Elementary Group Theory



|Unit-1|Elementary Group Theory
1.(a) If a * b = 3a+2b for a, b ∈ z, the set of integers, show that * is a binary operation on z.
Solution:
Given: a * b =3a+2b for a, b ∈ z
Closure property 
If a,b ∈ z then a+b ∈ z  eg.If a=1 and b=2,thena * b=3a+2b=3⨯1+2⨯2=3+4=7 ∈ z. It is unique.∴ * is a binary operation on z.
(b) Define Abelian group
Abelian group :A group (G,*) is said to be an abelian group if a * b = b * a for all a,b ∈ G.
Permutation group: Let S be finite and has n elements, the group A (S) is also denoted by Sₙ. It has n! elements and is also called group of permutations of degree n or symmetric group of degree n.

(2) Given the algebraic structure (G, *) with G={1,ω,ω²} where ω represents the cube root of unity and * stands for the binary operation of ordinary multiplication of complex numbers. Show that (G, *) is a group.
Solution:
The Cayley's table is given below:
x1ωω²
11ωω²
ωωω²1
ω²ω²1ω
(i) Since 1×1= 1, 1 × ω=ω×1=ω and 1×ω²=ω²×1=ω²
So, 1 is the multiplicative identity.
(ii) Since 1×1=1, so, 1is the multiplicative inverse of itself.
(iii)Since ω×ω²=ω²×ω=1
So, ω is the inverse of ω² ans ω² is the inverse of ω.
(iv) ω×ω²=1∈G and 1 × ω= ω∈G .It shows that the product of any two elements of G is also an element of G. So the operation * is closed.
(v) (1×ω)×ω²=ω×ω²=1 and 1 ×(ω×ω²)=1×1=1
∴(1×ω)×ω²=1×(ω×ω²)=1
∴ It also satisfies the associative, property
Thus,(G, *) is a group.

(3) Prepare the composition table of the addition modulo 3 and the multiplication modulo 3 for the set G={0,1,2}
Solution:
The composition tables are given below:








+₃012
0012
1120
2201










x₃012
0000
1012
2021

(4).Show that the set of integers Z forms a group under the operation of addition.
Solution:
a+b ∈ Z and is unique
∴Z is closed under addition
a+(b+c) = (a+b)+c
∴It holds associative law.
0∈G is identity,Since 0+a = a=a+0,∀a ∈ G
For any a ∈ G,we have -a ∈ G and
a+(-a)=0=(-a)+a
This all shows that G is a group under addition.

(5) If a,b∈ (G, *), then (i) (a⁻¹)⁻¹ =a, (ii) (a, b)⁻¹=b⁻¹a⁻¹
Proof : (i) We have a⁻¹ * a=e
Multiplying both sides on the left by (a⁻¹)⁻¹,we get
(a⁻¹)⁻¹ * (a⁻¹ * a)= (a⁻¹)⁻¹ * e
or, [(a⁻¹)⁻¹ * a⁻¹] * a =(a⁻¹)⁻¹ (by associativity law) e * a= (a⁻¹)⁻¹=a=(a⁻¹)¹
(ii) We have, (a * b) (b⁻¹ * a⁻¹) = a * (b * b⁻¹) *a⁻¹(by Associative law)
=a * e * a⁻¹ (by defⁿ of inverse) a * a⁻¹ (by defⁿ of identity)
=e (by defⁿ of inverse) and similarly (b⁻¹ * a⁻¹) (a * b)
=e. So, b⁻¹ * a⁻¹ is the inverse of a * b, i.e. (a,b)⁻¹= b⁻¹ * a⁻¹

(6) Show that the multiplication is a binary operation on the set S={-1,0,1} but the subtraction is not.
Solution:
Let us consider the operation of multiplication.
Let x,y,∈ S or not ∀ x,y ∈ S.
Since,-1 * 0= 0 ∈ S, 0 * 1 = 0 ∈ S
1 *(-1)= -1 ∈ S, 1* 1 = 1 ∈ S and -1 * (-1) =1∈ S, 0 * 0= 0
So, multiplication is the binary operation.
Again,the operation '-' , Let x-y ∈ S or not for all x,y ∈ S.
Since, -1-0=-1∈ S,1-0=1 ∈ S
but-1-1 =2 doesn't belongs to S•
∴ Subtraction is not the binary operation•

(7) Given a set G ={ 0,1} and a binary operation + defined by :
0 +0 =0;    0+1 = 1
1+0= 1;     1+ 1 =0
Find the additive identity and additive inverse of 0 and 1.

+01

001

110
Solution:
(a) Since 0+0=0; 0+1 =1+0, so 0 is the additive identity,
(b) Since 0+0= 0; 1+1 =0, so, 0 is the additive inverse of 0, and 1 is the additive inverse of 1. In other words , every element of G is the additive inverse of itself.
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