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Evidently, $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ has order $4$, but I think it’s infinite.

The four cosets are listed as $(0,0) + \langle (2,2) \rangle$, $(0,1) + \langle (2,2) \rangle$, $(1,0)+ \langle (2,2) \rangle$ and $(1,1) + \langle (2,2) \rangle$. However, $(2,0)$ doesn’t appear to be in any of these cosets. Maybe the answer I’m being told is wrong.

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You are right, $(\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle$ is infinite. You can embed $\mathbb{Z}$ via $k\mapsto (k,0)$ (and in other ways) into it. The quotient is not cyclic, because it contains elements of finite order, $(1,1)$ for example.

Probably it was meant to be $(\mathbb{Z} \oplus \mathbb{Z})/ (\langle 2\rangle\oplus \langle 2 \rangle)$ which indeed is a group of order $4$ (a Klein $4$-group).

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