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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$\begingroup$ It would help if you supplied some motivation and explained what $[N:N]$ means (typo?) $\endgroup$– GuntramCommented Apr 8, 2010 at 7:36

$\begingroup$ [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 $F/N$. $\endgroup$– Sagar KolteCommented Apr 8, 2010 at 8:15

2$\begingroup$ So you're asking about finite index normal subgroups of an infiniterank free group. Can you explain what kind of answer you are looking for? $\endgroup$– Tom ChurchCommented Apr 8, 2010 at 13:22

1$\begingroup$ 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. $\endgroup$– Ben Webster ♦Commented Apr 8, 2010 at 13:27

$\begingroup$ 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! $\endgroup$– Steve DCommented Apr 8, 2010 at 21:35

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1 Answer
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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.

$\begingroup$ 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... $\endgroup$ Commented Apr 9, 2010 at 3:55