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EDIT: In this question I forgot to put one of the assumptions, and the question was easier than it should be. Here is the revised question. Please vote to close this question as it is no longer relevant.


Question. Let $G$ be a finitely generated non-amenable discrete group, and $H$ be a subgroup of $G$ of infinite index. Can it happen that the index of the normalizer $N(H)$ of $H$ in $G$ is finite greater than $1$, and the Schreier graph of $G/H$ has subexponential growth?

If the answer is yes, I would very much like to see an example. It would be especially nice if $G$ could be taken to be a property $(T)$ group.

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what is the motivation of this question? –  Kate Juschenko Mar 22 '11 at 11:20
    
out of such an example one can construct a faithful probability measure preserving action of a non-ameanable group, with an ameanable equivalence relation. Besides, I'd be happy to learn what techniques can say something about "amenability" of the quotient G/H when H is not normal. –  Łukasz Grabowski Mar 22 '11 at 12:22
    
Maybe you mean normal closure instead of normalizer? Normalizer of the normal group is the whole group, in that case the answer to your question is $G=H\times \mathbb{Z}$, since $N(H)=G$ in this case. –  Kate Juschenko Mar 22 '11 at 14:34
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I assume "non-zero" is code for "greater than one"?! –  Igor Rivin Mar 22 '11 at 15:32
3  
Voting to close as no longer relevant per request of OP. –  Lucia Apr 16 at 16:44
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closed as off-topic by Yves Cornulier, Lucia, Stefan Kohl, Yemon Choi, Jeremy Rickard Apr 16 at 21:04

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1 Answer

up vote 9 down vote accepted

The answer to the question is yes. Here are two examples, the first one answers the question as it is posted and in the second one I assume that $N(H)$ is the normal closure of $H$ in $G$. The second example seems to be more interesting, in the sense that one have to do more computation on Schreier graph, compering to the first example, where the computation of the growth of the Schreier graph of $G/H$ is straightforward.

  • If $N(H)$ is normalizer of $H$, then the following example works. $G:=\mathbb{Z}\times\mathbb{F}_2$ and $H$ is a subgroup $(e,H_0)$, where $H_0$ is of finite index but not normal in $\mathbb{F}_2$. Then index of $N(H)=\mathbb{Z}\times N_{\mathbb{F}_2} (H_0)$ is finite and not equal to $1$, since $N_{\mathbb{F}_2}(H_0)\neq \mathbb{F}_2$, because $H_0$ is not normal in $\mathbb{F}_2$.

  • If $N(H)$ is normal closure of $H$ in $G$, then the following example works. Let $\mathbb{F}_2$ be the free group on two generators $a$ and $b$. Define a homomorphism $\phi:\mathbb{F}_2\rightarrow Aut(\mathbb{Z})=\mathbb{Z}_2$ by $\phi(a)=0,\phi(b)=1$. Then the question is valid for the semidirect product $G=\mathbb{F}_2\ltimes_{\phi} \mathbb{Z}$. The Schreier graph of $G/H$ has polynomial growth and the normal closure of $H$ is $\mathbb{F}_2\ltimes_{\phi} 2\mathbb{Z}$.

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Thanks, I'll have to edit the question though: I forgot to add that H should have no normal subgroup of finite index. Otherwise the application I have in mind (faithful pmp action with amenable equivalence relation) trivialize, because the action is in fact free. And yes, N(H) is the normalizer. –  Łukasz Grabowski Mar 23 '11 at 15:21
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