Let $k$ be order of $a$ mod $m$ and $r$ be order of $a$ mod $n$. Then it means $a^k\equiv 1(\bmod m)$ and $a^r\equiv 1(\bmod n)$. But since $m|n$ it means $a^r\equiv 1(\bmod m)$. Since $k$ is the order mod $m$ it means $k|r$.