Section 13 Homomorphisms Instructor: Yifan Yang Fall 2006 - - PowerPoint PPT Presentation

Section 13 – Homomorphisms

Instructor: Yifan Yang Fall 2006

Instructor: Yifan Yang Section 13 – Homomorphisms

Homomorphisms

Definition A map φ of a group G into a group G′ is a homomorphism if φ(ab) = φ(a)φ(b) for all a, b ∈ G.

Instructor: Yifan Yang Section 13 – Homomorphisms

Examples

1

Let φ : G → G′ be defined by φ(g) = e′ for all g ∈ G. Then clearly, φ(ab) = e′ = e′e′ = φ(a)φ(b) for all a, b ∈ G. This is called the trivial homomorphism.

2

Let φ : Z → Z be defined by φ(n) = 2n for all n ∈ Z. Then φ is a homomorphism.

3

Let Sn be the symmetric group on n letters, and let φ : Sn → Z2 be defined by φ(σ) =

• 0,

if σ is an even permutation, 1, if σ is an odd permutation. Then φ is a homomorphism. (Check case by case.)

Instructor: Yifan Yang Section 13 – Homomorphisms

Examples

1

Let φ : G → G′ be defined by φ(g) = e′ for all g ∈ G. Then clearly, φ(ab) = e′ = e′e′ = φ(a)φ(b) for all a, b ∈ G. This is called the trivial homomorphism.

2

Let φ : Z → Z be defined by φ(n) = 2n for all n ∈ Z. Then φ is a homomorphism.

3

Let Sn be the symmetric group on n letters, and let φ : Sn → Z2 be defined by φ(σ) =

• 0,

if σ is an even permutation, 1, if σ is an odd permutation. Then φ is a homomorphism. (Check case by case.)

Instructor: Yifan Yang Section 13 – Homomorphisms

Examples

1

Let φ : G → G′ be defined by φ(g) = e′ for all g ∈ G. Then clearly, φ(ab) = e′ = e′e′ = φ(a)φ(b) for all a, b ∈ G. This is called the trivial homomorphism.

2

Let φ : Z → Z be defined by φ(n) = 2n for all n ∈ Z. Then φ is a homomorphism.

3

Let Sn be the symmetric group on n letters, and let φ : Sn → Z2 be defined by φ(σ) =

• 0,

if σ is an even permutation, 1, if σ is an odd permutation. Then φ is a homomorphism. (Check case by case.)

Instructor: Yifan Yang Section 13 – Homomorphisms

example

1

Let GL(n, R) be the set of all invertible n × n matrices over

• R. Define φ : GL(n, R) → R× by φ(A) = det(A). Then φ is a

homomorphism since det(AB) = det(A) det(B).

2

Let F be the additive group of all polynomials with real

• coefficients. For a given real number a, the function

φa : F → R defined by φ(f) = f(a) is a homomorphism, called an evaluation homomorphism.

3

Let n be a positive integer. Define φn : Z → Zn by φn(r) = ¯

• r. Then φn is a homomorphism.

4

Let G = G1 × G2 × . . . × Gn be a direct product of groups. The projection map πi : G → Gi defined by πi(a1, a2, . . . , ai, . . . , an) = ai is a homomorphism.

Instructor: Yifan Yang Section 13 – Homomorphisms

example

1

Let GL(n, R) be the set of all invertible n × n matrices over

• R. Define φ : GL(n, R) → R× by φ(A) = det(A). Then φ is a

homomorphism since det(AB) = det(A) det(B).

2

Let F be the additive group of all polynomials with real

• coefficients. For a given real number a, the function

φa : F → R defined by φ(f) = f(a) is a homomorphism, called an evaluation homomorphism.

3

Let n be a positive integer. Define φn : Z → Zn by φn(r) = ¯

• r. Then φn is a homomorphism.

4

Let G = G1 × G2 × . . . × Gn be a direct product of groups. The projection map πi : G → Gi defined by πi(a1, a2, . . . , ai, . . . , an) = ai is a homomorphism.

Instructor: Yifan Yang Section 13 – Homomorphisms

example

1

Let GL(n, R) be the set of all invertible n × n matrices over

• R. Define φ : GL(n, R) → R× by φ(A) = det(A). Then φ is a

homomorphism since det(AB) = det(A) det(B).

2

Let F be the additive group of all polynomials with real

• coefficients. For a given real number a, the function

φa : F → R defined by φ(f) = f(a) is a homomorphism, called an evaluation homomorphism.

3

Let n be a positive integer. Define φn : Z → Zn by φn(r) = ¯

• r. Then φn is a homomorphism.

4

Let G = G1 × G2 × . . . × Gn be a direct product of groups. The projection map πi : G → Gi defined by πi(a1, a2, . . . , ai, . . . , an) = ai is a homomorphism.

Instructor: Yifan Yang Section 13 – Homomorphisms

example

1

Let GL(n, R) be the set of all invertible n × n matrices over

• R. Define φ : GL(n, R) → R× by φ(A) = det(A). Then φ is a

homomorphism since det(AB) = det(A) det(B).

2

Let F be the additive group of all polynomials with real

• coefficients. For a given real number a, the function

φa : F → R defined by φ(f) = f(a) is a homomorphism, called an evaluation homomorphism.

3

Let n be a positive integer. Define φn : Z → Zn by φn(r) = ¯

• r. Then φn is a homomorphism.

4

Let G = G1 × G2 × . . . × Gn be a direct product of groups. The projection map πi : G → Gi defined by πi(a1, a2, . . . , ai, . . . , an) = ai is a homomorphism.

Instructor: Yifan Yang Section 13 – Homomorphisms

Properties of homomorphisms

Definition Let φ be a mapping of a set X into a set Y. Let A ⊂ X and B ⊂ Y. The image φ[A] of A under φ is {φ(a) : a ∈ A}. The set φ[X] is the range of φ. The inverse image φ−1[B] of B in X is {x ∈ X : φ(x) ∈ B}.

Instructor: Yifan Yang Section 13 – Homomorphisms

Properties of homomorphisms

Theorem (13.12) Let φ be a homomorphism of a group G into a group G′.

1

φ(e) = e′.

