Cyclic Groups

Abstract

The simplest type of groups are group of integers modulo n and cyclic groups. Cyclic groups are groups in which every element is a power of some fixed element. In this chapter, we examine cyclic groups in detail and determine their important characteristics. We observe that a cyclic subgroup with generator a is the smallest subgroup containing the set $A = \{a\}$. Can we extend subgroups generated by sets with more than one element? We answer this question in this chapter too.

The simplest type of groups are group of integers modulo n and cyclic groups. Cyclic groups are groups in which every element is a power of some fixed element. In this chapter, we examine cyclic groups in detail and determine their important characteristics. We observe that a cyclic subgroup with generator a is the smallest subgroup containing the set $A = \{a\}$. Can we extend subgroups generated by sets with more than one element? We answer this question in this chapter too.

4.1 Group of Integers Modulo n

Let $a,b\in \mathbb {Z}$ and n be a positive integer. Already, we defined

$$ \begin{array}{l} a\equiv b (\mathrm{mod} \ n) \ \Leftrightarrow \ n|a-b. \end{array} $$

This relation is named as congruence modulo n. In Lemma 1.62, we proved that this relation is an equivalence relation. The equivalence class of $a\in \mathbb {Z}$ is the set

$$ \begin{array}{l} \overline{a}=\{b\in \mathbb {Z} \ | \ a\equiv b (\mathrm{mod} \ n)\}.\end{array} $$

The integers modulo n partition $\mathbb {Z}$ into n different equivalence classes. We denote the set of these equivalence classes by $\mathbb {Z}_n$. Thus,

$$ \begin{array}{l} \mathbb {Z}_n =\{\overline{0},\ \overline{1}, \ldots , \overline{n-1}\}. \end{array} $$

For instance, if $n=5$, then

$$ \begin{array}{l} \overline{0}=\{\ldots , -15,\ -10,\ -5,\ 0,\ 5,\ 10,\ 15, \ldots \}, \\ \overline{1}=\{\ldots , -14, \ -9,\ -4,\ 1,\ 6,\ 11,\ 16, \ldots \}, \\ \overline{2}=\{\ldots , -13,\ -8,\ -3,\ 2,\ 7,\ 12,\ 17, \ldots \}, \\ \overline{3}=\{\ldots , -12, \ -7,\ -2,\ 3,\ 8,\ 13,\ 18, \ldots \}, \\ \overline{4}=\{\ldots , -11, \ -6,\ -1,\ 4,\ 9,\ 14,\ 19, \ldots \}, \end{array} $$

and

$$ \begin{array}{l}\mathbb {Z}_5=\{\overline{0},\ \overline{1}, \ \overline{2}, \ \overline{3}, \ \overline{4}\}. \end{array} $$

Theorem 4.1

$(\mathbb {Z}_n, +)$ is an abelian group, where $+$ is the addition modulo n, i.e., $\overline{a}+\overline{b}=\overline{a+b}$, for all $\overline{a}, \overline{b}\in \mathbb {Z}_n$.

Proof

First we show that $+$ is well defined on $\mathbb {Z}_n$. If $\overline{a_1}=\overline{a_2}$ and $\overline{b_1}=\overline{b_2}$, then there exist integers $q_1$ and $q_2$ such that $a_1-a_2=q_1n$ and $b_1-b_2=q_2n$. So, we have

$$ \begin{array}{l} (a_1+b_1)-(a_2+b_2)=a_1-a_2+b_1-b_2=q_1n+q_2n=(q_1+q_2)n. \end{array} $$

This implies that $a_1+b_1\equiv a_2+b_2 (\mathrm{mod} \ n)$, or equivalently $\overline{a_1+b_1}=\overline{a_2+b_2}$. Clearly, addition on $\mathbb {Z}_n$ is associative and commutative. $\overline{0}$ is the additive identity for $\mathbb {Z}_n$ and each $\overline{a}\in \mathbb {Z}_n$ has an additive inverse $\overline{-a}$ in $\mathbb {Z}_n$. Note that $\overline{-a}=\overline{n-a}$. $\blacksquare$

Sometimes, when no confusion can arise, we use $0,\ 1, \ldots , n-1$ to indicate the equivalence classes $\overline{0},\ \overline{1}, \ldots , \overline{n-1}$.

Example 4.2

For $n=12$, we have $3+6=9$, $8+9=5$, $6+7=1$ and $5+10=3$. A familiar use of modular arithmetic is in the 12-hour clock (see Fig. 4.1, in which the day is divided into two 12-hour periods).

Fig. 4.1

5+10=3

Example 4.3

The Cayley table for $(\mathbb {Z}_5, +)$ is as follows:

$$ \begin{array}{|c|ccccc|}\hline \phantom {\Big |} \ + \ \ &{} \ \ \overline{0} \ \ &{} \ \ \overline{1} \ \ &{} \ \ \overline{2} \ \ &{} \ \ \overline{3} \ \ &{} \ \ \overline{4} \ \ \\ \hline \phantom {\Big |} \overline{0} &{} \overline{0} &{} \overline{1} &{} \overline{2} &{} \overline{3} &{} \overline{4} \\ \phantom {\Big |} \overline{1} &{} \overline{1} &{} \overline{2} &{} \overline{3} &{} \overline{4} &{} \overline{0} \\ \phantom {\Big |} \overline{2} &{} \overline{2} &{} \overline{3} &{} \overline{4} &{} \overline{0} &{} \overline{1}\\ \phantom {\Big |} \overline{3} &{} \overline{3} &{} \overline{4} &{} \overline{0} &{} \overline{1} &{} \overline{2} \\ \phantom {\Big |} \overline{4} &{} \overline{4} &{} \overline{0} &{} \overline{1} &{} \overline{2} &{} \overline{3} \\ \hline \end{array} $$

We can also multiply elements of $\mathbb {Z}_n$. Given two elements $\overline{a}$ and $\overline{b}$ in $\mathbb {Z}_n$, we define

$$ \begin{array}{l} \overline{a}\cdot \overline{b}=\overline{ab}. \end{array} $$

This is well defined and with respect to it, $\mathbb {Z}_n$ becomes a commutative monoid.

