# If $G\to\mathbb{Z}$ and $H\to\mathbb{Z}$ are surjective homomorphisms prove that exists $G\times H\to\mathbb{Z}$ with finitely generated kernel.

Let $$G$$ and $$H$$ be two finitely generated groups. If $$\varphi_G:G\to\mathbb{Z}$$ and $$\varphi_H:H\to\mathbb{Z}$$ are surjective homomorphisms prove that exists $$\varphi:G\times H\to\mathbb{Z}$$ with finitely generated kernel. I read this fact on a paper but without proof, so I thought that this must be either easy or well-known, however I failed.

The idea I had was that $$G=\ker(\varphi_G)\rtimes\mathbb{Z}$$ and $$H=\ker(\varphi_H)\rtimes\mathbb{Z}$$ so $$K=(\ker(\varphi_G)\times\ker(\varphi_H))\rtimes \mathbb{Z}$$ is finitely generated. So I was trying to find a map $$\varphi$$ with $$K$$ as its kernel, but I failed. The ideas of maps I thought were: $$\varphi(g,h)=\varphi_G(g)+\varphi_H(h),\varphi(g,h)=\varphi_G(g)-\varphi_H(h)$$ or $$\varphi(g,h)=\varphi_G(g)\varphi_H(h)$$ but none of them seemed to work and I don't know any other possibilty for $$\varphi$$.

• @SeanEberhard Maybe is obvious, but the part where I'm struggling is in proving that $K$ is actually the kernel of this map. Commented Jan 25 at 9:17
We have $$G \cong K_1 \rtimes_\alpha \mathbb Z$$ and $$H \cong K_2 \rtimes_\beta \mathbb Z$$, so $$G \times H \cong (K_1 \times K_2) \rtimes_{\alpha \times \beta} (\mathbb Z \times \mathbb Z)$$ . Let $$\varphi(g, h) = \varphi_G(g) - \varphi_H(h)$$ and let $$K = \ker(\varphi)$$. Then $$K = ((K_1 \times K_2) \rtimes (\mathbb Z \times \mathbb Z)) \cap K = (K_1 \times K_2) \rtimes ((\mathbb Z \times \mathbb Z) \cap K)$$ (this is the so-called "modular law of groups"). The group $$(\mathbb Z \times \mathbb Z) \cap K$$ is $$\langle(1,1)\rangle \cong \mathbb Z$$, so $$K \cong (K_1 \times K_2) \rtimes_{(\alpha,\beta)} \mathbb Z.$$ As you have presumably noted already, $$K$$ is generated by isomorphic copies of $$G$$ and $$H$$, so it is finitely generated.