GROUP AND SUBGROUP SET - ABSTRACT ALGEBRA ( TOPIC 5)

Binary Operation
A Binary Operation * on a set S is a
function mapping S X S into S. For each
(a,b) ϵ S X S, in short we denote the element
* ((a.b)) of S by a*b.

Example 1:
+ is a binary operation on a set S of Natural
numbers (N)
Let S = {1, 2}
S x S = {1, 2} x {1, 2}
= {(1, 1), (1, 2), (2, 1), (2,2)}

Example 1:
numbers (N)
To check if + is a binary operation to set S?
We’re going to operate it to every ordered pairs.
S x S = {1, 2} x {1, 2}
= {(1, 1), (1, 2), (2, 1), (2,2)}
{(1 + 1 = 2), (1 + 2 = 3), (2 + 1 = 3), (2+2 = 4)}

Example 1:
numbers (N)
> Since 2, 3, and 4 are natural numbers, our
⸫ + is
a Binary operation.
 Same process in -, x, /.
As long as it these operation satisfies the condition of
binary operation it will be a binary operation.

Binary Operation
Another thing about Binary Operation is
a way of combining two quantities/values
using any operation to create new one.
Common Binary operation:
+, -, x, /.

Example 2:
For every real numbers a and b:
a & b = 5a – 2b + 1.
Supposed the value for a = 3; b = 5
3 & 5 = 5(3) – 2(5) + 1
= 15 – 10 + 1
3 & 5= 6

Example 3 :
For every real numbers a and b:
a # b = a2 + b.
Supposed the value for
a = 1/2; b = (-2)
a # b = (½)2 + (-2)
= ¼ - 2
= -7/4

Closure or Induced
Operation/Method

Closure or Induced Property:
Given a ϵ S and b ϵ S, a binary
operation is closed ↔ a * b ϵ S.
If both a and b are from set S, then the result
of their binary operation should also be a
member of set S.

Example 1:
For every real numbers a and b, a binary
operation ~ is defined a ~ b = 2a + b. Is close
under real numbers?
Supposed the value of a = 4; b = (-3).
a ~ b = 2a + b
4 ~ (-3) = 2(4) + (-3)
= 8 – 3
= 5 ϵ R
⸫ a ~ b is closed under R.

Example 2:
Let V = {1, 2, 3, 4}. Is the operation Ø defined by a
Ø b = ab – 1. Is closed in W?
3 Ø 4 = 3(4) – 1
= 7 – 1
= 6 not belong to V
⸫ a Ø b is not closed under set V.

Example 2:
Let V = {1, 2, 3, 4}. Is the operation Ø defined by a
Ø b = ab – 1. Is closed in W?
1 Ø 3 = 1(3) – 1
= 3 – 1
= 2 ϵ V
⸫ a Ø b is closed under set V.

Closure or Induced Property:
REMEMBER
If a, b ϵ S, then a*b ϵ S.

Let’s try!
+ is a binary operation on the set of Real
Numbers (R*). {n|n > 0, is a set of R}. Is under
closure method?
Note: We all know that n > 0 means a
nonzero real numbers.
Proving this, knowing that 2 ϵ R* as well as -
2 ϵ R*, using the condition of Closure
method that a,b ϵ R* we can draw or operate
this situation “ 2 + (-2) = 0 is not

Let’s try!
an element of R* (non zero real numbers)., we
⸫
can say that closure method for this binary operation is
NOT SATISFIED.

Let’s try!
+ is a binary operation for 4Z. Is 4Z closed under
addition? Prove?
Proof:
Z = {. . .,-4, -3, -2, -1, 0, 1, 2, 3, 4, …}
4Z = {.., -16, -12, -8, -4, 0, 4, 8, 12, 16, …}
So, now we going to use the condition of Closure
Operation/Method: a, b ϵ 4Z, 16 + 12 = 28 ϵ 4Z; -16 + 12 = -4 ϵ 4Z,
we can conclude that closure operation/method is satisfied.
⸫

Associative Property/Method:
A binary operation * is associative if and
only if (a * b) * c = a * (b * c)
∀a, b, c ϵ S

Example 1:
For every real number a and b is the operation Ω
(omega) defined by a Ω b = (2a)(3b) associative?
(a Ω b) Ω c = a Ω (b Ω c)
Take note that c is also a real number

Example 1:
(a Ω b) Ω c = a Ω (b Ω c)
(2a) (3b) Ω c = a Ω (2b)(3c)
(6ab) Ω c = a Ω (6bc)
[2 (6ab)] (3c) = (2)(a) [3(6bc)]
(12ab) (3c) = 2a (18bc)
36abc = 36abc
⸫ a Ω b is associative.
This becomes your first
term.
This becom
es your 2nd
term
.
These things are based on the definition of
binary operations a,b ϵ S

Example 2:
For every real number a and b is the operation ♠
(spade) defined by a ♠ b = a b
ꜛ + b associative?
(a ♠ b) ♠ c = a ♠ (b ♠ c)
(a b
ꜛ + b) ♠ c = a ♠ (b c
ꜛ ♠ c)
(a b
ꜛ + b) c
ꜛ + c = a b vc
ꜛ + b c + c
ꜛ
⸫,♠ is not an associative.

