The Sylow Theorems

We prove Cauchy's theorem on the existence of elements of prime order, then establish the three Sylow theorems on the existence, conjugacy, and number of Sylow p-subgroups. Applications to classifying groups of small order illustrate the power of these results.

Folio Official

March 1, 2026

1 Introduction

Lagrange's theorem tells us that if $H$ is a subgroup of a finite group $G$ , then $∣ H ∣$ divides $∣ G ∣$ . But it says nothing about the converse: given a divisor $d$ of $∣ G ∣$ , must $G$ contain a subgroup of order $d$ ? In general, the answer is no — $A_{4}$ has order $12$ but no subgroup of order $6$ . The Sylow theorems provide a powerful affirmative answer in the case when $d$ is a prime power $p^{k}$ .

Definition 1 (Sylow

p

-subgroup).

Let

G

be a finite group and

p

a prime. Write

∣ G ∣ = p^{n} m

where

p ∤ m

. A subgroup of

G

of order

p^{n}

is called a Sylow $p$ -subgroup of

G

. We write

Syl_{p} (G)

for the set of all Sylow

p

-subgroups of

G

, and

n_{p} = ∣ Syl_{p} (G) ∣

for their number.

2 Cauchy's Theorem

Before turning to the Sylow theorems proper, we establish the foundational result of Cauchy, which provides a partial converse to Lagrange's theorem.

Theorem 2 (Cauchy's Theorem).

Let

G

be a finite group and

p

a prime. If

p ∣ ∣ G ∣

, then

G

contains an element of order

p

Proof.

We proceed by strong induction on

∣ G ∣

. The base case

∣ G ∣ = 1

is vacuously true.

Suppose

∣ G ∣ \geq 2

and the theorem holds for all groups of order less than

∣ G ∣

Case 1: There exists a proper subgroup

H < G

with

p ∣ ∣ H ∣

. Since

∣ H ∣ < ∣ G ∣

, the induction hypothesis gives an element of order

p

H

, which is also an element of

G

Case 2: For every proper subgroup

H < G

, we have

p ∤ ∣ H ∣

. Consider the class equation

∣ G ∣ = ∣ Z (G) ∣ + i \sum [G : C_{G} (a_{i})]

where the sum runs over representatives

a_{i}

of the non-central conjugacy classes. For each such

a_{i}

, the centralizer

C_{G} (a_{i})

is a proper subgroup, so

p ∤ ∣ C_{G} (a_{i}) ∣

by our assumption. Since

p ∣ ∣ G ∣

, we have

p ∣ [G : C_{G} (a_{i})]

for each

i

. The class equation then gives

∣ Z (G) ∣ = ∣ G ∣ - i \sum [G : C_{G} (a_{i})] \equiv 0 (mod p)

p ∣ ∣ Z (G) ∣

Since

Z (G)

is abelian and

p ∣ ∣ Z (G) ∣

, we can find an element of order

p

Z (G)

as follows. Pick any

z \in Z (G)

with

z \neq = e

, and set

m = ord (z)

. If

p ∣ m

, then

z^{m / p}

has order

p

and we are done. If

p ∤ m

, consider the quotient

Z (G) / ⟨ z ⟩

. This is an abelian group of order

∣ Z (G) ∣/ m

, and

p ∣ ∣ Z (G) ∣/ m

since

p ∣ ∣ Z (G) ∣

and

p ∤ m

. Moreover

∣ Z (G) / ⟨ z ⟩ ∣ < ∣ Z (G) ∣ \leq ∣ G ∣

, so by the induction hypothesis there exists an element

\overset{w}{ˉ} \in Z (G) / ⟨ z ⟩

of order

p

. Lifting to

w \in Z (G)

, we have

w^{p} \in ⟨ z ⟩

, so

ord (w)

is a multiple of

p

, and

w^{ord (w) / p}

is an element of order

p

G

. □

Remark 3.

Cauchy's theorem is the special case of the first Sylow theorem for

p^{1} ∣ ∣ G ∣

. Besides the inductive proof above, there is an elegant combinatorial proof due to McKay: one considers the set

{(a_{1}, \dots, a_{p}) \in G^{p} ∣ a_{1} a_{2} \dots a_{p} = e}

and lets

Z / p Z

act on it by cyclic permutation.

