group cohomology of abelianization

Question

Is it true that, for a finite or compact group $G$, $$H^3(G,\mathbb{Z})=H^3(G/[G,G],\mathbb{Z})\times H^3([G,G],\mathbb{Z})~?$$ It is clearly true for abelian and perfect $G$. I have checked a few other examples, and it has held.

Answer 1

In general, a short exact sequence of groups $1 \to N \to G \to G/N \to 1$ leads to a spectral sequence of group cohomology, with $$ E_2^{pq} = H^p(G/N, H^q(N, A)) \implies H^{p+q}(G,A).$$ This is known as the Lyndon-Hochschild-Serre spectral sequence. This tells us roughly that $H^3(G)$ will be determined by the cohomologies in pairs of degrees which add to $3$, modulo relations arising from total degrees $2$ and $4$.

Note here that $H^q(N,A)$ is a $G/N$-module, and may be nontrivial even if $A$ is a trivial $G$-module. So, this cohomology depends on the way $N$ and $G/N$ interact in $G$.