algebric structure_UNIT 4.ppt

Unit-III
Algebraic Structures
 Algebraic systems Examples and general
properties
 Semi groups
 Monoids
 Groups
 Sub groups

Algebraic systems
 N = {1,2,3,4,….. } = Set of all natural numbers.
Z = { 0,  1,  2,  3,  4 , ….. } = Set of all integers.
Q = Set of all rational numbers.
R = Set of all real numbers.
 Binary Operation: The binary operator * is said to be a binary
operation (closed operation) on a non empty set A, if
a * b  A for all a, b  A (Closure property).
Ex: The set N is closed with respect to addition and multiplication
but not w.r.t subtraction and division.
 Algebraic System: A set ‘A’ with one or more binary(closed)
operations defined on it is called an algebraic system.
Ex: (N, + ), (Z, +, – ), (R, +, . , – ) are algebraic systems.

Properties
 Commutative: Let * be a binary operation on a set A.
The operation * is said to be commutative in A if
a * b= b * a for all a, b in A
 Associativity: Let * be a binary operation on a set A.
The operation * is said to be associative in A if
(a * b) * c = a *( b * c) for all a, b, c in A
 Identity: For an algebraic system (A, *), an element ‘e’ in A is said to
be an identity element of A if
a * e = e * a = a for all a  A.
 Note: For an algebraic system (A, *), the identity element, if exists, is
unique.
 Inverse: Let (A, *) be an algebraic system with identity ‘e’. Let a be
an element in A. An element b is said to be inverse of A if
a * b = b * a = e

Semi group
 Semi Group: An algebraic system (A, *) is said to be a semi group if
1. * is closed operation on A.
2. * is an associative operation, for all a, b, c in A.
 Ex. (N, +) is a semi group.
 Ex. (N, .) is a semi group.
 Ex. (N, – ) is not a semi group.
 Monoid: An algebraic system (A, *) is said to be a monoid if the
following conditions are satisfied.
1) * is a closed operation in A.
2) * is an associative operation in A.
3) There is an identity in A.

Monoid
 Ex. Show that the set ‘N’ is a monoid with respect to
multiplication.
 Solution: Here, N = {1,2,3,4,……}
1. Closure property : We know that product of two natural numbers is
again a natural number.
i.e., a.b = b.a for all a,b  N
 Multiplication is a closed operation.
2. Associativity : Multiplication of natural numbers is associative.
i.e., (a.b).c = a.(b.c) for all a,b,c  N
3. Identity : We have, 1  N such that
a.1 = 1.a = a for all a  N.
 Identity element exists, and 1 is the identity element.
Hence, N is a monoid with respect to multiplication.

Subsemigroup & submonoid
Subsemigroup : Let (S, * ) be a semigroup and let T be a
subset of S. If T is closed under operation * , then (T, * ) is
called a subsemigroup of (S, * ).
Ex: (N, .) is semigroup and T is set of multiples of positive
integer m then (T,.) is a sub semigroup.
Submonoid : Let (S, * ) be a monoid with identity e, and let T
be a non- empty subset of S. If T is closed under the
operation * and e  T, then (T, * ) is called a submonoid of
(S, * ).

Group
 Group: An algebraic system (G, *) is said to be a group if
the following conditions are satisfied.
1) * is a closed operation.
2) * is an associative operation.
3) There is an identity in G.
4) Every element in G has inverse in G.
 Abelian group (Commutative group): A group (G, *) is
said to be abelian (or commutative) if
a * b = b * a a, b G.

Algebraic systems
Abelian groups
Groups
Monoids
Semi groups
Algebraic systems

Theorem
 In a Group (G, * ) the following properties hold good
1. Identity element is unique.
2. Inverse of an element is unique.
3. Cancellation laws hold good
a * b = a * c  b = c (left cancellation law)
a * c = b * c  a = b (Right cancellation law)
4. (a * b) -1 = b-1 * a-1
 In a group, the identity element is its own inverse.
 Order of a group : The number of elements in a group is called order
of the group.
 Finite group: If the order of a group G is finite, then G is called a
finite group.

