Dear GAP Forum,

Olivier Cormier asked

Thomas Breuer wrote:

In the example mentioned,
I suspect the desired action of G on H is the natural
symplectic one.
There are several ways to construct the group in question.

2. An alternative is to use the library of perfect groups in GAP.

That's a good method, but how do you know that the group in question is perfect?

An elementary way to see this is the following.

Let $p > 3$ be a prime,
and $G$ the split extension of the extraspecial group $E$
of order $p^3$ and exponent $p$ by the group $SL(2,p)$,
w.r.t. the natural action of $SL(2,p)$.

The factor group $G/E$ is perfect.
Its unique nontrivial normal subgroup is of order $2$,
with factor group isomorphic to the nonabelian simple group
$PSL(2,p)$.

If $G$ would have a normal subgroup $N$ of prime index then
this prime would be either $2$ or $p$.
The first case is impossible since then $N$ would contain $E$.
In the second case, $N \cap E$ would be a normal subgroup of
order $p^2$ inside $E$.
But such a normal subgroup would contain the derived subgroup
$E^\prime$ of $E$,
and this is impossible since $G$ acts irreducibly on the factor
$E / E^\prime$.

Thomas Breuer wrote:

3. A third possibility is the construction of the semidirect
product as a group of 4 by 4 matrices over the field with
5 elements.

I can't see why this construction gives the group in question.Can you say a bit
more or give some references about that?

A description of this construction -for the general case of
extensions of extraspecial groups in odd characteristic by
symplectic groups- can be found in Section 2 of

Paul Gerardin,
Weil Representations Associated to Finite Fields,
Journal of Algebra 46, 54-101 (1977).

I hope this helps.

Kind regards,
Thomas

Miles-Receive-Header: reply