# Math4302 Modern Algebra (Lecture 2)

## Recall from last lecture

### Binary operations

A binary operation that is not associative but commutative:

Consider $(\mathbb{Z},*)$ where $a*b=|a-b|$.

This is trivially commutative.

But $a=4,b=3,c=1$ gives $(a*b)*c=(4*3)*1=1*1=0$. and $a*(b*c)=4*(3*1)=4*2=2$.

#### Definition for identity element

An element $e\in X$ is called identity element if $a*e=e*a=a$ for all $a\in X$.

### Group

#### Definition of group

A group is a set $G$ with a binary operation $*$ that satisfies the following axioms:

1. Closure: $\forall a,b\in G, a* b\in G$ (automatically guaranteed by definition of binary operation).
2. Associativity: $\forall a,b,c\in G, (a* b)* c=a* (b* c)$.
3. Identity element: $\exists e\in G, \forall a\in G, a* e=e* a=a$.
4. Inverses: $\forall a\in G, \exists a^{-1}\in G, a* a^{-1}=a^{-1}* a=e$.

> [!NOTE]
>
> The inverse of $a$ is unique: If there is $b'\in G$ such that $b'*a=a*b'=e$, then $b=b'$.
>
> Proof:
>
> $b'=b'*e=b'*(a*b)=(b'*a)*b=e*b=b$.
>
> apply the definition of group.

<details>
<summary>Example of group</summary>

$(\mathbb{Z},+)$ is a group.

$(\mathbb{Q},+)$ is a group.

$(\mathbb{R},+)$ is a group.

with identity $0$ and all abelian groups.

---

$(\mathbb{Z},\cdot)$, $\mathbb{Q},\cdot)$, $(\mathbb{R},\cdot)$ are not groups ($0$ has no inverse).

---

We can fix this by removing $0$.

$(\mathbb{Q}\setminus\{0\},\cdot)$, $(\mathbb{R}\setminus\{0\},\cdot)$ are groups.

---

$(\mathbb{Z}\setminus\{0\},\cdot)$ is not a group.

$(\mathbb{Z}_+,+)$ is not a group.

---

Consider $S$ be the set of all functions from $\mathbb{R}$ to $\mathbb{R}$.

$(S,+)$

- Identity: $f(x)=0$
- Associativity: $(f+g)(x)=f(x)+g(x)$
- Inverse: $f(x)=-f(x)$

This is a group.

$(S,\circ)$

- Identity: $f(x)=x$
- Associativity: $(f\circ g)(x)=f(g(x))$
- Inverse: not all have inverse...... (functions which are not bijective don't have inverses)

This is not a group.

---

$\operatorname{GL}_(n,\mathbb{R})$: set of $n\times n$ invertible matrices over $\mathbb{R}$.

$(\operatorname{SL}_(n,\mathbb{R}),\cdot)$ where $\cdot$ is matrix multiplication.

- Identity: $I_n$
- Associativity: $(A\cdot B)\cdot C=A\cdot (B\cdot C)$
- Inverse: $(A^{-1})^{-1}=A$

This is a group.

**Matrix multiplication is not generally commutative**, therefore it's not abelian.

</details>

#### Definition of abelian group

A group $(G,*)$ is called abelian if $a* b=b* a$ for all $a,b\in G$. ($*$ is commutative)

#### Properties of group

1. $(a*b)^{-1}=b^{-1}* a^{-1}$

<details>
<summary>Proof</summary>

$(b^{-1}* a^{-1})*(a*b)=b^{-1}* a^{-1}*a*b=b^{-1}* e*b=b*b^{-1}=e$

$(a*b)* (b^{-1}* a^{-1})=a* b*b^{-1}* a^{-1}=a* e*a^{-1}=a*a^{-1}=e$

</details>

2. Cancellation from right and left:

$$
a*b=a*c\implies b=c
$$

$$
b*a=c*a\implies b=c
$$

<details>
<summary>Proof</summary>

$$
\begin{aligned}
    a*b&=a*c\\
    a^{-1}*(a*b)&=a^{-1}*(a*c)\\
    e*b&=e*c\\
    b&=c
\end{aligned}
$$

right cancellation are the same

</details>

> [!NOTE]
>
> This also implies that every row/column of the table representation of the binary operation is distinct.
>
> _If not, suppose $a,b$ have the same row/column, then we can prove $a=b$ using cancellation from right and left._

3. We can solve equations $a*x=b \text{ and } x*a=b
$ uniquely.

$x=a^{-1}* b$, similarly $x=b* a^{-1}$.

### Finite groups

Group with 1 element $\{e\}$.

Group with 2 elements $\{e,a\}$. (example is $(\{-1,1\},\times)$)

And

|*|e|a|
|-|-|-|
|e|e|a|
|a|a|e|

Group with 3 elements $\{e,a,b\}$.

And the possible ways to fill the table are:

|*|e|a|b|
|-|-|-|-|
|e|e|a|b|
|a|a|b|e|
|b|b|e|a|
