Showing $\mathrm{ord}_d(a) \mid \mathrm{ord}_m(a)$ if $d \mid m$. Also, let $1\le d$, $1 \le m$, and $\gcd(a,d)=1$.
What I have so far is:
Let $x=\mathrm{ord}_m(a)$.
Then we have $a^x\equiv 1 \pmod m\implies a^x=mk+1$ for some $k\in\mathbb{Z}$.
Let $m=m'd$. Then $a^x = mk +1 = d(m'k)+1$
I'm not sure where to go from here.
Thanks,
Brooke