Let’s try!
+ is a binary operation for 4Z. Is 4Z an associative
under addition? Prove?
Proof:
Z = {. . .,-4, -3, -2, -1, 0, 1, 2, 3, 4, …}
4Z = {.., -16, -12, -8, -4, 0, 4, 8, 12, 16, …}
So, now we going to use the condition of Associative
Operation/Method: a, b, c ϵ 4Z, a + b + c ϵ 4Z, 16 + (4 +8) = (16 +
4) + 8 = we can conclude that associative operation/method is
⸫
satisfied.

Associative Property:
REMEMBER
∀ a, b, c ϵ S, we have
a*(b*c) = (a*b)*c.

Identity Element Property:
A binary operation * is if there exists just
one element e such that: e*a = a*e = a,
where a,e ϵ R (S)

Example :
For common identity elements
For addition + e = 0
3 + 0 = 0 + 3 = 3
For multiplication x e = 1
3 x 1 = 1 x 3 = 3

Example :
Identity elements in tables
* a b c d
a c a d b
b a b c d
c d c d a
d b d a c
Identity = b
a*b=b*a = a

Inverse Property:
An element a -1 is called an inverse
⌃
under the binary operation * if a* a -1 = a
⌃
-1 * a = e, where e is the identity element
⌃
under operation *.

Example :
For inverse elements
For addition + e = 0
3 + ? = ? + 3 = 0 ? = Inverse ? = -3
⸫
element
For multiplication x e = 1
3 x ? = ? x 3 = 1
⸫ ? = 1/3

Example :
Find the inverse of each element in the table
* a b c d
a c a d b
b a b c d
c d c d a
d b d a c
Identity = b
a*b=b*a = a

Example :
* a b c d
a c a d b
b a b c d
c d c d a
d b d a c
Identity = b
a*b=b*a = a
So, a* a -1 = a -1 = b
⌃ ⌃
⸫ a -1 = d
⌃
b -1 = b inverse itself, c -1 = c, d -1 = a
⌃ ⌃ ⌃

Definition of a Group
A group consist of a nonempty set G
together with a binary operation *, such that
the following properties hold:

 CLOSURE
 ASSOCIATIVITY
 IDENTITY
 INVERSE
If a, b ϵ G, then a * b ϵ
G.
INSHORT, the sum, or
product, or result of a
and b must be an
element of G.
For all a,b,c ϵ G, we
have (a*b)*c = a*
(b*c), where a, b, and
c ϵ G.
There exist an element e
ϵ G such that, for any a ϵ
G, a*e = e*a = a.
For any a ϵ G, there
exist an element a-1 ϵ G
such that a* a-1 = a-1 * a =
e

How many elements are there
in ℤ?
INFINITE
(ℤ, +) is an example of an infinite group.

How many elements are there
in ℤ3?
3
Hence, (ℤ3, +3) is an example of finite group.

ORDER OF A GROUP
The order of a group G,
denoted by |G| is the number
of elements present in the
group or it’s cardinality.
The order of |ℤ| is infinite
thus the group ( , +)
ℤ has
an infinite order.
|ℤ3| = 3
(ℤ3, +3) could also generate 3
elements so its order is 3. this is an
example of finite order.

EXERCISE
Determine the order of the group:
1. ( ,
ℝ +)
2. (ℤ5
#
, ∙)
3. ( ,
ℚ ∙)
QUESTIONS?

REVIEW Is (ℤ4, +4) a group?
+4 0 1 2 3
0 0 1 2 3
1 1 2 3 0
2 2 3 0 1
3 3 0 1 2
Remember: ℤ4 = {0, 1, 2, 3}
i. Closure
From the Cayley table, we could see that ∀
a, b ∈ℤ4, a +4 b ∈ ℤ4.
Thus, closure holds.
ii. Associativity
∀ a, b, c ∈ ℤ4, it is known that (a
+4 b) +4 c = a +4 (b +4 c). Thus,
associativity holds.

REVIEW Is (ℤ4, +4) a group?
+4 0 1 2 3
0 0 1 2 3
1 1 2 3 0
2 2 3 0 1
3 3 0 1 2
Remember: ℤ4 = {0, 1, 2, 3}
iii. Identity
We could see that 0 ∈ ℤ4 and for any
a ∈ ℤ4, we could see that a +4 0 = 0 +4 a
=a. hence, identity property is satisfied.
iv. Inverse
We could see that each element in ℤ4
has an inverse. Thus, inverse property is
satisfied.

REVIEW
+4 0 1 2 3
0 0 1 2 3
1 1 2 3 0
2 2 3 0 1
3 3 0 1 2
Since all 4 properties
of a group are satisfied,
hence (ℤ4, +4) is a
group.

