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.

I think you are right. The set $X$ is the set of all two element subsets of $G$. For the action of $G$ on $X$, first note that $G$ has a regular action on itself. Use this action to define an action on $X$. More precisely, if $g_i \mapsto h_i$ and $g_j \mapsto h_j$ under the regular action of $G$ on itself $(i \neq j)$, then the action of $G$ on $X$ is defined as $\{g_i, g_j\} \mapsto \{h_i, h_j\}.$
