When does a $p$-group split as semidirect product with its commutator?

Question

Given a (finite) group $G$ and its commutator subgroup $G'$, when is it the case that $$1\to G'\to G\to G/G'\to 1$$ splits?

Specifically, can we say anything if we add the assumption that $G'$ is abelian ($G$ is nilpotent of class 2), or more generally if $G$ is nilpotent?

EDIT It is most interesting to consider this question when $G$ is a $p$-group. We may furthermore simplify this by supposing that $G$ is a $p$-group of exponent $p$ (ie. $g^p=1$ for every $g\in G$), and therefore both $G'$ and $G/G'$ are both elementary abelian.

Answer 1

If $G$ is nilpotent and $G' \ne 1$, then $G'$ does not have a complement in $G$.

Because if it had a complement $Q$, say, then $Q \cong G/G'$ would be abelian, so $$G' = [QG',QG'] = [Q,Q][Q,G'][G',G'] = [Q,G'][G',G'] \le [G,G'].$$

But if $G$ is nilpotent with $G' \ne 1$ then $[G,G']$ is the next term in the lower central series of $G$, so it is strictly contained in $G'$, contradiction.

Answer 2

The Schur-Zassenhaus Theorem states that if $G$ is a finite group, and $N$ is a normal subgroup whose order is coprime to the order of the quotient group $G/N$, then the short exact sequence $$ 1\to N\to G\to G/N\to 1 $$ splits. In particular, this applies for the normal subgroup $N=G'$, if the orders are coprime.

If $N$ is abelian, we can use the second group cohomology $H^2(G/N,N)$. If it is trivial, then the extension splits.