SOLVABILITY 
OF 
GROUPS OF ODD ORDER 
by 
WALTER FEIT (Cornell University) 
and 
. THOMPSON (University of Chicago) 
 TABLB OF CONTBNTS 
CHAPTER I l . 
. Introduction . . . 
. Notation and Definitions...... . . 
3. Quoted Results...... . . 
. Elementary Results...... . . 
. Numerical Results. . . . 
CHAPTER II l . . . . . 
. Preliminary Lemmas of Lie Type . . . 
. Preliminary Lemmas of Hall-Higman Type . . . 
. Miscellaneous Preliminary Lemmas..... . . . 
CHAPTER III . . 
. Tamely Imbedded Subsets of a Group . . . 
. Coherent Sets of Characters . r . 
. Some Applications of Theorem 10.1 . . . 
. Further Results about Tamely Imbedded Subsets...... . . 
. Self Normalizing Cyclic Subgroups . . . 
CHAPTER IV l . 
. Statement of Results Proved in Chapter IV . . 
. A Partition of . . 
. Lemmas about Commutators. . . . 
. A Domination Theorem and Some Consequences. . . 
. Configurations...... . . 
. An -theorem . . . 
. An -theorem for r . 
21. A -theorem for and a -theorem for r . 
22. Linking Theoremsl . . 
. Preliminary Results about the Maximal Subgroups of . 
. Further Linking Theorems . . . 
. The Isolated Prime...... . . 
. The Maximal Subgroups of . . 
CHAPTER l . 
. Statement of the Result Proved in Chapter . . 
. Characters of Subgroups of Type I l . 
. Characters of Subgroups of Type III and IV l . 
. Characters of Subgroups of Type II, III and IV l . 
. Characters of Subgroups of Type . . 
. Subgroups of Type . . 
. Subgroups of Type I . . 
. The Subgroups and . . 
. Further Results about and . . 
. The Proof of Theorem 27.1 . r 
CHAPTER VI l ....1011 
37. Statement of the Result Proved in Chapter VI . ....1011 
. The Sets and l ....1011 
39. The Proof of Theorem 37.1 . r ....1017 
BIBLIOGRAPHY . ....1029 
 SOLVABILITY OF . OF ODD ORDER 
WALTER FEIT AND JOHN G. THOMPSON 
CHAPTER I 
1. Introduction 
The purpose of this paper is to prove the following result: 
THEOREM. All finite groups of odd order are solvable. 
Some consequences of this theorem and a discussion of the proof 
may be found in [11]. 
The paper contains six chapters, the first three being of a general 
nature. The first section in each of Chapters IV and summarizes 
the results proved in that chapter. These results provide the starting 
point of the succeeding chapter. Other than this, there is no cross 
reference between Chapters IV. and VI. The methods used in Chapter 
IV are purely group theoretical. The work in Chapter relies heavily 
on the theory of group characters. Chapter VI consists primarily of 
a study of generators and relations of a special sort. 
2. Notation and Definitions 
Most of the following lengthy notation is familiar. Some comes 
from a less familiar set of notes of P. Hall , while some has arisen 
from the present paper. In general, groups and subsets of groups are 
denoted by German capitals, while group elements are denoted by 
ordinary capitals. Other sets of arious kinds are denoted by English 
script capitals. All groups considered in this paper are finite, except 
when explicitly stated otherwise. 
Ordinary lower case letters denote numbers or sometimes elements 
of sets other than subsets of the group under consideration. Greek 
letters usually denote complex valued functions on groups. However, 
- 
Received November 20, 1962. While working on this paper the first author was at 
. A-i30ou- arndted NbaytiffionealU S.cSSie.nceAmc FyResoundaetionarch gOffirancte (D-urha9504504504; mt)he onsecondtract nauthorumber 
by the Esso Education Foundation, the Sloan Foundation and the Institute for Defense 
Analyses. Part of this work was done at the 1960 Summer Conferenc.e on Group Theory 
in Pasadena. The authors wish to thank Professor A. A. Albert of the University of 
Chicago for making it possible for them to spend the year 1960-61 there. The authors 
are grateful to Professor E. C. Dade whose careful study of a portion of this paper 
disclosed several blunders. Special thanks go to Professor L. J. Paige who has expedited 
the publication of this paper. 
775 
 776 SOLVABILITY OF GROUPS OF ODD ORDER 
and are reserved for field automorphisms, permutations or other 
mappings, and is used with or without subscripts to denote a root of 
unity. Bold faced letters are used to denote operators on subsets of 
groups. 
The rational numbers are denoted by , while . denotes the 
field of nth roots of unity over . 
Set theoretic union is denoted by U. If and are sets, 
denotes the elements of which are not in means that 
is a proper subset of . 
the group generated by . . such that . 
will be identified with 1. 
the set of . . such that . . 
the group defined by the generators. . with 
the relations . . 
the number of elements in the set X. 
the set of non identity elements in the set X. 
a set of primes. If we customarily 
identify with . 
the complementary set of primes. 
-number a non zero integer all of whose prime factors 
are in . 
the largest -number dividing the non zero 
integer . 
-group a group with . 
-element a group element such that is a -group. 
-subgroup of a subgroup . of with . 
-subgroup of a -subgroup of for suitable . 
Hall subgroup of a -subgroup of X. 
is a normal subgroup of X. 
char is a characteristic subgroup of X. 
( mod ) the invetSC image in of . Here 
, and is a function from groups to 
subgroups. 
the maximal normal -subgroup of . 
( mod ). 
-closed group we say that is -closed if and only if has 
a normal -subgroup. 
the Fitting subgroup of , the maximal normal 
nilpotent subgroup of X. 
the Frattini subgroup of the intersection 
of all maximal subgroups of X. 
the nth term in the ascending central series 
of , defined inductively by: 
 2. NOTATION AND DEFINITIONS 777 
center of ( mod ). 
the smallest normal subgroup of such 
that is a -group. 
. 
[. ] [[. ] ]. 
and being subsets 
of a group. 
. 
. If is called 
the normal closure of in . 
, the commutator subgroup of X. 
the nth term of the descending central series 
of , defined inductively by: , 
. 
the subgroup of the -group generated by 
the elements of order at most . 
the subgroup of the -group generated by 
the powers of elements of X. 
the minimal number of generators of X. 
being a -subgroup of X. 
c1 (X) the class of nilpotency of the nilpotent group 
, that is, the smallest integer such that 
. 
the largest subset of commuting element- 
wise with and being subsets of a 
group X. In case there is no danger of 
confusion, we set . 
the largest subset of which normalizes 
and being subsets of a group . In case 
there is no danger of confusion, we set 
. 
ker the kernel of the homomorphism of the 
group into the group will often be 
suppressed. 
being a subset of X. 
, the weak closure of 
in with respect to the group X. 
Here and are subgroups of I. If 
, we say that is weakly closed 
in with respect to X. 
the set of primes which divide . 
the by matrix with 1 in 
and , zero 
elsewhere. 
 778 SOLVABILITY OF GROUPS OF ODD ORDER 
SL(2, p) 
special p-group 
extra special p-group 
self centralizing sub 
group of X 
self normalizing sub 
group of X 
(X) the group of 2 by 2 matrices of determinant one with coefficients in GF(p), the field of p elements. an elementary abelian p-group, or a non abelian p-group whose center, commutator subgroup and Frattini subgroup coincide and are elementary. a non abelian special p-group whose center is of order p. a subgroup 31 of I such that 21 = C(2I). Notice that self centralizing subgroups are abelian. a subgroup 21 of X such that 21 = JV(3l). the set of self centralizing normal subgroups 
of X. 
{21 1 21 e <9*t?^r (X), m(«) ^ m}. 
the set of subgroups of X which 21 normalizes 
and which intersect 21 in the identity only. 
In case there is no danger of confusion, we 
set M2(«) = M(2l). If M(2I) contains only the 
identity subgroup, we say that M(2t) is trivial, 
the 7r-subgroups in M(2l). 
if > and 5 are subgroups of the group X, 
and > < $, then $/Ł> is called a section, 
if > and $ are normal subgroups of X and 
Ł S $, then ffi/ is called a factor of X. 
if $/Ł> is a factor of X and a minimal normal 
subgroup of X/Ł>, it is called a chief factor 
of X. 
If $/$ and 8/3K are sections of X, and if each coset of ® in Ł has 
a non empty intersection with precisely one coset of 3Ji in 2 and each 
coset of 3Jt in 8 has a non empty intersection with precisely one coset 
of $ in >, then >/$ and 8/2ft are incident sections. 
If >/$ is a section of X and 8 is a subgroup of X which contains 
at least one element from each coset of 5? in , we say that 8 covers 
>/$. We say that 8 dominates the subgroup ® provided 8 covers the 
section ^V3E(^)/C'3e(^). The idea to consider such objects stems from [17]. 
If g = >/$ is a factor of X, we let C%(%) denote the kernel of 
the homomorphism of X into Aut g induced by conjugation. Similarly, 
we say that X in X centralizes g (°r ac^s trivially on gf) provided 1^(21; 7T) section factor chief factor 
We say that X has a Sylow series if X possesses a unique SPl,...,Pi- 
subgroup for each i = 1, • • , n, where ?r(X) = {plt • • , pw}. The ordered 
 2. NOTATION AND DEFINITIONS 779 
-tuple is called the complexion of the series [18]. 
A set of pairwise permutable Sylow subgroups of one for each 
prime dividing , is called a Sylow system for . This definition 
differs only superficially from that given in [16]. 
P.. Hall [18] introduced and studied the following propositions: 
contains at least one -subgroup. 
satisfies , and any two -subgroups of are conjugate 
in . 
satisfies , and any -subgroup of I is contained in a 
-subgroup of X. 
contains a nilpotent -subgroup. 
In [19], P. Hall studied the stabi lity group of the chain 
, that is, the group of all automorphisms of 
such that for all in and each . If 
and are subgroups of a larger group, and if normalizes , we 
say that stabilizes provided is a subgroup of the stability 
group of . 
By a character of we always mean a complex character of 
unless this is precluded by the context. A linear character is a 
character of degree one. An integral linear combimtion of characters 
is a linear combination of characters whose coefficients are rational 
integers. Such an integral linear combination is called a generalized 
character. If is a collection of generalized characters of a group, 
let be respectively the set of all integral (complex) 
linear combinations of elements in . Let . be the 
subsets of respectively consisting of all elements with 
. 
If and are complex valued class functions on , then the 
inner product and weight are denoted by 
. 
. 
The subscript is dropped in cases where it is clear from the context 
which group is involved. 
The principal character of is denoted by ; the character of 
the regular representation of I is denoted by . If is a complex 
valued class function of a subgroup of , then denotes the class 
function of induced by . 
The kernel of a character is the kernel of the representation with 
the given character. 
A generalized character is -rational if the field of its values is 
 780 SOLVABILITY OF GROUPS OF ODD ORDER 
linearly disjoint from . 
A subset of the group is said to be a trivial intersection set 
in . or a T. I. set in if and only if for every in , either 
or 
. 
If is a normal subgroup of the group and is a character 
of denotes the inertial group of , that is 
{ for all }. 
Clearly, for all characters of . 
A group is a Frobenius group with Frobenius kernel and 
only if is a proper normal subgroup of which contains the centralizer 
of every element in . It is well known (see 3.16) that the Frobe- 
nius kernel of is also characterized by the conditions 
1. . 
2. for every non principal irreducible character of . 
We say that is of Frobenius type if and only if the following 
conditions are satisfied: 
(i) If is the maximal normal nilpotent -subgroup of , then 
. 
(ii) If is a complement for in X. then contains a normal 
abelian subgroup such that for every non principal 
irreducible character of . 
(iii) contains a subgroup of the same exponent as such 
that is a Frobenius group with Frobenius kernel . 
In case I is of Frobenius type, the maximal normal nilpotent 
-subgroup of I will be called the Frobenius kernel of X. 
A group is a three step group if and only if 
(i) . where is a cyclic -subgroup of , 
and . 
(ii) contains a non cyclic normal -subgroup such that 
is nilpotent and is the maximal normal 
nilpotent -subgroup of . 
(iii) contains a cyclic subgroup such that for in 
. 
3. 
For convenience we single out various published results which are 
of use. 
 3. QUOTED RESULTS 781 
3.1. ([19] Lemma 1, Three subgroups lemma). If >, , % are 
subgroups of the group X and 
[Ł>, , 2] = [®, 8, &] = 1, then [8, $, ft] = 1 . 
3.2. [20] F(X) = n Cs(®)f the intersection being taken over all 
chief factors S> o/ the group X. 
3.3. [20] // X is solvable, tAen C(F(X)) = Z(F(X)). 
3.4. Le p be an odd prime and X a p-group. If every normal 
abelian subgroup of X is cyclic, then X is cyclic. If every normal 
abelian subgroup of X is generated by two elements, then X is isomorphic 
to one of the following groups: 
( i ) a central product of a cyclic group and the non abelian group 
of order p* and exponent p. 
(ii) a metacyclic group. 
(iii) gp <A, B | [B, A] = C, [C, A] = B"-\ C* = [B, C] = A- = &* = 
1, n > 1, (r, p) = 1>. 
(iv) a 3-group. 
A proof of this result, together with a complete determination of the 
relevant 3-groups, can be found in the interesting papers [1] and [2], 
3.5. [20] If X is a non abelian p-group, p is odd, and if every 
characteristic abelian subgroup of X is cyclic, then Ii is a central 
product of a cyclic group and an extra special group of exponent p. 
3.6. ([22] Hilfssatz 1.5). If a is a p' -automorphism of the p-group 
X, p is odd, and a acts trivially on &i(X), then a = 1. 
3.7. [20] If 21 and S3 are subgroups of a larger group, then 
[SI, 93] < <2I, S3>. 
3.8. If the Sp-subgroup ty of the group X is metacyclic, and if 
p is odd, then ty n OP(X) is abelian. 
This result is a consequence of ([23] Satz 1.5) and the well known 
fact that subgroups of metacyclic groups are metacyclic. 
3.9. [28] If 21 is a normal abelian subgroup of the nilpotent 
group X and SI is not a proper subgroup of any normal abelian subgroup 
of X, then 21 is self centralizing. 
3.10. If ^ is a Sp-subgroup of the group X, and 21 e 
 782 SOLVABILITY OF GROUPS OF ODD ORDER 
then where is a -group. The proof of Lemma 5.7 
in [27] is valid for all finite groups, and yields the preceding statement. 
3.11. Let and be subgroups of a group , where is a 
-group and is a -group normalized by . Suppose is a subgroup 
of which does not centralize . If is a subgroup of of least 
order subject to being normalized by not centralized by then 
is a special -group for some prime acts trivially on 
and acts irreducibly on . This statement is a paraphrase 
of Theorem of Hall and Higman [21]. 
3.12. ([3] Lemma 1). Let A be a nonsingular matrix let 
be a permutation of the elements of A. Suppose that can be 
derived from A by permuting the columns of A and can also be 
derived from A by permuting the rows of A. Then the number of 
rows left fixed by is equal to the number of columns left fixed by . 
The next two results follow from applying 3.12 to the character 
table of a group X. 
3.13 (Burnside). A group of odd order has no non principal real 
irreducible characters. 
3.14. If is an automorphism of the group then the number 
of irreducible characters fixed by is equal to the number of conjugate 
classes fixed by . 
3.15. ([8] Lemma 2.1). Let be a -group for some prime 
and let be an irreducible character of with Then 
(mod ), where the summation ranges over all irreduci ble 
characters of with . 
Let be a Frobenius group with Frobeni us kernel . Then 
3.16. (i). ([7], [26]). is a nilpotent -subgroup of and 
for some subgroup of with . 
3.16. (ii). ([4] . ). If are primes every subgroup 
of of order is cyclic. If then a -subgroup of is cyclic. 
3.16. (iii). ([7] Lemma 2.1 or [10] Lemma 2.1). A non principal 
irreducible character of induces an irreducible character of . 
Furthermore every irreducible character of which does not have 
in its kernel is induced by a character of ,. Thus in particular 
any complex representation of , wh ich does not have in its 
 4. ELEMENTARY RESULTS 
contains the regular representahon of as a constituent when 
restricted to . 
We will often use the fact that the last sentence of 3-16 (iii) is 
valid if "complex representation of " is replaced by "representation 
of over a field of characteristic prime to . 
4. Elementary Results 
LEMMA 4.1. Let be a group with center and let be an 
irreducible character of X. Then . 
Proof. For . Therefore 
' 
LEMMA 4.2. Let be a generalized character of the group . 
Suppose that are commuting elements of and the order of 
is a power of a prime . Let be an algebraic number field which 
contains the roots of unity and let be a prime ideal in the 
ring of integers of which divides Then 
(mod ). 
Proof. It is clearly sufficient to prove the result for a generalized 
character . and thus for every irreducible character, of the abelian 
group . If is an irreducible character of then 
and (mod ). This implies the required congruence. 
LEMMA 4.3. Let be a normal subgroup of the group and let 
be an irreducible ch aracter of which does not contain in its 
kernel. If and . then . 
Proof. Let r be all the irreducible characters of . 
Let r be all the remaining irreducible characters of . If 
. then is mapped isomorphically into where 
is the image of in . Consequently 
. 
This yields the required result. 
Lemma 4.3 is of fundamental importance in this paper. 
LEMMA 4.4. Let a normal subgroup of the group . Assume 
that if is any irreducible character of then is 
 784 SOLVABILITY OF GROUPS OF ODD ORDER 
a sum of irreducible characters of X. all of which have the same 
degree and occur with the same multiplicity in . For an y integer 
let be the sum of all the irreducible characters of of degree 
which do not have in th eir kernel. Then , where is a 
rational number and is a generalized character of . 
Proof. Let r be all the distinct characters of which 
are induced by non principal irreducible characters of . and which are 
sums of irreducible characters of of degree . Suppose that 
where is an irreducible character of I for all values of . 
It is easily seen that r form a set of pairwise orthogonal 
characters. Hence . This proves the lemma. 
If is a normal subgroup of the group . and is a character 
of then is defined by . 
LEMMA 4.5. Let be a normal subgroup of the group and let 
be an irreducible character of . Suppose contains a normal 
subgroup such that and such that . is abelian. Then 
a sum of irreducible characters of which have the same degree 
and occur with the same multiplicity in . This common degree 
is a multiple of . If furthermore is a -subgroup of , 
then is a sum of distinct irreducible characters of degree 
. 
Proof. Let be the character of induced by . Let 
be an irreducible Constituent of and let . be all the 
characters . Choose the notation so that 
if and only if Since , we get that 
for some integer . Thus, 
(4.1) . 
Hence, every irreducible constituent of is of the form so all 
irreducible constituents of have the same degree. The characters 
r form a group which permutes the constituents 
of transitively by multiplication. Hence for every of there 
are exactly values of such that . If now , are 
the distinct irreducible characters which are constituerits of , then 
(4.1) implies that . 
Suppose is a complement to . in being a -subgroup of . 
We must show that is a sum of distinct irreducible characters 
of . For any subgroups of with , and any character 
of , let denote .the of induced by . 
has the property that. . sum of distinct 
 4. ELEMENTARY RESULTS 785 
irreducible characters of , where . Let be the multi- 
plicative group of linear characters of which have in their kernel, 
and let be an irreducible constituent of . Then is 
prime to . and it follows from Lemma 4.2 that does not 
vanish on any element of of prime power order. This in turn 
implies that 
. 
If . we are done. Otherwise, let contain as a subgroup 
of prime index. It suffices to show that is reducible. or equivalently. 
that for every in . This is immediate. since , 
so that for some in Since is abelian, it follows 
that , as required. 
To complete the proof of the lemma (now that the necessary 
properties of have been established), it suffices to show that if 
. , 
where the are distinct irreducible characters of , then each is 
irreducible, and for . For if this is proved, the normality 
of in implies the lemma. The definition of implies that is 
a sum of distinct irreducible characters of . Furthermore, 
is the only irreducible constituent of whose restriction to is not 
orthogonal to Thus, if then Since vanishes 
outside , a simple computation yields that . Therefore 
is irreducible. The proof is complete. 
LEMMA 4.6. Let be an odd prime and.let be a normal 
subgroup of the group . Assume that is a Frobenius group with 
Frobenius kernel is a -group and . 
(i) If then . . 
(ii) If is for all elements then is a prime 
or . 
(iii) If , then either is cyclic or is not 
yclic. 
Proof. is represented on Suppose that . 
By 3.16 (iii) has a fixed point on . and thus on . This 
proves (i). If is not a prime, let . Then 3.16 (iii) implies 
that has a non-cyclic fixed point set on and thus on . 
This proves (ii). 
As for (iii), let be the largest integer such hat has a non 
trivial fixed point on . It follows that has a non trivial 
fixed point on S If is not cyclic then since 
 786 SOLVABILITY OF GROUPS OF ODD ORDER 
completely reducible on implies (iii) by 8.16 (iii). 
Suppose that is cyclic. If , then by [10] Lemma 1.4, is 
cyclic. Since is of class 1 or 2. is of exponent . As 
is not cyclic neither is . Thus it may be assumed that 
is non cyclic of exponent and class at most 2. If 
is abelian then (iii) follows from (i). If is of class 2 then by (i) 
has a fixed point on and on . As has exponent this implies 
that is not cyclic as required. 
5. Numerical Results 
In this section we state some elementary number theoretical results 
and some inequalities. The inequalities can all be proved by the methods 
of elementary calculus and their proof is left to the reader. 
LEMMA 5.1. If are primes and 
. 
then . 
Proof. Let . Since . Hence 
. 
Reading (mod ) yields . Therefore 
. 
If then the right hand side of the previous equation is strictly 
larger than the left hand side. Thus as required. 
The first statement of the following lemma is proved in [5]. The 
second can be proved in a similar manner. 
LEMMA 5.2. Let be odd primes and let . 
(i) If divides then . 
(ii) If divides then . 
If , then 
(5.1) , 
(5.2) , 
(5.8) . 
If then 
(5.4) , 
 5. NUMERICAL RESULTS 787 
(5.5) 
. 
(5.7) for . 
(5-8) for . 
(5-9) for or . 
(5.10) for . or . . 
(5.11) for . 
(5.12) for . 
 CHAPTER II 
6. Preliminary Lemmas of Lie Type 
Hypothesis 6.1. 
(i) is a prime, is a normal -subgroup of and is 
a non identity cyclic . 
(ii) . 
(iii) is elementary abelian and . 
(iv) is . 
Let . , and . Let the Lie ring 
associated to (. ). Then where and 
correspond to and respectively. Let . For 
, let be the linear transformation induced by on . 
LEMMA 6.1. Assume that Hypothesis 6.1 is satisfied. Let . , 
be the characteristic roots of Then the characteristic roots of 
are found among the elements with . 
Proof. Suppose the field is extended so as to include . . 
Since is a -group, it is possible to find a basis of 
such that . . Therefore, . As 
induces an automorphism of , this yields that 
. 
Since the vectors with span , the lemma follows. 
By using a method which differs from that used below, M. Hall 
proved a variant of Lemma 6.2. We are indebted to him for showing 
us his proof. 
LEMMA 6.2. Assume that Hypothesis 6.1 is satisfied, and that 
acts irreducibly . Assume further that is an odd 
prime and that and have the same characteristic polynomial. 
Then and 
Proof. Let be the characteristic roots of . . By 
Lemma 6.1 there exist integers such that . Raising 
this equation to a suitable power yields the existence of integers a 
and with such that . By Hypothesis 6-1 (ii). 
the preceding equality implies (mod ). Since acts 
irreducibly, we also have (mod ). Since is a -group, 
789 
 790 SOLVABILITY OF GROUPS OF ODD ORDER 
. Consequently, 
(mod ), 
(mod ), . 
Let be the resultant of the polynomials and 
. Since is a prime, the two polynomials are relatively prime, 
so is a nonzero integer. Also, by a basic property of resultants, 
(6.2) 
for suitable integral polynomials and . 
Let be a primitive qth root of unity over . so that we also 
have 
. 
For , this yields that , so that . 
Since is odd (6.1) and (6.2) imply that . This is not the case, 
so . 
Each term on the right hand side of (6.8) is non negative. As 
the geometric mean of non negative numbers is at most the arith- 
metic mean, (6.8) implies that 
. 
The algebraic trace of a primitive qth root of unity is -1. hence 
. 
Now (6.1) and (6.2) imply that 
. 
Since is irrational, equality cannot hold. 
LEMMA 6.8. If is a -group and , then 
is elementary abelian for all . 
Proof. The assertion follows from the congruence 
[. ] (mod ). 
valid for all in . 
LEMMA 6.4. Suppose that is a fixed point free pp-automorphisR 
 7. PRELIMINARY LEMMAS OF HML-HIGMAN TYPE 791 
of the -group . (mM ) for some integer 
independent of A. Then is of expment . 
Proof. Let so that is in for all A in . 
Then 
[. ][. ] 
[. ][. ] (mM ). 
Since is regular on is also regular on each . As the 
order of divides the above congruences now imply that c1 
and so is a regular -group. If . then the mapping 
induces a non zero linear map of to 
for suitable . Namely, choose so that but 
. and use the regularity of to guarantee linearity. Notice 
that , since by hypothesis . We find that (mod ). 
and so and has a fixed point on . contrary to 
assumption. Hence, . 
7. Preliminary Lemmas of Hall-Higman Type 
Theorem of Hall and Higman [21] is used frequently and will 
be referred to as (B). 
LEMMA 7.1. If is a -solvable linear group of odd order over 
a field of characteristic , then contains every element wh ose 
minimal polynomial is . 
Proof. Let be the space on which I acts. The hypotheses 
of the lemma, together with (B), guarantee that either or 
contains no element whose minimal polynomial is . 
Let be an element of with minimal polynomial . Then 
, and the subspace which is elementwise fixed by 
is proper and is I-invariant. Since is a -group, . Let 
. 
By induction on dim ( mod ). . Since 
( mM )( mod ) 
is a -group, the lemma follows. 
LEMMA 7.2. Let I be a -solvable roup of odd order, and a 
-su bgroup of X. Any one of the following guarantees 
that : 
 r92 SOLVABILITY OF GROUPS OF ODD ORDER 
1. 21 is abelian and \ X: -W(2l)| is prime to p. 
2. p^5 and [*P, 21, 21, 21, 21] = 1 for some Sp-subgroup ^ of X. 
3. [?p, 21, SI] = 1 for some SP-subgroup sp of X. 
4. 21 acts trivially on the factor OP/,P/P,(Ł)/OP,,P(X). 
Proof. Conditions 1, 2, or 3 imply that each element of 21 has a 
minimal polynomial dividing (x — I)""1 on Op/fp(X)/3>, where S = 
Z)(0p,,p(X)mod 0P,(X)). Thus (B) and the oddness of |X| yield 1, 2, and 
3. Lemma 1.2.3 of [21] implies 4. 
LEMMA 7.3. // X is p-solvable, and ^ is a Sp-subgroup of X, 
then M(^P) is a lattice whose maximal element is OP>(Ł). 
Proof. Since 0P,(X) < X and ^P n 0P>(X) = 1, O9,(H) is in 
Thus it suffices to show that if ft e MOP), then ftS0P'(X). Since 
is a group of order I^PHftl and ty is a Sp-subgroup of X, ft is a p'- 
group, as is ft0p/(X). In proving the lemma, we can therefore assume 
that OP/(X) = 1, and try to show that ft = 1. In this case, ft is faith 
fully represented as automorphisms of 0P(X), by Lemma 1.2.3 of [21]. 
Since 0p(X)g^p, we see that [Ł>, Op(X)]Sft D $P, and $ = 1 follows. 
LEMMA 7.4. Suppose ^ is a SP-subgroup of X araJ 21 
Then M(2I) contains only p'-groups. If in addition, X is p-solvable, 
then M(2I) is a lattice whose maximal element is 0P,(X). 
Proof. Suppose 21 normalizes § and 21 n & = <!> Let 21* be a 
Sp-subgroup of 2IŁ> containing 21. By Sylow's theorem, SR = 21* n & 
is a Sp-subgroup of >. It is clearly normalized by 21, and 21 n %. =<!>. 
If ?i ^ <1>, a basic property of p-groups implies that 21 centralizes 
some non identity element of $plf contrary to 3.10. Thus, ^ = <1> 
and § is a p'-group. Hence we can assume that X is p-solvable and 
that OP,(X) = <1> and try to show that § = <1>. 
Let Xa = OP(X)Ł2I. Then OP(X)2I is a Sp-subgroup of Xlf and 
21 6 <P^^~(0*>(X)2l). If XxcX, then by induction ^SO,^) and so 
[0P(X), ] gOp(X) n 0P,(XO = 1 and ft = 1. We can suppose that Xx = 
X. 
If 21 centralizes >, then clearly 21 < X, and so ker (X - > Aut 21) = 
2t x \, by 3.10 where &S&. Hence, ^ char 21 x & < X, and 
1 < X, so that & = 1. We suppose that 21 does not centralize >, 
and that § is an elementary g-group on which 21 acts irreducibly. 
Let S3 = 0P(X)/Z)(01,(X)) = ^ x SS3, where ^ = C^ft) and 332 = [55, ft]. 
Let Fe932 and Xe F, so that [JST,2l]S2l. Hence, [JCf 21] maps into 
SSi, since [[.XT, 21], ft] S ^ n OP(X) = 1. But 5S2 is X-invariant, so [X, 21] 
maps into 2^ n 93a = 1. Thus, 21 S ker (X - > Aut S32), and so [21, ft] 
 7. PRELIMINARY LEMMAS OF HALL-HIGMAN TYPE 793 
centralizes . As acts irreducibly on , we have , so 
. Thus, centralizes and so centralizes , so , as 
required. 
LEMMA 7.5. Suppose and are -subgroups of th e solvable 
group . If , then . 
Proof. We proceed by induction on . We can suppose that 
has no non identity normal subgroup of order prime . Suppose 
that possesses a non identity normal -subgroup . Then 
. 
Let . By induction, , 
so ( mod ), and we are done. Hence, we can assume 
that . In this case, is a -group, and . 
By hypothesis, . and so centralizes . By 3.3, we 
see that , so as desired. 
The next two lemmas deal with a -subgroup of the -solvable 
group and with the set 
is a subgroup of X. 
2. . 
8. The -length of is at most two . 
4. is not divisible by three distinct primes .} 
LEMMA 7.6. . 
Proof. Let . It suffices to show that 
for every prime . This is clear if , so suppose . Since 
is -solvable, satisfies , so we can suppose that is a , q- 
group. By induction, we can suppose that contains every proper 
subgroup of which contains . Since . we see that 
. If , then . Since 
. we have . Thus, we can assume that 
. Since , we see that . If 
, we are done, so suppose not. Then , 
so that contains , as required. 
LEMMA 7.7. Suppose are subgroups of which contain 
such that . for all in . Then . 
Proof. It suffices to show that for every prime . 
This is clear if , so suppose . Let be a -subgroup of 
 SOLVABILITY OF GROUPS OF ODD ORDER 
permutable with , which exists by in . Since 
satisfies , there is a -subgroup of which contains and 
is permutable with , Set . We next show that 
. 
If , this is the case by hypothesis, so we can suppose the p- 
length of is at least 8. Let , and . 
Then is a proper subgroup of so by induction on . we have 
. Let . Since is in 
. we have . Furthermore, by Sylow's theorem, 
. Let . Then with . Then , 
with in . in . Also, in in . 
and so . Since (. ). we have 
with in in . Hence, 
with in . in . 
Since . we have 
. 
By construction, Furthermore, 
and 
. 
completing the proof. 
LEMMA 7.8. Let be a finite group and a -su bgroup of 
Which is normalized by the -subgroup of Set . 
Suppose is a -solvable subgroup of containing and . 
there is a -solvable subgroup of which contains 
and . 
Proof. Let . Then does not centralize . 
Let be a subgroup of which is minimal with respect to being 
-invariant and not centralized by . Then , and 
, while . Hence, . and so 
. Since . centralizes . Since is a sub- 
group of . we have for suitable . As is a 
-group and is a -group, we can find an -invariant -subgroup 
of incident with . Hence, centralizes . Set 
. 
As is -solvable so is . If . then 
 8. MISCELLANEOUS PRELIMINARY LEMMAS 795 
[<P0, ] c S0 n 0P,(®) S 0P,(3) 
and > centralizes S3, contrary to construction. Thus, >g;Op/(®), as 
required. 
LEMMA 7.9. Let !Q be a p-solvable subgroup of the finite group 
X, and let ^ be a Sp-subgroup of >. Assume that one of the follow 
ing conditions holds: 
(a) |X| is odd. 
(b) p^5. 
( c ) p = 3 and a S^-subgroup of $ is abelian. 
Let SP0 = OP',P(!Q) D ^P and let ^P* be a p-subgroup of X containing ty. 
If ty is a Sp-subgroup of Ny($$0), then ^P0 contains every element of 
Proof. Let WeŁ«^^W*). By (B) and (a), (b), (c), it follows 
that 21 n %> = 21 0 % = i, say. If 2^ c 21, then there is a ^-invariant 
subgroup S3 such that 2^ c 33 S 51, 1 33 : 8^ | = p. Hence, [%, S3] S ^ g 
%, so SJVa(SPb)nSp*. Hence, <S3, ^p> is a p-subgroup of ^(^Po), 
so S3 C sp. Hence, S3 S a n ^P = 2^, which is not the case, so 21 = 2^, 
as required. 
8. Miscellaneous Preliminary Lemmas 
LEMMA 8.1. If X is a re-group, and <& is a chain X = X0 3 
Xj 2 • • • 3 Xn = 1, then the stability group 21 of ^ is a n-group. 
Proof. We proceed by induction on n. Let A 6 2t. By induction, 
there is a 7r-number m such that B = Am centralizes 3^. Let JTeX; 
then JT* = XY with F in Xlf and by induction, X* = J^F". It fol 
lows that B^ = 1. 
LEMMA 8.2. If ty is a p-group, then ?$ possesses a characteristic 
subgroup Ł sucfe that 
( i ) cl (&) ^ 2, and S/Z(S) is elementary. 
(ii) ker (Aut ^3-^->AutS) is a p-group. (res is fee homomor- 
phism induced by restricting A in Aut ^P ^o .) 
(iii) [^P,e] 
Proof. Suppose E can be found to satisfy (i) and (iii). Let 
5? = ker res. In commutator notation, [®, S] = 1, and so [®, S, ^5] = 1. 
Since [E, ^P] SS, we also have [(Ł, sp, ] = 1 and 3.1 implies [^P, , E] = 
1, so that [^P, ^]gZ(e). Thus, ffl stabilizes the chain ^P 2 21 so 
is a p-group by Lemma 8.1. 
 796 SOLVABILITY OF GROUPS OF ODD ORDER 
If now some element of is characteristic in , then 
(i) and (iii) are satisfied and we are done. Otherwise, let be a 
maximal characteristic abelian subgroup of , and let be the group 
generated by all subgroups of such that . , 
( mod ). . By construction, , and is seen 
to be characteristic. The maximal nature of implies that . 
Also by construction , so in particular, 
and c1 . By construction, is elementary. 
We next show that . This statement is of course 
equivalent to the statement that . Suppose by way of con- 
tradiction that . Let be a subgroup of of minimal 
order subject to (a) , and (b) Since satisfies (a) 
and (b), exists. By the minimality of , we see that 
and . Since centralizes , so do and . so we 
have and . The minimal nature of guarantees 
that is of order . Since is of 
order . so is of order . By construction of , we find 
, so , in conflict with (b). Hence, , and (i) and 
(iii) are proved. 
LEMMA 8.3. Let I be a -group, odd, and among all elements 
of , choose to maximize . Then . 
REMARK. The oddness of is required, as the dihedral group 
of order 16 shows. 
Proof. We must show that whenever an element of I of order 
centralizes . then the element lies in . 
If and , let , and let 
be an ascending chain of subgroups, each 
of index in its successor. We wish to show that . Suppose 
for some . Then is generated by its normal 
abelian subgroups and , so is of class at most two, so 
is regular. Let . of order . Then A in an 
integer. Since is regular, is of order 1 or . Hence, 
and . Hence, char . and 
follows. In particular, stabilizes the . 
It follows that if , then centralizes . Since 
. We next show that is of exponent . 
Since , we see that , and so 
, 
and c1 . If . then is regular, and being generated by 
 8. MISCELLANEOUS PRELIMINARY LEMMAS 797 
elements of order p, is of exponent p. It remains to treat the case 
p = 3, and we must show that the elements of 3) of order at most 
3 form a subgroup. Suppose false, and that (X, Y> is of minimal 
order subject to X 3 = Y2 = 1, (XY)* * 1, X and Y being elements 
of 3>. Since <F, r*>c<JT, F>, [Y, X] = y-1. JT^KST is of order 
three. Hence, [X, Y] is in fi^Sl), and so [Y, X] is centralized by 
both X and Y. It follows that (JTF)3 = X3Y*[Y, X]3 = 1, so ® is of 
exponent p in all cases. 
If ^(SOc®, let @<X, <gŁ3>, 18:^(21)1=0. Since ^(§l)S 
Z(<2), e is abelian. But m(@) = m(Sl) + 1 > m(2l), in conflict with 
the maximal nature of 21, since @ is contained in some element of 
by 3.9. 
LEMMA 8.4. Suppose p is an odd prime and X is a p-group. 
(i) // &*&^l (X) is empty, then every abelian subgroup of X 
is generated by two elements. 
(ii) // ^^^f (X) is empty and A is an automorphism of X of 
prime order q, p= q, then q divides p2 — 1. 
Proof. ( i ) Suppose 51 is chosen in accordance with Lemma 8.3. 
Suppose also that X contains an elementary subgroup @ of order p3. 
Let ! = Cft(0i(2l)), so that @x is of order p2 at least. But by Lemma 
8.3, (^SS^Sl), a group of order at most p\ and so ^ = ^(51). But 
now Lemma 8.3 is violated since ( centralizes L 
(ii) Among the A-invariant subgroups of X on which A acts non 
trivially, let > be minimal. By 3.11, § is a special p-group. Since 
p is odd, § is regular, so 3.6 implies that § is of exponent p. By 
the first part of this lemma, § contains no elementary subgroup of 
order p3. It follows readily that w(Ł>) ^ 2, and (ii) follows from the 
well known fact that q divides |Aut >//)(Ł>) |. 
LEMMA 8.5. If 3, is a group of odd order, p is the smallest 
prime in TT(Ł), and if in addition H contains no elementary subgroup 
of order p3, then X has a normal p-complement. 
Proof. Let ty be a SP-subgroup of X. By hypothesis, if § is a 
subgroup of sp, then ^^^l(^) is empty. Application of Lemma 8.4 
(ii) shows that JV3e(^))/C3E(^) is a p-group for every subgroup > of ^5. 
We apply Theorem 14. 4. 7 in [12] to complete the proof. 
Application of Lemma 8.5 to a simple group ® of odd order im 
plies that if p is the smallest prime in TT(®), then ® contains an 
elementary subgroup of order p3. In particular, if 3e;r(®), then ® 
contains an elementary subgroup of order 27. 
 798 SOLVABILITY OF GROUPS OF ODD ORDER 
LEMMA 8.6. Let be subgroups of a group and suppose 
that for every permutatim of 
Then is a subgroup of X. 
Proof. , as re- 
quired. 
LEMMA 8.7. If is a -group of automorph isms of the -group 
, if has no fixed points and acts trivially on . 
then . 
Proof. In commutator notation, we are assuming . 
and . Hence, Since . 
we also have . By the three subgroups lemma, we 
have . Since , the lemma follows. 
LEMMA 8.8. Suppose is a -group, is odd. A is an auto- 
morphism of of prime order . (mod ), and contains a 
subgroup of index such that is empty. Then 
and is elementary of order . 
Proof. Since (mod ) and is odd, does not divide . 
Since , Lemma 8.4 (ii) implies that A acts trivially on . 
Suppose that A has a non trivial fixed point on . We can 
then find an A-invariant subgroup of index in such that A 
acts trivially on . In this case, A does not act trivially on . 
and so , and is of index in . By induction, 
and is elementary of order Since A acts trivially 
on . it follows that is abelian of order If were elemen- 
tary, would not exist. But if were not elementary, then A 
would have a fixed point on . which is not possible. Hence 
A has no fixed points on . so by Lemma 8.7, . 
Next, suppose that A does not act irreducibly on Let 
be an irreducible constituent of A on . By induction, 
is of order . and . Since . is a 
proper A-invariant subgroup of . The only possibility is . 
and follows from the existence of . 
If , then follows from Lemma 5.1. 
we can suppose that , and that A acts irreducibly on , 
and try to derive a contradiction. We see that must be non 
abelian. This implies that . Let Since 
 8. MISCELLANEOUS PRELIMINARY LEMMAS 799 
p = 1 (mod q), and qn = 1 (mod p), n^3. Since Z)(O) = Z(O), ^ is 
even, O/Z(O) possessing a non singular skew-symmetric inner product 
over integers mod q which admits A. Namely, let E be a subgroup 
of order q contained in O' and let &! be a complement for & in O'. 
This complement exists since O' is elementary. Then Z(^?modS1) is 
A-invariant, proper, and contains Z)(O). Since A acts irreducibly on 
O/Z)(O), we must have Z)(O) = Z(Omod G^), so a non singular skew- 
symmetric inner product is available. Now O is regular, since cl(O) = 
2, and q is odd, so 1^(0)1 = 10:^(0)1, by [14]. Since cl(O) = 2, 
^(O) is of exponent q. Since 
|0 : <W)I ^IQ: 1*0)1 ^fl4, 
we see that | ^(O) | ^ q4. Since O0 exists, ^(Q) is non abelian, of 
order exactly q*, since otherwise O0 n ^i(O) would contain an elemen 
tary subgroup of order q3. It follows readily that A centralizes 
and so centralizes O, by 3.6. This is the desired contradiction. 
LEMMA 8.9. If ty is a p-group, if <$*S^fŁ($P) is non empty and 
21 is a normal abelian subgroup of ty of type (p, p), then 21 is con 
tained in some element of 
Proof. Let @ be a normal elementary subgroup of ty of order p3, 
and let d = 67^(21). Then ^ < *P, and <2l, ^> = g is abelian. If 
| g| = p2, then 21 = @! = g ^ , and we are done, since @ is contained 
in an element of ^^^(^P). If |g| ^ i?3, then again we are done, 
since g is contained in an element of 
If X and 2) are groups, we say that ?) is involved in X provided 
some section of X is isomorphic to 2) [18]. 
LEMMA 8.10. Let ty be a Sp-subgroup of the group X. Suppose that 
Z(*$) is cyclic and that for each subgroup 21 in ty of order p which 
does not lie in Z(ty), there is an element X = -X"(2l) of ty which 
normalizes but does not centralize <2l, ^(Z(^P))>. Then either SL(2, p) 
is involved in X or Qi(Z(ty)) is weakly closed in ty. 
Proof. Let S> = A(Z(^)). Suppose @ = 3)* is a conjugate of © 
contained in $p, but that @ ^= . Let 3) = <Ł>>, C = <Ł'>. By hypo 
thesis, we can find an element X = -X"(G?) in $p such that -X" normalizes 
<Ł', Z)> = 7$, and with respect to the basis (E, D) has the matrix 
(J J). Enlarge § to a SP-subgroup ^5* of CS(<Z). Since ( = S)0, 
^SC^B), so ^P* is a SP-subgroup of X, and @SZOP*). Since Z0P*) 
is cyclic by hypothesis, we have © = ?i(Z($P*)). By hypothesis, there 
is an element Y= F(S)) in ^5* which normalizes $ and with respect 
 800 SOLVABILITY OF GROUPS OF ODD ORDER 
to the basis (E, D) has the matrix 
Q J). Now (J J) and 
generate SL(2, p) [6, Sections 262 and 263], so SL(2, p) is involved in 
Ngffi), as desired. 
LEMMA 8.11. If 31 is a p-subgroup and S3 is a q-subgroup of X, 
p = g, and 31 normalizes S3 tfcera [S3, 31] = [S3, 21, SI]. 
Proo/. By 3.7, [St, S3] < 31S3. Since SlS3/[St, S3] is nilpotent, we 
can suppose that [31, S3] is elementary. With this reduction, [S3, 31, 31] < 
3133, and we can assume that [S3, 31, 31] = 1. In this case, 31 stabilizes 
the chain S3 2 [S3, 31] 31, so [S3, 81] = 1 follows from Lemma 8.1 and 
LEMMA 8.12. Let p be an odd prime, and @ an elementary sub 
group of the p-group $p. Suppose A is a pf -automorphism of ty which 
centralizes fli(C(Gf)). Then A = 1. 
Proof. Since < S 0i(CĄ(®))f A centralizes . Since <g is A-invari- 
ant, so is C$(6). By 3.6 A centralizes C^(6), so if @ S Z(SP), we are done. 
If C^(C) c $p, then C?(®)Z7($P) c ?p, and by induction A centralizes 
Now [SP, ] S JD(SP) and so [Sp, <S, <A>] = 1. Also, [@, <A>] - l, 
so that [Gc, <A>, ^?] = 1. By the three subgroups lemma, we have 
[<A>, p, 6] = 1, so that [Sp, <A>] S C^(@), and A stabilizes the chain 
5P3CĄ(C)=)1. It follows from Lemma 8.1 that A = 1. 
LEMMA 8.13. Suppose ^ is a Sp-subgroup of the solvable group 
@, *&i4^(ty) is empty and @ is of odd order. Then @' centralizes 
every chief p-factor of @. 
Proof. We assume without loss of generality that Op/(@) = 1. 
We first show that ^P < @. Let § = Op(@), and let S be a subgroup 
of § chosen in accordance with Lemma 8.2. Let SB = ^(C). Since 
p is odd and cl(E) ^ 2, SB is of exponent p. 
Since Op/(@) = 1, Lemma 8.2 implies that ker (@ - > Aut K) is a 
p-group. By 3.6, it now follows that ker (@ — ^-> Aut SB) is a p-group. 
Since Sp has no elementary subgroup of order p\ neither does SB, and 
so | SB : 77(3B) | ^ p2. Hence no p-element of @ has a minimal poly 
nomial (x - l)p on 3B//?(SB). Now (B) implies that ^?/kera < @/kera. 
and so ^5 < @, since ker a: S ^?. 
Since ^P < @, the lemma is equivalent to the assertion that if S 
is a Sp/-subgroup of @, then S' = 1. If S' = 1, we can suppose that 
8' 
centralizes every proper subgroup of ty which is normal in @. Since 
S is completely reducible on 5p/D($P)f we can suppose that [^5, S'] = ^P 
 8. MISCELLANEOUS PRELIMINARY LEMMAS 801 
and [Z)(*P), 8'] = 1. By Lemma 8.7 we have DW)SZ(W) and so 
i(?$) = $ is of exponent p and class at most 2. Since *$ has no 
elementary subgroup of order p\ neither does . If 5B is of order p, 
8' 
centralizes $ and so centralizes $p by 3.6, thus 8' = 1. Otherwise, 
| $ : D(®) | = p2 and 8 is faithfully represented as automorphisms of 
&ID(Sl). Since |8| is odd, 8' = 1. 
LEMMA 8.14. // <2> is a solvable group of odd order, and 
*$*K^J($P) is empty for every Sp-subgroup ty of & and every prime 
p, then @' is nilpotent. 
Proof. By the preceding lemma, @' centralizes every chief factor 
of @. By 3.2, @'SF(@), a nilpotent group. 
LEMMA 8.15. Let <Ł> be a solvable group of odd order and suppose 
that @ does not contain an elementary subgroup of order p3 for any 
prime p. Let ^ be a Sp-subgroup of @ and let & be any character 
istic subgroup of ty. Then ( fl ^P' < @. 
Proo/. We can suppose that ( S ^ , since E n ^ char sp. By 
Lemma 8.14 F(@) normalizes E. Since F(@)?P < @, we have @ = 
F(@)JV(5P). The lemma follows. 
The next two lemmas involve a non abelian p-group ^P with the 
following properties: 
( 1 ) p is odd. 
(2) $P contains a subgroup ^30 of order p such that 
where % is cyclic. 
Also, 21 is a p'-group of automorphisms of ^P of odd order. 
LEMMA 8.16. With the preceding notation, 
( i ) 21 is abelian. 
( ii ) JVb element of SI1 centralizes Ol(C(^0)). 
(iii) // 21 is cyclic, then either |2l| divides p l*o 
is empty. 
Proof, (ii) is an immediate consequence of Lemma 8.12. 
Let 93 be a subgroup of ^P chosen in accordance with Lemma 8.2, 
and let SB = fi^SB) so that 21 is faithfully represented on 2B. If ^P0g 
SB, then $02B is of maximal class, so that with 3B0 = SB, 3B<+1 = [SB,, SJJ], 
we have |SB,:S»<+1| = p, i = 0, 1, -..fn-l, |3B| = p«, and both (i) 
and (iii) follow. If %S3B, then m(3B) = 2. Since [SB, P] -g Z(SB), 
 SOLVABILITY OF GROUPS OF ODD ORDER 
it follows that . By Lemma 8.9, is empty. 
The lemma follows readily from 3.4. 
LEMMA 8.17. In the preceding notation, assume in additim that 
is a prime, that does not divide , that 
and that is cyclic. Then . 
Proof. Since centralizes and so - 
Since is cyclic, is not of type . Hence, 
. Since every automorphism of which is the identity 
on is inner, it follows that . where 
Since is cyclic, so is , and so , 
by virtue of and . 