2

φ(a−1) = φ(a)−1 for all a ∈ G.

3

If H is a subgroup of G, then φ[H] is a subgroup of G′.

4

If K ′ is a subgroup of G′, then φ−1[K ′] is a subgroup of G.

Instructor: Yifan Yang Section 13 – Homomorphisms

Properties of homomorphisms

Theorem (13.12) Let φ be a homomorphism of a group G into a group G′.

1

φ(e) = e′.

2

φ(a−1) = φ(a)−1 for all a ∈ G.

3

If H is a subgroup of G, then φ[H] is a subgroup of G′.

4

If K ′ is a subgroup of G′, then φ−1[K ′] is a subgroup of G.

Instructor: Yifan Yang Section 13 – Homomorphisms

Properties of homomorphisms

Theorem (13.12) Let φ be a homomorphism of a group G into a group G′.

1

φ(e) = e′.

2

φ(a−1) = φ(a)−1 for all a ∈ G.

3

If H is a subgroup of G, then φ[H] is a subgroup of G′.

4

If K ′ is a subgroup of G′, then φ−1[K ′] is a subgroup of G.

Instructor: Yifan Yang Section 13 – Homomorphisms

Properties of homomorphisms

Theorem (13.12) Let φ be a homomorphism of a group G into a group G′.

1

φ(e) = e′.

2

φ(a−1) = φ(a)−1 for all a ∈ G.

3

If H is a subgroup of G, then φ[H] is a subgroup of G′.

4

If K ′ is a subgroup of G′, then φ−1[K ′] is a subgroup of G.

Instructor: Yifan Yang Section 13 – Homomorphisms

Properties of homomorphisms

Theorem (13.12) Let φ be a homomorphism of a group G into a group G′.

1

φ(e) = e′.

2

φ(a−1) = φ(a)−1 for all a ∈ G.

3

If H is a subgroup of G, then φ[H] is a subgroup of G′.

4

If K ′ is a subgroup of G′, then φ−1[K ′] is a subgroup of G.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of φ(e) = e′. Consider φ(a), where a ∈ G. We have φ(a) = φ(ae) = φ(a)φ(e). By the cancellation law, φ(e) must equal to the identity e′. Proof of φ(a−1) = φ(a)−1. We have φ(a)φ(a−1) = φ(aa−1) = φ(e) = e′. Thus, φ(a−1) = φ(a)−1.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of φ(e) = e′. Consider φ(a), where a ∈ G. We have φ(a) = φ(ae) = φ(a)φ(e). By the cancellation law, φ(e) must equal to the identity e′. Proof of φ(a−1) = φ(a)−1. We have φ(a)φ(a−1) = φ(aa−1) = φ(e) = e′. Thus, φ(a−1) = φ(a)−1.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(3). We need to prove

1

Closed: Suppose that a′, b′ ∈ φ[H]. Then there exist a, b ∈ H such that φ(a) = a′ and φ(b) = b′. Thus, a′b′ = φ(a)φ(b) = φ(ab). Since H is a subgroup, ab ∈ H. Therefore, a′b′ is in φ[H].

2

identity: By Part (1), e′ = φ(e) ∈ φ[H].

3

inverse: Suppose that a′ ∈ φ[H]. Then a′ = φ(a) for some a ∈ H. By Part (b), (a′)−1 = φ(a)−1 = φ(a−1), and thus (a′)−1 ∈ φ[H].

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(3). We need to prove

1

Closed: Suppose that a′, b′ ∈ φ[H]. Then there exist a, b ∈ H such that φ(a) = a′ and φ(b) = b′. Thus, a′b′ = φ(a)φ(b) = φ(ab). Since H is a subgroup, ab ∈ H. Therefore, a′b′ is in φ[H].

2

identity: By Part (1), e′ = φ(e) ∈ φ[H].

3

inverse: Suppose that a′ ∈ φ[H]. Then a′ = φ(a) for some a ∈ H. By Part (b), (a′)−1 = φ(a)−1 = φ(a−1), and thus (a′)−1 ∈ φ[H].

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(3). We need to prove

1

Closed: Suppose that a′, b′ ∈ φ[H]. Then there exist a, b ∈ H such that φ(a) = a′ and φ(b) = b′. Thus, a′b′ = φ(a)φ(b) = φ(ab). Since H is a subgroup, ab ∈ H. Therefore, a′b′ is in φ[H].

2

identity: By Part (1), e′ = φ(e) ∈ φ[H].

3

inverse: Suppose that a′ ∈ φ[H]. Then a′ = φ(a) for some a ∈ H. By Part (b), (a′)−1 = φ(a)−1 = φ(a−1), and thus (a′)−1 ∈ φ[H].

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(3). We need to prove

1

Closed: Suppose that a′, b′ ∈ φ[H]. Then there exist a, b ∈ H such that φ(a) = a′ and φ(b) = b′. Thus, a′b′ = φ(a)φ(b) = φ(ab). Since H is a subgroup, ab ∈ H. Therefore, a′b′ is in φ[H].

2

identity: By Part (1), e′ = φ(e) ∈ φ[H].

3

inverse: Suppose that a′ ∈ φ[H]. Then a′ = φ(a) for some a ∈ H. By Part (b), (a′)−1 = φ(a)−1 = φ(a−1), and thus (a′)−1 ∈ φ[H].

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(3). We need to prove

1

Closed: Suppose that a′, b′ ∈ φ[H]. Then there exist a, b ∈ H such that φ(a) = a′ and φ(b) = b′. Thus, a′b′ = φ(a)φ(b) = φ(ab). Since H is a subgroup, ab ∈ H. Therefore, a′b′ is in φ[H].

2

identity: By Part (1), e′ = φ(e) ∈ φ[H].

3

inverse: Suppose that a′ ∈ φ[H]. Then a′ = φ(a) for some a ∈ H. By Part (b), (a′)−1 = φ(a)−1 = φ(a−1), and thus (a′)−1 ∈ φ[H].

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(3). We need to prove

1

Closed: Suppose that a′, b′ ∈ φ[H]. Then there exist a, b ∈ H such that φ(a) = a′ and φ(b) = b′. Thus, a′b′ = φ(a)φ(b) = φ(ab). Since H is a subgroup, ab ∈ H. Therefore, a′b′ is in φ[H].

