Finite index subgroup isomorphic to $\Bbb{Z}$ inside the free product $\Bbb{Z}_{2}*\Bbb{Z}_{2}$

Question

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?

Answer 1

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 $H$).

Answer 2

Your conclusion is correct.

Indeed, it checks out in GAP:

gap> F:=FreeGroup(2);
<free group on the generators [f1, f2] >
gap> rels:=[(F.1)^2, (F.2)^2];
[ f1^2, f2^2 ]
gap> G:=F/rels;
<fp group on the generators [ f1, f2 ]>
gap> H:=Subgroup(G, [(G.1)*(G.2)]);
Group([ f1*f2 ])
gap> Index(G, H);
2

The first command defines the free group F on two elements.

The second command defines a set rels of relators, which the third command uses to define G, which is the quotient of F by the subgroup generated by rels.

The fourth command defines H as the subgroup of G generated by the set [ (G.1)*(G.2) ], which corresponds to your $H$.

The final command checks the index of H in G.