Past exam of the building infinite groups course of the University of Oxford / 2011 / 3

Let $B=H_1\times \cdots \times H_m$, with each $H_i$ a copy of $H$. The permutation group $F\le S_m$ acts on $B$ byDefine the permutation wreath product to be the semidirect productExplicitly,Writing $h_i$ for the element with entry $h$ in coordinate $i$ and identity elsewhere, and writing $\sigma$ for $(1,\sigma )$, conjugation satisfiesThus the top permutation group permutes the coordinate subgroups in exactly its given action. Distinct coordinate subgroups commute, including when $H$ itself is nonabelian.

Put $A=A_m$ and let $T$ be the rooted tree of finite words over $\{1,\dots ,m\}$. View $a\in A$ as a rooted automorphism: it permutes the first letter by $a$ and fixes the suffix. Denote this rooted copy by $A$ again.

For $a\in A$, define a directed tree automorphism $b(a)$ by putting local permutation $a$ at every vertexand identity local permutation at every other vertex. These marked vertices are pairwise incomparable, so the prescribed actions are on disjoint rooted subtrees and define a unique automorphism. Equivalently its root permutation is trivial and its sections areThe map $a\mapsto b(a)$ is an injective homomorphism: it is the same copy of $a$ acting on each of those disjoint subtrees. Let $B=b(A)$ and defineBoth $A$ and $B$ are finite, so $H$ is finitely generated. For $m\ge 5$, $A_m$ is a nonabelian simple group and hence perfect. Thus $A,B\le H^{\prime }$, and because they generate $H$, $H$ is perfect.

Every section of either defining generator belongs to $H$, so sections of every element of $H$ belong to $H$. Its root permutation belongs to $A$. For a tree automorphism $g$, write $g(iw)=\sigma (i)g_i(w)$, where $g_i$ is its section at input letter $i$. In the semidirect-product convention of part (a), defineComposition of tree automorphisms gives exactly the wreath-product multiplication: this is why $\theta$ is a homomorphism. It is injective because an automorphism is determined by its first-level permutation and all its first-level sections. On the generators,It remains to prove surjectivity; merely writing these recursions would not establish it. Choose $\sigma =(234)\in A$, which fixes $1$ and sends $2$ to $3$. For $a,c\in A$, the two base tuples $\theta (b(a))$ and $\theta (\sigma b(c){\sigma }^{-1})$ overlap only in their first coordinate. Their commutator is thereforeSince $A$ is perfect, products of these commutators give every element of the copy of $B$ localized in coordinate $1$. Canceling this first coordinate from $\theta (b(a))$ then gives $(1,a,1,\dots ,1;1)$, the rooted copy of $A$ localized in coordinate $2$. Conjugating by the transitive rooted group $A_m$ supplies both localized $A$ and localized $B$ in every coordinate. Since $H=⟨A,B⟩$, the image contains the full base $H^m$. It also contains the top $A_m$. Therefore it is the entire wreath product andFinally $H$ is infinite: it contains the nontrivial rooted $A_m$, and a finite order $h=|H|$ would satisfy $h=h^m|A_m|$, which is impossible for $m\ge 5$.

Let $L$ be the subgroup of $H\mathrm{wr}{A}_m$ generated by all its elements of order $3$. The top group $A_m$ is generated by 3-cycles, also when $m=3$, so $A_m\le L$.

Take $\sigma =(123)$. For any $h\in H$, the conjugate $h_1\sigma h_1^{-1}$ has order $3$, as does ${\sigma }^{-1}$. Thus their product belongs to $L$ and equalsUsing ${\sigma }^{-1}$ instead supplies $d_{13}(h)=h_1{h}_3^{-1}$. Since distinct coordinate groups commute,The hypothesis that $H$ is perfect says that products of its commutators generate $H$, so $H_1\le L$. Transitivity of $A_m$ now gives $H_i\le L$ for every coordinate. Hence $H^m\le L$ and $A_m\le L$, provingThe proof does not require $H$ itself to be finitely generated, nor does it require $A_m$ to be perfect for $m=3,4$.