EXERCISE
Determine the following set if it’s a group or not:
1. ( ,
ℝ +)
2. (ℤ7
#
, +7 )
3. (ℝ+
, ∙)

KLEIN V
- read as “KLEIN 4”
- It was named Vierergruppe
(meaning four-group) by Felix
Klein in 1884.
- It is also called the Klein
group and is often symbolized
by letter V.

There are 4 elements
|V | = 4
The 4 elements are
{e, a, b, c}
We define the
operation * such
that;
e * a = a
a * b = c
a * a = e
REMEMBER!
Is V, * is a group?

* e a b c
e e a b c
a a e c b
b b c e a
c c b a e
Let V = {e, a, b, c} with
operation * defined earlier.
Is (V, *) a group?
KLEIN V
i. Closure
Let x, y ∈ V. We
could see that x * y
∈
V. Hence,

* e a b c
e e a b c
a a e c b
b b c e a
c c b a e
Is (V, *) a group?
KLEIN V
ii. Associativity
Let x, y, z ∈ V. We
could see that
(x * y) * z = x * (y * z).
Hence, associativity holds.

* e a b c
e e a b c
a a e c b
b b c e a
c c b a e
Is (V, *) a group?
KLEIN V
iii. Identity
We could see that e is
the identity element of V
and e ∈ V. Thus, identity
property is satisfied.

* e a b c
e e a b c
a a e c b
b b c e a
c c b a e
operation * defined earlier. Is (V,
*) a group?
iv. inverse
KLEIN V
e-1 = e
a-1 =
a
b-1 = b
c-1 = c
We could see that every
element of V has an
inverse that is also in V.
Hence, inverse property
is satisfied.

* e a b c
e e a b c
a a e c b
b b c e a
c c b a e
KLEIN V
Since all 4 properties of
a group are satisfied, hence
(V, *) with V = {e, a, b, c} and
operation * as defined is a
group.
QUESTIONS?

(V, *)
* e a b c
e e a b c
a a e c b
b b c e a
c c b a e
+4 0 1 2 3
0 0 1 2 3
1 1 2 3 0
2 2 3 0 1
3 3 0 1 2
(ℤ4, +4)

REVIEW
Consider the matrix A
=
1
4
5
3
. Compute for det
(A).
det (A) = 1(5) – 4(3) = -7
Does the matrix have an inverse?
YES
How do we know?
det (A) ≠ 0
A =
1
3
4
∈
GL2(ℝ)

GENERAL LINEAR GROUP
The general linear group
GLn(ℝ) is the group of invertible
n × n matrices with entries in ℝ
under matrix multiplication.
For our class our focus will
be

Let GL2(ℝ) = {𝑥 | 𝑥 ∈𝑎
𝑐
𝑑
𝑏 , where a, b, c, d ∈
ℝ
and ad – bc ≠ 0}. Prove that (GL2(ℝ), ∙) is a
group.
i. Closure
Let 𝑥
∈
𝑎
𝑏
𝑐
𝑑
and 𝑦
∈
𝑒
𝑓
𝑔 h
wher
e
𝑎, 𝑏, 𝑐, … , 𝑕 ∈ ℝ. Multiplying
them yields
𝑎
𝑏
𝑐
𝑑
𝑒
𝑓
𝑔 h
=
𝑎𝑒 + 𝑏𝑔 𝑎𝑓
+ 𝑏h
𝑐𝑒 + 𝑑𝑔 𝑐𝑓 +
𝑑h
2
∈ GL (ℝ). Since all entries
of
the resulting matrix are real numbers, then
closure holds.

Let GL2(ℝ) = {𝑥 | 𝑥 ∈
𝑎
𝑐
𝑑
ℝ
group.
ii. Associativity
Please see Kolman (Linear Algebra).
Hence, associativity holds true.

Let GL2(ℝ) = {𝑥 | 𝑥
∈
�
�
𝑐
𝑑
ℝ
group.
iii. Existence of an Identity Element
= =
𝑎 𝑏 1
0
𝑐 𝑑 0
1
1 0 𝑎
𝑏
0 1 𝑐
=
𝑎 1 +
𝑏(0)
𝑐 1 +
𝑑(0) 1
𝑎
+
0(𝑐)
𝑎 0 +
𝑏(1)
𝑐 0
+ 𝑑(1) 1
𝑏
+
0(𝑑)
=
𝑎
𝑏
𝑐
𝑑
𝑎
𝑏
𝑐
We could see that e
=
1
0
0
∈ GL2(ℝ).
Hence,
identity property is
satisfied.

Let GL2(ℝ) = {𝑥 | 𝑥
∈
�
�
𝑐
𝑑
𝑏 , where a, b, c, d ∈ ℝ
and
ad – bc ≠ 0}. Prove that (GL2(ℝ), ∙) is a
group.
iv. Inverse Property
For every matrix A
=
𝑎
𝑐
𝑑
𝑏
, ∃ A-1 considering ad –
bc
≠ 0 and hence A A-1 = A-1A = e
= property is satisfied.
1
0
0
1
. Hence,
inverse
Since all 4 properties of a group are satisfied,
hence (GL2(ℝ), ∙) is a group.

