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%%% Dear Ulf,
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%%% Authors: L.Di Martino, M.C.Tamburini and A.Zalesskii
%%% 
%%% Title: On Hurwitz Groups of low rank
%%% 
%%% E-mail: dimartino@vmimat.mat.unimi.it
%%% c.tamburini@dmf.bs.unicatt.it
%%% a.zalesskii@uea.ac.uk
%%% 
%%% Subject Classification: 20G40, 20F05.
%%% Key words: Hurwitz groups, classical groups, representations.
%%% 
%%% Abstract: A group G is said to be (2,3,7)-generated, or Hurwitz (when
%%% finite), if it can be generated by an element of order 2 and an element of
%%% order 3 whose product has order 7. Recent constructive results of
%%% A.Lucchini, M.C.Tamburini and J.S.Wilson show that the class of
%%% (2,3,7)-generated groups is very wide. In particular it contains most
%%% finite simple classical groups of high rank. By contrast, in this paper, we
%%% use a formula due to L.Scott to show that the Hurwitz linear  groups of low
%%% rank are very few.
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\footnote{}{{\sevenrm
\hskip-.5truecm

AMS 1991 Subject Classification: 20G40, 20F05. \endgraf

Key words: Hurwitz groups, classical groups, representations.}}

\centerline {\bf\tit On Hurwitz groups of low rank}

\vskip30pt
\centerline{L. Di Martino}
\medskip
\centerline {Dipartimento di Matematica e Applicazioni}

\centerline {Universit\`a degli Studi di Milano-Bicocca}

\centerline {Via Bicocca degli Arcimboldi 8, Ed. U7}

\centerline {20126 Milano, Italy.}

\centerline {e-mail: dimartino@vmimat.mat.unimi.it}

\bigskip
\centerline{M.C. Tamburini}
\medskip
\centerline {Dipartimento di Matematica e Fisica}

\centerline {Universit\`a Cattolica del Sacro Cuore}

\centerline {Via Trieste 17 , 25121, Brescia, Italy.}

\centerline {e-mail: c.tamburini@dmf.bs.unicatt.it}

\bigskip
\centerline {A.E. Zalesskii}
\medskip
\centerline{School of Mathematics}

\centerline {University of East Anglia}

\centerline {Norwich NR4 7TJ, UK.}

\centerline {e-mail: a.zalesskii@uea.ac.uk }

\vskip30pt
{\bf 1.\enskip Introduction}

\bigskip
Let $\Delta(2,3,7)$ denote
the infinite triangle group, defined by the presentation
$\langle X,Y\mid X^2=Y^3=(XY)^7=1\rangle.$
A non-trivial group $G$ is said to be $(2,3,7)$-generated
(or a {\it Hurwitz} group, when finite) if
it is an epimorphic image of
$\Delta(2,3,7).$ Hurwitz groups are particularly
interesting for the
theory of Riemann surfaces.
Namely, if $H$ is the
automorphism group of a compact
Riemann surface of genus $g>1,$ then  $H$ is finite of order
not exceeding $84(g-1),$ and the Hurwitz
groups are exactly those for which $|H|= 84(g-1).$
As a quotient of a
Hurwitz group is again Hurwitz, it is crucial
to determine the simple Hurwitz
groups.  The values of $q$ for which
the groups $\PSL(2,q)$ are Hurwitz were determined by
A. M. Macbeath in 1969 ([Mac]), whereas
J. Cohen in 1981 ([Coh]) showed that $PSL(3,2)$ is the only Hurwitz
group in the series $\PSL(3,q),$ and none of the groups in the series
${\rm PSU}(3,q^2)$ are Hurwitz. M. Conder in 1980
([Con]) proved that the alternating
groups $A_n$ are Hurwitz, provided $n>167 .$
More recently G. Malle
has shown
that the exceptional simple groups $G_2(q),\ q\geq 5 ;$
$^2G_2(q),\ q> 3 ;$
$^3D_4(q)\ {\rm for}\ q=p^m,\ p\not=3,\ q\not=4,$
and $^2F_4(2^{2m+1})^\prime$ for $m\equiv 1\ (3)$
are Hurwitz ( cfr. [Mal1], [Mal2]).
The sporadic Hurwitz groups
are also known, with the
exception of the Monster.
(For a bibliography and related comments, see [Jon]).

Recent constructive results of
Lucchini, Tamburini and Wilson ([LTW], [Lu], [Wil] )
show that the family of $(2,3,7)$-generated  groups,
and more generally of $(2,3,k)$-generated groups,
is very large. In fact,
for every $k\geq 7$  there are $2^{\aleph_0}$
isomorphism classes of simple $(2,3,k)$-generated
groups. In particular,
most finite classical groups of
large rank are Hurwitz ([LT]).
In this paper we prove, by contrast,
that most quasi-simple classical groups of small rank are
not $(2,3,7)$-generated.

Note that, whenever the centre of a quasi-simple group is non-trivial,
the problem of deciding whether the corresponding
simple central quotient is Hurwitz requires  further
analysis. This is why a general treatment of projective classical
groups is not envisaged in this paper.
Our first result, however,
deals with projective groups.
Namely, in Theorem 1 {\it ii})
we determine the (2,3,7)-generated subgroups
of $\PSL(3,\overline F),$ where $\overline F$
is the algebraic closure
of an arbitrary field $F.$ This
extends the results of [Mac] and [Coh], which
deal with the case $F=F_p,$
the Galois field of order $p .$
Our approach essentially amounts to bringing
the infinite triangle groups $\Delta(2,3,k),\ k\geq 7,$
into the picture. While referring to [Mag] II.5
for details and proofs, we recall here that the group
$\PSL(2,\R)$ over the reals acts naturally
on the complex upper half-plane
${\cal U}= \{\alpha\in \C \mid {\rm Im}\thinspace \alpha > 0\},$
preserving the hyperbolic metric which
turns ${\cal U}$ into a model of the Poincar\'e plane.
In this realization, the subgroup of $\PSL(2,\R)$ generated
by rotations of angles
${2\pi\over 2},\ {2\pi\over 3},\ {2\pi\over k}$
around the vertices of a hyperbolic triangle having
angles of size ${\pi\over 2},\ {\pi\over 3},\ {\pi\over k}$
is isomorphic to $\Delta(2,3,k).$
\medskip
Throughout the paper $F$ denotes a
field of characteristic $p\geq 0$
and $\overline F $ denotes the algebraic closure of $F.$
Moreover, we observe explicitly that, for any
prime $k\geq 7,$
a $(2,3,k)$-generated group $G$ is a perfect group.
It follows that, for any representation
$f: G\to \GL(n,F),$ $f(G)$ is a subgroup
of $\SL(n,F).$ We start with the following:

\medskip
{\bf Theorem 1}\enskip {\it Let $k\geq 7$ be a prime, and
$\epsilon \in \overline F$  be a primitive
$k^{\rm th}$ root of unity} ($\epsilon=1$ {\it if}
$p=k$).
\smallskip
{\it i}) {\it Up to isomorphism, a $(2,3,k)$-generated subgroup
$H$ of $\PSL(2,\overline F)$ is the projective image
of the group
$$\langle \pmatrix {0&-1\cr 1&0\cr}\ , \
\pmatrix {\epsilon&1\cr 0&\epsilon^{-1}\cr}\rangle .$$
If $p=0,\ H$ is isomorphic to $\Delta (2,3,k).$
If $p>0,$ $H$ is a subgroup
of $\PSL(2,p^{k-1}).$ In particular $H$ is finite and,
for $k=7,$ by $[Mac]$
$H\simeq {\rm PSL}(2,p)$ if $p\equiv 0, \pm1\ ({\rm mod} \ 7),$
$H\simeq {\rm PSL}(2,p^3)$ if $p\not\equiv 0,\pm 1\
({\rm mod}\ 7).$

\medskip
{\it ii}) {\it If $p\not= 2,$
the image of $H$ under the orthogonal embedding
$\varphi: \PSL(2,\overline F)\to {\rm SO}(3,\overline F)$ defined by
$$\pm\pmatrix{a&b\cr c&d\cr}\mapsto
\pmatrix{a^2 &2ab&b^2\cr
              ac&ad+bc&bd\cr
              c^2&2cd&d^2 \cr}$$
is irreducible for all primes $k\geq 7 .$
Assuming $k=7,$ let $K$ be an irreducible
$(2,3,7)$-generated subgroup of $\SL(3,\overline F).$ Then either
$K$ is isomorphic to $\PSL(2,7),$ or $p\not=2 $ and $K$ is isomorphic
to $\varphi (H).$
The irreducible $(2,3,7)$-generated subgroups
of $\PSL(3,\overline F)$ are the projective images of
those of $\SL(3,\overline F) .$ }
\medskip
{\it iii}) {\it If $p=0$ or $p\not\equiv 0,\pm 1\ ({\rm mod} \ k),$
the group $H$ embeds irreducibly into $\SL(4,\overline F),$
for all primes $k\geq 7 .$ The embedding is orthogonal if $p\not=2 .$

\medskip
{\it iv}) {\it If $p\not= 3,$ the group $H$
embeds irreducibly into a symplectic subgroup of $\PSL(4, \overline
F),$
for all primes $k\geq 7 .$}}
}
\medskip
We have included items
{\it iii}) and {\it iv}) in order to
emphasize the occurrence of orthogonal or symplectic representations of
Hurwitz groups in dimension 4 (cfr. Theorem 2).
We are aware that they are special cases of much more general
embeddings in tensor (or twisted tensor) product spaces.
\medskip
Our subsequent results only concern linear groups, and
are based on the application of a formula due
to L.L. Scott (see Section 2) to
appropriate representations of a group,
and the possible
contradictions that one may obtain assuming that the
group is $(2,3,7)$-generated.
Scott himself observed
that his formula could
be used for proving that certain linear groups
are not $(2,3,7)$-generated (he considered,
as examples, the groups ${\rm SL}(6,3)$ and ${\rm SL}(9,3)$).
A more systematic application of Scott's formula
in the context of Hurwitz groups appears in
[TV], where it is
shown that the groups $\SL(4,q)$ and $\SU(4,q^2),\ q$
odd, are not Hurwitz.
Although these techniques
work in principle over an arbitrary field,
we mainly concentrate on
classical groups over finite fields (Theorems 2,3,4),
%
and on the field of rational numbers (Theorem 5). We prove the
% and on subfields of $\C$ (Theorem 5). We prove the
following \lq\lq negative\rq\rq\ results.
\medskip
{\bf Theorem 2} \enskip {\it
Let $H$ denote an absolutely irreducible subgroup of
${\rm SL}(n,F),\ n\in\{4,5,6,7,10\}.$
Assume that $H$ is not contained in
an orthogonal group if
$p\not=2 ,$ and that $H$ is not
contained in a symplectic group if $p=2.$
If $n=6$ and $p=2,$ assume further
that $F$ is finite and $H=\SL(6,F)$ or $SU(6,F).$
Then $H$ is not $(2,3,7)$-generated.
In particular, if $n\in \{4,5,6,7,10\},$ then:}
\smallskip
{\it i}) {\it the groups\enskip
${\rm SL}(n,q),\ {\rm Sp}(n,q),\ {\rm SU}(n,q^2)$
are not Hurwitz, with the only possible exception of

\quad ${\rm Sp}(n,2^t)$ ;}
\smallskip
{\it ii}) {\it every complex irreducible character
of degree $n$ of a $(2,3,7)$-generated group, is real.
}
\medskip

Both Theorem 1 and Theorem 2 may be rephrased as statements about
representations of a (2,3,7)-generated group $G.$
Clearly, if $p\not= 2,$
$G$ has no absolutely irreducible representation of degree 2,
since the only involution
of $\SL(2,\overline F)$ is $-I.$
In particular $G$ has no complex irreducible representation
of degree 2.
It follows from Theorem 1
that, if $G$ has  a complex irreducible
representation of degree 3, then it has a quotient
isomorphic to 
$\PSL(2,7).$
In particular,
${\rm PSL}(2,7)$
does have complex irreducible representations
of degree 3. The remaining irreducible representations
of ${\rm PSL}(2,7)$ over $\C$ have degrees
6,7 and 8, with real characters. Similarly, the next Hurwitz
group ${\rm SL}(2,8)$
has non-real irreducible characters of
degree 9 and real characters of degree 7 and 8
(e.g. cfr. [Isa], p. 289).
Theorem 2 states that, under the given
assumptions on $F$ and for the stated values of $n,$ every irreducible
representation $f: G\to \SL(n,\overline F)$ is
orthogonal (symplectic if $p=2$).
It is worth to recall
that, in the natural characteristic, the
groups of type $G_2$ do have orthogonal representations
of degree 7 if $p\not= 2 ,$ and symplectic representations
of degree 6 if $p=2;$ whereas the groups of type $^2G_2$
and $^3D_4$ have orthogonal representations
of degree 7 and 8, respectively. Finally, note that
Holt, Plesken and Souvignier [HPS] have constructed
a 7-dimensional complex representation of $\Delta(2,3,7),$
whose kernel is generated by the element
$[X,Y]^{11}.$

\bigskip
{\bf Theorem 3}\enskip {\it Let $H$ be one of the groups:
\smallskip
a)\enskip
$\SL(n,q),\ q\not=3^6$ and, if
$q\equiv 1\ ({\rm mod}\ 7),$ $q\not=p,p^3;$
\enskip
$\Sp(n,q),\ q$ odd;\enskip
$\SU(n,q^2),\ q\not=3^3;$

\quad $n\in\{8,9,11\};$
\medskip
b)\enskip $\SL(12,q),\ q\not\equiv 1\ ({\rm mod}\ 7);$\enskip
$\Sp(12,q),\ q$ odd;\enskip $\SU(12,q^2),\ q\not\equiv -1\ ({\rm mod}\ 7).$
\medskip
Then $H$ is not Hurwitz .}
\bigskip
{\bf Theorem 4} \enskip {\it Assume
$13\leq n\leq 19,$ and let $H$ be
one of the groups
$\SL(n,q),\ q\not\equiv 1\ ({\rm mod}\ 3)\ ;$
\enskip $\Sp(n,q),\ q\ odd;$ $\SU(n,q^2),\
q\not\equiv -1\ ({\rm mod}\ 3)\ .$
Then $H$ is not Hurwitz, except possibly for the following cases}:
\smallskip
{\it a})\enskip
${\rm SL}(16,q),\ q\equiv 8\ ({\rm mod}\  21),\  {\rm SL}(16, 3^{6h}),
\enskip {\rm SU}(16,q^2),\ q\equiv -8\ ({\rm mod}\  21),\
{\rm SU}(16,3^{6h+3}).$
\smallskip
{\it b})\enskip ${\rm SL}(18,q),\ q\ odd\ \equiv 8\ ({\rm mod}\  21),\
{\rm SL}(18,3^{6h});$
${\rm SU}(18,q^2),\ q\ odd\ \equiv -8\ ({\rm mod}\  21),\
{\rm SU}(18,3^{6h+3}).$

\medskip
{\it c})\enskip ${\rm SL}(n,q),\ n\in \{15,17,19\},\
q\equiv 2,8,-10\ ({\rm mod}\  21)\ {or} \
q\in \{3^{6h},3^{6h+2},3^{6h+4}\}.$
\smallskip
{\it d})\enskip
${\rm SU}(n,q^2)\ n\in
\{15,17,19\},\ q\equiv 10,-2,-8\ ({\rm mod}\  21)\ or\
q\in \{3^{6h+1},3^{6h+3},3^{6h+5}\}.$
\smallskip
{\it e})\enskip
${\rm SL}(n,2^t),\ n\in\{14,18\},\ 2^t\equiv 2,8,-10\ ({\rm mod}\  21).$
\bigskip
{\bf Theorem 5}\enskip {\it
% Let $F$ be a subfield of $\C ,$
% and suppose that
% $\vert F(\theta):F\vert =12,$
% where $\theta\in \C$ is a primitive
% $21^{st}$ root of $1.$ Let $H$ denote an
% absolutely irreducible subgroup of
% ${\rm SL}(n,F),\ n\leq 19$ or $n=22 .$
% If $H$ is not contained in an orthogonal group, then
% $H$ is not $(2,3,7)$-generated.
% In particular, for these values of $n,$ the groups
% $\SL(n,\Z),\ \SL(n,\Q),\ \SL(n,\Q(\lambda )),$ where
% $\lambda$ is an indeterminate, are
% not $(2,3,7)$-generated.}
Let $\Q$ be the field of rational numbers, and
let $H$ denote an absolutely irreducible subgroup of
${\rm SL}(n,\Q),\ n\leq 19$ or $n=22 .$
If $H$ is not contained in an orthogonal group, then
$H$ is not $(2,3,7)$-generated.
In particular, for these values of $n,$ the group
$\SL(n,\Z)$ is not $(2,3,7)$-generated.}

\medskip
We are grateful to the Referee for the accurate reading of the
manuscript.


