Let S be a nonempty set then an operation * on a set S is said to be a binary operation on S if a * b belongs to S for every a,b in S.
ie: * from S x S to S defined by * ( a,b ) = a * b for every ( a, b ) in S x S.
Eg: Let R be the set of all real numbers then +, -, x are binary operations on R . But division is not a binary operation on R because division of any real number by zero is not defined.
Algebraic structure:
A non- empty set G together with one or more binary operations is called an algebraic structure. It is denoted by
Eg: ( G , * ),( G , *, o )
Group:
An algebraic structure ( G , * )
G1: * is associative in G
ie: a * ( b * c ) = ( a * b ) * c for every a,b,c in G
G2: There is a unique element e in G such that a * e = a = e * a for every ‘ a ‘ in G.
(Existence of identity element)
G3: For every element ‘ a ‘ in G there is an element ‘ a1 ‘ in G such that
a * a1 = e = a1 * a. (Existence of inverse element)
Eg: ( R , + ), ( Z , + )
Abelian Group:
A group
ie: a * b = b * a for every a, b in G
Eg: ( R , + ), ( Z , + )
Semi group:
A set together with a binary operation which is associative is called a semi group.
Eg: ( R , + ), ( Z , + ) is a semi group.
Monoid:
A set together with a binary operation which is associative and identity element exists is called a monoid.
Eg: ( R , + ), ( Z , + )
Finite Group and Infinite group: A group ( G, * ) is said to be a finite group if the under lying set G itself is finite otherwise G is infinite.
The number of elements in a group is called order of the group.
Addition modulo n: Let a, b be any two integers then addition modulo n of a and b is denoted by a +n b, is defined as the smallest positive integer r, is the remainder which when obtained by dividing ( a + b ) divided by 4.
Eg: 5 +3 4 = 2
- 20 +5 6 = 0
Multiplication modulo n: Let a, b be any two integers then Multiplication modulo n of a and b is denoted by a x n b, is defined as the smallest positive integer r, is the remainder which when obtained by dividing ( a b ) divided by 4.
Eg: 5x4 7 = 3
SOME EXAMPLES FOR GROUP:
Q1. Prove that G is the set of all rational numbers with the operation * defined below is a group.
Let a, b belongs to G then a * b = (ab / 2)
Proof:
(i) It is clear that if a, b belongs to G then a * b = (ab / 2) , belongs to G.
There fore * is closed in G.
(ii) Let a, b, c belongs to G then
a*( b * c ) = a*( bc / 2) = ( abc / 4 ) and
( a* b )* c = (ab / 2) * c = ( abc / 4 )
ie: a*( b * c ) = ( a* b )* c
Therefore * is associative in G.
(iii) Let x be the identity element in G then a * x = a = x * a
ie: a * x = a implies ( ax / 2 ) = a
implies x = 2 , belongs to G.
There fore x = 2 will act as the identity element in G.
(iv) Let y be the inverse of the non zero element ‘ a’ in G
Then a * y = 2 = y * a
Therefore a * y = 2 implies ( ay / 2 ) = 2
implies y = ( 4 / a ), belongs to G.
There fore y = ( 4 / a ) will act as the inverse of ‘ a ‘ in G.
Hence ( G , * ) is a group.
Example for an infinite non abelian group:
Let G be the set of all n x n non-singular matrices with real or complex entries with operation as matrix multiplication.
Verification:
(i) We have if A, B belongs to G then AB belongs to G.
Therefore matrix multiplication is a binary operation in G.
(ii) Since matrix multiplication is associative then here matrix multiplication is associative in G.
(iii) We have I, the n x n unit matrix belongs to G and AI = A = IA for every A in G.
Therefore I will act as the multiplicative identity in G.
(iv) Since every member in G is non-singular then clearly its inverse exists and belongs to G.
Therefore G is a group under matrix multiplication.
(v) Since matrix multiplication is commutative, G is abelian.
