Let $\Bbb{Z_{2}*Z_{2}}$ be the free product then I want to prove that there is a finite indexed subgroup isomorphic to $\Bbb{Z}$.
My attempt:
We denote the free product by $\langle a\rangle*\langle b\rangle$ for more clarity such that $a^{2}=e$ and $b^{2}=e$.
Then the subgroup $H=\langle ab\rangle $ is a subgroup isomorphic to $\Bbb{Z}$. Now as for the index I think it has index $2$ as the cosets are $\{H,bH\}$.
Am I correct in thinking this?
Best Answer
Every element $g$ in your group has a well defined length modulo $2$. Namely, when $g$ is written as a word in $a$ and $b$, the number of occurences of $a$’s and $b$’s (combined) can either be even or odd, but not both. Then the mapping $\mathbb{Z}_2* \mathbb{Z}_2 \rightarrow \mathbb{Z}_2$ which sends elements of even length to $0$ and elements of odd length to $1$ is a homomorphism. The kernel of this homomorphism is isomorphic to $\mathbb{Z}$ (it coincides with your $H$).