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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 $m=3,4$.

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 $B=H_1\times \cdots \times H_m$, 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.

First establish a stronger form of the coset-counting argument. If $K$ is a finitely generated infinite-index subgroup of a finitely generated group $L$ of growth at most $q$, choose a finite generating set $T$ for $K$, and let $M$ bound its lengths in a generating set $S$ of $L$. The Schreier graph supplies $r+1$ distinct left-coset representatives of $K$ of $S$-length at most $r$. The corresponding translates of $B_T(r)$ are disjoint and lie in $B_S((M+1)r)$. HenceThus an infinite-index finitely generated subgroup has growth at most $q-1$ whenever the ambient upper bound is $q$.

Every finitely generated $K\le H$ has infinite index in $G$. Applying this result with $L=G$ gives growth at most $n-1$. Minimality of $n$ among the supposed counterexamples therefore makes each such $K$ virtually nilpotent. Here $n\ge 1$, since a group with a uniform bound on its finite word balls is finite, and finite groups are virtually nilpotent.

For each finitely generated $K\le H$, let $d(K)$ be the least nonnegative integer giving a polynomial growth upper bound. It exists and belongs to $\{0,\dots ,n-1\}$. Choose $K$ for which this integer is maximal. If $K\le L\le H$ and $L$ is finitely generated, then $d(L)\le d(K)$. Were $[L:K]$ infinite, the strengthened bound would make $K$ have growth at most $d(L)-1\le d(K)-1$, contradicting the definition of $d(K)$. If $d(K)=0$, the same inequality would force the nonempty balls of $K$ to have cardinality tending to zero, also impossible. Hence every such overgroup has $[L:K]<\infty$.

Since $K$ is virtually nilpotent, choose a nilpotent finite-index subgroup $N\le K$. It is finitely generated by Schreier's lemma. For any $h\in H$, the subgroup $L=⟨K,h⟩$ is finitely generated and $[L:K]<\infty$. Pigeonholing the cosets $h^{j}K$ gives a positive $a$ with $h^a\in K$. Pigeonholing the cosets of $N$ in $K$ then gives a positive $b$ with $(h^a{)}^b\in N$. ConsequentlyNo normality of $K$ in $L$ or of $N$ in $K$ is required. Using least integer upper bounds also avoids needing any further structural theorem about the exact growth degree of a virtually nilpotent group.

Choose finite symmetric generating sets $S_i$ for all $G_i$, and let $L\ge 1$ bound the $S$-length of their elements. If $[G_{i-1}:G_i]=\infty$, the connected locally finite Schreier graph of left cosets of $G_i$ in $G_{i-1}$ is infinite. Its radius-$r$ ball has at least $r+1$ vertices: otherwise a finite-diameter locally finite connected graph would be finite. Equivalently, we can choose $r+1$ distinct left-coset representatives $t_{i,0},\dots ,t_{i,r}$ of $G_i$, each of $S_{i-1}$-length at most $r$, and hence $S$-length at most $Lr$.

The $(r+1{)}^n$ productsare distinct. Indeed, equality first forces equality of the $G_1$-cosets of the first factors, and hence their indices; cancellation then forces equality of the $G_2$-cosets of the second factors, and so on. Each product has $S$-length at most $nLr$. ThereforeLetting $r\to \infty$ yieldsFor $n=0$ the assertion is automatic.

Let $|g{|}_S$ be the least length of a word in $S\cup S^{-1}$ representing $g$, and write $B_S(r)=\{g:|g{|}_S\le r\}$. Polynomial growth of a group of degree at most $d$ means that, for some constant $C$, In particular these balls are finite. For the usual convention $S$ is finite; finiteness of $B_S(1)$ would force that in any case.

For a finite subset $T$, set $L=\max (1,{\max }_{t\in T}|t{|}_S)$. Replacing each letter of a $T$-word by an $S$-word givesThus $|B_T(r)|\le C(1+Lr{)}^d\le C{L}^d(1+r{)}^d$. This proves the same degree bound for $⟨T⟩$. If $T$ is empty the subgroup is trivial and the assertion is immediate.

All commutators of $H$ with $K$ lie in the central subgroup $Y$ of $HK$. In the standard commutator identities, conjugation by elements of $H$ or $K$ therefore acts trivially on these commutators. Since $Y$ is abelian, their order can also be reversed. ConsequentlyThus the map is a homomorphism in each variable separately; this does not assert that it is a homomorphism from the direct product jointly.

It factors through $H/C_H(K)$ in the first variable and $K/C_K(H)$ in the second. If these finite quotient groups have representatives $h_1,\dots ,h_a$ and $k_1,\dots ,k_b$, then the finite collection $[h_i,k_j]$ generates $Y$. Let $e$ be an exponent of $H/C_H(K)$. For every $h,k$,Hence $Y$ is a finitely generated abelian group of bounded exponent, and $Y$ is finite. For instance its order is at most $e^{ab}$.

