The Fundamental Theorem of Galois Theory and Normal Subgroups · Normal Subgroups Fundamental...

Normal Subgroups Fundamental Theorem of Galois Theory The Alternating Group

The Fundamental Theorem of GaloisTheory and Normal Subgroups

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Fundamental Theorem of Galois Theory and Normal Subgroups

Introduction

1. The Fundamental Theorem of Galois Theory tells when, ina nested sequence of field extensions F⊆ D⊆ E we havethat D is a normal extension of F.

2. The statement of the Fundamental Theorem of GaloisTheory will make it clear why normal subgroups areimportant for us.

3. But to competently work with normal subgroups in theproof of the Fundamental Theorem of Galois Theory, weshould start by investigating normality.

4. Let G be a group and let H be a subgroup. A simple way toview normal subgroups is to consider the equivalencerelation a∼ b iff ab−1 ∈ H.

5. We want to do algebra with the equivalence classes of thisrelation (similar to arithmetic modulo m).

Introduction1. The Fundamental Theorem of Galois Theory tells when, in

a nested sequence of field extensions F⊆ D⊆ E we havethat D is a normal extension of F.

5. We want to do algebra with the equivalence classes of thisrelation

(similar to arithmetic modulo m).

Definition.

Let G be a group. For any two sets A,B⊆ G, wedefine AB := {ab : a ∈ A,b ∈ B}. If A = {a} we also write aBfor {a}B, and if B = {b} we also write Ab for A{b}.

Definition. Let G be a group and let H ⊆ G be a subgroup.Then the sets gH = {gh : h ∈ H} are called the left cosets of Hand the sets Hg = {hg : h ∈ H} are called the right cosets of H.

Definition. Let G be a group and let N ⊆ G be a subgroup. Thesubgroup N is called a normal subgroup of G, also denotedN CG, iff for all g ∈ G we have that gN = Ng.

Definition. Let G be a group.

For any two sets A,B⊆ G, wedefine AB := {ab : a ∈ A,b ∈ B}. If A = {a} we also write aBfor {a}B, and if B = {b} we also write Ab for A{b}.

Definition. Let G be a group. For any two sets A,B⊆ G, wedefine AB := {ab : a ∈ A,b ∈ B}.

If A = {a} we also write aBfor {a}B, and if B = {b} we also write Ab for A{b}.

Definition. Let G be a group. For any two sets A,B⊆ G, wedefine AB := {ab : a ∈ A,b ∈ B}. If A = {a} we also write aBfor {a}B

, and if B = {b} we also write Ab for A{b}.

Definition. Let G be a group. For any two sets A,B⊆ G, wedefine AB := {ab : a ∈ A,b ∈ B}. If A = {a} we also write aBfor {a}B, and if B = {b} we also write Ab for A{b}.

Definition.

Let G be a group and let H ⊆ G be a subgroup.Then the sets gH = {gh : h ∈ H} are called the left cosets of Hand the sets Hg = {hg : h ∈ H} are called the right cosets of H.

Definition. Let G be a group and let H ⊆ G be a subgroup.

Then the sets gH = {gh : h ∈ H} are called the left cosets of Hand the sets Hg = {hg : h ∈ H} are called the right cosets of H.

Definition. Let G be a group and let H ⊆ G be a subgroup.Then the sets gH = {gh : h ∈ H} are called the left cosets of H

and the sets Hg = {hg : h ∈ H} are called the right cosets of H.

Definition.

Let G be a group and let N ⊆ G be a subgroup. Thesubgroup N is called a normal subgroup of G, also denotedN CG, iff for all g ∈ G we have that gN = Ng.

Definition. Let G be a group and let N ⊆ G be a subgroup.

Thesubgroup N is called a normal subgroup of G, also denotedN CG, iff for all g ∈ G we have that gN = Ng.

Lemma.

Let G be a group and let N ⊆ G be a subgroup. Thenthe following are equivalent.

1. N is a normal subgroup of G.2. For all g ∈ G we have that gNg−1 = N.3. For all g ∈ G we have that gNg−1 ⊆ N.

Proof. Good exercise.

Theorem. Let G be a group and let N CG be a normalsubgroup. Then the operation gN ◦ kN := (g◦ k)N turns the setof cosets of N into a group. This group is typically denotedG/N, and it is called the quotient group or factor group.

Proof. Simple exercise (maybe too simple).

Lemma. Let G be a group and let N ⊆ G be a subgroup.

Thenthe following are equivalent.

Lemma. Let G be a group and let N ⊆ G be a subgroup. Thenthe following are equivalent.

1. N is a normal subgroup of G.

2. For all g ∈ G we have that gNg−1 = N.3. For all g ∈ G we have that gNg−1 ⊆ N.

1. N is a normal subgroup of G.2. For all g ∈ G we have that gNg−1 = N.

3. For all g ∈ G we have that gNg−1 ⊆ N.

Proof.

Good exercise.

Theorem.

Let G be a group and let N CG be a normalsubgroup. Then the operation gN ◦ kN := (g◦ k)N turns the setof cosets of N into a group. This group is typically denotedG/N, and it is called the quotient group or factor group.

Theorem. Let G be a group and let N CG be a normalsubgroup.

Then the operation gN ◦ kN := (g◦ k)N turns the setof cosets of N into a group. This group is typically denotedG/N, and it is called the quotient group or factor group.

Theorem. Let G be a group and let N CG be a normalsubgroup. Then the operation gN ◦ kN := (g◦ k)N turns the setof cosets of N into a group.

This group is typically denotedG/N, and it is called the quotient group or factor group.

Proof.

Simple exercise (maybe too simple).

Proof. Simple exercise

(maybe too simple).

Why Do We Need Normality of N?

(gN)(hN) = {gn : n ∈ N}{hm : m ∈ N}= {gnhm : n,m ∈ N}= {ghpm : m,p ∈ N}= {gha : a ∈ N}= ghN.

or (gN)(hN) = g(Nh)N = ghNN = ghN. So for normalsubgroups the multiplication of the cosets is a group operation.

(gN)(hN)

= {gn : n ∈ N}{hm : m ∈ N}= {gnhm : n,m ∈ N}= {ghpm : m,p ∈ N}= {gha : a ∈ N}= ghN.

(gN)(hN) = {gn : n ∈ N}{hm : m ∈ N}

= {gnhm : n,m ∈ N}= {ghpm : m,p ∈ N}= {gha : a ∈ N}= ghN.

(gN)(hN) = {gn : n ∈ N}{hm : m ∈ N}= {gnhm : n,m ∈ N}

= {ghpm : m,p ∈ N}= {gha : a ∈ N}= ghN.

(gN)(hN) = {gn : n ∈ N}{hm : m ∈ N}= {gnhm : n,m ∈ N}= {ghpm : m,p ∈ N}

= {gha : a ∈ N}= ghN.

(gN)(hN) = {gn : n ∈ N}{hm : m ∈ N}= {gnhm : n,m ∈ N}= {ghpm : m,p ∈ N}= {gha : a ∈ N}

= ghN.

(gN)(hN) = g(Nh)N = ghNN = ghN. So for normalsubgroups the multiplication of the cosets is a group operation.

or (gN)(hN)

= g(Nh)N = ghNN = ghN. So for normalsubgroups the multiplication of the cosets is a group operation.

or (gN)(hN) = g(Nh)N

= ghNN = ghN. So for normalsubgroups the multiplication of the cosets is a group operation.

or (gN)(hN) = g(Nh)N = ghNN

= ghN. So for normalsubgroups the multiplication of the cosets is a group operation.

or (gN)(hN) = g(Nh)N = ghNN = ghN.

So for normalsubgroups the multiplication of the cosets is a group operation.

Corollary

(to the characterization of splitting fields/normalextensions). Let (F,+, ·) be a field of characteristic 0, let E bea normal extension of F and let D be an intermediate field, thatis, a field so that F⊆D⊆ E. Then E is a normal extension of D.

Proof. E is the splitting field of a polynomial f ∈ F[x]. BecauseF[x]⊆ D[x], E is the splitting field of a polynomial f ∈ D[x].Hence E is a normal extension of D.

(Characteristic 0 was needed so we could apply thecharacterization of normal extensions. We need characteristic 0in our version of the Fundamental Theorem of Galois Theorybelow, because we need the characterization of normalextensions and the representation of splitting fields as F(θ). Sofar, our need for characteristic 0 all traces back to us using therepresentation of splitting fields as F(θ).)

Corollary (to the characterization of splitting fields/normalextensions).

Let (F,+, ·) be a field of characteristic 0, let E bea normal extension of F and let D be an intermediate field, thatis, a field so that F⊆D⊆ E. Then E is a normal extension of D.

Corollary (to the characterization of splitting fields/normalextensions). Let (F,+, ·) be a field of characteristic 0, let E bea normal extension of F and let D be an intermediate field, thatis, a field so that F⊆D⊆ E.

Then E is a normal extension of D.

Corollary (to the characterization of splitting fields/normalextensions). Let (F,+, ·) be a field of characteristic 0, let E bea normal extension of F and let D be an intermediate field, thatis, a field so that F⊆D⊆ E. Then E is a normal extension of D.

Proof.

E is the splitting field of a polynomial f ∈ F[x]. BecauseF[x]⊆ D[x], E is the splitting field of a polynomial f ∈ D[x].Hence E is a normal extension of D.

Proof. E is the splitting field of a polynomial f ∈ F[x].

BecauseF[x]⊆ D[x], E is the splitting field of a polynomial f ∈ D[x].Hence E is a normal extension of D.

Proof. E is the splitting field of a polynomial f ∈ F[x]. BecauseF[x]⊆ D[x], E is the splitting field of a polynomial f ∈ D[x].

Hence E is a normal extension of D.

(Characteristic 0 was needed so we could apply thecharacterization of normal extensions.

We need characteristic 0in our version of the Fundamental Theorem of Galois Theorybelow, because we need the characterization of normalextensions and the representation of splitting fields as F(θ). Sofar, our need for characteristic 0 all traces back to us using therepresentation of splitting fields as F(θ).)

(Characteristic 0 was needed so we could apply thecharacterization of normal extensions. We need characteristic 0in our version of the Fundamental Theorem of Galois Theorybelow, because we need the characterization of normalextensions and the representation of splitting fields as F(θ).

Sofar, our need for characteristic 0 all traces back to us using therepresentation of splitting fields as F(θ).)

Theorem.

Fundamental Theorem of Galois Theory. Let F be afield of characteristic 0 and let E be a normal extension of F.

1. D 7→ G(E/D) is a bijective correspondence between the fields Dwith F⊆ D⊆ E and the subgroups of G(E/F).

2. Let D be an intermediate field between F and E. Then

[E : D] =∣∣G(E/D)

∣∣ and [D : F] =

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

3. Let D1, D2 be intermediate fields between F and E. ThenF⊆ D1 ⊂ D2 ⊆ E (where the central inclusion is proper)iff {id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where thecentral inclusion is proper).

4. Let D be an intermediate field between F and E. Then D isa normal extension of F iff G(E/D) is a normal subgroupof G(E/F). In this case G(D/F) is isomorphic toG(E/F)/G(E/D).

Theorem. Fundamental Theorem of Galois Theory.

Let F be afield of characteristic 0 and let E be a normal extension of F.

[E : D] =∣∣G(E/D)

∣∣ and [D : F] =

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

Theorem. Fundamental Theorem of Galois Theory. Let F be afield of characteristic 0 and let E be a normal extension of F.

