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The exact problem asks to Identify the isomorphism class of the quotient group $\mathbb{Z}\times\mathbb{Z}/\langle(2,2)\rangle$ within the classification of finitely generated abelian groups.

I'm trying to find the cosets, but I can't understand what they might look like outside a finite group. I know that $\langle(2,2)\rangle = \{..., (-2,-2), (0,0), (2,2), (4,4), ...\}$ in $\mathbb{Z}\times\mathbb{Z}$, so I started with $(1,n)$ and $(0,m)$ for any $n,m$ as coset representatives.

Using those two, I tried to solve for the finite order cosets $(xi, ni)=(2j,2j)$ where $x=0,1$ and I only found the identity coset for $x=1$ and the coset $(1,1)+\langle(2,2)\rangle$ for $x=1$. However it appears to me that this has order $2$.

My intuitive guess is the quotient group is isomorphic to $\mathbb{Z}\times\mathbb{Z_2}$, especially since the other finite coset seems to have order $2$. I really can't grasp these concepts around infinite groups. I don't know where I'm going wrong, but ultimately a general explanation of what is going on would be preferred so I can attempt to apply it myself in this context.

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    $\begingroup$ Try thinking like you would for vector spaces, and write down a different "basis" for $\mathbb{Z} \times \mathbb{Z}$, which plays well with the element $(2, 2)$. A strong hint: every element can be uniquely written as $a(1, 1) + b(1, 0)$ for $a, b \in \mathbb{Z}$. $\endgroup$ – Joppy Oct 1 '20 at 22:52
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    $\begingroup$ $2$ is prime... $\endgroup$ – Qiaochu Yuan Oct 1 '20 at 22:53
  • $\begingroup$ @QiaochuYuan oh yea. so $\mathbb{Z}\times\mathbb{Z_2}$ does work then, so would that be the answer or am I still doing something wrong. $\endgroup$ – Jack Oct 1 '20 at 22:56
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A presentation for $\Bbb Z\times \Bbb Z$ is

$$\langle a,b\mid ab=ba\rangle,\tag{1}$$

where $a\mapsto (1,0)$, say, and $b\mapsto (0,1)$; in which case $(2,2)$ corresponds so $(ab)^2$. The quotient by $\langle (2,2)\rangle$ is, in effect, the same as killing $(ab)^2$ in $(1)$, like so: let $c=ab$; then:

$$\begin{align} \Bbb Z\times \Bbb Z/\langle (2,2)\rangle&\cong \langle a,b, c\mid (ab)^2, c=ab=ba\rangle\\ &\cong \langle a,b, c\mid c^2, c=ba\rangle\\ &\cong \langle a,b,c\mid c^2, cb=bab\rangle\\ &\cong \langle b,c\mid c^2, cb=bc\rangle\\ &\cong \Bbb Z\times \Bbb Z_2, \end{align}$$

as you suspected.

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    $\begingroup$ In the right hand sides of your isomorphisms, the generators should be $a,b,c$ for the first three, and $b,c$ for the final one. $\endgroup$ – Derek Holt Oct 2 '20 at 7:42
  • $\begingroup$ Ah, yes, of course! Thank you, @DerekHolt. $\endgroup$ – Shaun Oct 2 '20 at 11:43

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