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\title{\large\bf Crossed Product Conditions for Division Algebras of Prime Power Degree}	
\author{D. Kiani, M. Mahdavi-Hezavehi\\	
 kiani@mehr.sharif.edu\\	
 mahdavih@sharif.edu \\	
 Department of Mathematical Sciences\\ Sharif University	
of Technology\\P. O. Box 11365-9415\\Tehran, Iran}	
\date{}	
\begin{document}	
\maketitle	
\begin{abstract}	
Let $ D $ be an $ F $-central division algebra of degree $ p^r $,	
$ p $ a prime. A set of criteria is given for $ D $ to be a	
crossed product in terms of irreducible soluble or	
abelian-by-finite subgroups of the multiplicative group $ D^* $	
of $ D $. Using the Amitsur's classification of finite subgroups	
of $ D^* $ and the Tits Alternative, it is shown that $ D $ is a	
crossed product if and only if $ D^* $ contains an irreducible	
soluble subgroup. Further criteria are also presented in terms of	
irreducible abelian-by-finite subgroups and irreducible subgroups	
satisfying a group identity.  Using the above results, it is	
shown that if $ D^* $ contains an irreducible finite subgroup,	
then $ D $ is a crossed product.	
	
\end{abstract}	
	
\section{Introduction}	
	
Let $D$ be an $ F $-central division algebra of degree $n$. The	
algebra $D$ is called a \emph{crossed product} if it contains a	
maximal subfield $ K $ such that $ K/F $ is Galois. We shall say	
that $ D $ is a \emph{nilpotent} crossed product if $ Gal(K/F) $	
is nilpotent. A subgroup $ G $ of $ D^* $ is said to be	
\emph{irreducible} if $ F[G] = D $. When $ n = p $, a prime, it	
is shown in \cite{tig} that $ D $ is cyclic if and only if $ D^*	
$ contains a nonabelian soluble subgroup.  A criterion is also	
given in \cite{nil} for $ D $ to be a supersoluble (nilpotent)	
crossed product division algebra in terms of subgroups $D^*$.	
More precisely, it is shown that $D$ is a supersoluble (nilpotent)	
crossed product if and only if $D^*$ contains an	
abelian-by-supersoluble (abelian-by-nilpotent) irreducible	
subgroup. The aim of this paper is to generalize some of these	
results to a division algebra of a prime power degree $ p^r $. In	
fact, we present a set of criteria for $ D $ to be a crossed	
product in terms of irreducible soluble or abelian-by-finite	
subgroups of $ D^* $. To be more precise, it is shown that $ D $	
is a nilpotent crossed product if and only if $ D^* $ contains an	
irreducible soluble subgroup. In addition, it is shown that,	
except for the case $ Char F = 0 $, $ p = 2 $ and $ r>1 $, $ D $	
is a crossed product if and only if either of the following	
conditions holds: (i) $ D^* $ contains an irreducible	
abelian-by-finite subgroup, or (ii) $ D^* $ contains an	
irreducible subgroup satisfying a group identity. Furthermore, it	
is proved that these conclusions also hold for the above excluded	
case provided that $ D^* $ contains no finite subgroup isomorphic	
to $ SL_2(Z_5) $. Finally, given a non-commutative $F$-central	
division algebra $ D $ of index $p^r$, $ p $ a prime, using the	
above mentioned results, it is shown that if $ D^* $ contains an	
irreducible finite subgroup $ G $, then $D$ is a crossed product.	
We note that soluble subgroups of the multiplicative group of a	
division ring were first studied by Suprunenko in \cite{su}.	
	
\section {Notations and conventions}	
	
 We now recall some	
notations and conventions that are used throughout. Let $ D $ be	
an $ F $-central division algebra and $ G $ be a subgroup of $	
D^* $. The $ F $-linear hull of $ G $, i.e., the $ F $-algebra	
generated by elements of $ G $ over $ F $, is denoted by $ F[G]	
$. $ G $ is called {\it irreducible} if $ D = F[G] $. For any	
group $ G $ we denote its center by $ Z(G) $. Given a subgroup $	
H $ of $ G $, $ N_G (H) $ means the {\it normalizer} of $ H $ in	
$ G $, and $\langle H,K \rangle $ the group generated by $ H $	
and $ K $, where $ K $ is a subgroup of $ G $. We shall say that	
$ H $ is {\it abelian-by-finite} if there is an abelian normal	
subgroup $ K $ of $ H $ such that $ H/K $ is finite. Let $ S $ be	
a subset of $ D $, then the {\it centralizer} of $ S $ in $ D $	
is denoted by $ C_{D} (S) $. For notations and results, used in	
the text, on central simple algebras see \cite{ro}.	
	
	
\section {Irreducible soluble subgroups}	
	
Let $ D $ be an $ F $-central division algebra of degree $ p^r $,	
$ p $ a prime. This section investigates the structure of $ D $	
under the condition that $ D^* $ has an irreducible soluble	
subgroup. To be more precise, it is shown that $D$ is a crossed	
product if and only if $D^*$ contains an irreducible soluble	
subgroup.  We begin our study with the following:	
	
\begin{lem}\label{if} {\it Let $D$ be a finite dimensional $ F $-central division	
algebra. If $ D $ is a soluble crossed product, then $D^*$	
contains an irreducible soluble subgroup.}	
 \end{lem}	
	
{\sc Proof.} Let $K$ be a maximal subfield of $D$ such that $K/F$	
is soluble Galois. By Skolem-Noether Theorem, for any $\sigma\in	
Gal(K/F)$ there exists an element $x\in N = N_{D^*}(K^*)$ such	
that $\sigma(k)=xkx^{-1}$, for all $k\in K$. Hence	
$N_{D^*}(K^*)/C_{D^*}(K^*)\simeq Gal(K/F)$. Since $K$ is a	
maximal subfield of $D$, we have $C_{D^*}(K^*)=K^*$. Therefore,	
$N_{D^*}(K^*)$ is a soluble subgroup of $ D^* $. To complete the	
proof of the lemma, it is enough to show that $ N $ is	
irreducible, i.e., $F[N]=D$. Put $D_1=F[N]$. We have	
$C_D(D_1)\subseteq C_D(K)=K$, and hence $C_D(D_1)$ is an	
intermediate field of the Galois extension $K/F$. By the fact	
that every element of $Gal(K/F)$ is the restriction of an inner	
automorphism of $N$ we conclude that $C_D(D_1)\subseteq	
Fix(Gal(K/F))$. Therefore $C_D(D_1)=F$. Now, by Centralizer	
Theorem, we obtain $D=C_D(F)=C_D(C_D(D_1))=D_1$, which completes	
the proof.\hfill$\Box$ \vspace{0.2 cm}	
	
The following lemma is used in many proofs below, its idea is due	
to Suprunenko \cite{su}.	
	
