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 ℤ

_{2}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 G and the trivial subgroup.

Some examples are **ℤ _{p}**, cyclic group of prime order, and

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} = gg^{-1}h = 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, 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.

- e ∈ N, e ∈ H because N and H are both subgroups ⇒ e ∈ N ∩ H.
- ∀g, h ∈ N ∩ H, gh ∈ N, gh ∈ H because N and H are both subgroups so they are both closed under the group operation, hence gh ∈ N ∩ H.
- ∀g ∈ N ∩ H, g
^{-1}∈ N, g^{-1}∈ H because N and H are both closed under taking inverses, and therefore g^{-1}∈ N ∩ H. Therefore, N ∩ H ≤ G.

Claim: **N ∩ H ◁ H** ↭ ∀ g ∈ N ∩ H, h ∈ H, gHg^{-1} = H?

g ∈ N, h ∈ H ⊆ G ⇒ [N ◁ G] hgh^{-1} ∈ H ∎

Theorem. Let A_{n}, 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 ∈ S_{n} can be written as a product of transpositions.

Since A_{n} 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) 😄)

- (ab)(ab) = 1 = (abc)(acb)
- (ab)(ac) = (acb)
- (ab)(cd) = (acb)(acd) ∎

Theorem. Let A_{n}, the alternating group, that is the group of even permutation of a finite set, for n ≥ 5. Then, all 3-cycles are conjugates in A_{n}.

Proof.

Consider the two 3-cycles (abc) and (xyz) in A_{n.}.

Let’s consider the following options:

- {a, b, c} = {x, y, z} ⇒ (xyz) = (abc) or (acb). If (xyz) = (abc) ⇒ e(abc)e
^{-1}= abc,**(abc)~(abc)**-Anyway ~ is reflexive-. If (xyz) = (acb). Since n ≥ 5 ⇒ ∃ i, j ∈ {1, 2, ···, n}\{a, b, c}, then:

(bc)(ij)(abc)[(bc)(ij)]^{-1} =[The Shoe-socks property] (bc)(ij)(abc)(ij)(bc) = (acb) ⇒ **(abc) ~ (acb)**

- {a, b, c} ≠ {x, y, z}. Without the loss of generality, let’s suppose that a ∉ {x, y, z} and x ∉ {a, b, c} ↭ (abc) fixes x and (xyz) fixes a.

Since n ≥ 5 ⇒ ∃ i ∈ {1, 2, ···, n}\{a, b, c, x}, then:

(axi)(abc)(axi)^{-1} = (axi)(abc)(ixa) = (xbc) ⇒ **(abc) ~ (xbc)** 🚀

- If {x, b, c} = {x, y, z}, then we return to the original case, so we have just proven 🚀 that (abc) ~ (xbc) and since {x, b, c} = {x, y, z}, then (xbc) ~ (xyz) ⇒ (abc) ~ (xyz) ∎
- Otherwise, if {x, b, c} ≠ {x, y, z}, then we could repeat the previous argument and we will obtain (abc) ~
**(xbc) ~ (xyz)**for some i ∈ {1, 2, ···, n}\{a, b, c,**x, y**}. In either case, the transitivity of conjugacy (it is an equivalent relation) guaranties that (a, b, c) ~ (x, y, z) ∈ A_{n}.

Theorem. Let A_{n}, the alternating group, that is the group of even permutation of a finite set, for n ≥ 5. Let N ◁ A_{n}. If N contains a 3-cycle, then N = A_{n}.

Proof.

N ◁ A_{n} ⇒ 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 A_{n}, and N is closed under conjugation ⇒ [A_{n}, n ≥ 3 is generated by the set of 3-cycles] N = A_{n} ∎

Theorem. A_{5} is a simple group.

