Sylow theorem : Sylow theorems 

Definition

Let p be a prime number; then we define a Sylow p-subgroup of G to be a maximal p-subgroup of G (i.e., a subgroup which is a p-group, and which isn't a proper subgroup of any other p-subgroup of G). The set of all Sylow p-subgroups for a given prime p is sometimes written Syl_p(G).

Collections of subgroups which are each maximal in one sense or another are not uncommon in group theory. The surprising result here is that in the case of Syl_p(G), all members are actually isomorphic to each other; and this property can be exploited to determine other properties of G.

Sylow theorems

The following facts were first proposed and proven by Norwegian mathematician Ludwig Sylow[?] in 1872. Given a finite group G and a prime p which divides the order of G, we can write the order of G as (pⁿ · s), where n > 0 and p doesn't divide s. Then:

• There exists a Sylow p-subgroup of G, of order pⁿ.
• All Sylow p-subgroups of G are conjugate to each other (and therefore isomorphic), i.e. if H and K are Sylow p-subgroups of G, then there exists an element g in G with g^-1H'g = K.
• Let n_p be the number of Sylow p-subgroups of G.
  • n_p divides s.
  • n_p = 1 mod p.

In particular, the above implies that every Sylow p-subgroup is of the same order, pⁿ; and conversely, if a subgroup has order pⁿ, then it is a Sylow p-subgroup, and so is isomorphic to every other Sylow p-subgroup. Because of the maximality condition, if H is any p-subgroup of G, then H is a subgroup of a p-subgroup of order pⁿ.

Example applications

Let G be a group of order 15 = 3 · 5. We have that n₃ must divide 5, and n₃ = 1 mod 3. The only value satisfying these constraints is 1; therefore, there is only one subgroup of order 3, and it must be normal (since it has no distinct conjugates). Similarly, n₅ divides 3, and n₅ = 1 mod 5; thus it also has a single normal subgroup of order 5. Since 3 and 5 are coprime, the intersection of these two subgroups is trivial, and so G must be a cyclic group. Thus, there is only 1 group of order 15 (up to isomorphism); namely Z₁₅.

For a more complex example, we can show that there are no simple groups of order 350. If |G| = 350 = 2 · 5² · 7, then n₅ must divide 14 ( = 2 · 7), and n₅ = 1 mod 5. Therefore n₅ = 1 (since neither 6 nor 11 divides 14), and thus G must have a normal subgroup of order 5², and so cannot be simple.

Proofs

The proofs of the Sylow theorem exploit the notion of group action in various creative ways. The group G acts on itself or on the set of its p subgroups in various ways, and each such action can be exploited to prove one of the Sylow theorem.

The proofs rely on several facts about the conjugacy classes of elements and subsets of G:

• |Cl(S)| = [G:N(S)]
  • where S is any subset of G, Cl(S) is the conjugacy class of sets T = g ^-1Sg for some g in G, and N(S) is the normalizer of S in G.
• |Cl(x)| = [G:C(x)]
  • where x is some element of G, Cl(x) is the conjugacy class of elements y = g ^-1xg for some g in G, and C(x) is the centralizer of x in G.
• |G| = Z(G) + ∑_i {Cl(x_i)}
  • the class equation of G, where Z(G) is the center of G, and the sum is taken over a representative set {x_i} of the conjugacy classes of G.

First, given a group of order pⁿ·s, we can prove that G has a subgroup K of order pⁿ using induction.

• Induct on the order of G. If G has a subgroup whose order is divisible by pⁿ then, by the inductive hypothesis, that subgroup has a subgroup of K of order pⁿ; and, therefore, so does G.
• So assume the contrary. Then for any element x which isn't in the center of G, Z(G), we find that the centralizer of x, C(x), is a subgroup of G; and so will not be divisible by pⁿ.
• By Lagrange's theorem, |G| = |C(x)|·[G:C(x)]; since |G| is divisible by pⁿ, [G:C(x)] is divisible by p; thus as noted above, |Cl(x)| = [G:C(x)] is divisible by p.
• Thus, each |Cl(x_i)| in the class equation is divisible by p, as is |G|; therefore, |Z(G)| is divisible by p. |Z(G)| is an abelian group; so since its order is divisible by p, it has a subgroup H of order p (see finitely generated abelian group). In fact, H (being in the center of G) is a normal subgroup of G.
• Since H is normal, the factor group G/H exists and will have order divisible by p^n-1. By the inductive hypothesis, G/H will then have a subgroup L which has order p^n-1.
• By the lattice theorem[?], there is a unique subgroup K of G such that K/H = L; since H has order p, then K has order pⁿ.