2

identity: By Part (1), e′ = φ(e) ∈ φ[H].

3

inverse: Suppose that a′ ∈ φ[H]. Then a′ = φ(a) for some a ∈ H. By Part (b), (a′)−1 = φ(a)−1 = φ(a−1), and thus (a′)−1 ∈ φ[H].

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(3). We need to prove

1

Closed: Suppose that a′, b′ ∈ φ[H]. Then there exist a, b ∈ H such that φ(a) = a′ and φ(b) = b′. Thus, a′b′ = φ(a)φ(b) = φ(ab). Since H is a subgroup, ab ∈ H. Therefore, a′b′ is in φ[H].

2

identity: By Part (1), e′ = φ(e) ∈ φ[H].

3

inverse: Suppose that a′ ∈ φ[H]. Then a′ = φ(a) for some a ∈ H. By Part (b), (a′)−1 = φ(a)−1 = φ(a−1), and thus (a′)−1 ∈ φ[H].

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(3). We need to prove

1

Closed: Suppose that a′, b′ ∈ φ[H]. Then there exist a, b ∈ H such that φ(a) = a′ and φ(b) = b′. Thus, a′b′ = φ(a)φ(b) = φ(ab). Since H is a subgroup, ab ∈ H. Therefore, a′b′ is in φ[H].

2

identity: By Part (1), e′ = φ(e) ∈ φ[H].

3

inverse: Suppose that a′ ∈ φ[H]. Then a′ = φ(a) for some a ∈ H. By Part (b), (a′)−1 = φ(a)−1 = φ(a−1), and thus (a′)−1 ∈ φ[H].

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(4). We need to show

1

Closed: Suppose that a, b ∈ φ−1[K ′]. We have φ(a), φ(b) ∈ K ′. Then φ(ab) = φ(a)φ(b) ∈ K ′ because φ(a), φ(b) ∈ K ′ and K ′ is a subgroup of G′.

2

Identity: By Part (1), we have φ(e) = e′ ∈ K ′. It follows that e ∈ φ−1[K ′] since K ′ is a subgroup.

3

Inverse: Let a ∈ φ−1[K ′]. We have φ(a) ∈ K ′. By Part (2), φ(a−1) = φ(a)−1. Since K ′ is a subgroup, φ(a) ∈ K ′ implies φ(a)−1 ∈ K ′.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(4). We need to show

1

Closed: Suppose that a, b ∈ φ−1[K ′]. We have φ(a), φ(b) ∈ K ′. Then φ(ab) = φ(a)φ(b) ∈ K ′ because φ(a), φ(b) ∈ K ′ and K ′ is a subgroup of G′.

2

Identity: By Part (1), we have φ(e) = e′ ∈ K ′. It follows that e ∈ φ−1[K ′] since K ′ is a subgroup.

3

Inverse: Let a ∈ φ−1[K ′]. We have φ(a) ∈ K ′. By Part (2), φ(a−1) = φ(a)−1. Since K ′ is a subgroup, φ(a) ∈ K ′ implies φ(a)−1 ∈ K ′.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(4). We need to show

1

Closed: Suppose that a, b ∈ φ−1[K ′]. We have φ(a), φ(b) ∈ K ′. Then φ(ab) = φ(a)φ(b) ∈ K ′ because φ(a), φ(b) ∈ K ′ and K ′ is a subgroup of G′.

2

Identity: By Part (1), we have φ(e) = e′ ∈ K ′. It follows that e ∈ φ−1[K ′] since K ′ is a subgroup.

3

Inverse: Let a ∈ φ−1[K ′]. We have φ(a) ∈ K ′. By Part (2), φ(a−1) = φ(a)−1. Since K ′ is a subgroup, φ(a) ∈ K ′ implies φ(a)−1 ∈ K ′.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(4). We need to show

1

Closed: Suppose that a, b ∈ φ−1[K ′]. We have φ(a), φ(b) ∈ K ′. Then φ(ab) = φ(a)φ(b) ∈ K ′ because φ(a), φ(b) ∈ K ′ and K ′ is a subgroup of G′.

2

Identity: By Part (1), we have φ(e) = e′ ∈ K ′. It follows that e ∈ φ−1[K ′] since K ′ is a subgroup.

3

Inverse: Let a ∈ φ−1[K ′]. We have φ(a) ∈ K ′. By Part (2), φ(a−1) = φ(a)−1. Since K ′ is a subgroup, φ(a) ∈ K ′ implies φ(a)−1 ∈ K ′.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(4). We need to show

1

Closed: Suppose that a, b ∈ φ−1[K ′]. We have φ(a), φ(b) ∈ K ′. Then φ(ab) = φ(a)φ(b) ∈ K ′ because φ(a), φ(b) ∈ K ′ and K ′ is a subgroup of G′.

2

Identity: By Part (1), we have φ(e) = e′ ∈ K ′. It follows that e ∈ φ−1[K ′] since K ′ is a subgroup.

3

Inverse: Let a ∈ φ−1[K ′]. We have φ(a) ∈ K ′. By Part (2), φ(a−1) = φ(a)−1. Since K ′ is a subgroup, φ(a) ∈ K ′ implies φ(a)−1 ∈ K ′.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(4). We need to show

1

Closed: Suppose that a, b ∈ φ−1[K ′]. We have φ(a), φ(b) ∈ K ′. Then φ(ab) = φ(a)φ(b) ∈ K ′ because φ(a), φ(b) ∈ K ′ and K ′ is a subgroup of G′.

2

Identity: By Part (1), we have φ(e) = e′ ∈ K ′. It follows that e ∈ φ−1[K ′] since K ′ is a subgroup.

3

Inverse: Let a ∈ φ−1[K ′]. We have φ(a) ∈ K ′. By Part (2), φ(a−1) = φ(a)−1. Since K ′ is a subgroup, φ(a) ∈ K ′ implies φ(a)−1 ∈ K ′.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(4). We need to show

1

Closed: Suppose that a, b ∈ φ−1[K ′]. We have φ(a), φ(b) ∈ K ′. Then φ(ab) = φ(a)φ(b) ∈ K ′ because φ(a), φ(b) ∈ K ′ and K ′ is a subgroup of G′.

