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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

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

  2. 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.

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    $\begingroup$ By an ideal do you mean a two-sided ideal? Or you just want to speak about a cyclic module? In the second case, consider the following example: $G=S_3$, and $Z[G]/R = M = \mathbb{Z}/6$, where the action of $\tau = (123)$ is trivial, and the action of $\sigma = (12)$ is given by acting with -1. Then the extension $1\to M\to A\to G\to 1$ in which $s(\sigma)^2 = 3\in M$, $s(\tau)^3 = 2\in M$ and $s(\sigma)s(\tau)s(\sigma)^{-1} = s(\tau)^{-1}$ will give you an element of order 6 in $H^2(G,M)$. $\endgroup$
    – Ehud Meir
    Commented Jul 13, 2018 at 8:13
  • $\begingroup$ Thanks for counterexample! (Yes, just a cyclic module, I was just trying to avoid cramming two slightly different "cyclic" in one sentence.) $\endgroup$
    – Denis T
    Commented Jul 13, 2018 at 8:24

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