Group Theory and Linear Algebra

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

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 $$S\times T=\left\{(s,t)\hspace{0.17em}\right|\hspace{0.17em}s\in S,t\in T\}\text{.}$$

If $S=\{1,2,3\}$ then $$S\times S=\left\{\begin{array}{ccc}(1,1),& (1,2),& (1,3)\\ (2,1),& (2,2),& (2,3)\\ (3,1),& (3,2),& (3,3)\end{array}\right\}\text{.}$$

Let $S$ be a set. A relation on $S$ is a subset of $S\times S\text{.}$

$<$ is a relation on $\mathbb{Z}\text{.}$ $a<b$ if there exists $x\in {\mathbb{Z}}_{>0}$ such that $a+x=b\text{.}$ $a<b$ means $(a,b)$ is in the relation $<\text{.}$

Let $m\in \mathbb{Z}\text{.}$ Let $a,b\in \mathbb{Z}\text{.}$ Define $=\text{mod}\hspace{0.17em}m,$ a relation on $\mathbb{Z},$ by $$a=b\hspace{0.17em}\text{mod}\hspace{0.17em}m\text{if}{r}_a=r_b,$$ where $a=q_{a}m+r_a$ and $b=q_{b}m+r_b$ with $q_a,q_b\in \mathbb{Z}$ and $0\le r_a<\left|m\right|$ and $0\le r_b<\left|m\right|\text{.}$

Define $a\hspace{0.17em}\text{mod}\hspace{0.17em}m$ to be $r_a,$ where $a=q_{a}m+r_a$ with $q_a\in \mathbb{Z}$ and $0\le r_a<\left|m\right|\text{.}$

Let $S$ be a set. Let $\sim$ be a relation on $S\text{.}$ Write $s_1\sim s_2$ if $(s_1,s_2)$ is in the relation $\sim \text{.}$

The relation $\sim$ is reflexive if $\sim$ satisfies: if $s\in S$ then $s\sim s\text{.}$

The relation $\sim$ is symmetric if $\sim$ satisfies: if $s_1,s_2\in S$ and $s_1\sim s_2$ then $s_2\sim s_1\text{.}$

The relation $\sim$ is transitive if $\sim$ satisfies: if $s_1,s_2,s_3\in S$ and $s_1\sim s_2$ and $s_2\sim s_3$ then $s_1\sim s_3\text{.}$

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

Let $S$ be a set. Let $\sim$ be an equivalence relation on $S\text{.}$ Let $s\in S\text{.}$ The equivalence class of $s$ is the set $$\left[s\right]=\{x\in S\hspace{0.17em}|\hspace{0.17em}x\sim s\}\text{.}$$

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

(a)	$\bigcup _{Y\in 𝒮}Y=S\text{.}$
(b)	If $X,Y\in 𝒮$ and $X\ne Y$ then $X\cap Y=\varnothing \text{.}$

Let $m=7\text{.}$ Then $$\begin{array}{rcl}{\displaystyle 36\hspace{0.17em}\text{mod}\hspace{0.17em}7}& =& {\displaystyle 1,\text{since}36=5·7+1,}\\ {\displaystyle -6\hspace{0.17em}\text{mod}\hspace{0.17em}7}& =& {\displaystyle 1,\text{since}-6=-1·7+1,}\\ {\displaystyle 1\hspace{0.17em}\text{mod}\hspace{0.17em}7}& =& {\displaystyle 1,\text{since}1=0·7+1\text{.}}\end{array}$$ The equivalence class of $36$ is $$\begin{array}{ccc}{\displaystyle \left[36\right]=\{\dots ,-13,-6,1,8,15,22,29,36,\dots \}}& =& {\displaystyle \left[1\right],}\\ {\displaystyle \{\dots ,-12,-5,2,9,16,23,30,37,\dots \}}& =& {\displaystyle \left[2\right],}\\ {\displaystyle \{\dots ,-11,-4,3,10,17,24,\dots \}}& =& {\displaystyle \left[3\right],}\\ {\displaystyle \{\dots ,-10,-3,11,18,25,\dots \}}& =& {\displaystyle \left[4\right],}\\ {\displaystyle \{\dots ,-9,-2,5,12,19,26,\dots \}}& =& {\displaystyle \left[5\right],}\\ {\displaystyle \{\dots ,-8,-1,6,13,20,27,\dots \}}& =& {\displaystyle \left[6\right],}\\ {\displaystyle \{\dots ,-7,0,7,14,21,28,\dots \}}& =& {\displaystyle \left[7\right]\text{.}}\end{array}$$ Recall that $\raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$m\mathbb{Z}$}\right.=\{1,2,3,4,5,6,7\}\text{.}$

