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.
Table of Contents
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, …, an-1}
o(b)=n and G’ = {b0, b1, b2, …, bn-1}.
Define a mapping φ: G → G’ by
φ(ak) = bk.
Claim that φ(ak) = bk 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 am = ar and bm = br. Thus, φ(am) = bm, 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]
FAQs
Answer: Both Q and R have the different cardinality. As two isomorphic cyclic groups must have the same cardinality, it follows that the additive groups (Q, +) and (R, +) are not isomorphic.