[E : D] =∣∣G(E/D)

∣∣ and [D : F] =

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

[E : D] =∣∣G(E/D)

∣∣ and [D : F] =

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

[E : D] =∣∣G(E/D)

∣∣ and [D : F] =

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

[E : D] =∣∣G(E/D)

∣∣ and [D : F] =

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

3. Let D1, D2 be intermediate fields between F and E. ThenF⊆ D1 ⊂ D2 ⊆ E

(where the central inclusion is proper)iff {id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where thecentral inclusion is proper).

[E : D] =∣∣G(E/D)

∣∣ and [D : F] =

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

3. Let D1, D2 be intermediate fields between F and E. ThenF⊆ D1 ⊂ D2 ⊆ E (where the central inclusion is proper)

iff {id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where thecentral inclusion is proper).

[E : D] =∣∣G(E/D)

∣∣ and [D : F] =

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

3. Let D1, D2 be intermediate fields between F and E. ThenF⊆ D1 ⊂ D2 ⊆ E (where the central inclusion is proper)iff {id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F)

(where thecentral inclusion is proper).

[E : D] =∣∣G(E/D)

∣∣ and [D : F] =

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

[E : D] =∣∣G(E/D)

∣∣ and [D : F] =

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

4. Let D be an intermediate field between F and E. Then D isa normal extension of F iff G(E/D) is a normal subgroupof G(E/F).

In this case G(D/F) is isomorphic toG(E/F)/G(E/D).

[E : D] =∣∣G(E/D)

∣∣ and [D : F] =

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

Proof (part 1, injectivity).

First recall that if D is anintermediate field so that F⊆ D⊆ E, then G(E/D) is asubgroup of G(E/F).Now let D1 and D2 be intermediate fields, that is, F⊆ D1 ⊆ Eand F⊆ D2 ⊆ E, so that G(E/D1) = G(E/D2). Then E is anormal extension of D1 and of D2. Hence D1 is the fixed fieldof G(E/D1) and D2 is the fixed field of G(E/D2). FromG(E/D1) = G(E/D2) we infer that D1 = D2. Hence thecorrespondence is one to one.

Proof (part 1, injectivity). First recall that if D is anintermediate field so that F⊆ D⊆ E, then G(E/D) is asubgroup of G(E/F).

Now let D1 and D2 be intermediate fields, that is, F⊆ D1 ⊆ Eand F⊆ D2 ⊆ E, so that G(E/D1) = G(E/D2). Then E is anormal extension of D1 and of D2. Hence D1 is the fixed fieldof G(E/D1) and D2 is the fixed field of G(E/D2). FromG(E/D1) = G(E/D2) we infer that D1 = D2. Hence thecorrespondence is one to one.

Proof (part 1, injectivity). First recall that if D is anintermediate field so that F⊆ D⊆ E, then G(E/D) is asubgroup of G(E/F).Now let D1 and D2 be intermediate fields

, that is, F⊆ D1 ⊆ Eand F⊆ D2 ⊆ E, so that G(E/D1) = G(E/D2). Then E is anormal extension of D1 and of D2. Hence D1 is the fixed fieldof G(E/D1) and D2 is the fixed field of G(E/D2). FromG(E/D1) = G(E/D2) we infer that D1 = D2. Hence thecorrespondence is one to one.

Proof (part 1, injectivity). First recall that if D is anintermediate field so that F⊆ D⊆ E, then G(E/D) is asubgroup of G(E/F).Now let D1 and D2 be intermediate fields, that is, F⊆ D1 ⊆ Eand F⊆ D2 ⊆ E

, so that G(E/D1) = G(E/D2). Then E is anormal extension of D1 and of D2. Hence D1 is the fixed fieldof G(E/D1) and D2 is the fixed field of G(E/D2). FromG(E/D1) = G(E/D2) we infer that D1 = D2. Hence thecorrespondence is one to one.

Proof (part 1, injectivity). First recall that if D is anintermediate field so that F⊆ D⊆ E, then G(E/D) is asubgroup of G(E/F).Now let D1 and D2 be intermediate fields, that is, F⊆ D1 ⊆ Eand F⊆ D2 ⊆ E, so that G(E/D1) = G(E/D2).

Then E is anormal extension of D1 and of D2. Hence D1 is the fixed fieldof G(E/D1) and D2 is the fixed field of G(E/D2). FromG(E/D1) = G(E/D2) we infer that D1 = D2. Hence thecorrespondence is one to one.

Proof (part 1, injectivity). First recall that if D is anintermediate field so that F⊆ D⊆ E, then G(E/D) is asubgroup of G(E/F).Now let D1 and D2 be intermediate fields, that is, F⊆ D1 ⊆ Eand F⊆ D2 ⊆ E, so that G(E/D1) = G(E/D2). Then E is anormal extension of D1 and of D2.

Hence D1 is the fixed fieldof G(E/D1) and D2 is the fixed field of G(E/D2). FromG(E/D1) = G(E/D2) we infer that D1 = D2. Hence thecorrespondence is one to one.

Proof (part 1, injectivity). First recall that if D is anintermediate field so that F⊆ D⊆ E, then G(E/D) is asubgroup of G(E/F).Now let D1 and D2 be intermediate fields, that is, F⊆ D1 ⊆ Eand F⊆ D2 ⊆ E, so that G(E/D1) = G(E/D2). Then E is anormal extension of D1 and of D2. Hence D1 is the fixed fieldof G(E/D1)

and D2 is the fixed field of G(E/D2). FromG(E/D1) = G(E/D2) we infer that D1 = D2. Hence thecorrespondence is one to one.

Proof (part 1, injectivity). First recall that if D is anintermediate field so that F⊆ D⊆ E, then G(E/D) is asubgroup of G(E/F).Now let D1 and D2 be intermediate fields, that is, F⊆ D1 ⊆ Eand F⊆ D2 ⊆ E, so that G(E/D1) = G(E/D2). Then E is anormal extension of D1 and of D2. Hence D1 is the fixed fieldof G(E/D1) and D2 is the fixed field of G(E/D2).

FromG(E/D1) = G(E/D2) we infer that D1 = D2. Hence thecorrespondence is one to one.

Proof (part 1, injectivity). First recall that if D is anintermediate field so that F⊆ D⊆ E, then G(E/D) is asubgroup of G(E/F).Now let D1 and D2 be intermediate fields, that is, F⊆ D1 ⊆ Eand F⊆ D2 ⊆ E, so that G(E/D1) = G(E/D2). Then E is anormal extension of D1 and of D2. Hence D1 is the fixed fieldof G(E/D1) and D2 is the fixed field of G(E/D2). FromG(E/D1) = G(E/D2) we infer that D1 = D2.

Hence thecorrespondence is one to one.

Proof (part 1, injectivity). First recall that if D is anintermediate field so that F⊆ D⊆ E, then G(E/D) is asubgroup of G(E/F).Now let D1 and D2 be intermediate fields, that is, F⊆ D1 ⊆ Eand F⊆ D2 ⊆ E, so that G(E/D1) = G(E/D2). Then E is anormal extension of D1 and of D2. Hence D1 is the fixed fieldof G(E/D1) and D2 is the fixed field of G(E/D2). FromG(E/D1) = G(E/D2) we infer that D1 = D2. Hence thecorrespondence is one to one.

Proof (part 1, surjectivity).

Let H be a subgroup of G(E/F)and let DH be the fixed field of H. Clearly H ⊆ G

(E/DH) is a

subgroup of G(E/DH). To prove equality, we show that both

have the same number of elements. E is also a normal extensionof DH . So there is a θ ∈ E so that E = DH(θ) and we have∣∣G(E/DH)∣∣ =

[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h = |H| ≤∣∣G(E/DH)∣∣ =

[E : DH] .

For the reverse inequality, consider the polynomial

p(x) =h

∏j=1

(x−σj(θ)

). For all σ ∈ H, we have

Proof (part 1, surjectivity). Let H be a subgroup of G(E/F)

and let DH be the fixed field of H. Clearly H ⊆ G(E/DH) is a

[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h = |H| ≤∣∣G(E/DH)∣∣ =

[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

Proof (part 1, surjectivity). Let H be a subgroup of G(E/F)and let DH be the fixed field of H.

Clearly H ⊆ G(E/DH) is a

[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h = |H| ≤∣∣G(E/DH)∣∣ =

[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

Proof (part 1, surjectivity). Let H be a subgroup of G(E/F)and let DH be the fixed field of H. Clearly H ⊆ G

(E/DH) is a

subgroup of G(E/DH).

To prove equality, we show that bothhave the same number of elements. E is also a normal extensionof DH . So there is a θ ∈ E so that E = DH(θ) and we have∣∣G(E/DH)∣∣ =

[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h = |H| ≤∣∣G(E/DH)∣∣ =

[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

(E/DH) is a

have the same number of elements.

E is also a normal extensionof DH . So there is a θ ∈ E so that E = DH(θ) and we have∣∣G(E/DH)∣∣ =

[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h = |H| ≤∣∣G(E/DH)∣∣ =

[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

(E/DH) is a

have the same number of elements. E is also a normal extensionof DH .

So there is a θ ∈ E so that E = DH(θ) and we have∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h = |H| ≤∣∣G(E/DH)∣∣ =

[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

(E/DH) is a

have the same number of elements. E is also a normal extensionof DH . So there is a θ ∈ E so that E = DH(θ)

and we have∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h = |H| ≤∣∣G(E/DH)∣∣ =

[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

(E/DH) is a

have the same number of elements. E is also a normal extensionof DH . So there is a θ ∈ E so that E = DH(θ) and we have∣∣G(E/DH)∣∣

=[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h = |H| ≤∣∣G(E/DH)∣∣ =

[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

(E/DH) is a

[E : DH]

=[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h = |H| ≤∣∣G(E/DH)∣∣ =

[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

(E/DH) is a

[E : DH] =

[DH(θ) : DH] .

LetH = {σ1, . . . ,σh}. Then h = |H| ≤

∣∣G(E/DH)∣∣ =[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

(E/DH) is a

[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}.

Then h = |H| ≤∣∣G(E/DH)∣∣ =

[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

(E/DH) is a

[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h

= |H| ≤∣∣G(E/DH)∣∣ =

[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

(E/DH) is a

[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h = |H|

≤∣∣G(E/DH)∣∣ =

[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

(E/DH) is a

[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h = |H| ≤∣∣G(E/DH)∣∣

=[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

(E/DH) is a

[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h = |H| ≤∣∣G(E/DH)∣∣ =

[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

(E/DH) is a

[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h = |H| ≤∣∣G(E/DH)∣∣ =

[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

For all σ ∈ H, we have

(E/DH) is a

[E : DH] =

[DH(θ) : DH] . Let

H = {σ1, . . . ,σh}. Then h = |H| ≤∣∣G(E/DH)∣∣ =

[E : DH] .

p(x) =h

∏j=1

(x−σj(θ)

Proof (part 1, surjectivity, concl.).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

because {σ ◦σj : j = 1, . . . ,h}= H. But then the coefficients ofp are fixed by H. So p ∈ DH[x]. Hence θ is a zero of apolynomial of degree h, and

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] ≤ h.

Hence∣∣G(E/DH)∣∣= |H|, which implies H = G

(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] ≤ h.

(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] ≤ h.

(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] ≤ h.

(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] ≤ h.

(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] ≤ h.

(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

because {σ ◦σj : j = 1, . . . ,h}= H.

But then the coefficients ofp are fixed by H. So p ∈ DH[x]. Hence θ is a zero of apolynomial of degree h, and

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] ≤ h.

(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

because {σ ◦σj : j = 1, . . . ,h}= H. But then the coefficients ofp are fixed by H.

