Transcript Math 3121 Lecture 11

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
– 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
1) Show h is a homomorphism
2) 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 welldefined 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).