A multiplicative inverse for $\overline{a}\in \mathbb {Z}_n$ is an element $\overline{b}\in \mathbb {Z}_n$ such that $\overline{a}\cdot \overline{b}=\overline{1}$. An element $\overline{a}\in \mathbb {Z}_n$ is a unit if it has a multiplicative inverse in $\mathbb {Z}_n$. In other words, the integer a is a unit modulo n, meaning that $ab\equiv 1 (\mathrm{mod} \ n)$ for some integer b.

By the above operation we do not obtain a group. For instance, $\overline{0}$ does not have a multiplicative inverse.

Lemma 4.4

$\overline{a}$ is a unit in $\mathbb {Z}_n$ if and only if $(a,n)=1$.

Proof

If $\overline{a}$ is a unit, then $ab-1=qn$ for some integers b and q, and so $ab+qn=1$. Now, by Theorem (1.51) we conclude that $(a,n)=1$.

Conversely, if $(a,n)=1$, then by Theorem (1.51) there exist $x,y\in \mathbb {Z}$ such that $1=ax+by$. Thus, $\overline{x}$ is a multiplicative inverse of $\overline{a}$. $\blacksquare$

Let $U_n$ denote the set of units of $\mathbb {Z}_n$.

Theorem 4.5

For each integer $n\ge 1$, the set $U_n$ forms an abelian group under multiplication modulo n, with identity element $\overline{1}$.

Proof

We first show that the product of two units $\overline{a}$ and $\overline{b}$ is also a unit. If $\overline{a}$ and $\overline{b}$ are units, then they have inverses $\overline{c}$ and $\overline{d}$, respectively, such that

$$ \begin{array}{l} \overline{a}\cdot \overline{c}=\overline{ac}=\overline{1} \ \mathrm{and} \ \overline{b}\cdot \overline{d}=\overline{bd}=\overline{1}. \end{array} $$

Hence, we obtain that

$$ \begin{array}{l} \overline{ab}\cdot \overline{cd}=\overline{abcd}=\overline{acbd}= \overline{ac}\cdot \overline{bd}=\overline{1}\cdot \overline{1}=\overline{1}. \end{array} $$

Consequently, $\overline{ab}$ has the inverse $\overline{cd}$, and is a unit. Moreover, for each $\overline{a}, \overline{b},\overline{c}\in U_n$, we have

$$ \begin{array}{l} \overline{a}\cdot \overline{b}=\overline{ab}=\overline{ba}= \overline{b}\cdot \overline{a},\\ \overline{a}\cdot ( \overline{b}\cdot \overline{c})=\overline{a}\cdot \overline{bc}= \overline{a(bc)}=\overline{(ab)c}= \overline{ab}\cdot \overline{c}=(\overline{a}\cdot \overline{b})\cdot \overline{c}. \end{array} $$

The identity element of $U_n$ is $\overline{1}$, since $\overline{a}\cdot \overline{1}=\overline{a}=\overline{1}\cdot \overline{a}$, for all $\overline{a}\in U_n$. Finally, if $\overline{a}\in U_n$, then by the definition of $U_n$, there exists $\overline{c}\in \mathbb {Z}_n$ such that $\overline{a}\cdot \overline{c}=\overline{1}=\overline{c}\cdot \overline{a}$. This yields that $\overline{c}\in U_n$ and $\overline{c}$ is the inverse of $\overline{a}$. $\blacksquare$

Definition 4.6

Let n be a positive integer. The Euler function $\varphi$ is defined to be the number of positive integers not exceeding n which are relatively prime to n.

Table 4.1 is a short table of values $\varphi (n)$:

Table 4.1 A short table of values $\varphi (n)$

Corollary 4.7

For each integer $n\ge 1$, $|U_n|=\varphi (n)$.

Proof

It follows from Lemma 4.4 and Theorem 4.5. $\blacksquare$

Example 4.8

In $\mathbb {Z}_{14}$, the group of units is $U_{14}=\{1,3,5,9,11,13\}$. The operation in $U_{14}$ is multiplication modulo 14. The following is Cayley table for $U_{14}$:

$$\begin{array}{|c|cccccc|}\hline \phantom {\Big |} \ \ \cdot \ \ &{} \ \ \overline{1} \ \ &{} \ \ \overline{3} \ \ &{} \ \ \overline{5} \ \ &{} \ \ \overline{9} \ \ &{} \ \ \overline{11} \ \ &{} \ \ \overline{13} \ \ \\ \hline \phantom {\Big |} \overline{1} &{} \overline{1} &{} \overline{3} &{} \overline{5} &{} \overline{9} &{} \overline{11} &{} \overline{13} \\ \phantom {\Big |} \overline{3} &{} \overline{3} &{} \overline{9} &{} \overline{1} &{} \overline{13} &{} \overline{5} &{} \overline{11}\\ \phantom {\Big |} \overline{5} &{} \overline{5} &{} \overline{1} &{} \overline{11} &{} \overline{3} &{} \overline{13} &{} \overline{9}\\ \phantom {\Big |} \overline{9} &{} \overline{9} &{} \overline{13} &{} \overline{3} &{} \overline{11} &{} \overline{1} &{} \overline{5}\\ \phantom {\Big |} \overline{11} &{} \overline{11} &{} \overline{5} &{} \overline{13} &{} \overline{1} &{} \overline{9} &{} \overline{3}\\ \phantom {\Big |} \overline{13} &{} \overline{13} &{} \overline{11} &{} \overline{9} &{} \overline{5} &{} \overline{3} &{} \overline{1}\\ \hline \end{array} $$

Example 4.9

If p is prime, then all positive integers smaller than p are relatively prime to p. Hence, $U_p=\{\overline{1},\ \overline{2}, \ldots , \overline{p-1}\}$.

