Sylow Theorems — Mathematics Knowledge System

Overview

The Sylow theorems, proved by Peter Ludwig Mejdell Sylow in 1872, are a partial converse to Lagrange's theorem: while Lagrange only says subgroup orders must divide |G|, the Sylow theorems guarantee that subgroups of certain sizes actually exist, and describe how many there are. Specifically, they concern Sylow p-subgroups — subgroups whose order is the largest power of a prime p dividing |G|. These theorems are the single most powerful tool for classifying finite groups of a given order and are proved using group actions on sets of subsets or on the set of Sylow subgroups by conjugation.

Intuition

If |G| = 12 = 2² × 3, Lagrange only tells you a subgroup's order must divide 12; the Sylow theorems go further and guarantee subgroups of order 4 (the full power of 2) and order 3 (the full power of 3) actually exist. The counting constraint n_p ≡ 1 (mod p) is a strong sieve: for |G| = 12, n_3 must be 1 or 4 (divisors of 4 that are ≡ 1 mod 3), immediately narrowing down the possible structures of G and often forcing a unique, hence normal, Sylow subgroup — the standard first move in classifying groups of a given small order.

Formal Definition

Definition

Let G be a finite group with |G| = pᵏm, where p is prime and p does not divide m. Let n_p denote the number of Sylow p-subgroups (subgroups of order pᵏ).

\exists\, P \leq G \text{ with } |P| = p^k \quad \text{(Sylow I: existence)}

Sylow I

\text{All Sylow } p\text{-subgroups are conjugate in } G \quad \text{(Sylow II)}

Sylow II

n_p \equiv 1 \pmod{p} \quad \text{and} \quad n_p \mid m \quad \text{(Sylow III)}

Sylow III

n_p = [G : N_G(P)] \quad \text{for any Sylow } p\text{-subgroup } P

n_p equals the index of the normalizer

Worked Examples

n_5 divides m = 3 and n_5 ≡ 1 (mod 5). Divisors of 3 are 1, 3; only 1 satisfies ≡1 mod 5.
$n_5 \mid 3,\ n_5 \equiv 1 \pmod 5 \implies n_5 = 1$
Similarly for p=3: n_3 divides 5 and n_3 ≡ 1 (mod 3). Divisors of 5 are 1, 5; only 1 works (5 ≡ 2 mod 3).
$n_3 \mid 5,\ n_3 \equiv 1 \pmod 3 \implies n_3 = 1$
A unique Sylow subgroup is automatically normal (Sylow II: conjugates permute among the n_p subgroups, but n_p=1 means it's fixed).
$P_5 \trianglelefteq G,\ P_3 \trianglelefteq G$
P_5 ∩ P_3 = {e} (orders 5,3 coprime), so G ≅ P_5 × P_3 ≅ ℤ_5 × ℤ_3 ≅ ℤ_15 (since gcd(3,5)=1).
$G \cong \mathbb{Z}_5 \times \mathbb{Z}_3 \cong \mathbb{Z}_{15}$

Answer: Both Sylow subgroups are unique hence normal, forcing G ≅ ℤ₅ × ℤ₃ ≅ ℤ₁₅ — every group of order 15 is cyclic.

Practice Problems

Difficulty 6/10

For a group of order 45 = 3² × 5, find n_5 and n_3, and conclude both Sylow subgroups are normal.

Difficulty 7/10

Explain why a group of order 6 must have a normal Sylow 3-subgroup, but need not have a normal Sylow 2-subgroup (cite S₃ as the relevant example).

Difficulty 7/10

Use the Sylow theorems to show there is no simple group of order 30.

Quiz

Summary

Sylow I: for |G|=pᵏm (p∤m), a subgroup of order pᵏ (a Sylow p-subgroup) always exists.
Sylow II: all Sylow p-subgroups are conjugate; Sylow III: their count n_p satisfies n_p ≡ 1 (mod p) and n_p | m.
These constraints are the standard tool for proving groups of a given order aren't simple or must be cyclic/direct products, by forcing n_p = 1 (hence normal).

References

WebsiteWikipedia — Sylow theorems