# Difference between revisions of "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.

## 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 $n$

In this case, $K \subseteq S$. Let $n = p^jm$ and $|G| = g = p^kl$ where $m,l$ are relatively prime to $p$. Clearly, $j \le k$ and $m|l$.

1. Let $u = p^{\max(j-i,0)}m$. Then, the order of $G/K$ is divisible by $u$: 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.
2. $up$ divides $n$: This follows essentially from the fact that exponent of $p$ in $u$ is strictly less than that in $n$.
3. There exists some natural number $r$ such that $G/K$ has $ru$ elements of order dividing $u$: This follows from fact (2).
4. 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).
5. 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 $p$.
6. If $\varphi(x)^u = 1$, we have $x^n = 1$: This follows from the previous step and step (2), which says that $up|n$.
7. $r = 1$, $j \ge i$ 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 $x^n = 1$.
8. 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.
9. $S$ is a subgroup: It is the inverse image in $G$ of a subgroup of $G/K$.