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2. ALGEBRAIC STRUCTURES AND
GRAMMARS DEFINITIONS:
Introduction
•
ALGEBRAIC SYSTEM
A System consisting of a set and one or more n-ary operations on the
set is called an algebraic system. An algebraic system is denoted by:
< S. f1,. f 2 , . .. > where S is a non empty set and f1,. f 2 , .. . are operations on S.
If relation on S is also included the operations and relation on the set
define a structure on the elements of S called an algebraic structure.
•
SEMIGROUP
Let S be a non empty set and o be a binary operation on S. The
algebraic system < S, o > is called a semi group if the operation o is associative. In
other words
< S. o > is a semi group if for any x, y. z  S.
(x o y) o z = x o (y o z)
The semi group may or may not have an identity element with
respect to the operation o. If there is an identity element then we have the
following definitions.
•
MONOID
A semi group < M., o > with an identity with respect to the operation
o is called monoid.
The algebraic system < M., o > is called a monoid, if for any x. y, z  M.
(x o y) o z = x o ( y o z ) and there exists an element
e  M such that for any x  M,
eo x = xoe= x
The identity element, if it exists is unique. Monoid is also
represented as < M., o.,e >
If in a semi group (monoid), < S, * >. the operation * is commutative, then
the semi group (monoid) is called commutative.
If in a monoid < M,.* ,e.> every elements invertible. then the monoid is
called a group.
2.1 GROUP
2.1.1 GROUPS
A group is denoted by < G. * >
A group < G, * > is an algebraic system in which the binary relation on * on
a G statisfies three conditions.
1. The operation is associative. That is for any three elements x. y. z  G. we
have
(x * y) * z = x * (y * z)
2. There is an identity element e  G such that x * e = e * x = x for all x  G
3. Every element in G is invertible. That is an element t. s * t = e = t * s is called
the inverse of s and is written as s-1
COMMUTATIVE GROUP
A group < G, * > in which the operation is commutative group of an
abelian group,
Thus a group is a semi-group with an identity and inverses.
2.1.2 ORDER OE A GROUP
The order of a group < G, * > denoted by |G| is the number of elements of G.
when G is finite.
2.1.3 The following remarks are made pertaining to the group < G, * >
• From the associativity of the operation*, it follows that inverse of every element
of G must be unique. The existence of a unique inverse of every element of G. also
guarantees the unique solvability of any equation of the type
The solution is given by
x = a-1* b
a * x = b.
a, b  G.
• Also the existence of inverse of every element implies that the cancellation
property holds i.e.
a*b= a*c => b=c
b * a = c * a => b = c for any a, b, c  G
• A group cannot have a zero element because even' element in a group
is invertible.
• Also a group cannot have any element, which is idempotent except the identity
element. To set this, let us assume that a  G is idempotent. Then
a * a = a and e = a-1 * a
= a - l * (a * a)
= (a -1 * a) * a
=e* a
=a
• Also the existence of the identity guarantees that no rows or columns in the
composition table of < G. * > arc identical.
2.1.4 THEOREMS ON GROUPS
• For any Group < G , * >
a) The identity element is unique
b) The inverse of each element of G is unique
c) If a. b, c  G and ab =ac, then b = c (Left - Cancellation property)
d) If a, b, c  G and ba = ca. then b = c (Right
Cancellation property)
e) If a. b  G, then ( a * b ) -1 = b -1 * a -1
Proof:
a) if e1 , e2 are both identities in G. then
e1 = e1 * e2 = e2
[ (e * a = a * e = a) for a  G ]
b) Let a  G and suppose that b. c are both inverses of a. Then
b=b*e=
b*(a*c)
= ( b * a) * c
=e*c
=c
c) c * a = c * b implies
c -1 *( c * a ) = c -1 *( c * b )
(c -1 * c) * a) = ( c -1 * c) * b)
e*a=e*b
a=b
d) Similarly a * c = b * c implies
(a * c) * c-1 = (b * c) * c-1
a * (c * c -1) = b * (c *c-1) a
*e=b*e
a=b
e) We shall show that b-1 * a-1 has the two properties required of (a*b) -1.
Then, because inverses are unique. (b -1 *a-1) must be (a*b) -1
(a * b) *( b-1 *a-1) = a * (b * b-1) *a-1
= a * e *a-1
= a * a-1 = e
Similary ( b-1 *a-1) *(a*b) =1
[Note how associativity and the meaning of e and inverse all come into play
in this proof]
THEOREM:
Let <G,*> be a finite cyclic group generated by an element a  G. If
G is of order n, that is | a | = n, then an = e. so that
G = {a, a2, a3, .... an = e }
Further mode, n is the least positive integer for which an = e.
Proof: Let us assume that for some positive integer m<n, a m =e. Since G is a
cyclic group, any element of G can be written as a k for some k  I. Now from
Euclid's algorithm, we can write k = mq + r. where q is some integer and 0 ≤ r < m,
this means
ak = amq+r = ( am )q * ar
so that every element of G can be expressed as a r. for some 0 ≤ r < m, implying that
G has at most m distinct elements, that is | G | = m < n which is a contradiction.
Hence a m = e for m < n is not possible. As a next step, we show that the elements a,
a2 , a3,…,
an are all distinct, where a n = e. Assume to the contrary that ai = aj for i
<j .< n. This means that a j-1 = e where j - i < n, which is a again a contradiction.
2.2 CYCLIC GROUP
A group <G,*> is said to be cyclic, if there exists an element a  G such
that every element of G can be written as some power of a. that is a n for some
integer n. The cyclic group is then said to be generated by a or a is a generator of
the group.
A cyclic group is abelian, because for any p. q  G. p= ar, q =as for some
r, s  I and p * q = ar + as = ar+s= as+r = as * ar = q * p.
2.2.2 Theorem
Let <G, * > be a finite cyclic group generated by an element a  G . If
G is of order n, that is |G| = n, then a n = e, so that G = {a. a 2 ....an-1, an = e}
Further more n is the least positive integer for which a n = e.
Proof: Let us assume that for some positive integer m < n. a m = e. Since G is a
cyclic group, any element of G can be written as a' for some k  l. Now from
Euclidean algorithm, we can write k = mq + r, where q is some integer and
0 <= r < m. This means ak = amq+r = (am)q * ar = ar
so that every element of G can be expressed as a r for some o< r < m, thus implying
that G has at most m distinct elements, that is |G| = m< n.
which is a contradiction. Hence a n = e for m<n is not possible. As a next step, we
show that the elements a. a2, a3... .an are all distinct, where an = e. Assume to the
contrary that ai = aj for i<j<n. This means that a j-i = e where j-i<n. which is again a
contradiction. Hence the theorem.
EXERCISES
1. Show that the group G = < Z4, +> is cyclic and that [1] and [3] generate G.
Since the operator "addition modulo + 4" is commutative, the group G is abelian.
Consider [1], we have [1] =[1]; [1]+[l]=[2], [1]+[1] + [l]=[3], [1]+[1]+[1] +[1] = [4]=[0]
.Hence < [1] > is a generator of the group G.
Note [0] and [2] do not generate G.
Hence G = < [1] > = <[3]>
2. Consider the group of complex numbers under multiplication given by
<G= { 1 , - 1 , + i, -i}, *>
show that “i” as well as "-i" generates G.
ie G = < i >, and G = < -i >
3. Proof that any sub group of a cyclic group is cyclic.
4. Consider the multiplicative group < U9= {1, 2, 4, 5, 7,8}. x9>
Show that <U9, x9> is a cyclic group 21 =2. 22 = 4. 23 = 8. 24 = 16 = "'7",
25 = 32 = “5”, 26 = 64= ”1”
.'. The element 2  U9 generates all the elements of U9. Similarly the element 5 also
generates all the elements of U9. Hence < U9. x9> is cyclic a group of order 6.
Hence V9 = < [2] >; V9 = < [5] >
How ever the elements 4 or 7 generates only the subgroup of V9 given by
<S ={1,4,7), x17> of order 3 and the element 8 generates [<8>={1, 8}] the subgroup
<H ={ 1, 8} , x9> of order 2 and the element <1> ={ 1} generates the identity element as a
subgroup of < V9 , x9 >of order 1.
The entire group < V9. x9 > is also a subgroup generated by all the elements of V9 and
is order 6.
Here there is a trivial group of order 1; and proper subgroups of order 2 and 3 and a
group of order 6 also a trivial group (the entire group < V9, x9>).
Hence Lagrange's theorem is satisfied, namely, the order of a subgroup divides the
order of group G. < S = {1, 4, 7}, x9> and < H = {1, 8}, x9} >are also cyclic. See the
Composition table's.
X9
1
4
7
X9
1
8
1
1
4
7
1
1
8
4
4
7
1
8
8
1
7
7
1
4
Also since < V9. x9> is a cyclic group, the subgroups.
2.2 CYCLIC GROUP DEFINITION
A group <a,*> is called cyclic if there is an element x  G such that for each
a  G, a = xn for some n  z.
• As an example consider the group < G,*>, where G = {1,-1.i,-i)
and is multiplication.
Here i1 = i , i 2 = -1, i3 = -i, i4 = 1, so every element of G is a power of i and we
say that i generates G. This is denoted by G =< i >: it is also true that G = < -i >
• Also the group <Z.4 , x4> is cyclic. Here we have multiples instead of
powers.
Nowz 4 ={[0],[l],[2],[3]}
Consider the element [1], we have
1. [1] = [1]. [1]+[1]=2. [1] = [2]. 3. [1]=[3], 4.[1]= [0]
Therefore [1] generates Z 4 ={[0],[l],[2],[3]}
Similarly the element [3]. we have
l . [3] = [3],
2.[3] = [2].
3. [3]= [l], 4.[1]=[0]
Therefore [3] generates Z 4 = {[0],[l].[2],[3]}
Here G = < Z4. +4 > = < [3] > = < [1 ] >
2.2.4
DEFINITION
For any n  Z+ where n>l, if Un = (a  <Zn, +,>| a is a unit} = {a  Z+| l<a<n-1
and gcd (a, n) = 1}, then<un, o>is an abelian group under the (closed) binary
operation of multiplication modulo n.
As an example consider < V 9, x9>
V9= {1, 2, 4, 5, 7. 8}.l is the identity element
2 x9 5= 10 (modulo 9) =1
.'.5 is the inverse of 2 and 7 is the inverse of 4. Also 1.2 = 2.1. etc
.'. < V9, x9 > is an abelian group of order 6.
More over for the multiplicative group V9= {1,2. 4, 5. 7, 8}.
We have 21=2. 22 =4, 23=8, 24=16 (modulo 9)=7
25 =32 (modulo 9) = 5; 26 = 64 ( modulo 9) = 1
So V9 is a cyclic group of order 6.
.Hence V9 = <2> ; Also V9 = <5>
2.2.1 DEFINITION
Given a group G, If a  G. consider the set S={a k|k  z} All the powers of
a,.. ..a-3, a-2, a-1 a° = e, a, a2 a3.... form a subgroups of G called cyclic group
generated by a. This subgroup S of G called the subgroup generated by G.
2.2.5 DEFINITION
If G is group and a  G, the order of a, denoted by O(a) is |<a>|
For the group G of complex numbers {l,-l, i, -i} under multiplication.
O(1)=1, O(-1) =2, O(i)= 4, O(-i) = 4
{1} is the subgroup of G generated by <l>and O(l)= |<1>| = 1, {1,-1} is the
subgroup of G generated by <-l> and O(-l) = |<-1>|
{l,-l,i,-i} is the trivial subgroup generated by <i>
O<-i> and O(i)=|<i>|=4 and O(-i)= |<-i>|= 4.
2.2.6
THEOREM
Let G be a cyclic group. If |a|= n. where n>l, then G is isomorphic to
<Zn, + 9 >
Proof:
If G = <a> = {a, a2,a3,…, an-1, an = e } then the function f:G—>Zn defined by
k
f(a ) =[k] is an isomorphism.
2.2.7
THEOREM
Any subgroup of a cyclic group is cyclic. Verify that <Zp, x p> is cyclic for the
primes 5, 7 and 11.
Z5 ={[1], [2], [3], [4]}
1. [2] =[2], 2. [2]=[4],
3. [2]= [1],
4. [2] = [3]
Therefore [2] generates or is a generator of the group <Z 5, x5>
Similarly 1. [3] =[3], 2. [3]=[1],
3. [3]= [4],
4. [3] = [2]
Therefore [3] generates or is a generator of the group <Z 5, x5>
<Z5, x5> where Z5 =<2> =<3> is a cyclic group.
Also Z 7 =<3>=<5> and Z 11 =<2>=<6>=<7>=<8>
EX: Find all the generator of <Z 12 , + 12 >, <Z 16 , + 16 >, <Z 24 , + 24 >
Ans: <Z 12 , + 12 >=<1>=<5>=<7>=<11>
<Z 16 , + 16 >= <1>=<3>=<5>=<7>=<9>=<11>=<13>=<15>
<Z 24 , + 24 >=<1>=<5>=<7>=<11>=<13>=<17>=<19>=<23>
2.3 PERMUTATION GROUP:
DEFINITION:
Any one-to-one and onto (bijection) mapping of a set S onto S is called
a permutation of S.
THEOREM:
(a) Every row or column is composition table of a group <G. *> is a
permutation of the elements of G.
(b) Sn :sets of permutations of the elements of a group <G, *> with |G| = n
and with a binary operation 0 on S n, representing the right composition of a
permutations is called the permutation group denoted by < S n . 0>, also called the
symmetric group of order n! and degree n.
The dihedral groups <D n, 0> are obtained from the symmetric permutation
group <S n . 0> by considering the symmetries of regular polygons of n sides.
With n = 3, 4, 5,…., n under the composition  . Dn is of order 2n and of a permutation
group of degree n.
Since <Dn,  > is a permutation group of degree, it can be shown that <Dn,  > is
sub group of <Sn,  >.
For n=3, the order of both <S3,  > and <D3,  > are the same (3! =6” for S3 and
2 x 3 = 6 for D3) and of degree 3.
For n = 4, D4 is order of 4! and D4 is of order 2 x 4=8
Considering an equilateral triangle shown here with no rotation 120○ and 240○ rotations
commencing with vertex1, in the counter clockwise
direction.
2
2
A
C
o
1
3
B
About the center O, perpendicular to the plane of the triangle and with reflections about
the central axis 3C, 1A, 1B respectively. We get 6 permutations  0 ,  1 ,  2 ,V1 ,V2 ,V3
respectively.
The set of rigid motions (in space) of the equilateral triangle forms the dihedral group
<D3,  > using the right composition for permutations.
The six permutations are respectively as follows.
1 2 3
0 = 