3 The First Sylow Theorem (Existence)

Theorem 4 (First Sylow Theorem).

Let

G

be a finite group and

p

a prime. Write

∣ G ∣ = p^{n} m

with

p ∤ m

and

n \geq 1

. Then

G

possesses a Sylow

p

-subgroup.

Proof.

Let

G

act on the collection

X = {S \subseteq G ∣ ∣ S ∣ = p^{n}}

of all subsets of

G

of size

p^{n}

by left multiplication:

g \cdot S = g S = {g s ∣ s \in S}

We have

∣ X ∣ = (p ^{n} ∣ G ∣) = (p ^{n} p ^{n} m)

. We claim that

p ∤ (p ^{n} p ^{n} m)

. Indeed, writing

(p ^{n} p ^{n} m) = i = 0 \prod p^{n} - 1 \frac{p ^{n} m - i}{p ^{n} - i},

one checks that for

0 \leq i < p^{n}

, the

p

-adic valuation of

p^{n} m - i

equals that of

p^{n} - i

, so the product is not divisible by

p

By the orbit decomposition

X = ⨆_{j} Orb (S_{j})

, and since

∣ X ∣

is not divisible by

p

, there must exist an orbit

Orb (S_{0})

with

p ∤ ∣ Orb (S_{0}) ∣

By the orbit-stabilizer theorem,

∣ Orb (S_{0}) ∣ = [G : Stab (S_{0})]

, so

∣ Stab (S_{0}) ∣ = ∣ G ∣/∣ Orb (S_{0}) ∣

. Since

p ∤ ∣ Orb (S_{0}) ∣

and

∣ G ∣ = p^{n} m

, we get

p^{n} ∣ ∣ Stab (S_{0}) ∣

On the other hand, let

H = Stab (S_{0})

. Then

h S_{0} = S_{0}

for all

h \in H

. Fixing any

s_{0} \in S_{0}

, the map

h \mapsto h s_{0}

embeds

H

into

S_{0}

, giving

∣ H ∣ = ∣ H s_{0} ∣ \leq ∣ S_{0} ∣ = p^{n}

Combining

p^{n} ∣ ∣ H ∣

and

∣ H ∣ \leq p^{n}

, we conclude

∣ H ∣ = p^{n}

. Thus

H = Stab (S_{0})

is a Sylow

p

-subgroup of

G

. □

4 The Second Sylow Theorem (Conjugacy)

Theorem 5 (Second Sylow Theorem).

Let

G

be a finite group and

p

a prime. Any two Sylow

p

-subgroups of

G

are conjugate. That is, if

P, Q \in Syl_{p} (G)

, then there exists

g \in G

such that

Q = g P g^{- 1}

Proof.

Fix

P \in Syl_{p} (G)

and let

Q \in Syl_{p} (G)

be arbitrary. Let

Q

act on the set

G / P

of left cosets by left multiplication:

q \cdot (g P) = (q g) P

We have

∣ G / P ∣ = [G : P] = m

where

p ∤ m

. Since

Q

is a

p

-group, the fixed-point theorem for

p

-groups gives

∣ (G / P)^{Q} ∣ \equiv ∣ G / P ∣ \equiv m \neq \equiv 0 (mod p)

. In particular,

(G / P)^{Q} \neq = \emptyset

Let

g P

be a fixed point. Then

q g P = g P

for all

q \in Q

, which means

g^{- 1} q g \in P

for all

q \in Q

. Hence

g^{- 1} Q g \leq P

. Since

∣ g^{- 1} Q g ∣ = ∣ Q ∣ = p^{n} = ∣ P ∣

, we conclude

g^{- 1} Q g = P

, i.e.,

Q = g P g^{- 1}

. □

Theorem 6 (Corollary).

P

is the unique Sylow

p

-subgroup of

G

, then

P ⊴ G

Proof.

For any

g \in G

, the conjugate

g P g^{- 1}

is also a Sylow

p

-subgroup (it has the same order

p^{n}

). By uniqueness,

g P g^{- 1} = P

. □

5 The Third Sylow Theorem (Counting)

Theorem 7 (Third Sylow Theorem).

Let

G

be a finite group with

∣ G ∣ = p^{n} m

where

p ∤ m

. The number

n_{p}

of Sylow

p

-subgroups satisfies:

$n_{p} ∣ m$ .
$n_{p} \equiv 1 (mod p)$ .