GENERAL LINEAR GROUP
What is the order of
GL2(ℝ)?
GL2(ℝ) is infinite
QUESTIONS?

CHECK THIS OUT
It is known that (ℤ, +)
is a group. Let us pick any
2 elements, say 4 and 7
What is 4 + 7?
11
How about 7 + 4?
11
What have you
noticed?
4 + 7 = 7 + 4
Such a group is an
example of an
ABELIAN GROUP.

ABELIAN GROUP
Let (G, *) be a group. If a, b ∈ G such
that a * b = b * a, then (G, *) is an ABELIAN
GROUP.
It is also called COMMUTATIVE
GROUP.
NON-ABELIAN GROUP
Let (G, *) be a group. If a, b ∈ G such
that a * b ≠ b * a, then (G, *) is a NON-
ABELIAN GROUP.
It is also called
NONCOMMUTATIVE GROUP.

EXERCISE
Classify the following groups as Abelian or Non-
Abelian: 1. (ℝ, +)
2. (ℤ5
#
, ∙5)
3. (ℚ#, ∙)
4. (GL2(ℝ), ∙)
QUESTIONS?
ABELIAN
ABELIAN
ABELIAN
NON ABELIAN

DEFINITION: ORDER OF AN ELEMENT OF A FINITE GROUP
It is the smallest positive power of the
element which gives you the identity
element.
Let (G, *) be a finite group and a ∈ G.
The order of an element <a> is n given
that an = e.

EXAMPLE
Let us consider
ℤ4 = {0, 1, 2, 3}. We
know that (ℤ4, +4) is
a finite group.

EXAMPLE
Let us consider
ℤ4= {0, 1, 2, 3}. We
know that (ℤ4, +4) is a
finite group.
<0> = {0}
<1> = {1, 2, 3, 0}
<2> = {2, 0}
<3> = {3, 2, 1, 0}
Let us determine
the order of each
element:
<0>= 1
<1>= 4
<2>= 2
<3>= 4
QUESTIONS?

DEFINITION: GENERATOR OF A FINITE GROUP
It is a set of group elements such that
possibly repeated application of the
generators on themselves and each other
is capable of producing all the elements in
the group.
It could be said that <a> is a generator
of G iff
<a> = G.

EXAMPLE
Let us consider
ℤ4= {0, 1, 2, 3}. We
know that (ℤ4, +4) is a
finite group.
<0> = {0}
<1> = {1, 2, 3, 0}
<2> = {2, 0}
<3> = {3, 2, 1, 0}
We could see that
ℤ4 = <1> =
<3>
Hence, the elements 1
and 3 are generators
of (ℤ4, +4) while
elements 0 and 2 are
NOT generators.

ADDITIONAL NOTE:
Speaking of a
finite group (G, *)
with G = n and if a
∈ G is a generator,
then
G = <a>

DEFINITION: FINITE CYCLIC GROUP
A certain finite group is
CYCLIC if it has at least 1
generator. If it has no
generator, then the group is
NOT CYCLIC.
Is (ℤ4, +4) a cyclic
group?
QUESTIONS?
Yes, because it has
generators 1 and 3.

EXERCISE
Determine the orders of the
elements of (ℤ8, +8), identify generators
if they exist, and if the group is cyclic or
not.
Solution:

EXERCISE
Determine the orders of the elements of (ℤ8, +8),
identify generators if they exist, and if the group is cyclic
or not.
Solution:
ℤ8 = {0, 1, 2, 3, 4, 5, 6, 7}
<0> = {0}
<1> = {1, 2, 3, 4, 5, 6, 7, 0}
<2> = {2, 4, 6, 0}
<3> = {3, 6, 1, 4, 7, 2, 5, 0}
<4> = {4, 0}
<5> = {5, 2, 7, 4, 1, 6, 3, 0}
<6> = {6, 4, 2, 0}
<7> = {7, 6, 5, 4, 3, 2, 1, 0}
We could see that ℤ8 = <1> = <3> = <5> = <7>
Generators: 1, 3, 5, 7
Since there is at least 1
generator, then (ℤ8, +8) is a
cyclic group.

EXERCISE
Determine the orders of the
elements of (ℤ5
#
, ∙5), identify
generators if they exist, and if
the group is cyclic or not.

EXERCISE
Determine the orders of the elements of (ℤ5
#
, ∙5),
identify generators if they exist, and if the group is
cyclic or not.
ℤ5
#
= {1, 2,
3, 4, 5}
<1> = {1}
<2> = {2, 4, 3, 1}
<3> = {3, 4, 2, 1}
<4> = {4, 1}
<5> = {0}
We could see that ℤ5
#
= <2>
= <3> Generators: 2, 3
Since there is at least 1
generator, then (ℤ5
#
, ∙5) is a cyclic
group.

EXERCISE
Determine the orders of the elements of
Klein V under operation * or (V, *), identify
generators if they exist, and if the group is
cyclic or not.
V = {e, a, b, c}
QUESTIONS?

