Degree of a permutation group What can we say about the set that a group can act on it as a permutation group? we know that this set is not unique. For example the alternative group $A_4$ acts on the sets of sizes 4 and 6.
 A: There is nothing you can say about the set as a whole, any group of any size can act on any set of any size because there is always the trivial action.
A more meaningful question would be to break a set up into orbits and ask what you can say about the orbits.  If $G$ acts on $X$ then an orbit is any set of the form
$$\mathcal O(x) = \{y \in X \ | \ gx = y \ \text{for some} \ g \in G\}$$
The orbits of a set partition it into pieces such that the $G$ action on each piece is completely independent of the other pieces.  The set as a whole does not have any size restrictions because it is possible for an orbit to have size $1$, hence putting any number of those together gives a set of any desired size.  But the orbits do have a restriction on size given by a theorem called the orbit stabilizer theorem.  This theorem says that the size of the orbit $\mathcal O(x)$ is exactly the index in $G$ of the stabilizer
$$G_x = \{g \in G \ | \ gx = x\}$$
So, in particular, the size of the orbit must divide the order of $G$.
A: This is in some sense just an elaboration on the answer by Jim:
An action of a group $G$ on a set $X$ is called $n$-transitive for some natural number $n$ is for any two $n$-tuples $(x_1,\dots,x_n)$ and $(y_1,\dots,y_n)$ of elements from $X$ such that for all $i\neq j$ $x_i\neq x_j$ and $y_i\neq y_j$ there is some $g$ in $G$ such that for all $i$ $g.x_i = y_i$ (in particular, if this holds for $n = 1$ we just call the action transitive, and the consition is that for any pair of elements $x$ and $y$ in $X$, there is an element $g\in G$ with $g.x = y$).
To unify this with Jim's answer, the action of $G$ on any of the orbits is transitive, and if the action is transitive, then there is just one orbit. So that answer shows that if the action is transitive, then the size of $X$ divides the order of $G$.
In more generality, one can show that if the action of $G$ on $X$ is transitive and $x$ is some element of $X$, then the action is $2$-transitive if and only if the action of $G_x$ (as defined in Jim's answer) is transitive on $X\setminus \{x\}$, so we see inductively that if the action is $n$-transitive and $|X| = k$ then $k(k-1)\cdots (k-(n-1))$ divides the order of $G$.
Added: Another condition we can put on the action is the following (assume the action is transitive): For each pair of elements $x$ and $y$ in $X$ there is a unique $g\in G$ such that $g.x = y$. If this is the case, we call the action faithful. Note that if the action is faithful, then for any $x\in X$, we have $G_x = \{1\}$. Let us more generally call an $n$-transitive action $n$-faithful if for each pair of $n$-tuples $(x_1,\dots,x_n)$ and $(y_1,\dots,y_n)$ of elements from $X$ such that for all $i\neq j$ $x_i\neq x_j$ and $y_i\neq y_j$ there is a unique $g\in G$ with $g.x_i = y_i$ for all $i$ (since we assume the action to be $n$-transitive, such a $g$ exists, and we now require it to be unique).
Just as with transitivity, we can show that if the action is $n$-transitive, then it is $n$-faithful if and only if for $x\in X$ we have that the action of $G_x$ on $X\setminus \{x\}$ (which is $(n-1)$-transitive) is $(n-1)$-faithful.
If the action of $G$ on $X$ is transitive and faithful, then we can see that since the size of $X$ is the index of the stabilizer of any element of $X$, this actually means that $|X| = |G|$ and more generally (using a similar argument as with the transitivity), if $|X| = k$ and the action is $n$-transitive and $n$-faithful, then $|G| = k(k-1)\cdots (k-(n-1))$
