Page 37 - DMTH502_LINEAR_ALGEBRA

Unit 1: Vector Space over Fields

Let us consider the following (p – 1) products: Notes
1 × s, 2 × s, 3 × s, … (p – 1) × s.
p p p p
All these are elements of G. Also no two of these can be equal as shown below:
Let i and j be two unequal integers such that

1 i p 1, 1 j p 1 and i j

Then i p s j p s
i s and j s leave the same least non-negative remainder when divided by p

i s – j s is divisible by p
(i – j) s is divisible by p.
Since 1 (i – j) < p – 1; 1 s p – 1 and p is prime, therefore (i – j) s cannot be divided by p.

i × s × j × s.
p p
Thus 1 × s, 2 × s, … (p – 1) × S are (p – 1) distinct elements of the set G. Therefore one of these
p p p
elements must be equal to 1.
Let s × s = 1. The s is the left inverse of s.
p
Commutative Law: The composition ‘Xp’ is commutative, since
a × b = least non-negative remainder when ab is divisible by p
p
= least non-negative remainder when ba is divided by p
= b × a
p
(G, X ) is a finite abelian group of order p – 1.
p
Theorem 8: The residue classes modulo form a finite group with respect to addition of residue
classes

Proof: Let G be the set of residue classes (mod m), then
G = { {0}, {1}, {2}, … {r }, … {r }, … {m – 1} }
1 2
or G = {0, 1, 2, … {r } … {r } … m – 1 (mod m)}
1 2
Closure axiom: (r ) + {r } = {r + r }
1 2 1 2
= {r} G where r is the least positive integer obtained as
remainder when r + r is divided by m (0 r m).
1 2
Thus the closure axiom is satisfied.
Associative axiom: The addition is associative.
Identity axiom: {0} G) and {0} + {r} = {r}. Hence the identity for addition is {0}.
Inverse axiom: Since {m – r} + {r} = {m} = {0}, the additive inverse of the element {r} is {m – r}.
Hence G is a finite group with respect to addition modulo m.

Theorem 9: The set of non-zero residue classes modulo p, where p is a prime, forms a group with
respect to multiplication of residue classes.
Proof: Let I = {…, –3, –2, –1, 0, 1, 2, 3, …} be the set of integers. Let a then {a} is residue class
modulo p of I,

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