G-set without fixed point question This is a qual exam question: Suppose $G$ is a group of order $pq$ with $p < q$ both prime. Prove that if $m ≥ q(p-1)$, then there exists a $G$-set $A$ with $m$ elements
and without fixed points.
Is the fixed point in the problem defined as "$x$ in $A$ such that $gx = x$ for all $g$ in $G$"? Any hint to answer this question? Thanks!
 A: You have stated the definition of fixed point correctly. Another way of thinking about it is as an orbit whose size is one. With the orbit-stabilizer theorem, we know that any orbit must have size a divisor of $pq$, so the sizes will be elements of $\{p,q,pq\}$ if there are no fixed points. A $G$-set is a disjoint union of its orbits, so the size of the whole set is the sum of the sizes of the orbits involved, and you can use this fact to turn the problem into one of elementary number theory a bit like the so-called Frobenius Problem with the case $d=2$. The Frobenius Problem actually makes a somewhat stronger arithmetic statement than is being made here.
For your purposes, you can simply prove a part of Frobenius, which states that every integer greater than or equal to $(p-1)(q-1)$ can be written as $ap+bq$ for some $a,b\ge0$. (You don't need the part where this is the tightest possible bound, and this bound is still noticeably better than the given one of $(p-1)q$.) To do this, consider Bézout's identity for relatively prime numbers [if two integers $p$ and $q$ are relatively prime then there exist integers $a$ and $b$ for which $ap+bq=1$] and combine with the fact that $(-q)p+(p)q=(q)p+(-p)q=0$ in order to manage the size of either $a$ or $b$ in this identity, and then employ the identity for an induction proof.
