Showing that $|X| \equiv |\text{Fix}(G)|\pmod{p}$.

Question

$\newcommand{\fix}{\text{Fix}}$ $\newcommand{\stab}{\text{stab}}$ $\newcommand{\orb}{\text{Orb}}$ Definition: Let $G$ act on a set $X$. We define $\fix(G) =\{x \in X : g \ast x = x, \forall g \in G\}$; i.e all elements of $X$ fixed by every $g \in G$.

Proposition: Let $G$ be a $p$-group acting on a finite set $X$. Then $|\fix(G)| \equiv |X|\pmod{p}$.

Attempt: To start, we know that $X$ is partitioned into it's orbits and since $G$ is finite we can use orbit-stabilizer theorem to assert that $$ \frac{|G|}{|\stab(x)|} = |\orb(x)|, \; (\forall x \in X). $$ Because $\stab(x)$ is a subgroup of $G$ and $G$ is finite; Lagrange's theorem implies that $|\stab(x)|$ divides the order of $G$, but since $G$ is a $p$-group this means that $|\stab(x)|$ is a power of $p$, and further that for each $x \in X$, $|\orb(x)| = p^\alpha$ for some $\alpha \geq 0$. Next let us consider what it means to be an element of the fixed set of $G$. By definition, $x \in \fix(G)$ means that $g \ast x = x$ for all $g \in G$; which with some thought, means that $\orb(x) = \{x\}$, and moreover $|\orb(x)| = 1$. In light of this, we may write $$ \fix(G) = \{x \in X : |\orb(x)| = 1\}, $$ and so $|\fix(G)|$ is just the number of elements of $X$ with just themselves in their orbit. Again as a corollary of Orbit-Stabilizer and that $X$ is a finite set, we know $$ |X| = \sum_{x \in X} |\orb(x)|. $$ But this gives more, reducing this sum modulo $p$ will annihilate any term such that $|\orb(x)| = p^\alpha$ where $\alpha>0$, and leave only terms whose orbits have $|\orb(x)| = 1$ (there are no other possible remainders because the orbits each have order a power of $p$ by the Orbit-Stabilizer theorem). But there are exactly $|\fix(G)|$ of those terms and hence $|X| \equiv |\fix(G)|\pmod{p}$.

Is this argument alright? Particularly is there a need for a more rigorous way to articulate the last paragraph? It felt slightly hand-wavy to me. Any recommendations? Thanks in advance.

Answer 1

To write the end "better" or in another way at least, I guess you can introduce, say, $r_i$ the number of orbits of cardinal $p^i$, and note that from a previous observation: $r_0=|\operatorname{Fix}(G)|$. Then, by what you said earlier, those are the only possible cardinals for our orbits, and thus, if we note $|G| =: p^\beta$: $$|X| = \sum_{i = 0}^{\beta} r_i p^i \equiv r_0 \mod p$$ Hence the desired result: $$|X| \equiv |\operatorname{Fix}(G)| \mod p$$

Answer 2

Your proof seems correct to me. Shortly, the orbit equation reads: $$|X|=\left|\operatorname{Fix}(G)\right|+\sum_i[G:\operatorname{Stab}(x_i)]$$ where the $x_i$'s are representatives of the non-sigleton orbits. Since $[G:\operatorname{Stab}(x_i)]$ is a proper power of $p$ for every $i$, then $|X|\equiv\left|\operatorname{Fix}(G)\right|\pmod p$.