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How is $\Bbb Z$ x $\Bbb Z$/$<(1,1)>$ Isomorphic to $\Bbb Z$? The book im reading says that we must choose the representatives $···,(-2,0), (-1,0), (0,0), (1,0), (2,0), ···$ for the cosets of $<(1, 1)>$ and doesnt give further explanation on why.

How are the representatives chosen and how do they form the factor group?

Mallacan
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  • Representatives from cosets can be chosen arbitrarily. It's sometimes convenient to choose particular elements for this purpose, but it's not required. Do you understand how to find the cosets? – CyclotomicField Aug 27 '22 at 15:29
  • Yes, $aH = ${$ah | h \in H$}, $Ha = ${$ha | h \in H$} – Mallacan Aug 27 '22 at 15:32
  • Ok so you have the cosets and you know $(1,1)$ is in the kernel, meaning it's the coset that contains the identity, which is $(0,0)$ in this case. Since this group is abelian the left and right cosets are the same. Can you now calculate all the cosets of this specific group? I recommend picking some small elements and figuring out which coset they belong to like $(3,5)$ and $(-2,1)$. After that you should be able to put the pieces together. – CyclotomicField Aug 27 '22 at 15:35
  • Oh, I get it. The representatives $⋅⋅⋅,(−2,0),(−1,0),(0,0),(1,0),(2,0),⋅⋅⋅$ will "fill up" all the $\Bbb Z x Z$ space. So we might as well choose $⋅⋅⋅,(0,-2),(0,-1),(0,0),(0,1),(0,2),⋅⋅⋅$. Thanks – Mallacan Aug 27 '22 at 15:41

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Hint

Take $\varphi :\mathbb Z\times \mathbb Z\to \mathbb Z$ defined by $$\varphi (m,n)=m-n.$$ One can prove that it's a surjective group homomorphism s.t. $\ker(\varphi )=\left<(1,1)\right>.$

Surb
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  • Yeah, this makes sense but how would you do it using cosets and representatives? – Mallacan Aug 27 '22 at 15:20
  • It's exactly the same idea. Indeed, $$(m,n)+\left<(1,1)\right>=(m-n,0)+\left<(1,1)\right>,$$ for all $(m,n)+\left<(1,1)\right>\in \mathbb Z\times \mathbb Z/\left<(1,1)\right>$. Thus define $\psi :\mathbb Z\times \mathbb Z/\left<(1,1)\right>\to \mathbb Z$ by $\psi\left((m,n)+\left<(1,1)\right>\right)=m-n.$ You can prove that $\psi$ is a well defined group homomorphism that is an isomorphism. @Mallacan – Surb Aug 27 '22 at 15:29
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    I get it, thanks – Mallacan Aug 27 '22 at 15:41