I'm following along the proof of Cauchy's Theorem using group actions, from https://en.wikipedia.org/wiki/Cauchy%27s_theorem_(group_theory) (proof 2).
The theorem states:
If $G$ is a finite group and $p$ is a prime number dividing the order of $G$ (the number of elements in G), then $G$ contains an element of order $p$.
The proof mentions:
[…] As remarked, orbits in $X$ under this action either have size 1 or size $p$. The former happens precisely for those tuples $(x,x,\ldots,x)$ for which $x^p=e$. Counting the elements of $X$ by orbits, and reducing modulo $p$, one sees that the number of elements satisfying $x^p=e$ is divisible by $p$. […]
(1) I'm not sure what "Counting the elements of $X$ by orbits" refers to. From what I think it means:
- For each element $x$ of $X$, I find the orbit of $x$
- Then I count the number of elements that have the same orbit?
(2) Then, how can I get from here to the conclusion that the number of elements satisfying $x^p=e$ is divisible by $p$?
Best Answer
Lemma: Let $G$ be a $p$-group acting on a finite set $X$, then $|X^G|\equiv |X|\pmod p$, where $X^G$ is the set of fixed points of this group action, i.e., set of $x\in X$ such that $gx = x$ for all $g\in G$.
Proof: Since $X$ is finite, then say $X^G = \{x_1,\ldots, x_n\}$ has $n$ elements (notice that each element has orbit size $1$), and pick representatives $x_{n+1},\ldots, x_{n+m}$ for orbits of size more than $1$. Since $X$ can be partitioned into disjoint union of orbits, we know $X = \coprod\limits_{i=1}^{n+m}\operatorname{orb}(x_i)$. For $i > n$, we have $|\operatorname{orb}(x_i)| = \frac{|G|}{|\operatorname{stab}(x_i)|}$ by the orbit-stabilizer theorem, but since the orbit is greater than $1$ and $G$ is a $p$-group, then $p$ must divide $\operatorname{orb}(x_i)$ for $i > n$. By the partition, we have $|X| = \sum\limits_{i=1}^{n+m}|\operatorname{orb}(x_i)|\equiv \sum\limits_{i=1}^n |\operatorname{orb}(x_i)| = |X^G|\pmod p. $