Order of Subgroup of Cyclic Group

Theorem
Let $C_n = \left \langle {g} \right \rangle$ be the cyclic group of order $n$ which is generated by $g$ whose identity is $e$.

Let $a \in C_n: a = g^i$, and let $H = \left \langle {a} \right \rangle$.

Then:
 * $\left|{H}\right| = \dfrac n {\gcd \left\{{n, i}\right\}}$

Proof
The fact that $H$ is cyclic follows from Subgroup of Cyclic Group is Cyclic.

We need to show that $H$ has $n / d$ elements.

Let $\left|{H}\right| = k$.

By Non-Trivial Group has Non-Trivial Cyclic Subgroup:
 * $\left|{H}\right| = \left|{a}\right| = k$

thus $a^k = e$.

Now $a = g^i$, so:

We now need to calculate the smallest $k$ such that $n \mathop \backslash i k$.

That is, the smallest $t \in \N$ such that $n t = i k$.

Let $d = \gcd \left\{{n, i}\right\}$.

Thus $t = \dfrac {k \left({i / d}\right)} {n / d}$

From Divide by GCD for Coprime Integers, $n / d$ and $i / d$ are coprime.

Thus from Euclid's Lemma, $\left({n / d}\right) \mathop \backslash k$.

As $a \mathop \backslash b \implies a \le b$, the smallest value of $k$ such that $k / \left({n / d}\right) \in \Z$ is $n / d$.

Hence the result.