Sylow Theorem II and III.

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Definitions. Two subgroups H and K of a group G are termed conjugate in G if there is an element g ∈ G such that H = gKg^-1. The normalizer of P, written as N(P), is {x ∈ G: xPx^-1 = P}.

1. If G is a group, the normalizer of a subset P ⊆ G is a subgroup of G, that is, N(P)≤ G.
2. A subgroup H ≤ G is a normal subgroup of its normalizer (∀g ∈G, g·H=H·g), H ◁ N(H).

A p-group is a group all of whose elements have an order equal to a power of the prime number p. A finite group is a p-group if and only if its order (the number of its elements) is a power of p.

Definition. Let G be a finite group, and let p be a prime dividing the order of G. Then, a Sylow p-subgroup of G is a maximal p-subgroup of G, that is, p^k divides |G| and p^k+1 does not divide |G|.

Lemma. Let P be a Sylow p-subgroup of a group G. Let x ∈ G be an element whose order is a power of p. If x ∈ N(P), then x ∈ P.

Proof.

Suppose P be a Sylow p-subgroup of a group G, x ∈ G, x ∈ N(P), an element whose order is a power of p, x ∈ P?

Since P ◁ N(P) = {x ∈ G: xPx^-1 = P}, i.e., P is a normal subgroup of its normalizer, we can consider the quotient group N(P)/P. Since x ∈ N(P), let K = ⟨xP⟩ a cyclic subgroup of N(P)/P.

As |x| is a power of p ⇒ |x|=p^a for some a∈ ℕ ⇒ (xP)^pa = x^paP = eP = P, hence |xP| must divide p^a, but p is prime ⇒ |xP| is a power of p, too

[By the Correspondence Theorem, N ◁ G, every subgroup of the quotient group G/N is of the form H/N, where N ≤ H ≤ G. Conversely, if N ≤ H ≤ G, then H/N ≤ G/N] Since, P ◁ N(P), K = ⟨xP⟩ ≤ N(P)/P, there exists a unique subgroup H such that P ≤ H ≤ N(P) and H/P = K.

And then, |H| = |P|·|K| ⇒[|xP| is a power of p, K = ⟨xP⟩, the order of K is a power of p, i.e., K is a p-group (a group in which the order of every element is a power of p), and, by assumption, P is a Sylow p-subgroup] H is a p-subgroup of G, and we already know that P ≤ H. Since P is a Sylow p-subgroup and hence a maximal p-subgroup ⇒ H = P ⇒ K = H/P = P/P is the trivial subgroup ⇒[K = ⟨xP⟩ = P/P] xP = P, the identity of N(P)/P ⇒ x ∈ P ∎

Theorem. Let H and K be subgroups of a group G. The number of distinct H-conjugates of K, that is, (hKh^-1) is [H: N(K)∩H].

Proof

The group G acts naturally on the set of its subgroups (Let G be a group and S denote the set of subgroups of G, including obviously K, by conjugation ·:G x S → S, g·T = gTg^-1, T ≤ G, where gTg^-1 is a subgroup for every g ∈ G.

We can restrict this action to its subgroup H ≤ G (·:H x S → S, h·T = hTh^-1, h ∈ H).

We need to compute the size of the orbit of this H-action containing K [By the Fundamental Counting Principal, |O_x| = [G:G_x]] |O_K| = [H: H_K] where H_K = { h ∈ H| h·K = hKh^-1 = K} is the stabilizer of K via the H-conjugation action. We claim that H_K = N(K) ∩ H and then, we are done.

• H_K ⊆ N(K) ∩ H. We know that the stabilizer is a subgroup of the group that is acting, H_K ≤ H. Besides, H_K ≤[We are using a restriction of a larger conjugation action by G, therefore, the stabilizer of K with respect to the H-action is a subgroup of the stabilizer of K with respect to the G-action, that is bigger, that is, H_K = { h ∈ H| h·K = hKh^-1 = K} ≤ G_K = { g ∈ G| g·K = gKg^-1 = K}] G_K = N(K), thus H_K ≤ N(K) ∩ H.

  H_K is the set of elements of H that fixes K, hKh^-1 = K, G_K is the set of elements that fixes K, and G_K = [Because we are using the conjugation action] N(K) = {g ∈ G: gKg^-1 = K}

• N(K) ∩ H ⊆ H_K. Conversely, if x ∈ N(K) ∩ H ⇒ x ∈ H, and x can act on K by conjugation (·:H x S → S, h·T = hTh^-1, h ∈ H), x·K = xKx^-1.

