every prime has a primitive root

every prime has a primitive root

Let

p

be a prime. Recall that an

integer

g

is said to be a

primitive root

for

p

(or more concretely for

(

ℤ

/

p

⁢

ℤ

)

×

) if the

multiplicative order

of

g

modulo

p

is

ϕ

⁢

(

p

)

=

p

-

1

. In other words,

g

is a

generator

of the

cyclic group

(

ℤ

/

p

⁢

ℤ

)

×

, i.e.

(

ℤ

/

p

⁢

ℤ

)

×

=

{

1

,

2

,

3

,

…

,

p

-

1

}

=

{

1

,

g

,

g

2

,

…

,

g

p

-

2

}

.

Before we go into the proof of the main

theorem

, we define the term “least

universal

exponent

” and prove a lemma.

Definition

.

Let

N

>

2

be a

natural number

. We say that

n

is the

least universal exponent

for

Z

/

N

⁢

Z

if

a

n

≡

1

mod

N

for all

a

∈

(

Z

/

N

⁢

Z

)

×

and

n

is the least

positive integer

with this property.

It follows from Euler’s theorem that the least universal exponent

n

of

ℤ

/

N

⁢

ℤ

is less or equal to the value of Euler’s phi function

ϕ

⁢

(

n

)

, and in fact, the least universal exponent should divide

ϕ

⁢

(

n

)

.

Lemma

.

Let

p

be a prime and let

n

be the least universal exponent of

Z

/

p

⁢

Z

. Then there is a number

a

∈

(

Z

/

p

⁢

Z

)

×

whose order is precisely

n

.

Proof.

In this proof we will use the properties of the multiplicative order of an integer. We shall prove that (a) the least universal exponent is the least common

multiple

of all the orders of all the elements of

(

ℤ

/

N

⁢

ℤ

)

×

, and (b) there is an

element

whose order is exactly the least common multiple. In order to show (a) and (b), it suffices to show that if

a

and

b

have orders

ord

⁡

(

a

)

=

h

and

ord

⁡

(

b

)

=

k

then there is an element

c

whose order is

ord

⁡

(

c

)

=

lcm

⁡

(

h

,

k

)

. This is clear: if

q

is a prime and

q

i

divides

h

=

ord

⁡

(

a

)

then, there is an element whose order is

p

i

(namely

a

h

/

q

i

has such order); and if

h

′

=

ord

⁡

(

c

)

and

k

′

=

ord

⁡

(

d

)

are relatively prime then

ord

⁡

(

c

⁢

d

)

=

h

′

⁢

k

′

. Finally, suppose that

lcm

⁡

(

h

,

k

)

=

q

1

i

1

⁢

⋯

⁢

q

t

i

t

, for some distinct primes

q

1

,

…

,

q

t

and positive integers

i

1

,

…

,

i

t

. Then, each

q

j

i

j

divides either

h

or

k

and there is an element

c

j

of order exactly

q

j

i

j

. Thus, the element

c

=

c

1

⁢

⋯

⁢

c

t

has exact order

lcm

⁡

(

h

,

k

)

.
∎

Theorem

.

Every prime

p

has a primitive root.

The following proof is due to Legendre.

Proof.

If

p

=

2

then

g

=

1

is a primitive root. Let us assume that

p

>

2

is prime and let

n

be the least universal exponent for

p

, i.e.

n

is the smallest positive integer such that

x

n

≡

1

mod

p

, for all non-zero

x

∈

ℤ

/

p

⁢

ℤ

. Notice that, in particular by the Lemma, there is some element

g

∈

ℤ

/

p

⁢

ℤ

such that

g

n

≡

1

but

g

m

≠

1

mod

p

for any

m

<

n

, i.e. the multiplicative order of

g

is precisely

n

. Also, notice that by Fermat’s little theorem,

n

≤

p

-

1

.

Now, the

polynomial

f

⁢

(

x

)

=

x

n

-

1

has at most

n

roots over the field

ℤ

/

p

⁢

ℤ

(see

http://planetmath.org/node/APolynomialOfDegreeNOverAFieldHasAtMostNRoots

this entry), and

f

⁢

(

x

)

≡

0

mod

p

for all non-zero

x

mod

p

. Thus

n

≥

p

-

1

. Hence,

n

=

p

-

1

and

g

is of exact order

p

-

1

, therefore

g

is a primitive root.
∎

Title

every prime has a primitive root

Canonical name

EveryPrimeHasAPrimitiveRoot

Date of creation

2013-03-22 16:20:41

Last modified on

2013-03-22 16:20:41

Owner

alozano (2414)

Last modified by

alozano (2414)

Numerical id

6

Author

alozano (2414)

Entry type

Theorem

Classification

msc 11-00

Related topic

APolynomialOfDegreeNOverAFieldHasAtMostNRoots

Related topic

ExistenceOfPrimitiveRootsForPowersOfAnOddPrime

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