Order of Powers

After proving the divisibility criterion for powers equal to the identity, we now determine the order of a power

a^m

. If

a

has finite order

n

, then the order of

a^m

is controlled by the greatest common divisor of

m

and

n

. This result is one of the most frequently used formulas in cyclic groups and finite group computations. It explains when a power of an element has the same order as the element and when the order becomes smaller. In this lesson, students will prove the formula

o(a^m)=\dfrac{n}{\operatorname{gcd}(m,n)}

and apply it in examples.

Let

(G,\circ)

be a group with identity element

e

and let

a\in G

. If

o(a)=n

, then for every positive integer

m

\begin{equation*} o(a^m)=\frac{n}{\operatorname{gcd}(m,n)}. \end{equation*}

After proving the divisibility criterion for powers equal to the identity, we now determine the order of a power

a^m

. If

a

has finite order

n

, then the order of

a^m

is controlled by the greatest common divisor of

m

and

n

o(a^m)=\dfrac{n}{\operatorname{gcd}(m,n)}

and apply it in examples.

Let

(G,\circ)

be a group with identity element

e

and let

a\in G

. If

o(a)=n

, then for every positive integer

m

\begin{equation*} o(a^m)=\frac{n}{\operatorname{gcd}(m,n)}. \end{equation*}

Given that

(G,\circ)

is a group with identity element

e

and

a\in G

. Given that

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

Let

m\in\mathbb{N}

. To prove that

\begin{equation*} o(a^m)=\frac{n}{\operatorname{gcd}(m,n)}. \end{equation*}

Let

\begin{equation*} d=\operatorname{gcd}(m,n). \end{equation*}

Then there exist relatively prime positive integers

r

and

s

such that

\begin{equation*} m=dr,\quad n=ds,\quad \operatorname{gcd}(r,s)=1. \end{equation*}

We first prove that

(a^m)^s=e

. Now

\begin{align*} (a^m)^s &=a^{ms}\\ &=a^{drs}\\ &=a^{rn}\\ &=(a^n)^r\\ &=e. \end{align*}

Therefore

a^m

has finite order and

\begin{equation*} o(a^m)\mid s. \end{equation*}

Let

\begin{equation*} o(a^m)=t. \end{equation*}

Then

\begin{equation*} (a^m)^t=e. \end{equation*}

Thus

\begin{align*} (a^m)^t=e &\implies a^{mt}=e. \end{align*}

Since

o(a)=n

, the divisibility criterion gives

\begin{equation*} n\mid mt. \end{equation*}

Using

m=dr

and

n=ds

, we get

\begin{align*} ds\mid drt &\implies s\mid rt. \end{align*}

Since

\operatorname{gcd}(r,s)=1

, Euclid's lemma gives

\begin{equation*} s\mid t. \end{equation*}

We already have

t\mid s

. Therefore

\begin{equation*} t=s. \end{equation*}

Since

\begin{equation*} s=\frac{n}{d}=\frac{n}{\operatorname{gcd}(m,n)}, \end{equation*}

we get

\begin{equation*} o(a^m)=\frac{n}{\operatorname{gcd}(m,n)}. \end{equation*}

Hence,

\begin{equation*} o(a^m)=\frac{n}{\operatorname{gcd}(m,n)}. \end{equation*}

\square

The formula shows two important cases. If

\operatorname{gcd}(m,n)=1

, then

o(a^m)=n

, so

a^m

has the same order as

a

. If

m

divides

n

, then

o(a^m)=\dfrac{n}{m}

. For example, if

o(a)=12

, then

o(a^5)=12

because

\operatorname{gcd}(5,12)=1

, while

o(a^4)=3

because

\operatorname{gcd}(4,12)=4

Let

(G,\circ)

be a group and let

a\in G

. If

o(a)=n

and

\operatorname{gcd}(m,n)=1

, then

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

Given that

(G,\circ)

is a group,

a\in G

o(a)=n

, and

\operatorname{gcd}(m,n)=1

. To prove that

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

Using the order of a power formula,

\begin{align*} o(a^m) &=\frac{n}{\operatorname{gcd}(m,n)}\\ &=\frac{n}{1}\\ &=n. \end{align*}

