# Math4302 Modern Algebra (Lecture 15)

## Group

### Normal subgroup

Suppose $H\leq G$, then the following are equivalent:

1. $aH=Ha$ for all $a\in G$
2. $aHa^{-1}= H$ for all $a\in G$
3. $aha^{-1}\subseteq H$ for all $a\in G$

Then $H\trianglelefteq G$

> [!TIP]
>
> If $H\leq G$ and if $aH$ is a right coset, then $aH=Ha$.
>
> Reason: If $aH=Hb$ for some $b\in G$, then $a\in aH$, so $a\in Hb$ but $a\in Ha$, so $Hb=Ha$.

<details>
<summary>Example</summary>

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

For example, if $\det:GL(n,\mathbb{R})\to \mathbb{R}-\{0\}$ is a homomorphism, then 

$$
H=\ker(\det)=\{A\in GL(n,\mathbb{R})|\det(A)=0\}=SL(n,\mathbb{R})\trianglelefteq GL(n,\mathbb{R})
$$

</details>

### Factor group

Consider the operation on the set of left coset of $G$, denoted by $S$. Define

$$
(aH)(bH)=abH
$$

#### Condition for operation

The operation above is well defined if and only if $H\trianglelefteq G$.

<details>
<summary>Proof</summary>

First, suppose $H\trianglelefteq G$, and $aH=a'H$m and $bH=b'H$, we want to show that $abH=ab'H$.

It is enough to show that $(ab)^{-1}a'b'=b^{-1}a^{-1}a'b'\in H$.

$aH=a'H\implies a^{-1}a'\in H$, and $bH=b'H\implies b^{-1}b'\in H$. Note that by proposition of normal group, $gHg^{-1}\subseteq H$ for any $g\in G$, so let $g=b^{-1}$, $h=a^{-1}a$.

Therefore $b^{-1}(a^{-1}a')(b^{-1})^{-1}=b^{-1}a^{-1}a'b\in H$, since $b^{-1} b'\in H$, then $b^{-1}a^{-1}a'b'\in H$.

---

Conversely, suppose this operation is well defined, then we show that $ghg^{-1}\in H$ for any $g\in G, h\in H$.

Note that $hH=eH$, the well-defineness implies that $(hH)(g^{-1}H)=(eH)(g^{-1}H)=g^{-1}H$. So $ghg^{-1}\in H$. (add $g$ on the left)

> $aH=bH\iff a^{-1}b\in H$, or equivalently $aH=bH\iff b^{-1}a\in H$.

</details>

#### Theorem for operation over left coset

If $H\trianglelefteq G$, the set of left coset of $G$ is a group under the operation defined above.

<details>
<summary>Proof</summary>

This operation is well defined by condition above.

- Identity: $eH=H$
- Inverse: $(aH)^{-1}=a^{-1}H$
- Associativity: $(aH bH)cH=aH(bH cH)=abcH$

</details>

Such group is called the factor group of $G$ by $H$.

<details>
<summary>(Non) Example of factor group</summary>

Recall from previous lectures, $G=S_3$ with $H=\{e,\tau_1\}$, with $\tau_1=(12), \tau_2=(23), \tau_3=(13)$.

- $\{e,\tau_1\}=\tau_1 H=H$
- $\{\tau_2,\rho_2\}=\tau_2 H=\rho_2 H$
- $\{\tau_3,\rho\}=\tau_3 H=\rho H$

And $(\tau_2 H)(\tau_3 H)=\tau_2 \tau_3 H=\rho H$.

However, if we take $\rho^2\in \tau_2 H$, and $\rho\in \tau_3 H$, $\rho^2\rho =e$. This is not in $\rho H$.

This is not well defined since $H$ is not normal.

</details>

#### Definition of factor (quotient) group

If $H\trianglelefteq G$, then the set of cosets with operation:

$$
(aH)(bH)=abH
$$

is a group denoted by $G/H$. This group is called the quotient group (or factor group) of $G$ by $H$.

<details>
<summary>Example</summary>

$5\mathbb{Z}\trianglelefteq \mathbb{Z}$, the cosets are $5\mathbb{Z}, 1+5\mathbb{Z}, 2+5\mathbb{Z}, 3+5\mathbb{Z}, 4+5\mathbb{Z}$.

Here $5\mathbb{Z}$ is the identity in the factor group.

And $\mathbb{Z}/5\mathbb{Z}\simeq \mathbb{Z}_5$

</details>
