# Simple Groups

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 G and the trivial subgroup.

Some examples are p, 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 = 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, 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 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) 😄)

1. (ab)(ab) = 1 = (abc)(acb)
2. (ab)(ac) = (acb)
3. (ab)(cd) = (acb)(acd) ∎

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:

• {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) 🚀

1. 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) ∎
2. 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) ∈ An.

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:

1. σ is a 3-cycle, we can assume without the lose of generality, σ = (123) ∈ N, then we are done.
2. 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-1-1 ∈ N, (τστ-1-1 = (12)(35)(12)(34)(35)(12)(34)(12) =[Notice that (12)(34)(12) = (12)(12)(34) = (34)] (35)(34)(35)(34) = (354)∈ N.
3. 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-1-1 ∈ N, (τστ-1-1 = (132)(12345)(123)(15432) = (134)∎

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 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.

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:

1. Case A. σ ∈ N has a fixed point. Up to renaming, we could suppose without losing generality that σ(k) = k. Then σ ∈ Ak-1 and we are going to be able to use our induction hypothesis ⇒ Ak-1 is simple and [Th. G -in our case, Ak-1- is a group, N ◁ G, and H ≤ G (in our case, Ak-1 ≤ Ak-1). Then, N ∩ H ◁ H] N ∩ Ak-1 ◁ Ak-1, σ ∈ N ∩ Ak-1, i.e., N ∩ Ak-1 is a nontrivial, normal subgroup of a simple group ⇒ N ∩ Ak-1= Ak-1 ⇒ Ak-1 ⊆ N, Ak-1 is generated by the set of 3-cycles (Recall: 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), so N contains a 3-cycle ⇒ N = Ak
2. 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-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 (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 = Ak

Finally, let’s address the case that α ∈ N is a product of disjoint 3-cycles in A6, 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)

# Bibliography

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