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 $$\phi: G_1 \to G_2: \phi \left({a^n}\right) = b^n$$. We would now like 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 $$G_1 = \left \langle {a} \right \rangle$$, $$\exists s, t \in \Z: x = a^s, y = a^t$$.

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.