Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2015 / Paper 2 / 8 / a / Solution

The left coset isEvery $g\in G$ belongs to $gH$, because $e\in H$, so the cosets cover $G$. If $gH$ and $kH$ intersect, choose $g{h}_1=k{h}_2$ in the intersection. Then $k^{-1}g=h_2{h}_1^{-1}\in H$, 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.