Math 3121 Abstract Algebra I

1 / 25

# Math 3121 Abstract Algebra I - PowerPoint PPT Presentation

Math 3121 Abstract Algebra I. Lecture 11 Finish Section 13 Section 14. Next Midterm. Midterm 2 is Nov 13. Covers sections: 7-14 (not 12) Review on Thursday. Section 13. Homomorphisms Definition of homomorphism (recall) Examples Properties Kernel and Image Cosets and inverse images

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about 'Math 3121 Abstract Algebra I' - jeanette-stewart

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

### Math 3121Abstract Algebra I

Lecture 11

Finish Section 13

Section 14

Next Midterm
• Midterm 2 is Nov 13.

Covers sections: 7-14 (not 12)

Review on Thursday

Section 13
• Homomorphisms
• Definition of homomorphism (recall)
• Examples
• Properties
• Kernel and Image
• Cosets and inverse images
• Monomorphisms
• Normal Subgroups
Images and Inverse Images
• Let X and Y be sets, and let f: X  Y
• Define f[A] and f-1[B] for subsets A of X and B of Y:

f[A] = { b in Y | b = f(a), for some a in A}

f-1[B] = { a in X | f(a) is in B}

Properties of Homomorphisms

Theorem: Let h be a homomorphism from a group G into a group G’. Then

1) If e is the identity in G, then h(e) is the identity in G’.

2) If a is in G, then h(a-1) = (h(a))-1

3) If H is a subgroup of G, then f[H] is a subgroup of G’.

4) If K’ is a subgroup of G’, then h-1[K’] is a subgroup of G.

Proof: Straightforward – in class and in the book

Kernel

Definition: Let h be a homomorphism from a group G into a group G’. The kernel of h is the inverse image of the trivial subgroup of G’:

Ker(h) = { x in G | h(x) = e’}

Examples of Kernels
• Modulo n: Z  Zn, x ↦ x + nZ
• Parity: Sn Z2
• Multiply by m: Zn Zn, x ↦ mx

n = 6, m = 1, 2, 3

Cosets of the kernel are inverse images of elements

Theorem: Let h be a homomorphism from a group G into a group G’. Let K be the kernel of h. Then

a K = {x in G | h(x) = h(a)} = h -1[{h(a)}]

and also

K a = {x in G | h(x) = h(a)} = h -1[{h(a)}]

Proof: h -1[{h(a)}] = {x in G | h(x) = h(a)} directly from the definition of inverse image.

Now we show that: a K = {x in G | h(x) = h(a)} :

x in a K ⇔ x = a k, for some k in K

⇔ h(x) = h(a k) = h(a) h(k) = h(a) , for some k in K

⇔ h(x) = h(a)

Thus, a K = {x in G | h(x) = h(a)}.

Likewise, K a = {x in G | h(x) = h(a)}.

Equivalence Relation
• Suppose: h: X  Y is any map of sets. Then h defines an equivalence relation ~h on X by:

x ~h y ⇔ h(x) = h(y)

The previous theorem says that when h is a homomorphism of groups then the cosets (left or right) of the kernel of h are the equivalence classes of this equivalence relation.

Monomorphisms and Epimorphisms
• Recall:

A homomorphism h: G  G’ is called a monomorphism if it is 1-1.

A homomorphism h: G  G’ is called an epimorphism if it is onto.

Monomorphism Test

Theorem: A homomorphism h is 1-1 if and only if Ker(h) = {e}.

Proof: Let h: G  G’ be a homomorphism.

Then h(x) = h(a) ⇔ x  a Ker(h).

If Ker(h) = {e}, then a Ker(h) = {a} and

h(x) = h(a) ⇔ x = a.

If Ker(h) is larger, then there is an k different from e in Ker(h), then ak ≠ a and h(ak) = h(a). So h is not 1-1.

Isomorphism Test

To show h : G  G’ is an isomorphism

• Show h is a homomorphism
• Show Ker(h) = {e}
• Show h is onto.
Normal Subgroups

Definition: A subgroup H of a group G is said to be normal if a H = H a, for all a in G.

Kernel is Normal
• Theorem: Let h: G  G’ be a group homomorphism, then Ker(h) is normal:
• Proof: By previous theorem, a Ker(h) = Ker(h) a, for all a in G. By the previous definition, Ker(h) is normal.
HW
• Not to hand in:

Page 133: 1, 3, 5, 7, 17, 19, 27, 29, 33, 35

• Hand in (due Thurs Nov 18)

Page 133: 44, 45, 49

Section 15
• Section 15: Factor Groups
• Multiplication of cosets
• Definition: Factor Group
• Theorem: The image of a group homomorphism is isomorphic to the group modulo its kernel.
• Properties of normal subgroups
• Theorem: For a subgroup of a group, left coset multiplication is well-defined if and only if the subgroup is normal.
• Theorem: The canonical map is a homomorphism.
Multiplication of Cosets
• Let H be a subgroup of a group G. When is

(a H) (b H) = a b H?

