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



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> **Question 1.** Is it possible to classify finite $\Gamma$-modules $A$ with $H^2(\Gamma,A)=0$ ?
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*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 $H^2(\Gamma,A)=0$ ?
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Finite $\Gamma$-modules with trivial $H^2$, where $\Gamma$ is a group of order 2
The answer to question 2 is yes. Take $A=\mathbb Z/8\mathbb Z$ where $\gamma$ acts by multiplication by $5$. Then $A^\Gamma= \ker( \cdot 4)= 2\mathbb Z/8\mathbb Z$ but $(\mathrm{id} +\gamma)(3)=6\cdot 3=2$, thus $H^2(\Gamma,A)$=0 (and obviously $A$ doesn't contain any induced module).


For the question 1 I do not really know what to say except that if $A$ is killed by 2, then $H^2(\Gamma,A)=0$ forces $A$ to be induced. However as you see in the example above, the "$H^2(\Gamma,A)=0$"-property does not pass to the 2-torsion subgroup $A[2]$ or the subquotients $A[2^i]/A[2^{i+1}]$, so I do not think this gives you anything when considering a general $A$. I also think that asking $H^2(\Gamma,A)=0$ doesn't really simplify the classification (so if we restrict to $2$-groups we just get finite $\mathbb Z\_2[x]/(x^2-1)$-modules for which $\ker(x-1)=\mathrm{im}(1+x)$).


