Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last updated: 17 September 2014

Lecture 3: Equivalence relations

A set is a collection of elements.

Let S and T be sets. The product of S and T is the set S×T= { (s,t)| sS,tT } .

If S={1,2,3} then S×S= { (1,1), (1,2), (1,3) (2,1), (2,2), (2,3) (3,1), (3,2), (3,3) } .

Let S be a set. A relation on S is a subset of S×S.

< is a relation on . a<b if there exists x>0 such that a+x=b. a<b means (a,b) is in the relation <.

Let m. Let a,b. Define =modm, a relation on , by a=bmodmif ra=rb, where a=qam+ra and b=qbm+rb with qa,qb and 0ra<|m| and 0rb<|m|.

Define amodm to be ra, where a=qam+ra with qa and 0ra<|m|.

Let S be a set. Let be a relation on S. Write s1s2 if (s1,s2) is in the relation .

The relation is reflexive if satisfies: if sS then ss.

The relation is symmetric if satisfies: if s1,s2S and s1s2 then s2s1.

The relation is transitive if satisfies: if s1,s2,s3S and s1s2 and s2s3 then s1s3.

An equivalence relation on S is a relation on S that is reflexive, symmetric and transitive.

Let S be a set. Let be an equivalence relation on S. Let sS. The equivalence class of s is the set [s]= {xS|xs}.

A partition of S is a collection 𝒮 of subsets of S such that

(a) Y𝒮Y=S.
(b) If X,Y𝒮 and XY then XY=.

Let m=7. Then 36mod7 = 1,since 36=5·7+1, -6mod7 = 1, since-6=-1·7+1, 1mod7 = 1,since 1=0·7+1. The equivalence class of 36 is [36]= { ,-13,-6,1,8,15 ,22,29,36, } = [1], { ,-12,-5,2,9, 16,23,30,37, } = [2], { ,-11,-4,3,10 ,17,24, } = [3], { ,-10,-3,11, 18,25, } = [4], { ,-9,-2,5,12, 19,26, } = [5], { ,-8,-1,6,13, 20,27, } = [6], { ,-7,0,7,14,21 ,28, } = [7]. Recall that m={1,2,3,4,5,6,7}.

Note that { [1], [2], [3], [4], [5], [6], [7] } is a partition of since

(a) [1][2][3][4][5][6][7] and
(b) if i,j{1,,7} and ij then [i][j]=.

Let m. Then =modm is an equivalence relation on .


To show:
(a) =modm is reflexive.
(b) =modm is symmetric.
(c) =modm is transitive.
To show:
(a) If a then a=amodm.
(b) If a,b and a=bmodm then b=amodm.
(c) If a,b,c and a=bmodm and b=cmodm then a=cmodm.
Assume a,b,c and a=bmodm and b=cmodm.
Let a=qa·m+ra, b=qb·m+rb, c=qc·m+rc with 0ra<m, 0rb<m, 0rc<m.
Since a=bmodm and b=cmodm then ra=rband rb=rc. Since = is an equivalence relation on , ra=ra, rb=raand ra=rc. So a=amodm, b=amodm and a=cmodm.
So =modm is an equivalence relation on .

Notes and References

These are a typed copy of Lecture 3 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on July 29, 2011.

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