Number of nth roots is a multiple of n

Statement
Suppose $$G$$ is a finite group and $$n$$ is any natural number dividing the order of $$G$$. Then, the size of the set:

$$\{g \in G \mid g^n = e \}$$

is a multiple of $$n$$.

Note that since the identity element is itself in this set, the size of the set is at least $$n$$.

Stronger facts

 * Number of nth roots of any conjugacy class is a multiple of n

Other related facts

 * Exactly n elements of order dividing n in a finite solvable group implies the elements form a subgroup
 * At most n elements of order dividing n implies every finite subgroup is cyclic
 * Number of nth roots of a subgroup is divisible by order of subgroup

Conjectures

 * Frobenius conjecture on nth roots

Facts used

 * 1) uses::Number of nth roots of any conjugacy class is a multiple of n

Proof
The proof follows from fact (1); in fact, the given statement is a special case of fact (1).