# Difference between revisions of "Exactly n elements of order dividing n in a finite solvable group implies the elements form a subgroup"

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## 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

The table below lists key facts used directly and explicitly in the proof. Fact numbers as used in the table may be referenced in the proof. This table need not list facts used indirectly, i.e., facts that are used to prove these facts, and it need not list facts used implicitly through assumptions embedded in the choice of terminology and language.
Fact no. Statement Steps in the proof where it is used Qualitative description of how it is used What does it rely on? Difficulty level Other applications
1 Minimal normal implies elementary abelian in finite solvable Setting up the elementary abelian normal subgroup $K$ of $G$, happens before we split into cases. The proof is inductive, and the goal is to induct from $G/K$ to $G$. Basic/intermediate group theory click here
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. Step (4) of the $p$ divides $n$, Step (1) of the other case We use it to compute the order of the quotient group $G/K$, from which we ultimately induct. The idea is to try to show that the inductive hypothesis conditions apply to $G/K$. Basic group theory 2 click here
3 Number of nth roots is a multiple of n (when $n$ divides the group order) Step (6) of $p|n$ case, Step (2) of the other case Applied to the quotient group to show that the number of roots of a certain kind of $G/K$ is at least a certain amount, which is then played off against it being at most a certain amount. Basic/intermediate group theory click here
4 Solvability is quotient-closed Step (13) of $p|n$ case, Step (6) of other case Show conditions prevail to apply inductive hypothesis to quotient group $G/K$ Basic group theory click here
5 Solvability is subgroup-closed Step (8) of $p$ does not divide $n$ case Show solvability of a subgroup of $G$ Basic group theory click here
6 Hall subgroups exist in finite solvable Step (8) of $p$ does not divide $n$ case Find a subgroup of order $n$ in a subgroup where the $p'$-part is $n$ Advanced click here
7 Order of element divides order of group Step (9) of $p$ does not divide $n$ case After finding subgroup of order $n$, show it is precisely the subgroup we want Basic group theory click here

## 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 $n$

No. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $n$ divides the order of $G/K$ Fact (2) $n$ divides the order of $G$, $K$ has order $p^i$, $p$ does not divide $n$ By Fact (2), the order of $G/K$ is $|G|/|K|$ which is $N/p^i$. We're given that $n$ divides $N$, so since $n$ is relatively prime to $p$, we must also have $n$ divides $N/p^i$.
2 The number of elements $z \in G/K$ such that $z^n$ is the identity is $rn$ for some natural number $r$ Fact (3) Step (1) Previous step+Fact direct
3 For every coset $z = xK$ of $K$ in $G$ such that $z^n$ is the identity in $G/K$, there exists an element $y \in xK$ such that $y^n$ is the identity in $G$ $p$ does not divide $n$ [SHOW MORE]
4 There are at least $rn$ solutions to $x^n$ being the identity in $G$ Steps (2), (3) Step-combination direct
5 $r = 1$, and there are exactly $n$ solutions to $z^n$ being the identity in $G/K$ There are exactly $n$ elements of order dividing $n$ Steps (2), (4) [SHOW MORE]
6 The set of elements in $G/K$ of order dividing $n$ forms a subgroup of order $n$ in $G/K$ Fact (4) $G$ is solvable, induction hypothesis Step (5) [SHOW MORE]
7 The inverse image in $G$ of the subgroup of $G/K$ obtained in Step (6) is a subgroup of $G$, say $H$. It has order $p^in$. Fact (2) $K$ has order $p^i$. [SHOW MORE]
8 $H$ has a $p$-complement, i.e., a Hall subgroup of order $n$. Call this $p$-complement $L$. Facts (5), (6) $p$ does not divide $n$, $G$ is solvable Step (7) [SHOW MORE]
9 $L = S$, i.e., $L$ is precisely the set of elements of order dividing $n$. Since it's a subgroup, this completes the proof. Fact (7) There are exactly $n$ elements of order dividing $n$. <toggledisplay>Since $L$ has order dividing $n$, all elements in there also have order dividing $n$ by Fact (7). So $L \subseteq S$. By size considerations, we have $L = S$.

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