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


1 Answer 1


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


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