\vspace{0.2 cm}	
	
\begin{lem}\label{index} {\it Let $D$ be a finite dimensional $ F $-central division	
algebra. Suppose that $G$ is a subgroup of $D^*$ such that	
$F^*\subseteq Z(G)$. If $K=Z(G)\cup \{0\}$ is a subfield of $D$	
and $G/K^*$ is abelian, then we have $[K[G]:K]=|G/K^*|$ and hence	
$G/K^*$ is a finite group.}	
 \end{lem}	
	
{\sc Proof.} Let $g_1,\ldots ,g_t$ be a set of linearly	
independent elements of $G$ over $ K $. It is clearly seen that	
$g_1K^*,\ldots ,g_tK^*$ are distinct elements of $G/K^*$. On the	
other hand, if $g_1K^*,\ldots ,g_tK^*$ are distinct elements of	
$G/K^*$, we shall show that $g_1,\ldots ,g_t$ are linearly	
independent over $K$. To see this, since $ G/K^* $ is abelian, for	
every $g \in G$ we have $gg_ig^{-1}=k_ig_i$ with $1 \leq i \leq t	
$, where $k_i\in K^*$. We claim that for each pair $i\neq j$ we	
can find an element $g$ in $G$ such that $k_i\neq k_j$. For	
suppose that for each $g$ in $G$ we have	
$gg_ig^{-1}g_i^{-1}=k_i=gg_jg^{-1}g_j^{-1}=k_j$. Therefore, we	
conclude that $[g,g_j^{-1}g_i]=1$, and hence $g_j^{-1}g_i \in	
K^*$. This contradicts the choice of $g_i' s$, and so the claim is	
established. Now, suppose that $g_1,\ldots ,g_t$ are linearly	
dependent over $K$ and consider a relation	
$$\lambda_1 g_1+\ldots +\lambda_tg_t= 0. ~~~~~~~(\ast)$$	
Of all relations of the form $ (\ast) $, there must be at least	
one for which the number of nonzero terms is least. Let $ (\ast)$	
be such a relation. Now, we may assume that $\lambda_1\neq 0 ,	
\lambda_2\neq 0$ and choose $g$ in $ G $ such that	
$k_1=gg_1g^{-1}g_1^{-1}\neq	
gg_2g^{-1}g_2^{-1}=k_2 $. From the relation $ (\ast) $ we obtain \\	
$k_1(\lambda_1 g_1+\ldots +\lambda_tg_t)-g(\lambda_1 g_1+\ldots	
+\lambda_tg_t)g^{-1}=\lambda_1 k_1 g_1+\ldots \lambda_tk_1g_t-	
(\lambda_1 k_1 g_1+\ldots \lambda_tk_tg_t)=	
\lambda_2(k_1-k_2)g_2+\ldots +\lambda_t(k_1-k_t)g_t=0$.\\	
Now, the last equation contradicts the choice of the relation	
$(\ast)$. Therefore, $g_1,\ldots ,g_t$ are a linearly independent	
subset of $G$ over $K$, and this completes the proof.	
\vspace{0.2cm}	
	
To prove our next lemma, we shall need the following results from	
\cite{nil}. \vspace{0.2cm}	
	
{\sc Lemma A.} {\it Let $D$ be a finite dimensional $ F $-central	
division algebra. Suppose that $K$ is a subfield of $D$	
containing $F$. If $G$ is an irreducible subgroup of $D^*$ such	
that $K^*\lhd G$, then $K/F$ is Galois and $G/C_G(K^*)\simeq	
Gal(K/F)$.} \vspace{0.2cm}	
	
{\sc Lemma B.} {\it Let $D$ be a finite dimensional $ F $-central	
division algebra and let $G$ be an irreducible subgroup of $D^*$.	
If $K$ is a subfield of $D$ containing $F$ such that	
$[G:C_G(K^*)]=[K:F]$, then $C_D(K)=F[C_G(K^*)]$.}	
 \vspace{0.2cm}	
	
{\sc Theorem C.} {\it Let $D$ be a noncommutative finite	
dimensional $ F $-central division algebra. Then $D$ is a	
nilpotent crossed product if and only if there exist an	
irreducible subgroup $ G $ of $ D^* $ and an abelian normal	
subgroup $ A $ of $ G $ such that $ G/A $ is nilpotent.}	
	
\begin{lem}\label{quad} {\it Let $ D $ be a finite dimensional $ F $-central	
division algebra of index $ n $. Assume that $ D^* $ contains an	
irreducible soluble subgroup. Then we have the following:\\	
(i) there is an irreducible soluble subgroup $ G $ and a maximal	
abelian normal subgroup $ K^* $ of $ G $ such that $ K = K^* \cup	
\{0\} $	
is a subfield of $ D $ and  $ G/K^* $ is finite. Furthermore, setting	
 $ H := C_G(K^*) $, then the derived group $ H' $ of $ H $ is also finite.\\	
(ii) Assume the notation of (i). If $ H' $ is abelian and $ n =	
q^r $, $ q $ a prime, then $	
G/K^* $ is a $ q $-group.\\	
(iii) Keep the notation of (i). If $ H' $ is nonabelian and $ n =	
q^r $, $ q $ a prime, then $ H^{t-2} $ is a $ q $-group, where $	
H^i $ denotes the $i$-th term of the derived series of $ H $.\\	
(iv) If $ D $ is a non-crossed product with index $ i(D) = 2^r $,	
then $ D^* $ contains the finite quaternion subgroup $ Q_8 $, or	
$ SL_2(Z_3) $, or the binary octahedral group of order $ 48 $.}	
 \end{lem}	
	
{\sc Proof.} (i) Let $ G_0 $ be an irreducible soluble subgroup of	
$ D^* $. By Lemma 3 of \cite{free}, we know that $G_0$ is	
abelian-by-finite, i.e., there is an abelian normal subgroup $A$	
in $G_0$ of finite index. Take $ A $ maximal in $ G_0 $, and set	
$K=F(A)$. One may easily show that $G_0\subseteq N_{D^*}(A)$ and	
that $K^*G_0$ is an irreducible soluble subgroup of $D^*$. Set $	
G = K^*G $. Then, it is easily seen that $K^*$ is maximal abelian	
normal in $ G $ and $G/K^*$ is a finite group. Furthermore, we	
know that $H/Z(H)$ is finite and hence, by Theorem 15.1.13 of	
\cite{sc}, the derived group $H'$ is a finite group.	
	
