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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? :(

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When you are asked to prove that something does not exist, usually you would be arguing by contradiction. Try assuming $T: \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ is an injective homomorphism and attempt to construct two elements which map to the same thing. –  Ben Passer Jul 14 at 14:07
You should be precise about what type of objects/homomorphisms these are. Abelian groups under addition, rings, what? –  Ragib Zaman Jul 14 at 14:09
sorry about that. i missed the word 'group' –  user134076 Jul 14 at 14:10
@Ben Passer, ok i'll try that 'contradiction' method! –  user134076 Jul 14 at 14:13
Another note: a homomorphism being injective is the same as insisting that its kernel contains only the identity element (make sure you can prove this fact!). So one of your two elements might as well be the identity, which must map to the identity of $\mathbb{Z}$. That means your task is to find another element which maps to the identity of $\mathbb{Z}$. –  Ben Passer Jul 14 at 14:17
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4 Answers 4

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 $h(n,m)=0?$

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Three proofs.

  1. 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 $\mathbb{Z}$.

  2. 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.

  3. 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.

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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.

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Hint : $\mathbb{Z} \times \mathbb{Z} $ is not cyclic but all subgroups of $\mathbb{Z}$ are cyclic.

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