# Math4302 Modern Algebra (Lecture 1)

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## Group and subgroups

### Group

#### Definition of binary operations

A binary operation (usually denoted by $*$) on a set $X$ is a function from $X\times X$ to $X$.

<details>
<summary>Example of binary relations</summary>

$+$ is a binary operation on $\mathbb{Z}$ or $\mathbb{R}$.

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$\cdot$ is a binary operation on $\mathbb{Z}$ or $\mathbb{R}$.

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division is not a binary operation on $\mathbb{Z}$ or $\mathbb{R}$. (Consider 0)

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Generally, we can define a binary operation over sets whatever we want.

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Let $X=\{a,b,c\}$ and we can define the (Cayley) table for binary operation as follows:

|*| a | b | c |
|---|---|---|---|
|a| a | b | b |
|b| b | c | c |
|c| a | b | c |

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If we let $X$ be the set of all functions from $\mathbb{R}$ to $\mathbb{R}$.

then $(f+g)(x)=f(x)+g(x)$,

$(f g)(x)=f(x)\circ g(x)$,

$(f\circ g)(x)=f(g(x))$, are also binary operations.

</details>

#### Definition of Commutative binary operations

A binary operation $*$ in a set $X$ is commutative if $a*b=b*a$ for all $a,b\in X$.


> [!TIP]
>
> Commutative basically means the table is symmetric on diagonal.

<details>
<summary>Example of non-commutative binary operations</summary>

$(f\circ g)(x)=f(g(x))$, is not generally commutative, consider constant functions $f(x)=1$ and $g(x)=0$.

</details>

#### Definition of Associative binary operations

A binary operation $*$ in a set $X$ is associative if $(a*b)*c=a*(b*c)$ for all $a,b,c\in X$.

$$
\begin{aligned}
a*((b*c)*d)&=a*(b*(c*d))\quad\text{apply the definition to b,c,d}\\
&=a*(b*(c*d))\quad \text{apply the definition to a,b, (c*d)}\\
&=(a*b)*(c*d)
\end{aligned}
$$

<details>
<summary>Example of non-associative binary operations</summary>

Suppose $X=\{a,b,c\}$

|*| a | b | c |
|---|---|---|---|
|a| a | b | b |
|b| b | c | c |
|c| a | b | c |

is not associative, take $a,b,c$ as examples.

$$
a*(b*c)=a*c=b\neq (a*b)*c=b*c=c
$$

</details>

#### Theorem for Associativity of Composition

(Associativity of Composition) Let S be a set and let $f,g$ and $h$ be functions from S to S. Then $(f\circ g)\circ h=f\circ(g\circ h)$.

> [!NOTE]
>
> There exists binary operation that is associative but not commutative.
>
> Consider $(f\circ g)$ where $f,g$ are functions over some set $X$.
> 
> $(f\circ g)(x)=f(g(x))$ is generally not commutative but always associative.
>
> There exists binary operation that is commutative but not associative.
>
> Consider operation defined belows:
>
> $S=\{a,b,c\}$
>
> |*| a | b | c |
> |---|---|---|---|
> |a| a | b | b |
> |b| b | b | c |
> |c| b | c | c |
>
> Note that this operation is commutative since the table is symmetric on diagonal.
>
> This operation is not associative, take $a,b,c$ as examples.
>
> $a*(b*c)=a*c=b\neq (a*b)*c=b*c=c$

#### Definition of Identity element

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

#### Uniqueness of identity element

If $X$ has an identity element, then it is unique.

<details>
<summary>Proof</summary>

Suppose $e_1$ and $e_2$ are identity elements of $X$. Then $e_1*e_2=e_2*e_1=e_1=e_2$.

</details>

<details>
<summary>Example of identity element</summary>

$0$ is the identity element of $+$ on $\mathbb{Z}$ or $\mathbb{R}$.

$1$ is the identity element of $\cdot$ on $\mathbb{Z}$ or $\mathbb{R}$.

identity zero $f(x)=0$ is the identity element of $(f+g)(x)=f(x)+g(x)$.

identity one $f(x)=1$ is the identity element of $(f\circ g)(x)=f(g(x))$.

identity function $f(x)=x$ is the identity element of $(f\circ g)(x)=f(g(x))$.

</details>

> [!WARNING]
>
> Not all binary operations have identity elements.
>
> Consider
>
> Suppose $X=\{a,b,c\}$
> |*| a | b | c |
> |---|---|---|---|
> |a| a | b | b |
> |b| b | c | c |
> |c| a | b | c |
>
> No identity element exists for this binary operation.