2

Identity: By Part (1), we have φ(e) = e′ ∈ K ′. It follows that e ∈ φ−1[K ′] since K ′ is a subgroup.

3

Inverse: Let a ∈ φ−1[K ′]. We have φ(a) ∈ K ′. By Part (2), φ(a−1) = φ(a)−1. Since K ′ is a subgroup, φ(a) ∈ K ′ implies φ(a)−1 ∈ K ′.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.12

Proof of Theorem 13.12(4). We need to show

1

Closed: Suppose that a, b ∈ φ−1[K ′]. We have φ(a), φ(b) ∈ K ′. Then φ(ab) = φ(a)φ(b) ∈ K ′ because φ(a), φ(b) ∈ K ′ and K ′ is a subgroup of G′.

2

Identity: By Part (1), we have φ(e) = e′ ∈ K ′. It follows that e ∈ φ−1[K ′] since K ′ is a subgroup.

3

Inverse: Let a ∈ φ−1[K ′]. We have φ(a) ∈ K ′. By Part (2), φ(a−1) = φ(a)−1. Since K ′ is a subgroup, φ(a) ∈ K ′ implies φ(a)−1 ∈ K ′.

Instructor: Yifan Yang Section 13 – Homomorphisms

Why are group homomorphisms important in group theory?

Let G = Z12 and G′ = Z7. How many homomorphisms from G to G′ are there?

• Solution. Let φ be a homomorphism from Z12 to Z7. On the one

hand, we have φ(0) = 0, by Theorem 13.11. On the other hand, we have 0 = 12 in Z12, and thus φ(0) = φ(12) = φ(1) + · · · + φ(1) = 12φ(1). Since 12 is relatively prime to 7, φ(1) must be equal to 0 in Z7. It follows that φ(n) = nφ(1) = 0 mod 7 for all n ∈ Z12. In other words, there is only one homomorphism, the trivial homomorphism, from Z12 to Z7.

Instructor: Yifan Yang Section 13 – Homomorphisms

Why are group homomorphisms important in group theory?

Let G = Z12 and G′ = Z7. How many homomorphisms from G to G′ are there?

• Solution. Let φ be a homomorphism from Z12 to Z7. On the one

hand, we have φ(0) = 0, by Theorem 13.11. On the other hand, we have 0 = 12 in Z12, and thus φ(0) = φ(12) = φ(1) + · · · + φ(1) = 12φ(1). Since 12 is relatively prime to 7, φ(1) must be equal to 0 in Z7. It follows that φ(n) = nφ(1) = 0 mod 7 for all n ∈ Z12. In other words, there is only one homomorphism, the trivial homomorphism, from Z12 to Z7.

Instructor: Yifan Yang Section 13 – Homomorphisms

Why are group homomorphisms important in group theory?

Let G = Z12 and G′ = Z7. How many homomorphisms from G to G′ are there?

• Solution. Let φ be a homomorphism from Z12 to Z7. On the one

hand, we have φ(0) = 0, by Theorem 13.11. On the other hand, we have 0 = 12 in Z12, and thus φ(0) = φ(12) = φ(1) + · · · + φ(1) = 12φ(1). Since 12 is relatively prime to 7, φ(1) must be equal to 0 in Z7. It follows that φ(n) = nφ(1) = 0 mod 7 for all n ∈ Z12. In other words, there is only one homomorphism, the trivial homomorphism, from Z12 to Z7.

Instructor: Yifan Yang Section 13 – Homomorphisms

Why are group homomorphisms important in group theory?

Let G = Z12 and G′ = Z7. How many homomorphisms from G to G′ are there?

• Solution. Let φ be a homomorphism from Z12 to Z7. On the one

hand, we have φ(0) = 0, by Theorem 13.11. On the other hand, we have 0 = 12 in Z12, and thus φ(0) = φ(12) = φ(1) + · · · + φ(1) = 12φ(1). Since 12 is relatively prime to 7, φ(1) must be equal to 0 in Z7. It follows that φ(n) = nφ(1) = 0 mod 7 for all n ∈ Z12. In other words, there is only one homomorphism, the trivial homomorphism, from Z12 to Z7.

Instructor: Yifan Yang Section 13 – Homomorphisms

Why are group homomorphisms important in group theory?

Let G = Z12 and G′ = Z7. How many homomorphisms from G to G′ are there?

• Solution. Let φ be a homomorphism from Z12 to Z7. On the one

hand, we have φ(0) = 0, by Theorem 13.11. On the other hand, we have 0 = 12 in Z12, and thus φ(0) = φ(12) = φ(1) + · · · + φ(1) = 12φ(1). Since 12 is relatively prime to 7, φ(1) must be equal to 0 in Z7. It follows that φ(n) = nφ(1) = 0 mod 7 for all n ∈ Z12. In other words, there is only one homomorphism, the trivial homomorphism, from Z12 to Z7.

Instructor: Yifan Yang Section 13 – Homomorphisms

Why are group homomorphisms important in group theory?

Let G = Z12 and G′ = Z7. How many homomorphisms from G to G′ are there?

• Solution. Let φ be a homomorphism from Z12 to Z7. On the one

hand, we have φ(0) = 0, by Theorem 13.11. On the other hand, we have 0 = 12 in Z12, and thus φ(0) = φ(12) = φ(1) + · · · + φ(1) = 12φ(1). Since 12 is relatively prime to 7, φ(1) must be equal to 0 in Z7. It follows that φ(n) = nφ(1) = 0 mod 7 for all n ∈ Z12. In other words, there is only one homomorphism, the trivial homomorphism, from Z12 to Z7.

Instructor: Yifan Yang Section 13 – Homomorphisms

Why are group homomorphisms important in group theory?

Now consider G = G′ = Z12. It can be shown that for all a ∈ Z12, the function φa : Z12 → Z12 defined by φa(r) = ar mod 12 is a homomorphism. From these two examples, we see that group homomorphisms are closely related to group

• structures. Thus, group homomorphisms are very important in

studying structural properties of groups.

Instructor: Yifan Yang Section 13 – Homomorphisms

Why are group homomorphisms important in group theory?

