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.

##### Example4.1

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 \}.\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.

##### Example4.2

If $H = \{ 2^n : n \in {\mathbb Z} \}$, then $H$ is a subgroup of the multiplicative group of nonzero rational numbers, ${\mathbb Q}^*$. If $a = 2^m$ and $b = 2^n$ are in $H$, then $ab^{-1} = 2^m 2^{-n} = 2^{m-n}$ is also in $H$. By Proposition 3.31, $H$ is a subgroup of ${\mathbb Q}^*$ determined by the element 2.

##### Remark4.4

If we are using the “+” notation, as in the case of the integers under addition, we write $\langle a \rangle = \{ na : n \in {\mathbb Z} \}$.

For $a \in G$, we call $\langle a \rangle$ the cyclic subgroup generated by $a$. If $G$ contains some element $a$ such that $G = \langle a \rangle$, then $G$ is a cyclic group. In this case $a$ is a generator of $G$. If $a$ is an element of a group $G$, we define the order of $a$ to be the smallest positive integer $n$ such that $a^n= e$, and we write $|a| = n$. If there is no such integer $n$, we say that the order of $a$ is infinite and write $|a| = \infty$ to denote the order of $a$.

##### Example4.5

Notice that a cyclic group can have more than a single generator. Both 1 and 5 generate ${\mathbb Z}_6$; 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 $\langle 2 \rangle = \{ 0, 2, 4 \}$.

The groups ${\mathbb Z}$ and ${\mathbb Z}_n$ are cyclic groups. The elements 1 and $-1$ are generators for ${\mathbb Z}$. We can certainly generate ${\mathbb Z}_n$ with 1 although there may be other generators of ${\mathbb Z}_n$, as in the case of ${\mathbb Z}_6$.

##### Example4.6

The group of units, $U(9)$, in ${\mathbb Z}_9$ is a cyclic group. As a set, $U(9)$ is $\{ 1, 2, 4, 5, 7, 8 \}$. 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. \end{align*}

##### Example4.7

Not every group is a cyclic group. Consider the symmetry group of an equilateral triangle $S_3$. The multiplication table for this group is Table 3.7. The subgroups of $S_3$ are shown in Figure 4.8. Notice that every subgroup is cyclic; however, no single element generates the entire group.

# SubsectionSubgroups of Cyclic Groups¶ permalink

We can ask some interesting questions about cyclic subgroups of a group and subgroups of a cyclic group. If $G$ is a group, which subgroups of $G$ are cyclic? If $G$ is a cyclic group, what type of subgroups does $G$ possess?

##### Example4.15

Let us examine the group ${\mathbb Z}_{16}$. The numbers 1, 3, 5, 7, 9, 11, 13, and 15 are the elements of ${\mathbb Z}_{16}$ that are relatively prime to 16. Each of these elements generates ${\mathbb Z}_{16}$. For example, \begin{align*} 1 \cdot 9 & = 9 & 2 \cdot 9 & = 2 & 3 \cdot 9 & = 11\\ 4 \cdot 9 & = 4 & 5 \cdot 9 & = 13 & 6 \cdot 9 & = 6 \\ 7 \cdot 9 & = 15 & 8 \cdot 9 & = 8 & 9 \cdot 9 & = 1 \\ 10 \cdot 9 & = 10 & 11 \cdot 9 & = 3 & 12 \cdot 9 & = 12\\ 13 \cdot 9 & = 5 & 14 \cdot 9 & = 14 & 15 \cdot 9 & = 7. \end{align*}