Download 1 = a

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
Detailed Study of groups is a
fundamental concept in the study of
abstract algebra. To define the notion of
groups,we require the concept of binary
operation or composition which is a type
of function that associates two elements
of the set to a unique element of that set.
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A binary operation on a set is a rule for combining
two elements of the set. More precisely, if S iz a
nonempty set, a binary operation on S iz a mapping f
: S  S  S. Thus f associates with each ordered pair
(x,y) of element of S an element f(x,y) of S.
IN OTHER WORDS,
An operation which combine two elements of a set to
give another elements of a set to give another
elements of the same set is called a binary operation
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
1. Ordinary addition ‘+’ is a binary operation
on Z
Consider
+: (Z*Z) Z
+: (3,7)=3+7=10 ∈Z
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

A non-empty set G equipped with one or more
binary operation defined on it is called an
algebraic structure.
Suppose ‘*’ is a binary operation on G , then (G
, *) is an algebraic structure.
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A group (G, ・) is a set G together with a binary
operation ・ satisfying the following axioms.
(i) Closure property:
a.b ∈ G for all a,b ∈ G.
(ii) The operation ・ is associative; that is,
(a ・ b) ・ c = a ・ (b ・ c) for all a, b, c ∈ G.
(iii) There is an identity element e ∈ G such that
e ・ a = a ・ e = a for all a ∈ G.
(iv) Each element a ∈ G has an inverse element a−1 ∈
G such that a-1・ a = a ・ a−1 = e.
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A group (G , .) is said to be abelian or
commutative if in addition to the above four
postulates, the following postulate is also
satisfied:
COMMUTATIVE PROPERTY:
If the operation is commutative, that is,
if a ・ b = b ・ a
for all a, b ∈ G,
the group is called commutative or abelian group
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Example:
 Let G be the set of complex numbers {1,−1, i,−i} and let ・
be the standard multiplication of complex numbers. Then
(G, ・) is an abelian group.


The product of any two of these elements is an element
of G; thus G is closed under the operation.
Multiplication is associative and commutative in G
because multiplication of complex numbers is always
associative and commutative.
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


The identity element is 1, and
The inverse of each element a is the element
1/a.
Hence
1−1 = 1 , (−1)−1 = −1 , i−1 = −I , and (−i)−1 = i.
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Solution.
(i)
CLOSURE PROPERTY: since the sum of two
integers is also an integer i.e. ,
a+b ∈ Z
for all a,b ∈ Z
Therefore the set Z is closed w.r.t. addition.
Hence closure property is satisfied.
(ii) ASSOCIATIVE LAW: since addition of
integers obey associative law, therefore
a+(b+c)=(a+b)+c for all a, b, c ∈ Z
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Thus addition is an associative composition.
(iii) EXISTENCE OF IDENTITY: the number
0 ∈ Z and a+0 = 0+a =a for all a ∈ Z
The integer 0 is the identity for (Z , +)
(iv) EXISTENCE OF INVERSE: for each a ∈ Z,
There exists a unique element -a ∈ Z such that
a+(-a)=0=(-a)+a
Thus each integer possesses an additive inverse.
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(v) COMMUTATIVE LAW: the commutative law
holds good for addition of integers
i.e.
a+b=b+a for all a,b ∈ Z
Thus (Z , +) is an abelian group. Also Z contains
an infinite number of elements.therefore (Z , +)
is an abelian group of infinite order.
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
An algebraic structure (G , *) is called a semigroup, if only the first two postulates, i.e.,
closure and associative law are satisfied.
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
The algebraic structure (N ,+), (W ,+), (Z, +),
(R, +) and (C, +) are semi-groups, where N , W
, Z , R and C have usual meanings.
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
Definition:
If the number of elements in the group G are
finite, then the group is called a finite group
otherwise it is an infinite group
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
Example of Finite Group:
{1,−1, i,−i} is an example of finite group.
Example of Infinite Group:
(Z , +) is an example of infinite group.
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
Definition:
The number of elements in a finite group is
called the order of the group. An infinite group
is said to be of infinite order.
The order of a group G is denoted by the
symbol o(G).
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If (G , .) is a group, then
(i) the identity element of G is unique.

