From e69362ce3c19aa6566306cdfb55b7215560f4a38 Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Mon, 2 Feb 2026 13:50:03 -0600
Subject: [PATCH] Create Math4302_L9.md

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+# Math4302 Modern Algebra (Lecture 9)
+
+## Groups
+
+### Non-cyclic groups
+
+#### Dihedral groups
+
+The dihedral group $D_n$ is the group of symmetries of a regular $n$-gon.
+
+(Permutation that sends adjacent vertices to adjacent vertices)
+
+$D_n<S_n$
+
+$|S_n|=n!, |D_n|=2n$
+
+We can classify dihedral groups as follows:
+
+$\rho \in D_n$ as the rotation of a regular $n$-gon by $\frac{2\pi}{n}$.
+
+$\phi\in D_n$ as a reflection of a regular $n$-gon with respect to $x$-axis.
+
+We can enumerate the elements of $D_n$ as follows:
+
+$$
+D_n=\langle \phi,\rho\rangle=\{e,\rho,\rho^2,\cdots,\rho^{n-1},\phi,\phi\rho,\phi\rho^2,\cdots,\phi\rho^{n-1}\}
+$$
+
+We claim these elements are all distinct.
+
+<details>
+<summary>Proof</summary>
+
+Consider the first half, clearly $\rho_i\neq \rho_j$ if $0\leq i<j\leq n-1$.
+
+Also $\phi\rho_i\neq \phi\rho_j$ if $0\leq i<j\leq n-1$. otherwise $\rho_i=\rho_j$
+
+Also $\rho^i\neq \rho^j\phi$ where $0\leq i,j\leq n-1$.
+
+Otherwise $\rho^{i-j}=\phi$, but reflection (with some point fixed) cannot be any rotation (no points are fixed).
+
+</details>
+
+In $D_n$, $\phi\rho=\rho^{n-1}\phi$, more generally, $\phi\rho^i=\rho^{n-i}\phi$ for any $i\in\mathbb{Z}$.
+
+### Group homomorphism
+
+#### Definition for group homomorphism
+
+Let $G,G'$ be groups.
+
+$\phi:G\to G'$ is called a group homomorphism if $\phi(g_1g_2)=\phi(g_1)\phi(g_2)$ for all $g_1,g_2\in G$ (Note that $\phi$ may not be bijective).
+
+This is a weaker condition than isomorphism.
+
+<details>
+<summary>Example</summary>
+
+$GL(2,\mathbb{R})=\{A\in M_{2\times 2}(\mathbb{R})|det(A)\neq 0\}$
+
+Then $\phi:GL(2,\mathbb{R})\to (\mathbb{R}-\{0\},\cdot)$ where $\phi(A)=\det(A)$ is a group homomorphism, since $\det(AB)=\det(A)\det(B)$.
+
+This is not one-to-one but onto, therefore not an isomorphism.
+
+---
+
+$(\mathbb{Z}_n,+)$ and $D_n$ has homomorphism $(\mathbb{Z}_n,+)\to D_n$ where $\phi(k)=\rho^k$
+
+$\phi(i+j)=\rho^{i+j\mod n}=\rho^i\rho^j=\phi(i)+\phi(j)$.
+
+This is not onto but one-to-one, therefore not an isomorphism.
+
+---
+
+Let $G,G'$ be two groups, let $e$ be the identity of $G$ and let $e'$ be the identity of $G'$.
+
+Let $\phi:G\to G'$, $\phi(a)=e'$ for all $a\in G$.
+
+This is a group homomorphism,
+
+$$
+\phi(ab)=\phi(a)\phi(b)=e'e'=e'
+$$
+
+This is generally not onto and not one-to-one, therefore not an isomorphism.
+
+</details>
+
+#### Corollary for group homomorphism
+
+Let $G,G'$ be groups and $\phi:G\to G'$ be a group homomorphism. $e$ is the identity of $G$ and $e'$ is the identity of $G'$.
+
+1. $\phi(e)=e'$
+2. $\phi(a^{-1})=(\phi(a))^{-1}$ for all $a\in G$
+3. If $H\leq G$, then $\phi(H)\leq G'$, where $\phi(H)=\{\phi(a)|a\in H\}$.
+4. If $K\leq G'$ then $\phi^{-1}(K)\leq G$, where $\phi^{-1}(K)=\{a\in G|\phi(a)\in K\}$.
+
+<details>
+<summary>Proof</summary>
+
+(1) $\phi(e)=e'$
+
+Consider $\phi(ee)=\phi(e)\phi(e)$, therefore $\phi(e)=e'$ by cancellation on the left.
+
+---
+
+(2) $\phi(a^{-1})=(\phi(a))^{-1}$
+
+Consider $\phi(a^{-1}a)=\phi(a^{-1})\phi(a)=\phi(e)$, therefore $\phi(a^{-1})$ is the inverse of $\phi(a)$ in $G'$.
+
+---
+
+(3) If $H\leq G$, then $\phi(H)\leq G'$, where $\phi(H)=\{\phi(a)|a\in H\}$.
+
+- $e\in H$ implies that $e'=\phi(e)\in\phi(H)$.
+- If $x\in \phi(H)$, then $x=\phi(a)$ for some $a\in H$. So $x^{-1}=(\phi(x))^{-1}=\phi(x^{-1})\in\phi(H)$. But $x\in H$, so $x^{-1}\in H$, therefore $x^{-1}\in\phi(H)$.
+- If $x,y\in \phi(H)$, then $x,y=\phi(a),\phi(b)$ for some $a,b\in H$. So $xy=\phi(a)\phi(b)=\phi(ab)\in\phi(H)$ (by homomorphism). Since $ab\in H$, $xy\in\phi(H)$.
+
+---
+
+(4) If $K\leq G'$ then $\phi^{-1}(K)\leq G$, where $\phi^{-1}(K)=\{a\in G|\phi(a)\in K\}$.
+
+- $e'\in K$ implies that $e=\phi^{-1}(e')\in\phi^{-1}(K)$.
+- If $x\in \phi^{-1}(K)$, then $x=\phi(a)$ for some $a\in G$. So $x^{-1}=(\phi(x))^{-1}=\phi(x^{-1})\in\phi^{-1}(K)$. But $x\in G$, so $x^{-1}\in G$, therefore $x^{-1}\in\phi^{-1}(K)$.
+- If $x,y\in \phi^{-1}(K)$, then $x,y=\phi(a),\phi(b)$ for some $a,b\in G$. So $xy=\phi(a)\phi(b)=\phi(ab)\in\phi^{-1}(K)$ (by homomorphism). Since $ab\in G$, $xy\in\phi^{-1}(K)$.
+
+</details>
+
+#### Definition for kernel and image of a group homomorphism
+
+Let $G,G'$ be groups and $\phi:G\to G'$ be a group homomorphism.
+
+$\operatorname{ker}(\phi)=\{a\in G|\phi(a)=e'\}=\phi^{-1}(\{e'\})$ is called the kernel of $\phi$.
+
+Facts: 
+
+- $\operatorname{ker}(\phi)$ is a subgroup of $G$. (proof by previous corollary (4))
+- $\phi$ is onto if and only if $\operatorname{ker}(\phi)=\{e\}$ (the trivial subgroup of $G$). (proof forward, by definition of one-to-one; backward, if $\phi(a)=\phi(b)$, then $\phi(a)\phi(b)^{-1}=e'$, so $\phi(a)\phi(b^{-1})=e'$, so $ab^{-1}=e$, so $a,b=e$, so $a=b$)