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# Cyclic Group – Theorem of Cyclic Group

July 20, 2022
written by Azhar Ejaz

A cyclic group is defined as an A group G is said to be cyclic if every element of G is a power of one and the same element ‘a’ of G.

i.e G= {ak|k∈Z}

Such an element ‘a’ is called the generator of G.

## Finite Cyclic Group

If the order of ‘a’ is finite if the least positive integer n such that an=e than G is called finite cyclic Group of order n.

It is written as G=< a:an=e>

Read as G is a cyclic group of order n generator by ‘a’

If G is a finite cyclic group of order n.

Than a,a2,a3,a4…an-1,an =e are the distinct elements of G.

For example: let G={1,-1,i,-i} prove that G is cyclic group under multiplication.

Solution:

G={1,-1,i,-i}

Since

(i)1=i

(i)2=-1

(i)3=-i

(i)4=1

Each element of G is generated by i

G=[i]={1,-1,i,-i}

So G is a cyclic group.

## Theorem: Every cyclic group is abelian.

Proof:-let G be a cyclic group generated by ‘a’

i.e G=<a|an=e>

Every element of G are the power of ‘a’

Let x,y∈G than

∃ Integer k, m such that

x=ak and y=am

To prove G is abelian, we prove xy=yx

Now

Xy=ak.am

xy=ak+m

xy=am+k      k,m are integer k+m=m+k

xy=am.ak

xy=yx

∀ x,y∈G

G is abelian.

Hence proved

Every cyclic group is abelian.

## Theorem: Every subgroup of a cyclic group is cyclic.

Proof:-let  G be a cyclic group .let generator be ‘a’

Every element of G is the power of ‘a’

Let H be a subgroup of G

Every element of H is also the power of a suppose k is the smallest positive integer.

Such that ak ∈H

To prove H is cyclic. we show that every element of H is the power of ak. it means H is generated by ak.

Let am∈H here we show that

am=(ak)i    i∈Z

k<m than by division algorithm

m=qk+r

0≤r<k

am=aqk+r

am=aqk.ar

am=(ak)q.ar

ar=am.(ak)-qà(1)

ak ∈H,   (ak)q∈H     ∴H is a subgroup

((ak)q)-1∈H               ∴H is a subgroup

So,  am , ((ak)q)-1∈H

am . ((ak)q)-1∈H

by(1)

ar∈H

∴ 0≤r<k which is a contradiction to k is the smallest positive integer s.that ak∈H

It is possible only if r=0

Thus

am=akq+r

am=akq+0

am=akq

am=( ak)q    q∈Z

am∈H generator by a k

So H is a cyclic subgroup.

Hence proved:-

Every subgroup of a cyclic group is cyclic.

## Theorem: Let G be a cyclic group of order n. let d be a positive divisor of n, then there is a unique subgroup of G of order d.

Proof:- let G=<a:an=e>

Let d be positive divisor of n

There are three possibilities

• d=1
• d=n
• 1<d<n
• If  d=1 than subgroup of G is of order 1 which is {e}
• If d=n than subgroup of G is of order n which is G.

Since {e} and G are trivial subgroup which are unique

• If 1<d<n then

n/d=qà(1)

Where q is integer

by (1)  n=qd

∴ e=an

n=qd

e=aqd=(aq)d

let aq=b

e=(b)d

So H=<b:bd=e>

To prove H is unique

Suppose K is another subgroup of G of order d

Let K<ak:(ak)d=e>

akd=e=an

kd=n

k=n/d=q

Hence q=k

Thus H=K

## Statement: Let G be any group .let a∈G has order n, then for any integer k, ak=e if and only if k=qn where q is an integer.

Proof:-let G be a group, a∈G and order of a=n

Here we prove

ak=e if and only if k=qn (where q is integer)

Suppose ak=e

Order of a=n

n is the smallest positive integer such that an=e

k>n

By division algorithm ∃ integer q,r

Such that

K=nq+rà(i)

ak=anq+r

ak= (an)q.ar

ak=(e)q.ar          an=e

ak=e.ar

e= ar        ak=e

∴ 0≤r<n and ar=e

r=0

Put in (i)

k=nq+0

k=nq

Conversely:

Suppose k=nq

ak=anq

ak=(an)q

ak=(e)q

ak=e

### What is a cyclic group?

A cyclic group is a group that is generated by a single element. A monogenous group is a special case of a cyclic group where the element that generates the group also acts as its own inverse.

### Is every cyclic group Abelian?

All cyclic groups are Abelian.

### Is every subgroup of a cyclic group cyclic?

every subgroup of a cyclic group is cyclic

### What are the properties of the cyclic groups?

All cyclic groups are also Abelian groups.
If G is a cyclic group with generator g and order n,
then every subgroup of G is also cyclic.
if G is a finite cyclic group with order n, the order of every element in G must divide n.

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