So p ∈ DH[x]. Hence θ is a zero of apolynomial of degree h, and

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] ≤ h.

(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

because {σ ◦σj : j = 1, . . . ,h}= H. But then the coefficients ofp are fixed by H. So p ∈ DH[x].

Hence θ is a zero of apolynomial of degree h, and

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] ≤ h.

(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

because {σ ◦σj : j = 1, . . . ,h}= H. But then the coefficients ofp are fixed by H. So p ∈ DH[x]. Hence θ is a zero of apolynomial of degree h

, and{

1,θ , . . . ,θ h}

is DH-linearlydependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] ≤ h.

(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

{1,θ , . . . ,θ h

}is DH-linearly

dependent.

Therefore∣∣G(E/DH)∣∣ =

[E : DH] =

[DH(θ) : DH] ≤ h.

(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣

=[E : DH] =

[DH(θ) : DH] ≤ h.

(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH]

=[DH(θ) : DH] ≤ h.

(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH]

≤ h.Hence

∣∣G(E/DH)∣∣= |H|, which implies H = G(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] ≤ h.

(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] ≤ h.

Hence∣∣G(E/DH)∣∣= |H|

, which implies H = G(E/DH).

σ(p(x)

∏j=1

(x−σj(θ)

∏j=1

((x−σj(θ)

∏j=1

(x−σ

(σj(θ)

∏j=1

(x−σj(θ)

{1,θ , . . . ,θ h

}is DH-linearly

dependent.Therefore

∣∣G(E/DH)∣∣ =[E : DH] =

[DH(θ) : DH] ≤ h.

(E/DH).

Proof (part 2).

First note that E is also a normal extension ofD. Thus

∣∣G(E/D)∣∣= [E : D] and

∣∣G(E/F)∣∣= [E : F]. Now

[D : F] =[E : F][E : D]

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

Proof (part 2). First note that E is also a normal extension ofD.

Thus∣∣G(E/D)

∣∣= [E : D] and∣∣G(E/F)

∣∣= [E : F]. Now

[D : F] =[E : F][E : D]

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

Proof (part 2). First note that E is also a normal extension ofD. Thus

∣∣G(E/D)∣∣= [E : D]

and∣∣G(E/F)

∣∣= [E : F]. Now

[D : F] =[E : F][E : D]

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

∣∣G(E/D)∣∣= [E : D] and

∣∣G(E/F)∣∣= [E : F].

[D : F] =[E : F][E : D]

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

∣∣G(E/D)∣∣= [E : D] and

∣∣G(E/F)∣∣= [E : F]. Now

[D : F]

=[E : F][E : D]

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

∣∣G(E/D)∣∣= [E : D] and

∣∣G(E/F)∣∣= [E : F]. Now

[D : F] =[E : F][E : D]

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

∣∣G(E/D)∣∣= [E : D] and

∣∣G(E/F)∣∣= [E : F]. Now

[D : F] =[E : F][E : D]

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .

Proof (part 3).

“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”:

Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper).

Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field.

Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}

= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)

⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)

⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)

⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F).

Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).

“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”:

Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper).

E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2.

Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1)

and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2).

Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1)

, everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2).

HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2.

Finally, by part 1, we must have D1 6= D2.

Proof (part 3).“⇒”: Let F⊆ D1 ⊂ D2 ⊆ E (where the central inclusion isproper). Every automorphism that fixes a field also fixes thesubfields of said field. Hence{id}= G(E/E)⊆ G(E/D2)⊆ G(E/D1)⊆ G(E/F). Thecontainment of G(E/D2) in G(E/D1) is proper because, by part1, D1 6= D2 implies G(E/D2) 6= G(E/D1).“⇐”: Let D1 and D2 be intermediate fields so that{id} ⊆ G(E/D2)⊂ G(E/D1)⊆ G(E/F) (where the centralinclusion is proper). E is also a normal extension of D1 and ofD2. Hence D1 is the fixed field of G(E/D1) and D2 is the fixedfield of G(E/D2). Because G(E/D2)⊂ G(E/D1), everyelement of D1 is fixed by every element of G(E/D2). HenceD1 ⊆ D2. Finally, by part 1, we must have D1 6= D2.

Proof (part 4, “⇒”).

Let D be a normal extension of F. Wemust prove that for all σ ∈ G(E/D) and all γ ∈ G(E/F) wehave that γσγ−1 ∈ G(E/D). By definition of G(E/D), this isequivalent to showing that for all σ ∈ G(E/D) and allγ ∈ G(E/F) the automorphism γσγ−1 fixes D.Let d ∈ D and let ν ∈ G(E/F). Then d ∈ E is a zero of an

irreducible polynomial p(x) =n

∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j = p(ν(d)

that is, ν(d) is a zero of p, too. Because D is a normal extensionof F, ν(d) ∈ D. Hence for all d ∈ D and all ν ∈ G(E/F), wehave that ν(d) ∈ D.

Proof (part 4, “⇒”). Let D be a normal extension of F.

Wemust prove that for all σ ∈ G(E/D) and all γ ∈ G(E/F) wehave that γσγ−1 ∈ G(E/D). By definition of G(E/D), this isequivalent to showing that for all σ ∈ G(E/D) and allγ ∈ G(E/F) the automorphism γσγ−1 fixes D.Let d ∈ D and let ν ∈ G(E/F). Then d ∈ E is a zero of an

∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j = p(ν(d)

Proof (part 4, “⇒”). Let D be a normal extension of F. Wemust prove that for all σ ∈ G(E/D) and all γ ∈ G(E/F) wehave that γσγ−1 ∈ G(E/D).

By definition of G(E/D), this isequivalent to showing that for all σ ∈ G(E/D) and allγ ∈ G(E/F) the automorphism γσγ−1 fixes D.Let d ∈ D and let ν ∈ G(E/F). Then d ∈ E is a zero of an

∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j = p(ν(d)

Proof (part 4, “⇒”). Let D be a normal extension of F. Wemust prove that for all σ ∈ G(E/D) and all γ ∈ G(E/F) wehave that γσγ−1 ∈ G(E/D). By definition of G(E/D), this isequivalent to showing that for all σ ∈ G(E/D) and allγ ∈ G(E/F) the automorphism γσγ−1 fixes D.

Let d ∈ D and let ν ∈ G(E/F). Then d ∈ E is a zero of an

∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j = p(ν(d)

Proof (part 4, “⇒”). Let D be a normal extension of F. Wemust prove that for all σ ∈ G(E/D) and all γ ∈ G(E/F) wehave that γσγ−1 ∈ G(E/D). By definition of G(E/D), this isequivalent to showing that for all σ ∈ G(E/D) and allγ ∈ G(E/F) the automorphism γσγ−1 fixes D.Let d ∈ D and let ν ∈ G(E/F).

Then d ∈ E is a zero of an

∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j = p(ν(d)

Proof (part 4, “⇒”). Let D be a normal extension of F. Wemust prove that for all σ ∈ G(E/D) and all γ ∈ G(E/F) wehave that γσγ−1 ∈ G(E/D). By definition of G(E/D), this isequivalent to showing that for all σ ∈ G(E/D) and allγ ∈ G(E/F) the automorphism γσγ−1 fixes D.Let d ∈ D and let ν ∈ G(E/F). Then d ∈ E is a zero of an

∑j=0

pjxj in F[x].

0 = ν(0) = ν

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j = p(ν(d)

∑j=0

pjxj in F[x]. Now

= ν(0) = ν

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j = p(ν(d)

∑j=0

pjxj in F[x]. Now

0 = ν(0)

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j = p(ν(d)

∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j = p(ν(d)

∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

∑j=0

ν (pj)ν(dj)

∑j=0

pjν (d)j = p(ν(d)

∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j

= p(ν(d)

∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j = p(ν(d)

∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j = p(ν(d)

that is, ν(d) is a zero of p, too.

Because D is a normal extensionof F, ν(d) ∈ D. Hence for all d ∈ D and all ν ∈ G(E/F), wehave that ν(d) ∈ D.

∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j = p(ν(d)

that is, ν(d) is a zero of p, too. Because D is a normal extensionof F

, ν(d) ∈ D. Hence for all d ∈ D and all ν ∈ G(E/F), wehave that ν(d) ∈ D.

∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j = p(ν(d)

that is, ν(d) is a zero of p, too. Because D is a normal extensionof F, ν(d) ∈ D.

Hence for all d ∈ D and all ν ∈ G(E/F), wehave that ν(d) ∈ D.

∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

∑j=0

ν (pj)ν(dj) =

∑j=0

pjν (d)j = p(ν(d)

Proof (part 4, “⇒”, concl.).

But then for all γ ∈ G(E/F) andσ ∈ G(E/D) we obtain the following for every d ∈ D. Becauseγ−1(d) ∈ D, we obtain σγ

−1(d) = σ

(γ−1(d)

−1(d), and

then γσγ−1(d) = γ

−1(d))

(γ−1(d)

)= d. Because

d ∈ D was arbitrary, γσγ−1 fixes D, so γσγ−1 ∈ G(E/D) andhence G(E/D) is normal in G(E/F).

Proof (part 4, “⇒”, concl.). But then for all γ ∈ G(E/F) andσ ∈ G(E/D)

we obtain the following for every d ∈ D. Becauseγ−1(d) ∈ D, we obtain σγ

−1(d) = σ

(γ−1(d)

−1(d), and

−1(d))

(γ−1(d)

)= d. Because

Proof (part 4, “⇒”, concl.). But then for all γ ∈ G(E/F) andσ ∈ G(E/D) we obtain the following for every d ∈ D.

Becauseγ−1(d) ∈ D, we obtain σγ

−1(d) = σ

(γ−1(d)

−1(d), and

−1(d))

(γ−1(d)

)= d. Because

Proof (part 4, “⇒”, concl.). But then for all γ ∈ G(E/F) andσ ∈ G(E/D) we obtain the following for every d ∈ D. Becauseγ−1(d) ∈ D, we obtain σγ

−1(d)

(γ−1(d)

−1(d), and

−1(d))

(γ−1(d)

)= d. Because

−1(d) = σ

(γ−1(d)

= γ−1(d), and

−1(d))

(γ−1(d)

)= d. Because

−1(d) = σ

(γ−1(d)

−1(d)

−1(d))

(γ−1(d)

)= d. Because

−1(d) = σ

(γ−1(d)

−1(d), and

then γσγ−1(d)

−1(d))

(γ−1(d)

)= d. Because

−1(d) = σ

(γ−1(d)

−1(d), and

−1(d))

(γ−1(d)

)= d. Because

−1(d) = σ

(γ−1(d)

−1(d), and

−1(d))

(γ−1(d)

= d. Because

−1(d) = σ

(γ−1(d)

−1(d), and

−1(d))

(γ−1(d)

Because

−1(d) = σ

(γ−1(d)

−1(d), and

−1(d))

(γ−1(d)

)= d. Because

d ∈ D was arbitrary, γσγ−1 fixes D

, so γσγ−1 ∈ G(E/D) andhence G(E/D) is normal in G(E/F).

−1(d) = σ

(γ−1(d)

−1(d), and

−1(d))

(γ−1(d)

)= d. Because

d ∈ D was arbitrary, γσγ−1 fixes D, so γσγ−1 ∈ G(E/D)

andhence G(E/D) is normal in G(E/F).

−1(d) = σ

(γ−1(d)

−1(d), and

−1(d))

(γ−1(d)

)= d. Because

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D)).