\vskip20pt
{\bf 2.\enskip Scott' s formula}
\par\nobreak\bigskip

We denote shortly by $M$ the algebra ${\rm Mat}(n,F)$
of $n\times n$ matrices over $F,$ and view
$M$ as acting on the left on the vector
space $V=F^n$ of column vectors. Given a representation
$f: H\to {\rm GL}(n,F)$ of a group $H$ over $F,$ for any
subset $X$ of $H$ we define
$d^X_V$ as the dimension of the subspace of $V$ fixed
pointwise by $f(X).$ In a similar way we define $\hat d^X_V,$
with respect to the dual representation
$\hat f: H\to {\rm GL}(n,F)$
given by $\hat f(h):=(f(h^{-1}))^t.$
The following formula, which is a special
case of a result of L.L.Scott ([Sc], Theorem 1), will
be crucial.

\medskip
{\bf Theorem 2.1} \enskip {\it Let
$f: H\to {\rm GL}(n,F)$ be a representation of
$H=\langle x,y\rangle .$
Then :}
$$d^{x}_V+d^{y}_V+d^{xy}_V\leq {\rm dim}\ V+ d^H_V+\hat d^H_V . $$

\bigskip
The above formula was conceived by Scott
as a generalization of the well-known {\it genus formula}
$\left[{n\over 2}\right]+2\left[{n\over 3}\right]+
6\left[{n\over 7}\right] \geq 2n-2$
for the index $n$ of a subgroup $S$ of a
$(2,3,7)$-generated group $G.$
Indeed, the genus formula is a consequence of Theorem 2.1.
For, let $G=\langle x,y\rangle$ with $x^2=y^3=(xy)^7=1,$
and consider the transitive permutation representation
$f:G\to \GL(n,\C)$ given by the action on the cosets of $S.$
Since the multiplicity in $f$
of the trivial representation is 1, $d^G_V=\hat d^G_V =1.$
Scott' s formula then gives:
$d^x_V\thinspace +d^y_V\thinspace +d^{xy}_V\leq n+2 .$
The genus formula follows easily. Indeed,
for every  $g\in G$ of prime order  $k ,$ whose
cyclic structure admits $a$ cycles of length $k ,$
$n-d^g_V=(k -1)a\leq (k -1)\left[{n\over k}\right].$
\medskip
Scott's formula will be applied essentially to the following
representations of a subgroup $H$ of $\SL(n,F),$
with generating pair $\{x,y\}.$
\smallskip
1) The representation of $H$ by conjugation on $M$
(also called the 'adjoint' representation of $H$) given by
$f(h):=  m \mapsto hmh^{-1}$  ($h \in H,\ m \in M$).
In this case, the fixed-points subspace of $M$ is the centralizer
of $H$ in $M,$ and therefore, if $H$ is supposed to be absolutely
irreducible, by Schur's Lemma it consists of scalar matrices. 
It follows $d_M^H = \hat d^H_M= 1,$ hence
Scott's formula reads

$$ (*)\qquad     d_M^x+d_M^y+d_M^{xy}\leq n^2 + 2.$$

Failing of the above inequality
means that $H$ is reducible over $\overline F$(and to some extent
our problem is reduced to linear groups of smaller
degree). Dealing with $(*)$ requires detailed information on
minimal values affordable by $d_M^x,\ d_M^y$ and $d_M^{xy}.$
These values will be worked out in Section 3.
\smallskip
2) The representation of $H$ on the symmetric square $S$ of
$\overline F^{n}.$ If $H$ is absolutely irreducible, then Scott's
formula takes the shape

$$(\ast\ast )\qquad d_S^x+d_S^y+d_S^{xy}\leq  {\rm dim}\thinspace S\quad ,
\quad {\rm or}\qquad  d_M^x+d_M^y+d_M^{xy}\leq  {\rm dim}\thinspace S + 2 $$
(the latter occurring only when $H$ is orthogonal if
$p\not=2$, symplectic if $p=2$). This will be shown in Section 4.


\vskip20pt
{\bf 3.\enskip The defect ${\rm def}_{\overline M}$}
\par\nobreak\bigskip
We first recall a well-known formula, due to Frobenius
(e.g. cfr. [Ja], p. 207).
Namely, let $B\in M:={\rm Mat}(n,F)$ and let
$n_1\leq\dots \leq n_s$ be the degrees of
the non-trivial invariant factors of $\lambda I-B$
(also called the similarity invariants of $B$).
The dimension
$d^B_M$ over $F$ of the centralizer of $B$ in
$M$ is given by the formula
$$d^B_M\ =\ \sum_{j=1}^s(2s-2j+1)
\thinspace n_j\ =\ (2s+1)n-2\sum_{j=1}^sjn_j.$$
In particular\enskip
$d^B_M\ \geq\ n+s^2-s $, since
$$\sum_{j=1}^sjn_j\ =\
\sum_{j=1}^sn_j+\sum_{j=2}^sn_j+\dots +
(n_{s-1}+n_s)+n_s\ \leq$$
$$\leq\  n+(n-1)+\dots +(n-s+2)+(n-s+1)
\leq\ sn-{s(s-1)\over 2} .$$

Moreover, considering the algebraic closure $\overline F$
and denoting by
$d^B_{\overline M}$ the dimension over
$\overline F$ of the centralizer of $B$ in
$\overline M={\rm Mat}(n,\overline F),$
clearly $d^B_M=d^B_{\overline M},$ since
the similarity invariants of $B$
remain unchanged in $\overline M.$
Now, for a fixed prime $k,$ define
$d^{k}_{\overline M}$ as the minimal
dimension of the centralizer of an
element of order $k$ in $\SL(n,\overline F),$ i.e.
$$d^{k}_{\overline M}\ :=\ \min\ \big\{d^A_{\overline M}\ \mid\
A\in {\rm SL}(n,\overline F)\ ,\ A^k=I.\big\}$$

\bigskip
{\bf Lemma 3.1}\enskip {\it Write $n=k\ell+r,\ 0\leq r<k .$

\nobreak{\medskip}
{\it i}) If ${k}=2$ and $n\equiv 2\ ({\rm mod}\ 4),$ assume
$p= 2.$ Let $B\in \SL(n,\overline F)$ have order $k,$
and let

\quad $n_1\leq\dots \leq n_s$ be the degrees
the non-trivial invariant factors of $\lambda I-B.$
Then $d^B_{\overline M}=d^k_{\overline M}$ iff

\quad $n_i={k}$ for every $i>1, $ and \enskip
$d^k_{\overline M}\ =\ (2\ell+1)r+{k}\ell^2 .$

\medskip
{\it ii}) If ${k}=2,$ $n\equiv 2\ ({\rm mod}\ 4)$ and
$p\not= 2,$ then\enskip
$d^2_{\overline M}\ =\ 2(\ell^2+1).$
}
\medskip
{\it Proof}
\par\nobreak\medskip
Let $A\in \GL(n,\overline F)$ be a matrix of order $k,$
such that $d^A_{\overline M}\leq d^C_{\overline M}$ for all
$C$ of order $k$ in $\GL(n,\overline F),$ and
denote by $n_1\leq\dots \leq n_s$ the degrees of
the non-trivial invariant
factors of $\lambda I-A.$ Suppose
that $n_i<{k}$ for some $i>1$ and
let $m$ be the maximum of such $i$'s.
If $n_1>1,$ there exists
$C\in \GL(n,\overline F)$ of order $k$
such that $\lambda I-C$ has $s$ invariant factors $\not= 1$ of
degrees $\overline n_1=n_1-1,\
\overline n_m=n_m+1,\ \overline n_i=n_i ,\ i\not=1,m.$
In this case we get :
$$d^{C}_{\overline M}\ =\
\sum_{j=1}^{s}(2s-2j+1)\overline n_j\ =\
2(1-m)+\sum_{j=1}^{s}(2s-2j+1)n_j\ <\ d^{A}_{\overline M},$$
a contradiction. On the other hand, if $n_1=1,$ there exists
$C\in \GL(n,\overline F)$ of order $k$
such that $\lambda I- C$ has $s-1$
invariant factors $\not= 1$
with degrees $\overline n_i=n_{i+1}$
for $i\not= m-1,$ $\overline n_{m-1}=n_{m}+1.$
A similar calculation gives again the contradiction
$d^{C}_{\overline M}\ <\ d^{A}_{\overline M}.$
It follows $n_i=k$ for all $i>1$ and $n_1=r$ if $r>0,$
$n_1=k$ if $r=0.$ Hence
$d^A_{\overline M}\ =\ (2\ell+1)r+{k}\ell^2\ \leq d^k_{\overline M}.$
\smallskip
{\it i}) We need to show that there exists
$B\in \SL(n,\overline F)$ such that
$B^k=I$ and $d^B_{\overline M}=d^A_{\overline M}.$
If $k=p$ or $r=0$ we may choose $B=A.$ Otherwise, we write
$r=2h+h_0\ (h_0\leq 1)$ and choose $B$ such that
$\lambda I- B$ has non-trivial
invariant factors
$$d_1(\lambda) = (\lambda \mp 1)^{h_0}\thinspace
\prod_{j=1}^h(\lambda -\alpha^j)
(\lambda -\alpha^{-j})\quad ,\quad
d_i(\lambda)=\lambda^k-1\ ,\ 2\leq i\leq \ell +1 ,$$
where $\alpha$ is a primitive $k$-th root of 1 in $\overline F .$
We conclude $(2\ell+1)r+{k}\ell^2\ =\  d^k_{\overline M}\ .$
Finally, assume $d^B_{\overline M}\ =\ d^k_{\overline M}\ .$
If $n_i<k$ for some $i>1 ,$ by the above argument there exists
$C\in \GL(n,\overline F)$ such that
$d^C_{\overline M}\ <\ \ d^B_{\overline M}\ \leq\ d^C_{\overline M}\ ,$
a contradiction.
\medskip
{\it ii}) Let $B\in {\rm SL}(n,F)$ be an
involution such that $d^B_M=d^2_{\overline M}\ .$ It is easily seen
that the non-trivial invariant factors of $\lambda I-B$ must have degrees
$n_1=n_2=1,\ n_3=\dots =n_s=2,$
and the formula can be computed directly. \qed

\bigskip
%%% FROM PROBLEM:
%>From the formulae given in Lemma 3.1,
From the formulae given in Lemma 3.1,
we obtain Table 1.
The last two columns list the values of the {\it defect} of
${\overline M},$ defined by
$${\rm def}_{\overline M}:=n^2+2-(d^2_{\overline M}+
d^3_{\overline M}+d^7_{\overline M}).$$
\vskip20pt
{\bf 4.\enskip The symmetric square action}
\par\nobreak\bigskip
Let $\overline V=\overline F^n,$ with basis
$\{ v_i\mid 1\leq i\leq n\}.$
We denote by $S$ the symmetric square of
$\overline V,$ i.e. the subspace
of $\overline V\otimes\overline V$
generated by the set:
$$\{ v_i\otimes v_i\mid 1\leq i\leq n\}\
\cup\  \{v_i\otimes v_j+v_j\otimes v_i\mid  1\leq i<j\leq n\} ,$$
and denote by $E$ the exterior square, i.e. the subspace
generated by the set:
$$\{v_i\otimes v_j-v_j\otimes v_i\mid  1\leq i<j\leq n\}.$$
Clearly ${\rm dim}\ (S)=n(n+1)/2$ and
${\rm dim}\ (E)=n(n-1)/2.$
If $p\not=2,$
$\overline V\otimes \overline V\ =\ S\dot + E$
where each of the summands is invariant under the
action $g(u\otimes v):= g(u)\otimes g(v)$ for every
$g\in \GL(n,\overline F)$ (diagonal action).
\bigskip
{\bf Lemma 4.1}\enskip {\it Let $H$ be an
irreducible subgroup of
${\rm GL}(n,\overline F)$ which fixes
a non-zero vector of $S.$ Then:
\smallskip
{\it i}) if $p\not =2,$ $H$ is contained in an orthogonal
group;

\smallskip
{\it ii}) if $p=2,$ $H$ is contained in  a symplectic
group;
\smallskip
{\it iii}) the identical representation of $H$
is equivalent to the dual one.
In particular $d^H_S=\hat d^H_S=1.$
}
\vfill\eject
\centerline{\bf Table 1: the defect of ${\overline M}$}
\par\nobreak\bigskip
{\sevenrm
$$\vbox{\settabs
\+11111111&11111111111111&11111111111111&11111111111111&111111111111111111&
11111111111111&11111111111111\cr
\+$n$&$d^2_{\overline M}$& $d^2_{\overline M}$& $d^3_{\overline M}$&
$d^7_{\overline M}$& ${\rm def}_{\overline M}$ &
${\rm def}_{\overline M}$\cr
\bigskip
\hrule
\bigskip
\+&$p\not=2$& $p=2$&
& &$p\not=2$ &$p=2$\cr
\bigskip
\hrule
\medskip
\+4&\quad 8&\quad 8&\quad 6&\quad  4&\quad 0
&\quad 0\cr
\bigskip
\hrule
\medskip
\+5&\quad 13&\quad 13&\quad 9&\quad 5&\quad 0
&\quad 0\cr
\bigskip
\hrule
\medskip
\+6&\quad 20&\quad 18&\quad 12&\quad 6&\quad 0
&\quad 2\cr
\bigskip
\hrule
\medskip
\+7&\quad 25&\quad 25&\quad 17&\quad 7&\quad 2
&\quad 2\cr
\bigskip
\hrule
\medskip
\+8&\quad 32&\quad 32&\quad 22&\quad 10&\quad 2
&\quad 2\cr
\bigskip
\hrule
\medskip
\+9&\quad 41&\quad 41&\quad 27&\quad 13&\quad 2
&\quad 2\cr
\bigskip
\hrule
\medskip
\+10&\quad 52&\quad 50&\quad 34&\quad 16&\quad 0
&\quad 2\cr
\bigskip
\hrule
\medskip
\+11&\quad 61&\quad 61&\quad 41&\quad 19&\quad 2
&\quad 2\cr
\bigskip
\hrule
\medskip
\+12&\quad 72&\quad 72&\quad 48&\quad 22&\quad 4
&\quad 4\cr
\bigskip
\hrule
\medskip
\+13&\quad 85&\quad 85&\quad 57&\quad 25&\quad 4
&\quad 4\cr
\bigskip
\hrule
\medskip
\+14&\quad 100&\quad 98&\quad 66&\quad 28&\quad 4
&\quad 6\cr
\bigskip
\hrule
\medskip
\+15&\quad 113&\quad 113&\quad 75&\quad 33&\quad 6
&\quad 6\cr
\bigskip
\hrule
\medskip
\+16&\quad 128&\quad 128&\quad 86&\quad 38&\quad 6
&\quad 6\cr
\bigskip
\hrule
\medskip
\+17&\quad 145&\quad 145&\quad 97&\quad 43&\quad 6
&\quad 6\cr
\bigskip
\hrule
\medskip
\+18&\quad 164&\quad 162&\quad 108&\quad 48&\quad 6
&\quad 8\cr
\bigskip
\hrule
\medskip
\+19&\quad 181&\quad 181&\quad 121&\quad 53&\quad 8
&\quad 8\cr
\bigskip
\hrule
\medskip
\+20&\quad 200&\quad 200&\quad 134&\quad 58&\quad
10&\quad 10\cr
\bigskip
\hrule
\medskip
\+21&\quad 221&\quad 221&\quad 147&\quad 63&\quad
12&\quad 12\cr
\bigskip
\hrule
\medskip
\+22&\quad 244&\quad 242&\quad 162&\quad 70&\quad
10&\quad 12\cr
\medskip
\hrule
}$$}
\vfill\eject

{\it Proof} ({\it of Lemma} 4.1)
\par\nobreak\medskip
{\it i}) and {\it ii})\enskip Let $\{v_1,\dots ,v_n\}$ be the standard basis of
$\overline V=\overline F^n.$ The map
$v_i\otimes v_j\mapsto v_i\thinspace v_j^t$ extends to an
isomorphism $\overline V\otimes\overline V\to
M(n,\overline F)$ of $\overline F$-vector spaces.
Via this isomorphism, the
elements of $S$ are identified with the symmetric matrices.
Clearly, for every $g\in {\rm GL}(n,\overline F)$
and every $v_i,v_j\in V,$
$$g v_i\otimes g v_j
 \ \mapsto\ g\thinspace  v_iv_j^t\thinspace  g^t.$$
Thereby, the diagonal action of ${\rm GL}(n,\overline F)$ on
$\overline V\otimes\overline  V$ is identified with
the action of ${\rm GL}(n,\overline F)$ on $M(n,\overline F)$ given by
$g\mapsto \gamma ,$ where $\gamma$ is the linear transformation
$B\mapsto gBg^t,$ for every $B\in M(n,\overline F).$
As $H$ fixes a non-zero vector of $S,$
there exists a non-zero symmetric matrix $B$
such that $hBh^t=B,$ for every $h\in H.$
Note that $B\overline V$ is a non-zero $H$-invariant
subspace of $\overline V .$
Hence, as $H$ is irreducible, 
$B$ is non-degenerate.
Since
${(h^t)}^tBh^t=B$ implies ${(h^{-1})}^tB^{-1}h^{-1}=B^{-1}$
for every $h\in H,$
it follows that, if $p\not=2,$ $H$ is orthogonal.
Thus {\it i}) holds.
If $p=2,$ {\it ii}) will be proved if we show that
$v^tBv=0$ for every $v\in \overline V.$ To this purpose let
$\overline V_0:= \{v\in \overline V\mid v^tBv=0\}.$ From
$$(w+\alpha u)^tB(w+\alpha  u)=w^tBw+2\alpha  w^tBu+u^tBu=
w^tBw+u^tBu\quad {\rm for} \ \alpha\in \overline F ,$$
it follows that $\overline V_0$ is a subspace of
$\overline V.$ As it is
$H$-invariant and $\not=\{0\},$ we obtain $\overline V_0=\overline V.$

\smallskip
{\it iii})\enskip As $B^{-1}hB=(h^{-1})^t$ for every $h\in H,$
the identical representation of $H$ is equivalent to the dual one.
It follows immediately that $d^H_S=\hat d^H_S=1,$ since absolutely
irreducible groups can fix only one form, up to a scalar multiple
(e.g. see [KL], Lemma 2.10.3).
\bigskip
For a matrix
$A\in M(n,\overline F), $ define $d^A_S$ as
the dimension of the subspace of the
symmetric square of $\overline V\otimes \overline V$
fixed pointwise
by $A\otimes A.$
Define $d^A_E$ in a similar way.
The following result is straightforward.