Proof.

Fix a Sylow

p

-subgroup

P

. The group

G

acts on

Syl_{p} (G)

by conjugation:

g \cdot Q = g Q g^{- 1}

. By the second Sylow theorem, this action is transitive, so

n_{p} = ∣ Syl_{p} (G) ∣ = [G : N_{G} (P)]

where

N_{G} (P) = {g \in G ∣ g P g^{- 1} = P}

is the normalizer of

P

. Since

P \leq N_{G} (P) \leq G

, Lagrange's theorem gives

n_{p} = [G : N_{G} (P)] ∣ [G : P] = m

To show

n_{p} \equiv 1 (mod p)

, let

P

act on

Syl_{p} (G)

by conjugation. The fixed points are those

Q \in Syl_{p} (G)

with

pQ p^{- 1} = Q

for all

p \in P

, i.e.,

P \leq N_{G} (Q)

. If

Q

is a fixed point, then both

P

and

Q

are Sylow

p

-subgroups of

N_{G} (Q)

, so they are conjugate in

N_{G} (Q)

Q = g P g^{- 1}

for some

g \in N_{G} (Q)

. But

Q ⊴ N_{G} (Q)

, so

g Q g^{- 1} = Q = g P g^{- 1}

, giving

P = Q

Thus the only fixed point is

P

itself, and the fixed-point theorem gives

n_{p} \equiv 1 (mod p)

. □

6 Applications: Determining Group Structure

The power of the Sylow theorems lies in the stringent constraints they impose on $n_{p}$ , which often suffice to pin down the structure of a group.

Example 8 (Groups of order

15

Let

∣ G ∣ = 15 = 3 \cdot 5

$n_{3} ∣ 5$ and $n_{3} \equiv 1 (mod 3)$ , so $n_{3} \in {1, 5}$ . Since $5 \equiv 2 (mod 3)$ , we must have $n_{3} = 1$ .
$n_{5} ∣ 3$ and $n_{5} \equiv 1 (mod 5)$ , so $n_{5} = 1$ .

The unique Sylow

3

-subgroup

P_{3}

and Sylow

5

-subgroup

P_{5}

are both normal. Since

P_{3} \cap P_{5} = {e}

(their orders are coprime) and

∣ P_{3} P_{5} ∣ = ∣ P_{3} ∣∣ P_{5} ∣/∣ P_{3} \cap P_{5} ∣ = 15 = ∣ G ∣

, we have

G = P_{3} \times P_{5} ≅ Z /3 Z \times Z /5 Z ≅ Z /15 Z

. Every group of order

15

is cyclic.

Example 9 (Groups of order

12

Let

∣ G ∣ = 12 = 2^{2} \cdot 3

$n_{3} ∣ 4$ and $n_{3} \equiv 1 (mod 3)$ , so $n_{3} \in {1, 4}$ .
$n_{2} ∣ 3$ and $n_{2} \equiv 1 (mod 2)$ , so $n_{2} \in {1, 3}$ .

n_{3} = 1

, there is a normal Sylow

3

-subgroup. The case

n_{3} = 4

also occurs (e.g., in

A_{4}

). Up to isomorphism, there are exactly five groups of order

12

Z /12 Z

Z /2 Z \times Z /6 Z

A_{4}

D_{6}

, and

Dic_{12}

(the dicyclic group).

Example 10 (Groups of order

pq

(

p < q

primes)).

Let

∣ G ∣ = pq

. Then

n_{q} ∣ p

and

n_{q} \equiv 1 (mod q)

. Since

p < q

, the only possibility is

n_{q} = 1

(for

p \equiv 1 (mod q)

would force

p \geq q + 1 > q > p

, a contradiction). Hence the Sylow

q

-subgroup

Q

is normal.

For the Sylow

p

-subgroup,

n_{p} ∣ q

and

n_{p} \equiv 1 (mod p)

, so

n_{p} \in {1, q}

. If

q \equiv 1 (mod p)

, then

n_{p} = q

is possible and a non-abelian group exists. If

q \neq \equiv 1 (mod p)

, then

n_{p} = 1

and

G ≅ Z / pq Z

7 Properties of Sylow Subgroups

Theorem 11.

