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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 questionrevised 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.

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.

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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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.

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.

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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Example of a non-normalan infinite index subgroup of a non-amenable group with certain properties.whose normalizer is of non-zero finite index, and such that the Schreier graph is of subexponential growth

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 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.

Remark: Answer of Kate below is valid for a previous version of the question, where I forgot to add that $H$ should have no finite-index subgroups which are normal in $G$.

Example of a non-normal infinite index subgroup of a non-amenable group with certain properties.

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.

Remark: Answer of Kate below is valid for a previous version of the question, where I forgot to add that $H$ should have no finite-index subgroups which are normal in $G$.

Example of an infinite index subgroup of a non-amenable group whose normalizer is of non-zero finite index, and such that the Schreier graph is of subexponential growth

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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Charles Matthews
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