$G$ is a group with subgroup $H$. For $a_i \in G$ there is a decomposition of the group into left cosets $G = a_1H \bigcup a_2H … \bigcup a_sH$
Show that $Ha_1^{-1} \bigcup Ha_2^{-1} … \bigcup Ha_s^{-1}$ is a decomposition of the group into right cosets.
I know that there are the same number of left and right cosets but I don't know how to proceed.
Best Answer
The subsets $H a_j^{-1}$ are certainly right cosets. The only thing to show is that they are disjoint since, as you observed, the number of left and right cosets are the same. If $H a_i^{-1} \cap H a_j^{-1} \neq \emptyset$, for $i \neq j$, then for some $h_1, h_2 \in H$ we must have $h_1 a_i^{-1} = h_2 a_j^{-1}$ which implies $a_i^{-1} = h_1^{-1} h_2 a_j^{-1}$. Then $a_i = a_j h_2^{-1} h_1$ and therefore $a_i \in a_j H$. But by assumption, $a_i H \cap a_j H = \emptyset$ which contradicts the conclusion that $a_i \in a_j H$. This completes the proof that the right cosets $H a_i^{-1}$ are pairwise disjoint.