Page 40 - DMTH502_LINEAR_ALGEBRA

Linear Algebra

Notes Commutative Property: The composition is commutative because the elements equidistant
from principal diagonal are equal each to each.
The set G has 6 elements. Hence (G, X ) is a finite abelian group of order 6.
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Self Assessment

7. Show that the set {1, 2, 3, 4} does not form a group under ‘addition modulo 5’, but it forms
a group under ‘multiplication modulo 5’.
8. Prove that the set {0, 1, 2, 3} is a finite abelian group of order 4 under addition modulo 4 as
composition.

1.3 Rings

The concept of a group has its origin in the set of mappings or permutations, of a set onto itself.
So far we have considered sets with one binary operation only. But rings are the outcome of the
motivation which arises from the fact that integers follow a definite pattern with respect to the
addition and multiplication. Thus we now aim at studying rings which are algebraic systems with
two suitably restricted and related binary operations.
Definition: An algebraic structure (R, +,.) where R is a non-empty set and + and . are two defined
operations in R, is called a ring if for all a, b, c in R, the following axioms are satisfied:
R . (R, +) is an abelian group, i.e.,
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(R ) a + b R (closure law for addition)
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(R ) (a + b) + c = a + (b + c) (associative law for addition)
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(R ) R has an identity, to be denoted by O, with respect to addition,
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i.e., a + 0 = a a R (Existence of additive identity)
(R ) There exists an additive inverse for every element in R, i.e., there exists an element –a in R
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such that
a + (–a) = 0 a R (Existence of additive inverse)
(R ) a + b = b + a (Commutative law for addition)
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R (R, .) is a semigroup, i.e.,
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(R ) a . b R (Closure law for multiplication)
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(R ) (a . b) . c = a . (b . c) (associative law for multiplication)
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R Multiplication is left as well as right distributive over addition, i.e.,
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a . (b + c) = a . b + a . c
and (b + c) . a = b . a + c . a

Elementary Properties of a Ring

Theorem 10: If R is a ring, then for all a, b R.
(a) a . 0 = 0 . a = 0
(b) a (–b) = (–a) b = – (ab)

(c) (–a) (–b) = ab

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