Publish On: 2019-05-24

# Ben joy

Total Post: 542

## Question: Describe the multiplication modulo p

Reply On: 2013-10-07

# Mjay Jollay

Total Post: 0

## ANS: Describe the multiplication modulo p

The multiplication modulo p, p being a positive integer, of any two integers a and b is denoted by a ×

a ×

where r is a least non-negative remainder when ab (i.e., the product of a and b) is divided by p. In other words, we find an ordinary product of two integers a and b, viz. ab and from this product we remove the multiples of p in such a way that the remainder r left out is such that either r = 0 or r<p and r is a positive integer.

For example,

8 ×

9 ×

Example: Prove that the set {1, 2, 3, 4, 5, 6} is a finite Abelian group of order 6 under multiplication modulo 7.

Solution: Let G = {1, 2, 3, 4, 5, 6} and let us form the composition table X

G

G

aX

= least non-negative remainder obtained on dividing a(bc) by 7.

= least non-negative remainder obtained on dividing (ab)c by 7. [∵ (ab)c = a. (bc) ∀ a, b c I]

= (cb) ×

≡ (a ×

G

1 × 7 a = a = aX7 1

G

G

Surely, the set G has 6 elements.

Hence <G, ×

_{p}b and is defined asa ×

_{p}b = r, 0 ≤ r p,where r is a least non-negative remainder when ab (i.e., the product of a and b) is divided by p. In other words, we find an ordinary product of two integers a and b, viz. ab and from this product we remove the multiples of p in such a way that the remainder r left out is such that either r = 0 or r<p and r is a positive integer.

For example,

8 ×

_{5}7 = 1 as 8 × 7 = 56 = 5.11 + 19 ×

_{4}7 = 3 as 9 × 7 = 63 = 15.4 + 3.Example: Prove that the set {1, 2, 3, 4, 5, 6} is a finite Abelian group of order 6 under multiplication modulo 7.

Solution: Let G = {1, 2, 3, 4, 5, 6} and let us form the composition table X

_{7},G

_{1}: Closure property: From the composition table, it is clear that all the entries are the element of G and, therefore, closure property holds good.G

_{2}: Associative law: For all a, b, c G,aX

_{7}(bX_{7}c) = a ×_{7}(b.c) as b ×_{7}c≡bc (mod 7)= least non-negative remainder obtained on dividing a(bc) by 7.

= least non-negative remainder obtained on dividing (ab)c by 7. [∵ (ab)c = a. (bc) ∀ a, b c I]

= (cb) ×

_{7}c≡ (a ×

_{7}b) X_{7}c [∵ aX_{7}b≡ab mod 7]G

_{3}: Identity element: 1 G is the identity element as1 × 7 a = a = aX7 1

G

_{4}: Existence of inverse: The inverses of 1, 2, 3, 4, 5, 6 are 1, 4, 5, 2, 3, 6 respectively.G

_{5}: Commutative law: The corresponding rows and columns in the composition table are identical, and as such the commutative law holds good.Surely, the set G has 6 elements.

Hence <G, ×

_{7}> is a finite albelian group of order 6.Like Us On Facebook for All Latest Updates