# Number of fixed points of a $p$-group action

If a $p$-group $G$ acts on a finite set $S$, show that the number of fixed points of the action is congruent mod $p$ to the cardinality $|S|$ of the set $S$.

Let $F$ be the collection of fixed points. Since I know that the cardinality of the orbit $O$ for any given $s \in S$ equals $|G: Stab_{G}(s)|$, I can say all the fixed points, by definition, have orbits of cardinality 1. Because of the order of $G$, each non-trivial index must be a power of $p$. So, summing up cardinalities of the non-trivial orbits will give me some multiple of $p$. Is this correct or only partway there?

• Yep, that's the idea.
– anon
Aug 20 '17 at 23:16
• Thanks, for some reason I wasn't sure what I was getting when I added all those orbits up, but I should be getting precisely the non-fixed elements by the very definition of an orbit.
– BMac
Aug 20 '17 at 23:19
• – Jess
Dec 17 '19 at 1:13
• math.stackexchange.com/questions/201297/…
– Jess
Dec 17 '19 at 1:13
• math.stackexchange.com/questions/506505/…
– Jess
Dec 17 '19 at 1:13