(ii)  Because $H'\subseteq C_G(K^*)$, $K^*H'$ is an abelian	
normal subgroup of $G$. Hence, by maximality of $K^*$, we have	
$H'\subseteq K^*=Z(H)$. Therefore, $H/K^*$ is abelian. Now, by	
Lemma~\ref{index}, we conclude that $[K[H]:K]=|H/K^*|$. Since	
$[D:F]=q^{2r}$ and $F^*\subseteq K^*$ we conclude that $[K[H]:K]$	
divides $q^{2r}$, i.e., there exists a natural number $s$ such	
that $|H/K^*|=[K[H]:K]=q^s$. Now, by Lemma A, we have $G/H\simeq	
Gal(K/F)$ and $K/F$ is a Galois extension. Since $i(D)=q^r$ there	
exists a natural number $t$ such that $|Gal(K/F)|=[K:F]=q^t$.	
Thus, $|G/H|=q^t$ and hence $|G/K^*|=q^{s + t}$, i.e., $G/K^*$ is	
a $ q $-group.	
	
(iii) Suppose that $ H' $ is nonabelian. Then, the soluble length	
of $H$ is $t = l(H)\geq 3$. Now, consider the derived chain	
$\langle e \rangle =H^t \subset H^{t-1} \subset \ldots \subset H'	
\subset H$. It is clear that $H^{t-1}$ is abelian and $H^{t-2}$	
is a nonabelian subgroup of $H'$. Now, we know that $H^{t-1}\lhd	
G$ and $H^{t-1}\subset C_G(K^*).$ Thus, $ H^{t-1}K^* $ is an	
abelian normal subgroup of $ G $. Hence, by maximality of $ K^*	
$, we conclude that $ H^{t-1} \subseteq K^* $. Therefore,	
$H^{t-2}K^*/K^*$ is an abelian subgroup of $G/K^*$. Set	
$N=H^{t-2}K^*$. We note that $N$ is normal in $G$, and hence	
$Z(N)$ is an abelian normal subgroup of $G$ containing $K^*$. By	
maximality of $K^*$, we have $Z(N)=K^*$. Now, $N$ is a subgroup	
of $D^*$ such that $N/K^*$ is abelian and $Z(N)=K^*$. By Lemma	
\ref{index}, we obtain $[K[N]:K]=|N/K^*|$. By our assumption, we	
know that $[K[N]:K]$ divides $q^{2r}$. Therefore, $N/K^{*} \simeq	
H^{t-2}/K^{*} \cap H^{t-2}$ is a $q$-group. Now, by Lemma B, we	
have $F[H]=K[H]=C_D(K)$. Put $D_1=C_D(K)$. We know that	
$Z(D_1)=K$. We now claim that $H^{t-2}$ is a $q$-group. To see	
this, let $x\in H^{t-2}$. Then, there exists a natural number $s$	
such that $x^{q^s}\in K^*\cap H^{t-2}$. On the other hand,	
$x^{q^s}\in H'\subset D_1'$, and hence $RN_{D_1/K}(x^{q^s})=1$.	
Since $x^{q^s}\in K^*=Z(D_1)^*$ we obtain	
$RN_{D_1/K}(x^{q^s})=(x^{q^s})^{i(D_1)}=x^{q^{s+u}}$, where $	
i(D_1) = q^u $. Therefore, $x^{q^{s+u}}=1$ and so $H^{t-2}$ is a	
finite $q$-group.	
	
(iv) Let $ G $ be the irreducible soluble subgroup obtained by	
(i) and keep the notations of the above cases. If $ H' $ is	
abelian, then by the case (ii), $G/K^*$ is a $ 2 $-group and hence	
nilpotent and so $G$ is an irreducible abelian-by-nilpotent	
subgroup of $D^*$. Now, by Theorem C, we conclude that $D$ is a	
nilpotent crossed product, which contradicts our assumption that	
$ D $ is a non-crossed product. Therefore, $ H' $ is nonabelian.	
By (iii) we conclude that $H^{t-2}$ is a finite $2$-group. Now,	
by a result of~\cite[p.45]{shir}, we conclude that $H^{t-2}$ is	
cyclic or a (generalized) quaternion group. Since $H^{t-2}$ is	
nonabelian we conclude that $H^{t-2}$ is a (generalized)	
quaternion group. We recall that the (generalized) quaternion	
group of order $2^{u}$, $u\geq 3$, is defined with the	
presentation $Q_{2^{u}}= \langle ~x,y~ |~ x^{2^{u-2}}=y^2,~	
y^4=1,~ yxy^{-1}=x^{-1} \rangle.$ It is clear that $\langle	
x\rangle \lhd Q_{2^{u}}$ and $Q_{2^u}^{(2)}=~\langle x^2 \rangle	
$. Thus, $\langle x^2 \rangle $ is a characteristic subgroup of	
$H^{t-2}$ and hence $\langle x^2 \rangle $ is an abelian normal	
subgroup of $G$. We note that $\langle x^2 \rangle \subset	
C_G(K^*) $ and so $ K^*\langle x^2 \rangle $ is an abelian normal	
subgroup of $ G $. Therefore, by maximality of $ K^* $, we have	
$\langle x^2 \rangle \subset K^*$. Thus, $x^2 \in Z(Q_{2^u})$,	
and hence $x^{-2}=yx^2y^{-1}=x^2$. Therefore, $x^4=1$. On the	
other hand, we have  $x^{2^{u-1}}=1$, and so $u=3$, i.e.,	
$H^{t-2}\simeq Q_8$. Now, assume that $N$ is a maximal normal	
$2$-subgroup of $H'$. For every $g\in G$, set $N_1=gNg^{-1}$. $	
N_1 $ is a normal $2$-subgroup of $H'$. Hence $NN_1$ is a normal	
$2$-subgroup of $H'$. By maximality of $ N $, we obtain $N_1	
\subseteq N$ and so we have $N \lhd G$. We note that $ H' $ is	
finite and hence $ N $ is a finite $ 2 $-group. Thus, as in the	
case of $ H^{t-2} $ above, we conclude that $N\simeq Q_8$.	
Therefore, by a result of~\cite[p.54]{shir}, we have either	
$H'\simeq Q_8 \times M$, where $M$ is a group of odd order, or	
$H'\simeq SL_2(Z_3)\times M$, where $M$ is a group of order $m$	
coprime to $6$, or $H'$ is isomorphic to the binary octahedral	
group. In the first case, $M$ is a characteristic subgroup of	
$H'$. Therefore, $M$ is a normal subgroup of $G$. Let $l$ be the	
soluble length of $M$. We know that $M^{l-1}$ is a nontrivial	
abelian normal subgroup of $G$. Thus, $M^{l-1}\subseteq K^*$, and	
hence for every $x\in M^{l-1}$ we have	
$x^{2^\alpha}=RN_{C_D(K)/K}(x)=1$, which contradicts the fact	
that $M$ is of odd order. Therefore, $M$ is trivial and so	
$H'\simeq Q_8$. One may easily show that other cases are also	
true by similar arguments and this proves the case (iv).	
\vspace{0.2 cm}	
	
