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 $k \neq \pm 1$ ?
has index 3 in SL(2,Z) =, for any k > 2 the subgrouphas 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.