Let G(E/D) be normal in G(E/F).Let µ ∈ G(D/F) be fixed. Then µ can be extended to anautomorphism γµ ∈ G(E/F) (good exercise) so that µ = γµ |D.(This extension is not unique.) Let γµ ,δµ ∈ G(E/F) be so thatµ = γµ |D = δµ |D. Then δ−1

µ γµ

∣∣D = µ−1µ = idD, that is,

δ−1µ γµ ∈ G(E/D). Hence

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

)γµG(E/D) = γµG(E/D).

Therefore Φ : G(D/F)→ G(E/F)/G(E/D), which maps eachµ ∈ G(D/F) to γµG(E/D) ∈ G(E/F)/G(E/D), whereγµ |D = µ , is well-defined. We claim that this function is thedesired isomorphism.

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D)). Let G(E/D) be normal in G(E/F).

Let µ ∈ G(D/F) be fixed. Then µ can be extended to anautomorphism γµ ∈ G(E/F) (good exercise) so that µ = γµ |D.(This extension is not unique.) Let γµ ,δµ ∈ G(E/F) be so thatµ = γµ |D = δµ |D. Then δ−1

µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D)). Let G(E/D) be normal in G(E/F).Let µ ∈ G(D/F) be fixed.

Then µ can be extended to anautomorphism γµ ∈ G(E/F) (good exercise) so that µ = γµ |D.(This extension is not unique.) Let γµ ,δµ ∈ G(E/F) be so thatµ = γµ |D = δµ |D. Then δ−1

µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D)). Let G(E/D) be normal in G(E/F).Let µ ∈ G(D/F) be fixed. Then µ can be extended to anautomorphism γµ ∈ G(E/F)

(good exercise) so that µ = γµ |D.(This extension is not unique.) Let γµ ,δµ ∈ G(E/F) be so thatµ = γµ |D = δµ |D. Then δ−1

µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D)). Let G(E/D) be normal in G(E/F).Let µ ∈ G(D/F) be fixed. Then µ can be extended to anautomorphism γµ ∈ G(E/F) (good exercise)

so that µ = γµ |D.(This extension is not unique.) Let γµ ,δµ ∈ G(E/F) be so thatµ = γµ |D = δµ |D. Then δ−1

µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D)). Let G(E/D) be normal in G(E/F).Let µ ∈ G(D/F) be fixed. Then µ can be extended to anautomorphism γµ ∈ G(E/F) (good exercise) so that µ = γµ |D.

(This extension is not unique.) Let γµ ,δµ ∈ G(E/F) be so thatµ = γµ |D = δµ |D. Then δ−1

µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D)). Let G(E/D) be normal in G(E/F).Let µ ∈ G(D/F) be fixed. Then µ can be extended to anautomorphism γµ ∈ G(E/F) (good exercise) so that µ = γµ |D.(This extension is not unique.)

Let γµ ,δµ ∈ G(E/F) be so thatµ = γµ |D = δµ |D. Then δ−1

µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D)). Let G(E/D) be normal in G(E/F).Let µ ∈ G(D/F) be fixed. Then µ can be extended to anautomorphism γµ ∈ G(E/F) (good exercise) so that µ = γµ |D.(This extension is not unique.) Let γµ ,δµ ∈ G(E/F) be so thatµ = γµ |D = δµ |D.

Then δ−1µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D)). Let G(E/D) be normal in G(E/F).Let µ ∈ G(D/F) be fixed. Then µ can be extended to anautomorphism γµ ∈ G(E/F) (good exercise) so that µ = γµ |D.(This extension is not unique.) Let γµ ,δµ ∈ G(E/F) be so thatµ = γµ |D = δµ |D. Then δ−1

µ γµ

∣∣D

= µ−1µ = idD, that is,δ−1

µ γµ ∈ G(E/D). Hence

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

µ γµ

∣∣D = µ−1µ

= idD, that is,δ−1

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

µ γµ

∣∣D = µ−1µ = idD

, that is,δ−1

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

µ γµ

δ−1µ γµ ∈ G(E/D).

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

µ γµ

δµG(E/D)

= δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

)γµG(E/D)

= γµG(E/D).

µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

Therefore Φ : G(D/F)→ G(E/F)/G(E/D)

, which maps eachµ ∈ G(D/F) to γµG(E/D) ∈ G(E/F)/G(E/D), whereγµ |D = µ , is well-defined. We claim that this function is thedesired isomorphism.

µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

Therefore Φ : G(D/F)→ G(E/F)/G(E/D), which maps eachµ ∈ G(D/F)

to γµG(E/D) ∈ G(E/F)/G(E/D), whereγµ |D = µ , is well-defined. We claim that this function is thedesired isomorphism.

µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

Therefore Φ : G(D/F)→ G(E/F)/G(E/D), which maps eachµ ∈ G(D/F) to γµG(E/D) ∈ G(E/F)/G(E/D), whereγµ |D = µ

, is well-defined. We claim that this function is thedesired isomorphism.

µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

Therefore Φ : G(D/F)→ G(E/F)/G(E/D), which maps eachµ ∈ G(D/F) to γµG(E/D) ∈ G(E/F)/G(E/D), whereγµ |D = µ , is well-defined.

We claim that this function is thedesired isomorphism.

µ γµ

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

δµδ−1µ

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D), cont.).

First note that for all µ,ν ∈ G(D/F)we have γµ◦νG(E/D) = (γµγν)G(E/D), becauseγµ◦ν |D = µ ◦ν = γµ |D ◦ γν |D = (γµγν)|D. Therefore

Φ(µ ◦ν) = γµ◦νG(E/D)= (γµγν)G(E/D)= γµγνG(E/D)G(E/D)= γµG(E/D)γνG(E/D)= Φ(µ)Φ(ν).

Hence Φ is a homomorphism.

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D), cont.). First note that for all µ,ν ∈ G(D/F)we have γµ◦νG(E/D) = (γµγν)G(E/D)

, becauseγµ◦ν |D = µ ◦ν = γµ |D ◦ γν |D = (γµγν)|D. Therefore

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D), cont.). First note that for all µ,ν ∈ G(D/F)we have γµ◦νG(E/D) = (γµγν)G(E/D), becauseγµ◦ν |D = µ ◦ν = γµ |D ◦ γν |D = (γµγν)|D.

Therefore

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D), cont.). First note that for all µ,ν ∈ G(D/F)we have γµ◦νG(E/D) = (γµγν)G(E/D), becauseγµ◦ν |D = µ ◦ν = γµ |D ◦ γν |D = (γµγν)|D. Therefore

Φ(µ ◦ν)

= γµ◦νG(E/D)= (γµγν)G(E/D)= γµγνG(E/D)G(E/D)= γµG(E/D)γνG(E/D)= Φ(µ)Φ(ν).

Φ(µ ◦ν) = γµ◦νG(E/D)

= (γµγν)G(E/D)= γµγνG(E/D)G(E/D)= γµG(E/D)γνG(E/D)= Φ(µ)Φ(ν).

Φ(µ ◦ν) = γµ◦νG(E/D)= (γµγν)G(E/D)

= γµγνG(E/D)G(E/D)= γµG(E/D)γνG(E/D)= Φ(µ)Φ(ν).

Φ(µ ◦ν) = γµ◦νG(E/D)= (γµγν)G(E/D)= γµγνG(E/D)G(E/D)

= γµG(E/D)γνG(E/D)= Φ(µ)Φ(ν).

Φ(µ ◦ν) = γµ◦νG(E/D)= (γµγν)G(E/D)= γµγνG(E/D)G(E/D)= γµG(E/D)γνG(E/D)

= Φ(µ)Φ(ν).

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D), cont.).

Let γ ∈ G(E/F) and let σ ∈ G(E/D)and d ∈ D be arbitrary. Because G(E/D) is normal in G(E/F),there is a ρ ∈ G(E/D) so that σγ = γρ . Henceσ(γ(d)

)= γρ(d) = γ(d), that is, γ(d) is fixed by every

σ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D), cont.). Let γ ∈ G(E/F) and let σ ∈ G(E/D)and d ∈ D be arbitrary.

Because G(E/D) is normal in G(E/F),there is a ρ ∈ G(E/D) so that σγ = γρ . Henceσ(γ(d)

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D), cont.). Let γ ∈ G(E/F) and let σ ∈ G(E/D)and d ∈ D be arbitrary. Because G(E/D) is normal in G(E/F),there is a ρ ∈ G(E/D) so that σγ = γρ .

Henceσ(γ(d)

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D), cont.). Let γ ∈ G(E/F) and let σ ∈ G(E/D)and d ∈ D be arbitrary. Because G(E/D) is normal in G(E/F),there is a ρ ∈ G(E/D) so that σγ = γρ . Henceσ(γ(d)

= γρ(d) = γ(d), that is, γ(d) is fixed by everyσ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

)= γρ(d)

= γ(d), that is, γ(d) is fixed by everyσ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

)= γρ(d) = γ(d)

, that is, γ(d) is fixed by everyσ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

σ ∈ G(E/D).

But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

σ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.

Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

σ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D.

Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

σ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D

, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

σ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.

Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

σ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D).

Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

σ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ .

Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

σ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D

= γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

σ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D

= γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

σ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D.

Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

σ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F)

that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F) is well-defined.

σ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D)

to γ|D ∈ G(D/F) is well-defined.

σ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D

∈ G(D/F) is well-defined.

σ ∈ G(E/D). But the fixed field of G(E/D) is D, so γ(d) ∈ D.Because d ∈ D was arbitrary, γ maps D to D. Because the sameargument applied to γ−1 shows that γ−1 maps D to D, we obtainthat for every γ ∈ G(E/F), the restriction γ|D is anautomorphism of D that fixes F.Now let δ ∈ γG(E/D). Then there is a σ ∈ G(E/D) so thatδ = γσ . Therefore δ |D = γσ |D = γ|D. Hence the functionΨ : G(E/F)/G(E/D)→ G(D/F) that maps eachγG(E/D) ∈ G(E/F)/G(E/D) to γ|D ∈ G(D/F)

is well-defined.

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D), concl.).

Moreover, for allγG(E/D),δG(E/D) ∈ G(E/F)/G(E/D) we have

Ψ(γG(E/D)δG(E/D)

(γδG(E/D)

)= γδ |D= γ|Dδ |D= Ψ

(γG(E/D)

)Ψ(δG(E/D)

Hence Ψ is a homomorphism.It is now easy to verify that Φ and Ψ are inverses of each other,which proves that Φ is the desired isomorphism.

Proof (part 4, “⇐”, G(D/F) is isomorphic toG(E/F)/G(E/D), concl.). Moreover, for allγG(E/D),δG(E/D) ∈ G(E/F)/G(E/D) we have

Ψ(γG(E/D)δG(E/D)

= Ψ(γδG(E/D)

(γG(E/D)

)Ψ(δG(E/D)

Ψ(γG(E/D)δG(E/D)

(γδG(E/D)

= γδ |D= γ|Dδ |D= Ψ

(γG(E/D)

)Ψ(δG(E/D)

Ψ(γG(E/D)δG(E/D)

(γδG(E/D)

)= γδ |D

= γ|Dδ |D= Ψ

(γG(E/D)

)Ψ(δG(E/D)

Ψ(γG(E/D)δG(E/D)

(γδG(E/D)

)= γδ |D= γ|Dδ |D

= Ψ(γG(E/D)

)Ψ(δG(E/D)

Ψ(γG(E/D)δG(E/D)

(γδG(E/D)

(γG(E/D)

)Ψ(δG(E/D)

Ψ(γG(E/D)δG(E/D)

(γδG(E/D)

(γG(E/D)

)Ψ(δG(E/D)

Hence Ψ is a homomorphism.

