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 = p^an₀, (n₀,p) = 1.

Definition 1: A subgroup of G of order p^a 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 Z_G, is defined as

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

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

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

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

Then N_G(S) < G. Moreover, if S is a subgroup of G, then S N_G(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,...,C_h are conjugacy classes of G, then

G = C₁ ... C_h

and C_i C_j = if i j.

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

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

G = g₁N_G(S) ... g_hN_G(S)

and

g_iN_G(S) g_jN_G(S) =    (i j).

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

However,

N_G(S) = {x G | xgx^-1 = g}

= {x G | xg = gx}.

But, by (3), g_j^-1g_i = m'm^-1 N(S), so that

g_j^-1g_ig = gg_j^-1g_i g_igg_i^-1 = g_jgg_j^-1,

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

g_i^-1xgx^-1g_i = g (g_i^-1x)g(g_i^-1x)^-1 = g

g_i^-1x N_G(S)

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

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

Case C: |Z_G| = n₀ or 1. Let C₁,...,C_t denote the conjugacy classes of G. Note that C_i = {g_i} if and only if g_i Z_g. Thus, let Z_G = {g₁,...,g_s} and number the C_i so that C_i = {g_i} (1 < i < s). Then, for i > s, C_i contains more than one element. Assume that p|(|C_i|) for all i > s. Then

n n₀ or 1 (mod p)

n₀ or 1 0 (mod p) (since p|n).

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

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

G = S₁g₁S₂ ... S₁g_tS₂

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

n = (p^a · p^a)/u_i,

where u_i = the order of U_i = g_i^-1S₁g_i S₂. But since n = p_an₀, we see that

n₀ = p^a/u_i.

Since U_i < S₂, u_i|p^a, and thus p^a/u_i = p^b for some b > 0 or p^a/u_i = 1. Since pn₀, there exists i for which p^a/u_i = 1. But then p^a = u_i and thus, g_i^-1S₁g_i S₂ = S₂, and therefore

g_i^-1S₁g_i S₂.

But both g_i^-1S₁g_i and S₂ have order p^a, so that we conclude that

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 = N_G(S). Then the number of p-Sylow subgroups of G is [G:N].

Proof: Let

G = g₁N g₂N ... g_iN,   t = [G:N]

be a coset decomposition. Then

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

we see that (g_j^-1g_i)S(g_j^-1g_i)^-1 = S, so that g_j^-1g_i N. Thus, g_i g_jN and g_iN = g_jN, 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 = g_in 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 = N_G(S) and S is a fixed p-Sylow subgroup. In particular, n = t|N|, so that t|n. Let

G = Sg₁N Sg₂N ... Sg_vN

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 = p^am₀, (m₀,p) = 1. By Theorem 1 of the section on decomposition with respect to two subgroups, we have

n = p^an₀ = p^ar/c_i

where c_i = |g_i^-1Sg_i N|, Note that since g₁ = 1, c₁ = |S N| = |S| = p^q. Therefore,

n₀ = r 1/c_i = m₀ p^a/c_i

= m₀(1 + p^a/c_i)

n₀/m₀ = 1 + p_a/c_i

t = n/r = 1 + p^a/c_i

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

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 Z_p Z_q.

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.