Third Sylow Theorem

Introduction

The Third Sylow Theorem provides detailed information about the number of Sylow $p$ -subgroups in a finite group. It gives precise constraints on this number, making it a powerful tool for group classification.

Statement

Third Sylow Theorem (Number): Let $n_{p}$ be the number of Sylow $p$ -subgroups of $G$ . Then:

$n_{p}$ divides $m$ (the part of the group order not divisible by $p$ )
$n_{p} \equiv 1 (\mod p)$
$n_{p} = [G : N_{G} (P)]$ , where $P$ is any Sylow $p$ -subgroup and $N_{G} (P)$ is its normalizer in $G$

Let $G$ be a finite group of order $| G | = p^{n} m$ , where $p$ is a prime and $p$ does not divide $m$ .

Proof Sketch

The proof uses group actions:

Define the action: Let $G$ act on the set of all its Sylow $p$ -subgroups by conjugation.
Transitivity: By the Second Sylow Theorem, this action is transitive, so there is only one orbit.
Count the orbit: The size of this orbit is $n_{p}$ .
Modular constraint: Restrict the action to a single Sylow $p$ -subgroup $P$ and count fixed points to get $n_{p} \equiv 1 (\mod p)$ .

Key Steps

The action of $G$ on the set of Sylow $p$ -subgroups is transitive
The size of the orbit is $n_{p} = [G : N_{G} (P)]$
When $P$ acts on the set of Sylow $p$ -subgroups, it fixes itself and possibly others
The number of fixed points is congruent to $n_{p}$ modulo $p$

Examples

Example 1: Groups of Order 12

Let $G$ be a group of order 12. We have $12 = 2^{2} \cdot 3$ .

Sylow 2-subgroups: $n_{2}$ must divide 3 and $n_{2} \equiv 1 (\mod 2)$ . So $n_{2} = 1$ or $n_{2} = 3$ .

Sylow 3-subgroups: $n_{3}$ must divide 4 and $n_{3} \equiv 1 (\mod 3)$ . So $n_{3} = 1$ or $n_{3} = 4$ .

Example 2: Groups of Order 30

Let $G$ be a group of order 30. We have $30 = 2 \cdot 3 \cdot 5$ .

Sylow 2-subgroups: $n_{2}$ must divide 15 and $n_{2} \equiv 1 (\mod 2)$ . So $n_{2} = 1, 3, 5, 15$ .

Sylow 3-subgroups: $n_{3}$ must divide 10 and $n_{3} \equiv 1 (\mod 3)$ . So $n_{3} = 1$ or $n_{3} = 10$ .

Sylow 5-subgroups: $n_{5}$ must divide 6 and $n_{5} \equiv 1 (\mod 5)$ . So $n_{5} = 1$ or $n_{5} = 6$ .

Example 3: Groups of Order 15

A group of order 15 must have $n_{3} = 1$ and $n_{5} = 1$ . The unique Sylow 3-subgroup and Sylow 5-subgroup are both normal, and it can be shown that the group must be the cyclic group $Z_{15}$ .

Applications

Application 1: Group Classification

The Third Sylow Theorem is crucial for classifying groups of a given order by constraining the number of Sylow subgroups.

Application 2: Simplicity Tests

If $n_{p} = 1$ for some prime $p$ , the unique Sylow $p$ -subgroup is normal, so the group is not simple.

Application 3: Structure Analysis

The constraints on $n_{p}$ help determine the possible structure of finite groups.

Consequences

Divisibility Constraints

The theorem provides strong constraints on the number of Sylow subgroups, which are essential for group classification.

Normalizer Information

The relationship $n_{p} = [G : N_{G} (P)]$ connects the number of Sylow subgroups to the structure of normalizers.