It is now easy to verify that Φ and Ψ are inverses of each other,which proves that Φ is the desired isomorphism.

Ψ(γG(E/D)δG(E/D)

(γδG(E/D)

(γG(E/D)

)Ψ(δG(E/D)

Hence Ψ is a homomorphism.It is now easy to verify that Φ and Ψ are inverses of each other

,which proves that Φ is the desired isomorphism.

Ψ(γG(E/D)δG(E/D)

(γδG(E/D)

(γG(E/D)

)Ψ(δG(E/D)

Proof (part 4, “⇐”, D is a normal extension of F).

It sufficesto prove that F is the fixed field of G(D/F). Let d ∈ D be fixedby all automorphisms µ ∈ G(D/F) and let γ ∈ G(E/F). By theabove, γ|D ∈ G(D/F). Hence γ(d) = γ|D(d) = d. Because E isnormal over F we conclude that d ∈ F. Therefore the fixed fieldof G(D/F) is F, and D is normal over F.

Proof (part 4, “⇐”, D is a normal extension of F). It sufficesto prove that F is the fixed field of G(D/F).

Let d ∈ D be fixedby all automorphisms µ ∈ G(D/F) and let γ ∈ G(E/F). By theabove, γ|D ∈ G(D/F). Hence γ(d) = γ|D(d) = d. Because E isnormal over F we conclude that d ∈ F. Therefore the fixed fieldof G(D/F) is F, and D is normal over F.

Proof (part 4, “⇐”, D is a normal extension of F). It sufficesto prove that F is the fixed field of G(D/F). Let d ∈ D be fixedby all automorphisms µ ∈ G(D/F)

and let γ ∈ G(E/F). By theabove, γ|D ∈ G(D/F). Hence γ(d) = γ|D(d) = d. Because E isnormal over F we conclude that d ∈ F. Therefore the fixed fieldof G(D/F) is F, and D is normal over F.

Proof (part 4, “⇐”, D is a normal extension of F). It sufficesto prove that F is the fixed field of G(D/F). Let d ∈ D be fixedby all automorphisms µ ∈ G(D/F) and let γ ∈ G(E/F).

By theabove, γ|D ∈ G(D/F). Hence γ(d) = γ|D(d) = d. Because E isnormal over F we conclude that d ∈ F. Therefore the fixed fieldof G(D/F) is F, and D is normal over F.

Proof (part 4, “⇐”, D is a normal extension of F). It sufficesto prove that F is the fixed field of G(D/F). Let d ∈ D be fixedby all automorphisms µ ∈ G(D/F) and let γ ∈ G(E/F). By theabove, γ|D ∈ G(D/F).

Hence γ(d) = γ|D(d) = d. Because E isnormal over F we conclude that d ∈ F. Therefore the fixed fieldof G(D/F) is F, and D is normal over F.

Proof (part 4, “⇐”, D is a normal extension of F). It sufficesto prove that F is the fixed field of G(D/F). Let d ∈ D be fixedby all automorphisms µ ∈ G(D/F) and let γ ∈ G(E/F). By theabove, γ|D ∈ G(D/F). Hence γ(d)

= γ|D(d) = d. Because E isnormal over F we conclude that d ∈ F. Therefore the fixed fieldof G(D/F) is F, and D is normal over F.

Proof (part 4, “⇐”, D is a normal extension of F). It sufficesto prove that F is the fixed field of G(D/F). Let d ∈ D be fixedby all automorphisms µ ∈ G(D/F) and let γ ∈ G(E/F). By theabove, γ|D ∈ G(D/F). Hence γ(d) = γ|D(d)

= d. Because E isnormal over F we conclude that d ∈ F. Therefore the fixed fieldof G(D/F) is F, and D is normal over F.

Proof (part 4, “⇐”, D is a normal extension of F). It sufficesto prove that F is the fixed field of G(D/F). Let d ∈ D be fixedby all automorphisms µ ∈ G(D/F) and let γ ∈ G(E/F). By theabove, γ|D ∈ G(D/F). Hence γ(d) = γ|D(d) = d.

Because E isnormal over F we conclude that d ∈ F. Therefore the fixed fieldof G(D/F) is F, and D is normal over F.

Proof (part 4, “⇐”, D is a normal extension of F). It sufficesto prove that F is the fixed field of G(D/F). Let d ∈ D be fixedby all automorphisms µ ∈ G(D/F) and let γ ∈ G(E/F). By theabove, γ|D ∈ G(D/F). Hence γ(d) = γ|D(d) = d. Because E isnormal over F we conclude that d ∈ F.

Therefore the fixed fieldof G(D/F) is F, and D is normal over F.

Proof (part 4, “⇐”, D is a normal extension of F). It sufficesto prove that F is the fixed field of G(D/F). Let d ∈ D be fixedby all automorphisms µ ∈ G(D/F) and let γ ∈ G(E/F). By theabove, γ|D ∈ G(D/F). Hence γ(d) = γ|D(d) = d. Because E isnormal over F we conclude that d ∈ F. Therefore the fixed fieldof G(D/F) is F

, and D is normal over F.

Proof (part 4, “⇐”, D is a normal extension of F). It sufficesto prove that F is the fixed field of G(D/F). Let d ∈ D be fixedby all automorphisms µ ∈ G(D/F) and let γ ∈ G(E/F). By theabove, γ|D ∈ G(D/F). Hence γ(d) = γ|D(d) = d. Because E isnormal over F we conclude that d ∈ F. Therefore the fixed fieldof G(D/F) is F, and D is normal over F.

Proposition.

Let σ ∈ Sn be a permutation. Then σ cannot berepresented both as a product of an even number oftranspositions and as a product of an odd number oftranspositions.

Proof. Suppose the opposite. Then the identity can berepresented as the product of 2m+1 transpositions. WLOG letm ∈ N0 be as small as possible. Let the transpositionsν1, . . . ,ν2m+1 be so that id = ν1 · · ·ν2m+1, so that, among allrepresentations of the identity with 2m+1 transpositions, thelargest number b ∈ {1, . . . ,n} that occurs in any of the νj is assmall as possible, and so that the number of transpositions inwhich this number occurs is as small as possible, too.Because ν1 · · ·ν2m+1 = id, there must be at least twotranspositions νj that contain b.

Proposition. Let σ ∈ Sn be a permutation.

Then σ cannot berepresented both as a product of an even number oftranspositions and as a product of an odd number oftranspositions.

Proposition. Let σ ∈ Sn be a permutation. Then σ cannot berepresented both as a product of an even number oftranspositions and as a product of an odd number oftranspositions.

Proof.

Suppose the opposite. Then the identity can berepresented as the product of 2m+1 transpositions. WLOG letm ∈ N0 be as small as possible. Let the transpositionsν1, . . . ,ν2m+1 be so that id = ν1 · · ·ν2m+1, so that, among allrepresentations of the identity with 2m+1 transpositions, thelargest number b ∈ {1, . . . ,n} that occurs in any of the νj is assmall as possible, and so that the number of transpositions inwhich this number occurs is as small as possible, too.Because ν1 · · ·ν2m+1 = id, there must be at least twotranspositions νj that contain b.

Proof. Suppose the opposite.

Then the identity can berepresented as the product of 2m+1 transpositions. WLOG letm ∈ N0 be as small as possible. Let the transpositionsν1, . . . ,ν2m+1 be so that id = ν1 · · ·ν2m+1, so that, among allrepresentations of the identity with 2m+1 transpositions, thelargest number b ∈ {1, . . . ,n} that occurs in any of the νj is assmall as possible, and so that the number of transpositions inwhich this number occurs is as small as possible, too.Because ν1 · · ·ν2m+1 = id, there must be at least twotranspositions νj that contain b.

Proof. Suppose the opposite. Then the identity can berepresented as the product of 2m+1 transpositions.

WLOG letm ∈ N0 be as small as possible. Let the transpositionsν1, . . . ,ν2m+1 be so that id = ν1 · · ·ν2m+1, so that, among allrepresentations of the identity with 2m+1 transpositions, thelargest number b ∈ {1, . . . ,n} that occurs in any of the νj is assmall as possible, and so that the number of transpositions inwhich this number occurs is as small as possible, too.Because ν1 · · ·ν2m+1 = id, there must be at least twotranspositions νj that contain b.

Proof. Suppose the opposite. Then the identity can berepresented as the product of 2m+1 transpositions. WLOG letm ∈ N0 be as small as possible.

Let the transpositionsν1, . . . ,ν2m+1 be so that id = ν1 · · ·ν2m+1, so that, among allrepresentations of the identity with 2m+1 transpositions, thelargest number b ∈ {1, . . . ,n} that occurs in any of the νj is assmall as possible, and so that the number of transpositions inwhich this number occurs is as small as possible, too.Because ν1 · · ·ν2m+1 = id, there must be at least twotranspositions νj that contain b.

Proof. Suppose the opposite. Then the identity can berepresented as the product of 2m+1 transpositions. WLOG letm ∈ N0 be as small as possible. Let the transpositionsν1, . . . ,ν2m+1 be so that id = ν1 · · ·ν2m+1

, so that, among allrepresentations of the identity with 2m+1 transpositions, thelargest number b ∈ {1, . . . ,n} that occurs in any of the νj is assmall as possible, and so that the number of transpositions inwhich this number occurs is as small as possible, too.Because ν1 · · ·ν2m+1 = id, there must be at least twotranspositions νj that contain b.

Proof. Suppose the opposite. Then the identity can berepresented as the product of 2m+1 transpositions. WLOG letm ∈ N0 be as small as possible. Let the transpositionsν1, . . . ,ν2m+1 be so that id = ν1 · · ·ν2m+1, so that, among allrepresentations of the identity with 2m+1 transpositions, thelargest number b ∈ {1, . . . ,n} that occurs in any of the νj is assmall as possible

, and so that the number of transpositions inwhich this number occurs is as small as possible, too.Because ν1 · · ·ν2m+1 = id, there must be at least twotranspositions νj that contain b.

Proof. Suppose the opposite. Then the identity can berepresented as the product of 2m+1 transpositions. WLOG letm ∈ N0 be as small as possible. Let the transpositionsν1, . . . ,ν2m+1 be so that id = ν1 · · ·ν2m+1, so that, among allrepresentations of the identity with 2m+1 transpositions, thelargest number b ∈ {1, . . . ,n} that occurs in any of the νj is assmall as possible, and so that the number of transpositions inwhich this number occurs is as small as possible, too.

Because ν1 · · ·ν2m+1 = id, there must be at least twotranspositions νj that contain b.

Proof (concl.).

It is possible to rewrite the product so that twotranspositions that contain b are adjacent, so that the number offactors is still 2m+1, and so that the number of occurrences ofb does not increase: Go through the product right to left untilyou find the second transposition that contains b. Use(bc)(ac) = (abc) = (ca)(ab) as needed (disjoint transpositionscommute) to push the second occurrence of b rightwards until itis adjacent to the first occurrence of b.By choice of m, the two adjacent transpositions that contain bcannot be equal, because then the identity could be representedas a product of 2m−1 transpositions. Hence the two adjacenttranspositions that contain b are (ab) and (bc) for some a 6= c.But now, because (ab)(bc) = (abc) = (ca)(ab), the twoadjacent transpositions that contain b can be rewritten so thatthe total number of bs in the product ν1 · · ·ν2m+1 is reduced by1. Contradiction.

