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I've recently encountered the following problem. Given a group $G$, a subgroup $H$ and a sequence $g_n\in G$, let $$ \liminf_{j\to\infty}H^{g_j} :=\bigcup_{n\ge 1} \bigcap_{j\ge n} H^{g_j}.$$ Here $$ H^g=g^{-1}Hg$$ denotes conjugation by the element $g$. The question is whether or not the above subgroup, denote it by $H_\infty$, is conjugated to a subgroup of $H$, i.e. if $$H_\infty\le H^g$$ for some $g\in G$.

This can quite easily be seen to be true for Noetherian groups $G$: the subgroups $$H_n:= \bigcap_{j\ge n} H^{g_j}$$ form an ascending chain of subgroups of $G$, therefore eventually becoming constant ($=H_\infty$). Since each $H_n$ satisfies $H_n\le H^{g_n}$ we find that for large enough $n$, $H_n=H_\infty\le H^{g_n}$.

I'm however interested in hyperbolic groups, where my understanding is that the Noetherian property is rather rare (I'm not really in group theory so I might be wrong) and so I'm wondering under what sort of hypotheses can we find a positive answer to the question (for an arbitrary sequence $(g_n)$ and subgroup $H\le G$. If needed one can assume $H$ to be finitely generated.)

I appreciate any help, thanks!

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    $\begingroup$ actually you can say that if $H_\infty$ is f.g. then it's contained in a conjugate of $H$ (same trivial argument). Of course this applies in the noetherian case, but not only. $\endgroup$ – YCor Jan 20 '15 at 0:37
  • $\begingroup$ @YCor yes you're right. It's just that in my particular case I have no way of checking if that is the case, so I kind of "have to" assume the Noetherian property. $\endgroup$ – Teri Jan 20 '15 at 8:08
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The answer is negative in general, even if you restrict to finitely generated subgroups. Indeed, in a hyperbolic group a f.g. subgroup can be conjugate to a proper subgroup of itself. For example, let $H$ be the free group of rank $2$ and let $\phi:H \to H$ be a non-surjective monomorphism. Define the group $G$ to be the ascending HNN-extension of $H$ with respect to $\phi$: $$G=\langle H, t \mid tft^{-1}=\phi(f) \mbox{ for all } f \in H \rangle.$$ It is easy to choose $\phi$ so that $G$ is hyperbolic (in fact it can satisfy the small cancellation condition $C'(1/6)$).

Now, clearly for any $n \in \mathbb{N}$, $t^{-n}Ht^n$ is properly contained in $t^{-n-1} H t^{n+1}$, so $$\bigcap_{j \ge n} H^{t^j}= H^{t^n},$$ hence $$H_\infty=\liminf_{j \to \infty} H^{t^j}=\bigcup_{n \in \mathbb{Z}} H^{t^n}.$$ It is easy to see that the latter is an infinitely generated normal subgroup of $G$.

Again, by choosing $\phi$ appropriately we can make sure that $$\bigcap_{j\in \mathbb{N}} \phi^j(H)=\bigcap_{j \in \mathbb{N}} H^{t^{-j}}=\{1\},$$ so $H$ cannot contain any normal subgroup of $G$. Thus $H_\infty$ is not contained in any conjugate of $H$ in $G$.

If you don't need $H$ to be finitely generated, then it should be easy to construct an example even when $G$ is the free group of rank $2$. Take $G=\langle a, t \rangle$ and $H=\langle t^{n} a t^{-n} \mid n\in \mathbb{N}\rangle$. Then the same argument shows that $\liminf_{j \to \infty} H^{t^j}= \langle t^n a t^{-n} \mid n\in \mathbb{Z}\rangle$. The fact that this is not contained in any conjugate of $H$ should not be difficult to show.

However, I think that the statement you want is true if $H$ is quasiconvex in $G$. This can quickly be deduced from the fact that quasiconvex subgroups of hyperbolic groups have finite height, i.e. the number of essentially distinct conjugates of the subgroup having infinite intersection is uniformly bounded from above. See the paper "WIDTHS OF SUBGROUPS" by RITA GITIK, MAHAN MITRA, ELIYAHU RIPS, AND MICHAH SAGEEV (TAMS 350, no. 1 (1998)).

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  • $\begingroup$ Thank you, this is indeed helpful! Pertaining to your comment on the other answer, ideally I'm interested in groups $G\times G$ where $G$ is the fundamental group of a compact CAT(0) space and $H=diag(G)$. Hyperbolic, however, is somewhat satisfying as well. $\endgroup$ – Teri Jan 19 '15 at 23:15
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    $\begingroup$ CAT(0) spaces are contractible. Did you mean to say that $G$ is the fund. group of a non-positively curved space (i.e., a space whose universal cover is CAT(0))? $\endgroup$ – Ashot Minasyan Jan 20 '15 at 11:46
  • $\begingroup$ Ahh yes, that's indeed what I meant, thanks! $\endgroup$ – Teri Jan 20 '15 at 12:03
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    $\begingroup$ The diagonal subgroup $H$ is a very special subgroup in $G \times G$. In particular, $H \cap H^{(g_1,g_2)}=\{(x,x) \mid x \in C_G(g_1^{-1}g_2)\}$. So, I think that your question may reduce to the question about intersections of centralizers in a CAT(0) group. The book of Bridson and Haefliger could help you with this. $\endgroup$ – Ashot Minasyan Jan 20 '15 at 17:22
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Let me give a simple counterexample. Let $G=Sym(\mathbb{Z})$ and let $H$ be the group of all permutations $f$ of $\mathbb{Z}$ such that $f(0)=0$. Let $g_{n}(x)=x-n$ for all $n,x$. Then $f\in H^{g_{n}}$ if and only if $f(n)=n$. Therefore $H_{\infty}$ is the group of all permutations that fix sufficiently large elements. However, $f\in H^{g}$ iff $f(g^{-1}(0))=g^{-1}(0)$. Therefore $H_{\infty}\not\subseteq H^{g}$ for any $g\in Sym(\mathbb{Z})$.

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  • $\begingroup$ So? The trivial subgroup is certainly conjugated to a subgroup of $H$. $\endgroup$ – Johannes Hahn Jan 19 '15 at 15:49
  • $\begingroup$ I guess I misread the question. $\endgroup$ – Joseph Van Name Jan 19 '15 at 15:55
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    $\begingroup$ Johannes Hahn. I changed the answer so that it actually answers the question now. $\endgroup$ – Joseph Van Name Jan 19 '15 at 19:57
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    $\begingroup$ As far as I understand, the question asks about the case when the ambient group is word hyperbolic. The group $G=Sym(\mathbb{Z})$ certainly isn't hyperbolic. $\endgroup$ – Ashot Minasyan Jan 19 '15 at 20:23

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