1 2 3
1 2 3

2 1 3
V1 = 
1 2 3

3 1 2
1 = 
1 2 3

1 3 2
V2 =
1 2 3

2 3 1
2=
1 2 3

3 2 1
V3=
The following table gives the dihedral group <D3,  > of the order 6 and degree 3

1
1
2
0
2
2
0
1
V1
V2
V3
0
1
2
0
0
1
2
V1
V2
V3
V3
V1
V2
V2
V3
V1
V1
V1
V2
V3
V2
V2
V3
V1
V3
V3
V1
V2
0
2
1
1
0
2
2
1
0
EXAMPLE:
Under this binary operation  composition of permutation <D3,  > is a group with
identity  0 .Moreover it is closed, since
 1
1 2 3
1 2 3 1 2 3
 

=
= V3
3 1 2
2 1 3 3 2 1
V1= 
Also
V1 

1 
V1
 1 
1=
1 2 3
1 2 3 1 2 3
2 1 3  3 1 2 = 1 3 2 = V 2



 

 V1   1
1=
1 2 3
3 1 2


Hence <D3,  > is not an abelian, Moreover
1 2 3 1 2 3
2 3 1 1 2 3 = 



0
Hence the inverse of inverse of  1 is  2 and the inverse of  2 is  1 or
 11   2 ,  21   1
Every other element is it’s own inverse
1 2 3 1 2 3 1 2 3
V1  V1= 
 
=
 =
2 1 3 2 1 3 1 2 3
Hence V11  V1 similarly V21  V2 , V31  V3
0
The associative property follows the association property of the composition functions in
the reverse order.
V1  V2  V3 = V3  V2  V1
We shall obtain the dihedral group <D4,  > by considering the symmetries of a square
shown here .There are 4 notations about the point O, the center of the square given by
A
1
B
4
O
2
B
3
A
1 2 3 4 
1 2 3 4
1 2 3 4
1 2 3 4
V2 =
V3=
V4 =




2 3 4 1 
3 4 1 2
1 2 3 4
1 2 3 4
V1= 
which are obtained by rotating the square counter clockwise through 90, 180, 270and
other multiples of 90  (also include clockwise rotations of the same type). Other rigid
rotations are obtained by rotating the square about the lines AA, BB and the diagonals 13
and 24. Note that these rotations are reflections and are given by,
V5=
1 2 3 4 
1 2 3 4 
1 2 3 4
4 3 2 1 V 6 = 2 1 4 3 V 7 = 1 4 3 2






D4 = { V1, V2, V3, V4, V5, V6, V7, V8}
1 2 3 4

3 2 1 4
V8 = 

v1
v2
v3
v4
v5
v6
v7
v8
v1
v2
v3
v4
v1
v8
v7
v5
v6
v2
v3
v4
v1
v2
v6
v5
v8
v7
v3
v4
v1
v2
v3
v7
v8
v6
v5
v4
v1
v2
v3
v4
v5
v6
v7
v8
v5
v7
v6
v8
v5
v4
V2
v1
v3
v6
v8
v5
v7
v6
v2
v4
v3
v1
v7
v6
v8
v5
v7
v3
v1
v4
v2
v8
v5
v7
v6
v8
v1
v3
v2
v4
D4 is given in the table and <D4,  > is the group of order 8 and permutation group of
degree 4.
EXAMPLES:
1. Find the group of rigid rotation of the rectangle, which is not a square. Show that this
is the sub group of dihedral group <D4,  > of the square consisting of the set of rotation
of the square given by
D4 = { r1, r2, r3, r4, r5, r6, r7, r8}
2. Show that <1, 4, 13, 16> is a sub group of <Z4, x17 >
3. For the dihedral group of <D4,  > of order 8.Show that each of the elements V2, V3,
V5, V7 and V8 , generates 2 elements sub groups < {V2, V4},  > and < {V3, V4},  >,
< {V5, V4},  > , < {V7, V4},  > <{V8, V4},  > and that V1 and V3 are both
generators of the 4 elements sub groups < {V1, V2, V3, V4},  >, besides the trivial
subgroups of < V4,  > and < {V1, V2,V3, V4, …, V8},  >
NOTE:
1. From Lagrange’s theorem, the dihedral group <D4,  > of order 8 has sub
groups of order 1, 2 , 4 and 8 which are divisions of 8.
2. 8 is not the prime and hence the group of order 8 has divisions other than 1 and
as proper sub groups.
3. Given the group 4
<S5,  > , let
1 2 3 4 5