 CHAPTER III 
9. Tamely Imbedded Subsets of a Group 
The character ring of a group has a metric structure which is 
derived from the inner product. Let be a subgroup of the group 
X. The purpose of this chapter is to state conditions on and which 
ensure the existence of an isometry that maps suitable subsets of 
the character ring of into the character ring of and has certain 
additional properties. If is in the character ring of and is 
defined then these additional properties will yield information con- 
cerning for some elements of . Once the existence of is 
established it will enable us to derive information about certain 
generalized characters of provided we know something about the 
character ring of . In this way it is possible to get global infor- 
mation about I from local information about . 
There are two stages in establishing the existence of . First we 
will require that is in some sense "nicely" imbedded in X. When 
this requirement is fulfilled it is possible to define for certain 
generalized characters of with . In this situation is 
explicitly defined in terms of induced characters of various subgroups 
of X. Secondly it is necessary that the character ring of have 
certain special properties. These properties make it possible to extend 
the definition of to a wider domain. In particular it is then possible 
to define for some generalized characters of with . 
The precise conditions that the character ring of needs to satisfy 
will be stated later. In this section we are concerned with the 
imbedding of in X. The following definition is appropriate. 
DEFINITION 9.1. Let be a subset of the group such that 
(9-1) . 
Let be the set of elements in such that . and let 
. 
We say that is tamely imbedded in if the following conditions 
are satisfied: 
(i) If two elements of are conjugate in , they are conjugate 
. 
(ii) If is non empty, then there are non identity subgroups 
of , with the following properties: 
 SOLVABILITY OF GROUPS OF ODD ORDER 
(a) for . 
(b) is a -subgroup of ; 
(c) and . 
(d) for ; 
(e) For . define 
. 
Then . is a non empty T. I. set in and . 
(iii) If . then there is a conjugate of in and an 
index such that 
. 
If is a tamely imbedded subset of then for , each of 
the groups : is called a supporting subgroup of . The collection 
is called a system of supporting subgroups of . 
In one important special case, the definition of tamely imbedded 
subset of I is fairly easy to master. Namely, if is empty, the 
reader can check that is a T. I. set. 
If is a tamely imbedded subset of with then in 
this section denotes the set of generalized characters of which 
vanish outside and denotes the complex valued class functions 
of which vanish outside . Similarly, is the subset of 
vanishing at 1. R. Brauer and M. Suzuki noted that if 
is a T. I. set in then the mapping from into the ring 
of class functions of . defined by 
is an isometry ([24], 662). They were then able to extend this 
isometry to certain subsets of . Several authors have since then 
used this technique and it has played an important role in recent 
work in group theory. 
In this chapter these results will be generalized in two ways. 
First we will consider tamely imbedded subsets of rather than T. I. 
sets in X. Secondly we will show that under a variety of conditions 
can be extended to various large subsets of The results 
proved in this chapter are important for the proof of the main theorem 
of this paper. However it is unnecessary in general to assume that 
has odd order or that is a minimal simple group. 
The following notation will be used throughout this section. 
For a tamely imbedded subset of let and for 
 9. TAMELY IMBEDDED SUBSETS OF A GROUP 805 
let and have the same meaning as in Definition 9.1. 
Define and 
for . 
For let 
(9.2) {, }. 
Since is tamely imbedded in . it follows from (9.2) and Definition 
9.1 that for 
. 
For and define 
. 
Let be the class function of : which satisfies 
. 
Let be the class function of induced by . Define 
(9.4) . 
If then (9.4) implies that is a generalized character of X. 
It is an immediate consequence of the definition of induced characters 
that for 
for 
for 
for . 
LEMMA 9.1. Suppose that is a tamely imbeWed subset of . 
If let be defined by (9.4). Then if is not 
conjugate to an element of for any l while 
for . 
Proof. If then a complement of : in is solvable. 
Thus ([28] . ) for every element of is conjugate to 
an element of the form with l Suppose 
 806 SOLVABILITY OF GROUPS OF ODD ORDER 
that is not conjugate to an element of ; then since . 
(9.4) implies that . This implies that unless 
is conjugate to an element of for some . 
Let A for some with . Suppose that 
for some and some with . Then 
Thus and . Furthermore 
. By assumption and . 
Thus contrary to the choice of L. Since . 
vanishes on . Consequently (9.4) implies that 
for 
for . 
Since is a T. I. set in X. with we get that 
. 
Thus (9.6) yields that 
(9.7) for . 
Assume first that . Then . Hence 
(9.5), (9.6) and (9.7) yield that . If , then 
and . Thus (9.5) and (9.7) yield that also in this case 
. The proof is complete in all cases. 
LEMMA 9.2. Suppose that is a tamely imbedded subset of I. 
If let be defined by (9.4). Then for 
for . 
Furthermore S is a linear combination of characters of . 
Proof. If then by Lemma 9.1 and the definition of 
. 
If . and , then is conjugate to an element A of 
for some l Thus by (9.5) and Lemma 9.1 as 
required. 
Let be an irreducible character of which does not have : 
in its kernel. Then 
 9. TAMELY IMBEDDED SUBSETS OF A GROUP 807 
(9.8) . 
By Lemma 4.3 vanishes on hence (9.8) and the first 
part of the lemma yield that 
. 
Since is a linear combination of characters of : this yields 
that . The lemma is proved. 
LEMMA 9.3. Suppose that is a tamely imbeMed subset of X. 
If and is defined by (9.4) then 
. 
Proof. Let . r be all the conjugate classes of which 
contain elements of . Let r be elements in such that 
. The number of elements in which are conjugate to 
an element of is easily seen to be 
. 
Thus by Lemma 9.1 and (9.3) 
(9.9) 
. 
By assumption . r are the conjugate classes of which 
contain elements of . Since this yields that 
. 
Therefore (9.9) implies the desired equality. 
LEMMA 9.4. Suppose that is a tamely imbedded subset of X. 
 808 SOLVABILITY OF GROUPS OF ODD ORDER 
Let be a generalized character of such that for . is 
constant on . If and if are defined by (9.4) 
then 
. 
Proof. Since is constant on for it follows from 
Lemma 9.1 that 
. 
Thus by Lemma 9.3 
. 
By Lemma 9.1 is a generalized character of V. which is constant 
on for . If now is replaced by in the first equation 
of the lemma the second equation follows. 
LEMMA 9.5. Suppose that is a tamely imbedded subset of 
I. Let be a class function of which is con stant on for 
. Let be the set of all elements in X. which are conjugate 
to some element of with Then 
. 
Proof. Define by 
if . 
if . 
By Lemma 9.1 
if 
if . 
Consequently Lemma 9.3 implies that 
 10. COHERENT SETS OF CHARACTERS 809 
. 
Lemma 9.5 is of great importance. Even the special case in 
which is of considerable interest and plays a role in section 
26. In this special case, Lemma 9.5 asserts simply that 
. 
. Coherent Sets of Characters 
Throughout this section let be a tamely imbedded subset of the 
group X. Let and let be the set of generalized char- 
acters of which vanish outside . Let be defined by (9.4). 
DEFINITION 10.1. A set of generalized characters of is 
coherent if and only if 
(i) . 
(ii) It is possible to extend from to a linear isometry 
mapping into the set of generalized characters of . 
(iii) . 
It is easily seen that if is a coherent set and with 
then also is a coherent set. It is more difficult to 
decide whether the union of two coherent subsets of is coherent. 
Examples are known in which consists of irreducible characters of 
and is not coherent though . In these examples 
is even a T. I. set in X. The main purpose of this section is to give 
some sufficient conditions which ensure that a subset of is 
coherent. 
LEMMA 10.1. Suppose that is a tamely imbedded subset of X. 
Let with . Assume that for 
is an character of . Furthermore for 
. Then is coherent. Furthermore, if are 
extensions of to then either or and . 
. 
Proof. For let , then . 
Thus since . Furthermore is defined. Since 
is an isometry this yields that 
(10.1) . 
 810 SOLVABILITY OF GROUPS OF ODD ORDER 
In particular (10.1) implies that if then . By Lemma 
9.1 , therefore is the difference of two irreducible charac- 
ters of X. 
If . then it follows from equation (10.1) that 
if and . It is now a simple consequence of (10.1) that 
there exists a unique irreducible character of which is not orthogonal 
to any for . Furthermore if is chosen to be plus or 
minus this character then it may be assumed that 
for . 
Now define by 
. 
This implies that 
. 
Hence (10.1) yields that the generalized characters are 
pairwise orthogonal and that they each have weight one. It is easily 
shown that a rational integral linear combination of the characters 
of degree zero is a rational integral linear combination of the 
generalized characters . Hence if is the set of generalized 
characters . then the linear mapping sending into is 
an isometry. Thus, is coherent and the extension of to is 
unique in this case. 
If , define for by , where has 
weight one. Any rational integral linear combination of and of 
degree zero is a multiple of . Thus, if is any extension of to 
or for . The proof is complete. 
Before proving the main result of this section, another definition 
is needed. The following notation is introduced temporarily. 
Let be a subset of which consists of pairwise orthogonal 
characters. If , let denote the smallest weight of any 
character in of minimum degree. If and are coherent 
subsets of and and are extensions of to and re- 
spectively, define 
(i) . 
(ii) . where 1 
(a) , 
(b) is not orthogonal to . 
(c) . 
 10. COHERENT SETS OF CHARACTERS 811 
DBFINITION 10.2. Let be a coherent subset of and let 
be an extension of to . The pair is subcoheren t in 
if the following conditions are satisfied: If is any coherent sub- 
set of which is orthogonal to and if and are extensions 
of to and respectively, then 
(i) is orthogonal to . 
(ii) If , then is a sum of two generalized 
characters, one of which is orthogonal to and the other is in 
. 
If is subcoherent in . we also say that is subcoherent 
in , which causes no confusion in case has been designated. 
Hypothesis 10.1. 
(i) is a tamely imbeMed subset of the group X. 
(ii) For is a subset of . 
(iii) consists of pairwise orthogonal characters. 
(iv) For any wi th is coherent with isometry . 
is partitioned into sets such that each . either consists of 
irreducible characters of the same degree and or 
is subcoherent in where on . 
(v) For l there exist integers such that 
, 
. 
(vi) is an irreducible character of . 
(vii) For any in teger wi th . 
(10.2) . 
THEOREM 10.1. Suppose that Hypothesis 10.1 is satisfied. Then 
is coherent. There is an extension of to such that 
either agrees with on or . for 
. 
Proof. The proof is by induction on . If the theorem 
follows by assumption. 
It is easily seen that satisfies the assumption of the theorem. 
Hence by induction it may be assumed that is coherent. Let 
denote an extension of to , with the property that for 
 812 SOLVABILITY OF GROUPS OF ODD ORDER 
IS S It agrees with on l or and , 
. 
Choose the notation so that has minimum weight among the 
characters in of degree . Let be the subset which 
contains . For define 
.. 
Thus and is defined. Define the integer by 
(10.8) . 
If and I S , then by (10.8) 
(10.4) 
. 
Since is irreducible and is an isometry on 
(10.5) for . 
By (10.4) 
(10.6) 
where for I S S l Equations (10.5) 
and (10.6) now yield that 
(10.7) . 
If then since is an integer (10.2) and (10.7) imply that 
||'-|| '. 
Therefore 
(I0.8) ' if' . 
We will show that . By Hypothesis l0.I (iv), can be ex- 
tended from to a linear isometry on . For 1 S 
let . be the image of under this extension. If is sub- 
coherent in , then is orthogonal to . Suppose that 
consists of irreducible characters of the same degree. If 
is not orthogonal to , then there exists and 
for some and with 1 S , such that . Assume 
first that consists of irreducible characters of the same degree. 
 10. COHERENT SETS OF CHARACTERS 813 
Then it may be assumed that and 
. Thus for suitable . Hence 
. 
Hence . Therefore 
. 
which is not the case as . Suppose now that is sub- 
coherent in . Then . is orthogonal to by definition. There- 
fore, for , 
(10.9) 
Thus, is not orthogonal to . If consists of irreducible 
characters this yields that . Hence, by (10.8). Suppose 
that is subcoherent in . If implies that 
(10.10) 
where and is orthogonal to . By changing notation 
if necessary it may be assumed that 
(10.11) 
by (10.9). Now (10.9), (10.10) and (10.11) yield that 
(10.12) . 
Hence, (10.8) and (10.12) imply that in all cases. Thus, (10.3) 
becomes 
(10.13) . 
For , 
Therefore, (10.13) implies that 
(10.14) . 
For . define by 
(10.15) . 
and extend the definition to by linearity. This implies that 
or and for it . Hence, is 
 814 SOLVABILITY OF GROUPS OF ODD ORDER 
orthogonal to . and thus is an isometry on . The 
proof is complete. 
If is a coherent subset of , then will be used to denote 
an extension of to . 
Hypothesis 10.2. 
(i) is a tamely imbedded subset of and : is a supporting 
subgroup of . 
(ii) If is any non-principal irreducible character of and 
is the character of induced by , then is a sum of irreducible 
characters of , all of which have the same degree and occur with 
the same multiplicity in . 
LEMMA 10.2. Suppose that Hypothesis 10.2 is satisfied. For any 
character of : let be the set of irreducible characters of 
whose restriction to coincides with . If is a generalized 
character of which is orthogonal to for all with 
then is constant on the cosets of which lie in 
. 
Proof. We first remark that by Lemma 4.3 characters in 
vanish on . and so generalized characters in 
vanish on Suppose that are distinct characters in 
with By a ssumption . Thus by the 
Frobenius reciprocity theorem Hence by ypothesis 
10.2 . where is a class unction of induced by a class 
function of : and is a generalized character of Thus 
for . The proof is complete. 
LEMMA 10.3. Suppose that HypotheSi is satisfied. Let 
be a coherent subset of which consists of pairwise orthogonal 
characters of . Assume further that contains at least two 
irreducible characters. Then if is constant on the cosets 
of wh ich lie in . 
Proof. Suppose that are distinct irreducible characters of 
which do not contain : in their kernel such that We 
will show that 
(10.16) . 
By Lemma 4.3 and vanish on Since is a 
T. I. set in and the mapping sending into 
 10. COHERENT SETS OF CHARACTERS 815 
defines an isometry on . By Lemma 10.1 this 
can be extended to an isometry of . Let be the 
respective images of under this isometry. By assumption 
contains two irreducible characters and . Since 
for , Lemma 9.2 implies that if (10.16) is violated then 
for . 
Thus by the Frobenius reciprocity theorem 
for . 
Thus by changing notation if necessary it may be assumed that 
for , where the sign is independent of . Hence 
(10.17) . 
Since Lemma 9.2 implies that 
. 
Thus by the Frobenius reciprocity theorem 
contrary to (10.17). Therefore (10.16) must hold. The result now 
follows from Lemma 10.2. 
LEMMA 10.4. Suppose that the assumptions of Lemma 10.3 are 
satisfied. Let be the least common multiple of all the orders of 
elements in . If is an irreducible character in , then 
contains all the values assumed by . 
Proof. By assumption contains another irreducible character 
. Let be any automorphism of whose fixed field contains . 
Then since it follows directly from (9.4) that 
. 
Therefore 
. 
As , this implies that . As may be an 
arbitrary automorphism of whose fixed field contains the result 
is proved. 
 816 SOLVABILITY OF GROUPS OF ODD ORDER 
LEMMA 10.5. Suppose that is a tamely imbedded subset of . 
Let have the same meaning as in (9.2) and let be a generalized 
character of which is constant on for . Let be a 
coherent subset of consisting of irreducible characters. Then 
there exist rational numbers and generalized characters of 
which are orthogonal to such that if then 
if is orthogonal to and if . . 
Proof. It is an immediate consequence of Lemma 9.4 that if 
is orthogonal to and if where ranges over . 
then 
(10.18) for . 
where are rational numbers and is a generalized character of 
which is orthogonal to . If , then Lemma 9.4 yields that 
(19.19) for 
where are rational numbers and is a generalized character of 
which is orthogonal to . There exists a generalized character 
of which is orthogonal to such that 
. 
Since for (10.18) and (10.19) imply respectively that 
. 
The lemma follows by a suitable change in notation. 
It is worth noting that if the hypotheses of Lemma 10.3 are 
satisfied for every subgroup in a system of supporting subgroups of 
. then that lemma implies that satisfies the hypotheses of Lemma 
10.5. This fact will be used later in this paper. 
11. Some Applications of Theorem lO-l 
In this section we are concerned with the problem of finding 
conditions under which it is possible to apply Theorem 10.1. That 
theorem will then allow us to conclude that certain sets of characters 
are coherent. To clarify matters the main Hypothesis is stated 
separately. This also serves to introduce the notation. 
 11. SOME APPLICATIONS OF THEOREM 10.1 817 
Hypothesis 11.1. 
(i) is a tamely imbedded subset of th e group 
has odd order. and is a union of cosets of . Let 
and let be the image of in . 
(ii) and are normal subgroups of such that is nilpotent 
and 
(11.1) . 
(iii) is the set of all characters of which are by 
non principal irreducible characters of , each of which vanishes 
outsid e Then consists of pairwise orthogonal characters. 
(iv) There exists an integer such that for . 
Furthermore contains an irreducible character of degree . 
(v) Define an equivalence relation on by the condition that 
two characters in are equivalent if and only if they have th e 
same degree and the same weight. Then each equi valence class of 
is ei ther subcoherent in or consists of irreducible characters. 
(vi) For any subgroup of which is normal in let 
be the subset of consisting of those characters which are equiva- 
len t to some character in that has in its kernel. 
In the application to the main theorem of this paper (11.1) will 
always be augmented by one of the following conditions. 
(11.2) . 
(11.8) . 
(11.4) . 
THEOREM 11.1. Suppose that Hypothesis 11.1 is satisfied. Let 
be a normal subgroup of which is contained in such that 
(11.5) . 
If is coherent and contains an irreducible character of degree 
then is coh erent. 
Proof. Let be a normal subgroup of which is contained in 
and is minimal with the property that is coherent. Suppose 
that . Choose such that is a chief factor of 53. 
Let , where is irreducible and 
. Let . be the subsets of 
consisting of all characters of a given weight and a given degree. For 
let be the common degree of the characters in . 
 818 SOLVABILITY OF GROUPS OF ODD ORDER 
By Hypothesis 11.1 all the assumptions of Theorem 10.1, except possibly 
inequality (10.2), are satisfied for . We will now verify that 
also inequality (10.2) is satisfied. 
Let r be all the irreducible characters of which do not 
have . in their kernel. Let denote the character of induced by 
. Then each is in by Lemma 4.3. Furthermore if ranges 
only over characters of then 
. 
Therefore 
(11-6) . 
If then Suppose that for a given there are 
values of such that . Then (11.6) implies that 
(11.7) 
where the summation in (11.7) ranges over the distinct ones among 
the . Since 
, 
(11.7) yields that 
. . 
where or equivalently 
(11.8) . . 
Since is nilpotent is in the center of . Every irreducible 
character of is a constituent of a character induced by an irreducible 
character of . Thus for , Lemma 4.1 implies that 
, 
or equivalently 
(11.9) l 
Suppose now that inequality (10.2) is violated for some value of . 
Then (11.8) and (11.9) yield that 
. 
 11. SOME APPLICATIONS OF THEOREM 10.1 819 
Thus 
. 
or 
. 
Since every term is an integer this implies that 
(11.10) . 
However , thus . Now (11.5) and (11.10) are 
incompatible. Therefore inequality (10.2), and thus all the assumptions 
of Theorem 10.1, are satisfied. Hence by that theorem is co- 
herent contrary to the minimal nature of . This finally implies 
that . Therefore is coherent. The proof is 
complete. 
The remainder of this section consists of applications of Theorem 
11.1. Lemmas 11.1 and 11.2 are closely related to Theorem 2 of [8]. 
By using the argument of that theorem the assumption that is 
odd in the following lemmas can be replaced by suitable weaker as- 
sumptions. However the stronger results are not relevant to this 
paper and will not be proved here. 
Hypothesis 11.2. 
(i) Hypothesis 11.1 equation (11.2) are satisfied. Thus 
. 
(ii) is odd is a Frobenius group with Frobenius 
kernel . 
LEMMA 11.1. Suppose that Hypothesis 11.2 is satisfied. If 
then is coherent. 
Proof. By Lemma 10.1 and 3.16 (iii) is coherent. The 
result now follows from Theorem 11.1. 
LEMMA 11.2. Suppose th at Hypothesis 11.2 is satisfied. Then 
is coherent except possibly if is a non abelian -group for some 
prime and 
. 
 820 SOLVABILITY OF GROUPS OF ODD ORDER 
Proof. If , where and are proper normal sub- 
groups of . then 
(mM ) for . 
Since is odd, this implies that 
for . 
Hence and is coherent by Lemma 11.1. As 
is nilpotent this implies that is coherent if is not a -group 
for any prime . Since is odd 
. 
Thus by Lemma 10.1 is coherent if is abelian. The result now 
follows directly from Lemma 11.1. 
LEMMA 11.3. Suppose that Hypothesis 11.2 is satisfied and 
a Frobenius group with Frobenius kernel . Assume that is a 
-group for some prime and . Then is coherent. 
Proof. If . is abelian Lemma 11.2 implies that is coherent. 
If . is not abelian then the second term of the descending central 
series modulo the third is cyclic. Thus 
(mM ). 
Therefore as is Md. Hence 
and the result follows from Lemma 11.1. 
LEMMA 11.4. Suppose that Hypothesis 11.2 is satisfied and is 
a Frobenius group wi th Frobenius kernel . Assume that is a 
-group for some prime and . If 
(11.11) 
then is coherent. 
Proof. If is abelian Lemma 11.2 implies that is coherent. 
If is non-abelian let be a subgroup of such that 
is a chief factor of . As is nilpotent is in the center of 
- Thus by Lemma 4.1 the degree of any irreducible character of 
is either 1 or . Hence the degree of any character in 
 11. SOME APPLICATIONS OF THEOREM 10.1 821 
is either or . Let be the subsets of con- 
sisting of all the characters of degree respectively. 
Let . By (11.11) 
. . 
Thus by Theorem 10.1 is coherent. 
If or , then or 
(mM ). 
As this yields that in either case 
(mod ). 
Therefore . Hence 
and is coherent by Lemma 11.1. Suppose that . 
Then by (11.11) 
. 
Since is coherent the result now follows from Theorem 11.1. 
The next two lemmas involve the following situation: 
Hypothesis 11.3. 
(i) Hypothesis 11.2 is satisfied. 
(ii) There exist primes and positive integers such that 
. Thus is a power of . 
LEMMA 11.5. Suppose that Hypothesis 11.3 is satisfied and 
is even. Then is coherent except possibly if is 
the smallest degree of any non linear irreducible character of whose 
kernel contains and for no subgroup of with , 
is a Frobenius group. 
Proof. Suppose that is not coherent. Then by Lemma 11.1 
. As it follows that or 
. If this implies that contrary 
to what has been proved above. Therefore . 
Let . be the set of non principal irreducible characters 
of of degree . Lemma 4.1 implies that is empty for 
. Let be the set of characters in of degree . 
 822 SOLVABILITY OF GROUPS OF ODD ORDER 
Since . it follows from Hypothesis 11.1 and 
Theorem 10.1 that is coherent- Suppose is non 
empty. Then 3.15 implies that 
. 
Therefore 
. 
Thus by Theorem 10.1 
is coherent. Since 
. 
Theorem 11.1 implies that is coherent. Thus it may be assumed' 
that is the smallest degree of any non linear irreducible character 
of . 
Suppose now that contains a subgroup such that 
is a Frobenius group. Then may be chosen so that is a chief 
factor of . Thus and by the earlier part of the lemma 
every irreducible character of has degree either 1 or As. 
is the smallest power of which satisfies (mod ). 
Since is a chief factor of this implies that is in the 
center of and . If is an irreducible character of 
of degree , then the orthogonality relations yield that 
for . As every non linear character of has degree 
the orthogonality relations may once again be used. They 
that 
(11.13) for . 
However 
which contradicts (11.13). Thus contains no subgroup such 
that is a Frobenius group. All statements in the lemma are 
proved. 
LEMMA 11.6. Suppose that Hypothesis 11.3 is satisfied. 
further that is odd and . Then is coherent. 
 12. FURTHER RESULTS ABOUT TAMELY IMBEDDED SUBSETS 823 
Proof. As is Md and (mod 8). it follows that (mod 8). 
Define the integer by 
(mM ) (mod ). 
If . then . Thus if and is 
Soherent by Lemma 11.1. If , then is cyclic. Therefore 
is coherent by Lemma 10.1. 
Suppose now that . Then since (mod ) we must 
have for some integer . Therefore 
. 
Since (mod ). this yields that 
. 
If 4.8" then is coherent by Lemma 11.1. Thus if is 
not coherent (11.14) implies that 
. 
Therefore . Hence or . In either case this implies 
that . As (mod 8) we get that . How- 
ever (mod 8). This contradiction arose from assuming that 
is not coherent. The proof is complete. 
12. Further Results about Tamely Imbedded Subsets 
In this section a fairly special situation is studied. Our purpose 
here is to get some information about certain sets of characters which 
may not be coherent. 
Hypothesis 12.1. 
(i) Let be a prime and let be a -subgroup of the group 
X. Assume that is tamely imbedded in and 
has odd order. Let and let . 
(ii) is the set of all characters of which are induced by 
-principal irreducible characters of . Define as equ ivalence 
on by the condition that two characters are equivalent 
if and only if they have the same degree and the same weight. Then 
each equivalence class of is either subcoherent in or consists 
of irreducible characters. 
(iii) Let r be all the integers which are degrees 
of irreducible characters of . Let be a fixed integer. For 
let . be the set of all characters in of degree 
 824 SOLVABILITY OF GROUPS OF ODD ORDER 
. Assume that each consists of irreducible characters. 
Let be an equivalence class in consisting of characters of 
degree . Let . 
In case Hypothesis 12.1 is satisfied the following notation will be 
used. 
(12.1) . 
Since is odd, and for . Thus by 
Lemma 10.1 is coherent for . 
For let be the number of non principal irreducible 
characters of of degree . By Hypothesis 12.1 acts regularly 
as a permutation group on the non principal irreducible characters of 
degree : for . Since is odd, no non principal irreducible 
character of is real. Thus even. Therefore 
(12.2) (mod ) for . 
Let . Define inductively to be the largest integer not 
exceeding such that is coherent. Suppose that 
. 
For , define 
and let . Define 
(12.4) for . 
where ranges from . to . Define 
(12.5) for . 
Then by Theorem 10.1 applied to 
for . 
By (12.2) 
(12.7) - (mod ) for . 
By 8.15 
(12.8) (mod ) for . 
 12. FURTHER RESULTS ABOUT TAMELY IMBEDDED SUBSETS 825 
LEMMA 12.1. Suppose that Hypothesis 12.1 is satisfied. Assume 
that 
. 
Them 
for . 
Furthermore if is odd, and (mod ). then 
. 
Proof. We will first prove that 
(12.9) for . 
This is true if since Suppose that Then by (12.5) 
and (12.6) 
. 
Assume now that the lemma is false and choose minimum to 
violate the result. Let . 
By (12.8) and (12.9) 
. 
Hence by (12.5) 
(12.10) . 
Inequalities (12.6) and (12.10) yield that or . 
This implies that 
. 
Conseq uently 
(12.11) . 
Suppose that 
 826 SOLVABILITY OF GROUPS OF ODD ORDER 
. 
Then by (12.9) 
. 
Hence by (12.7) . Thus 
since . This contradicts the choice of . Hence 
. 
As is even for . implies that 
(12.12) . 
The group contains a normal subgroup- of index . Every 
irreducible character of has degree strictly less than and 
the sum of the squares of the degrees of these characters is equal 
to . Hence (12.12) implies that every character of whose 
degree is strictly less than has in its kernel. Thus is a 
normal subgroup of and is a Frobenius group with Frobenius 
kernel . Therefore 
(12.18) (mod ), 
and the center of has order at least . Thus by Lemma 4.1 
. This yields that 
(12.14) . 
Define the integer by 
(12.15) . 
By (12.10) and by (12.12) . Thus 
(12.16) . 
Define the integer by 
(12.17) (mod ) . 
Equations (12.7), (12.18), (12.15) and (12.17) imply that 
(12.18) (mod ). 
 12. FURTHER RESULTS ABOUT TAMELY IMBEDDED SUBSETS 827 
If . then by (12.16) and (12.18) - Thus by (12.15) 
. hence by (12.7) 
. 
If and is odd, then 
. 
Thus by (12.18) . However this is impossible as is odd. 
Assume now that . As is a power of . (12.14) and (12.18) 
imply that either or . Since . the latter case 
leads to 
. 
Hence contrary to (12.11). Thus 
(12.19) . 
The inequality follows from (12.17) and the fact that is even. 
Now (12.11) and (12.19) yield that 
. 
Thus . (12.16) and (12.18) imply that 
(12.20) . . 
Equation (12.15) now becomes . Hence 
(mod ). 
Furthermore by (12.19) 
. 
Consequently 
. 
THEOREM 12.1. Suppose that Hypoth esis 12.1 is satisfied. Assume 
that for some with and Define 
. 
Suppose that and 
 828 SOLVABILITY OF GROUPS OF ODD ORDER 
Where and or.thogonal to Then 
. 
Furthermore if is . and (mod ) then 
. 
Proof. Let . If then and 
. Thus the result is trivial in this case. Hence it may be assumed 
that . In particular, is not coherent, hence is not 
coherent, so by Lemma 11.1 Consequently Lemma 
12.1 may be applied. Furthemore . and . Thus 
consists of irreducible characters. Let , where 
the notation is chosen so that and for . 
Suppose that . Define the integer by . Then 
since Lemma 9.4 implies that 
for . . 
Then 
. 
Therefore 
. 
where . is defined by (12.4). Let be defined by (12.5). 
Since and (12.21) yields that 
(12.22) . 
As a function of assumes its maximum at . 
Thus (12.22) implies that 
(12.23) . 
As is an integer Lemma 12.1 and (12.23) imply that 
. Furthermore if is odd, and (mM ) then 
. 
 13. SELF NORMALIZING CYCLIC SUBGROUPS 829 
The proof is complete. 
13. Self Normalizing Cydic Subgroups 
Hypothesis 13.1. 
(i) is a cyclic subgroup of the group with odd. 
Suppose that . where and for . 
Let 
. 
For any non empty subset of 
. 
(ii) Let be faithful irreducible characters of 
respecti vely. Define 
. 
for . 
If in Hypothesis 13.1 are both primes then (13.1) follows from 
the assumption that . Thus the situation described above 
is a generalization of this case. 
LEMMA 13.1. Suppose that Hypothesis 13.1 is satisfied. Then 
is T. I. set in X. There exists an orthonormal set 
of generalized characters of such that for 
, the values assumed by lie 
in respectively. and 
for . 
. 
Furthermore every irreducible character of distinct from all 
vanishes on . 
Proof. It follows directly from Hypothesis 13.1 that is a T. I. 
set in X. Define the generalized character of by 
. 
Clearly vanishes on . Thus 
for W e . 
(13-2) 
 830 SOLVABILITY OF GROUPS OF ODD ORDER 
for . Therefore and 
for . It follows directly from the definition 
of that the values of lie in . 
For any algebraic number field and any generalized character 
of a group let denote the field generated by and all the 
values assumed by . Since we see that 
for some with . If then , where and 
for . By (18.2) 
, 
where are distinct irreducible characters of X. 
Suppose that for . Let 
. 
Let be the Galois group of over . For let be 
the subgroup of whose fixed field is . 
Assume first that . By (18.2) for 
. If for some then contrary to 
assumption. Let and for . Then it may 
be assumed that . Since 
we must have . Therefore 
, . 
Hence 
. 
Therefore is not cyclic as it is the union of proper sub- 
groups. Hence is the non cyclic group of order 4 and 
for 8. As is odd this implies that . For 
let . where the notation is chosen so that . 
Therefore . Hence . Consequently 
as is abelian. This implies that which is not the 
case. Thus . 
If , then by (18.2) Hence by 
choosing the notation suitably it may be assumed that . If 
then replacing by and by if necessary 
we get that 
. 
By (18.2) . Hence also . Therefore 
 13. SELF NORMALIZING CYCLIC SUBGROUPS 831 
since and are all characters. This contradiction 
establishes that . Since we see that 
(13.3) . 
Furthermore and if then By 
definition for . Therefore 
for . 
Suppose that for some automorphism of . Then 
and (13.3) implies that . Thus by (13.2) 
. Consequently and so 
(13.4) . 
If now , then (13.3) yields that 
. 
Therefore and . Thus . Since 
and we get that . If then Thus 
(13.2) and (13.3) yield that Since is 
even we get that . Thus either and or . 
In the latter case (13.2), (13.3) and (13.4) imply that . Thus 
or equivalently contrary to assumption. 
Suppose now that . Thus and the previous argument 
with and interchanged yields that for or . 
Thus by (13.4) By (13.2) and (13.3) Thus 
since . Let be in the Galois group of 
over . Then and can be chosen so that 
. 
Hence (13.3) yields that 
. 
Since is not conjugate to this implies that 
contrary to the fact that and are all characters. 
Thus in any case there exists a non principal irreducible character 
 SOLVABILITY OF GROUPS OF ODD ORDER 
of such that and - Suppose that 
. Since is odd 
. 
Therefore 
. 
Hence 
. 
Since and are characters this yields that for . 
Hence and so . Consequently . 
which is not the case. Therefore 
(13.5) . 
Similarly there exists an irreducible character of with 
and . Thus by (13.5) . Now (13.2) yields 
that 
(13.6) 
for . The are distinct irreducible 
characters of whose values lie in the required field. Suppose now 
that 
with . Then by the Frobenius reciprocity theorem it follows 
from (13.6) that 
, 
S 
. 
Consequently for 
. 
In a similar way it can be shown that . Then it 
follows from (13.6) that for . 
This implies that if then 
 13. SELF NORMALIZING CYCLIC SUBGROUPS 833 
. 
The orthogonality relations for the irreducible characters of now 
yield that every irreducible character of distinct from all 
vanishes on . This completes the proof of the lemma. 
LEMMA 13.2. Suppose that Hypothesis 13.1 is satisfied. If is 
a generalized character of which vanish es on then 
. 
where for . 
Proof. Let 
, 
where for . By Lemma 13.1 
for . 
Hence 
for . 
This implies the desired result. 
Hypothesis 13.2. 
(i) The group satisfies Hypothesis 13.1. 
(ii) contains a normal subgroup such that 
and if is a non empty subset of then 
. 
Since is a -subgroup of , Hypothesis 13.2 (ii) implies that 
is a -subgroup of . Also, if ', then . 
LEMMA 13.3. Suppose that satisfies Hypothesis 13.2. Then 
is T. I. set in . For there 
exist irreducible characters of such that 
 834 SOLVABILITY OF GROUPS OF ODD ORDER 
. 
where is a set of integers depending on the sign depends 
only on . 
Proof. Hypothesis 13.2 implies that is a T. I. set in . 
For vanishes on . Define 
for . 
Then by Lemma 10.1 is coherent for . Let 
where the sign is chosen so that . Then 
for . 
The Frobenius reciprocity theorem now implies the required result 
since vanishes on . 
LEMMA 13.4. Suppose that satisfies Hypothesis 13.2. Let be 
an irreducible character of . Then there esists an integer such 
that 
. 
for some with . 
Proof. Let be the characters defined in Lemma 13.3. If 
for some with then the 
result follows from Lemma 13.3. Furthermore Lemma 13.3 implies 
that 
for . 
Hence if for all we have that for . This 
completes the proof of the lemma. 
We will use the fact that Lemma 13.4 is valid over fields of 
characteristic prime to , provided that is absolutely irreducible. 
LEMMA 13.5. Sttppose that satisfies Hypothesis 13.2. For 
 13. SELF NORMALIZING CYCLIC SUBGROUPS 835 
let be the characters defined by 
Lemma 13.3. Define 
for . 
Then is iruluced by an irreducible character of R. Further- 
more 
for . 
Proof. By Lemma 13.4 the characters 
are the only irreducible characters of which do not 
vanish on . Since each agrees on with a suitable linear 
character of it follows from Lemma 13.1 that 
is the set of irreducible characters of . Therefore agrees 
with on . Hence Lemma 13.1 implies that . Con- 
sequently if then 
for . 
Thus the Frobenius reciprocity theorem implies that is a constituent 
of for all values of . Since 
the lemma is proved. 
LEMMA 13.6. Suppose that satisfies Hypothesis 13.2, is a 
prime, is an estra special -group with . Let 
. Then divides either or . 
Proof. It is easily seen that a faithful irreducible character of 
has degree . Thus by Lemmas 13.4 and 13.5 
. 
This proves the result. 
LEMMA 13.7. Suppose that satisfies Hypothesis 13.2. Let 
be defined by Lemma 13.5. Then an irreducible character of 
either induces an irreducible character of or it irulruluces for some 
with . 
Proof. The group acts as a permutation group on the conjugate 
classes of R. If and leaves some conjugate class of fixed, 
 836 SOLVABILITY OF GROUPS. OF ODD ORDER . 
then since is a Hall subgroup of must centralize. some element 
of this conjugate class. Hence by assumption ffie only 
classes of which are fixed by any ' are those containing an 
element of . There are at most of these. The group also 
acts as a permutation group on the irreducible characters of . There- 
fore by 3.14 there are at most irreducible characters of which 
are fixed by any element . By Lemma 13.5 the distinct 
characters are fixed by every and these induce 
. Thus every other irreducible character of induces 
an irreducible character of . The proof is complete. 
Hypothesis 13.3. 
(i) is a tamely imbedded subset of the group and 
has odd order. 
(ii) satisfies Hypothesis 13.2, and satisfies Hypothesis 13.1 
with the same group . 
(iii) contains a normal ni lpotent subgroup such that 
. 
. 
. 
(iv) There exist subgroups such that 
is a system of supporting subgroups of and . Let 
for . 
(v) For let be defined 
respecti vely by Lemmas 13.1, 13.3 and 13.5. 
(vi) Let be the set of characters of which are induced by 
non principal irreducible characters of , each of which vanishes 
outside . 
LEMMA 13.8. Suppose that Hypothesis 13.3 is satisfied. Assume 
that for some with 
. Then vanishes in 
. 
Proof. By Lemma 13.3 . do not contain in their kernel, 
thus they do not contain in their kernel. Hence by Lemma 4.3 
vanish on . By Lemma 13.3 . Thus 
vanishes on . Hence . By Lemmas 
9.1 and 13.8 
 13. NORMALIZ.ING C.YCLIC 8.UBGROUPS 837 
. for . . 
the from Lemma . . . 
LEMMA 13.9. Suppose that Hypothesis 13.3 is satisfied. Choose 
with . Let. 
. 
Then is coherent and 
is an extension of to where either or . 
Proof. Since is odd Hence By Lemma 
13.5 
. 
Hence Lemma 13.8 yields that 
. 
By Lemma 9.1 vanishes on . Thus Lemma 13.2 implies 
that 
(13.7) . 
Now define 
Wh.ere the sign is the same as in (13.7). It is easily seen that is a 
linear isometry on . Thus is coherent. 
LEMMA 13.10. Suppose that Hypothesis 13.3 is satisfied. Let 
have the same meaning as in Lemma 13.9. Then is sub- 
coherent in where is defined as in Lemma 13.9. 
Proof. By Lemma 13.9 is coherent. Let be a coherent 
subset of which is orthogonal to . Let be an extension of 
to . 
Every generalized character in vanishes on . Thus by emma 
9.1 every generalized character in vanishes on If is 
 838 SOLVABILITY OF GROUPS OF ODD ORDER 
an irreducible character in . then as is odd. Further- 
more and thus vanishes on . Hence 
for . Therefore is orthogonal 
. If . then since is a linear combinatiort 
of and . with . Hence is orthogonal to 
Consequently is orthogonal to . 
Suppose now that with , where . 
is not orthogonal to and . Let . 
where is a linear combination of the generalized characters and 
or . Let be the set of 
integers such that . Lemma 13.8 implies that every gener- 
alized character in is orthogonal to for . 
Let , where is a linear combination of . with and 
for . Then 
(13.8) . 
By changing notation it may be assumed that and 
. By Lemma 9.4 
. 
Hence is a non zero integral multiple of . By (13.8) 
. 
Therefore 
(13.9) . 
By Lemma 13.2 
, 
where is as in Lemma 13.9 and where for 
. Now (13.9) yields that . Thus (13.8) and 
(13.10) imply 
. 
Every term in the second summation is non zero. Thus for 
. Hence or . Hence (13.8) and (13.10) 
yield that or . This shows that is subcoherent 
in and completes the proof of the lemma. 
In the proof of the main theorem of this paper we will reserve 
the letter to denote the extension of to defined by Lemma 
13.9. Thus will always be subcoherent in . 
 13. SELF NORMALIZING CYCLIC SUBGROUPS 839 
DEFINITION. A Z-group is a group all of whose Sylow subgroups 
are cyclic. 
Hypothesis 13.4. 
(i) with and solvable. Furthermore 
is a cyclic S-subgroup of is odd. 
(ii) For . Furthermore is a Z-group 
and . 
(iii) is faithfully irreducibly represented on a vector space 
over a field of characteristic not dividing contains a 
vector space of dimension at most 1 such that if 
then if anti only if . 
LEMMA 13.11. Suppose that Hypothesis 13.4 is satisfied. Then 
is nilpotent. Furthermore is a prime anti the representatim 
of on is absolutely irreducible. 
Proof. Let be the character of the representation of on . 
Let be a -subgroup of which is normalized but not centralized 
by . Then either or satisfies Hypothesis 13.2. Thus 
by Lemma 13.4 only one absolutely irreducible constituent of is 
not linear. Hence is absolutely irreducible. Furthermore Lemma 
13.4 and 3.16 (iii) imply that has , as a constituent. Thus 
is a prime. 
The nilpotence of is proved by induction on . We assume 
without loss of generality that the underlying field is algebraically 
closed. If then contrary to assumption. Thus by 
3.3 . Let be a minimal nilpotent normal subgroup of 
which is not centralized by . Then is a -group for some prime 
. Furthermore and . By Lemma 13.4 there 
is exactly one non linear irreducible constituent of . Let 
. 
where each is a linear character of . Assume first that . 
If is an irreducible constituent of . then Since 
. for we have Since is a sum of 
conjugate characters this implies that is abelian and the are 
distinct. Thus . where and is a Frobenius 
group. For let . If such that 
for some then since is the kernel of each Since 
permutes the constituents of transitively this implies that 
acts transitively on {. } Hence is odd. Thus 
 GROUPS OF ODD ORDER 
is even contradicting the absolute irreducibility of X. Therefore 
and is irreducible. 
By Lemma 13.4 this implies that or . 
If then is reducible since . As is a 
prime this implies that is a sum of linear characters and is 
abelian. Thus we can suppose that . By Lemma 13.4 
is irreducible. Thus if is any proper -invariant subgroup of 
with then satisfies the induction assumption and is 
nilpotent. If with then since is irreducible, 
. If is not a -subgroup of then is a proper sub- 
group of where is a -invariant -complement in . Thus 
and is nilpotent. Suppose now that is a -subgroup 
of . 
Since is cyclic. Let be the subgroup of 
index in . Then is a -group of class 2 and hence is a 
regular -group. If does not have exponent then there exists 
a characteristic subgroup of of index which is normal in but 
is not centralized by contrary to the minimality of . Thus 
has exponent . Therefore acts without fixed points on as 
is cyclic and . 
Let be a chief factor of with . Suppose first that 
does not centralize . Then is a Frobenius group which 
is represented on . As has no fixed points on Lemma 
4.6 implies that acts trivially on . Thus is 
nilpotent. Assume now that is abelian. Then for 
some prime . If is represented faithfully on . the 
minimal nature of implies that is represented irreducibly on 
. Let . Then acts without fixed points on 
Since is irreducible, . Thus . 
Hence . We will now reach a contradiction from the fact 
that . Let . Then . Thus is abelian. 
Let be the linear character of such that for 
. Let . Then and is an 
irreducible character of . The group . satisfies Hypothesis 13.2 
where is the normal subgroup. Thus by Lemma 13.4 no 
irreducible character of has degree . This completes 
the proof of the lemma in all cases. 
DEFINITION. Let and be subgroups of a roup with. 
. We say that is prime on if . 
. . . . for . 1 
I . . . . . . . .. 
. If is a then is necess.arily prime on 
. 
 13. SELF NORMALIZING CYCLIC SUBGROUPS 841 
LEMMA. 13.12. Let with. solvable, cyclic, 
and odd. that is prime 
is a Z-group. If then is nilpotent. If 
furthermore is not a prime then is nilpotent. 
Proof. Let be a counter example to the esult for which 
has minimum order. Since the hypotheses are satisfied 
by all -invariant factor groups of.U. 
Suppose that is not a prime. Let be a minimal normal 
subgroup of . Then is a -group for some prime and . 
By induction is nilpotent. If is a -invariant -group of 
for then and has no fixed points on 
. If is not nilpotent then it is possible to choose so that 
is not nilpotent. Let . Then is faithfully 
represented on . Hypothesis 13.4 is satisfied with in the role 
of . Thus by Lemma 13.11 is a prime contrary to assumption. 
Assume now that is a prime. Suppose that contains two 
distinct minimal normal subgroups and . For let 
be the inverse image of in . By induction is nilpotent 
for . The result now follows from the fact that 
. Thus it may be assumed that contains a unique minimal 
normal subgroup . Then for some prime . Let 
. Then is a -group. Thus the result follows 
by induction if . Assume now that . Then . 
Let be a -invariant -subgroup of . Then is faith- 
fully represented of . Hypothesis 13.4 is satisfied with in place 
of unless . Thus by Lemma 13.11 is nilpotent or 
. 
Let and let be -invariant -group of . If 
then is nilpotent since it is a -group and the result 
is proved. Assume that . By induction is nilpotent. 
Hence does not centralize by assumption. 
Let be a -group in which is minimal with the property 
that normalizes but does not centralize . Since is a 
group there is a prime such that contains no normal -sub- 
group, where jg a -group of Thus acts faithfully on 
Let As is faithfully represented on Lemmas 
4.6 and 13.4 imply that . Let . As is rep- 
resented faithfully on Lemmas 4.6 and 13.4 imply th.at 
. . Thus is a Z-group. and . 
. . . . 
. 
 842 SOLVABILITY OF GROUPS OF ODD ORDER 
By 3.11 is a special -group and . Thus is 
cyclic. By Lemma 13.11 the representation of on has a 
unique faithful irreducible constituent and this constituent is absolutely 
irreducible. Let be the character of this constituent. If 
then by Lemma 13.4 remains absolutely irreducible. Hence 
(mod ) contrary to (13.11). Therefore is an elementary 
abelian group and is a Frobenius group. Thus is a 
prime. If is not cyclic then is reducible in the field of 
elements as is faithful. Thus (mod ) contrary to (13.11). 
Therefore is a cyclic group of order and . is a Frobenius group. 
Hence 
(13.12) (mod ). 
Let be a invariant subgroup of which is minimal sub- 
ject to Thus the represenffition of on is 
irreducible. Therefore Since is lementary and 
we get that the hypotheses of the lemma are satisfied. 
Thus the minimal nature of implies that and . 
Therefore the representation of on is irreducible. Let 
be a minimal normal subgroup of which is not centralized by 
. Thus . Then and . Hence 
is cyclic. Let be the character of the representation of on . 
By Lemma 13.4 has exactly one irreducible cons ituent which does 
not have in its kernel. Let be this constituent and let 
. 
Since each is a character of a group of exponent it follows 
from (13.11) and (13.12) that each is absolutely irreducible. Thus 
for . By Lemma 13.11 is absolutely irreducible 
in the field of elements. By Lemma 13.4 . Since 
is odd (B) and (13.12) yield that 
(13-13) . 
Thus Let , where each is an irreducible char- 
acter of Thus 
(13.14) . 
Since is a set of conjugate characters- Since 
they are a 11 linear. Thus Hence . 
where is a Frobenius group and . Furthermore 
 13. SELF NORMALIZING CYCLIC SUBGROUPS 843 
(13-15) . 
Since for we see that for all 
Since for we get that no constituent of occurs 
with multiplicity greater than one. Since is a set of distinct 
linear characters of we et that Now (13-13). (13.14) and 
(13.15) yield that 
. 
This contradicts (13.11) and (13.12) since is odd. The proof is 
complete. 
 . . .. 
. 
. Statement of Results Provecl in Chapter IV 
In this chapter, we begin the proof of the main theorem of this 
paper. The proof is by contradiction. If the theorem is false, a 
minimal counterexample is seen to be a non cyclic simple group all 
of whose proper subgroups are solvable. Such a group is called a 
minimal simple group. Throughout the remainder of this chapter, 
is a minimal simple group of odd order. We will eventually derive 
a contradiction from the assumed existence of . 
. In this section, the results to be proved in this chapter are summar- 
ized. Several definitions are required. 
Let be the subset of consisting of all primes such that 
if is a -subgroup of . then either is empty or 
contains a subgroup of order such that where . 
is cyclic. Let be the subset of of those such that 
if is a -subgroup of and is the order of a cyclic subgroup 
of , then one of the following possibilities occurs: 
(i) divides . 
(ii) is abelian and divides . 
(iii) and divides . 
We now define five types subgroups of . The basic property 
shared by these five types is that they are all maximal subgroups of 
. Thus, for II, III, IV, , any group of type is by definition 
a maximal subgroup of . The remaining properties are more detailed. 
We say that is of type I provided 
(i) is of Frobenius type with Frobenius kernel . 
(ii) One of the following conditions is satisfied: 
(a) is a T. I. set in . 
(b) . 
(c) is abelian and . 
(iii) If . then and a of is 
abelian. 
