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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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1) 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
    
The additive group of rational numbers does not have a finite index subgroup. –  user29743 May 27 '12 at 15:14
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Is it finitely generated?I don't think so. –  User2040 May 27 '12 at 15:17
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@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

1 Answer 1

up vote 4 down vote accepted

No.

I suspect there are easier and more elegant ways to answer this question, but the following argument is one way to see it:

  1. There are finitely generated infinite simple groups:
  2. 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.

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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 4 '12 at 21:46
    
@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 4 '12 at 21:54
    
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 4 '12 at 22:20
    
No, I meant what I wrote. $H$ can very well permute the $G/H$-cosets. –  t.b. Jun 4 '12 at 22:40
    
@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 4 '12 at 22:57

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