• This is true for abelian groups, but not always when G is nonabelian.
• Consider S3: Let H = {ρ0, μ1}. The left cosets are

{ρ0, μ1}, {ρ1, μ3}, {ρ2, μ2}.

If we multiply the first two together, then

{ρ0, μ1}, {ρ1, μ3} = {ρ0 ρ1, ρ0 μ3, μ 1 ρ1, μ 1 μ3}

= {ρ1, μ3, μ2, ρ 2}

This has four distinct elements, not two!

Sometimes it does work.
• Consider S3: Let H = {ρ0, ρ1 , ρ2}. The left cosets are

{ρ0, ρ1 , ρ2}, {μ1, μ2 , μ3}

If we multiply the first two together, then

{ρ0, ρ1 , ρ2} {μ1, μ2, μ3} = {ρ0 μ1, ρ0 μ2, ρ0 μ3, ρ1 μ1, ρ1 μ2, ρ1 μ3, ρ2 μ1, ρ2 μ2, ρ2 μ3} = {μ1, μ2, μ3, μ3, μ1, μ2, μ2, μ3, μ1} = {μ1, μ2, μ3}

This is one of the cosets. Likewise,

{ρ0, ρ1 , ρ2} {ρ0, ρ1 , ρ2} = {ρ0, ρ1 , ρ2}

{μ1, μ2 , μ3}{ρ0, ρ1 , ρ2} = {μ1, μ2 , μ3}

{μ1, μ2 , μ3 }{μ1, μ2 , μ3} = {ρ0, ρ1 , ρ2}

Note that the cosets of {ρ0, ρ1 , ρ2} with this binary operation form a group isomorphic to ℤ2.

Canonical Homomorphism
• Note that there is a natural map from S3 from {{ρ0, ρ1 , ρ2}, {μ1, μ2 , μ3}} that takes any element to the coset that contains it. This gives a homomorphism called the cannonical homomorphism.
Theorem

Theorem: Let h: G  G’ be a group homomorphism with kernel K. Then the cosets of K form a group with binary operation given by (a K)(b K) = (a b) K. This group is called the factor group G/K. Additionally, the map μ that takes any element x of G to is coset xH is a homomorphism. This is called the canonical homomorphism.

Proof: Let (a K)(b K) = { a k1 b k2 | k1,k2 in K}. We show this is equal to (a b) K.

Clearly, a b K  (a K)(b K) (just consider what happens when k1 = e)

To prove the reverse apply h:

h[(a K)(b K)] = { h(a k1 b k2 )| k1,k2 in K}

But h(a k1 b k2)= h(a) h( k1) h(b) h(k2 )

= h(a) e’ h(b) e’= h(a) h(b) = h(a b)

Then h[(a K)(b K)] = {h(a b)| k1,k2 in K}= {h(a b)}

Thus (a K)(b K)  h-1[{h(a b)}] = a b K

So (a K)(b K) = a b K.

Associativity of Coset Multiplication

Proof continued:

This operation is associative:

((a K) (b K)) (c K) = (a b K) (c K) = a b c K

(a K)((b K) (c K)) = (a K) (b c K) = a b c K

Thus ((a K) (b K)) (c K) = (a K)((b K) (c K))

Identity and Inverse

Proof continued:

The coset e K = K is an identity:

(e K) (a K) = (e a) K = a K

For each coset a K, the coset a-1 K is an inverse:

(a-1 K) (a K) = (a-1 a) K = e K

(a K) (a-1 K) = (a a-1) K = e K

Canonical Map

Proof continued:

Let μ(a) = a K. Then

μ(a b) = a b K

and

μ(a) μ (b) = (a K)(b K) = a b K

Thus

μ(a b) = μ(a) μ (b)

Terminology
• Let H be a subroup of a group G. When the cosets satisfy the rule

(a H) (b H) = ( a b) H

We call the set of cosets the factor group and denote it by G/H. This is read G modulo H.

Note that for finite groups

order(G/H) = order(G)/order(H)

Coset Multiplication is equivalent to Normality

Theorem: Let H be a subgroup of a group G. Then H is normal if and only if

(a H )( b H) = (a b) H, for all a, b in G

Proof: Suppose (a H )( b H) = (a b) H, for all a, b in G.

We show that a H = H a, for all a in H.

We do this by showing: a H  H a and H a  a H, for all a in G.

a H  H a: First observe that a H a-1  (a H )( a-1 H) = (a a-1) H = H.

Let x be in a H. Then x = a h, for some h in H. Then x a-1 = a h a-1, which is in = a H a-1 , thus in H. Thus x a-1 is in H. Thus x is in H a.

H a  a H: H a  H a H = (e H )( a H) = (e a) H = a H.

This establishes normality.

For the converse, assume H is normal.

(a H )( b H)  (a b) H: For a, b in G, x in (a H )( b H) implies that x = a h1 b h2, for some h1 and h2 in H. But h1 b is in H b, thus in b H. Thus h1 b = b h3 for some h3 in H. Thus x = a b h3 h2 is in a b H.

(a b) H  (a H )( b H): x in (a b) H ⇒that x = a e b h, for some h in H.

Thus x is in (a H) (b H).