Futhermore, as x ∈ N(K) = G_K ⇒ xKx^-1 = K ⇒ x·K = xKx^-1 = K ⇒[x ∈ H] x ∈ H_K ∎

Sylow’s Second Theorem. Let G be a finite group and p a prime dividing |G|. Then, all Sylow p-subgroups are conjugates, that is, if P and Q are Sylow p-subgroups, then there exists some g ∈ G such that gPg^-1 = Q.

Proof.

Suppose G be a finite group and p a prime dividing |G| ⇒ ∃r ∈ ℕ: |G| = p^rm where p ɫ m. By Sylow’s First Theorem, there exist a p-subgroup P of order p^r, that is, P is a maximal p-subgroup of G, P ∈ Syl_p(G).

Let $\mathbb{S}$={gPg^-1: g ∈ G} be the set of all distinct G-conjugates of P, and let h * S be the conjugacy action, i.e., P x $\mathbb{S}$ → $\mathbb{S}$, ∀h ∈ P, ∀S ∈ $\mathbb{S}$, the conjugacy action h * S = h·S·h^-1 is a group action.

To show it is closed, ∀ S∈$\mathbb{S}$ ⇒ S = gPg^-1 ⇒ h*S = h·S·h^-1 = h(gPg^-1)h^-1 = (hg)P(hg)^-1 ∈ $\mathbb{S}$, hg ∈ G.

Let O = {P₁ = P, P₂, ···, P_k} be the complete orbit of all G-conjugates of P. By the previous lemma, the number of G-conjugates of P is k = [G: N(P)].

|G| = p^rm = [By Lagrange’s Theorem, recall that N(P) ≤ G] |N(P)|·|G:N(P)| = |N(P)|·k

Notice that as P ≤ N(P) ⇒ [By Lagrange’s Theorem] |P| = p^r | |N(P)| and since p^rm = |N(P)|·k, there’s no p’s left that can divide k, and therefore and p ɫ |G:N(P)|=k

Let Q ∈ Syl_p(G), an arbitrary Sylow’s p-subgroup. We claim that Q ∈ O = {P, P₂, P₃, ···, P_k} which would prove the theorem.

Since G acts on the set O = {P, P₂, P₃, ···, P_k} by conjugation ·:G x O → O, g·P_i = gPg^-1, we could restrict it to any subgroup, in particular, Q, that is, ·:Q x O → O, q·P_i = qPq^-1.

Besides, this will partition the set O into smaller sets. What is the number of distinct Q-conjugates of P_i is, by the previous theorem, [Q:N(P_i)∩Q], and by Lagrange’s Theorem |Q| = [Q:N(P_i)∩Q]|N(P_i)∩Q]| ⇒ [Q:N(P_i)∩Q]| must a divisor of |Q| = p^r ⇒ all the indices [Q:N(P_i)∩Q]| are powers of p or 1. Thus, k can be written as a sum of powers of p, which shows that p | k.

Since p ɫ k, then at least one of these powers is 1, that is, there must exist some conjugate P_i∈ O: [Q:N(P_i)∩Q]| = 1 ⇒ Q = N(P_i) ∩ Q ⇒ ∀x ∈ Q (|Q| = p^r), x ∈ N(P_i) ⇒ [Lemma. Let P be a Sylow p-subgroup of a group G. Let x ∈ G be an element whose order is a power of p. If x ∈ N(P), then x ∈ P.]x ∈ P_i ⇒ Q ⊆ P_i. Since Q is a maximal p-subgroup ⇒ P_i = Q ∎ Besides, k = [G:N(P)] = |Q| = [Q:N(P_i)∩Q]|N(P_i)∩Q]| = 1·|N(P_i)∩Q]| = |N(P_i)|

Notice that all Sylow p-subgroups are conjugates of each other. As conjugation is an inner automorphism on G, conjugation induces a group isomorphism between Sylow subgroups, and in particular all Sylow p-subgroups are isomorphic to each other and they all have the same order necessarily p^r if |G|= p^rm and p ɫ m

If |G| = p^rm, that is, p^r+1 ɫ |G| ⇒ ∃P ≤ G. By the first Sylow Theorem, |P| = p^r that is maximal, so that’s why all Sylow p-subgroups have order p^r.

Sylow’s Third Theorem. Let G be a finite group and p a prime dividing |G|. Then, the number of Sylow p-subgroups of G is congruent to 1 (mod p) and divides |G|, n_p ≡ 1 (mod p), n_p | |G|.

If |G| = p^rm and p ɫ m, the number of Sylow p-subgroups must divide m.

Proof.

