I'm trying to prove the following proposition from textbook Groups, Matrices, and Vector Spaces - A Group Theoretic Approach to Linear Algebra by James B. Carrell.
Let $G$ be a finite group. Then the number of elements of prime order $p$ is divisible by $p − 1$.
Could you please verify if my attempt is fine or contains errors? Thank you so much for your help!
My attempt:
Let $H$ be the set of all elements of prime order $p$.
For $x \in H$, $\langle x \rangle := \{x, x^2, \ldots, x^{p-1}, 1\}$ is a sub-group. Together with $p$ is prime, we get $\langle x \rangle$ is the only sub-group of itself. Consequently, either $\langle x \rangle = \langle y \rangle$ or $\langle x \rangle \cap \langle y \rangle = \emptyset$ for all $x,y \in H$.
It follows from $p$ is prime that if $1 \neq y \in \langle x \rangle$ then $y \in H$. Let $k$ be the cardinality of $\{\langle x \rangle \mid x \in H\}$. Then $|H|= |\cup_{x \in H} \left (\langle x \rangle \setminus \{1\} \right )| = k(p-1)$.
This is essentially correct, except that $\langle x\rangle$ has two subgroups, including the trivial subgroup, and $\langle x\rangle \cap \langle y\rangle$ is never empty, but rather equal to the trivial subgroup if $y\notin\langle x\rangle$.
You could rephrase it as saying that a subgroup of order $p$ has $p-1$ elements of order $p$, and two subgroups of order $p$ that are different intersect trivially, thus if $k$ is the number of subgroups of order $p$ we obtain $k(p-1)$ nonidentity elements, hence this is the number of elements of order $p$. I tend to favor less notation though, and your way is also fine.