Prime Order Cyclic Groups

After proving the greatest common divisor test for generators, the prime-order case becomes especially simple. If a cyclic group has prime order

p

, then every non-identity element is a generator. This happens because every exponent

1,2,\ldots,p-1

is relatively prime to

p

. Prime-order cyclic groups are important because they have the simplest possible nontrivial generator structure. In this lesson, students will prove the prime-order generator theorem and apply it to concrete finite cyclic groups.

Let

(G,\circ)

be a cyclic group and let

p

be a prime number. If

\begin{equation*} |G|=p, \end{equation*}

then every non-identity element of

G

is a generator of

G

After proving the greatest common divisor test for generators, the prime-order case becomes especially simple. If a cyclic group has prime order

p

, then every non-identity element is a generator. This happens because every exponent

1,2,\ldots,p-1

is relatively prime to

p

Let

(G,\circ)

be a cyclic group and let

p

be a prime number. If

\begin{equation*} |G|=p, \end{equation*}

then every non-identity element of

G

is a generator of

G

Given that

(G,\circ)

is a cyclic group,

p

is a prime number, and

|G|=p

. To prove that every non-identity element of

G

is a generator of

G

. Since

G

is cyclic, there exists

g\in G

such that

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

Let

x\in G

such that

\begin{equation*} x\neq e. \end{equation*}

Since

G=\langle g\rangle

, there exists

r\in\mathbb{Z}

such that

\begin{equation*} x=g^r. \end{equation*}

Because

x\neq e

, the exponent

r

is not divisible by

p

. Therefore

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

By the generator criterion for finite cyclic groups,

\begin{equation*} G=\langle g^r\rangle. \end{equation*}

Since

x=g^r

, we get

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

Hence, every non-identity element of

G

is a generator of

G

\square

This proof avoids an unnecessary detour. Since the group is already cyclic, every element is some power of a fixed generator. A non-identity element cannot correspond to an exponent divisible by

p

, because such a power would be the identity. Prime-ness then forces the exponent to be relatively prime to

p

, so the element is a generator.

Let

G=\langle a\rangle

and

|G|=5

. Then

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

Since

5

is prime, each of

\begin{equation*} a,\;a^2,\;a^3,\;a^4 \end{equation*}

is a generator of

G

. The only element that is not a generator is

e

In the additive group

(\mathbb{Z}_7,+_7)

, every nonzero residue class is a generator:

\begin{equation*} \overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6}. \end{equation*}

For instance,

\begin{equation*} \langle \overline{3}\rangle=\mathbb{Z}_7 \end{equation*}

because

\operatorname{gcd}(3,7)=1

Let

(G,\circ)

be a cyclic group and let

p

be a prime number. If

|G|=p

, then

G

has exactly

p-1

generators.

Given that

(G,\circ)

is a cyclic group,

p

is a prime number, and

|G|=p

. To prove that

G

has exactly

p-1

generators. By the previous theorem, every non-identity element of

G

is a generator. The group

G

has

p

elements, and exactly one of them is the identity element. Therefore the number of non-identity elements is

\begin{equation*} p-1. \end{equation*}

Hence,

G

has exactly

p-1

generators.

\square

Let

G

be a cyclic group of order

11

. How many generators does

G

have?

Let

G

be a cyclic group of order

11

. To find the number of generators of

G

. Since

11

is prime, every non-identity element of

G

is a generator. The group has

11

elements and one identity element. Therefore the number of generators is

\begin{align*} 11-1&=10. \end{align*}

Hence,

G

has

10

generators.

\square

Determine all generators of

(\mathbb{Z}_{13},+_{13})

Let

(\mathbb{Z}_{13},+_{13})

be the additive group modulo

13

. To determine all generators. Since

13

is prime, every nonzero residue class is a generator. Therefore all generators are

\begin{equation*} \overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10},\overline{11},\overline{12}. \end{equation*}

Hence, these are all generators of

\mathbb{Z}_{13}

\square

The theorem assumes the group is cyclic. Later, after Lagrange’s theorem is available, one proves the stronger result that every group of prime order is cyclic. At this stage, we use only the cyclic hypothesis and the generator criterion already proved.

Use the calculator below to list all generators of

\mathbb{Z}_p

when

p

is prime. Enter a prime number

p

. The calculator returns all nonzero residue classes, because each of them generates the additive group modulo

p

[1] State the prime-order generator theorem for cyclic groups. [2] How many generators does a cyclic group of order

17

have? [3] List all generators of

(\mathbb{Z}_5,+_5)

. [4] Why is the identity element not a generator when

|G|>1

? [5] If

G=\langle a\rangle

and

|G|=7

, determine whether

a^4

is a generator.

[1] If

G

is cyclic of prime order

p

, then every non-identity element of

G

is a generator. [2] It has

16

generators. [3] The generators are

\overline{1},\overline{2},\overline{3},\overline{4}

. [4] The identity generates only

\{e\}

, not the whole group when

|G|>1

. [5] Yes, because

\operatorname{gcd}(4,7)=1

Every non-identity element is a generator.

It has

p-1

generators.

No. The identity generates only the trivial subgroup.

Every integer from

1

p-1

is relatively prime to

p

BMLabs Mathematics

BMLabs Mathematics

Prime Order Cyclic Groups

Prime Order Cyclic Groups

PRIME CYCLIC GENERATORS

EVEN CYCLIC GROUPS

Coordinates and Locus

Invariants under Orthogonal Transformation