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: 16 September 2014

Lecture 2: gcd and Euclid's algorithm

Number systems – $\mathbb{Z},$ the integers

$$\mathbb{Z}=\{\dots ,(-1)+(-1)+(-1),(-1)+(-1),-1,0,1,1+1,1+1+1,\dots \}$$ with $(-1)+1=0,$ $1+(-1)=0,$ $0+1=1,$ $1+0=1,$ $0+(-1)=-1,$ $(-1)+0=-1\text{.}$

Let $d\in \mathbb{Z}\text{.}$ The multiples of $d$ is $$d\mathbb{Z}=\{\dots ,(-d)+(-d)+(-d),(-d)+(-d),-d,0,d,d+d,d+d+d,\dots \}\text{.}$$

Let $a,d\in \mathbb{Z}\text{.}$ The integer $d$ divides $a,$ $d|a,$ if $a\in d\mathbb{Z}\text{.}$

Let $x,m\in \mathbb{Z}\text{.}$ The greatest common divisor of $x$ and $m,$ $\text{gcd}(x,m),$ is $d\in {\mathbb{Z}}_{>0}$ such that

(a)	$d|x$ and $d|m,$
(b)	If $\ell \in {\mathbb{Z}}_{>0}$ and $\ell |x$ and $\ell |m$ then $\ell |d\text{.}$

Let $a,b\in \mathbb{Z}\text{.}$ Define $a<b$ if there exists $x\in {\mathbb{Z}}_{>0}$ such that $a+x=b\text{;}$
$a\le b$ if $a<b$ or $a=b\text{.}$

(Euclidean algorithm) Let $a,b\in \mathbb{Z}\text{.}$ There exist unique $q,r\in \mathbb{Z}$ such that

(a)	$a=bq+r,$
(b)	$0\le r<\left|b\right|,$ where $\left|b\right|=\{\begin{array}{cc}b,& \text{if}\hspace{0.17em}b\in {\mathbb{Z}}_{>0},\\ 0,& \text{if}\hspace{0.17em}b=0,\\ -b,& \text{if}\hspace{0.17em}-b\in {\mathbb{Z}}_{>0}\text{.}\end{array}$

If (a) and (b) hold write $a=r\hspace{0.17em}\text{mod}\hspace{0.17em}b\text{.}$

The 15th row of the multiplication table for $\raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$36\mathbb{Z}$}\right.$ is $$\begin{array}{ccccccccccccccccccccc}·& 1& 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12& 13& 14& 15& 16& 17& 18& 19& \cdots \\ 15& 15& 30& 9& 24& 3& 18& 33& 12& 27& 6& 21& 36& 15& 30& 9& 24& 3& \mathrm{<mi>\cdots </mi>}& \mathrm{}\end{array}$$ Notice that

(a)	$$\begin{array}{ccc}{\displaystyle 15·10}& =& {\displaystyle 150\text{in}\mathbb{Z},}\\ {\displaystyle 150}& =& {\displaystyle 4·36+6,\text{and}}\\ {\displaystyle 15·10}& =& {\displaystyle 6\text{in}\raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$36\mathbb{Z}$}\right.\text{.}}\end{array}$$
(b)	The numbers in row $15$ of the multiplication table for $\raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$36\mathbb{Z}$}\right.$ are $$3,6,9,12,15,18,21,24,27,30,33,36$$ (all multiples of $3$ in $\raisebox{1ex}{$\mathbb{Z}$}\!\left/ \!\raisebox{-1ex}{$36\mathbb{Z}$}\right.\text{).}$
(c)	$3=15·17+12(-7)\text{.}$

Let $x,m\in \mathbb{Z}\text{.}$ There exists $\ell \in {\mathbb{Z}}_{>0}$ such that $$\ell \mathbb{Z}=x\mathbb{Z}+m\mathbb{Z}\text{.}$$

Let $x,m\in \mathbb{Z}\text{.}$ Let $\ell \in {\mathbb{Z}}_{>0}$ such that $\ell \mathbb{Z}=x\mathbb{Z}+m\mathbb{Z}\text{.}$ Let $g=\text{gcd}(x,m)\text{.}$ Then $d=\ell \text{.}$

(Euclidean algorithm). Let $a,b\in \mathbb{Z}\text{.}$ There exist unique $q,r\in \mathbb{Z}$ such that

(a)	$a=bq+r,$
(b)	$0\le r<\left|b\right|,$ where $\left|b\right|=\{\begin{array}{cc}b,& \text{if}\hspace{0.17em}b\in {\mathbb{Z}}_{>0},\\ 0,& \text{if}\hspace{0.17em}b=0,\\ -b,& \text{if}\hspace{0.17em}-b\in {\mathbb{Z}}_{>0}\text{.}\end{array}$

		Proof.
	Assume $a,b\in \mathbb{Z}\text{.}$ To show: (a) There exist $q,r\in \mathbb{Z}$ such that (1) $a=bq+r,$ (2) $0\le r<\left|b\right|\text{.}$ (b) $q,r\in \mathbb{Z}$ such that $a=bq+r$ and $0\le r<\left|b\right|$ are unique. (a) Let $bq=$ the smallest integer in $b\mathbb{Z}$ less than or equal to $a$ and $r=a-qb\text{.}$ To show: (aa) $a=bq+r,$ (ab) $0\le r<\left|b\right|\text{.}$ (aa) Since $r=a-qb$ then $a=bq+r\text{.}$ (ab) Since $bq\le a$ and $b(q+1)>a,$ $0\le a-bq$ and $b>a-bq\text{.}$ So $0\le r$ and $b>r\text{.}$ (b) Assume $q_1,r_1,\in \mathbb{Z}$ and $a=b{q}_1+r_1$ and $0\le r_1<\left|b\right|$ and assume $q_2,r_2\in \mathbb{Z}$ and $a=b{q}_2+r_2$ and $0\le r_2<\left|b\right|\text{.}$ $q_1=q_2$ and $r_1=r_2\text{.}$ Since $a-r_1=b{q}_1$ and $0\le r_1<\left|b\right|,$ $b{q}_1$ is the largest integer in $b\mathbb{Z}$ which is $\le a\text{.}$ Since $a-r_2=b{q}_2$ and $0\le r_2<\left|b\right|,$ $b{q}_2$ is the largest integer in $b\mathbb{Z}$ which is $\le a\text{.}$ So $b{q}_1=b{q}_2$ and $q_1=q_2\text{.}$ So $r_1=a-b{q}_1=a-b{q}_2=r_2\text{.}$ $\square$

