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content/Math4302/Math4302_L1.md

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 # Math4302 Modern Algebra (Lecture 1)
 
+_Skip section 0_
+
+## 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 $\mathbb{R}$.
+
+$\cdot$ is a binary operation on $\mathbb{Z}$ or $\mathbb{R}$.
+
+division is not a binary operation on $\mathbb{Z}$ or $\mathbb{R}$.
+
+Generally, we can define a binary operation over sets whatever we want.
+
+Let $X=\{a,b,c\}$ and we can define the table for binary operation as follows:
+
+|*| a | b | c |
+|---|---|---|---|
+|a| a | b | b |
+|b| b | c | c |
+|c| a | b | c |
+
+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.
+
+</details>
+
+#### Theorem forAssociativity of Composition
+
+(Associativity of Composition) Let S be a set and let $f,g$ and $h$ be functions from S to S. Then $(f\circ g)\circ h=f\circ(g\circ h)$.
+
+#### 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 |