Page 19 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes So, f o g = I and g o f = I . This proves that g = f .
–1
B A
Conversely, suppose f has an inverse and that g = f . We must prove that f is one-one and onto.
–1

Note g o f is 1 – 1  f is 1 – 1

Suppose f(a ) = f(a ). Then g(f(a )) = g(f(a )).
1
1
2
2
 g o f(a ) = g o f(a )
1
2
 a = a , because g o f = I .
 2 A
So, f is one-one.

Note g o f is onto  g is onto.

Next, given b  B, we have f o g = I , So that f o g(b) = I (b) = b,
B
B
i.e., f(g(b)) = b. That is, f is onto.
Hence, the theorem is proved.
1.5 Some Number Theory

In this section we will spell out certain factorisation properties of integers that we will use
throughout the course. For this we first need to present the principle of finite induction.

1.5.1 Principle of Induction

We will first state an axiom of the integers that we will often use implicitly, namely, the
well-ordering principle. We start with a definition.
Definition: Let S be a non-empty subset of Z. An element a  S is called a least element (or a
minimum element) of S if a  b V b  S. For example, N has a least element, namely, 1. But Z
has no least element. In fact, many subsets of Z, like 2Z, (-1, -2, -3, ...... ), etc., don’t have least
elements.
The following axiom tells us of some sets that have a least element.
Well-ordering Principle: Every non-empty subset of N has a least element.

You may be surprised to know that this principle is actually equivalent to the principle of finite
induction, which we now state.
Theorem 5: Let S  N such that

(i) 1  S, and
(ii) whenever k  S, then k + l  S.
Then S = N.
This theorem is further equivalent to:

12 LOVELY PROFESSIONAL UNIVERSITY