The remaining types are by definition three groups. If 
is a three step group, we use the following notation: 
: .. 
. .. 
. : . 
. . 
. . . . . . 
-. 
denotes normal S.-subgrolup of 
. By definition, so we let be a complement for in , 
 SOLVABILITY OF GROUPS OF ODD ORDER 
In addition to being a three step group, each of the remaining four 
types has the property that if is any non empty subset of 
. then , by definition. The remaining 
properties are more detailed. 
We say that is of type II provided 
(i) and is abelian. 
(ii) . 
(iii) for every non empty subset of such that 
. 
(iv) is a prime. 
(v) For every prime , if are cyclic -subgroups of 
which are conjugate in but are not conjugate in , then either 
or . 
(vi) is a T. I. set in . 
We say that is of type III provided (ii) in the preceding defi- 
nition is replaced by 
, 
and the remaining conditions hold. 
We say that is of type IV provided (i) and (ii) in the definition 
of type II are replaced by 
, 
. 
and the remaining conditions hold. 
We say that is of type provided 
(i) . 
(ii) One of the following statemiments is true: 
(a) is a T. I. set in . 
(b) , where is cyclic and is a -subgroup of 
with . 
THEOREM 14.1. Let be a minimal simple group of odd order. 
Two elements of a nilpotent S-subgroup of are conjugate in 
if only if they are conjugate in . Either (i) or (ii) is true: 
(i) Every maimal subgroup of is of type I. 
(ii) (a) contains a cyclic subgroup with the 
property that for every non empty subset of . 
Also, . 
(b) contains maximal subgroups not of type I 
such that 
. ., . 
 14. STATEMENT OF RESULTS PROVED IN CHAPTER IV 
Every maximal subgroup of is either conjugate to or 
or is of type I. 
(d) Ei th er or is of type . 
(e) Both and are of type , or V. (They are 
not necessarily of the same type.) 
In order to state the next theorem we need further notation. If 
is of type , let 
, 
where is the Frobenius kernel of . 
If is of type II, III, IV, or V. we write . . 
Let be the maximal normal nilpotent -subgroup of . let be a 
complement for in and set . . 
. 
If is of type II, let 
. 
If is of type III. IV. or V. let 
. 
If is of type II, III, IV, or , let 
. 
We next define a set of subgroups associated to . 
Namely. if and only if is a maximal subgroup of and 
there is an element in such that and . Let 
be a subset of X. which is maximal with the property 
that and are not conjugate if . For , let be 
the maximal normal nilpotent -subgroup of . 
THEOREM 14.2. If is of type , or , then 
are tamely subsets of wi th 
. 
If is empty, are T. I. sets in . If is non 
empty, the subgroups . are a system of supporting subgroups 
for and for . 
The purpose of Chapter IV is to provide proofs for these two 
theorems. 
 848 SOLVABILITY OF , GROUPS OF ODD ORDER 
15* A Partition of TC(®) 
We partition TT(®) into four subsets, some of which may be empty: 
7T! = {p | A Sp-subgroup of © is a non identity cyclic group.} 
K* = {P 1 1. A Sp-subgroup of © is non cyclic. 
2. © does not contain an elementary subgroup of order p3.} 
;r3 '= {p\ 1. ® contains an elementary subgroup of order p3. 
2. If ^5 is a Sp-subgroup of , then kl(^P) contains a non 
identity subgroup.} 
7T4 = {p 1 1. ® contains an elementary subgroup of order p3. 
2. If ^P is a Sp-subgroup of , then M(^P) contains only 
It is immediate that the sets partition TT(©). The purpose of Lemma 
8.4 (i) is that condition 2 defining 7T2 is equivalent to the statement 
that &i4Ł(ty) is empty if ^P is a Sp-subgroup of . Lemma 8.5 
implies that 3 g ^ U ?r2. 
16. Lemmas about Commutators 
Following P. Hall [19], we adopt the notation 75133 = [21, 33], 
7w+12I33n+1 = [7tt2N3n, ], n = 1, 2, • • • , and 722W3E = [21, 33, (Ł]. 
If X is a group, ^3^(Ł) denotes the set of normal abelian 
subgroups of X. 
The following lemmas parallel Lemma 5.6 of [27] and in the 
presence of (B) absorb much of the difficulty of the proof of Theorem 
14.1. 
LEMMA 16.1. Let ty be a Sp-subgroup of ® and 2t an element of 
cx43^(^P). If % is a subgroup of ® such that 
(i) <2l»3> is a p-groupt 
(ii) g centralizes some element of Z(ty) n 21*, 
then 73g2l3 = 
Proof. Let Ze C(f$) n Z($) n 21*, and let E = C(Z). By Lemma 
7.2 (1) we have 21 s Op.,p(S) = . As ^P is a Sp-subgroup of (Ł, ^ = 
^? 0 ^> is a Sp-subgroup of >. Since 2t < ^P, so also 21 < ^3lf and since 
21 is abelian, we see that 72Ł>2I2 Ł Op>((Ł). Since § < E, we have 
7g2t s ^> and so 73g2l3 S O,,((E). Since <2I, g> is assumed to be a p- 
group, the lemma follows. 
If ^P is a non cyclic p-group, we define ^(ty) as follows: in case 
Z(^P) is non cyclic, <&(?$) consists of all subgroups of Z(^P) of type 
(p, p)\ in case Z(^P) is cyclic, ^(^P) consists of all normal abelian 
subgroups of ^P of type (p, p). 
LEMMA 16.2. Let ^ be a non cyclic SP-subgroup of , 21 e 
 LEMMAS ABOUT COMMUTATORS 
amdiet be a subgr0pp such that ..11 
(i ) is p-group, 
(ii) comtaims a subgr0up of such that . 
If , while if , them . Also, 
if , them . 
pro0f. If , the lemma follows from Lemma 16.1. If 
. then is of index in so is of index at most 
in a suitable of . In particular, . 
Let , and . Since , 
so . Hence . and SO 
being in . If . we conclude from (B) that . and 
SO . Since the lemma follows in this case. 
(Since centralizes . we have r.) 
Suppose now that . If . then and SO 
by (B). and the lemma follows. If , then . 
since . In this case, letting , we 
see that and so . that is, 
. Hence, and since , we see that . 
and SO . Continuing, we see that and SO 
, from which the lemma follows. 
LEMMA 16.8. Let be of let D. 
Let be a subgr0up 0f such that 
(i ) is 8-gr0up. 
. 
If , . 
proof. First suppose . Let . Since 
is of (B) implies that . Setting 
, we have . If . then 
and SO , since is a 8-group. If 
, then the definition of implies that . 
SO if , we must have for suitable . in 
. By definition of it that . 
We can suppose now that . In this case, the 
of implies that where 
and let be of containing and 
. Since is of in.dex at most 8 in and since 
centralizes , we have , and SO - If , 
it follows and we are done. Hence. 
we can suppose In it follows . 
 850 SOLVABILITY OF GROUPS OF ODD ORDER 
since | <P* : ^0 1 = 3. We also have D(W) S ^Po, and so < Ł C 
@. If e s ^Po, we have ( Ł Z((Z)f and since Z(8) char 6 char W, it 
follows that 72ge2 Ł <V($) 0 <Ł, g> = <1> and we are done. We can 
therefore suppose that ( Ł Z(C). Choose # in 6 - C@(<Ł). Since ^P* 
centralizes G^ it follows that E does not centralize S) = <D>. Consider 
[A #] = F * 1. Now < Ł Z(^P0) < $*, and so Fe Z(*P0). On the 
other hand, F lies in Z>(W) since both # and D are in W- Since 
Ee(S9 it follows that E centralizes F. Since <$p0f E> = *p*, it follows 
that F is in Z(?P*). But F is of order 3 and ^ = ^(ZOP*)), since 
Z(SP*) is cyclic. It follows that <F> = C^, and so 7 normalizes E and 
with respect to the basis (D, F) of ( has the matrix (A j). On the 
other hand, $P possesses an element which normalizes ( and with 
respect to the basis (D, F) has the matrix (| ?) Since these two 
matrices generate a group of even order, we have the desired con 
tradiction which completes the proof of this lemma. 
17* A Domination Theorem and Some Consequences 
In view of other applications, Theorem 17.1 is proved in greater 
generality than is required for this paper. 
Let ^P be a Sp-subgroup of the minimal simple group X and let 
21 be an element of &^^(?$). Let q be a prime different from p. 
THEOREM 17.1. Let Q, jQj be maximal elements of M(2l; 
(i) Suppose that Q is not conjugate to Ox by any element of 
Then for each element A in 21*, either CD(A) = 1 or C^(A) = 1. 
(ii) If We&W^C^), then D and X are conjugate by an 
element of C(2l). 
Proof. The proof of (i) proceeds by a series of reductions. If 
21 = 1, the theorem is vacuously true, so we may assume 21 = 1. 
Choose Z in Z(?P), and let O* be any element of M(2I; q) which 
is centralized by Z. By Lemmas 7.4 and 7.8, if S is any proper 
subgroup of X containing 21Q*, then O* ^ OP/(S). 
Now let D* denote any element of M(2t; q) and let S be a proper 
subgroup of 3 containing 21Q*. We will show that Q* a OP<(8). First, 
suppose Z($P) is non cyclic. Then D* = <CO*(Z) |ZeZOP)*>, so by 
the preceding paragraph, Q* a OP>(8). We can suppose that Z($P) is 
cyclic. Let Z be an element of Z(^P) of order p. We only need to 
show that [O*, Z] Ł OP>(8), by the preceding paragraph. Replacing 
O* 
by [Q*, Z], we may suppose that Q* = [Q*, Z]. Furthermore, 
we may suppose that 21 acts irreducibly on n*//)(Q*). 
Suppose Ze (V.F(S). Then Q* = [Q*, Z] Ł O,,.,(8) n Q* S <M2> 
and we are done. If 21 is cyclic, then Z is necessarily in <V,P(8), 
 17. A DOMINATION THEOREM AND SOME CONSEQUENCES 851 
since 21 n Op>,p(2) = 1. Thus, we can suppose that 21 is non cyclic. 
Let 2TX = Ca(Q*) = Ca(O*/iJ(n*)), so that 21/21! is cyclic and 
Zg2Ii. We now choose W of order p in 2Ij such that <Z, T7> <5p. 
Suppose by way of contradiction that O* g OP>(2). Then by Lemma 
7.8, we can find a subgroup $ of 21(7(210 which contains 2IO* and 
such that O*gOP,(S). In particular, JQ* g O,,(C(TF)). Thus, we 
suppose without loss of generality that 2 = C(W). Let ^?* be a Sp- 
subgroup of 8 which contains $P = $PnC(W). If ^P* = SJS, then 
ZeOp>,p(&), by Lemma 1.2.3 of [21], which is not the case. Hence, 
$ is of index p in <P*. Clearly, 21 E $ and ZeZ(?P). Hence, 
HP*, ^] S Z($) g 21. Let W = $P* n O,,.,(8) so that W is a SP-subgroup 
of 0,,.,(8). Then R$*, <Z>, Q*] s [21, O*] n <V.,(8) S Q* n O,,.,(8), so 
that [SRf, <Z>, O*] Ł <V(Ł). Let S3 = O,,.,(8)/<M8) and let ^ = C^O*). 
The preceding containment implies that [53, <Z)>] g S^. Let 332 = 
JV^CSSi). Then Z acts trivially on the Q*2l-admissible group Sy^. 
Hence, so does [<Z>, Q*] = JD*, that is, 5?2 S SB^ This implies that 
S3 = $B! is centralized by O* so O* g OP'(S). We have succeeded in 
showing that if D* is in M(2l; g') and 8 is any proper subgroup of X 
containing 2IO*, then D* S O,,(8). 
Now let ^J, • • • , ^ be the orbits under conjugation by C(2l) of 
the maximal elements of M(2t; q). We next show that if O 6 ?if 
QI e ^ and i ^ ^', then D n QI = 1. Suppose false and i, j, Q, Q! are 
chosen so that | D n Oa | is maximal. Let D* = JVJQ(O n Oi) and O* = 
JVDl(D Pi DI). Since jQ and C^ are distinct maximal elements of M(2t; q), 
O n QI is a proper subgroup of both Q* and Qf. Let S = JV(Q n Oi). 
By the previous argument, <D*, Qx*> Ł OP/(S). Let 91 be a S,-subgroup 
of Op>(%) containing D* and permutable with 21 and let 9^ be a Sq- 
subgroup of Op>(8>) containing D* and permutable with 21. The groups 
91 and 9^ are available by Dp>q in 210^(8). By the conjugacy of Sylow 
systems, there is an element C in OP<(2)21 such that W = 21 and 
9^ = 9tj. As 21 has a normal complement in Op/(8)2t, it follows that 
C centralizes 21. Let Q be a maximal element of M(2l; q) containing 
9*!. Then OndiSQfDJDnQi, and so 6 e ^-. Also, Q n Oa 2 
D**7 
D (Q n Oi)0 so that O e ^ and i = j. 
To complete the proof of (i), let Q, X be maximal elements of 
M(2t; <?) with Q e <^f d e ^. Suppose A 6 21* and CC(A)^1, (7ni(A)^l. 
Let 8 '= C(A), let 9t be a S9-subgroup of Op>(2) containing C^A) and 
permutable with 21, and let 9^ be a S9-subgroup of 0P/(8) containing 
CCl(A) and permutable with 21. Then 91" = 9tx for suitable C in C'(2t). 
Let Q* be a maximal element of M(2l; #) containing 9^. Then 
d* 
n D! 2 Cc^A) ^ 1 so O* e ^-. Also, Q* n & 2 (Ca(A))a ^= 1 so 
O* 
e <^ and i = j. This completes the proof of (i). 
As for (ii), if 21 6 ^K^(Sp)f then there is an element A in 21* 
 852 SOLVABILITY OF GROUPS OF ODD ORDER 
such that CQ(A) = 1 and CDl(A) = 1. By (i), } and Qj are conjugate 
under C(2l). 
COROLLARY 17.1. If p e 7rs U 7r4, $p is a Sp-subgroup of ® and 
21 6 i^^L^OP), /&ew /or eac/& prime q ^ p and each maximal element 
Q o/ H(2l; q), there is a Sp-subgroup of JV(2t) which normalizes Q. 
Proof. Let G e JV(2l). Then & is a maximal element of M(2l; g), 
since any two maximal elements of M(2t; q) have the same order, so 
&? = &c for suitable C = C(G) in C(2l). Hence, GC-1 normalizes Q. 
Setting 3 = JV(Q) n #(«), we see that 3 covers JV(2l)/C(2l), that 
is, JV(Q) dominates 51. Now we have 3fC(5l) = JV(5l) and $ contains 
21. Since C(2I) = 21 x ® where 3) is a p'-group, we have JV(2l) = 
= 321 = 32), and 3 contains a SP-subgroup of JV(3t) as required. 
COROLLARY 17.2. If p e 7r3 U 7r4, ^5 is a Sp-subgroup of 
, 2l€^g^^(^P) and q is a prime different from p, then ty 
normalizes some maximal element Q o/ M(2l; #). Furthermore if G 
is an element of ® such that 21* S $P, ffeen 21^ = 2F /or some JV m 
Proof. Applying Corollary 17.1, some Sp-subgroup ^3* of 
normalizes Dx, a maximal element of M(2I; g). Since ^? is a Sp-subgroup 
of JV(3l), ^5 = ^*^ for suitable -3T in JV(2I), and so ?p normalizes O = 
}f , a maximal element of M(2l; g). 
Suppose G e ® and 21* C sp. Then 21^ normalizes Q since 5p does, 
so 21 normalizes D61"1. Now d61"1 is a maximal element of M(2l; g) 
since any two such have the same order. Hence, Q?"1 — D^ for some 
C in C(2t), by Theorem 17.1 and so CG = N is in JV(Q). Since 2F - 
Slw = 21^, the corollary follows. 
COROLLARY 17.3. Ifpe n*> ^ is a Sp-subgroup of © and 
21 6 i$*if^OP), ^en M(2l) is trivial. 
Proof. Otherwise, M(2l; g) is non trivial for some prime q ^ p, 
by Lemma 7.4, and so M(^?; q) is non trivial, contrary to the definition 
of 7T4. 
Hypothesis 17.1. 
(i ) p 6 7T3, ^5 is a SP-subgroup of ® and 21 e ^ifl^OP). 
(ii) q is a prime different from p, M(2l; g) is won trivial and O 
is a maximal element of M(2t; <j) normalized by ^5. 
 17. A DOMINATION THEOREM AND SOME CONSEQUENCES 853 
REMARK. Most of Hypothesis 17.1 is notation. The hypothesis 
is that p e 7T3, for in this case a prime q is available such that (ii) is 
satisfied. Furthermore, we let 
93 = F(cca); $), 92 = #(Q) , and % 
LEMMA 17.1. t/nder Hypothesis 17.1 i/ G e ® and 21* Ł sp, 
eiement N in 
Proo/. By Corollary 17.2, 21* = a* for some element X in 5R. 
Since 5ft is solvable, Lemma 7.2 (1) and Corollary 17.2 imply that 
5ft = OP,(5R) • JV^(93), so we can write X = A/iN where Nt e O,/0R) and 
Ne A^S). Now a* is in SB, so in particular is in sp. Also a** = 2P 
is in p. Hence, if A is in 21, then A~N • A** = [A, N,}N is in ?p, 
and in particular is a ^-element. Since [A, Ai] is a p'-element, we 
,see that N,e C(2I). Hence 21^^ = 2P, and the lemma follows. 
LEMMA 17.2. Under Hypothesis 17.1, 3^ = O^CT?!). 
Proof. Since Z(3S) char S3, and S3 is weakly closed in ^P, 9^ 
contains JV(^P), so Theorem 14.4.1 of [12] applies. We consider the 
double cosets SR^Sp distinct from 9^. Denote by (X) the kernel of 
the homomorphism of $P onto the permutation representation of ^5 on 
the cosets of 9^ in ^Xty. Let P = P(X) be an element of ^P such 
that St(X)P is of order p in Z0P/ffl(JO). 
Suppose we are able to show that P can always be taken to lie 
in 21. In this case, we have [U, P, P] = 1 for all U in sp. Since 
3? 2^ 3 and © is simple we conclude from Theorem 14.4.1 in [12] that 
We now proceed to show that P can always be taken to lie in 21. 
The only restriction on the element X is that X&$119 that is, we 
must have St(X) * $p. 
Now 21 Ł S3, so Z(S3) centralizes 21. Since 21 e ^^>^(5P), we 
have Z(8S) S 21. It follows that 9^ contains C(2I). 
It suffices to show that 21 g ^(JT). For if a g (-?), choose A 
in 21 so that (St(X) n 2l)A is of order p in Z(^/^(^T) n 21). It follows 
that (X)A is of order p in Z(W®(X)). 
Suppose by way of contradiction that 21 C St(X). Then 21 s 5Rf 
so 21 S ^?*x for ^?* a suitable SP-subgroup of ^. But ^P* = ^PF for 
some Y in %. Setting Xl = YX, we have ^X^ = to^XW and 21 Ł SJ^i. 
Bence, 2Ixr1 S 5p, so by Lemma 17.1, aT1 = a^ for some JF in 
IB n JVflB). Since JV(S3) s ^, we have a = Sl^'i and We 31 n ^L Let 
i - -ST2. Since We3llf we have SR^fp = ^X^. 
Since -3T2 normalizes a, a normalizes QT1. By Theorem 17.1, 
 854 SOLVABILITY OF GROUPS OF ODD ORDER 
for some in . Hence (this defines r 
normalizes . Since and normalize , we see that . 
Since centralizes and , we have . 
We now write . where and 
Such a representation is possible since . Consider the equation 
. Since , we have . If . 
then [A. ] is a -element since . But [A. ] 
, an identity holding in all groups. Since . 
. Since , so [A. ] & 
-group. Hence 
. 
Since A is an arbitrary element of , we have . Now 
however , we have 
so , contrary to assumption. 
LEMMA 17.3. Uncler Hypothesis 17.1, . and; 
. 
Proof. We must show that contains at least one element from 
each coset W. from which the lemma follows directly. 
Let , and being 
a -group. Notice that by Lemma 7.4 together with 
. (This was the point in taking in place of ) 
By Sylow's theorem, contains some element of , so suppose 
. Since is contained in by Lemma 7.2 (1), we have 
, and normalizes . Hence, normalizes and 
by Theorem 17.1, for some in . Write . 
where . . so that , since normalizes . Hence, 
normalizes . But , since , so 
and contains an element of . 
LEMMA 17.4. Uncler Hypothesis 17.1, if is a subgroup of 
which contains , then . 
Proof. Let . Since normalizes so does . Hence, 
normalizes . But and contains . so normalizes. 
. By Theorem 17.1, for some in . Let 
. Now where and . Consider 
the equation . Let . 
We have . where is in 
 18. CONFIGURATIONS 855 
hence, is a -element 
of , since , so that . But 
. Hence, . Since is 
an arbitrary element of . it follows that centralizes , so 
is contained in . But now the elements and normalize 
. Since . the lemma follows. 
LEMMA 17.5. Under Hypothesis 17.1, if is a proper subgroup 
of which con tains , then . 
Proof. If , and . it suffices to show 
that . By Lemma 7.2 (1), we have . and so by Lemma 
17.4, . Thus it suffices to show that . By Lemma 
17.3 it suffices to show that . However, this last contain- 
ment follows from Lemma 7.2 (1) and Corollary 17.1. 
LEMMA 17.6. Under Hypothesis 17.1, if is a proper subgroup 
of and is a -subgroup of , th en . 
Proof. Suppose false, and that is chosen to maximize and 
with this restriction to minimize . Let . By 
minimality of we have . By maximality of is a 
-subgroup of . We assume without loss of generality that 
. In this case, Lemma 7.9 implies that . Since . 
by Lemma 17.4 we have . by Lemma 17.5, . so in 
particular, as required. 
18. Configurations 
The necessary -theorems emerge from a study of the following 
objects: 
1. A proper subgroup of . 
2. A -subgroup of . 
. A -subgroup of . 
4. . . . 
DEFINITION 18.1. A configuration is any 6-tuple 
satisfying (C). The semi-colon indicates that are determined 
when are given. 
DEFINITION 18.2. 
 856 SOLVABILITY OF GROUPS OF ODD ORDER 
(i) 2t is a ^-subgroup of . 
(ii) for every configuration (®, $p, 21; 93, 2ft, 9B), 
(a) 2tt centralizes Z(2B). 
(b) If Z(25) is cyclic, then 2ft centralizes Z2(3B)/Z(2B).} 
DEFINITION 18.3. 
ranging over all Sp-subgroups of ® in both unions. 
LEMMA 18.1. // p ^ 5, fc«n ^(p) U &W&lt;sK(p) S <if (p). 
Proo/. Let 21 e ^(p) U "ifi<K(p), and let (ft, $p, 51; S3, 2JJ, SB) be 
a configuration. Suppose by way of contradiction that either 2ft faila 
to centralize Z(2B) or Z(3B) is cyclic and 2ft fails to centralize 
Za(2B)/Z(2B). Since Op,,p(ft) centralizes both Z(SB) and Z2(9B)/Z(98), it 
follows that some element of 2ft induces a non identity p'-automorphism 
of either Z(2B) or Z2(9B)/Z(2B), so in both cases, some non identity 
p'-automorphism is induced on Z2(9B) by some element of 9JZ. By 3.6, 
some non identity p'-automorphism is induced on ^i(Z2(3B)) = SH^ by 
some element of 3R. Let 2B0 = 0i(Z(2B)) S SO^ and let 2B_! = <1>. 
Let aJi0 = ker(OP,,P>p,(^)->Aut2!Bo), aJ^==ker(OJ,,i,,,,(ft)-^Aut(SB1/a9o)). 
By definition of 3Jt, 3Ji is contained in 3Jti if and only if 93 acts trivially 
on Op',p,p/(ft)/2R*, t = 0 or 1. Suppose that 93 does not act trivially on 
O,'fP.p>(ft)/3Ri. Let 33 = 2l0 be a conjugate of 21 which lies in 93 and 
does not centralize Op'p,p'(ft)/9W< (93 depends on i). In accordance with 
3.11, we find a subgroup & of <V,p.p/(ft) such that 9^/2)^ is a special 
<7-group, is 93-admissible, and such that 95 acts trivially on 3y2Ji<, 
irreducibly and non trivially on 9J»/S)i, where 3), = D^iTnodyHi). Let 
95, = ker (95 — Aut (9^/2^)), so that 95, acts trivially on 9^/2^ and 
95/95, is cyclic. 
Let S, be a subgroup of 9B,/2B,_1 of minimal order subject to being 
952Ji-admissible and not centralized by 21, . The minimal nature of X, 
guarantees that 95, acts trivially on X,. If 95,5, is a generator for 
95/95,, then (B) guarantees that the minimal polynomial of B{ on X, is 
(x - l)r where r = r, = 1 95 : 95, |. 
Suppose i = 0. Since X0 is a p-group, while Op/(ft) is a p'-group, 
we can find a p-subgroup >0 of S? such that >0 and X0 are incident, 
and such that >0 is 95-admissible. In particular, 950 centralizes v 
Let ^?* be a Sp-subgroup of JV(95), so that ^P* is a SP-subgroup of . 
If 950 n Z(^?*)* is non empty, we apply Lemma 16.1 and have a contra 
diction. Otherwise, Lemma 16.2 gives the contradiction. 
We can now suppose that Z(9B) is cyclic. In particular, 9B0 is of 
order p. Since Xx is of the form 2)i/9B0 where Si is a suitable subgroup 
 19. AN ^-THEOREM 857 
of 3^!, we can find a p-subgroup & of $ incident with 2)i and 33- 
admissible. 
Choose B in 33^ Since SBi centralizes ?)!/3B0 and since 2B0 is of 
order p, it follows that >2 = C^(B) is of index 1 or p in &. If 
33i n Z($P*)f is non empty, application of Lemma 16.1 gives 73Ł>2333 = <1>, 
and so T^iSB4 = <1>, the desired contradiction. Otherwise, we apply 
Lemma 16.2 and conclude that o^B4 = <1>, and so T^i^B5 = <1>, from 
which we conclude that 1 33 : 33X | = 5. In this case, however, setting 
3 = Z(ty*) fl 33, we have 93 = <SBlf 3>, and so the extra push comes 
from Lemma 16.2 which asserts that 73Ł>233 = <1>, and so T4^4 = <1>, 
completing the proof of the lemma. 
19. An ^'theorem 
It is convenient to assume Burnside's theorem that groups of 
order paqb are solvable. The interested reader can reword certain of 
the lemmas to yield a proof of the main theorem of this paper with 
out using the theorem of Burnside. 
If p, q e 7T3 U 7T4, we write p ~ q provided © contains elementary 
subgroups 6? and g of orders p3 and q3 respectively such that <Gr, g>c®. 
Clearly, ~ is reflexive and symmetric. 
Hypothesis 19.1. 
(i ) p 6 7T3 U ?T4, q € TT(®) and p ^ q. 
(ii) A Sp-subgroup ty of ® centralizes every element of M(^P; q). 
LEMMA 19.1. Under Hypothesis 19.1, if 33 e ^(p), then 33 central 
izes every element of M(33; g). 
Proof. Suppose false, and that Q is an element of M(33; q) minimal 
with respect to 7S3D = <!)>. From 3.11 we conclude that 33 centralizes 
D(D) and acts irreducibly and non trivially on D/jD(Q), so in particular, 
O - 7Q33 and 330 = ker (33 — Aut Q) = <1>. Let ( = C(330), let ^ be 
a SP-subgroup of JV(33), and let ^50 = C(») n 5p. Since 33 e ^(?>), % 
is of index at most p in a Sp-subgroup ?& of E, and so ^50 < SR. Hence 
7^33 Ł ^Po. Since ^0 centralizes 33, we have 72^332 = <1>, so 
33 S Op'^E) = St. Let 8 = Op,(<Ł). Since 33 e ft < (E, 7Q33 Ł SB, so 
7O33 S *St n O G 8. Since jQ = 7QS3, we have Q Ł 8. 
By Lemma 8.9, 33 is contained in an element 21 of t$*^L^($P). 
Since 21 centralizes 33, we have 21 Ł 5p0.. Let S) = 218, and observe 
that 8 is a normal p-complement for 2t in S). By Hypothesis 19.1 
(ii), Theorem 17.1, Corollary 17.2, and DPiq in , 2t centralizes a Sg- 
subgroup of , so S) satisfies EŁq and every p, g-subgroup of S> 
 SOLVABILITY OF GROUPS OF ODD ORDER 
is nilpotent. But , and , so is not 
nilpotent. This contradiction completes the proof of this lemma. 
Hypothesis 19.2. 
(i) and . 
(ii) . 
(iii) A -subgroup of centralizes every element of 
a of centralizes every element of . 
THEOREM 19.1. Under Hypothesis 19.2, satisfies . 
We proceed by way of contradiction, proving the theorem by a 
sequence of lemmas. Lemmas 19.2 through 19.14 all assume Hypothesis 
19.2. We remark that Hypothesis 19.2 is symmetric in and . 
LEMMA 19.2. , whenever . 
Proof. Suppose , where . , and 
is minimal. By in , it follows that is a -group. 
By the previous lemma centralizes and centralizes 
. Since and are abelian, and . are abelian, so 
centralizes . Hence by 3.3. 
Let be a chief series for . one of whose terms is , and 
let be a chief factor of . If . then is obviously a 
central factor. If . and is a -group, then centralizes 
, and since is a chief factor, must also centralize . so 
is a central factor. The situation being symmetric in and . 
every chief factor of is central, and so is nilpotent, and 
. 
Let , let be a -subgroup of with Sylow system 
being a -subgroup of . since . By in 
for suitable in . Let be a maximal -subgroup of 
containing , with Sylow system . where . Let 
be a -subgroup of containing . Finally, let . 
and observe that . By Hypothesis 19.2, centralizes . 
By the previous lemma, centralizes . 
We next show that . Consider and let 
. Since centralizes , so does . so 
is nilpotent. But now centralizes and so Lemma 
1.2.3 of [21] implies that . It follows that . Since 
is weakly closed in a -subgroup of , it follows that is a 
-subgroup of . 
Again, centralizes , and now centralizes both 
assertions being a consequence of Hypothesis 19.2 (iii). It follows 
 19. AN -THEOREM 
readily that every chief factor of is central, and so is nilpotent. 
Since we are advancing by way of contradiction, we accept this lemma. 
LEMMA 19.3. If . then either is a -group or a 
of is of order and has the property that it 
does not centralize any . 
Proof. Let be a -subgroup of . and suppose . 
By Lemma 19.2, no element of centralizes any . Let 
be a -subgroup of containing and let . Then 
is of index 1 or in and is disjoint from follows. 
Lemmas 19.2 and 19.3 remain valid if and are interchanged 
throughout. In Lemmas 19.4 through 19.14 this symmetry is destroyed 
by the assumption that (which is not an assumption but a choice 
of notation). 
We now define a family of subgroups of . First, 
is the set theoretic union of the subfamilies , where 
ranges over the -subgroups of . Next, is the set theoretic 
union of the subfamilies , where ranges through the 
elements of . We proceed to build up . Form 
. Consider the collection 
of all -subgroups of which have the following properties: 
1. . 
(K) 2. . 
3. Every characteristic abelian subgroup of is cyclic. 
If is empty, we define to consist of all 
subgroups of of type . If is non empty, we define 
to consist of all subgroups of of type together with 
a11 subgroups of of type which contain . 
and ranges over . 
Notice that depends on too, but we write to 
emphasize that its elements are -subgroups of . The nature of 
is somewhat limited by 
LEMMA 19.4. If is a -subgroup of 
and are non empty, and if . 
then . 
Proof. Let . Then . 
From 3.5 and the definition of . we have . 
Hence and . From (B), we conclude that 
and as required. 
Using Lemma 8.9 and Lemma 19.4, we arrive at an alternative 
definition of being a -subgroup of . If is empty 
 860 SOLVABILITY OF GROUPS OF ODD ORDER 
for all a€^gl^(Sp)f J^OP) is the set of all subgroups 35 of <$ of 
type (p, p) such that 33^ is abelian. If JTX51) is non empty for some 
^e<5*if^OP) and fte jr(Sl), then .^(^P) consists of all subgroups 
of type (p, p) in Oq P(®) n ^P which contain Q1(Z(Oqtp(R) n $)), together 
with all subgroups 33 of Off,P(S) n ^P of type (p, p) such that 33^ i& 
abelian. Here we are also using (B) to conclude that Oq,9(St) n ^P- 
contains every element of 
LEMMA 19.5. Let St e JT(2l), wfeere 31 e ,Ł"g^7($p) and ^P is a 
Sp-subgroup of . Let ^P0 = ^P fl Off>p(A). // 3Ji is any proper subgroup 
of ® containing $p, tfeen ^P0OP.(9K) < 2W. 
Proo/. Since 73Wo=l, it follows from (B) that ^P0S^P 0 (W2») = 
tyl9 say. By Sylow's theorem, 2Ji = Op/(2»)JV^0Pi), so it suffices to 
show that ^Po < Ay^Pi) = 91. Choose N in 91. Then [W, %, ^Po, %] = !. 
Since SR, G SR Ł $p* G SV, it follows from (B) applied to W that 
% S W, so that *P0 = W, as required. 
LEMMA 19.6. Let ffi 6 JST(2l), 21 e ^^L^7(5P), ^P being a SP-subgroup 
of , and let % be a subgroup of index p in $P0 = Oq,p(&) n 
93= 
Proof. Since ^^^ (^P) is non empty, (B) implies that S is non 
abelian. Now &i(Z(^P)) is of order p and is contained in S. By 3.5 
S/IWZCSP)) is abelian. 
Let 8* = Si be a conjugate of 8 contained in $, G e . First, 
suppose that (Ł1(Z(^P)))6( = 3 is contained in $P0. Then C?0(3) = EI is 
of index 1 or p in $P0- Set (Ł2 = CX3). By Lemma 19.5, with S2 in 
the role of 2ft, ^ in the role of $, ^ in the role of ^?0, we see that 
T^g; = <1>, and it follows that 7*%8{ = <1>, so by (B), 2, G Sft. 
(Recall that p ^ 5.) 
Thus, if Sx g 5p0, but 8i G sp, then 3 g ^P0- But 2, normalizes ^P0r 
so SP0 PI Sj < SL Since Sx is of index p in sp?f any non cyclic normal 
subgroup of Si contains 3- Hence, % n 81 is cyclic and disjoint from 
3. If now fiiC^Po) is extra special of order p2r+1, we see that #1(81) 
contains an extra special subgroup 2 of order p*r~l which is disjoint 
from ^P0. 
Consider now the configuration (ffl, sp, 8; 53, 501, 28), and observe 
that 2B = ^P0. 2 is disjoint from $P0, so is faithfully represented on 
%=Oq>P g(St)IOg,p(St)9 a g-group. Furthermore, g is faithfully represented 
on fl1(2B)/fl1(Z(2B)), which makes sense, since Off,,(J8) acts trivially on 
Let ! be the subgroup of g which acts trivially on 
which also makes sense, since OQtp(St) acts trivially on 
Then /& is cyclic and 2 acts trivially on g/gx since p > g. 
 19. AN -THEOREM 861 
Since is a -group, acts faithfully on , so acts faithfully on 
If then divides , 
and so , by Lemma 5.2. 
On the other hand, acts faithfully on , and trivially 
on , so is isomorphic to a subgroup of the symplectic group 
. Hence, divides . 
so by Lemma 5.2 (ii). . Combining this with the previous 
paragraph, we have a contradiction, 
completing the proof of the lemma. 
We can now translate this information about to the general 
-subgroup of . To do this, we let be the set theoretic 
union of sets ranging over the -subgroups of 
is the set of all subgroups which can occur in the previous lemma. 
Formally, is the set of all subgroups of index in 
where , and . 
LEMMA 19.7. If and is a -subgroup of then 
. 
Proof. Let be a configuration. The lemma is 
clearly equivalent to the statement that . Let be a 
-subgroup of containing and let be a conjugate of 
contained in . Since , we have for some 
subgroup of . Now for some in , and so - 
By Lemma 19.6, we have . and so in 
particular, so (B) and imply this lemma. 
LEMMA 19.8. If then for every con- 
figuration for whi ch is a -group. 
Proof. Suppose false, and that is chosen to maximize , and, 
with this restriction to minimize . It follows readily that 
is a -subgroup of and that is a -subgroup of every 
-subgroup of which contains . 
By Lemma 18.1 and the isomorphism 
we conclude that centralizes . By minimality of . we 
also have . 
If is a -subgroup of containing , we see that 
centralizes , and so , by maximality of . It, 
now follows that centralizes , and maximality of yields 
: : 
Since does not act trivially on , and since , 
it follows that contains an elementary subgroup of order . But, 
 862 SOLVABILITY OF GROUPS OF ODD ORDER 
3Ji centralizes Z(OP(®)) = 3 and if 3 is non cyclic, then 3 contains 
an element of ^(^5), in violation of Lemma 19.3. Hence, 3 is cyclic. 
In this case, we conclude from Lemma 18.1 that a Sff-subgroup of 9JJ 
centralizes Z2(OP(®)) = -82- But 32 contains an element of ^(^P), so 
once again Lemma 19.3 is violated. This contradiction completes the 
proof of this lemma. 
LEMMA 19.9. // 8 e f(p) U *if~<K(p), then 2 Ł OP(ft) /or every 
p, q-subgroup ® o/ ® which contains 2. 
Proof. By Lemmas 19.7 and 19.8, it suffices to show that 8 
centralizes Oq(ft). If 8 e &*i^K(p)9 Theorem 17.1, Corollary 17.2 and 
Hypothesis 19.2 imply that 2 centralizes Off(ft). If 8 e f(p), then 
« e f(<$) for some Sp-subgroup p of . In this case, if 21 e &&^1W), 
the definition of .Sf ($P) implies that 21 n 8 = 2I0 is non cyclic. Hence, 
O,(®) is generated by its subgroups C(A) fl Off(ft) as A ranges over 
2IJ. By the preceding argument, 21 is contained in Op(®0) for every 
p, g-subgroup $0 of C(A) which contains 2t. Lemma 7.5 implies that 
% centralizes O,(ft). In particular, fl^Z^)) centralizes O,(ft). 
Consider C(^(Z(^))) 2 <$, Of(ft)>. Since 2 S Op(^) for every 
#, g-subgroup ^ of <^5, O7(S)> which contains S by (B) and Hypothesis 
19.2, a second application of Lemma 7.5 shows that 2 centralizes O9(®), 
as required. 
LEMMA 19.10. // 33e^(;p), then 33 centralizes every element o/ 
Proof. Suppose false, and O is chosen minimal subject to 
> e M(33; g) and 7&33 ^ <1>, so that we have Q = 7QS3 and S30 = 
ker (S3 — Aut O) = <1>. Let E = C(350). Since 33 e J^(p), we have 
33 6 ^~OP) for a suitable SP-subgroup ^P of . By definition of J^~OP), 
either C(33) contains an element ^ of S^<^H?$) or else C(33) contains 
a subgroup S& of index p in ^5 n Of ,„(Ł), ft e JT(2l) and 21 e ^^L^(^). 
Let § be a Sp,9-subgroup of ( containing 2IX in the first case, and % 
in the second case. Lemma 19.9 implies that 2It Ł O,($) in the first 
case and ^ Ł OP($) in the second case. In both cases, we have 
S3 Ł Op($). Now let fe be a SP,,-subgroup of (E containing S3Q. By 
Lemma 7.5, we have S3 E Of(*i) and so 7O33 Ł O,(&) n O = <1>, 
contrary to assumption. 
LEMMA 19.11. // » e jr(p)f 21 e &(q), then ® = <«, S3>. 
Proo/. Suppose <2l, S3> = ft c , and 21 and S3 are chosen to 
minimize S. By the minimal nature of ft, ft is a p, g-group. By the 
 19. AN ^-THEOREM 863 
previous lemmas, 21 centralizes OP($), and S3 centralizes Oq(®). It 
follows readily that $ is nilpotent, so ® = 21 x S3. But now C(2l) 
contains S3 in violation of Lemma 19.3, with p and q interchanged. 
This interchange is permissible since Lemma 19.3 was proved before 
we discarded the symmetry in p and q. 
LEMMA 19.12. If 3) is a p, q-subgroup of © and if ® possesses 
an elementary subgroup of order p3, then a Sp-subgroup of ® is 
normal in 3X 
Proof. Case 1. 3) contains a Sp-subgroup ^? of . Let Q be a 
Srsubgroup of , let d = > n Op.,(2>)f let 5 be a Sg-subgroup of ® 
containing Q, let S3 e ^(O), and Q, = (7^(33). Then Q, is of index 
1 or g in Q!. 
Next, let $ = Op(®), and assume by way of contradiction that 
Sc^P. By the preceding lemmas, ® contains F(ccŁ@(2l); $P) for every 
21 e ,Ł*gL^(5p). By the preceding lemma, no element of &$ centralizes 
any element of ^(p). 
If $ contains a non cyclic characteristic subgroup E, then every 
subgroup of ® of type (p, p) belongs to ^OP), and so C§(Q) is cyclic 
for Q 6 jQj. This implies that ^^^(D,) is empty, and if Oz possesses 
a subgroup of type (q, g), then 2? = 1 (mod q). However, if $ does 
not contain any non cyclic characteristic abelian subgroup, then every 
subgroup of $ of type (p, p) which contains Q^Z^)) lies in ^~(ty)9 
and we again conclude that &*&i4^(Qt) is empty, and if Qa is non 
cyclic, then p = 1 (mod q). 
Now Q! = Op,9(®)/5? admits a non trivial ^-automorphism since 
® c ^5, so i$*g5L/f7(Q1) is non empty, by Lemma 8.4 (ii) and p > q. 
Hence, Q, is non cyclic, being of index at most q in QI, and this 
yields p = 1 (mod q). We apply Lemma 8.8 and conclude that p = 
1 + q + q2, and Ox is elementary of order g3. This implies that any 
two subgroups of X of the same order are conjugate in . Since 
at least one subgroup of Oj of order # centralizes S3, every subgroup 
of Q! of order # centralizes some element of ^(g). Since at least 
one subgroup of Ox of order q centralizes some element of 
every subgroup of DX of order # centralizes some element of 
This conflicts with Lemma 19.11. 
Case 2. ® does not contain a Sp-subgroup of . Among all S> 
which satisfy the hypotheses but not the conclusion of this lemma, 
choose ® so that 1 3> n &i(2t) | is a maximum, where 21 ranges over all 
elements of ^^^(p), and with this restriction, maximize |®|p. 
Let 3>! be a Sp-subgroup of S>, and let ^P be a SP-subgroup of © 
containing SV 
 864 SOLVABILITY OF GROUPS OF ODD ORDER 
First, assume 3^ centralizes O9(3)). In this case, O,(3)) is a Sp- 
subgroup of 0g,P(3)). By maximality of | ® |p, 3\ is a Sp-subgroup of 
JV(0p(5»). This implies that x contains every element of ^g^^(^). 
To see this, let 21 e *S*g^0P), and let 2lx = 21 n i. Since OP(®) is 
a Sp-subgroup of O«,p(3)), it follows that 21 n 3\ S OP(S). If Six were 
a proper subgroup of 51, then 3^ would be a proper subgroup of 
Since this is not possible, we have 21 = 2lx. But now, 
; 3)0 < 2), and by maximality of | 3) |p, 3)t = $P follows, and 
we are in the preceding case. 
We can now assume that 3\ does not centralize Og(3)). Suppose 
3>! contains some element S3 of ^~(p). By Lemma 19.10, S3 centralizes 
Og(®). Since x does not centralize O«(5))f \ Og(S)) | > tf , and so Lemma 
19.11 is violated in C(Q), Q being a suitable element of Off(3)). Thus, 
we can suppose that i does not contain any element of J^~(p). In 
particular, 3) n fii(2l) is of order 1 or p for all 21 e ^K(p)m Let 
S3 € ^"($P), and 3>2 = (7^(33). Since ^^^(3)^ is non empty by 
hypothesis, 3)2 is non cyclic. Let Gf be a subgroup of 2 of type 
(P, P)- Since S3 Ł 3)x, <@, S3> is elementary of order at least p3. If 
@ does not centralize O«(3)), then there is an element # in * such 
that © does not centralize C(E) n Off(3>). But in this case, a SPf9- 
subgroup of C(E) is larger than ® in our ordering since S3 Ł C(^), 
CC^) possesses an elementary subgroup of order p3, and a Sp-subgroup 
of a Sp,g-subgroup g of C(JS') is not normal in g. This conflict forces 
every subgroup of 3)2 of type (p, p) to centralize Og(3)). Thus, ?i(3)2) = 
* 
centralizes O9(®), since * is generated by its subgroups of type 
(p, p). However, we now have JV(®*) a <3\, S3, O,(®)> and a Sp ,- 
subgroup gi of JV(®*) is larger than 3) in our ordering, possesses an 
elementary subgroup of order p3, and has the additional property that 
its SP-subgroups are not normal in %lm This conflict completes the 
proof of this lemma. 
Lemma 19.12 gives us a fairly good idea of the structure of the 
p, g-subgroups of . The remaining analysis is still somewhat detailed, 
but the moves are more obvious. 
For the remainder of this section, ty denotes a Sp-subgroup of , 
D a Sg-subgroup of JV($P), and Q a Sg-subgroup of © which contains Q. 
LEMMA 19.13. ^^L^(O) is non empty. 
Proof. We apply Hypothesis 19.2 (ii) and let 3) be a maximal 
p, g-subgroup of © which contains elementary subgroups of order p3 
and q\ By Lemma 19.12, p < 3), p being a SP-subgroup of 3>. 
Since 2) is a maximal p, g-subgroup of , p is a Sp-subgroup of , 
so p = sp* and the lemma follows. 
 19. AN ^-THEOREM 865 
We now choose 33 in ^(O) and set d = Cc(93). 
LEMMA 19.14. 
( i ) ^^^(Q!) is empty. 
(ii) D contains ^(ZCQ)). 
(iii) p = 1 (mod q). 
(iv) O* contains an element Y which centralizes an element of 
and has the additional property that C^(Y) contains an 
elementary subgroup of order q*. 
(v) If Xe&* and X centralizes an element of ^~(ty), then X 
does not centralize any element of ^(O), and C(X) does not contain 
an elementary subgroup of order q4. 
Proof. Let Gf be an elementary subgroup of Q of order q\ and 
choose Gf in Q if possible. If ty possesses a non cyclic characteristic 
abelian subgroup (Ł, then some element of G? has a non cyclic fixed 
point set on E. Since every subgroup of Ł of type (p, p) lies in 
.^~OP), (iv) is established in this case. 
If every characteristic abelian subgroup of ^? is cyclic, then some 
non cyclic subgroup G^ of Gf centralizes Z(ty). Since any non cyclic 
subgroup of ^? which contains Qi(Z(ty)) is normal in ^?, by 3.5, some 
element of G^ centralizes an element of ^OP), so (iv) is proved. 
If S S Q!, then Lemma 19.11 is violated in C(E), Ee @», E central 
izing an element of ^OP). Hence, (i) is proved. 
On the other hand, ^^^^(D) is non empty, so Ox possesses a 
subgroup & of type (q, q). If p & 1 (mod q), then some element of 
5i is seen to centralize an element of ^"(ty). Since this is forbidden 
by Lemma 19.11, (iii) follows. 
We now turn attention to (v). In view of Lemma 19.11, we only 
need to show that if X in }* centralizes an element of ^~OP), then 
C(X) does not contain an elementary subgroup of order #4. 
Let 21 be an element of ^($P) centralized by X, let § be a Sp>q- 
subgroup of C(X) and let ft be a maximal p, g-subgroup of ® containing 
>. By Dp>q in C(X), 8^ = W S & Ł $, for some G in C(X). Suppose 
by way of contradiction that C(X) contains an elementary subgroup 
of order q*. By DptQ in C(X), § contains an elementary subgroup of 
order <j4; thus, ® contains such a subgroup. 
We first show that a Sp-subgroup of ft is not normal in 5J. Suppose 
false. In this case, since $ is a maximal p, g-subgroup of , a Sp- 
subgroup of $ is conjugate to ^P, and so $ is conjugate to $PQ. 
However, (i) implies that Q does not contain an elementary subgroup 
-of order q\ since | D : QI | = tf, so $ does not contain one either. 
We now apply Lemma 19.12 and conclude that 5? does not possess 
 866 OF GROUPS OF ODD ORDER 
an elementary subgroup of order . It follows directly from Lemma 
8.13 that has -length one. Let be a -subgroup of containing 
, and let . By Lemma 19.10, centralizes . 
Since has -length one, . But then contains 
subgroups of larger order than . and also contains , contrary 
to the assumption that is a maximal -subgroup of . This 
contradiction proves (v). 
We now turn to (ii). Choose to satisfy (iv) and let be an 
elementary subgroup of of order . If were 
not contained in then would contain an elementary subgroup 
of order , and (v) would be violated. This completes the proof of 
this lemma. 
We remark that Lemma 19.2 and Lemma 19.14 (ii) imply that 
is cyclic. 
Theorem 19.1 can now be proved fairly easily. We again denote 
by an elementary subgroup of of order . and we let be an 
element of which centralizes an element of . Let . 
