# 抽象代数代写| Cyclic subgroups

1. Lecture 8

Reference: Judson, Chapter $4.1$

2. Subgroups

Recall the following fundamental example of a subgroup.

3. Example 1

Consider group $(\mathbb{Z},+) .$ Then for any fixed nonzero $m \in \mathbb{Z}$ there is a subgroup of elements divisible by $m$

$$m \mathbb{Z}:={k \cdot m \mid k \in \mathbb{Z}} \subset \mathbb{Z}$$

It turns out that the above example provides an exhaustive list of subgroups of $(\mathbb{Z},+) .$ Specifically, we have the following theorem:

4. Theorem 1

Let $H \subset \mathbb{Z}$ be a subgroup with respect to addition. Then either $H$ is trivial:

$$H={0}$$

or $H$ is of the form

$$H=m \mathbb{Z}$$

for some fixed nonzero $m \in \mathbb{Z}$.

Proof. Assume that $H$ is nontrivial. Then we have some nonzero integer $a \in H$. By subgroup property, we also have $-a \in H$, hence there is at least one positive integer in $H$.

Let $m$ be the smallest positive integer in $H$. We claim that $H=m \mathbb{Z}$.

1. $m \mathbb{Z} \subset H:$ Since $m \in H$, by subgroup properties, we have $-m \in H, 2 m=m+m \in H$, and, more generally, for any $k \in \mathbb{Z}$ we have $k \cdot m \in H .$ This proves $m \mathbb{Z} \subset H .$
2. $H \subset m \mathbb{Z}$. Let us take any element $a \in H$. We claim that $a$ is divisible by $m$ without remainder. Indeed, let us divide $a$ by $m$ with remainder:

$$a=m \cdot q+r, \quad r \in{0,1, \ldots, m-1}$$

Since $a \in H$ and $m \in H$, by subgroup properties we have

$$a-k \cdot m \in H$$

for any $k \in \mathbb{Z}$. In particular taking $k=q$ we conclude that

$$r \in H .$$

But $m$ is the smallest positive element of $H$, therefore $r$ must be 0 .

5. Cyclic subgroups

As usual, let $(G, *)$ be any abstract group. Before we introduce the notion of a cyclic subgroup let us make a trivial remark:

6. Remark 1

If we have a subgroup $H \subset G$ and an element $a \in H$, then for any $k \in{1,2,3, \ldots}$ we also have

$$a^{k}:=\underbrace{a * \cdots * a}_{k \text { times }} \in H .$$

Similarly, since necessarily $a^{-1} \in H$, we also have

$$a^{-k} \in H$$

Also $a^{0}=e \in H$, therefore we can write

$$\left{a^{k} \mid k \in \mathbb{Z}\right} \subset H$$

and this is true for any subgroup $H \subset G$ containing $a$.

Problem 1: Subset $\left{a^{k} \mid k \in \mathbb{Z}\right} \subset G$ is itself a subgroup.

This remark motivates the following definition.

7. Definition 1: Cyclic subgroup

A subgroup $H$ of $(G, *)$ is called cyclic, if there exists an element $a \in G$ such that

$$H=\left{a^{k} \mid k \in \mathbb{Z}\right} \subset G$$

Element $a$ is called a generator of a cyclic subgroup.

8. Remark 2

Generator of a cyclic subgroup is not unique. For example, both [1] and [2] in $\mathbb{Z}{3}$ generate the whole group $\left(\mathbb{Z}{3},+\right)$

NB 1: In the presentation $\left{a^{k} \mid k \in \mathbb{Z}\right}$ some elements might coincide, so that possibly $a^{l}=a^{n}$ for some pairs $l, n \in \mathbb{Z}$.

9. Example 2

Consider group $\left(\mathbb{Z}{7}^{\times}, \times\right)$and an element $3 \in \mathbb{Z}{7}^{\times} .$Consider elements $3,3^{2}, 3^{3}, \ldots$ in $\mathbb{Z}_{7}^{\times}$:

$$\begin{array}{cccccccccc} 3^{0}, & 3, & 3^{2}, & 3^{3}, & 3^{4}, & 3^{5}, & 3^{6}, & 3^{7}, & 3^{8}, & \ldots \ 1, & 3, & 2, & 6, & 4, & 5, & 1, & 3, & 2, & \ldots \end{array}$$

As we can see, elements in the bottom row repeat cyclically with period $6 .$

Problem 2: Do the same calculation for $2 \in \mathbb{Z}{7}^{\times}$and for $6 \in \mathbb{Z}{7}^{\times} .$What are the cyclic periods in this case?