3
$\begingroup$

Let $T=\operatorname{PSL}_n(q)$ with $n$ a prime number. Then the $\mathscr{C}_3$ subgroup $M=\langle x\rangle{:}\langle\sigma\rangle$ of $T$ is isomorphic to $\mathbb{Z}_{\frac{q^n-1}{(q-1)(n,q-1)}}{:}\mathbb{Z}_n$, where $x$ comes from the Singer cycle.

Note that $\sigma$ has a matrix which is a permutation matrix corresponding to $(1,2,\dots,n)$ in $\operatorname{SL}_n(q)$. It follows that $\langle\sigma\rangle$ is preserved by any outer automorphism of $T$.

My question is: is $\langle x\rangle$ preserved by any outer automorphism of $T$? It is true for the diagonal automorphism and the $\mathscr{C}_3$ subgroup of $\operatorname{PGL}_n(q)$ is isomorphic to $\mathbb{Z}_{\frac{q^n-1}{q-1}}{:}\mathbb{Z}_n$. How about the field automorphism and the graph automorphism? How can we find a matrix form of $x$ in this case?

That is to say, if $o\le\operatorname{Out}(T)$, then is the $\mathscr{C}_3$ subgroup of $T.o$ just $M.o$? And how about the case when $T=\operatorname{PSU}_n(q)$ and $M$ is isomorphic to $\mathbb{Z}_{\frac{q^n+1}{(q+1)(n,q+1)}}{:}\mathbb{Z}_n$, where $n$ is still a prime number? In this case we only need to consider the field automorphism.

Improve this question
$\endgroup$
4
  • $\begingroup$ There is no uniform shape for $x$. But there is the way to construct it in a particular situation (for given $n$ and $q$), see M. W. Short, "The primitive soluble permutation groups of degree less than 256," page 15. But it is going to be the matrix in not the same basis in which the matrix of $\sigma$ is a permutational matrix most probably. $\endgroup$
    – Anton B
    Commented May 25, 2020 at 6:01
  • 2
    $\begingroup$ For other questions see "The subgroup Structure of the Finite Classical Group" by P. Kleidman and M. Liebeck, it looks like $\S 4.3$ has some answers. Equation (4.3.8) shows that there is a field automorphism of $T$ normalising $\langle x \rangle$. $\endgroup$
    – Anton B
    Commented May 25, 2020 at 6:07
  • $\begingroup$ So the answer to the question is yes, this maximal subgroup extends to a maximal subgroup of ${\rm Aut}(T)$. $\endgroup$
    – Derek Holt
    Commented May 25, 2020 at 7:44
  • $\begingroup$ @DerekHolt I can now see that a field automorphism normalises $\langle x\rangle$ by Anton's comment. But how about the graph automorphism (even though the answer seems to be yes)? $\endgroup$
    – Groups
    Commented May 25, 2020 at 8:27

1 Answer 1

Reset to default
1
$\begingroup$

This is true by Proposition 4.3.6.(I) of Kleidman and Liebeck's book "The Subgroup Structure of the Finite Classical Groups", which says that, in all cases for the linear and unitary groups, there is a unique conjugacy colass of maximal ${\mathscr C}_3$-subgroups.

In fact in your situation it is easy to prove it directly using Zsigmondy's theorem: with a few exceptions, if $q$ is a prime power and $n>1$, then there is a prime $r$ dividing $q^n-1$ that does not divide $q^m-1$ for any $m<n$. So your maximal subgroup is in fact the normalizer of a (cyclic) Sylow $r$-subgroup of ${\rm PSL}(n,q)$, and then the result is clear.

The only exceptions to Zsigmondy's theorem that apply here are with $n=2$ and $q$ a Fermat prime, in which case the groups are the normalizers of Sylow $2$-subgroups.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .