Past exam of the building infinite groups course of the University of Oxford / 2011 / 2

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.

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.

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.