For every non-identity g∈G, there is exactly one element of S fixed by g. Prove that |S|mod k = 1.

Question

Suppose that a group G of order k acts on a finite set S. For every non-identity g∈G, there is exactly one element of S fixed by g. Prove that |S|mod k = 1.

I'm a little stuck here. I know that |S| is equal to the sum of order of all orbits and that the order of a orbit divides |G|. But I don't see how the condition of exactly one element of S fixed by g in G works here and don't know if this gives me the order of Stab(x) or Fix(g). Can someone help me with this proof?

Answer 1

The magic words are Burnside's Lemma, which says that the number orbits equals $$\frac{1}{k}\sum_{g\in G} Fix(g).$$ This means that the sum is divisible by $k,$ but the sum equals $|S| + (k-1).$