# Statement(s) > [!NOTE] Lemma (Coset Space Forms Partition) > Let $G$ be [[Groups|group]] and $H$ a [[Subgroup|subgroup]]. Let $g_{1},g_{2}$ be elements of $G.$ Then cosets $g_{1}H$ and $g_{2}H$ are non-empty and either equal or disjoint: that is, [[Coset space|(left) coset space]] forms a [[Partition of a Set|partition]] of $G.$ # Proof(s) **Proof:** Suppose $g_{1}H\cap g_{2}H \neq \emptyset,$ i.e. there exists $g_{3}\in G,$ such that $g_{3}\in g_{1}H$ and $g_{3}\in g_{2}H.$ Then for some $h_{1}\in H$ and $h_{2}\in H,$ $g_{1}h_{1}=g_{3}=g_{2}h_{2}.$ Now $g_{2}^{-1}g_{1}=h_{2}h_{1}^{-1}\in H$ so by [[Necessary Condition for Equality of Cosets]], $g_{1}H=g_{2}H.$ By definition, $g_{1}H \subset G$ so $\bigcup_{g\in H}gH \subset G.$ For all $g\in G,$ $1\in H$ gives $g\in gH$ so $gH$ is non-empty. Let $g\in G.$ Then $g\in gH$ which is a coset hence $G=\bigcup_{g\in H}gH.$ $\blacksquare$ # Application(s) **Consequences**: **Examples**: # Bibliography