updates?

content/Math4202/Math4202_L18.md

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+# Math4202 Topology II (Lecture 18)
+
+Revisit the idea of proof from previous lecture, using brick construction for lifting maps.

content/Math4302/Math4302_L16.md

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+# Math4302 Modern Algebra (Lecture 16)
+
+## 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 $a\in G$
+4. the operation $(aH)(bH)=abH$ is well defined for all $a,b\in G$, on the set of left coset of $H\leq G$
+
+Then $H\trianglelefteq G$
+
+If $H\trianglelefteq G$, then the set of left coset of $H\leq G$ is a group under the operation $(aH)(bH)=abH$
+
+$G/H$ is factor (or quotient) group of $G$ by $H$
+
+identity $H$=$eH$
+
+<details>
+<summary>Example</summary>
+
+If $|H|=\frac{1}{2}|G|$, then $H$ is a normal subgroup of $G$, then $G/H$ is isomorphic to $\mathbb{Z}_2$
+
+---
+
+Let $\phi:G\to G'$ be a homomorphism, then $\ker(\phi)\trianglelefteq G$
+
+$\mathbb{Z}/5\mathbb{Z}\trianglelefteq \mathbb{Z}$
+
+And $\mathbb{Z}/5\mathbb{Z}$ is isomorphic to $\mathbb{Z}_5$
+
+---
+
+$G/G$ is isomorphic to trivial group
+
+---
+
+$G/\{e\}$ is isomorphic to $G$
+
+---
+
+$\mathbb{R}/\mathbb{Z}$ is isomorphic to $S^1$
+
+---
+
+$\mathbb{Z}_3\times\mathbb{Z}_6/\langle (1,1)\rangle$ is isomorphic to $\mathbb{Z}_3$
+
+$\langle (1,1)\rangle=\{(1,1),(2,2),(0,3),(1,4),(2,5),(0,0)\}$
+
+</details>
+
+Recall
+
+1. The lagrange theorem, if $G$ is finite and $H\leq G$, then $|H| | |G|$.
+2. If $G$ is finite, abelian, $d||G|$, then $G$ has a subgroup of order $d$.
+
+We will show that 2. is not true if $G$ is not abelian. (consider $A_4$ with order 12, have no subgroup of order 6)
+
+<details>
+<summary>Proof</summary>
+
+Suppose $H\leq A_4$, and $|H|=6$. Then $A_4/H$ is normal in $A_4$, (since $|A_4/H|=2$), and $A_4/H$ is isomorphic to $\mathbb{Z}_2$.
+
+In other words, every element in $A_4/H$ has either order 1 or 2.
+
+So for any $\sigma\in A_4$, $(\sigma H)(\sigma H)=\sigma^2 H$. Therefore $\sigma^2=H$.
+
+But $\sigma=(1,3)(1,2)\in A_4$ and $\sigma^2=(1,3,2)\in H$.
+
+Similarly, $(1,3,2)(1,2,4)\dots(1,4,3)$ are all even permutations, making $|H|\geq 8$, that is a contradiction.
+
+</details>
+
+#### Fundamental homomorphism theorem (first isomorphism theorem)
+
+If $\phi:G\to G'$ is a homomorphism, then the function $f:G/\ker(\phi)\to \phi(G)$, ($\phi(G)\subseteq G'$) given by $f(a\ker(\phi))=\phi(a)$, $\forall a\in G$, is an well-defined isomorphism.
+
+<details>
+<summary>Proof</summary>
+
+First we will prove the well definedness and injectivity of $f$.
+
+We need to check the map will not map the same coset represented in different ways to different elements.
+
+Suppose $a\ker(\phi)=a'\ker(\phi)$, then $a^{-1}b\in \ker(\phi)$, this implies $\phi(a^{-1}b)=e'$ so $\phi(a)=\phi(b)$.
+
+Reverse the direction to prove the converse.
+
+Other properties are trivial.
+
+</details>

content/Math4302/Math4302_L17.md

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+# Math4302 Modern Algebra (Lecture 17)
+
+## Subgroup

content/Math4302/_meta.js

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     Math4302_L13: "Modern Algebra (Lecture 13)",
     Math4302_L14: "Modern Algebra (Lecture 14)",
     Math4302_L15: "Modern Algebra (Lecture 15)",
+    Math4302_L16: "Modern Algebra (Lecture 16)",
+    Math4302_L17: "Modern Algebra (Lecture 17)",
 }