Cyclic Groups of Same Order are Isomorphic

Theorem
Two cyclic groups of the same order are isomorphic to each other.

Proof
Let $G_1$ and $G_2$ be cyclic groups, both of finite order $k$.

Let $G_1 = \left \langle {a} \right \rangle, G_2 = \left \langle {b} \right \rangle$.

Then, by the definition of a cyclic group:
 * $\left|{a}\right| = \left|{b}\right| = k$

Also, by definition:
 * $G_1 = \left\{{a^0, a^1, \ldots, a^{k-1}}\right\}$

and:
 * $G_2 = \left\{{b^0, b^1, \ldots, b^{k-1}}\right\}$

Let us set up the obvious bijection:
 * $\phi: G_1 \to G_2: \phi \left({a^n}\right) = b^n$

The next task is to show that $\phi$ is an isomorphism.

Note that $\phi \left({a^n}\right) = b^n$ holds for all $n \in \Z$, not just where $0 \le n < k$, as follows:

Let $n \in \Z: n = q k + r, 0 \le r < k$, by the Division Theorem.

Then, by Element to the Power of Remainder:
 * $a^n = a^r, b^n = b^r$

Thus:
 * $\phi \left({a^n}\right) = \phi \left({a^r}\right) = b^r = b^n$

Now let $x, y \in G_1$.

Since $G_1 = \left \langle {a} \right \rangle$, it follows that:
 * $\exists s, t \in \Z: x = a^s, y = a^t$

Thus:

So $\phi$ is a homomorphism.

As $\phi$ is bijective, $\phi$ is an isomorphism from $G_1$ to $G_2$.

Thus $G_1 \cong G_2$ and the result is proved.