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

logo1

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

logo1


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).



logo1


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.







logo1







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

(similar to arithmetic modulo m).



logo1










logo1


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.



logo1


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}.





logo1


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}.





logo1


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}.





logo1


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}.





logo1



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.




logo1



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.




logo1



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.




logo1







logo1




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.



logo1




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.



logo1







logo1


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).



logo1


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

Thenthe following are equivalent.







logo1


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







logo1



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.






logo1



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.






logo1









logo1




Proof.

Good exercise.





logo1









logo1





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.




logo1





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.




logo1





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.




logo1









logo1






Proof.

Simple exercise (maybe too simple).



logo1






Proof. Simple exercise

(maybe too simple).



logo1









logo1


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.



logo1



(gN)(hN)

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




logo1



(gN)(hN) = {gn : n ∈ N}{hm : m ∈ N}

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




logo1



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

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




logo1



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

= {gha : a ∈ N}= ghN.




logo1



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

= ghN.




logo1







logo1




or

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



logo1




or (gN)(hN)

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



logo1




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

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



logo1




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

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



logo1




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

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



logo1







logo1


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).)



logo1


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.





logo1


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.





logo1


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.





logo1



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.




logo1



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.




logo1



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.




logo1







logo1




(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).)



logo1




(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).)



logo1







logo1


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).



logo1


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)∣∣ .





logo1


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)∣∣ .





logo1





[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).




logo1





[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).




logo1





[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).




logo1





[E : D] =∣∣G(E/D)

∣∣ and [D : F] =

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .





logo1





[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).



logo1





[E : D] =∣∣G(E/D)

∣∣ and [D : F] =

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .





logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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



logo1


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)




logo1


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)




logo1


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)




logo1



(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)




logo1



(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)




logo1



(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)




logo1



(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)




logo1



(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)




logo1



(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)




logo1



(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)




logo1



(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)




logo1



(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)




logo1



(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)




logo1



(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)




logo1



(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



logo1



(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)




logo1


Proof (part 1, surjectivity, concl.).

σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)

= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))

=h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))

=h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))

=h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1



σ(p(x)

)= σ

(h

∏j=1

(x−σj(θ)

))=

h

∏j=1

σ

((x−σj(θ)

))=

h

∏j=1

(x−σ

(σj(θ)

))=

h

∏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).



logo1


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)∣∣ .



logo1


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)∣∣ .



logo1


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)∣∣ .



logo1



∣∣G(E/D)∣∣= [E : D] and

∣∣G(E/F)∣∣= [E : F].

Now

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

=

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .



logo1



∣∣G(E/D)∣∣= [E : D] and

∣∣G(E/F)∣∣= [E : F]. Now

[D : F]

=[E : F][E : D]

=

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .



logo1



∣∣G(E/D)∣∣= [E : D] and

∣∣G(E/F)∣∣= [E : F]. Now

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

=

∣∣G(E/F)∣∣∣∣G(E/D)∣∣ .



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj) =

n

∑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.



logo1


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

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj) =

n

∑j=0

pjν (d)j = p(ν(d)

),




logo1


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

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj) =

n

∑j=0

pjν (d)j = p(ν(d)

),




logo1


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

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj) =

n

∑j=0

pjν (d)j = p(ν(d)

),




logo1


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

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj) =

n

∑j=0

pjν (d)j = p(ν(d)

),




logo1


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

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj) =

n

∑j=0

pjν (d)j = p(ν(d)

),




logo1




∑j=0

pjxj in F[x]. Now

0

= ν(0) = ν

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj) =

n

∑j=0

pjν (d)j = p(ν(d)

),




logo1




∑j=0

pjxj in F[x]. Now

0 = ν(0)

= ν

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj) =

n

∑j=0

pjν (d)j = p(ν(d)

),




logo1




∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

(n

∑j=0

pjdj

)

=n

∑j=0

ν (pj)ν(dj) =

n

∑j=0

pjν (d)j = p(ν(d)

),




logo1




∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj)

=n

∑j=0

pjν (d)j = p(ν(d)

),




logo1




∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj) =

n

∑j=0

pjν (d)j

= p(ν(d)

),




logo1




∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj) =

n

∑j=0

pjν (d)j = p(ν(d)