EXERCISE
Determine the orders of the elements of
Klein V under operation * or (V, *), identify
generators if they exist, and if the group is cyclic
or not.
V = {e, a, b, c}
<e> = {e}
<a> = {a, e}
<b> = {b,
e}
<c> = {c, e}
Generators: None
Since there is no generator,
then (V, *) is not cyclic.
QUESTIONS?

ORDER OF AN INFINITE CYCLIC
GROUP
An infinite group has an infinite order.
An infinite group is CYCLIC if it has at
least 1 generator. If it has no generator,
then the group is NOT CYCLIC.
If G is an infinite cyclic group
generated by a ∈ G, then a is an
element of infinite order, and all the
powers of a are different.

ORDER OF AN INFINITE CYCLIC GROUP
Is (ℤ, +) an infinite cyclic
group?
Is/Are there
generator(s)?
YES
-1 AND 1

ORDER OF AN INFINITE CYCLIC
GROUP
Are there generators in (ℝ, +)?
No
Is (ℝ, +) an infinite cyclic
group?
No

TRUE OR FALSE
If a group is cyclic, then it is Abelian.
If a group is Abelian, then it is cyclic.
TRUE
FALSE

EXERCISE
Categorize (ℤ, +),
(V, *), GL2(ℝ), (ℤ5
#
,
∙5)
(ℝ, +), (ℚ#,
∙), (ℤ4, ∙4), (10ℤ, +).
Finite
Group
Infinite
Group
Abelian (V, *) (ℤ, +)
Group
(ℤ5
#
,∙5) (ℝ, +)
(ℚ#, ∙)
(10ℤ, +)
Non-
Abelia
n
GL2(ℝ)

PERMUTATION
Given a finite set S, a permutation
of A is a function f : S → S which is
a BIJECTIVE FUNCTION- that is
both 1-1 (injective) and onto
(surjective).
Remember a set with finite
number of

PERMUTATION
We will focus specifically on the case
when S = {1, 2, 3, ... , n} for some fixed
integer n.
This means each group element will
permute this set. For example if A = {1, 2, 3}
then a permutation α might have α (1) = 2,
α (2) = 1, and α(3) = 3. We can write this as:
α
1 2
3
2 1
=

PERMUTATION
The first row is the domain (pre-
image) and the second row is your
codomain (image).
α
=
1 2
3
2 1
3
If we say α(1) = 2, we mean that
the element 1 (in the domain) is
mapped to 2 (in the codomain).

EXERCISE
Consider f
= 1. f(1)
2.
f(2)
3.
1 2 3 4
3 2 5 1
4
5 . What
is
4. f-1(1)
5. f-1(4)

PERMUTATION GROUP
Show that the function composition ( ͦ ) is a binary
operation on the collection of all permutation on a
set A. We call this operation permutation
multiplication. Let A be a set, and let and be
𝛼 𝛽
the permutation of A so that and are both one-
𝛼 𝛽
to-one functions mapping A to A. The composite
function of 𝛼 ͦ 𝛽 defined schematically by
A( )
𝛽 A(
→ )
𝛼 A.
→

PERMUTATION GROUP
gives mapping of A to A. Rather than keep the
symbol ( ͦ ) for permutation multiplication, we will
denote 𝛼 ͦ 𝛽 by the juxtaposition 𝛼𝛽, as we have
done for general groups. Now 𝛼𝛽 will be a
permutation if it is one to one and onto A.
REMEMBER: The action of 𝛼𝛽 on A must be read in
right-to-left.

EXERCISE
Suppose that
A = (1, 2, 3, 4, 5)
Let 𝛼 = 1 2 3 4
5
4 2 5 3
1
and 𝛽
=
1 2 3 4 5 .
3 5 4 2 1
𝛼𝛽 =
=
1 2 3 4
5
4 2 5 3
1
1 2 3 4 5 .
3 5 4 2 1
1 2 3 4 5 .
5 1 3 2 4

EXERCISE
Consider 2 permutations of length 5, say A
and B. We define AB = A ° B as the permutation
multiplication or mapping composition of B to A.
Note you can only do this if the permutations
involved have equal lengths.
Let A
=
1 2 3 4
5
2 3 5 1
4
and B
=
1 2 3 4 5 .
2 1 5 3 4
What is the value of AB or A °
B?
Let’s do this!

EXAMPLE
A =
1 2 3 4
5
2 3 5 1
4
and B
=
1 2 3 4
5
2 1 5 3
4
What is the value of AB or A °
B?
AB = A ° B
=
1 2 3
4
3 2 4
5
5
1

EXAMPLE
A =
1 2 3 4
5
2 3 5 1
4
and B
=
1 2 3 4
5
2 1 5 3
4
How about BA or B °
A?
BA = B ° A
=
1 2 3
4
1 5 4
2
5
3

REMEMBER:
Based on what we did, is AB = BA?
NO
Thus, mapping composition of
permutations is NOT COMMUTATIVE.