Since centralizes does not centralize any element 
of . by Lemma 19.2, and so does not centralize . Thus, we 
can find an element in with the property that does not 
centralize Consider We see that contains both 
and . Since does not centralize . is a non abelian 
group of order , with center . Let be a -subgroup of 
which contains ; since contains does not contain an 
elementary subgroup of order . Since is contained in the derived 
group of is contained in . We apply Lemma 8.13 and 
conclude that centralizes every chief -factor of . It follows 
that for suitably large . and so . But now 
if is any -subgroup of which contains . we have , 
by Lemma 7.5, and so is both a -group and q-group. so 
is , contrary to construction. This completes the proof of Theorem 
19.1. 
COROLLARY 19.1. If and then either 
or . 
Proof. If satisfies then both and are in . Other- 
wise, Hypothesis 19.2 is violated and the corollary follows. 
20. An Atheorem for 
Hypothesis 20.1 and . 
 20. AN -THEOREM FOR 867 
THEOREM 20.1. Under Hypothesis 20.1, satisfies . 
The proof of this theorem is by contradiction. The following 
lemmas assume that Hypothesis 20.1 is satisfied but does not satisfy 
. 
LEMMA 20.1. If is a -subgroup of and is a -subgroup 
of then either normalizes but does not centralize some -subgroup 
of , or normalizes but does not centralize some p-subgroup of . 
Proof. This lemma is an immediate consequence of Hypothesis 
20.1, Theorem 19.1, and the assumption that does not satisfy . 
We assume now that notation is chosen so that , a -subgroup 
of , does not centralize a maximal element of . Let 
be a -subgroup of permutable with , and let be a 
subgroup of containing . 
LEMMA 20.2. . 
Proof. Suppose false. Let be an element of . By 
Lemma 7.9, we have . We apply Lemma 17.4 and conclude 
that . where . . and so 
satisfies , contrary to assumption. 
Let . 
LEMMA 20.3. is a -subgroup of every proper subgroup 
of which contains . 
Proof. Let be a -subgroup of with Sylow system 
where and and let . where , 
. 
We first show that . Suppose by way of contradiction 
that . Since and both normalize and both 
normalize , setting , we see that is a 
group, and that normalizes . Let be a chief factor 
of with . Since it follows that central- 
izes , that is . In particular normalizes 
. Now is a maximal -subgroup of by Lemma 7.3, so 
is a -subgroup of . A second application of Lemma 7.3 
yields that is a -subgroup of . But is normalized 
by , so a third application of Lemma 7.3 yields 
so , contrary to our choice of . Thus, . 
We next show that . To do this, it suffices to show that 
 868 SOLVABILITY OF GROUPS OF ODD ORDER 
3$i centralizes 2, for if this is the case, then 2 Ł 
and so 2Q* is a g-subgroup of JVOPi)- Since G* is a ^-subgroup of 
2:2 Ł &* follows. 
To show that % centralizes 2, we first show that % centralizes 
By definition, Q* is a S,-subgroup of JV(Oi), and since 
is a g-subgroup of JV(Qi), we have C^GO Ł O*. Hence, 
] S Z, n [0*, %] Ł Zf n % = <1>. Suppose that ^ does 
not centralize Z2 and that 3 is a ^Grinvariant subgroup of 2, 
minimal subject to the condition 72^ = <1>. By minimality of 3, 
we have 3 = 72^. Since 3 is a g-group, 72^ c 3, and so 
72Ł3Q1?1 = <1>. Since 70^ = <1>, we also have 72G1$12:3 = <1>. The 
three subgroups lemma now yields 72^P1Ł3Q1 = <1>, so ^ centralizes 
7^PiŁ3 = 3. By what we have already shown this implies that $& 
centralizes 3. This conflict forces 7^ZZ = <!> 
We next show that ^ Ł 2. To do this, consider Cx(2y = ® < 2:. 
Since 2;j Ł SR, we see that Oi Ł . On the other hand, Z(^) central 
izes both 2; and 2,, so Z(^) Ł 2;, by 3.3. Hence, (E Ł C3:(Z(^1)) Ł 
C(Z(^)) S JV(Z(^)). Since d = Of(?iQ*), Lemma 7.5 implies that 
QI S Og(<Ł) char Ł < 2:, and so S^ Ł 2;,. 
Consider finally Cj(2y . Since ^ Ł S:2, we have ^(S,) Ł C^Qi) Ł 
C(Qa) Ł N(S\). Since % = OP(^PQ*), Lemma 7.5 implies that 
S& Ł OP(CX(2:2)) char Cx(2:2) < Z9 and so ^ Ł Z^ Since we have 
already shown that Z1 Ł %, we have Z1 = ^ < X, and so D* is a 
S?-subgroup of 2;, as required. 
To prove Theorem 20.1 recall that Q is a S^-subgroup of © con 
taining D*. Choose 21 in ^^Lx^(O), and let 21* = 51 n O*. We first 
show that 21* c 21. Suppose by way of contradiction that 21* = 21. 
Then 21 normalizes $&. Lemma 7.3 and the previous lemma imply 
that 51$! is a maximal element of M(2I; p). By Corollary 17.1, JVOPi) 
contains a S9-subgroup of , and ® satisfies Ep>q. Since we are 
advancing by contradiction, we have 21* c 21. 
We next show that 21* n Ox = <1>. To do this, we observe that 
21* 
n d < O*, so if 21* n >! # <1>, then 21* n k n Z(O*) ^ <l>. In 
this case, however, C(2l* n &i n Z(O*)) contains % and also contains 
1*21, contrary to the previous lemma. Thus, 21* n QI = <1>. Since 
3t* 
and f^ are both normal in O*, we have 721*^ = <1>. 
Let 2tx = JV^(O*), so that 21* c 2rx Ł 21. Observe that 72I1O* Ł 
i* 
n 21 c 2IX and so Q* normalizes 21^ Let 33 be any subgroup of 2^ 
which contains 21* properly. Since [S3, ^21*] iS 21*, we see that 33 
normalizes jQjSl* = DI x 21*. Since O* normalizes 33, D* also normal 
izes C0l(33) = 2), say. If 3) ^ <1>, then 3) n Z(O*) = <1>. But then 
the previous lemma is violated in C(S) n Z(d*)). Hence, 2) = <1>. 
Since CW n S2&* ^ 21*, we have C(33) n ^21* = 21*. 
 20. AN -THEOREM FOR 869 
Since normalizes also normalizes . 
Since has no fixed points on by the above argument, is 
abelian. But now and are normal abelian subgroups of , 
so is of class two, so is regular. It follows that if , 
, then . But is an arbitrary subgroup 
of which contains properly, so we can choose such that 
. For such a . the element centralizes . It now 
follows that is elementary. 
We take a different approach for an instant. does not centralize 
the elementary abelian group , and has no normal subgroup 
of index . by Lemma 17.3. It follows that is not of order . 
Returning to the groups and , since has no fixed points 
on , if . . then the mapping defined by 
in is an isomorphism of onto a subgroup of 
. Hence, is not cyclic. 
From the definition of , we see that contains . We 
wish to show that contains an element of . This is immediate 
if is non cyclic, so suppose is cyclic, If does not 
.contain any element of , then the element above can be taken 
to lie in some element of . However, . so 
.could not map onto a subgroup of order exceeding . We conclude 
that contains and also some element of . 
We will now show that for each element of . we can find 
-subgroup in which is not centralized by Z. Namely, 
is faithfully represented on , since and is a 
normal abelian subgroup of . We first consider the case in which 
is non cyclic. Let be a subgroup of of type which 
has non trivial intersection with , that is let contain . 
.Since acts non trivially on . acts non trivially on for 
.suitable in . Let and let be a -subgroup of 
permutable with . It is easy to see that does not centralize 
. 
If is cyclic, we use the fact that contains an element 
of . We can find an element in such that 
does not centralize . Let . By (B), it follows that 
, and so . Thus, contains an element 
which does not centralize. 
It now follows from Theorem 17.1 and the preceding argument 
that if is a maximal element of , then is faithfully 
represented on . If is a -subgroup of permutable with 
, then Lemma 20.2 is violated with and interchanged. This 
the proof of Theorem 20.1. 
 870 SOLVABILITY OF GROUPS OF ODD ORDER 
21. A C* 'theorem for 7T3 and a C-theorem for 7Ta 
It is convenient to introduce another proposition which is "between" 
CT and Dx. 
C?: X satisfies Cx, and if Z is a 7r-subgroup of Ł with the 
property that \Z\9 = \ X \p for at least one prime p in TT, then Z is- 
contained in a S^-subgroup of 36. 
THEOREM 21.1 If p, q e 7T3 and p ~ q, then & satisfies Cp*< 
Proof. We can suppose p ^ q. We first show that © satisfies 
Cp,g. By Theorem 20.1, © satisfies p,g. Let § be a Sp,,-subgroup> 
of © with Sylow system ^5, D, where ty is a Sp-subgroup of . We 
assume notation is chosen so that |$P| > |Q|. Then Op(Ł>) = <1> by 
Lemma 5.2. Lemma 7.3 implies that Op(Ł>) is a maximal element of 
H(D; p). If >i is another Sp,g-subgroup of © containing }, then Op(^)1> 
is also a maximal element of M(G; 2)). From Section 17 we conclude 
that Op(&) = G^Op^G for suitable G in @. Hence, G^G~l and & 
both normalize Op(Ł) so are conjugate in N(Op($Q)). 
Turning to C*,qy we drop the hypothesis |$P| > |Q|, and let 2 
be a maximal p, g-subgroup of © containing ^?. Let § be a Sp,g-subgroup^ 
of © containing ty. 
First, assume that Oq(Z) = 1. In this case, Og(Ł) is a maximal 
element of M0P; g). If O,(§) ^ 1, then Oq(&) is also a maximal element 
of M(^?; q). Thus, Theorem 17.1 yields that > is conjugate to Z. 
(Here, as elsewhere, we are using the fact that every maximal element 
of H(^P; q) is also a maximal element of M(2l; q) for all SI in ^^x^(^).> 
Thus, suppose Off($) = 1. In this case, if 21 e ^g^^(^P), then S3 < v 
g? =. F(ccy§l); ?P), by Lemma 17.5, so | © |f = | N(%) : C(5S) |f . But 
'N(Oq(Z)) dominates 95, so | N(Oq(Z)) \q > \ ® |ff, which is absurd. 
We can now suppose that Oq(Z) = 1. We apply Lemma 17.5 and 
conclude that 53 < Z, where 93 = F(ccy§l); $), and 2t 6 ^g=lx^(^). 
Let DO be a S7-subgroup of Z. Since Z is a maximal p, (/-subgroup- 
of , Do is a S,-subgroup of JV(9S). 
Let $ be a Sp,,-subgroup of © containing ^? and let D be a Sq- 
subgroup of . Let d = Of($). If Oj = <1>, then ^ Ł N(%), by 
Lemma 17.5, and we are done. Otherwise, § = D1JV^>(9S), again by 
Lemma 17.5, and we assume without loss of generality that JV^(9S) Ł Z.. 
Assume that N$(%>) n D! ^ <1>. Then in particular, 2 n 
contrary to Og(Z) = <1>. Hence, N$(%) n QI = <1>. 
We will now show directly that JV^(9S) = 2. Since N 
it suffices to show that | A^SB) |. ^ | Z |f . Now A(OO = Op 
(A(Qi) 0 A(SB)), by Lemma 17.1, and since JV^(9S) n d = <1>, it follows 
easily that | JV^(SS) |f = | N(&J n 
 21. A -THEOREM FOR AND A -THEOREM FOR 871 
Let . By Lemma 17.3 we have . 
Let . Since contains . By 
Lemma 17.5, we now have , which yields 
. Now contains and 
, since . Thus, is mapped isomorphically into 
. and it follows 
that , as required. 
Since it follows that , proving the theorem. 
THEOREM 21.2. Let be a subset of . Assume that satisfies 
for all . in . Then satisfies . 
Proof. By the preceding theorem, we can assume that contains 
at least three elements. By induction on . we assume that 
satisfies for every proper subset of . 
Let and let . , 
. . Let be a -subgroup of . . Then 
the S-subgroups of are conjugate to the S-subgroups of . 
For , let . Note that by dependS 
only on and and not on the particular -subgroup of we choose. 
Fix . let be a S-subgroup of , let 
groupof-subgroup 
which is possible by . and . 
Let Suppose that . 
With this assumption, we will show that . We can assume 
that . that , and try to show that 
. 
Let be a Sylow system for and let 
and be Sylow systems for and 
respectively. Here is a -subgroup of . 
Since is the -subgroup of , the condition 
says that is faithfully represented as autom. orphisms of . Now 
( .), 
where . Since and are disjoint normal subgroups 
of centralizes . If , then 
is clearly contained in and so and are disjoint 
normal subgroups of and so commute elementwise. But is 
faithfully represented as automorphisms of . so is faithfully 
represented as automorphisms of . It follows from Lemma 
5.2 that . 
Returning to the general situation, if . 
whenever . and is a -subgroup of then 
 872 SOLVABILITY OF GROUPS OF ODD ORDER 
. Permuting cyclically, we would have 
. The integers being pairwise relatively prime, 
we would find for all . This is not possible since a 
S-subgroup of is solvable. 
Returning to the groups and , we suppose without loss of 
generality that . Since 
commutes elementwise with Similarly. commutes element- 
wise with . Hence is a proper sub- 
group of normalizing . By Lemma 7.5, both and 
are -subgroups of . in particular, has a normal -complement. 
Since has a normal -complement, we can find an element in 
such that is permutable with . For such an 
element , let . 
We will now show directly that for each in . contains 
a -subgroup of . This is trivially true if so suppose 
that . Let . be a Sylow system for which is 
normalized by . where . is an -subgroup of . . 
We remark that by . each is a maximal element of . 
Let : and let By Lemma 17.5 and S we 
see that , where 
and . 
Let . Let be a coset of in . Then 
contains an element of by Lemma 17.5. Hence, 
where all lie in Let 
. Since , and since centralizes , 
we have . Let so that by Lemmas 
7.3 and 7.4. Hence, contains both and , and since 
normalizes . normalizes both and . By . 
is a -subgroup of . By the conjugacy of Sylow systems in . 
there is an element in such that . Since 
has a normal -complement, . so contains 
. 
Thus, if , we have . Since , 
we have . Hence by Lemma 17.5, 
so that . Thus maps onto Since 
centralizes . it follows that 
. Hence . as required. 
If now , then and so satisfies . 
We now treat the possibility that . In this case, both 
and are -groups. By (B), both groups contain . By 
Lemma 17.4, both and are contained in so once again 
f.fltisfies . 
It remains to prove . given and for every proper subset. 
 22. LINKING THEOREMS 873 
T of a. 
Let & and >x be two S^-subgroups of ® with Sylow systems 
^Pi • • • 9 ^n and D!, • • • , On respectively, ^< and O* being SPi-subgroups 
of , l^i^n. 
If F(Ł>) and F($i) are Prgroups, we apply Lemma 17.4 and conclude 
that Ł and & are conjugate in N(Z(%)), where 5? = 
21 e ^^^7(^1) and we have normalized by taking ^ = 
If F(Ł>) is a ^-group, then CPl,Pt for i — 2, • .-, n imply that 
is a prgroup. Thus, we can assume that neither F(Ł>) nor F(^) is 
a p-group for any prime p. 
Let m< = | Op<($) |, m- = | O,4(fe) |, 1 ^ t ^ n. For each i, we can 
choose Gi in (8 so that Of* = S&, 1 ^ j ^ n, i ^ j. Let ^ = >?% 
i — 1, • • , n, so that § n » contains a S^-subgroup of . 
Suppose 0,/ftJ 0 OPj(Ł>) = <1> for some i, j, i * j. Then Op/ft*) 
is faithfully represented on F(Ł>), since Opj(^i) ^ >. But in this case, 
OPJ(&i) centralizes OPJ($>) and also centralizes OPk($Q) for fe ^ i. Hence, 
Opj(&i) is faithfully represented on OPi(§>), and so m] ^ m^ by Lemma 
5.2. For the same reasons, nij ^ m-, since OPJ($) is faithfully repre 
sented on F(Ri). If for all i, j, l^i, j ^n,i^ j, OPj(^) n Op/Ł>) = 
<1>, we find m] ^ m< ^ m^, and so m'j — nit = 1. This is not possible 
since § and & are solvable. 
Hence, we assume without loss of generality that 2)12 = 
OPl(®2) n OPl(&) ^ <1>. We will now show that Op>(®2) is conjugate to 
Opj(Ł>). To see this, we first apply Lemma 7.4 and CP]L.P4 to conclude 
that Opj(S2) and OPi(Ł>) have the same order. Since 12 centralizes 
both OPi(®2) and O,{($), it follows that S = <SR, O,j(JW, OPi(^)> c . 
By Lemma 7.4, it follows that <OPi(^3), OPi($>)> g OPi(S). By Theorem 
17.1 and Cp*,Pi, OPi(^2) and OPl(^) are S-subgroups "of OPi(8), so are 
conjugate in S, being of the same order. Since OPi(Ł>) = <1>, C^ 
follows immediately. 
22. Linking Theorems 
One of the purposes of this section is to clarify the relationship 
between 7T3 and 7r4. 
Hypothesis 22.1. 
(i) pe7T3, ?e7r(®). 
(ii) A Sp-subgroup ty of ® does not centralize every element of 
M(^; q). 
THEOREM 22.1. Under Hypothesis 22.1, if X is a maximal 
element of M(^?; q) and Q is an element of C^ of order q, then C&^Q) 
contains an elementary subgroup of order q3. In particular t q 6 TTS U 7T4. 
 874 SOLVABILITY OF GROUPS OF ODD ORDER 
Proof. Choose char in accordance with Lemma 8.2, and set 
. From 3.6 and Lemma 8.2, it follows that does not 
centralize . Since is of exponent . 
Since , Lemma 17.3 implies that . 
Since has odd order, this in turn implies that is not generated 
by two elements. Consider the chain r 
Since does not centralize . does not stabilize . so we can 
find an integer and subgroups such that 
and such that is a chief factor of and 
with the additional property that does not centralize . Since 
, we also have , where 
. Since is odd it follows that 
Since . it follows that . If 
did not contain an elementary subgroup of order , then we would 
necessarily have since is of exponent Since . 
the only possibility is that is the non abelian group of order 
and exponent . But in this case . and 
contains an elementary subgroup of order since does, by Lemma 
8.13, Lemma 8.1, and the equation . 
Hypothesis 22.2. 
(i) is a -subgroup of and . 
(ii) does not centralize eveW element of 
and does not centralize every element of . 
THBORBM 22.2. Under Hypothesis 22.2, . 
The proof of this theore m is by contradiction. The following 
lemmas assume that . 
Since Hypothesis 22.2 is symmetric in and we can assume 
that , thereby destroying the symmetry. 
Let . Let be maximal ele.nents in . 
respectively. 
LBMMA 22.1. If , is an -invariant -subgroup of and 
if a of is non cyclic, then . 
Proof. Let be a -subgroup of normalized by . Since 
, either or is empty. If 
is empty, application of Lemma 8.5 to yields this lemma. 
Suppose is non empty. Then is empty, 
so has -length one. Thus, it suffices to how that centralizes 
. We suppose without loss of generality that normalizes . 
 22. LINKING THEOREMS 875 
Then by Corollary 17.2 is contained in a conjugate of . so 
possesses an elementary subgroup of order for in . of order 
, by Theorem 22.1. We will show that centralizes . Since 
is assumed non cyclic, is generated by its subgroups which 
are elementary of order . so it suffices to show that each such 
centralizes . If does not centralize . then does not 
centralize for suitable in . By Lemma 8.4, 
is non empty for such an E. so is violated 
in . 
Since centralizes , it follows that is 
empty, since . Hence, centralizes by Lemma 8.4, as 
required. 
We define as the set of -subgroups of which have 
the additional property that no or -subgroup is centralized by . 
LEMMA 22.2. is non empty. 
Proof. Suppose that . If we also had , 
then would be violated in . Hence, , and we 
can find such that and such that 
. Consider . By Lemma 17.6, 
and it follows readily that possesses a normal comple- 
ment to We can then find in such that 
is. a -group. By Lemma 22.1 and the fact that is a maximal 
element of . we have . But now . 
Since . if is a -subgroup of . then is 
empty. By Lemma 8.13 centralizes every chief -factor of 
. It follows that centralizes , contrary to construction, so 
we can assume that . 
Suppose . Since possesses an elementary subgroup 
of order we can find A in such that is non cyclic. 
Consider . By Lemma 17.6 we can assume that 
is a -group. Then Lemma 22.1 implies that 
. being a -subgroup of . Enlarge to , a maximal 
-invariant -subgroup with Sylow system . Lemma 17.6, 
Lemma 22.1 and maximality of imply that is a maximal element 
of . contrary to . 
We can now assume that and . 
Let be an -invariant subgroup of of minimal order subject 
to . Let be an -invariant subgroup of of minimal 
order subject to . Let 
ker . Since acts irreducibly on and on . 
it follows that is cyclic, . Since . 
 876 SOLVABILITY OF GROUPS OF ODD ORDER 
. An -invariant -subgroup of satisfies 
the conditions defining , by Lemma 17.6 and in . 
Let be a maximal element of , with Sylow system 
chosen so that normalizes both and being a -subgroup 
of . 
LEMMA 22.3. is cyclic and . 
Proof. Suppose is non cyclic. Then Lemma 22.1 yields 
The maximal nature of , together with Lemma 17.6, imply that 
is a maximal element of so is conjugate to . 
By Lemma 17.3, . Since does 
not possess an elementary subgroup of order . Now and 
Lemma 8.13 imply that , contrary to construction. Hence, 
is cyclic. 
If . then . The maximal nature 
of now conflicts with Lemma 17.6 and Theorem 22.1 proving this 
lemma. 
We choose in so that ; since is also a maximal 
element of . we assume without loss of generality that . 
LEMMA 22.4. 
(i) is non abelian. 
(ii) No non identity weakly closed subgroup of is contained 
in . 
(iii) contains an element of being any -subgroup 
of containing a -subgroup of . 
Proof. We first prove (ii). Suppose is weakly closed 
in . and . Then , so the maximal nature of 
together with Lemma 17.6 imply that . 
Since , so also . Since 
Lemma 8.13 implies , contrary to construction, proving (ii). 
If were abelian, then and Lemma 1.2.3 of 
imply that , in violation of (ii). This proves (i). 
Suppose . In this case, implies , and since 
it is clear that contains an element of . Since 
it follows that and so by (B), 
. It now follows that . and so , again by (B). 
Next, suppose that . In this case, since contains 
an element of being a -subgroup of . Since 
centralizes . by Lemma 19.1, we have . Since , 
and so by (B), . it follows 
. This proves (iii). 
 22. LINKING THEOREMS 877 
To prove Theorem 22.2 we will now show that centralizes 
. Suppose by way of contradiction that this is not the 
case. We can choose such that but . Since 
is cyclic is of order . From (B), we then have 
. 
If , we apply Lemma 16.2 and conclude that 
contrary to the above statement. Hence , and by Lemma 16.3 
we have ; in particular, . Now apply Lemma 16.3 
again, this time with in the role of , and conclude that 
. 
Let . By Lemma 8.11, we have . and so 
. Hence by (B), acts trivially on , and this 
implies that , so that is elementary. 
The equality and (B) i mply that an element of 
induces an automorphism of with matrix Since divides . 
we have . 
By definition of we ave . Since 
we have a contradiction, completing the proof that centralizes 
in all cases. 
Now centralizes , so by maximality of , we have 
and (B) implies that . Hence, 
. But and since does not possess an elementary 
subgroup of order . Lemma 8.13 implies that , contrary 
to construction, completing the proof of Theorem 22.2. 
For in , let be the se t, of all subgroups of 
of type such that every element of centralizes an element 
of . We allow to depend on W. 
Hypothesis 22.3. 
(i) . 
(ii) . 
THEOREM 22.3. Unter Hypothesis 22.3, if is a -subgroup 
of and if contains an element of , then a -subgroup of 
is normal in . 
Proof. Let be the set of subgroups of satisfying the 
hypotheses but not the conclusion of this theorem and let be the 
subset of all in which contain at least one element of . 
We first show that is empty. Suppose false and in is 
chosen to maximize . Let be a -subgroup of , and let 
where and . 
Since , Hypothesis 22.1 does not hold. Hence, Hypothesis 
 878 SOLVABILITY OF GROUPS OF ODD ORDER 
19.1 holds. Apply Lemma 19.1 and conclude that 93 centralizes Oq(&). 
Suppose ftp is a Sp-subgroup of . Then p centralizes Oq(®). 
By Lemma 17.5 and Hypothesis 22.3 (i), if 91 € <$*g^(ftp), and ^ = 
F(ccy2l); ftp), then ^ S O,,p(ft). Since Sp centralizes Og(ft), it follows 
that SBiEO^ft), and so SB^ft. By Lemma 17.2, JV(Z(9S1))=O'(JV(Z(SB1))). 
Since p^ q, JV(Z(SS1)) does not possess an elementary subgroup of 
order q\ so Lemma 8.13 implies that p < ft, contrary to the definition 
of J%?. Hence,, in showing that 3% is empty, we can suppose that 
$P is not a Sp-subgroup of . 
Since 55 centralizes Og(ft), we have p • Off(ft) S N($). Since S3 is 
weakly closed in % and p is not a Sp-subgroup of , p is not a 
Sp-subgoup of JV(aS). Maximality of | ft |p implies that p < ftp • O,(ft), 
and so Op(ft) is a Sp-subgroup of Oq p($). 
Let ^P be a Sp-subgroup of ® containing ftp, and let 21 e ^^L^(^P). 
Since Op(&) is a Sp-subgroup of Offtp(ft), it follows from (B) that 
?I n ftp = 21 n Op($). By maximality of | ft |p, p is a Sp-subgroup of 
JV(Op(ft)) and it follows readily that 21 S OP(ft). But in this case, 
SS2 = F(ccZ@(2I); ftp) < ^, by Lemma 17.5. Since ftp is not a Sp-subgroup 
of , it is not a Sp-subgroup of JV(5S2), and the maximality of p is 
violated in a Sp,9-subgroup of JV(9S2). This contradiction shows that 
^ is empty. 
Now let ® be in 3ff with | ft |p maximal. Let 2B E ftp, 3B e 5^(p). 
If 722O9(^) ^= <1>, then 2B does not centralize C(W) n O,(ft) for suitable 
TF in 2B*. But in this case a Sp,g-subgroup of C(W) contains an 
element of ^/(p) and also contains non normal Sp-subgroups, and 3% 
is non empty. Since this is not the case, 2B centralizes Og($), and 
so SBx = y(ccZ@(2B); ftp) centralizes Off(ft), 2B being an arbitrary element 
of ^(p) contained in $p. Since ^p is not a Sp-subgroup of , it is 
not a Sp-subgroup of JV(SBi), so maximality of | ft |p implies that ^p 
centralizes Oq(&). Hence, OP(ft) is a Sp-subgroup of O,,p(^). Since 
<SP is a Sp-subgroup of JV(Op(ft)) in this case, Z($P) Ł OP(ft) for every 
Sp-subgroup ^P of ® which contains ftp. It follows that p contains 
an element of ^(p). This contradiction completes the proof of this 
theorem. 
If p e 7T3 U ?T4, we define n(p) to be the set of primes q such that 
p ~ q, and we set 7T3(p) = n(p) n 7T3. 
THEOREM 22.4. // p, q € 7T3 and p — <?, then n3(p) = ^3(^). 
Proo/. We only need to show that if r e 7T3 and p — r, then r~q. 
Apply Theorem 21.1, let ft be a SP,a-subgroup of ® with Sylow 
system $, Q, and let S be a Sp,r-subgroup of ® with Sylow system 
W, • 
 22. LINKING THEOREMS 879 
If Hypothesis 22.2 is satisfied, Theorem 22.2 applies and yields 
this theorem. Hence, we suppose without loss of generality that 
centralizes . 
Let . Apply Lemma 17.5 and 
conclude that . 
If also centralizes , then we also have . and 
follows from consideration of . We can suppose that does not 
centralize . 
Suppose we are able to show that contains a -subgroup 
of . Apply Lemma 17.3 and conclude that 
, where . Let be a -subgroup of 
which is contained in . Since centralizes , it follows that 
is a -subgroup of . Let be a -subgroup of , 
so that . Hence, 
, 
since . Since contains a -subgroup of , so does 
. But contains a -subgroup of as well, and so . 
Thus, in proving this theorem, it suffices to show that 
contains a -subgroup of . 
We wish to show first that some element A of centralizes a 
subgroup of with . If is non cyclic, 
then every subgroup of of type is in and since 
possesses an elementary subgroup of order , an element A is available. 
Suppose then that is cyclic. If , then of course centralizes 
. If , then and once again centralizes 
. It now follows that contains an element A whose fixed-point 
set on is non cyclic, and this implies that 
contains an element of . 
For such an element A. let be a -subgroup of which 
is -invariant and contains . Then Lemma 17..5 implies that . 
contains an element of . Apply Theorem 22.3 and conclude that 
being a -subgroup of . If is a maximal element 
of containing , then Theorem 22.3 implies that 
being a -subgroup of . By maximality of is a maximal 
element of . Since contains a maximal element of , 
namely, , so does . It follows that contains a maximal 
element of where is a suitable -subgroup of . 
But. . and so for some in . and so 
is a maximal element of normalizing . By 
Lemma 17.4, is a maximal element of . 
Now centralizes , and is a maximal element of . 
It follows that is a -group. Since and 
 880 SOLVABILITY OF GROUPS OF ODD ORDER 
are conjugate by Theorem 17.1, it follows that JV(D2)/C(Q2) is a p'- 
group, and so ty centralizes G2. By C*q, it follows that Q2 is a Sq- 
subgroup of C(^P), completing the proof of this theorem. 
THEOREM 22.5. // pex^, then © satisfies C^(p). 
Proof. By Theorem 22.4, if q, r e rc3(p), then q ~ r. By Theorem 
20.1, ® satisfies Eq r for q, r e 7T8(p). By Theorem 21.2, ® satisfies C^,,. 
Hypothesis 22.4. 
(i) P€7r3, g€7rsU7T4. 
(ii) If ty is a Sp-subgroup of , feen ^5 contains a normal 
subgroup @ o/ type (p, p) which centralizes at least one maximal 
element of M(^5; q). 
LEMMA 22.5. Under Hypothesis 22.4, 6? centralizes every element 
of M(g; q). 
Proof. Suppose false and D is an element of M(Gc; q) minimal 
with respect to 7&@ * <1>. Then Q = 7Q and @0 = CC(O) = <1>. 
Let © = C(@0). Then § contains an element 21 of ^^^K^) with 
@ Ł 21. By Lemma 17.5, 21 Ł OP,|P(&), and so O = 7O@ is contained 
in Op/($). If O* is an 2l-invariant S9-subgroup of (M$), it follows 
readily that 7d*@ ^ <1>. If O is a maximal element of M(2I; g) 
containing Q*, then @ does not centralize O. Let Q0 be a maximal 
element of M(^5; (7) centralizing @. Since Q0 is also a maximal element 
of M(2I; g), we have G0 = ^ for suitable C in C(2l) S C(C). Since © 
does not centralize Q, ^ = @ does not centralize OQ. This contradiction 
completes the proof of this lemma. 
The next theorem is fairly delicate and brings ;r4 into play ex 
plicitly for the first time. 
Hypothesis 22.5. 
(i) pe7T3, <?e7T4. 
(ii) p ~ q. 
THEOREM 22.6. Under Hypothesis 22.5, if ty is a Sp-subgroup 
of © and D! is a maximal element of M(^P; q), then Ox = <1>. If 
D2 is a Sq-subgroup of JV(Di) permutable with ty and Q3 ts a Sg- 
subgroup of © containing GS, tfe^n Dx contains every element of 
Furthermore, 
Proof. By Theorem 19.1, ^P does not centralize d, so in particular 
 22. LINKING THEOREMS 881 
Suppose that Oa contains an element 33 of ^(Os). By Lemma 
19.1, 93 centralizes OP(^^) and since S3 is a normal abelian subgroup 
of Oj, (B) implies that 93 g O^SX). Let 91 be an element of 
^^^(Qs) containing 33. Let & = JV(33) 3 <Q3, O,0POi)>. Since 
^ e 7T4, O«,($) = <1>, and (B) implies that 21 S Off($). Hence, 
[a n 0,, 0P(m>] S Of(fc) 0 OP(m) = <1>, and by (B), 2t n O, S 
O^OPO,), and so 21 n Q, C O.0PO,), that is, a n O, = « n O^ If 
"21 n QI c 21, then 21 n $3* c A^OJ, contrary to JV^Qx) C 21 n O2 = 
^ n G!. Hence, a S QI. Since g e 7r4, Corollary 17.3 implies that 
W(a; p) is trivial, so OpCPO,) = <1>. By Lemma 7.9, it follows that 
&! contains every element of ^WL^(O3), and not merely 2t. This 
proves the theorem in this case. 
We can now assume that O2 does not contain any element of 
^<(Qs)i and try to derive a contradiction. 
Since G2 is a S?-subgroup of the normalizer of every non identity 
normal subgroup of ^G,, if D(Ł^ = <1>, then Ah800(Łli)) contains an 
element of ^(O3), and since A^/Z^Oi)) = Qa in this case, Oa contains 
an element of ^(Q3), contrary to assumption. Hence, D(Qi) = <1>. 
Let D* = 0P g(^Da) n Q2. Since [Q*, DJ = [O, ,(m>, ^1 < ?PQ,, 
and since every element of ^(Ł1$) normalizes [Df, C^], we conclude 
that Q! S Z(Q*). Since D(DŁ) n Oi is normalized by every element 
of ^(Ds) and also by <pp(WQj, ^O, 0 JV(Qf)> = ^Q2, we have 
D(Of) n Qi = <1>. This implies that Of = Oi x g for a suitable 
-subgroup g of Of. 
Since Z(Qs) S O2, we have Z(O3) S Of, by (B). Since O2 contains 
no element of ^(O3), Z(Og) is cyclic. For the same reason, Z(Os) nOi = 
-<!>, since otherwise, ^(ZCDg)) S Oi and every element of ^(O3) 
normalizes Oi. In particular, Oi is a proper subgroup of Of. This 
implies that OP(^S^) * <1>. More exactly, Qx = CoffOpOPQ,)). 
Let 33 e ^<(Q3) and let ^ = CQl(»), so that | Q! : &, \ = q. 
Suppose OP(^?Q2) is non cyclic. In this case, a basic property of 
p-groups implies that OP(^Q2) contains a subgroup © of type (p, p) 
which is normal in ^5. Since Oi is a maximal element of M(^p; g), 
Hypothesis 22.4 is satisfied. Since Oi is of index g in Oi, Theorem 
22.1 implies that C^ ^ <l>. Hence, <6, @> is a proper subgroup of 
® centralizing OL Choose ^ e cc^(33) and @! e cc^(©) so that ^ = 
<33!, i> is minimal. By Dptq in ^, it follows that 5 is a p, tf-group. 
By Lemma 19.1, SB? centralizes OP(ft) and by Lemma 22.5, @? centralizes 
Og(Ł). It follows that ffi = ! x @L Let STC = JV^). Since g e 7T4, F(5«) 
is a g'-group. By Lemma 22.5, @! centralizes F(5ft) so 3.3 is violated. 
This contradiction shows that OP(^?O2) is cyclic. 
Since Oi = Cd*(Op(^O2)), it follows that g is cyclic of an order 
'dividing p — 1. 
 882 SOLVABILITY OF GROUPS OF ODD ORDER 
Let 33 = fl^Qf ) = d x fl^g), and let & = J%>2(33). We see that 
m = &op(m), & n op(m> = <i>. Let aii = jy(33), an, = c(33). it 
is clear that 2^ n ^O2 = Of, since ^ n O,0PO^ = <1>, and since Of 
is a Sg-subgroup of OpgOPQj). 
Let S = Og(2ft mod 2JIO. We see that 8 n $0, = Of, again since 
Of is a Sg-subgroup of Op gOPO2). We observe that since Of contains 
Z(Os), 2Ji contains every element of ^(O3), and so contains 33. By 
Lemma 7.1, 8 contains S3. Hence, 8i)2Wi. 
We next show that 8/23^ = S is elementary. If /)(8) ^ <1>, then 
by a basic property of g-groups, C^(/)(8)) is of order at least q\ 
Hence, Oi = CfeU>(8)) n Oi = <1>. But in this case, Ox is normalized 
by ^OPO,), &,> = ^Os, and is centralized by /> (8 mod 23^), and so 
O2 is not a Sg-subgroup of JV(Oi). This is not possible, so /)(8) = <1>. 
We have in fact shown that if 2, < 2Ji, and 2^ c 8i S 8, then CJB(SI) 
is of order q. 
Since 8 is abelian, 8 normalizes [5?, S3] = ^(ZCO,)). It follows 
that CsB(8) = fl^ZCOs)). 
Let 8i = <»TC,3tti>, and let ^ = 33^, Me2«. Since 33 and 93, 
are conjugate in 2Ji, [5S,33J is of order g_and is centralized by 8. It 
follows that PB,J8J = O1(Z(S\)). Since 8 is abelian, and since 93^ 
covers 81/2Ji1 = 8U it follows that [S3, SJ =J?1(Z(Q8)). Let | Oi | = gnf 
and 1 8j : 2Jij | = qm. Since each element of 8} determines a non trivial 
homomorphism of 2$/Ł?i(Z(Q3)) into 0i(Z(Os)), it follows that m g n 
Since CJB(SI) = ^(ZCOs)), it also follows that m ^ n. Hence, m = n. 
This implies that 8! = 8, since any (j-element of Aut 33 which centralizes 
8j is in J?lf by 3.10. Here we are invoking the well known fact that 
Si is normal in a Sg-subgroup O of Aut 33 and is in fact in ^^^(O). 
(This appeal to the "enormous" group Aut 33 is somewhat curious.) 
Returning to 8, let 33* be a Sg-subgroup of 8, and let 2B = ^(33*). 
Since 0l(®*) S Z(33*), and Z(33*) is cyclic, it is easy to see that 
^(ZfSB)) = Ol(Z(Łi9))9 and that SEB/f?1(Z(SB)) is abelian. Hence, SB is 
an extra special group of order qzn+1 and exponent q. 
We next show that 23^ is a p'-group. Since 23^ C C(Qi), it suffices 
to show that no non identity p-element of JV(Qi) centralizes 33. This 
is clear by Dp>q in JVXOj), together with the fact that no non identity 
p-element of ^pOa centralizes 33. 
Since 23^ is a p'-group, so is 8. Since 8 < 2Ji, we assume with 
out loss of generality that JV^(33) normalizes 33*. 
Let < e ^^L^(^5), and set ^ = E n ^(33). Since ?p = 
Op(SpOj) • 2J^(33) and OP(?pO2) is cyclic, ex is non cyclic. Since (^ is. 
faithfully represented on 33, it is faithfully represented on SB = ^(33*). 
Since p > q, (^ centralizes 
 22. LINKING THEOREMS 883 
We can now choose C in &{ so that Sx does not centralize SS^ = 
Cgg(C). Let 2Ł2 = [Sd, SJ. We will show that 2B2 is non abelian. 
To do this, we first show that SS^ is extra special. Let We SH^ — 
fi^ZCSB)). Since C centralizes TF, C normalizes Cn(W). Since p > g, 
C acts trivially on SB/CggC^), and so C centralizes some element of 
SB-C»(TF). It follows that Z^) = Z(SB), so that 2^ is extra 
special. We can now find 3B3 S 2Bi so that 2^ = SK22B3 and SB2 n 2B3 S 
Z(2B); in fact, we take 2B3 = (7^(28,). By the argument just given, 
2Ł3 is extra special. Since SBj is, too, it follows that 3K2 is extra special, 
hence is non abelian. 
For such an element C, let Z = C(C) 2 <(Ł, 2B2>. By Lemma 17.5, 
( S Op/,p(3:). Since 3B2 = [S82, (EJ, by Lemma 8.11, it follows that 
2B2 C Opf(Z). It follows now that M((Ł; g) contains a non abelian group. 
But now Theorem 17.1 implies that the maximal elements of M((Ł; q) 
are non abelian. Since C^ is a maximal element of M(E; q) and X 
is elementary, we have a contradiction, completing the proof of this 
theorem. 
THEOREM 22.7. // p, q$ ;r3, and p — q, then u(p) = n(q). 
Proof. Suppose p ~ r. By Theorem 22.4, we can suppose that 
re7T4. Proceeding by way of contradiction, we can assume that a 
S9-subgroup O of © centralizes every element of M(Q; r), by Theorem 
22.1. By Theorem 19.1, a Sp-subgroup ^3 of © does hot centralize 
every element of M(^P; r). Applying Theorem 22.2, we can suppose 
that ^P centralizes every element of M0P; q). 
Let Oi be a maximal element of M0P; q) and let 9^ be a maximal 
element of M(^$; r). Let 5R2 be a Sr-subgroup of N(^i) permutable with 
<$ and let 5R3 be a Sr-subgroup of ® containing 5R2. Let 21 e ^^^l(^). 
By Theorem 22.6, OP(^5R2) = <1>, so 2t does not centralize 9^. We 
can then find A in 21* such that SRf = [C^^A), 21] ^ <1>. 
Suppose Q! is non cyclic. Then by Cp*><z, di contains an element 
of J*r(q). Let ^ = C(A)3<a,ai1*,OI>=>a:, and let ft be an 21- 
invariant S, r-subgroup of OP/(X) with Sylow system Sr, O^ By Theorem 
22.3, Dx < S. Since N(Łk) = 09(N(&d), it follows that 72Iftr - <1> 
by Lemma 8.11 and the fact that JV(Oi) does not contain an elementary 
subgroup of order r3. This violates the fact that SRf = 75Rf2I ^ <1>, 
by Dpr in . Hence, Oj is cyclic. 
Since Td^ = <1>, ^ = OP(^Q2) ^ <1>, where O2 is a Sg-subgroup 
of ® permutable with ty and containing SX, which exists by C*>g. 
Since JV^Pi) = O9(JV(^Pi)), it follows that C^ Ł Z(Q2), X being a S7- 
subgroup of <V(Am». 
Let S3 = V(ccy2l); 5P), and 9^ = JV(Z(93)). By Lemma 17.3, 5^ = 
 884 SOLVABILITY OF GROUPS OF ODD ORDER 
'Op^) • #! n N(MJ. Since k is a Srsubgroup of O,/^), it follows 
readily that 9^ n JV(3ti) contains an element of ^*(g). In particular, 
JV(9ti) contains an element of ^(q). If S is a Sq r-subgroup of JV(Sfti) 
with Sylow system 29, 8r, then 2q < S, by Theorem 22.3. By Theorem 
22.6, Sfti contains an element ( of &*g^(ftj. By Corollary 17.3, 
M((Ł) is trivial. Since 2g € M(E), we have a contradiction, completing 
the proof of this theorem. 
23. Preliminary Results about the Maximal Subgroups of © 
Hypothesis 23.1. 
(i) & is a non empty subset of 7T3. 
(ii) For at least one p in itf, ttr = n(p). 
We remark that by Theorem 22.7, Hypothesis 23.1 (ii) is equivalent 
to 
(ii)' 
n(p) = tar for all pin-tf. 
Under Hypothesis 23.1, Theorem 22.5 implies that @ contains a 
Stf-subgroup >. Since > also satisfies E&1 for all subsets tarx of tsr, § 
is a proper subgroup of © by P. Hall's characterization of solvable 
groups [15]. This section is devoted to a study of § and its normalizer 
2Jt = JV(Ł>). All results of this section assume that Hypothesis 23.1 
holds. Let -or = {plf • • , pn}, n^l, and let $&,•••,$* be a Sylow 
system for >. 
LEMMA 23.1. W is a maximal subgroup of © and is the unique 
maximal subgroup of © containing >. 
Proof. Let $ be any proper subgroup of ® containing >. We 
must show that $ g 3Ji. Since ® is solvable we assume without loss 
of generality that & is a -cr, tf-group for some # g tar. Let SR, • • , $pn, 
} be a Sylow system for $. It suffices to show that S& < ^O. 
Since g g tsr, PI^ q. Theorem 22.1 implies that ^i centralizes 
By Lemma 17.5, S3 < SRO, where S3 = F(ccZ@(Sl); ^) and 
By Lemma 17.2, 3^ = JV(Z(»)) = OPl(9^). Since % 
does not contain an elementary subgroup of order #3, Lemma 8.13 
implies that spx centralizes every g-factor of ^Q and so ^ < 
completing the proof of this lemma. 
LEMMA 23.2. // p< e n(F(§)), and ^ 6 ^^x^(^), then 
Proof. We can assume that i = 1. By C*ltPJ > contains a SPJ- 
subgroup of CC&i) for each ^* = 2, • • • , n. Thus, it suffices to show 
that if q&'tf, and O is a Sg-subgroup of C(SIi) permutable with SR, 
 23. PRELIMINARY RESULTS ABOUT THE MAXIMAL SUBGROUPS OF ® 885 
then G S 2JJ. 
By the preceding argument, 5R < 5&Q. Since S& normalizes 0(31!) = 
! x 3), ® being a pi-group, it follows that 5RQ = ?& x Q. 
Since F($) n SR = <1>, it follows that 3K = AXTO) n %), since 
F(i&) fl $Pi char > < 2ft and 2Ji is the unique maximal subgroup of © 
containing >. The lemma follows since N(F($) n %) 3 
LEMMA 23.3. Le l^i^n, and let 3l e 
S 2ft. 
Proof. We can assume that i = 1. If F($) is a Prgroup, then 
Lemma 17.5 implies that ^ < § and so S^ < 2ft, since ^ is weakly 
closed in F(Ł>) n 5R. In this case, JV(SSi) = 3K and we are done. 
We can suppose that F(iQ) is not a Pi-group, and so Z = Op/(^) = 
<1>. Let Z,, • • • , Zn be a ^-invariant Sylow system for Z, where f 
is a S^-subgroup of Z and we allow Zt = <1>. By C*l§Pi, Z, is a 
maximal element of M^; p{). 
Let Ne N(^). Then by Theorem 17.1, Zf = Z? where C2, • • - , Crt 
are in 0(31,) S SK. Since 2 char $ < 3M, each Z?< is contained in 2 
.and so 2* = Z. Since 2 ^ <1>, 5K = JV(S) 3 ^(550, as required. 
LEMMA 23.4. Let 1 ^ t^ n, 21, e ^K^(Sp4), 5S = F(crfa(3r4); sp,)- 
./jf <C(2Ii), JV(SSi)> E 3Ji, ^feen 2Ji is the unique maximal subgroup of 
<$ containing $&. 
Proof. We can assume that Ł = 1. Let 1 be a g-subgroup of 
<S permutable with $&. It suffices to show that O S 2W. 
Since D = Og(^O) • ^(SSi), it suffices to show that S^ = Og(^O) S 
UK. If D is centralized by 5plf then by hypothesis O Ł 3Ji. Otherwise 
Tve apply Theorem 22.1 and conclude that getsr. By Theorem 17.1, 
JO? C $ for suitable C 6 CO^) S SK, and the lemma follows. 
LEMMA 23.5. For each i = l, ,^, i/ Sli 6 
S SW, and 9Ji is ^fee unique maximal subgroup of © containing 
Proo/. First, suppose p{ e 7r(F(^)). Then (7(2^) E 2W, by Lemma 
23.2. Then by Lemma 23.3, #(«<) S 9R, 93, - F^c^SI,); ^), and then 
by Lemma 23.4, this lemma follows. We can suppose that p< g n(F(iQ)). 
We assume that i = l. Let C(2ti) = 2^ x , where S) is a pi-group. 
It suffices to show that for each q in 7r(S)), 2Ji contains a S9-subgroup 
JO of . If q € -or , this is the case by Cp*,g, so we can suppose that 
Since pl < 7r(F(Ł>)), ^ does not centralize F($). If F(§) were cyclic, 
p = max {P!, • • • , pw}, then a Sp-subgroup of § would be contained 
in F(!Q) and so be cyclic. Since this is not the case, F(Ł>) is non cyclic, 
 SOLVABILITY oF GROUPS oF oDD ORDER 
SO we can assume that is non cyclic. We can then find A 
in SO that contains an element of , say . 
Let ., and let be of' 
with Sylow system , where and . Since 
by Lemma 17.5, it follows that is a 
central factor of . Hence, is of and SO 
. 
We apply Theorem 22.3 and conclude is a 
elernent of it follows . 
is a maximal element of . By 
contains . By Theorem 17.1, there is an element in such 
that . Since normalizes , it follow that 
contains of . But , SO by what is. 
already proved, we have , and SO contains . 
We apply Lemmas 23.3 and 23.4 and complete the proof of this lemma.. 
24. Further Linking Theorems 
LEMMA 24.1. If then . 
pr00f. If , there is nothing to prove, so suppose 
Corollary 19.1 implies that . Let . We must 
show that . 
If and is of , then does not centralize 
every element of and does not centralize every element of 
. By Theorem 22.2, we have . 
If . then since also , we have . by Theorem 22.7. 
This completes the proof of this lemma. 
If let By Theorem| 
22.7 and Lemma 24.1. . It from 
that . By Theorem 22.5, satisfies . Let. 
be of . Clearly, since . 
It is easy to see that is non cyclic. Choose SO that the 
of is non cyclic. Let be a Sylow system 
for being a of . Thus, 'Jet is non cyclic, 
SO that contains a subgroup of type which is normal 
in . Let be an element of which contains . Let 
be a maximal element of . By Lemma 24.1 and Theorem 
22.6, . Let being . 
THEOREM 24.1. is . 
pr00f. Let be the set of -invariant subgroups of such 
 24. FURTHER LINKING THEOREMS 
Since . it follows that 
Suppose , and . Since normalizes 
By Lemma 17.6. . 
