From bdf0ff9f064c00931ea289e50f19f428c62b7956 Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Tue, 10 Feb 2026 10:08:34 -0600
Subject: [PATCH] Update Math4303_L12.md

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 # Math4303 Modern Algebra (Lecture 12)
 
-## Groups
\ No newline at end of file
+## Groups
+
+### Direct products
+
+$\mathbb{Z}_m\times \mathbb{Z}_n$ is cyclic if and only if $m$ and $n$ have greatest common divisor $1$.
+
+More generally, for $\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}\times \cdots \times \mathbb{Z}_{n_k}$, if $n_1,n_2,\cdots,n_k$ are pairwise coprime, then the direct product is cyclic.
+
+<details>
+<summary>Proof</summary>
+
+For the forward direction, use $\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}=\mathbb{Z}_{n_1n_2}$. if $n_1, n_2$ are coprime. 
+
+
+For the backward, suppose to the contrary that for example $\gcd(n_1,n_2)=d>1$, then $G=\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}\times H$, where any element in $H$ has order $\leq |H|$ and any element in $\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}$ has order $<\frac{n_1n_2}{d}$, therefore, all the elements in $G$ will have order strictly less than the size $n_1n_2\ldots n_k$ of the group.
+
+</details>
+
+#### Corollary for composition of cyclic groups
+
+If $n=p_1^{m_1}\ldots p_k^{m_k}$, where $p_i$ are distinct primes, then the group 
+
+$$
+G=\mathbb{Z}_n=\mathbb{Z}_{p_1^{m_1}}\times \mathbb{Z}_{p_2^{m_2}}\times \cdots \times \mathbb{Z}_{p_k^{m_k}}
+$$
+
+is cyclic.
+
+<details>
+<summary>Example for product of cyclic groups and order of element</summary>
+
+$$
+\mathbb{Z}_{8}\times\mathbb{Z}_8\times \mathbb{Z}_12
+$$
+
+the order for $(1,1,1)$ is 24.
+
+What is the maximum order of an element in this group?
+
+Guess:
+
+$8*3=24$
+
+</details>
+
+### Structure of finitely generated abelian groups
+
+#### Theorem for finitely generated abelian groups
+
+Every finitely generated abelian group $G$ is isomorphic to 
+
+$$
+Z_{p_1}^{n_1}\times Z_{p_2}^{n_2}\times \cdots \times Z_{p_k}^{n_k}\times\underbrace{\mathbb{Z}\times \ldots \times \mathbb{Z}}_{m\text{ times}}
+$$
+
+<details>
+<summary>Example</summary>
+
+If $G$ is abelian of size $8$, then $G$ is isomorphic to one of the following:
+
+- $\mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2$ (non cyclic)
+- $\mathbb{Z}_2\times \mathbb{Z}_4$ (non cyclic)
+- $\mathbb{Z}_2$ (cyclic)
+
+And any two of them are not isomorphic
+
+---
+
+Find all abelian group of order $72$.
+
+Since $72=2^3*3^2$, There are 3 possibilities for the $2^3$ part, and there are 2 possibilities for the $3^2$ part.
+
+Note that $\mathbb{Z}_8\times\mathbb{Z}_9$, where $8,9$ are coprime, $\mathbb{Z}_8\times\mathbb{Z}_9=\mathbb{Z}_{72}$, is cyclic.
+
+There are 6 possibilities in total.
+
+</details>
+
+#### Corollary for divisor size of abelian subgroup
+
+If $g$ is abelian and $|G|=n$, then for every divisor $m$ of $n$, $G$ has a subgroup of order $m$.
+
+> [!WARNING]
+>
+> This is not true if $G$ is not abelian.
+>
+> Consider $A_4$ (alternating group for $S_4$) does not have a subgroup of order 6.
+
+
+<details>
+<summary>Proof for the corollary</summary>
+
+Write $G=\mathbb{Z}_{p_1}^{n_1}\times \mathbb{Z}_{p_2}^{n_2}\times \cdots \times \mathbb{Z}_{p_k}^{n_k}$ where $p_i$ are distinct primes.
+
+Therefore $n=p_1^{m_1}\ldots p_k^{m_k}$.
+
+For any divisor $d$ of $n$, we can write $d=p_1^{m_1}\ldots p_k^{m_k}$, where $m_i\leq n_i$.
+
+Now for each $p_i$, we choose the subgroup $H_i$ of size $p_i^{m_i}$ in $\mathbb{Z}_{p_i}^{n_i}$. (recall that every cyclic group of size $r$ and any divisor $s$ of $r$, there is a subgroup of order $s$. If the group is generated by $a$, then use $a^{\frac{r}{s}}$ to generate the subgroup.)
+
+We can construct the subgroup $H=H_1\times H_2\times \cdots \times H_k$ is the subgroup of $G$ of order $d$.
+</details>
+
+### Cosets
+
+#### Definition of Cosets
+
+Let $G$ be a group and $H$ its subgroup.
+
+Define a relation on $G$ and $a\sim b$ if $a^{-1}b\in H$.
+
+This is an equivalence relation.
+
+- Reflexive: $a\sim a$: $a^{-1}a=e\in H$
+- Symmetric: $a\sim b\Rightarrow b\sim a$: $a^{-1}b\in H$, $(a^{-1}b)^{-1}=b^{-1}a\in H$
+- Transitive: $a\sim b$ and $b\sim c\Rightarrow a\sim c$ : $a^{-1}b\in H, b^{-1}c\in H$, therefore their product is also in $H$, $(a^{-1}b)(b^{-1}c)=a^{-1}c\in H$
+
+So we get a partition of $G$ to equivalence classes.
+
+Let $a\in G$, the equivalence class containing $a$
+
+$$
+aH=\{x\in G| a\sim x\}=\{x\in G| a^{-1}x\in H\}=\{x|x=ah\text{ for some }h\in H\}
+$$
+
+This is called the coset of $a$ in $H$.
+
+<details>
+<summary>Example</summary>
+
+Consider $G=S_3$
+
+</details>
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