\bigskip
{\bf Lemma 4.2}\enskip {\it Let $g\in {\rm GL}(n,F)$ be an element
of finite order coprime to $p.$ If $\nu$ is an eigenvalue of $g$,
let $m_\nu$ denote
the multiplicity of $\nu .$ Then:
$$d^g_S={m_1(m_1+1)+m_{-1}(m_{-1}+1)\over 2}+
\sum m_{\nu}m_{\nu^{-1}}$$
where the summation runs over all pairs
$\nu,\nu^{-1}$ of eigenvalues of $g$ in $\overline F,$
with $\nu \not=\nu^{-1}.$} \qed
\bigskip
In the sequel, for every $r\in \N$ we denote by $\pi_r$ the
companion matrix of $\lambda^r-1$ and
by $J_r$ the Jordan unipotent block of size $r.$
In particular, if $r=p ,$\enskip $\pi_p$ and $J_p$ are conjugate, and
in this case we will usually write $\pi_p.$
\medskip
{\bf Lemma 4.3}
\par\nobreak{\smallskip}
{\it i})\enskip {\it Assume $p=2$ and let
$g\sim \pmatrix {\underbrace {1,\dots ,1}_{k_1}\ ,\
\underbrace {\pi_2,\dots ,\pi_2}_{k_2}
}\in \GL(n,F).$  Then}
$$d^g_S={k_1^2+k_1\over 2}+k_2^2+k_2+k_1 k_2\ .$$
\smallskip
{\it ii})\enskip {\it Assume $p=3$ and let
$g\sim \pmatrix {\underbrace {1,\dots ,1}_{k_1}\ ,\
\underbrace {J_2,\dots ,J_2}_{k_2}\ ,\
\underbrace {\pi_3,\dots ,\pi_3}_{k_3}
}\in \GL(n,F).$
Then}
$$d^g_S={k_1^2+k_1\over 2}+k_2^2+
{3k_3^2+k_3\over 2}+k_1k_2+k_1k_3+2k_2k_3\ .$$
\smallskip
{\it ii})\enskip {\it Assume $p=7$ and let $g\in \GL(n,F),$
$$g\sim \pmatrix{
\underbrace {1,\dots ,1}_{k_1}\ ,\
\underbrace {J_2,\dots ,J_2}_{k_2}\ ,\
\underbrace {J_3,\dots ,J_3}_{k_3}\ ,\
\underbrace {J_4,\dots ,J_4}_{k_4}\ ,\
\underbrace {J_5,\dots ,J_5}_{k_5}\ ,\
\underbrace {J_6,\dots ,J_6}_{k_6}\ ,\
\underbrace {\pi_7,\dots ,\pi_7}_{k_7}
}.$$
Then} $d^g_S=$
$${k_1^2+k_1\over 2}+k_2^2+{3k_3^2+k_3\over 2}+2k_4^2+
{5k_5^2+k_5\over 2}+3k_6^2+
{7k_7^2+k_7\over 2}+$$
$$k_1\sum_{j=2}^7 k_j+
2k_2\sum_{j=3}^7 k_j+
3k_3\sum_{j=4}^7 k_j+
4k_4\sum_{j=5}^7 k_j+
5k_5\sum_{j=6}^7 k_j+
6k_6k_7.$$
\medskip
{\it Proof}
\par\nobreak\medskip
All these formulae can be computed directly. For $p$ odd, they can also
be obtained from Lemma 1.2.3 in [GM], where a corresponding formula is
given for $d_S^g,$ $g$ a unipotent in $\GL(n,F).$ \qed

\vskip20pt
{\bf 5.\enskip Constraints on (2,3,7)-generating triples}

\par\nobreak\bigskip
Our aim, in this section,  is to provide an explanation for the data
collected in the Tables 2, 3, 4, $4^\prime$ and 5,
which will play a crucial role in the proof of our results.
For this purpose we assume that $x,y,z=xy$ are elements
of order 2, 3 and 7 respectively, which
generate an absolutely irreducible
subgroup of $\SL(n,F).$ It follows that these
generators satisfy condition $(\ast)$ of Section 2.
In particular Tables 2, 3, 4 and $4^\prime$
describe the conjugacy classes to which $x,y$ and $z$
are constrained to belong in order to satisfy the bounds:
$$d^x_M\leq d^2_{\overline M}+def_{\overline M}\quad ,\quad
d^y_M\leq d^3_{\overline M}+def_{\overline M}\quad ,\quad
d^z_M\leq d^7_{\overline M}+def_{\overline M}$$
which follow from $(\ast).$
Note that explicit values for the right-hand sides
are deducible from Table 1.

\bigskip
{\bf A)\enskip Constraints on similarity invariants}

\nobreak{\bigskip}
In this subsection, as a consequence of the above mentioned bounds,
we obtain constraints on the number and the degrees of
the similarity invariants of $x,y,z.$

\bigskip
{\bf Lemma 5.1}\enskip {\it Let $g\in \GL(n,F)$ be an involution
with $s$ similarity invariants,
and set
$\delta:= d^2_{\overline M}+def_{\overline M}.$
Then $d^g_M\leq \delta$ iff
$${n\over 2}\leq s
\leq {n+\sqrt{2\delta -n^2}\over 2}.$$}
{\it Proof}

\nobreak\medskip
Clearly $g$ has has $2s-n$ invariants
of degree 1 and  $n-s$ of degree 2. Hence, by the Frobenius formula,
$d^g_M=2s^2-2ns+n^2 .$ Our claim follows immediately. \qed

\medskip
%%% FROM PROBLEM:
%>From the previous
From the previous
Lemma we obtain Table 2, which lists representatives
of all the conjugacy classes to which
$x$ may belong, and the
corresponding values of $d^x_S,$ for
every $n\leq 22.$
\bigskip

{\bf Lemma 5.2}\enskip {\it Let $g\in \GL(n,F)$ have order $3,$
$s$ be the number of its similarity invariants,
and denote by $k_1$ be the multiplicity
of the invariant of degree $1.$
Setting $\delta:=d^3_{\overline M}+def_{\overline M},$
$d^g_M\leq \delta$ iff
the following conditions are
satisfied:}
$$
k_1\ \leq\ \min \Big\{{3s-n\over 2}\thinspace ,
\thinspace \sqrt{6\delta-2n^2\over 3}\Big\}\quad ;\quad
{n\over 3}\ \leq s\
\leq \thinspace{3k_1+2n+\sqrt{6\delta-2n^2-3{k_1}^2}\over 6}.$$


\medskip
{\it Proof}

\nobreak\medskip
It is easy to see that $g$ has $k_2=3s-2k_1-n$ invariants
of degree 2 and $k_3=-2s+k_1+n$ of degree 3.
Hence, by the Frobenius formula,
$d^g_M=6s^2-6sk_1-4sn+2k_1n+2k_1^2+n^2.$
Our claim follows, noting that
$k_3\leq {n-k_1\over 3}$ implies
$k_1\ \leq\ {3s-n\over 2}.$
\qed
\bigskip

%%% FROM PROBLEM:
%>From Lemma 5.2 we obtain Table 3, which lists representatives
From Lemma 5.2 we obtain Table 3, which lists representatives
of all the conjugacy classes to which
$y$ may belong, and the corresponding
values of $d^y_S,$ for every
$n\leq 22.$ In this table we set $d_2:=\pmatrix {0&-1\cr 1&-1}.$
If $p\not= 3,\thinspace$ $d_2$ is conjugate to
$\pmatrix {\omega , \omega^{-1}},$ where
$\omega$ is a primitive cubic root of 1.
If $p=3,\thinspace$ $d_2$ is conjugate to
$J_2 .$ If $k_\ell >0$ set $d^\ell(\lambda)$ for the
similarity invariant
of degree $\ell $ of $y$ and note that,
by the condition ${\rm det}\ y=1:$
\medskip
{\sevenrm
$\cases{k_1\not\equiv 0\ ({\rm mod}\ 3)\cr
k_2\equiv 0\ ({\rm mod}\ 3)\cr}\enskip \Rightarrow\enskip
d^1(\lambda) = \lambda-1\qquad ;\qquad\qquad
k_2\not\equiv 0\ ({\rm mod}\ 3)\enskip \Rightarrow\enskip
d^1(\lambda) = \lambda-\omega^{\pm 1};$
\smallskip
$k_1\not\equiv 0\ ({\rm mod}\ 3)
\enskip \Rightarrow\enskip
d^2(\lambda) = (\lambda - 1)(\lambda-\omega^{\pm 1})\ ;\
\cases{ k_1\equiv 0\ ({\rm mod}\ 3)\cr
k_2\not\equiv 0\ ({\rm mod}\ 3)}\enskip \Rightarrow\enskip
d^2(\lambda) = (\lambda - \omega)(\lambda-\omega^{-1}).$}

\bigskip
{\bf Lemma 5.3}\enskip {\it Let $g\in \GL(n,F)$ be an element of order $7,$
with $s$ similarity invariants of
degrees $n_1\leq \dots \leq n_s\leq 7.$ Set
$t_1:=n_1-1\ ,\ t_i:=n_i-n_{i-1}\ (2\leq i\leq s-1)\ ,\
\delta:=
{1\over 2}\thinspace (d^7_{\overline M}+{\rm def}_{\overline M}-n)\ .$
For the stated values of $s,$
$d^g_M\leq d^7_{\overline M}+def_{\overline M}$
iff:

\bigskip
{\sevenrm
$s=3$ \enskip $\cases {
3t_1+t_2\ \leq\ \delta -3\cr
\cr
2t_1+t_2\geq n-9\cr
\cr
3t_1+2t_2\leq n-3
}$\qquad\qquad\qquad\qquad\qquad
$s=4$ \enskip $\cases {
6t_1+3t_2+t_3\ \leq\ \delta -6\cr
\cr
3t_1+2t_2+t_3\geq n-10\cr
\cr
4t_1+3t_2+2t_3\leq n-4
}$

\bigskip
$s=5$ \enskip $\cases {
10t_1+6t_2+3t_3+t_4\ \leq\ \delta -10\cr
\cr
4t_1+3t_2+2t_3+t_4\geq n-11\cr
\cr
5t_1+4t_2+3t_3+2t_4\leq n-5
}$\qquad
$s=6$ \enskip $\cases {
15t_1+10t_2+6t_3+3t_4+t_5\ \leq\ \delta -15\cr
\cr
5t_1+4t_2+3t_3+2t_4+t_5\geq n-12\cr
\cr
6t_1+5t_2+4t_3+3t_4+2t_5\leq n-6.
}
\ $}}
\bigskip
{\it Proof}
\par\nobreak{\medskip}
Note that $n_s=n-\sum_{i<s}n_i\ ,\ n\leq 7s\ ,\
n+s^2-s\leq d^g_M.$
For each of the stated values of $s,$ the first inequality follows
from the Frobenius formula, whereas the second
and the third one follow
from the conditions $7\geq n_s\geq n_{s-1}.$ \qed
\medskip
Using the above Lemma (and ruling out greater values
of $s$ via similar mutually incompatible inequalities)
we obtain Table 4, which lists representatives
of all the conjugacy classes to which
$z$ may belong in the modular case,
and the corresponding values of $d^z_S,$ for every
$n\leq 22.$   

\bigskip
{\bf B)\enskip Constraints on eigenvalue multiplicities}
\bigskip
An element $g\in {\rm GL}(n,F)$ of finite order $o(g)$
is called {\it modular} if $o(g)$  is a power of $p,$
{\it semisimple} if $o(g)$ is coprime to $p.$
In the semisimple case, $g$ is called
{\it real} if it is conjugate to $g^{-1}.$
Now, assume that $g$  is a semisimple element
of order 7. Denote by $\epsilon$
a primitive 7-th root of 1 in $\overline F$
and let $m_{\pm i}$ be the multiplicity of
$\epsilon^{\pm i},\ 0\leq i\leq 3,$
in the Jordan canonical form of $g$
over $\overline F.$ 
We note that $g$ is real whenever
$m_1=m_{-1},\  m_2=m_{-2},\  m_3=m_{-3}.$
Furthermore, we say that $g$ is
{\it rational} when it is conjugate to all its non-trivial
powers, i.e. when $m_1=m_2=m_{3}=m_{-1}=m_{-2}=m_{-3};$
and we say
that $g$ is {\it semirational} when it is conjugate
to $g^2,$ i.e when $m_1=m_2=m_{-3},\ m_{-1}=m_{-2}=m_{3}.$

Let us remark that, if $g$ has $s$ similarity invariants,
then obviously $m_{\pm i}\leq s$ for every $i,$ and
w.l.o.g. we may assume $m_1\geq m_{\pm i},$ for every $i>0.$
With the above notations, and assuming ${\rm det }\ g=1:$
$$d^g_M=m_0^2+\sum_{i>0}(m_i^2+m_{-i}^2) \quad ;\quad
m_1+2m_2+3m_3\equiv m_{-1}+2m_{-2}+3m_{-3}\ ({\rm mod}\ 7).$$
Using these conditions one may easily obtain
representatives
of the conjugacy classes to which the generator
$z$ may belong, listed in Table $4^\prime .$

\bigskip
{\bf Lemma 5.4}\enskip {\it Let $g\in \GL(n,F)$ have order $7.$
If $g$ is real, then
$2d_S^g\geq d^7_{\overline M}\ .$ If $g$ is semirational, then:}
$$\cases{
m_0+3m_1+3m_{-1}=n\cr
\cr
m_0^2+3{m_1}^2+3m^2_{-1}=d^g_M\cr
\cr
m_0^2+m_0+6m_1m_{-1}=2d^g_S}.$$

\medskip
{\it Proof}
\par\nobreak\medskip
Assume that $g$ is real. Then $d^g_M=m_0^2+2m_1^2+2m_2^2+2m_3^2$ and,
by Lemma 4.2 , $2d^g_S=m_0^2+m_0+2m_1^2+2m_2^2+2m_3^2 .$
It follows $2d^g_S-d^g_M=m_0\geq 0,$ hence
$2d_S^g\geq d^g_M\geq d^7_{\overline M}\ .$
The semirationality conditions are obvious.\ \qed
\bigskip
The next three lemmas are devoted to obtain Table 5
\medskip
{\bf Lemma 5.5}\enskip {\it Let $h,g\in \SL(n,q)$ be elements
of order $3$ and $7,$ respectively.

