**1.** Write a program to compute the number of conjugacy classes in *Sn*. What is the

largest *n*for which your program will work?

**References and Suggested Reading**

**[1]** De Bruijin, N. G. “Pólya’s Theory of Counting,” in*Applied 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 *n*divides *m, then* *G* does not necessarily possess a subgroup of order *n. For*
example, *A*4 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 for*G*to 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 *x*1*, . . . , 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*(*x*1)] +*· · ·*+ [*G*:*C*(*xk*)]*,*

where *Z*(*G*) = *{g* *∈* *G* : *gx* =*xg* for all*x* *∈* *G}* is the center of *G* and *C*(*xi*) = *{g* *∈* *G*:

*gxi*=*xig}* is the centralizer subgroup of*xi*.

We begin our investigation of the Sylow Theorems by examining subgroups of order *p,*
where*p*is prime. A group *G*is a*p-group* if every element in*G*has as its order a power of
*p, wherep* is a prime number. A subgroup of a group*G*is a*p-subgroup* if it is a*p-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*
divides*k, has an element of order* *p. Assume that* * _{|}G|*=

*n*and

*p|n*and consider the class equation of

*G:*

*|G|*=*|Z*(*G*)*|*+ [*G*:*C*(*x*1)] +*· · ·*+ [*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 of*G*and *p*divides*|C*(*xi*)*|*,*C*(*xi*) must contain an element of order*p.*

Hence,*G*must 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 *A*5. We know that *|A*5*|* = 60 = 22*·*3*·*5. By

Cauchy’s Theorem, we are guaranteed that *A*5 has subgroups of orders 2, 3 and 5. The

Sylow Theorems will give us even more information about the possible subgroups of*A*5.

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 *G*is *n*with *n > p*and that the theorem is true for all groups of
order less than*n, wherep* divides*n. We shall apply the class equation once again:*

*|G|*=*|Z*(*G*)*|*+ [*G*:*C*(*x*1)] +*· · ·*+ [*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 *p*divides[*G*:*C*(*xi*)]for all *i. Since* *p*divides*|G|*, the class

equation says that*p*must 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 in*G. Now consider the factor groupG*/*N* of order*|G|*/*p. By*
the induction hypothesis, *G*/*N* contains a subgroup *H* of order*pr−*1. The inverse image of
*H* under the canonical homomorphism*ϕ*:*G→G*/*N* is a subgroup of order*pr* in*G.*

A**Sylow***p-subgroupP* of a group*G*is a maximal*p-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*subgroup

*H,*is a

_{S}*H-set, whereH*acts on

*by conjugation. That is, we have an action*

_{S}*H× S → S*
defined by
*h·K* *7→hKh−*1
for*K* in* _{S}*.
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

*G*in 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−*1

*P x*=

*P, then*

*x∈P.*

Proof. Certainly*x∈N*(*P*), and the cyclic subgroup,* _{⟨}xP⟩ ⊂N*(

*P*)/

*P, has as its order a*power of

*p. By the Correspondence Theorem there exists a subgroupH*of

*N*(

*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*)*∩H*by *h−*1*Kh7→*(*N*(*K*)*∩H*)*h. To show that this map is a bijection, leth*1*, h*2 *∈H*

and suppose that (*N*(*K*)*∩H*)*h*1 = (*N*(*K*) *∩H*)*h*2. Then *h*2*h−*11 *∈* *N*(*K*). Therefore,

*K* = *h*2*h−*11*Kh*1*h−*21 or *h−*11*Kh*1 = *h−*21*Kh*2, 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* in*H.*

**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* *P*1 *and* *P*2 *are two Sylow*

*p-subgroups, there exists a* *g∈G* *such that* *gP*1*g−*1=*P*2*.*

Proof. Let *P* be a Sylow *p-subgroup of* *G* and suppose that * _{|}G|*=

*prm*with

*=*

_{|}P|*pr*. Let

*S*=*{P* =*P*1*, P*2*, . . . , Pk}*

consist of the distinct conjugates of*P* in*G. 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 divide}_{k.}

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 containing*Pi* 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 divide*k, one of these equivalence classes must contain only a*
single Sylow *p-subgroup, sayPj*. In this case,*x−*1*Pjx*=*Pj* for all *x∈Q. By Lemma*15.5,

*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 (mod*p*)

*and divides* *|G|.*

Proof. Let*P* be a Sylow *p-subgroup acting on the set of Sylowp-subgroups,*
*S* =*{P* =*P*1*, P*2*, . . . , Pk},*

by conjugation. From the proof of the Second Sylow Theorem, the only *P-conjugate of* *P*
is itself and the order of the other*P*-conjugacy classes is a power of *p. Each* *P*-conjugacy
class contributes a positive power of *p* toward * _{|S|}* except the equivalence class

*. Since*

_{{}P}*|S|*is the sum of positive powers of

*p*and1,

*1 (mod*

_{|S| ≡}*p*).

Now suppose that *G* acts on *S* by conjugation. Since all Sylow*p-subgroups are conju-*
gate, there can be only one orbit under this action. For*P* *∈ 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