This is a question following my post here: $2:1$ coverings of $S^1 \vee S^1 \vee S^1$. In this post, it turns out the number of 2:1 coverings of $S^1 \vee S^1 \vee S^1$ is the same as the number of index 2 subgroups of $\mathbb{Z}*\mathbb{Z}*\mathbb{Z}$ (free product on three generators).
So, I am asking how to find index 2 subgroups of $\mathbb{Z}*\mathbb{Z}*\mathbb{Z}$. I was told that these subgroups can be realizes as kernels of homomorphisms from $\mathbb{Z}*\mathbb{Z}*\mathbb{Z} \to \mathbb{Z}_2$. I was also told that any homomorphism factors through the abelianization $\mathbb{Z}\times \mathbb{Z}\times \mathbb{Z}$.
This makes me wonder: to find homomorphisms of $\mathbb{Z}*\mathbb{Z}*\mathbb{Z} \to \mathbb{Z}_2$ does it suffice to consider homomorphisms from the $\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z} \to \mathbb{Z}_2$ (in particular, the number of the the former is equal to the number of the latter)?
More generally, if $G,H$ are groups with $H$ abelian is there a bijective correspondence between homomorphisms from $G \to H$ and homomorphisms $G_{ab} \to H$ ($G_{ab}$ denotes abelianization of $G$)?
Best Answer
Yes: every homomorphism from $G$ to an abelian group factors through $G^{\rm ab}$ (the more common notation); this is the universal property of $G^{\rm ab}$, and in fact can be used to define $G^{\rm ab}$; and any morphism from $G^{\rm ab}$ to any group (in particular, to an abelian group) yields a homomorphism from $G$ by pre-composing with the projection $\pi\colon G\to G^{\rm ab}$).
While you can construct your morphisms by looking at $\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$, in fact there is no need here. A morphism from a free product corresponds to families of morphisms from the factors. Thus, every morphism $\mathbb{Z}\ast\mathbb{Z}\ast\mathbb{Z}$ to $G$ corresponds to a triple of morphism $(f,g,h)$, with $f,g,h\colon \mathbb{Z}\to G$, with each morphism acting on the corresponding copy of $\mathbb{Z}$. Here, you only have to decide whether each of the copies of $1$ go to $\mathbf{0}$ or to $\mathbf{1}$ in $\mathbb{Z}_2$, and make sure that at least one goes to $\mathbf{1}$ to ensure you get a surjective map.