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.
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 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.
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 (p^{n} · s), where n > 0 and p does not divide s. Then:
- There exists a Sylow p-subgroup of G, of order p^{n}.
- 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^{-1}H'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, p^{n}; and conversely, if a subgroup has order p^{n}, 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 p^{n}.
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.
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^{ -1}Sg 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^{ -1}xg 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 p^{n}·s, we can prove that G has a subgroup K of order p^{n} using induction.
- Induct on the order of G. If G has a subgroup whose order is divisible by p^{n} then, by the inductive hypothesis, that subgroup has a subgroup of K of order p^{n}; 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 p^{n}.
- By Lagrange's theorem, |G| = |C(x)|·[G:C(x)]; since |G| is divisible by p^{n}, [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 p^{n}.
K is necessarily maximal by Lagrange, so it is a Sylow p-subgroup; and therefore Syl_{p}(G) is not 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^{ -1}Mh = 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^{ -1}La = b^{ -1}Lb 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 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, 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| = p^{n}, 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 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 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 is not 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} does not divide p^{n}; 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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