)

,




logo1




∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj) =

n

∑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.



logo1




∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj) =

n

∑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.



logo1




∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj) =

n

∑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.



logo1




∑j=0

pjxj in F[x]. Now

0 = ν(0) = ν

(n

∑j=0

pjdj

)=

n

∑j=0

ν (pj)ν(dj) =

n

∑j=0

pjν (d)j = p(ν(d)

),




logo1


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)

)= γ

−1(d), and

then γσγ−1(d) = γ

(σγ

−1(d))

= γ

(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).



logo1


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)

)= γ

−1(d), and


(σγ

−1(d))

= γ

(γ−1(d)

)= d. Because




logo1


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)

)= γ

−1(d), and


(σγ

−1(d))

= γ

(γ−1(d)

)= d. Because




logo1


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)

)= γ

−1(d), and


(σγ

−1(d))

= γ

(γ−1(d)

)= d. Because




logo1



−1(d) = σ

(γ−1(d)

)

= γ−1(d), and


(σγ

−1(d))

= γ

(γ−1(d)

)= d. Because




logo1



−1(d) = σ

(γ−1(d)

)= γ

−1(d)

, and


(σγ

−1(d))

= γ

(γ−1(d)

)= d. Because




logo1



−1(d) = σ

(γ−1(d)

)= γ

−1(d), and

then γσγ−1(d)

= γ

(σγ

−1(d))

= γ

(γ−1(d)

)= d. Because




logo1



−1(d) = σ

(γ−1(d)

)= γ

−1(d), and


(σγ

−1(d))

= γ

(γ−1(d)

)= d. Because




logo1



−1(d) = σ

(γ−1(d)

)= γ

−1(d), and


(σγ

−1(d))

= γ

(γ−1(d)

)

= d. Because




logo1



−1(d) = σ

(γ−1(d)

)= γ

−1(d), and


(σγ

−1(d))

= γ

(γ−1(d)

)= d.

Because




logo1



−1(d) = σ

(γ−1(d)

)= γ

−1(d), and


(σγ

−1(d))

= γ

(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).



logo1



−1(d) = σ

(γ−1(d)

)= γ

−1(d), and


(σγ

−1(d))

= γ

(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).



logo1



−1(d) = σ

(γ−1(d)

)= γ

−1(d), and


(σγ

−1(d))

= γ

(γ−1(d)

)= d. Because




logo1


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.



logo1


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





logo1


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





logo1


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





logo1


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





logo1


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





logo1


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





logo1


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





logo1


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





logo1



µ γµ

∣∣D = µ−1µ

= idD, that is,δ−1


δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

=(

δµδ−1µ





logo1



µ γµ

∣∣D = µ−1µ = idD

, that is,δ−1


δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

=(

δµδ−1µ





logo1



µ γµ


δ−1µ γµ ∈ G(E/D).

Hence

δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

=(

δµδ−1µ





logo1



µ γµ



δµG(E/D)

= δµ

(δ−1µ γµ

)G(E/D)

=(

δµδ−1µ





logo1



µ γµ



δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

=(

δµδ−1µ





logo1



µ γµ



δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

=(

δµδ−1µ

)γµG(E/D)

= γµG(E/D).




logo1



µ γµ



δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

=(

δµδ−1µ





logo1



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.



logo1



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.



logo1



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.



logo1



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.



logo1



µ γµ



δµG(E/D) = δµ

(δ−1µ γµ

)G(E/D)

=(

δµδ−1µ





logo1


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.



logo1


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





logo1


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





logo1


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).




logo1



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).




logo1



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).




logo1



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).




logo1



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).




logo1







logo1


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.



logo1


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)





logo1


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)





logo1


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.



logo1



)= 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.



logo1



)= 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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1







logo1


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)

)= 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.



logo1


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)

).




logo1



Ψ(γG(E/D)δG(E/D)

)= Ψ

(γδG(E/D)

)

= γδ |D= γ|Dδ |D= Ψ

(γG(E/D)

)Ψ(δG(E/D)

).




logo1



Ψ(γG(E/D)δG(E/D)

)= Ψ

(γδG(E/D)

