Some applications of Sylow’s Theorems

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Recall

Cauchy’s Theorem. Let G be a finite group and let p be a prime that divides the order of G. Then, G has an element of order p, so also a subgroup with order p.

Sylow’s First Theorem. Let G be a finite group and p be a prime. If p^k | |G| for some integer k ∈ ℤ, then G has a subgroup of order p^k, i.e., ∃H ≤ G such that |H| = p^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.

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|

Theorem. Let G be a group. If G has an unique subgroup H (H ≤ G) of a given order, then that subgroup is normal, H ◁ G.

Proof.

Let H ≤ G be the only subgroup of G of a given order. ∀g ∈ G, [Note 2 below. The conjugate of a subgroup is a subgroup] gHg^-1 ≤ G

[Set equivalence of Regular Representations, If S is a finite subset of a group G, then |a·S| = |S| = |S·a| because the left (λ_a: S → S, ∀x∈S: λ_a(x) = a·x) and right regular representations are permutations, and therefore bijections, Note 1] Besides, |H^a| = |H|, that is because |H^a| = |aHa^-1| = |(aH)·a^-1| =[|a·S| = |S| = |S·a| ] |aH| =[All cosets have the same size] |H|

However, by assumption, H is the only subgroup of G of order |H| ⇒ gHg^-1 = H ⇒ [A subgroup N of G is normal if and only if N is self-conjugate, that is, ∀g∈G: g·N·g^-1 = N, g^-1·N·g = N ] H is normal∎

Note 1: Let G be a group and N be a normal subgroup N ◁ G (∀g ∈ G: g·N = N·g where g·N denotes the subset product of g with N = {a·n: n ∈ N}.

Note 2: Let G be a group, let H ≤ G be a subgroup of G, then the conjugate of H by a is a subgroup, i.e., ∀H ≤ G, a ∈ G: H^a ≤ G where H^a = aHa^-1.

Corollary. A Sylow p-subgroup of a finite group G is a normal subgroup of G if and only if it is the only Sylow p-subgroup of G.

Proof

⇐) If G has only one Sylow p-subgroup P ⇒ [Let G be a group. If G has an unique subgroup H (H ≤ G) of a given order, then that subgroup is normal, H ◁ G.] P must be normal, P ◁ G.

⇒) Conversely, suppose a Sylow p-subgroup P is normal ⇒ [Second Sylow Theorem. All the Sylow p-subgroups of a finite group are conjugates] since P is normal, it equals its conjugates, so it is the only one.

Theorem. Let G be a group with H, K ◁ G, H ∩ K = {e}, and HK = G. Then, G ≋ H x K.

Proof.

• Let’s start by showing that ∀h ∈ H, k ∈ K, hk = kh.

Since H ◁ G, ∀h ∈ H, k∈ K, khk^-1∈ H ⇒ ∃h’∈ H such that khk^-1 = h’ ⇒ [*k both sides of the equation] kh = h’k

Similarly, since K ◁ G, ∀h ∈ H, k∈ K, h^-1kh ∈ K, ∃k’∈ K such that h^-1kh = k’ ⇒ [*h both sides of the equation] kh = hk'

Therefore, kh = h’k = hk’ ⇒ hk’ = h’k ⇒ [*h’^-1] (h’)^-1hk’ = k ⇒ [*k’^-1] (h’)^-1h = k(k’)^-1

h^’-1h = kk^’-1 [h^’-1h ∈ H and kk^’-1 ∈ K] ∈ H ∩ K = {e}

Thus, h^’-1h = e ⇒ h’ = h and kk^’-1 = e ⇒ k’ = k. Finally, kh = h’k = hk’ = hk, kh = hk.

Then, we can define Φ:H x K → G, Φ((h, k)) = hk. We claim that Φ is a group isomorphism.

