Show that no group homomorphism $h : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ is injective.

Question

Show that no group homomorphism $h : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ is injective.

If $h$ is injective, then $\ker(h)= \{(a,b) \mid h(a,b)=0 \} =0$. Now the map $h$ is completely determined by where it sends the generators $(1,0)$ and $(0,1)$ so can we use these two to show a contradiction?

Answer 1

If there were an injective homomorphism, that would embed $\mathbb Z^2$ as a subgroup of the cycllic group $\mathbb Z$. But every subgroup of a cyclic group is cyclic, whereas $\mathbb Z^2$ is not cyclic.

Answer 2

If $h(1,0)=n,h(0,1)=m$, then $h(m,0)=mn=h(0,n)$. This example demonstrates $h$ is not injective, unless $m=n=0$. But in this case, $h(g)=0$ for all elements $g$ and is not injective.

This is Arturo’s hint. No Bézout!