Hence,

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

\square

Let

(G,\circ)

be a group and let

a\in G

. If

o(a)

is infinite, then

o(a^m)

is infinite for every positive integer

m

Given that

(G,\circ)

is a group,

a\in G

, and

o(a)

is infinite. Let

m\in\mathbb{N}

. To prove that

o(a^m)

is infinite. If possible let

o(a^m)

be finite. Then there exists

k\in\mathbb{N}

such that

\begin{equation*} (a^m)^k=e. \end{equation*}

Therefore

\begin{align*} (a^m)^k=e &\implies a^{mk}=e. \end{align*}

Since

mk\in\mathbb{N}

, this contradicts that

a

has infinite order. A contradiction. Hence,

o(a^m)

is infinite for every positive integer

m

\square

Let

o(a)=12

. Then

\begin{align*} o(a^4) &=\frac{12}{\operatorname{gcd}(4,12)}\\ &=\frac{12}{4}\\ &=3. \end{align*}

Thus

a^4

has order

3

Let

o(a)=12

. Then

\begin{align*} o(a^5) &=\frac{12}{\operatorname{gcd}(5,12)}\\ &=\frac{12}{1}\\ &=12. \end{align*}

Thus

a^5

has the same order as

a

Let

(G,\circ)

be a group and let

a\in G

. If

o(a)=18

, find

o(a^6)

Let

(G,\circ)

be a group and let

a\in G

. Given that

\begin{equation*} o(a)=18. \end{equation*}

To find

o(a^6)

. Using the order of a power formula,

\begin{align*} o(a^6) &=\frac{18}{\operatorname{gcd}(6,18)}\\ &=\frac{18}{6}\\ &=3. \end{align*}

Hence,

\begin{equation*} o(a^6)=3. \end{equation*}

\square

Let

(G,\circ)

be a group and let

a\in G

. If

o(a)=20

, determine all positive integers

m

with

1\leq m<20

such that

o(a^m)=20

Let

(G,\circ)

be a group and let

a\in G

. Given that

\begin{equation*} o(a)=20. \end{equation*}

To determine all positive integers

m

with

1\leq m<20

such that

o(a^m)=20

. By the formula,

\begin{equation*} o(a^m)=\frac{20}{\operatorname{gcd}(m,20)}. \end{equation*}

Thus

o(a^m)=20

if and only if

\begin{equation*} \operatorname{gcd}(m,20)=1. \end{equation*}

The positive integers

m

with

1\leq m<20

and

\operatorname{gcd}(m,20)=1

are

\begin{equation*} 1,3,7,9,11,13,17,19. \end{equation*}

Hence, the required values are

\begin{equation*} 1,3,7,9,11,13,17,19. \end{equation*}

\square

Use the calculator below to compute the order of

a^m

from

o(a)=n

. Enter

n

and

m

. The calculator applies the formula

o(a^m)=n/\operatorname{gcd}(m,n)

[1] State the formula for

o(a^m)

when

o(a)=n

. [2] If

o(a)=15

, find

o(a^5)

. [3] If

o(a)=15

, find

o(a^4)

. [4] If

o(a)=24

, find all

m

with

1\leq m<24

such that

o(a^m)=24

. [5] Prove that if

a

has infinite order, then

a^m

has infinite order for every

m\in\mathbb{N}

[1]

o(a^m)=\dfrac{n}{\operatorname{gcd}(m,n)}

. [2]

o(a^5)=15/\operatorname{gcd}(5,15)=3

. [3]

o(a^4)=15/\operatorname{gcd}(4,15)=15

. [4] The values are

1,5,7,11,13,17,19,23

. [5] If

(a^m)^k=e

, then

a^{mk}=e

, contradicting that

a

has infinite order.

It is

n/\operatorname{gcd}(m,n)

When

\operatorname{gcd}(m,n)=1

Yes. For example, if

o(a)=12

, then

o(a^4)=3

Every positive power

a^m

also has infinite order.

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