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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 congruent to 1 (mod p). If P and Q are any two p-Sylow subgroups of G, then there exists g element of 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 element of G such that P = gQg-1.

Theorem 4 (SIII): Suppose that G has r p-Sylow subgroups. Then r|n and r congruent to 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 element of G | gx = xg for all x element of G}.

Then ZG normal 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 element of G |gSg-1 subset of S}.

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

Let g,g' element of G. We say that g is conjugate to g' if there exists x element of 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 element of C, then

C = {xgx-1 | x element of G}.

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

G = C1 union ...union Ch

and Ci intersection Cj = empty set if i not equal j.

Proposition 5: Let g element of 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) union ... union ghNG(S)

and

(2)
giNG(S) intersection gjNG(S) = empty set   (i not equal 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 element of giNG(S) intersection gjNG(S) (inot equal j), then there exist m,m' element of NG(S) such that

(3)
x = gim = gjm'.

However,

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

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

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

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

gi-1xgx-1gi = g implies (gi-1x)g(gi-1x)-1 = g
implies gi-1x element of NG(S)
implies x element of 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 normal G since x element of 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: pdoes not divide(|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 element of 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|
implies n congruent to n0 or 1 (mod p)
implies n0 or 1 congruent to 0 (mod p) (since p|n).

But (n0,p) = 1, so that pdoes not divide|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 pdoes not divide|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 union ... union 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 = summ of i to t (pa · pa)/ui,

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

n0 = sum over all i pa/ui.

Since Ui < S2, ui|pa, and thus pa/ui = pb for some b > 0 or pa/ui = 1. Since pdoes not dividen0, there exists i for which pa/ui = 1. But then pa = ui and thus, gi-1S1gi intersection S2 = S2, and therefore

gi-1S1gi superset of 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 SnormalG.

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

backwards implication Let S' be any p-Sylow subgroup of G. Then by SII, S' = gSg-1 for some g element of G. But since S normal 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 union g2N union ... union 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 element of N. Thus, gi element of 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 element of G. Now g = gin for some n element of 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 union Sg2N union ... union 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 = sum over i to v par/ci

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

n0 = rsum over i to v 1/ci = m0 sum over i to v pa/ci
(5)
= m0(1 + sum over i=2 to v pa/ci)
implies n0/m0 = 1 + sum over i=2 to v pa/ci
implies t = n/r = 1 + sum over i=2 to v pa/ci

Note that since gi-1Sgi intersection 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 sum over i=2 to v pa/ci congruent to 0 (mod p) and t congruent to (mod p). Assume, on the contrary, that i > 2 and ci = pa. Then gi-1Sgi intersection N is a p-Sylow subgroup of N. But S is a p-Sylow subgroup of N and SnormalN, since N = NG(S). Therefore, by Corollary 6, gi-1Sgi intersection N = S and thus gi-1Sgi superset of S. However, both gi-1Sgi and S contain pa elements, so that gi-1Sgi = S. Thus, gi element of N and SgiN = SN = N. However, since SgiN intersection SgjN = empty set if i not equal 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 not congruent to 1(mod q), q not congruent to 1(mod p), and let G be any abelian group of order pq. Then G isomorphic to Zp cross 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 intersection 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 not congruent to 1(mod p), and p,pqcongruent to 0(mod p), we see that P is the only p-Sylow subgroup of G and that PnormalG by Corollary 6. Similarly, QnormalG. Therefor by Theorem 6 of the section on group isomorphisms, PQ is a subgroup of G of order pq/|PintersectionQ| = pq. Thus, every element of G can be written in the form xy (x element of P, y element of Q). Finally, if x element of P, y element of Q, then (xyx-1)y-1 element of Q, since QnormalG. Moreover, xyx-1y-1 = x(yx-1y-1) element of P since PnormalG, so that xyx-1y-1 element of P intersection Q = {1} and thus xy = yx. By Theorem 3 of the section on direct products, we see that G isomorphic to P cross Q. Thus, G is abelian, since P and Q are cyclic.

Thus, for example, since 5 not congruent to 1(mod 3) and 3 not congruent to 1(mod 5), we see that the only group of order 15 is Z15 isomorphic Z5 cross 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 PnormalG and G is not simple.