Theorem 4.10

Let $p>1$ be an integer. Then the following statements are equivalent.

• (1) p is prime;

• (2) If $\overline{a}\cdot \overline{b}=\overline{0}$ in $\mathbb {Z}_p$, then $\overline{a}=\overline{0}$ or $\overline{b}=\overline{0}$.

Proof

($1\Rightarrow 2$) Since p is prime, it follows that $U_p=\mathbb {Z}_p-\{0\}$ is a group. Assume that $\overline{a}\cdot \overline{b}=\overline{0}$ in $\mathbb {Z}_p$. If $\overline{a}=\overline{0}$, then there is nothing to prove. If $\overline{a}\not =\overline{0}$, then $\overline{a}\in U_p$. So, there exists $\overline{x}\in U_p$ such that $\overline{x}\cdot \overline{a}=\overline{1}$. Therefore, we get

$$ \begin{array}{l} \overline{0}=\overline{x}\cdot \overline{0}=\overline{x}\cdot (\overline{a}\cdot \overline{b})=(\overline{x}\cdot \overline{a})\cdot \overline{b}=\overline{1}\cdot \overline{b}=\overline{b}. \end{array} $$

($2\Rightarrow 1$) Suppose that $p=ab$ for some integers a and b. We show $a=\pm 1$ or $\pm p$. Since $p=ab$, it follows that $\overline{ab}=\overline{p}=\overline{0}$. This implies that $\overline{a}\cdot \overline{b}=\overline{0}$ in $\mathbb {Z}_p$. Now, by (2) we conclude that $\overline{a}=\overline{0}$ or $\overline{b}=\overline{0}$. If $\overline{a}=\overline{0}$, then p|a, or equivalently $a=kp$ for some integer k. Hence, we have $p=ab=kpb$. This implies that $1=kb$. Since k and b are integers, it follows that the only possibilities are $b=\pm 1$, and so $a=\pm p$. If $\overline{b}=\overline{0}$, then a similar argument shows that $a=\pm 1$. Therefore, p is prime. $\blacksquare$

Exercises

• 1. Solve equation $x^2+\overline{4}x+\overline{8}=\overline{0}$ in $\mathbb {Z}_{11}$.

• 2. Find the number of distinct solutions to the equation $x^2+\overline{(-1)}=\overline{0}$ in $\mathbb {Z}_{5}$, $\mathbb {Z}_{13}$, and in $\mathbb {Z}_{9}$.

• 3. Find, if possible, a multiplicative inverse for $\overline{8}$ in each of $\mathbb {Z}_5$, $\mathbb {Z}_{21}$, and $\mathbb {Z}_{264}$.

• 4. Find a positive integer n and three elements $\overline{a}$, $\overline{b}$, and $\overline{b}$ in $\mathbb {Z}_n$ such that none of $\overline{a}\cdot \overline{b}$, $\overline{b}\cdot \overline{c}$, and $\overline{c}\cdot \overline{a}$ is equal to $\overline{0}$, yet $\overline{a}\cdot \overline{b}\cdot \overline{c}=\overline{0}$.

• 5. Given $\overline{a},\ \overline{b}\in \mathbb {Z}_n$, we say that $\overline{b}$ is a square root  of $\overline{a}$ if $\overline{b}\cdot \overline{b}=\overline{a}$.

  • (a) Find all square roots of elements in $\mathbb {Z}_{17}$, if exist;

  • (b) If p is any prime, show that a has at most two square roots modulo n;

  • (c) Give an example that shows that it is possible for a number to have more than two square roots.

4.2 Cyclic Groups

This section contains a few observations about cyclic groups. The structure of cyclic groups is relatively simple. We examine cyclic groups in detail and determine many of their important characteristics.

If G is a group, $a\in G$ and $k\in \mathbb {Z}$, then we define

$$ a^n=\left\{ \begin{array}{ll} \underbrace{a a\ldots a}_{k\ \mathrm{times}} &{} \ \mathrm{if} \ k>0 \\ e&{} \ \mathrm{if} \ k=0 \\ \underbrace{a^{-1} a^{-1}\ldots a^{-1}}_{-k\ \mathrm{times}} &{} \ \mathrm{if} \ k<0. \end{array} \right. $$

Lemma 4.11

If G is a group and $a\in G$, then $H=\{a^n \ | \ n \in \mathbb {Z}\}$ is a subgroup of G.

Proof

It is straightforward. $\blacksquare$

Definition 4.12

Let G be a group. A subgroup H of G is cyclic if $H=\{a^n \ | \ n \in \mathbb {Z}\}$, for some $a\in H$. In this case we say that H is the cyclic subgroup generated by a.

When this happens, we write $H=\langle a \rangle$.

Definition 4.13

A group G is called cyclic if there is an element $a\in G$ such that $G=\langle a \rangle$. Such an element a is called a generator of G.

Example 4.14

The set of integers under addition is an example of an infinite group which is cyclic and is generated by both 1 and $-1$.

Example 4.15

If $K_4=\{e,\ a, \ b, \ c\}$ is the Klein’s 4-group, then $H=\{e, \ a\}$ is a cyclic subgroup of $K_4$; however, $K_4$ is not a cyclic group.

Example 4.16

The set $\mathbb {Z}_n=\{0, 1, \ldots , n-1\}$ for $n\ge 1$ is a cyclic group under addition modulo n. Both 1 and $-1$ are generators.

Depending on which n we are given, $\mathbb {Z}_n$ may have many generators.

Example 4.17

$\mathbb {Z}_{12} =\langle 1 \rangle =\langle 5 \rangle =\langle 7 \rangle = \langle 11 \rangle$. To verify for instance that $\mathbb {Z}_{12}=\langle 5 \rangle$ we can use a string art. Figure 4.2 is an example of string art that illustrates how 5 generates $\mathbb {Z}_{12}$. Twelve tacks are placed along a circle and numbered. A string is tied to tack 0 and then looped around every fifth tack. As a result, the numbers of the tacks that are reached are exactly the ordered numbers of 5 modulo 12. Note that if every seventh tack were used, the same artwork would be obtained. If every third tack were connected, as in Fig. 4.3, the resulting loop would only use four tacks, and so 3 does not generate $\mathbb {Z}_{12}$.

