Garis-garis Besar Perkuliahan
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Sets and Relations
Definitions and Examples of Groups
Subgroups
Lagrange’s Theorem
Mid-test 1
Homomorphisms and Normal Subgroups 1
Homomorphisms and Normal Subgroups 2
Factor Groups 1
Factor Groups 2
Mid-test 2
Cauchy’s Theorem 1
Cauchy’s Theorem 2
The Symmetric Group 1
The Symmetric Group 2
22/5/10
Final-exam
Definitions and
Examples of Groups
Section 1
The Set of 1-1 mappings of
S onto itself, A(S)
Lemma 1. A(S) satisfies the following:
a) f, g  A(S) implies that f ∘ g A(S).
b) f, g, h  A(S) implies that (f ∘ g)∘h = f ∘ (g ∘ h)
c) There exists an element –the identity mapping,
i –such that i∘f = f∘i = f for every f  A(S).
d) Given f  A(S), there exists a g  A(S) (g = f -1)
such that g∘f = f∘g = i.
The Set of 1-1 mappings of
S onto itself, A(S)
Lemma 2. If S has n elements, then A(S) has n! elements.
Let f  A(S), where S  x1 , x2 ,
, xn .
f has n choices of a place to send x1.
But now f is not free to send x2 anywhere, since f is 1-1.
Thus f can send x2 anywhere except to f(x1) in n-1
choices.
Proceed this way, we see that f can send xi into n-(i-1)
different places.
The Set of 1-1 mappings of
S onto itself, A(S)
Consider A(S) = S3, where S  x1 , x2 , x3.
List of all the elements of S3:
i : x1  x1 , x 2  x 2 , x 3  x 3
f : x1  x 2 , x 2  x 3 , x 3  x1
g : x1  x 2 , x 2  x1 , x 3  x 3
g f : x1  x1 , x 2  x 3 , x 3  x 2
f g : x1  x 3 , x 2  x 2 , x 3  x1
f f : x1  x 3 , x 2  x1 , x 3  x 2
Problems
1. If f  Sn, show that there is some positive
integer k, depending on f, such that f k = i.
2. If S has three or more elements, show that
we can find f, g  A(S) such that fg  gf.
3. For f  A(S), let C(f) = {g  A(S) | fg = gf}.
Prove that:
a) g, h  C(f) implies that gh  C(f).
b) g  C(f) implies that g -1  C(f).
c) C(f) is not empty.
Definitions and Examples of
Groups
Definition. A nonempty set G is said to be a group if in G
there is defined an operation  such that:
a) a, b  G implies that a  b G. (closure)
b) Given a, b, c  G, then (a  b)  c = a  (b  c)
(associativity)
c) There exists a special element e  G such that e  a =
a  e = a for every a  G.
(e is called the identity or unit element of G)
d) For every a  G there exists an element b  G such
that a  b = b  a = e.
(b is written as a-1 and is called the inverse of a in G.
Examples
Group
Not a Group
(R, +)
(R, •)
(Z, +)
(N, +)
(Q, +)
(Q, •)
(R\{0}, •)
(R\{0}, +)
({0}, -)
(Z, -)
(R+, •)
(R-, •)
(all polynomials, +)
(Q \{0}, •)
(all polynomials, •)
(Q \{0}, +)
Finite Groups
A group G is said to be a finite group if it has a
finite number of elements. The number of
elements in G is called the order of G and
is denoted by |G|.
Let a be an element of a group G. If there is a
smallest positive integer n such that an = e
then n is the order of a. Denoted |a|.
Abelian Groups
A group G is said to be abelian if a  b =
b  a for all a, b  G.
The word abelian derives from the name
of the great Norwegian mathematician
Niels Henrik Abel (1802-1829), one
of the greatest scientists Norway has
ever produced.
Examples (Nonabelian)
1) Let G be the set of all mappings Ta,b: R  R
defined by Ta,b(x) = ax + b for any real
number x, where a, b are real numbers and
a  0. Under the composition, G forms a
nonabelian group. Verify the following
formula
Ta,bTc,d = Tac,ad+b
2. Let G = {f  A(S) | f(s)  s for only a finite
number of s  S}, where S is supposed to
be an infinite set. Under the product in
A(S), G is a nonabelian group.
The Cancellation Law
Lemma 3. Let (G, •) be a group.
If a, b, c  G, then a•b = a•c  b = c and
b•a = c•a  b = c .
Proof:
Let e be the identity of G. Then
a•b = a•c
 a-1•(a•b) = a-1•(a•c) [Existence of inverses]
 (a-1•a)•b = (a-1•a) •c [Associative]
 e•b = e•c [Def. of inverse]
 b = c [Def. of identity]
 a•b = a•c  b = c. The proof of the second part is
virtually identical.
Simple Remarks
Lemma 4. If G is a group, then:
a) Its identity element is unique.
b) Every a  G has a unique inverse a-1  G.
c) If a  G, (a-1)-1 = a.
d) For a, b  G, (ab)-1 = b-1 a-1.
Proof of Lemma 2 (b)
Let G be a group and aG. Let e be the
identity of G. Suppose b, c G are both
inverses of a. Then ab = e and ac = e. So
ab = ac. Then by cancellation law, b = c.
Thus the inverse of a is unique.
Existence and Uniqueness of a
solution to ax = b
Theorem 1. Let a, b G. The equation ax = b
has a unique solution in G.
Proof: Let e be the identity of G. Solving the equation:
ax = b
 a-1(ax) = a-1b [Existence of inverses]
 (a-1a)x = a-1b [Associative Law]
 ex = a-1b [Def. of inverse]
 x = a-1b [Def. of identity]
 IF there is a solution, it must be a-1b. So we have shown
uniqueness.
Note that a-1b G due to existence of inverses and closure.
Now, note that a(a-1b) = (aa-1)b = eb = b. This proves that a
solution exists. 
Problems
For each of the rules 1-5, either prove that the rule
is true in any group, or give a counterexample:
1. If x2 = e then x = e.
2. If x2 = a2 then x = a.
3. (ab)2 = a2b2
4. If x2 = x then x = e.
5. For every x  G there is y  G such that x = y2.
Problems
6. Assume that a, b commute in G, i.e., ab = ba.
Prove: a-1 commutes with b-1; a commutes
with ab; a2 commutes with b2.
7. Find all multiplication tables of groups of order
4 with headlines and sidelines labeled 1, a, b,
c. Hint: You can assume that 1 is the neutral
element and a has order 2. You will then find
only 2 possible multiplication tables.
Problems
8. If G is a group in which a2 = e for all a  G,
show that G is abelian.
9. Show that a group of order 5 must be
abelian.
10. If G is a finite group, prove that, given a 
G, there is an integer n > 0, depending on
a, such that an = e.
11. In Problem 10, show that there is an integer
m > 0 such that am = e for all a  G.
Question?
If you are confused like this kitty is,
please ask questions =(^ y ^)=