\par\nobreak\smallskip
{\it i}) If $q\equiv 2\ ({\rm mod}\ 3),$ then $h$
is real;
\smallskip
{\it ii}) if $q\equiv 3,5\ ({\rm mod}\ 7),$ then $g$ is rational;
\smallskip
{\it iii}) if $q\equiv 2,4\ ({\rm mod}\ 7),$ then $g$ is semirational;
\smallskip
{\it iv}) if $q\equiv 6\ ({\rm mod}\ 7),$ then $g$ is real.
}
\medskip
{\it Proof}
\par\nobreak\medskip
{\it i}) As $\vert F_q(\omega):F_q\vert =2,$
the minimum polynomial of $\omega$
over $F_q$ has degree 2. It follows that the similarity
invariants of $h$ can only be
$$\lambda-1\quad , \quad\prod_{k=1}^2(\lambda -\omega^k)\quad {\rm or} \quad
\prod_{k=0}^2(\lambda -\omega^k).$$
\smallskip
{\it ii}) As 3 and 5 have order 6 (mod 7),
$[F_q(\epsilon):F_q]=6.$ Thus
the minimum polynomial of $\epsilon$
over $F_q$ has degree 6. It follows that the similarity
invariants of $g$ can only be
$$\lambda-1\quad , \quad\prod_{k=1}^6(\lambda -\epsilon^k)
\quad {\rm or} \quad
\prod_{k=0}^6(\lambda -\epsilon^k).$$
{\it iii}) and {\it iv})   are proved with similar arguments. \qed

\bigskip
{\bf Remark}\enskip
We recall here that any companion matrix is conjugate to its transpose:
in fact they have the same minimum and characteristic polynomial,
which coincide. It follows immediately that any matrix
in ${\rm GL}(n,F)$ is conjugate to its transpose.

\medskip
{\bf Lemma 5.6} \enskip {\it Let $g$ be a semisimple element
of a symplectic or orthogonal group
over $F.$

\smallskip
{\it i}) $g$ is real;

\smallskip
{\it ii}) if $F=F_q$ with $q\equiv 2,3,4,5\ ({\rm mod}\ 7)$ and $g$
has order $7,$ then $g$ is rational.
}
\medskip
{\it Proof}
\par\nobreak\medskip
{\it i}) Let $J$ be the matrix of the
non-degenerate symplectic or orthogonal form defining
the relevant group.
%%% FROM PROBLEM:
%>From  $g^tJg=J$ it follows that
From  $g^tJg=J$ it follows that
$g$ is conjugate to $(g^{-1})^{t},$
hence to $g^{-1}$ by the remark above.

\smallskip
{\it ii}) $q\equiv 3,5\ ({\rm mod}\ 7)$ implies  $g$ rational,
$q\equiv 2,4\ ({\rm mod}\ 7)$ implies $g$ conjugate to $g^2.$
On the other hand, by {\it i}),  $g$ is
conjugate to $g^{-1}.$ Then it is readily seen that $g$
is conjugate to all its non-trivial powers, hence is rational.

\medskip
{\bf Lemma 5.7} \enskip {\it Let
$h,g\in \SU(n,q^2)$ be elements of order $3$
and $7,$ respectively.
\smallskip
{\it i}) If $q\equiv 1\ ({\rm mod}\ 3),$ then $h$ is real;
\smallskip
{\it ii}) if $q\equiv 2,4\ ({\rm mod}\ 7),$ then $g$ is rational;
\smallskip
{\it iii}) if $q\equiv 1\ ({\rm mod}\ 7),$ then $g$ is real;
\smallskip
{\it iv}) if $q\equiv 3,5\ ({\rm mod}\ 7),$ then $g$ is semirational.
}
\medskip
{\it Proof}
\par\nobreak\medskip
Let $A\in \SU(n,q^2)$ be semisimple. W.l.o.g. we may assume that
$A^t\overline A=I,$
where $\overline A$ is obtained from
$A$ applying
the field automorphism $\alpha\mapsto \alpha^q$ to all
its entries. By the above remark, $A$ is conjugate to
$(\overline A^{-1}).$ Moreover, considering the Jordan canonical
form of $A$ over $\overline F,$ it is clear that $\overline A$ is
conjugate to $A^q.$ Hence $A$ is conjugate to $A^{-q}.$

{\it i}) $h$ is conjugate to $h^{-q}=h^{-1}$
by the previous observation.\enskip
{\it ii}) $g$ is semirational and
conjugate to $g^{-2}$ or to $g^{-4}.$ It follows $g$ rational.
\enskip {\it iii}) $g$ is conjugate to $g^{-1}.$\enskip
{\it iv}) In this case
$q^2\equiv 2,4\ ({\rm mod}\ 7).$ Therefore $g$ is semirational.\qed

\vfill\eject
\centerline {\bf Table 2: admissible conjugacy classes for $x$}
\bigskip
{\sevenrm
$$\vbox{\settabs
\+111111&11111111111111&
11111111111111111111111111111111111111111111111111111111111&11111111111\cr
\+$n$&\quad $s$&\quad Representatives&$\quad d^x_S$\cr
\medskip
\hrule
\medskip
\+4&\quad 2&\quad $\pmatrix{(\pi_2)^2}$&\quad 6\cr
\smallskip
\hrule
\smallskip
\+5&\quad 3&\quad $\pmatrix{1,(\pi_2)^2}$&\quad 9\cr
\smallskip
\hrule
\smallskip
\+6&\quad 4&\quad $\pmatrix{1^2,(\pi_2)^2}$&\quad 13\cr
\smallskip
\+&\quad 3\ ,\ p= 2&\quad $\pmatrix{(\pi_2)^3}$\ p= 2&\quad 12\cr
\smallskip
\hrule
\smallskip
\+7&\quad 4&\quad $\pmatrix{-1,(\pi_2)^3}$&\quad 16\cr
\smallskip
\hrule
\smallskip
\+8&\quad 4&\quad $\pmatrix{(\pi_2)^4}$&\quad 20\cr
\smallskip
\+&\quad 5\ ,\ p= 2&\quad $\pmatrix{1^2,(\pi_2)^3}$\ p= 2&\quad 21\cr
\smallskip
\hrule
\smallskip
\+9&\quad 5&\quad $\pmatrix{1,(\pi_2)^4}$&\quad 25\cr
\smallskip
\hrule
\smallskip
\+10&\quad 6&\quad $\pmatrix{1^2,(\pi_2)^4}$&\quad 31\cr
\smallskip
\+&\quad 5\ ,\ p= 2&\quad $\pmatrix{(\pi_2)^5}$\ p= 2&\quad 30\cr
\smallskip
\hrule
\smallskip
\+11&\quad 6&\quad $\pmatrix{-1,(\pi_2)^5}$&\quad 36\cr
\smallskip
\hrule
\smallskip
\+12&\quad 6&\quad $\pmatrix{(\pi_2)^6}$&\quad 42\cr
\smallskip
\+&\quad 7\ ,\ p= 2&\quad $\pmatrix{1^2,(\pi_2)^5}$\ p= 2&\quad 43\cr
\smallskip
\hrule
\smallskip
\+13&\quad 7,8&\quad $\pmatrix{1,(\pi_2)^6}\ ,\
\pmatrix{(-1)^3,(\pi_2)^5}$&\quad 49, 51\cr
\smallskip
\hrule
\smallskip
\+14&\quad 8&\quad $\pmatrix{1^2,(\pi_2)^6}$&\quad 57\cr
\smallskip
\+&\quad 7\ ,\ p= 2&\quad $\pmatrix{(\pi_2)^7}$\ p= 2&\quad 56\cr
\smallskip
\hrule
\smallskip
\+15&\quad 8,9&\quad $\pmatrix{-1,(\pi_2)^7}\ ,\
\pmatrix{1^3,(\pi_2)^6}$&\quad 64, 66\cr
\smallskip
\hrule
\smallskip
\+16&\quad 8&\quad $\pmatrix{(\pi_2)^8}$&\quad 72\cr
\smallskip
\+&\quad 9\ ,\ p= 2&\quad $\pmatrix{1^2,(\pi_2)^7}$\ p= 2&\quad 73\cr
\smallskip
\hrule
\smallskip
\+17&\quad 9,10&\quad $\pmatrix{1,(\pi_2)^8}\ ,\
\pmatrix{(-1)^3,(\pi_2)^7}$&\quad 81, 83\cr
\smallskip
\hrule
\smallskip
\+18&\quad 10&\quad $\pmatrix{1^2, (\pi_2)^8}$&\quad 91\cr
\smallskip
\+&\quad 9,11\ ,\ p= 2&\quad $\pmatrix{(\pi_2)^9}\ ,\
\pmatrix{1^4,(\pi_2)^7}$\ p=2&\quad 90,94\cr
\smallskip
\hrule
\smallskip
\+19&\quad 10,11&\quad
$\pmatrix{-1,(\pi_2)^9}\ ,\
\pmatrix{1^3,(\pi_2)^8}$&\quad 100,102\cr
\smallskip
\hrule
\smallskip
\+20&\quad 10,12&\quad
$\pmatrix{(\pi_2)^{10}}\ ,\ \pmatrix{1^4, (\pi_2)^8}$&\quad 110,114\cr
\smallskip
\+&\quad 11\ ,\ p= 2&\quad $\pmatrix{1^2,(\pi_2)^9}$\ p=2&\quad 111\cr
\smallskip
\hrule
\smallskip
\+21&\quad 11,12,13&\quad$\pmatrix{1,(\pi_2)^{10}}\ ,\
\pmatrix{(-1)^3,(\pi_2)^9}\ ,\ \pmatrix{1^5,(\pi_2)^8}$&\quad
121,123,127 \cr
\smallskip
\hrule
\smallskip
\+22&\quad 12&\quad$\pmatrix{1^2,(\pi_2)^{10}}$&\quad 133\cr
\smallskip
\+&\quad 11,13,\ p= 2&\quad $\pmatrix{(\pi_2)^{11}}\ ,\
\pmatrix{1^4,(\pi_2)^9}$\ p=2&\quad 132,136\cr
\smallskip
\hrule
}$$}
\vfill\eject
\centerline {\bf Table 3: admissible conjugacy classes for  $y$}
{\sevenrm
$$\vbox{\settabs
\+111111&11111111111111&
11111111111111111111111111111111111111111111111111111111111&11111111111\cr
\+$n$&\quad $s$&\quad Representatives &$\quad d^y_S$\cr
\medskip
\hrule
\medskip
\+4&\quad 2&\quad $\pmatrix{1,\pi_3}$&\quad 4\cr
\smallskip
\hrule
\smallskip
\+5&\quad 2&\quad
$\pmatrix {d_2,\pi_3}$&\quad
5\cr
\smallskip
\hrule
\smallskip
\+6&\quad 2&\quad $\pmatrix{(\pi_3)^2}$&\quad 7\cr
\smallskip
\hrule
\smallskip
\+7&\quad 3&\quad $\pmatrix{1,(\pi_3)^2}\ ,\
\pmatrix{(d_2)^2,\pi_3}$&\quad 10\cr
\smallskip
\hrule
\smallskip
\+8&\quad 3, 4&\quad $\pmatrix{d_2,(\pi_3)^2}\ ,\
\pmatrix{(1)^2,(\pi_3)^2}$&\quad 12,14\cr
\smallskip
\hrule
\smallskip
\+9&\quad 3,4&\quad $\pmatrix{(\pi_3)^3}$&\quad 15\cr
\smallskip
\+&&\quad $\pmatrix{J_1,J_2,(\pi_3)^2}\ p=3$&\quad 16\cr
\smallskip
\hrule
\smallskip
\+10&\quad 4&\quad $\pmatrix{1,(\pi_3)^3}$&\quad 19\cr
\smallskip
\+&&\quad $\pmatrix{(d_2)^2,(\pi_3)^2}\ p=2$&\quad 19\cr
\smallskip
\hrule
\smallskip
\+11&\quad 4,5&\quad $\pmatrix{d_2,(\pi_3)^3}\ ,\
\pmatrix{(1)^2,(\pi_3)^3}$&\quad 22,24\cr
\smallskip
\hrule
\smallskip
\+12&\quad 4,5&\quad $\pmatrix{(\pi_3)^4}$&\quad 26\cr
\smallskip
\+&&\quad $\pmatrix{J_1,J_2,(\pi_3)^3}\ p=3$&\quad 27\cr
\smallskip
\hrule
\smallskip
\+13&\quad 5,6&\quad $\pmatrix{1,(\pi_3)^4}\ ,\
\pmatrix{(d_2)^2,(\pi_3)^3}$&\quad 31\cr
\smallskip
\+&&\quad $\pmatrix{(\omega)^2 ,(1,\omega) ,(\pi_3)^3}\ p\not=3$
&\quad 28\cr
\smallskip
\+&&\quad $\pmatrix{(J_1)^2,J_2,(\pi_3)^3}\ p=3$
&\quad 33\cr
\smallskip
\hrule
\smallskip
\+14&\quad 5,6&\quad $\pmatrix{d_2,(\pi_3)^4}\ ,\
\pmatrix{(1)^2,(\pi_3)^4}$&\quad 35,37\cr
\smallskip
\+&&\quad $\pmatrix{\omega , (1,\omega)^2 ,(\pi_3)^3}\ p\not=3$
&\quad 33\cr
\smallskip
\+&&\quad $\pmatrix{J_1,(J_2)^2,(\pi_3)^3}\ p=3$&\quad 37\cr
\smallskip
\hrule
\smallskip
\+15&\quad 5,6,7&\quad $\pmatrix{(\pi_3)^5}\ ,\
\pmatrix{d_2^3,(\pi_3)^3}\ ,\ \pmatrix{(1)^3,(\pi_3)^4}$&
\quad 40, 42, 44\cr
\+&&\quad $\pmatrix{(\omega )^3,(\pi_3)^4}\ ,\
\pmatrix{(1,\omega)^3,(\pi_3)^3}\ p\not=3$&\quad 38, 39\cr
\+&&\quad $\pmatrix{J_1,J_2,(\pi_3)^4}\ p=3$&\quad 41\cr
\smallskip
\hrule
\smallskip
\+16&\quad 6,7&\quad $\pmatrix{1,(\pi_3)^5}\ ,\
\pmatrix{(d_2)^2,(\pi_3)^4}$&\quad 46\cr
\smallskip
\+&&\quad $\pmatrix{(\omega)^2, (1,\omega), (\pi_3)^4}\ p\not=3$
&\quad 43\cr
\smallskip
\+&&\quad $\pmatrix{(J_1)^2,J_2, (\pi_3)^4}\ p=3$
&\quad 48\cr
\smallskip
\hrule
\smallskip
\+17&\quad 6,7&\quad $\pmatrix{d_2,(\pi_3)^5}\ ,\
\pmatrix{(1)^2,(\pi_3)^5}$&\quad 51, 53\cr
\smallskip
\+&&\quad $\pmatrix{\omega ,(1,\omega)^2, (\pi_3)^4}\ p\not=3$
&\quad 49\cr
\smallskip
\+&&\quad $\pmatrix{J_1,(J_2)^2, (\pi_3)^4}\ p=3$&\quad 53\cr
\smallskip
\hrule
\smallskip
\+18&\quad 6,7,8&\quad $\pmatrix{(\pi_3)^6}\ ,\
\pmatrix{(d_2)^3,(\pi_3)^4}\ ,\
\pmatrix{(1)^3,(\pi_3)^5}$&\quad 57,59,61\cr
\smallskip
\+&&\quad $\pmatrix{(\omega)^3,(\pi_3)^5}\ ,\
\pmatrix{(1,\omega)^3,(\pi_3)^4}\ p\not=3$
&\quad 55,56\cr
\smallskip
\+&&\quad $\pmatrix{(J_1,J_2,(\pi_3)^5}\ p=3$&\quad 58\cr
\smallskip
\hrule
\smallskip
\+19&\quad 7,8&\quad $\pmatrix{1,(\pi_3)^6}\ ,\
\pmatrix{d_2^2,(\pi_3)^5}$&\quad 64\cr
\smallskip
\+&&\quad
$\pmatrix{1,(1,\omega )^3,(\pi_3)^4}\ ,\
\pmatrix{(\omega )^2,(1,\omega), (\pi_3)^5}\ p\not=3$&\quad 64,61\cr
\smallskip
\+&&\quad
$\pmatrix{J_1,(J_2)^3,(\pi_3)^4}\ ,\
\pmatrix{(J_1)^2,J_2,(\pi_3)^5}\ p=3$&\quad 67,66\cr
\smallskip\hrule
}$$}
\bigskip
\centerline {\bf Table 3\ (continuation)}
{\sevenrm
$$\vbox{\settabs
\+1111&11111111&111111111111111111
11111111111111111111111111111111111111111111111111111111111&11111111111\cr
\+$n$&\quad $s$&\qquad\qquad $y$&$\quad d^y_S$\cr
\medskip
\hrule
\medskip
\+20&\quad 7,8,9&\quad $\pmatrix{d_2,(\pi_3)^6}\ ,\
\pmatrix{(d_2)^4,(\pi_3)^4}\ ,\
\pmatrix{(1)^2,(\pi_3)^6}$&\quad 70,74,72\cr
\smallskip
\+&&\quad $\pmatrix{\omega,(\omega ,1)^2,(\pi_3)^5}\ ,\
\pmatrix{(\omega )^3 , d_2 ,(\pi_3)^5}
\ p\not=3$&\quad 68,69 \cr
\smallskip
\+&&\quad $\pmatrix{J_1,(J_2)^2,(\pi_3)^5}\ ,\
\pmatrix{(J_1)^3,J_2, (\pi_3)^5}\ p=3$&\quad 72,75\cr
\smallskip
\hrule
\smallskip
\+21&\quad 7,8,9&\quad $\pmatrix{(\pi_3)^7}\ ,\
\pmatrix{(d_2)^3,(\pi_3)^5}\ ,\
\pmatrix{(1)^3,(\pi_3)^6}$&\quad 77,79,81\cr
\smallskip
\+&&\quad $\pmatrix{(\omega)^3,(\pi_3)^6}\ ,\
\pmatrix{(\omega,1)^3,(\pi_3)^5}
\ p\not=3$&\quad 75,76 \cr
\smallskip
\+&&\quad $\pmatrix{J_1,J_2,(\pi_3)^6}\ ,\
\pmatrix{(J_1)^2,(J_2)^2,(\pi_3)^5}
\ p=3$&\quad 78,81\cr
\smallskip
\hrule
\smallskip
\+22&\quad 8,9,10&\quad $\pmatrix{1,(\pi_3)^7}\ ,\
\pmatrix{(d_2)^2,(\pi_3)^6}\ ,\
\pmatrix{(1)^4,(\pi_3)^6}$&\quad 85,91\cr
\smallskip
\+&&\quad $\pmatrix{(\omega)^2,(1, \omega),(\pi_3)^6}\ ,\
\pmatrix{1,(1, \omega)^3,(\pi_3)^5}
\ p\not=3$&\quad 82,85 \cr
\smallskip
\+&&\quad $\pmatrix{(J_1)^2,J_2,(\pi_3)^6}\ ,\
\pmatrix{J_1,(J_2)^3,(\pi_3)^5}
\ p=3$&\quad 87,88\cr
\smallskip
\+&&\quad $\pmatrix{(\omega)^3, (d_2)^2,(\pi_3)^5}
\ p=2$&\quad 85\cr
\smallskip
\hrule
}$$}

