# Math4302 Modern Algebra (Lecture 14)

## Group

### Cosets

Left cosets:

$$
aH=\{x|a\sim x\}=\{x\in G|a^{-1}x\in H\}=\{x|x=ah\text{ for some }h\in H\}
$$

Right cosets:

$$
Ha=\{x|x\sim'a\}=\{x\in G|xa^{-1}\in H\}=\{x|x=ha\text{ for some }h\in H\}
$$

And $G=\sqcup_{a\in G}aH=\sqcup_{a\in G}Ha$ (all sets are disjoint)

And $H$ is both a left and right coset of $G$

<details>
<summary>Example of left and right cosets</summary>

$G=S_3=\{e,\rho,\rho^2,\tau_1,\tau_2,\tau_3\}$ with $H=\{e,\rho, \rho^2\}$, $\tau_1=(12), \tau_2=(23), \tau_3=(13)$.

Number of distinct coset is $|G|/|H|=2$.

The (left and right) cosets are:

$$
\tau_1 H=\tau_2 H=\tau_3 H=\{\tau_1,\tau_2,\tau_3\}\\
H=\rho H=\rho^2 H=\{e,\rho,\rho^2\}
$$

For this case, left and right cosets are the same (gives the same partition of $G$).

---

$H=\{e,\tau\}$

Left cosets:

$$
e H=H=\tau_1 H\\
\rho H=\{\tau_3,\rho\}=\tau_3 H\\
\rho^2 H=\{\tau_2,\rho^2\}=\tau_2 H
$$

Right cosets:

$$
H=H e=H\tau_1\\
H\tau_2=\{\tau_2,\rho\}=H\rho \\
H\tau_3=\{\tau_3,\rho^2\}=H\rho^2
$$

</details>

#### Definition of Normal Subgroup

A subgroup $H\leq G$ is called a normal subgroup if $aH=Ha$ for all $a\in G$. We denote it by $H\trianglelefteq G$

<details>
<summary>Example of normal subgroup</summary>

Every subgroup of an abelian group is a normal subgroup.

Prove using direct product of cyclic groups.

---

If $G$ is finite, and $|H|=\frac{|G|}{2}$, then $H\trianglelefteq G$.

> there are exactly two cosets, and one of them must be $H$, then the left coset $G\setminus H$ will always be the same as the right $G\setminus H$.

$A_n\trianglelefteq S_n$

---

If $\phi:G\to G'$ is a homomorphism, then $\ker(\phi)\trianglelefteq G$

We will use the equivalent definition of normal subgroup. ($aha^{-1}\in H$ for all $a\in G, h\in H$)

$\phi(aha^{-1})=phi(a)\phi(h)\phi(a)^{-1}=\phi(a)e'\phi(a)^{-1}=e'$, so $aha^{-1}\in \ker(\phi)$

---

Consider $\operatorname{GL}(n,\mathbb{R})$ be all the invertible matrices of size $n\times n$

Let $H=\{A\in \operatorname{GL}(n,\mathbb{R})|\det(A)=1\}$.

$H\trianglelefteq \operatorname{GL}(n,\mathbb{R})$

$\phi:\operatorname{GL}(n,\mathbb{R})\to (\mathbb{R}-\{0\},\cdot)$ where $\phi(A)=\det(A)$

Then $H=\ker(\phi)$

</details>

#### Lemma for equivalent definition of normal subgroup

The following are equivalent:

1. $H\trianglelefteq G$
2. $aHa^{-1}=H$ for all $a\in G$
3. $aHa^{-1}\subseteq H$ for all $a\in G$, that is $aha^{-1}\in H$ for all $a\in G$

<details>
<summary>Proof</summary>

We first show that $1\implies 2$.

$aHa^{-1}\subseteq H$:

If $aH=Ha$, for every $h\in H$, $ah=h'a$ for some $h'$, so $aha^{-1}=h'\in H$.

$H\subseteq aHa^{-1}$:

we have $Ha=aH$, so for every $h\in H$, $ha=ah'$ for some $h'$, so $h=ah'a^{-1}\in aHa^{-1}$.

$2\implies 3$: clear

$3\implies 1$: 

$aH\subseteq Ha$. for any $h\in H$, $\forall aha^{-1}\in H$, so $aha^{-1}=h'\in H$, so $ah=h'a\in Ha$ so $aH\subseteq Ha$.

$Ha\subseteq aH$: apply previous part to $a^{-1}$., and $a^{-1}H\subseteq Ha^{-1}$, so $\forall h\in H$ $a^{-1}h=h'a^{-1}\in Ha^{-1}$, so $ha=ah'$.
