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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?

Ricardo
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    No contradition. Just show that there must be some $(k,0)$ that maps to the same thing as some $(0,m)$, with $k$ and $m$ not both zero. Use the fact that given two integers $a$ and $b$, they must have a common multiple. – Arturo Magidin Oct 16 '22 at 21:18
  • Is that the lemma by Bezout? @ArturoMagidin – Ricardo Oct 16 '22 at 21:41
  • If $h(1,0) = m$ and $h(0,1) = n$ then for any $(a,b)$ we have $h(a,b) = h((1,0)a + (0,1)b) = ma + nb$. – Ricardo Oct 16 '22 at 21:52
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    No, Bezout's Lemma tells you about the gcd. You don't need Bezout, or even the least common multiple, for this. If $h(1,0)=a$ and $h(0,1)=b$, and not both are zero, then $h(b,0)=h(0,a)$. – Arturo Magidin Oct 16 '22 at 22:38

2 Answers2

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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!

FShrike
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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.

Anne Bauval
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