Looking at this Example of group cohomology not annihilated by exponent of $G$? I stumbled upon one question I couldn't solve (probably because it's hard), so I post it here.

Using Lyndon resolvent, we can always find $G$-module $M$ of rank equal to number of relators in some presentation such that $H^2(G, M)$ has maximal exponent (and is equal to $\Bbb Z/|G|$). Because one-relator finite group is cyclic, I asked myself

Is it true that if there's ideal $R$ with $H^2(G, \Bbb Z[G]/R)$ of exponent $|G|$, then $G$ cyclic?

If there's $2$-generated $G$-module $M$ with $H^2$ of exponent $|G|$, what can we say about $G$?

Of course, we can look not only on second cohomology, and obtain series of filtrations on category of finite groups: $G \in LC_{n, k}$ (LC for "large cohomology") if exists a $k$-generated module s. t. $exp(H^n(G, M) = |G|$. (or maybe $H^{\leq n}$, or $H^{\geq n}$,..). Looks interesting.