Let be the image of under the projection of onto 
Since , we see that is a self centralizing sub- 
group of and it follows readily that is 
centralized by and . By Lemma 1.2.3 of [Z1], we have 
and hence is a . 
Let . be an -invariant Sylow system of being a 
-subgroup of . If {. } it follows from that 
is a aximal element of Since. . this implies that 
is a -subgroup of . If {. }, then Theorem 
22.1 implies that centralizes . so that . Finally, if , 
then there is an element of such that , by Theorem 17.1. 
Let be a fixed -subgroup of . By the preceding 
paragraph, is a -subgroup of . and is a -subgroup of 
. Since , it follows that is permu-. 
table with so that is a proper -subgroup of . This means 
that contains a unique maximal element. Since is A-in- 
variant for each . since , and since 
. the theorem follows. 
THEOREM 24.2. Let and . Then 
contains is a maximal subgroup of and is the only 
maximal su bgroup of containing . 
Proof. Since is a proper subgroup of . We first 
show that contains . Let be an -invariant -subgroup of 
, so that is a maximal element of . either by virtue of 
, or by virtue of so that centralizes . For in 
for some in by Theorem 17.1 together with 
. Since is a -subgroup of . Hence, . and 
so . Thus, . 
To show that . we use the fact that , 
where Since , it suffices to show that 
We will in fact show that . Let be a 
invariant -subgroup of . If . then . so that 
normalizes . Hence, normalizes , and we see that 
. Thus, contains and . 
Let be a maximal subgroup of conffiining . It is easy 
to see that by Lemma 7.3, so that , and is a 
maximal subgroup of . 
 888 SOLVABILITY OF GROUPS OF ODD ORDER 
Let be any proper subgroup of containing . To show that 
, it suffices to treat the case that is a -group. Let 
be a -subgroup of permutable with . Since , it suffices 
to show that . This is clear by if . If 
, this is also clear, by Theorem 17.1, since and 
If . then : centralizes by Theorem 22.1, and 
we are done, since . 
If , and is a -subgroup of . we define 
{ contains some element of }. 
contains a subgroup of type 
such that for each in . 
LEMMA 24.2. If and is a -subgroup of then 
every subgroup of which contains a subgroup of type 
is in . 
Proof. Let , so that is non cyclic. 
Let be a subgroup of of type . If . then . 
Since is contained in an element of . it follows that 
is in . 
THEOREM 24.3. If is a -subgroup of , am) is 
cmtained in a unique maximal subgroup of , then each element of 
is contained in a unique maximal subgroup of . 
Proof. Let be the unique maximal subgroup of containing 
. We remark that if this theorem is proved for the pair . 
then it will also be proved for all pairs where . This 
prompts the following definition: is the set of all subgroups 
of such that contains for some in and 
some . Clearly . 
Suppose some element of is contained in a maximal sub- 
group of different from . Among all such elements of 
let be maximal. By hypothesis, . Let be a maximal 
subgroup of different from which contains and let be a 
-subgroup of which contains . If , then . 
Since . maximality of implies that , so 
that for some in Since , so also . 
But , and maximality of is violated. Hence, 
is a -subgroup of . 
Let be chosen so that for some . 
Since every element of is contained in , every element of 
 24. FURTHER LINKING THEOREMS 889 
is contained in . Hence . If 
, then , so , by maximality of 
Since by Lemma 17.6, we find that . 
contrary to assumption. The theorem is proved. 
THEOREM 24.4. Let and let be a -subgroup of . 
If each element of is contained in a unique maximal subgroup 
of . then for each each element of 
is the unique maximal subgroup of containing . 
Proof. For , let be the set of subgroups 
of such that contains a subgroup of type such that 
contains an element of for some and all . Here 
denotes the set of . Suppose . or 4 
is minimal with the property that some element of is con- 
tained in at least two maximal subgroups of . This implies that 
does not contain any elements which are contained in two 
maximal subgroups of being an arbitrary element of . Choose 
in with maximal subject to the condition that is 
contained in a maximal subgroup of with . We see 
that is a of . Let be a subgroup of of type 
with the property that contains an element of 
for suitable in , and each in . (We allow to depend on 
Q.) Since is generated by its subgroups , 
it follows that . 
Let be an element of . Then . or we are 
done. Let Since by (B). it 
follows that . Hence, , by maximality of . 
Since , we have . completing the proof 
of this theorem, since . 
THEOREM 24.5. If and is a -subgrmp of then 
is contained in a unique maximal subgroup of . 
Proof. If , this theorem follows from Lemma 23.5. 
Suppose . Let , where and 
let . be a -subgroup of containing . If is non 
cyclic, we are done by Theorem 24.2, so we suppose that 
is cyclic. 
Let be the unique maximal subgroup of containing . Suppose 
we are able to show that for some in . Since 
is cyclic, is also cyclic. Hence, If 
. then normalizes , by Theorem 17.1, 
 SOLVABILITY oF GROUPS oF oDD ORDER 
Logether with . Since for every proper 
subgroup of which contains . it suffices to show that every 
element of is contained in . This follows readily by , 
Theorem 22.1 and . 
Thus, it suffices to show that . Choose such that , 
of , is non cyclic, and let : be of 
permutable with . It suffices to show that for 
some , by Theorems 24.8 and 24.4 together with the fact that 
is the unique maximal subgroup of containing . 
Let , so that is a maximal element of . 
By Lemma 17-8, . Since is contained in and no 
other maximal subgroup, . Thus, if is generated by 
two elements, then centralizes and we are done. If 
is non cyclic, then every subgroup of of type is contained 
in . Since contains a subgroup of type 
is non cyclic for some in , and we are done in this case. There 
remains the possibility that is cyclic, while is not 
generated by two elements. Since every subgroup of of 
type which contains is contained in , by Lemma 
24.2, and since contains such a subgroup for some in ,we 
are done. 
The preceding theorems give precise information regarding the 
of the maximal subgroups of for in . 
THEOREM 24.6. Let let be a maximal of 
. If is of and is n0t of . 
then contains a cyclic subgroup of index at . 
Pr00f. Let be of containing . let 
and let so that |--1 or . If is non cyclic, 
then . and so is contained in a unique maximal sub- 
group of , which must be , since . But , a con- 
tradiction, so is cyclic, as required. 
Theorem 24.6 is of interest in its own right, and plays an important 
role in the study of to which all the preceding results are now 
turned. 
24.1. 
1. . 
2. is 0f . 
8. is a proper subgroup of such that 
( i ) . 
(ii) If . there is a subgr0up 0f in 
 24. FURTHER LINKING THEOREMS 891 
accordance with Lemma 8.2 such that is generated by two ele- 
ments. 
THEOREM 24.7. Uruder Hypothesis 24.1, is contained a 
unique maximal subgroup of and centralizes . 
Proof. Let be any proper subgroup of containing . We 
must show that centralizes . 
By Lemma 8.2, ker is a 8-group, so is contained in 
. It follows that and in particular . 
Suppose . Then , so is 
generated by two elements. Since is odd, a -subgroup of 
centralizes , so centralizes . Since also centralizes . 
we have . 
Suppose . Since is a normal abelian subgroup of 
we have . Since , we conclude that 
. By the preceding paragraph, centralizes 
. Thus, it suffices to show that centralizes . 
8ince , and since normalizes . it 
suffices to show that centralizes . Let . 
.so that contains . Since is a normal abelian subgroup of . 
(B) implies that . Hence, , which implies that 
induces only 8-automorphisms on , and suffices to complete 
the proof. 
Hypothesis 24.2. 
1. . 
2- is a -subgroup of . 
8. If is any proper subgroup of containing and if 
then every subgroup of chosen in accordance with Lemma 
8.2 satisfies . 
REMARK. If , then Hypothesis 24.1 and Hypothesis 24.2 
exhaust all possibilities. 
LEMMA 24.8. Und er Hypothesis 24.2, an element 
of such that the normal closure of in is abelian. 
Proof. If is non cyclic, every element of satisfies 
this l.emma. Otherwise, set , and let be a non 
cyclic normal abelian subgroup of R. Since contains an 
element of which meets the demands of this lemma. 
THEOREM 24.8. Let and let be a -su bgroup of . If 
assume that contains an element such that th e normal 
 892 SOLVABILITY OF GROUPS OF ODD ORDER 
closure 0f in is abelian. If , let be any ele- 
ment of . If is any pr0per subgr0up of such that 
and, if is a 0f , then , where- 
' . and is the largest normal subgr0up of which- 
centralizes . 
Pr0of. 0bSVSer that contains . 
Since mod mM . maximality of guaran- 
tees that mod . If , then Sylow's theorem yields 
this theorem since is weakly closed in . 
Suppose by way of contradiction that . Let mod . 
By Lemma 1-2.8 of [21]. . 
Let , and let . Let be the normal 
closure of in . Suppose . Since 
and since by Sylow's theorem, we see that 
Maximality of implies that . In particular, . Since- 
. , by Sylow's theorem we have mod , which is not 
the case. Hence, . Since , we also have 
. Since . the identity 
implies that contains a conjugate of such 
that . Since application of Theorem C of 
[21] to yields a special 
such that acts irreducibly and non trivially on . Since 
is , and does not centralize does not centralize 
. Furthermore, . where and 
. and is invariant under . 
Since is a -group and . we have . where 
and . If , Lemma 18.1 gives 
an immediate contradiction. If . and , we also have 
a contradiction with (B), since . If , Lemma 
16.8 implies that is cyclic, and that . However, 
the normal closure of in is abelian, and so 
the desired contradiction, completing the proof of this theorem. 
REMARK. Except for the case , and the side conditions 
and Theorem 24.8 is a repetition of Lemma 18.1. 
Hypothesis 24.8. 
1. . 
2. is of is a maximal element 0f , 
is a 0f permutable with . 
8. is a of containing . where 
 24. FURTHER LINKING THEOREMS 893 
for . the normal closure of in is abelian. 
4. . 
THEOREM 24.9. Unxter Hypothesis 24.3, ei th er contains as 
element of or contains an element of . Furth er- 
more, and satisfies . 
Proof. Let be the largest normal subgroup of which 
centralizes . Then . by Theorem 24.8. Since , 
. If is non cyclic, then . 
Suppose is a non identity cyclic group. By Lemma 17.6, 
. Since a Sylow -subgroup of is cyclic, it follows that 
centralizes . where . and so . If 
centralizes . then contains a -subgroup of . 
In this case, normalizes for some in . Let be a 
-subgroup of containing , with . By the conjugacy 
of Sylow systems in . we have for suitable 
in . Hence, normalizes and . Since is weakly 
closed in and we are done. If does not centralize 
, then is a -group, since is cyclic. In 
this case the factorization, , together with . 
yields that , for some subgroup of . This in 
turn implies that every non cyclic subgroup of is in . 
Since and is cyclic, the -subgroups of 
are non cyclic. Hence, contains a non cyclic subgroup such that 
normalizes for some in . By the conjugacy of Sylow 
systems, we can find in such that and . Since 
is weakly closed in , and we are done, since every non 
cyclic subgroup of is contained in . 
Suppose . Then is a -group. From , 
we conclude that normalizes for some in and the conjugacy 
of Sylow systems, together with the fact that is weakly closed in 
imply that normalizes . This completes the proof of the first 
assertion of the theorem. 
If then S].nce every element of is 
contained in a unique aXlmal subgroup of by Theorem 24.3, 
if contains an element of , then is not a -subgroup 
of . But is a maximal -subgroup of , by Lemma 7.3. If 
contains an element of , then since by 
(B) and Theorem 22.7, we see that contains an element of 
. Hence, . Thus, in all cases, Since also 
contains a -subgroup of , satisfies Since is contained 
in and no other maximal subgroup of satisfies as required. 
 894 SOLVABILITY OF GROUPS OF ODD ORDER 
Hypothesis 24.4. 
1. . 
2- is a -subgroup of . 
8. contains a subgroup which is elementary of order 27 
with the property that for all . 
Hypothesis 24.5. 
1. . 
2. A -subgroup of is contained in at least two maximal 
subgroups of . 
LEMMA 24.4. Assume that Hypothesis 24.5 is satisfied and that 
if Hypothesis 24.4 is also satisfied. If , let be an 
arbitrary element of . If , let be the subgroup 
given in Hypothesis 24.4. Let be the weak closure of in . and 
let be the subgroup of generated by its subgroups such that 
and is cyclic for suitable in . Let be a proper 
subgroup of coutaining . with the properties that is a q- 
group for some prime and has -length at most two. Let 
be any one of the pairs Then 
, where normalizes and is a -group, and 
normalizes . 
Proof. Let be a -subgroup of . and let . Then 
. The lemma will follow immediately if we can show that 
normalizes and induces only -automorphisms on X. 
Suppose by way of contradiction that either some element of 
induces a non trivial -automorphism on , or does not normalize 
I. If , we can find such that either some element 
of induces a non trivial -automorphism of or else does 
not normalize X. Similarly, if , we can find and in 
such that is cyclic and such that either some element 
of induces a non trivial -automorphism of or else 
does not ormalize . 
Let , so that Since jg gener- 
ated by the subgroups which h. ave the property that acts 
irreducibly and non trivially on we can find 
such that either does not normalize or some element of 
induces a non trivial -automorphism on , and with the additional 
property that acts irreducibly on . 
Let . so that 
is cyclic. Let , and Since and since 
 24. FURTHER LINKING THEOREMS 895 
it follows that . Also, since is a normal 
abelian subgroup of , we have . 
Suppose that . If . then since is generated 
by two elements, it follows that . Hence, . 
Since the normal closure of in is abelian, we have 
, and (B) implies that of centralizes . so 
centralizes . 
Suppose . We first treat the case that for some 
e and is 
of index at most in . If is a of containing , 
then . Hence, , and so 
. It follows that . (Note that since ) 
Hence, . and so , so that 
of centralizes and so centralizes . 
We can now suppose that for all such that 
. In this case, since is generated by two ele- 
ments, there is a normal elementary subgroup of of order such 
that . Hence, . Since we can 
find in . Consider . Since is 
cyclic, if and , then . Let . 
Let be a of containing , so that 
or . We have . and so 
. This implies that . Let ; then . 
, so that Since it 
follows that . Since . and since 
, we have . This shows 
that of centralizes and so centralizes . 
Suppose now that , so that . In this case, 
. Hence, since is cyclic. Since is con- 
tained in , if denotes the normal closure of in , then 
centralizes being a subgroup of . 
Let so that is normal in . If , 
we have , since , and it follows that 
group of centralizes . Namely, if r is part 
of a chief series for , then centralizes each , so that a 
of centralizes each . so centralizes . If . 
then for some . and we have 
. and we are done. 
THEOREM 24.10. Umcler Hyp0thesis 24.5, . 
Furtherm0re, Hypothesis 24.4 is m0t satisfied,. 
Pr00f. Suppose that either or Hypothesis 24.4 is satisfied. 
Let be any element of in case and let be the 
 896 SOLVABILITY OF GROUPS OF ODD ORDER 
subgroup given by Hypothesis 24.4 in case p = 3. Let 2B, 2B* be as 
in Lemma 24.4. Let ^ = JV(Z(SP)), 9?2 = JV(2B), 5K3 = <Y(Z(2B*))f and 
let § be any proper subgroup of © containing $p. Then by Lemma 
24.4 and Lemma 7.7, we have $ = ($ n 9*i)($ n %) = ($ n 9W(ft 0 913) = 
($ n 5K2)(€> n Si,). Taking ft = %, we get 8^ S 9W, MI E 9W. Taking 
& = 5tt2, we get S^E^Ws^SS^. Taking Ł = 9t3, we get 9l3g 
9*1%, 9^3 ^ 9^1- By Lemma 8.6, we conclude that 9^% is a group 
and so ^SS^SRa for every proper subgroup > of © containing $p. If 
UyRa = , then Op(SRi) is contained in every conjugate of 912, against 
the simplicity of . Hence, SyJla is the unique maximal subgroup of 
© containing $p. 
We can now suppose that p = 3 and that Hypothesis 24.4 is not 
satisfied. Suppose q € ?r(3), q = 3. Let O be a S9-subgroup of ® 
permutable with sp and let 2Jt be the unique maximal subgroup of ® 
containing Q. If § = O3(3K) and E is a subgroup of $ chosen in 
accordance with Lemma 8.2, then Theorem 24.7 yields that m(Z(E)) ^ 3. 
Let © be a subgroup of Q of type (q, q, q) and let QV(Z(&)) = 
<E! x • • • x (Łr, each (Ł; being a minimal -invariant subgroup. If Gr 
centralizes Z(€), then any subgroup of Z((Ł) of type (3, 3, 3) will 
serve as 21. This is so, since in this case, C(A) Ł 9J1 for all A in 21*. 
Otherwise, | E< | ^ 27 for some i, and since Gr/C@(Łi) is cyclic, 
C@((Ł<) e J*I(Q), so we let 21 be any subgroup of <Ł< of type (3, 3, 3). 
The proof is complete. 
25. The Isolated Prime 
Hypothesis 25.1. 
1. 3 e 7T4. 
2. A S3-subgroup ^ of ® is contained in at least two maximal 
subgroups of . 
THEOREM 25.1. Under Hypothesis 25.1, there is a q-subgroup 
O of ® permutable with ^ such that if$ = $PO and if Sjj, O are 
the images of ^3, Q respectively in >/O3(Ł>), ^Aew ^ ^ 1 is cyclic, ^ 
ts faithfully and irreducibly represented on S/D(Ł}), and Q does not 
centralize S3 = 
Proof. There is at least one proper subgroup of © containing ty 
and not normalizing ZCP)9 since otherwise JV(Z($P)) is the unique 
maximal subgroup of © containing ?p. Let & be minimal with these 
two properties. Then $ = $PQ for some g-group O. Since 3 e :r4, 
O,(Ł>) = 1. Since ^^x^(D) is empty, § has g-length 1. Hence, 
 25. THE ISOLATED PRIME 897 
O3(&)Ł1 < ft. By Lemma 8.13, ^ is abelian. By minimality of ft, $ 
acts faithfully and irreducibly on Q//)(Q). If SJJ = 1, then ?p < ft, 
and } normalizes Z(^P), which is not the case. 
Since O does i^ot normalize Z(^P), Q does not centralize Z(O3(ft)) 
so does not centralize #i(Z(O3(Ł>))). The proof is complete. 
We will now show that Hypothesis 24.4 is satisfied. ^D is rep 
resented on 33 = 0i(Z(O8(ft))), and it follows from (B) that the minimal 
polynomial of a generator of ty is (x — I)1*1. Hence, there is an ele 
mentary subgroup 21 of S3 of order 27 on which ty acts indecomposably. 
Let <P0 = C^(2I) and let @ - JWZflR,)) so that 21 Ł @. Choose Ae 21*. 
and set & = C(A). Let ^P* be a S3-subgroup of ( containing ^0- (It 
may occur that ^5 = ^P* but this makes no difference in the following 
argument.) If S& = $*, then 72&2l2 = 1. Suppose | <P* : ^0 1 = 3. Then 
<$P, ^P*> Ł N(yQ), so that <<P, $P*> normalizes C. Since p and ^P* are 
conjugate in JV(^P0)> any element of ^P* — $P0 has minimal polynomial 
{x - I)3 on e. 
Let A = O3(<Ł). Then | A : A n % I - 1 or 3, so that 7®@ s ^Po, 
and 72^®2 = 1. By (B), @ S ft. If • ft S ^Po, then @ Ł Z(ft), and 
72e2t2 = 1. Suppose f ft : ft n % I = 3. Then Z)(ft) S ^P0, so that 
G Ł Cft(jD(ft)). If Cft(D(ft)) Ł %, then © S Z(C^(D(St)))9 and once 
again 72S2l2 = 1. Hence we can suppose that C%(D($t)) contains an 
element K of ® — ft n % Since ft C ^P*, it follows from the preceding 
paragraph that the class of C$(D(&)) is at least three. On the other hand, 
if X and Fare in C$(D(Sl)), then [X, Y\ e CK(D(St)) n SV. Since & Ł Z)(^), 
we have [JT, Y, Z] = 1 for all ^T, F, Z in Cfi(-D(ft)). This contradiction 
^shows that 72E2I2 = 1 for all A in 21*. Combining this result with 
the results of Section 24 yields the following theorem. 
THEOREM 25.2. // pe7T4, and ^ is a SP-subgroup of , then ty 
is contained in a unique maximal subgroup of . 
THEOREM 25.3. Let pen* and let ty be a Sp-subgroup of . Then 
each element of J^ffi) is contained in a unique maximal subgroup of . 
Proof. First, assume that if p = 3, then ^($P) contains an ele 
ment S3 whose normal closure in C(Z(^P)) is abelian, while if p ^ 5, 
53 is an arbitrary element of ^(^P). 
Let 2Ji be the unique maximal subgroup of ® containing $p. Let 
J^*(^p) be the set of subgroups ^50 of ^P such that ^?0 contains S^ for 
suitable E in *9^l^J(^P), M in 9JI. Suppose by way of contradiction 
that some element % of J^*(^P) is contained in a maximal subgroup 
^ of © different from 2Ji, and that |$P0| is maximal. It follows 
readily that ^?0 is a Sp-subgroup of 2)^. Since % contains E^ for 
 898 SOLVABILITY OF GROUPS OF ODD ORDER 
suiffible in in Thus the 
of Theorem 24.8 are satisfied, playing the role of and the 
role of Since and since 
( being self centralizing), we conclude from the factorization 
given in Theorem 24.8 and from the maximality of that . 
There remains the possibility that for every in , the 
normal closure of in is non abelian, and . 
Let . If contains a on cyclic characteristic abelian 
subgroup then contains an element of and is ab e- 
lian. Since we are assuming there are no such elements, every 
characteristic abelian subgroup of is cyclic. The structure of is 
given by 3.5. If is any element of then by 
(B). so . 
As before, let be chosen so that is contained in 
a maximal subgroup of different from , with maximal. 
Then is a -subgroup of and . 
Let . Since implies that . 
Since by maximality of , we conclude that We 
need to show that . Consider . Since 
, we conclude that and maximality of 
implies that so it suffices to show that 
and it follows readily from 
that it suffices to show that . Since , we have 
, so that , and induces only 3-auto- 
morphisms on , so centralizes , and follows in case 
. 
Suppose . If . then , and since 
implies that . In this case, 
. so . There remains the possibility 
that . 
If , then and (B) is violated. Hence, . so 
that . Hence, . If , then 
and we are done. If , we conclude that centralizes 
, since . This is absurd, since 
by (B) applied to , completing the proof of this theorem 
Before combining all these results, we require an additional result 
about . 
THEOREM 25.4. Let , let be a -subgroup of let 
be the unique maximal subgroup of containing . Then . 
Proof. Let , and suppose in has the property 
that Then . By Theorem 25.3, we have 
 25. THE ISOLATED PRIME 899 
so that . Hence . By (B) 
and for each in . Hence, so maxi- 
mality of implies . By uniqueness of (or because 
is weakly closed in , we have . Furthermore, 'by Theorem 
25.8. if then '. Thus, is not in the kernel 
of the permutation representation of on the cosets of in . 
We can then find in such that has order in . 
so Theorem 14.4.1 in [12] yields this theorem. 
We are now in a position to let and coalesce, that is, we 
set . 
THEOREM 25.5. Let be a matcimal subgroup 0f . If 
is a 0f , then either is a 0f 
or has a cyclic mbgroup 0f inlex at most 
f0r every 0f . If is the largest subset 0f with 
the property that c0ntains 0f then ';He, 
. 
Pr00f. Let be a of containing . Suppose 
. Then . by Theorems 24.8, 24.5, and 25.8. Thus, 
if , then is cyclic. Since 
or the first assertion follows. 
Let be of for in W. (If W is empty there 
is no more to prove.) If , then by uniqueness of 
and Lemma 17.2. If . then by uniqueness of and 
Theorem 25.5. Hence, . If , then central- 
izes every chief -factor of , by Lemma 8.18. Since we 
conclude that . 
THEOREM 25.6. is rtiti0ne1, into non empty subsets r 
, with the f0ll0wing pr0perties: 
( i ) If , then satisfies if 0nly if f0r 
some . 
(ii ) If is of , then is a maximal 
subgroup of . .. is 0f square free for 
each . 
(iii) If is of , if 
for some ., then is of 0rcler 
, where is cyclic, . 
Pr0of. By Lemma 8.5, is non empty. By Corollary 19.1, 
Theorems 24.8, 24.4, 24.5, 25.2 and is an equivalence relation on 
and if are the equivalence classes of under ' then (i) 
holds. 
 900 SOLVABILITY OF GROUPS OF ODD ORDER 
Let : be a -subgroup of and let be a 
subgroup of for . By Theorem 25.5, is a maxi- 
mal subgroup of , and . 
Suppose and . If is any non identity 
characteristic subgroup of . then either or 
by Theorems 24.3, 24.4, 24.5, 25.5 and 25.3. Since 
contains every element of both and , we con- 
clude that is elementary of order or . Suppose . If 
contains then contains an element of , so that 
. If does not contain then 
contains an elementary subgroup of order , so once again 
. The same argument applies to . so that . 
Hence for some in . Hence . so . 
contrary to hypothesis. Hence, is of order . 
If . then , so that . contrary 
to the fact that has order 1 or for all in , by the 
preceding paragraph. Hence, . If , and 
, then is of index at most in and is 
disjoint from . since . Hence, . This 
proves (iii), the cyclicity of following from . The proof 
is complete. 
26. The Maximal Subgroups of 
The purpose of this section is to use the preceding results, notably 
Theorems 25.5 and 25.6, to complete the proofs of the results stated 
in Section 14. 
LEMMA 26.1. If is a -subgroup of then 
. 
Proof. If is abelian, the lemma follows from theorem 
and the simplicity of . If is non abelian, is not metacyclic, 
by 3.8. Also, . as already observed several times. Thus, from 
3.4 we see that is a non abelian group of order . The hypo- 
theses of Lemma 8.10 are satisfied, so by Theorem 
14.4.2 in [12] and the simplicity of . Since and 
since has -length one, the lemma follows. 
LEMMA 26. 2. If and is a -subgroup of , then is 
abelian or is a central product of a cyclic group and a non abelian 
group of order and exponent . 
Proof. We only need to show that is not isomorphic to (iii) 
 26. THE MAXIMAL SUBGROUPS OF 
in 3.4. Suppose false. Let . and let be a fixed 
subgroup of Set . The oddness of guar- 
antees that is abelian. 
Let be a chief series for , one of whose terms is and 
which is -admissible. Let be the character of on the ith term 
of modulo the , where , and . 
Since is cyclic, . From 3.4, we see that . 
Furthermore, . and . Combining these 
equalities yields , so , and Lemma 26.1 is violated. 
If normalizes we say that is prime on provided any 
two elements of have the same fixed points on . if is a 
prime, is necessarily prime on . If is solvable, then is prime 
on if and only if for each prime , there is a -subgroup of 
which is normalized by and such that is prime on . 
The next two lemmas are restatements of Lemma 13.12. 
LEMMA 26.3. Suppose is a sol vable -group, is a cyclic 
pp-subgroup of Aut which is prime . Assume also that 
is odd. If is not a prime, if the centralizer of in 
is a Z-group, if has no fixed points on , then is 
milpotent. 
LEMMA 26.4. Suppose is a solvable -group is a 
subgroup of Aut of prime order. Assume also that is 
. If th e centralizer of in is a Z-group, if has 
fixed points on then is nilpotent. 
denotes the set of all proper subgroups of denotes 
those subgroups of such that, for all does not contain 
an element of for any -subgroup of . 
denotes the set of maximal subgroups of 
. 
If , then does not contain an elementary subgroup of order 
for any prime , so is nilpotent. Furthermore, if 
then has a Sylow series of complexion (. ). 
Suppose and is a subgroup of type with . 
Let . be the distinct -subgroups of which contain . 
Since . , it follows that . and that 
contains an element of order centralizing . 
Since is -closed, this implies that . 
. This fact is very important, since it shows that the 
subgroups of of order are contained in two conjugate 
slasses in , one class containing , the remaining subgroups 
 902 SOLVABILITY OF GROUPS OF ODD ORDER 
lying in a single conjugate class. 
If denotes the largest normal nilpotent -subgroup 
of . Note that by Lemma 8.5, . More explicitly, 
contains the largest prime in . Note also that is a S-sub- 
group of . 
If denotes the unique -subgroup of , where 
is the equivalence class of under associated with . 
That is, if and only if and contains a -subgroup of 
. Or again, if and only if contains an elementary subgroup 
of order . Or again, if and only if and contains an 
element of for some -subgroup of . 
Suppose . and a -subgroup of 
centralizes . Since is the unique maximal subgroup of 
containing , it follows that , so that is a -sub- 
group of . Then by Lemma 26.1, . Since the derived group. 
of is nilpotent, we have . Thus, if is the largest. 
subset of such that some -subgroup of centralizes 
, then contains a unique -subgroup is a 
normal nilpotent -subgroup of is a -subgroup of 
and the structure of the -subgroups of is given by Lemma 
26.2. We set Since 
and centralizes , and since is the unique maximal 
subgroup of containing , it follows that is a T.I. set 
in . 
If and is a -subgroup of . then the definitions; 
of and imply that is of type . In this case, wa 
set , and remark that char . 
. Furthermore, if is an element of order in . then 
contains an elementary subgroup of order . If , set 
being any -subgroup of . The relevance of 
lies in the fact that if is any element of of order . then 
is contained in only one maximal subgroup of . namely, the 
one that contains . This statement is an immediate consequence of. 
the theorems proved about , explicitly stated in Theorem 25.5.. 
If , then is contained in a unique maximal subgroup 
of . so we set . The existence of the mapping from 
to is naturally crucial. 
If , set . If let consist of. 
a11 elements in with the property that some power of H. 
say is either in or is in for some -subgroup 
of with . 
Let and let be a -subgroup of with let 
denote the set of subgroups of of type such that 
 26. THE MAXIMAL SUBGROUPS OF 903 
for some element in . If , then 
. Furthermore, if and is a subgroup of of type 
, and if is contained in at least two maximal subgroups of 
, then for every -subgroup of which contains . 
LEMMA 26.5. 
(i) If then is a . set in . 
(ii) If . then is a . set in . 
Proof. 
(i) is cyclic and normal in , by Lemma 26.2. Hence, 
if for some in , then , 
so , as required. 
(ii) It is immediate from the definition that is a normal 
.subset of , so is a T.I. set in . Suppose and 
. Choose so that is in either or 
for some -subgroup of . and such that is of prime 
order. If , then since , it follows 
that . Hence , and so . 
Suppose . Then and so . This 
implies that contains non cyclic -subgroups. By 
Theorem 25.6 (ii), we again have . The lemma is proved. 
With Lemma 26.5 at hand, it is fairly clear that the one remaining 
obstacle in this chapter is . In dealing with . we will repeatedly 
use the assumption that is odd. 
LEMMA 26.6. Let let be a -subgroup of let 
. If is any non identity subgroup of is 
in the -subgroup of , then . 
Proof. In any case, , by Theorem 25.6 (iii). If is non 
cyclic, then contains an element of for some 
subgroup of and we are done. Otherwise, . so 
contains an element of , and we are done. 
LEMMA 26.7. Suppose is a -subgroup 
of and is a -subgroup of . If is cyclic, th en is 
prime on . 
Proof. Suppose false. Then (mod ). and every q- 
subgroup of is -closed. Also for some -subgroup 
of , by Lemma 26.2 and . If is cyclic, or if is 
non abelian, then , by Lemma 26.1. Since every chief 
 904 SOLVABILITY OF GROUPS OF ODD ORDER 
-factor of is centralized by , it follows that 
centralizes and we are done. 
If is abelian and non cyclic, then normalizes some 
of . Since the lemma is assumed false, 
, so . If is a maximal -subgroup of 
containing , then is -closed, so contains a -subgroup of . 
This violates the hypothesis of this lemma. 
LEMMA 26.8. Let and suppose that or 
. If is any -subgroup of and contains an elemenb 
of for some -subgroup of , then is p-closed. 
Proof. Let . The hypotheses imply that 
and . The lemma follows. 
LEMMA 26.9. Let . and suppose that or 
. If is a -subgroup of which is normalized by the cyclic 
-subgroup of , then is prime on . 
Proof. If , the lemma is trivial. Otherwise, the lemma 
follows fr.om Lemma 26.8, since contains an element of 
for some -subgroup of . 
LEMMA 26.10. Let . let be a -subgroup of for 
some prime . If is non abelian , th en does not 
contain a cyclic subgroup of index . 
Proof. We can suppose that for if then 
and by Theorem 25.6 (ii). Hence, proceeding by vvay 
of contradiction we can suppose that 
, where . Note that . 
If is nilpotent, then , so by maximality of 
. This implies that is a -subgroup of which is not the case. 
Hence, is not nilpotent. In particular, . It follows that 
for all in . 
We first show that . For centralizes . so if 
is an element of containing . then normalizes some 
-subgroup of with . It follows from Lemma 8.16 that 
centralizes . If , then , which is not the 
case, so . 
Choose in and let be a -subgroup of normalized 
by . We can now choose such that is normalized by 
, is centralized by some non identity element of . but is 
not centralized by . For otherwise, centralizes , and 
 26. THE MAXIMAL SUBGROUPS OF 
, which is not the case. For such a choice of and 
P. let be a -subgroup of which contains . By Lemma 
26.7, there is a -subgroup of which contains and is contained 
in . Since does not centralize , and since . a 
subgroup of is contained in . by Lemma 26.8. 
We wish to show that . This is clear if contains an 
element of for some -subgroup of , by Lemma 26.6. 
Otherwise, Lemma 8.5 implies that , since . By Lemma 
26.6, , so contains a -subgroup of . This implies that 
Since the subgroups of of order different from 
are conjugate in , and since is a normal subset of , 
we can suppose that . 
Let be a -subgroup of containing and let , 
so that , or else . It follows that central- 
izes for some in . But contains a -subgroup of , 
so contains an element of order equal to that of W. 
Since and have the same order, a -subgroup of 
has exponent , which is not the case. The proof is complete. 
LEMMA 26.11. Let let be a -subgroup of for 
some prime . If is non abelian, then . 
Proof. First, suppose . If , we are done. Other- 
wise, contains a cyclic subgroup of index and we are done by 
the preceding lemma. 
We can now suppose that . If is nilpotent, the lemma 
follows readily from Lemmas 26.1 and 26.2. We can suppose that 
is not nilpotent and that . Since is non abelian, Lemma 
26.2 implies that is of order , or else is metacyclic. In the 
second case, we are done by the preceding lemma. 
We first show that . Since centralizes . 
it follows readily that dominates , by Sylow's theorem. 
If . then . and so , by Lemma 26.1, 
and we are done. 
Let be a -subgroup of which is normalized by , with 
. 
We show that . For otherwise, centralizes , by 
Lemma 8.16, so that . By Lemmas 26.1 and 26.2, 
, contrary to . Hence, . 
Let We next show that has no fixed points on 
Let . and suppose by way of contradiction that 
. Let . and let be the maximal normal subgroup 
of of order prime to . Let be permutable Sylow subgroups of 
. . Since , it follows that is not contained 
 SOLVABILITY OF GROUPS OF ODD ORDER 
in any conjugate of This implies that This in urn 
implies that centralizes every chief -factor of by Lemma 8.13. 
Hence, and it follows that covers Since 
by Lemma 26.6, we have a contradiction. Hence. . 
We next show that if then This is 
clear if is non cyclic, since , so suppose 
is cyclic. We remark that an easy consequence of the 
preceding paragraph. 
Let be a maximal subgroup of containing , and let 
a -subgroup of containing . If is non cyclic, then 
is contained in a nique maximal subgroup of , 
and since , we have Since and 
since we have . Thus, we can suppose that 
is cyclic. 
Since acts regularly on . we can suppose that a -subgroup 
normalizes and that . 
If is nilpotent, then Since . we 
have Hence, we can suppose that is not nilpotent. 
Choose in , and let be a -subgroup of normalized 
by Since cyclic, Since does not cen- 
tralize It follows from that by Lemma 
8.16. Since is a Frobenius group, it follows that 
. Let . 
Let be a -subgroup of which contains and , and 
let be a -subgroup of containing . If is non cyclic, then 
so If is cyclic, then in any case . since 
Let be a -subgroup of . If does not centralize 
then , and so and once again . If cen- 
tralizes and then Since the structure of is 
determined by Lemma 26.2, and since centralizes it follows 
that centralizes , so that , and once again . 
Thus, in any case, we see that . This implies that . 
so centralizes every chief -factor of . This is absurd, since 
is a Frobenius group. We conclude that for every 
in . 
We will now show directly that . Choose . 
Then normalizes and Since has no fixed points on 
is generated by its subgroups . By 
the preceding paragraph, we COnclude that . Since is the 
unique maximal subgroup of containing , we have . 
so . By Lemma 26.1, . so . The proof is 
complete. 
LEMMA 26.12. Suppose and is an non cyclic 
 26. THE MAXIMAL SUBGROUPS OF 
-subgroup of for some prime further that a 
subgroup of is non abelian. Then , where , 
centralizes is a Frobenius group with Frobenius 
kernel aM contains for every of 
which contains . 
Proof. Let be a -subgroup of containing . If , 
then , and if is any automorphism of of prime 
order . then , by Lemma 8.16. The same inequality clearly 
holds if . 
Choose in and let be a -subgroup of normalized 
by . 
Let . We will show that is a Frobenius group. 
Let and suppose by way of contradiction that . 
First consider the case that . Let . and let be a 
-subgroup of normalized by with . Then 
, so centralizes Since , it follows 
that centralizes . Thus, if or . we con- 
clude that , which is contrary to hypothesis. Otherwise, 
or so that , or centralizes . But in 
these cases, we at least have . so . which yields 
, and so a -subgroup of is non cyclic, and centralizes 
. Again we conclude that , which is not the case. Hence, 
we can suppose that . 
Let be a -subgroup of containing . . 
and let be a maximal -subgroup of containing . . 
, where is a -subgroup of and is a -subgroup 
of Since is a -subgroup of is a -subgroup of 
. If contains an elementary subgroup of order , then . 
and maximality of implies that is contained in a conjugate of 
. contrary to hypothesis. If does not contain an elementary 
subgroup of order , then either or centralizes . If , 
then . so once again for some . If . then 
. and since centralizes a centralizes , by Lemma 
26.2. In this case, . If is non cyclic, then . 
either by Lemma 26.6, in case . or because in case . 
If is cyclic, then . In this case is conjugate 
to a subgroup of . since is a -subgroup of . Since 
it follows that centralizes so that centralizes 
some -subgroup of . If this is not possible. But 
if , then is non cyclic, so . Thus, 
in all these cases, contains a -subgroup of . Since this is not 
possible, is a Frobenius group, and so is a Frobenius 
group. 
 908 SOLVABILITY OF GROUPS OF ODD ORDER 
Suppose . We will show that if is any subgroup of 
of order with , then . Let 
with . First consider the case , for some in 
. Let be a non identity -subgroup of and let 
be a -subgroup of containing . If . then 
Lemma 26.2 implies that is a -subgroup of . In this case, 
since and are conjugate and since is a -subgroup of . 
contains a subgroup of order which centralizes the -subgroup of 
. Since is a Frobenius group, this implies that if is 
any subgroup of of order . then either is a Frobenius 
group, or centralizes , the -subgroup of . This violates 
the choice of . Hence, . If a -subgroup of is abelian, 
then so . If some -subgroup of contains 
in its center, then by Lemma 8.10, . Hence, we can 
suppose that is of order and . In this case, 
is of index in and is of index in 
and contains . 
Let . If is contained in a conjugate of , then 
so , since . Similarly, 
, and we are done. If is contained in an element of . then 
since is a Frobenius group, we see that 
and . 
Hence, in showing that , we can suppose that 
is contained in an element of . Since is a 
Frobenius group, this implies that . Since is a -subgroup 
of . we conclude that is a -subgroup of . By what we have 
already proved, is a Frobenius group. This implies that 
is nilpotent, so centralizes . 
Since is the unique maximal subgroup of containing . it 
follows that centralizes , so that . which is absurd 
since . We conclude that . 
We next show that if and contains an element of 
, then . Here, as above, is a subgroup of of 
order . Let be a -invariant -subgroup of with 
. From Lemma 26.7, we conclude that 
contains a -subgroup of , and we can assume that . 
Let . If , then , so 
. If is nilpotent, then by Lemma 26.7, we see that 
contains a -subgroup of which is -invariant. Since 
is a Frobenius group, and so . We can suppose 
that is not nilpotent, and that . In particular, . 
It follows that is a -subgroup of , so that is a Fro- 
benius group, and so centralizes and follows. 
 26. THE MAXIMAL SUBGROUPS OF 
Thus. in all cases. 
Suppose now that contains two distinct subgroups such 
that and We can choose in 
such that . If . we get an easy contradiction. 
Namely, , and so and so 
that contrary to assumption. 
If . then contains . If contains 
an abelian with then , 
and , which is the desired contradiction. Otherwise, if is 
a -subgroup of with , then is of 
index in and is of index in . while both 
and are -subgroups of . Further- 
more, since a -subgroup of is -closed, it follows that 
and are -subgroups of . 
Furthermore, has a normal complement in since and 
no element of centralizes . By the conjugacy of Sylow 
systems in , we can therefore find such 
that Since 
and we conclude that 
SO , which is not the case, since is in and is not. 
Hence, there is exactly one subgroup of of order which 
has a fixed point on , so centralizes Since , 
where . the lemma follows. 
Lemma 26.12 is quite important because, given , (and the 
hypothesis of Lemma 26.12) it produces a unique factorization of . 
Namely, exactly one subgroup of of order is in the center of 
a -subgroup of , and exactly one subgroup . of of order 
centralizes . and . This is a critical point in dealing 
with tamely imbedded subsets. Furthermore, Lemma 26.12 shows 
that is nilpotent, a useful fact. 
LEMMA 26.13. Suppose is an abelian, non cyclic 
-subgroup of for some prime Suppose ftrther that a 
subgroup of is abelian. Then the following statements are true: 
(i) is a -subgroup of . 
(ii) . 
(iii) If and are elements of which are conjugate in 
but are not conjugate in , eith er or . 
(iv) Either dminates or . 
(v) One of th e following coruli tions holds: 
(a) . 
(b) for every non identity of such 
 910 SOLVABILITY OF GROUPS OF ODD ORDER 
that . 
Proof. If , then and all parts of the lemma follow 
immediately. We can suppose that . 
In proving this lemma, appeal to Lemmas 8.5 and 8.16 will be 
made repeatedly. 
If centralizes , then and all parts of 
the lemma follow immediately. We can suppose that does not 
centralize . This implies that . 
We first prove an auxiliary result: if is any -subgroup of 
containing and if , then is -closed. To see 
this, let be a -subgroup of . and let be a -subgroup 
of which contains . Let be a -subgroup of con- 
taining and let be a -subgroup of containing . If , 
then for some in and so If then 
not contain elementary subgroups of order or . so either or 
. If and . then . Suppose . Then 
centralizes the -subgroup of . There is no loss of generality 
in supposing that is a maximal . -subgroup of . It follows 
from this normalization that is a -subgroup of , and 
. Hence, we can suppose . Since . If 
is not of order , then is contained in a conjugate of . by 
Lemma 26.7, and we are done. Hence, we can suppose that 
is of order . But now contains -subgroups of 
order exceeding . so that -subgroups of are -closed. Since 
. is -closed 
(i) is an immediate application of the preceding paragraph, since 
some element of centralizes an element of . 
We turn next to (iv). Suppose , and is 
a non identity -invariant -subgroup of . Let 
be a -subgroup of permutable with . By the first 
paragraph of the proof, normalizes , so by Sylow's theorem 
dominates . Suppose for some normalizes 
and dominates . Let be a -subgroup of 
permutable with . Then normalizes and so dominates 
. Since , we see that some -subgroup of 
dominates and is normalized by . It follows that the normal- 
izer of every -subgroup of dominates for some in . 
and so dominates (iv) is proved. 
Notice that if , then by (iv)| elements of 
are conjugate in if and only if they are conjugate in . Thus, 
in the case, it only remains to prove (ii). We emphasize that in any 
case (i) and (iv) are proved. 
 26. THE MAXIMAL SUBGROUPS OF 
Since , if . then and the lemma follows. 
We can suppose that . Let and let be a 
invariant -subgroup of . If centralizes then (ii) 
follows immediately. Thus, we can choose in such that 
does not centralize . If . then , 
so that (ii) holds. If . then is cyclic, by Lemma 8.16, and 
the containment . Hence . where is a 
Frobenius group. 
Let . If is nilpotent, then , so by Lemma 
26.7, , and (ii) follows. Suppose is not nilpotent. Hence, 
contains an elementary subgroup of order for some prime . If 
then for some in . Since 
we have and (ii) follows. Suppose . In this case, 
normalizes a -subgroup of . Since centralizes 
and is a Frobenius group, and since , it follows that 
. Let . By (iv) applied to , we get 
. Since , and since the derived group of 
is nilpotent, centralizes , which is a contradiction. 
Hence, , and (ii) holds. The lemma is proved in case 
, and (i) is proved in all cases. 
Throughout the remainder of the proof, we assume 
(26.1) 1 
Suppose is a non identity subgroup of and 
. 
There are three cases: 
(a) and 
(b) and 
(c) . 
In each of these cases, we will show that 
(26.8) 
Case is nilpotent. 
Choose so that contains an element of order . 
and let be a -invariant -subgroup of . By (26.1), 
, so . If we conclude that 
, by Lemma 26.7. If , then 
centralizes for some in , and so . 
and follows. 
Case is not nilpotent. 
In this case, contains an elementary subgroup of order 
 912 SOLVABILITY OF GROUPS OF ODD ORDER 
for some prime . If , then , for some 
in . Since . we have . If , let 
be a -subgroup of normalized by , where is a 
non identity -subgroup of , as in Case . Let 
so that is a Frobenius group by (26.1). If is a 
Frobenius group, then centralizes . and . This is not the 
case, since for all . Hence, has a fixed point 
on , so has a fixed point on . By (iv) applied to 
, it follows that , and so centralizes . 
which is not the case. Thus (26.3) holds in case (a). 
In analysing case (b), we use the fact that , and 
that if is any subgroup of which is disjoint frorn . 
then is of square free order and for every in . 
Let be a non identity -invariant -subgroup of . 
so that . Suppose that (26.3) does not hold. 
We will show that is contained in a maximal subgroup 
of such that is not nilpotent, and such that is not conjugate 
to . 
Case for some in . 
Consider . Since and have non cyclic 
-subgroups, and since , it follows that is contained in no 
conjugate of . Let be a -invariant -subgroup of . 
If is nilpotent, then and so by 
Lemma 26.7. This is not the case, since has non cyclic 
-subgroups. Hence, is not nilpotent, so we take 
. 
Case is nilpotent, but is not contained in any 
conjugate of . 
Since . . If is not of order . 
then for some in . Suppose that is 
of order . Let , so that has non cyclic -subgroups 
and . Since is contained in no conjugate of 
neither is . If is nilpotent, then a -subgroup of is con- 
tained in , by (26.1) and so . which is not the case. 
We apply (iv) to . If . then so 
that centralizes , which is not the case. Hence, (26.1) holds 
with replacing . Let be any subgroup of of order 
different from . Then is a Frobenius group. Choose 
and let be a -subgroup of invariant under . 
If does not centralize , then so that case 
holds with replacing . replacing . 
Suppose then that centralizes . Then so a 
-subgroup of is contained in . We suppose without 
 26. THE MAXIMAL SUBGROUPS OF 913 
loss of generality that normalizes . If now is any subgroup 
of of order which does not centralize , then since does 
not centralize , we conclude that . 
Thus, in all cases, if are the distinct subgroups 
of of order which have fixed points on . then , so 
that or . 
Choose . Then there are indices not necessarily 
distinct, such that . If , then , by (a). If 
then , so that and . 
Hence, , so . and centralizes , which 
is not the case. Hence, (b) implies (26.3). 
We will now complete the proof of this lemma in case . 
Since some element of has a fixed point on (ii) holds 
by (26.3). Also, by (26.3), alternative holds. It remains to prove 
(iii). Suppose are elements of which are conjugate in , 
but are not conjugate in , and that . 
Theorem 17.1 is violated. 
We next verify (26.3) under hypothesis (c). 
Suppose by way of contradiction that (26.3) does not hold. Let 
be a non identity -invariant -subgroup of . We 
will produce a subgroup of such that is not nilpotent, and 
such that . Once this is done, then it will follow as in case 
that of the subgroups of of order have fixed points 
on , and (26.3) will follow. 
Suppose is a maximal subgroup of containing . If 
is nilpotent, then . If is non abelian, then 
for some in . Furthermore, from (26.1) and the fact 
that is not a -subgroup of , we conclude that . 
Hence, contains . Let be a maximal subgroup of 
containing . If is nilpotent, then and (26.3) holds. 
Hence, is not nilpotent, so we can take . If is 
abelian, then and (26.3) holds. Thus, (26.3) holds in all cases. 
The completion of the proof that (26.3) implies this lemma is a 
application of Theorem 17.1. 
LEMMA 26.14. Suppose and is a non abelian -subgroup 
Then . Furthermore, one of the following 
is true; 
(a) centralizes . 
(b) for every non subgroup of . 
(c) . 
Proof. Suppose . If , then . and so 
 914 SOLVABILITY OF GROUPS OF ODD ORDER 
Since , the lemma is proved. If 
then contains a cyclic subgroup of index Since is assumed to 
be non abelian, is a non abelian metacyclic group, so , by 
3.8. Lemma 26.10 is violated. 
Through the remainder of the proof, we assume . 