Ex. Show that, the set of all integers is a group with
respect to addition.
 Solution: Let Z = set of all integers.
Let a, b, c are any three elements of Z.
1. Closure property : We know that, Sum of two integers is again an
integer.
i.e., a + b  Z for all a,b  Z
2. Associativity: We know that addition of integers is associative.
i.e., (a+b)+c = a+(b+c) for all a,b,c  Z.
3. Identity : We have 0  Z and a + 0 = a for all a  Z .
 Identity element exists, and ‘0’ is the identity element.
4. Inverse: To each a  Z , we have – a  Z such that
a + ( – a ) = 0
Each element in Z has an inverse.

Contd.,
 5. Commutativity: We know that addition of integers is commutative.
i.e., a + b = b +a for all a,b  Z.
Hence, ( Z , + ) is an abelian group.

Ex. Show that set of all non zero real numbers is a group with respect to
multiplication .
 Solution: Let R* = set of all non zero real numbers.
Let a, b, c are any three elements of R* .
1. Closure property : We know that, product of two nonzero real
numbers is again a nonzero real number .
i.e., a . b  R* for all a,b  R* .
2. Associativity: We know that multiplication of real numbers is
associative.
i.e., (a.b).c = a.(b.c) for all a,b,c  R* .
3. Identity : We have 1  R* and a .1 = a for all a  R* .
 Identity element exists, and ‘1’ is the identity element.
4. Inverse: To each a  R* , we have 1/a  R* such that
a .(1/a) = 1 i.e., Each element in R* has an inverse.

Contd.,
 5.Commutativity: We know that multiplication of real numbers is
commutative.
i.e., a . b = b . a for all a,b  R*.
Hence, ( R* , . ) is an abelian group.
 Ex: Show that set of all real numbers ‘R’ is not a group with respect
to multiplication.
 Solution: We have 0  R .
The multiplicative inverse of 0 does not exist.
Hence. R is not a group.

Example
 Ex. Let (Z, *) be an algebraic structure, where Z is the set of integers
and the operation * is defined by n * m = maximum of (n, m).
Show that (Z, *) is a semi group.
Is (Z, *) a monoid ?. Justify your answer.
 Solution: Let a , b and c are any three integers.
Closure property: Now, a * b = maximum of (a, b)  Z for all a,b  Z
Associativity : (a * b) * c = maximum of {a,b,c} = a * (b * c)
 (Z, *) is a semi group.
Identity : There is no integer x such that
a * x = maximum of (a, x) = a for all a  Z
 Identity element does not exist. Hence, (Z, *) is not a monoid.

Example
 Ex. Show that the set of all strings ‘S’ is a monoid under the
operation ‘concatenation of strings’.
Is S a group w.r.t the above operation? Justify your answer.
 Solution: Let us denote the operation
‘concatenation of strings’ by + .
Let s1, s2, s3 are three arbitrary strings in S.
Closure property: Concatenation of two strings is again a string.
i.e., s1+s2  S
Associativity: Concatenation of strings is associative.
(s1+ s2 ) + s3 = s1+ (s2 + s3 )

Contd.,
 Identity: We have null string ,   S such that s1 +  = S.
  S is a monoid.
 Note: S is not a group, because the inverse of a non empty string
does not exist under concatenation of strings.

Example
 Ex. Let S be a finite set, and let F(S) be the collection of all functions
f: S  S under the operation of composition of functions, then
show that F(S) is a monoid.
Is S a group w.r.t the above operation? Justify your answer.
 Solution:
Let f1, f2, f3 are three arbitrary functions on S.
Closure property: Composition of two functions on S is again a function
on S.
i.e., f1o f2  F(S)
Associativity: Composition of functions is associative.
i.e., (f1 o f2 ) o f3 = f1 o (f2 o f3 )

Contd.,
 Identity: We have identity function I : SS
such that f1 o I = f1.
 F(S) is a monoid.
 Note: F(S) is not a group, because the inverse of a non bijective
function on S does not exist.

Ex. If M is set of all non singular matrices of order ‘n x n’.
then show that M is a group w.r.t. matrix multiplication.
Is (M, *) an abelian group?. Justify your answer.
 Solution: Let A,B,C  M.
1.Closure property : Product of two non singular matrices is again a non
singular matrix, because
AB = A . B  0 ( Since, A and B are nonsingular)
i.e., AB  M for all A,B  M .
2. Associativity: Marix multiplication is associative.
i.e., (AB)C = A(BC) for all A,B,C  M .
3. Identity : We have In  M and A In = A for all A  M .
 Identity element exists, and ‘In’ is the identity element.
4. Inverse: To each A  M, we have A-1  M such that
A A-1 = In i.e., Each element in M has an inverse.