(ii) every element has a unique inverse.
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Proof::
If possible let e1 and e2 be two identities in the
group (G, .)
Since e1 is identity and e2 ∈ G
therefore
e1 . e2 = e2 = e2. e1
…..(1)
Also since e2 is the identity and e1 ∈ G
therefore
e1 . e2 = e1 = e2 . e1
…..(2)
therefore from (1) and (2), e1 = e2
Hence the identity is unique
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PROOF:
Let a be any element of the group ( G , .)
If possible , let b1 and b2 be two inverses of a
under the binary operation ‘ . ‘ and let e be the
identity element in G. Then
a . b1 = e = b 1 . a
And a . b2 = e = b2 . a
Now b1 = b1 . e
=b1 . ( a . b2)
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b1=(b1 . a) . b2 [by associativity]
=e . b2 = b2
Therefore b1 = b2.
hence the inverse is unique.
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Proposition . If a, b are elements of a group G ,
then
(i) (a−1)−1 = a.
(ii) (ab)−1 = b−1a−1.
i.e. , the inverse of the product of two elements of a
group is the product of their inverses in the reverse
order.

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Proof:
for each a ∈ G, we have
a . a−1 = e = a−1 . a

Inverse of a−1 is a.
Therefore
(a−1)−1 = a
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Proof:
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Subgroups:: It often happens that some subset of a
group will
also form a group under the same operation.Such
a group is called a subgroup. If (G, ・) is a
group and H is a nonempty subset of G, then
(H, ・) is called a subgroup of (G, ・) if the
following conditions hold:
(i) a ・ b ∈ H for all a, b ∈ H. (closure)
(ii) a−1 ∈ H for all a ∈ H. (existence of inverses)


