About groups act faithfully on a set Suppose group $G$ act faithfully on a set $X$ of $5$ elements, and there are $2$ orbits, of order $2$ and $3$ respectively. Then what should the group $G$ be like?
Note: A group $G$ acts faithfully on a set $\Leftrightarrow$ $gx=x$ for all $x\in X$ iff $g=e$.
My attempt:
Suppose $X=\{a_1,a_2,a_3,a_4,a_5\}$. Since there is an orbit of order $2$ (suppose it is $G\cdot a_1$), and the group must have an unit element $e$, then there must be another element $g$ in $G$ to make sure the order of $G\cdot a_1$ is $2$. But how to reduce the number of orbits (like making $G\cdot a_2$ same to $G\cdot a_1$)? I meet confusion here.
How will the condition "act faithfully" affect this problem?
Or whether I make some mistakes on understanding or thinking?
 A: Too long for a comment, but more of a hint and head-start:
Some things to consider:
If $G\cdot a_1$ is an orbit of order $2$, it includes $e\cdot a_1$, so that orbit is $\{a_1, a_j\}$ for some $j$. This orbit must also be the same as $G\cdot a_j$, because whatever the element $g$ is that takes $a_1$ to $a_j$, $g^{-1}$ takes $a_j$ to $a_1$.
Orbits have to be disjoint (do you see why?), so there are just the two orbits $\{a_1,a_j\}$ and the rest of $X$. You can suppose without loss of generality that $j=2$ by renaming the elements of $X$.
Every element of $G$, then, consists of a permutation of $\{a_1,a_2\}$ that fixes the other elements of $X$ composed with a permutation of $\{a_3,a_4,a_5\}$ that fixes $a_1$ and $a_2$. How many such things can there be, and how can you form a group from elements like this? (Acting faithfully means that two different group elements can’t do the same thing to $X$.)
A: If $G$ acts faithfully on a set of $5$ elements, then $G$ is isomorphic to a subgroup of $S_5$ (i.e. it embeds into $S_5$). The subgroups of $S_5$ act naturally (i.e. as a group of permutations) on $\{1,2,3,4,5\}$ and give rise under this action to the orbit setup as in the OP if and only if they are of the form $\langle\sigma\rangle$, where $\sigma\in S_5$ has cycle structure $(2,3)$. Therefore, $G\cong \langle\sigma\rangle$, where $\sigma\in S_5$ has cycle structure $(2,3)$.
