1. Math 3121 Abstract Algebra I
Lecture 11
Finish Section 13
Section 14

2. Next Midterm Midterm 2 is Nov 13. Covers sections: 7-14 (not 12)
Review on Thursday

3. Section 13 Homomorphisms Definition of homomorphism (recall) Examples
Properties
Kernel and Image
Cosets and inverse images
Monomorphisms
Normal Subgroups

4. 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}

5. 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

6. 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’}

7. 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

8. 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)}.

9. 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.

10. 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.

11. 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.

12. Isomorphism Test To show h : G  G’ is an isomorphism
Show h is a homomorphism
Show Ker(h) = {e}
Show h is onto.

13. 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.

14. 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.

15. HW Not to hand in: Hand in (due Thurs Nov 18)
Page 133: 1, 3, 5, 7, 17, 19, 27, 29, 33, 35
Hand in (due Thurs Nov 18)
Page 133: 44, 45, 49

16. 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.

17. 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!

18. 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.

19. 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.

20. 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.

21. 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))

22. 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

23. 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)

24. 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)

25. 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).