Now consider G = G′ = Z12. It can be shown that for all a ∈ Z12, the function φa : Z12 → Z12 defined by φa(r) = ar mod 12 is a homomorphism. From these two examples, we see that group homomorphisms are closely related to group

• structures. Thus, group homomorphisms are very important in

studying structural properties of groups.

Instructor: Yifan Yang Section 13 – Homomorphisms

Why are group homomorphisms important in group theory?

Now consider G = G′ = Z12. It can be shown that for all a ∈ Z12, the function φa : Z12 → Z12 defined by φa(r) = ar mod 12 is a homomorphism. From these two examples, we see that group homomorphisms are closely related to group

• structures. Thus, group homomorphisms are very important in

studying structural properties of groups.

Instructor: Yifan Yang Section 13 – Homomorphisms

Why are group homomorphisms important in group theory?

Now consider G = G′ = Z12. It can be shown that for all a ∈ Z12, the function φa : Z12 → Z12 defined by φa(r) = ar mod 12 is a homomorphism. From these two examples, we see that group homomorphisms are closely related to group

• structures. Thus, group homomorphisms are very important in

studying structural properties of groups.

Instructor: Yifan Yang Section 13 – Homomorphisms

Kernel

Since {e′} is a subgroup of G′, Theorem 13.11 shows that φ−1[{e′}] is a subgroup of G. This subgroup is of great importance. Definition Let φ : G → G′ be a group homomorphism. The subgroup φ−1[{e′}] = {g ∈ G : φ(g)} is called the kernel of φ, and denoted by Ker(φ).

Instructor: Yifan Yang Section 13 – Homomorphisms

Kernel

Theorem (13.15) Let φ : G → G′ be a group homomorphism, and let H = Ker(φ). Let a ∈ G. Then the set {x ∈ G : φ(x) = φ(a)} = φ−1[{φ(a)}] is the left coset aH, and is also the right coset Ha. Consequently, the partition of G into left cosets is the same as the partition into right cosets. Corollary (13.18) A group homomorphism φ : G → G′ is one-to-one if and only if Ker(φ) = {e}.

Instructor: Yifan Yang Section 13 – Homomorphisms

Kernel

Theorem (13.15) Let φ : G → G′ be a group homomorphism, and let H = Ker(φ). Let a ∈ G. Then the set {x ∈ G : φ(x) = φ(a)} = φ−1[{φ(a)}] is the left coset aH, and is also the right coset Ha. Consequently, the partition of G into left cosets is the same as the partition into right cosets. Corollary (13.18) A group homomorphism φ : G → G′ is one-to-one if and only if Ker(φ) = {e}.

Instructor: Yifan Yang Section 13 – Homomorphisms

Example

Question Are there any non-trivial homomorphisms from A4 to Z2? Solution Suppose that φ : A4 → Z2 is a non-trivial homomorphism. Then there is a σ in A4 such that φ(σ) = 1. By Theorem 13.15, Ker(φ) and σKer(φ) are left cosets, and form a partition of A4. In other words, (A4 : Ker(φ)) = 2, and |Ker(φ) = 6|. However, we see earlier that A4 does not have a subgroup of order 6. Thus, there is no non-trivial homomorphism from A4 to Z2.

Instructor: Yifan Yang Section 13 – Homomorphisms

Example

Question Are there any non-trivial homomorphisms from A4 to Z2? Solution Suppose that φ : A4 → Z2 is a non-trivial homomorphism. Then there is a σ in A4 such that φ(σ) = 1. By Theorem 13.15, Ker(φ) and σKer(φ) are left cosets, and form a partition of A4. In other words, (A4 : Ker(φ)) = 2, and |Ker(φ) = 6|. However, we see earlier that A4 does not have a subgroup of order 6. Thus, there is no non-trivial homomorphism from A4 to Z2.

Instructor: Yifan Yang Section 13 – Homomorphisms

Example

Question Are there any non-trivial homomorphisms from A4 to Z2? Solution Suppose that φ : A4 → Z2 is a non-trivial homomorphism. Then there is a σ in A4 such that φ(σ) = 1. By Theorem 13.15, Ker(φ) and σKer(φ) are left cosets, and form a partition of A4. In other words, (A4 : Ker(φ)) = 2, and |Ker(φ) = 6|. However, we see earlier that A4 does not have a subgroup of order 6. Thus, there is no non-trivial homomorphism from A4 to Z2.

Instructor: Yifan Yang Section 13 – Homomorphisms

Example

Question Are there any non-trivial homomorphisms from A4 to Z2? Solution Suppose that φ : A4 → Z2 is a non-trivial homomorphism. Then there is a σ in A4 such that φ(σ) = 1. By Theorem 13.15, Ker(φ) and σKer(φ) are left cosets, and form a partition of A4. In other words, (A4 : Ker(φ)) = 2, and |Ker(φ) = 6|. However, we see earlier that A4 does not have a subgroup of order 6. Thus, there is no non-trivial homomorphism from A4 to Z2.

Instructor: Yifan Yang Section 13 – Homomorphisms

Example

Question Are there any non-trivial homomorphisms from A4 to Z2? Solution Suppose that φ : A4 → Z2 is a non-trivial homomorphism. Then there is a σ in A4 such that φ(σ) = 1. By Theorem 13.15, Ker(φ) and σKer(φ) are left cosets, and form a partition of A4. In other words, (A4 : Ker(φ)) = 2, and |Ker(φ) = 6|. However, we see earlier that A4 does not have a subgroup of order 6. Thus, there is no non-trivial homomorphism from A4 to Z2.

Instructor: Yifan Yang Section 13 – Homomorphisms

Example

Question Are there any non-trivial homomorphisms from A4 to Z2? Solution Suppose that φ : A4 → Z2 is a non-trivial homomorphism. Then there is a σ in A4 such that φ(σ) = 1. By Theorem 13.15, Ker(φ) and σKer(φ) are left cosets, and form a partition of A4. In other words, (A4 : Ker(φ)) = 2, and |Ker(φ) = 6|. However, we see earlier that A4 does not have a subgroup of order 6. Thus, there is no non-trivial homomorphism from A4 to Z2.

