updates

content/Math4302/Math4302_L2.md

@@ -0,0 +1,176 @@
+# 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.
+
+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 $\{e,a\}$.
+
+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|