Let , so that is of order . by Lemma 26.2 and 
Lemma 26.10. 
If ' is nilpotent, then , and all parts of the lemma follow. 
We can suppose that is not nilpotent. In particular, . We 
can further assume that . 
Since is non abelian, centralizes . 
Choose and let be a -invariant -subgroup of . 
If then centralizes . 
Thus, if , then centralizes a 
subgroup of . If , all parts of the lemma follow. 
Let and let be a -subgroup of normalized 
by , and such that does not centralize . If there are no such 
primes . we are done. 
Let be any subgroup of of order different from . We 
will show that . 
Since does not centralize . Set 
If , then . Otherwise, is a non trivial cyclic 
subgroup of , and is a Frobenius group. 
Let be a maximal subgroup of containing . If is 
nilpotent, then . so , by Lemma 26.6. We can 
suppose that is not nilpotent and that is not conjugate to . 
If a -subgroup of is non then centralizes . which is 
not the case. Hence, a -subgroup of is abelian and non cyclic. 
We can apply Lemma 26.12 to and a -subgroup of which 
contains . We conclude that is a Frobenius group. Since 
is a Frobenius group, centralizes , and so . We 
conclude that contains in all cases. 
Now let be the distinct subgroups of of order 
different from . Here . Let be any proper subgroup 
of containing . Let . Since is generated by its 
subgroups we have . Let 
and choose in . We can then find indices , not necessarily 
distinct, such that . Hence, . Since 
contains an element of , we have . Hence, 
so in particular, . 
Let be any non identity subgroup of . If is non cyclic, 
then . If is cyclic, then . The 
proof is complete. 
 26. THE MAXIMAL SUBGROUPS OF 915 
LEMMA 26.15. Suppose is a cyclic S-subgroup of 
and . Then is prime on and is a 
Z-group. 
Proof. Suppose is prime on , but that is a non cyclic 
-subgroup of . Choose and let be the 
-subgroup of . Since . it follows that . Thus, 
if is a -subgroup of , while if is also a -subgroup 
of , by Lemma 8.12. Since . we have , so that . 
Let be a maximal subgroup of containing . If a 
-subgroup of is cyclic, then dominates . which is 
not the case, since . Hence, . Let be a 
-subgroup of permutable with . If is a -subgroup of . 
then normalizes . Otherwise, normalizes since , 
and Lemma 8.5 applies to . 
Let be a maximal -subgroup of containing and let 
be a -subgroup of R. Then , so that is a -subgroup 
of . Let be a maximal subgroup of containing . 
If were non abelian, then by Lemma 26.14, which is 
not the case. Hence, is abelian. If . then by Lemma 26.13, 
we have since centralizes . Since this is 
impossible, we see that . 
If , then by Lemma 26.1, together with the fact that 
covers , we see that , contrary to 
hypothesis. Hence, . Since . this implies that 
is a non abelian group of order and exponent . 
Since some element of has a non identity fixed point on . 
and since centralizes . we see that by Lemma 26.13. 
Since and since , it follows that . the 
desired contradiction. 
Thus, in proving this lemma, it suffices to show that is prime 
on . 
First, suppose that is a -group for some prime . We can 
clearly suppose that , and that . 
Case 1. . Let , so that Lemma 26.9 
applies. Let . Then since . Lemma 26.10 
applies. If , Lemma 26.9 applies. 
Case 2. and a -subgroup of is abelian. 
If . or and . Lemma 26.7 applies. 
Let . and let be an -invariant -subgroup of . 
If centralizes . we have an immediate contradiction. Hence, 
does not centralize . 
We can suppose by way of contradiction that . 
If contains a -subgroup of , which is not the 
 916 SOLVABILITY OF GROUPS OF ODD ORDER 
case. Otherwise, , so every -subgroup of is -closed, and 
contains a -subgroup of , which is not the case. 
Case 8- and a -subgroup of is non abelian. 
Here, , by Lemma 26.2. Since . 
the lemma follows. 
Case 4. . In this case, also, we have . and 
the lemma follows. 
Next, suppose that , where is a non identity . 
. Suppose by way of contradiction that is an -invariant 
-subgroup of and that is not prime on . We can suppose 
that does not centralize . 
Let be a maximal subgroup of containing . Then 
is not conjugate to , either because is not a -subgroup of 
, or because . Let be a -subgroup of which 
contains and is -invariant. 
Suppose . Then , since . and 
. Furthermore, is non cyclic. Suppose In this 
case. so a -subgroup of normalizes some -subgroup 
of , and it follows that normalizes some -subgroup of . This 
implies that is a -subgroup of . But in this case 
so that centralizes and so . Suppose . If , 
then , which is not the case. Hence, so that 
. Once again we get that . Hence, we necessarily have 
in all cases. 
Since is prime on , from the first part of the lemma, we 
conclude that is cyclic. 
We next assume that is nilpotent. 
Suppose . Since is a -subgroup of , 
it follows that and is a -subgroup of , so that . 
Since , we can suppose that is non cyclic. 
In this case, however, is a -subgroup of and is conjugate 
to , which is not the case. 
We can now suppose that is not nilpotent. 
Suppose . Let be a complement for in 
which contains . Then is nilpotent and so . 
Case 1. . In this case, is a -subgroup of . and 
dominates . This violates . 
Case 2. , and a -subgroup of is abelian. In this case, 
, so once again is a -subgroup of and 
dominates . 
Case 8. and a -subgroup of is non abelian. Since 
i-s cyclic, we have , so some -subgroup of normalizes 
some -subgroup of . But now dominates since every . q- 
 26. THE MAXIMAL SUBGROUPS OF 917 
subgroup of is -closed, and dominates . 
Case 4. . If , then every -subgroup of which 
contains a .-subgroup of is -closed, so once again dominates 
and is a -subgroup of . Hence, Since is not 
tonjugate to . it follows that if is a -subgroup of containing 
, then . which implies that is cyclic, and . Hence, 
, since centralizes . But now , so . 
Thus, once again is a -subgroup of and dominates . 
A11 these possibilities have led to a contradiction. We now get 
to the heart of the matter. Suppose . 
We will show that . 
Let be a -subgroup of containing and invariant 
under . Suppose that 
We will derive a contradiction from the assumption that (26.4) holds. 
If is an absurdity, since . If , 
then a -subgroup of is non cyclic, so , as already 
remarked. If . then centralizes a -subgroup of , so 
is a -subgroup of . In this case, however, , an absurdity, 
by (26.4). Thus, if (26.4) holds, then and . 
Since (26.4) is assumed to hold, it follows that is a -subgroup 
-of . Hence, is non cyclic. We have already shown 
that is cyclic. We conclude that (26.4) does not hold. 
If , then centralizes , by Lemma 8.16 (ii), so 
(26.4) holds. Hence, . 
Since (26.4) does not hold, and since is 
. It follows that is non cyclic. This implies that 
. since . Since , and since . 
it follows that a -subgroup of is -closed. This in turn 
implies that some -subgroup of normalizes some -subgroup of 
Since is a -subgroup of is forced to be a -subgroup 
-of . But , and , so centralizes . The 
proof of the lemma is complete in case . 
If , the lemma follows immediately by applying the 
preceding result to all pairs of elements of . 
LEMMA 26.16. Suppose and is not ni lpotent. Then 
is a prime and is a S-subgroup of . 
Proof. Let and let be a -subgroup of . By 
Lemma 26.11, is abelian. Suppose is non cyclic. If a -subgroup 
is non abelian, then is nilpotent, by Lemma 26.12. Hence, 
 918 SOLVABILITY OF GROUPS OF ODD ORDER 
we can suppose that a -subgroup of is abelian. By Lemma 
is a -subgroup of . By theorem, the simplicity of , 
and Lemma 26.15, contains elements which are conjugate in 
but are not conjugate in . If and if 
has a fixed point on , then . so that and 
are conjugate in Since this is not the case, is a 
Frobenius group, and so is nilpotent, contrary to assumption. 
Hence, . By Lemma 26.13, either or 
is a Frobenius group, which is not the case. Hence, 
is cyclic. 
Let be a complement to in so that is a cyclic -subgroup 
of . 
By Lemma 26.15, is prime on and is a Z-group. 
Let and suppose that is not a prime. By 
Lemma 26.3, is nilpotent. By 3.7, Hence , 
so that is a -group. It follows that . the 
desired contradiction. 
LEMMA 26.17. Suppose and 
. Let and 
Set . Then a -subgroup of 
has a Sylow series of complexion . Furthermore, 
if has -length 1. 
Proof. We first show that has -length 1 for each in . 
If , this is clear, so suppose . Let be a 
-subgroup of and let be a subgroup of of order such that 
where is cyclic. 
Let . and . It suffices to show that 
has -length one, since . Let be a subgroup of 
chosen in accordance with Lemma 8.2, and set Then 
ker If then . and we 
are done. We can suppose that . This implies that , 
since has order and is of exponent . We are assuming 
by way of contradiction that has -length , so by (B), we have 
. Hence, . 
Set and let . Then maps. 
onto a -subgroup of . Hence has a normal series 
where and are -groups and . 
Since is abelian. Also is faithfully represented 
on and since . 
By Lemma 26.16, is a prime, and is a -subgroup 
of . We let be a -subgroup of , so that is of order . 
Since , it follows that maps onto . Let. 
 26. THE MAXIMAL SUBGROUPS OF 919 
denote the image of in and let denote the image of in 
. Since is a -group and a -group, we assume without loss 
of generality that normalizes . 
Let be the linear character of on . so that . Let 
be the linear character of on . Since divides , 
is non cyclic. Hence, is not a Z-group. contrary 
to Lemma 26.15. 
Thus, has -length one for each . Since a -subgroup of 
has a Sylow series of complexion (. ) and since a -subgroup 
of is -closed, it suffices to show that a -subgroup of has a 
Sylow series of complexion . 
Let be a .-subgroup of with Sylow system where 
. By Lemma 8.16, centralizes Hence is -closed, 
since has -length one. The lemma follows. 
LEMMA 26.18. Let let be a complement for 
in . Then there is at most one prime in wi th the following 
properties: 
(i) A -subgroup of is a non cyclic abelian group. 
(ii) A -subgroup of is non abelian. 
Furthermore, if contains a prime satisfying (i) and (ii). then 
a subgroup of is a Z-group. 
Proof. Suppose a.nd both and satisfy (i) 
and (ii). Let be a -subgroup of and let be a -subgroup 
of permutable with . 
Let , where centralizes 
is a Frobenius group and for some S-subgroup of 
. Assume without loss of generality that . Then 
normalizes . It follows that centralizes , and this 
implies that centralizes . It follows that satisfies . 
By Lemma 26.17, contains a -subgroup of By 
Lemma 8.16, centralizes . so centralizes Since , 
we see that . By Lemma 26.2, and Lemma 26.10, now centralizes 
. This is a contradiction, proving the first assertion. 
Now suppose satisfies (i) and (ii), is a -subgroup of and 
is a non cyclic -subgroup of permutable with p- 
Case 1. is non abelian. 
In this case. is a -subgroup of and by Lemma 26.14. 
Since normalizes Write , where centralizes 
is a Frobenius group, and for some 
subgroup of of with . Then centralizes . 
If centralizes then satisfies as can be seen by considering 
 920 SOLVABILITY OF GROUPS OF ODD ORDER 
. 
We now show that does not satisfy . Otherwise, since 
, we see that normalizes some -subgroup of . 
Then centralizes by Lemma 26.2, Lemma 26.14, and Lemma 8.16, 
This is not possible since is abelian. 
Hence, does not satisfy , so does not centralize and 
. This implies that . Hence 
is of order . 
Consider . Since a -subgroup of has order 
, it follows that a -subgroup of is q-closed. Let be a 
-subgroup of containing . If is not of order then 
contains a -subgroup of a -subgroup of is 
and a -subgroup of has larger order than . As 
, this is not possible. Hence has order . But now a 
-subgroup of contains and . so a -subgroup of 
is -closed. This in turn implies that a -subgroup of 
has order larger than . which is a contradiction. 
Case 2. is a non cyclic abelian group. 
By the first part of the proof, and by Lemma 26.13, is a 
subgroup of . Since centralizes , and since 
. it follows that satisfies . This implies that a 
-subgroup of is -closed, by Lemma 26.2. Hence, centralizes 
the center of some -subgroup of . since centralizes , 
(where , as in Case 1). To obtain the relation 
, we have used Lemma 26.13 to conclude that there are at. 
least 2 subgroups of of order which have no fixed points on , or 
else in which case normalizes and so centralizes . 
But now dominates , so centralizes some -subgroup 
of , contrary to . The proof is complete. 
LEMMA 26.19. Let . Suppose is . 
pose further that ei ther is nilpotent or is not 
prime. Then is of type I or V. 
Proof. Let be a complement for . Since 
by hypothesis (we always have ). is abelian. 
Case 1. is cyclic. 
We wish to show that is nilpotent, so suppose is not. 
a prime. Since is not a prime, since is prime on . since 
has no fixed points on , and since is a 
Z-group, it ollows from Lemma 26.3 that is nilpotent, so that. 
is cyclic. 
Case . 
In this case, is a Frobenius group with Frobenius kernel 
 26. THE MAXIMAL SUBGROUPS oF 921 
. so condition (i) in type I holds. If is a T.I. set 
in , then is of type I, since (ii) (a) holds, so suppose is 
not a T.I. set in . Let where is the 
subgroup of and . If , then clearly 
. If - then also , since is a Frobenius 
group. Similarly, if and X. is non abelian, then . 
Suppose . Then either and is abelian, or 
. We will show that the second possibility cannot occur. 
Choose in such that . and let 
an element of of prime order and , 
then , and , contrary to assumption. Hence, 
. In this case, , and since 
, both and are in . so . 
Hence, . 
Thus, if , then contains a prime such 
that the of is abelian and . Since does 
not divide or . but does divide we can find 
e such that and . Let b e t he 
subgroup of . Then , where is normalized by and 
is cyclic, . Since , it follows that and are 
isomorphic . Hence, normalizes every subgroup of . 
Once again, choose in so that . 
Then , so is not contained in any conjugate of 
. Let . We apply Lemma 26.13 to and . 
Since , we have . 
Suppose were not abelian. Let 'be a non abelian 
subgroup of . Apply Lemma 26.16 to and , and conclude 
that , and so 'Ht., which is not the case. Thus, 
alternative (ii) (c) in the definition of type I holds, so is of type 
I.(Since is generated by two elements.) 
Case . 
Since , we have . It follows 
that for every non empty subset of . Let 
. If is any non empty subset of . then each element of 
is of the form . Thus, if , 
then . Since is a three step 
group with in the role of in the role of in the 
role of . Since , we take , so that (i) in the 
definition of type V holds. If (ii) (a) holds, then is of type V, 
SO suppose (ii) (a) does not hold. 
Since is non abelian. Let . 
where is a non abelian of (there may be 
several). 
 922 SOLVABILITY OF GROUPS OF ODD ORDER 
We will show that is a T.I. set in Suppose 
and is a maximal intersection, so that is contained 
in no conjugate of - Let . with . Apply Lemma 
26.14 to and and conclude that . a contradiction. 
Hence, is a T.I. set in . 
Since is not a T.I. set in . choose so that 
is a maximal intersection. Since is a T.I. set 
in , we see that and is con- 
tained in no conjugate of , while Since jg a T-I. 
set in . and since is cyclic. By construction, 
is non abelian, so . It only remains to show that . 
Apply Lemma 8.16 to and . If does not centralize 
then divides and we are done. Suppose that centralizes 
. Then is faithfully represented on , so if 
, we are done. Otherwise. we let be an 
element of of order such that where is 
cyclic. Since we have . so 
. By Lemma is empty. By 
Lemma 26.2, is a central product of a cyclic group and , 
with Since and since centralizes . we 
have is faithfully represented on , and since 
centralizes , each element of induces a linear transformation of 
of determinant 1. Thus, divides either or , 
since is isomorphic to a cyclic -subgroup of (. ). Hence, 
and is of type V. 
Case 2. is non cyclic. 
Case . There is an element such that the -subgroup 
of is non cyclic and a -subgrou of is non abelian. In 
this case, Lemma 26.18 implies that where is cyclic. 
Let , with . and with 
a Frobenius group. Also is a cyclic -subgroup of . 
We will show that is a Frobenius group. If . 
this is the case, so suppose . By Lemma 26.16, is prime on 
. Let , and suppose . Then is 
a Frobenius group. Let be a maximal subgroup of containing 
being a fixed subgroup of of prime order. Then is 
not conjugate to . Hence, . Since is a Frobenius 
group, , so a -subgroup of is abelian. By Lemma 
26.12, is a Frobenius group, so centralizes . 
Since , we see that , which is not the 
case. Hence, , so is a Frobenius group, as is . 
itself is a group of Frobenius type. 
Suppose is not a T.I set in and . It follows readily 
 26. THE MAXIMAL SUBGROUPS oF 923 
that is abelian and is generated by two elements. is of type I. 
Case 2b. Whenever a of is non cyclic, a 
subgroup of is abelian. 
Let be the set of primes 'In such that a 
of is non cyclic. Let , where is the of 
. Thus is a cyclic -subgroup of , and . By Lemma 
26.13, is a -subgroup of . 
We first show that if and is the of . then 
(26.6) 
This is an immediate consequence of Lemma 26.13 (iv) and 
theorem, since 'He' . 
We next show that either or is a Frobenius group. 
Suppose . By Lemma 26.15 is prime on . Suppose 
. Let be the of for some 
, and let be a maximal subgroup of containing . 
Then is not conjugate to . 
By Lemma 26.13 (ii), Logether with , there is some 
element of . which has no fixed points on , so is 
cyclic. By construction . Suppose is non 
cyclic for some in . Let be a non cyclic of 
. If a of is abelian, then by 
Lemma 26.13 (i) and (ii). Since , we have , which 
is not the case. Hence, a of is non abelian. If 
were non abelian, then for some in . by Lemma 26.14 
with in the role of . Hence, is abelian. By Lemma 26.13, 
centralizes and is a Frobenius 
. By (26.6), , so . Since . we can 
find a of which is normalized by . Since 
and are not conjugate, so does not 
lie in . and does not centralize . There are at least 
subgroups of with the property that is a Fro- 
benius group, by (26.6). Each of these has a fixed point on . 
It follows from Lemma 26.13 (iii) that dominates . This is 
absurd, by (26.6) and Lemma 8.13. Hence, is cyclic for 
a11 in . In particular, is cyclic. This implies that 
is faithfully represented on so is faithfully 
represented on . By (26.6), at least subgroups of or order 
have fixed poinLs on , so dominates , which violates 
by Lemma 8.13. is a Frobenius group. Thus, in the defi- 
nition of a group of Frobenius type, the primes in are taken care 
of. Let . with , and where is cyclic, 
. If . then char . By Lemma 26.14 
 SOLVABILITY oF GROUPS oF oDD ORDER 
(v), it follows that is a Frobenius group. If 
then by Lemma 26.14 (iii), there is some element of order in 
such that is a Frobenius group. Thus, contains a sub- 
group of the same exponent as with the property that 
is a Frobenius group. is of Frobenius type. 
If is not a T.I. set in , and , it follows. 
readily that is abelian and is generated by two elements. The. 
proof is complete. 
LEMMA 26.20. Let let be the subset of primes p- 
in such that a of is non cyclic abelian 
group and, a 0f is abelian. Let be a c0mplement f0r 
in . Then of is a n0rma1 abelian sub- 
gr0up of 0r . 
Pr00f. We can suppose . Let and let be a 
subgroup of . We first show that . Let and let. 
be of permutable with . If is non abelian, 
then , by Lemma 26.14. If centralizes. 
, then so that centralizes . We can suppose. 
that does not centralize . Since centralizes. 
, and since , it follows that so 
that centralizes . 
If is a non cyclic abelian group, then by Lemma 
If . then normalizes and centralizes. 
. If . then dominates . 
SO centralizes . If , then domi- 
nates . so that dominates and once again centralizes . 
Suppose is cyclic. We can suppose that normalizes 
Then centralizes . If . then centralizes 
since . We can suppose and that a 
of is in . In this case, however, for all 
. so which is absurd. Hence, , so that is 
a normal abelian subgroup of . 
Suppose contains a non abelian for some prime 
. Then , which implies that , since 
dominates each Sylow subgroup of . 
Thus, in showing that or . we can suppose that. 
every Sylow subgroup of is abelian. By Lemma 26.18 and the defi- 
nition of , this implies that of is a -group. 
This in turn implies that is a -subgroup of . Let be 
a complement for in . Then is cyclic. If , then 
is abelian and we are done. We can suppose . 
Suppose is not of prime order. Let . By 
 26. THE MAXIMAL SUBGROUPS OF 925 
Lemma 26.3, and Lemma 26.16 is nilpotent. If then 
for some -subgroup of . Hence, domi- 
nates every Sylow subgroup of Since can be assumed 
non cyclic, , and we are done. If , then 
, and we are done. 
We can now suppose that is of prime order . We can now 
write . where and , and we 
suppose by way of contradiction that . 
Choose so that , where is the -subgroup of - 
If centralizes , then by Lemma 
26.13, since . Since 
contrary to construction. Hence we can assume that 
is a Frobenius group, where . Let be a maxi- 
mal subgroup of containing . Since it, 
follows that , since dominates . Since is 
not conjugate to . it follows that . so that 
Since . both and 
are in , so commute elementwise. Thus contrary 
to the above argument. The lemma is proved. 
LEMMA 26.21. Let aM suppose contains 
prime such that a -subgroup of is non cyclic. Then is 
of type I. 
Proof. 
Case 1. A -subgroup of is abelian. 
Case 2. A -subgroup of is non abelian. 
In Case 1, let be the subset of those in such that, 
a -subgroup of is an abelian non cyclic -subgroup of . Then 
and if is a complement for in , then a -subgroup of 
is an abelian direct factor of by Lemma 26.20. Let . 
If were not a then some Sylow subgroup of would 
be non abelian, by Lemma 26.18 and the definition of . But then 
by Lemma 26.14. Since dominates every Sylow sub- 
group of we would find , which is not the case. Hence, 
is a Z-group. 
Let be a complement for in , and let be the 
subgroup of . Let . Since . is a Z-group, 
and since dominates every Sylow subgroup of central- 
izes . By Lemma 26.13, . Hence is a Frobenius 
group. 
Let . be the -subgroup of , and let . 
If is a Z-group. then as in the preceding paragraph. If 
 926 SOLVABILITY OF GROUPS OF ODD ORDER 
not a Z-group, then since dominates every Sylow 
subgroup of we find , which is not the case. Hence, 
is a Frobenius group. 
If is non abelian, then for every -subgroup 
of , so that is a T.I. set in . By Lemma 26.18, is 
of Frobenius type, so is of type I. If is abelian, is abelian, 
so is of type I by Lemma 26.19. 
In Case 2, let be a complement for i n . let a 
group of and let be a -subgroup of . Let be a complement 
for in . Then is a -subgroup of . and is a possi- 
bility. We can suppose is permutable with , so that normalizes 
. since by Lemma 26.18, is a Z-group. and . 
Let . where centralizes is a Frobenius 
group, normalizes both and , and for some 
subgroup of . By hypothesis, . 
Suppose . Let , and suppose that 
. Let be a fixed subgroup of of prime order. Then 
is a -group normalized by Since 
is non abelian, . Hence is a Frobenius group, 
Since is prime on is a Frobenius group. In par- 
ticular, every subgroup of of prime order centralizes . 
Let . and suppose that . so that our running 
assumptions are: is not 
a Frobenius group, and let be a subgroup of prime order such 
that . It follows that . But 
centralizes so is not a -subgroup of . Hence 
is a Frobenius group, in case . Hence, is of 
Frobenius type in this case. If is non abelian, then 
for every -subgroup of , so is a T.I. set 
in and is of type I. If is abelian, and is not a T.I 
set in , and , then and is abelian. 
is of type I in this case. 
Suppose now that . In this case . Since 
is a Frobenius group and centralizes . it follows 
readily that is a Frobenius group, and that is of type I. 
Next suppose . Since is prime on is 
of prime order. Since does not centralize does centralize 
. Let . so that . Since centralizes 
centralizes . Since is non abelian and is a 
Frobenius group, it follows that is a T.I. set in and that 
is cyclic. 
Let be a maximal subgroup of containing . Then 
is not conjugate to . Let be a complement to which con- 
 26. THE MAXIMAL SUBGROUPS OF 927 
tains . If , then since centralizes a non 
cyclic p-group, which is not the case. Hence, , and we can 
suppose that . 
Since for every non empty subset of . it follows 
that . is prime on . Let , so that . and 
is prime on . Since . it follows that . 
If is not a -subgroup of . then for some 
in . But then is a Frobenius group, as is . 
so that centralizes . which is absurd. Hence is a 
subgroup of . 
If , then either or a -subgroup of is 
abelian. But in the first case, dominates , contrary to 
while in the second case normalizes some 
of with . and is a robenius group. 
As is prime on and is not a prime, it follows that 
centralizes If is a -subgroup of , then 
dominates so which is not the case. Otherwise. a 
subgroup of is non and is contained in the 
center of some -subgroup of . But . and a 
-subgroup of is abelian. Hence, . 
We next show that is a complement to in . 
Namely, turning back to the definition of . we have . 
But , and centralizes . Hence, or is a 
Frobenius group with Frobenius kernel . Now, since 
, it follows that . This implies that 
has a normal complement in . If then is a Frobenius 
group with Frobenius kernel and . This is absurd 
since is prime on . and is not a prime. Thus 
is a complement to in . Now, however, is nilpotent. 
Since has no fixed points on , it follows that 
. 
Since centralizes , it follows that . We next 
show that is a T.I. set in . Namely, divides 
since . Hence . since is a prime, , 
so is a T.I. set in . 
We now turn to . Let be a maximal subgroup of 
which contains . Then is not conjugate to either or 
, since the -subgroups of these three maximal subgroups are 
pairwise non isomorphic. Let be a -subgroup of containing 
and normalized by . If , then does not map onto 
, since centralizes . But then covers 
. This is not the case since and . 
Hence, , so , and . 
 928 SOLVABILITY OF GROUPS OF ODD ORDER 
Since and since 
. 
it follows that . Hence, normalizes 
But . (This turns the tide.) Suppose 
. Then contains a non identity subgroup 
of . But contains , and we find that 
which is not the case. Hence . 
By Lemma 26.17, has -length one. Let , so that 
. Then is a Frobenius group whose 
Frobenius kernel is of index , or else . In any case, 
by Lemma 8.16, centralizes . But now which is a 
contradiction to . 
We have now exhausted all possibilities under the assumption 
that . 
Suppose . In this case, is cyclic and is nor- 
malized by . Since is a Frobenius group, centralizes 
, so centralizes . This implies that is a Frobenius 
group, or . In both cases, is of Frobenius type. If , 
then is non abelian, so for every -subgroup of 
, and is a T.I. set in . If . then 
is abelian, and the lemma follows from Lemma 26.19. 
LBMMA- 26.22. Let be the set of Z-subgroups of with 
the following properties: 
(i) If are primes, every subgroup of of order is 
cyclic. 
(ii) and for any non empty 
subset of . 
Then is empty or consists of a unique conjugate class of 
subgroups. 
Proof. If , and satisfies (i) and . then 
contains elements. Since is a 
Z-group, . is clearly a normal subset of so 
. Suppose and . Then there is a power of , say 
such that and such that has order where 
. is a prime divisor of . Then and so 
. Thus, the number of elements of which are conjugate 
to an element of is 
. 
 26. THE MAXIMAL OF 929 
Suppose another subgroup of and satisfies 
and (ii). Set We can assume 
. by (26.7), and it follows that . The proof is complete. 
LEMMA 26.23. Let . suppose is a -subgroup of 
is not a and is cyclic. Then is of 
lype I or or has the following properties: 
(i) is a n ilpotent . set in . 
(ii) If is a complemen t for in then 
(a) is a non abelian -group and every subgroup of 
of order is cyclic, primes. 
(b) is prime on and is a non 
identity cyclic group. 
(iii) satisfies the hypotheses of Lemma 26.22. 
Proof. If . the lemma follows from Lemma 26.19. 
We can therefore suppose that . Let be a complement 
for in , let be a complement to in Then 
is a cyclic -subgroup of , and is not a prime: 
If is a Frobenius group, then for every non 
identity -subgroup of , so is a T.I. set in and 
we are done. We can suppose that is not a Frobenius group. 
Suppose is a Frobenius group with Frobenius kernel . 
With this hypothesis, we will show that is of type I. 
Let be a cyclic -subgroup of Suppose 
. Then normalizes Consider . 
Since is not a prime and is a Frobenius group, it follows 
that . Hence, is a -subgroup of Since 
does not centralize , it follows that every subgroup of of 
prime order centralizes Since is not square free, 
.and contains a -subgroup such that . Consider 
. If . then . If or and a 
rsubgroup of is non abelian, then , so once again 
If and a -subgroup of is abelian, then 
contains a -subgroup of , contrary to . 
Hence and is a Frobenius group. 
Since is not a Frobenius group, . contains a non cyclic 
.subgroup for some prime If is abelian, and a -subgroup 
.of is non abelian, then , and is a -group. In this 
case, is a group, and so is of type I. If 
is abelian, and a -subgroup of is abelian, then is a -subgroup 
of . In this case, every subgroup of of prime order centralizes 
so centralizes for some non identity subgroup 
of . Since and a -subgroup of is abelian, it follows 
 930 SOLVABILITY OF GROUPS OF ODD ORDER 
that if is a -subgroup of which does not centralize , then 
a -subgroup of is abelian, and is normalized by a 
subgroup of with Since . 
Since is of type , there is a direct factor of which 
normalizes every subgroup of . Hence, is this direct factor. 
Hence, divides , so we have where is 
normalized by . It follows that is a Frobenius group for 
. 
Suppose every Sylow subgroup of is abelian. Let be the 
subset of in such that a -subgroup of is non cyclic, and 
let be a -subgroup of . By Lemma 26.18 and the preceding 
paragraph, is a normal abelian subgroup of . Hence, is of 
Frobenius type. Since is non abelian. is a T.I. set in . so 
is of type I. 
Thus, if is a Frobenius group and every Sylow subgroup 
of is abelian, then is of type I. 
Suppose is a Frobenius group, and is a non abelian 
-subgroup of . Then is a -subgroup of and Since 
every subgroup of of prime order centralizes 
and since Lemma 26.9 implies that centralizes 
. This violates the containment . Hence, if 
is a Frobenius group, is of type I. 
Suppose now that is not a Frobenius group. Let 
. By Lemma 26.15, is a -group. By Lemma 26.3, 
is nilpotent so is cyclic. Since every subgroup of of 
prime order centralizes , it follows that normal.izes 
, so centralizes since Aut is abelian. Hence, . 
Since every subgroup of of prime order centralizes 
, it follows that is abelian. Suppose were non 
cyclic. Let be a non cyclic -subgroup of . By Lemma 26.12, 
together with is a -subgroup of . 
Let be a -subgroup of which does not centralize 
, and let . Then 
. It follows now from 
that either is not a -subgroup of or , both of which 
are false. Hence, is cyclic. This yields that every subgroup of 
of order is cyclic, being primes. 
We next show that is prime on Since 
for all , it suffices to show that 
for all Suppose false and is a -subgroup 
of such that . Since is a Fro- 
benius group, it follows that is a -subgroup of and 
. In this case, let be a -subgroup of which does not 
 26. THE MAXIMAL SUBGROUPS OF 931 
centralize and consider . If , Lemma 26.9 
is violated; if . then so ; If 
is not a -subgroup of , contrary to . Hence, is 
prime on , and so for all . Since is 
non abelian, is a T.I. set in . 
Let . and let . By construction, . 
, and . Since 
for every non empty subset of . Since , this 
implies that and for every non empty subset 
of . Thus, satisfies the hypotheses of Lemma 26.22. The proof 
is complete. 
LEMMA 26.24. Suppose is of type V. Then 
is tamely imbedded in . 
Proof. We can suppose that is not a T.I. set in . Let 
, where is a complement to in . Then . 
and . Hence, is non abelian. Let , where 
is a non abelian -subgroup of , and is the -subgroup of 
for some prime (there may be several). 
We show that is a T.I. set in . If , this is the case. 
Suppose , and . Then . Let 
be a maximal subgroup of containing . By Lemma 26.14, 
. Hence . so and so 
and . 
Since is not a T.I. set in , it follows that is cyclic. 
Suppose . and . Since every subgroup 
of is normal in , it follows that . Furthermore, 
, so is of order , and , 
where is a non identity cyclic subgroup of , and 
(Notice that since ) 
Let be a maximal subgroup of containing . Then a 
-subgroup of is abelian, by Lemma 26.14, so is a 
-subgroup of . by Lemma 26.6. By Lemma 26.12 is a 
Frobenius group. 
Let 52 be a complement to in which contains 
Since is a Frobenius group, it follows that . 
This implies that , so that . 
We next show that . This is equivalent to 
showing that . Suppose false and is a prime 
divisor of . Since divides or . 
Since divides , and is a Frobenius group, 
. Thus, if is any element of of order , then 
 932 SOLVABILITY OF GROUPS OF ODD ORDER 
is contained in a unique maximal subgroup of . Let be an 
element of of order and let . Then . Since 
. is conjugate to in . Since is a Frobenius 
group or is a -group. We can thus find in such 
that . and we can suppose that is a -group. 
This implies that , so that . Since is a 
subgroup of . we have . Since and , 
. and so . But now . 
contrary to and a Frobenius group. Hence, 
. 
By construction, We next show that is a 
complement to in . Since it follows that 
, since and . Thus, . 
We next show that two elements of are conjugate in if 
and only if they are conjugate in . Let , and 
. Since is a T.I. set in . we can suppose M. . If 
then , so is non cyclic, and so . 
We can suppose . In this case is a -subgroup of 
. Now . so we can find 
so that . As observed earlier, this implies that 
. Since . and 
. we have . Then , so and are 
conjugate in , namely, by , since . 
Let be a set of representatives for the conjugate 
classes of elements in which are in and satisfy 
. As we saw in the preceding paragraph, 
is contained in a unique maximal subgroup of , for each , in fact, 
is the unique maximal subgroup of which contains 
. Let , and suppose notation is 
chosen so that are non conjugate in , while is con- 
jugate to some with , if . Set . 
. so that if . 
Let 
. 
Since . it follows that . Also, 
and . If , then is abelian, and in 
fact , where is a cyclic subgroup of . 
Since : is a Frobenius group, 
(26.8) . 
. 
so is a T.I. set in . 
 26. THE MAXIMAL SUBGROUPS OF 
Suppose . Then . and 
where , and is a Frobenius group so 
that divides . Now normalizes can be so 
chosen.) If is abelian, then is a Frobenius group by 
Lemma 26.21, (together with a Frobenius group), and is a 
T.I. set in . If is non abelian, then since is prime on 
. and is prime on is prime on . If is not a 
prime, then centralizes . Since is cyclic and every 
subgroup of is normal in , we have . But 
since . Thus, we can suppose is a prime. If 
centralizes . Lemma 26.21 implies that is of type I. Thus, we 
can suppose that is a Frobenius group. Hence is a 
Frobenius group, as is . Since is a Frobenius group 
and is also a Frobenius group, : is a nilpotent T.I. set in . 
Hence is a non identity cyclic subgroup and satisfies 
the hypotheses of Lemma 26.22 with the obvious factorization 
. But also satisfies the hypotheses of Lemma 26.22, so 
and are isomorphic. In particular, divides , so 
divides . This is absurd, since divides and is a 
Frobenius group with Frobenius kernel . Hence, this case 
cannot arise. Hence, is a T.I. set in , and in fact (26.8) hold.s. 
Since is a -subgroup of . we have . 
Since and . are not conjugate in , by 
construction, we have if . The factorization of 
is now immediate, . We have already shown that 
. Thus, is tamely imbedded in . 
Hypothesis 26.1. 
(i) and is a S-subgroup of . 
(ii) is a prime and is a complement to 
in . 
(iii) is not nilpotent. 
(iv) . 
LEMMA 26.25. Under HypotheSis 26.1, is cyclic and 
satisfies the hypotheses of Lemma 26.22 with the factorization 
is contained in a unique maximal subgroup 
. every element of . is of type I or 
is conjugate to or . 
Proof. Since is not nilpotent, . Let be any maximal 
subgroup of containing . 
Let consist of those in such that either or 
or , and let be a -invariant -subgroup of 
 OF GROUPS OF ODD ORDER 
and let be a -subgroup of We will show is 
nilpotent and that . 
Choose and let be a -invariant -subgroup of . If 
then has -length one, by Lem:ma 26.17. 
Hence, centralizes , so has a normal -complement. 
If then by 3.16 (i) or Lemma 13.4, 
so in this case, too, has a normal -complement. 
Hence, is nilpotent and Since is not nilpotent, . 
Furthermore, . By construction, , so 
for every non empty subset of . Thus, is a T.I. 
set in Since , Lemma 26.14 implies that 
for every non empty subset of . Thus satis- 
fies Hypothesis (ii) in Lemma 26.22. 
Let is not conjugate to . either because 
is not a -subgroup of or because Thus, 
1. If , then since [', ]. But in that case,. 
some -subgroup of normalizes , so is a -subgroup of . 
But in that case, Hence, . 
Since for every non empty subset CIf it. 
follows that has a normal complement in , say a]ld is . 
-subgroup of . Suppose . Then is disjoint from 
is a Frobenius group, and . Further- 
more, a -invariant of has a normal complement 
in . and is abelian, by Lemmas 26.10 and 26.11. Thus is a 
direct factor of , and , since and . If a 
subgroup of is abelian, then dominates , so . 
which is not the case. If a -subgroup of is non abelian, then since: 
is nilpotent, is contained in the center of some -subgroup. 
of . This is absurd, since and is an abelian -subgroup. 
of Hence, . 
Again, let be a -subgroup of normalized by , and let. 
be a -subgroup of normalized by . Then either or 
is a Frobenius group. In both these case, we conclude 
. If does not centralize , then by Lemma 
, so is the unique maximal subgroup of If 
centralizes . then , so if is the unique maximal 
subgroup of containing . But if . then , so of" 
course is the unique maximal subgroup of containing . 
Thus, in all cases, is the unique maximal subgroup of 
. 
We next see that if are primes then every subgroup of" 
of order is cyclic. We next show that . 
Suppose false and where is the -subgroup 
 26. THE MAXIMAL SUBGROUPS oF 935 
of . If . then since urnS', it follows that and 
is the non abelian group of order and exponent , so that 
. Since has a normal complement in and every subgroup 
of of order is cyclic, . Thus, we can suppose that 
e .. By definition of , we also have . Apply Lemma 8.17 
and conclude that divides . Since is a -group. Lemma 
18.4 applied to acting on the of implies that 
centralizes the -subgroup of .. since , it follows once again 
that . Hencet with cyclic . 
, then is cyclic. Suppose is non cyclic. 
Since is nilpotent and since is nilpotent by Lemma 26.4, 
it follows that contains a prime such that 
of is non abelian. Hence, contains a non abelian 
. By construction, , so . This implies 
that is a T.I. set in . 
Since is assumed non cyclic, hence non abelian, and since 
every subgroup of of order is cyclic, it follows that 
is not a prime. By Lemma 26.28 is a nilpotent T.I. set in . 
Set . If . 
,the sets have pair- 
wise empty intersections. Hence, 
fig, 
SO that 
. 
Since , the last inequality is not possible. Hence, 
is cyclic. 
Let be a maximal subgroup of which is not conjugate to 
either S or . If is not a -subgroup of , then Lemmas 26.10, 
26.11 and 26.21 imply that is of type I. If is a -subgroup of 
but is non cyclic, Lemma 26.21 implies that is of type I. 
If is a -subgroup of is cyclic, and is not a prime, 
then by Lemma 26.28. is of type I or contains a subgroup 
which satisfies the hypotheses of Lemma 26.22. But 
also satisfies the hypotheses of Lemma 26.22, so is conjugate to 
. Since can be assumed, either , ' , or 
. The first case yields the second case 
yields and we are done in this case. Lemmas 26.22 
and 26.28 complete the proof. 
 936 SOLVABILITY OF GROUPS OF ODD ORDER 
LEMMA 26.26. Under Hypothesis 26.1 is either of type , or 
(i) is a prime. 
(ii) 
(a) . anti is a -subgroup of . 
(b) is not ni lpotent. 
Proof. By Lemma 26.25, and is cyclic. As 
and , it follows that for every 
non empty subset of . Since , this implies that 
has as a complement. If is not a prime, is nilpotent, 
by Lemma 26.3. This implies directly that is of type , condition 
(ii) in the definition of type following easily, since is non abelian. 
We can suppose that is not of type V. Hence, (i) is satisfied. 
Since is not nilpotent, (ii) (a) and (ii) (b) also hold. 
Lemma 26.26 is important, since if is not of type , then 
satisfies Hypothesis 26.1, as does . 
LEMMA 26.27. Uruder Hypothesis 26.1, one of the following hold s: 
(i) (ii) is a tamely imbedded subset of is a 
-subgroup of . 
Proof. Suppose . If is a T.I. set in we are done. 
Hence, we can suppose that is not a T.I. set in . 
Since is not a T.I. set in and since is a T.I. set in 
(, so Lemma 26.5 (ii) applies), . We first treat 
the case in which is non abelian. Let , where is a 
non abelian -subgroup of . and is the subgroup of We 
show that is the unique maximal subgroup of containing . 
Suppose it. By Lemma 26.1, . 
In particular, , so is a -subgroup of . If 
. then by Sylow's theorem, is conjugate to in 
, so that , and . Hence, we can 
suppose is not conjugate to . Clearly, is not conjugate to , 
since . Hence, is of type I. But then so 
, contrary to assumption. Hence, is contained in 
and no other maximal subgroup of . This implies that is a S- 
subgroup of . 
Choose . There are such elements and 
since is not a T.I. set in . If is not a -element, then 
for some integer , contrary to the fact that is 
a T.I. set in . Hence is a -element and we can suppose that 
. If , then for some , and con- 
tains a -subgroup of both and , which is not the case. Hence, 
 26. THE MAXIMAL SUBGROUPS OF 937 
. Since was any non abelian Sylow subgroup of , it follows 
that is abelian. 
Let . A -subgroup of is non cyclic. Let 
be a -subgroup of containing . If . then by Lemma 
26.7, , so . If , the same equality holds 
by Lemma 26.14 and the containment Thus. 
is not conjugate to . Since is non cyclic, is not conjugate 
to . Hence, is of type I. and this implies directly that 
. Since a -subgroup of is non abelian, 
Lemmas 26.12 and 26.18 imply that 
, 
and it is obvious that is a T.I. set in with as its 
normalizer. We have verified all the properties in the definition of 
a tamely imbedded subset except the conjugacy condition for and 
the coprime conditions. By definition of , together with the 
fact that is a -subgroup of , it follows that . 
If then is conjugate to This is not the case, as 
is non cyclic. Thus, if is a set of representatives for the 
conjugate classes of maximal subgroups of which contain for 
some in and are different from . it follows that 
for . It remains only to verify the conjugacy condition for 
elements of . Let be elements of which are conjugate 
in . We can suppose that and have order and are in . 
otherwise it is immediate that and are conjugate in . Let 
, then . Since 
, it follows that and are conjugate in . (It is at this point 
that we once again have made use of the fact that the subgroups in 
have two conjugate classes of subgroups of order Thus, 
is a tamely imbedded subset of in this case. 
We now assume that is abelian. We first show that is a 
-subgroup of . Otherwise, is not a -subgroup of for 
some non identity -subgroup of 0. Let . Then 
is not conjugate to , since . Suppose is conjugate 
to . Since is a Frobenius group, we have . Thus is 
not nilpotent, since by hypothesis . Hence, is not of 
type V. By Lemma 26.26, is a prime. Since is 
also a prime, it follows that if is a -subgroup of . normalized 
by , then is a Frobenius group, , since is not 
nilpotent). If . then ince . it follows that is 
conjugate to . But divides , and so , which 
is not the case. Hence . But 
 938 SOLVABILITY OF GROUPS OF ODD ORDER 
so , which is absurd since is a -group. Hence, 
is not conjugate to ither or so is of type I. Since is 
of prime order and is a Frobenius group, Since 
we have Hence , which is absurd. 
Hence, is a -subgroup of . This implies directly that 
for all non identity Sylow subgroups of . 
Since is an abelian -subgroup of , and is a T.I. set in , 
the condition implies that two elements of are conjugate 
in if and only if they are conjugate in . 
Suppose , and . Then is a -element, and we 
can suppose . Let . . Since is an abelian S- 
subgroup of and since , it follows that is not con- 
jugate to or . It is now straightforward to verify that is 
tamely imbedded in . 
LEMMA 26.28. Hypothesis 26.1, either or is of type 
II. If is of type II, then 
is T.I. set in Both and are of type II, III, IV or V. 
Proof. First, suppose is of type , but that is not of type 
II. Suppose . By Lemma 26.27. is a tamely imbedded 
subset of . As is a -subgroup of in this case, we have 
. By Lemma 26.24, is a tamely imbedded subset 
of . We now use the notation of section 9. Suppose 
and some element of is conjugate to some element of This 
implies the existence of it such that divides 
, which is not the case. Setting it 
follows that no element of is conjugate to an element of or 
. We find, with . that by Lemma 9.5, 
(26.9) , 
which is not the case. Hence . If were a non abelian 
-subgroup of , then . by Lemma 26.14. Since 
, this is impossible. Hence is abelian, and 
. Thus, is of type II in this case, since the above in- 
formation implies directly that is nilpotent. 
Suppose now that is not of type V. Then from Lemma 26.26 
we have , where is a normal -subgroup of is 
 26. THE MAXIMAL SUBGROUPS OF 939 
a Frobenius group with Frobenius kernel , and is a non identity 
-group. Since is of prime order . it follows from 3.16 that 
contains a subgroup such that is elementary of 
order . and centralizes . 
We next show that centralizes . This is an immediate 
application of 3.16. If , then is of type III or IV accord- 
ing as is abelian or non abelian. If neither nor is of type 
II, then both and are tamely imbedded subsets of , by Lemma 
26.27, since both and satisfy Hypothesis 26.1. Once again, 
(26.9) yields a contradiction. 
If is of type II. then is a T.I. set in . Suppose 
and . Choose Then 
. If then since is a T.I. set in We 
can suppose , and without loss of generality, we assume 
that has prime order . If a -subgroup of is non 
cyclic, then by Lemmas 26.12 and 26.13, . We can suppose 
that the -subgroup of is cyclic, so that . Since 
. it follows that . Choose with 
. If , it follows readily that . so 
we can suppose . In this case, is a Frobenius 
group, and this implies that , which is not the case. 
The proof is complete. 
LEMMA 26. 29. If and is of type then 
is a tamely imbedded subset of . 
Proof. We first show that is tamely imbedded in . 
If is a T.I. set in we are done. If is abelian, 
the conjugacy property for elements of holds. Suppose is 
abelian, , and . Let with . 
Suppose is of type I. Then is disjoint from , since 
. Let be a complement for in which contains 
. Lemmas 26.12 and 26.13 imply that . 
If is a set of representatives for the conjugate classes 
of maximal subgroups of constructed in this fashion, then , 
for . Also, . Suppose . 
for some . and some We can suppose that 
 940 SOLVABILITY OF GROUPS OF ODD ORDER 
has prime order . Let be a prime divisor of . 
so that , Since is of type , this implies that a 
-subgroup of is non cyclic so that . Since does not 
centralize a -subgroup of . But now Lemma 8.16 implies 
that the -subgroup of centralizes a S.-subgroup of . which 
is not the case. Hence, for every . 
By construction 
contains a non identity element. From Lemma 26.13 we have 
and is a T.I. set in . Thus, if is abelian and every 
with the property that and for some . is 
of type , then is tamely imbedded in . 
Suppose is not of type I. Since it is obvious that 
is not of type V. It is equally obvious that is not of type III 
or IV. Hence, is of type II. Since is a -subgroup of . it 
is a -subgroup of , and it follows that is a complement to 
Since is relatively prime to and to each , 
we only need to show that is relatively prime to , 
. Let , so that is a prime and contains 
a -subgroup of . Since , it follows that if 
is a -subgroup of . either , or is 
a Frobenius group. Thus for and 
is a tamely imbedded subset of . Since for every element 
of 
, 
by Lemmas 26.12 and 26.13, the lemma is proved if is abelian. 
We can now suppose that is non abelian, and is not a T.I. 
set in . Let be a non abelian -subgroup of and let 
. Since is not a T.I. set in Lemmas 26.14 and 
26.13 imply that is a cyclic T.I. set in . It follows directly 
from Lemma 26.12 that is a tamely imbedded subset of . 
It remains to show that is a tamely imbedded subset of . 
This is an immediate consequence of Lemmas 26.12 and 26.13. 
LEMMA 26.30. If is a nilpotent S-subgroup of , then two 
elements of are conjugate in if and only if they are conjugate 
in . . 
Proof. Let . If and is of type , 
 26. THE MAXIMAL SUBGROUPS OF 941 
we are done. If and is not of type I. we are done. If 
then If is of type I. is abelian, and 
we are done. If is not of type I. then is of type III or IV. and 
we are done. 
We now summarize to show that the proofs of Theorems 14.1 
and 14.2 are complete. By Lemma 26.30, the conjugacy property for 
nilpotent -subgroups holds. If every element of . is of type I. we 
are done by Lemma 26.29. We can therefore suppose that con- 
tains an element not of type I. Choose not of type I. 
