# Special automorphisms of extraspecial groups

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 $H=Aut_{SL(V)}(G)$ ?

• 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)? Jul 25, 2013 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 ${\rm SL}(3,p)$. Apr 4, 2015 at 7:49

$\DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\SL}{SL} \DeclareMathOperator{\Aut}{Aut}$ 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.

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