# Proving that there is a homomorphism $f : \mathbb{Z}^2 \ast \mathbb{Z}^2 \to \mathbb{Z}^2$ with kernel free and not finitely generated

I want to find an example of a homomorphism $$f : \mathbb{Z}^2 \ast \mathbb{Z}^2 \to \mathbb{Z}^2$$ such that $$\ker f$$ is free and not finitely generated.

My idea is to define $$f$$ on each copy of $$\mathbb{Z}^2$$ as the identity map, then use the universal property of free products to get a map $$f : \mathbb{Z}^2 \ast \mathbb{Z}^2 \to \mathbb{Z}^2$$.

The kernel is free since conjugates of $$\mathbb{Z}^2$$ map isomorphically to their images: $$g \mathbb{Z}^2 g^{-1}$$ maps to $$\mathbb{Z}^2$$ isomorphically since in the image, we can move the $$g^{-1}$$ past the elements of $$\mathbb{Z}^2$$ and cancel with $$g$$. Then the kernel intersects trivially with all conjugates, so acts freely on the Bass-Serre tree of $$\mathbb{Z}^2 \ast \mathbb{Z}^2$$ (which has vertex stabilisers given by conjugates of $$\mathbb{Z}^2$$), so $$\ker f$$ is free.

I can't think of any finite generating set for $$\ker f$$, but I am having a tough time thinking about why it can't be finitely generated. Certainly there are finitely generated subgroups with not finitely generated subgroups (e.g. the rank 2 free group $$F_2$$ has the countably infinite rank free group $$F_\infty$$ as a subgroup), so thinking about subgroups won't help. Maybe constructing an isomorphism to $$F_\infty$$?

It's a classical and not hard theorem that if a finitely generated group $$G$$ has a finitely generated subgroup $$N$$ such that $$N$$ is normal and both $$N$$ and $$G/N$$ are infinite, then $$G$$ has a single end.
There's also a more explicit approach. Namely, consider the action of $$G\ast G$$ on its Bass-Serre tree (which has valency $$|G|$$). Let $$N$$ be the kernel of the homomorphism onto $$G$$ that is identity on each factor. It acts freely, hence is free. It's not hard to see that $$N$$ has the same vertex orbits (namely 2). But all directed edges emanating from a single vertex are in distinct orbits. Hence, $$N$$ is infinitely generated as soon as $$G$$ is infinite.