1. Write a program to compute the number of conjugacy classes in Sn. What is the
largest nfor which your program will work?
References and Suggested Reading
[1] De Bruijin, N. G. “Pólya’s Theory of Counting,” inApplied Combinatorial Mathemat- ics, Beckenbach, E. F., ed. Wiley, New York, 1964.
14.5. PROGRAMMING EXERCISE 171 [2] Eidswick, J. A. “Cubelike Puzzles—What Are They and How Do You Solve Them?”
American Mathematical Monthly 93(1986), 157–76.
[3] Harary, F., Palmer, E. M., and Robinson, R. W. “Pólya’s Contributions to Chem- ical Enumeration,” in Chemical Applications of Graph Theory, Balaban, A. T., ed. Academic Press, London, 1976.
[4] Gårding, L. and Tambour, T. Algebra for Computer Science. Springer-Verlag, New York, 1988.
[5] Laufer, H. B.Discrete Mathematics and Applied Modern Algebra. PWS-Kent, Boston, 1984.
[6] Pólya, G. and Read, R. C.Combinatorial Enumeration of Groups, Graphs, and Chem- ical Compounds. Springer-Verlag, New York, 1985.
[7] Shapiro, L. W. “Finite Groups Acting on Sets with Applications,”Mathematics Mag- azine, May–June 1973, 136–47.
15
The Sylow Theorems
We already know that the converse of Lagrange’s Theorem is false. If G is a group of order m and ndivides m, then G does not necessarily possess a subgroup of order n. For example, A4 has order12 but does not possess a subgroup of order6. However, the Sylow
Theorems do provide a partial converse for Lagrange’s Theorem—in certain cases they guarantee us subgroups of specific orders. These theorems yield a powerful set of tools for the classification of all finite nonabelian groups.
15.1 The Sylow Theorems
We will use what we have learned about group actions to prove the Sylow Theorems. Recall for a moment what it means forGto act on itself by conjugation and how conjugacy classes are distributed in the group according to the class equation, discussed in Chapter 14. A group G acts on itself by conjugation via the map (g, x) 7→ gxg−1. Let x1, . . . , xk be
representatives from each of the distinct conjugacy classes of G that consist of more than one element. Then the class equation can be written as
|G|=|Z(G)|+ [G:C(x1)] +· · ·+ [G:C(xk)],
where Z(G) = {g ∈ G : gx =xg for allx ∈ G} is the center of G and C(xi) = {g ∈ G:
gxi=xig} is the centralizer subgroup ofxi.
We begin our investigation of the Sylow Theorems by examining subgroups of order p, wherepis prime. A group Gis ap-group if every element inGhas as its order a power of p, wherep is a prime number. A subgroup of a groupGis ap-subgroup if it is ap-group. Theorem 15.1 Cauchy. Let G be a finite group and p a prime such that p divides the order of G. Then Gcontains a subgroup of order p.
Proof. We will use induction on the order of G. If |G|= p, then clearly G itself is the required subgroup. We now assume that every group of order k, where p ≤ k < n and p dividesk, has an element of order p. Assume that |G|=n andp|nand consider the class equation ofG:
|G|=|Z(G)|+ [G:C(x1)] +· · ·+ [G:C(xk)].
We have two cases.
Case 1. Suppose the order of one of the centralizer subgroups, C(xi), is divisible by p for
some i, i = 1, . . . , k. In this case, by our induction hypothesis, we are done. Since C(xi)
is a proper subgroup ofGand pdivides|C(xi)|,C(xi) must contain an element of orderp.
Hence,Gmust contain an element of order p.
Case 2. Suppose the order of no centralizer subgroup is divisible by p. Then p divides
[G: C(xi)], the order of each conjugacy class in the class equation; hence, p must divide
15.1. THE SYLOW THEOREMS 173 the center of G, Z(G). Since Z(G) is abelian, it must have a subgroup of order p by the Fundamental Theorem of Finite Abelian Groups. Therefore, the center of G contains an element of order p.
Corollary 15.2. Let G be a finite group. Then Gis a p-group if and only if |G|=pn. Example 15.3. Let us consider the group A5. We know that |A5| = 60 = 22·3·5. By
Cauchy’s Theorem, we are guaranteed that A5 has subgroups of orders 2, 3 and 5. The
Sylow Theorems will give us even more information about the possible subgroups ofA5.
We are now ready to state and prove the first of the Sylow Theorems. The proof is very similar to the proof of Cauchy’s Theorem.
Theorem 15.4 First Sylow Theorem. Let G be a finite group and p a prime such that pr divides|G|. Then Gcontains a subgroup of order pr.
Proof. We induct on the order of G once again. If |G| = p, then we are done. Now suppose that the order of Gis nwith n > pand that the theorem is true for all groups of order less thann, wherep dividesn. We shall apply the class equation once again:
|G|=|Z(G)|+ [G:C(x1)] +· · ·+ [G:C(xk)].
First suppose that p does not divide [G : C(xi)] for some i. Then pr | |C(xi)|, since pr
divides|G|=|C(xi)| ·[G:C(xi)]. Now we can apply the induction hypothesis to C(xi).