Fig. 4.2

An examples of string art to show that 5 is a generator of $\mathbb {Z}_{12}$

Fig. 4.3

An example of string art to show that 3 is not a generator of $\mathbb {Z}_{12}$

Example 4.18

An nth root of unity is a complex number z which satisfies the equation $z^n = 1$ for some positive integer n. Let $\omega _n = e^{2\pi i/n}$ be an nth root of unity. All the nth roots of unity form a group under multiplication. It is a cyclic group, generated by $\omega _n$, which is called a primitive root of unity. The term “primitive” exactly refers to being a generator of the cyclic group, namely an nth root of unity is primitive when there is no positive integer k smaller than n such that $\omega _n^k=1$. See Fig. 4.4 for $n=5$.

Fig. 4.4

5th roots of unity, where $\alpha =\displaystyle \frac{2\pi }{5}$

Definition 4.19

The order of an element a in a group G is the least positive integer n such that $a^n=e$. In additive notation this would be $na=0$. If no such integer exist, then we say a has infinite order. The order of an element a is denoted by o(a).

Note that the critical part of Definition 4.19 is that the order is the least positive integer with the given property. The terminology order is used both for groups and group elements, but it is usually clear from the context which one is considered.

So, to find the order of an element a we need only compute the series of products $a, \ a^2, \ a^3, \ldots$ until we first reach the identity. If the identity never appears in the series, then a has infinite order.

Definition 4.20

A torsion group is a group all of whose elements have finite order. On the other hand, a group is said to be torsion-free if all its non-identity elements have infinite order.

Lemma 4.21

Let $a\not =e$ be an element of a group G. If n is the order of a and $a^k=e$ for some integer k, then n divides k.

Proof

By the Division algorithm, there exist integers q and r such that $k=qn+r$ and r is strictly smaller than n. So, we have

$$ \begin{array}{l} e=a^k=a^{nq+r}=a^{nq}a^r. \end{array} $$

Since a has order n, it follows that $a^{nq}=(a^n)^q=e$. This implies that $a^r=e$, which contradicts our choice of n as the smallest positive integer for which a power of a is the identity, unless $r=0$. In this case $k=nq$ and we conclude that n must divide k. $\blacksquare$

Theorem 4.22

If $a\in G$ has order n, then

$$ \begin{array}{l} o(a^k)=\displaystyle \frac{n}{(n,k)}. \end{array} $$

Proof

Suppose that $(n,k)=d$. Then, there exist integers u and v such that $k=ud$, $n=vd$ and $(u,v)=1$. Now, we have

$$ \begin{array}{l} (a^k)^v=(a^{ud})^v=a^{uvd}=a^{un}=(a^n)^u=e. \end{array} $$

Moreover, if $(a^k)^m=e$, then n|km. So, vd|udm, or equivalently v|um. Since $(u,v)=1$, by Euclid’s lemma we conclude that v|m. Therefore, we obtain $o(a^k)=n/d$. $\blacksquare$

We now turn to some results about cyclic groups.

Theorem 4.23

Every cyclic group is abelian.

Proof

Suppose that G is a cyclic group. By definition, there exists $a\in G$ such that $G=\langle a \rangle$. We need to show that $xy=yx$, for all $x,y \in G$. Now, x and y are of the form $x^m$ and $y^n$, for some integers m and n, respectively. Moreover, we have

$$ \begin{array}{l} xy=a^m a^n=a^{m+n}=a^{n+m}=a^na^m=yx, \end{array} $$

as desired. $\blacksquare$

Theorem 4.24

Every subgroup of a cyclic group is also cyclic.

Proof

Let $G=\langle a \rangle$ be our cyclic group and suppose that H is a subgroup of G. We must show that H is cyclic. If it consists of the identity alone, then it is trivial subgroup and there is nothing to prove. So, we may assume that $H\not = \{e\}$. We consider the set

$$ \begin{array}{l} S=\{s\in \mathbb {N} \ | \ a^s\in H\}. \end{array} $$

Since H is non-trivial, it follows that S is non-empty. But being a non-empty subset of positive integers, S must have the least positive integer, call it k. We next claim that $H=\langle a^k\rangle$. In order to prove this claim, it suffices to let h be an arbitrary element of H and show that $h\in \langle a^k\rangle$. Since $h\in G=\langle a\rangle$, it follows that $h=a^m$, for some integer m. We now invoke the Division algorithm to m and k to obtain integers q and r such that $m=kq+r$, where $0\le r<k$. Therefore, we have

$$ \begin{array}{l} h=a^m=a^{kq+r}=a^{kq} a^r. \end{array} $$

This implies that

$$ \begin{array}{l} a^r=a^{-kq}a^m. \end{array} $$

As $a^k$ belongs to H, so does $a^{-kq}$. By closure property of a subgroup, we know that $a^{-kq}a^m=a^r$ belongs to H. As r is strictly smaller than k, it contradicts the choice of k as the smallest element of S, unless $r=0$. This proves that $h=a^{qk}=(a^k)^q\in \langle a^k\rangle =H$. $\blacksquare$

Remark 4.25

Theorem 4.24 tells us that all subgroups of additive group $\mathbb {Z}$ are of the form $n\mathbb {Z}=\{nk \ | \ k\in \mathbb {Z}\}$.

Theorem 4.26

Let G be a finite cyclic group generated by a.

• (1) The generator a has finite order;

• (2) If $o(a)=n$, then the elements $a^k$ with $k=0,1, \ldots , n-1$ are all distinct and

  $$ \begin{array}{l} G=\{e, \ a, \ a^2, \ \ldots , a^{n-1}\}. \end{array} $$

Proof

Since G is cyclic, it follows that

$$ \begin{array}{l} G=\langle a \rangle =\{ \ldots ,\ a^{-2},\ a^{-1}, \ e, \ a, \ a^2, \ldots \}. \end{array} $$

Since G is finite, it follows that for some integers k and m we have $a^m=a^k$. This implies that $a^{m-k}=e$ and so a has finite order.

