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.
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.