# Classify $\mathbb{Z}\times\mathbb{Z}/\langle(0,3)\rangle$ according to the fundamental theorem of finitely generated abelian groups.

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!

• The element $(0,1)$ is of order $3$ and so the group can't be $\mathbb{Z}$. – Yanko Nov 8 '18 at 19:14
• It goes like for this question. – Dietrich Burde Nov 8 '18 at 19:48

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 $$\mathbb{Z}\times\mathbb{Z}_3$$.

• 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 at 18:48
• @JonathanRayner $G/N$ need not be isomorphic to $G/N'$ even if $N$ and $N'$ are isomorphic. That conclusion is wrong. – freakish Apr 13 at 18:54
• Oh true, okay thanks! – Jonathan Rayner Apr 13 at 19:43

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 $$(x,y) \in \mathbb{Z}\times\mathbb{Z} \mapsto (x, y \bmod 3) \in \mathbb{Z}\times\mathbb{Z_3}$$.