• Φ is homomorphism Φ((h₁, k₁)(h₂, k₂)) = Φ((h₁h₂, k₁k₂)) = h₁h₂k₁k₂

Φ((h₁, k₁))Φ((h₂, k₂)) = h₁k₁h₂k₂ = [∀h ∈ H, k ∈ K, hk = kh] h₁h₂k₁k₂

• Ker(Φ) = {(h, k) ∈ H x K: Φ((h, k)) = e}

Φ((h, k)) = e ⇒ hk = e ⇒ h = k^-1 ∈ H ∩ K ⇒[By assumption, H ∩ K = {e}] h = e ⇒ [hk = e] k = e, i.e., Ker(Φ) = {(e, e)}, and by the First Group Isomorphism Theorem, (H x K)/Ker(Φ) ≋ Φ(H x K).

• However, Φ is surjective since by assumption G = HK. ∀g ∈ G, ∃h ∈ H, k ∈ K: g = hk. Therefore, ∃(h, k) ∈ H x K such that Φ(h, k) = hk = g, so Φ(H x K) = G, and finally (H x K)/Ker(Φ) = [Ker(Φ) = {(e, e)}] = H x K ≋ Φ(H x K) = [Φ is surjective] G, therefore H x K ≋ G ∎

Theorem. Let G be a group of order pq, p, q be distinct primes such that p < q, then it has a normal subgroup of order q. In particular, no group of order pq can be simple. Besides, if p does not divide q -1 (q ≆ 1 (mod p)), then G is cyclic and G ≋ ℤ_pq ≋ ℤ_pxℤ_q

Proof.

Let |G| = pq, p < q, n_q be the number of Sylow q-subgroups.

By Sylow’s Third Theorem, n_q | |G| and n_q ≡ 1 (mod q) ⇒ The divisors of pq are 1, p, q, and pq. However, q and pq are ≡ 0 (mod q). And as p < q, p ≇1 (mod q) -0, 1, 2, ···, p (p ≡ p (mod q) ),.·· q-1-. Therefore, n_q = 1 ⇒ There is exactly one Sylow q-subgroup of G, say H ⇒[H Sylow p-Subgroup is unique ↭ H ◁ G] H must be normal.

Next, suppose that q ≆ 1 (mod p). Let n_p be the number of Sylow p-subgroups of G. Again, by Sylow’s Third Theorem, n_p | pq (so n_p = 1, q, p, and pq) and n_p ≡ 1 (mod p). However, q and pq are ≡ 0 (mod q) and the assumption above (q ≆ 1 (mod p)) prevents n_q = q, and therefore n_p =1 ⇒ there is a unique Sylow p-subgroup, and then as a unique Sylow p-subgroup is normal.

Recall that every group of order p prime is a cyclic group and isomorphic to ℤ_p. Considering this, these Sylow subgroups H, K must be cyclic and isomorphic to ℤ_p and ℤ_q respectively.

H ∩ K ≤ H, H ∩ K ≤ K ⇒ |H ∩ K| divides both p and q, and therefore is the trivial subgroup, {e}.

It follows that |HK| = ^|H||K|⁄_|H∩K| = pq = |G| ⇒ HK = G. Then, by the previous theorem [Recall. Theorem. Let G be a group with H, K ◁ G, H ∩ K = {e}, and HK = G. Then, G ≋ H x K.], G ≋ ℤ_pxℤ_q ≋ [The direct product of two cyclic group of coprime order is itself cyclic] ℤ_pq

Examples

• Let G be a group, |G| = 40 = 2³·5, n₅ = 1 (1 | 40 and 1 ≡ 1 (mod 5)), so G has exactly one subgroup of order 5, and therefore is normal. It also has 1 or 5 2-subgroup of order 8 (8 | 40 and 1 ≡ 1 (mod 2), 5 ≡ 1 (mod 2)). If there is only one subgroup of order 8, then it is obviously normal. Otherwise, none of the all five subgroups are normal, and they can be obtained with any particular one, say H, and conjugating xHx^-1.

