From f3c54c4dc795610f8ebdd65a462815236f50d504 Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Fri, 30 Jan 2026 13:58:22 -0600
Subject: [PATCH] Update Math4302_L8.md

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-#  Math4302 Modern Algebra (Lecture 8)
\ No newline at end of file
+# Math4302 Modern Algebra (Lecture 8)
+
+## Subgroups
+
+### Cyclic group
+
+#### Subgroup of cyclic group is cyclic
+
+Every subgroup of a cyclic group is cyclic.
+
+#### Order of subgroup of cyclic group
+
+If $a\in G$ and $|\langle a\rangle|$ be the smallest positive $n$ such that $a^n=e$, then $\langle a\rangle=\{e,a,a^2,\cdots,a^{n-1}\}$ and $a^{m_1}=a^{m_2}\iff m_1=m_2\mod n$. ($n$ divides $m_1-m_2$)
+
+#### Size of subgroup of cyclic group
+
+Let $G=\langle a\rangle$ and $H=\langle a^m\rangle$. Then $|H|=\frac{|G|}{d}$ where $d=\operatorname{gcd}(|G|,|H|)$. In particular, $\langle a^m\rangle=G\iff \operatorname{gcd}(n,m)=1$.
+
+#### GCD decides the size of subgroup
+
+Suppose $G=\langle a\rangle$, $|G|=n$.
+
+Then $\langle a^{m_1}\rangle=\langle a^{m_2}\rangle\iff \operatorname{gcd}(n,m_2)=\operatorname{gcd}(n,m_1)$.
+
+<details>
+<summary>Proof</summary>
+
+$\implies$:
+
+$\langle a^{m_1}\rangle=\langle a^{m_2}\rangle\implies \operatorname{gcd}(n,m_1)=\operatorname{gcd}(n,m_2)$
+
+$\impliedby$:
+
+Suppose $d=\operatorname{gcd}(n,m_1)=\operatorname{gcd}(n,m_2)$.
+
+Enough to show $a^{m_1}\in \langle a^{m_2}\rangle$. (then we conclude $\langle a^{m_1}\rangle=\langle a^{m_2}\rangle$ and by symmetry $\langle a^{m_2}\rangle=\langle a^{m_1}\rangle$.)
+
+Equivalent to show that $a^{m_1}=(a^{m_2})^k$ for some integer $k$. That is $n$ divides $m_1-km_2$ for some $k\in \mathbb{Z}$.
+
+From last lecture, we know that $d$ can be written as $d=nr+m_2 s$ for some $r,s\in \mathbb{Z}$.
+
+Multiply by $\frac{m_1}{d}$, (since $d$ divides $m_1$, this is an integer).
+
+So $m_1=nr\frac{m_1}{d}+m_2s\frac{m_1}{d}$.
+
+Therefore $n$ divides $m_1-(\frac{m_1}{d}s)m_2$, so $k=\frac{m_1}{d}s$. works.
+
+</details>
+
+#### Corollaries for subgroup of cyclic group
+
+Let $G=\langle a\rangle$ be a cyclic group of finite order.
+
+1. If $H\leq G$, then $|H|$ is a divisor of $|G|$. (More generally true for finite groups.)
+2. For any $d$ divides $|G|$, there is exactly one subgroup of $G$ of order $d$. $\langle a^m\rangle$ where $m=\frac{|G|}{d}$.
+
+<details>
+<summary>Examples</summary>
+
+$(\mathbb{Z}_18,+)$.
+
+The subgroup with size $6$ is $\langle 3\rangle=\{0,3,6,9,12,15\}=\langle 15\rangle$.
+
+Note that $\operatorname{gcd}(18,3)=3=\operatorname{gcd}(18,15)$.
+
+$\langle 6\rangle=\{0,6,12\}$.
+
+$\langle 9\rangle=\{0,9\}$.
+
+$\langle 2\rangle=\{0,2,4,6,8,10,12,14,16\}$ (generators are $2,4,8,10,14,16$ since they have gcd $2$ with $18$).
+
+</details>
+
+### Non-cyclic groups
+
+Let $G$ be a group and $a,b\in G$, then we use $\langle a,b\rangle$ to mean the subgroup of $G$ generated by combination of $a$ and $b$.
+
+$$
+\langle a,b\rangle\coloneqq \{e,a,b,ab,ba,a^{-1},b^{-1},(ab)^{-1},(ba)^{-1},\ldots\}
+$$
+
+This is a subgroup of $G$ since it is closed and $e=a^0$.
+
+#### Klein 4 group
+
+Klein 4 group is abelian but not cyclic.
+
+|*|e|a|b|c|
+|--|---|---|---|---|
+|e|e|a|b|c|
+|a|a|e|c|b|
+|b|b|c|e|a|
+|c|c|b|a|e|
+
+The subgroups are
+
+$\langle e\rangle=\{e\}$
+
+$\langle a\rangle=\{e,a\}$
+
+$\langle b\rangle=\{e,b\}$
+
+$\langle c\rangle=\{e,c\}$
+
+Therefore $G$ is **not cyclic** and **not isomorphic** to $\mathbb{Z}_4$.
+
+Here $G=\langle a,b\rangle=\{e,a,b,ab=c\}$.
+
+More generally, if we have $a_i\in G$, where $i\in I$, then $\langle a_i,i\in I\rangle=$ all possible combinations of $a_i$ with their inverses. Is a subgroup of $G$.
+
+Another way to describe is that $\langle a_i,i\in I\rangle=\bigcap_{H\leq G, a_i\in H,i\in I}H$.
+
+#### Definition of finitely generated group
+
+If $G$ is a group and if there is a finite set $a_1,\ldots, a_n\in G$ such that $G=\langle a_1,\ldots, a_n\rangle$, then $G$ is called finitely generated.
+
+<details>
+<summary>Examples</summary>
+
+Any finite group is finitely generated.
+
+---
+
+$(\mathbb{Q},+)$ is not finitely generated.
+
+Suppose for the contrary, there is a finite set $\frac{a_1}{b_1},\ldots,\frac{a_n}{b_n}\in \mathbb{Q}$ such that 
+
+$$
+\mathbb{Q}=\langle \frac{a_1}{b_1},\ldots,\frac{a_n}{b_n}\rangle=\{t_1\frac{a_1}{b_1},\ldots,t_n\frac{a_n}{b_n}|t_1,t_2,\ldots,t_n\in \mathbb{Z}\}
+$$.
+
+Pick prime $p$ such that $p>|b_1|,\ldots,|b_n|$. Then $\frac{1}{p}\in \mathbb{Q}$.
+
+$$
+\frac{1}{p}=t_1\frac{a_1}{b_1}+t_2\frac{a_2}{b_2}+\cdots+t_n\frac{a_n}{b_n}=\frac{A}{b_1b_2\cdots b_n}
+$$
+
+This implies that $pA=b_1b_2\cdots b_n$.
+
+Since $p$ is prime, $p|b_i$ for some $i$.
+
+However, by our construction, $p>|b_i|$ and cannot divide $b_i$.
+
+Contradiction.
+
+</details>