Two Cyclic Groups of Same Order are Isomorphic

Cyclic groups of same order are isomorphic. So there is only one cyclic group of order n, up to isomorphism. In this article, we will prove that two cyclic groups of same order are isomorphic.

Proof

Let us consider two cyclic groups G and G’ of same order generated by a and b respectively. Thus, |G| = |G’| = o(a) = o(b). We need to show that G ≅ G’.

We will consider two cases.

G and G’ are Finite:

Let G and G’ have order n. Therefore, we have that

o(a)=n and G = {a⁰, a¹, a², …, a^n-1}

o(b)=n and G’ = {b⁰, b¹, b², …, b^n-1}.

Define a mapping φ: G → G’ by

φ(a^k) = b^k.

Claim that φ(a^k) = b^k holds for any integer k. To prove this, consider an integer m. By division algorithm, m = nq+r for some integers q and r with 0 ≤ r < n-1.

Then as both a and b have the same order n, so we deduce that a^m = a^r and b^m = b^r. Thus, φ(a^m) = b^m, As m is arbitrary, our claim follows.

φ is bijective: By definition, φ is bijective.

φ is a homomorphism:

Take x, y ∈ G. So x = au and y=av for some integers u and v. Now, φ(xy) = φ(au av) = φ(au+v) = bu+v ⇒ φ(xy) = bu bv ⇒ φ(xy) = φ(x) φ(y). This proves that φ is a homomorphism

So by the first isomorphism theorem, φ is an isomorphism, that is, G ≅ G’. This proves that any two finite cyclic group of same order n are isomorphic.

Main Topic: Cyclic Group: Definition, Examples, Properties, Orders

Order of a Group | Subgroups of Cyclic Groups

G and G’ are Infinite

Now, we consider that both G and G’ are infinite cyclic groups. Using the fact infinite cyclic groups are isomorphic to Z, we conclude that both G and G’ are isomorphic to (ℤ, +).

Therefore, we deduce that G ≅ G’. This proves that any two infinite cyclic groups are isomorphic.

Read Also: Prove that Groups of Prime Order are Cyclic

Infinite Cyclic Group is Isomorphic to ℤ [With Generators]

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