Let be $G$ finitely generated; My question is: Does always exist $H\leq G,H\not=G$ with finite index? Of course if G is finite it is true. But $G$ is infinite?
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- There are finitely presented infinite simple groups, see e.g. here, Theorem 5.5 for an interesting family. 2) It is easy to show that every finite index subgroup contains a finite index normal subgroup. 3) Combine 1. and 2.
– t.b. May 27 '12 at 14:15 -
Thanks!I thought that was easier. – User2040 May 27 '12 at 15:14
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The additive group of rational numbers does not have a finite index subgroup. – May 27 '12 at 15:14
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2Is it finitely generated?I don't think so. – User2040 May 27 '12 at 15:17
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2@Lima: No, $\mathbb{Q}$ is quasicyclic (every finitely generated subgroup is cyclic), but not cyclic, so it is not finitely generated. – Arturo Magidin May 27 '12 at 16:20
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ah sorry! dropped a hypothesis. – May 27 '12 at 16:28
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No.
I suspect there are easier and more elegant ways to answer this question, but the following argument is one way to see it:
- There are finitely generated infinite simple groups:
- In 1951, Higman constructed the first example in A Finitely Generated Infinite Simple Group, J. London Math. Soc. (1951) s1-26 (1), 61–64.
- Very popular are Thompson's groups.
- I happen to like the Burger–Mozes family of finitely presented infinite simple torsion-free groups, described in Lattices in product of trees. Publications Mathématiques de l'IHÉS, 92 (2000), p. 151–194 (full disclosure: I wrote my thesis under the direction of M.B.).
- See P. de la Harpe, Topics in Geometric Group Theory, Complement V.26 for further examples and references.
- If a group $G$ has a finite index subgroup $H$ then $H$ contains a finite index normal subgroup of $G$, in particular no infinite simple group can have a non-trivial finite index subgroup.
See also Higman's group for an example of a finitely presented group with no non-trivial finite quotients. By the same reasoning as above it can't have a non-trivial finite index subgroup.
t.b.
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Is point 2. well known? Is it obvious with sufficient thought? (by well-known I mean to students, rather than professionals/experts) – Ben Millwood Jun 04 '12 at 21:46
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@benmachine: Yes, I think it's well known in your sense. It's easy to prove: $G$ acts by on the set of cosets $G/H$. Thus you get a homomorphism $G \to \operatorname{Aut}(G/H)$ and you can take its kernel $N \subset H$. Then $G/N$ is isomorphic to a subgroup of $\operatorname{Aut}(G/H)$ and the latter has order dividing $n!$ if $H$ has index $n$. – t.b. Jun 04 '12 at 21:54
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Yeah, that sounds reasonable. I'd never heard the result before but I guess it just hasn't come up. Shouldn't $H$ be inside $N$, though, rather than the other way around? – Ben Millwood Jun 04 '12 at 22:20
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@t-b: hmm, perhaps we're thinking of different actions: it would possibly help if you weren't missing the verb in the action of $G$ on the cosets :P – Ben Millwood Jun 04 '12 at 22:57
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I implement $G/H = {aH,:,a \in G}$ as left cosets and act with $G$ on the left $g(aH) = (ga)H$ (action on the right won't be well-defined!). Just cross out the "by" in my first comment to you or add the noun bijections... :) See example 2.9 on p. 4 of Keith Conrad's blurb. – t.b. Jun 04 '12 at 23:05
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Oh, we're thinking of the same action and I'm just being silly. Sometimes it helps to be literally staring at the definition! Anyway, thank you for clearing that up. – Ben Millwood Jun 04 '12 at 23:16