Is group cohomology killed by exponent of group?

Question

Let $G$ be a finite group, the exponent $e(G)$ is defined to be the lcm of order of elements in $G$. Let $M$ be a $G$ module, we know by restriction corestriction that $H^i(G,M)$ is annihilated by $|G|$, for positive $i$. Is there examples that $H^2(G,M)$ is not annihilated by $e(G)$?

Answer 1

Yep. For every finite group, there's $M$ such that $H^2(G, M) = \Bbb Z/|G|$.

For every group (not necessary finite) augmentation ideal $I$ can be covered by free module of rank equal to rank of group: suppose $G$ generated by $g_i$, then map goes like $$\Bbb Z[G]^{\mathrm{rk} G} \to I, x_i \mapsto (g_i -1)$$

So we have short exact sequence $M \to \Bbb Z[G]^{\mathrm{rk} G} \to I$ where $M$ is kernel; by cohomological long exact sequence $$H^2(G, M) = H^1(G, I) = \Bbb Z/|G|$$

It's noteworthy that usually this module $M$ will have pretty big rank (we can bound it above by number of relations for some presentation of $G$ (see Lyndon, Schupp, Combinatorial group theory, Ch. II.3 on Fox calculus). It's an interesting question which conditions are implied on group by existense of cyclic (or $k$-generated) module which has $|G|$-torsion; I don't know answer to it.