We are now prepared to prove the following. \vspace{0.2 cm}	
	
\begin{thm}\label{prime} {\it Let $D$ be an $ F $-central division	
algebra of index $ q^r$, $ q $ a prime. If $D^*$ contains an	
irreducible soluble subgroup, then $D$ is a crossed product.}	
 \end{thm}	
	
{\sc Proof.} We may consider the following two cases:	
	
 Case 1. $Char F = p > 0 $. By Lemma \ref{quad}, we know that there is	
an irreducible soluble subgroup $ G $ and abelian normal subgroup	
$ K^* $ of $ G $ such that $ K = K^* \cup \{0\} $ is a subfield of	
$ D $ and  $ G/K^* $ as well as $ H' $ is finite, where	
$H=C_G(K^*)$. Since $CharF=p>0$, by a result of	
\cite[p.215]{lam}, we conclude that $H'$ is cyclic. Now, by	
Lemma~\ref{quad}, $G/K^*$ is a $ q $-group and hence it is	
nilpotent. Thus, $G$ is an irreducible abelian-by-nilpotent	
subgroup of $D^*$. Now, by Theorem C, we conclude that $D$ is a	
nilpotent crossed product, which completes the proof of this case.	
	
Case 2. $ Char F = 0 $. We keep to the notations of the case 1.	
If $ H' $ is abelian, then as in the above case we obtain the	
result. So, we may assume that $ H' $ is nonabelian. By	
Lemma~\ref{quad}, we know that $H^{t-2}$ is a finite $q$-group.	
If $ q $ is odd, then, by a result of \cite[p.45]{shir}, we	
conclude that $H^{t-2}$ is cyclic, which contradicts the fact that	
$H^{t-2}$ is nonabelian. So, we may assume that $ q = 2 $. We now	
proceed by induction on $ r $. If $ r = 1 $, then it is clear	
that $ D $ is cyclic. Assume that the result holds for all $ n<r	
$. Now, by a result of~\cite[p.45]{shir} again, we conclude that	
$H^{t-2}$ is cyclic or a (generalized) quaternion. Since	
$H^{t-2}$ is nonabelian we conclude that $H^{t-2}$ is a	
(generalized) quaternion. As in the proof of Lemma~\ref{quad}, one	
may easily show that $H^{t-2}\simeq Q_8 $. Therefore, $ H^{t-2} $	
is normal in $ G $. Set $ D_1 = F[H^{t-2}] $. It is clear that $	
i(D_1) = 2 $ and $ Z(D_1) = F $ and $ D_1 $ is a crossed product.	
Now, by the Double Centralizer Theorem, we have $ D \simeq D_1	
\otimes_F C_D(D_1) $. Since $ G $ normalizes $ D_1 $ we see that	
for any $ g \in G $ we may define a natural homomorphism $ f_g :	
D_1 \rightarrow D_1 $, given by the rule $ f_g(x) = gxg^{-1} $ for	
any $ x \in D_1 $. Hence, by Skolem-Noether Theorem there is an	
element $ a_g \in D_1^* $ such that $ f_g =f_{a_g} $. If $ u,v	
\in D_1 $ satisfy $ f_u =f_v $, then for any $ x\in D_1 $ we have	
$ uxu^{-1}=vxv^{-1} $. Therefore, $ u^{-1}v \in Z(D_1)=F $, which	
shows that $u,v $ are equal modulo $ F^* $, i.e., $ F^*u=F^*v $.	
Now, for any $ x\in D_1$ we have $ gxg^{-1}=a_gxa_g^{-1} $, and	
hence $ b_g :=a_g^{-1}g \in C_D(D_1) $. The fact that $ b_g $	
commutes with $ a_g $ implies that $ a_g, g $, and $ b_g $	
pairwise commute. Set $ A=\cup_{g \in G} F^*a_g $ and $ B=\cup_{g	
\in G} F^*b_g $. We claim that $ A, B $ are groups. To see this,	
it is enough to show that for any $ g,h \in G $ we have $	
F^*a_{g^{-1}} =F^*a_g^{-1}, F^*a_ha_g =F^*a_{hg}, F^*b_{g^{-1}}	
=F^*b_g^{-1}, F^*b_hb_g =F^*b_{hg} $. For any $ x\in D_1,	
f_{a_{g^{-1}}}(x) = f_{g^{-1}}(x) =g^{-1}xg =	
(a_gb_g)^{-1}x(a_gb_g)=a_g^{-1}xa_g = f_{a_g^{-1}}(x) $.	
Therefore, $ F^*a_{g^{-1}} =F^*a_g^{-1} $. Also, we have $	
f_{a_{hg}}(x) = f_{hg}(x) $. Hence $ f_{a_{hg}}(x) =	
hgxg^{-1}h^{-1} =ha_gxa_g^{-1}h^{-1} = a_ha_gxa_g^{-1}a_h^{-1}	
=(a_ha_g)x(a_ha_g)^{-1} =f_{a_ha_g}(x) $. Therefore, $ F^*a_ha_g	
=F^*a_{hg}$ which shows that $ A $ is a group. Next, considering	
the fact that $ a_g \in D_1 $ and $ b_h \in C_D(D_1) $ we obtain	
$ b_hb_g = b_ha_g^{-1}g =a_g^{-1}b_hg =	
a_g^{-1}a_h^{-1}hg=(a_ha_g)^{-1}hg $. Thus, since $ A $ is a	
group we conclude that $ F^*b_hb_g = F^*(a_ha_g)^{-1}hg	
=F^*a_{hg}^{-1}hg =F^*b_{hg}, F^*b_g^{-1} = F^*a_gg^{-1}	
=F^*a_{g^{-1}}^{-1}g^{-1} = F^*b_{g^{-1}} $. Therefore, $ B $ is	
also a group. We claim that $ B $ is soluble that is normalized	
by $ G $. To see this, consider the epimorphism $ \theta : G	
\rightarrow B/F^* $ given by $ \theta(g) = F^*b_g $ for all $ g\in	
G $. Hence $ B/F^* $ as a homomorphic image of a soluble group is	
soluble, and so is $ B $. Set $ D_2 =C_D(D_1) $. If we show that	
$ D_2 = F[B] $, then by induction $ D_2 $ is a crossed product. To	
prove this, put $ D_3 = F[B] $. Now, for all $ g \in G $ we have $	
g=a_gb_g=(a_g \otimes 1)(1 \otimes b_g) = a_g \otimes b_g $.	
Therefore, we conclude that $ G \subset D_1 \otimes D_3 $ and	
hence $ F[G] = D = D_1 \otimes D_3 = D_1 \otimes D_2 $. Finally,	
one may easily see that $ [D_3:F] = [D_2:F] $, and so $ D_3	
\subseteq D_2 $, i.e., $ D_3 = D_2 $ and so the result follows.	
\vspace{0.2 cm}	
	
