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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

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