From the answer to another question I asked ([Projective representations of a finite abelian group](https://mathoverflow.net/questions/452161/projective-representations-of-a-finite-abelian-group/452166#452166)) and from the structure theorem of finite abelian groups it follows that if $A$ is a finite abelian group with $H^2(A,U(1))=0$ then for every subgroup $B\subset A$ we also have $H^2(B,U(1))=0$ (namely $U(1)$ is a cohomologically trivial $A$-module).


1. Is there a way to prove this without using the structure theorem and computing explicitly the groups? For instance, is it the case that the map $H^2(A,U(1))\rightarrow H^2(B,U(1))$ induced by $K(B,1)\rightarrow K(A,1)$ is surjective? If yes, why?
2. Depending on the answer to the previous question this might be trivial or not: is it true that for any $A$-module $M$ and any $n$ such that $H^n(A,M)=0$, then $H^n(B,M)=0$? If not, are there conditions on $M$ (and maybe $n$) for which this is true?



Trivial group cohomology induces trivial cohomology of subgroups
For any abelian group $A$ we have a canonical isomorphism $\bigwedge^2A\to H\_2(A,\mathbb{Z})$, given by the (anti-symmetric) Pontrjagin product $H\_1(A,\mathbb{Z})\times H\_1(A,\mathbb{Z}) \to H\_2(A,\mathbb{Z})$, see for example Section 6 of Breen, "[On the functorial homology of abelian groups](https://doi.org/10.1016/S0022-4049(98)00140-6)". It follows that if $A$ is finite and $B$ is a subgroup of $A$ then $H\_2(B,\mathbb{Z}) \to H\_2(A,\mathbb{Z})$ is injective. Now $H^2(A,U(1))$ is the Pontrjagin dual abelian group to $H\_2(A,\mathbb{Z})$, and so for $B$ a subgroup of $A$ the restriction map $H^2(A,U(1))\to H^2(B,U(1))$ is surjective.


**Edit.** And now I see that I was too fast in writing this. It is not true that if $B \to A$ is an inclusion of finite abelian groups then $\bigwedge^2 B \to \bigwedge^2 A$ is injective. Let $A=\mathbb{Z}/p \times \mathbb{Z}/p^2$, generated by elements $u$ and $v$, and let $B$ be the subgroup $\mathbb{Z}/p\times \mathbb{Z}/p$ generated by $u$ and $pv$. Then $u\wedge pv$ is non-zero in $B\wedge B$ but zero in $A \wedge A$ because it equals $pu\wedge v$. So the answer to question 1 is no, for this inclusion.


Nonetheless, it is true for an inclusion of finite abelian groups, that if $\bigwedge^2A=0$ then $\bigwedge^2B=0$. Thus $H^2(A,U(1))=0$ implies $H^2(B,U(1))=0$.


But beware that for infinite abelian groups, even this is false. Look for example, at the inclusion $\mathbb{Z}/p\times\mathbb{Z}/p\to\mathbb{Z}/p\times\mathbb{Q}/\mathbb{Z}$; in this case $\bigwedge^2A=0$ but $\bigwedge^2B=\mathbb{Z}/p$.


