Given a subgroup of a group, the numbers of left and right cosets are equal

Let G be a group and H \leq G. Prove that the map x \mapsto x^{-1} sends each left coset to a right coset (of H), and hence |H \backslash G| = |G/H|.

(This is a slightly different approach.)

Define \psi : G \rightarrow H \backslash G by \psi(x) = Hx^{-1}. Suppose y^{-1}x \in H; then y^{-1} \in Hx^{-1}. and we have Hy^{-1} = Hx^{-1}, so \psi(x) = \psi(y). By a lemma to a previous exercise, this induces a mapping \varphi : G/H \rightarrow H \backslash G given by \varphi(xH) = Hx^{-1}. \psi is clearly surjective, so \varphi is surjective. Now suppose \varphi(x) = \varphi(y); then Hx^{-1} = Hy^{-1}, so that in particular y^{-1} \in Hx^{-1} and hence y^{-1}x \in H. Thus xH = yH. Hence \varphi is a bijection.

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