A5 is simple. Primitive element theorem.

No pressure, no diamonds, Thomas Carlyle.

For every problem there is always, at least, a solution which seems quite plausible. It is simple and clean, direct, neat and nice, and yet very wrong, #Anawim, justtothepoint.com

Recall:

• In a group G, two elements g, h ∈ G are called conjugate when h = xgx^-1 for some x ∈ G.
• Let a ∈ G, its conjugacy class [a] is the set of elements conjugate to it, i.e., [a] = {xgx^-1: x ∈ G}, e.g., [(12)] = {(12), (23), (13)}. S₃ has three conjugacy classes, namely [(1)], [(12)] = {(12), (23), (13)}, and [(123)] = {(123), (132)}

  (13)(12)(13) = (23), notice that (13)^-1 = (13). (23)(12)(23) = (13)

Theorem. For each cycle (i₁i₂···i_k) in S_n and each σ ∈ S_n, σ(i₁i₂···i_k)σ^-1 = (σ(i₁)σ(i₂)···σ(i_k))

Proof.

Let π = σ(i₁i₂···i_k)σ^-1.

1. π(σ(i₁)) = σ(i₁i₂···i_k)σ^-1σ(i₁) = σ(i₁i₂···i_k)(σ^-1σ)(i₁) = σ(i₁i₂···i_k)(i₁) = σ(i₂). Therefore, π sends σ(i₁) to σ(i₂), σ(i₂) to σ(i₃), …, σ(i_k) to σ(i₁)

2. Let’s show that π does not move anything other that σ(i₁), σ(i₂)···σ(i_k). Let pick a that is not among σ(i₁), σ(i₂)···σ(i_k). Claim: π(a) = a ↭ σ(i₁i₂···i_k)σ^-1(a) = a.

Since a ≠ σ(i_j) for j = 1, 2, ···, k ⇒ σ^-1(a) ≠ i_j for j = 1, 2, ···, k ⇒ The cycle (i₁i₂···i_k) does not move it (“it” means σ^-1(a)) ⇒ π(a) = σ(i₁i₂···i_k)σ^-1(a) = σ(i₁i₂···i_k)(σ^-1(a)) = σ(σ^-1(a)) = a∎

Theorem. All cycles of the same length in S_n are conjugate.

Proof.

Let’s pick two cycles of the same length in S_n, say (a₁a₂···a_k), (b₁b₂···b_k). Let’s select σ ∈ S_n such that σ(a₁) = b₁, σ(a₂) = b₂, ··· σ(a_k) = b_k, and let σ be an arbitrary bijection from the complement of {a₁, a₂, ···, a_k} to the complement of {b₁, b₂, ···, b_k}.

Then, using the previous theorem, we see conjugation by σ carries the first k-cycle to the second. 

Lemma 1. For all n ≥ 3, the alternating group A_n is generated by 3 cycles.

Proof.

The identity (1) is (123)(132), which is a product of 3-cycles. Now, let’s take a non-identity element of A_n, say σ and write it as a product of transpositions, σ = τ₁τ₂···τ_r.

By definition, A_n = {σ ∈ S_n : σ is even, sgn(σ) = 1}, so the left side has sign 1 (Recall: sgn(σ) = (-1)^m where m is the number of transpositions in the decomposition of σ), and the right side has sign (-1)^r, so r is even. Therefore, we can collect the products on the right side of the equation into successive transpositions τ_iτ_i+1, i = 1, 3, and so on.

We will now show that every product of two transpositions in S_n is a product of two 3-cycles, hence σ is a product of 3-cycles. There are three possibilities:

1. If both transpositions are equal, τ_iτ_i+1, then their product is the identity, τ_iτ_i+1 = (1) = (123)(132).
2. If τ_i and τ_i+1 have exactly one element in common, say τ_i = (ab) and τ_i+1 = (bc) ⇒ τ_iτ_i+1 = (ab)(ac) = (acb) = (abc)(abc), and therefore, we can replace τ_iτ_i+1 with a product of two 3-cycles.
3. If τ_i and τ_i+1 have no elements in common, so they are disjoint, say τ_i = (ab) and τ_i+1 = (cd) ⇒ τ_iτ_i+1 = (ab)(cd) = [(bc)(bc) = (id)] ((ab)(bc))((bc)(cd)) = (bca)(bcd) = [This last step is done exclusively for aesthetically purposes] (abc)(bcd) and we can replace τ_iτ_i+1 with a product of two 3-cycles.

Lemma 2. For n ≥ 5, all 3-cycles are conjugate in A_n

Proof.

We will show every 3-cycle in A_n is conjugate within A_n to (123). We are not really losing any generality over here.

Let σ be a 3-cycle in A_n ⇒ [Previous Theorem. All cycles of the same length in S_n are conjugate. This may no be true in subgroups of S_n] σ can be conjugated to (123) in S_n: ∃τ ∈S_n such that τστ^-1 = (123)

If τ ∈ Aⁿ, we are done, 🎉

Otherwise, τ ∉ A_n, let τ’ = (45)τ, so τ’∈ A_n and τ’στ’^-1 = [(45)^-1 = (45), Shoes and sock Theorem] (45)τστ^-1(45) = [τστ^-1 = (123)] (45)(123)(45) = (123) ∎

Theorem. A₅ is simple.

A_n is simple ∀n ≥ 5, and S_n, A_n are not solvable ∀n ≥ 5.

Proof

Let’s suppose for the sake of contradiction that there exist N ◁ A₅, a non-trivial normal subgroup of A₅, |N| > 1. So we can pick σ ∈ N, σ ≠ (1).