2 3 1 4 5
1 2 3 4 5

2 3 1 4 5
 
Determine
 
   ,    ,  -1 ,  -1 ,  3,  3 , (    )-1 and  -1   -1.
Let  and  be the following elements of the symmetric group S6 .
1
2
3
4
5
3 1
5
4
6
 =
Find
6

2
1 2 3 4 5 6

5 3 1 6 2 4
 
  ,   ,  2,  -1
There are 8 symmetries as follows
For 0 0 , 9 0 0 , 18 0 0 , 27 0 0 , .Let  (  ) be the symmetry obtained by rotating s about its
center  degrees and let  (  )be the symmetry obtained by reflecting s about the y
axis and then rotating s about it’s center  degrees. Any symmetry of s is completely
determined by it’s effect on the vertices of s and hence  can be represented as
permutation in s4.They are
1 2 3 4

1 2 3 4
 (0 0 ) = 
1 2 3 4 

 2 1 4 3
 0 0   
 90 0   

2 3 4 1 
1 2 3 4 
 90 0 = 
1 2 3 4

3 4 1 2
 180 0 = 
1 2 3 4 

 4 1 2 3
 270 0  = 
 180 0  = 
 270 0 = 
5.
1 2 3 4

3 2 1 4
1 2 3 4 

4 3 2 1 
1 2 3 4

1 4 3 2
For the equilateral triangle with three rotations and three reflections about the attitudes,
let G ={  0 ,  1 ,  2 , V1, V2, V2} be the set of regid notations in space of the equilateral
triangle , the group <G,  > is the group of permutations of the set {1, 2, 3}.The group G
is closed under the binary operation  with identity  0 Also  1   2 =  0
 21
=
 11 =  2
 1 and every other element is its own inverse;
Since V1  V1 =  0 =  2   2 = V3  V3
Moreover  1  V1 =V3 and V1   2 =V2 with

1
1
2
0
2
2
0
1
V1
V2
V2
0
1
2
0
0
1
2
V1
V2
V3
V3
V1
V2
V2
V3
V1
V1
V1
V2
V3
V2
V2
V3
V1
V3
V3
V1
V2
0
2
1
1
0
2
2
1
0
1 
V1 =V3
 V2  V1 1
It follows that G is abelian. This is denoted by <D3,  > called the dihedral group of
order 6 and degree3.
This group can also be obtained as the group of all permutations of the set S3 ={1, 2, 3}
denoted by <S3,  > under the operation of function composition . This group
<S3,  > is the symmetric group of order 3.
Definition
If H is a subgroup of G then for any a  G the set aH ={ ah / h  H}is called the left
coset in G, the set Ha ={ha / h  H} is the right coset of H in G.
Permutation Group
1
A one to one mapping P = 
 j1
2
3
...
j2
j3 ...
n
onto itself is called a permutation as
j n 
denoted by P where j i =p(i).The set of all permutations is denoted by Sn and there are
n!=1,2,3…n of them. The composition and the inverse of permutations in Sn belongs to
Sn and the identity function “  ” belongs to Sn. Thus Sn forms a group under
composition of function with  as a binary operation on Sn representing the (right)
composition of permutations and  as a left composition of the function related by
P i  P j = P j  P i for i, j=1, 2,3….n
Accordingly | P i  P j | (ar) = | P i  P j | (ar) = P j Pi (a r ) and similarly for other
elements of the set S={a 1 ,a 2 ,….a n }.
This group< Sn,  >is called the symmetric group of order n! and of a degree n.
The symmetric group of n=3.With S =1, 2, 3 given by <S 3 ,  > has 3!=6 elements as
follows
1 2 3
e= 
 = 0;
1 2 3
 1= 
;
3 1 2
1 2 3
 2 =

2 3 1
1 2 3
1 2 3
v1= 
;
2 1 3
1 2 3
v2 =
;
1 3 2
1 2 3
v3=

3 2 1
Similarly considering the symmetries of (regular polygons of n sides) an equilateral
triangle whose vertices are denoted by 1, 2, 3 by taking all the possible notations and
reflections which leave the final position of the triangle unchanged from it’s original
position except for renaming the vertices. We get certain other group for general n,
known as  dihedral group <Dn,>,where Dn is the order of  degree n, for n=  <S 3 ,>
and <D 3 ,> the order is both 6 and degree 3. But for n = 4, D 4 is of order 8 and S 4 is of
order 4!=16.But both are of degree 4.
2.4 SUBGROUPS
DEFINITION:
Let <G. *> be a group and s  G be such that it satisfies the following conditions:
(1) e  S .where e is the identity of <G. *>
(2) For any a  s, a-1  s
(3) For a, b  s, a * b  s Then <s, *> is called a subgroup of <G, *>
THEOREMS
(a) A subset S   is a sub group of <G, *> if for any pair of elements a, b  S.
a* b - 1  S
Proof:
Assuming that S is a subgroup it is clear that if a, b  S, then b - 1  S and
a* b - 1  S . To prove the converse, let us assume that a, b  S and a* b - 1  S for
any pair a, b.
Taking b =a. a*a-1 = e S .From e, a, b S, we have e*a-1 = a -1= S. similarly
b - 1  S. Finally because a and b - 1 are in S, we have a* b  S. hence <S. *> is a subgroup
of <G, *>.
(b) If <G, *> is a group and   S  G , with S finite then S is a subgroup of
G if and only if S is closed under the binary operation * of G
Proof:
If S is a subgroup of a G. then S is closed under the binary operation of G.
[due to the theorem (a)].
Conversely, let S be a finite nonempty subset of G that is closed, if a  S. then
S={as /s  S}  S because of the closure condition.
By cancellation in G, as1= a s2 => s1 = s2.
So |aS| = |S| with aS  S and |aS | = |S|, it follows from S being finite that aS= S.
As a  S there exists b  S with ab = a. But (in G) a b = a = a e.
So b= e and S contains the identity.
Since e  S = a. there is an element e  S much that a e = e
Then (ea) 2 = ea * ea = e*(a*e)*a ; e*a=e and e = a-1
Consequently S is a subgroup of G.
(c)
If S is a non empty subset of a group G. then S is a subgroup of G if and only if (1)
For all a, b.  S, ab  S and (2), for all a  S,
a -1  S
Proof:
If S is a subset group of G. then by definition S. then by definition. S is a group
under the same binary operation. Hence it satisfies all the group conditions, including the
two mentioned here.
Currently let   S  G with S satisfying the condition (1) and(2).
For all a, b, c  S, (ab)c = a(bc) in G. so (ab)c= a(bc) in S, [S inherits the
associative property from G]. Finally G, as S   . let G  S by condition (2) a-1  S and
by condition (1) a. a-1 =e  S so S contains the Identity element and is a group.
DEFINITION:
Let <H, * > be a subgroup of <G. * >, for any for any a e G, the set aH defined by
aH= (a*h/h  H)
is called the left cosset of H in G determined by the element a  G. The element a is
called the representative element of the left cosset aH. Thus set Ha= {h*a/h  H} is the
right coset of H in G.
(d) THEOREM
Let <H, * > be a subgroup of <G. * >. The set of left cosets of H in G form
a partition G, Every element of G belongs to one and only left coset of H in G.
Define a left coset relation modulo H, denoted by the symbol  such that for
a, b  G a  b or more precisely a  b (mod H) if b -1 * a  H We shall show
that this relation is an equivalence relation.
Since H is a subset of G. ( eG  H. For any a  G
Therefore a

a-1 *a = eG  H
b (mod H). Also if b-1*a  H, then (b -1 * a)-1 = (a-1 * b)  H.
implying that a  b (mod H)
then b  a (mod H). Similarly we can see that if b -1 * a  H and e-1*a H, then
e-1*a =( e-1* b) (b-1*a)  H, implying that from a
have a


b (mod H) and b  c (mod H) we
c (mod H).
Hence a left coset modulo H relation in an equivalence relation on G. obviously
the left coset equivalence relation defined here partitions G into equivalence classes.
For any a  H denote the equivalence class containing G by [a] that is
[a] ={x  G /x  a(modH)}
={x  G /a"'*x  H) "
={x  G h  H) where a-1 * x =h say or x-1= a * h
={a* h  G| h  H}
={a*h| h  H)
This is the definition of a left coset.
Hence the set of left coset of H in G form a partition of G. As a is an arbitrary
element of G, every element of G belongs to one and only one left coset of H in G.
R
Similarly defining a right cosets equivalence relation by a
the right cosets of H in G again partition G.

b(mod H). if a * b-1  H.
(e)
Let H be a finite subgroup of G. show that H and left coset a * H (or the right
coset H * a) have the same number of elements.
Proof:
Let H = {h1, h2, h3,…hk] where H has k elements. Then
aH = {a*h1, a*h2 ........ a*hk}. However a*hi = a* hj implies hj = hi, hence the k elements
listed in a*H are distinct. Thus H and aH have the same number of elements │H│= aH
Similarly │H│= Ha .
We have already seen that the left or right coset relation defined by a subgroup is an
equivalence relation.
(f) We shall find conditions on the subgroup much that the left coset relation
determined by it is a congruence relation [once a congruence relation is a
available, the existence of a homomorphism follows].
Let <H, *> be a subgroup of <G. *> and a ,b, p, q  G .G be such that
a

p(mod H) and b

q(mod H) ..... (1)
We wish to determine whether condition (1) implies
(a*b)

