# Statement(s) > [!NOTE] Statement 1 (....) > Let $G$ be a [[Groups|group]] and $H$ a [[Subgroup|subgroup]]. Let $g\in G.$ Then there exists a [[Bijection|bijection]] between the [[Coset|left coset]] of $H$ modulo $g,$ $gH$ and $H.$ # Proof(s) **Proof:** Define $\begin{align} \phi: H\to gH \\ h \mapsto gh \end{align}$Let $h_{1},h_{2} \in H,$ such that $\phi(h_{1})=\phi(h_{2}).$ Then $gh_{1}=gh_{2}.$ Premultiplying both sides by $g^{-1}$ gives $h_{1} = h_{2}$ and so $\phi$ is injective. Let $x\in gH.$ Then by definition of $gH,$ $x=gh$ for some $h\in H$ and so $\phi$ is surjective. $\blacksquare$ **Proof:** Define $\begin{align} \phi: H\to gH \\ h \mapsto gh \end{align}$ and Define $\begin{align} \theta: gH\to H \\ x \mapsto g^{-1} x \end{align}$ Then $\phi$ and $\theta$ are inverses of each other. $\blacksquare$ # Application(s) **Consequences**: **Examples**: # Bibliography