From c663850ee225c78a465e4ab8af57f64c0a7ff741 Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Sat, 7 Mar 2026 11:19:20 -0600
Subject: [PATCH] Create Math4302_L22.md

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+# Math4302 Modern Algebra (Lecture 22)
+
+## Groups
+
+### Group acting on a set
+
+Let $X$ be a $G$-set, recall that the orbit of $x\in X$ is $\{g\cdot x|g\in G\}$.
+
+#### The orbit-stabilizer theorem
+
+For any $x\in X$, ,$G_x=\{g\in G|g\cdot x=x\}\leq G$.
+
+Let $(G:G_x)$ denote the index of $G_x$ in $G$, then $(G:G_x)=\frac{|G|}{|G_x|}$, which equals to the number of left cosets of $G_x$ in $G$.
+
+<details>
+<summary>Proof</summary>
+
+Define $\alpha:gG_x\mapsto g\cdot x$.
+
+$\alpha$ is well-defined and injective.
+
+$$
+gG_x=g'G_x\iff g^{-1}g'\in G_x\iff (g^{-1}g')\cdot x=x\iff g^{-1}\cdot(g'\cdot x)=x\iff g'\cdot x=g\cdot x
+$$
+
+$\alpha$ is surjective, therefore $\alpha$ is a bijection.
+
+</details>
+
+<details>
+<summary>Example</summary>
+
+Number of elements in the orbit of $x$ is $1$ if and only if $g\cdot x=x$ for all $g\in G$.
+
+if and only if $G_x=G$.
+
+</details>
+
+#### Theorem for orbit with prime power groups
+
+Suppose $X$ is a $G$-set, and $|G|=p^n$ for some prime $p$. Let $X_G$ be the set of all elements in $X$ whose orbit has size $1$. (Recall the orbit divides $X$ into disjoint partitions.) Then $|X|\equiv |X_G|\mod p$.
+
+<details>
+<summary>Examples</summary>
+
+Let $G=D_4$ acting on $\{1,2,3,4\}=X$.
+
+$X_G=\emptyset$ since there is no element whose orbit has size $1$.
+
+---
+
+Let $G=\mathbb{Z}_{11}$ acting on a set with $|X|=20$ if the action is not trivial, then what is $|X_G|$?
+
+Using the theorem we have $|X_G|\equiv 20\mod 11=9$. Therefore $|X_G|=9$ or $20$, but the action is not trivial, $|X_G|=9$.
+
+An instance for such $X=\mathbb{Z}_{11}\sqcup\{x_1,x_2,\ldots,x_9\}$, where $\mathbb{Z}_{11}$ acts on $\{x_1,x_2,\ldots,x_9\}$ trivially. and $\mathbb{Z}_{11}$ acts on $x_1$ with addition.
+
+</details>
+
+<details>
+<summary>Proof</summary>
+
+If $x\in X$ such that $|Gx|\geq 2$, then $\frac{|G|}{|G_x|}=|Gx|\geq 2$.
+
+So $|G|=|G_x||Gx|\implies |Gx|$ divides $|G|$.
+
+So $|Gx|=p^m$ for some $m\geq 1$.
+
+Note that $X$ is the union of subset of elements with orbit of size $1$, and distinct orbits of sizes $\geq 2$. (each of them has size positive power of $p$)
+
+So $p|(|X|-|X_G|)$.
+
+this implies that $|X_G|\equiv |X_G|\mod p$.
+
+</details>
+
+#### Corollary: Cauchy's theorem
+
+If $p$ is prime and $p|(|G|)$, then $G$ has a subgroup of order $p$.
+
+> This does not hold when $p$ is not prime.
+>
+> Consider $A_4$ with order $12$, and $A_4$ has no subgroup of order $6$.
+
+<details>
+<summary>Proof</summary>
+
+It is enough to show, there is $a\in G$ which has order $p$: $\{e,a,a^2,\ldots,a^{p-1}\}\leq G$.
+
+Let $X=\{(g_1,g_2,\ldots,g_p)|g_i\in G,g_1g_2g_3\ldots g_p=e\}$.
+
+Then $|X|=|G|^{p-1}$ since $g_p$ is determined uniquely by $g_p=(g_1,g_2,\ldots,g_{p-1})^{-1}$.
+
+Therefore we can define $\mathbb{Z}_p$ acts on $X$ by shifting.
+
+$i\in \mathbb{Z}_p$ $i\cdot (g_1,g_2,\ldots,g_p)=(g_{i+1},g_2,\ldots,g_p,g_1,\ldots,g_i)$.
+
+$X$ is a $\mathbb{Z}_p$-set.
+
+- $0\cdot (g_1,g_2,\ldots,g_p)=(g_1,g_2,\ldots,g_p)$.
+- $j\cdot (i\cdot (g_1,g_2,\ldots,g_p))=(i+j)\cdot (g_1,g_2,\ldots,g_p)$.
+
+By the previous theorem, $|X|\equiv |X_G|\mod p$.
+
+Since $p$ divides $|G|^{p-1}$, $p$ also divides $|X_G|$. Therefore $(e,e,e,\ldots,e)\in X_G$. Therefore $|X_G|\geq 1$.
+
+So $|X_G|\geq 2$, we have $(a,a,\ldots,a)\in X_G$, $a\neq e$, but $a^p=e$, so $ord(a)=p$.
+
+</details>
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