Suppose G is a group, and p is prime. Then the number of elements of G of order p is multiple of p-1.

Question

I need help.

Suppose $G$ is a group, and $p$ is prime. Then the number of elements of $G$ of order $p$ is multiple of $p-1$.

Give me any advise or note.

Answer 1

Hints:

1) Every element of order $p$ generates a cyclic subgroup of order $p$ isomorphic to $\mathbb Z_p$.

2) In that subgroup, every element except the identity has order $p$.

3) Any two subgroups of order $p$ intersect either at $e$, or they are identical.

4) The collection of all elements of order $p$ is a union of such subgroups of order $p$.

5) Count...

Answer 2

This can be generalized for arbitrary $n \geq 1$.

The number of elements of order $n$ is $k \cdot \varphi(n)$, where $k$ is the number of cyclic subgroups of order $n$ and $\varphi$ is the Euler totient function. This follows from the fact that a cyclic subgroup of order $n$ has exactly $\varphi(n)$ generators.