What are finite groups $H$ such that $H^n(H,\mathbb{Q/Z}) \cong H_n(G,\mathbb{Z})$?

Question

Let $G$ be a finite group and $G^{\prime}$ be its commutator subgroup. Let $\mathbb{Z}$ and $\mathbb{Q}$ denote the integers and rationals. $\mathbb{Z}$ and $\mathbb{Q/Z}$ treated as trivial $G$-module. We know that $H^1(G,\mathbb{Z})Hom(G,\mathbb{Q/Z}) \cong Hom(G/G^{\prime},\mathbb{Q/Z}) \cong G/G^{\prime} \cong H_1(G,\mathbb{Z})$ and $H^2(G,\mathbb{Q/Z}) \cong H_2(G,\mathbb{Z})$, where $H_n$ and $H^n$ denote the homolgy and cohomology resp.

What are finite groups $H$ such that $H^n(H,\mathbb{Q/Z}) \cong H_n(G,\mathbb{Z})$ ?

I thought Universal coefficient theorem will help. But it do not seem to help.

Answer 1

It's true for any finite group (for $n>0$).

The long exact sequence of cohomology for the short exact sequence $0\to\mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}\to0$ gives $H^n(G,\mathbb{Q}/\mathbb{Z})\cong H^{n+1}(G,\mathbb{Z})$, and the Universal Coefficient Theorem gives $\operatorname{Ext}^1(H_n(G,\mathbb{Z}),\mathbb{Z})\cong H^{n+1}(G,\mathbb{Z})$. Finally, $H_n(G,\mathbb{Z})$ is a finite abelian group, and $A\cong\operatorname{Ext}^1(A,\mathbb{Z})$ (non-naturally) for any finite abelian group $A$.