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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](http://www.numdam.org/item?id=PMIHES_2000__92__151_0), 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 2.2012-05-27
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    Thanks!I thought that was easier.2012-05-27
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    The additive group of rational numbers does not have a finite index subgroup.2012-05-27
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    Is it finitely generated?I don't think so.2012-05-27
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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.2012-05-27
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    ah sorry! dropped a hypothesis.2012-05-27

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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:

  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)2012-06-04
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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 $n$.2012-06-04
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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?2012-06-04
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    No, I meant what I wrote. $H$ can very well permute the $G/H$-cosets.2012-06-04
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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 :P2012-06-04
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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](http://www.math.uconn.edu/~kconrad/blurbs/grouptheory/gpaction.pdf).2012-06-04
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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.2012-06-04