Suppose that gcd(a,n)=1, m|n, then order of a in m divides order of a in n.
Let $\displaystyle k$ be order of $\displaystyle a$ mod $\displaystyle m$ and $\displaystyle r$ be order of $\displaystyle a$ mod $\displaystyle n$. Then it means $\displaystyle a^k\equiv 1(\bmod m)$ and $\displaystyle a^r\equiv 1(\bmod n)$. But since $\displaystyle m|n$ it means $\displaystyle a^r\equiv 1(\bmod m)$. Since $\displaystyle k$ is the order mod $\displaystyle m$ it means $\displaystyle k|r$.