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content/Math4302/Math4302_L6.md

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+# Math4302 Modern Algebra (Lecture 6)
+
+## Subgroups
+
+### Dihedral group
+
+The dihedral group $D_n$ is the group of all rotations and reflections about the center of the regular polygon of $n$ sides.
+
+$|S_n|=n!, |D_n|=2n$
+
+### Cyclic group
+
+$G=\langle a\rangle=\{1,a,a^2,\cdots\}$ for some $a\in G$
+
+<details>
+<summary>Example of cyclic group</summary>
+
+$(\mathbb{Z}_n,+)$ is cyclic and $\mathbb{Z}_n=\langle 1\rangle=\{0,1,2,\cdots,n-1\}$
+
+---
+
+$(\mathbb{Z},+)$ is cyclic and $\mathbb{Z}=\langle 1\rangle=\langle -1 \rangle$
+
+---
+
+$S_3$ is not cyclic
+
+$\langle e\rangle=\{e\}$
+$\langle (1,2)\rangle=\{e,(1,2)\}$
+$\langle (1,3)\rangle=\{e,(1,3)\}$
+$\langle (2,3)\rangle=\{e,(2,3)\}$
+$\langle (1,2,3)\rangle=\{e,(1,2,3),(1,3,2)\}$
+$\langle (1,3,2)\rangle=\{e,(1,3,2),(1,2,3)\}$
+
+</details>
+
+#### Every cyclic group is abelian
+
+Every cyclic group is abelian
+
+<details>
+<summary>Proof</summary>
+
+Let $G=\langle a\rangle$ be a cyclic group, then $\forall g_1,g_2\in G$ we have $g_1g_2=g_2g_1$ since $g_1g_2=a^k_1a^k_2=a^{k_1+k_2}$ and $g_2g_1=a^k_2a^k_1=a^{k_1+k_2}$
+
+</details>
+
+#### Definition for order of element
+
+Let $G$ be a group, then the order of $g\in G$ is defined to be the size of the smallest subgroup containing $g$.
+
+If $|\langle g\rangle|$ is infinite, then we say that $g$ has infinite order.
+
+<details>
+<summary>Example of order of element</summary>
+
+$5$ in $(\mathbb{Z},+)$ has infinite order.
+
+---
+
+$5$ in $(\mathbb{Z}_{10},+)$ has order $2$.
+
+$\langle 5\rangle=\{0,5\}$.
+
+---
+
+$5$ in $(\mathbb{Z}_{6},+)$ has order $6$.
+
+$\langle 5\rangle=\{0,5,4,3,2,1\}$.
+
+</details>
+
+#### Lemma for order of element
+
+Let $G$ be a group, then $a\in G$ has order $n$ if $n$ is the smallest positive integer such that $a^n=e$.
+
+<details>
+<summary>Proof</summary>
+
+There are 2 cases:
+
+Case 1:
+
+There is no positive $n$ such that $a^n=e$.
+
+Then $a^i\neq a^j$ if $i\neq j, i,j\in \mathbb{N}$.
+
+Reason: if $a^i=a^j$, then $a^{i-j}=e$.
+
+Then the order of group is infinite.
+
+Case 2:
+
+There is a positive $n$ such that $a^n=e$.
+
+Let $n$ be the smallest such positive integer. Then we claim $\langle a^n\rangle=\{e,a^1,a^2,\cdots,a^{n-1}\}$.
+
+We claim they are all distinct.
+
+Suppose not, then we can have $a^i=a^j$ for $i\neq j$, $0\leq i,j\leq n-1$.
+
+Then $a^{i-j}=e$ but $i-j\leq n-1$. Therefore $n$ is not the smallest positive integer such that $a^n=e$.
+
+</details>
+
+#### Theorem for cyclic group up to isomorphism
+
+Suppose $G$ is a cyclic group,
+
+- If $|G|=n$, then $|G|\simeq \mathbb{Z}_n^+$
+- If $|G|=\infty$, then $|G|\simeq \mathbb{Z}$.
+
+<details>
+<summary>Proof</summary>
+
+Case 1:
+
+If $|G|=\infty$, then we can map $G$ to $(\mathbb{Z},+)$, where $G=\langle a\rangle$. $\phi(n)=a^n$. This gives a bijection between $G$ and $(\mathbb{Z},+)$.
+
+where $\phi(n+m)=a^{n+m}=a^n a^m=\phi(n)\phi(m)$.
+
+Case 2:
+
+If $|G|=n$, then we can map $G$ to $(\mathbb{Z}_n,+)$, where $G=\langle a\rangle$. $\phi(n)=a^n$. This gives a bijection between $G$ and $(\mathbb{Z}_n,+)$.
+
+where $\phi(n+m)=\phi(r)=a^{n+m}=a^n a^m=\phi(n)\phi(m)$.
+</details>
+
+<details>
+<summary>Example</summary>
+
+Let $H=\langle (12)(345)\rangle\subseteq S_5$. Then $H\simeq \mathbb{Z}_6^+$.
+
+Let $\tau=(12)(345)$
+
+All the elements of $H$ are:
+
+- $\tau^0=(12)(345)$
+- $\tau^1=(453)$
+- $\tau^2=(12)(534)$
+- $\tau^3=(345)$
+- $\tau^4=(12)(453)$
+- $\tau^5=(534)$
+
+</details>
+
+#### GCD and order
+
+If $G=\langle a\rangle$, then $H=\langle a^k\rangle$, $|H|=\frac{n}{d}$ where $d=\operatorname{gcd}(n,k)$.

content/Math4302/_meta.js

@@ -8,4 +8,5 @@ export default {
     Math4302_L3: "Modern Algebra (Lecture 3)",
     Math4302_L4: "Modern Algebra (Lecture 4)",
     Math4302_L5: "Modern Algebra (Lecture 5)",
+    Math4302_L6: "Modern Algebra (Lecture 6)",
 }