Order of Subgroup of Cyclic Group

Theorem
Let $$G$$ be a cyclic group whose identity is $$e$$ and whose order is $$n$$.

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

Then $$\left|{H}\right| = \frac n {\gcd \left\{{n, i}\right\}}$$.

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

Let $$\left|{H}\right| = k$$. By 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 $$n \backslash i k$$.

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

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

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

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

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

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

Thus concludes the proof.