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In mathematics, specifically in the field of finite group theory, the Sylow theorems are a collection of theorems named after the Norwegian mathematician Ludwig Sylow.[1] The three main Sylow theorems are the Sylow E theorem, the Sylow D theorem, and the Sylow C theorem, sometimes referring to the "existence", "development" and "conjugate" theorem respectively. Another theorem gives the number of Sylow p-subgroups of a group for fixed prime p; this is sometimes referred to as the Sylow "counting" theorem. Given any prime number and any finite group, the Sylow E theorem asserts the existence of a subgroup of this finite group, having order the power of this prime. The Sylow D theorem asserts the equivalence between maximality of a p-subgroup of a finite group, and that it is a Sylow subgroup of the group. Finally, the Sylow C theorem asserts that all Sylow p-subgroups of a group (for fixed prime p) are conjugate to each other.

The Sylow theorems and the notion of a Sylow p-subgroup (for fixed prime p) form a fundamental part of finite group theory and have very important applications in the Classification of finite simple groups. In some sense, they assert a partial converse to Lagrange's theorem and give detailed information about the number of subgroups of fixed order a group contains.

Sylow p-subgroups

Given a prime number p, a p-group (also p-primary group) is a group such that for each element g of the group there exists a nonnegative integer n such that g to the power pn is equal to the identity element. (In other words, g has order pn). A Sylow p-subgroup (sometimes p-Sylow subgroup) of a group G is 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).

Sylow theorems

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 have the largest possible order: if |G| = pnm with n > 0 where p does not divide m, then any Sylow p-subgroup P has order |P| = pn. That is, P is a p-group and gcd(|G:P|, p) = 1. These properties can be exploited to further analyze the structure of G.

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

Theorem 1: For any prime factor p with multiplicity n of the order of a finite group G, there exists a Sylow p-subgroup of G, of order pn.

The following weaker 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: Given a finite group G and a prime number p, 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 p be a prime factor with multiplicity n of the order of a finite group G, so that the order of G can be written as pn · m, where n > 0 and p does not divide m. Let np be the number of Sylow p-subgroups of G. Then the following hold:

• np divides m, which is the index of the Sylow p-subgroup in G.
• np ≡ 1 mod p.
• np = |G : NG(P)|, where P is any Sylow p-subgroup of G and NG denotes the normalizer.

Consequences

The Sylow theorems imply that for a prime number p 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 np = 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.)

Sylow theorems for infinite groups

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 np = |Cl(K)| is finite, then every Sylow p-subgroup is conjugate to K, and np ≡ 1 mod p, where Cl(K) denotes the conjugacy class of K.

Example applications

Let G be a group of order 15 = 3 · 5 and n3 be the number of Sylow 3-subgroups. Then $n_3\mid 5$ and $n_3\equiv1\pmod3$. 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 must divide 3, and n5 must equal 1 (mod 5); thus it must also have 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 the internal direct product of groups of order 3 and 5, that is the cyclic group of order 15. Thus, there is only one group of order 15 (up to isomorphism).

A more complex example involves the order of the smallest simple group which is not cyclic. Burnside's paqb theorem states that if the order of a group is the product of two prime powers, then it is solvable, and so the group is not simple, or is of prime order and is cyclic. This rules out every group up to order 30 (= 2 · 3 · 5).

If G is simple, and |G| = 30, then n3 must divide 10 ( = 2 · 5), and n3 must equal 1 (mod 3). Therefore n3 = 10, since neither 4 nor 7 divides 10, and if n3 = 1 then, as above, G would have a normal subgroup of order 3, and could not be simple. G then has 10 distinct cyclic subgroups of order 3, each of which has 2 elements of order 3 (plus the identity). This means G has at least 20 distinct elements of order 3. As well, n5 = 6, since n5 must divide 6 ( = 2 · 3), and n5 must equal 1 (mod 5). So G also has 24 distinct elements of order 5. But the order of G is only 30, so a simple group of order 30 cannot exist.

Next, suppose |G| = 42 = 2 · 3 · 7. Here n7 must divide 6 ( = 2 · 3) and n7 must equal 1 (mod 7), so n7 = 1. So, as before, G can not be simple.

On the other hand for |G| = 60 = 22 · 3 · 5, then n3 = 10 and n5 = 6 is perfectly possible. And in fact, the smallest simple non-cyclic group is A5, the alternating group over 5 elements. It has order 60, and has 24 cyclic permutations of order 5, and 20 of order 3.

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 Helmut Wielandt published in 1959. In the following, we use a | b as notation for "a divides b" and a $\nmid$ b for the negation of this statement.

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

Proof: Let |G| = pkm, and let Ω denote the set of subsets of G of size pk. Clearly:

$|\Omega | ={p^km \choose p^k}\mathrm{.}$

Let νp(|Ω|)p(m)= r and G act on Ω by left multiplication. Then,

$|\Omega | =\sum_{[o],\ o\in\Omega}|Go|\mathrm{.}$

where the sum is taken over the equivalence classes of Ω under the action of G. Thus, there exists an element A∈ Ω, such that νp(|GA|) ≤ νp(|Ω|). Let its orbit θ = GA, and by construction pr+1 $\nmid$ |θ|. Now |θ| = |GA| = [G : GA] where GA denotes the stabilizer subgroup of the set A, hence pk | |GA| so pk ≤ |GA|. On the other hand, fix an element a ∈ A . For any two g,h ∈ GA we have that ga = ha implies g=h, because a ∈ A ⊆ G means that a-1 ∈ G, whereupon one may cancel on the right. Thus the function [gga] maps GA to A injectively; therefore |A| ≥ |GA|, hence |GA| = pk and GA 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|/|Gx| (where Gx denotes the stabilizer), which is a multiple of p by assumption. The result follows immediately.

Theorem 2: If H is a p-subgroup of G and P is a Sylow p-subgroup of G, then there exists an element g in G such that g−1HgP. 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 $\nmid$ [G : P] by definition so p $\nmid$0|, hence in particular |Ω0| ≠ 0 so there exists some gP ∈ Ω0. It follows that for some gG and ∀ hH we have hgP = gP so g−1hgPP and therefore g−1HgP. Now if H is a Sylow p-subgroup, |H| = |P| = |gPg−1| so that H = gPg−1 for some gG.

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

Proof: By Theorem 2, np = [G : NG(P)], where P is any such subgroup, and NG(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 xP so that PNG(Q). By Theorem 2, P and Q are conjugate in NG(Q) in particular, and Q is normal in NG(Q), so then P = Q. It follows that Ω0 = {P} so that, by the Lemma, |Ω| ≡ |Ω0| = 1 mod p.

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).

Notes

1. ^ L. Sylow, "Théorèmes sur les groupes de substitutions," Mathematische Annalen, vol. 5, pages 584-594 (1872). Available on-line (in French): http://www.springerlink.com/content/u4v5u2273530123r/fulltext.pdf .

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.