Problem - 4180
Show that if integer $a$ has multiplicative order of $hk$ modulo $n$, then $a^h$ has order of $k$ modulo $n$.
By the given condition, we have $a^{hk}\equiv 1\pmod{n}$, but $a^m\not\equiv 1\pmod{n}$ for any integer $0 < m < hk$.
Because $\left(a^h\right)^k\equiv a^{hk}\equiv 1\pmod{n}$, we find the order of $a^h$ modulo $n$ is at most $k$. If there exists a positive integer $r$ less than $k$ such that $a^r\equiv 1\pmod{n}$, we shall have $a^{rk}\equiv 1\pmod{n}$ and $rk < hk$. This is impossible. Therefore the given claim must hold.