Contd.,
  M is a group w.r.t. matrix multiplication.
We know that, matrix multiplication is not commutative.
Hence, M is not an abelian group.

Ex. Show that the set of all positive rational numbers forms an abelian
group under the composition * defined by
a * b = (ab)/2 .
 Solution: Let A = set of all positive rational numbers.
Let a,b,c be any three elements of A.
1. Closure property: We know that, Product of two positive rational
numbers is again a rational number.
i.e., a *b  A for all a,b  A .
2. Associativity: (a*b)*c = (ab/2) * c = (abc) / 4
a*(b*c) = a * (bc/2) = (abc) / 4
3. Identity : Let e be the identity element.
We have a*e = (a e)/2 …(1) , By the definition of *
again, a*e = a …..(2) , Since e is the identity.
From (1)and (2), (a e)/2 = a  e = 2 and 2  A .
 Identity element exists, and ‘2’ is the identity element in A.

Contd.,
 4. Inverse: Let a  A
let us suppose b is inverse of a.
Now, a * b = (a b)/2 ….(1) (By definition of inverse.)
Again, a * b = e = 2 …..(2) (By definition of inverse)
From (1) and (2), it follows that
(a b)/2 = 2
 b = (4 / a)  A
 (A ,*) is a group.
 Commutativity: a * b = (ab/2) = (ba/2) = b * a
 Hence, (A,*) is an abelian group.

Theorem
 Ex. In a group (G, *) , Prove that the identity element is
unique.
 Proof :
a) Let e1 and e2 are two identity elements in G.
Now, e1 * e2 = e1 …(1) (since e2 is the identity)
Again, e1 * e2 = e2 …(2) (since e1 is the identity)
From (1) and (2), we have e1 = e2
 Identity element in a group is unique.

Theorem
 Ex. In a group (G, *) , Prove that the inverse of any element is
unique.
 Proof:
 Let a ,b,c G and e is the identity in G.
 Let us suppose, Both b and c are inverse elements of a .
 Now, a * b = e …(1) (Since, b is inverse of a )
 Again, a * c = e …(2) (Since, c is also inverse of a )
 From (1) and (2), we have
 a * b = a * c
  b = c (By left cancellation law)
 In a group, the inverse of any element is unique.

Theorem
 Ex. In a group (G, *) , Prove that
(a * b)-1 = b-1 * a-1 for all a,b G.
 Proof :
 Consider,
 (a * b) * ( b-1 * a-1)
 = (a * ( b * b-1 ) * a-1) (By associative property).
 = (a * e * a-1) ( By inverse property)
 = ( a * a-1) ( Since, e is identity)
 = e ( By inverse property)
 Similarly, we can show that
 (b-1 * a-1) * (a * b) = e
 Hence, (a * b)-1 = b-1 * a-1 .

Ex. If (G, *) is a group and a  G such that a * a = a ,
then show that a = e , where e is identity element in G.
 Proof: Given that, a * a = a
  a * a = a * e ( Since, e is identity in G)
  a = e ( By left cancellation law)
 Hence, the result follows.

Ex. If every element of a group is its own inverse, then show that
the group must be abelian .
 Proof: Let (G, *) be a group.
 Let a and b are any two elements of G.
 Consider the identity,
 (a * b)-1 = b-1 * a-1
  (a * b ) = b * a ( Since each element of G is its own
 inverse)
 Hence, G is abelian.

Note: a2 = a * a
a3 = a * a * a etc.
 Ex. In a group (G, *), if (a * b)2 = a2 * b2 a,b  G
then show that G is abelian group.
 Proof: Given that (a * b)2 = a2 * b2
  (a * b) * (a * b) = (a * a )* (b * b)
  a *( b * a )* b = a * (a * b) * b ( By associative law)
  ( b * a )* b = (a * b) * b ( By left cancellation law)
  ( b * a ) = (a * b) ( By right cancellation law)
 Hence, G is abelian group.