)= γδ |D

= γ|Dδ |D= Ψ

(γG(E/D)

)Ψ(δG(E/D)

).




logo1



Ψ(γG(E/D)δG(E/D)

)= Ψ

(γδG(E/D)

)= γδ |D= γ|Dδ |D

= Ψ(γG(E/D)

)Ψ(δG(E/D)

).




logo1



Ψ(γG(E/D)δG(E/D)

)= Ψ

(γδG(E/D)


(γG(E/D)

)Ψ(δG(E/D)

).




logo1



Ψ(γG(E/D)δG(E/D)

)= Ψ

(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.



logo1



Ψ(γG(E/D)δG(E/D)

)= Ψ

(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.



logo1



Ψ(γG(E/D)δG(E/D)

)= Ψ

(γδG(E/D)


(γG(E/D)

)Ψ(δG(E/D)

).




logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.




logo1


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.




logo1



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.



logo1



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.



logo1



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.



logo1



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.



logo1



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.



logo1



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.



logo1



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.



logo1






logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.



logo1


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.







logo1


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.







logo1



Proof.

Simple exercise (maybe too simple).






logo1



Proof. Simple exercise

(maybe too simple).






logo1









logo1




Proposition.

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





logo1









logo1





Proof.

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




logo1






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.



logo1






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.



logo1






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.



logo1






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.



logo1






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.



logo1









logo1


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).



logo1


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





logo1



Lemma.

Any even permutation is a product of 3-cycles.




logo1







logo1




Proof.

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



logo1




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

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



logo1




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

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



logo1







logo1


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.



logo1


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




logo1



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




logo1



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




logo1



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




logo1



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




logo1




γ :=(

1 2 3 4 5 · · · na b c x y · · ·


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




logo1




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




logo1




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




logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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.



logo1




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




logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

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

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.



logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

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

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.



logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

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

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.



logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

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

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.



logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

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

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.



logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

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

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.



logo1



N 3 α

(βα

−1β−1)

=(

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

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

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.



logo1



N 3 α

(βα

−1β−1)

=(

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

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

)(kji)

= (jkl)(kji)

= (ilj)

Hence N = An.



logo1



N 3 α

(βα

−1β−1)

=(

αβα−1)

β−1

=((ijkl . . .)γ(ijk)γ−1(. . . lkji)

)(kji)

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

)(kji)

= (jkl)(kji)= (ilj)

Hence N = An.



logo1



N 3 α

(βα

−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


logo1


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

(βα

−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.



logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)






logo1



N 3 α

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)






logo1



N 3 α

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)


)(lji)

= (jkm)(lji) = (ilkmj)




logo1



N 3 α

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)


)(lji) = (jkm)(lji)

= (ilkmj)




logo1



N 3 α

(βα

−1β−1)

=(

αβα−1)

β−1


)(lji)






logo1


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

(βα

−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.



logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(kmi)


)(kmi)

= (jlα(m))(kmi) =






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(kmi)


)(kmi)

= (jlα(m))(kmi) =






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(kmi)


)(kmi)

= (jlα(m))(kmi) =






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(kmi)


)(kmi)

= (jlα(m))(kmi) =






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(kmi)


)(kmi)

= (jlα(m))(kmi) =






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(kmi)


)(kmi)

= (jlα(m))(kmi) =






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(kmi)


)(kmi)

= (jlα(m))(kmi) =






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(kmi)


)(kmi)

= (jlα(m))(kmi) =






logo1


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

(βα

−1β−1)

=(

αβα−1)

β−1


)(kmi)


)(kmi)

= (jlα(m))(kmi) =






logo1



N 3 α

(βα

−1β−1)

=(

αβα−1)

β−1


)(kmi)


)(kmi)

= (jlα(m))(kmi) =






logo1



N 3 α

(βα

−1β−1)

=(

αβα−1)

β−1


)(kmi)


)(kmi)

= (jlα(m))(kmi)

=






logo1



N 3 α

(βα

−1β−1)

=(

αβα−1)

β−1


)(kmi)


)(kmi)

= (jlα(m))(kmi) =






logo1



N 3 α

(βα

−1β−1)

=(

αβα−1)

β−1


)(kmi)


)(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


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

Documents

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