$\mathbb{Z} \times \mathbb{Z}$ 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 $h((x,y) + (m,n)) = h((x,y)) + h((m,n))$.
 A: I think that you are possibly confusing  two different subgroups.
(i) $G=\mathbb{Z}\times\mathbb{Z}$ has a subgroup $H_1=\{(2k,3k)| k\in\mathbb{Z}\}$, and then $G/H_1\simeq\mathbb{Z}$.
(ii) $G=\mathbb{Z}\times\mathbb{Z}$ has a subgroup $H_2=\{(2k,3\ell)| k,\ell\in\mathbb{Z}\}$, and then $G/H_2\simeq\mathbb{Z}_2\times\mathbb{Z}_3\simeq\mathbb{Z}_6$. 
Of course $H_1\leqslant H_2$, but they are not equal. 
A: No. If $G/H$ was isomorphic to both $\mathbb Z$ and $\mathbb{Z}_2\times\mathbb{Z}_3$, then $\mathbb{Z}\simeq\mathbb{Z}_2\times\mathbb{Z}_3$. But this is not true. For instance, $\mathbb Z$ is an infinite group, whereas $\mathbb{Z}_2\times\mathbb{Z}_3$ has order $6$.
A: You have that $(2,3)H$ and $(4,3)H$ are both send to the same element using the homomorphism. However it's not true that $(2,3)H = (4,3)H$, as this would mean that $(2,0) \in H$, which isn't the case.
Hence $G/H$ isn't isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_3$.
A: 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.))
A: Your $h$ is simply an epimorphism then the only thing you get is that 
$$\frac{G/H}{\ker h}\quad\mbox{isomorphic to}\quad \Bbb Z_2\times\Bbb Z_3$$ by a theorem of homomorphisms.