Let $ D $ be an $ F $-central division algebra of degree $ p^r $,	
$ p $ a prime. Using the above result one may conclude that if $	
D^* $ contains an irreducible finite subgroup $ G $, then $D$ is a	
crossed product. To see this, by a result of~\cite[p.51, Thm	
2.1.11]{shir}, we know that either $G$ is soluble or $G\simeq	
SL_2(Z_5)$. If the first case happens, then the result follows	
from Theorem \ref{prime}. If the second case occurs, then as in	
the course of the proof of Theorem 2.1.11 of~\cite[p.51]{shir},	
we have $[\mathbb{Q}(G):\mathbb{Q}]\leq 8$. Since $\mathbb{Q}	
\subseteq F $ we clearly have $[F[G]:F]\leq 8$ and hence $ [D:F]	
= 4 $ because $ G $ is irreducible. Therefore, $ D $ is cyclic	
and so the result also follows for this case. Later on we shall	
present a different proof of this fact which may be of some	
interest. Now, combining Lemma~\ref{if} and Theorem~\ref{prime},	
we are able to obtain one of our main results in the following	
form.	
	
\begin{cor}\label{primezero} {\it Let $D$ be an $F$-central division	
algebra of index $p^r$, $ p $ a prime. Then,  $D$ is a crossed	
product if and only if $D^*$ contains an irreducible soluble	
subgroup.}	
\end{cor}	
	
\section {Irreducible abelian-by-finite subgroups}	
	
This section turns to the case where the multiplicative group $	
D^* $ contains an irreducible abelian-by-finite subgroup. Let $D$	
be an $F$-central division algebra of index $p^r$, $ p $ a prime.	
It is proved that except when $ Char F = 0 $ and $ p=2, r>1 $,	
$D$ is a crossed product if and only if $D^*$ contains an	
irreducible abelian-by-finite subgroup. Furthermore, the	
conclusion also holds for the excluded case provided that $ D^* $	
contains no finite subgroup isomorphic to $ SL_2(Z_5) $. Using	
the above result, and the Tits Alternative which asserts that a	
finitely generated linear group either contains a non-cyclic free	
subgroup or it is soluble-by-finite \cite{zal}, we are able to	
show that $D$ is a crossed product if and only if $D^*$ contains	
an irreducible subgroup satisfying a group identity. Furthermore,	
the conclusion also holds for the above excluded case provided	
that $ D^* $ contains no finite subgroup isomorphic $ SL_2(Z_5)	
$. To prove our results, we shall need the following lemma.	
	
\begin{lem}\label{quad1} {\it Given a field $ F $ of characteristic zero,	
 let $D$ be an $ F $-central division	
algebra of index $ 2^r, r>1 $. Assume that $ D^* $ contains an	
irreducible abelian-by-finite subgroup. If $ D $ is a non-crossed	
product, then $ D^* $ contains a copy of the finite group	
$SL_2(Z_5)$.}	
 \end{lem}	
	
{\sc Proof.} Suppose that $G$ is an irreducible abelian-by-finite	
subgroup of $D^*$ and $A$ is a  maximal abelian normal subgroup	
of $G$ such that $G/A$ is finite. Set $K=F(A)$. It is clear that	
$G\subseteq N_D{^*}(K^*)$, and hence $G_1=GK^*$ is an irreducible	
subgroup of $D^*$ so that $G_1/K^*$ is finite. One may easily	
show that $K^*$ is a maximal abelian normal subgroup of $G_1$.	
Put $H=C_{G_1}(K^*)$. By maximality of $K^*$, we have $Z(H)=K^*.$	
Now, we know that $H/Z(H)$ is finite, and so by Theorem	
of~\cite[p.443, Thm. 15.1.13]{sc}, the derived group $H'$ is a	
finite group. We claim that $H'$ is nonabelian. For if $ H' $ is	
abelian, then $ H $ is soluble. Now, by Lemma A, we have	
$G_1/H\simeq Gal(K/F)$ and $K/F$ is a Galois extension. Thus,	
$G_1/H$ is a $2$-group and hence $G_1$ is soluble. We note that	
$G_1$ is an irreducible soluble subgroup of $D^*$. By	
Theorem~\ref{prime}, we conclude that $D$ is a crossed product	
which is a contradiction. Thus, $ H' $ is nonabelian as claimed.	
Therefore, by a result of~\cite[p.51]{shir}, this implies that	
either $H'$ is a soluble group or $H'\simeq SL_2(Z_5)$. If the	
first case occurs, then $H$ is soluble and hence as above $ G $	
is soluble. Therefore, by Theorem~\ref{prime}, we conclude that	
$D$ is a crossed product which is a contradiction. So, we have	
$H'\simeq SL_2(Z_5)$, and the result follows. \vspace{0.2 cm}	
	
\begin{thm}\label{ap} {\it Let $D$ be an $F$-central division	
algebra of index $q^r$, $ q $ a prime. If $ D^* $ contains an	
irreducible abelian-by-finite subgroup, then, except for the case	
$ Char F = 0 $ and $ q=2, r>1 $, $D$ is a crossed product.	
Furthermore, the conclusion also holds for the above excluded	
case provided that $ D^* $ contains no finite subgroup isomorphic	
to $ SL_2(Z_5) $.}	
 \end{thm}	
	
