# Is group cohomology with the inversion action order two?

Let $\mathbb{Z}^\text{inv}$ denote the $\mathbb{Z}/2$-module defined by the inversion action on $\mathbb{Z}$. Let $G_0$ be a finite group that acts trivially on $\mathbb{Z}^\text{inv}$. Then one can compute via the Künneth Theorem and UCT for cohomology,\begin{align}H^3(G_0\times\mathbb{Z}/2;\mathbb{Z}^\text{inv})&=\text{Tor}(H^3(G_0;\mathbb{Z}),\mathbb{Z}/2)\times H^2(G_0;\mathbb{Z})\otimes_\mathbb{Z}\mathbb{Z}/2\times \mathbb{Z}/2\\&=H^2(G_0;\mathbb{Z}/2)\times \mathbb{Z}/2.\end{align}We find that every non-zero element in the cohomology group is of order $2$.

Now let $G_0\rightarrow G\xrightarrow{x} \mathbb{Z}/2$ be a possibly non-trivial extension of $\mathbb{Z}/2$ by $G_0$. A $G$-action on $\mathbb{Z}^\text{inv}$ is defined by $g:n\mapsto (-1)^{x(g)}n$. Is it true that every non-zero element in $H^3(G;\mathbb{Z}^\text{inv})$ is of order $2$?

## migrated from math.stackexchange.comOct 17 '17 at 19:26

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• Just to clarify, here $\mathbb{Z}/2$-module means $\mathbb{Z}[\mathbb{Z}/2]$-module (i.e., one considers $\mathbb{Z}/2$ as a group, not as a ring). – YCor Oct 17 '17 at 19:28
• Yes, that's what I mean. – Alex Turzillo Oct 17 '17 at 19:44