= Solution
The left <coset> is
$$
\boxed{gH=\{gh:h\in H\}.}
$$
Every $g\in G$ belongs to $gH$, because $e\in H$, so the cosets cover $G$. If $gH$ and $kH$ intersect, choose $gh_1=kh_2$ in the intersection. Then $k^{-1}g=h_2h_1^{-1}\in H$, so $g=kh$ for some $h\in H$. Thus $gH=khH=kH$.
Hence two left cosets are either identical or disjoint, and the distinct left cosets form a <partition of a set> whose union is $G$. Finiteness is unnecessary for this partition argument. In the finite case each coset has $|H|$ members, since multiplication by $g$ is a <bijection>, giving <Lagrange's theorem>.
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