Let

G

be a finite group and

P \in Syl_{p} (G)

. If

H \leq G

is a

p

-group, then

H \leq g P g^{- 1}

for some

g \in G

. In particular, every maximal

p

-subgroup of

G

is a Sylow

p

-subgroup.

Proof.

Let

H

act on

G / P

by left multiplication. Since

∣ G / P ∣ = m

with

p ∤ m

and

H

is a

p

-group, the fixed-point theorem gives

∣ (G / P)^{H} ∣ \equiv m \neq \equiv 0 (mod p)

. In particular, there is a fixed point

g P

. Then

H g P = g P

, so

g^{- 1} H g \leq P

, and therefore

H \leq g P g^{- 1}

. □

Theorem 12 (Frattini's Argument).

Let

N ⊴ G

and

P \in Syl_{p} (N)

. Then

G = N \cdot N_{G} (P)

Proof.

Let

g \in G

be arbitrary. Since

N ⊴ G

, we have

g P g^{- 1} \leq g N g^{- 1} = N

, so

g P g^{- 1}

is a Sylow

p

-subgroup of

N

. By the second Sylow theorem applied within

N

, there exists

n \in N

with

g P g^{- 1} = n P n^{- 1}

. Then

n^{- 1} g \in N_{G} (P)

, so

g = n (n^{- 1} g) \in N \cdot N_{G} (P)

. □

8 Summary

The material covered in this chapter:

Cauchy's theorem: if $p ∣ ∣ G ∣$ , then $G$ has an element of order $p$ .
First Sylow theorem: Sylow $p$ -subgroups exist.
Second Sylow theorem: all Sylow $p$ -subgroups are conjugate.
Third Sylow theorem: $n_{p} ∣ m$ and $n_{p} \equiv 1 (mod p)$ .
Applications to determining the structure of groups of orders $15$ , $12$ , and $pq$ .
Maximality of Sylow subgroups and Frattini's argument.

The Sylow theorems are the single most powerful tool for analyzing the structure of finite groups concretely. They allow us to systematically answer the question: to what extent does the order of a group determine its structure?

Group Theory Algebra Textbook Sylow Theorems p-groups

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The Sylow Theorems

Folio Official

March 1, 2026

1 Introduction

Definition 1 (Sylow

p

-subgroup).

Let

G

be a finite group and

p

a prime. Write

∣ G ∣ = p^{n} m

where

p ∤ m

. A subgroup of

G

of order

p^{n}

is called a Sylow $p$ -subgroup of

G

. We write

Syl_{p} (G)

for the set of all Sylow

p

-subgroups of

G

, and

n_{p} = ∣ Syl_{p} (G) ∣

for their number.

2 Cauchy's Theorem

Before turning to the Sylow theorems proper, we establish the foundational result of Cauchy, which provides a partial converse to Lagrange's theorem.

Theorem 2 (Cauchy's Theorem).

Let

G

be a finite group and

p

a prime. If

p ∣ ∣ G ∣

, then

G

contains an element of order

p

Proof.

We proceed by strong induction on

∣ G ∣

. The base case

∣ G ∣ = 1

is vacuously true.

Suppose

∣ G ∣ \geq 2

and the theorem holds for all groups of order less than

∣ G ∣

Case 1: There exists a proper subgroup

H < G

with

p ∣ ∣ H ∣

. Since

∣ H ∣ < ∣ G ∣

, the induction hypothesis gives an element of order

p

H

, which is also an element of

G

Case 2: For every proper subgroup

H < G

, we have

p ∤ ∣ H ∣

. Consider the class equation

∣ G ∣ = ∣ Z (G) ∣ + i \sum [G : C_{G} (a_{i})]

where the sum runs over representatives

a_{i}

of the non-central conjugacy classes. For each such

a_{i}

, the centralizer

C_{G} (a_{i})

is a proper subgroup, so

p ∤ ∣ C_{G} (a_{i}) ∣

by our assumption. Since

p ∣ ∣ G ∣

, we have

p ∣ [G : C_{G} (a_{i})]

for each

i

. The class equation then gives

∣ Z (G) ∣ = ∣ G ∣ - i \sum [G : C_{G} (a_{i})] \equiv 0 (mod p)

p ∣ ∣ Z (G) ∣

Since

Z (G)

is abelian and

p ∣ ∣ Z (G) ∣

, we can find an element of order

p

Z (G)

as follows. Pick any

z \in Z (G)

with

z \neq = e

, and set

m = ord (z)

