structure of (ℤ/n⁢ℤ)^× as an abelian group

structure of

(

ℤ

/

n

⁢

ℤ

)

×

as an abelian group

The

automorphism group

of the

cyclic group

C

n

≅

ℤ

/

n

⁢

ℤ

is

(

ℤ

/

n

⁢

ℤ

)

×

. This article determines the

structure

of

(

ℤ

/

n

⁢

ℤ

)

×

as an

abelian group

.

Theorem 1

.

Let

n

≥

2

be an

integer

whose

factorization

is

n

=

p

1

a

1

⁢

p

2

a

2

⁢

…

⁢

p

r

a

r

where the

p

i

are distinct primes. Then:

1.

(

ℤ

/

n

⁢

ℤ

)

×

≅

(

ℤ

/

p

1

a

1

⁢

ℤ

)

×

×

(

ℤ

/

p

2

a

2

⁢

ℤ

)

×

×

…

×

(

ℤ

/

p

r

a

r

⁢

ℤ

)

×

2.

(

ℤ

/

p

k

⁢

ℤ

)

×

is a cyclic group of order

p

k

-

1

⁢

(

p

-

1

)

for all

odd primes

p

.

3.

(

ℤ

/

2

k

⁢

ℤ

)

×

is the

direct product

of a cyclic group of order

2

and a cyclic group of order

2

k

-

2

for

k

≥

2

.

Corollary 2

.

Aut

⁡

(

C

n

)

≅

(

ℤ

/

n

⁢

ℤ

)

×

is cyclic if and only if

n

=

2

,

4

,

p

k

, or

2

⁢

p

k

for

p

an odd prime and

k

≥

0

an integer.

Proof.

(of

theorem

)

(1): This is a restatement of the

Chinese Remainder Theorem

.

(2): Note first that the result is clear for

k

=

1

, since then

(

ℤ

/

p

⁢

ℤ

)

×

is the

multiplicative group

of the

finite field

ℤ

/

p

⁢

ℤ

and thus is cyclic (any

finite subgroup

of the multiplicative group of a field is cyclic). Also,

|

(

ℤ

/

p

k

⁢

ℤ

)

×

|

=

ϕ

⁢

(

p

k

)

=

p

k

-

1

⁢

(

p

-

1

)

. Since

(

ℤ

/

p

k

⁢

ℤ

)

×

is

abelian

, it is the direct product of its

q

-

primary components

for each prime

q

∣

ϕ

(

p

k

)

; we will show that each of those

q

-primary components is cyclic. For

q

=

p

, it suffices to find an element of

(

ℤ

/

p

k

⁢

ℤ

)

×

of order

p

k

-

1

.

1

+

p

is such an

element

; see Lemma

3

below. For

q

≠

p

, consider the map

ℤ

/

p

k

⁢

ℤ

→

ℤ

/

p

⁢

ℤ

:

a

+

(

p

k

)

↦

a

+

(

p

)

i.e. the reduction-by-

p

map. This is a

ring homomorphism

; restricting it to

(

ℤ

/

p

k

⁢

ℤ

)

×

gives a

surjective

group homomorphism

π

:

(

ℤ

/

p

k

⁢

ℤ

)

×

→

(

ℤ

/

p

⁢

ℤ

)

×

. Since

|

(

ℤ

/

p

⁢

ℤ

)

×

|

=

p

-

1

, it follows that the kernel of

π

has order

p

k

-

1

. Thus for

q

≠

p

, the

q

-primary component of

(

ℤ

/

p

k

⁢

ℤ

)

×

must map isomorphically into

(

ℤ

/

p

⁢

ℤ

)

×

by order considerations. But

(

ℤ

/

p

⁢

ℤ

)

×

is cyclic, so the

q

-primary component is as well.

Thus each

q

-primary component of

(

ℤ

/

p

k

⁢

ℤ

)

×

is cyclic and thus

(

ℤ

/

p

k

⁢

ℤ

)

×

is also cyclic.

(3): The result is true for

k

=

2

, when

(

ℤ

/

2

2

⁢

ℤ

)

×

≅

V

4

, the Klein

4

-group. So assume

k

≥

3

.

5

has exact order

2

k

-

2

in

(

ℤ

/

2

k

⁢

ℤ

)

×

(see Lemma

4

below). Also by that lemma,

5

2

k

-

3

≠

-

1

is

(

ℤ

/

2

k

⁢

ℤ

)

×

, so that

5

2

k

-

3

and

-

1

are two distinct elements of order

2

. Thus

(

ℤ

/

2

k

⁢

ℤ

)

×

is not cyclic, but has a cyclic subgroup of order

2

k

-

2

; the result follows.
∎

Proof.

