 # PreTeXt Sample Book: Abstract Algebra (SAMPLE ONLY)

## Section4.1Cyclic groups

Often a subgroup will depend entirely on a single element of the group; that is, knowing that particular element will allow us to compute any other element in the subgroup.
Suppose that we consider $$3 \in {\mathbb Z}$$ and look at all multiples (both positive and negative) of 3. As a set, this is
\begin{equation*} 3 {\mathbb Z} = \{ \ldots, -3, 0, 3, 6, \ldots \}\text{.} \end{equation*}
It is easy to see that $$3 {\mathbb Z}$$ is a subgroup of the integers. This subgroup is completely determined by the element 3 since we can obtain all of the other elements of the group by taking multiples of 3. Every element in the subgroup is “generated” by 3.
If $$H = \{ 2^n : n \in {\mathbb Z} \}\text{,}$$ then $$H$$ is a subgroup of the multiplicative group of nonzero rational numbers, $${\mathbb Q}^*\text{.}$$ If $$a = 2^m$$ and $$b = 2^n$$ are in $$H\text{,}$$ then $$ab^{-1} = 2^m 2^{-n} = 2^{m-n}$$ is also in $$H\text{.}$$ By Proposition 3.3.8, $$H$$ is a subgroup of $${\mathbb Q}^*$$ determined by the element 2.
The identity is in $$\langle a \rangle$$ since $$a^0 = e\text{.}$$ If $$g$$ and $$h$$ are any two elements in $$\langle a \rangle \text{,}$$ then by the definition of $$\langle a \rangle$$ we can write $$g = a^m$$ and $$h = a^n$$ for some integers $$m$$ and $$n\text{.}$$ So $$gh = a^m a^n = a^{m+n}$$ is again in $$\langle a \rangle \text{.}$$ Finally, if $$g = a^n$$ in $$\langle a \rangle \text{,}$$ then the inverse $$g^{-1} = a^{-n}$$ is also in $$\langle a \rangle \text{.}$$ Clearly, any subgroup $$H$$ of $$G$$ containing $$a$$ must contain all the powers of $$a$$ by closure; hence, $$H$$ contains $$\langle a \rangle \text{.}$$ Therefore, $$\langle a \rangle$$ is the smallest subgroup of $$G$$ containing $$a\text{.}$$

### Remark4.1.4.

If we are using the “+” notation, as in the case of the integers under addition, we write $$\langle a \rangle = \{ na : n \in {\mathbb Z} \}\text{.}$$
For $$a \in G\text{,}$$ we call $$\langle a \rangle$$ the cyclic subgroup generated by $$a\text{.}$$ If $$G$$ contains some element $$a$$ such that $$G = \langle a \rangle \text{,}$$ then $$G$$ is a cyclic group. In this case $$a$$ is a generator of $$G\text{.}$$ If $$a$$ is an element of a group $$G\text{,}$$ we define the order of $$a$$ to be the smallest positive integer $$n$$ such that $$a^n= e\text{,}$$ and we write $$|a| = n\text{.}$$ If there is no such integer $$n\text{,}$$ we say that the order of $$a$$ is infinite and write $$|a| = \infty$$ to denote the order of $$a\text{.}$$
Notice that a cyclic group can have more than a single generator. Both 1 and 5 generate $${\mathbb Z}_6\text{;}$$ hence, $${\mathbb Z}_6$$ is a cyclic group. Not every element in a cyclic group is necessarily a generator of the group. The order of $$2 \in {\mathbb Z}_6$$ is 3. The cyclic subgroup generated by 2 is $$\langle 2 \rangle = \{ 0, 2, 4 \}\text{.}$$
The groups $${\mathbb Z}$$ and $${\mathbb Z}_n$$ are cyclic groups. The elements 1 and $$-1$$ are generators for $${\mathbb Z}\text{.}$$ We can certainly generate $${\mathbb Z}_n$$ with 1 although there may be other generators of $${\mathbb Z}_n\text{,}$$ as in the case of $${\mathbb Z}_6\text{.}$$
The group of units, $$U(9)\text{,}$$ in $${\mathbb Z}_9$$ is a cyclic group. As a set, $$U(9)$$ is $$\{ 1, 2, 4, 5, 7, 8 \}\text{.}$$ The element 2 is a generator for $$U(9)$$ since
\begin{align*} 2^1 & = 2 \qquad 2^2 = 4\\ 2^3 & = 8 \qquad 2^4 = 7\\ 2^5 & = 5 \qquad 2^6 = 1\text{.} \end{align*}
Not every group is a cyclic group. Consider the symmetry group of an equilateral triangle $$S_3\text{.}$$ The subgroups of $$S_3$$ are shown in Figure 4.1.8. Notice that every subgroup is cyclic; however, no single element generates the entire group. Figure 4.1.8. Subgroups of $$S_3$$
Let $$G$$ be a cyclic group and $$a \in G$$ be a generator for $$G\text{.}$$ If $$g$$ and $$h$$ are in $$G\text{,}$$ then they can be written as powers of $$a\text{,}$$ say $$g = a^r$$ and $$h = a^s\text{.}$$ Since
\begin{equation*} g h = a^r a^s = a^{r+s} = a^{s+r} = a^s a^r = h g\text{,} \end{equation*}
$$G$$ is abelian.