Cyclic Groups of the Same Order are Isomorphic

Theorem

Any two Cyclic Groups of the same Order (Groups) are isomorphic. More specifically,

If $n \in Z^{+}$ and $⟨ x ⟩$ and $⟨ y ⟩$ are both cyclic groups of order $n$ then the map $φ : ⟨ x ⟩ \to ⟨ y ⟩$ where $x^{k} \mapsto y^{k}$ is well defined and is an Isomorphism.
If $⟨ x ⟩$ is an infinite cyclic group, the map $φ : Z \to ⟨ x ⟩$ where $k \mapsto x^{k}$ is well defined an is an isomorphism.

Proof

Suppose $⟨ x ⟩, ⟨ y ⟩$ are both cyclic groups of order $n$ . Let $φ : ⟨ x ⟩ \to ⟨ y ⟩$ be defined by:

φ (x^{k}) = y^{k}

We must first prove that $φ$ is well defined. That is:

x^{r} = x^{s} \Rightarrow φ (x^{r}) = φ (x^{s})

Since $x^{r - s} = 1$ then via The GCD of any two orders is the same order then $n | r - s$ . Write $r = t n + s$ so:

\begin{aligned} φ (x^{r}) & = φ (x^{t n + s}) \\ = y^{t n + s} \\ = (y^{n})^{t} y^{s} \\ = 1^{t} y^{s} \\ = y^{s} \\ = φ (x^{s}) \end{aligned}

thus $φ$ is well-defined. It is immediate from the laws of exponents that $φ (x^{a} x^{b}) = φ (x^{a}) φ (x^{b})$ since:

\begin{aligned} φ (x^{a} x^{b}) & = φ (x^{a + b}) \\ = y^{a + b} \\ = y^{a} y^{b} \\ = φ (x^{a}) φ (x^{b}) \end{aligned}

that is $φ$ is a Homomorphism (or Group Morphism). Since the element $y^{k} \in ⟨ y ⟩$ is the image of $x^{k}$ under $φ$ then this map is surjective.

Since both groups have the same finite order, any surjection from one to the other is a bijection, so $φ$ is specifically an isomorphism.

If $⟨ x ⟩$ is an infinite cyclic group, let $φ : Z \to ⟨ x ⟩$ be defined by $φ (k) = x^{k}$ . Note that this map is already well defined since there is no ambiguity in the representation of elements in the domain.

Using the Orders of Cyclic Subgroups, then $x^{a} \neq x^{b}$ for all distinct $a, b \in Z$ , then $φ$ is injective. By the definition of a Cyclic Group then $φ$ is surjective. As above, the laws of exponents ensure $φ$ is homomorphic, so then $φ$ is an isomorphism.

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