As we have studied previously, G acts on Syl_P(G) by conjugation. By Sylow's Second Theorem, this action has a single orbit (there must exist some conjugate P_i∈ O = {P, P₂, P₃, ···, P_k}: [Q:N(P_i)∩Q]| = 1 ⇒ P_i = Q, one of the equivalence classes contains only a single Sylow p-subgroup), O = {P₁, P₂, ···, P_k}, and |O| = k = [G:N(P_i)], [G:N(P_i)] | |G|, therefore the number of Sylow p-subgroups (they are all conjugate of P) of G divides |G|.

Let P ∈ Syl_P(G). Then P also acts on Syl_P(G) by conjugation. Notice that {P} is a trivial orbit of this action (x ∈ P, xPx^-1 = P, because P is a subgroup, so is close under multiplication). Claim: This action has no other fixed points, P is the only element of $\mathbb{S}$ whose orbit has length 1.

[Theorem. Let P be a Sylow p-subgroup of a group G. Let x ∈ G be an element whose order is a power of p. If x ∈ N(P), that is, xPx^-1=P ⇒ x ∈P] Suppose that there is another {Q}, xQx^-1 = Q ⇒ x ∈ N(Q), but as x ∈ P a Sylow p-subgroup, then [By Lagrange’s Theorem, all elements have order of power of p] ∃a ∈ ℕ such that |x|=p^a ⇒ x ∈ Q ⇒ Q = P. Thus, this action has no other fixed points.

So, therefore {P}, {P₂, P₃···} and the class equation applies where X = Syl_P(G), |X_P| = 1, there is only one fixed point.

|X| = |X_P| + $\sum_{i=k}^n |O_{x_i}| = 1 + \sum_{i=k}^n p^{e_i} ≡ 1 (mod~ p)$ Notice that |X:C(x_i)| divides |X| = |Syl_P(G)| = p^r.

Example. Consider n = 6 = 2·3

n_k = # of Sylow k-subgroup = |Syl_p(G)|, then n_k | |G| and n_k ≡ 1 (mod p)

By Sylow’s Third Theorem, n₃|6, n₃ ≡ 1 (mod 3) ⇒ n₃ = 1, and therefore any group of order 6 must have a unique (and normal) Sylow 3-subgroup but [n₂|6, n₂ ≡ 1 (mod 2) ⇒ n₂ = 1 or 3] can have 1 or 3 Sylow 2-subgroups, e.g., ℤ₆ has a unique Sylow 2-subgroup (Fig 3, ℤ₃), ℤ₆ has a unique Sylow 2-subgroup (Fig 3, ℤ₂) and S₃ has 3 Sylow 2-subgroups (Fig 4).

Consider one of the Sylow 2-subgroups of S₃, say {(1), (12)}. By Sylow’s Third Theorem, we should be perfectly able to get the other two Sylow 2-subgroups from this one by conjugation: (13){(1), (12)}(13)^-1 = {(13)(1)(31), (13)(12)(31)} = {(1), (23)}, and (23){(1), (12)}(23)^-1 = {(23)(1)(32), (23)(12)(32)} = {(1), (13)}

• Consider n = 12 = 2²·3. By Sylow’s Third Theorem any group of order 12 must have n₂ = 1 or 3 Sylow 2-subgroups (1 and 3 | 12, and 1≡1 (mod 2), 3≡1 (mod 2)) and n₃ = 1 or 4 Sylow 3-subgroups. Examples:

1. ℤ₁₂ has a unique Sylow 2-subgroup (ℤ₄) and a unique Sylow 3-subgroup (ℤ₃) -Fig 1.a.-, i.e., n₂ = n₃ = 1.
2. The alternating group A₄, the group of even permutations of 4 elements, has a unique Sylow 2-subgroup (V₄), but 4 Sylow 3-subgroups (ℤ₃) -Fig 1.b-, i.e., n₂ = 1, n₃ = 4.
3. D₆, the dihedral group, has 3 Sylow 2-subgroup (V₄) and a unique Sylow 3-subgroup (ℤ₃).

   Consider one of the Sylow 3-subgroups of A₄, say {(), (234), (243)}. By Sylow’s Third Theorem, we should be perfectly able to get the other three Sylow 3-subgroups from this one by conjugation:

4. (134){(), (234), (243}(134)^-1 = {(134)()(431), (134)(234)(431), (134)(243)(431)} = {(), (124), (142)}.
5. (421){(), (234), (243)}(124) = {(), (132), (123)}.
6. (123){(), (234), (243)}(123)^-1 = {(123)()(321), (123)(234)(321), (123)(243)(321)} = {(), (143), (134)}

If σ = (a₁a₂···a_k), then a₁ → ₂→···→a_k→a₁. Then, σ^-1 = a₁ ← ₂←···←a_k←a₁, this is the cycle written backwards (a_ka_k-1···a₁). If it is a 2-cycle: (ab)^-1 = b^-1a^-1

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