Instructor: Yifan Yang Section 13 – Homomorphisms

Example

Question Are there any non-trivial homomorphisms from A4 to Z2? Solution Suppose that φ : A4 → Z2 is a non-trivial homomorphism. Then there is a σ in A4 such that φ(σ) = 1. By Theorem 13.15, Ker(φ) and σKer(φ) are left cosets, and form a partition of A4. In other words, (A4 : Ker(φ)) = 2, and |Ker(φ) = 6|. However, we see earlier that A4 does not have a subgroup of order 6. Thus, there is no non-trivial homomorphism from A4 to Z2.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.15

We will show that

1

aH ⊂ φ−1[{φ(a)}]: Let x = ag ∈ aH. Since φ is a homomorphism, we have φ(x) = φ(a)φ(h). By assumption that H = Ker(φ), it follows that φ(x) = φ(a), and x ∈ φ−1[{φ(a)}].

2

φ−1[{φ(a)}] ⊂ aH: Suppose that x ∈ φ−1[{φ(a)}]. We have φ(x) = φ(a), and thus φ(a)−1φ(x) = e′. By Theorem 13.12(2), φ(a)−1 = φ(a−1). It follows that φ(a−1x) = e′, and a−1x ∈ H. We see that x ∈ aH. The proof of φ−1[{φ(a)}] = Ha is similar.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.15

We will show that

1

aH ⊂ φ−1[{φ(a)}]: Let x = ag ∈ aH. Since φ is a homomorphism, we have φ(x) = φ(a)φ(h). By assumption that H = Ker(φ), it follows that φ(x) = φ(a), and x ∈ φ−1[{φ(a)}].

2

φ−1[{φ(a)}] ⊂ aH: Suppose that x ∈ φ−1[{φ(a)}]. We have φ(x) = φ(a), and thus φ(a)−1φ(x) = e′. By Theorem 13.12(2), φ(a)−1 = φ(a−1). It follows that φ(a−1x) = e′, and a−1x ∈ H. We see that x ∈ aH. The proof of φ−1[{φ(a)}] = Ha is similar.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.15

We will show that

1

aH ⊂ φ−1[{φ(a)}]: Let x = ag ∈ aH. Since φ is a homomorphism, we have φ(x) = φ(a)φ(h). By assumption that H = Ker(φ), it follows that φ(x) = φ(a), and x ∈ φ−1[{φ(a)}].

2

φ−1[{φ(a)}] ⊂ aH: Suppose that x ∈ φ−1[{φ(a)}]. We have φ(x) = φ(a), and thus φ(a)−1φ(x) = e′. By Theorem 13.12(2), φ(a)−1 = φ(a−1). It follows that φ(a−1x) = e′, and a−1x ∈ H. We see that x ∈ aH. The proof of φ−1[{φ(a)}] = Ha is similar.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.15

We will show that

1

aH ⊂ φ−1[{φ(a)}]: Let x = ag ∈ aH. Since φ is a homomorphism, we have φ(x) = φ(a)φ(h). By assumption that H = Ker(φ), it follows that φ(x) = φ(a), and x ∈ φ−1[{φ(a)}].

2

φ−1[{φ(a)}] ⊂ aH: Suppose that x ∈ φ−1[{φ(a)}]. We have φ(x) = φ(a), and thus φ(a)−1φ(x) = e′. By Theorem 13.12(2), φ(a)−1 = φ(a−1). It follows that φ(a−1x) = e′, and a−1x ∈ H. We see that x ∈ aH. The proof of φ−1[{φ(a)}] = Ha is similar.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.15

We will show that

1

aH ⊂ φ−1[{φ(a)}]: Let x = ag ∈ aH. Since φ is a homomorphism, we have φ(x) = φ(a)φ(h). By assumption that H = Ker(φ), it follows that φ(x) = φ(a), and x ∈ φ−1[{φ(a)}].

2

φ−1[{φ(a)}] ⊂ aH: Suppose that x ∈ φ−1[{φ(a)}]. We have φ(x) = φ(a), and thus φ(a)−1φ(x) = e′. By Theorem 13.12(2), φ(a)−1 = φ(a−1). It follows that φ(a−1x) = e′, and a−1x ∈ H. We see that x ∈ aH. The proof of φ−1[{φ(a)}] = Ha is similar.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.15

We will show that

1

aH ⊂ φ−1[{φ(a)}]: Let x = ag ∈ aH. Since φ is a homomorphism, we have φ(x) = φ(a)φ(h). By assumption that H = Ker(φ), it follows that φ(x) = φ(a), and x ∈ φ−1[{φ(a)}].

2

φ−1[{φ(a)}] ⊂ aH: Suppose that x ∈ φ−1[{φ(a)}]. We have φ(x) = φ(a), and thus φ(a)−1φ(x) = e′. By Theorem 13.12(2), φ(a)−1 = φ(a−1). It follows that φ(a−1x) = e′, and a−1x ∈ H. We see that x ∈ aH. The proof of φ−1[{φ(a)}] = Ha is similar.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.15

We will show that

1

aH ⊂ φ−1[{φ(a)}]: Let x = ag ∈ aH. Since φ is a homomorphism, we have φ(x) = φ(a)φ(h). By assumption that H = Ker(φ), it follows that φ(x) = φ(a), and x ∈ φ−1[{φ(a)}].

2

φ−1[{φ(a)}] ⊂ aH: Suppose that x ∈ φ−1[{φ(a)}]. We have φ(x) = φ(a), and thus φ(a)−1φ(x) = e′. By Theorem 13.12(2), φ(a)−1 = φ(a−1). It follows that φ(a−1x) = e′, and a−1x ∈ H. We see that x ∈ aH. The proof of φ−1[{φ(a)}] = Ha is similar.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.15

We will show that

1

aH ⊂ φ−1[{φ(a)}]: Let x = ag ∈ aH. Since φ is a homomorphism, we have φ(x) = φ(a)φ(h). By assumption that H = Ker(φ), it follows that φ(x) = φ(a), and x ∈ φ−1[{φ(a)}].

