# Absolute irreducibility of a symmetric square?

This is a question I received today by email, which somebody more experienced with finite group representations can probably answer directly. Take $F:=\mathbb{F}_q$ for some prime power $q$, so $G:=\mathrm{GL}(V)$ acts naturally on both $V:=F^n$ and its symmetric square in $V \otimes V$.

If $H$ is an absolutely irreducible subgroup of $G$, does it act absolutely irreducibly on the symmetric square of $V$ (and if so, what is a reference for this)?

Maybe it's enough here just to start with any absolutely irreducible representation of a finite group $H$ in $G$, not necessarily faithful? Also, I'm unclear about what happens in characteristic 2. Otherwise the statement looks reasonable but I can't readily find a reference in standard textbooks.

• Magaard and Malle have some work on related questions. mathematik.uni-kl.de/~malle/download/symclass.pdf newton.ac.uk/preprints/NI97035.pdf – NAME_IN_CAPS Aug 25 '14 at 18:52
• It's not always true: If $H = SO_n(\mathbb{F}_q)$ then $\mathrm{Sym}^2 V$ has an invariant vector. What sort of answer is your correspondent looking for? – David E Speyer Aug 25 '14 at 18:55
• There are lots of small counterexamples. For example if $V$ is a $2$-dimensional module for a dihedral group in coprime characteristic, then its symmetric square has dimension $3$ and splits as $1 \oplus 2$. – Derek Holt Aug 25 '14 at 19:20
• Thanks to all of you for your helpful comments, which confirm my vague impression that the question (apparently from a European student) was poorly formulated. Possibly the subgroup $H$ is assumed to act irreducibly on the symmetric square, but that wasn't how the question was stated. I'll try to reply appropriately, without suggesting any of you as likely correspondents. – Jim Humphreys Aug 25 '14 at 22:17
• I tried to write this earlier, but somehow the system kicked me off: if $H$ is a finite group with an absolutely irreducible module of dimension greater than $2,$ and we take an absolutely irreducible module $V$ of maximal dimension for $H,$ then the action of $H$ on the symmetric square of $V$ can't be absolutely irreducible. – Geoff Robinson Aug 25 '14 at 23:27

Certainly not in general : For an extreme case, consider the case that $n = q-1,$ where $q$ is a power of $2$. Let $H$ be a subgroup of order $(q-1)^{3}$ of ${\rm GL}(V),$ generated by a scalar matrix of order $(q-1)$ and the permutation matrix associated to the $n$-cycle $(12 \ldots n).$ Then $H$ is nilpotent of class $2$ and acts absolutely irreducibly on $V.$ The $H$-module $V \otimes V$ is a direct sum of $n$ absolutely irreducible $H$-modules of dimension $n,$ each of which is isomorphic to the first Frobenius twist of $V.$