Proof (concl.). It is possible to rewrite the product so that twotranspositions that contain b are adjacent

, so that the number offactors is still 2m+1, and so that the number of occurrences ofb does not increase: Go through the product right to left untilyou find the second transposition that contains b. Use(bc)(ac) = (abc) = (ca)(ab) as needed (disjoint transpositionscommute) to push the second occurrence of b rightwards until itis adjacent to the first occurrence of b.By choice of m, the two adjacent transpositions that contain bcannot be equal, because then the identity could be representedas a product of 2m−1 transpositions. Hence the two adjacenttranspositions that contain b are (ab) and (bc) for some a 6= c.But now, because (ab)(bc) = (abc) = (ca)(ab), the twoadjacent transpositions that contain b can be rewritten so thatthe total number of bs in the product ν1 · · ·ν2m+1 is reduced by1. Contradiction.

Proof (concl.). It is possible to rewrite the product so that twotranspositions that contain b are adjacent, so that the number offactors is still 2m+1

, and so that the number of occurrences ofb does not increase: Go through the product right to left untilyou find the second transposition that contains b. Use(bc)(ac) = (abc) = (ca)(ab) as needed (disjoint transpositionscommute) to push the second occurrence of b rightwards until itis adjacent to the first occurrence of b.By choice of m, the two adjacent transpositions that contain bcannot be equal, because then the identity could be representedas a product of 2m−1 transpositions. Hence the two adjacenttranspositions that contain b are (ab) and (bc) for some a 6= c.But now, because (ab)(bc) = (abc) = (ca)(ab), the twoadjacent transpositions that contain b can be rewritten so thatthe total number of bs in the product ν1 · · ·ν2m+1 is reduced by1. Contradiction.

Proof (concl.). It is possible to rewrite the product so that twotranspositions that contain b are adjacent, so that the number offactors is still 2m+1, and so that the number of occurrences ofb does not increase:

Go through the product right to left untilyou find the second transposition that contains b. Use(bc)(ac) = (abc) = (ca)(ab) as needed (disjoint transpositionscommute) to push the second occurrence of b rightwards until itis adjacent to the first occurrence of b.By choice of m, the two adjacent transpositions that contain bcannot be equal, because then the identity could be representedas a product of 2m−1 transpositions. Hence the two adjacenttranspositions that contain b are (ab) and (bc) for some a 6= c.But now, because (ab)(bc) = (abc) = (ca)(ab), the twoadjacent transpositions that contain b can be rewritten so thatthe total number of bs in the product ν1 · · ·ν2m+1 is reduced by1. Contradiction.

Proof (concl.). It is possible to rewrite the product so that twotranspositions that contain b are adjacent, so that the number offactors is still 2m+1, and so that the number of occurrences ofb does not increase: Go through the product right to left untilyou find the second transposition that contains b.

Use(bc)(ac) = (abc) = (ca)(ab) as needed (disjoint transpositionscommute) to push the second occurrence of b rightwards until itis adjacent to the first occurrence of b.By choice of m, the two adjacent transpositions that contain bcannot be equal, because then the identity could be representedas a product of 2m−1 transpositions. Hence the two adjacenttranspositions that contain b are (ab) and (bc) for some a 6= c.But now, because (ab)(bc) = (abc) = (ca)(ab), the twoadjacent transpositions that contain b can be rewritten so thatthe total number of bs in the product ν1 · · ·ν2m+1 is reduced by1. Contradiction.

Proof (concl.). It is possible to rewrite the product so that twotranspositions that contain b are adjacent, so that the number offactors is still 2m+1, and so that the number of occurrences ofb does not increase: Go through the product right to left untilyou find the second transposition that contains b. Use(bc)(ac) = (abc) = (ca)(ab) as needed

(disjoint transpositionscommute) to push the second occurrence of b rightwards until itis adjacent to the first occurrence of b.By choice of m, the two adjacent transpositions that contain bcannot be equal, because then the identity could be representedas a product of 2m−1 transpositions. Hence the two adjacenttranspositions that contain b are (ab) and (bc) for some a 6= c.But now, because (ab)(bc) = (abc) = (ca)(ab), the twoadjacent transpositions that contain b can be rewritten so thatthe total number of bs in the product ν1 · · ·ν2m+1 is reduced by1. Contradiction.

Proof (concl.). It is possible to rewrite the product so that twotranspositions that contain b are adjacent, so that the number offactors is still 2m+1, and so that the number of occurrences ofb does not increase: Go through the product right to left untilyou find the second transposition that contains b. Use(bc)(ac) = (abc) = (ca)(ab) as needed (disjoint transpositionscommute)

to push the second occurrence of b rightwards until itis adjacent to the first occurrence of b.By choice of m, the two adjacent transpositions that contain bcannot be equal, because then the identity could be representedas a product of 2m−1 transpositions. Hence the two adjacenttranspositions that contain b are (ab) and (bc) for some a 6= c.But now, because (ab)(bc) = (abc) = (ca)(ab), the twoadjacent transpositions that contain b can be rewritten so thatthe total number of bs in the product ν1 · · ·ν2m+1 is reduced by1. Contradiction.

Proof (concl.). It is possible to rewrite the product so that twotranspositions that contain b are adjacent, so that the number offactors is still 2m+1, and so that the number of occurrences ofb does not increase: Go through the product right to left untilyou find the second transposition that contains b. Use(bc)(ac) = (abc) = (ca)(ab) as needed (disjoint transpositionscommute) to push the second occurrence of b rightwards until itis adjacent to the first occurrence of b.

By choice of m, the two adjacent transpositions that contain bcannot be equal, because then the identity could be representedas a product of 2m−1 transpositions. Hence the two adjacenttranspositions that contain b are (ab) and (bc) for some a 6= c.But now, because (ab)(bc) = (abc) = (ca)(ab), the twoadjacent transpositions that contain b can be rewritten so thatthe total number of bs in the product ν1 · · ·ν2m+1 is reduced by1. Contradiction.

Proof (concl.). It is possible to rewrite the product so that twotranspositions that contain b are adjacent, so that the number offactors is still 2m+1, and so that the number of occurrences ofb does not increase: Go through the product right to left untilyou find the second transposition that contains b. Use(bc)(ac) = (abc) = (ca)(ab) as needed (disjoint transpositionscommute) to push the second occurrence of b rightwards until itis adjacent to the first occurrence of b.By choice of m, the two adjacent transpositions that contain bcannot be equal

, because then the identity could be representedas a product of 2m−1 transpositions. Hence the two adjacenttranspositions that contain b are (ab) and (bc) for some a 6= c.But now, because (ab)(bc) = (abc) = (ca)(ab), the twoadjacent transpositions that contain b can be rewritten so thatthe total number of bs in the product ν1 · · ·ν2m+1 is reduced by1. Contradiction.

Proof (concl.). It is possible to rewrite the product so that twotranspositions that contain b are adjacent, so that the number offactors is still 2m+1, and so that the number of occurrences ofb does not increase: Go through the product right to left untilyou find the second transposition that contains b. Use(bc)(ac) = (abc) = (ca)(ab) as needed (disjoint transpositionscommute) to push the second occurrence of b rightwards until itis adjacent to the first occurrence of b.By choice of m, the two adjacent transpositions that contain bcannot be equal, because then the identity could be representedas a product of 2m−1 transpositions.

Hence the two adjacenttranspositions that contain b are (ab) and (bc) for some a 6= c.But now, because (ab)(bc) = (abc) = (ca)(ab), the twoadjacent transpositions that contain b can be rewritten so thatthe total number of bs in the product ν1 · · ·ν2m+1 is reduced by1. Contradiction.

Proof (concl.). It is possible to rewrite the product so that twotranspositions that contain b are adjacent, so that the number offactors is still 2m+1, and so that the number of occurrences ofb does not increase: Go through the product right to left untilyou find the second transposition that contains b. Use(bc)(ac) = (abc) = (ca)(ab) as needed (disjoint transpositionscommute) to push the second occurrence of b rightwards until itis adjacent to the first occurrence of b.By choice of m, the two adjacent transpositions that contain bcannot be equal, because then the identity could be representedas a product of 2m−1 transpositions. Hence the two adjacenttranspositions that contain b are (ab) and (bc) for some a 6= c.

But now, because (ab)(bc) = (abc) = (ca)(ab), the twoadjacent transpositions that contain b can be rewritten so thatthe total number of bs in the product ν1 · · ·ν2m+1 is reduced by1. Contradiction.

Proof (concl.). It is possible to rewrite the product so that twotranspositions that contain b are adjacent, so that the number offactors is still 2m+1, and so that the number of occurrences ofb does not increase: Go through the product right to left untilyou find the second transposition that contains b. Use(bc)(ac) = (abc) = (ca)(ab) as needed (disjoint transpositionscommute) to push the second occurrence of b rightwards until itis adjacent to the first occurrence of b.By choice of m, the two adjacent transpositions that contain bcannot be equal, because then the identity could be representedas a product of 2m−1 transpositions. Hence the two adjacenttranspositions that contain b are (ab) and (bc) for some a 6= c.But now, because (ab)(bc) = (abc) = (ca)(ab)

, the twoadjacent transpositions that contain b can be rewritten so thatthe total number of bs in the product ν1 · · ·ν2m+1 is reduced by1. Contradiction.

Proof (concl.). It is possible to rewrite the product so that twotranspositions that contain b are adjacent, so that the number offactors is still 2m+1, and so that the number of occurrences ofb does not increase: Go through the product right to left untilyou find the second transposition that contains b. Use(bc)(ac) = (abc) = (ca)(ab) as needed (disjoint transpositionscommute) to push the second occurrence of b rightwards until itis adjacent to the first occurrence of b.By choice of m, the two adjacent transpositions that contain bcannot be equal, because then the identity could be representedas a product of 2m−1 transpositions. Hence the two adjacenttranspositions that contain b are (ab) and (bc) for some a 6= c.But now, because (ab)(bc) = (abc) = (ca)(ab), the twoadjacent transpositions that contain b can be rewritten so thatthe total number of bs in the product ν1 · · ·ν2m+1 is reduced by1.

Contradiction.

Proof (concl.). It is possible to rewrite the product so that twotranspositions that contain b are adjacent, so that the number offactors is still 2m+1, and so that the number of occurrences ofb does not increase: Go through the product right to left untilyou find the second transposition that contains b. Use(bc)(ac) = (abc) = (ca)(ab) as needed (disjoint transpositionscommute) to push the second occurrence of b rightwards until itis adjacent to the first occurrence of b.By choice of m, the two adjacent transpositions that contain bcannot be equal, because then the identity could be representedas a product of 2m−1 transpositions. Hence the two adjacenttranspositions that contain b are (ab) and (bc) for some a 6= c.But now, because (ab)(bc) = (abc) = (ca)(ab), the twoadjacent transpositions that contain b can be rewritten so thatthe total number of bs in the product ν1 · · ·ν2m+1 is reduced by1. Contradiction.

Proposition.

The permutations that can be represented as acomposition of an even number of transpositions form asubgroup of Sn, called the alternating group An on n elements.

Proposition. For n > 1, the alternating group An is a normalsubgroup of the symmetric group Sn.

Proof. For every σ ∈ Sn, the functions ν 7→ σν and ν 7→ νσ

are bijections on Sn. If σ is even, these bijections map evenpermutations to even permutations and odd permutations to oddpermutations. If σ is odd, they map odd permutations to evenpermutations and even permutations to odd permutations. Butthen, if σ is even, we have σAn = An = Anσ . If σ is odd wehave σAn = Sn \An = Anσ . Thus An is normal in Sn.

