# 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, pk divides |G| and pk+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.

Since P ◁ N(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|=pa, (xP)pa = xpaP = eP = P, |xP| must divide pa, 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 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 (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 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, |Ox| = [G:Gx]] |OK| = [H: HK] where HK = { h ∈ H| h·K = hKh-1 = K} is the stabilizer of K via the H-conjugation action. We claim that HK = N(K) ∩ H.

• HK ⊆ N(K) ∩ H. We know that the stabilizer is a subgroup of the group that is acting, HK ≤ H. Besides, HK[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.] GK = N(K), thus HK ≤ N(K) ∩ H.

HK is the set of elements of H that fixes K, xKx-1 = K, GK is the set of elements that fixes K, and GK = [Because we are using the conjugation action] N(K) = {x ∈ G: xKx-1 = K}

• N(K) ∩ H ⊆ HK. Conversely, if x ∈ N(K) ∩ H ⇒ x ∈ H, and x can act on K by conjugation, x·K = xKx-1.

Futhermore, as x ∈ N(K) = GK ⇒ xKx-1 = K ⇒ x·K = xKx-1 = K ⇒ x ∈ HK

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.

Let |G| = prm where p ɫ m. By Sylow’s First Theorem, there exist a p-subgroup P of order pr, P ∈ Sylp(G), that is, P (maximal) ≤ 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 for S∈$\mathbb{S}$ ⇒ S = gPg-1 ⇒ h*S = h·S·h-1 = h(gPg-1)h-1 = (hg)P(hg)-1 ∈ $\mathbb{S}$

Let O = {P1 = P, P2, ···, Pk} 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| = prm = [By Lagrange’s Theorem] |N(P)|·|G:N(P)| = |N(P)|·k

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

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

Since G acts on the set O = {P, P2, P3, ···, Pk} by conjugation, we could restrict it to any subgroup, in particular, Q. Besides, this will partition the set O into smaller sets. What is the number of distinct Q-conjugates of Pi is, by the previous theorem, [Q:N(Pi)∩Q], and by Lagrange’s Theorem |Q| = [Q:N(Pi)∩Q]|N(Pi)∩Q]| ⇒ [Q:N(Pi)∩Q]| must a divisor of |Q| = pr ⇒ the number of conjugates in every equivalence class of the partition is a power of p.

However, since p does not divide k, one of these equivalence classes (x, y ∈ X -O-, x ~ y ↭ ∃g ∈ G -Q-: g·x = y ↭ Pj*x = x-1Pjx = Pj) must contain only a single Sylow p -subgroup, say Pj. In other words, x-1Pjx = Pj, i.e., x ∈ N(Pj) [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 ∈ Q, x ∈ Pj ⇒ Q ⊆ Pj. Since Q is a maximal p-subgroup ⇒ Pj = Q ∎

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 pr if |G|= prm and p ɫ m

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|, np ≡ 1 (mod p), np | |G|

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

Proof.

As we have studied previously, G acts on SylP(G) by conjugation. By Sylow’s Second Theorem, this action has a single orbit (one of the equivalence classes contains only a single Sylow p-subgroup), O = {P1, P2, ···, Pk}, and the size of this orbit is the index of the normalizer of any p-subgroup Pi, i.e., k = [G:N(Pi)] | |G|, therefore the number of Sylow p-subgroups of G divides |G|.

Let P ∈ SylP(G). Then P also acts on SylP(G) by conjugation. Notice that {P} is a trivial orbit of this action (x ∈ P, xPx-1 = P, because P is close). 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 p ∈ 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 ⇒ |x|=pa ⇒ x ∈ Q ⇒ Q = P.

So, therefore {P}, {P2, P3···} and the class equation applies where X = SylP(G)

|X| = |XP| + $\sum_{i=k}^n |O_{x_i}| = 1 + \sum_{i=k}^n |p^{e_i}| ≡ 1 (mod~ p)$

Example. Consider n = 6 = 2·3

nk = # of Sylow k-subgroup = |Sylp(G)| | |G| and nk ≡ 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) 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), (12)}(31) = {(1), (23)}, and (23){(1), (12)}(23)-1 = (23){(1), (12)}(32) = {(1), (13)}

• Consider n = 12 = 22·3. By Sylow’s Third Theorem any group of order 12 must have 1 or 3 Sylow 2-subgroups (1 and 3 | 12, and 1≡1 (mod 2), 3≡1 (mod 2)) and 1 or 4 Sylow 3-subgroups. Examples:
1. 12 has a unique Sylow 2-subgroup (ℤ4) and a unique Sylow 3-subgroup (ℤ3) -Fig 1.a.-.
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-.
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){(), (234), (243)}(431) = {(), (124), (142)}.
5. (421){(), (234), (243)}(124) = {(), (132), (123)}.
6. (123){(), (234), (243)}(123)-1 = (123){(), (234), (243)}(321) = {(), (143), (134)}

If σ = (a1a2···ak), then a12→···→ak→a1. Then, σ-1 = a12←···←ak←a1, this is the cycle written backwards (akak-1···a1). If it is a 2-cycle: (ab)-1 = b-1a-1

# Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.
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