If $o(a)=n$, then we claim that $G=\{e, \ a , \ a^2, \ \ldots , a^{n-1}\}$. We see that for $0\le m<k\le n-1$, the elements $a^m$ and $a^k$ are distinct as otherwise $a^{m-k}=e$ with $|m-k|<n$. Consequently, $\{e, \ a , \ a^2, \ \ldots , a^{n-1}\}$ is a subgroup of G containing n elements. Now, if x is an arbitrary element of G, then $x=a^j$, for some integer j. If $j>0$, then applying the Division algorithm, we obtain $a^j=a^r$, where $0\le r<n-1$. If $j<0$ and $s=-j$, then we can make $a^s=a^r$, for some $r<n$ and $a^j=a^{-s}=(a^s)^{-1}=(a^r)^{-1}=a^{n-r}$. This yields that $G=\{e, \ a , \ a^2, \ \ldots , a^{n-1}\}$. Therefore, we conclude that the cyclic group generated by an element of order n has precisely n elements. $\blacksquare$

Corollary 4.27

Let G be a group and a be an element of G. Then,

$$ \begin{array}{l} o(a)=|\langle a \rangle |. \end{array} $$

Corollary 4.28

Let G be a finite group of order n. Then, G is cyclic if and only if there exists an element of G of order n.

Corollary 4.29

If G is an infinite cyclic group, then G has an infinite number of subgroups.

Proof

If $G=\langle a \rangle$, then $\langle a^n \rangle \not = \langle a^m \rangle$, for each integer $n\not =m$. $\blacksquare$

Theorem 4.30

Let G be a cyclic group of order n generated by a and let k be a positive integer. If $d=(n,k)$, then $x^k$ and $x^d$ generate the same cyclic subgroup of G.

Proof

If $d=(n,k)$, we show that $a^k$ and $a^d$ are both powers of each other. Since d|k, it follows that $a^k=(x^d)^{k/d}$. On the other hand, since $d=(k,n)$, it follows that $d=kr+ns$, for some $r,s\in \mathbb {Z}$. This implies that $a^d=a^{kr+ns}=a^{kr}a^{ns}=a^{kr}=(a^k)^r$. This yields that $a^k$ and $a^d$ generate the same subgroup of G. $\blacksquare$

Theorem 4.31

Let G be a cyclic group of order n. Then, for each m dividing n, G has a unique subgroup of order m, namely $\langle a^{n/m}\rangle$.

Proof

If $k={n}/{m}$, then $(a^k)^m=(a^{n/m})^m=a^n=e$ and no smaller positive integer power of $a^k$ could be e. So, $\langle a^k \rangle$ is a subgroup of order m. Next, we show that $\langle a^k \rangle$ is the only subgroup of order m. In order to do this, let H be an arbitrary subgroup of G of order m. By the proof of Theorem 4.24, we have $H=\langle a^d \rangle$, where d is the smallest positive integer such that $a^d$ is in H. We apply the Division algorithm to obtain integers q and r such that $n=dq+r$ and $0\le r<d$. Then, we have $e=a^n=a^{dq+r}=(a^d)^q a^r$. This implies that $a^r=(a^d)^{-q}\in H$. Consequently, $r=0$ and so $n=dq$. Moreover, we have

$$ \begin{array}{l} m=|H|=| \langle a^d \rangle |=\displaystyle \frac{n}{d}. \end{array} $$

Therefore, we obtain $d={n}/{m}=k$, i.e., $H=\langle a^d \rangle = \langle a^k\rangle$. This completes the proof. $\blacksquare$

Theorem 4.32

(Generators of a Finite Cyclic Group) Let G be a cyclic group of order n generated by a. Then, $G=\langle a^k\rangle$ if and only if k and n are relatively prime.

Proof

Suppose that k and n are relatively prime. Then, we may write $1=kx+ny$, for some integers x and y. Then, we have

$$ \begin{array}{l} a=a^{kx+ny}=a^{kx}a^{ny}=a^{kx}=(a^k)^x. \end{array} $$

This implies that a belongs to $\langle a^k \rangle$, and so all powers of a belong to $\langle a^k \rangle$. Thus, we conclude that $G=\langle a^k \rangle$ and $a^k$ is a generator of G.

Conversely, suppose that k and n are not relatively prime. Then, there exists an integer $d>1$ such that d|n and d|k. Hence, there exist integers m and r such that $n=md$ and $k=dr$. So, we get

$$ \begin{array}{l} (a^k)^m=a^{drm}=a^{rn}=(a^n)^r=e. \end{array} $$

Since $d>1$, it follows that $m<n$. This shows that $a^k$ is not a generator of G. $\blacksquare$

Corollary 4.33

Let G be a cyclic group of order n generated by a. Then, G has $\phi (n)$ generators.

Corollary 4.34

Let k be a positive integer. Then, k is a generator of $\mathbb {Z}_n$ if and only if k and n are relatively prime.

Theorem 4.35

Let G be a cyclic group of order n generated by a. If d is a positive divisor of n, then the number of elements of order d is $\phi (d)$.

Proof

By Theorem 4.31, there exists exactly one subgroup of order d, say $\langle b \rangle$. Then, every element of order d generates $\langle b \rangle$. On the other hand, by Theorem 4.32, an element $b^k$ generates $\langle b \rangle$ if and only if $(k,d)=1$. The number of such elements is $\varphi (d)$. $\blacksquare$

Theorem 4.36

Every infinite cyclic group has exactly two generators.

