Proof explanation for Order of Subgroups, by Fraleigh Theorem: $G$ a cyclic group, $H < G$, $H = \langle a^s \rangle$ then $|H|=\frac{G}{\operatorname{gcd}(s,n)}.$
Proof:
Let $|H|=| \langle a^s \rangle |=m$ for $m$ smallest such that $(a^s)^m=e$
So $(a^s)^m=e$ if and only if $n|ms$. 
How do we find $m$ smallest satisfying $n|ms$? Why does Fraleigh introduce Bezout's Identity, and how does the GCD help?

The proof in Fraleigh uses Bezout's identity to show that $n∣ms$ iff $\frac{n}{d}\mid m\frac{s}{d}$ where $d=\gcd(s,n)$, so that $m=\frac{n}{d}$ is the smallest such $m$. 
To see this another way, and perhaps see why the $\gcd$ is relevant, clearly we want the smallest $m$ such that $n\mid ms$. Now, $s$ contains some of the factors of $m$ (the product of those is $\gcd(n,s)$) and does not contain others. So to make $ms$ divisible by $n$, $m$ must be the product of those other factors. But this product is precisely $\frac{n}{\gcd(n,s)}$.
