$(\mathbb{Z}\times\mathbb{Z})/\langle (1,2) \rangle$ is isomorphic to what?

Question

$$(\mathbb{Z}\times\mathbb{Z})/\langle (1,2) \rangle$$

We can see that $\langle(1,2)\rangle = \{\cdots,(-1,-2),(0,0)(1,2),(2,4)(3,6),\cdots\}$

We have that $(a,b), (c,d)$ are in the same coset when $(a-c, b-d) = (k,2k)\implies a = c+k, b = d+2k$

Let's try a few c's and d's:

$c = 0, d = 0$, then $a = k, b = 2k \implies (0,0) $ is in the coset $(k,2k)$ which is $\{\cdots,(-1,-2),(0,0)(1,2),(2,4)(3,6),\cdots\}$

$c=1, d=0$ then $a = 1+k, d = 0+2k$ so the coset is of the form $(1+k,2k)$ which is $\{(1,0), (2, 2), (3, 4), (4, 6), \cdots\}$

$c = 0, d = 1$ then $a = k, b = 1+2k$ so the coset is of the form $(k,1+2k)$ which is $\{(0,1), (2,2), (4,3), (6,4), \cdots\}$

$c=1,d=1$ then $a = 1+k, b = 1+2k$ the coset is of the form $(1+k, 1+2k)$ which is $\{(1,1),(2,5), (3,7), (4,9), \cdots\}$

How can I see an isomorphism here?

In general, the cosets will be of the form $\{\{(c+k, d+2k), k\in \mathbb{Z}\}, c,d\in\mathbb{Z}\}$. I cannot see any isomorphism here.

Answer 1

Note that the onto homomorphism $\Bbb Z\times \Bbb Z\to\Bbb Z$, $(a,b)\mapsto 2a-b$ has kernel $\langle(1,2)\rangle$.

Answer 2

The usual trick here is to rely on the First Group Isomorphism Theorem. In particular: If we can find a surjective homomorphism $\phi:\Bbb Z \times \Bbb Z \to H$ whose kernel is $\langle (1,2) \rangle$, then we have $(\Bbb Z \times \Bbb Z)/ \langle (1,2) \rangle \cong H$.

In this case, the map $\phi(x,y) = 2x - y$ does just what we want.

---

On the other hand: keeping your analysis in mind, note that every coset can be written in the form $$ (0,d) + \langle (1,2) \rangle $$ for some $d \in \Bbb Z$. You can then show that $d \mapsto (0,d) + \langle (1,2) \rangle$ is an isomorphism.