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There is a well known fact says that if $H$ is a finite index subgroup of $G$ then $H$ has a finite index subgroup $N$ which is normal in $G$. If $H$ is not normal in $G$ then one can find $N$ proper, namely $N \neq H$. My question is as follows: If $H$ is normal in $G$, is there always a proper finite index subgroup $N$ in $H$ which is normal in $G$?

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Well, not if $H = \{e\}$, of course. Probably you want $G$ to be infinite? –  Pete L. Clark Aug 1 '11 at 23:55
    
It seems that we don't have the same interpretation of your question. I think that the answer will be “no” in all cases, but it might be worthwhile to rephrase it. –  PseudoNeo Aug 2 '11 at 0:01

3 Answers 3

No: it can very well happen that a group has only one nontrivial normal subgroup. The symmetric group $\mathfrak S(n)$, for $n \geq 5$, has this property.

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Thanks! If $H$ is infinite and finitely generated, is there any counter example? –  MTG Aug 2 '11 at 2:24

Not in general. Take an infinite simple group $H$ which admits an outer automorphism $a$, and let $G$ be the semdirect product $H \langle a \rangle$ (in fact, your question as stated does not exclude the possibility that $H =G $, in which case taking $H = G$ to be any infinite simple group does the trick, but as I indicate, this sort of thing can still happen when $H$ is a proper normal subgroup of $G$). An explicit example with $H$ not quite simple is to take $H = {\rm SL}(3,\mathbb{C})$, and $a$ to be the automorphism of $H$ given by $M \to (M^{t})^{-1}$ for each invertible matrix of determinant $1$. This is an outer automorphism of $H$. The only proper normal subgroup of $H$ is its center $Z(H)$, which consists of scalar matrices and has infinite index (and is normal in $G = H\langle a \rangle$).

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Thanks! If $H$ is infinite and finitely generated, is there any counter example? –  MTG Aug 2 '11 at 2:23

No -- take $G$ to be any infinite simple group, and take $H = G$. For an especially dramatic example, take a Tarski Monster group.

If you want an example where $H$ is actually proper in $G$, let $G_1$ be any infinite simple group, and take $G = G_1 \times G_2$, where $G_2$ is any finite simple group. For what you need to know about normal subgroups of a product of two nonisomorphic simple groups, see here.

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Thanks! If $H$ is infinite and finitely generated, is there any counter example? –  MTG Aug 2 '11 at 2:23
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The Tarski Monsters are 2-generated. There are even infinite finitely presented simple groups. –  Derek Holt Aug 2 '11 at 6:48
    
What about locally indicable $H$? If $H$ is locally indicable, i.e. there is a non trivial homomorphism from $H$ onto (Z,+), –  MTG Aug 2 '11 at 13:13
    
Thanks! What about locally indicable $H$? If $H$ is locally indicable, i.e. there is a non trivial homomorphism from $H$ onto (Z,+), and also for every its finitely generated subgroups, then $H$ is not simple and has a normal subgroup with infinite index. But I don't know about the existence of such group with finite (and greater than one) index! –  MTG Aug 2 '11 at 13:27
    
If $H$ maps on ${\mathbb Z}$, then it clearly has normal subgroups of index $n$ for all $n$. The answer to your original question is yes for finitely generated locally indicable groups. –  Derek Holt Aug 2 '11 at 17:47

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