Timeline for Does such an infinite index subgroup exist?
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| Apr 10, 2012 at 12:30 | comment | added | user6976 | @Steven: HNN is a free construction. So every group with a subgroup $H$, two isomorphic subgroups $A,B < H$ of different finite indices, and an element $g$ with $A^g=B$ is a homomorphic image of the HNN extension of $H$ with some kernel $N$. So one way to proceed is to understand where property 3 breaks and what is needed from $N$, and then to construct $H,A,B,N$. So consider my answer as a first step in the process. By the way, the result may be negative, that is no such $H,A,B,N$ exist at all. | |
| Apr 10, 2012 at 8:27 | comment | added | Steven Deprez | I suppose the easiest example of this kind is $\mathbb{Z}$ with subgroups $n\mathbb{Z}$ and $m\mathbb{Z}$. The corresponding HNN extension is the Baumslag-Solitar group $BS(m,n)$. Property 1 holds when $n\not=m$, property 2 holds since $H=\IZ$ is an almost normal subgroup of $G=BS(m,n)$. But i am quite sure property 3 does not hold. (we should have something like $[H:H\cap t^k H t^{-k}]=(\frac{m}{n})^k$) A similar problem should happen in general with HNN extensions. | |
| Apr 10, 2012 at 2:03 | history | edited | user6976 | CC BY-SA 3.0 |
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| Apr 9, 2012 at 22:45 | comment | added | user6976 | I see now. You can take the Thompson group F: it has subgroups of indices 2 and 3 isomorphic to F. I will think about a simpler example (lamplighter group).u | |
| Apr 9, 2012 at 22:13 | comment | added | Ian Agol | Property 1 is not true in your example. $H$ must have euler characteristic $=0$ to have two isomorphic subgroups of different index. | |
| Apr 9, 2012 at 18:42 | comment | added | user6976 | @Ian: You mean property 1 is not true or the other 2 properties? Both $U$ and $V$ are direct products of a free group of rank 11 and a free group of rank 25. Hence these subgroups are isomorphic. What's wrong? | |
| Apr 9, 2012 at 18:33 | comment | added | Ian Agol | I don't think this answer can possibly work. We have $[H:H\cap gHg^{-1}|= \chi(H\cap gHg^{-1})/\chi(H) = \chi(H \cap g^{-1}Hg)/\chi(H) = [H:H\cap g^{-1}Hg]$ (when the index is finite), so property (1) is not satisfied since $\chi(F_2\times F_3)=2$. Maybe there's some modification with groups with $\chi=0$. | |
| Apr 9, 2012 at 18:12 | comment | added | Dave Penneys | And aren't the indices the same as $[H\colon U]=(10)(12)=120=(24)(5)=[H\colon V]$? | |
| Apr 9, 2012 at 14:53 | comment | added | Dave Penneys | Thanks for your answer Mark! I've been working through the example, and 2 and 3 still seem non-trivial (especially 3). For example, $[H\colon H\cap t^2 H t^{-2}]$ seems highly dependent on how $t$ moves $U\cap V$ (and thus depends on the choice of isomorphism $U\cong V$). Are there any good references for the general properties of HNN extensions and why 2 and 3 should hold? Thanks again! | |
| Apr 6, 2012 at 8:24 | history | edited | user6976 | CC BY-SA 3.0 |
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| Apr 6, 2012 at 8:17 | history | answered | user6976 | CC BY-SA 3.0 |