\bigskip
\centerline {\bf Table 4 : admissible conjugacy classes for
$z$ modular}
{\sevenrm
$$\vbox{\settabs
\+111111&11111111111111&
11111111111111111111111111111111111111111111111111111111111&11111111111\cr
\+$n$&\quad $s$&\quad Representatives &$\quad d^z_S$\cr
\medskip
\hrule
\medskip
\+4&\quad 1&\quad $\pmatrix{J_4}$&\quad 2\cr
\smallskip
\hrule
\smallskip
\+5&\quad 1&\quad $\pmatrix{J_5}$&\quad 3\cr
\smallskip
\hrule
\smallskip
\+6&\quad 1&\quad $\pmatrix{J_6}$&\quad 3\cr
\smallskip
\hrule
\smallskip
\+7&\quad 1,2&\quad $\pmatrix{\pi_7}\ , \ \pmatrix{J_1,J_6}$&\quad 4,5\cr
\smallskip
\hrule
\smallskip
\+8&\quad 2&\quad $\pmatrix{J_1,J_7}\ , \ \pmatrix{J_2,J_6}$&\quad 6\cr
\smallskip
\hrule
\smallskip
\+9&\quad 2,3&\quad $\pmatrix{J_2,J_7}\ , \ \pmatrix{J_3,J_6}\ ,\
\pmatrix{(J_1)^2,J_7}$&\quad 7,8,9\cr
\smallskip
\hrule
\smallskip
\+10&\quad 2&\quad $\pmatrix{J_3,J_7}$&\quad 9\cr
\smallskip
\hrule
\smallskip
\+11&\quad 2,3&\quad $\pmatrix{J_4,J_7}\ , \ \pmatrix{J_5,J_6}\ ,\
\pmatrix{J_1,J_3,J_7}$&\quad 10,11,12\cr
\smallskip
\hrule
\smallskip
\+12&\quad 2,3&\quad $\pmatrix{J_5,J_7}\ , \ \pmatrix{(J_6)^2}$&\quad 12\cr
\smallskip
\+&&\quad $\pmatrix{J_1,J_4,J_7}\ , \ \pmatrix{J_1,J_5,J_6}\ ,\
\pmatrix{J_2,J_3,J_7}$&\quad 13,14\cr
\hrule
\smallskip
\+13&\quad 2,3&\quad $\pmatrix{J_6,J_7}\ ,\
\pmatrix{J_1,J_5,J_7}$&\quad 13,15\cr
\smallskip
\+&&\quad $\pmatrix{J_1,(J_6)^2}\ ,\ \pmatrix{J_2,J_4,J_7}
$&\quad 15 \cr
\smallskip
\hrule
\smallskip
\+14&\quad 2,3&\quad $\pmatrix{(J_7)^2}\ ,\
\pmatrix{J_1, J_6, J_7}\ ,\
\pmatrix{J_2, J_5, J_7}$&\quad 15,16,17\cr
\smallskip
\hrule
\smallskip
\+15&\quad 3,4&\quad $\pmatrix{J_1, (J_7)^2}\ ,\
\pmatrix{J_2,J_6, J_7}\ ,\
\pmatrix{J_3,J_5,J_7}\ ,\
\pmatrix{J_3,(J_6)^2}$&
\quad 18, 20\cr
\smallskip
\+&&\quad $
\pmatrix{(J_4)^2,J_7}\ ,\ \pmatrix{(J_1)^2,J_6, J_7}\ ,\
\pmatrix{J_1,J_2,J_5, J_7}$&
\quad 20, 21\cr
\smallskip
\hrule
\smallskip
\+16&\quad 3,4&\quad $\pmatrix{J_2, (J_7)^2}\ ,\
\pmatrix{J_3,J_6, J_7}\ ,\
\pmatrix{J_4,J_5,J_7}\ ,\
\pmatrix{J_4,(J_6)^2}$&
\quad 20, 21, 22 \cr
\smallskip
\+&&\quad $\pmatrix{(J_1)^2,(J_7)^2}\ ,\
\pmatrix{J_1,J_2, J_6,J_7}\ ,\
\pmatrix{J_1,J_3,J_5,J_7}$&\quad 22,24\cr
\smallskip
\hrule
}$$}
\bigskip
\centerline{\bf Table 4 \ (continuation)}
{\sevenrm
$$\vbox{\settabs
\+111&1111111&111111
111111111111111111111111111111111111111111111111111111111111111&11111111111\cr
\+$n$&\quad $s$&\qquad\qquad $z$&$\quad d^z_S$\cr
\medskip
\+17&\quad 3,4,5&\quad $\pmatrix{J_3, (J_7)^2}\ ,\
\pmatrix{J_4,J_6, J_7}\ ,\
\pmatrix{(J_5)^2,J_7}
\pmatrix{J_5,(J_6)^2}$&
\quad 23, 25\cr
\smallskip
\hrule
\smallskip
\+&&\quad $\pmatrix{J_1,J_2, (J_7)^2}\ ,\
\pmatrix{J_1,J_3, J_6,J_7}\ ,\ \pmatrix{J_1,J_4,J_5,J_7}$
&\quad 24,25\cr
\smallskip
\+&&\quad $
\pmatrix{(J_2)^2,J_6,J_7}\ ,\
\pmatrix{(J_1)^3,(J_7)^2}$&\quad 26,25,27\cr
\smallskip
\hrule
\smallskip
\+18&\quad 3,4,5&\quad $\pmatrix{J_4, (J_7)^2}\ ,\
\pmatrix{J_5,J_6, J_7}\ ,\
\pmatrix{(J_6)^3}\ ,\ \pmatrix{J_1,J_3,(J_7)^2}$&
\quad 25,26,27\cr
\smallskip
\+&&\quad $\pmatrix{J_1,(J_5)^2, J_7}\ ,\
\pmatrix{(J_2)^2,(J_7)^2}\ ,\
\pmatrix{J_2,J_3, J_6, J_7}$&\quad 29,27,28\cr
\smallskip
\+&&\quad $\pmatrix{J_1,J_4,J_6,J_7}\ ,\
\pmatrix{(J_1)^2,J_2, (J_7)^2}$&\quad 27,29\cr
\smallskip
\hrule
\smallskip
\+19&\quad 3,4,5&\quad $\pmatrix{J_5, (J_7)^2}\ ,\
\pmatrix{(J_6)^2, J_7}\ ,\
\pmatrix{J_1,J_4,(J_7)^2}\ ,\ \pmatrix{J_1,J_5,J_6,J_7}$&
\quad 28,29,30\cr
\smallskip
\+&&\quad $\pmatrix{J_1,(J_6)^3}\ ,\
\pmatrix{J_2,J_3, (J_7)^2}\ ,\
\pmatrix{J_2,J_4,J_6,J_7}$&\quad 31,30\cr
\smallskip
\+&&\quad $\pmatrix{J_2,(J_5)^2,J_7}\ ,\
\pmatrix{(J_3)^2,J_6,J_7}\ ,\
\pmatrix{(J_1)^2,J_3, (J_7)^2}$&\quad 32\cr
\smallskip
\+&&\quad $\pmatrix{(J_1)^2,J_4,J_6,J_7}\ ,\
\pmatrix{J_1,(J_2)^2,(J_7)^2}$&\quad 32\cr
\smallskip
\hrule
\smallskip
\+20&\quad 3,4,5&\quad $\pmatrix{J_6, (J_7)^2}\ ,\
\pmatrix{J_1, (J_6)^2, J_7}\ ,\
\pmatrix{J_1,J_5,(J_7)^2}\ ,\ \pmatrix{J_2,J_4,(J_7)^2}$&
\quad 30,32\cr
\smallskip
\+&&\quad $\pmatrix{J_2,(J_6)^3}\ ,\
\pmatrix{(J_3)^2, (J_7)^2}\ ,\
\pmatrix{J_3,J_4,J_6,J_7}$&\quad 34\cr
\smallskip
\+&&\quad $\pmatrix{J_2,J_5,J_6,J_7}\ ,\
\pmatrix{J_3,(J_5)^2,J_7}\ ,\
\pmatrix{(J_1)^2,J_4,(J_7)^2}$&\quad 33,36,34\cr
\smallskip
\+&&\quad $\pmatrix{(J_1)^2,J_5,J_6,J_7}\ ,\
\pmatrix{J_1,J_2,J_3,(J_7)^2}\ ,\
\pmatrix{J_1,J_2,J_4,J_6,J_7}$&\quad 35\cr
\smallskip
\hrule
\smallskip
\+21&\quad 3,4,5,6&\quad $\pmatrix{(J_7)^3}\ ,\
\pmatrix{J_1,J_6, (J_7)^2}\ ,\
\pmatrix{J_2, J_5, (J_7)^2}\ ,\
\pmatrix{J_2, (J_6)^2, J_7}$&\quad 33,34,35\cr
\smallskip
\+&&\quad $\pmatrix{J_3,J_4,(J_7)^2}\ ,\
\pmatrix{J_3,J_5,J_6,J_7}\ ,\
\pmatrix{(J_4)^2,J_6,J_7}\ ,\
\pmatrix{J_3,(J_6)^3}$&\quad 36,37,38\cr
\smallskip
\+&&\quad $\pmatrix{J_4,(J_5)^2,J_7}\ ,\
\pmatrix{(J_1)^2,J_5,(J_7)^2}\ ,\
\pmatrix{(J_1)^2,(J_6)^2,J_7}$&\quad 39,37\cr
\smallskip
\+&&\quad $\pmatrix{J_1,J_2,J_4,(J_7)^2}\ ,\
\pmatrix{J_1,J_2,J_5,J_6,J_7}\ ,\
\pmatrix{J_1,(J_3)^2,(J_7)^2}$&\quad 37,38,39\cr
\smallskip
\+&&\quad $\pmatrix{J_1,J_3,J_4,J_6,J_7}\ ,\
\pmatrix{(J_2)^2,J_3,(J_7)^2}\ ,\
\pmatrix{(J_1)^3,J_4,(J_7)^2}$&\quad 39,40\cr
\smallskip
\hrule
\smallskip
\+22&\quad 4,5,6&\quad $\pmatrix{J_1,(J_7)^3}\ ,\
\pmatrix{J_2,J_6, (J_7)^2}\ ,\
\pmatrix{J_3, J_5, (J_7)^2}\ ,\
\pmatrix{J_3, (J_6)^2, J_7}$&\quad 37,39\cr
\smallskip
\+&&\quad $\pmatrix{(J_4)^2,(J_7)^2}\ ,\
\pmatrix{(J_1)^2,J_6,(J_7)^2}\ ,\
\pmatrix{J_4,J_5,J_6,J_7}$&\quad 39,40\cr
\smallskip
\+&&\quad $\pmatrix{J_1,J_2,J_5,(J_7)^2}\ ,\
\pmatrix{J_1,J_2,(J_6)^2,J_7}\ ,\
\pmatrix{J_1,J_3,J_4,(J_7)^2}\ ,\
$&\quad 40,41\cr
\smallskip
\+&&\quad $\pmatrix{J_1,J_3,J_5,J_6,J_7}\ ,\
\pmatrix{(J_2)^2,J_4,(J_7)^2}\ ,\
\pmatrix{(J_1)^3,J_5,(J_7)^2}$&\quad 42,41,43\cr
\smallskip
\hrule
}$$}

