# Cyclic Group – Theorem of Cyclic Group

*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= {a^{k}|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:a^{n}=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,a^{2},a^{3},a^{4}…a^{n-1},a^{n} =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|a^{n}=e>

Every element of G are the power of ‘a’

Let x,y∈G than

∃ Integer k, m such that

x=a^{k} and y=a^{m}

To prove G is abelian, we prove xy=yx

Now

Xy=a^{k}.a^{m}

xy=a^{k+m}

xy=a^{m+k} k,m are integer k+m=m+k

xy=a^{m}.a^{k }

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 a^{k }∈H

To prove H is cyclic. we show that every element of H is the power of ak. it means H is generated by a^{k}.

Let a^{m}∈H here we show that

a^{m}=(a^{k})^{i} i∈Z

k<m than by division algorithm

m=qk+r

0≤r<k

a^{m}=a^{qk+r}

a^{m}=a^{qk}.a^{r}

a^{m}=(a^{k})^{q}.a^{r}

a^{r}=a^{m}.(a^{k})^{-q}à(1)

a^{k }∈H, (a^{k})^{q}∈H ∴H is a subgroup

((a^{k})^{q})^{-1}∈H ∴H is a subgroup

So, a^{m }, ((a^{k})^{q})^{-1}∈H

a^{m }. ((a^{k})^{q})^{-1}∈H

by(1)

a^{r}∈H

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

It is possible only if r=0

Thus

a^{m}=a^{kq+r}

^{ }a^{m}=a^{kq+0}

^{ }a^{m}=a^{kq}

^{ }a^{m}=( a^{k})^{q} q∈Z

^{ }a^{m}∈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:a^{n}=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=a^{n}

n=qd

e=a^{qd}=(a^{q})^{d}

let a^{q}=b

e=(b)^{d}

So H=<b:b^{d}=e>

To prove H is unique

Suppose K is another subgroup of G of order d

Let K<a^{k}:(a^{k})^{d}=e>

a^{kd}=e=a^{n}

kd=n

k=n/d=q

Hence q=k

Thus H=K

## Theorem:-

## Statement: Let G be any group .let a∈G has order n, then for any integer k, a^{k}=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

a^{k}=e if and only if k=qn (where q is integer)

Suppose a^{k}=e

Order of a=n

n is the smallest positive integer such that a^{n}=e

k>n

By division algorithm ∃ integer q,r

Such that

K=nq+rà(i)

a^{k}=a^{nq+r}

a^{k}= (a^{n})^{q}.a^{r}

a^{k}=(e)^{q}.a^{r} a^{n}=e

a^{k}=e.a^{r}

e= a^{r } a^{k}=e

∴ 0≤r<n and a^{r}=e

Which is contradiction

r=0

Put in (i)

k=nq+0

k=nq

Conversely:

Suppose k=nq

a^{k}=a^{nq}

a^{k}=(a^{n})^{q}

a^{k}=(e)^{q}

a^{k}=e

## Frequently Asked Question-FAQs

### 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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