Page 31 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Conversely, if we are given a table, we can define a binary operation on S. For example, we can
define the operation * on S = {1, 2, 3} by the following table.

* 1 2 3
1 1 2 3
2 3 1 2
3 2 3 1

From this table we see that, for instance, 1 * 2 = 2 and 2 * 3 = 2.

Now 2 * 1 = 3 and 1 * 2 = 2. 2 * 1  1 * 2. That is, * is not commutative.
Again, (2 * 1) * 3 = 3 * 3 = 1 and 2 * (1 * 3) = 2

 (2 * 1) * 3  2 * (1 * 3).  , * is not associative.
See how much information a mere table can give !

Now consider the following definition.
Definition: Let * be a binary operation on a non-empty set S and let a,, . . . . . .,a ,  S. We define
k+1
the product a, * . ... .. * a , as follows:
k+1
If k = 1, a, * a is a well defined element in S.
2
If a * ..... * a, is defined, then
l
a, * .... . . * a = (a, * . ..... * a,) * a k+1
k+l
We use this definition in the following result.

Theorem 2: Let a,, ...... ,a m+n be elements in a set S with an associative binary operation *. Then
(a, * ..,...* a,) * (a,,, * ...... * a m+n ) = a * ...... * a m+n .
1
Proof: We use induction on n. That is, we will show that the statement is true for n = 1. Then,
assuming that it is true for n – 1, we will prove it for n.
If n = 1, our definition above gives us
(a, .......* a,) * a,,, = a, *...... * a m+1 .

Now, assume that
(a, *.......* a,) * (a,,, ........ a m+n–1 ) = a, * ....... a,,,,
Then
(a, * ......... * a) * (a,,, .......... a)

= (a *.........* a,) * ((a m+1 * ..... * a m+n–1 ) * a m+n )
1
= ( (a, *......* a,,) * (a,,, *......* a,,,) ) * a, since * is associative
= (a ......... a m+n–1 ) * a m+n by induction
1
= a, * ....... a, by definition.
Hence, the result holds for all n.

We will use Theorem 2 quite often in this course, without explicitly referring to it.
Now that we have discussed binary operations let us talk about groups.

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