Also it is clear that number of elements in G is infinite.
Hence G is an infinite non abelian group.
1. Show that G = { a + bV2; a, b in Q } where V2 is the square root of 2, is a group under usual addition.
2. Show that V the set of all vectors in a space is an infinite abelian group under vector addition.
3.Show that G = {1, w, w2 }where w is the cube root of unity forms a finite abelian group of order 3 under the operation multiplication.
4. Show that G= { 1,-1,i,-i } is a finite abelian group under the operation multiplication of order 4
From this section on wards in a group ( G, * ) we take a * b = ab unless otherwise stated.
Theorem ( Cancellation laws): Let G be a group and a, b belongs to G then
(i) ab = ac implies b = c ( Left cancellation law )
(ii) ba – ca implies b = c ( Right cancellation law)
Proof:
Suppose ab = ac
Since G is a group and ‘ a ‘ belongs to G, a-1 belongs to G.
Multiply a-1 on the left of both sides then we get, a-1 (ab) = a-1(ac)
implies (a-1a)b = (a-1a) c
implies eb = ec
implies b = c.
Similarly, we can verify that ba – ca implies b = c.
Theorem: In any group G the identity element is unique.
Proof:
Let e and e’ be two identity elements in G
Then for any ’ a ‘ belongs to G we have a e = a = e a and a e’ = a = e’a
From this it is clear that a e = a = a e’
implies e = e’
Hence identity element is unique in G.
Theorem :Let G be a group then inverse of an element ‘ a ‘ in G is unique.
Proof:
Let b and c be the inverses of ‘ a ‘ in G
Then ab = e = ba and ac = e = ca ………………………….(1)
Claim: b = c
We have from (1) ab = e = ac
Then by Left cancellation law b = c.
Therefore claim.
Hence the inverse of the element ‘ a ‘ in G is unique.
Theorem: Let G be a group and a-1 is the inverse of ‘ a ‘ then (a-1) – 1= a.
Proof:
Since a-1 is the inverse of ‘ a ‘ then a (a-1) = e = (a-1) a
Since a belongs to G then (a-1) belongs to G implies (a-1) – 1 belongs to G
Therefore (a-1) (a-1) – 1 = e = (a-1) – 1(a-1)
We have , (a-1) (a-1) – 1 = e implies a(a-1) (a-1) – 1 = ea
implies (aa-1) (a-1) – 1 = a
implies e (a-1) – 1 = a
Theorem ( Invertibility): Let G be a group then (ab) –1 = b-1 a-1 for every a, b in G
Proof:
Claim: (ab) (b-1 a-1) = e =(b-1 a-1) (ab)
We have (ab) (b-1 a-1) = a(b b-1 )a-1
= a(e) a-1
= a a-1
= e
Similarly, we will get (b-1 a-1) (ab) = e
ie: (ab) (b-1 a-1) = e =(b-1 a-1) (ab).
There fore claim.
Hence (ab) –1 = b-1 a-1 for every a, b in G.
Note : The set of all remainders which when divide the sum of any two integers by ‘ n ‘ is {0,1,2,3,…, n-1}, denoted by Z n.
Theorem: The set Z n = {0,1,2,3,…, n-1} forms a group under the operation addition modulo n.
ORDER OF AN ELEMENT IN A GROUP
Let G be a group then order of an element ‘ a ‘ in G is the least positive integer ‘ n ‘ such that
a n = e.It is denoted by o(G).
Note: If G is an additive group then order of an element ‘ a ‘ in G is the least positive integer
‘ n ‘ such that na = e.
Eg: Consider the multiplicative group G ={1,w,w 2 }, where ‘w’ is the cube root of unity.
We have 1 2 = 1 implies o(1) = 1
w 3 = 1 implies o(w) = 3
w 6 = 1 implies o(w 2 ) = 3
Eg: Consider the multiplicative group G ={1,-1, I, -i .
