[Math] Let $p$ be prime number. In any finite group, the number of elements of order $p$ is multiple of $p-1$.

Question

This is the problem in the book that I want to prove, but it doesn't seem right.

For example let's say I have a group of $9$ elements. If this group is non-cyclic then every element (except identity) has an order of $3$ (prime) (because of Lagrange's theorem).
So there are actually $8$ elements of order $3$.

Theorem seems to work for cyclic groups.

Is there is something I am missing here or this only holds for cyclic groups?

Answer 1

Let $X$ be the set of elements of order $p$. The relation $$x \sim y :\Longleftrightarrow \langle x \rangle = \langle y \rangle$$ is obviously an equivalence relation and the equivalence class of $x \in X$ is $\langle x \rangle \setminus \{e\}$. Thus each equivalence class has size $p-1$, i.e. $|X|$ is a multiple of $p-1$.