Exactly n elements of order dividing n in a finite solvable group implies the elements form a subgroup

Statement
Suppose $$G$$ is a fact about::finite solvable group, and $$n$$ is a natural number dividing the order of $$G$$. If there are exactly $$n$$ elements of $$G$$ whose $$n^{th}$$ power is the identity element, then these elements form a subgroup. This subgroup is clearly a normal subgroup. In fact, it is a characteristic subgroup, and even better, is a fully characteristic subgroup, a homomorph-containing subgroup and a variety-containing subgroup.

Related facts

 * Number of nth roots is a multiple of n
 * Number of nth roots of any conjugacy class is a multiple of n
 * At most n elements of order dividing n implies every finite subgroup is cyclic

Related conjectures

 * Frobenius conjecture on nth roots: The conjecture states that the assumption of solvability can be dropped.

Proof
We do the proof by induction. Specifically, we assume that the statement holds true for all finite solvable groups of strictly smaller orders than $$N$$.

PROOF OF INDUCTIVE STEP:

Given: A finite solvable group $$G$$ of order $$N$$. A natural number $$n$$. $$S$$ is the set of $$n^{th}$$ roots of unity in $$G$$ (i.e., the set of elements of order dividing $$n$$, and $$S$$ has $$n$$ elements.

To prove: $$S$$ is a subgroup.

Proof: By fact (1), $$G$$ has a nontrivial elementary abelian normal subgroup $$K$$ of order $$p^i$$ for some prime power $$p^i$$ dividing the order of $$G$$. We split the proof into two cases, based on whether $$p$$ divides $$n$$ or not.

Proof that the subgroup is normal, characteristic and fully invariant
This follows from the fact that if $$\alpha$$ is a homomorphism from $$G$$ to any subgroup of $$G$$, then $$(\alpha(g))^n = \alpha(g^n)$$. Hence, if $$g^n = 1$$, so is $$(\alpha(g))^n$$.