{\sc Proof.} We consider the following three cases:\\	
 Case 1. $Char F =p>0$. Suppose that $G$ is an irreducible abelian-by-finite	
subgroup of $D^*$ and $A$ is a  maximal abelian normal subgroup	
of $G$ such that $G/A$ is finite. Set $K=F(A)$. It is clear that	
$G\subseteq N_D{^*}(K^*)$, and hence $G_1=GK^*$ is an irreducible	
subgroup of $D^*$ so that $G_1/K^*$ is finite. One may easily	
show that $K^*$ is a maximal abelian normal subgroup of $G_1$.	
Put $H=C_{G_1}(K^*)$. By maximality of $K^*$, we have $Z(H)=K^*.$	
Now, we know that $H/Z(H)$ is finite, and so by Theorem	
of~\cite[p.443]{sc}, the derived group $H'$ is a finite group.	
Thus, by a result of \cite[4, Cor. 13.3]{lam}, we conclude that	
$H'$ is cyclic. Therefore, $H$ is a soluble group. Now, by Lemma	
A, we have $G_1/H\simeq Gal(K/F)$ and $K/F$ is a Galois	
extension. Thus, $G_1/H$ is $q$-group and hence $G_1$ is soluble.	
We note that $G_1$ is an irreducible soluble subgroup of $D^*$. By	
Theorem~\ref{prime}, we conclude that $D$ is a crossed product,	
which completes the proof of this case.\\	
Case 2. $ Char F = 0 $. If $ q =2 $ and $ r= 1$, then it is clear	
that $ D $ is cyclic. So, we may assume that $ q $ is odd.	
Keeping to the notations of the above case, we know that $H/Z(H)$	
is finite as well as the derived group $H'$. Therefore, by a	
result of~\cite[p.51]{shir}, we know that either $H'$ is a	
soluble group or $H'\simeq SL_2(Z_5)$. In the first case $H$ is	
soluble and as in the above case we have that $G_1$ is also a	
soluble subgroup of $D^*$. Thus, by Theorem~\ref{prime}, we	
conclude that $D$ is a crossed product. We claim that the second	
case leads to a contradiction. So, we may assume that $H'\simeq	
SL_2(Z_5)$. In the course of the proof of Theorem 2.1.11	
of~\cite[p.51]{shir}, that the only finite insoluble subgroup of	
a division ring is $SL_2(Z_5)$ we obtain	
$[\mathbb{Q}(H'):\mathbb{Q}]\leq 8$. Since $\mathbb{Q}\subseteq	
K$ we conclude that $[K[H']:K]\leq 8$. On the other hand, we have	
$K\subseteq Z(K[H'])$. Set $D_1=K[H']$. Now, we know that $D_1$	
is a division algebra with $[D_1:Z(D_1)]\leq 8$. Therefore,	
$[D_1:Z(D_1)]=4$ and hence $2$ divides $q^{r}$, which	
contradicts our assumption that $q$ is odd.\\	
Case 3. Assume that $ Char F = 0, p=2$, and $ r>1 $. If $ D $ is	
not a crossed product, then, by Lemma \ref{quad1}, we conclude	
that $ D^* $ contains a copy of the finite group $ SL_2(Z_5) $,	
which is a contradiction. This completes the proof of the theorem.	
\vspace{0.2 cm}	
	
Combining Lemma~\ref{if} and Theorem~\ref{ap}, we obtain the	
following.	
	
\begin{cor}\label{apaz} {\it Let $D$ be an $F$-central division	
algebra of index $p^r$, $ p $ a prime. Then, except when $ Char F	
= 0 $ and $ p=2, r>1 $, $D$ is a crossed product if and only if	
$D^*$ contains an irreducible abelian-by-finite subgroup.	
Furthermore, the conclusion also holds for the above excluded case	
provided that $ D^* $ contains no finite subgroup isomorphic to $	
SL_2(Z_5) $.}	
\end{cor}	
	
Using the above result, and the Tits Alternative which asserts	
that a finitely generated linear group either contains a	
non-cyclic free subgroup or it is soluble-by-finite \cite{zal},	
we are able to prove the following criterion.	
 \vspace{0.2 cm}	
	
\begin{cor}\label{gi} {\it  Let $D$ be an $F$-central division	
algebra of index $p^r$, $ p $ a prime. Then, except when $ Char F	
= 0 $ and $ p=2, r>1 $, $D$ is a crossed product if and only if	
$D^*$ contains an irreducible subgroup satisfying a group	
identity. Furthermore, the conclusion also holds for the above	
excluded case provided that $ D^* $ contains no finite subgroup	
isomorphic to $ SL_2(Z_5) $.}	
\end{cor}	
	
{\sc Proof.} The ``only if `` part is clear by Lemma~\ref{if}.	
Assume that $G$ is an irreducible subgroup of $ D^* $ satisfying	
a group identity. Since $[D:F]<\infty$ we may view $G$ as a	
linear group. Let $ G_1 $ be a subgroup of $ G $ generated by the	
elements of a basis of $ D $ over $ F $. Thus, by Tits	
Alternative, we know that $G_1$ is soluble-by-finite, i.e., there	
is a soluble normal subgroup $ N $ of $ G_1 $ such that $ G_1/N $	
is finite. Now, by Lemma 3 of~\cite{free}, $N$ is	
abelian-by-finite. Thus, $G_1$ is abelian-by-finite. Therefore, by	
Theorem~\ref{apaz}, $D$ is crossed product.	
 \vspace{0.2 cm}	
	
Now, one may apply the above results to prove the following	
criterion for $ D $ to be cyclic. This is one of the main results	
of \cite{ekm}.	
	
\begin{cor}\label{pgi} {\it Let $D$ be an $F$-central division algebra	
of prime degree $p$. Then $D$ is cyclic if and only if $D^*$	
contains a nonabelian subgroup satisfying a group identity.}	
\end{cor}	
	
{\sc Proof.} The ``only if`` part is clear by Lemma~\ref{if}. If	
$p=2$, then $D$ is cyclic. Let $p$ be an odd prime. Now, by	
Corollary~\ref{gi}, one can easily show that $D$ is	
cyclic.\hfill$\Box$	
 \vspace{0.2 cm}	
	
Let $ D $ be an $ F $-central division algebra of finite index $	
i(D) =n $ and $ G $ be an irreducible subgroup of $ D^* $. Assume	
that $ A $ is a maximal abelian normal subgroup of $ G $. We	
conclude this section with some remarks concerning the relation	
between the cardinal of $ G/A $ and the dimension of $ D/F $.	
 \vspace{0.2 cm}	
	
Remark 1. Let $ D $ be a finite dimensional $ F $-central	
division algebra. If $ G $ is an irreducible subgroup of $ D^* $	
with maximal abelian normal subgroup $ A $ such that $ G/A $ is	
nilpotent, then $ |G/A| = i(D) $. To see this, set $ G_1 = K^*G	
$, where $ K = F[A] $. It is easily seen that $ G_1 $ is	
irreducible with maximal abelian normal subgroup $ K^* $ such that	
$ G_1/K^* \simeq G/A $ and so $ G_1/K^* $ is nilpotent. As in the	
proof of Theorem 3.4 of \cite{nil}, one may easily check that $	
K/F $ is Galois and $ K $ is a maximal subfield of $ D $.	
Therefore, we have $ C_{G_1}(K^*) = K^* $. Now, by Lemma B, we	
have $ G_1/K^* \simeq Gal(K/F) $, i.e., $ |G_1/K^*| = |G/A| =	
[K:F] = i(D) $. \vspace{0.2 cm}	
	
Remark 2. Let $D$ be an $F$-central division algebra of index	
$p^r$, $ p $ a prime. Assume that $ G $ is an irreducible	
subgroup of $ D^* $ with maximal abelian subgroup $ A $ such that	
$ G/A $ is finite. Then, except when $ Char F = 0 $ and $ p=2,	
r>1 $, we have $ |G/A	
 |=i(D) $. Furthermore, the conclusion also holds for the excluded	
case provided that $ D^* $ contains no finite subgroup isomorphic	
to $ SL_2(Z_5) $. To prove this, we may use Theorem \ref{prime},	
Theorem \ref{ap}, and the Remark 1 to obtain the result.	
	
