Is group cohomology with the inversion action order two?

Question

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 $2$?

Answer 1

This is false.

Let's work with $H^2(G,U(1))$ instead of $H^3(G,\mathbb{Z})$.

A counterexample is given by the group $\mathbb{Z}_3 \times (\mathbb{Z}_3 \rtimes \mathbb{Z}_2)$ with the map $x(a,b,c)=c$. There is an order $3$ cohomology class represented by the cocycle $$\omega((a,b,c),(a',b',c'))=e^{2\pi i\,a'b/3}~.$$ Another counterexample is $\mathbb{Z}_4 \rtimes \mathbb{Z}_4$ with $x(a,b)=b\text{ mod }2$, which has an order $4$ class represented by $$\omega((a,b),(a',b'))=e^{2\pi i\,a'b/4}~.$$

However the claim is true when there exists an element of $g$ in the center of $G$ with $x(g)=1$.

The details are worked out in Appendix C of our paper https://arxiv.org/abs/2311.18782.