Finite groups
 Ex. Show that G = {1, -1} is an abelian group under multiplication.
 Solution: The composition table of G is
 . 1 – 1
 1 1 – 1
 – 1 – 1 1
1. Closure property: Since all the entries of the composition table are
the elements of the given set, the set G is closed under
multiplication.
2. Associativity: The elements of G are real numbers, and we know that
multiplication of real numbers is associative.
3. Identity : Here, 1 is the identity element and 1 G.
4. Inverse: From the composition table, we see that the inverse elements
of
1 and – 1 are 1 and – 1 respectively.

Contd.,
Hence, G is a group w.r.t multiplication.
5. Commutativity: The corresponding rows and columns of the table are
identical. Therefore the binary operation . is commutative.
Hence, G is an abelian group w.r.t. multiplication..

Ex. Show that G = {1, , 2} is an abelian group under multiplication.
Where 1, , 2 are cube roots of unity.
 . 1  2
 1 1  2
   2 1
 2 2 1 
1. Closure property: Since all the entries of the composition table are the
elements of the given set, the set G is closed under multiplication.
2. Associativity: The elements of G are complex numbers, and we know that
multiplication of complex numbers is associative.
4. Inverse: From the composition table, we see that the inverse elements of
1 , 2 are 1, 2,  respectively.

Contd.,
 Hence, G is a group w.r.t multiplication.
 5. Commutativity: The corresponding rows and columns of the table
are identical. Therefore the binary operation . is commutative.
 Hence, G is an abelian group w.r.t. multiplication.

Ex. Show that G = {1, –1, i, –i } is an abelian group under multiplication.
 . 1 –1 i -i
 1 1 -1 i - i
 -1 -1 1 -i i
 i i -i -1 1
 -i -i i 1 -1
1. Closure property: Since all the entries of the composition table are
the elements of the given set, the set G is closed under
multiplication.
2. Associativity: The elements of G are complex numbers, and we know
that multiplication of complex numbers is associative.

Contd.,
 4. Inverse: From the composition table, we see that the inverse
elements of
1 -1, i, -i are 1, -1, -i, i respectively.
are identical. Therefore the binary operation . is commutative.
Hence, (G, .) is an abelian group.

Modulo systems.
 Addition modulo m ( +m )
 let m is a positive integer. For any two positive integers a and b
 a +m b = a + b if a + b < m
 a +m b = r if a + b  m where r is the remainder obtained
 by dividing (a+b) with m.
 Multiplication modulo p ( p )
 let p is a positive integer. For any two positive integers a and b
 a p b = a b if a b < p
 a p b = r if a b  p where r is the remainder obtained
 by dividing (ab) with p.
 Ex. 3 5 4 = 2 , 5 5 4 = 0 , 2 5 2 = 4

Ex.The set G = {0,1,2,3,4,5} is a group with respect to addition modulo 6.
 +6 0 1 2 3 4 5
 0 0 1 2 3 4 5
 1 1 2 3 4 5 0
 2 2 3 4 5 0 1
 3 3 4 5 0 1 2
 4 4 5 0 1 2 3
 5 5 0 1 2 3 4
 1. Closure property: Since all the entries of the composition table
are the elements of the given set, the set G is closed under +6 .

Contd.,
 2. Associativity: The binary operation +6 is associative in G.
for ex. (2 +6 3) +6 4 = 5 +6 4 = 3 and
2 +6 ( 3 +6 4 ) = 2 +6 1 = 3
 3. Identity : Here, The first row of the table coincides with the top
row. The element heading that row , i.e., 0 is the identity element.
 4. . Inverse: From the composition table, we see that the inverse
elements of 0, 1, 2, 3, 4. 5 are 0, 5, 4, 3, 2, 1 respectively.
are identical. Therefore the binary operation +6 is commutative.
 Hence, (G, +6 ) is an abelian group.

Ex.The set G = {1,2,3,4,5,6} is a group with respect to multiplication
modulo 7.
 7 1 2 3 4 5 6
 1 1 2 3 4 5 6
 2 2 4 6 1 3 5
 3 3 6 2 5 1 4
 4 4 1 5 2 6 3
 5 5 3 1 6 4 2
 6 6 5 4 3 2 1
 1. Closure property: Since all the entries of the composition table
are the elements of the given set, the set G is closed under 7 .

