From 5b103812b48592bdbb59ddff97ea941045a04088 Mon Sep 17 00:00:00 2001
From: Trance-0 <60459821+Trance-0@users.noreply.github.com>
Date: Fri, 6 Feb 2026 13:49:41 -0600
Subject: [PATCH] Update Math4302_L11.md

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-# Math4302 Modern Algebra (Lecture 11)
\ No newline at end of file
+# Math4302 Modern Algebra (Lecture 11)
+
+## Groups
+
+### Symmetric groups
+
+#### Definition of odd and even permutations
+
+$\sigma$ is an even permutation if the number of transpositions is even.
+
+$\sigma$ is an odd permutation if the number of transpositions is odd.
+
+#### Theorem for parity of transpositions
+
+The parity of the number of transpositions is unique.
+
+<details>
+<summary>Proof</summary>
+
+Prove using the determinant of a matrix, swapping the rows of the matrix multiply the determinant by $-1$.
+
+Consider the identity matrix $I_n$. Then the determinant is $1$, let $(ij)A$, where $i\neq j$ denote the matrix obtained from $A$ by swapping the rows $j$ and $i$, then the determinant of $(1j)A$ is $-1$.
+
+And,
+
+$$
+\det((a_1b_1)(a_2b_2)\cdots(a_nb_n)A)=(-1)^n\det(A)
+$$
+
+</details>
+
+$S_3$ has 6 permutations $\{e,(12),(13),(23),(12)(23),(13)(23)\}$, 3 of them are even $\{e,(12)(23),(13)(23)\}$ and 3 of them are odd $\{(13),(12),(23)\}$.
+
+#### Theorem for the number of odd and even permutations in symmetric groups
+
+In general, $S_n$ has $n!$ permutations, half of them are even and half of them are odd.
+
+<details>
+<summary>Proof</summary>
+
+Consider the set of odd permutations in $S_n$ and set of even permutations in $S_n$. Consider the function: $\alpha:S_n\to S_n$ where $\alpha(\sigma)=\sigma(12)$.
+
+$\sigma$ is a bijection,
+
+If $\sigma_1(12)=\sigma_2(12)$, then $\sigma_1=\sigma_2$.
+
+If $\phi$ is an even permutation, $\alpha(\phi(12))=\phi(12)(12)=\phi$, therefore the number of elements in the set of odd and even permutations are the same.
+</details>
+
+#### Definition for sign of permutations
+
+For $\sigma\in S_n$, the sign of $\sigma$ is defined by $\operatorname{sign}(\sigma)=1$ if sigma is even and $-1$ if sigma is odd.
+
+Then $\beta: S_n\to \{1,-1\}$ is a group under multiplication, where $\beta(\sigma)=\operatorname{sign}(\sigma)$.
+
+Then $\beta$ is a group homomorphism.
+
+#### Definition of alternating group
+
+$\ker(\beta)\leq S_n$, and $\ker(\beta)$ is the set of even permutations. Therefore the set of even permutations is a subgroup of $S_n$. We denote as $A_n$ (also called alternating group).
+
+and $|A_n|=\frac{n!}{2}$.
+
+### Direct product of groups
+
+#### Definition of direct product of groups
+
+Let $G_1,G_2$ be two groups. Then the direct product of $G_1$ and $G_2$ is defined as
+
+$$
+G_1\times G_2=\{(g_1,g_2):g_1\in G_1,g_2\in G_2\}
+$$
+
+The operations are defined by $(a_1,b_1)*(a_2,b_2)=(a_1*a_2,b_1*b_2)$.
+
+This group is well defined since:
+
+The identity is $(e_1,e_2)$, where $e_1\in G_1$ and $e_2\in G_2$. (easy to verify)
+
+The inverse is $(a_1,b_1)^{-1}=(a_1^{-1},b_1^{-1})$.
+
+Associativity automatically holds by associativity of $G_1$ and $G_2$.
+
+<details>
+<summary>Examples</summary>
+
+Consider $\mathbb{Z}_\1\times \mathbb{Z}_2$.
+
+$$
+\mathbb{Z}_\1\times \mathbb{Z}_2=\{(0,0),(0,1),(1,0),(1,1)\}
+$$
+
+$(0,0)^2=(0,0)$, $(0,1)^2=(0,0)$, $(1,0)^2=(0,0)$, $(1,1)^2=(0,0)$
+
+This is not a cyclic group, this is isomorphic to klein four group.
+
+---
+
+Consider $\mathbb{Z}_2\times \mathbb{Z}_3$.
+
+$$
+\mathbb{Z}_2\times \mathbb{Z}_3=\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2)\}
+$$
+
+This is cyclic ((2,3) are coprime)
+
+Consider:
+
+$$
+\langle (1,1)\rangle=\{(0,0),(1,1),(0,2),(1,0),(0,1),(1,2)\}
+$$
+
+</details>
+
+#### Lemma for direct product of cyclic groups
+
+$\mathbb{Z}_m\times \mathbb{Z}_n\simeq \mathbb{Z}_{mn}$ if and only if $m$ and $n$ have greatest common divisor $1$.
+
+<details>
+<summary>Proof</summary>
+
+First assume $\operatorname{gcd}(m,n)=d>1$
+
+Consider $(r,s)\in \mathbb{Z}_m\times \mathbb{Z}_n$.
+
+We claim that order of $(r,s)$ is at most $\frac{mn}{d}<mn$.
+
+Since $\frac{mn}{d}$ is integer, $\frac{mn}{d}=m_1dn_1$ where $m_1d$ is multiple of $m$ and $n_1d$ is multiple of $n$.
+
+Therefore $r$ combine with itself $\frac{mn}{d}$ times is $0$ in $\mathbb{Z}_m$ and $s$ combine with itself $\frac{mn}{d}$ times is $0$ in $\mathbb{Z}_n$.
+
+---
+
+Other direction:
+
+Assume $\operatorname{gcd}(m,n)=1$.
+
+Claim order of $(1,1)=mn$, so $\mathbb{Z}_m\times \mathbb{Z}_n=\langle (1,1)\rangle$.
+
+If $k$ is the order of $(1,1)$, then $k$ is a multiple of $m$ and a multiple of $n$.
+
+</details>
+
+Similarly, if $G_1,G_2,G_3,\ldots,G_k$ are groups, then
+
+$$
+G_1\times G_2\times G_3\times \cdots\times G_k=\{(g_1,g_2,\ldots,g_k):g_1\in G_1,g_2\in G_2,\ldots,g_k\in G_k\}
+$$
+
+is a group.
+
+Easy to verify by associativity. $(G_1\times G_2)\times G_3=G_1\times G_2\times G_3$.
+
+#### Some extra facts for direct product
+
+1. $G_1\times G_2\simeq G_2\times G_1$, with $\phi(a_1,a_2)=(a_2,a_1)$.
+2. If $H_1\leq G_1$ and $H_2\leq G_2$, then $H_1\times H_2\leq G_1\times G_2$.
+
+> [!WARNING]
+>
+> Not every subgroup of $G_1\times G_2$ is of the form $H_1\times H_2$.
+>
+> Consider $\mathbb{Z}_2\times \mathbb{Z}_2$ with subgroup $\{(0,0),(1,1)\}$, This forms a subgroup but not of the form $H_1\times H_2$.