Sixth rule A, Almost all properties apply to the empty set, most of them to finite sets, and dividing by zero is a terrible idea. 0 is not a number, 1 plus one equals zero in ℤ2 and $\sqrt{2}$ and $-\sqrt{2}$ are algebraically the same. Galois’ Theory and Partial differential equations are not for the faint of heart. Absolute numbers have no meaning, 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.
Some examples are ℤp, cyclic group of prime order, and An the alternating group, that is the group of even permutation of a finite set, for n ≥ 5. A4 is not simple, because V4 ◁ A4.
Theorem. Every subgroup of an Abelian group is normal.
Proof.
Let G be an Abelian group and let H ≤ G be a subgroup of G.
We claim that H is normal, i.e., H is invariant under conjugation by elements of G, that is, ∀g ∈ G, gHg-1 = H, but this is obvious ∀h ∈ H, ghg-1 =[G is Abelian] gg-1h = h.
Theorem. Let G be an Abelian group. Then, G is simple if and only if G is cyclic of primer order.
Proof.
⇒) Suppose G is Abelian and simple.
G is Abelian ⇒ Every subgroup of an Abelian group is normal. G is simple ⇒ It has no proper normal subgroups.
Let H be a subgroup, suppose x ∈ H, x ≠ e ⇒ [G is simple and Abelian, H is not a proper subgroup] H = G ⇒ G = H = ⟨x⟩ and G is cyclic ⇒ [A finite Abelian group of order n has a subgroup of order m for every divisor m of n] |G| must be prime. Therefore, G is cycle of prime order ∎
⇐) Conversely, if G is an Abelian group and cyclic of prime order ⇒[By Lagrange’s theorem, given a finite group G, the order of any subgroup divides the order of G] G does not have any nontrivial, proper subgroups, and therefore, any non-identity element would generate the whole group ⇒ G is simple ∎
Theorem. Let G be a group with N ◁ G and H ≤ G. Then, N ∩ H ◁ H.
Proof.
Claim: N ∩ H ◁ H ↭ ∀ g ∈ N ∩ H, h ∈ H, gHg-1 = H? ↭ ghg-1 ∈ H?
g ∈ N, h ∈ H ⊆ G ⇒ [N ◁ G] hgh-1 ∈ H ∎
Theorem. Let An, the alternating group, that is the group of even permutation of a finite set, for n ≥ 3 is generated by the set of 3-cycles.
Proof.
We have already demonstrated that every element of σ ∈ Sn can be written as a product of transpositions.
Since An consists of only even permutations, it suffices to show that every pair of transpositions is a product of 3-cycles. There are really three only options (consider that (ab)=(ba) 😄)
Theorem. Let An, the alternating group, that is the group of even permutation of a finite set, for n ≥ 5. Then, all 3-cycles are conjugates in An.
Proof.
Consider the two 3-cycles (abc) and (xyz) in An..
Let’s consider the following options:
(bc)(ij)(abc)[(bc)(ij)]-1 =[The Shoe-socks property] (bc)(ij)(abc)(ij)(bc) = (acb) ⇒ (abc) ~ (acb)
Since n ≥ 5 ⇒ ∃ i ∈ {1, 2, ···, n}\{a, b, c, x}, then:
(axi)(abc)(axi)-1 = (axi)(abc)(ixa) = (xbc) ⇒ (abc) ~ (xbc) 🚀
Theorem. Let An, the alternating group, that is the group of even permutation of a finite set, for n ≥ 5. Let N ◁ An. If N contains a 3-cycle, then N = An.
Proof.
N ◁ An ⇒ N is closed under conjugation. If N contains a 3-cycle, then it contains all the 3-cycles because all 3-cycles are conjugates in An, and N is closed under conjugation ⇒ [An, n ≥ 3 is generated by the set of 3-cycles] N = An ∎
Theorem. A5 is a simple group.
Proof. Let N be a nontrivial, normal subgroup of A5. By the previous theorem, it suffices to show that N contains a 3-cycle and therefore N = A5 ⊥ ⇒ N is not a nontrivial, proper normal subgroup ⇒ A5 is a simple group.
Let σ ∈ N, σ ≠ id = (). There are three options:
Theorem. Theorem. An, the alternating group, that is the group of even permutation of a finite set, n ≥ 5, is a simple group.
Proof.
Let’s proceed by strong induction, where n = 5 is our base case and it has already being demonstrated that A5 is a simple group.
Suppose that An is simple ∀5 ≤ n < k, is Ak simple?
Let N be a nontrivial, normal subgroup of Ak. Let σ ∈ N, σ ≠ 1. By a previous theorem, it suffices to show that N contains a 3-cycle and therefore N = An ⊥ N is not a trivial, proper normal subgroup.
There are two options:
Notice that στ(i) = σ(i) = j ≠ τ(j) = τ(σ(i)), thus στ ≠ τσ. Thus, α = [σ, τ] = στσ-1τ-1
The commutator of two elements, g and h ∈ G is [g, h] = ghg-1h-1. It is equal to the group’s identity if and only if g and h commute.
Since N is normal, and σ ∈ N ⇒ [σ-1 ∈ N (N is in particular a subgroup), then τσ-1τ-1 ∈N (normal) and σ(τσ-1τ-1) ∈N (subgroup)] α = σ(τσ-1τ-1) ∈ N. Besides, στσ-1 is a conjugate of τ (Recall: τ is a 3-cycle and All 3-cycles are conjugates in An), so it needs to be a 3-cycle, and α ∈ N is a product of two (not necessarily disjoint, α = (στσ-1)·τ-1 3-cycles, say (abc)(xyz), so it involves (at most) 6 letter of the alphabet, so it fixes k - 6 elements and we are done if k > 6 ↭ k -6 > 0 or k = 6 but they are not disjoint 3-cycles ⇒ Case A, α ∈ N, α fixes at least one point ⇒ N = Ak -we still have a problem if k = 6 and α ∈ N is a product of non-disjoint 3-cycles.
Finally, let’s address the case that α ∈ N is a product of disjoint 3-cycles in A6 (k = 6), we may assume α = (123)(456). Let β = (234). Then (15324) =🚀 [α, β] = α(βα-1β-1)∈ N (Recall, α ∈ N ⇒[N≤G] α-1 ∈ N, N◁G ⇒ βα-1β-1∈ N ⇒ α(βα-1β-1)∈ N). Thus, [α, β] fixes 6 and the above argument applies, and in either case, N = Ak ∎
🚀 αβα-1β-1 = (123)(456)(234)(132)(465)(243) = (15324)