let G be a group of order n, let p be a prime dividing n, and let n = pan0, (n0,p) = 1. A subgroup of G of order pa is called a p-Sylow subgroup.

Suppose that G has r p-Sylow subgroups. Then r|n and r 1 (mod p). If P and Q are any two p-Sylow subgroups of G, then there exists g G such that P = gQg-1. Let G be an abelian group whose order is divisible by a prime p. Then G contains an element of order p. G contains a p-Sylow subgroup A group is considered cyclic if it is generated by a single element. commutative

### Sylow's Theorems

We now come to the three theorems of Sylow, which are among the most important results in the study of finite groups. Recall that earlier we proved Lagrange's theorem, which asserts that if G is a finite group and H is a subgroup of G, then H divides the order of G. Note, however, that the converse of Lagrange's theorem is false. That is, if G has order n and if m|n, then there does not necessarily exist a subgroup of G of order m. For example, A4 is a group of order 12 which has no subgroups of order 6. Sylow's theorem comes about as close as one can come to providing a converse to Lagrange's theorem.

Throughout this section, let G be a group of order n, let p be a prime dividing n, and let n = pan0, (n0,p) = 1.

Definition 1: A subgroup of G of order pa is called a p-Sylow subgroup.

Our main results are the following theorems of the Norwegian mathematician Sylow, which we will denote SI, SII, and SII.

Theorem 2 (SI): G contains a p-Sylow subgroup.

Theorem 3 (SII): If P and Q are any two p-Sylow subgroups of G, then there exists g G such that P = gQg-1.

Theorem 4 (SIII): Suppose that G has r p-Sylow subgroups. Then r|n and r 1 (mod p).

In order to prove these remarkable results, a number of preliminaries must be introduced.

If G is any group, then the center of G, denoted ZG, is defined as

ZG = {g G | gx = xg for all x G}.

Then ZG G. Moreover, G is abelian if and only if ZG = G.

If S is any subset of G, then the normalizer of S, denoted NG(S), is defined by

NG(S) = {g G |gSg-1 S}.

Then NG(S) < G. Moreover, if S is a subgroup of G, then S NG(S).

Let g,g' G. We say that g is conjugate to g' if there exists x G such that g = xg'x-1. It is immediate that the relation of conjugacy is an equivalence relation on G. The equivalence classes of G with respect to conjugacy are called conjugacy classes. If C is a conjugacy class of G, and g C, then

C = {xgx-1 | x G}.

Moreover, if C1,...,Ch are conjugacy classes of G, then

G = C1 ... Ch

and Ci Cj = if i j.

Proposition 5: Let g G, S = {g}, hg = the number of elements in the conjugacy class containing g. Then

|NG(S)| = n/hg.

Proof: Let C denote the conjugacy class containing g, h = hg, C = {g1gg1-1,...,ghggh-1}. Let us show that

(1)
G = g1NG(S) ... ghNG(S)

and

(2)
giNG(S) gjNG(S) =    (i j).

This will suffice to prove the proposition. For giNG(S) contains |NG(S)| elements for i = 1,...,h and by (2), none of these sets overlap. Therefore, by (1), n = hg · |Ng(S)|. If x giNG(S) gjNG(S) (i j), then there exist m,m' NG(S) such that

(3)
x = gim = gjm'.

However,

NG(S) = {x G | xgx-1 = g}
= {x G | xg = gx}.

But, by (3), gj-1gi = m'm-1 N(S), so that

gj-1gig = ggj-1gi giggi-1 = gjggj-1,

which contradicts the choice of gi's. Thus (2) is proved. Let x G. Then xgx-1 = giggi-1 for some i, so that

gi-1xgx-1gi = g (gi-1x)g(gi-1x)-1 = g
gi-1x NG(S)
x giNG(S).

Thus (1) is proved.

Let us now give proofs of the Sylow theorems.

Proof of SI: The result is clearly true if n = 2, so let us proceed by induction on n. Let us consider three cases:

Case A: p|(|ZG|). In this case, by Proposition 7 of the section on abelian groups, ZG contains an element x of order p. Let P = [x]. Then P G since x ZG and thus G/P is defined and clearly of order pa-1n0. By induction G/P has a subgroup W of order pa-1. But W is of the form H/P for some subgroup H of G. Then H is a subgroup of G of order pa.

Case B: p(|ZG|) and 1 < |ZG| < n0. In this case, G/ZG is of order pan' for some n' < n0, (n',p) = 1. By induction, there exists a subgroup W of G/ZG of order pa But W is of the form H/ZG for some subgroup H of G. But the H is of order pan' < n, so that by induction, H has a subgroup H0 of order pa.

Case C: |ZG| = n0 or 1. Let C1,...,Ct denote the conjugacy classes of G. Note that Ci = {gi} if and only if gi Zg. Thus, let ZG = {g1,...,gs} and number the Ci so that Ci = {gi} (1 < i < s). Then, for i > s, Ci contains more than one element. Assume that p|(|Ci|) for all i > s. Then

n = |ZG| + |Cs+1| + ... + |Ct|
n n0 or 1 (mod p)
n0 or 1 0 (mod p) (since p|n).

But (n0,p) = 1, so that p|Ci| for some i > s. Let c be any element belonging to Ci. Then, if S = {c}, Proposition 5 implies that |NG(S)| = n/|Ci| = pan' [n' < n0,(n',p) = 1], since |Ci| > 1 and p|Ci|. Therefore, by induction, NG(S) has a subgroup of order pa.

