2
$\begingroup$

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?

Improve this question
$\endgroup$
15
  • 3
    $\begingroup$ The group$G$ that you have defined there is a free group on the generators $t,s_1,s_3,\ldots$, so there are certainly subgroups of index $m$ containing $s_1^m$. $\endgroup$
    – Derek Holt
    Commented Sep 24, 2019 at 11:01
  • $\begingroup$ @DerekHolt sorry, I should have written it correctly. It is possible for the $s_{i}$ to have relations between them, I wanted to say that $t$ only acts on $s_{1}$. $\endgroup$
    – J.L.
    Commented Sep 24, 2019 at 12:20
  • $\begingroup$ I think you need to make it clearer exactly what you are trying to achieve. It is unlikely that you can say very much without further assumptions. $\endgroup$
    – Derek Holt
    Commented Sep 24, 2019 at 12:52
  • $\begingroup$ The group is now just the free product of $\langle t\rangle$ with $\langle s_2,\dots,s_n\mid R\rangle$ with $s_1$ defined as $ts_2t^{-1}$. $\endgroup$
    – YCor
    Commented Sep 24, 2019 at 12:52
  • 1
    $\begingroup$ @Karen: I'm still a bit confused about your set-up. (You seem to be using the symbol "$*$" in a quite nonstandard way!) However, if all you want is a finite-index subgroup of $M$ that contains $s_1^m$ but not $s_1$, then this exists: $M$ is a 3-manifold group, and Hamilton proved that abelian subgroups of 3-manifold groups (eg $\langle s_1^m\rangle$) are separable; see Hamilton, Emily, Abelian subgroup separability of Haken 3-manifolds and closed hyperbolic $n$-orbifolds, PLMS. $\endgroup$
    – HJRW
    Commented Sep 25, 2019 at 8:29

0

Reset to default

You must log in to answer this question.