Let $p$ be a prime and let $M\_p$ be the $\mathrm{GL}\_2(\mathbb{F}\_p)$-module of $2 \times 2$ matrices over $\mathbb{F}\_p$ with trace $0$ (the action is by conjugation).


Is it true that for $p$ large enough we have $H^2(\mathrm{GL}\_2(\mathbb{F}\_p), M\_p) \neq \{0\}$ ?



Second cohomology of the adjoint representation
Yes it has dimension $1$ for $p \ge 5$.


This module is actually the natural module for ${\rm SO}(3,p)$, which is isomorphic to ${\rm PGL}(2,p)$ (the scalars of ${\rm GL}(2,p)$ act trivially), so the result is probably in the literature somewhere.


But it's easy to do it by direct calculation. A Sylow $p$-subgroup $P$ of $G = {\rm GL}(2,p)$, which you can take to be the upper unitriangular matrices, has order $p$, and acts indecomposably on the module, so it fixes a submodule of dimension $1$.


So in an extension by the module $M\_p$, the $p$-th power of a generator of $P$ must lie in this fixed submodule, and hence $H^2(G,M\_p)$ and so also $H^2(G,M\_p)$ have dimension at most $1$. Now for $p \ge 5$, it is not hard to see that taking this $p$-th power to be nontrivial defines a nonsplit extension of $P$ of $M\_p$. (It is a split extension when $p=3$, so $H^2(P,M\_3) = H^2(G,M\_3) = 0$.)


Since distinct $p$-Sylow subgroups have trivial intersection, to check that this element of $H^2(P,M\_p)$ is stable in $G$, we just need to check that it is stable in $N = N\_G(P)$, which is the group of upper triangular nonsingular matrices. Checking that the actions of $N$ on $M\_p$ and $P$ extend to give an action of $N$ on the nonsplit extension is a routine calculation. I could give details if you like.


