Timeline for Non-normal subgroups of triangle groups
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Dec 13, 2016 at 8:15 | comment | added | Sridhar Ramesh | Explicit examples would be good as well. If I at any point sound like I know anything, that is misleading. :) | |
| Dec 13, 2016 at 8:07 | comment | added | YCor | OK, I'll do this when I find time. Btw, the question actually sounded like you know the answer and you ask for a more explicit example. | |
| Dec 13, 2016 at 8:01 | comment | added | Sridhar Ramesh | Ah, ok, great; thanks for all your help! If you post your comments as an answer, I will accept it. | |
| Dec 13, 2016 at 7:44 | comment | added | YCor | Because hamiltonian groups have an abelian subgroup of index 2, by the structure theorem. Yet here you need a non-normal subgroup of finite index. So it's enough to have a finite quotient that is not hamiltonian. Actually if a f.g. residually finite group has the property that all its finite quotients have an abelian subgroup of index 2, then it itself has an abelian subgroup of index 2. | |
| Dec 13, 2016 at 7:29 | comment | added | Sridhar Ramesh | Ah, ok, I see now the residual finiteness, but why does N being non-abelian guarantee a non-normal subgroup? What prevents N from being a Hamiltonian group? [Or, rather, I guess, what's really of interest isn't necessarily subgroups of N which are non-normal in N, but rather subgroups of N which are non-normal in the ambient triangle group; still, why does non-abelianness guarantee such a subgroup of N?] | |
| Dec 13, 2016 at 3:58 | comment | added | YCor | However my claim there's a non-normal subgroup was a bit quick. It's true as soon as the finite index subgroup (intersected with von Dyck) is non-abelian. This holds unless we have a Coxeter group of affine type. Here the only exceptions are $AA_2$, $CC_2$, $GG_2$, and $II_1$ (here double letter means tilde). They correspond to $(a,b,c)=$ $(3,3,3)$, $(2,4,4)$, $(2,3,6)$, $(2,2,\infty)$. But actually it's OK in the first 3 cases because we get a normal subgroup isomorphic to $\mathbf{Z}^2$ with a noncentral action. However we get no normal subgroup in the $(2,2,\infty)$ case. | |
| Dec 13, 2016 at 3:51 | comment | added | YCor | This is a finitely generated Coxeter group, so it's linear over a field (see e.g. Bourbaki on Lie groups and Lie algebras) and hence residually finite. But I guess here one can use a more direct argument (since you need much less than just being residually finite). | |
| Dec 13, 2016 at 3:34 | comment | added | Sridhar Ramesh | In the infinite case, pardon my inability to see it right away, but why is it clear the triangle group is residually finite? And, supposing we take N as above (intersected with the von Dyck group if necessary), how do we know it has a non-normal subgroup? | |
| Dec 13, 2016 at 3:08 | comment | added | YCor | Then in the finite case it seems not to exist at all: if $a=b=2$, you have a dihedral group (times one of order 2). Then in the dihedral group, all subgroup of the von Dyck group will be normal. So the only remaining cases are the Coxeter groups of type $A_3,B_3, H_3$. In $A_3$ and $H_3$, the von Dyck subgroup has index 2, and index 4 in $B_3$. In $A_3=Sym_4$, it does not exist either, and also in the binary icosahedral (of order 120) $H_3$. I haven't checked $B_3$ (of order 48). | |
| Dec 13, 2016 at 3:00 | comment | added | YCor | Oh yes I forgot. In the infinite case it's no problem, just intersect with the von Dyck group first if necessary. | |
| Dec 13, 2016 at 2:47 | comment | added | Sridhar Ramesh | Ah, $\{1, z\}$ doesn't work; I want subgroups which contain only products of even numbers of generators, while $z$ is a product of an odd number of generators. (As for the latter case, I have to think about the residually finite reasoning some more.) | |
| Dec 13, 2016 at 2:32 | comment | added | YCor | At least you know that there are such subgroups (unless $(a,b,c)=(2,2,2)$ as $\Delta$ is then abelian). Indeed otherwise we can assume $c>2$ and then you can choose the subgroup $\{1,z\}$ if $\Delta$ is finite. If $\Delta$ is infinite, there is a normal subgroup of finite index $N$ such that in $\Delta/N$, the orders of $xy,yz,zx$ are $a,b,c$ respectively (it exists because $\Delta$ is residually finite). Then choose any non-normal subgroup of $N$, since you want a non-normal subgroup. | |
| Dec 13, 2016 at 1:39 | history | edited | Sridhar Ramesh | CC BY-SA 3.0 |
The period placement at the end of the second paragraph is weird on my computer, so with minor wording changes, I try to fix it…
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| Dec 13, 2016 at 1:25 | history | asked | Sridhar Ramesh | CC BY-SA 3.0 |