2

φ−1[{φ(a)}] ⊂ aH: Suppose that x ∈ φ−1[{φ(a)}]. We have φ(x) = φ(a), and thus φ(a)−1φ(x) = e′. By Theorem 13.12(2), φ(a)−1 = φ(a−1). It follows that φ(a−1x) = e′, and a−1x ∈ H. We see that x ∈ aH. The proof of φ−1[{φ(a)}] = Ha is similar.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.15

We will show that

1

aH ⊂ φ−1[{φ(a)}]: Let x = ag ∈ aH. Since φ is a homomorphism, we have φ(x) = φ(a)φ(h). By assumption that H = Ker(φ), it follows that φ(x) = φ(a), and x ∈ φ−1[{φ(a)}].

2

φ−1[{φ(a)}] ⊂ aH: Suppose that x ∈ φ−1[{φ(a)}]. We have φ(x) = φ(a), and thus φ(a)−1φ(x) = e′. By Theorem 13.12(2), φ(a)−1 = φ(a−1). It follows that φ(a−1x) = e′, and a−1x ∈ H. We see that x ∈ aH. The proof of φ−1[{φ(a)}] = Ha is similar.

Instructor: Yifan Yang Section 13 – Homomorphisms

Proof of Theorem 13.15

We will show that

1

aH ⊂ φ−1[{φ(a)}]: Let x = ag ∈ aH. Since φ is a homomorphism, we have φ(x) = φ(a)φ(h). By assumption that H = Ker(φ), it follows that φ(x) = φ(a), and x ∈ φ−1[{φ(a)}].

2

φ−1[{φ(a)}] ⊂ aH: Suppose that x ∈ φ−1[{φ(a)}]. We have φ(x) = φ(a), and thus φ(a)−1φ(x) = e′. By Theorem 13.12(2), φ(a)−1 = φ(a−1). It follows that φ(a−1x) = e′, and a−1x ∈ H. We see that x ∈ aH. The proof of φ−1[{φ(a)}] = Ha is similar.

Instructor: Yifan Yang Section 13 – Homomorphisms

Examples

1

Let n be a positive integer. Let φ : Z → Zn be defined by φ(a) = ¯ a, the residue class modulo n containing n. Then Ker(φ) = nZ. Also, for each ¯ b ∈ Zn, the set φ−1[{¯ b}] = {. . . , b − 2n, b − n, b, b + n, . . .} = b + nZ, which is indeed a coset.

2

Let U be the set of all complex number z of unit length, i.e., |z| = 1. Consider φ : R → U defined by φ(x) = eix. The kernel is Ker(φ) = {x : eix = 1} = {2nπ : n ∈ Z}, and the sets φ−1[{φ(a)}] are equal to {a + 2nπ : n ∈ Z} = a + Ker(φ).

Instructor: Yifan Yang Section 13 – Homomorphisms

Examples

1

Let n be a positive integer. Let φ : Z → Zn be defined by φ(a) = ¯ a, the residue class modulo n containing n. Then Ker(φ) = nZ. Also, for each ¯ b ∈ Zn, the set φ−1[{¯ b}] = {. . . , b − 2n, b − n, b, b + n, . . .} = b + nZ, which is indeed a coset.

2

Let U be the set of all complex number z of unit length, i.e., |z| = 1. Consider φ : R → U defined by φ(x) = eix. The kernel is Ker(φ) = {x : eix = 1} = {2nπ : n ∈ Z}, and the sets φ−1[{φ(a)}] are equal to {a + 2nπ : n ∈ Z} = a + Ker(φ).

Instructor: Yifan Yang Section 13 – Homomorphisms

Examples

1

Let n be a positive integer. Let φ : Z → Zn be defined by φ(a) = ¯ a, the residue class modulo n containing n. Then Ker(φ) = nZ. Also, for each ¯ b ∈ Zn, the set φ−1[{¯ b}] = {. . . , b − 2n, b − n, b, b + n, . . .} = b + nZ, which is indeed a coset.

2

Let U be the set of all complex number z of unit length, i.e., |z| = 1. Consider φ : R → U defined by φ(x) = eix. The kernel is Ker(φ) = {x : eix = 1} = {2nπ : n ∈ Z}, and the sets φ−1[{φ(a)}] are equal to {a + 2nπ : n ∈ Z} = a + Ker(φ).

Instructor: Yifan Yang Section 13 – Homomorphisms

Examples

1

Let n be a positive integer. Let φ : Z → Zn be defined by φ(a) = ¯ a, the residue class modulo n containing n. Then Ker(φ) = nZ. Also, for each ¯ b ∈ Zn, the set φ−1[{¯ b}] = {. . . , b − 2n, b − n, b, b + n, . . .} = b + nZ, which is indeed a coset.

2

Let U be the set of all complex number z of unit length, i.e., |z| = 1. Consider φ : R → U defined by φ(x) = eix. The kernel is Ker(φ) = {x : eix = 1} = {2nπ : n ∈ Z}, and the sets φ−1[{φ(a)}] are equal to {a + 2nπ : n ∈ Z} = a + Ker(φ).

Instructor: Yifan Yang Section 13 – Homomorphisms

Examples

1

Let n be a positive integer. Let φ : Z → Zn be defined by φ(a) = ¯ a, the residue class modulo n containing n. Then Ker(φ) = nZ. Also, for each ¯ b ∈ Zn, the set φ−1[{¯ b}] = {. . . , b − 2n, b − n, b, b + n, . . .} = b + nZ, which is indeed a coset.

2

Let U be the set of all complex number z of unit length, i.e., |z| = 1. Consider φ : R → U defined by φ(x) = eix. The kernel is Ker(φ) = {x : eix = 1} = {2nπ : n ∈ Z}, and the sets φ−1[{φ(a)}] are equal to {a + 2nπ : n ∈ Z} = a + Ker(φ).

Instructor: Yifan Yang Section 13 – Homomorphisms

Examples

1

Let n be a positive integer. Let φ : Z → Zn be defined by φ(a) = ¯ a, the residue class modulo n containing n. Then Ker(φ) = nZ. Also, for each ¯ b ∈ Zn, the set φ−1[{¯ b}] = {. . . , b − 2n, b − n, b, b + n, . . .} = b + nZ, which is indeed a coset.