Proof. Let N be a nontrivial, normal subgroup of A_{5}. By the previous theorem, it suffices to show that N contains a 3-cycle and therefore **N = A _{5} ⇒ N is not a nontrivial, proper normal subgroup ⇒ A_{5} is a simple group.**

Let σ ∈ N, σ ≠ id = (). There are three options:

- σ is a 3-cycle, we can assume without the lose of generality, σ = (123) ∈ N, then we are done.
- σ is a 2-2 cycle, we can assume without the lose of generality, σ = (12)(34) ∈ N, and let consider τ = (12)(35) ⇒ [N ◁ G] τστ
^{-1}∈ N, and [N is a subgroup, too] (τστ^{-1})σ^{-1}∈ N, (τστ^{-1})σ^{-1}= (12)(35)(12)(34)(35)(12)(34)(12) =[Notice that (12)(34)(12) = (12)(12)(34) = (34)] (35)(34)(35)(34) = (354)∈ N. - σ is a 5-cycle, we can assume without the lose of generality, σ = (12345) ∈ N, and let τ = (132) ⇒ [N ◁ G] τστ
^{-1}∈ N, and [N is a subgroup, too] (τστ^{-1})σ^{-1}∈ N, (τστ^{-1})σ^{-1}= (132)(12345)(123)(15432) = (134)∎

Theorem. Theorem. A_{n}, 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 induction, where n = 5 is our base case and it has already being demonstrated that A_{5} is a simple group.

Suppose that A_{n} is simple ∀5 ≤ n < k.

Let N be a nontrivial, normal subgroup of A_{k}. Let σ ∈ N, σ ≠ 1. By a previous theorem, it suffices to show that N contains a 3-cycle and therefore N = A_{n} ⊥ N is not a trivial, proper normal subgroup.

There are two options:

**Case A. σ ∈ N has a fixed point.**Up to renaming, we could suppose without losing generality that σ(k) = k. Then σ ∈ A_{k-1}and we are going to be able to use our induction hypothesis ⇒**A**and [Th. G -in our case, A_{k-1}is simple_{k-1}- is a group, N ◁ G, and H ≤ G (in our case, A_{k-1}≤ A_{k-1}). Then, N ∩ H ◁ H]**N ∩ A**, σ ∈ N ∩ A_{k-1}◁ A_{k-1}_{k-1}, i.e., N ∩ A_{k-1}is a nontrivial, normal subgroup of a simple group ⇒ N ∩ A_{k-1}= A_{k-1}⇒ A_{k-1}⊆ N, A_{k-1}is generated by the set of 3-cycles (Recall: Let A_{n}, 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), so N contains a 3-cycle ⇒ N = A_{k}- Case B. Suppose that σ ∈ N does not have any fixed points. ∃i, j: σ(i) = j ≠ i, Let τ be a 3-cycle such that τ(i) = i and τ(j) ≠ j.

Notice that στ(i) = σ(i) = j ≠ τ(j) = τ(σ(i)), thus στ ≠ τσ. Thus, α = [σ, τ] = στσ^{-1}τ^{-1}

The commutator of two elements, g and h ∈ G is [g, h] = ghg^{-1}h^{-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^{-1}τ^{-1} ∈N (normal) and σ(τσ^{-1}τ^{-1}) ∈N (subgroup)] α = σ(τσ^{-1}τ^{-1}) ∈ N. Besides, στσ^{-1} is a conjugate of τ (Recall: τ is a 3-cycle and All 3-cycles are conjugates in A_{n}), 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 (we are done if k > 6) points or α ∈ N is a product of non-disjoint 3-cycles ⇒ **Case A, α ∈ N, α fixes at least one point ⇒ N = A _{k}** ∎

Finally, let’s address the case that α ∈ N is a product of disjoint 3-cycles in A_{6}, we may assume α = (123)(456). Let β = (234). Then (15324) =🚀 [α, β] = α(βα^{-1}β^{-1})∈ N (Recall, α ∈ N ⇒[N≤G] α^{-1} ∈ N, N◁G ⇒ βα^{-1}β^{-1}∈ N ⇒ α(βα^{-1}β^{-1})∈ N). Thus, [α, β] fixes 6 and the above argument applies, and in either case, N = A_{k} ∎

🚀 αβα^{-1}β^{-1} = (123)(456)(234)(132)(465)(243) = (15324)

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.

- NPTEL-NOC IITM, Introduction to Galois Theory.
- Algebra, Second Edition, by Michael Artin.
- LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
- Field and Galois Theory, by Patrick Morandi. Springer.
- Michael Penn (Abstract Algebra), and MathMajor.
- Contemporary Abstract Algebra, Joseph, A. Gallian.
- Andrew Misseldine: College Algebra and Abstract Algebra.