I am trying to solve the following qual study problem:
Let $p$ and $q$ be distinct primes, and let $G$ be any group of order $pq$. Show that for every integer $n \geq pq$, there exists a set $\Omega$ of cardinality $n$ and a group action of $G$ on $\Omega$ without fixed points (meaning, for each $\alpha \in \Omega$, there exists $g \in G$ such that $g\alpha \neq \alpha$.
I was able to conclude that for such an action, all the orbits must have size $p$, $q$, or $pq$, and for any $\Omega$ of size $n \geq pq$, it is possible to partition $\Omega$ into disjoint sets of size either $p$ or $q$. But is it enough or even possible to define an action by simply specifying each orbit to be a part in this partition? I'm skeptical, but this answer seems to imply that it's sufficient.
If not, is there a better way to approach this problem?
Given a partition $\Omega = \coprod_{i=1}^n \Omega_i$ into pairwise disjoint subsets $\Omega_i \subseteq \Omega$ of size $|\Omega_i| \in \{p,q\}$, it sufficies to construct an action of $G$ on each $\Omega_i$ without fixed points. By combining these actions, we then get an action of $G$ on the whole of $\Omega$ without fixed points.
Instead of using the $\Omega_i$, it sufficies to show that there exists actions of $G$ on some sets of cardinalty $p$ and $q$ without fixed points.
It follows from $|G| = pq$ that $G$ contains a subgroup $H \subseteq G$ of order $|H| = p$ (one can use Sylow’s theorems or Cauchy’s theorem to see this). Then the action of $G$ on the set of left cosets $G/H$ given by $$ g \cdot g'H = (gg')H $$ is transitive, and because $|G/H| = |G|/|H| = q \neq 1$ it follows that this action has no fixed points.
By swapping the roles of $p$ and $q$ we can also construct an action of $G$ on a set with $p$ elements without fixed points.