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Let $G= (\mathbb{Z} \bigoplus \mathbb{Z}) \star (\mathbb{Z} \bigoplus \mathbb{Z})$, where $\star$ denotes the free product, let F be the commutator subgroup of G, it is free by a theorem of Kurosh. Find a proper normal subgroup of F (other than the trivial one) such that it is of infinite index.

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It would help if you supplied some motivation and explained what $[N:N]$ means (typo?) –  Guntram Apr 8 '10 at 7:36
    
[N:N] is the commutator subgroup of N. By the way of motivation: There is a theorem by Auslander and Lyndon which says that $F/[N:N]$ has non trivial center if and only if $F/N$ is a finite group. I am interested in the finiteness of $F/N$. –  Sagar Kolte Apr 8 '10 at 8:15
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So you're asking about finite index normal subgroups of an infinite-rank free group. Can you explain what kind of answer you are looking for? –  Tom Church Apr 8 '10 at 13:22
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I'm with Tom here. As you've written it this question doesn't have an answer other than "when it's not trivial." Maybe someone will come along and just know something about this problem, but it would be helpful, and lead to better answers if you said something about what sort of answers would help. Also, if you edited the question to clarify, rather than just leaving a comment. –  Ben Webster Apr 8 '10 at 13:27
    
I don't understand the point of this question. $F$ is free of infinite rank, so there are tons of normal subgroups of infinite index! –  Steve D Apr 8 '10 at 21:35

1 Answer 1

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The commutator subgroup $F' = [F:F]$ of $F$. It is normal. $F$ is not abelian, so $F'$ is nontrivial. The quotient $F/F'$ is a free abelian group of infinite rank, so $[F:F']$ is infinite.

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Along the lines of Steve D's comment above, I am surprised that this obvious observation has become the accepted answer. Admittedly, the competition was not especially fierce. Anyway, always happy to help, however little... –  Pete L. Clark Apr 9 '10 at 3:55

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