Abstract Algebra
Sylow Theorems
How many distinct groups of order 100 exist? Without the Sylow theorems this seems hopeless. With them, it becomes a few lines of reasoning. In 1872, Sylow gave us an X-ray machine for seeing the internal structure of finite groups.
- The Sylow theorems are used in cryptography: the order of the group of points on an elliptic curve over a finite field determines the Sylow subgroups, directly affecting the security of ECDH
- In physics: classifying crystallographic symmetry groups uses Sylow theory to analyze possible crystal structures
Предварительные знания
p-Groups
A **p-group** is a group whose order is a power of a prime p: |G| = pⁿ. Examples: Z₄ (p=2, n=2), Z₈, Z₂×Z₂, D₄ (all of order 8 = 2³). p-groups have a special property: their center Z(G) is nontrivial (|Z(G)| ≥ p). This is the fundamental lemma from which the Sylow theorems grow.
The **class equation** is the key tool: |G| = |Z(G)| + Σ |Cl(x)|, summed over representatives of non-central conjugacy classes. For a p-group every |Cl(x)| > 1 is divisible by p, and |G| = pⁿ is too. Therefore |Z(G)| is divisible by p - the center contains at least p elements.
**Corollary:** Every group of order p² is abelian. Proof: |Z(G)| is divisible by p. If |Z(G)| = p², then G = Z(G) is abelian. If |Z(G)| = p, then G/Z(G) ≅ Zₚ is cyclic, which forces G to be abelian (a paradox that rules out this case). Therefore |Z(G)| = p² and G is abelian.
A p-group is a group with p elements (prime order)
A p-group has order pⁿ for some n ≥ 1. A group of prime order is a special case (n=1).
Groups of prime order are cyclic and 'boring'. The real interest begins at n ≥ 2, for example D₄ of order 8 = 2³.
Group G has order 25 = 5². What can we conclude about G?
The Three Sylow Theorems
Let |G| = pⁿ · m where p ∤ m. A **Sylow p-subgroup** is a subgroup of order pⁿ (a maximal p-subgroup). The three great Sylow theorems: **Theorem I (Existence):** G contains at least one Sylow p-subgroup. **Theorem II (Conjugacy):** All Sylow p-subgroups are conjugate: if P, Q are Sylow p-subgroups, then Q = gPg⁻¹ for some g ∈ G. **Theorem III (Count):** The number nₚ of Sylow p-subgroups satisfies: - nₚ ≡ 1 (mod p) - nₚ | m
**Ludwig Sylow (1832-1918)** was a Norwegian mathematician who proved his theorems in 1872 in a paper of just 10 pages. He spent 40 years as a secondary school teacher - and during that time created one of the most fundamental theorems in algebra. He received a university position only at age 65. His 1872 paper remains one of the most influential in the history of group theory.
**Key insight:** nₚ = 1 means the unique Sylow p-subgroup is **normal** in G (since all Sylows are conjugate, and a subgroup is normal iff it is its own only conjugate). This is a powerful tool: showing nₚ = 1 forces the group to have a normal subgroup → a quotient group can be formed.
Group G has order 15 = 3 × 5. What is n₃ (the number of Sylow 3-subgroups)?
Applications: Classifying Groups of Order pq
**Theorem:** If |G| = pq where p < q are prime and p ∤ (q−1), then G ≅ Z_{pq} (the unique group up to isomorphism). **Proof sketch:** By the Sylow theorems: - nₚ | q and nₚ ≡ 1 (mod p) → nₚ ∈ {1, q}. If q ≢ 1 (mod p), then nₚ = 1. - nq | p and nq ≡ 1 (mod q) → nq = 1. If both nₚ = nq = 1, then G has normal H ≅ Zₚ and K ≅ Zq. Since H∩K = {e} and |HK| = pq = |G|, we get G = HK ≅ Zₚ × Zq ≅ Z_{pq}.
**When p | (q−1): semidirect products.** If p | (q−1), there is a nontrivial homomorphism φ: Zₚ → Aut(Zq) ≅ Z_{q-1}, giving the semidirect product Zq ⋊ Zₚ - a non-commutative group of order pq. Example: |G| = 21 = 3×7, and 3 | (7−1) = 6. There exist both Z₂₁ and Z₇⋊Z₃ - exactly 2 groups of order 21.
If nₚ > 1, the group cannot be classified
nₚ > 1 makes things harder but does not block classification. It means there is no normal p-subgroup and the structure may be more complex (semidirect products, etc.).
The Sylow constraints narrow down nₚ, and often both possible values are realized - each corresponding to a distinct group (as with groups of order pq when p|(q-1)).
How many non-isomorphic groups of order 35 = 5 × 7 exist?
Key Ideas
- p-group: |G| = pⁿ. The center of a p-group is nontrivial (from the class equation)
- Groups of order p² are always abelian
- Sylow p-subgroup: a subgroup of order pⁿ, where |G| = pⁿ·m, p∤m
- Theorem I: a Sylow p-subgroup exists
- Theorem II: all Sylow p-subgroups are conjugate
- Theorem III: nₚ ≡ 1 (mod p) and nₚ | m
- nₚ = 1 ⟺ the unique Sylow p-subgroup is normal
Further Paths
The Sylow theorems are the gateway to the theory of finite simple groups. The classification of finite simple groups (the enormous Monster theorem) relies on analyzing Sylow subgroups.
- Group Actions — The proof of the Sylow theorems uses the action of G on cosets - a natural motivation for the next lesson
- Solvable Groups — All p-groups are solvable; Sylow theorems help prove solvability via normal series
Вопросы для размышления
- Prove that every group of order p² is abelian. Use the fact: if G/Z(G) is cyclic, then G is abelian.
- How many non-isomorphic groups of order 21 are there? Find n₃ and n₇, then consider both cases.
- Why does conjugacy of Sylow subgroups (Theorem II) imply they are 'the same' from the perspective of internal structure?