Proof

Suppose that a is a generator of the infinite cyclic group G. Then, we have $G=\{ \ldots ,\ a^{-2},\ a^{-1}, \ e, \ a, \ a^2, \ldots \}$. Now, if $a^k\in G$ is another generator of G, then $G=\{ \ldots ,\ a^{-2k},\ a^{-k}, \ e, \ a^k, \ a^{2k}, \ldots \}.$ Since $a^{k+1}\in G$, it follows that there exists an integer m such that $a^{k+1}=a^{km}$. This implies that $a^{k(1-m)+1}=e$. Since G is infinite, it follows that $k(m-1)+1=0$, or equivalently $k(m-1)=1$. This yields that $k=\pm 1$. Consequently, if a is a generator of G, then another generator of G is only $a^{-1}$. $\blacksquare$

For any finite group G, let $\theta _G(d)$ be the number of elements of G of order d. Then, it is clear that

$$ \begin{array}{l} |G|=\displaystyle \sum _{d=1}^{|G|} \theta _G(d). \end{array} $$

Theorem 4.37

In a finite group (not necessary cyclic), $\theta _G(d)$ is divisible by $\varphi (d)$.

Proof

Let G be a finite group. If G has no element of order d, then $\varphi (d)|0=\theta _G(d)$ and we are done. So, suppose that there is $a\in G$ such that $o(a)=d$. By Theorem 4.35, $\langle a \rangle$ has $\varphi (d)$ elements of order d. Now, if all elements of order d belong to $\langle a \rangle$, we are done. Hence, suppose that $x\in G$ such that $o(x)=d$ and $x\in \langle a \rangle$. Then, $\langle x \rangle$ has $\varphi (d)$ elements of order d too. Next, if $\langle a \rangle$ and $\langle x \rangle$ have no elements of order d in common, we have found $2\varphi (d)$ elements of order d. Note that if there is $y\in \langle a \rangle \cap \langle x \rangle$ with $o(y)=d$, then we obtain $\langle a \rangle =\langle x \rangle =\langle y \rangle$, and this is a contradiction. Continuing, we conclude that the number of elements of order d is a multiple of $\varphi (d)$. $\blacksquare$

Exercises

• 1. Draw the Hasse diagram for a cyclic group of order 30.

• 2. Show that the elements of finite order in an abelian group G form a subgroup of G.

• 3. Let $U_{24}$ be the group of invertible elements in $\mathbb {Z}_{24}$. Find all cyclic subgroups of $U_{24}$.

• 4. Find an example of a non-cyclic group, all of whose proper subgroups are cyclic.

• 5. Find a collection of distinct subgroups $\langle a_1 \rangle$, $\langle a_2 \rangle$, $\ldots$, $\langle a_n \rangle$ with the property that $\langle a_1 \rangle \le \langle a_2 \rangle \le \ldots \le \langle a_n \rangle$ with n as large as possible.

• 6. Let G be a group and $a,b\in G$. Prove that

  • (a) $o(a)=o(a^{-1})$;

  • (b) $o(ab)=o(ba)$;

  • (c) $o(a)=o(g^{-1}ag)$, for all $g\in G$;

  • (d) If $ab=ba$, and o(a) and o(b) are relatively prime, then $o(ab)=o(a) o(b)$.

• 7. If a cyclic group has an element of infinite order, how many elements of finite order does it have?

• 8. Give an example of a group that has exactly 6 subgroups (including the trivial subgroup and the group itself). Generalize to exactly n subgroups for any positive integer n.

• 9. If G is an abelian group and contains cyclic subgroups of orders 4 and 5, what other sizes of cyclic subgroups must G contain? Generalize.

• 10. If G is an abelian group and contains a pair of cyclic subgroups of order 2, show that G must contain a subgroup of order 4. Must this subgroup be cyclic?

• 11. Let G be an abelian group. Show that the elements of finite order in G form a subgroup. This subgroup is called the torsion subgroup of G.

• 12. Find the torsion subgroup of the multiplicative group $\mathbb {R}^*$ of non-zero real numbers.

• 13. Find the torsion subgroup T of the multiplicative group $\mathbb {C}^*$ of non-zero complex numbers.

• 14. Let G be an abelian group of order mn such that m and n are relatively prime. If there exists $a,b\in G$ such that $o(a)=m$ and $o(b)=n$, prove that G is cyclic.

• 15. Show that both $U_{25}$ and $U_{27}$ are cyclic groups.

• 16. Prove that $U_{2^n}$ ($n\ge 3$) is not cyclic.

• 17. Let a and b be elements of a group. If $o(a)=m$, $o(b)=n$, and m and n are relatively prime, show that $\langle a \rangle \cap \langle b \rangle =\{e\}$.

• 18. Suppose that $o(x)=n$. Find a necessary and sufficient condition on r and s such that $\langle x^r\rangle \le \langle x^s\rangle$.

• 19. Suppose that $o(x)=n$. Show that $\langle x^r\rangle = \langle x^s\rangle$ if and only if $(n,r)=(n,s)$.

• 20. Suppose that a is a group element such that $o(a^{28})=10$ and $o(a^{22})=20$. Determine o(a).

• 21. Determine the Hasse diagram for subgroups of $\mathbb {Z}_{p^2q}$, where p and q are distinct.

• 22. Give an example of a group and elements $a,b\in G$ such that o(a) and o(b) are finite but o(ab) is not of finite order.

4.3 Generating Sets

We are interested in answering the following questions: What is the smallest subgroup of a group G containing elements $x_1, \ldots , x_n \in G$? How can we describe an arbitrary element in this subgroup? Or, more generally, what is the smallest subgroup of a group G containing a subset A of G and how can we describe an arbitrary element in this subgroup?

Definition 4.38

Let G be a group and A be a subset of G. Let $\{H_i\ | \ i\in I\}$ be the family of all subgroups of G which contain A. Then

$$ \begin{array}{l} \bigcap \limits _{i\in I}H_i \end{array} $$

is called the subgroup of G generated by the set A  and denoted by $\langle A\rangle$.