(p *q)(mod H)
……
(2)
If relation (2) is satisfied for any a, b, p q  G for which (1) holds, then the left
coset relation determined by H in a congruence relation on G.
Condition (1) can be written as
p-1 * a  H and q-1 * b  H
Let us denote b-1 *a = h1 and q-1 * b= h2 for some h1, h2 H. In a similar manner (2) can be
written as
(p*q)-1(a*b)  H
or
q(q-1 * p-1) * (a * b)  H
q-1*(p-1*a) *b  H,
ie q-1 *h1* b  H
……(3)
Condition (3) is clearly satisfied if h1* b = b* h3
when h3 is some element of H. In that case, q-1 *h1* b = q-1b * h3=h2 * h2  H
We can restate our last result by saying that a left coset relation determined by H is a
congruence relation provided that for any a  G and hi  H. there exists an element
h2  H
such that h1* a = a * h2
………..(4)
or equivally a-1 * hi*a  H
……..(4a)
Condition (4) or (4a) is automatically satisfied if G is an abelian group .
This means for an abelian group the left coset relation is a congruence relation no
matter what subgroup is used to generate the left cosets.
(g) In addition, we shall show also that conditions (4) or (4a) are equivalent
to the condition.
aH = Ha
….(4b)
When aH and Ha are left and right coset respectively of H in G determined by an
element a  G. If we assume (4b) and choose an element h2 H
we have a*h2  H and (4b) says then a * h2  Ha that is, there exists an element in H.
say h1 such that a*h2 = h1* a.
Hence (4b) =>(4)
Let us assume (4) and consider the different values of h1  H. then h1*a all the element of
Ha and from (4) we have Ha  Ha. because a * h2 or  H.
By considering a-1 in place in (4), we can see that (4) implies aH  HG.
Hence
aH = Ha and (4)
=>(4b).
2.5 Group Isomorphism:
Let < G, * > and < H,  > be two groups. A mapping g : G  H is called
a group homomorphism from < G, * > to < H,  > if for any a, b  G
g(a*b)=g(a)  g(b)
(1)
assumes that a group homomorphism preserves identities, inverses and
subgroups, If eG and eH are the identities of < G, * > and < H,  > respectively,
then
g(eG) = eH
(2)
Also for any a  G ,
g(a-1)=[g(a)]-1
(3)
If < s , k > is a subgroup of < G, * > and g(s) denotes the image set of s
under the homomorphism, then < g(s),  ) is a subgroup of < H,  ) We have
g(eG * eH) = g(eG) = g (eG)  g(eH).
so that g(eG) is idempotent. But the only idempotent element of a group is the
identify, hence g(eG) = eH.
Also for any a  G. a-1  G and hence
g(a * a-1 ) = g (eG) = eH = g(a)  g(a-1)
g(a-1 * a) = g(eG) = eH = g(a-1)  g(a)
which implies g(a-1) =:[g(a)] -1.
Finally, let < S. * > be a subgroup of <G, *>
Therefore g(eg) = eH  g(S).
For any a  S, a-1  S and g(a) as well as g(a -1) = [g(a)]-1 are in g(S).
Also for a , b  S, a * b  S
and consequently for g(a), g(b)  g(S),
g(a)  g(b)  g(S).
This guarantees that <g(S),  > is a subgroup of <H,  >. because every element of g(S)
can be written as g(b) for some b  S.
2.5.2 DEFINITION
A group homomorphism g is called monomorphism, epimorphism or
isomorphism depending upon whether g is one-to-one, onto or one-to-one and onto,
respectively.
A homomorphism from a group < G. * > to < G, * > is called endomorphism,
while an isomorphism of < G, * > to < G, * > is called an isomorphism.
Let g be group homomorphism from < G, * > to < H,  >. The set of elements of
G which are mapped into H , the identity of H is called the kernel of the homomorphism
of g and denoted by ker(g).
DEFINITION:
A one-to-one mapping of a set S onto S is called a permutation of S.
THEOREM:
Every row or column in the Composition table of a group < G. * > is a
permutation of the elements of G.
Proof : Firstly, we shall show that no row of column in the composition table can have
an element of G more than once.
Assume to the contrary that the row corresponding to the element a  G has two
entries which are both k, that is assume that
a * b 1= a * b1+ 2 = k.
b 1 , b2, k  G and b1  b2
From the cancellation property we have b1= b2
which is a contradiction.
A similar result holds for any column.
Secondly, we shall show that every elements of G appears in each row and
column of the table composition.
Again consider the row corresponding to the element a  G and let b be any
element of G. Since b = a * (a-1 * b ), b must appear in the row corresponding to the
element a  G.
A similar argument holds for every column of the table of composition. From the
above result and the fact that no two rows or columns are identical, it follows that every
row of the composition table is obtained by a permutation of the elements of G and that
each row is a distinct permutation.
The same result applies to the columns of the composition table.
THEOREM:
Every finite group of order n is isomorphic to a permutation group of degree n.
Proof: Let < G, * > be a group of order n. ft is shown that every row and column in the
composition table of < G, * > represents a permutation of the elements of G.
Corresponding to an element a  G. let pa denote the permutation given by the
column under a in the composition table. Thus
pa(c) = c * a for any c  G.
For every column we can define permutations of the elements of G. Let the set
of permutations be denoted by P. Obviously, P has n elements.
We shall show that < P, 0 > is a group where 0 denotes the right
composition of the permutations of P. Note that since e  G, p e  P. and
pe 0 pa = pa 0 Pe= pa for any a  G.
Also for any a  G, pa-1 0 pa = pe
Also for a, b  G, pa 0 pb = pa* b
(4)
This Equation follows from the fact that for any element c  G. pa (c)= c*a
so that ( p a 0 pb) (c) = p b0 pa (c), pb(pa (c)) = p b ( c * a ) = c * a * b
T
= c * ( a * b ) = pa*b (c)
Hence < P. 0 > is a group, the last step is sufficient to guarantee that < P. 0 > is a
group, because it shows that < P. 0 > is isomorphic to < G, * >
Consider a mapping f : G —> p given by f(a)= pa for any a  G. Natural f is oneto-one onto. Equation a, b  G, p a 0 pb= p a*b can be written as
f (a * b ) = f(a) * f(b) showing that f is an isomorphism.
2.5.3 SIMPLE EXAMPLES IN GROUPS:
•
The set {-1, 1} under multiplication is a group, the identity is 1 and each
element in its own inverse.
•
The set Z of integers is not a group under subtraction, since subtraction is not
an association binary operation for Z [(3-2)-4 = - 3  5 = 3-(2-4)]
•
Define the binary operation o or z by x o y = x+y+1 .Show that < z, o > is
an abelian group
•
For x, y  Z,
x + y+1 Z :.x o y = x+y+1 is a closed binary operation for z.
Also for any w, x, y  Z
wo(xoy)= wo [x+y+1] = w +(x+y+l)+l
=(w + x + l)+y+1 = [w o x] o y.
Hence the binary operation o is associative.
x o y = x+y+1 = y+x+1 = y o x for all x, y  z. So “o” is also commutative.
If x  Z. then x o (-1) = x + (-1) +1 = x[(-l) o x ]
[So -1 is the identity element for o]
Finally, for any x  z, we have -x-2  z and
x o (-x -2 ) = x + (-x -2 ) + 1 = -1 [= (-x -2 ) o x ].
[So -x-2 is the inverse for x under 0. Consequently < z. 0 > is an
abelian group.]
•
The set z of integers is an abelian group under addition. The identity element
is o and -a is two addictive universal a in z.
•
< z* = z - {o}; > is not a subgroup of Z
•
The nonzero rational numbers q / (o) form an abelian group under
multiplication. The number 1 is the identity element and q / p is the multiplication
inverse of the rational numbers p / q.
•
< Zm. +m >, [where Zm is the integer | or w congruent modulo j m given by
{ [o], [1], ... [m-l] } ] is a group under addition, but it is not a group under
multiplication.
As an example < Z12, + 12 > is a group under addition but it is not a group
under multiplication.
However, Let Um denote a reduced residue system modulo m which consists
of those integers relatively prime to m.
Then U m is a group under multiplication (modulo m)
As an example U 1 2 = {3, 5, 7. 11},
< U12, x 1 2 > is a group under multiplication x 12.
The composition table for < U12, x 1 2 > is given by
(i)
x12
1
5
7
11
1
1
5
7
11
5
5
1
11
7
7
7
11
1
5
11
11
7
5
1
Not that < U12. x 1 2 > is an abelian group, (a x12 = bx12 a for
a, b  U12 (ii)
element of U12 can
Also < U12, x12 > is not a cyclic group, because no
be represented in terms of the powers of any element
of U 12 , since each element in U12 is a prime number.
2.6 LAGRANGE'S THEOREM
(h) If G is a finite group of order n with H a subgroup of order m,
then m divides n
Proof:
If H = G the result follows. Otherwise m<n and there exists an element a
 G . Since a  H, it follows that
a H  H, so aH  H= φ . If G = aH  H, then
│G│ = │a H│ +│H│ = 2│H│ and the theorem follows.
If there is an element b  G-(H  a H) with bH  ( H  a H )= φ .
If G = bH  aH  H
we have | a | = 3 | H |.
Otherwise we choose an element c  G with c  b H  aH  H. The group G is
finite, so this process terminates and we find that
G = a l H  a 2 H…  ak H.
Therefore | G |= k | H | and m divides n.
Examples:
(i). Let G =<Z6, +> If (l) H = {0, 2, 4} then H is a non-empty subset of G. <H, +> is also a
group under the binary operation "+" of G, as shown the truth table.
+
0
2
4
0
0
2
4
2
2
4
0
4
4
0
2
Such a non empty subset H  G is called a subgroup under the same binary
operation "+" of G.
In general every group< G, * > has {e}. G as subgroups and they are the trivial
subgroups of G. The others are called proper subgroups of G.
(j). Let G = <Z6,+> be the groups chosen. The non-empty subsets. H={0, 2, 4}
(k). Find all the distinct subgroup of <Zn.+ 12>
[Ans:<{0},+ 12>; < Z12, + 12>: <{0,2,4,6,8,10} + 12>:<{0,4,8},+ 1 2 >
<{0,3,6,9), +12 >. <{0,6},.+
12>]
(1). Let < G, * > be a group and let< S, * >and < T. * > be subgroup of < G, * >
Show that < S  T, * > is a subgroup of < G, * >, while < S  T, * > is not a
subgroup < G, * >
[Ans :for the set S  T  G, establish the closure existence of identity and inverse.
In < Z12,+ 12 >
<{0,4,8} + 12>and <{0.6},+ 12>are subgroups of < Z12. + 12 >;
but <{0, 4, 6,8,+12 > is not a subgroup of < Z12,+ 12 >. as the union set is not closed]
(m). Find all the subgroups of <Z5 , +5 >
|Z5| =5
.'.by Lagrange's theorem, the order of a subgroup divides 5. Hence the
subgroup of G must be of order 1 or 5. The possible subgroups <z 5, +5>
are (1) <{0}.+ 5 >and (2) <Z 5 + 5 >.
2.7 GRAMMAR AND LANGUAGES
A language L is a set of strings or sentences over some finite alphabet VT
A language consists of a finite set or infinite set of sentences. Any device which
specifies a language should be finite.
One method of specification which satisfies this requirement uses a generation
device called a grammar.
A grammar consists of a finite set of rules or productions which specify the syntax
of the language. A grammar also implies a structure on the sentences of a language. The
study of grammars constitutes an important sub area of computer science called formal
languages (Noam chomsky)
A second method of language application is to have a machine, called an accepts
which determines whether a given sentence belongs to the language. There are very
interesting and important relationships that exist between grammars and acceptors.
A sentence in English has a structure described in terms of subject, phrase,
predicate noun and so on.
For a program, the structure is given in terms of procedures, statements,
expressions, etc.
The symbol → separates the syntactic class “sentence” from its definition.
The syntactic class and the arrow symbol along with the interpretation of a
production enable us to describe a language.
A system or language which describes another language is known as a meta
language.
The diagram of a sentence describes its syntax but not its meaning or semantic.
The device designed to give the syntactic definition of the language is called a
grammar.
2.7.1. Formal definition of a language
VT: is finite not empty set of symbols called alphabet; terminal symbols x, y, z…
The meta language which is used to generate strings in the language is assumed to
contain a set of syntactic classes or variables called non terminal symbols: VN, capital
letters.
The elements of VN are used to define the syntax (structure) of the language.
VN and VT are assumed to be disjoint.
VT  VN is called the vocabulary of the language: S1, S2, ….. are elements of the
vocabulary.
Strings of symbols over the vocabulary are given by α, β, γ….. The length of the
string α will be denoted by | α | .
2.7.2 GRAMMAR
A (phrase structure) grammar is defined by a 4-tuple G = <VN, VT, S, φ> where VN and
VT are sets of terminal and non terminal(syntactic class) symbols respectively: S is a
distinguished element of VN and therefore the vocabulary is called the starting symbol: φ
is a finite subset of the relation from (VN  VT)* to ( VN  VT)*
In general an element <  , 
 is written as    and is called a production rule or
rewriting rule.
G= <VN, VT, S, φ> in which VN = {I, L, D}
VT={a, b, c, …, w, x, y, z, 0,1, 2,…9}
S=I
Φ = {I  L, I  IL, I  ID, L  a, L  b, …L  z, D  0, D  1, …, D  9}
Definition:
Let G = <VN, VT, S, φ> be a grammar. For σ,  ( VN  VT)*, σ is set to be a
direct derivation of  written as   σ,if there are strings 1 and  2 (including
possibly empty strings) such that,  = 1   2 and σ = 1   2 and    is a
production of G.
If   σ, we may also that  directly produces σ or σ directly reduces to .
Definition: Let G = <VN, VT, S, φ> be a grammar. The string  produces σ, written as