EXERCISE
1 2 3
4
2 3 5
1
5
4
Consider A
=
what is
, B
=
1 2 3
4
2 1 5
3
4
5 , and C =
1 2 3
4
3 5 2
4
5
1
,
1. AC =
1 2 3 4
5
5 4 3 1
2
2. CB =
1 2 3 4
5
5 3 1 2
4
3. C-1 =
1 2 3 4
5
5 3 1 4
2
4. A-1 = 1 2 3 4
5

EXERCISE
Consider A
=
1 2 3 4
5
2 3 5 1
4
, B
=
1 2 3 4
2 1 5 3
4
5
, and C =
1 2 3 4
5
3 5 2 4
1
,
what is
5. ABC = A ° B ° C
=
1 2 3 4 5 1 2 3 4 5 1 2 3 4 5
2 3 5 1 4 2 1 5 3 4 3 5 2 4 1
=
1 2 3 4 5
4 1 2 5 3
6. BCA = B ° C ° A
=
7. A2 = AA = A ° A
=
1 2 3 4 5 1 2 3 4 5 =
1 2 3 4 5
2 3 5 1 4 2 3 5 1 4 3 5 4 2 1
8. B3 =
1 2 3 4 5
2 1 3 4 5
1 2 3 4 5
4 1 2 5 3

EXERCISE
Consider A
=
what is
1 2 3 4
5
2 3 5 1
4
, B = 1
2 3 4
2 1 5 3 4
5 , and C =
1 2 3 4
5
3 5 2 4
1
,
9. AB-3 =
10. A-1CB =
1 2 3 4
5
3 2 4 1
5
1 2 3 4
5
3 2 5 1
4

EXERCISE
Consider A
=
, B
=
1 2 3 4
2 1 5 3
4
5 , and C =
1 2 3 4
5
3 5 2 4
1
,
what is AB
=
1
2
3
2
1 2 3 4
5
2 3 5 1
4
3 4 5
4 5 1
11. A2B2 =
1 2 3 4
5
3 5 2 1
4
12. (AB)2 =
1 2 3 4
5
4 2 5 1

EXERCISE
Consider A
=
1 2 3
4
2 3 5
1
5
4 , B = 1 5 ,
and
C =
1 2 3
4
3 5 2
4
5
1
, C-1 =
1 2 3
4
5 3 1
4
2
5 , C3 =
2 3 4
2 1 5 3 4
1
2
3
4
5
3
5
2
what is
13. C-3 = C-1 ° C-1 ° C-1 =
14. (C-1)3 =
15. (C3)-1 = (xm)n = xmn = (xn)m
1 2 3 4
5
3 5 2 4
1
1 2 3 4
5
3 5 2 4
1
1 2 3 4
5
3 5 2 4

EXERCISE
Consider A
=
1 2 3 4
5
2 3 5 1
4
, B = 1
2 3 4
2 1 5 3 4
5 , and C =
1 2 3 4
5
3 5 2 4
1
1 2 3 4 5 1 2 3 4 5
3 2 4 5 1 3 5 2 4 1
16. (A ° B) ° C =
=
17. A ° (B ° C)
=
1
2
1 2 3 4 5
4 1 2 5 3
2 3 4 5 1 2 3 4 5
3 5 1 4 5 4 1 3 2
=
1 2 3 4
5

PERMUTATION
𝑖 =
1
1
2
2
3
3
𝛾 =
1
2
2
3
3
1
𝛼 =
1
1
2
3
3
2
𝛿 =
1
3
2
1
3
2
𝛽=
1
2
2
1
3
3
𝜀 =
1
3
2
2
3
1
i. Closure
∀𝑥, 𝑦 ∈ 𝑆3, it could be seen
that
𝑥 ° 𝑦 ∈ 𝑆3. Hence, closure holds.
Is (S3,°) a group?
° 𝑖 𝛼 𝛽 𝛾 𝛿 𝜀
𝑖 𝑖 𝛼 𝛽 𝛾 𝛿 𝜀
𝛼 𝛼 𝑖 𝛾 𝛽 𝜀 𝛿
𝛽 𝛽 𝛿 𝑖 𝜀 𝛼 𝛾
𝛾 𝛾 𝜀 𝛼 𝛿 𝑖 𝛽
𝛿 𝛿 𝛽 𝜀 𝑖 𝛾 𝛼
𝜀 𝜀 𝛾 𝛿 𝛼 𝛽 𝑖

PERMUTATION
𝑖 =
1
1
2
2
3
3
𝛾 =
1
2
2
3
3
1
𝛼 =
1
1
2
3
3
2
𝛿 =
1
3
2
1
3
2
𝛽=
1
2
2
1
3
3
𝜀 =
1
3
2
2
3
1
iii. Existence of an Identity Element
∀𝑥 ∈ 𝑆3, ∃𝑒 = 𝑖 such that 𝑥
° 𝑒 =
𝑒 ° 𝑥 = 𝑥. Hence, identity
property is satisfied.
Is (S3,°) a group?

