## Section4.1Cyclic Subgroups

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 $$3\text{.}$$ 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 $$3\text{.}$$ Every element in the subgroup is “generated” by $$3\text{.}$$

###### Example4.2.

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 3.31, $$H$$ is a subgroup of $${\mathbb Q}^*$$ determined by the element $$2\text{.}$$

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 $$a\text{.}$$

###### Remark4.4.

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 $$a\text{.}$$

###### Example4.5.

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 $$3\text{.}$$ 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 $${\mathbb Z}_6\text{.}$$

###### Example4.6.

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*}
###### Example4.7.

Not every group is a cyclic group. Consider the symmetry group of an equilateral triangle $$S_3\text{.}$$ The multiplication table for this group is Figure 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. Figure 4.8. 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.

### SubsectionSubgroups of Cyclic Groups

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?

The main tools used in this proof are the division algorithm and the Principle of Well-Ordering. Let $$G$$ be a cyclic group generated by $$a$$ and suppose that $$H$$ is a subgroup of $$G\text{.}$$ If $$H = \{ e \}\text{,}$$ then trivially $$H$$ is cyclic. Suppose that $$H$$ contains some other element $$g$$ distinct from the identity. Then $$g$$ can be written as $$a^n$$ for some integer $$n\text{.}$$ Since $$H$$ is a subgroup, $$g^{-1} = a^{-n}$$ must also be in $$H\text{.}$$ Since either $$n$$ or $$-n$$ is positive, we can assume that $$H$$ contains positive powers of $$a$$ and $$n \gt 0\text{.}$$ Let $$m$$ be the smallest natural number such that $$a^m \in H\text{.}$$ Such an $$m$$ exists by the Principle of Well-Ordering.

We claim that $$h = a^m$$ is a generator for $$H\text{.}$$ We must show that every $$h' \in H$$ can be written as a power of $$h\text{.}$$ Since $$h' \in H$$ and $$H$$ is a subgroup of $$G\text{,}$$ $$h' = a^k$$ for some integer $$k\text{.}$$ Using the division algorithm, we can find numbers $$q$$ and $$r$$ such that $$k = mq +r$$ where $$0 \leq r \lt m\text{;}$$ hence,

\begin{equation*} a^k = a^{mq +r} = (a^m)^q a^r = h^q a^r\text{.} \end{equation*}

So $$a^r = a^k h^{-q}\text{.}$$ Since $$a^k$$ and $$h^{-q}$$ are in $$H\text{,}$$ $$a^r$$ must also be in $$H\text{.}$$ However, $$m$$ was the smallest positive number such that $$a^m$$ was in $$H\text{;}$$ consequently, $$r=0$$ and so $$k=mq\text{.}$$ Therefore,

\begin{equation*} h' = a^k = a^{mq} = h^q \end{equation*}

and $$H$$ is generated by $$h\text{.}$$

First suppose that $$a^k=e\text{.}$$ By the division algorithm, $$k = nq + r$$ where $$0 \leq r \lt n\text{;}$$ hence,

\begin{equation*} e = a^k = a^{nq + r} = a^{nq} a^r = e a^r = a^r\text{.} \end{equation*}

Since the smallest positive integer $$m$$ such that $$a^m = e$$ is $$n\text{,}$$ $$r= 0\text{.}$$

Conversely, if $$n$$ divides $$k\text{,}$$ then $$k=ns$$ for some integer $$s\text{.}$$ Consequently,

\begin{equation*} a^k = a^{ns} = (a^n)^s = e^s = e\text{.} \end{equation*}

We wish to find the smallest integer $$m$$ such that $$e = b^m = a^{km}\text{.}$$ By Proposition 4.12, this is the smallest integer $$m$$ such that $$n$$ divides $$km$$ or, equivalently, $$n/d$$ divides $$m(k/d)\text{.}$$ Since $$d$$ is the greatest common divisor of $$n$$ and $$k\text{,}$$ $$n/d$$ and $$k/d$$ are relatively prime. Hence, for $$n/d$$ to divide $$m(k/d)$$ it must divide $$m\text{.}$$ The smallest such $$m$$ is $$n/d\text{.}$$

###### Example4.15.

Let us examine the group $${\mathbb Z}_{16}\text{.}$$ The numbers $$1\text{,}$$ $$3\text{,}$$ $$5\text{,}$$ $$7\text{,}$$ $$9\text{,}$$ $$11\text{,}$$ $$13\text{,}$$ and $$15$$ are the elements of $${\mathbb Z}_{16}$$ that are relatively prime to $$16\text{.}$$ Each of these elements generates $${\mathbb Z}_{16}\text{.}$$ 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\text{.} \end{align*}