2

Let U be the set of all complex number z of unit length, i.e., |z| = 1. Consider φ : R → U defined by φ(x) = eix. The kernel is Ker(φ) = {x : eix = 1} = {2nπ : n ∈ Z}, and the sets φ−1[{φ(a)}] are equal to {a + 2nπ : n ∈ Z} = a + Ker(φ).

Instructor: Yifan Yang Section 13 – Homomorphisms

Examples

1

Let n be a positive integer. Let φ : Z → Zn be defined by φ(a) = ¯ a, the residue class modulo n containing n. Then Ker(φ) = nZ. Also, for each ¯ b ∈ Zn, the set φ−1[{¯ b}] = {. . . , b − 2n, b − n, b, b + n, . . .} = b + nZ, which is indeed a coset.

2

Let U be the set of all complex number z of unit length, i.e., |z| = 1. Consider φ : R → U defined by φ(x) = eix. The kernel is Ker(φ) = {x : eix = 1} = {2nπ : n ∈ Z}, and the sets φ−1[{φ(a)}] are equal to {a + 2nπ : n ∈ Z} = a + Ker(φ).

Instructor: Yifan Yang Section 13 – Homomorphisms

Normal subgroups

Definition A subgroup H of a group G is normal if its left cosets and right cosets coincide, that is, if gH = Hg for all g ∈ G. (Alternatively, gHg−1 = H for all g ∈ G, or ghg−1 ∈ H for all h ∈ H and g ∈ G.) Corollary (13.20) If φ : G → G′ is a group homomorphism, then Ker(φ) is a normal subgroup.

Instructor: Yifan Yang Section 13 – Homomorphisms

Normal subgroups

Definition A subgroup H of a group G is normal if its left cosets and right cosets coincide, that is, if gH = Hg for all g ∈ G. (Alternatively, gHg−1 = H for all g ∈ G, or ghg−1 ∈ H for all h ∈ H and g ∈ G.) Corollary (13.20) If φ : G → G′ is a group homomorphism, then Ker(φ) is a normal subgroup.

Instructor: Yifan Yang Section 13 – Homomorphisms

Example

Let G = S3. There are 6 subgroups, namely, {e}, {e, (1, 2)}, {e, (1, 3)}, {e, (2, 3)}, {e, (1, 2, 3), (1, 3, 2)}, and G. Among them, it is easy to see that {e} and G are normal. The subgroup {e, (1, 2, 3), (1, 3, 2)} is normal since it is of index 2 in

• S3. The three subgroups of order 2 are not normal. For

example, we have (1, 3)(1, 2)(1, 3) = (2, 3) ∈ {e(1, 2)}. Thus, {e, (1, 2)} is not a normal subgroup.

Instructor: Yifan Yang Section 13 – Homomorphisms

Example

Let G = S3. There are 6 subgroups, namely, {e}, {e, (1, 2)}, {e, (1, 3)}, {e, (2, 3)}, {e, (1, 2, 3), (1, 3, 2)}, and G. Among them, it is easy to see that {e} and G are normal. The subgroup {e, (1, 2, 3), (1, 3, 2)} is normal since it is of index 2 in

• S3. The three subgroups of order 2 are not normal. For

example, we have (1, 3)(1, 2)(1, 3) = (2, 3) ∈ {e(1, 2)}. Thus, {e, (1, 2)} is not a normal subgroup.

Instructor: Yifan Yang Section 13 – Homomorphisms

Example

Let G = S3. There are 6 subgroups, namely, {e}, {e, (1, 2)}, {e, (1, 3)}, {e, (2, 3)}, {e, (1, 2, 3), (1, 3, 2)}, and G. Among them, it is easy to see that {e} and G are normal. The subgroup {e, (1, 2, 3), (1, 3, 2)} is normal since it is of index 2 in

• S3. The three subgroups of order 2 are not normal. For

example, we have (1, 3)(1, 2)(1, 3) = (2, 3) ∈ {e(1, 2)}. Thus, {e, (1, 2)} is not a normal subgroup.

Instructor: Yifan Yang Section 13 – Homomorphisms

Example

Let G = S3. There are 6 subgroups, namely, {e}, {e, (1, 2)}, {e, (1, 3)}, {e, (2, 3)}, {e, (1, 2, 3), (1, 3, 2)}, and G. Among them, it is easy to see that {e} and G are normal. The subgroup {e, (1, 2, 3), (1, 3, 2)} is normal since it is of index 2 in

• S3. The three subgroups of order 2 are not normal. For

example, we have (1, 3)(1, 2)(1, 3) = (2, 3) ∈ {e(1, 2)}. Thus, {e, (1, 2)} is not a normal subgroup.

Instructor: Yifan Yang Section 13 – Homomorphisms

Example

Let G = S3. There are 6 subgroups, namely, {e}, {e, (1, 2)}, {e, (1, 3)}, {e, (2, 3)}, {e, (1, 2, 3), (1, 3, 2)}, and G. Among them, it is easy to see that {e} and G are normal. The subgroup {e, (1, 2, 3), (1, 3, 2)} is normal since it is of index 2 in

• S3. The three subgroups of order 2 are not normal. For

example, we have (1, 3)(1, 2)(1, 3) = (2, 3) ∈ {e(1, 2)}. Thus, {e, (1, 2)} is not a normal subgroup.

Instructor: Yifan Yang Section 13 – Homomorphisms

Example

Let G = S3. There are 6 subgroups, namely, {e}, {e, (1, 2)}, {e, (1, 3)}, {e, (2, 3)}, {e, (1, 2, 3), (1, 3, 2)}, and G. Among them, it is easy to see that {e} and G are normal. The subgroup {e, (1, 2, 3), (1, 3, 2)} is normal since it is of index 2 in

• S3. The three subgroups of order 2 are not normal. For

example, we have (1, 3)(1, 2)(1, 3) = (2, 3) ∈ {e(1, 2)}. Thus, {e, (1, 2)} is not a normal subgroup.

Instructor: Yifan Yang Section 13 – Homomorphisms

Homework

Do Problems 18, 24, 27, 38, 40, 44, 47, 50, 51, 53 of Section 13.

Instructor: Yifan Yang Section 13 – Homomorphisms