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?
Finitely Generated Group
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abstract-algebra
group-theory
examples-counterexamples
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0ah sorry! dropped a hypothesis. – 2012-05-27
1 Answers
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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.
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0Oh, 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