Question

Prove that every finite cyclic group of order n is isomorphic to zn

Answer 1

A finite group is a group with a finite number of elements. ... Every cyclic group G of order n is isomorphic to Zn. Proof.

I hope it will help you

Answer 2

Answer:

Let (G,∘) be a finite group whose identity element is e.

Then (G,∘) is cyclic of order n if and only if (G,∘) is isomorphic with the additive group of integers modulo n (Zn,+n).

Proof

Let (G,∘) be a cyclic group of order n.

From List of Elements in Finite Cyclic Group:

G={a0,a1,a2,…,an}

where a0=e,a1=a

From the definition of integers modulo n, Zn can be expressed as:

Zn={[[0]]n,[[1]]n,…,[[n−1]]n}

where [[x]]n is the residue class of x modulo n.

Let ϕ:G→Zn be the mapping defined as:

∀k∈{0,1,…,n−1}:ϕ(ak)=[[k]]n

By its definition it is clear that ϕ is a bijection.

Also:

ϕ(ar∘as) = ϕ(ar+s) Powers of Group Elements: Sum of Indices

= [[r+s]]n Definition of ϕ

= [[r]]n+n[[s]]n Definition of Modulo Addition

= ϕ(ar)+nϕ(as) Definition of ϕ

Thus the morphism property of ϕ is demonstrated, and ϕ is thus a group homomorphism.

By definition, a group isomorphism is a group homomorphism which is also a bijection.

Now suppose G is a group such that ϕ:Zn→G is a group isomorphism.

Let a=ϕ([[1]]n).

Let g∈G.

Then g=ϕ([[k]]n) for some [[k]]n∈Zn.

Therefore:

g==ϕ([[k]]n=ϕ([[1]]n+n⋯(k)⋯+n[[1]]n)==ak

g==ϕ([[k]]n=ϕ([[1]]n+n⋯(k)⋯+n[[1]]n)==akSo every element of G is a power of a.

g==ϕ([[k]]n=ϕ([[1]]n+n⋯(k)⋯+n[[1]]n)==akSo every element of G is a power of a.So by definition G is cyclic.