Timeline for Finite index subgroup of HNN extension
Current License: CC BY-SA 4.0
20 events
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| Sep 26, 2019 at 11:10 | comment | added | HJRW | To do that, you need to check that the subgroup $\langle s_1^m,t\rangle$ is separable and doesn't contain $s_1$. If I've understood the set-up correctly, $\langle s_1,t\rangle\cong\mathbb{Z}^2$, in which case this follows again from Hamilton's theorem. | |
| Sep 25, 2019 at 14:47 | comment | added | J.L. | @HJRW ah ok yes! I forgot that tree groups are 3-manifold groups... However, I'd obtain that a power of $t$ lies in that finite index subgroup, but I would like $t$ to be a generator of that finite index subgroup... I want $t$ to have relations only with $s_{1}^m$ in the finite index subgroup, so that it is still an HNN extension with stable letter $t$. | |
| Sep 25, 2019 at 13:47 | comment | added | HJRW | Well, tree groups are 3-manifold groups (see, for instance, p. 234 of Behrstock & Neumann, "Quasi-isometric classification of graph manifold groups", Duke). Since $M$ is a finite-index subgroup of your tree group $G$, it follows that $M$ is the fundamental group of a finite cover of a 3-manifold, which is also a 3-manifold. | |
| Sep 25, 2019 at 12:45 | comment | added | J.L. | @HJRW sorry, why is $M$ a $3$-manifold group? | |
| Sep 25, 2019 at 8:34 | history | edited | J.L. | CC BY-SA 4.0 |
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| Sep 25, 2019 at 8:29 | comment | added | HJRW | @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. | |
| Sep 25, 2019 at 7:02 | history | edited | J.L. | CC BY-SA 4.0 |
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| Sep 25, 2019 at 6:23 | comment | added | J.L. | @HJRW I think that I have solved my misunderstanding. I will edit the post completely explaining it better, I'm very sorry. Thanks for your answers | |
| Sep 24, 2019 at 14:26 | comment | added | HJRW | @Karen, I guess $M$ is what I called $H$? Two things are certainly true: a finite-index subgroup of a tree group need not be a RAAG; and a finite-index subgroup of a tree group doesn't split freely. If I had to guess what the problem was, I would guess that, when passing to a finite-index subgroup, you may not have "remembered" all the lifts of some of your elements. It's quite easy to accidentally get a free product that way. | |
| Sep 24, 2019 at 14:17 | comment | added | YCor | @Karen it's hard to interpret your comment without knowing what you mean by $M$. Why don't you want $s_1$ to appear in $R$? | |
| Sep 24, 2019 at 14:04 | comment | added | J.L. | @HJRW Yes, $M$ is a finite index subgroup of a tree group (right-angled Artin group), but I don't know if $M$ is a RAAG again. I'm quite confused with the fact that I have obtained that $M$ is a free product, since otherwise my RAAG would be a free product, which is a contradiction... | |
| Sep 24, 2019 at 13:49 | comment | added | HJRW | The moral of my previous comment is that, at the very least, you probably need to assume that $H$ is residually finite and that $s_2$ has infinite order in $H$ (or at least that the order of $s_2$ is somehow compatible with $m$). In this case, what you want is probably true, but it would help if you could clarify what you mean by " $t$ only acts on $s_1^m$". Do you mean that $s_1^i\notin G_0$ for $1\leq i<m$? | |
| Sep 24, 2019 at 13:45 | comment | added | HJRW | The subgroup you are looking for may not exist. Let $H=\langle s_2,\ldots,s_n\mid R\rangle$. As @YCor mentioned, your group is just $G=H*\langle t\rangle$ (since the generator $s_1$ only appears once in one relator, and so can be eliminated). Since you haven't told us anything about $H$, it could be a subgroup without proper finite-index subgroups. (Indeed, it could be the trivial group!) You're looking for a finite-index subgroup $G_0$ of $G$ that contains $t$. But if $H$ doesn't have proper finite-index subgroups then $G_0$ must also contain $H$, so $G_0=G$. | |
| Sep 24, 2019 at 12:52 | comment | added | YCor | 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 $ts_2t^{-1}$. | |
| Sep 24, 2019 at 12:52 | comment | added | Derek Holt | 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. | |
| Sep 24, 2019 at 12:21 | history | edited | J.L. | CC BY-SA 4.0 |
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| Sep 24, 2019 at 12:20 | comment | added | J.L. | @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 $s_{1}$. | |
| Sep 24, 2019 at 11:01 | comment | added | Derek Holt | 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 $s_1^m$. | |
| Sep 24, 2019 at 10:40 | review | First posts | |||
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| Sep 24, 2019 at 10:36 | history | asked | J.L. | CC BY-SA 4.0 |