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J.L.
J.L.
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Let $GX$ be a tree group (a right-angled Artin group such that the graph is a tree), such that not all the subgroups of $GX$ are necessarily RAAGs, so the length of the tree is greater or equal than $4$.

Then, $GX= A\ast_{C} B$, where $C=\langle c \rangle$ is a cyclic group. Let $t$ be an hyperbolic element in $GX$. It is known that there is a finite index subgroup $M$ such that $M= B \ast_{C\cap M} \langle t \rangle$ ($M$ is an HNN extension). Hence, the generators of $$M= G \ast (M\cap A)\ast (M\cap B) \ast (M\cap A^g) \ast (M \cap B^g) \ast \cdots \ast \langle t \rangle$$$M$ are the generators of $$G, (M\cap A), (M\cap B), (M\cap A^g), (M \cap B^g), \cdots, \{ t \}$$ and the relations tell me (apart from the relations in the groups $G,\dots, M\cap B^g$) that $t^{-1}s_{1}t=s_{2}$ where $s_{1}$ is an element in a cyclic edge group. So up to conjugacy, I may assume that $s_{1}$ is $c^n$ for some $n\in \mathbb{N}$.

Now, I would like to find a finite index subgroup in $M$ such that $s_{1}^m$ is a generator and $s_{1}$ is not in the subgroup. Since $s_{1}$ lies in the edge group, is it enough to find a finite index subgroup in $A$ with that property?

Let $GX$ be a tree group (a right-angled Artin group such that the graph is a tree), such that not all the subgroups of $GX$ are necessarily RAAGs, so the length of the tree is greater or equal than $4$.

Then, $GX= A\ast_{C} B$, where $C=\langle c \rangle$ is a cyclic group. Let $t$ be an hyperbolic element in $GX$. It is known that there is a finite index subgroup $M$ such that $M= B \ast_{C\cap M} \langle t \rangle$ ($M$ is an HNN extension). Hence, $$M= G \ast (M\cap A)\ast (M\cap B) \ast (M\cap A^g) \ast (M \cap B^g) \ast \cdots \ast \langle t \rangle$$ and the relations tell me (apart from the relations in the groups $G,\dots, M\cap B^g$) that $t^{-1}s_{1}t=s_{2}$ where $s_{1}$ is an element in a cyclic edge group. So up to conjugacy, I may assume that $s_{1}$ is $c^n$ for some $n\in \mathbb{N}$.

Now, I would like to find a finite index subgroup such that $s_{1}^m$ is a generator and $s_{1}$ is not in the subgroup. Since $s_{1}$ lies in the edge group, is it enough to find a finite index subgroup in $A$ with that property?

Let $GX$ be a tree group (a right-angled Artin group such that the graph is a tree), such that not all the subgroups of $GX$ are necessarily RAAGs, so the length of the tree is greater or equal than $4$.

Then, $GX= A\ast_{C} B$, where $C=\langle c \rangle$ is a cyclic group. Let $t$ be an hyperbolic element in $GX$. It is known that there is a finite index subgroup $M$ such that $M= B \ast_{C\cap M} \langle t \rangle$ ($M$ is an HNN extension). Hence, the generators of $M$ are the generators of $$G, (M\cap A), (M\cap B), (M\cap A^g), (M \cap B^g), \cdots, \{ t \}$$ and the relations tell me (apart from the relations in the groups $G,\dots, M\cap B^g$) that $t^{-1}s_{1}t=s_{2}$ where $s_{1}$ is an element in a cyclic edge group. So up to conjugacy, I may assume that $s_{1}$ is $c^n$ for some $n\in \mathbb{N}$.

Now, I would like to find a finite index subgroup in $M$ such that $s_{1}^m$ is a generator and $s_{1}$ is not in the subgroup. Since $s_{1}$ lies in the edge group, is it enough to find a finite index subgroup in $A$ with that property?

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J.L.
J.L.
  • 321
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  • 6

Let $G$$GX$ be a HNN extensiontree group (a right-angled Artin group such that the graph is a tree), such that not all the subgroups of $GX$ are necessarily RAAGs, so the formlength of the tree is greater or equal than $\langle s_{1},s_{2},\dots,s_{n},t \mid R, t^{-1}s_{1}t=s_{2}\rangle$$4$.

Then, $GX= A\ast_{C} B$, where $R$$C=\langle c \rangle$ is a set of relationscyclic group. Let $t$ be an hyperbolic element in the alphabet $\{s_{2},\dots,s_{n}\}$$GX$.

I want to find It is known that there is a finite index subgroup of $G$ substituting$M$ such that $s_{1}$ by a power$M= B \ast_{C\cap M} \langle t \rangle$ $s_{1}^m$($M$ is an HNN extension). Hence, $$M= G \ast (M\cap A)\ast (M\cap B) \ast (M\cap A^g) \ast (M \cap B^g) \ast \cdots \ast \langle t \rangle$$ and suchthe relations tell me (apart from the relations in the groups $G,\dots, M\cap B^g$) that $t$ only acts on$t^{-1}s_{1}t=s_{2}$ where $s_{1}^m$$s_{1}$ is an element in a cyclic edge group. So up to conjugacy, I can't takemay assume that $\langle s_{1}^m,s_{2},\dots,s_{n},t \mid R, t^{-1}s_{1}^m t=s_{2}^m\rangle$ because$s_{1}$ is $c^n$ for some $n\in \mathbb{N}$.

