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Take an infinite finitely generated group $G$ with an infinite subgroup $N$ which has an infinite centraliser $Z = Z_G(N)$. Let $S$ be some [symmetric] generating set of $G$ and for $g \in G$, denote by $\mathrm{norm}(g)$ the combinatorial distance between $g$ and the identity in the Cayley graph w.r.t. $S$. Then $A= \mathrm{norm}(N)$ and $B=\mathrm{norm}(Z)$ are some subsets of the ($\geq 0$) integers.

Question: is it possible that $A$ and $B$ are asymptotically separated by large gaps, that is for every $C>1$ there is a $n_0$ such that

  • $\forall a \in A \cap [n_0,\infty[$, $B \cap [a/C,aC] = \emptyset$

  • $\forall b \in B \cap [n_0,\infty[$, $A \cap [b/C,bC] = \emptyset$

Note that if either $Z$ or $N$ contain an element $x$ of infinite order, then the answer is no (since $\mathrm{norm}(x^n) \to \infty$ and the triangle inequality implies $\mathrm{norm}(x^n) - \mathrm{norm}(x^{n-1}) \leq \mathrm{norm}(x)$, so the largest gap has length $< \mathrm{norm}(x)$).

This question is partially motivated by the fact that it is possible for the centre of an [infinite finitely generated] group $C = Z_G(G)$ to be so that $\mathrm{norm}(C)$ [is infinite] and has large gaps (in the sense that forall $C>1$ there is a $n > 2$ such that $\mathrm{norm}(C) \cap ]n/C,Cn[ = \emptyset$)

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    $\begingroup$ These conditions are trivially fulfilled if $N$ or $Z$ is finite. Also your argument that this is "no" if $Z$ or $N$ is not torsion also carries over the case when $Z$ or $N$ is locally finite. That is, your conditions imply that $Z$ and $N$ are locally finite. $\endgroup$
    – YCor
    Commented May 21, 2020 at 21:08
  • $\begingroup$ You are right: I forgot to add the adjective "infinite" before $N$ and $Z_G(N)$. $\endgroup$
    – ARG
    Commented May 21, 2020 at 21:11
  • $\begingroup$ Your remark on locally finiteness is essentially an "upgrade" on my remark on the fact that no element (of $N$ or $Z$) may have infinite order... or am I missing an important point? $\endgroup$
    – ARG
    Commented May 21, 2020 at 21:18
  • $\begingroup$ I said "your argument... carries over". It's essentially the same argument but it yields a much stronger conclusion. $\endgroup$
    – YCor
    Commented May 21, 2020 at 21:23
  • $\begingroup$ Thanks (i just was not sure if there was nonetheless a new ingredient required). So for eventual readers (which are as slow as I am): (1) If a subgroup $S$ is infinite, $\mathrm{norm}(S)$ is unbounded (because $G$ is fin.gen.) (2) In a fin. gen. subgroup, the norm of the generating elements bound the size of the gap (again by the triangle inequality) (3) Hence the $N$ and $Z$ [in the question] may not have infinite finitely generated subgroups (which is to say that they are locally finite) $\endgroup$
    – ARG
    Commented May 21, 2020 at 21:32

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