What is $\mathbb Z \oplus \mathbb Z / \langle (2,2) \rangle$ isomorphic to? This question came up after I'd solved the following exercise:

Determine the order of $\mathbb Z \oplus \mathbb Z / \langle (2,2) \rangle$. Is the group cyclic?

I had no trouble solving the exercise: The answer is the group is infinite and non-cyclic.
But as a bonus exercise I tried to figure out what it is isomorphic to and now I'm stuck. 
Here is what I have so far:
$[(1,0)]$ and $[(0,1)]$ generate infinite disjoint subgroups. 
Also, every element in the odd diagonal $[(2k+1, 2k+1)]$ has order $2$. Let's call the odd diagonal $D$. Then $D \cong \oplus_{k \in \mathbb Z} \mathbb Z_2$. 

Is this correct? And if so, is $\mathbb Z \oplus \mathbb Z / \langle
 (2,2) \rangle \cong \oplus_{k \in \mathbb Z} \mathbb Z_2 \oplus
 \mathbb Z \oplus \mathbb Z$?

And if not could someone help me by showing me what it is isomorphic to?

Let $G = \dfrac{\mathbb Z \oplus \mathbb Z}{ \langle (2,2) \rangle}$.


*

*$v_1=(1,1)$ is mapped to an element of order $2$ in $G$.

*$v_2=(2,1)$ is mapped to an element of infinite order in $G$.
This means that $G$ is infinite and non-cyclic.
Note that $\mathbb Z \oplus \mathbb Z = v_1 \mathbb Z \oplus v_2\mathbb Z$. Therefore
$$
G = \dfrac{\mathbb Z \oplus \mathbb Z}{ \langle 2v_1 \rangle}
 = \dfrac{v_1 \mathbb Z \oplus v_2\mathbb Z}{v_1 2\mathbb Z \oplus 0}
 \cong \dfrac{\mathbb Z}{2\mathbb Z} \times \mathbb Z
$$
