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Suppose that $F = \langle a,b \rangle$ is the free group on two generators and let $H=\langle X,Y\rangle$ be the subgroup of $F$ generated by $X = (ab)^k$, $k$ non-zero integer, and $Y = a$.

What is the index of $H$ in $F$ ?

We know that $H$ is a free group on two generators. When $k = \pm 1$, it is easy to see that $F = H$. But what about when $k \neq \pm 1$ ?

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    Since F is generated by a and ab, which have no relations, you could replace ab by b in your definition of X. That is, F = and H = = . The group SL(2,Z) has two standard generators S = [0 -1|1 0] and T = [1 1|0 1], where S has order 4 and ST has order 6. While has index 3 in SL(2,Z) = , for any k > 2 the subgroup has infinite index in SL(2,Z). The proof, shown to me by Vincentiu Pasol, uses the action of SL(2,Z) on primitive vectors in Z^2. Because of this result,[F:H] is infinite for k > 2.2011-06-20
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    The above seems worth writing as an answer instead of a comment.2011-06-20
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    As a general rule any proper subgroup of $F_2$ isomorphic to $F_2$ must have infinite index. A finite-index subgroup of a free group is always a free group of higher rank.2011-06-20
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    Jim: a subgroup of index 1 wouldn't have higher rank and your statement is also incorrect for a free group of rank 1.2011-06-21
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    @KCd: Jim said "proper", so index 1 doesn't count. Also, he was speaking of $F_2$ (and he's right), not of free groups of rank 1.2011-06-21
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    Thank you all for the wonderful insight. I appreciate your help. KCd: In your first comment, you mention "Since F is generated by a and ab, which have no relations, you could replace ab by b in your definition of X. That is, $F = \langle a,b \rangle$ and $H = \langle Y,X \rangle = \langle a,b^k \rangle$". Why is this true ? Jim: Could you please provide a reference for your statement (third comment) ? According to Swlabr's answer, the formula guarantees that the index must be 1 assuming it is finite. But this does not give answer the question posed completely. Why is $\langle a,(ab)^k \rangle$2011-06-23
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    Since $F = = ,$ we see that the quotient of $F$ by the normal closure of $H$ is equal to $/<(ab)^k>$, which is cyclic of order $k$. In particular, if $k > 1$ this is non-trivial, hence $H$ has proper normal closure in $F$, and so $H$ itself is a proper subgroup of $F$.2011-06-23
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    [Moderator's note, @user12484 's comment was converted from a misused "answer", which unfortunately resulted in the last sentence being cut off. The following is the full last sentence as written by the user.] Why is $\langle a,(ab)^k\rangle$ is a proper subgroup of $\langle a,b\rangle$ when $k\neq \pm 1$?2011-06-24

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