Note that $$\{\left[1\right],\left[2\right],\left[3\right],\left[4\right],\left[5\right],\left[6\right],\left[7\right]\}$$ is a partition of $\mathbb{Z}$ since

(a)	$\left[1\right]\cup \left[2\right]\cup \left[3\right]\cup \left[4\right]\cup \left[5\right]\cup \left[6\right]\cup \left[7\right]$ and
(b)	if $i,j\in \{1,\dots ,7\}$ and $i\ne j$ then $$\left[i\right]\cap \left[j\right]=\varnothing \text{.}$$

Let $m\in \mathbb{Z}\text{.}$ Then $=\hspace{0.17em}\text{mod}\hspace{0.17em}m$ is an equivalence relation on $\mathbb{Z}\text{.}$

		Proof.
	To show: (a) $=\hspace{0.17em}\text{mod}\hspace{0.17em}m$ is reflexive. (b) $=\hspace{0.17em}\text{mod}\hspace{0.17em}m$ is symmetric. (c) $=\hspace{0.17em}\text{mod}\hspace{0.17em}m$ is transitive. To show: (a) If $a\in \mathbb{Z}$ then $a=a\hspace{0.17em}\text{mod}\hspace{0.17em}m\text{.}$ (b) If $a,b\in \mathbb{Z}$ and $a=b\hspace{0.17em}\text{mod}\hspace{0.17em}m$ then $b=a\hspace{0.17em}\text{mod}\hspace{0.17em}m\text{.}$ (c) If $a,b,c\in \mathbb{Z}$ and $a=b\hspace{0.17em}\text{mod}\hspace{0.17em}m$ and $b=c\hspace{0.17em}\text{mod}\hspace{0.17em}m$ then $a=c\hspace{0.17em}\text{mod}\hspace{0.17em}m\text{.}$ Assume $a,b,c\in \mathbb{Z}$ and $a=b\hspace{0.17em}\text{mod}\hspace{0.17em}m$ and $b=c\hspace{0.17em}\text{mod}\hspace{0.17em}m\text{.}$ Let $a=q_a·m+r_a,$ $b=q_b·m+r_b,$ $c=q_c·m+r_c$ with $0\le r_a<m,$ $0\le r_b<m,$ $0\le r_c<m\text{.}$ Since $a=b\hspace{0.17em}\text{mod}\hspace{0.17em}m$ and $b=c\hspace{0.17em}\text{mod}\hspace{0.17em}m$ then $$r_a=r_b\text{and}{r}_b=r_c\text{.}$$ Since $=$ is an equivalence relation on $\mathbb{Z},$ $$r_a=r_a,r_b=r_a\text{and}{r}_a=r_c\text{.}$$ So $a=a\hspace{0.17em}\text{mod}\hspace{0.17em}m,$ $b=a\hspace{0.17em}\text{mod}\hspace{0.17em}m$ and $a=c\hspace{0.17em}\text{mod}\hspace{0.17em}m\text{.}$ So $=\hspace{0.17em}\text{mod}\hspace{0.17em}m$ is an equivalence relation on $\mathbb{Z}\text{.}$ $\square$

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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