Indeed, $\langle A\rangle$ is the smallest subgroup of G that contains A. The elements of A are the generators of the subgroup $\langle A \rangle$, which may also be generated by other subsets. If $A=\{x_1, \ldots , x_n\}$, then we write $\langle x_1, \ldots , x_n \rangle$ instead of $\langle A \rangle$. If $G=\langle x_1, \ldots , x_n \rangle$, then G is said to be finitely generated.

Remark 4.39

If $A=\{a\}$, then $G=\langle a \rangle$ is a cyclic group.

Theorem 4.40

If G is a group and A is a non-empty subset of G, then

$$ \begin{array}{l} \langle A \rangle =\{x_1^{\alpha _1} \ldots x_n^{\alpha _n} \ | \ x_i \in A, \ \alpha _i \in \mathbb {Z}, \ 1\le i\le n, \ n\in \mathbb {N}\}. \end{array} $$

Proof

Let

$$ \begin{array}{l} H =\{x_1^{\alpha _1} \ldots x_n^{\alpha _n} \ | \ x_i \in A, \ \alpha _i \in \mathbb {Z}, \ 1\le i\le n, \ n\in \mathbb {N}\}. \end{array} $$

Clearly, H is a non-empty subset of G. Suppose that x and y are two arbitrary elements of H. Then, there exist $x_1, \ldots , x_n,y_1, \ldots , y_m \in A$ and $\alpha _1, \ldots , \alpha _n, \beta _1, \ldots , \beta _m \in \mathbb {Z}$ such that

$$ \begin{array}{l} x=x_1^{\alpha _1} \ldots x_n^{\alpha _n} \ \ \mathrm{and} \ \ y=y_1^{\beta _1} \ldots y_m^{\beta _m}. \end{array} $$

Hence, we obtain

$$ \begin{array}{l} xy^{-1}=x_1^{\alpha _1} \ldots x_n^{\alpha _n}y_m^{-\beta _m} \ldots y_1^{-\beta _1}\in H. \end{array} $$

Thus, H is a subgroup of G. Moreover, $A\subseteq H$. Consequently, by the definition of $\langle A \rangle$, we conclude that $\langle A \rangle \subseteq H$.

On the other hand, since $\langle A \rangle$ is a subgroup of G, it follows that all elements of the form $x_1^{\alpha _1} \ldots x_n^{\alpha _n}$ (for $x_i\in A$ and $\alpha _i \in \mathbb {Z}$) belong to $\langle A \rangle$. This yields that $H\subseteq \langle A \rangle$. Therefore, we deduce that $\langle A \rangle =H$. $\blacksquare$

Remark 4.41

If G is an additive group and A is a non-empty subset of G, then

$$ \begin{array}{l} \langle A \rangle =\{\alpha _1 x_1+ \ldots \alpha _nx_n \ | \ x_i \in A, \ \alpha _i \in \mathbb {Z}, \ 1\le i\le n, \ n\in \mathbb {N}\}. \end{array} $$

Example 4.42

Every finite group is finitely generated since $G=\langle G \rangle$.

Example 4.43

The additive group $\mathbb {Q}$ is not finitely generated: a finite set of rational numbers has a common denominator, say m, and the subgroup of $\mathbb {Q}$ generated by these rational numbers (their integral multiples and sums thereof) will only give rise to rational numbers with denominators dividing m. Not all rationales have such denominators (try $\frac{1}{m+ 1}$), so $\mathbb {Q}$ does not have a finite set of generators as an additive group.

Example 4.44

A finitely generated group is at most countable, so an uncountable group is not finitely generated. For example, the group of real numbers under addition is not finitely generated.

If G is a group and $\{H_i\ | \ i\in I\}$ is a family of subgroups of G, then $\bigcup \limits _{i\in I}H_i$ is not a subgroup of G in general. The subgroup $\langle \bigcup \limits _{i\in I}H_i \rangle$ generated by the set $\bigcup \limits _{i\in I}H_i$ is called the subgroup generated by groups $\{H_i\ | \ i\in I\}$. If H and K are subgroups of G, then the subgroup $\langle H\cup K\rangle$ generated by H and K is called the join of H and K and is denote by $H\vee K$.

Theorem 4.45

If G is a group, A is a subgroup of Z(G), the center of G, and $z\in G$, then $\langle A\cup \{z\}\rangle$, the subgroup generated by A and z, is abelian.

Proof

Let $H=\langle A\cup \{z\}\rangle$ and y be an arbitrary element of H. First, we show that $y=az^k$, for some $a\in A$ and integer k. By Theorem 4.40, we can write $y=a_1a_2\ldots a_n$, where for each $1\le i\le n$, $a_i\in A$, $a_i=z$ or $a_i=z^{-1}$. Since A is a subgroup of Z(G), it follows that $a_1,a_2,\ldots ,a_n$ commute. We set all $a_i$s that belong to A on the left side and hence other elements will be z or $z^{-1}$. If a is the product of $a_i$s that are in A, then $y=az^k$ for some integer k.

Now, suppose that x and y are two arbitrary elements of H. Then, we have

$$ \begin{array}{l} x=a_1z^{k_1} \ \ \mathrm{and} \ \ y=a_2z^{k_2}, \end{array} $$

for some $a_1,a_2 \in A$ and integers $k_1, k_2$. Thus, we obtain

$$ \begin{array}{ll} xy &{} = (a_1z^{k_1})a_2z^{k_2} \\ &{} = a_2 (a_1z^{k_1})z^{k_2} \\ &{} = a_2a_1z^{k_2}z^{k_1} \\ &{} = a_2 z^{k_2} a_1 z^{k_1} \\ &{} = yx. \end{array} $$

This yields that H is an abelian subgroup of H. $\blacksquare$

Exercises

• 1. In $(\mathbb {Z}, +)$, determine the subgroup generated by

  • (a) $\{4,\ 6\}$;

  • (b) $\{4, \ 5\}$.

• 2. Let G be a group and X be a non-empty subset of G. For every $a\in G$, prove that $\langle a^{-1}Xa\rangle = a^{-1} \langle X \rangle a$.

