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Let $G$ be an extraspecial group of order $p^{2r+1}$ (where $p$ is an odd prime), and let $V$ be a faithful representation of $G$. Consider the normal subgroup $H$ of $Aut(G)$ consisting of all elements of $Aut(G)$ acting trivially on the center $Z(G)$.

Is every element of $H$ induced by $SL(V)$ ? More generally, is it true that $H=Aut_{SL(V)}(G)$ ?

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I am not clear on what is being asked. Is V supposed to be a module for G over the complex numbers? Is it supposed to be a simple module? Now G is being viewed as a subgroup of Aut(V) = GL(V). Are we being asked if each element of H can be obtained as conjugation by an element of the normalizer of G in SL(V)? –  Marty Isaacs Jul 25 '13 at 15:41
Well it's not true in characteristic $p$. You could take the image of the representation to be a Sylow $p$-subgroup of ${\rm SL}(3,p)$. –  Derek Holt Apr 4 at 7:49

1 Answer 1

Well, I do know that if $G$ has exponent $p$ then $H/Inn(G)\cong Sp(2r,p)$ and if $G$ has exponent $p^2$ then $H/Inn(G)$ is the semi-direct product of $Sp(2r-2,p)$ and another extraspecial group of order $p^{2n-1}$.

I believe I also see why $Aut_{SL(V)}G \subset H$, however the other direction is not entirely clear to me.

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I corrected my post- I was thinking about the exponent $p$ case. The other direction is OK, as I explained, because the whole of ${\rm Sp}(2r,p)$ is induced by the action of ${\rm SL}(V)$ since the representation does extend to the appropriate central extension of $H,$ which has a subgroup $Z(G) \times {\rm Sp}(2r,p).$ –  Geoff Robinson May 15 '13 at 7:03
@John, as soon as you get 50 points of reputation you will be able to add comments. Using the answer box to respond to someone's answer does not work at all. –  Mariano Suárez-Alvarez May 15 '13 at 7:19

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