Sixth rule B. Cherry picking on data, variables and metrics, perverse incentives, false causality, sampling bias, the observer effect, and over manipulating the data can lead you to any desired outcome with “rigorous” standards. Futhermore, there are many infinites, but the biggest of them all is human’s stupidity and sheer arrogance, Apocalypse, Anawim, #justtothepoint.
Definition. A group G is simple if it has no trivial, proper normal subgroups or, alternatively, if G has precisely two normal subgroups, namely G and the trivial subgroup.
Theorem. Theorem. A_{n}, the alternating group, that is the group of even permutation of a finite set, n ≥ 5, is a simple group.
Therefore, A_{5} is a simple group, but |A_{5}| = ^{5!}⁄_{2} = 60. 60 is big, isn’t it? Big is better, those who deny this fact have small penises. Ultimately, what is considered “big” is subjective, but whatever… The main question in this article is this: is there a “smaller” simple group?
{1, 2, 3, 4, ···, 60}
1 is the trivial subgroup and it is not simple by construction, it does not have proper subgroups.
|G| = p, prime ⇒[A group of order p where p is a prime number is cyclic and Abelian] G is Abelian and cyclic of prime order ⇒[G Abelian ⇒ G simple ↭ G is cyclic of primer order. ] G is simple.
|G| = p^{n}, p prime, n > 1.
Recall: Let G be a group. The center Z(G) of G is a normal subgroup of G, Z(G) = {g ∈ G : g·x = x·g, ∀x ∈ G} ◁ G.
Theorem. A finite p-group (|G| = p^{n}) cannot be simple unless it has order p.
Proof.
If G is Abelian ⇒ All subgroups are normal so it’s enough to find a non-trivial proper subgroup. If G is order p^{n} with n ≥ 2 ⇒[Cauchy’s Theorem for Abelian Groups. Let G be a finite Abelian group, |G| = n, and let p be a prime number dividing the order of G, then G contains an element of order p] then there is an element x of order p. If you let H = ⟨x⟩, the subgroup generated by x, then |H|=p, so H is a non-trivial proper normal subgroup ⇒ G is not simple.
If G is not Abelian ⇒ G - Z(G) ≠ ∅. By the class equation, $|G| = |Z(G)|+\sum_{i=1}^k |G:C(a_i)|$ where Z(G) is the center of G and C(a_{i}) is the centralizer of a_{i} and a_{1}, ···, a_{k} are representatives for the distinct conjugacy classes of G with more than one element (for otherwise they would be in the center).
Now, recall that |G:C(a_{i})|=|G|/|C(a_{i})| =[C(a_{i}) = {x ∈ G: x·a_{i}=a_{i}·x} ≤ G, |G| = p^{n}] |C(a_{i})| = p^{k} for some k < n. If k = n ⇒ C(a_{i}) = G ⇒ a_{i} ∈ Z(G)⊥] |G:C(a_{i})| = p^{n-k} ⇒[By the class equation, |C(a_{i})| = p^{k} for some k < n, and |G| = p^{n}] p | |Z(G)| ⇒ Z(G) is non-trivial, so Z(G) is a normal and proper subgroup ⇒ G is not simple.
Therefore, groups of orders 4 = 2^{2}, 8 = 2^{3}, 9, 16, 25, 27, 32 and 49 are not simple.
Theorem. Distinct Sylow p-subgroups of prime order intersect only at the identity.
Proof.
Suppose P and Q are distinct Sylow p-subgroups of prime order p. P ∩ Q ≤ P (and of Q) ⇒[By Lagrange, p prime] |P ∩ Q| = 1 or p (⊥ P ≠ Q) ⇒ |P ∩ Q| = 1 ∎
Therefore, it is not possible groups of order 6 (2·3), 10(2·5), 14(2·7), etc.
Possible orders left: 12, 20, 24, 28, 30, 36, 40, 42, 44, 48, 52, 56
|G| = 28 = 2^{2}·7 ⇒ n_{7}| 28 (so 1, 2, 4, and 7) and n_{7} ≡ 1 (mod 7) ⇒ n_{7} = 1 ⇒[Corollary. A Sylow p-subgroup of a finite group G is normal ↭ it is the only Sylow p-subgroup.] We have found a proper, normal, Sylow 7-subgroup ⇒ G is not simple. A similar argument removes 20 (2^{2}·5), 44 (2^{2}·11) and 52 (2^{2}·13).