Proof of SII: Let S1 and S2 be two p-Sylow subgroups of G, and let

G = S1g1S2 ... S1gtS2

be the decomposition of G relative to the two subgroups S1 and S2. Then since |S1| = |S2| = pa, Theorem 1 of the section on decomposition with respect to two subgroups asserts that

n = (pa · pa)/ui,

where ui = the order of Ui = gi-1S1gi S2. But since n = pan0, we see that

n0 = pa/ui.

Since Ui < S2, ui|pa, and thus pa/ui = pb for some b > 0 or pa/ui = 1. Since pn0, there exists i for which pa/ui = 1. But then pa = ui and thus, gi-1S1gi S2 = S2, and therefore

gi-1S1gi S2.

But both gi-1S1gi and S2 have order pa, so that we conclude that

gi-1S1gi = S2.

Corollary 6: Let S be a p-Sylow subgroup of G. Then S is the only p-Sylow subgroup of G if and only if SG.

Proof: Let g G. Then gSg-1 is a p-Sylow subgroup of G, so that gSg-1 = S. Thus, SG.

Let S' be any p-Sylow subgroup of G. Then by SII, S' = gSg-1 for some g G. But since S G, gSg-1 = S, so that S' = S.

Proposition 7: Let S be a p-Sylow subgroup of G, N = NG(S). Then the number of p-Sylow subgroups of G is [G:N].

Proof: Let

G = g1N g2N ... giN,   t = [G:N]

be a coset decomposition. Then

(4)
g1Sg1-1,g2Sg2-1,...,giSgi-1

are p-Sylow subgroups of G. These are distinct, since if

giSgi-1 = gjSgj-1,

we see that (gj-1gi)S(gj-1gi)-1 = S, so that gj-1gi N. Thus, gi gjN and giN = gjN, so that i = j. Thus, the p-Sylow subgroups (4) are distinct. Let S' be any p-Sylow subgroup. Then S' = gSg-1 for some g G. Now g = gin for some n N. Thus, S' is contained in the list (4), which must therefore include all p-Sylow subgroups of G.

Proof of SIII: Let t be the number of p-Sylow subgroups of G. Then, by Proposition 7, t = [G:N], where N = NG(S) and S is a fixed p-Sylow subgroup. In particular, n = t|N|, so that t|n. Let

G = Sg1N Sg2N ... SgvN

be the decomposition of G with respect to the two subgroups S and N. Without loss of generality, assume that g1 = 1, so that Sg1N = N. Let |N| = r = pam0, (m0,p) = 1. By Theorem 1 of the section on decomposition with respect to two subgroups, we have

n = pan0 = par/ci

where ci = |gi-1Sgi N|, Note that since g1 = 1, c1 = |S N| = |S| = pq. Therefore,

n0 = r 1/ci = m0 pa/ci
(5)
= m0(1 + pa/ci)
n0/m0 = 1 + pa/ci
t = n/r = 1 + pa/ci

Note that since gi-1Sgi N is a subgroup of gi-1Sgi, which has order pa, we see that ci = pb for some b(0 < b < a). We assert that ci < pa for i = 2,...,v. This fact in conjunction with (4), suffices to complete the proof, since then pa/ci 0 (mod p) and t (mod p). Assume, on the contrary, that i > 2 and ci = pa. Then gi-1Sgi N is a p-Sylow subgroup of N. But S is a p-Sylow subgroup of N and SN, since N = NG(S). Therefore, by Corollary 6, gi-1Sgi N = S and thus gi-1Sgi S. However, both gi-1Sgi and S contain pa elements, so that gi-1Sgi = S. Thus, gi N and SgiN = SN = N. However, since SgiN SgjN = if i j and since Sg1N = N, we see that i = 1, which is a contradiction to the choice of i. Thus, ci < pa.

Let us now give a few examples of applications of the Sylow theorems.

Proposition 8: Let p and q be distinct primes, p 1(mod q), q 1(mod p), and let G be any abelian group of order pq. Then G Zp Zq.

Proof: By the fundamental theorem of abelian groups, it suffices to show that G is abelian. Let P and Q, be respectively a p-Sylow and a q-Sylow subgroup of G. Since every element of P except the identity has order p and every element of Q except the identity has order q, we see that P Q = {1}. The number of p-Sylow subgroups divides pq and is therefore 1, p, q, pq. Moreover, this number is congruent to 1 (mod p). Since q 1(mod p), and p,pq 0(mod p), we see that P is the only p-Sylow subgroup of G and that PG by Corollary 6. Similarly, QG. Therefor by Theorem 6 of the section on group isomorphisms, PQ is a subgroup of G of order pq/|PQ| = pq. Thus, every element of G can be written in the form xy (x P, y Q). Finally, if x P, y Q, then (xyx-1)y-1 Q, since QG. Moreover, xyx-1y-1 = x(yx-1y-1) P since PG, so that xyx-1y-1 P Q = {1} and thus xy = yx. By Theorem 3 of the section on direct products, we see that G P Q. Thus, G is abelian, since P and Q are cyclic.

Thus, for example, since 5 1(mod 3) and 3 1(mod 5), we see that the only group of order 15 is Z15 Z5 Z3.

Example 1: No group of order 20 is simple.

Let G be a group of order 20. The number of 5-Sylow subgroups is one of 1,2,4,5,10,20. However, the number of 5-Sylow subgroups is congruent to 1(mod 5), so that this number is 1. Thus, if P is the unique 5-Sylow subgroup, then PG and G is not simple.