The strategy is to verify the two partition conditions: (1) every element lies in some coset, and (2) two cosets that share an element must be identical. The key technique in (2) is the double-inclusion argument $aH \subseteq cH \subseteq aH$. **Step 1: Every element lies in a coset.** For any $g \in G$, we have $g = ge \in gH$ (since $e \in H$). So $g$ lies in the coset $gH$. **Step 2: Overlapping cosets are equal.** [claim:Coset Equality From Shared Element] If $c \in aH \cap bH$, then $aH = bH$. [/claim] [proof] Since $c \in aH$, there exists $k \in H$ with $c = ak$. For any $h \in H$: \begin{align*} ch = akh \in aH, \end{align*} since $kh \in H$ by closure. Therefore $cH \subseteq aH$. Conversely, $a = ck^{-1}$ (since $k^{-1} \in H$), so for any $h \in H$: \begin{align*} ah = ck^{-1}h \in cH. \end{align*} Therefore $aH \subseteq cH$. Combining gives $aH = cH$. By the identical argument with $b$ in place of $a$, $bH = cH$. Hence $aH = cH = bH$. [/proof] **Step 3: Conclusion.** By Step 1, $\{gH : g \in G\}$ covers $G$. By Step 2, any two cosets are either disjoint or identical. Therefore the distinct left cosets of $H$ in $G$ form a partition of $G$.