Let G be a group of order 56 = 2^{3}·7. n_{7} | 8 (1, 2, 4, and 8), n_{7} ≡ 1 (mod 7), then the only options are 1 and 8. If G has a unique Sylow 7-subgroup, then it is normal and G is not simple.
Let’s suppose that n_{7} = 8, these subgroups have order 7, prime, so they are cyclic and isomorphic to ℤ_{7}. Let us consider two of these, P, Q ∈ Syl_{7}(G), P ≠ Q ⇒ [P ∩ Q is a subgroup of P and Q, its order must divide 7] P ∩ Q = 7 (P = Q ⊥) or 1.
Theorem. Distinct Sylow p-subgroups of prime order intersect only at the identity ↭ ∀P, Q ∈ Syl_{7}(G), P ≠ Q, P ∩ Q = {e} Therefore, G must have 8 subgroups = 8*(7 -1 {e}) = 8·6 = 48 elements of order 7 (each subgroup has also e the order 1). This leaves 56 - 48 = 8 elements unaccounted for in the group.
As a Sylow 2-subgroup of G necessarily has eight elements and at least one exists (Sylow’s First Theorem) ⇒ A single Sylow 2-subgroup accounts for the remaining eight elements ⇒[Corollary. A Sylow p-subgroup of a finite group G is normal ↭ it is the only Sylow p-subgroup.] normal ⇒ G is not simple.
The same argument applies to 12 = 2^{2}·3, n_{3} = 1 (then it is normal, G is not simple), or 4. 4 Sylow 3-subgroup that only share the identity, so there are 4*(3-1)= 8 elements of order 3. 12 -8 = 4 remaining elements and these four elements have to be in a “unique” Sylow 2-subgroup ⇒ normal ⇒ G is not simple.
n_{5} = 1 ⇒ There is a unique Sylow 5-subgroup ⇒ it would be normal and G would not be simple.
Let’s consider n_{5} = 6, these subgroups have order 5, prime, they are cyclic and isomorphic to ℤ_{5} = {1, a, a^{2}, a^{3}, a^{4}}.
By the same reasoning as before, distinct Sylow p-subgroups of prime order intersect only at the identity ↭ ∀P, Q ∈ Syl_{7}(G), P ≠ Q, P ∩ Q = {e}, so this will account for 6 subgroups · (5-1) elements = 24 elements of G of order 5, leaving 30 - 24 = 6 remaining unaccounted elements.
If G has a unique Sylow 3-subgroup (n_{3} = 1), then G is not simple.
Let’s assume that G has non-unique Sylow 3-subgroups ⇒ n_{3} | 10 (1, 2, 5, 10), n_{3} ≡ 1 (mod 3) ⇒[By assumption, n_{3} ≠ 1] n_{3} = 10 (10 ≡ 1 (mod 3)). Then, this account for 10 Sylow 3-subgroups (3 is prime, so they are cycle, ℤ_{3}={1, b, b^{2}}) · (3-1) elements = 20 elements of order 3, but we have only 6 elements left ⊥ n_{3} = 1 and G is not simple.
|G| = 42 = 2·3·7, n_{7} | 6 (1, 2, 3, 6), n_{7} ≡ 1 (mod 7) ⇒ n_{7} = 1, i.e., there is a unique Sylow 7-subgroup ⇒ G is not simple.
|G| = 24 = 2^{3}·3, n_{2}|3 & n_{2}≡1 (mod 2), n_{2} = 1 (normal ⇒ G is not simple) or 3. n_{3} | 8 (1, 2, 4, 8) and n_{3} ≡ 1 (mod 3) ⇒ n_{3} = 1 (normal ⇒ G is not simple) and 4 (4 ≡ 1(mod 3)).
Let’s suppose n_{2} = 3, n_{3} = 4. How is it? ↭ subgroups of order 8? There are five possibilities: ℤ_{8}, ℤ_{4}xℤ_{2}, (ℤ_{2})^{3}, D_{4}, and Q_{8}.
