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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    $\begingroup$ 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)? $\endgroup$ Jul 25, 2013 at 15:41
  • $\begingroup$ Well it's not true in characteristic $p$. You could take the image of the representation to be a Sylow $p$-subgroup of ${\rm SL}(3,p)$. $\endgroup$
    – Derek Holt
    Apr 4, 2015 at 7:49

2 Answers 2


$\DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\SL}{SL} \DeclareMathOperator{\Aut}{Aut}$ For clarity, let's assume that $V$ is a module on which $G$ acts by a faithful representation $D\colon G \to \GL(V)$. Then "every element of $H$ is induced by $\SL(V)$" means that for every $h\in H$ there is some $S\in \SL(V)$ such that $D(g^h) = D(g)^S$ for all $g\in G$.

The exact answer will depend on the assumptions on the module $V$ and the ground field:

When $V$ is an irreducibe, faithful module over the complex numbers (or any algebraically closed field of characteristic $\neq p$), then $H=\Aut_{\SL(V)}(G)$.

Proof. An irreducible, faithful character of $G$ vanishes outside the center, and is determined by its values on the center. Thus the representation $g\mapsto D(g^h)$ is similar to $D$ for $h\in H$, and so there is $S\in \GL(V)$ with $D(g^h) = D(g)^S$ for all $g$. As the field is algebraically closed, we can multiply with a suitable scalar to get $S\in \SL(V)$. This shows $H\subseteq \Aut_{\SL(V)}(G)$.

Conversely, when $V$ is absolutely irreducible, then every element in the normalizer of $D(G)$ in $\SL(V)$ (or $\GL(V)$) induces an element of $H$, since the center of $G$ is mapped to the center of $\GL(V)$ by $D$. Thus $\Aut_{\SL(V)}(G) \subseteq H$.

In all other cases, both inclusions may fail:
We still have $H\subseteq \Aut_{\SL(V)}(G)$, when $V$ is not irreducible, but all irreducible constituents are faithful. On the other hand, if we allow for non-faithful (=linear) constituents, then $h\in H$ may map these to other constituents, and so $D$ and $D^h$ can not be similar.
Also, when the field is not algebraically closed, but $V$ irreducible (even absolutely irreducible), then $D$ and $D^h$ are similar, but maybe not in $\SL(V)$. This happens in the exponent $p^2$ case, and also in the case where $G$ has order $3^3$ and exponent $3$.

If $V$ is not absolutely irreducible, then an element in $\Aut_{\SL(V)}(G)$ may induce a non-trivial action on the center of $G$. (Whether this can happen is determined by the restriction of $V$ to the center.)

And finally, in characteristic $p$ it's not true, as said by Derek Holt. (In characteristic $p$, the only simple module is the trivial module, so we would have to allow arbitrary modules here.)


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.

  • $\begingroup$ 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 $Z(G) \times {\rm Sp}(2r,p).$ $\endgroup$ May 15, 2013 at 7:03
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    $\begingroup$ @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. $\endgroup$ May 15, 2013 at 7:19

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