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 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.
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^{-1}kh ∈ K, ∃k’∈ K such that h^{-1}kh = k’ ⇒ [*h both sides of the equation] kh = hk'
Therefore, kh = h’k = hk’ ⇒ hk’ = h’k ⇒ [*h’^{-1}] (h’)^{-1}hk’ = k ⇒ [*k’^{-1}] (h’)^{-1}h = k(k’)^{-1}
h^{’-1}h = kk^{’-1} [h^{’-1}h ∈ H and kk^{’-1} ∈ K] ∈ H ∩ K = {e}
Thus, h^{’-1}h = 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((h_{1}, k_{1}))Φ((h_{2}, k_{2})) = h_{1}k_{1}h_{2}k_{2} = [∀h ∈ H, k ∈ K, hk = kh] h_{1}h_{2}k_{1}k_{2}
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 ≋ ℤ_{pq} ≋ ℤ_{p}xℤ_{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_{p}xℤ_{q} ≋ [The direct product of two cyclic group of coprime order is itself cyclic] ℤ_{pq}
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_{p}xℤ_{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 ℤ_{15} ≋ ℤ_{3}xℤ_{15}.
Let G be a group of order 14 = 2·7. It necessarily has a normal Sylow 7-group, which is isomorphic to ℤ_{7}. Besides, n_{2} = 1 or 7. If n_{2} = 1, then G ≋ ℤ_{14}, but it could have seven Sylow 2-subgroups, each one of them isomorphic to ℤ_{2} 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 ℤ_{7}, 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, (ℤ_{7} - {e}) and 7 (ℤ_{2}- {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 = r^{n} for some n ∈ ℤ ⇒ ssrss = sr^{n}s ⇒ [s^{2} = e] r = sr^{n}s = [srs=r^{n}⇒ (replace r by r^{n}) sr^{n}s = (r^{n})^{n}] = r^{n2}. Therefore, r = r^{n2}, since r has order 7, n^{2}≡1 mod 7 ⇒ n = ± 1.
Thus, there are only two groups, up to isomorphism, of order 14, namely ℤ_{14} and D_{7}.
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 ⇒ n_{2} = 1 or p. If n_{2} = 1, G ≋ ℤ_{2}xℤ_{p} ≋ ℤ_{2p}. Otherwise, n_{2} = p, G ≋ D_{p} as we have shown in the previous example.