# 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.

## Facts used

1. Minimal normal implies elementary abelian in finite solvable
2. Lagrange's theorem: We in particular are interested in the version which states that if $A$ is a subgroup of $B$, $|B/A| = |B|/|A|$, where $B/A$ is the quotient set. When $A$ is a normal subgroup, $B/A$ is the quotient group.
3. Number of nth roots is a multiple of n
4. Solvability is quotient-closed
5. Hall subgroups exist in finite solvable

## Proof

This proof uses the principle of mathematical induction in a nontrivial way (i.e., it would be hard to write the proof clearly without explicitly using induction).

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.

### Case that $p$ divides $n$

No. Assertion/construction Facts used Given data/assumptions used Previous steps used Explanation
1 The order of every element in $K$ is either $1$ or $p$ $K$ is elementary abelian of order $p^i$
2 $K \subseteq S$ $p$ divides $n$, $S$ comprises all elements whose $n^{th}$ power is the identity Step (1) [SHOW MORE]
3 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]
4 $G/K$ has order $p^{k-i}l$ Fact (2) $K$ has order $p^i$ Step (3) [SHOW MORE]
5 Let $u = p^{\max(j-i,0)}m$. Then, $u$ divides the order of $G/K$. Steps (3), (4) [SHOW MORE]
6 There exists some natural number $r$ such that $G/K$ has $ru$ elements of order dividing $u$ Fact (3) Step (5) [SHOW MORE]
7 Let $\varphi:G \to G/K$ be the quotient map. The set of elements $x \in G$ such that $\varphi(x)^u$ is the identity element of $G/K$ has size $rup^i$ $K$ has order $p^i$ Step (6) [SHOW MORE]
8 If $\varphi(x)^u$ is the identity element in $G/K$, then $x^{up}$ is the identity element in $G/K$. Step (1) [SHOW MORE]
9 $up$ divides $n$. $p$ divides $n$ Steps (3), (5) [SHOW MORE]
12 $r = 1, j \ge i$ and $x^n$ is the identity element iff $x^u \in K$. In other words, $S = \{ x \mid x^u \in K \}$. Steps (5), (11) [SHOW MORE]
13 The set of elements $y \in G/K$ such that $y^u$ is the identity element forms a subgroup. In particular, $S/K$ is a subgroup Fact (4) inductive hypothesis, $G$ is solvable Step (12) [SHOW MORE]
14 $S$ is a subgroup Step (13) [SHOW MORE]

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