. If

p ∣ m

, then

z^{m / p}

has order

p

and we are done. If

p ∤ m

, consider the quotient

Z (G) / ⟨ z ⟩

. This is an abelian group of order

∣ Z (G) ∣/ m

, and

p ∣ ∣ Z (G) ∣/ m

since

p ∣ ∣ Z (G) ∣

and

p ∤ m

. Moreover

∣ Z (G) / ⟨ z ⟩ ∣ < ∣ Z (G) ∣ \leq ∣ G ∣

, so by the induction hypothesis there exists an element

\overset{w}{ˉ} \in Z (G) / ⟨ z ⟩

of order

p

. Lifting to

w \in Z (G)

, we have

w^{p} \in ⟨ z ⟩

, so

ord (w)

is a multiple of

p

, and

w^{ord (w) / p}

is an element of order

p

G

. □

Remark 3.

Cauchy's theorem is the special case of the first Sylow theorem for

p^{1} ∣ ∣ G ∣

. Besides the inductive proof above, there is an elegant combinatorial proof due to McKay: one considers the set

{(a_{1}, \dots, a_{p}) \in G^{p} ∣ a_{1} a_{2} \dots a_{p} = e}

and lets

Z / p Z

act on it by cyclic permutation.

3 The First Sylow Theorem (Existence)

Theorem 4 (First Sylow Theorem).

Let

G

be a finite group and

p

a prime. Write

∣ G ∣ = p^{n} m

with

p ∤ m

and

n \geq 1

. Then

G

possesses a Sylow

p

-subgroup.

Proof.

Let

G

act on the collection

X = {S \subseteq G ∣ ∣ S ∣ = p^{n}}

of all subsets of

G

of size

p^{n}

by left multiplication:

g \cdot S = g S = {g s ∣ s \in S}

We have

∣ X ∣ = (p ^{n} ∣ G ∣) = (p ^{n} p ^{n} m)

. We claim that

p ∤ (p ^{n} p ^{n} m)

. Indeed, writing

(p ^{n} p ^{n} m) = i = 0 \prod p^{n} - 1 \frac{p ^{n} m - i}{p ^{n} - i},

one checks that for

0 \leq i < p^{n}

, the

p

-adic valuation of

p^{n} m - i

equals that of

p^{n} - i

, so the product is not divisible by

p

By the orbit decomposition

X = ⨆_{j} Orb (S_{j})

, and since

∣ X ∣

is not divisible by

p

, there must exist an orbit

Orb (S_{0})

with

p ∤ ∣ Orb (S_{0}) ∣

By the orbit-stabilizer theorem,

∣ Orb (S_{0}) ∣ = [G : Stab (S_{0})]

, so

∣ Stab (S_{0}) ∣ = ∣ G ∣/∣ Orb (S_{0}) ∣

. Since

p ∤ ∣ Orb (S_{0}) ∣

and

∣ G ∣ = p^{n} m

, we get

p^{n} ∣ ∣ Stab (S_{0}) ∣

On the other hand, let

H = Stab (S_{0})

. Then

h S_{0} = S_{0}

for all

h \in H

. Fixing any

s_{0} \in S_{0}

, the map

h \mapsto h s_{0}

embeds

H

into

S_{0}

, giving

∣ H ∣ = ∣ H s_{0} ∣ \leq ∣ S_{0} ∣ = p^{n}

Combining

p^{n} ∣ ∣ H ∣

and

∣ H ∣ \leq p^{n}

, we conclude

∣ H ∣ = p^{n}

. Thus

H = Stab (S_{0})

is a Sylow

p

-subgroup of

G

. □

4 The Second Sylow Theorem (Conjugacy)

Theorem 5 (Second Sylow Theorem).

Let

G

be a finite group and

p

a prime. Any two Sylow

p

-subgroups of

G

are conjugate. That is, if

P, Q \in Syl_{p} (G)

, then there exists

g \in G

such that

Q = g P g^{- 1}

Proof.

