The strategy is to combine two earlier results: the cosets of $H$ partition $G$ (so $G$ is a disjoint union of cosets), and each coset has the same size as $H$ (so the total count is $|H|$ times the number of cosets). **Step 1: Partition $G$ into cosets.** By the [Cosets Partition the Group](/theorems/781) theorem, $G$ is the disjoint union of, say, $k$ distinct left cosets of $H$: \begin{align*} G = g_1H \;\dot\cup\; g_2H \;\dot\cup\; \cdots \;\dot\cup\; g_kH, \end{align*} where we may take $g_1 = e$. **Step 2: Count.** By the [Coset Bijection](/theorems/780) theorem, $|g_iH| = |H|$ for each $i$. Since the cosets are pairwise disjoint and their union is $G$: \begin{align*} |G| = \sum_{i=1}^{k} |g_iH| = k \cdot |H|. \end{align*} Therefore $|H|$ divides $|G|$, and the quotient $k = |G|/|H| = |G : H|$ is the index of $H$ in $G$.