  σ, if there are strings  0 , 1 , …  n (>0) such that
 =  0  1 , 1   2 …  n1   n and  n = σ.
The relation  is the transitive closure of the relation  .
If we let n=0 then we can define the reflective transitive closure of  as.

*
     or  
We shall show that the string abc12 is derived from I by the following derivative
sequence:
I  ID  IDD  ILDD  ILLDD
 LLLDD  aLLDD  abLDD
 abcDD  abcID  abc12
As long as we have a non terminal character in the string, we can produce a new
string from it.
On the other hand, if a string contains only terminal symbols, then the derivation
is complete and we cannot produce any further string from it.
Definition: A sentential form is any derivation of the unique non-terminal symbol S.
The language L generated by a grammar G is the set of all sentential forms whose
symbols are terminals
L(G)= {σ / S σ and σ є VT}
Therefore, the language is merely a subset of the set of all terminal strings over VT.
Ex.
Let G2 = <{E, T, F}, {a, +, *, (1)}, E,  >
When  consists of production.
E→E+T
E→T
T→ T*F
T→ F
F→ (E)
F→ a
Where the variables E, T and F represents the names “expression”, “term” and “factor”
commonly used in conjunction with arithmetic expressions.
A derivation for the expression a*a + a is
E E + T T + T  T * F + T
F*F+Ta*F+T
 a * a + Ta * F + T
a*a+T
a*a+F
a*a+a
Ex.
The language L (G3) = {an bn cn / n1} is generated by the following grammar
G3 = <{S, B, C}, {a, b, c}, S,  >
When  consists of production.
S→ a S B C2, S→ a B C3, CB→BC
a B → a b, b B → b b ; b C→ b c, c C→ c C
SaSBC
aa BCBC
aaBBcC
aabBcC
aabbCC
aabbcC
aabbcc
 a2 b2 c2
In general,
San-1 S (BC)n-1
 an (Bc)n
 an Bn Cn
 an b Bn-1 Cn
 an bn Cn
 an bn c Cn-1
 an bn c cn-1
 an bn cn
EX: The language L (G4) = {an b am / n1} is generated by a grammar.
G4 = <{S, C}, {a, b}, S,  >
When  consists of production.
S→ a C a
c→ a C a
c→ b
A derivation for a2 b a2 consists of the following steps
S a C a  a a C a.a
aaCaa
 a a b a.a  a2 b a2
Ex.
The language L (G5) = {an b am n, m1} is generated by the following grammar
G5 =<{S, A, B, C}, {a, b}, S,  > , Where the set of production is
S→ a S , S→ a B, B→ b c, C → a C, C→ a
The sentence a2 b a3 has the following derivation.
S a S  a a B  a a b c
a abaa abaac
 a a b a a a  a2 b a3
Noam Chomsky classified grammars into four classes by imposing restrictions in the
productions.
Definitions:
A context sensitive grammar contains only productions of the form    where
   (the word  is at least as long as  )
This restriction on a production prevents  from being empty. Because of this
restriction, the set of sentences generated by a grammar is recursively solvable. The
restriction in the productions of a context sensitive grammar can be equivalently stated as
follows:
 and  in the production    can be expressed as   1   2 and
  1  2 (1 and / or  2 are possibly empty ) where  must be non empty.
The meaning of “context sensitive” becomes clearer with this reformulation.
The application of a production 1  2  1  2 to a sentential form means that A is
rewritten as  in the context 1 and  2 .
Context sensitive grammars are said to generate context sensitive languages.
[G3 is an example of a context sensitive grammar]
We now impose a further restriction on production to obtain a context-free grammar.
Definition:
A context free grammar contains productions of only the form    where
   and  VN ( is a single non termin al)
With such grammars do not have the power to represent even significant parts of the
English language, since context dependency is often required in order to properly analyze
the structure of a sentence context free grammars are not capable of specifying (or
determining) that a certain variable was declared when it is used in some expression in a
subsequent statement of a source program.
However these grammars can specify most of the syntax for computer or artificial
 languages, since these are by and large, simple in structure.
Context free grammars are said to generate context free languages.
Grammars G1, G2 and G4 are examples of context free grammars.
A final restriction leads to the definition of regular grammars.
Definition:
A regular grammars contains only productions of the form    where    ,
  VN and  has the form a B or a where a V T and  has the form a B or a where a
 V T and B  VN.
The set of languages generated by such grammars is said to be regular.
G5 is an example of a regular grammar.
Phase structure type:
Let the unrestricted, context sensitive, context free and regular grammars, denoted by the
class symbols T0, T1, T2 and T3 respectively. If L(Ti) represents the class of languages
that can be generated by the class Ti grammar, it can be show that
L(T3)  (L(T2)  (L(T1)  (L(T0) and therefore the four classes of grammar form a
hierarchy. Corresponding to each class of grammars, there is a class of machines
(acceptors) that will accept the class of languages generated by the former.
The meta variables are synthetic classes will be enclosed by the symbols < and >.
Using this terminology, the symbol.
<sentence> is a symbol of VN and the symbol “sentence” is an element of VT. In this way
no confusion or ambiguity, when attempting to distinguish the two symbols.
This meta language is known as “BACKUS NAUR FORM” (BNF) and has been used %
extensively on the formal definition of many programming languages.
A popular language described using BNF is ALGOL.
<Identifier> :: <letter> | <identifier>
<Letter> | <Identifier> <digit>
<Letter> :: = | a | b | c |……| y | z >
<digit> :: 0 |, 1 |, 2 |…..| 8 |, 9.
The symbol :: = replaces the symbol  in the grammar notation and | is used to separate
different right hand sides of productions corresponding to the same left hand side. The
symbol :: = is interpreted as “is defined as “and | as ”or”.
BNF gives a much more compact description of a language than could be achieved in the
previous meta languages.
Ex: consider the grammar G = <VN , VT, S, P > where VN = {S, A}, VT = {0, 1} and P
consists of the production.
{S  OS, SIA, SO, SI, AO, AO}
The possible derivations in G are
(1) S  O S  OOS  OOOS  OOOIA  OOOIOS  OOOIOI
(2) S IS IOS  IOIA  IOIOS  IOIOO
Ex: A close look at P reveals that L(G) = { w | w  {0, 1}}
adjacent 1s. This grammar G is a regular grammar.
+
, w contains no
(VN  VT)+ set of finite strings on {VN  VT}
(VN  VT)* = (VN  VT)+ {  }, is the empty string or the null string which is defined
as the string of length 0 (or denoted by  ).
w  =  w = w.
Ex: The following is an example of a context sensitive grammar.
G = <VN, VT, S , P> where P consists of productions:
VN = {S, B, C}, VT = {a, b, c}
P = {S  SB, SB  SC, C  abs}
Ex: The following is an example of a context free grammar.
G = <VN, VT, S , P>
P = {C  B, B  abS}
Ex: The following is an example of a regular grammar.
G = <VN, VT, S , P>, VN = {S, B, C},
P = { S  aB, S  C, B  as}, VT = {a, b, c}
S  an Bn Cn
 (a b)n Cn
 an (b C)n
 an bn cn
Ex: The grammar
G = <VN, VT, S , P> with VN = {S}, VT = {0, 1} and
with P = { S  OSI, S   }
It is a context free grammar. It generates the languages.
L(G) = {On 1n | n  0}
{S  OSI
 OOSII
 OOOSIII
 O3  I3
 O3  I3
 On  In
 On In , n  0 }
Formal definition of a Language
Let VT be a finite nonempty set of symbols called the alphabet. The symbols in VT are
called terminal symbols.
The meta language generates strings in the language containing a set of syntactic classess
or variables called nonterminal symbols. The set of nonterminal symbols is denoted by
VN and the elements of VN are used to define the syntax (structure) of the language.
Capital letters such as A, B, C…..X, Y, Z denote nonterminal symbols, while S 1, S2,….
Represent the elements of the vocabulary.
Lower Case letters such as x, y, z,… denote terminal symbols.
The strings of symbols over the vocabulary, are given by  ,  ,  ..... The length of the
string  is denoted by  .
Definition:
A phrase Structure Grammar
It is defined by a 4-tuple G = <VN, VT, S ,  > where VT and VN are sets of terminal and
non terminal synatic class symbols respectively. S, a distinguished element of V N and
therefore of the vocabulary, is called the starting symbol  is a finite subset of the
relation from (VT  VN (VT  VN)* to (VT  VN)*. In general, an element <  ,   is
written as    and is called a production rule or a rewriting rule.
Definition:
Let G = <VN, VT, S ,  > be a grammar. For    (V N  VT ) *  (},  denotes an
empty string  is said to be a direct derivative of written as   C if, there are strings
1 and  2 ( including possibly empty strings) such that   1  2 and   1   2
and    is a production of G.
Definition: Let G = < VN, VT, S ,  > be a grammar. The string  produces

