Cyclic Subgroup added (#83)

Author: minsubb13

content/Basic Algebra/Cyclic Subgroup.md

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+# Definition
+If $G$ is a [[Group]] and $a \in G$, then $H = \{a^n\; |\; n \in \mathbb{Z} \}$ is a [[Subgroup]] of $G$. This group is the **cyclic subgroup $\langle a \rangle$ of $G$ generated by** a. Also, given a group $G$ and an element $a$ in $G$, if $G = \{a^n\; |\; n \in \mathbb{Z} \}$, then $a$ is a **generator of** $G$ and the group $G = \langle a \rangle$ is **cyclic**.
+
+Let $a$ be an element of a group $G$. If the cyclic subgroup $\langle a \rangle$ of $G$ is finite, then the **order of** $a$ is the order $|\langle a \rangle|$ of this cyclic subgroup. Otherwise, we say that $a$ is of **infinite order**.
+
+## Example
+### Cyclic Subgroups of $\mathbb{Z}/6\mathbb{Z}$
+
+Let $\mathbb{Z}/6\mathbb{Z}$ be an additive group of integers modulo 6. For any element $a \in \mathbb{Z}/6\mathbb{Z}$, the cyclic subgroup generated by $a$, denoted $\langle a \rangle$, is defined as:
+$\langle a \rangle = \{na\; |\; n \in \mathbb{Z} \} = \{na\;(mod \space 6) \; |\; n \in \mathbb{Z} \}$
+
+To understand the generation of cyclic subgroups, we can use the following addition table:
+
+|   || 0 | 1 | 2 | 3 | 4 | 5 |
+|---|---|---|---|---|---|---|---|
+| 0 || 0 | 1 | 2 | 3 | 4 | 5 |
+| 1 || 1 | 2 | 3 | 4 | 5 | 0 |
+| 2 || 2 | 3 | 4 | 5 | 0 | 1 |
+| 3 || 3 | 4 | 5 | 0 | 1 | 2 |
+| 4 || 4 | 5 | 0 | 1 | 2 | 3 |
+| 5 || 5 | 0 | 1 | 2 | 3 | 4 |
+
+Case 1: Generator $a = 1$
+
+- $\langle 1 \rangle = \{0⋅1, 1⋅1, 2⋅1, 3⋅1, 4⋅1, 5⋅1\} \space (mod \space 6) = \{0, 1, 2, 3, 4, 5\} = \mathbb{Z}/6\mathbb{Z}$
+
+- $|\langle 1 \rangle| = 6$
+
+Case 2: Generator $a = 2$
+
+- $\langle 2 \rangle = \{0⋅2, 1⋅2, 2⋅2, 3⋅2, 4⋅2, 5⋅2\} \space (mod \space 6) = \{0, 2, 4\}$
+
+- $|\langle 2 \rangle| = 3$
+
+Case 3: Generator $a = 3$
+
+- $\langle 3 \rangle = \{0⋅3, 1⋅3, 2⋅3, 3⋅3, 4⋅3, 5⋅3\} \space (mod \space 6) = \{0, 3\}$
+
+- $|\langle 3 \rangle| = 2$
+
+Case 4: Generator $a = 4$
+
+- $\langle 4 \rangle = \{0⋅4, 1⋅4, 2⋅4, 3⋅4, 4⋅4, 5⋅4\} \space (mod \space 6) = \{0, 4, 2\} = \langle 2 \rangle$
+
+- $|\langle 4 \rangle| = 3$
+
+Case 5: Generator $a = 5$
+
+- $\langle 5 \rangle = \{0⋅5, 1⋅5, 2⋅5, 3⋅5, 4⋅5, 5⋅5\} \space (mod \space 6) = \{0, 5, 4, 3, 2, 1\} = \mathbb{Z}/6\mathbb{Z}$
+
+- $|\langle 5 \rangle| = 6$
+
+### Theorem 1
+The order of each cyclic subgroup of $\mathbb{Z}/6\mathbb{Z}$ divides the order of $\mathbb{Z}/6\mathbb{Z}$ (Lagrange's Theorem).
+
+Proof: $|\langle a \rangle| \in \{1, 2, 3, 6\}$ for all $a \in \mathbb{Z}/6\mathbb{Z}$
+
+### Theorem 2
+$\mathbb{Z}/6\mathbb{Z}$ is itself cyclic, generated by either $1$ or $5$.
+
+i.e., $\mathbb{Z}/6\mathbb{Z} = \langle 1 \rangle = \langle 5 \rangle$
+
+#### Summary of Distinct Cyclic Subgroups
+1. Trivial subgroup: $\langle 0 \rangle = \{0\}$, order 1
+2. Order 2 subgroup: $\langle 3 \rangle = \{0, 3\}$
+3. Order 3 subgroup: $\langle 2 \rangle = \langle 4 \rangle = \{0, 2, 4\}$
+4. Order 6 subgroup: $\langle 1 \rangle = \langle 5 \rangle = \mathbb{Z}/6\mathbb{Z}$
+
+### Conclusion 
+Cyclic subgroups of $\mathbb{Z}/6\mathbb{Z}$ are $\{0 \}, \langle 2 \rangle, \langle 3 \rangle, \mathbb{Z}/6\mathbb{Z}$.