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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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    ah sorry! dropped a hypothesis.2012-05-27

1 Answers 1

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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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    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