 ( reduces to  or  is the derivation of  ) written as    if the re are strings
 0 , 1 ,...,  n (n  0) suchthat    0  1 , 1   2 ,...,  n1   n and  n   . The



relation  is the transitive closure of  as      or    .
Definition: A sentential form is any derivative of the unique non terminal symbol S.
The language L generated by a grammar G is the set of all sentential forms whose
symbols are terminal, (i.e.)

L(G) = {  S  and   VT* }
Therefore the language is merely a subset of all terminal strings over VT .
Chomsky classified grammars into four classes by imposing restrictions on the
productions.
Context-sensitive text free grammar Regular grammar.
RELATIONS
The word “ relation” suggests some familiar examples of relations such as
the relation of father to son, mother to son, brother to sister etc. In arithmetic,
“greater than”, “lessthan” or that of equality between two real numbers are familiar
examples. There are relations like : between the area of a circle to its radius and
between the area of square to its side etc.
Definition:
(1)
Binary relation: Any set of ordered pairs defines a binary relation.
(2)
The ordered pair, say, <x, y>  R is a relation is written as xRy read as “x is in
relation R to y”
The symbol “>” denotes ““greater than” , or a > b. More preciously, the
relation > is
> = {<x, y> / x, y are real numbers and x > y}
The set R(S) of all objects y such that for some x, <x, y>  S is called the range
of S that is
R(S) = { y / (  x)(<x, y>  S )}
Any relation in X is a subset of the Cartesian product X x X. The set X x X itself
defines a relation in X and is called universal relation in X, while the empty set which is
also a subset of X x X is called a void relation in X.
Properties of Binary relations in a set
Definition:
 A binary relation R is a set X is reflexive, if for every x  X, x R x, that is
<x, x>  R or R is reflexive in X  (x)(x  Xx R x). The relation  is
reflexive in the set of real numbers, since of any x, we have x  x. The
relation of inclusion  is reflexive in the family of all subsets of a universal
set. [ <,  are not reflexive].
 A relation R in a set X is symmetric if, for every x and y in X, whenever
x R y, then y R x.
That is R is symmetric in X  (x)(y)(x  X and y  X and x R yy R x).
[ the relations  and < are not symmetric in the set of real numbers, while
the relation of equality is]
 A relation R is a set X is transitive if, for every x, y and z in x, whenever
x R y, and y R z, then x R z. That is, R is transitive in x 
(x)(y)(z)(x  X and y  X and z  X and x R y and y R z x R z )
[ the relations  , < and = are transitive in the set of real numbers. The
relations  ,  and equality are also transitive in the family of subsets of a
universal set. The relation of triangles in a plane is transitive while
the relation is being a mother is not]
 A relation R in a set X is irreflexive if, for every x  X, < x, x>  R
 A relation R in asset X is antisymmetric if, for every x and y in X. whenever
x R y and y R x then x = y. Symbolically R is antisymmetric in X iff
(x)(y)(x  X and y  X and x R y and y R x  x = y
Relation Matrix
A relation R from a finite set X to a finite set Y can also be represented by a
matrix of R.
Let X ={ x1, x2, …, xm},
Y = { y1, y2, …. ym}
And R be a relation from X to Y. The relation matrix of R can be obtained by
constructing table whose columns are preceded by a column consisting of the successive
element of X and whose rows are headed by a row consisting of the successive element
of Y. If xi R yj, then we enter a 1 in the i th row and jth column. otherwise we enter a zero
in the ith row and the jth column.
For m=3, n=2 and R is given by
R = {<x1, y1>, <x2, y1>,<x3, y2>, <x2, y2>}The table for the above is
Y1
Y2
x1
1
0
x2
1
1
x3
0
1
The relation matrix is
1 0
1 1


0 1
Equivalence Relations:
Definition
A relation R in a set X is called an equivalence relation if it is reflexive,
symmetric and transitive.
If R is an equivalence relative on A, then the sets R(a) are
traditionally called equivalence classes of R. R(a) is also denoted by [a]. The partition P
of A consists of all equivalence classes of R and thus partition is denoted by A/R this
Example:
Let A  1,2,3,4 and
R  (1,1), (1,2), (2,1), (2,2), (3,4), (4,3), (3,3), (4,4)
Now R(1)  [1]  {1,2}  R(2)  [2]
Also R(3)  [3]  {3,4}  R(4)  [4]
 A / R  {{1,2},{3,4}}
Examples of equivalence relation

Equality of numbers on a set of real numbers.

Equality of subsets of a universal set

Similarity of triangle as a set of triangles.

Relation of lines being parallel on a set of lines in a plane.

Relation of living in the same town on the set of persons living in
Tamil Nadu.

