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 $F_\infty$?


1 Answer 1


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.

Hence, in your case, the kernel being infinite with infinite quotient, it is infinitely generated.

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.


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