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Sylow theorem

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The Sylow theorems of group theory form a partial converse to the theorem of Lagrange, which states that if H is a subgroup of a finite group G, then the order of H divides the order of G. The Sylow theorems guarantee, for certain divisors of the order of G, the existence of corresponding subgroups, and give information about the number of those subgroups.

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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 is not 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 Sylp(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 Sylp(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 does not 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 np be the number of Sylow p-subgroups of G.
    • np divides s.
    • np = 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.


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(xi)}
    • the class equation of G, where Z(G) is the center of G, and the sum is taken over a representative set {xi} 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 is not 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(xi)| 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 pn-1. By the inductive hypothesis, G/H will then have a subgroup L which has order pn-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 Sylp(G) is not empty.

Next, we use the existence of K to prove that Cl(K), the conjugacy class of K, is in fact Sylp(G); so every Sylow p-subgroup is conjugate to K. As a bonus, we simultaneously find that np = 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'(Li)|), the sum of the order of those classes over some representative subset {Li} of Cl(K).
  • Given a, b in H, and some L in Cl(K), a -1La = b -1Lb if and only if aNH(L) = bNH(L), where NH(L) is the normalizer of L in H.
  • Therefore, |Cl'(L)| = [H:NH(L)]. Note that NH(L) = (H ∩ NG(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(Li))] = |Cl'(Li)| = pm for some m ≥ 0 for all Li in Cl(K)
  • At this point, it would be nice to know a bit more about (H ∩ N(Li)). Given any L in Cl(K), consider an element x in H, where x is not in L. Then x is not 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, xq = 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 HL, then |Cl'(L)| = [H:(H ∩ N(L))] = pm 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'(Li)s, |Cl(K)| = kp for some k if H is not 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 Sylp(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 is not in Cl(K) (as its order cannot be both 0 and 1 mod p).
  • Therefore, H is in Cl(K), and so Sylp(G) = Cl(K).
  • Thus every Sylow p-subgroup is conjugate to K; and np = |Cl(K)| = 1 mod p.

The other numerical fact about np follows almost immediately.

  • Since np = 1 mod p, np does not divide pn; since np = [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 np = |Cl(K)| is finite, then every Sylow p-subgroup is conjugate to K, and np = 1 mod p.

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