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+# Math4302 Modern Algebra (Lecture 14)
+
+## Group
+
+### Cosets
+
+Left cosets:
+
+$$
+aH=\{x|a\sim x\}=\{x\in G|a^{-1}x\in H\}=\{x|x=ah\text{ for some }h\in H\}
+$$
+
+Right cosets:
+
+$$
+Ha=\{x|x\sim'a\}=\{x\in G|xa^{-1}\in H\}=\{x|x=ha\text{ for some }h\in H\}
+$$
+
+And $G=\sqcup_{a\in G}aH=\sqcup_{a\in G}Ha$ (all sets are disjoint)
+
+And $H$ is both a left and right coset of $G$
+
+<details>
+<summary>Example of left and right cosets</summary>
+
+$G=S_3=\{e,\rho,\rho^2,\tau_1,\tau_2,\tau_3\}$ with $H=\{e,\rho, \rho^2\}$, $\tau_1=(12), \tau_2=(23), \tau_3=(13)$.
+
+Number of distinct coset is $|G|/|H|=2$.
+
+The (left and right) cosets are:
+
+$$
+\tau_1 H=\tau_2 H=\tau_3 H=\{\tau_1,\tau_2,\tau_3\}\\
+H=\rho H=\rho^2 H=\{e,\rho,\rho^2\}
+$$
+
+For this case, left and right cosets are the same (gives the same partition of $G$).
+
+---
+
+$H=\{e,\tau\}$
+
+Left cosets:
+
+$$
+e H=H=\tau_1 H\\
+\rho H=\{\tau_3,\rho\}=\tau_3 H\\
+\rho^2 H=\{\tau_2,\rho^2\}=\tau_2 H
+$$
+
+Right cosets:
+
+$$
+H=H e=H\tau_1\\
+H\tau_2=\{\tau_2,\rho\}=H\rho \\
+H\tau_3=\{\tau_3,\rho^2\}=H\rho^2
+$$
+
+</details>
+
+#### Definition of Normal Subgroup
+
+A subgroup $H\leq G$ is called a normal subgroup if $aH=Ha$ for all $a\in G$. We denote it by $H\trianglelefteq G$
+
+<details>
+<summary>Example of normal subgroup</summary>
+
+Every subgroup of an abelian group is a normal subgroup.
+
+Prove using direct product of cyclic groups.
+
+---
+
+If $G$ is finite, and $|H|=\frac{|G|}{2}$, then $H\trianglelefteq G$.
+
+> there are exactly two cosets, and one of them must be $H$, then the left coset $G\setminus H$ will always be the same as the right $G\setminus H$.
+
+$A_n\trianglelefteq S_n$
+
+---
+
+If $\phi:G\to G'$ is a homomorphism, then $\ker(\phi)\trianglelefteq G$
+
+We will use the equivalent definition of normal subgroup. ($aha^{-1}\in H$ for all $a\in G, h\in H$)
+
+$\phi(aha^{-1})=phi(a)\phi(h)\phi(a)^{-1}=\phi(a)e'\phi(a)^{-1}=e'$, so $aha^{-1}\in \ker(\phi)$
+
+---
+
+Consider $\operatorname{GL}(n,\mathbb{R})$ be all the invertible matrices of size $n\times n$
+
+Let $H=\{A\in \operatorname{GL}(n,\mathbb{R})|\det(A)=1\}$.
+
+$H\trianglelefteq \operatorname{GL}(n,\mathbb{R})$
+
+$\phi:\operatorname{GL}(n,\mathbb{R})\to (\mathbb{R}-\{0\},\cdot)$ where $\phi(A)=\det(A)$
+
+Then $H=\ker(\phi)$
+
+</details>
+
+#### Lemma for equivalent definition of normal subgroup
+
+The following are equivalent:
+
+1. $H\trianglelefteq G$
+2. $aHa^{-1}=H$ for all $a\in G$
+3. $aHa^{-1}\subseteq H$ for all $a\in G$, that is $aha^{-1}\in H$ for all $a\in G$
+
+<details>
+<summary>Proof</summary>
+
+We first show that $1\implies 2$.
+
+$aHa^{-1}\subseteq H$:
+
+If $aH=Ha$, for every $h\in H$, $ah=h'a$ for some $h'$, so $aha^{-1}=h'\in H$.
+
+$H\subseteq aHa^{-1}$:
+
+we have $Ha=aH$, so for every $h\in H$, $ha=ah'$ for some $h'$, so $h=ah'a^{-1}\in aHa^{-1}$.
+
+$2\implies 3$: clear
+
+$3\implies 1$: 
+
+$aH\subseteq Ha$. for any $h\in H$, $\forall aha^{-1}\in H$, so $aha^{-1}=h'\in H$, so $ah=h'a\in Ha$ so $aH\subseteq Ha$.
+
+$Ha\subseteq aH$: apply previous part to $a^{-1}$., and $a^{-1}H\subseteq Ha^{-1}$, so $\forall h\in H$ $a^{-1}h=h'a^{-1}\in Ha^{-1}$, so $ha=ah'$.
diff --git a/content/Math4302/_meta.js b/content/Math4302/_meta.js
index 77f25a3..f540043 100644
--- a/content/Math4302/_meta.js
+++ b/content/Math4302/_meta.js
@@ -16,4 +16,5 @@ export default {
     Math4302_L11: "Modern Algebra (Lecture 11)",
     Math4302_L12: "Modern Algebra (Lecture 12)",
     Math4302_L13: "Modern Algebra (Lecture 13)",
+    Math4302_L14: "Modern Algebra (Lecture 14)",
 }