Commutators in an unrestricted infinite wreath product Consider an unrestricted wreath product $G = \prod_X A \rtimes B$, where $A$ is a group and $B$ is some subgroup of $\mathrm{Sym}(X)$.  I am wondering what the circumstances are under which $\prod_X A$ is contained in (the closure of) the derived group $[G,G]$.


*

*If $X$ is finite, then $\prod_X A \nleq [G,G]$, because there is a nontrivial $G$-invariant homomorphism $\prod_X A \rightarrow A$ given by summing the entries.

*If there is $b \in B$ such that all orbits of $\langle b \rangle$ are infinite, then a standard argument shows that every element of $\prod_X A$ is a commutator of $G$.  (We represent $h \in \prod_X A$ as $[b,h']$, where the entries of $h$ along each $b$-orbit are given by differences of successive entries of $h'$.)
Edit: Suppose $\prod_X A$ has the product topology (with respect to some group topology on $A$).  Under what circumstances does $[G,G]$ contain a dense subgroup of $\prod_X A$?
 A: Then for the density question, the only obstruction is the existence of infinite orbits.
Proposition: let $B$ be a group, let $A$ be a nontrivial abelian group and $X$ a $B$-set. Then the intersection of $D(A^X\rtimes B)$ with $A^X$ is dense in $A^X$ (for the product topology, $A$ being discrete, and $D$ meaning derived subgroup) if and only if all $B$-orbits are infinite.
Lemma: Let $B$ be a group, let $A$ be an abelian group and $X$ a transitive $B$-set. Then the coinvariants of $B$ in the $\mathbf{Z}$-module $A^{(X)}$ consists of the subgroup $A^{(X)}_0$ of elements with sum zero.
[The coinvariants of a $\mathbf{Z}B$-module $M$ is the $\mathbf{Z}$-submodule of $M$ generated by the elements $bm-m$ when $(b,m)\in B\times M$. So the quotient is the largest quotient module with trivial $B$-action.]
Proof: clearly it is contained in $A^{(X)}_0$. Also $A^{(X)}_0$ is generated as $\mathbf{Z}$-module by the elements with support of cardinal two. Let $\delta_x(a)-\delta_y(a)$ be such an element (with self-explanatory notation). Since the action is transitive, there is $b\in B$ such that $y=bx$, and hence this is a commutator in an obvious way.
As a consequence, we get:
Lemma: Let $B$ be a group, let $A$ be an abelian group and $X$ a $B$-set. Then the coinvariants of $B$ in the $\mathbf{Z}$-module $A^{(X)}$ consists of the set $A^{(X)}_{B,0}$ of elements $m\in A^{(X)}$ such that for every $B$-orbit $Y\subset X$ we have $\sum_{y\in Y}m(y)=0$. $\Box$
Proof of the proposition: if there is a finite orbit $Y$, the mapping $(m,b)\mapsto \sum_Y m$ is a continuous surjective homomorphism $A^X\rtimes B\to A$. If $A$ is nontrivial, this homomorphism is nontrivial on $A^X$ and this implies failure of the density (this part was already noted by the OP). 
Conversely, suppose that every $B$-orbit is infinite. Let $Y$ be a finite subset of $X$, and let $m_0$ be an element of $A^Y$. Since all orbits are infinite, we can find $m\in A^{(X)}_{B,0}$ whose restriction to $Y$ is $m_0$. By the corollary, $m\in D(A^X\wr B)$. This proves the density.
Corollary: Let $B$ be a group, let $A$ be a group and $X$ a $B$-set. Then the intersection of $D(A^X\rtimes B)$ with $A^X$ is dense in $A^X$ if and only if


*

*either $A$ is a perfect group, or

*all $B$-orbits are infinite. $\Box$
