Math 3121Abstract Algebra I

Lecture 11Finish 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 ↦ mxn = 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 isomorphism1) Show h is a homomorphism2) Show Ker(h) = {e}3) 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.

TheoremTheorem: 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 KSo (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 KFor 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 KThus

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) HWe 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).