K is necessarily maximal by Lagrange, so it is a Sylow p-subgroup; and therefore Syl_p(G) isn't empty.

Next, we use the existence of K to prove that Cl(K), the conjugacy class of K, is in fact Syl_p(G); so every Sylow p-subgroup is conjugate to K. As a bonus, we simultaneously find that n_p = 1 mod p.

• For the following argument, let H be any Sylow p-subgroup (not necessarily distinct from K). We can further partition Cl(K) using inner automorphisms induced by H as follows:
  • Given members L and M of Cl(K), let M be in Cl'(L) iff there exists some h in H such that h ^-1Mh = L.
• It is easily seen that Cl'(L) is an equivalence class; and that Cl(K) is the disjoint union of these classes; so |Cl(K)| = ∑_i(|Cl'(L_i)|), the sum of the order of those classes over some representative subset {L_i} of Cl(K).
• Given a, b in H, and some L in Cl(K), a ^-1La = b ^-1Lb if and only if aN_H(L) = bN_H(L), where N_H(L) is the normalizer of L in H.
• Therefore, |Cl'(L)| = [H:N_H(L)]. Note that N_H(L) = (H ∩ N_G(L)); so |Cl(L)| = [H:(H ∩ N(L))]. Since H is a p-group, (H ∩ N(L)) is also a p-group (by Lagrange); and so [H:(H ∩ N(L_i))] = |Cl'(L_i)| = p^m for some m ≥ 0 for all L_i in Cl(K)
• At this point, it would be nice to know a bit more about (H ∩ N(L_i)). Given any L in Cl(K), consider an element x in H, where x isn't in L. Then x isn't in N(L), and so H ≠ (H ∩ N(L)):
  • Assume x is in N(L). Recalling that L is a normal subgroup of N(L), consider the element xL/L of the factor group N(L)/L.
  • Since H is a p-group, x^q = e, where q is some power of p. Therefore, we must have that xL/L has order dividing a power of p. But since |L| = pⁿ, p cannot divide |N(L)/L|; and therefore xL/L is the identity in N(L)/L. Thus x must be a member of L; and thus a member of N(L), contrary to assumption.
• Thus, if L in Cl(K) and H ≠ L, then |Cl'(L)| = [H:(H ∩ N(L))] = p^m for some m>0. On the other hand, if H = L in Cl(K), then |Cl'(L)| = |Cl'(H)| = [H:(H ∩ N(H))] = [H:H] = 1.
• Since Cl(K) is the disjoint union of the Cl'(L_i)s, |Cl(K)| = kp for some k if H isn't in Cl(K), and equals kp+1 if H is in Cl(K).
• We stipulated at the beginning that H was not necessarily distinct from K. Since K is in Syl_p(G), then we can take H = K to show that |Cl(K)| = 1 mod p; at the same time, this then provides a contradiction if we assume that H isn't in Cl(K) (as its order cannot be both 0 and 1 mod p).
• Therefore, H is in Cl(K), and so Syl_p(G) = Cl(K).
• Thus every Sylow p-subgroup is conjugate to K; and n_p = |Cl(K)| = 1 mod p.

The other numerical fact about n_p follows almost immediately.

• Since n_p = 1 mod p, n_p doesn't divide pⁿ; since n_p = [G : N(K)], it divides |G|, so it must divide s.

It is worth pointing out that the above argument as regards conjugacy is valid as long as |Cl(K)| = [G:N(K)] is finite; so we can state an infinite analog of the Sylow theorems:

• If K is a Sylow p-subgroup of G, and n_p = |Cl(K)| is finite, then every Sylow p-subgroup is conjugate to K, and n_p = 1 mod p.

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