Let’s H, K be two distinct Sylow 2-subgroups (|H|=|K| = 2^{3} = 8), then |HK| = ^{|H||K|}⁄_{|H∩K|} = ^{8·8}⁄_{|H∩K|} ⇒[HK ≤ G] |HK| ≤ 24 and |H ∩ K| ≤ 4 & |H ∩ K| | 8 (By assumption, H, K are two distinct subgroups, H ∩ K < H), therefore the only option is |H∩K| = 4 and |HK| = ^{64}⁄_{4} = 16 (e.g., |H∩K| = 2 and |HK| = ^{64}⁄_{2} = 32 ⊥)
Futhermore, [H] = 8 = [H: H∩K]·[H∩K] = [H: H∩K]·4 ⇒ [H: H∩K] = [K: H∩K] = 2 ⇒ H ∩ K ◁ H, K
A subgroup of index 2 is always normal. Suppose H ≤ G: [G : H] = 2 ⇒ ∃ Two left cosets in G. If g ∈ H (H ≤ G), then obviously gH = H = Hg. If g ∉ H, gH = G \ H (there are only two cosets) ⇒[For the same reason, there are only two right cosets and both cosets partition G] Hg = G \ H ⇒ gH = Hg ⇒ H is normal.
H ∩ K ◁ H, K ⇒ H, K ≤ N(H∩K) = {g ∈ G: g (H∩K) = (H∩K)g}, N(H∩K) because N(H∩K), that is the normalizer of H∩K, is the largest subgroup in G that (H∩K) is normal inside of it. It contains both H and K since H ∩ K is normal in both H and K, and every combination, in particular the set HK (|HK| = 16) ⇒ N(H∩K) ≤ G, |N(H∩K)| ≥ 16 ⇒ [By Lagrange’s Theorem, it should divide 24, but 16 ɫ 24] |N(H∩K)| = 24 ⇒ N(H∩K) = G ⇒[H ◁ G ↭ N(H) = G] H∩K ◁ G. G is not simple because there is a normal proper subgroup (order 4), namely H∩K.
Let’s assume n_{2} = 3, n_{3} = 4 ⇒ [By Sylow’s Second Theorem, All Sylow p-subgroups are conjugates, that is, if P and Q are Sylow p-subgroups ⇒ ∃g∈G: gPg^{-1} = Q], G acts non-trivially on the set X = Syl_{2}(G) -By assumption, n_{2}=3, |Syl_{2}(G)|=3- by conjugation ⇒ [By Strong Cayley’s Theorem. Let G be a group acting on X. Let S_{X} be the symmetric group on X. For g ∈ G, define the map σ_{g}: X → X (σ_{g}:Syl_{2}(G) → Syl_{2}(G)), σ_{g}(x) = gxg^{-1}. Then, σ_{g} ∈ S_{X} and the map Φ: G → S_{X} given by Φ(g) = σ_{g} is a group homomorphism.] Φ: G → S_{Syl2(G)} (≋ S_{3}, the symmetric group on 3 letters) is a non-trivial group homomorphism. In particular, Ker(Φ) ≠ G.
S_{Syl2(G)} ≋ S_{3} ⇒ |S_{Syl2(G)}| = 3! = 6, |G| = 48 =[G/Ker(Φ) ≋ Im(Φ)] |Im(Φ)| * |Ker(Φ)|, |Ker(Φ)| ≥ 8 (Otherwise, if |Ker(Φ)| < 8, then |Im(Φ)| > 6 ⊥) Therefore, [Recall ker(Φ) is always a normal subgroup of the domain of a group homomorphism, Ker(Φ) ≠ G, |Ker(Φ)| ≥ 8] Ker(Φ) is a non-trivial proper normal subgroup of G ⇒ G is not a simple group.
Theorem. Let G be a group of order n, n be a positive integer that is not prime, and p be a prime divisor of n. If 1 is the only divisor of n that is equal to 1 (mod) p, then G is not simple.
Proof:
There are two possibilities:
If n is a prime-power, say |G| = n = p^{k} where p is prime ⇒[Let p be prime number. The center of any p-group is nontrivial. A p-group is a group in which the order of every element is a power of p. A finite group is a p-group ↭ its order is a power of p] A group of order n has a nontrivial center. Since the center of any group G is a normal subgroup, G cannot be a simple group.
If n is not a prime-power, by Sylow’s Third Theorem, n_{p} | n and n_{p} ≡ 1 (mod p) ⇒ By assumption, n_{p} = 1, the Sylow p-subgroup is proper (n is not a prime-power) and unique, and therefore it is normal and G is not simple.