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 — A4 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 pk.
Let G be a finite group and p a prime. Write ∣G∣=pnm where p∤m. A subgroup of G of order pn is called a Sylowp-subgroup of G. We write Sylp(G) for the set of all Sylow p-subgroups of G, and np=∣Sylp(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.
Let G be a finite group and p a prime. If p∣∣G∣, then G contains an element of order p.
We proceed by strong induction on ∣G∣. The base case ∣G∣=1 is vacuously true.Suppose ∣G∣≥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 in 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∑[G:CG(ai)]
where the sum runs over representatives ai of the non-central conjugacy classes. For each such ai, the centralizer CG(ai) is a proper subgroup, so p∤∣CG(ai)∣ by our assumption. Since p∣∣G∣, we have p∣[G:CG(ai)] for each i. The class equation then gives ∣Z(G)∣=∣G∣−i∑[G:CG(ai)]≡0(modp)
so p∣∣Z(G)∣.Since Z(G) is abelian and p∣∣Z(G)∣, we can find an element of order p in Z(G) as follows. Pick any z∈Z(G) with z=e, and set m=ord(z). If p∣m, then zm/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)∣≤∣G∣, so by the induction hypothesis there exists an element wˉ∈Z(G)/⟨z⟩ of order p. Lifting to w∈Z(G), we have wp∈⟨z⟩, so ord(w) is a multiple of p, and word(w)/p is an element of order p in G. □
3 The First Sylow Theorem (Existence)
Let G be a finite group and p a prime. Write ∣G∣=pnm with p∤m and n≥1. Then G possesses a Sylow p-subgroup.
Let G act on the collection X={S⊆G∣∣S∣=pn} of all subsets of G of size pn by left multiplication: g⋅S=gS={gs∣s∈S}.We have ∣X∣=(pn∣G∣)=(pnpnm). We claim that p∤(pnpnm). Indeed, writing (pnpnm)=i=0∏pn−1pn−ipnm−i,
one checks that for 0≤i<pn, the p-adic valuation of pnm−i equals that of pn−i, so the product is not divisible by p.By the orbit decomposition X=⨆jOrb(Sj), and since ∣X∣ is not divisible by p, there must exist an orbit Orb(S0) with p∤∣Orb(S0)∣.By the orbit-stabilizer theorem, ∣Orb(S0)∣=[G:Stab(S0)], so ∣Stab(S0)∣=∣G∣/∣Orb(S0)∣. Since p∤∣Orb(S0)∣ and ∣G∣=pnm, we get pn∣∣Stab(S0)∣.On the other hand, let H=Stab(S0). Then hS0=S0 for all h∈H. Fixing any s0∈S0, the map h↦hs0 embeds H into S0, giving ∣H∣=∣Hs0∣≤∣S0∣=pn.Combining pn∣∣H∣ and ∣H∣≤pn, we conclude ∣H∣=pn. Thus H=Stab(S0) is a Sylow p-subgroup of G. □
4 The Second Sylow Theorem (Conjugacy)
Let G be a finite group and p a prime. Any two Sylow p-subgroups of G are conjugate. That is, if P,Q∈Sylp(G), then there exists g∈G such that Q=gPg−1.
Fix P∈Sylp(G) and let Q∈Sylp(G) be arbitrary. Let Q act on the set G/P of left cosets by left multiplication: q⋅(gP)=(qg)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∣≡∣G/P∣≡m≡0(modp). In particular, (G/P)Q=∅.Let gP be a fixed point. Then qgP=gP for all q∈Q, which means g−1qg∈P for all q∈Q. Hence g−1Qg≤P. Since ∣g−1Qg∣=∣Q∣=pn=∣P∣, we conclude g−1Qg=P, i.e., Q=gPg−1. □
If P is the unique Sylow p-subgroup of G, then P⊴G.
For any g∈G, the conjugate gPg−1 is also a Sylow p-subgroup (it has the same order pn). By uniqueness, gPg−1=P. □
5 The Third Sylow Theorem (Counting)
Let G be a finite group with ∣G∣=pnm where p∤m. The number np of Sylow p-subgroups satisfies:
np∣m.
np≡1(modp).
Fix a Sylow p-subgroup P. The group G acts on Sylp(G) by conjugation: g⋅Q=gQg−1. By the second Sylow theorem, this action is transitive, so np=∣Sylp(G)∣=[G:NG(P)]
where NG(P)={g∈G∣gPg−1=P} is the normalizer of P. Since P≤NG(P)≤G, Lagrange's theorem gives np=[G:NG(P)]∣[G:P]=m.To show np≡1(modp), let P act on Sylp(G) by conjugation. The fixed points are those Q∈Sylp(G) with pQp−1=Q for all p∈P, i.e., P≤NG(Q). If Q is a fixed point, then both P and Q are Sylow p-subgroups of NG(Q), so they are conjugate in NG(Q): Q=gPg−1 for some g∈NG(Q). But Q⊴NG(Q), so gQg−1=Q=gPg−1, giving P=Q.Thus the only fixed point is P itself, and the fixed-point theorem gives np≡1(modp). □
6 Applications: Determining Group Structure
The power of the Sylow theorems lies in the stringent constraints they impose on np, which often suffice to pin down the structure of a group.
Let ∣G∣=15=3⋅5.
n3∣5 and n3≡1(mod3), so n3∈{1,5}. Since 5≡2(mod3), we must have n3=1.
n5∣3 and n5≡1(mod5), so n5=1.
The unique Sylow 3-subgroup P3 and Sylow 5-subgroup P5 are both normal. Since P3∩P5={e} (their orders are coprime) and ∣P3P5∣=∣P3∣∣P5∣/∣P3∩P5∣=15=∣G∣, we have G=P3×P5≅Z/3Z×Z/5Z≅Z/15Z. Every group of order 15 is cyclic.
Let ∣G∣=12=22⋅3.
n3∣4 and n3≡1(mod3), so n3∈{1,4}.
n2∣3 and n2≡1(mod2), so n2∈{1,3}.
If n3=1, there is a normal Sylow 3-subgroup. The case n3=4 also occurs (e.g., in A4). Up to isomorphism, there are exactly five groups of order 12: Z/12Z, Z/2Z×Z/6Z, A4, D6, and Dic12 (the dicyclic group).
Let ∣G∣=pq. Then nq∣p and nq≡1(modq). Since p<q, the only possibility is nq=1 (for p≡1(modq) would force p≥q+1>q>p, a contradiction). Hence the Sylow q-subgroup Q is normal.For the Sylow p-subgroup, np∣q and np≡1(modp), so np∈{1,q}. If q≡1(modp), then np=q is possible and a non-abelian group exists. If q≡1(modp), then np=1 and G≅Z/pqZ.
7 Properties of Sylow Subgroups
Let G be a finite group and P∈Sylp(G). If H≤G is a p-group, then H≤gPg−1 for some g∈G. In particular, every maximal p-subgroup of G is a Sylow p-subgroup.
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∣≡m≡0(modp). In particular, there is a fixed point gP. Then HgP=gP, so g−1Hg≤P, and therefore H≤gPg−1. □
Let N⊴G and P∈Sylp(N). Then G=N⋅NG(P).
Let g∈G be arbitrary. Since N⊴G, we have gPg−1≤gNg−1=N, so gPg−1 is a Sylow p-subgroup of N. By the second Sylow theorem applied within N, there exists n∈N with gPg−1=nPn−1. Then n−1g∈NG(P), so g=n(n−1g)∈N⋅NG(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: np∣m and np≡1(modp).
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?