order of groups Question: Suppose $\operatorname{ord}(g)=20$. Find elements $h,k\in G$ such that $\operatorname{ord}(h)=4$, $\operatorname{ord}(k)=5$, and $hk^{-1}=k^{-1}h=g$.
I can't seem to find anything in my notes on how to complete this question. Can someone help hint how to find the solution to this question please?

Hint
Let $h=g^a$ then we have
$$o(h)=\frac{20}{\gcd(20,a)}:=\alpha$$
in fact
$$h^\alpha=g^{\operatorname{lcm}(20,a)}=e\Rightarrow o(h)|\alpha$$
and 
$$h^m=g^{ma}=e\iff 20|ma\iff \alpha|m \frac{a}{\gcd(20,a)}$$
hence we see that $\alpha\le o(h)$ and then $\alpha=o(h)$.
Now choose $a$ such that $o(h)=4$ and do the same thing for $k$.
