Prime Order Elements

After proving the formula for the order of a power, we now study the special case of prime order. Prime order elements are especially simple because a positive integer is either divisible by the prime or relatively prime to it. This gives a sharp dichotomy for powers: a power is either the identity, or it has the same prime order as the original element. This result is important later when studying cyclic groups of prime order and generators. In this lesson, students will prove the main prime-order consequences and solve standard examples.

Let

(G,\circ)

be a group with identity element

e

, let

a\in G

, and let

p

be a prime number. If

o(a)=p

, then

\begin{equation*} o(a^k)=p \quad \forall\, k\in\mathbb{N},\; 1\leq k<p. \end{equation*}

Let

(G,\circ)

be a group with identity element

e

, let

a\in G

, and let

p

be a prime number. If

o(a)=p

, then

\begin{equation*} o(a^k)=p \quad \forall\, k\in\mathbb{N},\; 1\leq k<p. \end{equation*}

Given that

(G,\circ)

is a group with identity element

e

a\in G

, and

p

is a prime number. Given that

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

To prove that

\begin{equation*} o(a^k)=p \quad \forall\, k\in\mathbb{N},\; 1\leq k<p. \end{equation*}

Let

k\in\mathbb{N}

such that

\begin{equation*} 1\leq k<p. \end{equation*}

Since

p

is prime and

1\leq k<p

, we get

\begin{equation*} \operatorname{gcd}(k,p)=1. \end{equation*}

Using the order of a power formula,

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

Hence,

\begin{equation*} o(a^k)=p \quad \forall\, k\in\mathbb{N},\; 1\leq k<p. \end{equation*}

\square

The theorem says that if

a

has prime order

p

, then all nontrivial powers

a,a^2,\ldots,a^{p-1}

also have order

p

. The identity appears at

a^p

. This is one of the reasons prime order groups are structurally simple. There is no smaller positive divisor of

p

available to create a smaller order for a non-identity power.

Let

(G,\circ)

be a group with identity element

e

, let

a\in G

, and let

p

be a prime number. If

o(a)=p

, then for every

k\in\mathbb{N}

, either

\begin{equation*} a^k=e \end{equation*}

\begin{equation*} o(a^k)=p. \end{equation*}

Given that

(G,\circ)

is a group with identity element

e

a\in G

, and

p

is a prime number. Given that

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

To prove that for every

k\in\mathbb{N}

, either

a^k=e

o(a^k)=p

. Let

k\in\mathbb{N}

. Since

p

is prime, either

\begin{equation*} p\mid k \end{equation*}

\begin{equation*} \operatorname{gcd}(k,p)=1. \end{equation*}

[1] Let

p\mid k

. Then there exists

q\in\mathbb{N}

such that

\begin{equation*} k=pq. \end{equation*}

Now

\begin{align*} a^k &=a^{pq}\\ &=(a^p)^q\\ &=e^q\\ &=e. \end{align*}

Therefore

a^k=e

. [2] Let

\operatorname{gcd}(k,p)=1

. Using the order of a power formula,

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

Therefore

o(a^k)=p

. Hence, for every

k\in\mathbb{N}

, either

a^k=e

o(a^k)=p

\square

Let

o(a)=5

. Then

\begin{equation*} o(a)=o(a^2)=o(a^3)=o(a^4)=5. \end{equation*}

Also,

\begin{equation*} a^5=e,\quad a^{10}=e,\quad a^{15}=e. \end{equation*}

Thus every positive power of

a

is either the identity or has order

5

In the group

\{1,\omega,\omega^2\}

under complex multiplication, suppose

\omega^3=1

and

\omega\neq1

. Then

\begin{equation*} o(\omega)=3. \end{equation*}

Since

3

is prime, the non-identity powers

\omega

and

\omega^2

both have order

3

. Indeed,

\begin{equation*} (\omega^2)^3=\omega^6=(\omega^3)^2=1 \end{equation*}

and

\omega^2\neq1

Let

(G,\circ)

be a group and let

a\in G

. If

o(a)=7

, find

o(a^3)

Let

(G,\circ)

be a group and let

a\in G

. Given that

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

To find

o(a^3)

. Since

7

is prime and

1\leq3<7

, the prime-order theorem gives

\begin{equation*} o(a^3)=7. \end{equation*}

Hence,

o(a^3)=7

\square

Let

(G,\circ)

be a group and let

a\in G

. If

o(a)=7

, determine whether

a^{21}=e

Let

(G,\circ)

be a group and let

a\in G

. Given that

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

To determine whether

a^{21}=e

. By the divisibility criterion,

\begin{equation*} a^{21}=e \iff 7\mid21. \end{equation*}

Since

\begin{equation*} 21=7\cdot3, \end{equation*}

we get

\begin{equation*} 7\mid21. \end{equation*}

Therefore

\begin{equation*} a^{21}=e. \end{equation*}

Hence,

a^{21}=e

\square

Let

(G,\circ)

be a group and let

a\in G

. If

o(a)=p

, where

p

is prime, prove that

a^k=e

if and only if

p\mid k

Let

(G,\circ)

be a group and let

a\in G

. Given that

o(a)=p

, where

p

is prime. To prove that

\begin{equation*} a^k=e \iff p\mid k. \end{equation*}

This follows directly from the divisibility criterion for powers equal to the identity. Since

o(a)=p

, for every

k\in\mathbb{N}

\begin{equation*} a^k=e \iff p\mid k. \end{equation*}

Hence,

\begin{equation*} a^k=e \iff p\mid k. \end{equation*}

\square

Prime order does not mean every power is different. The powers repeat after

p

steps:

\begin{equation*} a^{p+1}=a,\quad a^{p+2}=a^2. \end{equation*}

Prime order means that before reaching

a^p=e

, no positive smaller exponent gives the identity. Therefore

a,a^2,\ldots,a^{p-1}

are non-identity elements, and each has order

p

Use the calculator below for the prime-order case. Enter a prime

p

and a positive exponent

k

. The calculator decides whether

a^k=e

o(a^k)=p

under the assumption

o(a)=p

[1] If

o(a)=11

, find

o(a^4)

. [2] If

o(a)=11

, determine whether

a^{33}=e

. [3] If

o(a)=11

, determine whether

a^{35}=e

. [4] Prove that if

o(a)=p

and

1\leq k<p

, then

a^k\neq e

. [5] If

o(a)=p

and

p\nmid k

, find

o(a^k)

[1]

o(a^4)=11

. [2] Yes, because

11\mid33

. [3] No, because

11\nmid35

. [4] If

a^k=e

, then

p\mid k

, which is impossible for

1\leq k<p

. [5]

o(a^k)=p

The power

a^k

is either the identity, or it has order

p

Exactly when

p\mid k

Because

\operatorname{gcd}(k,p)

is either

1

p

BMLabs Mathematics

BMLabs Mathematics

Prime Order Elements

Prime Order Elements

PRIME ORDER TEST

PRODUCTS AND CONJUGATES

Coordinates and Locus

Invariants under Orthogonal Transformation