If H, K ≤ G subgroups of a finite group of order 40, K ◁ G, |K| = 5, |H| = 8 ⇒ |H∩K| needs to divide 8 and 5 (it is a subgroup of H and K), so |H∩K| = 1 ⇒ |HK| = ^|H||K|⁄_|H∩K| ⇒ G = HK. If H is normal, too, then by the previous theorem G ≋ H x K.

• Let G be a group of order 15 = 3·5, 5 ≆ 1 (mod 3) [Theorem. Let G be a group of order pq, p, q be distinct primes such that p < q, then it has a normal subgroup of order q. In particular, no group of order pq can be simple. Besides, if p does not divide q -1 (q ≆ 1 (mod p)), then G is cyclic and G ≋ ℤ_pq ≋ ℤ_pxℤ_q ] There is no simple group of order 15, because it has a normal Sylow 5-group and a normal Sylow 3-group. All groups of order 15 are cyclic, that is, isomorphic to ℤ₁₅ ≋ ℤ₃xℤ₁₅.

• Let G be a group of order 14 = 2·7. It necessarily has a normal Sylow 7-group, which is isomorphic to ℤ₇. Besides, n₂ = 1 or 7. If n₂ = 1, then G ≋ ℤ₁₄, but it could have seven Sylow 2-subgroups, each one of them isomorphic to ℤ₂ containing the identity and a element of order 2.

From Cauchy theorem: |G| < ∞, p is prime, p ∣ |G| ⇒ in G exists an element with order p, so also subgroup with order p. In our case, |S| = 2, |P| = 7.

Let’s suppose the case that we have seven Sylow 2-subgroups. G has a unique Sylow 7-subgroup which is isomorphic to ℤ₇, and six elements (without considering the identity) of order 7 in G. Therefore, we have that one element in G have order 1, namely the identity, ({e}), 6 have order 7, (ℤ₇ - {e}) and 7 (ℤ₂- {e}) corresponding to the subgroups of order 2, have order 2.

Let r, |r| = 7, |s| = 2, P be the unique Sylow 7-subgroup ⇒ P is normal and cyclic (P ◁ G, P = ⟨r⟩) ⇒ [P ◁ G] srs^-1 ∈ P ⇒ [P = ⟨r⟩] srs^-1 = [s=s^-1 because s has order 2] srs = rⁿ for some n ∈ ℤ ⇒ ssrss = srⁿs ⇒ [s² = e] r = srⁿs = [srs=rⁿ⇒ (replace r by rⁿ) srⁿs = (rⁿ)ⁿ] = rⁿ². Therefore, r = rⁿ², since r has order 7, n²≡1 mod 7 ⇒ n = ± 1.

1. n = 1, srs = r ⇒ sr = rs ⇒ G ≋ ⟨s⟩ x ⟨r⟩ ≋ ℤ₂ x ℤ₇ ≋ ℤ₁₄
2. n = -1 or 6 (1 + 6 ≡ 0 (mod 7), 6 = -1 in ℤ₇) ⇒ srs = r^-1. Therefore, we have r⁷=1, s²=1, and sr = r^-1s, G = ⟨r, s: r⁷ = e = s², sr = r^-1s⟩ this is the dihedral group D₇.

Thus, there are only two groups, up to isomorphism, of order 14, namely ℤ₁₄ and D₇.

Theorem. Let p be a odd prime. Then, there are only two groups of order 2p, up to isomorphism, namely ℤ_2p and D_p.

Proof.

It is the same argument as before, 2p is a semi-prime ⇒ There is a unique, normal Sylow p-subgroup, and cyclic (Every prime group is cyclic), isomorphic to ℤ_p, and G cannot be simple.

p ≡ 1 (mod 2) because p is an odd number ⇒ n₂ = 1 or p. If n₂ = 1, G ≋ ℤ₂xℤ_p ≋ ℤ_2p. Otherwise, n₂ = p, G ≋ D_p as we have shown in the previous example.

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