By Lemma 26.21, if , a -subgroup of is cyclic. This 
implies that is a -subgroup of . First, suppose is not a 
prime. Then by Lemma 26.23, is of type or satisfies the con- 
ditions listed in Lemma 26.23. Suppose that is not of type and 
is a complement to in . Let be the smallest prime such 
that a -subgroup of is not contained in and choose 
. . By Lemmas 26.12 and 26.13, is not of 
type I. Lemma 26.21 implies that is a -subgroup of and 
is cyclic. By construction, is not nilpotent, and also by construc- 
tion is not conjugate to . We will now show that is a 
prime. Otherwise, since is not of type I or satisfies the 
conditions of Lemma 26.23. In this case, both and are 
nilpotent T.I. sets in and satisfies the hypotheses of Lemma 
26.22. . . . , 
so that 
(26.10) , 
which is not the case. Hence is a prime, so that satisfies 
Hypothesis 26.1. But then Lemma 26.25 implies that is of type 
V. Thus, whenever . satisfies the hypotheses of Lemma 26.23, 
is of type I or V. 
Suppose every element of is of type I or , and there is 
an element of type V. Let , and let be a -subgroup 
of . Choose so that . Then is not of type I. 
Suppose is of type V. By Lemma and are tamely 
imbedded subsets of . Since (. ) it follows that and 
do not contain elements in the same conjugate class of , 
Setting . 
then (26.10) holds, by Lemma 9.5, which is not the case. 
We can now suppose that it contains an element not of type 
I or V. Lemmas 26.21, 26.23 and the previous reduction imply that 
is a -subgroup of . is not nilpotent, and is a prime. 
Lemmas 26.25 and 26.28 cowplete the proof of Theorem 14.1. 
 942 SOLVABILITY OF GROUPS OF ODD ORDER 
As for Theorem 14.2, Lemmas 26.28 and 26.29, together with 
Theorem 14.1, imply all parts of the theorem, since if is of type 
II, III, IV, or is any tamely imbedded subset of which 
satisfies , and is a cyclic subgroup of which satisfies 
the hypothesis of Lemma 26.22, then adjoining all L. 
, to does not alter the set of supporting subgroups for as 
for all . The proofs are complete. 
 CHAPTER 
27. Statement of the Result Proved in Chapter 
The following result is proved in this chapter. 
THEOREM 27.1. Let be a minimal simple group of odd order. 
Then satisfies the following conditions: 
(i) and are odd primes with contains elemen tary 
abelian subgroups and with . and are 
T-I. sets in . 
(ii) , where and are Frobenius groups with 
Frobenius kernels respectively. , 
and . 
(iii) If then and is a self-normal- 
izing cyclic subgroup of . Furthermore, . 
and . 
(iv) is a cyclic group which is T.I. set in . Further- 
more, is a cyclic group of order 
and is a Frobenius group with Frobenius kernel . 
In this chapter we take the results stated in Section 14 as our 
starting point. The notation introduced in that section is also used. 
There is no reference to any result in Chapter IV which is not con- 
tained in Section 14. The theory of group characters plays an es- 
sential role in the proof of Theorem 27.1. In particular we use the 
material contained in Chapter III. 
Sections 28-31 consist of technical results concerning the characters 
of various subgroups of . In Section 32 the troublesome groups of 
type are eliminated. In Section 33 it is shown that groups of 
type I are Frobenius groups. By making use of the main theorem 
of it is then easy to show that the first possibility in Theorem 
14.1 cannot occur. The rest of the chapter consists of a detailed 
study of the groups and until in Section 36 we are able to supply 
a proof of Theorem 27.1. 
28. Characters of Subgroups of Type I 
Hypothesis 28.1. 
(i) is of Frobenius type with Frobenius kernel and comple- 
ment . 
(ii) where is abelian. is cyclic. . 
 SOLVABILITY OF GROUPS OF ODD ORDER 
(iii) is a subgroup of with the same exponent as such 
that is a Frobenius group with Frobenius kernel . 
LEMMA 28.1. Under Hypothesis 28.1, has an irreducible charac- 
of degree which does not have in its kernel. 
Proof. If is cyclic, then I is a Frobenius group and the lemma 
is immediate. We may assume that is non cyclic. 
Let be a chief factor of with . Let 
. Then is cyclic. Since is of Frobenius type, the 
exponent of is the exponent of . Hence, Let 
be the normal closure of in . Then is abelian. Let be 
a non principal linear character of . Then so 
Lemma 4.5 completes the proof. 
LEMMA 28.2. Suppose is of type anxi satisfies Hypo- 
thesis 28.1. Suppose further that contains an element such 
that and . Then the set of irreducible 
eharacters of which do not have in their kernel is coherent. 
Proof. By Lemmas 28.1 and 4.5, it follows that Hypothesis 11.1 
and (11.4) are satisfied if we take . and let 
play the role of . 
Since is in the center of , it follows that Thus, 
by assumption, is not a chief factor of . Therefore, 
(28.1) . 
Let {. }, where the notation 
is chosen so that if and only if . and where 
. By (28.1) we get that (11.5) holds with 
and by Theorem 11.1 the lemma will follow as soon as it is shown 
that is coherent. 
for . Then each is an integer and 
. By Theorem 10.1, the coherence of will 
follow once inequality (10.2) is established. Suppose (10.2) does not 
hold. Then for some with 
. 
Every character in is a constituent of a character induced 
by a linear character of . Therefore, 
(28.8) . 
Let and let Thus, 
 29. CHARACTERS OF SUBGROUPS OF TYPE III AND IV 
and . If is the inverse image of : in , then 
is of Frobenius type and satisfies Hypothesis 28.1. Two applications 
of Lemma 28.1 imply that . Hence, (28.2) does not hold 
for any . . The proof is complete. 
29. Characters of Subgroups of Type III and IV 
The following notation will be used. 
is a subgroup of type II, III, or IV. plays the 
role of in the definition of subgroups of type II, III, and IV 
given in Section 14. , and have the same meaning as in these 
definitions. is a subgroup of type II. III, IV, or whose 
existence follows from Theorem 14.1 (ii) (b), (e). 
Let {. } and for . let : be the -sub- 
group of . Define 
, . 
. 
Let . and . 
By definition, is a prime. 
is the set of characters of which are induced by nonprincipal 
irreducible characters of . 
is the set of characters of which are induced by irreducible 
characters of that do not have . in their kernel. 
The purpose of this section is to prove the following result. 
THEOREM 29.1. 
(i) If is of type III then is coheren t except possibly 
if . for some prime and . 
(ii) If is of type IV, then is coheren t except possibly 
for some prime aM is not coherent. 
Hypothesis 29.1. 
is a subgroup of type III or IV. 
Throughout this section, Hypothesis 29.1 will be assumed. Thus, 
by Theorem 14.1 (ii) (d), is of type II. Consequently, has prime 
order . Let . and . Thus, by 3.16 (i), 
for . Since , this yields that . 
As does not act trivially on , Lemma 4.6 (i) implies that 
. 
For any subgroup of , let denote the set of characters 
in which have the same degree and the same weight as some 
character in that has in it kernel. 
 SOLVABILITY OF GROUPS OF ODD ORDER 
LEMMA 29.1. Hypothesis 11.1 is satisfied if in that hypothesis 
is replaced by is replaced by is taken as 
is replaced by . and are replaced by , and . 
Proof. By Theorem 14.2, Condition (i) is satisfied. Condition (ii) 
follows from the fact that is a three step group. Condition (iii) 
is immediate and Condition (vi) is simply definition (consistent with 
the present definition). Since is a Frobenius group, contains 
an irreducible character of degree . Hence, Condition (iv) is satisfied. 
The group satisfies Hypothesis 13.2. Hence, by Theorem 14.2, 
Hypothesis 13.3 is satisfied with , and , and 
with replaced by . By Lemmas 13.7, 13.9, and 13.10, 
Condition (v) of Hypothesis 11.1 is satisfied. The proof is complete. 
LEMMA 29.2. If is coherent, then is coherent. 
Proof. As It does not act trivially on . Since 
is a Frobenius group, 3.16 (iii) yields that . As 
either and or and yields that 
. 
Hence, (11.5) is satisfied with . By Lemma 29.1, Theorem 
11.1 may be applied. This implies the required result. 
LEMMA 29.3. If is not coherent, then . 
Proof. Let . We have , as and 
centralizes . Hence, is a Frobenius group. Let 
be all the degrees of characters in and let for 
. Then for each is an integer and . Every 
character of is a constituent of a character induced by a linear 
character of . Thus, for . There are at least 
irreducible characters of degree in . Thus, if is. 
not coherent, inequality (10.2) must be violated for some . In par- 
ticular, this implies that 
 29. CHARACTERS OF SUBGROUPS OF TYPE III AND IV 
Therefore, , so since As is a 
normal subgroup of the Frobenius group we have (mod ). 
Since and are both odd, this implies that as required. 
LEMMA 29.4. If is not coherent, then . 
and . 
Proof. By Lemma 29.8. If then 
so that Hence, and fol- 
lows directly from the fact that If , 
then since is cyclic, Lemma 4.6 (i) implies that some 
non identity element of is in the center of Thus, 
divides which is not the case. Since does not act 
trivially on (iii) now implies that Since 
has prime order and lies outside . we get that 
a Frobenius group. Hence, by 8.16 (i), is nilpotent. Conse- 
quently, is in the center of . As the fixed points of 
on are a direct factor of , and since has no fixed points 
on , we have . The lemma is proved. 
LEMMA 29.5. If is not coherent then is an elementary 
abelian -group of order . 
Proof. In view of Lemma 29.4 it suffices to show that . 
By 8.16 . Thus, if . there exists a subgroup . 
of such that and . If acts irreducibly on 
. then . Hence, is an extra special -group 
and for some integer contrary to Lemma 29.4. 
Suppose that acts reducibly on . Since the irreducible 
constituents of this representation are conjugate under the action of 
. all constituents have the same dimension. As and 
is a prime, this yields that they must all be one dimensional. There- 
fore, there exist elements in such that 
and 
where are linear characters of (mod ) with 
for and a suitably chosen generator of Since 
is odd, for any with . Hence, if 
are given, there exists such that For . 
let be an element of such that 
 SOLVABILITY OF GROUPS OF ODD ORDER 
. 
Since , we get that 
[Pl ] 
. 
Since . this yields that for . 
Since , we get th.at contrary to construction. 
Thus, as required. 
LEMMA 29.6. If is not coherent , then is 
not coherent. 
Proof. Suppose that . Assume that is coherent. 
Let . Let be the equivalence classes of 
chosen so that every character in has degree 
for . and . Suppose is not coherent. 
By Hypothesis 11.1, and Lemma 29.1, all parts of Hypothesis 10.1 
are satisfied except possibly inequality (10.2). Since is not 
coherent, inequality (10.2) must be violated for some . 
Every character in . is a constituent of a character induced 
by a linear character of . Thus for . Hence, 
violation of inequality (10.2) yields that 
Since (mod ) and . this implies that 
. 
Hence is coherent. Since , the proof is 
complete. 
The proof of Theorem 29.1 is now immediate. Lemmas 29.2, 29.4 
and 29.5 imply statement (i). Lemmas 29.2, 29.4, 29.5, and 29.6 
imply statement (ii). 
30. Characters of Subgroups of Type II, III and IV 
The notation introduced at the beginning of Section 29 is used 
in this section. The main purpose of this section is to prove the 
following result. 
 30. CHARACTERS oF SUBGROUPS oF TYPE II. III AND IV 
THEOREM 80.1. Let be a subgr0up 0f type II, III or IV. 
Them is c0heremt except p0ssibly if is 0f type II, is mom 
abeliam 8-gr0up, is a Frobemius group with Fr0bemius kermel 
is a subgr0up 0f type V. 
A11 lemmas in this section will be proved under the following as- 
sumption. 
Hypothesis 80.1. 
( i ) is a subgr0up 0f type II, III, or IV. 
(ii) is m0t c0heremt except possibly if is of type II. 
(iii) has exp0memt . 
For any subgroup of let be the set of characters in 
which have in their kernel. Notice that this notation differs 
from that used in Section 29. 
LEMMA 80.1. The degree of every character is 
by . 
Pr00f. Every character in is a constituent of a character of 
induced by a nonprincipal character of . For any character of 
let be the character of induced by . Set . 
Let . If is of type II 0t III. then by Lemma 4.5 it 
suffices to show that if , then . 
Let be the kernel of and let such that . 
Then and for . As . if , 
then centralizes some element in . Hence, . Let 
. Then char and . 
Suppose . If is of type II, then is a T.I. set in 
by Theorem 14.2. Hence, contrary t0 definition. 
If is of type III, then by Theorem 29.1, is represented irre- 
ducibly on . Since is in the kernel of this represen- 
tation. Thus, contrary to Theorem 29.1. Thus, . 
for and s0 in case is of type II 0t III. 
If is of type IV. we will show that Hypothesis 11.1 and (11.2) 
are satisfied with in that hypothesis being taken as our present 
being taken as and being taken as , and being 
taken as . Certainly (i) is satisfied. Since is a Frobenius group 
with Frobenius kernel (ii) and (11.2) are satisfied, and the 
,remaining conditions follow immediately from the fact that is a 
 950 SOLVABILITY OF GROUPS OF ODD ORDER 
Frobenius group. The present plays the role of in Hypothesis; 
11.1 (iii). 
Notice now that Hypothesis 11.2 is satisfied. By Lemma 11.2 
and the fact that is not coherent it follows that is a non 
abelian -group for some prime whose derived group and Frattini 
subgroup coincide. But . Since . is of exponent 
. so . As has no fixed points on . it follows readily that 
every non linear character of has degree divisible by , as required. 
LEMMA 30.2. For . anti has 
ponen t . 
Proof. If is of type III or IV, the result follows from Theo- 
rem 29.1. Suppose is of type II. Then is a T.I. set in by 
Theorem 14.2. Let be the exponent of for 
. Then and char . Thus, if 
then . contrary to definition of subgroups of type II. 
Suppose for some with Since 
is cyclic, this implies the existence of a subgroup with 
such that is a chief factor of . By 3.16 (i), is nilpotent. 
Thus, and , contrary to definition. 
LEMMA 30.3. For . either or (. ) and. 
. In the first case, is the direct product of 
groups of order each of which is normalized by . In the second 
case, is cyclic of order and acts irreducibly on . 
Proof. By Lemma 30.2. is represented irreducibly on . 
As , the restriction of this representation to breaks up. 
into a direct sum of irreducible representations all of which have the 
same degree . By Lemma 30.2, and so or . 
If , the order of every element in divides . 
Hence, by Lemma 30.2, . 
If then acts irreducibly on Thus, is. 
cyclic. By Lemma 30.2, . Therefore, . Let 
. Then the characteristic roots of are algebraically 
conjugate over . Hence, this is also the case for every power 
of U. If . then some power of has its charac- 
teristic roots in and thus is a scalar. This violates the fact 
that is a Frobenius group. 
LEMMA 30.4. Suppose for some . Let- 
 30. CHARACTERS OF SUBGROUPS OF TYPE II, III AND IV 951 
, 
and let . Then for some integer . Furth er- 
more, contains at least 
irreducible characters of degree at least characters 
of weight and degree . 
Proof. By Lemma 30.3, is cyclic. By Theorem 29.1, is 
not of type IV. so is abelian. Hence, is a Frobenius group. 
By Lemma 30.2, . Furthermore, since acts irreducibly 
on . is the direct product of cyclic groups of the 
same order . Thus, and . By 3.16 (iii) 
every non principal irreducible character of induces an irre- 
ducible character of of degree . Since is abelian, this 
implies that every irreducible character of which does not have 
in its kernel induces an irreducible character of of degree 
. Hence, has at least 
. 
distinct irreducible characters of degree . 
Since satisfies Hypothesis 13.2, Lemma 13.7 implies that all 
but non principal irreducible characters of induce irre- 
ducible characters of . The result now follows. 
LEMMA 30.5. Suppose that for some wi th . 
Let 
and let Then is an integer 
contains at least 
irreducible characters of degree , where . 
Proof. For any subgroup I of , let . By Lemma 
30.3, contains a cyclic subgroup which is normalized by such 
that 
 952 SOLVABILITY OF GROUPS OF ODD ORDER 
and such that for some subgroup which is normalized 
by . Since acts irreducibly on , it follows that 
. Let be the kernel of the representation of on . Then 
is cyclic and so . There are at least 
distinct linear characters of which do not have in their 
kernel. Each of these induces an irreducible character of of 
degree . Thus, by Lemma 30.1, and there are at 
least 
distinct irreducible characters of of degree which have in 
their kernel, and as characters of have in their kernel. If one 
of these induced a reducible character of or two of these induced 
the same character of , then would normalize , contrary to 
the fact that acts irreducibly on . 
LEMMA 30.6. If contains no irreducible character of degree 
then and . 
Furthermore, is coherent. 
Proof. By Lemmas 30.3 and 30.5, and divides 
for . Suppose that for some , 
. 
Then 
. 
Therefore, , and (. ). Thus, 
(30-1) . 
Now Lemma 30.4 implies that (30.1) holds for . Thus, . 
Hence, . Also, . and 
 30. CHARACTERS OF SUBGROUPS OF TYPE II, III AND IV 953 
so . 
If a character in is equivalent to a character in , 
then its degree is prime to , so . Thus, the equivalence 
relation in Hypothesis 11.1 has the property that the present set 
is a union of equivalence classes. Therefore, consists 
of reducible characters of degree . Theorem 14.2 implies 
that Hypothesis 13.3 is satisfied. Hence, Lemma 13.9 implies that 
is coherent. 
The remaining lemmas in this section will be proved under the 
following stronger assumption. 
Hypothesi s 30.2. 
(i) Hypothesis 30.1 is satisfied. 
(ii) is not coherent. 
LEMMA 30.7. If is not coherent, then . 
and . The degree of every 
character in is either or contains exactly 
irreducible characters of degree . 
Proof. Let be all the degrees of characters in 
. Define for . By Lemmas 13.10, 30.1 
and 30.6, all the assumptions of Theorem 10.1 are satisfied except 
possibly inequality (10.2). Every character in is a constituent 
of a character of which is induced by a linear character of . 
Hence, and so . 
Choose the notation so that for and 
for . If is not coherent then inequality 
(10.2) is violated. Lemmas 30.2 and 30.3 imply that for , 
. Thus by Lemmas 30.4 and 30.5, there exists with 
, such that 
. 
Therefore, 
(30-2) . 
For 
 SOLVABILITY OF GROUPS OF ODD ORDER 
. 
By Theorem 29.1, . Thus, (30.2) implies that 
. If , then . . 
Assume first that If , then since and 
(30.2) yields . Thus, every character in 
has degree . Therefore the definition of subcoherence 
implies directly that is coherent contrary to assumption. 
Suppose now that . Then (30.2) yields that 
. Therefore, 
(30.4) (mod ), . 
Further, (30.2) also implies that 
. 
It follows from (30.4) that 
(30.6) (mod ). 
Each term on the left of (30.5) is an integer. Hence, if . 
yields that 
contrary to (30.5). Consequently, . 
Now (30.2) and (30.3) imply that so that We also 
conclude that , so that . Furthermore, . 
and Since is odd, we have Finally 
we get that and so . If or if 
contains more than irreducible characters of degree , then (30.2) 
is replaced by a strict inequality which is impossible as . 
Thus, , and so and the degree of a character in 
is either or . If is of type II or III, then 
and the result is proved. 
Suppose that is of type IV. Since the degree of any character 
in is either or . is generated by two elements. 
Since is generated by two elements. Thus, if we set 
, replace and by , and replace by in Hypothesis 11.2, 
then by Lemma 29.1, Hypothesis 11.2 holds and by Lemma 11.3 and 
Theorem 29.1, we conclude that is coherent, contrary to 
 30. CHARACTERS OF SUBGROUPS OF TYPE III AND IV 955 
assumption. 
LEMMA 30.8. is coherent. 
Proof. By Lemma 30.7, it may be assumed that is a p- 
group for some prime . that , and that . Suppose 
that is not coherent. Let be the set of irreducible charac- 
ters in of degree . Then by Lemma 30.7 
(30.7) , . 
Let be the set of irreducible characters in of degree . The 
group satisfies Hypothesis 13.2. Hence, by Lemmas 13.5, 13.7 
and 30.7, there are reducible characters in of weight 
and degree which have in their kernel. As the sum of the 
squares of degrees of irreducible characters of is , we get 
that 
(30.8) . 
Since is abelian and is generated by two elements, we also have 
(30.9) . 
Now (30.7), (30.8) and (30.9) yield that 
(30.10) 
. 
Hence, by (5.8), is non empty. 
Let for . The character is in- 
duced by a linear character of some subgroup of index in . 
Define 
(30.11) , 
where is the character of induced by Since , it 
follows that induces on Since does not normalize 
. (30.11) is seen to imply that 
. 
Since is tamely imbedded in and vanishes on . we get 
that 
 SOLVABILITY OF GROUPS OF ODD ORDER 
. 
Furthermore, 
for all values of and . 
Suppose that for some . Then for all 
. Hence (30.10) and (30.12) imply that 
. 
Thus 
(30.13) 
. 
Therefore 
. . 
Hence 
. 
Thus by (5.1). Now (30.13) becomes 
. 
Thus 
. 
This implies that 
. 
 30. CHARACTERS OF SUBGROUPS OF TYPE II. III AND IV 957 
Therefore , or equivalently . 
Thus (80.7) yields t.hat 
. 
This is impossible since and both a and are odd. Thus' 
(80.14) for . 
Define . Suppose that . 
As is an isometry on . this yields that 
for all . 
Therefore, 
(80.15) , 
where is a linear combination of elements in and is orthogonal 
. Since it follows that . 
Since 
(80.16) . 
(80.7) and (80.16) yield 
. 
This implies that 
, 
0t . Since is an integer, or 1 and . 
Suppose . Then (80.15) becomes 
(80.17) . 
As vanish on , we have 
(80.18) a. 
Since , we get that 
 958 SOLVABILITY OF GROUPS OF ODD ORDER 
(30.19) , 
for some integer and some which is orthogonal to . Now 
(30.14), (30.17), (30.18) and (30.19) yield that 
. 
- 
Reading this equality mod , we get 
(mod ). 
Thus , contrary to Lemma 30.7. Hence, . Consequently 
, and so for some . Since , 
or- . This implies directly that is coherent. Lemma 
13.10 and Theorem 10.1 now yield that is coherent. The proof 
is complete. 
LEMMA 30.9. is of type II. 
Proof. If is of type III or IV. then Theorem 29.1 yields that 
. Thus, by Lemma 30.8, is coherent. Hence, Hypothesis 
30.2 implies that is of type II. 
LEMMA 30.10. If contains an irreducible character of degree 
. then Hypothesis 11.1 is satisfied with . 
and . 
Proof. By Theorem 14.2, Condition (i) is satisfied. Condition (ii) 
follows from the definition of three step group. Conditions (iii) and 
(vi) are immediate, while Condition (iv) holds by assumption. The 
group satisfies Hypothesis 13.2. Hence, by Theorem 14.2 Hypo- 
thesis 13.3 is satisfied with and By 
Lemmas and 13.10, Hypothesis 10.1 is satisfied. Thus, 
Lemma 10.1 yields that Condition (v) of Hypothesis 11.1 is satisfied. 
The proof is complete. 
LEMMA 30.11. If contains an irreducible character of degree 
then 
. 
Proof. By Hypothesis 30.2, is not coherent. Thus, Lemmas 
30.8, 30.9, and 30.10, together with Theorem 11.1 yield the result. 
 30. CHARACTERS OF SUBGROUPS OF TYPE II, III AND IV 959 
LEMMA 80.12. For . and is a 
Frobenius group. 
Proof. Suppose that for some . Then Lemmas 80.2 
and 80.11 yield that . Thus, . 
Therefore, (5.1) implies that . Hence, or 7. Thus, a 
divides 4 or 6. As is odd and , this implies that 
which is not the case. Therefore, by Lemma 80.8, is 
cyclic of order for . If were not a Frobenius 
group, then for some . { U ell} would lie in Since 
and char , this implies that , contrary 
to Lemma 80.9. 
LEMMA 80.18. . and . 
Proof. By Lemma 80.8, . Choose the notation so that 
. Let where is 
a chief factor of for . Thus, . is of class two 
and so is a regular -group. By Lemma 4.6 (i) centralizes an 
element of for Since is cyclic, this 
implies that , has exponent . Let . Then the 
regularity of , yields that has the same minimal polynomial 
on as on . Hence, by Lemma 6.2, . Now 
Lemma 80.11 implies that if , then 
(80.20) . 
Since , (80.20) implies that 
. 
Hence, by (5.9), and . Thus, (80.20) becomes 
(80.21) . 
If (80.21) implies that 
. 
Thus, and so by (5.1). Hence and (80.21) yields 
1000, which is not the case. If , then 
(80.21) and (5.6) imply that . Thus, or . If 
then 
 SOLVABILITY OF GROUPS OF ODD ORDER 
and . Since and this cannot be the case. 
If then 
is a prime. Thus 2801 . Suppose now that . 
Then by (5.7), . Thus, , or 11. Let be a prime 
factor of . Then and (mod ). Thus, (mod ). 
If , then (mod 6) and , which is impossible. If 
. then and (mod 14). Thus or 48. 
Since (mod 29) and (mod 48). these cases cannot 
occur. If . then and (mod 22). Thus, 
. , or 419. Since (mod ). the quad- 
ratic reciprocity theorem implies that so that (mod 5). 
Thus, or 419. Since (mod 89), 
(mod 199), (mod ) (mod 419). these cases cannot 
occur. Hence, , and the lemma is proved. 
If is not coherent, then Lemmas 80.8 and 80.12 imply that 
is not a prime. Hence, is of Type V. The other statements 
in Theorem 80.1 follow directly from Lemmas 80,9 and 80.18. 
31. Characters of Subgroups of Type 
In this section is a subgroup of type V. Let be the 
subgroup of which satisfies condition (ii) of Theorem 14.1. By 
Theorem 14.1 (ii) (d) is of type II. The notation introduced at 
the beginning of Section 29 will be used. 
is the set of all characters of which are induced by non 
principal irreducible characters of . For any class function of 
let be the class function of induced by . 
For let be the generalized charac- 
ters of defined by Lemma 18.1 and let be the characters of 
defined by Lemma 18.8. 
Hypothesis 18.2 is satisfied with . and replaced 
by . By Lemma 18.7 has exactly irreducible characters which 
induce reducible characters of . Denote these by for , 
where . Let for . Since is a prime 
the characters are algebraically conjugate for . 
Therefore 
for . 
 31. CHARACTERS OF SUBGROUPS OF TYPE 
LEMMA 31.1. contains an irreducible character of ex- 
cept possibly if is a prime and is a Frobenius group. 
Proof. If is not a Frobenius group then there are strictly 
more than classes of whose order is not relatively prime to 
. The result now follows from Lemma 13.7. 
Suppose that is a Frobenius group. By Lemma 6.2 and 3.16 
(iii) is abelian and if the result is false. Then Lemma 
13.7 implies that contains exactly conjugate classes which 
are in . Therefore 
. 
Hence 
. 
This implies that is an elementary abelian -group for some prime 
. Since is cyclic is a prime as required. 
LEMMA 31.2. Let 
. 
Then for . 
Proof. Lemma 10.3 implies that by Lemma 9.4 
. 
Since is rational on by Lemma 13.1, is independent of 
. Thus (31.1) implies that 
(31.2) . 
for some integer , where is an integral linear combination of 
irreducible characters of each of which vanishes on . 
Let . Let be a prime dividing . let be an element 
of order in and let be a prime divisor of in the ring of 
integers of . Let have the same meaning as in Hypothesis 
13.1. Thus by Lemmas 13.1 and 13.3 
(31.3) , 
where is independent of . Therefore 
 962 SOLVABILITY OF GROUPS OF ODD ORDER 
(31.4) . 
In view of Lemma 4.2 (31.3) and (31.4) imply that 
(mod ) 
(mod ) 
(mod ). 
Thus (31.2) and (31.5) yield that (mod ). Thus as 
required. 
The main purpose of this section is to prove that is coherent. 
Theorem 12.1 will play an important role in the proof of this fact. 
The lemmas in this section will from now on satisfy the following 
assumption. 
Hypothesis 31.1. 
is not coherent. 
By theorem is a Frobenius group. Hence by Lemma 
11.2 is a -group. Define 
(31.6) , . 
Let r be all the integers which are degrees of 
irreducible characters of . Let 
(31.7) , . 
By Lemma 13.10 Hypothesis 12.1 is satisfied. Let . be defined 
by (12.3) for . 
LEMMA 31.3. Sttppose th at for some integer . Then 
is not a prime power. 
Proof. Suppose that for some prime . Then by Lemma 
11.5 and contains a subgroup which is 
normal in and satisfies and . Therefore 
and contains irreducible characters r of 
degree . Define 
. . 
By Lemma 9.4 we have that 
(31.8) . 
Furthermore 
 31. CHARACTERS OF SUBGROUPS OF TYPE V 
, 
(81.9) 
. 
Suppose that for some with . Then 
(81.8) and (81.9) imply that 
. 
Hence , or which is not the case. Therefore 
(81.10) for . 
Equation (81.9) also yields that for some integer 
(81.11) , 
for . 
Furthermore Lemma 18.8 implies that for , 
(81.12) . 
Since vanishes on and Lemma 18.2 yields that 
(81-18) 
where for Now (81.12) and 
(81.18) imply that 
for . 
Define a2.. Then (81.18) implies that 
(81.14) 
'. 
For any value of the term in the last summation in (81-14) is non 
zero. Furthermore . Thus (81.14) implies that 
if there are exactly values of with , then 
(81.15) is even. 
The last statement follows from the fact that since 
and thus has its values in . By definition 
. 
Lemma 81.2 implies that for any value of with 
 SOLVABILITY OF GROUPS OF ODD ORDER 
or . 
Now (31.8), (31.11) and (31.15) yield that 
. 
or 
. 
Therefore 
. 
Suppose that , then . Now (31.8) and (31.11) imply 
that By (31.15) 
this implies that or Assume first that then 
(31.10) implies that . Hence by (31.16) 
. 
This implies that . Consequently (31.8), (31.10) and (31.11) 
yield that 
which is not the case. 
Assume now that . Choose with 
so that for . Thus and by 
(31.16) 
. 
Since has its values in and has its values in 
for any algebraic conjugate of . By Lemma 13.1 has at 
least algebraic conjugates. Hence , therefore . 
Since vanishes on Lemma 13.1 implies that for 
. 
Hence if then 
. 
Therefore . Hence 
. 
Consequently (31.8), (31.10) and (31.11) yield that 
 31. CHARACTBRS OF SUBGROUPS OF TYPE 965 
. 
The assumption that has led to a contradiction in all cases. 
Therefore (31.8), (31.11) and (31.15) imply that 
, . 
Thus for . Thus (31.14) implies that 
. 
Hence or and . Thus or 
and . In either case this implies that the set of charac- 
ters consisting of . and is coherent. 
This includes all characters in which have in their kernel. 
Since the result now follows from Theorem 
11.1 with - . and . 
LEMMA 31.4. is coheren t. 
Proof. By Theorem 30.1 is a power of 3 if is not coherent. 
By Lemma 31.3 is odd. Thus the lemma follows from Lemma 11.6. 
LEMMA 31.5. For let be as irretucible charac- 
of wi th . Let be the normal closure of in 
. Let be all the degrees of irreducible characters 
of . Then is a Frobenius group. For any value of with 
let be an irreclucible character of of tegree . 
Define 
. 
for 
for . 
Them 
for . . 
for . . 
Furthermore if is a prime then one of the following possibilities 
must occur: 
, 
. 
- 
 SOLVABILITY oF GROUR OF oDD ORDER 
with f0r . 
Pr00f. For let 
's0 , 
where each is a linear combination of the generalized characters 
and each is orthogonal to each of these generalized characters. 
Since for is orthogonal to and and 
a11 of these vanish on , Lemma 13.2 implies that 
(31.17) . 
where is a set of integers depending on . Since 
. Since , 
Theorem 12.1 implies that 
(31.18) for all . 
Assume first that . Then . Thus (31.17) and 
(31.18) imply that 
. 
If then for each value of or . Thus 
which is not the case. Hence and so 
(31.19) . let . 
As is orthogonal to Lemma 31.4 yields that for all 
for . 
By (31.19) . Hence 
. 
Suppose now that for some . Then for all with 
. Hence (31.18) and (31.19) imply that 
which is not the case. The result is proved in case . 
Let . Then . By assumption for 
. since is a prime. By (31.17) 
 31. CHARACTERS OF SUBGROUPS OF TYPE 967 
for 
where the sign is independent of . Since this yields 
that for . Hence (31.17) and (31.18) imply that 
. 
If this yields that and and the result follows. 
If then we get that 
. 
Hence and the result is proved also in this case. 
LEMMA 31.6. Let . have the same meaming as in Lemma 
31.5. Define 
. 
Then for . 
Proof. Let be the equivalence class in defined by (12.3) 
which contains X. If is in then the result follows from the 
coherence of . For any , let be the number of characters 
of degree in and define as in (12.4) by 
(31.20) , 
where is the minimum degree of any character in . 
Let 
(31.21) , 
where is an integral linear combination of the gener- 
alized characters and is orthogonal to and to every . 
Theorem 12.1 yields that 
(31.22) . 
vanishes on and . Furthermore for 
. Therefore Lemma 13.2 implies that 
(31.23) . 
where . 
Since is orthogonal to Lemma 31.4 yields that 
for . 
 OF GROUPS OF ODD ORDER 
By (81.28) 
, 
where the sign is independent of . Therefore 
for . . By (81.22) and (81.28) we see that 
(81.24) 
. 
If and then for each at most one of 
vanishes. Hence (81.24) yields that 
. 
This is impossible as either or is even. If then 
(81.24) implies that 
2 . 
If . then 2 for . Hence 
which is not the case. Thus or and . Thus 
we get 
for . 
Assume that the result is false. Then for some value of 
. We will next show that for . If this is false 
then there exists such that . If is any character in 
then (. ). Thus (81.25) implies that 
(81.26) . 
Thus for every in P. Let be the exponent of . By 
Lemmas 80.1, 80.4 and 80.5 . Thus is in the kernel of 
. Define 
. 
By (81.25) 
 31. CHARACTERS OF SUBGROUPS OF TYPE 
Thus (81.22), (81.28) and (81.26) yield that 
, 
where if and 16/25 if . and ranges over 
the irreducible characters in . By Lemma 18.7 there exist irre- 
ducible characters of which induce the characters for . 
Consequently 
where ranges over the irreducible characters of which are distinct 
from all and do not have in their kernel. Therefore . 
otherwise since is odd there are at least 2 characters I of 
degree at least . Furthermore 
. 
This implies that 
(81.27) , 
where if and otherwise. Let , where 
. Let and . 
Since is of type II and . Thus (81.27) implies that 
. Since for some prime dividing we get 
that . Thus by (5.1). Hence which is not 
the case as . Hence no such group exists. Thus is an 
elementary abelian -group for some prime. Therefore is a 
prime and for . Consequently for 
contrary to assumption. 
Returning to (81.24) we see that 
. 
Therefore for . Thus 
(81.28) for . 
Now (81.24) implies that 
(81.29) 
. 
 970 SOLVABILITY OF GROUPS OF ODD ORDER 
Suppose that . Thus and . Then (31.29) im- 
plies that . As this implies that 
or which is not the case. Therefore 
(31.30) . 
By (31.29) either or . . 
Now (31.23) and (31.28) imply that 
or 
. 
This is equivalent to 
(31.31) or 
. 
Since . is a real valued generalized character. Thus 
. By (31.31) , hence by (31.22) . Now 
(31.21) implies that 
(31.32) , 
where for ranges over the characters of degree 
in . 
Suppose that contains an irreducible character . Then by 
Lemma 31.4 
for . 
As is rational valued on elements of for all . Thus 
(31.31) and (31.32) imply that 
. 
Therefore contains no irreducible characters. Hence by Lemma 
31.1 
(31.33) a prime. 
Now Lemma 31.3 implies that is odd, where is defined in 
(31.6). As ThCOtCm 12.1 implies that if is 
 31. CHARACTERS OF SUBGROUPS OF TYPE 971 
defined in (31.20) then 
(31.34) (mM ) or . 
Assume first that in (31.32). Let be defined as in Lemma 
31.5. Suppose that 
. 
- 
Then (31.31) and (31.32) yield that 
. 
Thus by Lemma 31.5 
(31.35) . 
Then 
(31.36) . 
where for . Suppose that 
. Then (31.31), (31.32) and (31.36) yield that . 
Hence . Thus by (31.35) and (31.36) 
Thus (31.34) yields that 
(31.37) (mod ). 
Equations (31.31), (31.32), (31.35) and (31.36) imply that 
Hence (31.37) implies that (mod ). This contradiction arose 
from assuming . 
Assume now that . Then 
Hence (mM ). Thus (31.34) and 3.15 imply that 
(31.38) (mM ). 
Now (31.31) and (31.32) yield that 
 SOLVABILITY OF GROUPS OF ODD ORDER 
. 
Therefore 
(31.39) . 
By (31.38), . Thus (31.39) yields that 
. 
It is easily verified that is a monotone increasing function for 
and . Thus . By (31.39) . 
Hence . Now (31.39) becomes 
, 
or equivalently 
(31.40) . 
Therefore . hence 
Thus . As is even, (31.38) now yields that . 
Now (31.40) becomes 
, 
or 
. 
As is a prime one of the factors is and the other is . As 
the factors differ by 2 this implies that . Hence . 
Since implies that . This contradiction establishes 
the lemma in all cases. 
THEOREM 31.1. is coh erent. 
Proof. Suppose that is not coherent so that Hypothesis 31.1 
is assumed. Let have the same meaning as in Lemmas 
31.5 and 31.6. Choose . Then 
(31.41) . 
. 
for . 
 31. CHARACTERS OF SUBGROUPS OF TYPE 
Lemmas 31.2, 31.5 and 31.6 together with (31.41) imply that 
or 
and . If the latter possibility occurs then by Lemma 
10.1 it may be assumed after changing notation that in any case 
(31.44) . 
Now Lemma 31.5, (31.43) and (31.44) imply that 
(31.45) 
. for . 
Since and . 
implies that 
(31.46) for . 
Lemmas 31.2 and 31.5 and equations (31.42) and (31.44) yield that 
(31.47) (. ). 
By Lemma 13.10 is subcoherent in . Since 
it follows from (31.47) that 
(31.48) . 
Let have the same meaning as in Lemma 31.5. Then there 
exists a subgroup of such that is a chief factor of 
and . Let be the irreducible characters of of 
degree . . Then (31.46) implies directly that is 
coherent. Hypothesis 11.1 is satisfied with and 
. If is not coherent then Theorem 11.1 implies that 
. As is a Frobenius group this implies that . 
Therefore is an extra special -group. Thus for 
some integer . Define 
. 
Then consists of all characters in having the same weight 
and degree as some character in which has in its kernel. By 
(31.48) is coherent. Thus if is not coherent Theorem 11.1 
implies that 
 SOLVABILITY OF GROUPS OF ODD ORDER 
(31.49) . 
Lemma 13.6 applied to the group implies that or 
and . As is odd this yields that in 
any case. Thus by (31.49) 
. 
This contradiction suffices to prove Theorem 31.1. 
COROLLARY 31.1.1. If is an irreducible character of of 
degree then 
. 
Proof. Let and let . By Theorem 31.1 
(31.50) 
. 
As is rational on . . By Lemma 13.9 . 
Thus (31.50) implies that 
for . 
Hence by Lemma 31.2 for As 
and we get that 
. 
As vanishes on Lemma 13.2 now implies the required result. 
COROLLARY 31.1.2. is a Frobenius group and is a prime. 
Proof. Suppose that contains an irreducible character . 
Choose in . Then . If is not 
coherent may be chosen in by Theorem 30.1 and Lemma 31.1. 
Hence by Corollary 31.1.1 and Lemmas 13.9 and 30.8, 
. 
Therefore contains no irreducible characters. Lemma 31.1 now 
implies that is a Frobenius group and is a prime. 
 32. SUBGROUPS OF TYPE 
. Subgroups of Type 
THEOREM 32.1. contains no subgroup of type V. 
Proof. Suppose that the result is false and is a subgroup of 
type V. is tamely imbedded in by Theorem 14.2. For 
let have the same meaning as in Definition 9.1 and let be 
defined by (9.2). Let be the set of elements in which are 
conjugate to some element of for . By Lemma 9.5 
. 
Let be an irreducible character of degree in . By Theo- 
rem 31.1 and Lemmas 10.3 and 9.4 
(32.2) for 
where is independent of T. Now Theorem 31.1 and Corollary 31.1.1 
imply that in (32.2). Thus for . Hence 
Theorem 31.1 and Lemmas 10.3 and 9.5 imply that 
(32.3) . 
Let be defined by Theorem 14.1 (ii) (a) and let . 
Define 
. 
Thus Theorem 14.2 (ii) (a) implies that 
(32.4) . 
Let be the set of elements in which are conjugate to some 
element of . Since is a T.I. set in . 
(32.5) . 
Define 
. 
 SOLVABILITY OF GROUPS OF ODD ORDER 
Then (82.1), (82.4) and (82.5) imply that 
By (82.8) 
(82.7) . 
By Corollary 81.1.2 is a prime and is a Frobenius group. 
Hence by Lemma 18.1 . are algebraically conjugate 
characters whose values lie in . Every element whose order is 
divisible by lies in . Thus is a rational integer 
for and . Now Corollary 81.1.1 implies that 
for . Hence (mod 2) for 
. Therefore for . Now (82.6) and (82.7) imply 
that 
or 
(82.8) . 
Since yields that and . Thus, 
acts irreducibly on . Therefore is an extra special group. 
Let . Then by Lemma 18.6, Thus (82.8) 
implies that . Hence which is not the 
case. The proof is complete. 
COROLLARY 82.1.1. Let be a sttbgroup of type II. III or IV. 
Let have the same meamimg as Sectiom 29. Them is 
coheremt. 
Proof. This is an immediate consequence of Theorems 80.1 and 
82.1. 
33. Subgroups of Type I 
LEMMA 88.1. Let be a maximal subgroup of let have 
the same meamimg as sectiom 14. If is of type I with Frobemius 
 33. SUBGROUPS OF TYPE I 
kernel let be the set of all irreducible characters of which 
do not have in their kernel. If is of type II. III or IV let 
be the set of characters of each of which is induced by a 
principal irreducible character of which vanishes outside . Let 
have the same meaning as in section 9 and let be defined by 
(9.2). If then can be defined. Furthermore is constant 
on for . 
Proof. Since is odd Lemmas 10.1 and 13.9 imply that 
can always be defined as {X. } is coherent. 
If then and there is nothing to prove. If 
with let be a supporting subgroup of such that 
. If is of type I then the result follows from Lemmas 4.5 
and 10.3. By definition cannot be of type III or N. If is of 
type II then the result is a simple consequence of Corollary 32.1.1. 
The main purpose of this section is to prove 
THEOREM 33.1. Every subgroup of type I is a Frobenius . 
A11 the remaining lemmas in this section will be proved under 
the following assumption. 
Hypothesis 33.1. 
contains a subgroup of type I which is not a Frob enius . 
If Hypothesis 33.1 is satisfied the following notation will be used. 
is a set of primes defined as follows: if and only if 
contains a subgroup of type I with Frobenius kernel such that 
a S-subgroup of is not cyclic. 
is the smallest prime in . . 
is a -subgroup of . 
is a -subgroup of with . 
is a maximal subgroup of such that . 
has the same meaning as in Lemma 33.1. 
If is of type I let be the Frobenius kernel of . Let 
with . 
If is of type II, III or IV let be the maximal normal nilpotent 
-subgroup of . Let be a complement of in and let ; be 
a complement of in with . 
LEMMA 33.2. is the unique maximal subgroup of which 
contains . Furthermore is either a Frobenius or 
is type III or IV can be chosen to lie in . 
 978 SOLVABILITY OF GROUPS OF ODD ORDER 
Proof. By Theorem 32.1 is not of type V. If is of type II. 
III or IV then since is not cyclic. Since is a T.I. set 
in it may be assumed that . 
There exists such that . Thus e]..ther 
or is cyclic and . If a -subgroup of is abelian 
then is the -subgroup of . Hence char and so 
. Therefore is of type III or IV and . 
By definition is the unique maximal subgroup which contains 
. If the -subgroup of is not abelian then is of type 
IV and it may be assumed that . Then and in this 
case also is the unique maximal subgroup of contains 
. 
Suppose that is of type I. Let . be a -subgroup (of with 
. If then is abelian. Thus, and so 
. Hence, is an abelian -subgroup of . By construction, 
. Hence, . by Burnside's transfer theorem. Since 
is odd, if an element of induces an automorphiscn of of 
prime order , then . By the minimal nature of a -subgroup 
of is cyclic. Let . Since is of type is 
cyclic. We can now find a prime such that some element 
induces an automorphism of order on . Let be a -subgroup 
of permutable with . Since normalizes . and is 
cyclic. Since is a Frobenius group, centralizes . Let 
. Then , and . 
Let be a maximal subgroup containing . The minimal 
nature of implies that . Hence, by Lemma 8.13, centralizes 
every chief -factor of , so centralizes which is not the 
case. We conclude that . Therefore . Herpce . 
is not a T.I. set since is not a T.I. set in . This y:ields that 
either or . In either case this implies that every 
prime divisor of is less than . The minimal nature of now 
implies that is a Frobenius group. 
The previous parts of the lemma imply that if is maximal 
subgroup of which contains then is a Froben.ius group 
and divides the order of the Frobenius kernel of . If is abelian 
then and . If is non abelian then 
. The uniqueness of is proved. 
LEMMA 33.3. There exists an irreducible character which 
does not have in its hernel such that or . 
Proof. Let be a character of which does not have in its 
kernel and is induced by a linear character of if is a Frobenius 
group and by a linear character of if is of type III or IV. 
 33. SUBGROUPS OF TYPE 
Either and so , or is cyclic. In either case 
this implies that if then or . 
If is of type III or IV then is a prime and the result 
follows. Suppose that is a Frobenius group. If then 
has the required properties by assumption. If then is 
abelian since is not a T.I. set in . Thus and . 
Suppose that where and . Then an 
element of of order acts as a scalar on . There exists 
such that . Thus contains a Frobenius group of order 
which is not the case. Therefore every prime in divides 
or. every prime in divides . Since 
this yields that or . The lemma follows 
since . 
LEMMA 33.4. Let be the character defined in Lemma 33.3. 
Then 
for 
Proof. Set . Observe that if is a Frobenius group, 
then since . it follows that , so that . This 
equality also holds if is of type III or IV. 
Set so that , where is a gener- 
alized character of orthogonal to . Let . be the 
characters in of degree . Since divides or , 
it follows that , and so . 
We next show that is coherent. If is a Frobenius group, 
the coherence of follows from Lemma 11.1 and the fact that 
is of type I. 
Suppose is of type III or IV. Then Hypothesis 11.1 and (11.2) 
are satisfied with the present in the role of in the role of 
, and in the role of . By Lemma 11.1, we may assume that 
. Hence, and , so 
that . If is non abelian, then divides . Hence, 
we may assume that is abelian of order and is of type III. 
By Theorem 29.1 (i), no element of centralizes . This implies 
that if . are the characters in of degree , then . 
. . 
Let , so that , with 
. If the coherence of follows from Theorem 
30.1. As and it follows that . 
and . Hence, and . But now 
, so that . This is not 
the case as and are real valued generalized characters of 
 980 SOLVABILITY OF GROUPS OF ODD ORDER 
orthogonal to . The coherence of is proved in all cases. 
Since , the lemma follows from Lemmas 9.4 and 33.1. 
LEMMA 33.5. Let be the character defined Lemma 33.3. Then 
. 
Proof. Let be the set of all elements in which are conjugate 
to an element of 
ments in which are conjugate to an element of for some . 
No subgroup of can be a supporting subgroup for both and . 
If were a supporting subgroup of then would not be minimal 
in the set . Thus is disjoint from . Therefore by Lemmas 9.5, 
4.5, 10.3, 33.1 and 33.4 
' 
LEMMA 33.6. Let where . Then there exists 
in . such that . Furthermore satisfies 
Hypothesis 28.1. 
Proof. If is of type , then . Thus, is nilpotent and 
hence abelian. The result follows from 3.16 (ii) and the fact that 
is not cyclic. 
Suppose is not of type I. If then we may assume 
that . Then is a Frobenius group and . By 
3.16 (ii), centralizes an element of . Since is a prime, 
this contradicts the fact that contains a Frobenius group of order 
. Thus, . Let . Since is a T.I. set in , 
we get that is a cyclic normal -subgroup of . If , then 
is abelian and the result follows from 3.16 (ii). 
Assume now that . We may assume that . 
If does not centralize , then there exists such 
that is a Frobenius group. Hence, by 3.16 (ii). But 
in this case, lies i-n no normal abelian subgroup of contrary to 
the definition of groups of Frobenius type. Thus, centralizes . 
Since is abelian and . this implies that 
 33. SUBGROUPS OF TYPE 981 
. The lemma now follows from 3.16 (ii). 
LEMMA 33.7. Let be the set of all irreducible characters of 
which do not have in thei r k ernel. Let be the character 
defined in Lemma 33.3. If is coherent then is constant on . 
Proof. Let be a set of supporting subgroups of in 
, and let . By definition, 
. 
Suppose and We will show that . For 
otherwise, some power of is -conjugate to an element A of . 
