Permutation wreath product ($H{\wr }_{X}K=H^X⋊K$)

If a group $K$ acts on a finite set $X$, it acts on the direct product of groups $H^X$ by permuting its factors. The resulting semidirect product is the permutation wreath product. For $K=S_m$ acting on $m$ two-point blocks and $H=C_2$, it describes independent within-block swaps combined with a permutation of blocks, and has order $2^{m}m!$. The action on $X$ is part of the definition; it need not be the regular action used in a regular wreath product.

For any perfect group $H$ and $m\ge 3$, the permutation wreath product $H\mathrm{wr}{A}_m$ is generated by elements of order three. The top $A_m$ is generated by 3-cycles. With $\sigma =(123)$, the product of the conjugate $h_1\sigma h_1^{-1}$ and ${\sigma }^{-1}$ is $h_1{h}_2^{-1}$; using ${\sigma }^{-1}$ gives $h_1{h}_3^{-1}$. The commutator of $h_1{h}_2^{-1}$ and $k_1{k}_3^{-1}$ is $[h,k{]}_1$, since distinct coordinate groups commute. Perfectness gives the entire first coordinate group, and transitivity gives all coordinates. No finite-generation assumption on $H$ is needed.

For every $m\ge 5$, let $A=A_m$ act by rooted permutations on the rooted tree of finite words over $m$ letters. Define $b(a)$ to have local permutation $a\in A$ at the pairwise incomparable vertices $1^{r}2$ for $r\ge 0$, and identity elsewhere. Then $b$ is an embedding of $A$ and satisfies $b(a)=(b(a),a,1,\dots ,1)$ with trivial root permutation. The group $H=⟨A,b(A)⟩$ is finitely generated and perfect because it is generated by two finite perfect subgroups.

The root permutation and sections define an injective homomorphism $H\to H\mathrm{wr}{A}_m$. To prove surjectivity, conjugate $b(c)$ by the rooted cycle $(234)$ and take its commutator with $b(a)$. In wreath coordinates this leaves only $b([a,c])$ in the first coordinate. Perfectness of $A_m$ therefore supplies the entire localized copy of $b(A)$ there. Canceling it from the recursion for $b(a)$ supplies a localized rooted $a$ in the second coordinate. Transitivity moves both kinds of generators to every coordinate, yielding the full base $H^m$ and hence the full wreath product. The group is infinite, since a nontrivial finite order $h$ could not satisfy $h=h^m|A_m|$.