Sylow theorem

The Sylow theorems of group theory, 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.
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Definition
Let p be a prime number; then we define a Sylow psubgroup of G to be a maximal psubgroup of G (i.e., a subgroup which is a pgroup, and which is not a proper subgroup of any other psubgroup of G). The set of all Sylow psubgroups 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 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 (p^{n} · s), where n > 0 and p does not divide s.
Theorem 1: There exists a Sylow psubgroup of G, of order p^{n}.
Theorem 2: All Sylow psubgroups of G are conjugate to each other (and therefore isomorphic), i.e. if H and K are Sylow psubgroups of G, then there exists an element g in G with g^{−1}Hg = K.
Theorem 3: Let n_{p} be the number of Sylow psubgroups of G.
 n_{p} divides s.
 n_{p} = 1 mod p.
In particular, the above implies that every Sylow psubgroup is of the same order, p^{n}; conversely, if a subgroup has order p^{n}, then it is a Sylow psubgroup, and so is isomorphic to every other Sylow psubgroup. Due to the maximality condition, if H is any psubgroup of G, then H is a subgroup of a psubgroup of order p^{n}. There is also an infinite analog of the Sylow theorems:
Theorem: If K is a Sylow psubgroup of G, and n_{p} = Cl(K) is finite, then every Sylow psubgroup is conjugate to K, and n_{p} = 1 mod p, where Cl(K) denotes the conjugacy class of K.
Example applications
Let G be a group of order 15 = 3 · 5. We have that n_{3} must divide 5, and n_{3} = 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, n_{5} divides 3, and n_{5} = 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_{15}.
For a more complex example, we can show that there are no simple groups of order 350. If G = 350 = 2 · 5^{2} · 7, then n_{5} must divide 14 ( = 2 · 7), and n_{5} = 1 mod 5. Therefore n_{5} = 1 (since neither 6 nor 11 divides 14), and thus G must have a normal subgroup of order 5^{2}, and so cannot be simple.
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 psubgroups 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 <math>\nmid<math> 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^{k}m, and let p^{r} be chosen such that no higher power of p divides m. Let Ω denote the set of subsets of G of size p^{k} and note that Ω = <math>{p^km \choose p^k}\mathrm{,}<math> and furthermore that p^{r+1} <math>\nmid<math> <math>{p^km \choose p^k}<math> by the choice of r. Let G act on Ω by left multiplication. It follows that there is an element A ∈ Ω with an orbit θ = A^{G} such that p^{r+1} <math>\nmid<math> θ. Now θ = A^{G} = [G : G_{A}] where G_{A} denotes the stabilizer subgroup of the set A, hence p^{k}  G_{A} so p^{k} ≤ G_{A}. Note that the elements ga ∈ A for a ∈ A are distinct under the action of G_{A} so that A ≥ G_{A} and therefore G_{A} = p^{k}. Then G_{A} is the desired subgroup.
Lemma: Let G be a finite pgroup, let G act on a finite set Ω, and let Ω_{0} denote the set of points of Ω that are fixed under the action of G. Then Ω ≡ Ω_{0} 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/C_{G}(x) (where C_{G}(x) denotes the centralizer), which is a multiple of p by assumption. The result follows immediately.
Theorem 2: If H is a psubgroup of a finite group G and P is a Sylow psubgroup of G then there exists a g ∈ G such that H ≤ gPg^{−1}. In particular, the Sylow psubgroups for a fixed prime p are conjugate in G.
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 Ω_{0} ≡ Ω = [G : P] mod p. Now p <math>\nmid<math> [G : P] by definition so p <math>\nmid<math> Ω_{0}, hence in particular Ω_{0} ≠ 0 so there exists some gP ∈ Ω_{0}. It follows that hgP = gP so g^{−1}hgP = P, g^{−1}hg ∈ P, and thus h ∈ gPg^{−1} ∀ h ∈ H, so that H ≤ gPg^{−1} for some g ∈ G. Now if H is a Sylow psubgroup, H = P = gPg^{−1} so that H = gPg^{−1} for some g ∈ G.
Theorem 3: Let q denote the order of any Sylow psubgroup 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 psubgroups of G, and let P act on Ω by conjugation. Let Q ∈ Ω_{0} 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 Ω = {P} so that, by the Lemma, Ω ≡ Ω_{0} = 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 psubgroup can be found in polynomial time of the input (the degree of the group times the number of generators).
Reference
H. Wielandt. "Ein Beweis für die Existenz der Sylowgruppen." Archiv der Mathematik, 10:401402, 1959.fr:Théorèmes de Sylow pl:Twierdzenie Sylowa de:SylowSätze