\vfill\eject
In Table $4^\prime$ below we set $d_6:=\pmatrix
{\epsilon ,\epsilon^{2}, \epsilon^{3}, \epsilon^{-3},
\epsilon^{-2}, \epsilon^{-1}}.$
\vskip20pt
\centerline {\bf Table $4^\prime$ :
admissible conjugacy classes for  $z$ semisimple}
\medskip
\centerline {\bf ($p\not= 2$  if $n=14$)}
\bigskip
{\sevenrm
$$\vbox{\settabs
\+111&11111&1111111111111111111
111111111111111111111111111111111111111111111111111111111111111&11111111111\cr
\+$n$&\quad $s$&\quad Representatives &$\quad d^z_S$\cr
\medskip
\hrule
\medskip
\+4&\quad 1&\quad $\pmatrix
{\epsilon ,\epsilon^{3},\epsilon^{-3}, \epsilon^{-1} }\ ,\
\pmatrix
{\epsilon ,\epsilon^{2},\epsilon^{-2}, \epsilon^{-1}}\ ,\
\pmatrix
{1, \epsilon ,\epsilon^{2},\epsilon^{-3}}  $
&\quad 2,1\cr
\smallskip
\hrule
\smallskip
\+5&\quad 1&\quad $\pmatrix
{1,\epsilon ,\epsilon^{3}, \epsilon^{-3}, \epsilon^{-1}}\ ,\
\pmatrix
{1,\epsilon ,\epsilon^{2}, \epsilon^{-2}, \epsilon^{-1}}$&\quad 3\cr
\smallskip
\hrule
\smallskip
\+6&\quad 1&\quad $d_6$&\quad 3\cr
\smallskip
\+&\quad 2&\quad$\pmatrix
{\epsilon , 1, \epsilon ,\epsilon^{3}, \epsilon^{-3}, \epsilon^{-2}}\ ,\
\pmatrix
{\epsilon ,1, \epsilon ,\epsilon^{2}, \epsilon^{-3}, \epsilon^{-1}}$ \ p=2
&\quad 2,3\cr
\smallskip
\+&\quad 2&\quad$\pmatrix
{1 ,1, \epsilon ,\epsilon^{3}, \epsilon^{-3}, \epsilon^{-1}}\ ,\
\pmatrix
{1 ,1, \epsilon ,\epsilon^{2}, \epsilon^{-2}, \epsilon^{-1}}$\ p=2
&\quad 5\cr
\smallskip
\hrule
\smallskip
\+7&\quad 1&\quad $\pmatrix{\pi_7}$&\quad 4\cr
\smallskip
\hrule
\smallskip
\+10&\quad 2&\quad $\pmatrix
{\epsilon ,\epsilon^{2},\epsilon^{-3},\pi_7}\ , \ \pmatrix
{1,\epsilon,\epsilon^{-1} ,\pi_7}$&\quad 7,9\cr
\smallskip
\+&\quad 2&\quad $\pmatrix
{\epsilon,\epsilon^{2} ,\epsilon^{-2}, \epsilon^{-1}, d_6}
\ ,\
\pmatrix
{\epsilon,\epsilon^{3} ,\epsilon^{-3}, \epsilon^{-1}, d_6}$\ &\quad 9\cr
\smallskip
\+&\quad 2&\quad $\pmatrix
{\epsilon, \epsilon^{-3}, \epsilon^{-2}, \epsilon^{-1},
1, \epsilon ,\epsilon^{3}, \epsilon^{-3},
\epsilon^{-2}, \epsilon^{-1}}\ ,\
\pmatrix
{\epsilon, \epsilon^{3}, \epsilon^{-3}, \epsilon^{-2},
1, \epsilon ,\epsilon^{2},\epsilon^{3}, \epsilon^{-3},
\epsilon^{-2}}$&\quad 7\cr
\smallskip
\+&\quad 2&\quad $\pmatrix
{\epsilon, \epsilon^{2}, \epsilon^{3}, \epsilon^{-2},
1, \epsilon ,\epsilon^{2},\epsilon^{3},
\epsilon^{-2}, \epsilon^{-1}}\ ,\
\pmatrix
{\epsilon, \epsilon^{2}, \epsilon^{3}, \epsilon^{-1},
1, \epsilon ,\epsilon^{2},\epsilon^{3},
\epsilon^{-3}, \epsilon^{-1}}$ &\quad 7\cr
\smallskip
\+&\quad 3,2&\quad $
\pmatrix
{\epsilon,\epsilon, \epsilon^{-2}, \pi_7}\ ,\
\pmatrix
{1, \epsilon, \epsilon^{3}, \epsilon^{-2},
1, \epsilon ,\epsilon^{3},
\epsilon^{-3},\epsilon^{-2}, \epsilon^{-1}}$&\quad 7\cr
\smallskip
\+&\quad 2&\quad $
\pmatrix
{1, \epsilon, \epsilon^{-3}, \epsilon^{-2},
1, \epsilon ,\epsilon^{2},\epsilon^{-3},
\epsilon^{-2}, \epsilon^{-1}}\ ,\
\pmatrix
{1, \epsilon, \epsilon^{2}, \epsilon^{3},
1, \epsilon ,\epsilon^{2}, \epsilon^{3},\epsilon^{-3},
\epsilon^{-2}}$&\quad 7\cr
\smallskip
\hrule
\smallskip
\+13&\quad 2&\quad $\pmatrix
{d_6,d_7}$&\quad 13\cr
\smallskip
\+&\quad 3&\quad $\pmatrix
{\epsilon,\epsilon^{-3} , \epsilon, \epsilon^{-3},
\epsilon^{-2},\epsilon^{-1},\pi_7}\ ,\
\pmatrix
{\epsilon,\epsilon^{-2} , \epsilon, \epsilon^{-2},
\epsilon^{3},\epsilon^{-1},\pi_7}$&\quad 12\cr
\smallskip
\+&\quad 3&\quad $\pmatrix
{\epsilon,\epsilon^{-1} , \epsilon, \epsilon^{-1},
\epsilon^{3},\epsilon^{-3},\pi_7}\ ,\
\pmatrix
{\epsilon,\epsilon^{-1} , \epsilon, \epsilon^{-1},
\epsilon^{2},\epsilon^{-2},\pi_7}$&\quad 15\cr
\smallskip
\+&\quad 3&\quad $\pmatrix
{\epsilon,\epsilon^{3} , \epsilon, \epsilon^{3},
\epsilon^{2},\epsilon^{-3},\pi_7}\ ,\
\pmatrix
{\epsilon,\epsilon^{2} , \epsilon, \epsilon^{2},
\epsilon^{3},\epsilon^{-2},\pi_7}$&\quad 12\cr
\smallskip
\+&\quad 3&\quad $\pmatrix
{\epsilon,1,\epsilon, \epsilon^{3} , \epsilon^{-3},
\epsilon^{-2},\pi_7}\ ,\
\pmatrix
{\epsilon,\epsilon^{3},1, \epsilon, \epsilon^{3} , \epsilon^{-1},
\pi_7}$&\quad 12,13\cr
\smallskip
\+&\quad 3&\quad $\pmatrix
{\epsilon,\epsilon^{-2},1,\epsilon , \epsilon^{2},
\epsilon^{-2},\pi_7}\ ,\
\pmatrix
{\epsilon,1, \epsilon, \epsilon^{2} ,\epsilon^{-3},
 \epsilon^{-1},  \pi_7}$&\quad 13\cr
\smallskip
\+&\quad 3&\quad $\pmatrix
{\epsilon,1, \epsilon , \epsilon^{2}, \epsilon^{3},
\epsilon^{-2}, \epsilon^{-1},
1, \epsilon , \epsilon^{2}, \epsilon^{3},
\epsilon^{-2}, \epsilon^{-1}}\ ,\
\pmatrix
{1, \epsilon,1, \epsilon, \epsilon^{2}, \epsilon^{3},
\pi_7}$&\quad 13\cr
\smallskip
\+&\quad 3&\quad $\pmatrix
{1, 1, \epsilon, \epsilon^{3}, \epsilon^{-3},
\epsilon^{-1},\pi_7}\ ,\
\pmatrix
{1, 1, \epsilon, \epsilon^{2}, \epsilon^{-2},
\epsilon^{-1},\pi_7}$&\quad 15\cr
\smallskip
\hrule
\smallskip
\+14&\quad 3&\quad $\pmatrix
{\epsilon ,\epsilon^{-2},\epsilon, \epsilon^{2}, \epsilon^{3},
\epsilon^{-3}, \epsilon^{-2}, \pi_7}\ ,\
\pmatrix
{\epsilon, \epsilon^{3}, \epsilon, \epsilon^{2},\epsilon^{3},
\epsilon^{-2},\epsilon^{-1}, \pi_7} $&\quad 14\cr
\smallskip
\+&\quad 3&\quad $\pmatrix
{\epsilon, \epsilon^{-3}, \epsilon, \epsilon^{2},\epsilon^{3},
\epsilon^{-3},\epsilon^{-1}, \pi_7}\ ,\
\pmatrix
{\epsilon, \epsilon^{2}, \epsilon, \epsilon^{2},\epsilon^{-3},
\epsilon^{-2},\epsilon^{-1}, \pi_7}  $&\quad 15\cr
\smallskip
\+&\quad 3&\quad $\pmatrix {(\pi_7)^2}\ ,\ \pmatrix
{\epsilon, \epsilon^{-3},1, \epsilon, \epsilon^{-3},\epsilon^{-2},
\epsilon^{-1}, \pi_7}$ &\quad 15,14 \cr
\smallskip
\+&\quad 3&\quad $\pmatrix
{\epsilon, \epsilon^{-2},1, \epsilon, \epsilon^{3},\epsilon^{-2},
\epsilon^{-1}, \pi_7} $&\quad 14\cr
\smallskip
\+&\quad 3&\quad $\pmatrix
{\epsilon, \epsilon^{-1},1, \epsilon, \epsilon^{3},\epsilon^{-3},
\epsilon^{-1}, \pi_7}\ ,\
\pmatrix
{\epsilon, \epsilon^{-1},1, \epsilon, \epsilon^{2},\epsilon^{-2},
\epsilon^{-1}, \pi_7} $&\quad 17\cr
\smallskip
\+&\quad 3&\quad $\pmatrix
{\epsilon, \epsilon^{3},1, \epsilon, \epsilon^{2},\epsilon^{3},
\epsilon^{-3}, \pi_7}\ ,\
\pmatrix
{\epsilon, \epsilon^{2},1, \epsilon, \epsilon^{2},\epsilon^{3},
\epsilon^{-2}, \pi_7}  $&\quad 14\cr
\smallskip
\+&\quad 3&\quad $\pmatrix
{1, \epsilon,1, \epsilon, \epsilon^{3},\epsilon^{-3},
\epsilon^{-2}, \pi_7}\ ,\
\pmatrix
{1, \epsilon,1, \epsilon, \epsilon^{2},\epsilon^{-3},
\epsilon^{-1}, \pi_7}  $&\quad 15,16\cr
\smallskip
\hrule
}$$}
\vfill\eject
\centerline{\bf Table 5\enskip (Lemmas 5.5, 5.6, 5.7)}
\bigskip
{\sevenrm
$$\vbox{\settabs
\+111111111111111111&111111111111111111111111111&
11111111111111111111111&111111111111111111111111111\cr
\+$q$&\qquad $\SL(n,q)$&\qquad $\Sp(n,q)$&\qquad $\SU(n,q^2)$\cr
\medskip
\hrule
\medskip
\+$\equiv 1\ (21)$&&$y$ real, $z$ real&
$y$ real, $z$ real\cr
\medskip
\hrule
\medskip
\+$\equiv 2,-10\ (21)$&$y$ real, $z$ semirat. &$y$ real, $z$ rational&
\qquad\qquad $z$ rational\cr
\medskip
\hrule
\medskip
\+$\equiv 4,-5\ (21)$&\qquad \qquad $z$ semirat. &$y$ real, $z$ rational&
$y$ real, $z$ rational\cr
\medskip
\hrule
\medskip
\+$\equiv 5,-4\ (21)$&$y$ real, $z$ rational &$y$ real, $z$ rational&
\qquad\qquad $z$ semirat. \cr
\medskip
\hrule
\medskip
\+$\equiv 8\ (21)$&$y$ real&$y$ real, $z$ real&
\qquad\qquad $z$ real\cr
\medskip
\hrule
\medskip
\+$\equiv 10,-2\ (21)$&\qquad\qquad $z$ rational &$y$ real, $z$ rational&
$y$ real, $z$ semirat. \cr
\medskip
\hrule
\medskip
\+$\equiv -8\ (21)$&\qquad\qquad $z$ real &$y$ real, $z$ real&
$y$ real\cr
\medskip
\hrule
\medskip
\+$\equiv -1\ (21)$&$y$ real, $z$ real &$y$ real, $z$ real&
\cr
\medskip
\hrule
\medskip
\+$3^{6h}$&$y$ modular&$y$ modular, $z$ real&
$y$ modular, $z$ real\cr
\medskip
\hrule
\medskip
\+$3^{6h+1}$&$y$ modular, $z$ rational&$y$ modular, $z$ rational&
$y$ modular, $z$ semirat. \cr
\medskip
\hrule
\medskip
\+$3^{6h+2},\ 3^{6h+4}$&$y$ modular, $z$ semirat. &$y$ modular, $z$ rational&
$y$ modular, $z$ rational\cr
\medskip
\hrule
\medskip
\+$3^{6h+3}$&$y$ modular, $z$ real&$y$ modular, $z$ real&
$y$ modular\cr
\medskip
\hrule
\medskip
\+$3^{6h+5}$&$y$ modular, $z$ rational&$y$ modular, $z$ rational&
$y$ modular, $z$ semirat. \cr
\medskip
\hrule
\medskip
\+$7^t$&\qquad\qquad $z$ modular&$y$ real, $z$ modular&
$y$ real, $z$ modular\cr
\medskip
\hrule
\medskip
}$$}

\bigskip
{\bf 6.\enskip Proof of Theorem 1}
\par\nobreak\bigskip

We start with an easy but useful remark. Namely,
let $k\geq 7$ be a prime, $G$ be a $(2,3,k)$-generated group, and
$f: G\to \SL(n,F)$ be a representation of degree $n \leq 2.$
If $p\not=2,$
the only involution of $\SL(2,F)$ is $-I,$ and therefore
$f(G)$ is abelian. Hence $f(G)=\{1\}.$
This readily implies that
a $(2,3,k)$-generated subgroup $G$ of $\SL(3,F)$
or $\SL(4,F)$ which does not fix any
non-zero vector is absolutely irreducible.
For otherwise, by the above remark, $G$ necessarily
admits the eigenvalue $1.$
\medskip
{\it i}) \enskip Let
$\overline H\leq \SL(2,\overline F)$ be the
preimage of a $(2,3,k)$-generated subgroup of $\PSL(2,\overline F).$
Clearly we may assume $\overline H=\langle \overline x, z\rangle$ where
$\overline x^2=-I,\ z^k=I.$ Moreover
$\overline H$ is irreducible,
since it is not soluble.
Therefore, if $u$ is an eigenvector of $z,$
$u$ and $\overline xu$ are linearly independent.
It follows that, with respect to the basis
$\{u,\overline xu\} ,$
$$\overline x=\pmatrix {0&-1\cr 1&0\cr}\quad , \quad
z=z_i= \pmatrix {\epsilon^i&a\cr 0&\epsilon^{-i}\cr}\quad 1\leq i\leq k-1$$
for a suitable $a\in \overline F .$
Thus $y_i:=\overline x z_i$ has trace $a.$ On the other hand, $y_i$ must be
conjugate to $\pm\pmatrix{0&-1\cr 1&-1},$ which gives $a=\pm1 .$
Both choices for $z_i$ give rise to conjugate subgroups.
This is clear if $p=k,\ \epsilon =1.$
On the other hand, if $p\not= k,$
let {\fiverm $t:=\pmatrix{1&\epsilon^{-i}-\epsilon ^i \cr
-\epsilon^{-i}+\epsilon ^i &1}.$}\enskip $t$ is non-singular, centralizes
$\overline x$  and
{\fiverm $$t^{-1}\thinspace
\pmatrix {\epsilon^i&-1\cr 0&\epsilon^{-i}\cr}\thinspace t
\ =\ \pmatrix {\epsilon^i&1\cr 0&\epsilon^{-i}\cr}^{-1} .$$}
Hence we may assume $a=1 .$
Let $\overline H_i:=\langle \overline x, z_i\rangle .$
Note that, for every $i ,$ $(\overline x z_i)^3=- I$
and $\overline H_i\leq \SL(2,R),$ where $R$ is isomorphic to
$\Q(\epsilon)$ if $p=0,$ $R=F_{p^{k-1}}$ if $p>0.$
If $p=0,$ we may consider
the automorphism $\gamma_i$ of
$\GL(n,R)$ induced by the automorphism of
$R$ sending
$\epsilon$ to $\epsilon^i , \ 1\leq i\leq k-1.$
Since $\overline H_i=\gamma_{i}(\overline H_1),$
the projective image of $\overline H_1$ is, up to isomorphism,
the only $(2,3,k)$-generated subgroup of $\PSL(2,\overline F),$
and therefore (cfr. [Mag]) it is isomorphic to
$\Delta(2,3,k).$

Now assume $p>0$ and $k=7,$ hence
$\overline H_i\leq \SL(2,F_{p^6}).$
As $\overline H_i$ is perfect, it follows from the knowledge
of the  subgroups of $\PSL(2,p^6)$ that
the projective image $H_i$ of $\overline H_i$ is isomorphic
to $\PSL(2, p^{m_i})$ for some $m_i\in \{1,2,3,6\}.$
The results of [Mac] give
$H_i\simeq {\rm PSL}(2,p)$ if $p\equiv 0, \pm1$(mod 7)
and $H_i\simeq {\rm PSL}(2,p^3)$ if $p\not\equiv 0,\pm 1$(mod 7).