Relation of statements being equivalent in the set of statements.
Recurrence relation:
Definition:
A recurrence relation for the sequence a0, a1, …, an-1 is an equation that relates an to
certain of its predecessors a0, a1, …, an-1.
Initial conditions for the sequence a0, a1, …an-1
are explicity given values for a finite
number of the terms of the sequence.
EX: The FIBONACCI sequence {Fn} is defined by the recurrence relation.
Fn = Fn-1 +Fn-2. n  3 and initial conditions.
F1=1, F2= 2.
RELATIONAL DATABASE
A database is a collection of records that are manipulated by a computer. For
example, an air line database might contain records of passengers reservations, flight
schedules, equipment, and so an.
Database management systems are programs that help users access the
information in databases.
The relational database model is based on the concept of an n-ary relation.
Some terminology:
The columns of an n-ary relation are called attributes. The domain of an
attribute is asset to which all the elements in that attribute belong.
A single attribute or a combination of attributes for a relation is a KEY if the
values of the attributes uniquely define an n-tuple.
The ID number may be taken as a key. Each person has a unique
identification number.
A database management responds to queries. A query is request for information from
the database.
The selection operator chooses certain n-tuples from a relation.
Where as the selection operator choose rows of a relation, the projection operator choose
columns.
Join: The selection and projection operators manipulate a single relation.
Join: manipulates two relations. The join operation on relations R1 and R2 begins by
examining all pairs of tuples, one from R1 and one from R2.
The join condition specifies a relationship between one attribute in R1 and an attribute in
R2 .
Definition:
The set of integers modulo m, written Zm is the sum of distinct equivalence
classes under the relation of congruence modulo m in Z. So Zm is the set {[0] m,[1]m,
…,[m-1]m}
Examples:
1. For the equivalence relation R = { (a, a), (b, b) (c, c), (a, c), (c, a)}. What is the
set [a]? Does it have any other names?
[ ans: [a] = {a, c} = [c]]
2. Let A = {1, 2, 3, 4, 5} and let f and g belong to SA, compute to g . f and f .g for
the following.
f = (5, 2, 3) g = (3,4, 1) compute f.g and g.f
3. Let f= (1,2,3,4) g= (3,2,4,5) Write f.g and g.f in the answer in array form.
Ans:
g.f =
f.g =
1
4
2
2
3
5
4
1
5
3
1
2
2
5
3
1
4
4
5
2
3. The permutation function can be shown either in array form or in cycle form.
(a) Let A = { 1, 2, 3, 4, 5} and let f  SA be given in array form by
f=
1 2 3 4 5
4 2 3 5 1
In cyclic form f = (5,1, 4) = (4, 5,1) = (1, 4, 5)
DEFINITION:
1. A set of ordered pairs defines a relation.
2. A relation R in a set X is called an equivalence relation if it is reflexive,
symmetric and transitive.
3. Define the distance between words and weight w(a) of a word a.
4. A mapping f: X  Y is called onto (subjective ) if the range R1 = Y: otherwise it
is called into.
5.
A mapping f: X  Y is called one - to- one (injective ) if district elements of X
are mapped into distinct elements of Y.(i.e) if no member of T is the image under
f of two distinct elements of S.
6.
A mapping f: X  Y is called one - to- one (injective ) if it is both one - to- one
and onto.
7.
A mapping Ix: X  X is called identity map if Ix = {< x, x> / x  X }
8. Let f be a function f: X  Y. If there exists a function g: X  Y such that
g . f = Ix and f . g = Iy then g is called the inverse function of f, denoted of
f -1
9. Let f be a function f: S  T is an onto (surjective) function if the range of f
equals the co domain of f..
10. Let f be a function f: X  Y is bijective if it is both one to one and onto.
2.12 Exercises on Relations and Function
A) Say true or false
1. If the relation R on S={ 1,2,3} is antisymmetric and if (a, b) and (b, a) belong to R,
then a * b. (False)
2. The equivalence relation R on N given by xRy ↔ x + y is even , partitions N into
two equivalence classes. (True)
3. If 4 is the maximum integer that can be stored on a microcomputer, the integer that
will be stored for the values 3 + 4 if addition modulo 5 is used is 2. (True)
4. The relation R on Q defined by R= {x,y  Q: │x│  │y│ is reflexive, transitive.
(True)
5. h: N↔N where h is defined by h(x)=x-4 is a function . (False) [For values 0,1, 2, 3 of
the domains the corresponding values of h(x) fall outside the co-domain.]
6. The floor function and the ceilinf function are one-to-one. (False)
7. Let f: R→R be defined by f(x) = x2. Let g: R→R be defined by g(x) = x .The value of
(fog)(2.3) is the same as the value of (gof)(2.3). (False) [(fog)(x) = 4;(gof)(x)=5]
8. The entire truth table defines a function f such that f:{T.F)"→ {T,F} for n
statement variables. (True)
9. The function associated with a tautology maps {T.F}" →{T}. while the function
associated with a contradiction maps {T,F}n →{F}. (True)
10. n2 +2n is divisible by 2 for all n. (False)
11. n2 +2n is divisible by 3. (True)
12. An infinite subset of a denumberable set is denumberable. (True)
13. The set R of real numbers is denumberable. (False)
14. The power set of A is 2A. (True)
15. 2n+1 < 3n for all n. (False) [only for n > l it is true]
16. n3 < 3n for all n. (False) [true only for n > 4]
17. n! >n2 for n > 4. (True)
18. n! < nn for all n. (False) [True for n >2]
B) Say True or False
1. A ternary relation R on a set of real numbers is given by (a, b, c)  R if b is between
a and c.
(True)
2. If x  A  B, then (x-A)  (x-B) is a null set. (True)
3. The function f: N→ N , where f (n) = n2 is a subjection. (False)
4. On the set of positive integers the relation " y – x + 2 is a prime " is reflexive. (True)
5. On the set of positive integers the relation “x is a proper factor of y" is transitive.
(True)
6. On the set of positive integers the relation "(y-x)+2" is a prime number" is
symmetric. (True)
7. If a relation R on a set S is anti symmetric and if (a, b) and (b, a) belong to R,
then a  b must be true. (False)
8. On the set of relational numbers Q, the relation R defined by {(x, y)  R | x| < |y|} is
reflexive, symmetric and transitive. (False) [it is not symmetric]
9. The relation R=-={(9,7),(6,5),(3,6),(8,5)} on N X N, where N is the set of natural
numbers is one-to-one . (True)
10. If R is an asymmetric relative on set S, then R is irreflexive. (True)
11. On the set of all times in the plane, the relation defined by {(x. y)  R / x is parallel to
y or x coincides with y} is an equivalence relation. (True)
12. On the set of natural numbers N, the relation R defined by
R={(x, y)  R |x = y2 } is antisymmetric. (True)
13. The relation R on the set S={x | x is a student in your class} defined by
R ={(x, y)  S| x sits in the same row as y } is are equivalence relations. (True)
C) Say True (T) or False (F) in the following :
1. In a one-to-many binary relation, at least one first component is two different ordered
pairs-T
2. If an anti symmetric binary relation contains (x, y), then (y, x) will not belong to the
relations-F [(x, x) can belong]
3. A least element of a partially ordered set precedes all elements except itself-T
4. An equivalence relation can not also be a partial ordering-F
[The relation of equality is both a partial ordering and equivalence relation]
5. A partial ordering of a set determines a partition of that set -F
6. A binary relation on X * Y that is not one-to-many or many-to-one is a function from
X to Y.
-F [It may not have an unique element for each member of the domain]
7. To prove that a function is onto, begin with an arbitrary of the range and show that it
has a pre usage
-F [Every element of the range has a pre usage: begin with element
of the co domain]
8. To show that a function is one-to-one assume f(x1) = f(x2) for some x1 and x2 in the
domain and show that x1 = x2
-T
D) Say true or false in the following
1. The relation r = {(i, j) │i-j│ = 2} on the set {1, 2, 3. 4, 5. 6} is transitive.
2. The relation of being a mother is not transitive.
3. In general which is irreflexive and symmetric cannot be transitive.
4. The subset B = {{a}. {a. c}) of the set S = {a, b, c} is a partition of S.
5. For a finite set. the largest partition consists of the set itself.
6. The ordering on the factors of 24 extends divisibility.
7. In the set of human beings the relation x r y if x and y share, at least one parent is an
equivalence relation.
8. For a set x with n numbers, the number of relations on x which are symmetric is
2(n(n+l)/2).
E) Choose the appropriate alternative in the following
1. The number of relations R on A (a.b}is
(a) 16 (b )8 ( c ) 4
(d)2
Ans: (a)
2. The number of relations ""divides'" o n { l , 2, 3, 6} is
(a)3 (b)4 (c) 6
(d) 9
Ans: (d)
3. The relations "divides "' and "is less than or equal to" on the set {2 , 6,12}
are
(a) in general different
(b) the same
(c) not comparable
(d) such that ""divides" contains less number of pairs than '"is less than or
equal to". Ans: (b)
4. The number of functions f: x→x when x ={a, b} is
(a) 4 (b) 6 (c) 8
(d) 9
Ans: (d)
5. On the set of positive integers, the relation "x/y and y/x in lowest terms are both
fractions
with an odd numerator and denominator " is
(a) Transitive only (b) refexive only (c) symmetric only
(d) an equivalence
relation Ans: (d)
6. The number of functions f: S → T with |S|= m and |T|=n is
(a)nm
(b)mn (c) m!n!
(d) n!/m!
Ans: (a)
7. The number one to one functions f: S→S with |S|=n is
(a)n11 (b)n!
Ans: (b)
(c)n 2
(d) n
8. Let f: N→ N be defined by f(x) = x + 1 .Let g: N →N be defined by g(x)=3x
then gof(5) is equal to
(a)3x+3
(b)3x+l
(c)18 (d) 15
Ans: (c)
9. The distance between the code words a = ( 1 1 0 1 ) and b = ( 1000) is
(a) 3 (b) 2 (c) 1
(d) 0
Ans: (b)
10. For the codes {10000, 01010, 00001} in  5 , the minimum distance, the number
of errors which can be detected and connected are
(a) {2,2.2} (b) {2.0,1}
(c) {2,1,0}
(d) {3.2,1}
Ans: ( c)
11. N is the set of natural numbers including zero. The function f : N→N, f(j) = j2 + 2
is
(a) one to one (b) onto
(c) one to one onto
(d) none of them
Ans: (a)
12. N is the set of natural numbers including zero. The function f: N→ {0,l}.
0 if j is odd
f(j)=
1 j is even
(a) one to one
(b) onto(c) one to one onto
(d) none of them
Ans: (b)
13.
I4 ={0,1,2.3}.The function f: I4→ I4: f(x) = 3x(mod4) is
(a) one to one only (b) onto only
(c) one to one onto
(d) none of these
Ans: ( c)
14. Which of the ordered pairs belong to R on N x N defined by x R y ↔ x > y2
{(1,2),(2,1),(5,2),(6,4),(4,3)}
(a) {(2,1), (6,4)}
(b) {(5,2), (6,4)} (c){(2.1). (5.2)}
15. Let S={1, 2, 3} and R={(1, 1), (1,2), (1, 3), (3,1), (2, 3)}. Then R is
(a) an equivalence relation
(b) not reflexive, not symmetric and not transitive(c)
reflexive and anti symmetric (d) transitive only
Ans: (b)
16. Let S (n, k) denote the number of ways to partition a set of n elements into k blocks.
Then S (3,2) is equal to
(a) 8 (b) 6 (c) 3
(d)1
Ans: (c)
17. Let S (n, k) denote the number of ways to partition a set of n element into k blocks.
Then S (4.2) is equal to
(a) 6
(b)7 (c)8
(d)9
Ans: (b)
F)Choose the appropriate alternative in the following
1. The number of different reflexive symmetric relations on a set of three
elements is
(a) 3
(b)4
(c)23
(d)0
2. The relation on a set of positive integers given by r = {|y-x| +2 is a
prime number} is
(a) Reflexive
(b) symmetric
(c) transitive
(d) reflexive
and symmetric
4. The number of partitions of a set x = {a, b, c, d} into pair wise disjoint non
empty subsets of x with union x is
(a) 18
(b) 16
(c) 15
(d) 12
G) SHORT ANSWER QUESTIONS
1. The following are relations an A = (0.1,2,) R = {(i, j) / j = i + 1 j = i/2}:
S={(i, j)|i=j+2}
a) Determine the composite relations:
R º S; S º R; R º R º R
b) Construct the relation matrices for the relations in (a)
2. Given R = {(i, j) / i, j  I, the set of integers, j- i =1}
What is Rm ? (m is a positive integer)
3. The transitive closure of the relation R on a finite set. is denoted by R* and is a
relation on A given by R+ = Uk-1 Ri.
i=1
Find the graph of the relation R over (1, 2, 3, 4, 5, 6) given by {<1, 3>, <1, 5>,
<2, 5>, <3, 6>, <4, 5>, <5, 4>. <6, 6>) Draw the graph of the relations R5 and R8+:
find R+.
4. Let S be a set with n elements (finite). The number of ways that i distinct
elements can be selected from n elements is given by
n 
n!
 i   i!(n  I )!
 