Conditions (i) and (ii) are equivalent to the single
condition:
(iii) a ・ b−1 ∈ H for all a, b ∈ H.
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Prove that:
(i) The identity of the sub-group is same as that
of the group.
(ii)
The inverse of any element of a sub-group is
the same as the inverse of that element in the
group.
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let H be a sub-group of the group G.
if e is the identity element of G, then
ea = ae = a for all a ∈ G
(i)
As H is a sub-set of G
therefore
ea = ae= a for all a ∈ H
[ since a ∈ H => a ∈ G]
=> e is an identity element of H.
Hence identity of the sub-group is same as that of the
group.
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(ii) Let e be the identity of G as well as of H.
Let a be any element of H
 a is an element of group G
suppose b is the inverse of a in H and c is the inverse of a in G.
 ab = ba = e
….. (1)
ac = ca = e
…. (2)
From (1) & (2), ab = ac
Therefore b=c
Hence, the result.
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Solution.
Let G = {…. , -4 , -3 , -2 , -1 , 0 , 1 , 2 , 3 , 4 , …..}
i.e. G is additive group of integers.
Let H = {… , -3k , -2k , -k , 0 , k , 2k , 3k ,….}
Therefore H ≠ ф
Here H is a subset of G.
We have to show that H is a subgroup of G.
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Let ak , bk be any two elements of H such that a ,
b are integers.
Inverse of bk in G is –bk.
Now ak - bk = (a - b)k , which is an element of H
as
(a – b) is some integer.
Thus for ak , bk ∈ H , we have ak – bk ∈ H
Hence H is a subgroup of G.
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Solution :H is a subset of R but H is not a subgroup of R ,
The reason being that the composition in H is
different from the composition in R.
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
Definition. Let G be a group and let a  G ane
e be the identity element in G. If ak = e for some
k  1, then the smallest such exponent k  1 is
called the order of a; if no such power exists,
then one says that a has infinite order.
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
Definition.
If G is a group and a  G, write
<a > = {an : n Z} = {all powers of a } .
It is easy to see that <a > is a subgroup of G .
< a > is called the cyclic subgroup of G generated by
a.
A group G is called cyclic if there is some a  G with
G = < a >; in this case a is called a generator of G.
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Proof:Let G be a finite group and o(G)= s
Let a  G such that o(a)= s
If H = {an : n Z}
Then o(H) = s = o(a)
Therefore
H is a cyclic subgroup of G.
Also o(H) = o(G) implies G itself is a cyclic group
and a is a generator of G.
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Example:If G = {0 , 1 , 2 , 3 , 4 , 5 } and binary operation + 6
is
Then prove that G is a cyclic group.
Solution .
we see that
1
1=1
1 2 = 1+ 6 1 = 2
3
2
1=1 +61=3
4
3
1 = 1 +6 1 = 4
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5
4
1=1 +6 1=5
5
6
1 =1 +6 1=0
Therefore
G = { 0 , 1 , 2 , 3 , 4 , 5 } is a cyclic group.
1 is generator of given group.
Therefore
G=<1>
Hence proved
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Definition:
let G be a group and H be any subgroup of G. For
any
a  G , the set Ha = { ha : h  H} is called right coset
of H in G generated by a.
Similarly, the set aH = { ah : h  H} is called left coset
of H in G generated by a.
Obviously , Ha and aH are both subsets of G . H is
itself a right and left coset as eH = H =He , where e
is an identity of G.
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Find the right cosets of the subgroup { 1, -1 } of the
group { 1 , -1 , i, -i} w.r.t. usual multiplication.
Solution.
let G = { 1 , -1 , i, -i} be a group w.r.t. usual
multiplication
And
H = { 1, -1 } be a subgroup of G.
The right cosets of H in G are
H(1) = {1(1) , -1(1)} = {1 , -1}= H
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H(-1)={1(-1) , -1(-1)}={-1 , 1}=H
H(i)={1(i) , -1 (i)} ={i, -i}
H(-i) = {1(-i) , -1(-i)}= {-i , i}
thus we have only two distinct right cosets of H
in G.
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Define group. Show that Z (the set of all
integers) is an abelian group w.r.t. addition.
2. Show that the set of integers Z is an abelian
grop w.r.t. binary operation ‘ * ‘ defined as a * b
= a + b + 1 for a , b ∈Z.
3. If (G , .) is a group, then
(i) the identity element of G is unique.
1.
(ii) every element has a unique inverse.
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4. If a, b are elements of a group G , then
(i) (a−1)−1 = a.
(ii) (ab)−1 = b−1a−1.
i.e. , the inverse of the product of two elements of a group is
the product of their inverses in the reverse order.
5. If every element of a group is its own inverse, then show
that the group is abelian.
6.If a group has four elements , show that it must be
abelian
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7.The order of every element of a finite group is
finite and is less than or equal to the order of
the group.
8.Let G = { 0 , 1 , 2 , 3 , 4 , 5 }. Find the order of
elements of the group G under the binary
operation addition modulo 6.
9. Define subgroup.Let G be the additive group of
integers. Prove that the set of all multiples of
integers by a fixed integer k ig a subgroup of G.
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10. Prove that:
(i) the identity of the subgroup is same as that
of the group.
(ii) the inverse of any element of a subgroup is
the same as the inverse of that element in the
group.
11.Let H be the multiplicative group of all
positive real numbers and R the additive group
of all real numbers. Is H a subgroup of R?
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12.Let G be a group with binary operation denoted as
multiplication. The set
{h ∈ G : for all x ∈ G} is called the centre of the group G
. Show that the centre of G is a subgroup of G.
13. Define cyclic group. If a finite group ‘ s ‘ contains
an element of order ‘ s ‘ , then the group must be
cyclic.
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14. If G = { 0 , 1 , 2 , 3 ,4 , 5 } and binary operation
is addition modulo 6 , then prove that G is a
cyclic group.
15.Define cosets. Find the right cosets of the
subgroup { 1 , -1 } of the group { 1, -1 , i , -i}
w.r.t. usual multiplication.
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DO ANY THREE QUESTIONS.
1. If a, b are elements of a group G , then
(i) (a−1)−1 = a.
(ii) (ab)−1 = b−1a−1.
i.e. , the inverse of the product of two elements of a group
is the product of their inverses in the reverse order.

2. Define cyclic group. If a finite group ‘ s ‘ contains
an element of order ‘ s ‘ , then the group must be
cyclic.
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3. If G = { 0 , 1 , 2 , 3 ,4 , 5 } and binary operation is
addition modulo 6 , then prove that G is a
cyclic group.
4.Define cosets. Find the right cosets of the
subgroup { 1 , -1 } of the group { 1, -1 , i , -i}
w.r.t. usual multiplication.
5. Define subgroup.Let G be the additive group of
integers. Prove that the set of all multiples of integers
by a fixed integer k ig a subgroup of G.
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