{\it ii})\enskip A direct computation shows that the group
$$\varphi (H)= \langle
\pmatrix{0&0& 1\cr
              0&-1&0\cr
              1&0&0\cr}\ ,\
              \pmatrix{\epsilon^2 &2\epsilon&1\cr
              0&1&\epsilon^{-1}\cr
              0&0&\epsilon^{-2} \cr}\rangle $$
preserves the bilinear symmetric
form defined by the matrix
${\rm antidiag} (1,-2,1);$ moreover, if $p\not=2,$
it does not fix any non-zero vector.
Hence $\varphi (H)$
is an irreducible subgroup of ${\rm SO}(3,\overline F),$
for all $k\geq 7.$
%%% FROM PROBLEM:
%>From now on assume $k=7,$ and let $K=\langle x,z\rangle $ be an irreducible
From now on assume $k=7,$ and let $K=\langle x,z\rangle $ be an irreducible
subgroup of $\SL(3, \overline F),$ with
$x^2=z^7=(xz)^3=1.$
Clearly $x$ has a 2-dimensional eigenspace. Therefore,
by the irreducibility of $K,$ $z$ cannot have a
2-dimensional eigenspace. In particular, if $p\not=7,$
$z$ is either
conjugate to
$z_1={\rm diag} (\eta ,\eta^2 ,\eta^4)$ or to
$z_2={\rm diag} (\eta ,1,\eta^{-1}),$ where
$\eta =\epsilon ^i$ for some $i$ with $1\leq i\leq 6 .$
Notice that $z$ leaves invariant a 2-dimensional
subspace $\langle u,v\rangle .$ If $p\not= 7,$ we may assume
$zu= \eta u, zv=\eta^2 v$ if $z\sim z_1$ and
$zu= \eta u, zv=v$ if $z\sim z_2 .$
On the other hand, if $p=7,$ we may assume $zu=u.$
Since $K$ is irreducible,
$u,v,xu$ are linearly independent. Moreover, as
${\rm det}\ x=1 ,$
$xv=au-v+axu$ for some $a\in \overline F .$
It follows that, with respect to the basis
$\{u,v-au,xu\}$
{\fiverm $$x=\pmatrix{0&0& 1\cr
              0&-1&0\cr
              1&0&0\cr}\ ,\quad {\rm and}\quad
              z= z_1=\pmatrix{\eta &b&c\cr
              0&\eta^2&d\cr
              0&0&\eta^4\cr}\quad {\rm or}\quad
              z= z_2=\pmatrix{\eta &b&c\cr
              0&1&d\cr
              0&0&\eta^{-1}\cr}$$}
where $b,c,d$ are suitable elements of $\overline F$ (and $\eta = 1$ if
$p=7$).\enskip
Note that $b\not=0,$ again by the irreducibility of $K.$ Thus,
conjugating $x$ and $z_i$ by ${\rm diag}\thinspace (1,b^{-1},1),$
we may also assume $b=1.$
\medskip
1)\enskip Let $K_1=\langle x,z_1\rangle .$
As $y=xz_1$ must have trace 0, it follows
$c=\eta^2 .$
Moreover, equating the elements
in position (1,1) of $y^2$ and $y^{-1},$ one gets
$d= \eta^4+\eta^5.$
This gives
{\fiverm $$z_1=\pmatrix{\eta &1&\eta^2\cr
              0&\eta ^2&\eta^4+\eta^5\cr
              0&0&\eta^4 \cr}\quad
              y=\pmatrix{0&0&\eta^4 \cr
              0&-\eta^2&-\eta^4-\eta^5 \cr
              \eta&1&\eta^2\cr}$$}
and a direct calculation shows that $z_1^7=y^3=[x,y]^4=1.$
But the relations $x^2=y^3=(xy)^7=[x,y]^4=1$
correspond to a well-known presentation for
$\PSL(2,7)$ (see [CM], 7.5, p.96). We conclude
$K_1\simeq \PSL(2,7).$
Note that $K_1$ is irreducible, as it does not fix  any
non-zero vector.

\smallskip
2)\enskip Now let  $K_2:=\langle x,z_2\rangle ,$
Again $y = xz_2$ has trace 0, hence
$c=1 .$ Moreover, equating the elements
in position (1,1) of $y^2$ and $y^{-1},$ one gets
$d= 2.$
This gives
{\fiverm $$x=\pmatrix{0&0& 1\cr
              0&-1&0\cr
              1&0&0\cr}\ ,\quad
              z_2=\pmatrix{\eta &1&1\cr
              0&1&2\cr
              0&0&\eta^{-1} \cr}.$$}
If $p=2,\ K_2$ is reducible, whereas if $p\not= 2$
$K_2$ is irreducible, as it does not fix any non-zero vector.
In the latter case, conjugating $x$ and $z_2$ by
${\rm diag}\thinspace (1,2\eta^4,1),$ one gets the matrices
{$$\pmatrix{0&0& 1\cr
              0&-1&0\cr
              1&0&0\cr}\ =\
              \varphi \Big (\pm
              \pmatrix {0&-1\cr 1&0\cr}\Big )
              \quad ,\quad
              \pmatrix{\eta &2\eta^4&1\cr
              0&1&\eta^3\cr
              0&0&\eta^{6} \cr}\ =\ \varphi \Big (
              \pm\pmatrix {\eta^4&1\cr 0&\eta^{-4}\cr} \Big )$$}
where $\varphi$ is the embedding given in the
statement. It follows from {\it i}) that the projective image of
{\fiverm $$\langle \pmatrix {0&-1\cr 1&0\cr}\ , \
\pmatrix {\eta^4&1\cr 0&\eta^{-4}\cr}\rangle $$}
is isomorphic to $H,$ in other words
$K_2\simeq \varphi (H) .$
\medskip
Finally, let
$\langle x, z\rangle\leq \SL(3,\overline F)$
be the preimage of a
$(2,3,7)$-generated subgroup of
$\PSL(3,\overline F).$ We may assume $p\not= 3,$ so that
the centre of $\SL(3,\overline F)$ has order 3 and is generated
by the scalar matrix $\beta^3\thinspace I ,$ where
$\beta$ is a primitive $9^{th}$ root of 1.
We may also assume that $x$ and
$z$ have respective orders
2 and 7. In particular, $x$ is conjugate to
${\rm diag} (-1,-1,1)$ and has an eigenspace of dimension 2.
Assume, by way of contradiction, that
$y= x\thinspace z$
has order 9, hence
$y^3=\beta^3\thinspace I .$
Since $y$ must have pairwise distinct eigenvalues
by the irreducibility of $\langle x, z\rangle ,$
$y$ should be conjugate to
${\rm diag} (\beta ,\beta^4 , \beta^7),$ which has determinant
$\not=1 ,$ a contradiction.
It follows that $y$ has order 3, hence
$\langle x, z\rangle$  is conjugate
either to $K_1$ or to $K_2 .$
 
\medskip
{\it iii}) Under our assumptions we may consider the
automorphism $\gamma_i$ of $\GL(n,R)$
induced by $\epsilon \mapsto \epsilon^i ,$
for some $i\not\equiv 0,\pm 1\ ({\rm mod}\ k).$
The image of
$H$ under the map
$g\mapsto g\otimes \gamma_i(g)$
is the group
{\fiverm $$\langle
\pmatrix
{0&0&0&1\cr
0&0&-1&0\cr
0&-1&0&0\cr
1&0&0&0\cr}\ ,\
\pmatrix
{\epsilon^{i+1}&\epsilon^i&\epsilon&1\cr
0&\epsilon^{i-1}&0&\epsilon^{-1}\cr
0&0&\epsilon^{-i+1}&\epsilon^{-i}\cr
0&0&0&\epsilon^{-i-1}\cr}\rangle .$$}
As $1$ is not an eigenvalue of the second generator,
it follows, from the remark at the beginning of the proof, that this
group is irreducible unless possibly when $p=2$ and it leaves
invariant a 2-dimensional subspace. But
this possibility is ruled out, using a direct argument
(omitted here, but entirely similar to the one given
below, see {\it iv}).
Note that the above group preserves the symmetric bilinear
form defined by the matrix $J\otimes J,$ where $J={\rm antidiag}(-1,1),$
hence it is orthogonal if $p\not= 2 .$
\medskip
{\it iv}) The group
$\overline H=\langle \pmatrix {0&-1\cr 1&0\cr}\ , \
\pmatrix {\epsilon&1\cr 0&\epsilon^{-1}\cr}\rangle$
considered in {\it i}) acts on the
third symmetric power of $V:= \overline F^2,$ that is on the space of
homogeneous polynomials of degree 3 in two variables $t,u .$ With
respect to the basis $\{t^3,t^2u,tu^2,u^3\},$
$\overline H$ acts faithfully as
{\fiverm $$\langle
\pmatrix
{0&0&0&-1\cr
0&0&1&0\cr
0&-1&0&0\cr
1&0&0&0\cr}\ ,\
\pmatrix
{\epsilon^{3}&\epsilon^2&\epsilon&1\cr
0&\epsilon&2&3\epsilon^{-1}\cr
0&0&\epsilon^{-1}&3\epsilon^{-2}\cr
0&0&0&\epsilon^{-3}\cr}\rangle .$$}
$\overline H$ preserves the symplectic form
defined by the matrix ${\rm antidiag} (-3,1,-1,3),$
and the projective image of $\overline H$
is isomorphic to $H.$
If $p=3,$ this form is degenerate, thus
$\overline H$ is reducible. 
Let $p\not= 3.$
As $\overline H$ is perfect and 1 is not a common eigenvalue
of $\overline x$ and $z,$ it follows
easily that $\overline H$ cannot
leave invariant a subspace of dimension 1 or 3.
Assume that $\overline H$ leaves invariant a 2-dimensional
subspace $U=\langle u_1,u_2\rangle.$
If $p\not=k,$ $z_{|U}$ is diagonalizable and
must have determinant 1. Thus we may choose as $u_1$ an
eigenvector of $z$ with corresponding eigenvalue
$\epsilon^{3}$ or $\epsilon ,$ and set
$u_2=\overline x\thinspace u_1 .$
A direct calculation of the eigenvectors of $z$ gives
the two possibilities:
$$U=\langle e_1,\ e_4\rangle\qquad {\rm or}\qquad
U=\big \langle\ {\epsilon\over 1-\epsilon^2}\thinspace e_1\thinspace +
\thinspace e_2 \enskip ,\enskip
-e_3\thinspace +\thinspace {\epsilon\over 1-\epsilon^2}\thinspace
e_4\ \big\rangle\ ,$$
where $\{e_1,e_2,e_3,e_4\}$ is the standard basis of $\overline F^4.$
In both cases $U$ is not $z$-invariant. Finally let $p=k,$ hence
$\epsilon =1.$ Clearly $U$ contains the unique eigenspace
of $z ,$ generated by $e_1.$ Since $\overline x\thinspace e_1=e_4 ,$
we obtain $U=\langle e_1,e_4\rangle$. But, as noted above, this subspace
is not $z$-invariant. We conclude that
$\overline H$ is irreducible. \qed


\vskip20pt
{\bf 7.\enskip Case by case analysis}
\par\nobreak\bigskip
Our aim in this section is to prove Theorems 2,3,4 and 5.
In the statements of all these theorems, $H$
is an absolutely irreducible
subgroup of $\SL(n,F).$ Furthermore, if $p\not= 2,$
$H$ is not contained in an orthogonal group
and, if $p=2,$ $H$ is not contained in a symplectic group.
Hence, by Lemma 4.1, $d^H_S=0.$
Now, suppose that $H$ is generated
by $x,\ y,\ z=xy$ of order 2, 3 and 7 respectively.
By applying Scott' s formula to the
adjoint representation of $H$ on $M=\Mat(n,F)$
and to the representation of $H$ on the symmetric square $S,$
we obtain the necessary conditions:
$$(\ast)\quad d^x_M+d^y_M+d^z_M\leq n^2+2\qquad {\rm and}\qquad (\ast\ast)
\quad d^{z}_S\leq {\rm dim}\ S-
(d^x_S+d^y_S).$$
We conclude that $H$ is not $(2,3,7)$-generated
whenever the assumption that $x,y,z$ satisfy both
$(\ast )$ (or a refined version of it, cfr. Lemma 7.1 below)
and $(\ast\ast )$ leads to a
contradiction.

\medskip
We note that $x,y,z$ satisfy $(\ast)$
precisely when $x,y,z$ belong to the
\lq admissible \rq\ conjugacy classes obtained in Section 5.
These classes are listed in Tables 2 and 3
for $x$ and $y,$ and in Table 4 for $z$ modular.
For certain values of $n$ they are also listed in Table $4^\prime$
for $z$ semisimple. The tables also give the corresponding
values of $d^x_S,\ d^y_S,\ d^z_S.$

\bigskip
{\bf Proof of Theorem 2}

\par\nobreak\medskip
Let $H$ be as stated in the theorem, and assume
that it is (2,3,7)-generated. The assumption that
$x,y,z$ satisfy $(\ast )$ and $(\ast\ast )$ leads immediately
to a contradiction, except in the case $n=6,\ p=2 ,$ which requires
a little more work.

\medskip
$\underline{n=4}$.\enskip
$d^x_S=6,\ d^y_S=4$ and ${\rm dim}\ S=10$ imply $d^z_S\leq 0.$
A contradiction, as $d^z_S\geq 1.$
\medskip
$\underline{n=5}$. \enskip
$d^x_S=9,
\ d^y_S=5$ and ${\rm dim}\ S=15$ imply $d^z_S\leq 1 .$
A contradiction, as $d^z_S= 3.$
\medskip
$\underline{n=6}$.\enskip Assume first
$d^x_S=13.$ Then $d^y_S=7$
and ${\rm dim}\ S=21$ imply
$d^z_S\leq 1.$ A contradiction, as $d^z_S\geq 2.$
We are left with the case $d^x_S=12,$
$x$ conjugate to $\pmatrix {(\pi_2}^3),$
$p=2 .$ Now  $d^y_S=7$ implies $d^z_S\leq 2,$ $z$ conjugate to
$\pmatrix{\epsilon ,\epsilon ,1,\epsilon^{3} ,
\epsilon^{-3} ,\epsilon^{-2}}.$\enskip
The faithful action of $H$ on the exterior square $E,$ of dimension
15, gives rise to an embedding $i: H\to {\rm GL}(15,\overline F).$
Recall that in the present case we are assuming
$H=\SL(6,q)$ or $\SU(6,q^2).$ It is
well-known (e.g. cfr. [Lie]) that $i(H)$ is
irreducible. It is also easy to see that $i(x),$ $i(y)$
and $i(z)$ are respectively conjugate to the matrices
$$\pmatrix{1^3,(\pi_2)^6}\ ,\
\pmatrix{(\pi_3)^5}\ ,\
\pmatrix{(1),(\epsilon)^3,(\epsilon^2)^2,(\epsilon^3),
(\epsilon^4)^3,(\epsilon^5)^3,(\epsilon^6)^2}.$$
Hence, by applying Scott' s formula to the action of $i(H),$
by conjugation, on the space $M_E={\rm Mat}(15,\overline F),$
we get the contradiction:
$$d^{i(x)}_{M_E}+d^{i(y)}_{M_E}+d^{i(z)}_{M_E}=
117+75+37=229>227={\dim}(M_E)+2.$$
$\underline{n=7}$.\enskip
$d^x_S=16,\ d^y_S=10\ {\rm and \ dim}\ S=28$ imply
$d^z_S\leq 2.$ A contradiction, as $d^z_S\geq 4.$
\smallskip
$\underline{n=10}$.\enskip
$d^x_S\geq 30,\ d^y_S=19\ {\rm and \ dim}\ S=55$ imply $d^z_S\leq 6.$
A contradiction, as $d^z_S\geq 7.$\qed

\bigskip
The following lemma gives a refined version of $(\ast ),$ which will
be needed in order to prove Theorem 3.
\medskip
{\bf Lemma 7.1}\enskip {\it Let $H$ be one of
the groups $\SL(n,q),\ \Sp(n,q)$
or $\SU(n,q).$ Assume either
\smallskip
a)\ $q \equiv 1\ ({\rm mod}\ 7)$ and
$q\not\in \{p,p^3,3^6\};$ or
\smallskip
b)\ $q \equiv 3,5\ ({\rm mod}\ 7)$ and $q \not=p\ .$
\smallskip
If $(x,y,z)$ is a $(2,3,7)$-generating triple for $H$ and
$y$ is real\ or\ modular, then:}
$$ d^x_M\leq d^2_{\overline M}+{\rm def}_{\overline M}-2\quad ,\quad
d^y_M\leq d^3_{\overline M}+{\rm def}_{\overline M}-2\quad ,\quad
d^z_M\leq d^7_{\overline M}+{\rm def}_{\overline M}-2 .$$
\medskip
{\it Proof}
\par\nobreak\medskip
{\it a})\enskip Let $d$ be the multiplicative order of
$p$ (mod 7). Then $q =p^{dk},\ k\in \N.$
Assume first $k>1$
and let $\sigma$ be the automorphism
of $F_q$ defined by $\alpha\mapsto \alpha^{p^d}.$
Then $\sigma$ is non-trivial, as the subfield of the
elements fixed by $\sigma$ has order $p^d< q .$
Consider the representation
$\gamma_1:\ {\rm GL}(n,q)\to {GL}(n^2,q)$
such that, for every $g\in \GL(n,q),\ \gamma_1(g)$
is the automorphism
of ${\rm Mat}(n,q)$ defined by
$A\mapsto g^{\sigma} Ag^{-1}.$
By a well-known theorem of Steinberg,
$\gamma_1(H)$ and its dual are irreducible
(cfr. [St]). Note that $\gamma_1$ is the representation
afforded by the tensor product module $V\otimes V^{\ast\sigma},$
where $V$ is the natural module for $\GL(n,q)$ and
$V^{\ast\sigma}$ denotes the dual of
$V$ \lq twisted\rq \ by $\sigma$).
Scott' s formula then gives:
$$d^x_{M_1}+d^y_{M_1}+d^z_{M_1}\leq n^2,$$
where, as usual, $d^g_{M_1}$ denotes the
dimension of the space of fixed points of $\gamma_1(g).$
Under our assumptions, $y$ is conjugate
to a matrix over $F_p,$ while $z$ is conjugate
to a matrix over $F_{p^d}.$
As $x$ is clearly conjugate
to a matrix over $F_p,$ it follows that
$d^x_{M_1}=d^x_M,$  $d^y_{M_1}=d^y_M,$ $d^z_{M_1}=d^z_M$
and we are done.
Now assume $k=1,$ i.e. $q =p^d.$
%%% FROM PROBLEM:
%>From $q \not\in \{p,\ p^3\}$ it follows
From $q \not\in \{p,\ p^3\}$ it follows
$d\in \{2,6\}.$ In particular $p\not= 2,$ as $2$ has order 3 (mod 7),
and $p\not= 3$ as we exclude the case $q =3^6.$
Let $\sigma$ be the
automorphism of order 2 of $F_q ,$
and consider the representation
$\gamma_2:\ {\rm GL}(n,q)\to {GL}(n^2,q)$
such that, for every $g\in \GL(n,q),\ \gamma_2(g)$  is the automorphism
of ${\rm Mat}(n,q)$ defined by
$A\mapsto g^{\sigma} Ag^t.$
Again $\gamma_2(H)$ and its dual are irreducible (cfr. [St];
$\gamma_2$ is the representation
afforded by the module $V\otimes V^{\sigma}$).
Hence, with obvious notation,
$d^x_{M_2}+d^y_{M_2}+d^z_{M_2}\leq n^2.$
As $\sigma (\epsilon)=\epsilon^{-1},$
$\sigma (\omega)= \omega^{\pm 1}$ and
$x,y,z$ are diagonalizable, it follows that
$x^\sigma,\ y^\sigma,\ z^\sigma$
are respectively conjugate to
$x, y^{\pm 1}, z^{-1}.$
Our claim follows immediately.
\medskip
{\it b})\enskip In this case $q =p^t,\ t>1.$ Note that,
as $q \equiv 3,5\ ({\rm mod}\ 7),$ by Lemma 5.5 $z$ is rational,
hence is conjugate to a matrix over $F_p$. Now let $\sigma$ be
the Frobenius automorphism $\alpha\mapsto \alpha^p,$ and apply
the same argument used above for the case $k>1.$ \qed

