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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}.
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.
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|=p^{a}, (xP)^{pa} = x^{pa}P = eP = P, |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 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, |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.
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.] G_{K} = N(K), thus H_{K} ≤ N(K) ∩ H.
H_{K} is the set of elements of H that fixes K, xKx^{-1} = K, G_{K} is the set of elements that fixes K, and G_{K} = [Because we are using the conjugation action] N(K) = {x ∈ G: xKx^{-1} = K}
N(K) ∩ H ⊆ H_{K}. 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) = G_{K} ⇒ xKx^{-1} = K ⇒ x·K = xKx^{-1} = K ⇒ 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.
Let |G| = p^{r}m where p ɫ m. By Sylow’s First Theorem, there exist a p-subgroup P of order p^{r}, P ∈ Syl_{p}(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 = {P_{1} = P, P_{2}, ···, 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^{r}m = [By Lagrange’s Theorem] |N(P)|·|G:N(P)| = |N(P)|·k
Notice that as P ≤ N(P) ⇒ [By Lagrange’s Theorem] |P| = p^{r} | |N(P)|, 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, P_{2}, P_{3}, ···, P_{k}} which would prove the theorem.
Since G acts on the set O = {P, P_{2}, P_{3}, ···, P_{k}} 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 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} ⇒ 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 ↭ P_{j}*x = x^{-1}P_{j}x = P_{j}) must contain only a single Sylow p -subgroup, say P_{j}. In other words, x^{-1}P_{j}x = P_{j}, i.e., x ∈ N(P_{j}) [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 ∈ P_{j} ⇒ Q ⊆ P_{j}. Since Q is a maximal p-subgroup ⇒ P_{j} = 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 p^{r} if |G|= p^{r}m 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|, n_{p} ≡ 1 (mod p), n_{p} | |G|
If |G| = p^{r}m 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 (one of the equivalence classes contains only a single Sylow p-subgroup), O = {P_{1}, P_{2}, ···, P_{k}}, and the size of this orbit is the index of the normalizer of any p-subgroup P_{i}, i.e., k = [G:N(P_{i})] | |G|, therefore the number of Sylow p-subgroups 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 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|=p^{a} ⇒ x ∈ Q ⇒ Q = P.
So, therefore {P}, {P_{2}, P_{3}···} and the class equation applies where X = Syl_{P}(G)
|X| = |X_{P}| + $\sum_{i=k}^n |O_{x_i}| = 1 + \sum_{i=k}^n |p^{e_i}| ≡ 1 (mod~ p)$
Example. Consider n = 6 = 2·3
n_{k} = # of Sylow k-subgroup = |Syl_{p}(G)| | |G| and n_{k} ≡ 1 (mod p)
By Sylow’s Third Theorem, n_{3}|6, n_{3} ≡ 1 (mod 3) ⇒ n_{3} = 1, and therefore any group of order 6 must have a unique (and normal) Sylow 3-subgroup but [n_{2}|6, n_{2} ≡ 1 (mod 2) ⇒ n_{2} = 1 or 3] can have 1 or 3 Sylow 2-subgroups, e.g., ℤ_{6} has a unique Sylow 2-subgroup (Fig 3) and S_{3} has 3 Sylow 2-subgroups (Fig 4).
Consider one of the Sylow 2-subgroups of S_{3}, 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 one of the Sylow 3-subgroups of A_{4}, 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:
If σ = (a_{1}a_{2}···a_{k}), then a_{1} → _{2}→···→a_{k}→a_{1}. Then, σ^{-1} = a_{1} ← _{2}←···←a_{k}←a_{1}, this is the cycle written backwards (a_{k}a_{k-1}···a_{1}). If it is a 2-cycle: (ab)^{-1} = b^{-1}a^{-1}