Now, I would needlike to add conjugates offind a finite index subgroup such that $s_{1}^m$ is a generator and $s_{1}$ is not in orderthe subgroup. Since $s_{1}$ lies in the edge group, is it enough to havefind a finite index subgroup.

I have tried to visualize the problem in $A$ with Bass-Serre theory, but I can't convince myself that $t$ doesn't touch any other element.

Can someone help me, pleaseproperty?

Let $G$ be a HNN extension of the form $\langle s_{1},s_{2},\dots,s_{n},t \mid R, t^{-1}s_{1}t=s_{2}\rangle$, where $R$ is a set of relations in the alphabet $\{s_{2},\dots,s_{n}\}$.

I want to find a finite index subgroup of $G$ substituting $s_{1}$ by a power $s_{1}^m$ and such that $t$ only acts on $s_{1}^m$. I can't take $\langle s_{1}^m,s_{2},\dots,s_{n},t \mid R, t^{-1}s_{1}^m t=s_{2}^m\rangle$ because I would need to add conjugates of $s_{1}^m$ in order to have a finite index subgroup.

I have tried to visualize the problem with Bass-Serre theory, but I can't convince myself that $t$ doesn't touch any other element.

Can someone help me, please?

Let $GX$ be a tree group (a right-angled Artin group such that the graph is a tree), such that not all the subgroups of $GX$ are necessarily RAAGs, so the length of the tree is greater or equal than $4$.

Then, $GX= A\ast_{C} B$, where $C=\langle c \rangle$ is a cyclic group. Let $t$ be an hyperbolic element in $GX$. It is known that there is a finite index subgroup $M$ such that $M= B \ast_{C\cap M} \langle t \rangle$ ($M$ is an HNN extension). Hence, $$M= G \ast (M\cap A)\ast (M\cap B) \ast (M\cap A^g) \ast (M \cap B^g) \ast \cdots \ast \langle t \rangle$$ and the relations tell me (apart from the relations in the groups $G,\dots, M\cap B^g$) that $t^{-1}s_{1}t=s_{2}$ where $s_{1}$ is an element in a cyclic edge group. So up to conjugacy, I may assume that $s_{1}$ is $c^n$ for some $n\in \mathbb{N}$.

Now, I would like to find a finite index subgroup such that $s_{1}^m$ is a generator and $s_{1}$ is not in the subgroup. Since $s_{1}$ lies in the edge group, is it enough to find a finite index subgroup in $A$ with that property?

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J.L.
J.L.
  • 321
  • 1
  • 6

Let $G$ be a HNN extension of the form $\langle s_{1},s_{2},\dots,s_{n},t \mid t^{-1}s_{1}t=s_{2}\rangle$$\langle s_{1},s_{2},\dots,s_{n},t \mid R, t^{-1}s_{1}t=s_{2}\rangle$, where $R$ is a set of relations in the alphabet $\{s_{2},\dots,s_{n}\}$.

I want to find a finite index subgroup of $G$ substituting $s_{1}$ by a power $s_{1}^m$ and such that $t$ only acts on $s_{1}^m$. I can't take $\langle s_{1}^m,s_{2},\dots,s_{n},t \mid t^{-1}s_{1}^m t=s_{2}^m\rangle$$\langle s_{1}^m,s_{2},\dots,s_{n},t \mid R, t^{-1}s_{1}^m t=s_{2}^m\rangle$ because I would need to add conjugates of $s_{1}^m$ in order to have a finite index subgroup.

I have tried to visualize the problem with Bass-Serre theory, but I can't convince myself that $t$ doesn't touch any other element.

Can someone help me, please?

Let $G$ be a HNN extension of the form $\langle s_{1},s_{2},\dots,s_{n},t \mid t^{-1}s_{1}t=s_{2}\rangle$.

I want to find a finite index subgroup of $G$ substituting $s_{1}$ by a power $s_{1}^m$ and such that $t$ only acts on $s_{1}^m$. I can't take $\langle s_{1}^m,s_{2},\dots,s_{n},t \mid t^{-1}s_{1}^m t=s_{2}^m\rangle$ because I would need to add conjugates of $s_{1}^m$ in order to have a finite index subgroup.

I have tried to visualize the problem with Bass-Serre theory, but I can't convince myself that $t$ doesn't touch any other element.

Can someone help me, please?

Let $G$ be a HNN extension of the form $\langle s_{1},s_{2},\dots,s_{n},t \mid R, t^{-1}s_{1}t=s_{2}\rangle$, where $R$ is a set of relations in the alphabet $\{s_{2},\dots,s_{n}\}$.

I want to find a finite index subgroup of $G$ substituting $s_{1}$ by a power $s_{1}^m$ and such that $t$ only acts on $s_{1}^m$. I can't take $\langle s_{1}^m,s_{2},\dots,s_{n},t \mid R, t^{-1}s_{1}^m t=s_{2}^m\rangle$ because I would need to add conjugates of $s_{1}^m$ in order to have a finite index subgroup.

I have tried to visualize the problem with Bass-Serre theory, but I can't convince myself that $t$ doesn't touch any other element.

Can someone help me, please?

Source Link
J.L.
J.L.
  • 321
  • 1
  • 6