there is no injective group homomorphism from $\mathbb Z\times\mathbb Z$ into $\mathbb Z$ there is no injective group homomorphism from $\mathbb Z\times\mathbb Z$ into $\mathbb Z$
But i don't know why it is true.
should i investigate all group homomorphisms from $\mathbb Z\times\mathbb Z$ into $\mathbb Z$?
Do you have any idea to help me? :(
 A: Hint : $\mathbb{Z} \times \mathbb{Z} $ is not cyclic but all subgroups of $\mathbb{Z}$ are cyclic.
A: We have $$\begin{array}{rcl}h(h(0,1),0) &=& h(\underbrace{1+\cdots +1}_{h(0,1)\mbox{ times}},0)\\ &=& h(\underbrace{(1,0)+\cdots +(1,0)}_{h(0,1)\mbox{ times}})\\ &=& \underbrace{h(1,0)+\cdots +h(1,0)}_{h(0,1)\mbox{ times}}\\ &=& h(0,1).h(1,0).\end{array}$$
We also have $$\begin{array}{rcl}h(0,h(1,0)) &=& h(0,\underbrace{1+\cdots +1}_{h(1,0)\mbox{ times}})\\ &=& h(\underbrace{(0,1)+\cdots +(0,1)}_{h(1,0)\mbox{ times}})\\ &=& \underbrace{h(0,1)+\cdots +h(0,1)}_{h(1,0)\mbox{ times}}\\ &=& h(0,1).h(1,0).\end{array}$$
It follows that $h(h(0,1),0)=h(0,h(1,0))$ and hence if $h$ is injective then we must have $h(0,1)=0$ and $h(1,0)=0$ but then $h$ has non-trivial kernel which contradicts injectivity.
A: Suppose we have a group homomorphism $h: \mathbb{Z}\times \mathbb{Z} \to \mathbb{Z}.$ Our goal is to show $h$ is not injective, which for a group homomorphism is the same as showing that the kernel of $h$ is more than just $\{ (0,0)\}.$ 
The essence of $h$ is contained in where it sends $(0,1)$ and $(1,0).$ Suppose $h(1,0) = a$ and $h(0,1)=b.$ Then since $h$ is a group homomorphism, we must have $h(n,m) = an+bm.$ We want to observe why such a formula for $h$ implies that it is not injective. $(0,0)$ is in the kernel; can you find another $(n,m)$ such that $h(n,m)=0?$
A: Three proofs.


*

*From the classification of subgroups of $\mathbb{Z}$ we know that every two non-trivial subgroups of $\mathbb{Z}$ have a non-trivial intersection (in fact, $n \mathbb{Z} \cap m \mathbb{Z}$ contains $n \cdot m$). But in $\mathbb{Z} \times \mathbb{Z}$ we have two copies of $\mathbb{Z}$ with trivial intersection. Hence, $\mathbb{Z} \times \mathbb{Z}$ doesn't embed into $\mathbb{Z}$.

*If there was an injective homomorphism $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$, then by localization we would get an injective homomorphism $\mathbb{Q} \times \mathbb{Q} \to \mathbb{Q}$, which is impossible by linear algebra.

*If there was an injective homomorphism $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$, the image would be isomorphic to $\mathbb{Z}$, so that $\mathbb{Z} \times \mathbb{Z} \cong \mathbb{Z}$. After applying $A \mapsto A/2A$, we would get $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \cong \mathbb{Z}/2\mathbb{Z}$, which  contradicts the group orders.
