# 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 $H$.

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"volledig stelsetl van representanten" translates to "complete system/set of representatives". – Ittay Weiss Mar 14 '13 at 11:56

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.

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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!

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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 $ab^{-1}\in H \iff (b^{-1})^{-1} a^{-1}\in H$.

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