Claim: N = A₅. We will show that N contains a 3 cycle ⇒ [Lemma 2 and N ◁ A₅] all 3-cycles are conjugate in A₅ (n = 5) and they are all contained in N (normal) ⇒ [Lemma 1] N = A₅.

N ⊆ A₅, so only even permutations are allowed in A₅, that is, the cycle structure of σ is a three-cycle, say (abc) (🎉), (ab)(cd), and (abcde)

(ab)(cd) ∈ N ⇒ [N ◁ A₅] (abe)(ab)(cd)(abe)^-1 ∈ N, then (abe)(ab)(cd)(abe)^-1(ab)(cd) ∈ N, but (abe)(ab)(cd)(abe)^-1(ab)(cd) = (abe)(ab)(cd)(aeb)(ab)(cd) = (aeb) ∈ N

(abcde) ∈ N ⇒ [N ◁ A₅] (abc)(abcde)(abc)^-1∈ N ⇒ [N is a subgroup, too] ((abc)(abcde)(abc)^-1)(abcde)^-1 = ((abc)(abcde)(acb))(aedcb) = (adebc)(aedcb) = (abd)∈ N

Definition. An extension of fields K/F is called simple if K = F(α) for some element α. In this case, α is called a primitive element for K.

Primitive element theorem. Let K/F be a finite extension. Then, a finite extension of fields is primitive if and only if there exists only finitely many subfields of K containing F.

Proof:

⇒ Let K/F be a primitive extension, K = F(α), and let f ∈ F[x] be the irreducible polynomial of α over F. There is an injective map from the intermediate fields of K/F to the set of divisors of f in K[x] which is obviously a finite set, Φ: {intermediate fields of K/F} → {divisors of f in K[x]}

Let L be an intermediate field, and let g ∈ L[x] be the irreducible polynomial of α over L. Hence, g divides f in L[x]

Let L’ be the field generated over F by the coefficients of g(x). Obviously, g ∈ L’[x], and since K = F(α), then K = L(α) = L’(α).

[K : L] = degree of the irreducible polynomial of α over L = deg(g) = degree of the irreducible polynomial of α over L’ = [K : L’] ⇒ [K : L] = [K : L’] and L’⊆ L ⇒ L = L’ = F(coefficients of g), g | f in L[x] and K[x]. It follows that the subfields of K containing F are the subfields generated by the coefficients of the monic factors of f(x), hence there are only finitely many such subfields. This is one reasoning, let’s see the same result with another argument.

The map is defined as follows, Φ: {intermediate fields of K/F} → {divisors of f in K[x]}, Φ(L) = g where g is the irreducible polynomial of α over L, g divides f in L[x], and obviously in K[x], too.

Suppose Φ(L₁) = Φ(L₂) = g ⇒ L₁ = F(coefficients of g) = L₂ ⇒ Φ is one to one ⇒ Since {divisors of f in K[x]} is a finite set, K/F has only finitely many intermediate fields.

⇐ Suppose conversely that K/F has only finitely many intermediate fields.

• Case 1. If F is a finite field, then we have that K is a finite field ⇒ [Recall: The multiplicative group of non-zero elements of a finite field is cyclic.] K^x = K\{0} is a cyclic group, generated by, say α. Then, clearly K = F(α) 🎉
• Case 2. Suppose f is infinite. By assumption, [K : F] < ∞, then ∃α₁, α₂, ···, α_n ∈ K such that K = F(α₁, α₂, ···, α_n)

It suffices to show that ∀α, β ∈ K, ∃ γ ∈ K such that F(α, β) = F(γ)

Case base, n = 2, F(α, β) = F(γ). Induction hypothesis, F(α₁, α₂, ···, α_n-1) = F(γ’), therefore F(α₁, α₂, ···, α_n) = F(γ’, α_n) = [n = 2] F(γ’’)

∀c ∈ F, consider α + cβ ∈ F(α, β), and the subfields F(α +cβ), F ⊆ F(α + cβ) ⊆ F(α, β) ⊆ K. Therefore, since there are obviously infinitely many choices for c ∈ F and (by assumption) only finitely many such subfields ⇒ ∃c₁, c₂ ∈ F, c₁ ≠ c₂, such that F(α + c₁β) = F(α + c₂β).

Futhermore, α + c₁β and α + c₂β both lie in F(α + c₁β), and taking their differences show that (α + c₁β) - (α + c₂β) = (c₁ -c₂)β ∈ F(α + c₁β) ⇒ [c₁≠c₂, so c₁-c₂≠0] β ∈ F(α + c₁β), and then [c₁β ∈ F(α + c₁β)] also α = α + c₁β -c₁β ∈ F(α + c₁β) ⇒ F(α, β) ⊆ F(α + c₁β) ⊆_trivially F(α, β) ⇒ F(α, β) = F(α + c₁β)∎

Corollary. A finite separable extension K/F is primitive.

Proof.

Let L be the Galois closure of K over F. Since L/F is Galois, by Galois’ Main Theorem, L/F has only finitely many intermediate fields (there is a bijection between these intermediate fields and the subgroups of its Galois group which is indeed a finite set) ⇒ [F ⊆ K ⊆ L] K/F has only finitely many intermediate fields ⇒ [Primitive element theorem. Let K/F be a finite extension. Then, a finite extension of fields is primitive if and only if there exists only finitely many subfields of K containing F.] K/F is primitive ∎

Corollary. Any finite extension of fields of characteristic zero is simple.

Proof.

It follows from the previous corollary since any finite extension of fields in characteristic zero is separable∎