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We wish to classify the factor group $(\mathbb{Z} \times \mathbb{Z})/\langle (2, 2) \rangle$, that is, find a group to which it is isomorphic. (According to the fundamental theorem of finitely generated abelian groups. Initially, I thought the group had but two cosets, forcing an isomorphism to $\mathbb{Z}_2$. Obviously, this is wrong, due to the existence of cosets such as $(1, 0) + \langle (2, 2) \rangle$ However, I am unable to see how I am to find an isomorphism here.

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  • $\begingroup$ Andrew: my answer was total crap, sorry. $\endgroup$ – Ian Coley Mar 21 '14 at 8:28
  • $\begingroup$ No problem; I've provided this site with both crappy questions and answers :) $\endgroup$ – Andrew Thompson Mar 21 '14 at 16:27
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Hints:

  • Show that $\Bbb{Z}\times\Bbb{Z}$ is generated (freely) by the elements $u=(1,0)$ and $v=(1,1)$. IOW every element $(a,b)\in\Bbb{Z}\times\Bbb{Z}$ can be written as a linear combination of $u$ and $v$ with integer coefficients in a unique way.
  • Show that $mu+nv$ is in the subgroup generated by $(2,2)$ if and only if $m=0$ and $n$ is even.
  • Show that the mapping $mu+nv+\langle(2,2)\rangle \mapsto (m,\overline{n})$ is an isomorphism from $(\Bbb{Z}\times\Bbb{Z})/\langle(2,2)\rangle$ to $\Bbb{Z}\times(\Bbb{Z}/2\Bbb{Z})$.
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    $\begingroup$ In general there is a result aka the stacked bases theorem that helps understand quotient groups of free abelian groups like here. Usually it is explained at the same time as the Smith normal form and invariant factors of matrices with integer coefficients. $\endgroup$ – Jyrki Lahtonen Mar 21 '14 at 8:30
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    $\begingroup$ I wonder why so many very specific questions like this are asked. It would surely make more sense to teach students a general method (i.e. algorithm) for solving such problems, and the Smith normal form is the natural way to do it. $\endgroup$ – Derek Holt Mar 21 '14 at 12:21
  • $\begingroup$ We have not been told about such an algorithm, but I would highly appreciate it if anyone (you?) could provide an answer to this question, solving the tasking using said method, seeing as a straight-forward algorithm for such a task would be brilliant. (And actually help me understand the material, given that I understand why the algorithm works.) $\endgroup$ – Andrew Thompson Mar 21 '14 at 16:26
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The kernel of the surjective map $\mathbb Z\times\mathbb Z\to \mathbb Z\times\mathbb Z/2\mathbb Z$ given by $(m,n)\mapsto (m-n,n)$ is $\langle(2,2)\rangle$.

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