Originally Posted by

**ThePerfectHacker** Given positive integer $\displaystyle a$:

$\displaystyle \left< [a] \right> = \left< [d] \right> $ where $\displaystyle d=\gcd(a,n)$.

Now since $\displaystyle \gcd(b,n) = d$ it means $\displaystyle \left< [b] \right> = \left< [d] \right>$.

Thus, $\displaystyle \left< [a] \right> = \left< [b] \right>$.

So that means we need to prove $\displaystyle \left< [a] \right> = \left< [d] \right>$ when $\displaystyle \gcd(a,n) = d$. Note $\displaystyle [a]\in \left< [d] \right> $ because $\displaystyle d|a$, this means $\displaystyle \left< [a] \right> \subseteq \left< [d] \right>$. Note there exist $\displaystyle x,y\in \mathbb{Z}$ so that $\displaystyle ax+ny=d$. Thus, $\displaystyle [d] = [ax+ny] = [ax]$, so $\displaystyle [d] \in \left< [a] \right>$ which means $\displaystyle \left< [d] \right> = \left< [a] \right>$.