PERMUTATION
𝑖 =
1
1
2
2
3
3
𝛾 =
1
2
2
3
3
1
𝛼 =
1
1
2
3
3
2
𝛿 =
1
3
2
1
3
2
𝛽=
1
2
2
1
3
3
𝜀 =
1
3
2
2
3
1
ii. Associativity
∀𝑥, 𝑦, 𝑧 ∈ 𝑆3, it is known
that (x ° y) ° z = x ° (y ° z). Thus,
associativity holds.
Is (S3,°) a group?

PERMUTATION
𝑖 =
1
1
2
2
3
3
𝛾 =
1
2
2
3
3
1
𝛼 =
1
1
2
3
3
2
𝛿 =
1
3
2
1
3
2
𝛽=
1
2
2
1
3
3
𝜀 =
1
3
2
2
3
1
iv. Inverse
Property
𝑖-1 =
𝛼-1=
𝛽 -1 =
𝛾-1 =
𝛿-1=
𝜀-1=
Is (S3,°) a group?

PERMUTATION
𝑖 =
1
1
2
2
3
3
𝛾 =
1
2
2
3
3
1
𝛼 =
1
1
2
3
3
2
𝛿 =
1
3
2
1
3
2
𝛽=
1
2
2
1
3
3
𝜀 =
1
3
2
2
3
1
iv. Inverse
Property
𝑖-1 = 𝑖
𝛼-1= 𝛼
𝛽 -1 = 𝛽
𝛾-1
= 𝛿
𝛿-1= 𝛾
𝜀-1= 𝜀
Since ∀ 𝑥 ∈ S3, ∃ 𝑥 -1 ∈ S3 such
that 𝑥°𝑥 -1 = 𝑥 -1°𝑥 = e = 𝑖, thus,
inverse property is satisfied.
Is (S3,°) a group?

PERMUTATION
𝑖 =
1
1
2
2
3
3
𝛾 =
1
2
2
3
3
1
𝛼 =
1
1
2
3
3
2
𝛿 =
1
3
2
1
3
2
𝛽=
1
2
2
1
3
3
𝜀 =
1
3
2
2
3
1
Since all properties of a
group are satisfied, hence
(S3,°) is a group.
There is a special type of
terminology for this type of
Is (S3,°) a group?

SYMMETRIC GROUP
The symmetric group Sn is
the group of bijections from
any set of n objects, which we
usually call simply {1,2, 3, ..., n},
to itself.
An element of this group is
called a permutation of {1, 2,
3, ..., n}. The group operation in
Sn is composition of
S = stands for
symmetric
n = stands for
the number of
elements in a
set

REMEMBER:
Is symmetric group Abelian?
NO
Is symmetric group cyclic?
NO
The fact that it is not Abelian, then it is not
Cyclic.

p → q
If a group is cyclic, then it is Abelian.
~𝑞 → ~ p
If a group is not Abelian, then it is not
cyclic.

ORDER OF AN ELEMENT OF Sn
Recall:
The order of an element <a> is n given
that an = e.
The identity permutation
e =
1
2
1
2
3 …
𝑛
3 …
𝑛
Suppose A
=
1 2 3 4
5
3 2 5 1
∈ S5. Find
𝐴 .

LET’S DO THIS!
∈
S5
A = 1 2
3 4
5
3
2
5
1

LET’S DO THIS!
∈
S5
A1 = A
=
A2 =
A =
1 2
3 4
5
3
2
5
1
5 2 4 3 1
1 2 3 4 5
4 2 1 5 3
1 2 3 4 5
1 2 3 4 5
A3 =
A4 =
Thus, 𝐴

EXERCISE
Find the order of each element of a group in
Sn:
2. B = ∈
S5
1 2 3 4 5
5 3 4 2 1

EXERCISE
Find the order of each element of a group in
Sn:
2. B = ∈
S5
B6 =
1 2 3 4 5
5 3 4 2 1
1 2 3 4
5
1 2 3 4
5
𝐵 =
6

NOTE:
Finding the order of a
particular element of Sn
sometimes would take a lot
of time!
Sige lang deserve
ninyo ina!
QUESTIONS?

ORBITS, AND CYCLE, GROUPS
Each permutation σ of a set A determines a natural
partition of A into cells with the property that a, b ∈
A are in the same cell if and only if b = σ n (a) for
some n ∈ ℤ. We can establish a partition using an
appropriate equivalence relation:
ORBITS

For a, b A, a ~ b if and only if b =
∈ σ n (a)
for some n ∈ .
ℤ
Reflexive: Clearly a ~ a since a = i (a) = σ 0 (a).
Symmetric: If a ~ b then b = σ n (a) for some n ∈ ℤ.
But then a = σ -n (b) and –n ∈ ℤ.
Transitive: Suppose a ~ b and b ~ c, then b = σ n
(a) and c = σ m (b) for some n, m ∈ ℤ. Substituting,
we find that c = σ m (σ n (a)) = σ n+m (a), a ~ c.
ORBITS

Let σ be a permutation of a set A.
The equivalence classes in A determine
by the equivalence relation.
DEFINITION OF ORBITS OF σ.
Example:
Find the orbit of the permutation:
1 2 3 4 5 6 7
8
3 8 6 7 4 1 5
σ = ∈
S8

ORBITS OF σ.
Example:
Find the orbit of the permutation:
1 2 3 4 5 6 7
8
3 8 6 7 4 1 5
2
σ = ∈
S8
Solution:
To find the orbit containing 1, we
apply σ repeatedly, obtaining symbolically
1 → 3 → 6 → 1 → 3 → 6 → 1 → 3 → . . .

