Showing that any central group extension of a finite group $G$ by the trivial $G$-module $\mathbb Q$ is a semidirect product

Question

Suppose that $G$ is a finite group and view the additive group $\mathbb{Q}$ as a trivial $G$-module.

I want a concrete way of understanding why the second cohomology group $H^2(G, \mathbb{Q})$ is trivial.

A good start comes from the fact that this is the same as understanding why any central extension of $G$ by $\mathbb Q$ is (up to isomorphism) $G \ltimes \mathbb Q$, i.e., the extension splits.

Unfortunately, I cannot see why this is true and would really appreciate an explanation.

I believe that there are very general theorems/concepts from cohomology theory that could be used here, but I am looking for answers that are as concrete as possible.

Answer 1

You are only considering the case of trivial action, so you are trying to prove that the extension is isomorphic to the direct product $G \times {\mathbb Q}$.

Show it first for the case when $G$ is cyclic, using the divisibility of ${\mathbb Q}$. Then the required complement is just the torsion subgroup of the extension.

As you say, there are much more general results. If $M$ is an $KG$-module with $K$ a field of characteristic $0$, and $G$ is a finite group, then $H^n(G,M)=0$ for all $n \ge 1$.