MATH 415 Lecture 7

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Section 10: Cosets and Lagrange's Theorem

Given $H<G$ , where $G$ is finite, $\left|H\right|\mid \left|G\right|$ .

Define two Equivalence relations: $\sim _{L}$ (left) and $\sim _{R}$ (right)

Theorem 10.1

Let $H\leq G$ .

Let $a\sim _{L}b$ iff $a^{-1}\,b\in H$
Let $a\sim _{R}b$ iff $a\,b^{-1}\in H$

Then $\sim _{L}$ and $\sim _{R}$ are equivalence relations

Proof. For $\sim _{L}$

Reflexivity: $a^{-1}\,a=e\in H$
Symmetry: $a^{-1}\,b\in H$ implies $(a^{-1}\,b)^{-1}=b^{-1}\,a\in H$
Transitivity: $a^{-1}\,b\in H$ and $b^{-1}\,c\in H$ give $(a^{-1}\,b)\,(b^{-1}\,c)=a^{-1}\,c\in H$ .

For $\sim _{R}$

Reflexivity: $a\,a^{-1}=e\in H$
Symmetry: $a\,b^{-1}\in H$ implies $(a\,b^{-1})^{-1}=b\,a^{-1}\in H$
Transitivity: $a\,b^{-1}\in H$ and $b\,c^{-1}\in H$ give $(a\,b^{-1})\,(b\,c^{-1})=a\,c^{-1}\in H$ .

quod erat demonstrandum

Cosets

The defined relation means that if $a\sim _{L}b$ , then there exists a $h\in H$ such that $b=a\,h$ . Respectively, if $a\sim _{R}b$ , then there exists a $h\in H$ such that $b=h\,a$ . In shorthand, let $a\,H=\left\{a\,h~\mid ~h\in H\right\}$ and $H\,a=\left\{h\,a~\mid ~h\in H\right\}$ , then

{\begin{aligned}a\sim _{L}b&\iff b\in a\,H\\a\sim _{R}b&\iff b\in H\,a\end{aligned}}

$a\,H$ is called a left coset of $H$ containing $a$ , and
$H\,a$ is called a right coset of $H$ containing $a$ .

In general, $a\,H\neq H\,a$ . For abelian (commutative) groups, $a\,H=H\,a$ , so we will use $\sim$ for these groups.

Example 10.4

Take $n\,\mathbb {Z} <\mathbb {Z}$ . If we have $a\sim b$ (or $b-a\in \mathbb {Z}$ and $a-b\in \mathbb {Z}$ ), then we can say that $a$ and $b$ are congruent modulo $n$ .

Cosents correspond to elements of $\mathbb {Z} _{n}$ . For example, when $n=6$ and $H=\left\{0,3\right\}$ , we get

coset containing 0 is $\left\{0,3\right\}$
coset containing 1 is $\left\{1,4\right\}$
coset containing 2 is $\left\{2,5\right\}$

Take a look at the following addition table:

+₆	0	3	1	4	2	5
0	0	3	1	4	2	5
3	3	0	4	1	5	2
1	1	4	2	5	3	0
4	4	1	5	2	0	3
2	2	5	3	0	4	1
5	5	2	0	3	1	4

Example 10.7

Take group $S_{3}=\left\{\rho _{0},\rho _{1},\rho _{2},\mu _{1},\mu _{2},\mu _{3}\right\}$ , where each $\rho$ is a triangular rotation and each $\mu$ is a transposition. Note that $e=\rho _{0}$ , and note that $S_{3}$ is non-abelian. Let $H=\left\{\rho _{0},\mu _{1}\right\}$ . Then

left coset containing $\rho _{0}$ is $\rho _{0}\,H=\left\{\rho _{0},\mu _{1}\right\}$
left coset containing $\rho _{1}$ is $\rho _{1}\,H=\left\{\rho _{1}\,\rho _{0},\rho _{1}\,\mu _{1}\right\}=\left\{\rho _{1},\mu _{3}\right\}$
left coset containing $\rho _{2}$ is $\rho _{2}\,H=\left\{\rho _{2}\,\rho _{0},\rho _{2}\,\mu _{1}\right\}=\left\{\rho _{2},\mu _{2}\right\}$

Notice the differences between left and right cosets

right coset containing $\rho _{0}$ is $H\,\rho _{0}=\left\{\rho _{0},\mu _{1}\right\}$
right coset containing $\rho _{1}$ is $H\,\rho _{1}=\left\{\rho _{1},\mu _{2}\right\}$
right coset containing $\rho _{2}$ is $H\,\rho _{2}=\left\{\rho _{2},\mu _{3}\right\}$

The Theorem of Lagrange

Let $H<G$ where $G$ is a finite group. Then the order of $H$ is a divisor of the order of $G$ :

\left|H\right|\mid \left|G\right|\,\!

Proof. Given a finite group $G$ . Let $a_{1}\in G$ be the identity element, and let $H$ , $a_{2}\,H$ , $a_{3}\,H$ , … $a_{k}\,H$ be cosets.

Observe that either $a_{i}\,H=a_{j}\,H$ or $a_{i}\,H\cap a_{j}\,H=\emptyset$ .

Let $\phi$ be a function from $H$ to $a\,H$ with $\phi (h)=a\,h$ . $\phi$ a bijection.

We have $G$ is a union of distinct subsets $H$ , where all orders $|H_{i}|$ are equal. Thus $\left|G\right|=k\,\left|H\right|$ , so $|H|$ divides $|G|$ .

quod erat demonstrandum

We denote the number ${\frac {\left|G\right|}{\left|H\right|}}=\left(G:H\right)$ as the index of $H$ in $G$ .

Corollary. Every group of prime order is cyclic.

Suppose $G$ has prime order $p$ , and let $a\neq e$ be generator for $H=\left\langle a\right\rangle$ . Then either $|H|=1$ or $|H|=p$ . The first case is impossbile because $|H|\geq 2$ (namely, $e$ and $a$ must be elements). Thus $|H|=p$ , and its cyclic generator is $a$ . So $G$ must be cyclic.