$\mathbb{Z} \times \mathbb{Z}$ and application of isomorphism theorems Let $G = \mathbb{Z} \times \mathbb{Z}$ with group law given by addition. Let $H$ be the subgroup generated by the element $(2,3)$. Then $G/H$ is isomorphic to $\mathbb{Z}$. 
Is $G/H$ also isomorphic to $\mathbb{Z}_2\times \mathbb{Z}_3$
using the homomorphism $h((x,y)) = (x\mod 2,\ y\mod 3)$? (here, $h((0,0)) = (0,0)$ and $h((x,y) + (m,n)) = h((x,y)) + h((m,n))$.

For $a,b \in G$ to land in the same equivalence class modulo $H$, there must be an integer $k$ such that $a = b + k(2,3) = b + (2k, 3k)$ (which should look very similar to the definition of equivalence classes of residues in the integers modulo a non-zero integer).  As an example, the equivalence class of $G/H$ containing $(0,0)$ (which is $H$) also (and only) contains
$$  \dots, (-6,-9), (-4,-6), (-2,-3), (0,0), (2,3), (4,6), (6,9), (8,12), \dots $$
Your rule "$h((x,y) = (x \pmod{2}, y \pmod{3})$" puts more things into the equivalence class containing $(0,0)$, including 
$$
\begin{matrix}
\ddots & \vdots & \vdots & \vdots & \vdots & \vdots & \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} \\
\dots  & (-4, 6) & (-2, 6) & (0, 6) & (2, 6) & (4, 6) & \dots  \\
\dots  & (-4, 3) & (-2, 3) & (0, 3) & (2, 3) & (4, 3) & \dots  \\
\dots  & (-4, 0) & (-2, 0) & (0, 0) & (2, 0) & (4, 0) & \dots  \\
\dots  & (-4,-3) & (-2,-3) & (0,-3) & (2,-3) & (4,-3) & \dots  \\
\dots  & (-4,-6) & (-2,-6) & (0,-6) & (2,-6) & (4,-6) & \dots  \\
\kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.} & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots
\end{matrix}
$$
Notice that you can take this second mass of points of $\mathbb{Z} \times \mathbb{Z}$ and cover all of $\mathbb{Z} \times \mathbb{Z}$ with just six congruent copies: the copies containing each of $(0,0)$, $(0,1)$, $(0,2)$,  $(1,0)$, $(1,1)$, and $(1,2)$.  You can't cover all of $\mathbb{Z} \times \mathbb{Z}$ with finitely many congruent copies of $H$ (listed above).  (Proof?  No two points of the subgroup generated by $(3,-2)$ are in the same coset of $H$, so since there are infinitely many points in this subgroup, there must be infinitely many distinct cosets of $H$.  (Why that particular subgroup?  Although many different choices would work, draw a picture.))
