Finite and Infinite Cyclic Groups

After proving that cyclic groups are abelian, we now examine how the size of a cyclic group is determined by the order of its generator. If a cyclic group is generated by an element

a

, then all elements are powers of

a

. Therefore the group is finite exactly when the powers of

a

eventually return to the identity. This connects cyclic groups with the earlier section on order of an element. In this lesson, students will prove the finite and infinite cases and learn how the order of a generator controls the whole group.

Let

(G,\circ)

be a finite group and let

a\in G

such that

\begin{equation*} G=\langle a\rangle. \end{equation*}

|G|=n

, then

\begin{equation*} o(a)=n. \end{equation*}

After proving that cyclic groups are abelian, we now examine how the size of a cyclic group is determined by the order of its generator. If a cyclic group is generated by an element

a

, then all elements are powers of

a

. Therefore the group is finite exactly when the powers of

a

Let

(G,\circ)

be a finite group and let

a\in G

such that

\begin{equation*} G=\langle a\rangle. \end{equation*}

|G|=n

, then

\begin{equation*} o(a)=n. \end{equation*}

Given that

(G,\circ)

is a finite group,

a\in G

G=\langle a\rangle

, and

|G|=n

. To prove that

\begin{equation*} o(a)=n. \end{equation*}

Since

G=\langle a\rangle

, every element of

G

is an integral power of

a

. Since

|G|=n

, the distinct powers are

\begin{equation*} e,a,a^2,\ldots,a^{n-1}. \end{equation*}

Therefore

\begin{equation*} a^n=e. \end{equation*}

Also, if

1\leq r<n

, then

\begin{equation*} a^r\neq e, \end{equation*}

because otherwise the list of powers would repeat before producing

n

distinct elements. Thus the least positive integer

r

such that

a^r=e

n

. Hence,

\begin{equation*} o(a)=n. \end{equation*}

\square

Let

(G,\circ)

be a group and let

a\in G

such that

\begin{equation*} G=\langle a\rangle. \end{equation*}

\begin{equation*} o(a)=n, \end{equation*}

then

\begin{equation*} G=\{e,a,a^2,\ldots,a^{n-1}\} \end{equation*}

and

\begin{equation*} |G|=n. \end{equation*}

Given that

(G,\circ)

is a group,

a\in G

G=\langle a\rangle

, and

o(a)=n

. To prove that

\begin{equation*} G=\{e,a,a^2,\ldots,a^{n-1}\} \end{equation*}

and

|G|=n

. Since

o(a)=n

, we have

\begin{equation*} a^n=e. \end{equation*}

Also, the elements

\begin{equation*} e,a,a^2,\ldots,a^{n-1} \end{equation*}

are distinct. Let

x\in G

. Since

G=\langle a\rangle

, there exists

k\in\mathbb{Z}

such that

\begin{equation*} x=a^k. \end{equation*}

By the division algorithm, there exist

q,r\in\mathbb{Z}

such that

\begin{equation*} k=nq+r,\quad 0\leq r<n. \end{equation*}

Now

\begin{align*} x &=a^k\\ &=a^{nq+r}\\ &=(a^n)^q\circ a^r\\ &=e^q\circ a^r\\ &=a^r. \end{align*}

Thus every element of

G

belongs to

\{e,a,a^2,\ldots,a^{n-1}\}

. Therefore

\begin{equation*} G=\{e,a,a^2,\ldots,a^{n-1}\}. \end{equation*}

Since these

n

elements are distinct, we get

\begin{equation*} |G|=n. \end{equation*}

Hence,

\begin{equation*} G=\{e,a,a^2,\ldots,a^{n-1}\} \end{equation*}

and

|G|=n

\square

Let

(G,\circ)

be a group and let

a\in G

such that

G=\langle a\rangle

. Then

G

is infinite if and only if

a

has infinite order.

Given that

(G,\circ)

is a group,

a\in G

, and

G=\langle a\rangle

. To prove that

G

is infinite if and only if

a

has infinite order. [1] To prove that if

G

is infinite, then

a

has infinite order. If possible let

a

have finite order. Let

\begin{equation*} o(a)=n. \end{equation*}

Then by the previous theorem,

\begin{equation*} G=\{e,a,a^2,\ldots,a^{n-1}\}. \end{equation*}

Therefore

G

is finite. A contradiction. Hence, if

G

is infinite, then

a

has infinite order. [2] To prove that if

a

has infinite order, then

G

is infinite. If possible let

G

be finite. Since

G=\langle a\rangle

, two positive powers of

a

must be equal. Thus there exist

r,s\in\mathbb{N}

such that

\begin{equation*} r<s \end{equation*}

and

\begin{equation*} a^r=a^s. \end{equation*}

Then

\begin{align*} a^r=a^s &\implies a^{s-r}=e. \end{align*}

Since

s-r\in\mathbb{N}

, this contradicts that

a

has infinite order. A contradiction. Hence, if

a

has infinite order, then

G

is infinite. Hence,

G

is infinite if and only if

a

has infinite order.

\square

The group

(\mathbb{Z},+)

is cyclic with generator

1

. The element

1

has infinite order under addition because no positive multiple of

1

0

. Therefore

(\mathbb{Z},+)

is an infinite cyclic group.

The group

(\mathbb{Z}_{10},+_{10})

is cyclic with generator

\overline{1}

. Since

\begin{equation*} 10\overline{1}=\overline{0} \end{equation*}

and no smaller positive multiple gives

\overline{0}

, we get

\begin{equation*} o(\overline{1})=10. \end{equation*}

Therefore

|\mathbb{Z}_{10}|=10

Let

(G,\circ)

be a cyclic group generated by

a

. If

o(a)=7

, list all elements of

G

Let

(G,\circ)

be a cyclic group generated by

a

. Given that

\begin{equation*} o(a)=7. \end{equation*}

To list all elements of

G

. Since

G=\langle a\rangle

and

o(a)=7

, we get

\begin{equation*} G=\{e,a,a^2,a^3,a^4,a^5,a^6\}. \end{equation*}

Hence, the elements of

G

are

\begin{equation*} e,a,a^2,a^3,a^4,a^5,a^6. \end{equation*}

\square

Use the calculator below to list the elements of a finite cyclic group generated by

a

when

o(a)=n

. Enter

n

. The calculator displays the formal list

e,a,a^2,\ldots,a^{n-1}

. This reinforces the relation between the order of a generator and the size of a finite cyclic group.

[1] If

G=\langle a\rangle

and

o(a)=5

, list the elements of

G

. [2] If

G=\langle a\rangle

and

|G|=12

, find

o(a)

. [3] Give an example of an infinite cyclic group. [4] Give an example of a finite cyclic group of order

8

. [5] Prove that if

G=\langle a\rangle

and

a

has infinite order, then all powers

a^k

are distinct.

[1]

G=\{e,a,a^2,a^3,a^4\}

. [2]

o(a)=12

. [3]

(\mathbb{Z},+)

is an infinite cyclic group. [4]

(\mathbb{Z}_8,+_8)

is a finite cyclic group of order

8

. [5] If

a^r=a^s

with

r<s

, then

a^{s-r}=e

, contradicting infinite order.

It has

n

elements.

It is infinite exactly when its generator has infinite order.

They are

e,a,a^2,\ldots,a^{n-1}

when

o(a)=n

No. A finite cyclic group has a generator of finite order equal to the group size.

BMLabs Mathematics

BMLabs Mathematics

Finite and Infinite Cyclic Groups

Finite and Infinite Cyclic Groups

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Invariants under Orthogonal Transformation