Contd.,
 2. Associativity: The binary operation 7 is associative in G.
for ex. (2 7 3) 7 4 = 6 7 4 = 3 and
2 7 ( 3 7 4 ) = 2 7 5 = 3
 3. Identity : Here, The first row of the table coincides with the top
row. The element heading that row , i.e., 1 is the identity element.
 4. . Inverse: From the composition table, we see that the inverse
elements of 1, 2, 3, 4. 5 ,6 are 1, 4, 5, 2, 5, 6 respectively.
are identical. Therefore the binary operation 7 is commutative.
 Hence, (G, 7 ) is an abelian group.

More on finite groups
 In a group with 2 elements, each element is its own inverse
 In a group of even order there will be at least one element (other
than identity element) which is its own inverse
 The set G = {0,1,2,3,4,…..m-1} is a group with respect to addition
modulo m.
 The set G = {1,2,3,4,….p-1} is a group with respect to multiplication
modulo p, where p is a prime number.
 Order of an element of a group:
 Let (G, *) be a group. Let ‘a’ be an element of G. The smallest
integer n such that an = e is called order of ‘a’. If no such number
exists then the order is infinite.

Examples
 Ex. G = {1, -1, i, -i } is a group w.r.t multiplication.The order –i is
a) 2 b) 3 c) 4 d) 1
 Ex. Which of the following is not true.
 a) The order of every element of a finite group is finite and is a
divisor of the order of the group.
b) The order of an element of a group is same as that of its inverse.
 c) In the additive group of integers the order of every element
except
 0 is infinite
 d) In the infinite multiplicative group of nonzero rational numbers
the
 order of every element except 1 is infinite.
 Ans. d

Sub groups
 Def. A non empty sub set H of a group (G, *) is a sub group of G,
 if (H, *) is a group.
Note: For any group {G, *}, {e, * } and (G, * ) are trivial sub groups.
 Ex. G = {1, -1, i, -i } is a group w.r.t multiplication.
H1 = { 1, -1 } is a subgroup of G .
H2 = { 1 } is a trivial subgroup of G.
 Ex. ( Z , + ) and (Q , + ) are sub groups of the group (R +).
 Theorem: A non empty sub set H of a group (G, *) is a sub group of G
iff
 i) a * b  H  a, b  H
 ii) a-1  H  a  H

Theorem
 Theorem: A necessary and sufficient condition for a non empty
subset H of a group (G, *) to be a sub group is that
a  H, b  H  a * b-1  H.
 Proof: Case1: Let (G, *) be a group and H is a subgroup of G
Let a,b  H  b-1  H ( since H is is a group)
 a * b-1  H. ( By closure property in H)
 Case2: Let H be a non empty set of a group (G, *).
Let a * b-1  H  a, b  H
 Now, a * a-1  H ( Taking b = a )
 e  H i.e., identity exists in H.
 Now, e  H, a  H  e * a-1  H
 a-1  H

Contd.,
  Each element of H has inverse in H.
Further, a  H, b  H  a  H, b-1  H
 a * (b-1)-1  H.
 a * b  H.
 H is closed w.r.t * .
 Finally, Let a,b,c  H
 a,b,c  G ( since H  G )
 (a * b) * c = a * (b * c)
 * is associative in H
 Hence, H is a subgroup of G.

Ex. Show that the intersection of two sub groups of a group G is again a
sub group of G.
 Proof: Let (G, *) be a group.
 Let H1 and H2 are two sub groups of G.
 Let a , b  H1  H2 .
 Now, a , b  H1  a * b-1  H1 ( Since, H1 is a subgroup of G)
 again, a , b  H2  a * b-1  H2 ( Since, H2 is a subgroup of G)
  a * b-1  H1  H2 .
 Hence, H1  H2 is a subgroup of G .

Ex. Show that the union of two sub groups of a group G need not be
a sub group of G.
 Proof: Let G be an additive group of integers.
 Let H1 = { 0, 2, 4, 6, 8, …..}
 and H2 = { 0, 3, 6, 9, 12, …..}
 Here, H1 and H2 are groups w.r.t addition.
 Further, H1 and H2 are subsets of G.
  H1 and H2 are sub groups of G.
 H1  H2 = { 0, 2, 3, 4, 6, …..}
 Here, H1  H2 is not closed w.r.t addition.
 For ex. 2 , 3  G
 But, 2 + 3 = 5 and 5 does not belongs to H1  H2 .
 Hence, H1  H2 is not a sub group of G.

