About the intersection of two commutator subgroups I'm reading an example in Lennox & Stonehewer's book "Subnormal Subgroup of Group".
There (p.144/145) they construct the following example. Let $L$ be an infinite elemetary abelian $2$-group and let $K=\langle x\rangle$ be a cyclic group of order $2$. Form the wreath product $W=L\wr K$. Let $F$ be the field of two elements, $B=FW$ (the group ring), and form the split extension $G=B\rtimes W$ (the action is by multiplication), so essentially $G$ is $C_2\wr (L\wr C_2)$. They claim $$[B,L]\cap[B,L^x]=[B,L,L^x].$$ I cannot understand why the equality holds. It is clear that $[B,L]\cap[B,L^x]\geq[B,L,L^x]$, but I do not see the other inequality. I thought I could see $[B,L]$ as $B(1-L)$ in the group ring $B$, so as a left-ideal, but why the intersection should then be $B(1-L)(1-L^x)$? I think there must be some smart way to see this, but I cannot see it.
 A: This is an attempt at a reasonably simple-minded explanation (which may not be sufficiently rigorous).
The base group $B$ is freely generated (as a vector space over ${\mathbb F}_2$) by elements $b_{(l,m)}$ with $l \in L$ and $m \in L^x$. So elements  $g \in B$ are uniquely represented as sums $$g=\sum_{l \in L,\, m \in L^x}{a_{(l,m)} b_{(l,m)}},$$
where each $a_{(l,m)}=0$ or $1$, and all but finitely many
$a_{(l,m)}$ are $0$.
Writing all of the groups $B$, $L$, and $L^x$ additively, we have, for $l' \in L$, $[b_{(l,m)},l'] = b_{(l+l',m)} + b_{(l,m)}$. It follows from this that $[B,L]$ consists of all elements $g$ of the form above in which, for each fixed $m \in L^x$, the number of $l \in L$ for which $a_{(l,m)}$ is nonzero is even.
Similarly, $[B,L^x]$ consists of all sums of the form above in which, for each fixed $l \in L$, the number of $m \in L^x$ for which $a_{(l,m)}$ is nonzero is even.
So $[B,L] \cap [B,L^x]$ consists of the elements $g$ represented by sums that both of the above two conditions hold.
To prove the result we need to show that an element $g$ satisfying these two conditions lies in $[B,L,L^x]$. We do this by induction on the number of $m \in L^x$ for which there exists $l \in L$ with $a_{(l,m)} \ne 0$. In the base case, when there no such $m$, we have $b=0$.
Suppose that $m \in L^x$ is such that there exists $l \in L$ with $a_{(l,m)} \ne 0$. Then $h := \sum_{l \in L,}{a_{(l,m)} b_{(l,m)}}$ has an even number of nonzero $a_{(l,m)}$, and so it lies in $[B,L]$. By assumption, there exists some $m' \in L^x$ with $m \ne m'$ such that $a_{(l,m')} \ne 0$.
Now $h' := [h,m+m']$ lies in $[B,L,L^x]$, it has the same coefficients $a_{l,m}$ for that fixed value of $m$ as $g$ and all other nonzero coefficients of $h'$ are of the form  $a_{l,m'}$.
So the expression for $g + h'$ has no nonzero coefficients of the from $a_{l,m}$ for the chosen value of $m$, and we can apply induction to $g + h'$ to complete the proof.
