List all cosets of H and K 
Let $G = \mathbb Z_3 \times \mathbb Z_6$, $H = \langle (1,2)\rangle$ and let $K = \langle (1,3)\rangle$. List all cosets of $H$ and $K$.

Can somebody please explain me how to do this problem. First of all I'm having trouble how to obtain the set of $H$ and $K$.
Definition: Let H be a subgroup of the group $G$, and let a element of $G$. The set 
$aH$ = {x element of G | x = ah for some h element of H}
is called the left coset of $H$ in $G$ determined by $a$. Similarly, the right coset of $H$ in $G$ determined by $a$ is the set 
$Ha$ = {x element of G |x = ha for some h element of H}.
The number of left cosets of $H$ in $G$ is called the index of $H$ in $G$, and is denoted by $[G : H]$
I need some explanation on how to do this kind of problems I really want to understand. Any help will be appreciated.

Let's do this for $H$, and leave $K$ as a similar exercise :


*

*$H = \langle (1,2)\rangle = \{(1,2), (2,4), (0,0)\}$ since
$$
(3,6) \equiv (0,0)
$$
You need to check that there are no other elements. Hence, $|H| = 3$ and $[G:H] = 18/3 = 6$. Thus, there are 6 cosets you need to find.

*The cosets are all of the form $(i,j) + H$, where $(i,j) \in G$.
$$
(0,0) + H, (1,0) + H, \ldots, (2,4)+H, (2,5) + H
$$
There are 18 such elements; but there are overlaps :
$$
(i,j) + H = (k,l) + H \Leftrightarrow (i-k,j-l) \in H
$$
$$
\Leftrightarrow (i-k,j-l) \in \{(1,2), (2,4), (0,0)\}
$$

*Now we list the cosets :
$$
(0,0) + H = (1,2) + H = (2,4) + H
$$
$$
(1,0) + H = (2,2) + H = (3,4) + H
$$
$$
(2,0) + H = (0,2) + H = (1,4) + H
$$
$$
(0,1) + H = (1,3) + H = (2,5) + H
$$
$$
(1,1) + H = (2,3) + H = (3,5) + H
$$
$$
(2,1) + H = (0,3) + H = (1,5) + H
$$
These are all the cosets of $H$. Can you do the same thing with $K$?