• 3. Prove that group generated by $G\setminus H$ equals G, where H is a proper subgroup of G.

• 4. Let G be an abelian group and $A=\{a_1, \ldots , a_n\}$ be a finite subset of G. If the order of each element of A is finite, prove that $\langle A \rangle$ is a finite subgroup of G.

• 5. Suppose that G is a group containing subgroups H and K. Show that $H\cup K = H\vee K$ if and only if $H\cap K \in \{H,\ K\}$.

• 6. Show that there is a group G containing subgroups H, K, L such that $H\cup K\cup L = H\vee K\vee L$ but $H\cap K\cap L\not \in \{H,\ K,\ L\}$.

4.4 Worked-Out Problems

Problem 4.46

Let G be an abelian group and $x,y\in G$ be of finite orders.

• (1) Show that o(xy)|[o(x), o(y)], where [o(x), o(y)] stands for least common multiple;

• (2) Give an example to illustrate that $o(xy)\not =[o(x),o(y)]$, in general.

Solution

(1) Assume that x and y are two elements of orders m and n, respectively. Let $[m,n]=r$. Then, we have $r=mm^\prime =nn^\prime$, and hence

$$ \begin{array}{l} x^ry^r=(x^m)^{m\prime } (y^n)^{n\prime }=e. \end{array} $$

Now, suppose that $o(xy)=k$, then k is the smallest positive integer k such that $(xy)^k=e$. Since G is abelian, it follows that

$$ \begin{array}{l} (xy)^r=x^ry^r=e \ \mathrm{and} \ (xy)^k=x^ky^k=e. \end{array} $$

This shows that $(xy)^k=(xy)^r=e$. Since $o(xy)=k$, we conclude that k|r. This completes the proof.

(2) For this part, let x be a non-identity element of G and let $y=x^{-1}$. Then, it is clear that $1=o(e)=o(xx^{-1})=o(xy)\not =[o(x),o(y)]$. $\blacksquare$

Problem 4.47

Prove that a group is finite if and only if it has only finitely many subgroups.

Solution

Suppose that G is a finite group of order n. Since the number of subsets of G is $2^n$, it follows that the number of subgroups of G is less than $2^n$.

Conversely, suppose that the number of subgroups of G is finite. Let $a\in G$ be an arbitrary element. We claim that $\langle a \rangle$ is finite. If $\langle a \rangle$ is an infinite group, then the order of a is infinite. Now, for each positive integer k, let $H_k=\langle a^k \rangle$. We prove $H_k$s are distinct. Indeed, if $H_i=H_j$, for some positive integers m and n, then $a^i\in \langle a^j \rangle$ and $a^j\in \langle a^i \rangle$. So, there exist integers r and s such that $a^i=a^{jr}$ and $a^j=a^{is}$. This yields that $a^{i-jr}=e$ and $a^{j-is}=e$. Since the order of a is infinite, it follows that $i-jr=0$ and $j-is=0$, or i|j and j|i. Hence, $i=j$. Consequently, $H_k$s are distinct. This means that G has many infinitely subgroups, and it is a contradiction. Therefore, $\langle a \rangle$ is finite, for every $a\in G$. Since

$$ \begin{array}{l} G=\bigcup \limits _{a\in G}\langle a \rangle \end{array} $$

and the number of subgroups of G is finite, we conclude that

$$ \begin{array}{l} G=\langle a_1 \rangle \cup \langle a_2\rangle \cup \ldots \cup \langle a_n \rangle . \end{array} $$

Since the finite union of finite sets is finite, it follows that G is finite. $\blacksquare$

4.5 Supplementary Exercises

• 1. Let $G=\{a_1, \ a_2, \ \ldots , \ a_n\}$ be a finite abelian group of order n and $x=a_1a_2\ldots a_n$ be the product of all the elements in G. Show that $x^2=e$.

• 2. Suppose that the elements x, y in the group G satisfy the relation $xyx^{-1}=y^2$, where y is a non-identity element.

  • (a) Show that $x^5yx^{-5}=y^{32}$;

  • (b) If $o(x)=5$, compute o(y).

• 3. Let G be the collection of all rational numbers x which satisfy $0 \le x <1$. Show that the binary operation

  $$ x\oplus y=\left\{ \begin{array}{ll} x+y &{} \ \mathrm{if} \ 0\le x+y <1 \\ x+y-1 &{} \ \mathrm{if} \ x+y \ge 1. \end{array} \right. $$

  makes G into an infinite abelian group all of whose elements have finite order.

• 4. Let G be a group and $x\in G$ such that $o(x)=mn$, where m and n are positive integers and relatively prime. Show that one can write $x=ab$, where $o(a)=m$, $o(b)=n$ and $ab=ba$. Moreover, prove the uniqueness of such a representation.

• 5. Give an example of an infinite group that has exactly two elements of order 4.

• 6. For every integer n greater than 2, prove that the group $U_{n^2-1}$ is not cyclic.

• 7. Let G be an abelian and suppose that G has elements of orders m and n, respectively. Prove that G has an element whose order is the least common multiple of m and n.

• 8. Let G be a finite group. Prove that the following conditions are equivalent:

  • (a) G is cyclic and $|G|=p^n$, where p is prime and n is a non-negative integer;

  • (b) If A and B are subgroups of G, then $A\le B$ or $B\le A$.

• 9. Prove that the additive group of all rational numbers is a torsion-free group and, further, that it can be represented as a union of an ascending chain of cyclic subgroups.

• 10. Suppose that G is a finitely generated group and that $G =\langle Y\rangle$, where Y is not necessarily a finite set. Prove that there is a finite subset X of Y such that $G = \langle X\rangle$.

• 11. Prove that a group G is finitely generated if and only if any increasing sequence $H_1\le H_2\le \ldots$ of subgroups of G stabilizes, i.e., $H_i=H_j$ for $i,j\ge k$ starting from some k.