Page 36 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes = least non-negative remainder when a + (b + c) is divisible by m
= least non-negative remainder when (a + b) + c divided by m.
since a + (b + c) = (a + b) + c
+
= (a + b) + c [by definition of m]
m
= (a + b) + c [ a + b = a + b (mod m)]
m m m
‘+ ’ is an associative composition.
m
Existence of Identity Element: We have 0 G. Also, if a is any element of G, then 0 + a = a + m .
0
m
Therefore 0 is the identity element.
Existence of Inverse: The inverse of 0 is 0 itself. If r G and r 0, then m – r G. Also (m – r) + r
m

= 0 = r + m (m – r). Therefore (m – r) is the inverse of r.
Commutative Property: The composition ‘ m’ is commutative also.
+
Since
a + b = least non-negative remainder when a + b is divided by m
m
= least non-negative remainder when b + a is divided by m
= b + a.
m
The set G contains m elements.

Hence (G, m) is a finite abelian group of order m.
+
Multiplicative Group of Integers Modulo p where p is Prime

The set G of (p – 1) integers 1, 2, 3, …, p – 1, p being prime, is a finite abelian group of order
p – 1, the composition being multiplication modulo p.
Let G = {1, 2, 3, … p – 1} where p is prime.

Closure Property: Let a and b be any elements of G. Then 1 < a < p – 1, 1 < b < p – 1. Now by
definition a × b = r where r is the least non-negative remainder when the ordinary product a b
p
is divided by p. Since p is prime, therefore a b is not exactly divisible by p. Therefore r cannot be
zero and w shall have 1 r p – 1. Thus a × b G a, b G. Hence the closure axiom is satisfied.
p
Associative Law: a, b, c, be any arbitrary elements of G.
Then a p b p c a p (bc ) [ b × C = bc (mod p)]
p
= least non-negative remainder when a (bc) is divided by p
= least non-negative remainder when ab (c) is divided by p
= (ab) × C
p
= (a × b) × C [ ab = a × pb (mod p)]
p p
‘X’ is an associative composition.
p
Existence of left identity: We have 1 G. Also if a is any element of G, then 1 × a = a. Therefore
p
1 is the left identity.
Existence of left inverse: Let s be any member of G. Then 1 < s < p – 1.

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