Homomorphism and Isomorphism.
 Homomorphism : Consider the groups ( G, *) and ( G1, )
A function f : G  G1 is called a homomorphism if
f ( a * b) = f(a)  f (b)
 Isomorphism : If a homomorphism f : G  G1 is a bijection then f is
called isomorphism between G and G1 .
Then we write G  G1

Example
 Ex. Let R be a group of all real numbers under addition and R+ be a
group of all positive real numbers under multiplication. Show that
the mapping f : R  R+ defined by f(x) = 2x for all x  R is an
isomorphism.
 Solution: First, let us show that f is a homomorphism.
 Let a , b  R .
 Now, f(a+b) = 2a+b
 = 2a 2b
 = f(a).f(b)
  f is an homomorphism.
 Next, let us prove that f is a Bijection.

Contd.,
 For any a , b  R, Let, f(a) = f(b)
  2a = 2b
  a = b
  f is one.to-one.
 Next, take any c  R+.
 Then log2 c  R and f (log2 c ) = 2 log2 c = c.
  Every element in R+ has a pre image in R.
 i.e., f is onto.
  f is a bijection.
 Hence, f is an isomorphism.

Example
 Ex. Let R be a group of all real numbers under addition and R+ be a
group of all positive real numbers under multiplication. Show that
the mapping f : R+  R defined by f(x) = log10 x for all x  R is
an isomorphism.
 Solution: First, let us show that f is a homomorphism.
 Let a , b  R+ .
 Now, f(a.b) = log10 (a.b)
 = log10 a + log10 b
 = f(a) + f(b)
  f is an homomorphism.
 Next, let us prove that f is a Bijection.

Contd.,
 For any a , b  R+ , Let, f(a) = f(b)
  log10 a = log10 b
  a = b
  f is one.to-one.
 Next, take any c  R.
 Then 10c  R and f (10c) = log10 10c = c.
  Every element in R has a pre image in R+ .
 i.e., f is onto.
  f is a bijection.
 Hence, f is an isomorphism.

Theorem
 Theorem: Consider the groups ( G1, *) and ( G2, ) with identity
elements e1 and e2 respectively. If f : G1  G2 is a group
homomorphism, then prove that
a) f(e1) = e2
b) f(a-1) = [f(a)]-1
c) If H1 is a sub group of G1 and H2 = f(H1),
then H2 is a sub group of G2.
d) If f is an isomorphism from G1 onto G2,
then f –1 is an isomorphism from G2 onto G1.

Proof
 Proof: a) we have in G2,
e2  f(e1) = f (e1) ( since, e2 is identity in G2)
= f (e1 * e1) ( since, e1 is identity in G1)
= f(e1)  f(e1) ( since f is a homomorphism)
e2 = f(e1) ( By right cancellation law )
 b) For any a  G1, we have
f(a)  f(a-1) = f (a * a-1) = f(e1) = e2
and f(a-1)  f(a) = f (a-1 * a) = f(e1) = e2
 f(a-1) is the inverse of f(a) in G2
i.e., [f(a)]-1 = f(a-1)

Contd.,
 c) H2 = f (H1) is the image of H1 under f; this is a subset of G2.
 Let x , y  H2.
 Then x = f(a) , y = f(b) for some a,b H1
 Since, H1is a subgroup of G1, we have a * b-1  H1.
 Consequently,
 x  y-1 = f(a)  [f(b)]-1
 = f(a)  f(b-1)
 = f (a * b-1) f(H1) = H2
 Hence, H2 is a subgroup of G2.

Contd.,
 d) Since f : G1  G2 is an isomorphism, f is a bijection.
  f –1 : G2  G1 exists and is a bijection.
 Let x, y  G2. Then x  y  G2
 and there exists a, b  G1 such that x = f(a) and y = f(b).
  f –1 (x  y ) = f –1 (f(a)  f(b) )
 = f –1 (f (a* b ) )
 = a * b
 = f –1 (x) * f –1 (y)
 This shows that f –1 : G2  G1 is an homomorphism as well.
  f –1 is an isomorphism.

