If $G$ is a finite group, it is not true in general that $G$ is the semidirect product of a normal subgroup $N$ and the quotient group $G/N$. It is also not true in general that there is a subgroup of $G$ isomorphic to $G/N$.
But if $N=[G,G]$, the commutator subgroup of $G$, then I believe both these statements should be true. Are they?
My intuition is that, since the elements of $N$ are products of non-trivial commutators of $G$, whereas $G/N$ is abelian, then if $G/N$ were isomorphic to a subgroup of $G$, we would clearly have $N \bigcap G/N=\langle e \rangle$. Further, since $G/N$ is the "albelianization" of $G$, it seems intuitive that $G$ should contain a subgroup $H$ isomorphic to $G/N$. That would naturally lead to all elements of $G$ as products of elements of $N$ and $H$, which gives us $G$ as a semidirect product of $N$ and $H$. But I do not know how to prove that such an $H$ exists, or if it may not.
I ask this question because, if true, it would lead to a nice intuitive characterization of nonabelian solvable groups as nonabelian only with respect to the semidirect product of otherwise abelian factors. Thanks in advance!
EDIT: As I note in my comment, the counterexample $Q_8$ may not be valid, as $Q_8/\mathbb Z_2 \cong \mathbb Z_2 \times \mathbb Z_2$, not $\mathbb Z_4$. But please let me know if this is irrelevant.
EDIT 2: I have asked a related question about this as a group extension here.
It's false. Take the quaterions $Q_8$. Here $[G,G]=Z(G)=\langle -1\rangle \cong \mathbb Z_2$, but $Q_8$ is not the semidirect product of $\mathbb Z_2$ by $\mathbb Z_4$. To see this, simply note that every element of $Q_8$ of order $4$ squares to $-1$.
EDITED: I should have said $Q_8$ is not the semidirect product of $\mathbb Z_2$ by a group of order $4$.