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

Statement

Suppose $G$ is a 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.

Proof

Given: A group $G$. 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.

Case that $p$ divides $n$

No. Assertion/construction Facts used Given data/assumptions used Previous steps used Explanation
1 $K \subseteq S$ $p$ divides $n$, $K$ is elementary abelian of order $p^i$, $S$ comprises all elements whose $n^{th}$ power is the identity [SHOW MORE]
2 Let $n = p^jm$ and $|G| = N = p^kl$ where $m,l$ are relatively prime to $p$. Then, $j \le k$ and $m|l$. $n$ divides the order of $G$ [SHOW MORE]
3 Let $u = p^{\max(j-i,0)}m$. Then, the order of $G/K$ is divisible by $u$. <toggledisplay>This is clear from the fact that the order of $G/K$ equals $p^{k-i}l$, and since $m|l$ and $j \le k$, this follows.

(remaining stuff needs to be cleaned up. Numbers may be off).

1. $up$ divides $n$: This follows essentially from the fact that exponent of $p$ in $u$ is strictly less than that in $n$.
2. There exists some natural number $r$ such that $G/K$ has $ru$ elements of order dividing $u$: This follows from fact (2).
3. Let $\varphi:G \to G/K$ be the quotient map. The set of elements $x \in G$ such that $\varphi(x)^u = 1$ has size $rup^i$: This follows from step (3).
4. If $\varphi(x)^u = 1$, we have $x^{up} = 1$: This follows from the fact that the order of any element in $K$ is either $1$ or $p$.
5. If $\varphi(x)^u = 1$, we have $x^n = 1$: This follows from the previous step and step (2), which says that $up|n$.
6. $r = 1, j \ge i$ and $x^n = 1 \iff x^u \in K$: By steps (3) and (5), there are at least $rup^i$ elements $x$ for which $x^n = 1$. Thus, $rup^i \le n$. Simplifying, we obtain that $rp^{\max(i,j)} \le p^j$. This forces $r = 1$ and $j \ge i$. Further, since equality holds even under these assumptions, we conclude that the $rup^i$ solutions to $x^u \in K$ exhaust all solutions to $x^n = 1$.
7. The set of elements $y \in G/K$ such that $y^u = 1$ forms a subgroup. In particular, $S/K$ is a subgroup: By the previous step, $r = 1$, so there are exactly $u$ solutions in $G/K$ to $y^u = 1$. By fact (3), $G/K$ is solvable, so the induction hypothesis applies, and the set of solutions forms a subgroup.
8. $S$ is a subgroup: It is the inverse image in $G$ of a subgroup of $G/K$.

Case that $p$ does not divide $n$

1. $n$ divides the order of $G/K$: This follows because $n$ divides the order of $G$ and $n$ is relatively prime to $p$.
2. The number of elements $z \in G/K$ such that $z^n = 1$ is $rn$ for some natural number $r$: This follows from fact (2).
3. For every coset $z = xK$ of $K$ in $G$ such that $z^n = 1$, there exists an element $y \in xK$ such that $y^n = 1$: Start with some element $x$ in the coset. Let $a = x^n$. By definition, $a \in K$. Now, find an integer $m$ such that $p|mn - 1$ (this can be done since $p$ and $n$ are relatively prime). Let $b = a^m$ and let $y = xb^{-1}$. Note that since $b$ is a power of $a$, which in turn is a power of $x$, $x$ commutes with $b$. Thus, if $y = xb^{-1}$, then $y^n = (xb^{-1})^n = x^nb^{-n} = ab^{-n} = a^{1-mn} = 1$ since $1 - mn$ is a multiple of $p$, and $a^p = 1$. Further, $b^{-1} \in K$, so $y \in xK$. Thus, we have found an element of $xK$ whose order divides $n$.
4. There are at least $rn$ solutions to $x^n = 1$ in $G$: This follows from steps (2) and (3).
5. $r = 1$, and there are exactly $n$ solutions to $z^n = 1$ in $G/K$: This follows from the previous step and the assumption that there are exactly $n$ solutions.
6. The set of solutions to $z^n = 1$ in $G/K$ forms a subgroup, say $H/K$: By fact (3), $G/K$ is solvable, so we can apply the induction hypothesis to the conclusion of the previous step.
7. The inverse image, say $H$ of this subgroup $H/K$ of $G/K$ in $G$ is a group of order $p^in$, where $n$ is relatively prime to $p$.
8. $H$ has a $p$-complement, i.e., a Hall subgroup of order $n$: This follows from fact (4).
9. This Hall subgroup of order $n$ is precisely equal to the set of solutions to $x^n = 1$: All elements in the Hall subgroup of order $n$ are solutions to $x^n = 1$, and we know that there are exactly $n$ solutions. Hence, these must be the only solutions.

Proof that the subgroup is normal, characteristic and fully characteristic

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 $(\alpha(g))^n$.