Commutator subgroup having a complement

Question

Suppose we have a finite group $G$ such that $[G,G]$ has a complement in $G$. Then what are some "good" things we can conclude from this situation? I know that "good" is not well-defined, but I am sure there are people on here who would come up with some good facts. Thanks

Answer 1

Another observation, if $H$ is a complement of $G'$, then $H$ is also a complement of each member of the lower central series $\gamma_{i+1}(G)$ for any $i\geq 1$. This can be proved by induction: if $G=HG'$, then it follows that $G'=H'[G',G]$, and this can be substituted in $G=HG'$. In particular, $G$ cannot be nilpotent.

In addition to Mesel's remarks: if $H$ is normal, then $H \subseteq Z(G)$, the center of $G$, since $H \cap G'=1$. In that case, $G=G'Z(G)$.

Answer 2

A few observation;

Let $G=G'H$ where $G'\cap H=1$;

since $G/G'\cong H$ we must have $H$ is a abelian group.

If $H$ is normal in $G$ then $G'$ is a perfect group. (so, $G$ is not solvable)

since $G= G'\times H\implies G'= G''\times H'$ but since $H$ is abelian $H'=e$ So we have $G'=G''$ so $G'$ is a perfect group.

You can set such group as an example, $G=A_5\times C_2$.