Now suppose $G$ is nilpotent and $G/{\gamma }_2(G)$ is finite. Induct on the nilpotency class $c$. For $c=1$, $G$ is its own abelianization and is finite. If $c\ge 2$, the group $\bar{G}=G/{\gamma }_c(G)$ has smaller class andsince ${\gamma }_c(G)\le {\gamma }_2(G)$. By induction $\bar{G}$ is finite. Apply the preceding result with $H=G={\gamma }_1(G)$ and $K={\gamma }_{c-1}(G)$. Their commutator subgroup is $Y={\gamma }_c(G)\le Z(G)$. Both relevant centralizer quotients are finite, because their centralizers contain ${\gamma }_c(G)$ and $G/{\gamma }_c(G)$ is finite. Thus ${\gamma }_c(G)$ is finite. The extension of this finite normal subgroup by the finite quotient $\bar{G}$ is finite. The converse follows because a quotient of a finite group is finite. Therefore

Suppose inductively that $G=H{\gamma }_i(G)$. In the quotient $G/{\gamma }_{i+1}(G)$, the image of ${\gamma }_i(G)$ is central. Hence this quotient is generated by the image of $H$ together with a central subgroup, so its commutator subgroup lies in the image of $H$. Pulling this back givesThe original assumption now gives $G=H{\gamma }_2(G)\le H{\gamma }_{i+1}(G)\le G$. Starting with $i=2$, induction proves $G=H{\gamma }_i(G)$ for every $i\ge 2$. These products are subgroups because each ${\gamma }_i(G)$ is normal; no normality of $H$ is needed.

If $G$ is finitely generated, every quotient, including its abelianization $G/{\gamma }_2(G)$, is finitely generated. Conversely, lift a finite generating set of that quotient to $g_1,\dots ,g_t\in G$, and set $H=⟨g_1,\dots ,g_t⟩$. Then $G=H{\gamma }_2(G)$. For a nilpotent group, ${\gamma }_{c+1}(G)=1$ for some $c$, and the established identity gives $G=H$. Thus

Use the commutator convention $[x,y]=x^{-1}{y}^{-1}xy$, and let $[A,B]$ denote the subgroup generated by all such commutators with $x\in A,y\in B$. The lower central series isA central series has $[G_j,G]\le G_{j-1}$ for each $j\ge 1$. We prove the requested inclusion by induction. At $i=0$, ${\gamma }_1(G)=G_n$. If $i<n$ and ${\gamma }_{i+1}(G)\le G_{n-i}$, thenThereforeIn particular ${\gamma }_{n+1}(G)=1$. The proof uses the central-series condition directly; no equality of the intermediate terms is claimed.

A bounded holomorphic function of planar Brownian motion is a local martingale up to exit, because its real and imaginary parts are harmonic and Itô's formula has zero drift. Stop before the given $S<T_D$. Localization inside compact subdisks gives the local martingale $f(Z_{t\wedge S}^z)$, and $|f|<1$ makes it bounded. The bounded local martingale criterion therefore gives a true martingale.

Exit from the unit disk is finite almost surely, so $S<T_D$ is finite too. Taking $t\to \infty$ and using the dominated convergence theorem yieldsThe same proof works for any bounded holomorphic function, after rescaling if needed; it does not require the stopping time itself to be bounded.

The disk automorphism satisfies $f(w)=0$ and maps the unit circle bijectively to itself. Its derivative $(1-|w{|}^2)/(1-\overline{w}z{)}^2$ is bounded above and away from zero on the closed disk. By conformal invariance of planar Brownian motion, the image path up to the exit time is a time-changed planar Brownian motion started at zero. Its exit point is $f(Z_T)$; the clock changes when it exits, but not where.

The original exit time is finite almost surely. For example, $|Z_{t\wedge T}{|}^2-2(t\wedge T)$ is a martingale, so $\mathbb{E}(t\wedge T)\le (1-|w{|}^2)/2$. Boundedness of $f^{\prime }$ makes the image exit clock finite too. By rotational invariance of planar Brownian motion, the image exit angle is uniform on $[0,2\pi )$.

Let $\Theta =\arg Z_T$. Parts (a) and (b) show that $\varphi (\Theta )$ modulo $2\pi$ is uniform, where $\varphi$ is strictly increasing and increases by exactly $2\pi$ in one circuit. The one-dimensional change-of-variables formula therefore givesIts integral is one because the total angle change is $2\pi$. This harmonic measure density is the Poisson kernel for the unit disk divided by $2\pi$. It is uniform for $w=0$ and concentrates near the boundary point closest to $w$ as $|w|$ increases.

Differentiate the argument through the nonvanishing curve, using $g^{\prime }=\mathrm{Im}({\beta }^{\prime }/\beta )$. Since ${\beta }^{\prime }=-i{e}^{i\theta }$,Also $|w-e^{i\theta }{|}^2=1+|w{|}^2-2(w_1\cos \theta +w_2\sin \theta )$. Subtracting one therefore givesThus the lifted boundary angle $\varphi (\theta )=-\theta +2g(\theta )$ is strictly increasing. Its total change around the circle is $2\pi$, because the disk automorphism has boundary degree one. The same total change follows directly from the winding number one of $w-e^{i\theta }$. This is the boundary-angle derivative of a disk automorphism.

Let $\beta (\theta )=w-e^{i\theta }=|\beta (\theta )|e^{ig(\theta )}$. It is never zero, since $|w|<1$. On the unit circle,Dividing these identities givesAny other continuous lift of the argument differs by an integer multiple of $2\pi$, which leaves the exponential unchanged.