Let $\Gamma=\{1,\gamma\}$ be a group of order 2. Let $A$ be a finite $\Gamma$-module, that is, a finite abelian group on which $\Gamma$ acts. It is a hopeless problem to classify finite $\Gamma$-modules.

We consider $A^\Gamma=\{a\in A\mid{}^\gamma a=a\}$ and $$H^2(\Gamma,A)=A^\Gamma/\{a'+{}^\gamma a'\mid a'\in A\}.$$

Question 1.Is it possible to classify finite $\Gamma$-modules $A$ with $H^2(\Gamma,A)=0$ ?

*Remark.* If $A$ is of odd order, then $H^n(\Gamma,A)=0$ for $n>0$.

*Reduction 1.* Write $A=A_{\rm odd}\oplus A_2$, where $|A_{\rm odd}|$ is odd and $|A_2|=2^m$.
Then $H^2(\Gamma,A)=H^2(\Gamma,A_2)$.
Therefore, from now on we assume that $A$ is a 2-group.

*Remark.* Let $B$ be an *induced $\Gamma$-module*, that is, $B=C\oplus C$, where $C$ is an abelian group
and $\Gamma$ acts on $B$ by
$$^\gamma(c,c')=(c',c)\quad\text{for $(c,c')\in C\oplus C=B$.} $$
Then $H^n(\Gamma,B)=0$ for $n>0$.

*Reduction 2.* Let $B\subset A$ be a $\Gamma$-submodule that is induced.
Then the canonical epimorphism $A\to A/B$ induces an isomorphism $H^n(\Gamma,A)\cong H^n(\Gamma,A/B)$ for $n>0$. Therefore, from now on we assume that $A$ has no induced $\Gamma$-submodules.

Question 2.Does there exist a finite $\Gamma$-module $A$ of order $2^m$ for some $m>1$ without induced $\Gamma$-submodules and such that $H^2(\Gamma,A)=0$ ?