(of Corollary)

(

⇐

)

is clear, since

(

ℤ

/

C

2

⁢

ℤ

)

×

≅

{

1

}

(

ℤ

/

C

4

⁢

ℤ

)

×

≅

C

2

(

ℤ

/

C

p

k

⁢

ℤ

)

×

≅

C

p

k

-

1

⁢

(

p

-

1

)

(

ℤ

/

C

2

⁢

p

k

⁢

ℤ

)

×

≅

(

ℤ

/

C

2

⁢

ℤ

)

×

×

(

ℤ

/

C

p

k

⁢

ℤ

)

×

≅

C

p

k

-

1

⁢

(

p

-

1

)

(

⇒

)

: Assume

(

ℤ

/

n

⁢

ℤ

)

×

is cyclic. If

n

is a power of

2

, then by the theorem, it must be either

2

or

4

. Otherwise, if

n

has two distinct odd prime factors

p

,

q

, then

(

ℤ

/

n

⁢

ℤ

)

×

contains the direct product

(

ℤ

/

p

r

⁢

ℤ

)

×

×

(

ℤ

/

q

s

⁢

ℤ

)

×

. But the orders of these two

factor groups

are both even (they are

ϕ

⁢

(

p

r

)

=

p

r

-

1

⁢

(

p

-

1

)

and

ϕ

⁢

(

q

s

)

=

q

s

-

1

⁢

(

q

-

1

)

respectively), so their direct product is not cyclic. Thus

n

can have at most one odd prime as a factor, so that

n

=

2

m

⁢

p

k

for some integers

m

,

k

, and

(

ℤ

/

C

n

⁢

ℤ

)

×

=

(

ℤ

/

C

2

m

⁢

ℤ

)

×

×

(

ℤ

/

C

p

k

⁢

ℤ

)

×

But the order of

(

ℤ

/

C

p

k

⁢

ℤ

)

×

is even, so that (since the order of

(

ℤ

/

C

2

m

⁢

ℤ

)

×

is also even for

m

≥

2

) we must have

m

=

0

or

1

, so that

n

=

p

k

or

2

⁢

p

k

.

∎

The above proof used the following lemmas, which we now prove:

Lemma 3

.

Let

p

be an odd prime and

k

>

0

a

positive integer

. Then

1

+

p

has exact order

p

k

-

1

in the multiplicative group

(

Z

/

p

k

⁢

Z

)

×

.

Proof.

The result is obvious for

k

=

1

, so we assume

k

≥

2

. By the

binomial theorem

,

(

1

+

p

)

p

n

=

1

+

∑

i

=

1

p

n

(

p

n

i

)

⁢

p

i

Write

ord

p

⁡

(

m

)

for the largest power of a prime

p

dividing

m

. Then by a theorem on

divisibility of prime-power binomial coefficients

,

ord

p

⁡

(

(

p

n

i

)

⁢

p

i

)

=

n

+

i

-

ord

p

⁡

(

i

)

Now,

i

-

ord

p

⁡

(

i

)

is

1

if

i

=

1

, and is at least

2

for

i

>

1

(since

p

≥

3

). We thus get

(

1

+

p

)

p

n

=

1

+

p

n

+

1

+

r

⁢

p

n

+

2

,

r

∈

ℤ

Setting

n

=

k

-

1

gives

(

1

+

p

)

p

k

-

1

≡

1

(

p

k

)

; setting

n

=

k

-

2

gives

(

1

+

p

)

p

k

-

2

≡

1

+

p

k

-

1

≢

1

(

p

k

)

.
∎

Lemma 4

.

For

k

≥

3

,

5

has exact order

2

k

-

2

in the multiplicative group

(

Z

/

2

k

⁢

Z

)

×

(which has order

2

k

-

1

). Additionally,

5

2

k

-

3

≢

-

1

(

2

k

)

.

Proof.

The proof of this lemma is essentially identical to the proof of the preceding lemma.
Again by the binomial theorem,

5

2

n

=

(

1

+

2

2

)

2

n

=

∑

i

=

1

2

n

(

2

n

i

)

⁢

2

2

⁢

i

Then

ord

2

⁡

(

(

2

n

i

)

⁢

2

i

)

=

n

+

2

⁢

i

-

ord

2

⁡

(

i

)

Now,

2

⁢

i

-

ord

2

⁡

(

i

)

is

2

if

i

=

1

, and is at least

3

for

i

>

1

. We thus get

5

2

n

=

1

+

2

n

+

2

+

r

⁢

2

n

+

3

,

r

∈

ℤ

Setting

n

=

k

-

2

gives

5

2

k

-

2

≡

1

(

2

k

)

; setting

n

=

k

-

3

gives

5

2

k

-

3

≡

1

+

2

k

-

1

≢

±

1

(

2

k

)

. (Note that

1

+

2

k

-

1

≢

-

1

since

k

≥

3

).
∎

References

1

Dummit, D., Foote, R.M.,

Abstract Algebra

, Third Edition

, Wiley, 2004.

Title

structure of

(

ℤ

/

n

⁢

ℤ

)

×

as an abelian group

Canonical name

StructureOfmathbbZnmathbbZtimesAsAnAbelianGroup

Date of creation

2013-03-22 18:42:40

Last modified on

2013-03-22 18:42:40

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

4

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 20A05

Classification

msc 20E36

Classification

msc 20E34

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