Orbits in G = $Z_6$ by listing 2 element subsets in G. 1) Let $G = \mathbb{Z}_6$. List all 2-element subsets of $G$, and show that under the regular action of G (by left addition) there are 3 orbits, 2 of length 6, one of length 3. Deduce that the stabilizers of 2-element subsets of $G$ have order 1 or 2.
2) Carry out the same calculations as in the previous question for $S_3$,
showing that there are four orbits, of lengths 3, 3, 3 and 6.
These are the questions - I have the answers but what I am really struggling with is getting to grips with what it all means. 
Firstly I know the definition of orbit to be the equivalence class for $x\sim y$ if $\theta(g)(x)=y$ for g$\in$G and $x,y\in X$ with $G$ being the Group and $X$ the set on which $G$ acts. 
But what I am not seeing is how I can apply this to the questions. Is my "Set" the set of all two element subsets? And if so what is my group action $\theta$ and how can I apply it? 
Thanks in advance.

The two element subsets: 
$$\{0,1\},\{0,2\},\{0,3\},\{0,4\},\{0,5\},\{1,2\},\{1,3\},\{1,4\},\{1,5\},\{2,3\},\{2,4\},\{2,5\},\{3,4\},\{3,5\},\{4,5\}.$$
Left addition: Take $\left[a\right]_6 \in \mathbb{Z}_6$ and do $a+H$ (where $H$ is a two element subset of $G$).
For example: 
$$1+\{0,1\} =\{1,2\}, 1+\{1,2\} = \{2,3\}, 1+\{2,3\} = \{3,4\}, 1+\{3,4\} = \{4,5\}, 1+\{4,5\} = \{5,0\}, 1+\{5,0\} = \{0,1\} $$
Here is an example of an orbit of length $6$. 
$\textbf{Lastly}$: To do a problem like this with maximum efficiency think of these sets as  ordered pairs $(x,y)$. If you want to show that there are three orbits, think of all the ordered pairs $(x,y)$ in which you can add $1,2,3,4,5$ to and get back to the original ordered pair. 
