Suppose that $\gcd(a,m)=1$ and that $a$ has order $t$ modulo $m$. Then $a^k$ has order $t$ modulo $m$ if and only if $\gcd(k,t)=1$.
My question is about a step in the direction $\Rightarrow$:
Suppose that $a$ and $a^k$ both have order $t$ modulo $ m$ and that $\gcd(k,t)=r$. Then $1 \equiv a^t \equiv (a^t)^{\frac k r} \equiv (a^k)^{\frac t r} \mod m$. $\underline{\text{This implies that } r=1}$. I don't know where this very last step comes from. Any ideas?
Thanks a lot!
If $a^k$ has order $t$ mod m, then $(a^k)^t\equiv 1 \mod m$ and $t$ is minimal with this property. So if $(a^k)^{t/r}\equiv 1\mod m$, then $r=1$. Done.
$\ \color{#c00}{d\mid k,t} \,\Rightarrow\, (a^{\large k})^{\Large \color{#c00}{\frac{t}d}}\!= (a^{\large t})^{\Large \color{#c00}{\frac{k}d}}\! = 1^{\Large \frac{k}d}\! = 1,\ $ so $\,a^{\large k}\,$ has order $\,n\le \dfrac{t}{\color{#0a0}d}\,\ $ [so $\, n = t\,\Rightarrow \color{#0a0}{d=1}]$
Remark $ $ Generally $\ {\rm ord}\,a = t\,\Rightarrow\, {\rm ord}(a^{\large k}) = t/\gcd(t,k).\ $ See here for proofs.
