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This is an improved version of my previous question, where I forgot to put one of the assumptions.

Question. Let $G$ be a finitely generated non-amenable discrete group, and $H$ be a subgroup of $G$ of infinite index, such that no finite-index subgroup of $H$ is normal in $G$. Can it happen that the index of the normalizer $N(H)$ of $H$ in $G$ is finite, 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.

I hope I got everything right this time, but if the question is very easy then please point out the example in the comments, so that I can improve the question without starting a new thread.

The motivation for the question is that out of such an example one can construct an ergodic, faithful, non-free action of a non-amenable group whose equivalence relation is amenable. (Most likely such examples have been known before.)


EDIT: There's not enough space in the comments to answer Jesse's question below so I answer it here. I haven't thought the construction exactly through, but it should work like this: the Borel space is $X:=${0,1}${}^{G/H}$, and the action is induced by action of $G$ on $G/H$. There's a theorem (of Kaimanovich?) which says that if there is a graphing of a relation such that each component is of subexponential growth then the relation is amenable. In our case connected components are Schreier graphs so our relation is amenable, no matter what measure we put on the Borel space.

The measure is not the product measure. In $G/H$ we have (disjoint) images $C_i$ of $[G:N(H)]$ cosets of $N(H)$. Call $C_i$ "cosets" as well. The measure is supported on those sequences which are non-zero on at most one of the cosets $C_i$. So as a measure space $X$ is the union of $[G:N(H)]$ spaces {0,1}${}^{C_i}$. On each of these subspaces of $X$ the measure is defined to be the product measure normalized by $\frac{1}{[G:N(H)]}$.

The action is not essentially free, because H stabilizes {0,1}${}^C_0$, where $C_0$ is the trivial coset of $N(H)$.

I'm not sure about faithfulness, if I find a good argument why it's faithful I'll add it here. Also, I don't see anymore why I wanted $H$ to have no finite index subgroups which are normal in $G$, although it seemed important to us when we discussed it couple of days ago...

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I guess you need index strictly greater than 1 again? –  Kate Juschenko Mar 23 '11 at 16:13
    
@Kate: this time not, because I write "no finite-index subgroup of H is normal in G", so it's automatic :-) –  Łukasz Grabowski Mar 23 '11 at 16:16
    
How do you produce an amenable equivalence relation from such a subgroup? –  Jesse Peterson Mar 23 '11 at 18:31
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If $G/H$ has a $G$ invariant mean then this will imply that the action of $G$ on $\{ 0, 1 \}^{G/H}$ will not be strongly ergodic (in fact this is if and only if). But this actions will be essentially free as soon as it is faithful and so in this case should not give an amenable equivalence relation unless $G$ itself is amenable. –  Jesse Peterson Mar 25 '11 at 5:08
    
Sorry, I think now that essential freeness is actually equivalent to any non-identity element having infinitely many non-fixed points, and so is not equivalent to faithfulness, as I put in my last comment. But I still think that the invariant mean should give you only non-strong ergodicity, and not in general give hyperfinite. –  Jesse Peterson Mar 26 '11 at 0:22
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1 Answer

I think that this not possible if $G$ has property (T). Indeed in this case $N(H)$ would be itself finitely generated with property (T) (because of finite index), and the quotient $N(H)/H$ is amenable, having subexponential growth. By property (T), $N(H)/H$ is finite, so that $H$ has finite index in $G$, contrary to your assumption.

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