The number of distinct sub sets of S (including  ) is
n  n

 2
i 0  i 
n
5. Given a set S, the set of all subsets of S including the empty set  is called the power
set of S and is denoted by P (S) ={A/A  S}=2s.
a) What is the power set of (  , a, (a)}?
b) Let S= {a }, Find the power sets of S and of 2s?
b) Test whether the binary relation R={(l,2),(2,2),(3,3),(1.2),(2,l)}on S={1, 2, 3} is an
equivalent relation or not
Ans: yes
c) Let R be the Relation on the power set P(S) of a finite set S of cardinality n. Define R
on P(S) by (A,B)  R iff A  B= 
(i) Consider the specific case mo and determine the cardinality of R
[S={A, B, C}; P(S) = [{{  },{A},{B},{C},{A,B},{B,C}}]]
The cardinality of R = (1 + 1 + 1 )3 = 33 = 27
(ii) What is the cardinality of R for an arbitrary n.
Express your answer in terms of n. [Ans : 3n]
iii)
Not reflexive, not symmetric, not antisymmetric, not transitive
(e) Consider the following relation on S = {1, 2, 3, 4, 5, 6} and R - {<i, j>) | |i-j|=2}. Is
R reflexive, symmetric, transitive and draw the graph of R.
[Ans : R is not reflexive, it is symmetric; it is not transitive]
(f) R is a relation in S = {1,2,.. .9}defined by xRy iff x + y =10. Is R an equivalence
relation on S [no: R is not reflexive or transitive]
(g) S = {a,b,c} n=3
P(S) = [{  },{a},{b},{c}.{a.b}.{a,c},{b,c},{a,b,c)]. R on P(S) is (A.b)
eRiff A n B - O
R=
(  ,  ),(  ,a),(a,  ).(  .b),(  ,c),(c,  ),(  ,ab), (ab,  ),(  .ac). (ac,  ),
(  ,bc),(  .abc),(abc,  ),(a, b),(b. a).(a, c),(c, a). (b, c),(c, b),(a, bc),
(bc, a),(b, ac). (ac, b),(c, ab),(ab, c)
Cardinality of R=27 = 3 3
H) Short answer questions
1. Define: binary relation, universal relation, and void relation, reflexive,
symmetric, anti symmetric and transitive relation.
2. Define: equivalence relation.
3. Let x = {1, 2. 3. ..7} and r = {<x, y>/x-y is divisible by 3}. Show that r
is an equivalence relation. Draw the graph of r.
4. Draw the hasse diagram for the following
a. Let A = {1, 2, 3, 4} and r be the relation "|" on A.
b. Let A = {2, 3, 4, 6, 12, 36, 48} and r is the relation "/" on A.
c. Let A = {a, b, c}and r be the relation "|" on A.
d. Let A = {1,2,3,4,5,6} and write m/n in case m divides n.
6. Define the terms
a. Transitive closure.
b. Reflexive closure.
7. Let r be a relation on the set of positive real numbers so that its graphical
representation consists of the points in the first quadrant of the Cartesian plane. What
can you expect if r is
a. Reflexive
b. Symmetric and
c. Transitive?
8 Let r be a binary relation on the set of all positive integers such that r = ((a.b)/a=b~} is
relation r reflexive? Symmetric? Anti Symmetric? Transitive? Or equivalence
relation or a partial ordering relation?
9 Let A be a set of books. Let r be a binary relation on A such that (a. b) is in r if book
"a" costs more and contains fewer pages than book Is r reflexive? Symmetric? Anti
symmetric? Transitive?
2. State the principle of finite induction.
3. Let f: R-> R be given by f(x)=3x=4 .it is a bijiction what is f' 2
4. Let f: R 2 -> R 2 where f(x,y)=(y+l.x+l), what is f'
5. Let S={2.4,6.8}and T:{ 1.5.7}
8. The number of functions
f: S -»T with I S j =m and I TI =n is
(a)nm (b)mn (c) m! n! (d) n! An ................... Ans: (A)
9. The number of one-to-one functions f: S->s with I S I = n is
(a) n n (b) n! (c) n 2 (d) n .....................................Ans: (B)
10. Let f: N->N be defined by f(x) = x+1. Let g: N->N be defined by g(x) ^3x.Then
( go f) (5) is equal to
(a) 3x-t-5 (b)3x+l (c) 18 (d) 15 ........................ Ans :(c)
11. The distance between the code Words a=(l 101) and b=(1000) is
(a) 3 (b) 2 (c) 1 (d) 0 ............................. Ans :( b)
12. For the codes {10000,01010, 00001,}in V\ the minimum distance, the number of
errors which can be detected and corrected are
(a){2,2,2};(b){2, l,0};(c) {2, 1, 0), (d) {3.2.1} ................... Ans : (c)
13. N is the set of natural number including zero. The function f: N->N, f (j) =j"+2 is
(a) One-to-one (b) onto (c) one to one onto (d) none of them .......... (Ans): (a)
14. N is the set of natural number including Zero. The function f : N->{0.1}.
fOJ is odd
is
1, j is even
(a) one to one (b) onto (c) one to one onto (d) none of them .......... (Ans): (b)
15.1 4= {0, 1. 2, 3}. The function f: I 4-> I 4; f(x) -3x (mod 4) is
(a)one to one only (b)onto only (c) one to one onto (d) none of this ......... Ans: (c)
16. The floor function [x] associate with each real number x the greatest integer less
than to x. The ceiling function [x]associates with each real number x the smallest
integer greater than or equal to x.
(a) Sketch the graph of the function [x]
(b) Sketch the graph of the function[x]
17. A function f: S^T is an onto (suijietive) function if the range of f equals the
codomain of f.
A function f: S—>T is one-to-one (or injective) function, if no member of T is
theimage under f of two distinct elements of S.
A function f: S—>T is bijietive if it is both one to one and onto.
Draw the Hasse diagram for the partial during set <S-{2,3,5,7,21,42,105.210}. 1> where
x|y denotes x divides y. Name any least elements, minimal elements, greatest elements
and maximal elements.
[Least element is 2, minimal element is 2,greatest element is 210.maximal element
is 210
210
42
105
Example:
The relation R={(1,2),(2,1),(1,3)} on the set S={1,2,3} is neither symmetric (1,3) belongs but (3,1) does not-nor antisymmetric-(l,2) and (2,1) belong ,but 1^2.
It is possible to extend the relation R on the set S to a relation R on S so that R
will contain all of the ordered pairs in R plus the additional ordered pairs needed for the
desired property to hold. Thus RcR .If R is the smallest such set then R is called the
closure of R with respect to that property.
(d) Is D?4 a distributed lattice ........Explain
(e) Is D24 a Boolean lattice....... Explain
[Ans: (b) 4 has no complement; (c) it is not a complemented lattice; (d) it is a
distributed lattice (e) it is not a Boolean lattice]
J) Additional Examples on Relations and Function
Test whether the binary relation R=-{(1,2),(2,2)(3,3),(1,2).(2,1)) on S- {1,2,3} is
an equivalence relation or not
[Ans : yes]
Define the greatest element and maximal element in a partially ordered set <P,S>
[Ans : yeP is a greatest element if x<y for all x EP; yeP is a maximal element if
dt there is no xeP with y<x]