Cosets
 If H is a sub group of( G, * ) and a  G then the set
Ha = { h * a h  H}is called a right coset of H in G.
Similarly aH = {a * h  h  H}is called a left coset of H is G.
 Note:- 1) Any two left (right) cosets of H in G are either identical or
disjoint.
 2) Let H be a sub group of G. Then the right cosets of H form a
partition of G. i.e., the union of all right cosets of a sub group H is
equal to G.
3) Lagrange’s theorem: The order of each sub group of a finite group
is a divisor of the order of the group.
 4) The order of every element of a finite group is a divisor of the
order of the group.
 5) The converse of the lagrange’s theorem need not be true.

Example
 Ex. If G is a group of order p, where p is a prime number. Then the
number of sub groups of G is
 a) 1 b) 2 c) p – 1 d) p
 Ans. b
 Ex. Prove that every sub group of an abelian group is abelian.
 Solution: Let (G, * ) be a group and H is a sub group of G.
 Let a , b  H
  a , b  G ( Since H is a subgroup of G)
  a * b = b * a ( Since G is an abelian group)
 Hence, H is also abelian.

State and prove Lagrange’s Theorem
 Lagrange’s theorem: The order of each sub group H of a finite
group G is a divisor of the order of the group.
 Proof: Since G is finite group, H is finite.
 Therefore, the number of cosets of H in G is finite.
 Let Ha1,Ha2, …,Har be the distinct right cosets of H in G.
 Then, G = Ha1Ha2 …, Har
 So that O(G) = O(Ha1)+O(Ha2) …+ O(Har).
 But, O(Ha1) = O(Ha2) = ….. = O(Har) = O(H)
  O(G) = O(H)+O(H) …+ O(H). (r terms)
 = r . O(H)
 This shows that O(H) divides O(G).

Hass Diagram, Lattices and
Boolean Algebra

Hasse Diagram
• A Hasse diagram is a graphical
representation of a poset.
• Since a poset is by definition reflexive and
transitive (and antisymmetric), the
graphical representation for a poset can be
compacted.
• For example, why do we need to include
loops at every vertex? Since it’s a poset, it
must have loops there.

Constructing a Hasse Diagram
• Start with the digraph of the partial order.
• Remove the loops at each vertex.
• Remove all edges that must be present
because of the transitivity.
• Arrange each edge so that all arrows point
up.
• Remove all arrowheads.

Hasse Diagram Terminology
• Let (S, ≼) be a poset.
• a is maximal in (S, ≼) if there is no bS such that
a≼b. (top of the Hasse diagram)
• a is minimal in (S, ≼) if there is no bS such that
b≼a. (bottom of the Hasse diagram)

Which elements of the poset ({, 2, 4, 5, 10, 12, 20, 25}, | )
are maximal? Which are minimal?
The Hasse diagram
for this poset shows
that the maximal
elements are:
12, 20, 25
The minimal
elements are:
2, 5

• Let (S, ≼) be a poset.
• a is the greatest element of (S, ≼) if b≼a for all
bS…
– It must be unique
• a is the least element of (S, ≼) if a≼b for all bS.

• Does the poset represented by this Hasse
diagram have a greatest element? If so, what is
it? Does it have a least element? If so, what is it?
b c d
The poset represented by
this Hasse diagram does
not have a greatest element,
because the greatest
element must be unique.
It does have a least
element, a.

d e
c
this Hasse diagram has
neither a greatest element
nor a least element,
because they must be
unique.

d
c
this Hasse diagram does
not have a least element,
because the least element
must be unique.
It does have a greatest
element, d.

d
b c
this Hasse diagram has a
greatest element, d.
It also has a least element,
a.

• Let A be a subset of (S, ≼).
• If uS such that a≼u for all aA, then u is called an upper
bound of A.
• If lS such that l≼a for all aA, then l is called an lower
bound of A.
• If x is an upper bound of A and x≼z whenever z is an
upper bound of A, then x is called the least upper bound
of A.
• If y is a lower bound of A and z≼y whenever z is a lower
bound of A, then y is called the greatest lower bound of
A.

Lattices
• A lattice is a partially ordered set in which
every pair of elements has both a least
upper bound and greatest lower bound.
f
e
c d
b
a
h
e f g
b c d
a

Lattice example
• Are the following three posets lattices?

algebric structure_UNIT 4.ppt

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