\section {Irreducible finite subgroups}	
	
Let $ D $ be an $ F $-central division algebra of degree $ p^r $,	
$ p $ a prime. This section studies the structure of $ D $ under	
the condition that $ D^* $ has an irreducible finite subgroup.	
Using Amitsur's classification of finite multiplicative subgroups	
of a division ring, it is proved that if $ D^* $ contains an	
irreducible finite subgroup, then $D$ is a crossed product.	
	
\begin{thm}\label{finite} {\it Let $D$ be an $F$-central division	
algebra of index $q^r$, $ q $ a prime. If $ D^* $ contains an	
irreducible finite subgroup $ G $, then $D$ is a crossed product.}	
 \end{thm}	
	
{\sc Proof.} We first observe that $ Char F = 0 $. Since	
otherwise $ G $ is cyclic and since $ G $ is irreducible we obtain	
$ D = F $ which is a contradiction. If $ q $ is odd, then the	
result follows from Corollary \ref{gi}. So we may assume that $	
q=2 $. By a result of~\cite[p.51]{shir}, we know that either $G$	
is soluble or $G\simeq SL_2(Z_5)$. If the second case occurs,	
then as in the course of the proof of Theorem 2.1.11	
of~\cite[p.51]{shir}, that the only finite insoluble subgroup of	
a division ring is $SL_2(Z_5)$, we may obtain	
$[\mathbb{Q}(G):\mathbb{Q}]\leq 8$. Since $\mathbb{Q} \subseteq F	
$ we clearly have $[F[G]:F]\leq 8$ and hence $ [D:F] = 4 $	
because $ G $ is irreducible. Therefore, $ D $ is cyclic and so	
the result follows for this case. It remains to consider the case	
where $ G $ is soluble and $ q = 2 $. By Lemma 3 of \cite{free},	
we know that $G$ is abelian-by-finite, i.e., there is an abelian	
normal subgroup $A$ in $G$ of finite index. Take $ A $ maximal in	
$ G $, and set $K=F(A)$. One may easily show that $G\subseteq	
N_{D^*}(A)$ and that $G_1 =K^*G$ is an irreducible soluble	
subgroup of $D^*$ with maximal abelian normal subgroup $ K^* $.	
Set $ H=C_G(A) $, $ H_1 = C_{G_1}(K^*) $. It is clearly seen that	
$ H_1 = HK^* $. Since elements of $ K^* $ and $ H $ pairwise	
commute we conclude that $ H_1' = H' $. Now, by Lemma B, $ K/F $	
is Galois with $ G_1/H_1 \cong Gal(K/F) $. Therefore, $ G_1/H_1 $	
is a $ 2 $-group and hence it is nilpotent. Now, one may easily	
show that $ G \cap H_1 = H $ and $ G_1 = GK^* = GH_1 $. Thus, we	
have $ G_1/H_1 \cong G/H $ and hence $ G/H $ is a $ 2 $-group. If	
$ H_1' $ is abelian, then as in the proof of Lemma \ref{quad},	
one may easily show that $ H_1/K^* $ is a $ 2 $-group. Now, since	
$ A= H \cap K^* $ and $ H_1 = HK^* $ we conclude that $ H/A $ is a	
$ 2 $-group. This means that $ G/A $ is also a $ 2 $-group and	
hence $ G $ is abelian-by-nilpotent. Therefore, by Theorem C, we	
conclude that $ D $ is crossed product. Thus, we may assume that	
$ H_1'=H' $ is nonabelian. Let $ l(H)=t $ be the derived length	
of $ H $. As in the proof of Lemma \ref{quad}, one may easily	
show that $ H^{t-2} $ is a nonabelian $ 2 $-group and it is	
isomorphic to the quaternion group $ Q_8$. This means that $ H $	
contains a normal subgroup isomorphic to $ Q_8 $. Now, assume	
that $T=O_2(H)$ is a maximal normal $2$-subgroup of $H$. As in	
Lemma \ref{quad}, it is easily seen that $O_2(H)\simeq Q_8$. Now,	
by a result of~\cite[p.54]{shir}, we have either $H\simeq Q_8	
\times M$, where $M$ is a group of odd order, or $H\simeq	
SL_2(Z_3)\times M$, where $M$ is a group of order $m$ coprime to	
$6$, or $H$ is isomorphic to the binary octahedral group. We deal	
with these cases separately as follows:	
	
Case 1. $H\simeq Q_8 \times M$, where $M$ is a group of odd	
order. We claim that $ M $ is normal in $ G $. Since $ H $ is	
normal in $ G $ for each $ g\in G $ and $ m \in M $ we have $	
gmg^{-1} = (q,m_1) \in H $. Comparing the orders of both sides of	
the last relation, one may easily conclude that $ q = 1 $ and so	
the claim is established. Now, we show that $ M $ is abelian.	
Otherwise, $ M' $ is nontrivial. If $ l $ is the soluble length	
of $ M $, then $ l \geq 2 $. Thus, $ M^{l-1} \subseteq M' $ is a	
nontrivial abelian subgroup. This implies that $ AM^{l-1} $ is an	
abelian normal subgroup of $ G $ and hence by the choice of $ A $	
we obtain $ M^{l-1} \subseteq A $. By Lemma B, we know that $	
F[H_1] = C_D(K) $. Since $ M \subseteq H \subseteq H_1 $ we	
obtain $ M' \subseteq C_D(K)' $. Take an element $ x \in M^{l-1}	
\subseteq A \subseteq K^* $. We have $ 1 = RN_{C_D(K)/K}(x) =	
x^{2^s} $, where $ i(C_D(K))=2^s $. This shows that the order of	
$ x $ is a power of $ 2 $ which contradicts the fact that $ M $	
has odd order. Hence $ M' $ must be trivial and so $ M $ is	
abelian. It is clear that $ H/M \simeq Q_8 $ and $ G_1/H_1 \simeq	
G/H $ is a $ 2 $-group. Since $ M $ is normal in $ G $ we	
conclude that $ G/M $ is also a $ 2 $-group. This says that $ G $	
is abelian-by-nilpotent and hence, by Theorem C, $ D $ is a	
crossed product.	
	
