Cyclic Subgroup

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 $⟨a⟩$ 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 $G=⟨a⟩$ is cyclic.

Let $a$ be an element of a group $G$. If the cyclic subgroup $⟨a⟩$ of $G$ is finite, then the order of $a$ is the order $|⟨a⟩|$ 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 $⟨a⟩$, is defined as: $⟨a⟩=\{na|n\in \mathbb{Z}\}=\{na(mod6)|n\in \mathbb{Z}\}$

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 $a=1$

• $⟨1⟩=\{0\cdot 1,1\cdot 1,2\cdot 1,3\cdot 1,4\cdot 1,5\cdot 1\}(mod6)=\{0,1,2,3,4,5\}=\mathbb{Z}/6\mathbb{Z}$

• $|⟨1⟩|=6$

Case 2: Generator $a=2$

• $⟨2⟩=\{0\cdot 2,1\cdot 2,2\cdot 2,3\cdot 2,4\cdot 2,5\cdot 2\}(mod6)=\{0,2,4\}$

• $|⟨2⟩|=3$

Case 3: Generator $a=3$

• $⟨3⟩=\{0\cdot 3,1\cdot 3,2\cdot 3,3\cdot 3,4\cdot 3,5\cdot 3\}(mod6)=\{0,3\}$

• $|⟨3⟩|=2$

Case 4: Generator $a=4$

• $⟨4⟩=\{0\cdot 4,1\cdot 4,2\cdot 4,3\cdot 4,4\cdot 4,5\cdot 4\}(mod6)=\{0,4,2\}=⟨2⟩$

• $|⟨4⟩|=3$

Case 5: Generator $a=5$

• $⟨5⟩=\{0\cdot 5,1\cdot 5,2\cdot 5,3\cdot 5,4\cdot 5,5\cdot 5\}(mod6)=\{0,5,4,3,2,1\}=\mathbb{Z}/6\mathbb{Z}$

• $|⟨5⟩|=6$

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: $|⟨a⟩|\in \{1,2,3,6\}$ 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., $\mathbb{Z}/6\mathbb{Z}=⟨1⟩=⟨5⟩$

Summary of Distinct Cyclic Subgroups

1. Trivial subgroup: $⟨0⟩=\{0\}$, order 1
2. Order 2 subgroup: $⟨3⟩=\{0,3\}$
3. Order 3 subgroup: $⟨2⟩=⟨4⟩=\{0,2,4\}$
4. Order 6 subgroup: $⟨1⟩=⟨5⟩=\mathbb{Z}/6\mathbb{Z}$

Conclusion

Cyclic subgroups of $\mathbb{Z}/6\mathbb{Z}$ are $\{0\},⟨2⟩,⟨3⟩,\mathbb{Z}/6\mathbb{Z}$.