Show that the set $\{r^{-1} : r\in R \}$ contains exactly one element out of each right coset of $H$. I'm dutch and I'm not sure if I translated this right. If there are some more dutchies here, how could I translate: "volledig stelsel van representanten"
Let $H$ be a subgroup of of $G$ and consider the left cosets of $H$. Pick from every distinct left coset one element, and put them in the set $R$. Show that the set $\{r^{-1} : r\in R \}$ contains exactly one element out of each right coset of $H$.
 A: Well, suppose there were $r,s\in R$ such that $r^{-1}$ and $s^{-1}$ are both in the same right coset of $H$.  What could you conclude?
I don't speak Dutch, but another way of phrasing this would be "Show that $\{r^{-1} \mid r\in R\}$ is a right transversal for $H$ in $G$".  What you have written is fine, though.
A: You need to show that every $r\in R$, the element $r^{-1}$ is an element of precisely one right coset of $H$. Since any element in a group is an element of some right coset, all you need to do is show that no two distinct elements elements $r^{-1}$ and $s^{-1}$, with $r,s\in R$, are in the same coset of $H$. So, assume to the contrary that $r^{-1},s^{-1}\in gH$ for some $g\in G$. Then that means that you can write $r^{-1}$ and $s^{-1}$ as a certain product. Compute inverses and you'll find that $r$ and $s$ belong to the same left coset, but that is impossible for distinct elements in $R$. 
Succes!
A: By definition, $a,b\in G$ are in the same right coset of $H$ iff $ab^{-1}\in H$, and they are in the same left coset iff $a^{-1}b\in H$.
Hint: Show that $ab^{-1}\in H \iff (b^{-1})^{-1} a^{-1}\in H$.
