Page 13 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Definition: A relation R defined on a set S is said to be

(i) reflexive if we have aRa, V a  S.
(ii) symmetric if aRb  bRa V a, b  S.

(iii) transitive if aRb and bRc  aRc V a, b, c  S.
To get used to these concepts, consider the following examples.

Example: Consider the relation R on Z given by ‘aRb if and only if a > b’. Determine
whether R is reflexive, symmetric and transitive.
Solution: Since a > a is not true, aRa is not true. Hence, R is not reflexive.

If a > b, then certainly b > a is not true. That is, aRb does not imply bRa. Hence, R is not
symmetric.
Since a > b and b > c implies a > c, we find that aRb, bRc implies aRc. Thus, R is transitive.

Example: Let S be a non-empty set. Let (S) denote the set of all subsets of S, i.e.,
(S) = (A | A  S}. We call p (S) the power set of S.
Define the relation R on (S) by
R= { (A,B) | A, B  (S) and A  B ].

Check whether R is reflexive, symmetric or transitive.
Solution: Since A  A V A   (S), R is reflexive.
If A  B, B need not be contained in A. (In fact, A  B and B  A  A = B.) Thus, R is not symmetric.

If A  B and B  C, then A C V A, B, C  (S). Thus, R is transitive.
A very important property of an equivalence relation on a set S is that it divides S into a number
of mutually disjoint subsets, that is, it partitions S. Let us see how this happens.
Let R be an equivalence relation on the set S. Let a  S. Then the set { b  S | aRb } is called the
equivalence class of a in S. It is just the set of elements in S which are related to a. We denote it
by [a].
This is
[1] = { n | 1 R n ; n  N }

= { n 1 n N and 5 divides 1-n)
= { n I n  N and 5 divides n-1)
= { 1, 6, 11, 16, 21 ........ }.
Similarly,

[2] = { n | n  N and 5 divides n–2 }
= { 2, 7, 12, 17, 22. ......... },
[3] = { 3, 8, 13, 18, 23 .......... },
[4] = { 4 , 9, 14 , 19, 24, . ........ },

[5] = { 5, 10, 15, 20, 25, ....... },

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