Groups of Prime Order

We have seen that the order of every element divides the order of a finite group. When the order of the group is prime, this leaves very few possibilities. A non-identity element cannot have order

1

, so its order must be the prime number itself. This forces the whole group to be generated by that one element. Thus every group of prime order is cyclic. This result is one of the first examples where Lagrange's theorem describes the whole structure of a group.

Let

(G,\circ)

be a finite group. The group

G

is called a

\textbf{group of prime order}

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

for some prime number

p

We have seen that the order of every element divides the order of a finite group. When the order of the group is prime, this leaves very few possibilities. A non-identity element cannot have order

1

Let

(G,\circ)

be a finite group. The group

G

is called a

\textbf{group of prime order}

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

for some prime number

p

A group of prime order has only two possible subgroup sizes:

1

and the prime number

p

. In

\mathbb{Z}_p

under addition, every nonzero element repeatedly added to itself reaches every residue class. Choose a prime and a nonzero element to see the entire group appear in one cycle. The identity element is the only element that fails to generate the group.

Prime order leaves no room for a proper nontrivial cyclic subgroup. Once a non-identity element generates more than one element, its generated subgroup must have all

p

elements.

Let

(G,\circ)

be a finite group. If

|G|

is prime, then

G

is cyclic.

Given that

(G,\circ)

is a finite group and

|G|

is prime. To prove that

G

is cyclic. Let

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

where

p

is a prime number. Since

p>1

, the group

G

contains at least one element different from the identity element. Let

a\in G

such that

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

Then

\langle a\rangle

is a subgroup of

G

. By Lagrange's theorem,

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

Therefore

\begin{equation*} |\langle a\rangle|\mid p. \end{equation*}

Since

p

is prime, the only positive divisors of

p

are

1

and

p

. Therefore

\begin{equation*} |\langle a\rangle|=1 \end{equation*}

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

Since

a\neq e

, the subgroup

\langle a\rangle

contains at least

e

and

a

. Therefore

\begin{equation*} |\langle a\rangle|\neq 1. \end{equation*}

Thus

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

Since

\langle a\rangle\subseteq G

and both sets have

p

elements, we get

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

Hence,

G

is cyclic.

\square

Let

(G,\circ)

be a finite group. If

|G|

is prime, then every non-identity element of

G

is a generator of

G

Given that

(G,\circ)

is a finite group and

|G|

is prime. To prove that every non-identity element of

G

is a generator of

G

. Let

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

where

p

is prime, and let

a\in G

such that

a\neq e

. By Lagrange's theorem,

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

Since

p

is prime, we get

\begin{equation*} o(a)=1 \end{equation*}

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

Since

a\neq e

, we have

o(a)\neq 1

. Therefore

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

Thus

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

Since

\langle a\rangle\subseteq G

, we get

\langle a\rangle=G

. Hence, every non-identity element of

G

is a generator of

G

\square

This calculator checks how primality restricts element orders. Enter a proposed group order and choose whether the element is the identity or non-identity. When the group order is prime, the only possible element orders are

1

and

p

. For a non-identity element, order

1

is impossible, so the element must generate the whole group.

Let

(G,\circ)

be a finite group. If

|G|

is prime, then

G

is abelian.

Given that

(G,\circ)

is a finite group and

|G|

is prime. To prove that

G

is abelian. By the theorem on groups of prime order,

G

is cyclic. Every cyclic group is abelian. Therefore

G

is abelian. Hence,

G

is abelian.

\square

Let

G=\mathbb{Z}_7

under addition modulo

7

. Then

\begin{equation*} |G|=7. \end{equation*}

Since

7

is prime,

G

is cyclic. In fact, every nonzero element of

\mathbb{Z}_7

generates the group. For example,

\begin{equation*} \langle 3\rangle=\{0,3,6,2,5,1,4\}=\mathbb{Z}_7. \end{equation*}

Let

(G,\circ)

be a group with

|G|=11

. Prove that

G

is cyclic.

Let

(G,\circ)

be a group with

|G|=11

. Since

11

is prime, every group of order

11

is cyclic by the theorem on groups of prime order. Therefore

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

for every non-identity element

a\in G

. Hence,

G

is cyclic.

Let

(G,\circ)

be a group with

|G|=13

. If

a\in G

and

a\neq e

, find

o(a)

Let

(G,\circ)

be a group with

|G|=13

, and let

a\in G

with

a\neq e

. By Lagrange's theorem,

\begin{equation*} o(a)\mid 13. \end{equation*}

Since

13

is prime,

o(a)=1

o(a)=13

. Since

a\neq e

, we get

\begin{equation*} o(a)\neq 1. \end{equation*}

Therefore

\begin{equation*} o(a)=13. \end{equation*}

1. Prove that every group of order

5

is cyclic. 2. Let

G

be a group of order

17

. If

a\in G

and

a\neq e

, find

o(a)

. 3. Let

G

be a group of order

19

. Prove that

G

is abelian.

1. Since

5

is prime, every group of order

5

is cyclic by the theorem on groups of prime order. 2. Since

17

is prime and

a\neq e

, we get

\begin{equation*} o(a)=17. \end{equation*}

3. Since

19

is prime,

G

is cyclic. Every cyclic group is abelian. Therefore

G

is abelian.

A non-identity element generates a subgroup whose order divides the prime group order, so the generated subgroup must be the whole group.

Every non-identity element generates it. The identity element generates only the trivial subgroup.

Yes. They are cyclic, and every cyclic group is abelian.

BMLabs Mathematics

BMLabs Mathematics

Groups of Prime Order

Groups of Prime Order

OVERVIEW

GROUP OF PRIME ORDER

PRIME ORDER GENERATOR EXPLORER GUIDE

PRIME ORDER GENERATOR EXPLORER

PRIME ORDER GROUP CALCULATOR GUIDE

PRIME ORDER GROUP CALCULATOR

A GROUP OF ORDER SEVEN

A GROUP OF ORDER ELEVEN

SOLUTION

GENERATORS IN A GROUP OF PRIME ORDER

SOLUTION

EXERCISES

ANSWERS

Coordinates and Locus

Invariants under Orthogonal Transformation