\bigskip
{\bf Proof of Theorem 3}
\par\nobreak\medskip

Let $H$ be as stated in the theorem,
and suppose that it is (2,3,7)-generated. Our strategy is
as follows. Using Tables 2,3 and 4
and Lemma 5.4 we show that $z$ cannot be modular,
real or semirational. It follows from
Lemmas 5.5, 5.6 and 5.7
that $H$ can only
be either {\it i}) $\SL(n,q)$ with $q\equiv 1\ ({\rm mod}\ 7),$
or {\it ii}) $\SU(n,q)$ with $\sqrt q\equiv -1\ ({\rm mod}\ 7).$
This proves our statement when $H$ is a symplectic group
and when $H$ has degree $n=12.$
Thus we are left with the cases $H=\SL(n,q)$ or $\SU(n,q),$
$q\equiv 1\ ({\rm mod}\ 7),$ $n\in \{8,9,11\}.$
We assume, by way of contradiction,
$q \not\in \{p,p^3,3^6\}.$ Since $y$
is real or modular (cfr. Table 3) and
${\rm def}_{\overline M}=2 ,$
it follows from Lemma 7.1 that $z$ satisfies the condition
$d^z_M=d^7_{\overline M}.$
Hence, by Lemma 3.1, $z$ has two similarity invariants of respective
degrees $n-7$ and $7.$ We reach a contradiction
by showing that $z$ does not satisfy the condition
$(\ast\ast ) ,$ i.e. $d^z_S\leq {\rm dim}\ S-(d^x_S+d^y_S).$

\medskip
$\underline{n=8}$.\enskip
$d^x_S\geq 20,\ d^y_S\geq 12$ and ${\rm dim}\ S=36$ imply
$d^z_S\leq 4.$\enskip $z$ cannot be modular since, for $p=7,$
$d^z_S=6.$
By Lemma 5.4 $z$ cannot be real, since $2d^z_S<d^7_{\overline M}=10.$
Assume $z$ semirational. Then
$d^z_M\leq 12$ implies, by the same Lemma, $m_0=2,\ \ m_{1}=m_{-1}=1.$
But this in turn implies $z$ real,
a contradiction.
Finally, in cases {\it i}) and {\it ii}),
$d^z_M=d^7_{\overline M}$ implies
$z\sim \pmatrix{1,\pi_7},$
hence $d^z_S=6,$ a contradiction.

\medskip
$\underline{n=9}.$\enskip
$d^x_S=25,\ d^y_S\geq 15$ and
${\rm dim}\ S=45$ imply $d^z_S\leq 5.$
$z$ cannot be modular since, for $p=7,$
$d^z_S\geq 7.$
$z$ cannot be real, as
$2d^z_S<d^7_{\overline M}=13.$
Assume $z$ semirational, and non-real.
$d^z_M\leq d^7_{\overline M}+{\rm def}_{\overline M}=15$
implies $m_0=0,\ m_1=2,\ m_{-1}=1.$ But, in this case,
$d^z_S=6,$ against the bound $2d^z_S\leq 10.$
Finally, in cases {\it i}) and {\it ii}),
$d^z_M=d^7_{\overline M}$ implies
$z\sim \pmatrix{\epsilon ,\epsilon^{-1},\pi_7},$
hence $d^z_S=7,$ a contradiction.

\medskip
$\underline{n=11}.$\enskip
$d^x_S=36,\ d^y_S\geq 22$ and
${\rm dim}\ S=66$ imply $d^z_S\leq 8.$
$z$ cannot be modular since, for $p=7,$
$d^z_S\geq 10.$
$z$ cannot be real, as
$2d^z_S<d^7_{\overline M}=19.$
Assume $z$ semirational.
$d^z_M\leq d^7_{\overline M}+{\rm def}_{\overline M}=21$
implies $m_0=2,\ m_1=2,\ m_{-1}=1.$ But, in this case,
$d^z_S=9,$ against the bound $2d^z_S\leq 16.$
Finally, in cases {\it i}) and {\it ii}),
$d^z_M=d^z_{\overline M}$ implies
that $z$ is conjugate either to
$\pmatrix{\epsilon,\epsilon^2 , \epsilon^{-1},\epsilon^{-2},\pi_7}$
or to $\pmatrix{1,\epsilon , \epsilon^2 ,\epsilon^{-3},\pi_7}.$
In both cases $d^z_S\geq 9,$ a contradiction.

\medskip
$\underline{n=12}.$\enskip $d^x_S\geq 42,\ d^y_S\geq 26$ and
${\rm dim}\ S=78$ imply $d^z_S\leq 10.$
$z$ cannot be modular since, for $p=7,$
$d^z_S\geq 12.$
$z$ cannot be real, as
$2d^z_S<d^7_{\overline M}=22.$\enskip
Assume $z$ semirational, and non-real.
$d^z_M\leq d^7_{\overline M}+{\rm def}_{\overline M}=26$
implies $m_0=3,\ m_1=2,\ m_{-1}=1.$ But, in this case,
$d^z_S=12,$ against the bound
$2d^z_S\leq 20 .$
\qed

\bigskip
{\bf Proof of Theorem 4}
\par\nobreak{\medskip}
Let $H$ be as stated in the theorem,
and assume that it is (2,3,7)-generated.
Note that, under our assumptions,
$y$ is real or modular by Lemmas 5.5, 5.6 and 5.7.
This assumption is necessary in order to obtain contradictions
by playing condition $(\ast )$ against
$(\ast \ast ).$ Our claims follow from the arguments
below and Table 5.
\medskip
{\bf A)}\enskip $x,y,z$ cannot satisfy $(\ast )$ and
$(\ast \ast ).$ 
\smallskip
$\underline{n=13}.$\enskip
$d^x_S\geq 49,\ d^y_S\geq 31$ and
${\rm dim}\ S=91$ imply $d^z_S\leq 11,$ contradicting
Tables 4 and  $4^\prime .$
\smallskip
$\underline{n=14},\ p\not= 2.$\enskip
$d^x_S=57,\ d^y_S\geq 35$ and
${\rm dim}\ S=105$ imply $d^z_S\leq 13,$
contradicting Tables 4 and $4^\prime .$
For the case $p=2,$ see below.
\medskip
{\bf B)}\enskip If $x,y,z$ satisfy $(\ast )$ and
$(\ast \ast ),$ then $z$ is neither modular, nor real, nor semirational.
\smallskip
$\underline{n=16}.$\enskip
$d^x_S\geq 72,\ d^y_S\geq 46$ and
${\rm dim}\ S=136$ imply $d^z_S\leq 18.$
Thus $z$ cannot be modular by Table 4.\enskip
On the other hand, if $p\not=7,$
$2d^z_S\leq 36 < 38= d^7_{\overline M},$ hence $z$ is not real.
Finally, it is easy to see, using the condition $d^z_M\leq 44,$
that $z$ can be semirational (and not real) only if
$m_0=1,\ m_1=3,\ m_{-1}=2.$
But in this case $d^z_S=19,$ a contradiction.
\smallskip
$\underline{n=18},\ p\not=2.$\enskip
$d^x_S\geq 91,\ d^y_S\geq 57$ and
${\rm dim}\ S=171$ imply $d^z_S\leq 23.$
Using this condition and $d^z_M\leq 54,$
it is easy to check that $z$ cannot
be semirational. The cases $z$ real or modular are
dealt with below.

\medskip
{\bf C)}\enskip If $x,y,z$ satisfy $(\ast )$ and
$(\ast \ast ),$ then $z$ is neither modular nor real.
\smallskip
$\underline{n=14},\ p= 2 .$\enskip
$d^x_S\geq 56$ and $d^y_S\geq 35$ imply
$d^z_S\leq 14.$\enskip
Assume $z$ real. Using Lemma 5.4, we get
$28\geq 2d^z_S\geq d^z_M\geq d^7_{\overline M}=28,$ whence
$d^z_M=d^7_{\overline M}.$ By Lemma 3.1 $z$ is conjugate
to $\pmatrix{(\pi_7)^2}$ and $d^z_S= 15,$ a contradiction.
\smallskip
$\underline{n=18},\ p\geq 2.$\enskip
$d^x_S\geq 90$ implies $d^z_S\leq 24.$
$z$ cannot be modular by Table 4.\enskip
Assume $z$ real.
$48\geq 2d^z_S\geq d^z_M\geq d^7_{\overline M}=48$ implies
$d^z_M=d^7_{\overline M}.$ By Lemma 3.1 $z$ is conjugate
to $\pmatrix{(\epsilon,
\epsilon^{-1},(\epsilon^{i},\epsilon^{-i}),(\pi_7)^2}\ (i=2\ {\rm or}\ 3),$
and we get $d^z_S= 25,$ a contradiction.

\smallskip
$\underline{n=15}.$\enskip
$d^x_S\geq 64,\ d^y_S\geq 40$ and
${\rm dim}\ S=120$ imply $d^z_S\leq 16.$
Thus $z$ cannot be modular by Table 4.
On the other hand, if $p\not=7,$
$2d^z_S\leq 32 < 33=d^7_{\overline M},$ hence $z$ is not real.
\smallskip
$\underline{n=17}.$\enskip
$d^x_S\geq 81,\ d^y_S\geq 51$ and
${\rm dim}\ S=153$ imply $d^z_S\leq 21.$
Thus $z$ cannot be modular.\enskip
On the other hand, if $p\not=7,$
$2d^z_S\leq 42 < 43= d^7_{\overline M},$ hence $z$ is not real.
\smallskip
$\underline{n=19}.$\enskip
$d^x_S\geq 100,\ d^y_S\geq 64$ and
${\rm dim}\ S=190$ imply $d^z_S\leq 26.$
\enskip $z$ cannot be modular.\enskip
If $p\not=7,$
$2d^z_S\leq 52 < 53=d^7_{\overline M},$ hence $z$ is not real.
\qed
\bigskip
{\bf Proof of Theorem 5}
% \par\nobreak{\medskip}
% Let $H$ be as stated in the theorem,
% and assume that it is (2,3,7)-generated.
% As $\vert F(\theta):F\vert=12,$
% $y$ is real and $z$ is rational.
% Using Lemmas 5.3 and 4.2 it is easy to determine
% the admissible conjugacy classes of $z$
% and, when they exist, to compute $d^z_S.$
% (Alternatively this information can be
% deduced from Table 4.)
% Our claim follows since, for the values of $n$
% considered in the statement, conditions
% $(\ast )$ and
% $(\ast \ast )$ are incompatible. \qed
\par\nobreak{\medskip}
Let $H$ be as stated in the theorem,
and assume that it is (2,3,7)-generated.
As $\vert \Q(\omega):\Q\vert=2$ and
$\vert \Q(\epsilon):\Q\vert=6 ,$
$y$ is real and $z$ is rational.
Using Lemmas 5.3 and 4.2 it is easy to determine
the admissible conjugacy classes of $z$
and, when they exist, to compute $d^z_S.$
(Alternatively this information can be
deduced from Table 4.)
Our claim follows since, for the values of $n$
considered in the statement, conditions
$(\ast )$ and
$(\ast \ast )$ are incompatible. \qed

\vskip20pt
\centerline{\bf References}

\nobreak{\bigskip}
[Coh] J. COHEN, On non-Hurwitz groups and noncongruence
subgroups of the modular group, Glasgow Math. J. {\bf 22}
(1981), 1-7.
\smallskip
[Con] M. CONDER, Generators for alternating and symmetric
groups, {\it J. London Math. Soc.} (2) {\bf 22} (1980), 75-86.
\smallskip
[CM] H. COXETER and W. MOSER, {\it Generators and relations for discrete
groups}, Ergeb. Math. {\bf 14} Springer, Third Edition 1972.
\smallskip
[Fe] W.FEIT, {\it The representation theory of finite groups}, North-Holland
1982.

\smallskip
[GM] D. GLUCK and K. MAGAARD, Character and fixed point ratios in finite
classical groups, Proc. London Math. Soc. (3) {\bf 71} (1995), 547-584.

\smallskip
[HPS] D.F. HOLT, W. PLESKEN and B. SOUVIGNIER, Constructing a
representation of the group $(2,3,7,11)$, J. Symbolic Computation,
{\bf 24}(1997), 489 - 492.

\smallskip
[Isa] I.M. ISAACS, {\it Character Theory of Finite Groups}, Academic Press 1976.

\smallskip
[Ja] N. JACOBSON, {\it Basic Algebra} I, W.H.Freeman and Company 1985.
\smallskip
[Jon] G.A. JONES, Characters and surfaces: a survey. In: `{\it An atlas
of finite groups: $10$ years}', Cambridge Univ. Press, Cambridge, 1997, 96
-116.

\smallskip
[KL] P.B. KLEIDMAN and  M.W. LIEBECK, {\it The Subgroup Structure
of the Finite Classical Groups}, L.M.S.
Lecture Notes Series
{\bf 129}, Cambridge University Press 1990.
\smallskip
[Lie] M.W. LIEBECK, On the orders of maximal subgroups
of the finite classical groups, Proc. London Math. Soc. (3)
{\bf 50} (1985), 426-446.

\smallskip
[Lu] A. LUCCHINI, $(2,3,k)$-generated groups of large rank (to appear).

\smallskip
[LTW] A. LUCCHINI, M. C. TAMBURINI and J. S. WILSON,
Hurwitz groups of large rank\ (to appear).

\smallskip
[LT] A. LUCCHINI and M. C. TAMBURINI Classical groups of large
rank as Hurwitz groups\ (to appear).

\smallskip
[Mac] A.M. MACBEATH, Generators of the linear fractional
groups, Proc. Symp. Pure Math. {\bf 12} (1969), 14-32.

\smallskip
[Mag] W. MAGNUS, {\it Noneuclidean Tesselations and Their Groups},
Academic Press 1974.
\smallskip
[Mal1] G. MALLE, Hurwitz groups and $G_2(q),$ Canad. Math. Bull.
{\bf 33} (1990), 349-357.
\smallskip
[Mal2] G. MALLE, Small rank exceptional Hurwitz groups, {\it Groups
of Lie type and their geometries}, LMS Lecture Notes, vol. 207, Cambridge
University Press, 1995, 173-183.

\smallskip
[Sc] L.L. SCOTT, Matrices and cohomology, Ann. Math. {\bf 105} (1977),
473-492.

\smallskip
[St] R. STEINBERG, {Lectures on Chevalley groups}, Mimeographed
Notes, Yale 1967.

\smallskip
[TV] C. TAMBURINI, S. VASSALLO, (2,3)-generazione di $\SL(4,q)$
in caratteristica dispari e problemi collegati, Boll. Un. Mat. Ital.
({\bf 7}) 8B (1994), 121-134.

\smallskip
[Wil] J.S. WILSON, Simple images of triangle groups (to appear).

\bye

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