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: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…a^n-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=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)^-qà(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: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=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=aⁿ

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 aⁿ=e

k>n

By division algorithm ∃ integer q,r

Such that

K=nq+rà(i)

a^k=a^nq+r

a^k= (aⁿ)^q.a^r

a^k=(e)^q.a^r          aⁿ=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=(aⁿ)^q

a^k=(e)^q

a^k=e

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