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?
Your guess is too large; this group is generated by two elements, whereas your guess is countably generated. Your arguments for both the torsion-free and torsion parts are incorrect:
Torsion-free: “disjoint” is not a meaningful word to use here. Any two subgroups contain the identity in common. The meaningful thing to ask here is about independence, and $(1, 0)$ and $(0, 1)$ are not independent, because $2(1, 0) + 2(0, 1) = 0$ in the quotient.
Torsion: the elements on the odd diagonal are also not independent; in fact they are all equal in the quotient.
To figure out the correct quotient, it’s convenient to pick a different choice of generators. Hint: one of them should be $(1, 1)$.
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
$$
G = \dfrac{\mathbb Z \oplus \mathbb Z}{ \langle 2v_1 \rangle}
= \dfrac{v_1 \mathbb Z \oplus v_2\mathbb Z}{v_1 2\mathbb Z \oplus 0}
\cong \dfrac{\mathbb Z}{2\mathbb Z} \times \mathbb Z
$$