Proposition. The permutations that can be represented as acomposition of an even number of transpositions form asubgroup of Sn

, called the alternating group An on n elements.

Proposition. The permutations that can be represented as acomposition of an even number of transpositions form asubgroup of Sn, called the alternating group An on n elements.

Proof.

Simple exercise (maybe too simple).

Proof. Simple exercise

(maybe too simple).

Proposition.

For n > 1, the alternating group An is a normalsubgroup of the symmetric group Sn.

Proof.

For every σ ∈ Sn, the functions ν 7→ σν and ν 7→ νσ

are bijections on Sn.

If σ is even, these bijections map evenpermutations to even permutations and odd permutations to oddpermutations. If σ is odd, they map odd permutations to evenpermutations and even permutations to odd permutations. Butthen, if σ is even, we have σAn = An = Anσ . If σ is odd wehave σAn = Sn \An = Anσ . Thus An is normal in Sn.

are bijections on Sn. If σ is even, these bijections map evenpermutations to even permutations and odd permutations to oddpermutations.

If σ is odd, they map odd permutations to evenpermutations and even permutations to odd permutations. Butthen, if σ is even, we have σAn = An = Anσ . If σ is odd wehave σAn = Sn \An = Anσ . Thus An is normal in Sn.

are bijections on Sn. If σ is even, these bijections map evenpermutations to even permutations and odd permutations to oddpermutations. If σ is odd, they map odd permutations to evenpermutations and even permutations to odd permutations.

Butthen, if σ is even, we have σAn = An = Anσ . If σ is odd wehave σAn = Sn \An = Anσ . Thus An is normal in Sn.

are bijections on Sn. If σ is even, these bijections map evenpermutations to even permutations and odd permutations to oddpermutations. If σ is odd, they map odd permutations to evenpermutations and even permutations to odd permutations. Butthen, if σ is even, we have σAn = An = Anσ .

If σ is odd wehave σAn = Sn \An = Anσ . Thus An is normal in Sn.

are bijections on Sn. If σ is even, these bijections map evenpermutations to even permutations and odd permutations to oddpermutations. If σ is odd, they map odd permutations to evenpermutations and even permutations to odd permutations. Butthen, if σ is even, we have σAn = An = Anσ . If σ is odd wehave σAn = Sn \An = Anσ .

Thus An is normal in Sn.

Definition.

A group is called simple iff it has no nontrivialnormal subgroup.

Lemma. Any even permutation is a product of 3-cycles.

Proof. (ab)(ab) = id = (abc)(cba), (ab)(bc) = (abc),(ab)(cd) = (acb)(acd).

Definition. A group is called simple iff it has no nontrivialnormal subgroup.

Lemma.

Any even permutation is a product of 3-cycles.

Proof.

(ab)(ab) = id = (abc)(cba), (ab)(bc) = (abc),(ab)(cd) = (acb)(acd).

Proof. (ab)(ab) = id = (abc)(cba)

, (ab)(bc) = (abc),(ab)(cd) = (acb)(acd).

Proof. (ab)(ab) = id = (abc)(cba), (ab)(bc) = (abc)

,(ab)(cd) = (acb)(acd).

Theorem.

An is simple for all n≥ 5.

Proof. Let N CAn be a normal subgroup that is not equal to{id}. Assume N contains the 3-cycle (abc). Let x,y 6∈ {a,b,c}be two distinct elements of {1, . . . ,n}. Then one of

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

)(where we simply complete the

assignment beyond 5 to somehow get a permutation) or

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

N 3 γ−1(abc)γ = (123) or N 3 δ−1(abc)δ = (123). Now that(123) ∈ N, a similar argument proves that all 3-cycles are in N.So, if N contains a 3-cycle, then N = An.Thus, we are done if we can prove that N contains a 3-cycle.Let α ∈ N and consider the representation of α as a product ofdisjoint cycles.

Theorem. An is simple for all n≥ 5.

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

Proof.

Let N CAn be a normal subgroup that is not equal to{id}. Assume N contains the 3-cycle (abc). Let x,y 6∈ {a,b,c}be two distinct elements of {1, . . . ,n}. Then one of

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

Proof. Let N CAn be a normal subgroup that is not equal to{id}.

Assume N contains the 3-cycle (abc). Let x,y 6∈ {a,b,c}be two distinct elements of {1, . . . ,n}. Then one of

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

Proof. Let N CAn be a normal subgroup that is not equal to{id}. Assume N contains the 3-cycle (abc).

Let x,y 6∈ {a,b,c}be two distinct elements of {1, . . . ,n}. Then one of

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

Proof. Let N CAn be a normal subgroup that is not equal to{id}. Assume N contains the 3-cycle (abc). Let x,y 6∈ {a,b,c}be two distinct elements of {1, . . . ,n}.

Then one of

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

assignment beyond 5 to somehow get a permutation)

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

is even. But then

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even.

But then

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

3 γ−1(abc)γ = (123) or N 3 δ−1(abc)δ = (123). Now that(123) ∈ N, a similar argument proves that all 3-cycles are in N.So, if N contains a 3-cycle, then N = An.Thus, we are done if we can prove that N contains a 3-cycle.Let α ∈ N and consider the representation of α as a product ofdisjoint cycles.

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

N 3 γ−1(abc)γ

= (123) or N 3 δ−1(abc)δ = (123). Now that(123) ∈ N, a similar argument proves that all 3-cycles are in N.So, if N contains a 3-cycle, then N = An.Thus, we are done if we can prove that N contains a 3-cycle.Let α ∈ N and consider the representation of α as a product ofdisjoint cycles.

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

N 3 γ−1(abc)γ = (123)

or N 3 δ−1(abc)δ = (123). Now that(123) ∈ N, a similar argument proves that all 3-cycles are in N.So, if N contains a 3-cycle, then N = An.Thus, we are done if we can prove that N contains a 3-cycle.Let α ∈ N and consider the representation of α as a product ofdisjoint cycles.

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

N 3 γ−1(abc)γ = (123) or N

3 δ−1(abc)δ = (123). Now that(123) ∈ N, a similar argument proves that all 3-cycles are in N.So, if N contains a 3-cycle, then N = An.Thus, we are done if we can prove that N contains a 3-cycle.Let α ∈ N and consider the representation of α as a product ofdisjoint cycles.

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

N 3 γ−1(abc)γ = (123) or N 3 δ−1(abc)δ

= (123). Now that(123) ∈ N, a similar argument proves that all 3-cycles are in N.So, if N contains a 3-cycle, then N = An.Thus, we are done if we can prove that N contains a 3-cycle.Let α ∈ N and consider the representation of α as a product ofdisjoint cycles.

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

N 3 γ−1(abc)γ = (123) or N 3 δ−1(abc)δ = (123).

Now that(123) ∈ N, a similar argument proves that all 3-cycles are in N.So, if N contains a 3-cycle, then N = An.Thus, we are done if we can prove that N contains a 3-cycle.Let α ∈ N and consider the representation of α as a product ofdisjoint cycles.

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

N 3 γ−1(abc)γ = (123) or N 3 δ−1(abc)δ = (123). Now that(123) ∈ N, a similar argument proves that all 3-cycles are in N.

So, if N contains a 3-cycle, then N = An.Thus, we are done if we can prove that N contains a 3-cycle.Let α ∈ N and consider the representation of α as a product ofdisjoint cycles.

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

N 3 γ−1(abc)γ = (123) or N 3 δ−1(abc)δ = (123). Now that(123) ∈ N, a similar argument proves that all 3-cycles are in N.So, if N contains a 3-cycle, then N = An.

Thus, we are done if we can prove that N contains a 3-cycle.Let α ∈ N and consider the representation of α as a product ofdisjoint cycles.

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

N 3 γ−1(abc)γ = (123) or N 3 δ−1(abc)δ = (123). Now that(123) ∈ N, a similar argument proves that all 3-cycles are in N.So, if N contains a 3-cycle, then N = An.Thus, we are done if we can prove that N contains a 3-cycle.

Let α ∈ N and consider the representation of α as a product ofdisjoint cycles.

γ :=(

1 2 3 4 5 · · · na b c x y · · ·

δ := (xy)γ =(

1 2 3 4 5 · · · na b c y x · · ·

)is even. But then

Proof (cont.).

Case 1: The representation of α as a product ofdisjoint cycles contains an r-cycle (ijkl . . .) with r ≥ 4. Thenα = (ijkl . . .)γ , where γ is a product of cycles that do not shareelements with (ijkl . . .). Let β := (ijk). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

=((ijkl . . .)(ijk)(. . . lkji)

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.

Proof (cont.). Case 1: The representation of α as a product ofdisjoint cycles contains an r-cycle (ijkl . . .) with r ≥ 4.

Thenα = (ijkl . . .)γ , where γ is a product of cycles that do not shareelements with (ijkl . . .). Let β := (ijk). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

=((ijkl . . .)(ijk)(. . . lkji)

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.

Proof (cont.). Case 1: The representation of α as a product ofdisjoint cycles contains an r-cycle (ijkl . . .) with r ≥ 4. Thenα = (ijkl . . .)γ , where γ is a product of cycles that do not shareelements with (ijkl . . .).

Let β := (ijk). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

=((ijkl . . .)(ijk)(. . . lkji)

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.

Proof (cont.). Case 1: The representation of α as a product ofdisjoint cycles contains an r-cycle (ijkl . . .) with r ≥ 4. Thenα = (ijkl . . .)γ , where γ is a product of cycles that do not shareelements with (ijkl . . .). Let β := (ijk).

Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

=((ijkl . . .)(ijk)(. . . lkji)

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.

Proof (cont.). Case 1: The representation of α as a product ofdisjoint cycles contains an r-cycle (ijkl . . .) with r ≥ 4. Thenα = (ijkl . . .)γ , where γ is a product of cycles that do not shareelements with (ijkl . . .). Let β := (ijk). Then N contains theproduct βα−1β−1

and hence

N 3 α

−1β−1)

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

=((ijkl . . .)(ijk)(. . . lkji)

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.

Proof (cont.). Case 1: The representation of α as a product ofdisjoint cycles contains an r-cycle (ijkl . . .) with r ≥ 4. Thenα = (ijkl . . .)γ , where γ is a product of cycles that do not shareelements with (ijkl . . .). Let β := (ijk). Then N contains theproduct βα−1β−1 and hence

−1β−1)

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

=((ijkl . . .)(ijk)(. . . lkji)

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.

N 3 α

−1β−1)

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

=((ijkl . . .)(ijk)(. . . lkji)

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.

N 3 α

−1β−1)

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

=((ijkl . . .)(ijk)(. . . lkji)

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.

N 3 α

−1β−1)

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

=((ijkl . . .)(ijk)(. . . lkji)

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.

N 3 α

−1β−1)

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

=((ijkl . . .)(ijk)(. . . lkji)

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.

N 3 α

−1β−1)

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

=((ijkl . . .)(ijk)(. . . lkji)

)(kji)

= (jkl)(kji)

= (ilj)

Hence N = An.

N 3 α

−1β−1)

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

=((ijkl . . .)(ijk)(. . . lkji)

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.

N 3 α

−1β−1)

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

=((ijkl . . .)(ijk)(. . . lkji)

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.Bernd Schroder Louisiana Tech University, College of Engineering and Science

Proof (cont.).

