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 $H\le G$.

• 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 $b=ah$. Respectively, if $a{\sim }_{R}b$, then there exists a $h\in H$ such that $b=ha$. In shorthand, let $aH=\{ah\mid h\in H\}$ and $Ha=\{ha\mid h\in H\}$, then

$\begin{array}{rl}a{\sim }_{L}b& ⟺b\in aH\\ a{\sim }_{R}b& ⟺b\in Ha\end{array}$

• $aH$ is called a left coset of $H$ containing $a$, and
• $Ha$ is called a right coset of $H$ containing $a$.

In general, $aH\ne Ha$. For abelian (commutative) groups, $aH=Ha$, so we will use $\sim$ for these groups.

Example 10.4

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

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

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

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 $S_3=\{{\rho }_0,{\rho }_1,{\rho }_2,{\mu }_1,{\mu }_2,{\mu }_3\}$, 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 $H=\{{\rho }_0,{\mu }_1\}$. Then

• left coset containing ${\rho }_0$ is ${\rho }_{0}H=\{{\rho }_0,{\mu }_1\}$
• left coset containing ${\rho }_1$ is ${\rho }_{1}H=\{{\rho }_1{\rho }_0,{\rho }_1{\mu }_1\}=\{{\rho }_1,{\mu }_3\}$
• left coset containing ${\rho }_2$ is ${\rho }_{2}H=\{{\rho }_2{\rho }_0,{\rho }_2{\mu }_1\}=\{{\rho }_2,{\mu }_2\}$

Notice the differences between left and right cosets

• right coset containing ${\rho }_0$ is $H{\rho }_0=\{{\rho }_0,{\mu }_1\}$
• right coset containing ${\rho }_1$ is $H{\rho }_1=\{{\rho }_1,{\mu }_2\}$
• right coset containing ${\rho }_2$ is $H{\rho }_2=\{{\rho }_2,{\mu }_3\}$

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 $a_{i}H\cap a_{j}H=\mathrm{\varnothing }$.

For all $a,b$, we have $\left|aH\right|=\left|bH\right|=\left|H\right|$.

Let $\varphi$ be a function from $H$ to $aH$ with $\varphi (h)=ah$. $\varphi$ 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 $\frac{\left|G\right|}{\left|H\right|}=(G:H)$ as the index of $H$ in $G$.

Corollary. Every group of prime order is cyclic.

Suppose

$G$

has prime order

$p$

, and let

$a\ne e$

be generator for

$H=⟨a⟩$

. Then either

$|H|=1$

or

$|H|=p$

. The first case is impossbile because

$|H|\ge 2$

(namely,

$e$

and

$a$

must be elements). Thus

$|H|=p$

, and its cyclic generator is

$a$

. So

$G$

must be cyclic.

quod erat demonstrandum

Given $K\le H\le G$ with $(H:K)$ and $(G:H)$ are finite, then $(G:K)$ is finite and $(G:K)=(G:H)(H:K)$.