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content/Math4302/Math4302_L15.md

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+# Math4302 Modern Algebra (Lecture 15)
+
+## Group
+
+### Normal subgroup
+
+Suppose $H\leq G$, then the following are equivalent:
+
+1. $aH=Ha$ for all $a\in G$
+2. $aHa^{-1}= H$ for all $a\in G$
+3. $aha^{-1}\subseteq H$ for all $a\in G$
+
+Then $H\trianglelefteq G$
+
+> [!TIP]
+>
+> If $H\leq G$ and if $aH$ is a right coset, then $aH=Ha$.
+>
+> Reason: If $aH=Hb$ for some $b\in G$, then $a\in aH$, so $a\in Hb$ but $a\in Ha$, so $Hb=Ha$.
+
+<details>
+<summary>Example</summary>
+
+If $\phi:G\to G'$ is a homomorphism, then $\ker(\phi)\trianglelefteq G$
+
+For example, if $\det:GL(n,\mathbb{R})\to \mathbb{R}-\{0\}$ is a homomorphism, then 
+
+$$
+H=\ker(\det)=\{A\in GL(n,\mathbb{R})|\det(A)=0\}=SL(n,\mathbb{R})\trianglelefteq GL(n,\mathbb{R})
+$$
+
+</details>
+
+### Factor group
+
+Consider the operation on the set of left coset of $G$, denoted by $S$. Define
+
+$$
+(aH)(bH)=abH
+$$
+
+#### Condition for operation
+
+The operation above is well defined if and only if $H\trianglelefteq G$.
+
+<details>
+<summary>Proof</summary>
+
+First, suppose $H\trianglelefteq G$, and $aH=a'H$m and $bH=b'H$, we want to show that $abH=ab'H$.
+
+It is enough to show that $(ab)^{-1}a'b'=b^{-1}a^{-1}a'b'\in H$.
+
+$aH=a'H\implies a^{-1}a'\in H$, and $bH=b'H\implies b^{-1}b'\in H$. Note that by proposition of normal group, $gHg^{-1}\subseteq H$ for any $g\in G$, so let $g=b^{-1}$, $h=a^{-1}a$.
+
+Therefore $b^{-1}(a^{-1}a')(b^{-1})^{-1}=b^{-1}a^{-1}a'b\in H$, since $b^{-1} b'\in H$, then $b^{-1}a^{-1}a'b'\in H$.
+
+---
+
+Conversely, suppose this operation is well defined, then we show that $ghg^{-1}\in H$ for any $g\in G, h\in H$.
+
+Note that $hH=eH$, the well-defineness implies that $(hH)(g^{-1}H)=(eH)(g^{-1}H)=g^{-1}H$. So $ghg^{-1}\in H$. (add $g$ on the left)
+
+> $aH=bH\iff a^{-1}b\in H$, or equivalently $aH=bH\iff b^{-1}a\in H$.
+
+</details>
+
+#### Theorem for operation over left coset
+
+If $H\trianglelefteq G$, the set of left coset of $G$ is a group under the operation defined above.
+
+<details>
+<summary>Proof</summary>
+
+This operation is well defined by condition above.
+
+- Identity: $eH=H$
+- Inverse: $(aH)^{-1}=a^{-1}H$
+- Associativity: $(aH bH)cH=aH(bH cH)=abcH$
+
+</details>
+
+Such group is called the factor group of $G$ by $H$.
+
+<details>
+<summary>(Non) Example of factor group</summary>
+
+Recall from previous lectures, $G=S_3$ with $H=\{e,\tau_1\}$, with $\tau_1=(12), \tau_2=(23), \tau_3=(13)$.
+
+- $\{e,\tau_1\}=\tau_1 H=H$
+- $\{\tau_2,\rho_2\}=\tau_2 H=\rho_2 H$
+- $\{\tau_3,\rho\}=\tau_3 H=\rho H$
+
+And $(\tau_2 H)(\tau_3 H)=\tau_2 \tau_3 H=\rho H$.
+
+However, if we take $\rho^2\in \tau_2 H$, and $\rho\in \tau_3 H$, $\rho^2\rho =e$. This is not in $\rho H$.
+
+This is not well defined since $H$ is not normal.
+
+</details>
+
+#### Definition of factor (quotient) group
+
+If $H\trianglelefteq G$, then the set of cosets with operation:
+
+$$
+(aH)(bH)=abH
+$$
+
+is a group denoted by $G/H$. This group is called the quotient group (or factor group) of $G$ by $H$.
+
+<details>
+<summary>Example</summary>
+
+$5\mathbb{Z}\trianglelefteq \mathbb{Z}$, the cosets are $5\mathbb{Z}, 1+5\mathbb{Z}, 2+5\mathbb{Z}, 3+5\mathbb{Z}, 4+5\mathbb{Z}$.
+
+Here $5\mathbb{Z}$ is the identity in the factor group.
+
+And $\mathbb{Z}/5\mathbb{Z}\simeq \mathbb{Z}_5$
+
+</details>

content/Math4302/_meta.js

@@ -17,4 +17,5 @@ export default {
     Math4302_L12: "Modern Algebra (Lecture 12)",
     Math4302_L13: "Modern Algebra (Lecture 13)",
     Math4302_L14: "Modern Algebra (Lecture 14)",
+    Math4302_L15: "Modern Algebra (Lecture 15)",
 }