Fix

P \in Syl_{p} (G)

and let

Q \in Syl_{p} (G)

be arbitrary. Let

Q

act on the set

G / P

of left cosets by left multiplication:

q \cdot (g P) = (q g) P

We have

∣ G / P ∣ = [G : P] = m

where

p ∤ m

. Since

Q

is a

p

-group, the fixed-point theorem for

p

-groups gives

∣ (G / P)^{Q} ∣ \equiv ∣ G / P ∣ \equiv m \neq \equiv 0 (mod p)

. In particular,

(G / P)^{Q} \neq = \emptyset

Let

g P

be a fixed point. Then

q g P = g P

for all

q \in Q

, which means

g^{- 1} q g \in P

for all

q \in Q

. Hence

g^{- 1} Q g \leq P

. Since

∣ g^{- 1} Q g ∣ = ∣ Q ∣ = p^{n} = ∣ P ∣

, we conclude

g^{- 1} Q g = P

, i.e.,

Q = g P g^{- 1}

. □

Theorem 6 (Corollary).

P

is the unique Sylow

p

-subgroup of

G

, then

P ⊴ G

Proof.

For any

g \in G

, the conjugate

g P g^{- 1}

is also a Sylow

p

-subgroup (it has the same order

p^{n}

). By uniqueness,

g P g^{- 1} = P

. □

5 The Third Sylow Theorem (Counting)

Theorem 7 (Third Sylow Theorem).

Let

G

be a finite group with

∣ G ∣ = p^{n} m

where

p ∤ m

. The number

n_{p}

of Sylow

p

-subgroups satisfies:

$n_{p} ∣ m$ .
$n_{p} \equiv 1 (mod p)$ .

Proof.

Fix a Sylow

p

-subgroup

P

. The group

G

acts on

Syl_{p} (G)

by conjugation:

g \cdot Q = g Q g^{- 1}

. By the second Sylow theorem, this action is transitive, so

n_{p} = ∣ Syl_{p} (G) ∣ = [G : N_{G} (P)]

where

N_{G} (P) = {g \in G ∣ g P g^{- 1} = P}

is the normalizer of

P

. Since

P \leq N_{G} (P) \leq G

, Lagrange's theorem gives

n_{p} = [G : N_{G} (P)] ∣ [G : P] = m

To show

n_{p} \equiv 1 (mod p)

, let

P

act on

Syl_{p} (G)

by conjugation. The fixed points are those

Q \in Syl_{p} (G)

with

pQ p^{- 1} = Q

for all

p \in P

, i.e.,

P \leq N_{G} (Q)

. If

Q

is a fixed point, then both

P

and

Q

are Sylow

p

-subgroups of

N_{G} (Q)

, so they are conjugate in

N_{G} (Q)

Q = g P g^{- 1}

for some

g \in N_{G} (Q)

. But

Q ⊴ N_{G} (Q)

, so

g Q g^{- 1} = Q = g P g^{- 1}

, giving

P = Q

Thus the only fixed point is

P

itself, and the fixed-point theorem gives

n_{p} \equiv 1 (mod p)

. □

6 Applications: Determining Group Structure

The power of the Sylow theorems lies in the stringent constraints they impose on $n_{p}$ , which often suffice to pin down the structure of a group.

Example 8 (Groups of order

15

Let

∣ G ∣ = 15 = 3 \cdot 5

$n_{3} ∣ 5$ and $n_{3} \equiv 1 (mod 3)$ , so $n_{3} \in {1, 5}$ . Since $5 \equiv 2 (mod 3)$ , we must have $n_{3} = 1$ .
$n_{5} ∣ 3$ and $n_{5} \equiv 1 (mod 5)$ , so $n_{5} = 1$ .

The unique Sylow

3

-subgroup

P_{3}

and Sylow

5

-subgroup

P_{5}

are both normal. Since

P_{3} \cap P_{5} = {e}

(their orders are coprime) and

∣ P_{3} P_{5} ∣ = ∣ P_{3} ∣∣ P_{5} ∣/∣ P_{3} \cap P_{5} ∣ = 15 = ∣ G ∣

, we have

G = P_{3} \times P_{5} ≅ Z /3 Z \times Z /5 Z ≅ Z /15 Z

. Every group of order

15

is cyclic.

