Sylow theorems

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In mathematics, specifically group theory, the Sylow theorems, named after Ludwig Sylow, form a partial converse to Lagrange's theorem, 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.

Contents 1 Definition 2 Sylow theorems 3 Example applications 4 Proof of the Sylow theorems 5 Finding a Sylow subgroup 6 References

[edit] Definition

Let p be a prime number; then we define a Sylow p-subgroup (sometimes p-Sylow 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 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.

[edit] Sylow theorems

The following theorems were first proposed and proven by Norwegian mathematician Ludwig Sylow in 1872, and published in Mathematische Annalen. 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.

Theorem 1: There exists a Sylow p-subgroup of G, of order pⁿ.

The following less general version of theorem 1 was first proved by Cauchy.

Corollary: Given a finite group G and a prime number p dividing the order of G, then there exists an element of order p in G .

Theorem 2: If H is a subgroup of G and |H|=p^b, P is p-Sylow subgroup of G, then there exists an element a in G such that aHa^-1 is a subgroup of P. In particular, all Sylow p-subgroups of G are conjugate (and therefore isomorphic) to each other, i.e. if H and K are Sylow p-subgroups of G, then there exists an element g in G with g⁻¹Hg = K.

Theorem 3: Let n_p be the number of Sylow p-subgroups of G.

• n_p divides s.
• n_p = 1 mod p.
• n_p = |G : N_G(P)|, where P is any Sylow p-subgroup of G and N_G denotes the normalizer.

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

A very important consequence of Theorem 3 is that the condition n_p = 1 is equivalent to saying that the Sylow p-subgroup of G is a normal subgroup. (There are groups which have normal subgroups but no normal Sylow subgroups, such as S₄.)

There is an analogue of the Sylow theorems for infinite groups. We define a Sylow p-subgroup in an infinite group to be a p-subgroup (that is, every element in it has p-power order) which is maximal for inclusion among all p-subgroups in the group. Such subgroups exist by Zorn's lemma.

Theorem: 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, where Cl(K) denotes the conjugacy class of K.

[edit] 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 Z/15Z.

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.

[edit] Proof of the Sylow theorems

The proofs of the Sylow theorems 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 theorems. The following proofs are based on combinatorial arguments of H. Wielandt published in 1959. In the following, we use a | b as notation for "a divides b" and a b for the negation of this statement.

Theorem 1: A finite group G whose order |G| is divisible by a prime power p^k has a subgroup of order p^k.

Proof: Let |G| = p^km, and let p^r be the highest power of p that divides m. Let Ω denote the set of subsets of G of size p^k. Clearly:

and it is not hard to see ( e. g. using division properties of factorials ) that p^r is precisely the maximal power of p that divides |Ω|, in particular, p^r+1 . Let G act on Ω by left multiplication. It follows by the choice of r that there is an element A ∈ Ω with an orbit θ = GA such that p^r+1 |θ|. Now |θ| = |GA| = [G : G_A] where G_A denotes the stabilizer subgroup of the set A, hence p^k | |G_A| so p^k ≤ |G_A|. On the other hand, fix an element a ∈ A . Any g ∈ G_A must be equal to ba^-1 for some b ∈ A and therefore |A| ≥ |G_A|, hence |G_A| = p^k and G_A is a desired subgroup.

Lemma: Let G be a finite p-group, let G act on a finite set Ω, and let Ω₀ denote the set of points of Ω that are fixed under the action of G. Then |Ω| ≡ |Ω₀| mod p.

Proof: Write Ω as a disjoint sum of its orbits under G. Any element x ∈ Ω not fixed by G will lie in an orbit of order |G|/|G_x| (where G_x denotes the stabilizer), which is a multiple of p by assumption. The result follows immediately.

Theorem 2: If H is a subgroup of G and |H|=p^b, P is p-Sylow subgroup of G, then there exists an element a in G such that aHa^-1 is a subgroup of P. In particular, 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⁻¹Hg = K.

Proof: Let Ω be the set of left cosets of P in G and let H act on Ω by left multiplication. Applying the Lemma to H on Ω, we see that |Ω₀| ≡ |Ω| = [G : P] mod p. Now p [G : P] by definition so p |Ω₀|, hence in particular |Ω₀| ≠ 0 so there exists some gP ∈ Ω₀. It follows that hgP = gP so g⁻¹hgP = P, g⁻¹hg ∈ P, and thus h ∈ gPg⁻¹ ∀ h ∈ H, so that H ≤ gPg⁻¹ for some g ∈ G. Now if H is a Sylow p-subgroup, |H| = |P| = |gPg⁻¹| so that H = gPg⁻¹ for some g ∈ G.

Theorem 3: Let q denote the order of any Sylow p-subgroup of a finite group G. Then n_p | |G|/q and n_p ≡ 1 mod p.

Proof: By Theorem 2, n_p = [G : N_G(P)], where P is any such subgroup, and N_G(P) denotes the normalizer of P in G, so this number is a divisor of |G|/q. Let Ω be the set of all Sylow p-subgroups of G, and let P act on Ω by conjugation. Let Q ∈ Ω₀ and observe that then Q = xQx⁻¹ for all x ∈ P so that P ≤ N_G(Q). By Theorem 2, P and Q are conjugate in N_G(Q) in particular, and Q is normal in N_G(Q), so then P = Q. It follows that Ω₀ = {P} so that, by the Lemma, |Ω| ≡ |Ω₀| = 1 mod p.

[edit] Finding a Sylow subgroup

The problem of finding a Sylow subgroup of a given group is an important problem in computational group theory. In permutation groups, it has been proven by William Kantor that a Sylow p-subgroup can be found in polynomial time of the input (the degree of the group times the number of generators).

[edit] References

• Florian Kammüller and Lawrence C. Paulson. "A Formal Proof of Sylow's Theorem: An Experiment in Abstract Algebra with Isabelle HOL". University of Cambridge, UK. 2000. link
• H. Wielandt. "Ein Beweis für die Existenz der Sylowgruppen". Archiv der Mathematik, 10:401-402, 1959.