Using Euclid's algorithm find $$\text{gcd}(1288,1144)\text{.}$$ Hodgson says: If $a=bq+r$ with $0\le r<\left|b\right|$ then $\text{gcd}(a,b)=\text{gcd}(b,r)\text{.}$ $$\begin{array}{ccc}{\displaystyle 1288}& =& {\displaystyle 1144+144}\\ {\displaystyle 1144}& =& {\displaystyle 7·144+136}\\ {\displaystyle 144}& =& {\displaystyle 136+8}\\ {\displaystyle 136}& =& {\displaystyle 17·8+0}\end{array}\begin{array}{ccc}{\displaystyle 9·144}& =& {\displaystyle 1296}\\ {\displaystyle 8·144}& =& {\displaystyle 1152}\\ {\displaystyle 7·144}& =& {\displaystyle 1008}\end{array}$$ So $$\begin{array}{ccc}{\displaystyle \text{gcd}(1288,144)}& =& {\displaystyle \text{gcd}(1144,144)}\\ & =& {\displaystyle \text{gcd}(144,136)}\\ & =& {\displaystyle \text{gcd}(136,8)}\\ & =& {\displaystyle \text{gcd}(8,0)=8\text{.}}\end{array}$$ Note: $$\begin{array}{ccc}{\displaystyle 8}& =& {\displaystyle 144-136}\\ & =& {\displaystyle 144-(1144-7·144)=8·144-1144}\\ & =& {\displaystyle 8(1288-1144)-1144}\\ & =& {\displaystyle 8·1288-9·1144\text{.}}\end{array}$$

Let $x,m\in {\mathbb{Z}}_{>0}$ with $1\le x\le m\text{.}$ There exists $\ell \in {\mathbb{Z}}_{>0}$ such that $$\ell \mathbb{Z}=x\mathbb{Z}+m\mathbb{Z}\text{.}$$

		Proof.
	Let $\ell \in {\mathbb{Z}}_{>0}$ be minimal such that $\ell \in x\mathbb{Z}+m\mathbb{Z}\text{.}$ To show: $\ell \mathbb{Z}=z\mathbb{Z}+m\mathbb{Z}\text{.}$ To show: (a) $\ell \mathbb{Z}\subseteq x\mathbb{Z}+m\mathbb{Z}\text{.}$ (b) $x\mathbb{Z}+m\mathbb{Z}\subseteq \ell \mathbb{Z}\text{.}$ (a) Since $\ell \in x\mathbb{Z}+m\mathbb{Z},$ $$\ell \mathbb{Z}\subseteq x\mathbb{Z}+m\mathbb{Z}\text{.}$$ (b) Assume $y\in x\mathbb{Z}+m\mathbb{Z}\text{.}$ To show: $y\in \ell \mathbb{Z}\text{.}$ Since $\ell$ is minimal $y\ge \ell \text{.}$ So $y=q\ell +r$ with $0\le r<\ell \text{.}$ So $r=y-q\ell \in x\mathbb{Z}+m\mathbb{Z}\text{.}$ So $r=0,$ since $\ell$ is minimal positive integer in $x\mathbb{Z}+m\mathbb{Z}\text{.}$ So $y=q\ell \text{.}$ So $y\in \ell \mathbb{Z}\text{.}$ So $\ell \mathbb{Z}=x\mathbb{Z}+m\mathbb{Z}\text{.}$ $\square$

Let $x,m\in \mathbb{Z}\text{.}$ Let $\ell \in {\mathbb{Z}}_{>0}$ such that $\ell \mathbb{Z}=x\mathbb{Z}+m\mathbb{Z}\text{.}$ Let $d=\text{gcd}(x,m)\text{.}$ Then $d=\ell \text{.}$

		Proof.
	Let $d=\text{gcd}(x,m)\text{.}$ Let $\ell \in {\mathbb{Z}}_{>0}$ such that $\ell \mathbb{Z}=x\mathbb{Z}+m\mathbb{Z}\text{.}$ To show: $\ell =d\text{.}$ To show: (a) $\ell |d\text{.}$ (b) $d|\ell \text{.}$ (a) Since $x\in \ell \mathbb{Z},$ then $\ell |x\text{.}$ Since $m\in \ell \mathbb{Z},$ then $\ell |m\text{.}$ Since $d=\text{gcd}(x,m),$ then $\ell |d\text{.}$ (b) Since $d|x$ and $d|m,$ then $x\in d\mathbb{Z}$ and $m\in d\mathbb{Z}\text{.}$ So $x\mathbb{Z}+m\mathbb{Z}\subseteq d\mathbb{Z}\text{.}$ So $\ell \mathbb{Z}\subseteq d\mathbb{Z}\text{.}$ So $\ell \in d\mathbb{Z}\text{.}$ So $d|\ell \text{.}$ $\square$

Notes and References

These are a typed copy of Lecture 2 from a series of handwritten lecture notes for the class Group Theory and Linear Algebra given on July 27, 2011.

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