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
2
$\begingroup$
abstract-algebra
group-theory
examples-counterexamples
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61) 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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0Thanks!I thought that was easier. – 2012-05-27
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0The additive group of rational numbers does not have a finite index subgroup. – 2012-05-27
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2Is it finitely generated?I don't think so. – 2012-05-27
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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. – 2012-05-27
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0ah sorry! dropped a hypothesis. – 2012-05-27
1 Answers
4
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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0Is 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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0@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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0Yeah, 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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0No, I meant what I wrote. $H$ can very well permute the $G/H$-cosets. – 2012-06-04
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0@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 – 2012-06-04
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0I 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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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