We have 1 2 = 1 implies o(1) = 1
(-1) 2 = 1 implies o(1) = 1
i 4 = 1 implies o(i) = 4
( -i ) 4 = 1 implies o(-i ) = 4
Subgroup:
A non empty subset H of a group G is said to be subgroup of G if H itself be a group under the same operation defined in G.
Eg: Consider the multiplicative group G ={1,-1, i, -i }.Let H = { 1, - 1 } then H is a subgroup of G.
Theorem: A non empty subset H of a group G is a subgroup of G Û
(i) (ab) belongs to H for every a, b in H
(ii) a-1 exists and belongs to H for every ‘a’ in H.
Proof:
First suppose that H is a subgroup of G.
Then (ab) belongs to H for every a,b in H. ( By closure property in H )
Conversely, suppose that the two conditions in the statement holds.
To prove H is a subgroup of G.
By (i) closure property holds in H.
By (ii) for every ‘ a ‘ in H, a-1 exists and belongs to H.
Since a, a-1 belongs to H we get (aa-1) belongs to H……………….(1)
Since a, a-1 belongs to G we have aa-1 = e belongs to G
Then from (1) ‘e ‘ belongs to H
There fore H is a group under the same operation in G
Hence H is a subgroup of G.
Theorem: A non empty subset H of a group G is a subgroup of G Û
(ab - 1) belongs to H for every a, b in H
Proof:
First suppose that H is a subgroup of G.
Let a, b belongs to H
Since ‘ b ‘ belongs to H and H is a subgroup, b - 1 belongs to H.
Since a, b - 1 belongs to H and H is a subgroup, (ab - 1) belongs to H.
Conversely, suppose that (ab - 1) belongs to H for every a, b in H.
To prove H is a subgroup of G.
We have a, a belongs to H implies (aa - 1) belongs to H
implies e belongs to H.
there fore e, a belongs to H implies (ea - 1) belongs to H
implies a - 1 belongs to H.
So a, b belongs to H implies a, b - 1 belongs to H
implies a(b - 1) - 1 belongs to H ( By assumption )
implies (ab) belongs to H.
Hence H is a subgroup of G.
Theorem: Let H and K be subgroups of a group G then H intersection K is also a subgroup of G.
Proof:
Claim: (ab - 1) belongs to H intersection K for every a, b in H intersection K.
Let a, b belongs to H intersection K.
Then ‘ a ‘ belongs to H intersection K and ‘ b ‘ belongs to H intersectionK..
We have, a, b belongs to H and H is a subgroup implies (ab - 1) belongs to H and
a, b belongs to K and K is a subgroup implies (ab - 1) belongs to K.
implies (ab - 1) belongs to H intersection K..
implies H intersection K. is also a subgroup of G.
Theorem: For any group G the set H = { x in G / xa = ax for every ‘a’ in G} forms a subgroup of G.
Proof:
Let x and y be two elements in H then xa = ax and ya = ay for every ‘a ‘in G.
We have ya = ay implies y - 1 ( ya) y - 1 = y - 1 (ay) y – 1
implies (y - 1 y)(a y - 1 ) = (y - 1 a) (y y – 1)
implies e (a y - 1 ) = (y - 1 a)e
implies (a y - 1 ) = (y - 1 a) for every ‘ a ‘ in G
implies y- 1 belongs to H.
That is y belongs to H implies y- 1 belongs to H.
Claim: (x y- 1) belongs to H.
We have (x y- 1)a = x (y- 1a)
= x (a y - 1 )
= (x a) y - 1
= (a x ) y - 1
= a (x y- 1) for every a in G.
There fore (x y- 1) belongs to H.
Hence H is a subgroup of G.
Centre of a group:
For any group G the set H = { x in G / xa = ax for every ‘a’ in G} forms a group is called the centre of a group G . It is denoted by Z(G) .
That is, Z(G) = { x in G / xa = ax for every ‘a’ in G}