# Statements > [!NOTE] Lemma > Let $(G,\circ)$ be a [[Groups|group]] and let $H$ be a [[Subgroup|subgroup]] of $G.$ Let $x,y \in G.$ Let $xH$ denote the [[Coset|left coset]] of $H$ by $x.$ Then $xH= yH \iff x^{-1}y\in H$ **Note**: We may replace $x^{-1}y$ with $y^{-1} x.$ # Proof **Proof**: Suppose $xH=yH.$ Note that $y=y\circ 0 \in yH=xH$. So there exists $z \in I$ such that $y=x\circ z$. This gives $x^{-1}y= z \in H.$ Conversely, suppose $x^{-1}y \in H$. So $x^{-1}y=z$ for some $z \in H$ and $y=xz$. Let $s\in yH$ then $s=ys^{*}$ for some $s^{*}\in H.$ Thus $s=(xz)s^{*}=x(zs^{*})$. Since $zs^{*} \in H$, $s \in xH$ and $yH \subset xH.$ Similarly we have $y^{-1}x=z^{-1}\in H$ so as above $xH \subset yH$. The two inclusions give $xH=yH$. $\square$