I'm recently a bit indulged with basic number theory and investigating in the Carmichael $\lambda(n)$ function, which is defined as the smallest positive number that makes $a^x\equiv\ (\mathrm{mod}\ n)$ true for every $a$ co-prime to $n$.
On the other hand, the order of $a$ modulo $n$ (given $a$ is co-prime to $n$) is defined as the smallest number that makes $a^x\equiv1\ (\mathrm{mod}\ n)$ true, for the specific $a$. Denote order of $a$ modulo $n$ as $\mathrm{ord}_n(a)$, it's obvious that of $\mathrm{ord}_n(a)|\lambda(n)$.
My problem is, it seems that the order of all $a$ co-prime to $n$ seems to traverse all factors of $\lambda(n)$. That is to say, for every factor of $\lambda(n)$, there exists $a$ co-prime to $n$ whose order equals the factor. Or you can write it more formally: $$ \{\mathrm{ord}_n(a):\gcd(a,n)=1\}=\{d:d|\lambda(n)\}. $$ I concluded the conjecture from some numerical calculations based on python, but I have no idea why. How can I prove it?