 Solution

For a subgroup $H$ of $G$, the left coset represented by $g$ isSuppose two left cosets intersect. There are $h_1,h_2\in H$ such that $g_1{h}_1=g_2{h}_2$, soBecause $H$ is a subgroup, $h_2{h}_1^{-1}\in H$. Thus, for every $h\in H$, $g_{1}h=g_2(h_2{h}_1^{-1}h)\in g_{2}H$, proving $g_{1}H\subseteq g_{2}H$. Interchanging the two representatives gives the reverse inclusion. Therefore two left cosets are equal or disjoint.