[Self studying Robinson, ex 1.6.20]
Robinson asks us to prove that $G/[B,K] \cong(H/H') \times K$ (where $G=H \wr K$, $K\neq 1$ and $B$ is the base group.) As a hint, we are told to show first that $B'\leq [B,K]$.
I can prove the hint. First, the generators of $B'=[B,B]$ are elements of the form $g^{-1}h^{-1}gh$. Elements in different components of B commute, so w.l.o.g. g and h can be taken as elements of the same, single component of B, say the y component $H_y$ (where $y\in K$). It is therefore enough to show that this $[g,h]\in [B,K]$
Since K is not 1 we can choose a non-trivial k in K. Then $[g,k]$ is the element of B with $g$ as the $yk$ component and $g^{-1}$ as the $k$ component.
Then $[g,h] = [g,k][h,k][(gh)^{-1},k]$ and so lies in $[B,K]$. $\square$
But now I have no idea how to proceed to the conclusion. I need another hint, please.
Hint: Consider the following map $f:G\to (H/H')\times K$: $$((h_y)_{y\in K},\,k)\mapsto\left(\prod_y\overline{h_y},\,k\right)$$ where the elements $\overline{h_y}\in H/H'$ already commute.
Prove that this map is a surjective homomorphism and show that its kernel is just $[B,K]$, using your lemma for one direction.