# Finite $\Gamma$-modules with trivial $H^2$, where $\Gamma$ is a group of order 2

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 $$H^2(\Gamma,A)=0$$ ?

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 $$\ker(x-1)=\mathrm{im}(1+x)$$).