I am having an issue classifying $\mathbb{Z}\times\mathbb{Z}/\langle(0,3)\rangle$ according to the fundamental theorem of finitely generated abelian group (i.e. finding what $\mathbb{Z}\times\mathbb{Z}/\langle(0,3)\rangle$ is isomorphic to). I think It should $\mathbb{Z}$, but I am not sure why. Thanks!
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The element $(0,1)$ is of order $3$ and so the group can't be $\mathbb{Z}$. – Yanko Nov 08 '18 at 19:14
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It goes like for this question. – Dietrich Burde Nov 08 '18 at 19:48
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Well, it is not $\mathbb{Z}$. What is the order of (the coset) $[(0,1)]$?
You don't need the fundamental theorem at all. What you need is the following: if $G,G'$ are groups and $N\subseteq G$, $N'\subseteq G'$ are normal subgroups then $N\times N'$ is normal in $G\times G'$ and
$$(G\times G')/(N\times N')\simeq (G/N)\times (G'/N')$$
With that you can easily check that $\langle(0,3)\rangle=\{0\}\times 3\mathbb{Z}$ and so your group is $\mathbb{Z}\times\mathbb{Z}_3$.
freakish
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What stops me from using ${0} \times 3 \mathbb{Z} \cong {0} \times \mathbb{Z}$ and then concluding $(\mathbb{Z} \times \mathbb{Z})/ ({0} \times \mathbb{Z}) \cong \mathbb{Z}$? – Jonathan Rayner Apr 13 '19 at 18:48
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@JonathanRayner $G/N$ need not be isomorphic to $G/N'$ even if $N$ and $N'$ are isomorphic. That conclusion is wrong. – freakish Apr 13 '19 at 18:54
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We have
$\qquad \mathbb{Z}\times\mathbb{Z} = \mathbb{Z} e_1 \oplus \mathbb{Z} e_2 $
$\qquad \langle(0,3)\rangle = \mathbb{Z} (0 e_1) \oplus \mathbb{Z} (3e_2) $
Therefore,
$\qquad \mathbb{Z}\times\mathbb{Z}/\langle(0,3)\rangle \cong \mathbb{Z}\times\mathbb{Z_3}$
An explicit isomorphism is induced by $(x,y) \in \mathbb{Z}\times\mathbb{Z} \mapsto (x, y \bmod 3) \in \mathbb{Z}\times\mathbb{Z_3}$.
lhf
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