Calculate the Factor Group $(\mathbb{Z}_4 \times \mathbb{Z}_6)/\langle(0,2)\rangle$ 
I am attempting to understand and compute: $(\mathbb{Z}_4 \times \mathbb{Z}_6)/\langle(0,2)\rangle$

I know $(0,2)$ generates $H = \{(0,0),(0,2),(0,4)\}$, which has  order 3 because there are 3 elements. Now, I must find all the cosets  which there should be 8. The reason there should be 8 is because $4\cdot 6/3 = 8$.
This is where I run into the issue. I have a hard time generating the cosets, but furthermore, once I actually do get the cosets, I have a hard time analyzing them and knowing what it means. If someone could walk me through a step-by-step solution that would be great.
The answer is that the factor group is isomorphic to $\mathbb{Z}_4 \times \mathbb{Z}_2$. 
Reference: Fraleigh, A First 
Course in Abstract Algebra, p. 147, Example 15.10.  
 A: You are right, the order of the quotient is 8. You want to enumerate all the cosets of $H$. Each coset is of the form $(a,b)H$ for some $(a,b) \in G:= \mathbb{Z}_4\times \mathbb{Z}_6$. The question is, when are two cosets the same? Now
$$
(a,b)H = (c,d)H \Leftrightarrow (a-c, b-d) \in H \Leftrightarrow a=c, 2\mid (b-d)
$$
So the distinct left cosets would correspond to the elements of the set
$$
\{(0,1), (0,2), (1,1), (1,2), (2,1), (2,2), (3,1), (3,2)\}
$$
Now the next question is the group structure on $G/H$ : This is given by multiplication of cosets
$$
(a,b)H \cdot (c,d)H = (a+c,b+d)H
$$
Consider the multiplication
$$
(0,1)H \cdot (0,2)H = (0,3)H = (0,1)H
$$
since $(0,3)-(0,1) = (0,2) \in H$.
Similarly,
$$
(1,2)H \cdot (3,2)H = (4,4)H = (0,0)H
$$
since $4 \equiv 0$ in $\mathbb{Z}_4$ and $(0,4 \in H$; and so on. Try and see if you can write down the multiplication table for $G/H$.
Once you have done that, explain why
(i) $G/H$ is abelian
(ii) $G/H$ has a cyclic subgroup $H$ of order 4 and a cyclic subgroup $K$ of order 2 which are disjoint.
(iii) Define the "natural" map $f : H\times K \to G$ and show that it is an isomorphism.
Conclude that $G/H \cong \mathbb{Z}_4\times \mathbb{Z}_2$
