The strategy is to construct an explicit map from $H$ to $gH$ and verify it is a bijection using left cancellation in the [group](/page/Group) $G$. **Step 1: Define the map.** Consider the [function](/page/Function): \begin{align*} \varphi : H &\to gH \\ h &\mapsto gh. \end{align*} **Step 2: Injectivity.** Suppose $\varphi(h_1) = \varphi(h_2)$, i.e., $gh_1 = gh_2$. Pre-multiplying both sides by $g^{-1}$ gives $h_1 = h_2$. **Step 3: Surjectivity.** Every element of $gH$ has the form $gh$ for some $h \in H$, and $\varphi(h) = gh$, so $\varphi$ is surjective. Since $\varphi$ is a bijection, $|H| = |gH|$ whenever $H$ is finite.