Hence, we may assume that pdivides[G:C(xi)]for all i. Since pdivides|G|, the class
equation says thatpmust divide|Z(G)|; hence, by Cauchy’s Theorem,Z(G)has an element of order p, say g. Let N be the group generated by g. Clearly, N is a normal subgroup of Z(G) since Z(G) is abelian; therefore, N is normal in G since every element in Z(G)
commutes with every element inG. Now consider the factor groupG/N of order|G|/p. By the induction hypothesis, G/N contains a subgroup H of orderpr−1. The inverse image of H under the canonical homomorphismϕ:G→G/N is a subgroup of orderpr inG.
ASylow p-subgroupP of a groupGis a maximalp-subgroup ofG. To prove the other two Sylow Theorems, we need to consider conjugate subgroups as opposed to conjugate elements in a group. For a group G, let S be the collection of all subgroups of G. For any subgroupH,S is aH-set, whereH acts onS by conjugation. That is, we have an action
H× S → S defined by h·K 7→hKh−1 forK inS. The set N(H) ={g∈G:gHg−1=H}
is a subgroup of G called the the normalizer of H in G. Notice that H is a normal subgroup of N(H). In fact,N(H)is the largest subgroup of Gin which H is normal. Lemma 15.5. Let P be a Sylowp-subgroup of a finite groupG and let x have as its order a power of p. If x−1P x=P, then x∈P.
Proof. Certainlyx∈N(P), and the cyclic subgroup,⟨xP⟩ ⊂N(P)/P, has as its order a power ofp. By the Correspondence Theorem there exists a subgroupH ofN(P)containing P such that H/P = ⟨xP⟩. Since |H| = |P| · |⟨xP⟩|, the order of H must be a power of p. However, P is a Sylow p-subgroup contained in H. Since the order of P is the largest power of p dividing |G|,H =P. Therefore, H/P is the trivial subgroup and xP =P, or x∈P.
Lemma 15.6. Let H and K be subgroups of G. The number of distinct H-conjugates of K is [H :N(K)∩H].
Proof. We define a bijection between the conjugacy classes of K and the right cosets of N(K)∩Hby h−1Kh7→(N(K)∩H)h. To show that this map is a bijection, leth1, h2 ∈H
and suppose that (N(K)∩H)h1 = (N(K) ∩H)h2. Then h2h−11 ∈ N(K). Therefore,
K = h2h−11Kh1h−21 or h−11Kh1 = h−21Kh2, and the map is an injection. It is easy to
see that this map is surjective; hence, we have a one-to-one and onto map between the H-conjugates of K and the right cosets of N(K)∩H inH.
Theorem 15.7 Second Sylow Theorem. Let Gbe a finite group andp a prime dividing |G|. Then all Sylow p-subgroups of G are conjugate. That is, if P1 and P2 are two Sylow
p-subgroups, there exists a g∈G such that gP1g−1=P2.
Proof. Let P be a Sylow p-subgroup of G and suppose that |G|= prm with |P|=pr. Let
S={P =P1, P2, . . . , Pk}
consist of the distinct conjugates ofP inG. By Lemma 15.6,k= [G:N(P)]. Notice that |G|=prm=|N(P)| ·[G:N(P)] =|N(P)| ·k.
Since pr divides|N(P)|,p cannot dividek.
Given any other Sylow p-subgroup Q, we must show that Q ∈ S. Consider the Q- conjugacy classes of each Pi. Clearly, these conjugacy classes partition S. The size of the
partition containingPi is[Q:N(Pi)∩Q]by Lemma 15.6, and Lagrange’s Theorem tells us
that|Q|= [Q:N(Pi)∩Q]|N(Pi)∩Q|. Thus, [Q:N(Pi)∩Q]must be a divisor of|Q|=pr.
Hence, the number of conjugates in every equivalence class of the partition is a power of p. However, sincep does not dividek, one of these equivalence classes must contain only a single Sylow p-subgroup, sayPj. In this case,x−1Pjx=Pj for all x∈Q. By Lemma15.5,
Pj =Q.
Theorem 15.8 Third Sylow Theorem. Let G be a finite group and let p be a prime dividing the order of G. Then the number of Sylow p-subgroups is congruent to 1 (modp)
and divides |G|.
Proof. LetP be a Sylow p-subgroup acting on the set of Sylowp-subgroups, S ={P =P1, P2, . . . , Pk},
by conjugation. From the proof of the Second Sylow Theorem, the only P-conjugate of P is itself and the order of the otherP-conjugacy classes is a power of p. Each P-conjugacy class contributes a positive power of p toward |S| except the equivalence class {P}. Since |S| is the sum of positive powers ofp and1,|S| ≡1 (modp).
Now suppose that G acts on S by conjugation. Since all Sylowp-subgroups are conju- gate, there can be only one orbit under this action. ForP ∈ S,
|S|=|orbit of P|= [G:N(P)]
by Lemma 15.6. But [G : N(P)] is a divisor of |G|; consequently, the number of Sylow p-subgroups of a finite group must divide the order of the group.
15.2. EXAMPLES AND APPLICATIONS 175