Case 2: The representation of α as a product ofdisjoint cycles contains a 3-cycle (ijk) and all cycles in thisrepresentation have 2 or 3 elements. If α is equal to the 3-cycle(ijk), then there is nothing to prove. Otherwise there are distinctelements l,m 6∈ {i, j,k}, a cycle (lm·) that is either atransposition or a 3-cycle, and a permutation γ that consists ofcycles that contain none of the elements of (ijk) and (lm·) sothat α = (ijk)(lm·)γ . Let β := (ijl). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

=((ijk)(lm·)γ(ijl)γ−1(·ml)(kji)

)(lji)

=((ijk)(lm·)(ijl)(·ml)(kji)

)(lji) = (jkm)(lji) = (ilkmj)

By case 1, N contains a 3-cycle and hence N = An.

Proof (cont.). Case 2:

The representation of α as a product ofdisjoint cycles contains a 3-cycle (ijk) and all cycles in thisrepresentation have 2 or 3 elements. If α is equal to the 3-cycle(ijk), then there is nothing to prove. Otherwise there are distinctelements l,m 6∈ {i, j,k}, a cycle (lm·) that is either atransposition or a 3-cycle, and a permutation γ that consists ofcycles that contain none of the elements of (ijk) and (lm·) sothat α = (ijk)(lm·)γ . Let β := (ijl). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

Proof (cont.). Case 2: The representation of α as a product ofdisjoint cycles contains a 3-cycle (ijk)

and all cycles in thisrepresentation have 2 or 3 elements. If α is equal to the 3-cycle(ijk), then there is nothing to prove. Otherwise there are distinctelements l,m 6∈ {i, j,k}, a cycle (lm·) that is either atransposition or a 3-cycle, and a permutation γ that consists ofcycles that contain none of the elements of (ijk) and (lm·) sothat α = (ijk)(lm·)γ . Let β := (ijl). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

Proof (cont.). Case 2: The representation of α as a product ofdisjoint cycles contains a 3-cycle (ijk) and all cycles in thisrepresentation have 2 or 3 elements.

If α is equal to the 3-cycle(ijk), then there is nothing to prove. Otherwise there are distinctelements l,m 6∈ {i, j,k}, a cycle (lm·) that is either atransposition or a 3-cycle, and a permutation γ that consists ofcycles that contain none of the elements of (ijk) and (lm·) sothat α = (ijk)(lm·)γ . Let β := (ijl). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

Proof (cont.). Case 2: The representation of α as a product ofdisjoint cycles contains a 3-cycle (ijk) and all cycles in thisrepresentation have 2 or 3 elements. If α is equal to the 3-cycle(ijk), then there is nothing to prove.

Otherwise there are distinctelements l,m 6∈ {i, j,k}, a cycle (lm·) that is either atransposition or a 3-cycle, and a permutation γ that consists ofcycles that contain none of the elements of (ijk) and (lm·) sothat α = (ijk)(lm·)γ . Let β := (ijl). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

Proof (cont.). Case 2: The representation of α as a product ofdisjoint cycles contains a 3-cycle (ijk) and all cycles in thisrepresentation have 2 or 3 elements. If α is equal to the 3-cycle(ijk), then there is nothing to prove. Otherwise there are distinctelements l,m 6∈ {i, j,k}

, a cycle (lm·) that is either atransposition or a 3-cycle, and a permutation γ that consists ofcycles that contain none of the elements of (ijk) and (lm·) sothat α = (ijk)(lm·)γ . Let β := (ijl). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

Proof (cont.). Case 2: The representation of α as a product ofdisjoint cycles contains a 3-cycle (ijk) and all cycles in thisrepresentation have 2 or 3 elements. If α is equal to the 3-cycle(ijk), then there is nothing to prove. Otherwise there are distinctelements l,m 6∈ {i, j,k}, a cycle (lm·) that is either atransposition or a 3-cycle

, and a permutation γ that consists ofcycles that contain none of the elements of (ijk) and (lm·) sothat α = (ijk)(lm·)γ . Let β := (ijl). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

Proof (cont.). Case 2: The representation of α as a product ofdisjoint cycles contains a 3-cycle (ijk) and all cycles in thisrepresentation have 2 or 3 elements. If α is equal to the 3-cycle(ijk), then there is nothing to prove. Otherwise there are distinctelements l,m 6∈ {i, j,k}, a cycle (lm·) that is either atransposition or a 3-cycle, and a permutation γ that consists ofcycles that contain none of the elements of (ijk) and (lm·)

sothat α = (ijk)(lm·)γ . Let β := (ijl). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

Proof (cont.). Case 2: The representation of α as a product ofdisjoint cycles contains a 3-cycle (ijk) and all cycles in thisrepresentation have 2 or 3 elements. If α is equal to the 3-cycle(ijk), then there is nothing to prove. Otherwise there are distinctelements l,m 6∈ {i, j,k}, a cycle (lm·) that is either atransposition or a 3-cycle, and a permutation γ that consists ofcycles that contain none of the elements of (ijk) and (lm·) sothat α = (ijk)(lm·)γ .

Let β := (ijl). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

Proof (cont.). Case 2: The representation of α as a product ofdisjoint cycles contains a 3-cycle (ijk) and all cycles in thisrepresentation have 2 or 3 elements. If α is equal to the 3-cycle(ijk), then there is nothing to prove. Otherwise there are distinctelements l,m 6∈ {i, j,k}, a cycle (lm·) that is either atransposition or a 3-cycle, and a permutation γ that consists ofcycles that contain none of the elements of (ijk) and (lm·) sothat α = (ijk)(lm·)γ . Let β := (ijl).

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

Proof (cont.). Case 2: The representation of α as a product ofdisjoint cycles contains a 3-cycle (ijk) and all cycles in thisrepresentation have 2 or 3 elements. If α is equal to the 3-cycle(ijk), then there is nothing to prove. Otherwise there are distinctelements l,m 6∈ {i, j,k}, a cycle (lm·) that is either atransposition or a 3-cycle, and a permutation γ that consists ofcycles that contain none of the elements of (ijk) and (lm·) sothat α = (ijk)(lm·)γ . Let β := (ijl). Then N contains theproduct βα−1β−1

and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

Proof (cont.). Case 2: The representation of α as a product ofdisjoint cycles contains a 3-cycle (ijk) and all cycles in thisrepresentation have 2 or 3 elements. If α is equal to the 3-cycle(ijk), then there is nothing to prove. Otherwise there are distinctelements l,m 6∈ {i, j,k}, a cycle (lm·) that is either atransposition or a 3-cycle, and a permutation γ that consists ofcycles that contain none of the elements of (ijk) and (lm·) sothat α = (ijk)(lm·)γ . Let β := (ijl). Then N contains theproduct βα−1β−1 and hence

−1β−1)

αβα−1)

β−1

)(lji)

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

= (jkm)(lji) = (ilkmj)

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

)(lji) = (jkm)(lji)

= (ilkmj)

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

N 3 α

−1β−1)

αβα−1)

β−1

)(lji)

Proof (concl.).

Case 3: The representation of α as a product ofdisjoint cycles contains only transpositions. Then there aredistinct elements i, j,k, l ∈ {1, . . . ,n} and a permutation γ that isa product of cycles that contain none of i, j,k, l so thatα = (ij)(kl)γ . Moreover, there is an elementm ∈ {1, . . . ,n}\{i, j,k, l}. Let β := (ikm). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

=((ij)(kl)γ(ikm)γ−1(lk)(ji)

)(kmi)

=((ij)(kl)(ikα(m))(lk)(ji)

)(kmi)

= (jlα(m))(kmi) =

{(ikjlm) if α(m) = m,

(jlα(m))(kmi) if α(m) 6= m.

In either case N contains a 3-cycle and hence N = An.

Proof (concl.). Case 3:

The representation of α as a product ofdisjoint cycles contains only transpositions. Then there aredistinct elements i, j,k, l ∈ {1, . . . ,n} and a permutation γ that isa product of cycles that contain none of i, j,k, l so thatα = (ij)(kl)γ . Moreover, there is an elementm ∈ {1, . . . ,n}\{i, j,k, l}. Let β := (ikm). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

Proof (concl.). Case 3: The representation of α as a product ofdisjoint cycles contains only transpositions.

Then there aredistinct elements i, j,k, l ∈ {1, . . . ,n} and a permutation γ that isa product of cycles that contain none of i, j,k, l so thatα = (ij)(kl)γ . Moreover, there is an elementm ∈ {1, . . . ,n}\{i, j,k, l}. Let β := (ikm). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

Proof (concl.). Case 3: The representation of α as a product ofdisjoint cycles contains only transpositions. Then there aredistinct elements i, j,k, l ∈ {1, . . . ,n}

and a permutation γ that isa product of cycles that contain none of i, j,k, l so thatα = (ij)(kl)γ . Moreover, there is an elementm ∈ {1, . . . ,n}\{i, j,k, l}. Let β := (ikm). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

Proof (concl.). Case 3: The representation of α as a product ofdisjoint cycles contains only transpositions. Then there aredistinct elements i, j,k, l ∈ {1, . . . ,n} and a permutation γ that isa product of cycles that contain none of i, j,k, l

so thatα = (ij)(kl)γ . Moreover, there is an elementm ∈ {1, . . . ,n}\{i, j,k, l}. Let β := (ikm). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

Proof (concl.). Case 3: The representation of α as a product ofdisjoint cycles contains only transpositions. Then there aredistinct elements i, j,k, l ∈ {1, . . . ,n} and a permutation γ that isa product of cycles that contain none of i, j,k, l so thatα = (ij)(kl)γ .

Moreover, there is an elementm ∈ {1, . . . ,n}\{i, j,k, l}. Let β := (ikm). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

Proof (concl.). Case 3: The representation of α as a product ofdisjoint cycles contains only transpositions. Then there aredistinct elements i, j,k, l ∈ {1, . . . ,n} and a permutation γ that isa product of cycles that contain none of i, j,k, l so thatα = (ij)(kl)γ . Moreover, there is an elementm ∈ {1, . . . ,n}\{i, j,k, l}.

Let β := (ikm). Then N contains theproduct βα−1β−1 and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

Proof (concl.). Case 3: The representation of α as a product ofdisjoint cycles contains only transpositions. Then there aredistinct elements i, j,k, l ∈ {1, . . . ,n} and a permutation γ that isa product of cycles that contain none of i, j,k, l so thatα = (ij)(kl)γ . Moreover, there is an elementm ∈ {1, . . . ,n}\{i, j,k, l}. Let β := (ikm).

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

Proof (concl.). Case 3: The representation of α as a product ofdisjoint cycles contains only transpositions. Then there aredistinct elements i, j,k, l ∈ {1, . . . ,n} and a permutation γ that isa product of cycles that contain none of i, j,k, l so thatα = (ij)(kl)γ . Moreover, there is an elementm ∈ {1, . . . ,n}\{i, j,k, l}. Let β := (ikm). Then N contains theproduct βα−1β−1

and hence

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

Proof (concl.). Case 3: The representation of α as a product ofdisjoint cycles contains only transpositions. Then there aredistinct elements i, j,k, l ∈ {1, . . . ,n} and a permutation γ that isa product of cycles that contain none of i, j,k, l so thatα = (ij)(kl)γ . Moreover, there is an elementm ∈ {1, . . . ,n}\{i, j,k, l}. Let β := (ikm). Then N contains theproduct βα−1β−1 and hence

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi)

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

In either case N contains a 3-cycle and hence N = An.Bernd Schroder Louisiana Tech University, College of Engineering and Science

N 3 α

−1β−1)

αβα−1)

β−1

)(kmi)

= (jlα(m))(kmi) =

In either case N contains a 3-cycle and hence N = An.Bernd Schroder Louisiana Tech University, College of Engineering and Science

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