Number of nth roots is a multiple of n

This article states a result of the form that one natural number divides another.
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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 $n$.

Related facts

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. 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).