Case 2. $H\simeq SL_2(Z_3)\times M$. Since the order of $ M $ is	
prime to $ 6 $ and $ |SL_2(Z_3)|= 24 $, as in the case 1, we	
conclude that $ M $ is an abelian normal subgroup of $ G $. Now,	
$ M $ as an abelian normal subgroup of $ D^* $ is cyclic. Set $ M	
= <m> $ such that for each natural number $ s $ with $ (s,6) = 1	
$ we have $ m^s = 1 $. Since $ SL_2(Z_3) \subseteq G $ we have $	
2||G| $ and hence there exists $ g \in G $ such that $ g^2 = 1 $,	
i.e., $ -1 \in G $.	
	
If $ m \in F^* $, then $ m \in Z(G) $. Therefore, $ 1,m, \cdots,	
m^{s-1},-1,-m, \cdots,-m^{s-1} $ are distinct elements of $ Z(G)	
$ for if $ m^i = -m^j $ with $ 0 \leq i,j \leq s-1 $, then	
raising to the power of $ s $ we obtain $ 1=-1 $ which is a	
contradiction to the fact that $ Char F = 0 $. Thus, $ |Z(G)|>2s	
$. Now, $ G $ as an irreducible subgroup of $ D^* $ contains a	
basis $ g_1, g_2, \cdots, g_t $ with $ t= 2^{2r} $. Since $g_1,	
g_2, \cdots, g_t $ are linearly independent over $ F $ we	
conclude that $g_1Z(G), g_2Z(G), \cdots, g_tZ(G) $ are distinct	
elements of $ G/Z(G) $ and hence $ |G/Z(G)| \geq t $. Therefore,	
we have $ |G| \geq 2^{2r}\times 2s $. On the other hand, we have	
$ |M|=s $ and so $ |H|=2^3\times 3 \times s $ and also $ G/H	
\simeq Gal(K/F) $, where $ K/F $ is Galois. If $ [K:F] = 2^r $,	
then $ K $ is a maximal subfield of $ D $ and hence $ D $ is	
crossed product. So, we may assume that $ [K:F] \leq 2^{r-1} $.	
In this case we obtain $ |G| \leq 2^3 \times 3 \times s \times	
2^{r-1} $. Therefore, $ 2^{2r+1} \times s \leq 2^{r+2} \times 3	
\times s $ which implies that $ 2^{r-1} \leq 3 $, i.e., $ r=1 $	
or $ r = 2 $. If $ r = 1 $, then it is clear that $ D $ is	
cyclic. If $ r = 2 $, then, by a result of \cite[p. 183]{ro}, $ D	
$ is a crossed product.	
	
If $ m \notin F^* $, then $ MA $ is an abelian normal subgroup of	
$ G $. By maximality of $ A $, we conclude that $ M \subseteq A	
\subseteq K^* $. Since $ m $ is not in $ F $ we obtain $	
|Gal(K/F)|=2^u $ with $ u \geq 1 $. Since $ Z(H) $ is an abelian	
normal subgroup of $ G $, by maximality of $ A $ , we have $ Z(H)	
=A $. Therefore, $ A = <-1> \times M $ and hence $ |A| = 2s $.	
Since $ Q_8 $ is normal in $ G $ let $ O_2(G) = Q_{2^l} $. It is	
clearly seen that $ (Q_{2^l})^2 = <x^2> $, where $ Q_{2^l} = <x,y|	
x^{2^{l-1}} = y^4 =1, yxy^{-1} = x^{-1} > $. Now, one may easily	
show that $ N = <x^2> $ is normal in $ G $. Since the orders of $	
M $ and $ N $ are coprime we have $ M \cap N = 1 $. Therefore,	
each element of $ M $ commutes with each element of $ N $, i.e.,	
$ MN $ is abelian. Since $ -1 \in N $ we obtain $ A \subseteq MN	
$. But this contradicts the choice of $ A $ unless $ <x^2> = <-1>	
$, i.e., $ x^4 = 1 $ and $ l = 3 $. Thus, $O_2(G) = Q_8 $. Now, by	
a result of~\cite[p.54]{shir} again we have three subcases to	
consider as follows:	
	
Subcase 1. $ G \simeq Q_8 \times M_1 $, where the order of $ M_1	
$ is odd. If $ |M_1| = 2n+1 $, then $ |G| = 2^3 \times (2n + 1)	
$. Now, we have $ |H| = 2^3 \times 3 \times s $, where $ s $ is	
odd, and $ |G/H| = |Gal(K/F)| = 2^u $ with $ K \neq F $.	
Therefore, $ (2n+1)=3s \times 2^u $ which is not possible.	
	
Subcase 2. $ G \simeq SL_2(Z_3) \times M_1 $, where the order of	
$ M_1 $ is prime to $ 6 $. Since the order of $ M_1 $ is odd, as	
in the Subcase 1, we obtain a contradiction.	
	
Subcase 3. $G$ is isomorphic to the binary octahedral group of $	
48 $ elements. Then, $ |G| = 2^4 \times 3 $. Since $ -1 \in Z(G)	
$ we obtain $ |Z(G)| \geq 2 $. As before, because $ G $ is	
irreducible we have $ |G/Z(G)| \geq 2^{2r} $. Therefore, $	
2^{2r+1} \leq 2^4 \times 3 $. This means that either $ r = 1 $ or	
$ r = 2 $, and as above we conclude that $ D $ is a crossed	
product.	
	
Case 3.  $H$ is isomorphic to the binary octahedral group of $ 48	
$ elements. Then, $ |H| = 2^4 \times 3 $. As in the Subcase 3, we	
conclude that $ |G| \geq 2^{2r+1} $. In addition, as in the	
previous cases, we have $ |G/H| < 2^{r-1} $, and hence $ |G| \leq	
2^{r-1} \times 2^4 \times 3 $, i.e., $ 2^{2r+1} \leq 2^{r+3}	
\times 3 $. This implies that either $ r=1 $ or $ r=2 $ or $ r=3	
$. For the cases $ r=1 $ or $ r=2 $, as before, we conclude that	
$ D $ is crossed product. Assume that $ r=3 $. If $ D $ is not a	
crossed product, then, by Lemma \ref{quad}, $ D^* $ contains a	
copy of $ Q_8 $. It is clear that $ [F[Q_8]:F] = 4 $. Set $ B =	
F[Q_8] $. Then, by Centralizer Theorem, we have $ D \simeq B	
\otimes C_D(B) $. Since $ i(C_D(B)) = 4 $, by a result of	
\cite[p. 183]{ro}, $ C_D(B) $ is a crossed product. Therefore, $ D	
$ which is a tensor product of crossed products is a crossed	
product division algebra. This completes the proof of the	
theorem. \vspace{0.2 cm}	
	
The authors thank the Research Council of Sharif University of	
Technology for support.  \vspace{0.2 cm}	
	
	
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\end{thebibliography}	
\end{document}	
