TITLE: Derived subgroup of the base group of a standard wreath product
QUESTION [7 upvotes]: Let $G=H\wr K$ be the standard wreath product with $K\ne 1$. Prove
  that $B'\leq [B,K]$ where $B$ is the base group of $G$. Deduce that $G/[B,K]\cong (H/H')\times K$.

This is problem 1.6.20 from A Course on the Theory of Groups. When I initially looked at this problem, I quickly dismissed it thinking it would just require some simple commutator calculus. Maybe this is still the case, but after spending some time on this, I cannot find a reason why this is true. Here's an example of some of the things I've tried.
For $a,b\in B$, with $k^{-1}bk=c$, for some $c\in B$ and $k\in K$, 
\begin{align*}
[a,c]&=[a,k^{-1}bk]\\
&=[a,bk][a,k^{-1}]^{bk}\\
&=[a,k][a,b]^{k}[a,k^{-1}]^{bk}.
\end{align*}
This seems to go no where... Any help is appreciated, thank you.

REPLY [4 votes]: Here is an outline of how to do it. $B$ is a direct product $H_1 \times \cdots \times H_k$ of copies of $H$. If $h \in H$ and $k \in K$ conjugates $H_1$ to $H_2$, then $$[k,(h,1,\ldots,1)] = (h,h^{-1},1,\ldots,1) \in [B,K].$$ Now $[B,K]$ is normalized by $B$, so for any $h' \in H$, $$[(h',1,\ldots,1),(h,h^{-1},1,\ldots,1)] = ([h',h],1,\ldots,1) \in [B,K],$$ so $[B,K]$ contains $(H',1,\ldots,1)$. Since the same is true for the other direct factors of $B$, it contains $[B,B]$.