Since is a supporting subgroup of some tamely imbedded subset of 
. it follows that . Hence, is in . 
We next show that is of type I or II, . Suppose 
is not of type I. Then , and we assume that 
. Since is a supporting subgroup of we may 
choose in so that . By the first paragraph, 
. Hence, . If . then by a well known 
property of nilpotent groups, we have , so that , 
which is not the case. Hence, so is not of type 
III or IV; is of type II. 
Let be the least common multiple of the orders of all elements 
of . We will show that . If is 
of type I. then is a Frobenius group, so divides . and we only 
need to verify that is not conjugate to or . As 
none of the groups is a Frobenius group, this is clear. 
Suppose is of type III of so that . Since 
none of is of type III or IV, we have 
Since . it is trivial that 
. 
We appeal to Lemma 10.4 and conclude that is rational on 
and on every supporting subgroup of . 
Let be a supporting subgroup of and let be a character 
of : with . Let be irreducible characters of th 
. Then and no irreducible character 
appearing in is rational on . Thus, . 
If is of type I. then Hypothesis 10.2 is satisfied with our present 
in the role of . If is of type II, then a complement to in 
is abelian, and again Hypothesis 10.2 is satisfied. Hence, by Lemma 
10.2, is constant on the cosets of in , and in particular 
is constant on all the sets . As is assumed coherent, 
an appeal to Lemma 10.5 completes the proof of this lemma. 
 982 SOLVABILITY OF GROUPS OF ODD ORDER 
Theorem 33.1 will now be proved by showing that Hypothesis 
leads to a contradiction. 
Choose and . By Lemmas 33.1 and 33.4 
(33.2) . 
Let be a prime divisor of in . By Lemma 4.2 
(33.3) (mod ) 
(mod ). 
Now (33.2), (33.3) and (33.4) yield that 
(mM ). 
By Lemma 10.4 is rational. Thus 
(mod ). 
Since by Lemma 33.3, we get that 
(33.5) for . 
If every element in commutes with an element of then (33.5) 
implies that 
(33.6) for . 
If not every element in commutes with an element of then 
is constant on by Lemmas 28.2, 33.6 and 33.7. As (33.5) holds 
for at least one element in . we get that (33.6) holds in any case. 
Now Lemma 33.5 and (33.6) imply that 
. 
This can be written as 
(33.7) where . 
Since and is a complement to in 
yields that 
. 
Hence or . Thus and a -subgroup 
of is cyclic contrary to the simplicity of and the fact that 
 . THE SUBGROUPS AND 983 
is odd. This contradiction completes the proof of Theorem 33.1. 
THEOREM 33.2. c0ntains a subgroup 0f type II. 
Pr00f. Suppose false. Then by Theorems 14.1 and 33.1, every 
maximal subgroup of is a Frobenius group. Let be a maximal 
subgroup of and let be a complement to the Frobenius kernel of 
. We will show that is abelian. Suppose false. 
Let be the set of primes such that for some maximal subgroup 
with Frobenius kernel and complement . a of is 
not in . Let be the least prime in . We may suppose that 
a of is not contained in . Then . Let 
be a maximal subgroup of containing . Since 
'He.. If is contained in the Frobenius kernel of . 
then so is . This is impossible as does not centralize , 
while is nilpotent. Hence . Since , it follows 
that is not contained in , and that a of 'He. is cyclic. 
Hence, by Burnside's transfer theorem, is not simple. Since this 
is not possible, is abelian. 
Let . Let be a maximal subgroup of containing . 
It follows that is nilpotent. Hence, is solvable by the main 
theorem of [10]. The proof is complete. 
34. The Subgroups and 
By Theorems 32.1 and 33.2 contains two subgroups and . 
each of which is of type II, III or IV and which satisfy Condition 
(ii) (b) of Theorem 14.1. The following notation will be used 
the rest of this chapter. This differs slightly from that introduced 
previously. 
. 
Thus and are both primes. Let be the of and 
let be the of . Then . Let 
, . 
Let be a complement of in and let be a complement of 
in . By 3.16 (i) and are nilpotent, thus 
. 
if is of type II and 
, 
 SOLVABILITY OF GROUPS OF ODD ORDER 
if is of type II. Let 
. . 
If is of type III or IV let . If is of type II then 
a maximal subgroup which contains is not conjugate to 
since is not -closed. Hence by Theorem 33.1 is a Frobenius 
group. Let be the Frobenius kernel of . Thus . Define 
similarly. Let 
, 
. 
is the set of characters of which are induced by irreducible 
characters of which do ot have in their kernel. 
is the set of characters of which are induced by irreducible 
characters of which do not have in their kernel. 
The set as defined here is a subset of the as defined in 
Section 29. hus by Corollary 32.1.1 and are coherent. 
are the sets of irreducible characters of 
respectively which do not have respectively in their kernel. 
For are the generalized characters 
of defined by Lemma 13.1; are the characters of defined by 
Lemma 13.3; are the characters of defined by Lemma 13.3. For 
is the character of defined by Lemma 13.5. For 
is the character of defined by Lemma 13.5. 
If , where is a maximal subgroup of and if 
is a class function of then denotes the class function of induced 
by . Whenever this notation is used will be uniquely determined 
by the context. 
Throughout this section no distinction is made between and . 
Any result in this section about one of these groups is automatically 
for the other by symmetry. 
LEMMA 34.1. Either 
and is cyclic or is the product of at most cyclic groups 
and . For is induced by a linear 
character of . . Either is a Frobenius group with 
and 
 34. THE SUBGROUPS AND 985 
or contains an irreducible character of degree which is 
by a linear character of . 
Proof. If then by is nilpotent. Thus 
is nilpotent contrary to assumption. Hence contains a subgroup 
such that and is a chief factor of . Hence 
is represented on the elementary abelian group By 8.16 (i) 
is nilpotent. Therefore is faithfully and irreducibly 
represented on . By 8.16 (iii) . 
Let where . By Lemma 4.6 (i) 
. Thus . Hence any non principal 
linear character of induces for some with . 
As is a prime the characters are algebraically conjugate for 
Thus for . Let for 
a linear character of . 
Suppose that . Then contains a subgroup 
such that is a chief factor of . Let be a non 
principal linear character of . Then induces an irreducible 
character of of degree . 
Suppose that is represented reducibly on . Since 
the irreducible constituents of this representation all have the same 
dimension. This dimension is 1 since is a prime. Thus is the 
direct product of cyclic subgroups for some integer , each of which 
has order dividing . No element of is represented as a 
scalar as is a Frobenius group. Therefore and . 
The irreducible constituents of the representation of on are 
distinct since is irreducibly represented on . Let 
where for some generator of and 
such that normalizes each . Let 
with and . for 1. Sup- 
pose and centralizes for some . Let then 
. 
Then . If then is conjugate to . Hence 
is conjugate to which is impossible as is odd. Therefore 
. Then . for and so . This proves 
that no element of . leaves fixed- Let be a non principal 
linear character of with ker - Let . 
then induces an irreducible character of of degree 
. 
 986 SOLVABILITY OF GROUPS OF ODD ORDER 
Assume now that is irreducibly represented on . Then 
is cyclic since is abelian. If a subgroup of acts reducibly on 
then it is represented by scalar matrices. As is a Frobenius 
group every non identity subgroup of acts irreducibly on . 
Thus permutes the subgroups of order in and no element 
of leaves any such subgroup fixed. Hence 
. 
Suppose now that contains no irreducible character of degree . 
By an earlier part of the lemma this implies that . 
Thus and . Since , we must have 
. By 3.16 (i) is nilpotent. If then there exists 
a subgroup of such that . Hence is the center 
of since acts irreducibly on . Thus is an extra special 
-group. This implies that is even which is not the case. Thus 
. Hence is a Frobenius group. Consequently contains 
irreducible characters of degree . Lemma 13.7 now implies 
that 
LEMMA 34.2. Either is a Frobenius group with aM 
or is a Frobenius group with and 
. 
Proof. If the result is false then Lemma 34.1 implies that 
contains an irreducible character of degree and contains an 
irreducible character of degree . Every character in is rational 
valued on by Lemma 10.4. Since is odd this implies that every 
generalized character of weight 1 in is orthogonal to . Define 
. 
Then and . Thus 
(. ) 
. 
 34. THE SUBGROUPS AND 987 
This proves the lemma. 
LEMMA 34.3. For 
. 
Proof. Since is a T.I. set in and is coherent the Frobenius 
reciprocity theorem implies that for 
for , 
where is a generalized character of , and . Therefore 
. 
This implies that 
(34.1) 
. 
By Lemma 34.1, . thus 
. 
The result now follows from (34.1). 
LEMMA 34.4. For 
. 
Proof. Since is a T.I. set in the coherence of and the 
Frobenius reciprocity theorem imply that for , 
where is a generalized character of . Therefore for 
. 
If and is a prime divisor of in then by 
Lemma 4.2 
 SOLVABILITY OF GROUPS OF ODD ORDER 
(mM ). 
Thus the expression in (34.2) is non zero. The result now follows 
from the fact that 
(mod ). 
LEMMA 34.5. Suppose that contains an irreducible character 
of degree which is induced by a character of . Then 
. 
Proof. As is a T.I. set in the coherence of and the 
Frobenius reciprocity theorem imply that 
for , 
for some generalized character of . Therefore 
. . 
If then by Lemma 34.1 
. 
Hence the result follows from (34.3) in this case. If then 
thus (34.3) also implies the result in this case. 
LEMMA 34.6. Let be the set of elements in which are not 
conjugate to any element of or . Suppose that contains 
an irreducible character of degree Define 
{ (mM )}. 
Then 
. 
Proof. Suppose that . Let . Then 
and 
. 
 34. THE SUBGROUPS AND 989 
Since is rational. Thus for 
As we must have that 
(84.4) 
(mod ). 
Suppose that . Then since we must have that 
for . Hence by Lemma 18.1 
contradicting the fact that . Hence and by 
as required. 
LEMMA 84.7. 
(i) If then aM . 
(ii) If then and (18/20) . 
(iii) If and then . 
(iv) If then or Furthermore . 
(v) If then is an elementary abelian -group aM 
or aM . 
(vi) If aM then . 
Proof. If is a Frobenius group with 
then all the statements in the lemma are immediate. Suppose that 
this is not the case. Then by Lemma 84.1 contains an irreducible 
character which is induced by a linear character of By Lemma 
84.2 is a Frobenius group with . . 
. 
and are T.I. sets. Let have the same 
meaning as in Lemma 84.6. Then 
. 
Since is rational valued on by Lemma 10.4, Lemma 84.5 implies 
that 
' 
(84.6) 
. 
 SOLVABILITY OF GROUPS OF ODD ORDER 
If Lemma 34.3 is applied to then Lemmas 13.1 and 34.4 yield that 
. 
. 
. 
Lemmas 13.1 and 34.3 also imply that 
' 
. 
Lemma 34.6 and (34.5), (34.6), (34.7) and (34.8) now imply that 
. 
Since , this can be simplified to 
, 
. 
 THE SUBGROUPS AND 991 
By Lemma 84.1 and ; thus (84.9) implies that 
, 
. 
Let then 
(84.11) . 
Suppose first that . Then (84.10) implies hat 
. 
Hence by (5.2) 
. 
Therefore . 
(84.12) . 
Therefore 
(84.18) . 
Suppose that Then by (84.11) . Thus (84.12) implies 
that 
. 
Thus or which is not the case. Hence 
and (84.18) completes the proof of statement (ii) of the lemma. 
Suppose now that . Hence (84.10) yields that 
(84.14) . 
As this implies that 
. 
Hence by (5.8) 
 SOLVABILITY OF GROUPS OF ODD ORDER 
. 
Thus 
. 
This yields that 
Hence . 
Assume that . Then (84.11) implies that 
. 
SuPPose first then . Then (84.14) implies that 
. 
Hence 
. 
Therefore 
contrary to (84.16). Suppose that . Then by (84.16). 
Thus and since is a power of 8 and This 
contradicts (84.11). Hence . Thus by (84.16) Hence 
and since is a power of 8. Thus statement (i) of the 
lemma follows from (84.15) and statement (ii). If then 
and . By (84.15) 
. 
Hence by Lemma 84.1 Thus by (84.17). 
This completes the proof of statement (iii) of the lemma. 
Assume now that . Let not neces- 
sarily integral) Then (84.9) implies that 
 34. THE SUBGROUPS AND 
. 
- 
Therefore 
. 
or 
. 
Suppose that . Then (84.18) implies that 
. 
Therefore 52-19, or 
. 
Therefore, . Hence or . Since , Lemma 
34.1 now implies that . Thus if and 
if . Hence one of the following must occur: 
, , . 
or 
. . , . 
By (84.11) 
. or . 
If then by (84.18) 
. 
Hence , which is not the case. Therefore we have shown that 
either or and . If . then 
, and by (84.11) or . Thus in any case 
(84.19) or . 
This proves statement (iv) of the lemma. 
 994 SOLVABILITY OF GROUPS OF ODD ORDER 
If then (34.19) implies that and , hence 
and . Since , Lemma 34.1 implies that , thus 
. If then acts irreducibly on and centralizes 
. If is non abelian this implies that . Hence 
is an extra special -group contrary to the fact that . 
Thus is abelian. Hence . If this implies 
that is represented on and so acts irreducibly on 
contrary to and . Thus is elementary 
abelian. Statement (v) of the lemma is proved. 
Suppose that and ; then (34.18) implies that 
. 
Therefore, which is impossible. Hence if then . 
This proves statement (vi) of the lemma and completes the proof of 
Lemma 34.7. 
LEMMA 34.8. If then is a Frobenius group and 
. 
Proof. By Lemma 34.7 (i) . Thus if is not a 
Frobenius group then by Lemma 34.1 . Thus by Lemma 
34.7 (i) 
. 
Therefore which is not the case, since . 
LEMMA 34.9. If then and either 
or (mod ) and . 
Proof. By Lemma . Lemma 34.8 implies that 
and . Let . If (mod r 
then 
(mod ). 
Thus (mod ). If (mod ) then (mod ). 
Hence (mod ) as . Thus in any case if the 
result is false. Now Lemma 34.7 (ii) implies that 
. 
Hence 
 34. THE SUBGROUPS AND 
. 
Thus 18 contrary to the fact that . 
LEMMA 84.10. 
or if or 
or 15 or 88 if 
or if . 
Proof. Let be a. complement of in which contains 
. Every Sylow subgroup of is cyclic and every subgroup of prime 
order is normal in by 8.16 (ii) and Theorem 88.1. Thus 
. Hence or or . The result now 
follows from Lemma 84.7. 
By Theorem 88.1 is tamely imbedded in unless and 
. By Lemma 84.7 this can only happen if and . 
In that case let be the set of characters of which are induced 
by non principal irreducible characters of . In all other cases let 
. Define similarly. Then and are always 
defined. 
LEMMA 84.11. Suppose that is coherent and . If 
and 
then . If furthermore then 
. 
Proof. Let . Let with . Let 
. Then . Define 
. 
vanish on respectively. As and are T.I. 
sets in 
Furthermore by Lemma 18.8 
(84.21) , 
 996 SOLVABILITY OF GROUPS OF ODD ORDER 
where are real valued generalized characters of which are 
orthogonal to . The assumed inequalities and (34.20) imply that 
. Thus if then 
(mod 2) 
(mod 2). 
Since is rational valued on this implies that 
for . . Hence by Lemma 13.1 
(mod 2). 
Thus for . . Hence 
. 
Suppose now that then 
(34.22) . 
Let be the set of elements in which are conjugate to some 
element of with Since is coherent by assumption, 
(34.22) Lemmas 33.1 and 9.4 imply that for . 
. Furthermore Lemma 9.5 and (34.22) imply that 
(34.23) . 
By Lemma 9.5 
(34.24) . 
Let be the set of elements in which are not conjugate to 
any element of or . Now (34.22) implies that if 
then is rational and 
(mod 2). 
Thus for . Hence (34.23) implies that 
. 
 35. FURTHER RESULTS ABOUT AND 997 
Therefore 
. 
. and 
. 
thus the required inequality follows. 
LEMMA 84.12. If is cyclic then T.I. set in unless 
and . 
Proof. Since is a cyclic -subgroup in is a -subgroup 
-of . Suppose that is not a T.I. set in and let 
. Then . Since is a maximal 
ubgroup of this implies that . Thus 
and is a T.I. set in . 
35. Further Results About and 
The notation of Section 84 is used in this section. However we 
will destroy the symmetry of and by choosing the notation so that 
. 
The next three lemmas are restatements of Lemmas 84.7, 84.8, 
t84.9 and 84.10. 
LEMMA 85.1. If then 
. Either or (mod ) 
. Furthermore are Frobenius groups. 
or and or . 
LEMMA 85.2. Suppose that Then , 
and . is a Frobeniu s group with . Either or 
and . Furthermore and 
or . 
 998 SOLVABILITY OF GROUPS OF ODD ORDER 
LEMMA 85.8. Suppose that Then 
or if 
or 88 if . 
Furthermore one of the following possibilities occurs: 
(i) . is an elementary abelian -group 
With or . 
(ii) . is an elementary abelian -group 
with . 
LEMMA 85.4. Either or 
. 
Proof. By (5.12) 
. 
Therefore if then by Lemma 84.1 
. 
Suppose now that . Then by Lemma 85.1. By 
Lemma 85.2 and Thus if 
then by Lemma 84.2 
. 
Hence . Thus or . If then . 
As 1098 is a prime this implies that and the result follows 
from the first part of the lemma. If then . 
Thus and . Thus . The proof is complete. 
LEMMA 85.5. is coherent. 
Proof. Suppose that is not coherent. Then by Lemma 11.2 
is a power of some prime . As is cyclic (mod ). Thus 
 35. FURTHER RESULTS ABOUT AND 
. 
Let , then by Lemma 11.3. By Lemma 11.1 
. 
Suppose that . Then and (35.2) and (35.3) imply 
that If this yields that Then 
by (35.2) which is not the case as . Hence Thus Lemma 
11.4 implies that . Hence by (35.2) and so 
which is impossible. 
By Lemmas 35.1 and 35.2 we may assume now that 
. Thus (35.2) and (35.3) imply that 
. 
thus . Hence Lemma 11.4 implies that 
. 
This completes the proof in all cases. 
LEMMA 35.6. . If then or 
. 
Proof. If then . Hence by Lemma 
34.2. Assume now that . 
Assume first that . By Lemmas 35.1 and 35.2 
and . By Lemma 34.2 Thus 
and 
. 
Hence by Lemmas 35.5 and 34.11 . 
Thus 
. 
Therefore . 
Assume now that . Let . 
(mod 10) as (mod 10). If and , then 
. Thus 21.11. 
 SOLVABLITY OF GROUPS OF ODD ORDER 
and 
. 
Thus Lemmas 35.5 and 34.11 imply that Thus 
which is not the case. Therefore Ot 
, and we are done in this case. 
By Lemma 35.4 it may now be assumed that 
If then (mod ) since (mod ) Thus 
(35.4) if . 
Therefore 
. . 
Hence by Lemmas 35.5 and 34.11 Hence (35.4) and- 
Lemmas 35.1 and 35.2 imply that 
. 
Thus Hence by (5.4). Thus Hence 
and . By assumption , hence Thus 
. This completes the proof in all cases. 
LEMMA 35.7. 
or . 
Proof. This follows directly from Lemmas 35.1, 35.2, 35.3 and 35.6-. 
THEOREM 35.1. If is conjugate to then the conclusions: 
of Theorem 27.1 hold. 
Proof. By Lemma 35.6 if then and . 
Thus . Hence does not divide . Thus by Lemmas 35.1 
and 35.2, is cyclic. By Theorem 33.1 is a Frobeniua 
group with Frobenius kernel . Hence by Lemma 34.12 is a 
T.I. set in . Since and Lemma 35.T 
implies that is a cyclic group of order . Thus condition 
(iv) of Theorem 27.1 holds. Since is cyclic so is . Thus char 
. Hence if then which is not the case. Hence 
 36. THE PROOF OF THEOREM 27.1 
. By Lemma 35.6 . Thus and . 
Hence condition (iii) of Theorem 27.1 holds. If or , 
then or respectively. This implies that is 
elementary abelian of order and is elementary abelian of order 
. Hence condition (i) of Theorem 27.1 holds. 
Since is cyclic and and are Frobenius groups 
and . Since is cyclic every divisor of 
satisfies (mod ). Thus . Hence by Lemma 34.1 
. Let . Suppose that 
(mod ). Then (mod ) since 
(mod ). 
Thus if . then . Furthermore Lemma 35.1 implies that 
in this case . Thus by Lemma 35.3 (i) 
whichls impossible as Thus and Sup- 
pose that Then Hence 
since . As and (mod ) we see 
that . Thus if , Lemma 35.1 yields that 
. Since , Lemma 35.3 (i) implies that . 
This is impossible since . This verifies condition (ii) of 
Theorem 27.1 and completes the proof of the theorem. 
36. The Proof of Theorem 27.1 
In this section the study of the groups and is continued. All 
the lemmas in this section will be proved under the followi ng assumption. 
Hypothesis 36.1 
(i) . 
(ii) is not conjugate to . 
The following notation is used in addition to that introduced in 
Section 34. 
and 
. 
If is defined since is odd. Let . 
Then 
 1w2 SOLVABILITY OF GROUPS OF ODD ORDER 
if 
if 
(36.3) for . 
(36.4) for , 
where are in respectively, are orthogonal 
to respectively. are li near combinations of the 
generalized characters and are orthogonal to each . 
Then and are real valued generalized characters each of 
which is orthogonal to . Thus 
(36.5) (mod 2). 
(36.6) (mod 2). 
(36.7) (mod 2). 
It is a simple consequence of Lemma 13.1 that 
(36.8) (mod 2). 
(36.9) (mM 2). 
By Hypothesis 36.1 (ii) is orthogonal to . Thus 
(36.10) (mod 2). 
Since is an isometry (36.1), (36.2), (36.3) and (36.4) yield that 
(36.11) 
. 
LEMMA 36.1. is coherent. 
Proof. If is of type IV then by Lemmas 35.2 and 35.3 
. or 7 so by Lemma 11.1 the result follows from Theorem 29.1. If 
is of type III then is abelian and the result follows from 
Lemma 11.2. Suppose that is not coherent. Then and 
by Lemma 11.2 is an -group for some prime . Furthermore 
jg of type II. Let then by Lemmas 11.1, 11.3 and 
 36. THE PROOF OF THEOREM 27.1 1003 
11.4 , 
(36.15) with , 
(36.16) or and . 
Suppose first that is not cyclic. Then by Lemma 35.1 . 
If . then by Lemma 35.3 is cyclic and 
. 
Thus by Lemma 34.1 is cyclic. Hence is generated by two 
elements. If then Lemma 34.1 implies that is generated by 
two elements. Thus . As is of type II is a T.I. set in 
. Consequently there exists an element of order such that 
. Thus is cyclic. Hence (mod ). This contradicts 
(36.15) and (36.16). 
Suppose now that is cyclic. Thus (mod ). By (36.16) 
is irreducibly represented on . Thus acts as a 
group of scalar matrices on . Hence by Lemma 6.4 has 
prime exponent. Since is a cyclic subgroup of this implies that 
(36.17) . 
If then Lemmas 35.1, 35.7 and (36.15) and (36.16) imply that 
. 
Hence and so 
. 
Thus which is not the case. 
Suppose that : If then (36.16) and Lemmas 35.3 and 
35.7 ].mply that 
. 
Hence 
. 
Thus . Hence . If then (36.16) and Lemmas 
35.3 and 35.7 imply that 
. 
Hence 
 SOLVABILITY OF GROUPS OF ODD ORDER 
. 
Therefore in this case also. Thus or 11. By Lemma 
84.1 and (86.17) either or If 
then and since . However in this case 
which is impossible by Lemma 85.1. Thus . Hence by 
(86.17) if if and if then 
or . If then (86.16) and (86.17) imply that 
which is not the case. If then (86.16) and (86.17) imply that 
19' 86.49 1800 . 
Thus 19' which is not the case. If and then 
(86.16) and (86.17) imply that 
which is not the case. 
Assume now that and . Then (86.15) and 
(86.16) imply that 
(86.18) (mod ll) . 
Since 
we must have . However 
(mod 11) 
contrary to (86.18). The proof is complete. 
LEMMA 86.2. . 
Proof. Suppose that . Then by (86.10) either 
or . If , then (mod p). Hence 
by Lemmas 85.1, 85.5 and 86.1, 
or . 
Therefore by (5.11) . Hence 
 36. THE PROOF OF THEOREM 27.1 
which is impossible for . 
LEMMA 86.8. or . 
Proof. If then and by Lemma 85.8. 
Since Lemma 84.1 implies that . 
Thus . By Lemma 84.2 . 
Suppose first that Then by (86.8) . 
Thus . Hence by (86.5). Since is coherent 
(86.18) implies that 
, 
which is not the case. 
Suppose now that . Then by Lemma 85.7 
Let Then (mod ) 
since 
(mod ). 
As this implies that . Hence by Lemma 85.2 
and (86.12) 
(86.19) . 
Since is coherent this implies that . Thus by 
. 
Since is coherent (86.18) and (86.19) imply that . Thus 
by (86.5) 
(86.21) (mod 2). 
Since is coherent (86.11), (86.20) and (86.21) imply that 
. 
Hence by Lemma 85.2 
. 
Therefore . Hence by (5.5). Thus and 
. If the -subgroup of has order , then (mod 5). 
Thus . Therefore 
 1006 SOLVABILITY OF GROUPS OF ODD ORDER 
. 
Thus the coherence of implies that . Hence (36.7) yields 
that (mod 2). Therefore (36.8), (36.11) and (36.21) imply 
that 
contrary to (36.20). Thus and consequently 
or . 
LEMMA 36.4. . 
Proof. If the result is false then by Lemma 
35.7. Thus (36.8) implies that . Therefore by (36.5) 
and (36.10) and Since (36.13) implies 
that and 
. , 
(36.22) . 
where ranges over . Thus by (36.6) is odd. Hence by 
Lemma 36.3 and (36.12) 
. 
where is odd and is orthogonal to all . Therefore by (36.22) 
. 
Since this implies that . Hence . Thus 
by Lemma 35.3 (i) , hence . Hence 
or . In either of these cases by Lemma 34.1 
since . Thus or which is not the case. 
LEMMA 36.5. . 
Proof. If then as is a T.I. set in . This 
contradicts Lemma 36.4. 
LEMMA 36.6. is cyclic. 
Proof. By Lemma 34.1 if is not cyclic then , where 
each is cyclic and . Let for 
. If then Lemma 35.3 (i) implies that 
 36. THE PROOF OF THEOREM 27.1 1N7 
which is not the case. Thus . If then (mod 4) 
and so is even. If then (mM 8) and 
so which is not the case. Thus and . 
Therefore . Thus . Furthermore (mod ). 
Since (mod ) by Lemma 86.4 we have that and 
(mod ). By Lemma 84.2 . Hence Lemma 86.8 and 
(86.10), (86.11) and (86.12) imply that 
or . 
The first possibility implies that . Thus . 
Hence by (5.4). The second possibility in (86.28) yields that 
. 
Therefore 
. 
As this implies that 
(86.24) . 
Let for some integer . Then since (86.24) yields 
that . Furthermore 
(86.25) if if . 
As and or . If then by 
(86.25) which is impossible as (mod 8). If , then 
or 87. As (mod 6) this implies that . Then 
and so 87 (mM 11) by Lemma 86.4 which is not the case. 
LEMMA 86.7. or or 
. 
Proof. If then by Lemmas 84.1 and 86.6 . 
Thus by Lemma 85.8 (i) . Hence 
lip which implies that . Therefore by Lemma 
84.1 an integer. By Lemma 85.8 (i) . If 
is a prime such that (mod ) then either or 
 10o SOLVABILITY OF GROUPS OF ODD ORDER 
(mod 8). Hence or 9. If then 
(mM 9). Hence (mM 8). Thus or 7 (mM 9). In none 
of these cases is (mod 9). Hence or 7. 
LEMMA 86.8. . 
Proof. Let . Assume that (mod ) by 
Lemma 86.4. If , then (mM ). thus 
(mod ) and so . If , then 
(mod ). Furthermore (mod 6) since (mM 6) and 
(mod 6) since (mod 8). Thus if then 
(mod 6). Hence . If then 
(mod ). If then by Lemma 86.6 the -subgroup of 
is generated by two elements. Hence (mod ) by 
Lemma 86.4. However and . Thus . Let 
. Then (mod 6). Hence (mod 6). 
Thus (mod 6), hence (mod 6). Therefore 
. Thus in any case 
(86.26) , . 
Therefore . Hence by (86.18) and the coherence 
of 
. 
Assume first that , then by (86.12) and the coherence 
of 
. 
Suppose now that . Then by (86.10) . 
Hence the coherence of and (86.11) imply that 
(86.29) . 
By (86.27) and (86.5) (mod 2). If also were odd 
then by (86.8) for Thus 
by (86.11) contrary to what has been proved. Therefore 
(mod 2). Hence by (86.7) Thus by (86.14) 
and (86.29) 
. 
Now (86.28) implies that in any case 
 36. THE PROOF OF THEOREM 27.1 
. 
For any prime let be the -subgroup of . 
Suppose first that . then . Hence (36.30) 
.mplies that 
. 
Therefore which is impossible for . 
Suppose now that then by (36.26). 
Hence (36.30) implies that 
. 
Thus or . Since (mod 3) this yields that . 
or . 
If then . If then as 
(mod 7). Thus (36.30) implies that 19' . If then 
. Let , then as (mod 13). Hence 
(36.30) implies that 61' 27.13' . If then . Let 
, then as (mod 19). Hence (36.30) implies 
that 127' 27.19' . 
Assume finally that then . Thus 
(36.30) implies that 
. 
Therefore . so . Since (mod 7). 
(mod 7) or (mM 7). Thus or . 
If then then as 
(modll). Hence (36.30) implies that 19' 27.11' 287.11 . If 
then As (mod 23). . Hence (36.30) 
implies that 79' 27.23' . 
Therefore in all cases. Hence (mod ) by Lemmas 
36.4 and 36.5. Since (mod ) and (mod ). 
Hence by Lemma 36.7 . 
The proof of Theorem 27.1 under Hypothesis 36.1 is now im- 
mediate. 
Let and have the same meaning as in the earlier part 
of this section. By Lemma 35.2 . By Lemma 36.5 . 
The other properties of Condition (i) follow from the structure of 
and and Theorem 14.1. Thus Condition (i) is verified. By Lemma 
35.6 . Hence . By Lemma 36.3 . 
hence by Lemma 36.5. The other properties of Condi- 
 1010 SOLVABILITY OF GROUPS OF ODD ORDER 
tion (iii) follow from the structure of and . Thus Condition (iii) 
is verified. Lemmas 36.6 and 36.8 imply that is cyclic. By 
Lemmas 34.12 and 36.4 is a T.I. set in . Hence Lemma 
36.4 completes the verification of Condition (iv). 
Lemmas 34.1, 36.3, 36.5 and 36.8 imply that is a Frobenius 
group. Lemma 36.8 implies that . Lemmas 36.4, 
36.6 and 36.8 imply that if then (mod ). Thus 
. The other statements in Condition (ii) follow from 
the structure of and . 
By Theorem 35.1 this completes the proof of Theorem 27.1 in 
a11 cases. 
 CHAPTER VI 
37. Statement of the Result Proved in Chapter VI 
The purpose of this chapter is to prove the following result. 
THEOREM 37.1. There are no groups which satisfy conditions 
(i)-(iv) of Theorem 27.1. 
Once it is proved, Theorem 37.1 together with Theorem 27.1 will 
serve to complete the proof of the main theorem of this paper. In 
this chapter there is no reference to anything in Chapters II-V other 
than the statement of Theorem 27.1. The following notation is used 
throughout this chapter. 
is a fixed group which satisfies conditions (i)-(iv) of Theorem 
27.1. 
and . 
. . Thus 
so that . 
and are fixed elements of and respectively. 
For any integer . is the ring of integers mod . If 
is a prime power then is the field of elements. 
acts as a linear transformation on . Let be the minimal 
polynomial of on . Then is an irreducible polynomial of 
degree over . Let be a fixed root of in . Then 
is a primitive uth root of unity in and are all 
the characteristic roots of on . 
38. The Sets and 
LEMMA 38.1. There exists an element such that nor- 
malizes 
Proof. normalizes and is contained in a cyclic sub- 
group of of order . Hence some element of order in 
normalizes . Since every subgroup of order in 
is of the form for some . Hence it is possible 
to choose such that normalizes . Since , 
 1012 SOLVABILITY OF' GROUPS OF ODD ORDER 
does not normalize hence and normalizes . 
From now on let 
(38.1) 
where satisfies Lemma 38.1. Notice that normalizes , since 
normalizes and centralizes . Define by 
(38.2) 
LEMMA 38.2. If then 
unless and . 
Proof. . Hence acts trivially on 
. However if then is a Frobenius 
group with Frobenius kernel . Thus as required. 
LEMMA 38.3. Every element of has a unique representation 
in the form , where and is a polynomial of 
degree at most over . 
Proof. There are up' ordered pairs with and 
of degree at most over . Thus it is sufficient to show 
the uniqueness of in such a representation. 
If . Then reading mM yields that . 
Since is irreducible we get that (mod ). Thus 
as required. 
LEMMA 38.4. Every element of has a unique representa- 
tion in the form , where and . 
Proof. If and 
then reading mod we get that . Hence 
Since . . As we have that , and so 
. . The representation is unique. There are 
ordered triples with and . . Each triple 
gives rise to an element of and . The 
result now follows. 
LEMMA 38.5. Let Then 
 38. THE SETS AND 1013 
if and only if 
(i) 
(ii) . 
Proof. Let . Then 
. 
Thus by Lemma 38.3 if and only if 
(mod ). 
The first equation allows us to rewrite the second as 
(mod ). 
Thus the lemma is proved. 
DEFINITION 38.1. The set is defined to consist of all ordered 
triples such that 
(i) for . 
(ii) . 
(iii) . 
DEFINITION 38.2. is the set of all elements such that 
for suitable . 
LEMMA 38.6. . 
Proof. If then by Lemma 38.4 and are de- 
termined by . 
LEMMA 38.7. if and only if 
(i) for 
(ii) 
(iii) . 
Proof. By Lemma 38.5, 
if and only if and This implies 
the result. 
LEMMA 38.8. If then (. . ). 
 1014 SOLVABILITY OF GROUPS OF ODD ORDER 
Proof. If then by Lemma 88.7 . 
As this yields that 
. 
As the result follows from Lemma 88.7. 
LEMMA 88.9. For let be the conjugate class of 
which contains and let be the sum of the elements in 
in the group ring of over the integers. Let 
. 
If then . 
Proof. Let r be all the irreducible characters of 
and let r be all the other irreducible characters of . It 
is a well known consequence of the orthogonality relations ([4] r 
that 
. 
Since is cyclic, for all By 8.16 
for all . Thus 
(88.8) . 
By the orthogonality relations 
for . 
Therefore 
(88.4) . 
By (88.8) and (88.4) 
. 
Thus 
(88.5) . 
Since yields that 
. 
 38. THE SETS AND 1015 
As and is a prime we have , and the lemma follows. 
LEMMA 38.10- . 
Proof. Assume first that . Consider the set of polynomials 
of the form with . There are 
of these and none of them has 0 as a root. Thus if were re- 
ducible for every value of there would exist such that 
and have a common root . Then 
. 
Since this yields that , hence How- 
aver Thus there exists some polynomial which 
is irreducible over . Let be a root of in . Then 
. 
Therefore for some . . and for some 
. Furthermore Thus . 
Since . Hence by Lemmas 38.6 and 38.7 
. 
Assume now that . Then Lemma 38.9 implies the existence 
of . with or such that 
. 
Therefore 
. 
Let . . Then . If 
then (38.6) becomes ; as is a Frobenius group this 
implies contrary to the choice of and If then (38.6) 
implies that , hence . If then (38.6) 
yields that or commutes with Thus . 
hence also . Therefore are all non zero and by Definition 
38.1 and Lemma 38.6 . 
The following result about finite fields is of importance for the 
proof of Theorem 37.1. 
LEMMA 38.11. For define and for 
let If then for some . 
Proof. Assume that the result is false and for all . 
We will first prove by induction that 
 1016 SOLVABILITY OF GROUPS OF ODD ORDER 
(88.7) for r 
If follows from the definition of . Assume now 
holds for . Then 
. 
This establishes (88.7). 
Now (88.7) implies that for , 
. 
Therefore 
. 
Thus 
(88.8) for . 
Define by 
(88.9) . 
Thus has coefficients in and (88.8) yields that 
(88.10) 
for . . 
Let for . then . Hence (88.10) yields 
for . 
Therefore 
(88.11) for . 
 39. THE PROOF OF THEOREM 37.1 1017 
is a polynomial of degree at most . By (38.11) 
has at least roots. As we must 
have that . By (38.9) is a root of , hence 
contrary to the choice of . The proof is complete. 
39. The Proof of Theorem 37.1 
LEMMA 39.1. There exist functions and such that 
(i) and map into . 
(ii) maps into , 
(iii) . 
Furthermore for (iii) determines 
and uniquely anti . are all non- 
ze ro. 
Proof. By Lemma 38.4 the functions exist and are uniquely de- 
fined by 
provided that does not lie in . It is easily seen that if 
and does not lie in . 
Suppose that . Then . Then 
and . Therefore either or . 
Suppose that . Then . Thus 
-f-- and Hence or . 
Suppose that . Then . Hence 
, then . Thus or . This com- 
pletes the proof of the lemma. 
Throughout the rest of this section will denote the func- 
tions defined in Lemma 39.1. For as in Lemma 38.1, define 
. 
LEMMA 39-2. 
(i) 
(ii) 
(iii) , 
for . 
Proof. Since and is abelian, (i) is immediate. 
(iii) is a direct consequence of (i). By definition . 
Thus which implies (ii). 
LEMMA 39.3. For . " UP' . 
 1018 SOLVABILITY OF GROUPS OF ODD ORDER 
Proof. By (38.2) . By (38.1) . Hence 
. 
Conjugating both sides by . we get that 
. 
If both sides are raised to the power, the lemma follows. 
LEMMA 39.4. 
. 
Proof. Substitute (38.1) into (iii) of Lemma 39.1 to get 
. 
Conjugate by to get 
. 
Now use the results of Lemma 39.2 to derive that 
which implies the lemma. 
LEMMA 39.5. If then 
. 
Proof. In the definition of conjugate (iii) by . Then 
, 
or 
. 
Hence Lemma 39.3 yields that 
. 
Since is abelian, this implies that 
. 
Conjugating by implies the result by (38.1) and the fact that 
is abelian. 
LEMMA 39.6. For define 
 39. THE PROOF OF THEOREM 37.1 1019 
(. ) 
(. . ) 
(. ) 
( . ) 
. 
Then 
. 
Proof. Use Lemmas 89.4 and 89.5 to obtain 
. 
. 
Multiply on the left by and on the right by 
to get 
A 
where 
, 
or equivalently 
. 
The lemma follows. 
LBMMA 89.7. Let . Use the notation of Lemma 
39.6. If then there exist elements sueh that 
(i) 
(ii) 
(iii) Y. 
Proof. Conjugate (89.1) by Q. Since . this yields 
that 
 1020 SOLVABILITY OF GROUPS OF ODD ORDER 
. 
Taking inverses we get 
. 
Multiplying this by (39.1) on the left yields 
. 
Conjugating by yields 
. 
Use Lemma 39.2 (iii) and (38.1) to get 
. 
Conjugate this by to obtain . 
. 
Multiply on the left by and on the right by to obtain 
, 
(39.2) 
Y. 
Suppose that . Then (39.2) implies that 
By Hypothesis , hence by Lemma 38.2, . By Lemma 
39.1 . Thus the above equality cannot hold in the Frobenius 
group . Hence . This proves statement (i) of the 
lemma. 
Let . By (39.2) is a conjugate of 
which lies in . All conjugates of which lie in are of.' 
the form 
with . Hence 
(39.3) 
for some . Thus . Since , we get that 
. By (39.2) , thus . 
Hence 
. 
 39. THE PROOF OF THEOREM 37.1 1021 
However . Since N(II) , this yields that 
. Hence . Thus 
(39.4) 
for some . Now (39.2) and (39.4) show that 
. 
Since , we have N(II), thus 
. Therefore 
(39.5) 
for some . Consequently 
. 
If this is compared with (39.3) we see that 
(39.6) . 
Using (39.4) and (39.6) in (39.5) leads to 
(39.7) Y. 
Comparing (39.2) and (39.7), we get 
Y. 
Conjugating by gives 
(39.8) . 
If we substitute (39.7) into (39.2) we get 
. 
Multiply on the left by and on the right by 
to get 
. 
Since the right hand side is the left hand side conjugated by , we 
see that centralizes the left hand side. Hence 
(39.9) 
for some . Reading (39.9) mod yields that 
which proves (ii) of the lemma. Substituting (ii) of Lemma 39.2 
 1022 SOLVABILITY OF GROUPS OF ODD ORDER 
into (39.8) we get that 
(39.10) . 
Substituting (39.10) into (39.1) leads to 
. 
Multiply on the left by and on the right by . Then using 
Lemma 39.2 (ii) and (iii) this becomes 
. 
Use to get 
. 
Conjugate by and multiply on the left by to get 
(39.11) . 
Conjugate by and take inverses, then 
Y. 
Multiply by (39.11) on the right to get 
. 
Conjugate by to get 
. 
Using (39.2) and (39.3), this yields 
. 
Now by the second equation in (39.12) 
. 
Thus the first equation in (39.12) impliCS that 
. 
By (39.3) and (39.4), . Hence 
(39.13) 
for some . We wish to show that . To do this con- 
jugate (39.13) by to get 
 39. THE PROOF OF THEOREM 37.1 
by (39.7). Multiply (39.13) by the inverse of (39.14) on the right to 
get 
(39.15) . 
By Lemma 38.2 and have the same order. Since the left 
hand side of (39.15) is in , this implies that the order of divides 
thus . 
Multiply (39.13) on the left by and on the right by 
to get 
(39.16) Y. 
By (39.7) the right hand side is in , while the left hand side is 
in . Since this yields that 
(39.17) 
for some . Conjugate by to get 
. 
Comparing this with (39.9) yields that 
. 
so that 
. 
Using (39.16) and (39.17) this yields 
or 
. 
Hence by (39.7) 
Y. 
This immediately implies (iii) of the lemma and thus completes the 
proof. 
LEMMA 39.8. Let let have the same meaning 
as in Lemma 39.6. Then . 
Proof. Suppose that , so that Lemma 39.7 may be applied. 
Let 
 1024 SOLVABILITY OF GROUPS OF ODD ORDER 
( . ) 
. 
By Lemma 38.5 (i) 
( . ) 
( . ). 
Hence in the notation of Lemma 39.6 
. 
. 
Since this yields that 
. 
Thus 
or 
. 
By Lemma 39.7 (ii) this implies that . If , 
then by Lemma 38.2, has an inverse in . Thus 
contrary to Lemma 39.7 (i). Therefore . Now Lemma 39.7 
(iii) becomes 
(39.18) . 
Reading (39.18) mod implies that . Thus (39.18) yields that 
and commute. Since by Lemma 39.1, this implies 
that 
. 
Thus which is not the case. Therefore as 
required. 
LEMMA 39.9 Let . let and have the same 
meaning as in Lemma 39.6. Then . 
Proof. Since by Lemma 39.8, (39.1) becomes 
 39. THE PROOF OF THEOREM 37.1 1025 
. 
Conjugating by and using (38.2) we get that 
. 
Now (39.19) and (39.20) imply that 
. 
lTherefore 
. 
Suppose that . Then by Lemma 38.2 . As 
is a T.I. set in . (39.21) now implies that . 
As this implies that 
. 
commutes with Y. Hence . This is contrary to 
Lemma 39.1. Thus . 
Now (39.21) implies that . Therefore by Lemma 
.38.2 . 
LEMMA 39.10. Let and have the same meaning 
as in Lemma 39.6. Then . 
Proof. In view of Lemmas 39.8 and 39.9 equation (39.1) becomes 
. 
Reading (39.22):mod implies that 
.or using the definition of 
. 
Hence and (39.22) becomes 
. 
acts as a linear transformation on . It is convenient to use 
the exponential notation. Thus , so that . 
l39.24) can be rewritten as 
. 
In exponential notation this becomes 
 1026 SOLVABILITY OF GROUPS OF ODD ORDER 
(39.25) . 
Define 
(39.26) 
. 
Since is a Frobenius group with Frobenius kernel . is 
an invertible linear transformation on . By (39.25) A annihilates 
Y. Hence also A(1 ) annihilates Y. By (39.26) 
. 
Therefore 
. 
Thus 
(39.27) . 
By Lemma 39.3 
. 
By (39.27) this yields that 
(39.28) . 
Lemma 39.2 also implies that 
. 
Raising this to the th power we get that 
(39.29) . 
Now (39.28) and (39.29) yield that 
(39.30) . 
Another application of Lemma 39.3 gives 
(39.31) . 
Thus (39.30) and (39.31) imply that 
. 
Since . . Therefore 
 39. THE PROOF OF THEOREM 37.1 1027 
. 
As is a T.I. set in (39.32) now implies that 
. 
Therefore commutes with and so . Now (39.27) yields 
that . or 
. 
Consequently . Hence . Now (39.23) implies 
that as required. 
LEMMA 39.11. Let have the same meaning as in Definition 
38.2. If then . 
Proof. Let and suppose that . By 
Lemma 38.8 (. ). Let (. . ) play the role 
of . By Lemma 39.10 (. . ). Thus Lemmas 
38.5 and 39.1 imply that 
(39.33) (. ) 
. 
Let and ( . ). By 
Lemma 39.1 for . By (39.33) . Now 
it follows from (39.34) and Lemma 38.7 that . Thus 
. 
Since was an arbitrary element of we get that for any 
integer . . Thus in particular, . . Hence 
by (38.2). as was to be shown. 
It is now very easy to complete the proof of Theorem 37,1. 
Define the set by 
. 
Since . , Lemma 38.10 yields that is not empty. The 
definition of and Lemma 38.7 yield that and if 
and only if . Lemma 39.11 implies that if and only 
if Therefore if then Since - 
. we have for . Thus if has the 
same meaning as in Lemma 38.11 then there exists 
such that for all values of . This contradicts Lemma 
38.11, and completes the proof of the main theorem of this paper. 
 BIBLIOGRAPHY 
1. N. Blackburn, On a special class of -groups, Acta Math., 100 (1958), 45-92. 
2. -. Generalizations of certain elementary theorems on -groups, Proc. London 
Math. Soc. (3). 11 (1961). 1-22. 
3. R. Brauer, On the connection between the wdinary and the modular characters of 
finite groups, Ann. of Math. (2), 42 (1941), 926-935. 
4. W. Burnside, Theory of Groups of Finite Order, Cambridge, 1911. 
5. -, On groups of order . Proc. London Math. SOC., (2), 2 (1904), . 
6. L. E. Dickson, Linear Groups, New York, 1958. 
7. W. Feit, On the stmcture of Frobenius groups, Can. J. Math., 9 (1958), 587-596. 
8. -, On a class of doubly transitive groups, Ill. J. Math., 4 (1960), 
170-186. 
9. -, Exceptwnal Characters, Proceedings of the Symposium in pure mathematics, 
A.M.S., 6 (1962), 67-70. 
10. W. Feit, M. Hall, Jr. and J. G. Thompson, Finite groups in the centralizer 
of any non-ientity element is nilpotent, Math. Zeitschr., 74 (1960), 1-17. 
11. W. Feit and J. G. Thompson, A solvability for finite groups and some 
consequences, Proc. Nat. Acad. , 968-70. 
12. M. Hall, Jr., The Theory of Groups, New York, 1959. 
13. . Hall, A note on soluble groups, J. London Math. SOC., a (1928), 98-105. 
14. -, A contribution to the theory of groups of prime power order, Proc. London 
Math. Soc., (2) . 
15. -, A property of soluble groups, . London Math. SOC., 12 
(1937), 198-200. 
16. -. On the Sybw systems of a soluble group, Proc. London Math. SOC., (2) 43 
(1937), 316-323. 
17. -. On the system normalizers of a soluble group, Proc. London Math. SOC., 
(2), 43 (1937), 507-528. 
18. -. Theorems like Sylows, Proc. London Math. SOC., (3), 6 (1956), 286-304. 
19. -, Some conditwns for a group to be nilpotent, Ill. J. Math., 2 
(1958), 787-801. 
20. -, Lecture Notes. 
21. . Hall and G. Higman, The -length of a -soluble group, and rdution theorems 
problem, Proc. London Math. SOC., (3), 7 (1956), 1-42. 
22. B. Huppert, Subnormale untergruppen unxt Sylowgruppen, Acta Szeged., 22 (1961). 
46-61. 
23. -. mit modularer Sylow-Gruppe, Math. Zeitschr., 75 (1961), 140-153. 
24. M. Suzuki, On finite groups with cyclic Sylow subgroups all odd primes. Amer. 
. Math., 77 (1955), 657-691. 
25. A type of simple groups of finik order, Proc. Nat. Acad. Sci.. 
(1960), 868-870. 
26. G. Thompson, Finite groups with -free of prime order, 
Proc. Nat. Acad. Sci.. (1959), 578-581. 
27. -, Normal -complements for finite groups, Math. Zeitschr., 72 (1960), 332-354. 
28. H. Zassenhaus, The Theory of Groups, Second Edition, New York, 1958. 
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