4th Law. Life, economic collapse, and history are like a roll of toilet paper, the closer it gets to the end, the faster it goes, Apocalypse, Anawim, #justtothepoint.
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 pk | |G| for some integer k ∈ ℤ, then G has a subgroup of order pk, i.e., ∃H ≤ G such that |H| = pk.
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|, np ≡ 1 (mod p), np | |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 (λa: S → S, ∀x∈S: λa(x) = a·x) and right regular representations are permutations, and therefore bijections, Note 1] Besides, |Ha| = |H|, that is because |Ha| = |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: Ha ≤ G where Ha = 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.
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((h1, k1))Φ((h2, k2)) = h1k1h2k2 = [∀h ∈ H, k ∈ K, hk = kh] h1h2k1k2
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).
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 ≋ ℤpq ≋ ℤpxℤq
Proof.
Let |G| = pq, p < q, nq be the number of Sylow q-subgroups.
By Sylow’s Third Theorem, nq | |G| and nq ≡ 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, nq = 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 np be the number of Sylow p-subgroups of G. Again, by Sylow’s Third Theorem, np | pq (so np = 1, q, p, and pq) and np ≡ 1 (mod p). However, q and pq are ≡ 0 (mod q) and the assumption above (q ≆ 1 (mod p)) prevents nq = q, and therefore np =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 ℤp. Considering this, these Sylow subgroups H, K must be cyclic and isomorphic to ℤp and ℤq 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 ≋ ℤpxℤq ≋ [The direct product of two cyclic group of coprime order is itself cyclic] ℤpq
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 ≋ ℤpq ≋ ℤpxℤq ] 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.
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 ℤ2p and Dp.
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 ℤp, and G cannot be simple.
p ≡ 1 (mod 2) because p is an odd number ⇒ n2 = 1 or p. If n2 = 1, G ≋ ℤ2xℤp ≋ ℤ2p. Otherwise, n2 = p, G ≋ Dp as we have shown in the previous example.