Given an integer a and a positive integer n with gcd(a,n) = 1, the multiplicative order of a modulo n is the smallest positive integer k with $a^k \equiv 1 (mod\ n)$
There exist a lemma to this saying that
$ [a]^r$ for $0 <= r <= k-1$ has order k if and only if gcd$(r,k)=1$
I want to prove this by contradiction but i might not be doing the setup correctly.
$ [a]^r$ for $0 <= r <= k-1$ has a order less than k if and only if gcd$(r,k) !=1$
Note $\ (a^{\large r})^{\large x}\!=1 \!\iff\! k\mid rx\!\iff\! k\mid rx,kx\!\iff\! k\mid (rx,kx)\!=\! (r,k)x\!\iff\! k/(r,k)\mid x\ $
The first $\iff$follows by $\,k = {\rm\ ord}\, a,\,$ and the third by the $\rm gcd$ universal and distributive law.
Since $\,\ell = {\rm ord}\,a^{\large r}\,$ is the least such $\,x,\,$ clearly $\ \ell = k/(r,k),\ $ so $\ \ell = k\iff (r,k)= 1$
Remark $\ $ Alternatively, dually, instead of $\rm gcd$ we can use $\rm lcm$ and its universal property, i.e.
$\quad\ \ (a^{\large r})^{\large x} = 1\iff k\mid rx\iff r,k\mid rx\iff {\rm lcm}(r,k)\mid rx\iff {\rm lcm}(r,k)/r\mid x$
They agree since $\ {\rm lcm}(r,k)/r = k/\gcd(r,k)\,$ by $\ {\rm lcm}(r,k)\gcd(r,k) = rk,\,$ by here.
