Let $p$ and $q$ be primes, let $k$ be a positive integer such that $q^k\mid(p-1)$, and let $a$ and $b$ be reduced residues modulo $p$. Suppose that both $a$ and $b$ have order $q^k$ modulo $p$. Prove that exactly $q-2$ of the numbers $ab,ab^2,\ldots,ab^{q-1}$ have order $q^k$ modulo $p$, while the other number has order $q^j$ for some $0\le j<k$.
I'm using the following lemmas:
Let $a$ have order $r\pmod m$ and $b$ have order $s\pmod m$. Let $t$ be the order of $ab \pmod m$. Then $t$ divides $\frac{rs}{(r,s)}$ and $t$ is a multiple of $\frac{rs}{(r,s)^2}$.
If $a$ has order $h$ modulo $m$, then $a^k$ has order $h/(h,k)$ modulo $m$.
By the second lemma $b^j$ has order $q^k / (q^k, j)$ for all $1 \leq j \leq q-1$, and since $(q^k,j) = 1$ we have that $b^j$ has order $q^k$. Thus, by the first lemma the order $t$ of $ab^j$ divides $(q^k \cdot q^k) / q^k = q^k$. How do I continue? Some hints would really help, thanks!