Commutator of the semidirect product

Question

Let $B$ be a an abelian finite group and $A$ be a finite solvable group acting faithfully on $B$ by automorphisms. Let $G = B\ltimes A$ be the semidirect product of $B$ and $A$.

My question. Is it true that $[G,G] = B[A,A]$?

Of course $[G,G]\subseteq B[A,A]$ and it is not difficult to show that $[G,G] = B[A,A]$, when the order of $B$ is odd number or more generally $B = H[B,A]$, where $H=\{x^2: x\in B\}$, and $[B,A]$ is a commutator group (which can be considered as a standart commutator group in $G$).

But what can we say in the general case?

Answer 1

The answer to your question in general is no, since we do not necessarily have $B \le [G,G]$.

For a counterexample let $G$ be the dihedral group of order $8$, with $B$ cyclic of order $4$ and $A$ of order $2$.