Example 9 (Groups of order

12

Let

∣ G ∣ = 12 = 2^{2} \cdot 3

$n_{3} ∣ 4$ and $n_{3} \equiv 1 (mod 3)$ , so $n_{3} \in {1, 4}$ .
$n_{2} ∣ 3$ and $n_{2} \equiv 1 (mod 2)$ , so $n_{2} \in {1, 3}$ .

n_{3} = 1

, there is a normal Sylow

3

-subgroup. The case

n_{3} = 4

also occurs (e.g., in

A_{4}

). Up to isomorphism, there are exactly five groups of order

12

Z /12 Z

Z /2 Z \times Z /6 Z

A_{4}

D_{6}

, and

Dic_{12}

(the dicyclic group).

Example 10 (Groups of order

pq

(

p < q

primes)).

Let

∣ G ∣ = pq

. Then

n_{q} ∣ p

and

n_{q} \equiv 1 (mod q)

. Since

p < q

, the only possibility is

n_{q} = 1

(for

p \equiv 1 (mod q)

would force

p \geq q + 1 > q > p

, a contradiction). Hence the Sylow

q

-subgroup

Q

is normal.

For the Sylow

p

-subgroup,

n_{p} ∣ q

and

n_{p} \equiv 1 (mod p)

, so

n_{p} \in {1, q}

. If

q \equiv 1 (mod p)

, then

n_{p} = q

is possible and a non-abelian group exists. If

q \neq \equiv 1 (mod p)

, then

n_{p} = 1

and

G ≅ Z / pq Z

7 Properties of Sylow Subgroups

Theorem 11.

Let

G

be a finite group and

P \in Syl_{p} (G)

. If

H \leq G

is a

p

-group, then

H \leq g P g^{- 1}

for some

g \in G

. In particular, every maximal

p

-subgroup of

G

is a Sylow

p

-subgroup.

Proof.

Let

H

act on

G / P

by left multiplication. Since

∣ G / P ∣ = m

with

p ∤ m

and

H

is a

p

-group, the fixed-point theorem gives

∣ (G / P)^{H} ∣ \equiv m \neq \equiv 0 (mod p)

. In particular, there is a fixed point

g P

. Then

H g P = g P

, so

g^{- 1} H g \leq P

, and therefore

H \leq g P g^{- 1}

. □

Theorem 12 (Frattini's Argument).

Let

N ⊴ G

and

P \in Syl_{p} (N)

. Then

G = N \cdot N_{G} (P)

Proof.

Let

g \in G

be arbitrary. Since

N ⊴ G

, we have

g P g^{- 1} \leq g N g^{- 1} = N

, so

g P g^{- 1}

is a Sylow

p

-subgroup of

N

. By the second Sylow theorem applied within

N

, there exists

n \in N

with

g P g^{- 1} = n P n^{- 1}

. Then

n^{- 1} g \in N_{G} (P)

, so

g = n (n^{- 1} g) \in N \cdot N_{G} (P)

. □

8 Summary

The material covered in this chapter:

Cauchy's theorem: if $p ∣ ∣ G ∣$ , then $G$ has an element of order $p$ .
First Sylow theorem: Sylow $p$ -subgroups exist.
Second Sylow theorem: all Sylow $p$ -subgroups are conjugate.
Third Sylow theorem: $n_{p} ∣ m$ and $n_{p} \equiv 1 (mod p)$ .
Applications to determining the structure of groups of orders $15$ , $12$ , and $pq$ .
Maximality of Sylow subgroups and Frattini's argument.

Group Theory Algebra Textbook Sylow Theorems p-groups

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Mathematics "between the lines" — exploring the intuition textbooks leave out, written in LaTeX on Folio.

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The Sylow Theorems

1 Introduction

2 Cauchy's Theorem

3 The First Sylow Theorem (Existence)

4 The Second Sylow Theorem (Conjugacy)

5 The Third Sylow Theorem (Counting)

6 Applications: Determining Group Structure

7 Properties of Sylow Subgroups

8 Summary

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The Sylow Theorems

1 Introduction

2 Cauchy's Theorem

3 The First Sylow Theorem (Existence)

4 The Second Sylow Theorem (Conjugacy)

5 The Third Sylow Theorem (Counting)

6 Applications: Determining Group Structure

7 Properties of Sylow Subgroups

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