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.