ORBITS OF σ.
Solution:
To find the orbit containing 1, we
apply σ repeatedly, obtaining symbolically
1 → 3 → 6 → 1 → 3 → 6 → 1 → 3 → . . .
Since σ -1 would simply reverse the direction of
arrows in this chain, we can see that that the
orbit containing 1 is {1, 3, 6}. We now choose
an integer from 1 to 8 not in {1, 3, 6}, say 2,
and similarly find that the orbit containing 2

ORBITS OF σ.
Solution:
is {2, 8}. Finally , we find the orbit that
containing 4 is {4, 7, 5}. Since these three
orbits include all the integer from 1 to 8, we
can see that the complete list of orbits of σ is;
{1, 3, 6} {2, 8} {4, 7, 5}

ORBITS OF σ.
Example:
Find the orbits of the ff permutation.
1 2 3 4 5
5 3 4 2 1
1. A = ∈
S5
1 2 3 4
5
4 2 5 1
3
∈
S5
2. B =

A permutation σ ∈ Sn is a cyclic if it
has at most one orbit containing more
than 1 element. The length of a cycle is
the number if elements in its largest orbit.
CYCLES
Example:
Find the cycle of the permutation:
1 2 3 4
5
A =

CYCLE NOTATION
We now write down
a more compact
notation for Sn.
Consider the following
element in S5:
A =
1 2 3 4
5
3 2 5 1
4
You could
see 1 → 3 → 5 → 4
→ 1
2 → 2
1
3
5
4
2

CYCLE NOTATION
We now write down
a more compact
notation for Sn.
Consider the following
element in S5:
A =
1 2 3 4
5
3 2 5 1
4
You could
see
1 → 3
→ 5 →
A = (1 3 5 4) (2)
or
A = (1 3 5 4)
Note: (2) was no longer
written because it is
understood that it is
mapped to itself. Also,
writing cycle form of a
permutation is NOT
unique.

CYCLE NOTATION
A = (1 3 5 4)
or
A = (3 5 4 1)
or
A = (5 4 1 3)
or
A = (4 1 3 5)
1
3
5
4
2

EXERCISE
Write the following in cycle
notation:
1. A
=
1 2 3 4
5
2 4 5 1
3
A = (1 2 4) (3 5)
2. B =
1 2 3 4 5 6 7
1 3 7 6 4 5 2
B = (1) (2 3 7) (4 6 5)
B = (2 3 7) (4 6 5)

CYCLE NOTATION to 2-ROW FORM
Let A ∈ S7 such that A = (1 3 2)
(4 6).
What is A in 2-row form?

CYCLE NOTATION to 2-ROW FORM
Let A ∈ S7 such that A = (1 3 2)
(4 6).
What is A in 2-row form?
Solution:
1 → 3 → 2 → 1
4 → 6 → 4
5 →
5
7 →
7
Let us fill
in
A =
1 2 3 4 5 6 7
3 1 2 6 5 4 7

EXERCISE
Write the following in 2-row
form:
1. If length of A = 6 and A = (2 4)
(1 6)
Answer:
A =
1 2 3 4 5
6

EXERCISE
Write the following in 2-row form:
2. If length of B = 7 and B = (14 2 6) (3 7 5)
Answer:
B =
1 2 3 4 5 6 7
4 6 7 2 3 1 5

JOINT AND DISJOINT CYCLES
Two or more cycles
are DISJOINT if they do
not have any common
element.
Example:
A = (1 2)
B = (3 5)
A and B are disjoint
Two or more cycles
are JOINT if they have
at least 1 common
element.
Example:
C = (3 1)
D = (3 4 5)
C and D are joint cycles.

LET’S DO THIS
Consider A, B to be permutations of length 7
with A = (1 2 4) and B = (3 7). Write these products
in 2- row form.
a. AB =
b. BA =
1 2 3 4 5 6 7
2 4 7 1 5 6 3
1 2 3 4 5 6 7
2 4 7 1 5 6 3

WHAT HAVE YOU NOTICED?
A = (1 2 4) and B = (3 7)
What have you noticed with AB and
BA?
AB = BA =
1 2 3 4 5 6
7
2 4 7 1 5 6
3
What kind of cycles are A and B in relation
to each other?
Disjoint
Remember: Disjoint cycles
commute.

REMEMBER:
An ORBIT is a set, while a CYCLE is a
permutation of a set ( which permutes
its elements cyclically.

GROUP AND SUBGROUP SET - ABSTRACT ALGEBRA ( TOPIC 5)

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GROUP AND SUBGROUP SET - ABSTRACT ALGEBRA ( TOPIC 5)