[proofplan] We verify the three defining properties of an equivalence relation directly from the subgroup axioms for $H$. Reflexivity uses the identity element, symmetry uses closure of $H$ under inverses, and transitivity uses closure of $H$ under multiplication. Once the relation is known to be an equivalence relation, the equivalence class of $x$ is identified by rewriting the condition $x \sim_R y$ as the statement that $y$ has the form $hx$ with $h \in H$. [/proofplan] [step:Use the identity element of $H$ to prove reflexivity] Let $x \in G$ be arbitrary. Since $H \le G$, the identity element $e$ of $G$ belongs to $H$. By the inverse axiom in $G$, \begin{align*} xx^{-1} = e \in H. \end{align*} Therefore $x \sim_R x$. Since $x \in G$ was arbitrary, $\sim_R$ is reflexive. [guided] We first prove reflexivity, which means that every element of $G$ is related to itself. Let $x \in G$ be arbitrary. By the definition of $\sim_R$, the statement $x \sim_R x$ is equivalent to \begin{align*} xx^{-1} \in H. \end{align*} The group inverse axiom gives $xx^{-1}=e$, where $e$ is the identity element of $G$. Because $H \le G$ is a subgroup, $H$ contains the same identity element $e$. Hence \begin{align*} xx^{-1} = e \in H. \end{align*} Thus $x \sim_R x$. Since the choice of $x \in G$ was arbitrary, the relation $\sim_R$ is reflexive. [/guided] [/step] [step:Invert the defining subgroup element to prove symmetry] Let $x,y \in G$ satisfy $x \sim_R y$. By definition of $\sim_R$, \begin{align*} yx^{-1} \in H. \end{align*} Since $H$ is a subgroup, it is closed under inverses. Therefore \begin{align*} (yx^{-1})^{-1} \in H. \end{align*} Using the inverse-of-a-product identity in $G$, \begin{align*} (yx^{-1})^{-1} = xy^{-1}. \end{align*} Thus $xy^{-1} \in H$, so $y \sim_R x$. Hence $\sim_R$ is symmetric. [guided] We next prove symmetry. Let $x,y \in G$ and assume $x \sim_R y$. The definition of $\sim_R$ translates this assumption into the subgroup membership condition \begin{align*} yx^{-1} \in H. \end{align*} To prove $y \sim_R x$, we must prove \begin{align*} xy^{-1} \in H. \end{align*} The connection between these two expressions is inversion. Since $H \le G$ is a subgroup, every element of $H$ has its inverse in $H$. Applying this to the element $yx^{-1} \in H$, we get \begin{align*} (yx^{-1})^{-1} \in H. \end{align*} By the inverse-of-a-product identity in the group $G$, \begin{align*} (yx^{-1})^{-1} = (x^{-1})^{-1}y^{-1} = xy^{-1}. \end{align*} Therefore $xy^{-1} \in H$, which is exactly the condition $y \sim_R x$. Hence $\sim_R$ is symmetric. [/guided] [/step] [step:Multiply the two defining subgroup elements to prove transitivity] Let $x,y,z \in G$ satisfy $x \sim_R y$ and $y \sim_R z$. By definition of $\sim_R$, \begin{align*} yx^{-1} \in H \end{align*} and \begin{align*} zy^{-1} \in H. \end{align*} Since $H$ is a subgroup, it is closed under multiplication. Hence \begin{align*} (zy^{-1})(yx^{-1}) \in H. \end{align*} Associativity in $G$ and the inverse axiom give \begin{align*} (zy^{-1})(yx^{-1}) = z(y^{-1}y)x^{-1} = zex^{-1} = zx^{-1}. \end{align*} Thus $zx^{-1} \in H$, so $x \sim_R z$. Hence $\sim_R$ is transitive. [guided] We now prove transitivity. Let $x,y,z \in G$ and assume both $x \sim_R y$ and $y \sim_R z$. Unpacking the definition of $\sim_R$, these assumptions mean \begin{align*} yx^{-1} \in H \end{align*} and \begin{align*} zy^{-1} \in H. \end{align*} To prove $x \sim_R z$, we must prove \begin{align*} zx^{-1} \in H. \end{align*} The expression $zx^{-1}$ is obtained by multiplying the two known elements in the correct order: \begin{align*} (zy^{-1})(yx^{-1}). \end{align*} Because $H \le G$ is a subgroup, it is closed under the group operation. Since both factors lie in $H$, their product lies in $H$: \begin{align*} (zy^{-1})(yx^{-1}) \in H. \end{align*} Using associativity in $G$ and the inverse axiom $y^{-1}y=e$, we compute \begin{align*} (zy^{-1})(yx^{-1}) = z(y^{-1}y)x^{-1} = zex^{-1} = zx^{-1}. \end{align*} Therefore $zx^{-1} \in H$, which is exactly the statement $x \sim_R z$. Hence $\sim_R$ is transitive. [/guided] [/step] [step:Conclude that the relation is an equivalence relation] The relation $\sim_R$ is reflexive, symmetric, and transitive by the preceding three steps. By the definition of an [equivalence relation](/page/Equivalence%20Relation), $\sim_R$ is an equivalence relation on $G$. [/step] [step:Rewrite the equivalence class condition as membership in the right coset] Fix $x \in G$. Define the equivalence class of $x$ under $\sim_R$ by \begin{align*} [x]_{\sim_R} := \{y \in G : x \sim_R y\}. \end{align*} By definition of $\sim_R$, \begin{align*} [x]_{\sim_R} &= \{y \in G : yx^{-1} \in H\}. \end{align*} We prove equality with $Hx := \{hx : h \in H\}$ by double inclusion. First, let $y \in [x]_{\sim_R}$. Then $yx^{-1} \in H$. Define $h := yx^{-1} \in H$. Multiplying on the right by $x$ gives \begin{align*} hx = (yx^{-1})x = y(x^{-1}x) = ye = y. \end{align*} Hence $y = hx$ for some $h \in H$, so $y \in Hx$. Conversely, let $y \in Hx$. Then there exists $h \in H$ such that $y=hx$. Multiplying on the right by $x^{-1}$ gives \begin{align*} yx^{-1} = (hx)x^{-1} = h(xx^{-1}) = he = h \in H. \end{align*} Therefore $x \sim_R y$, so $y \in [x]_{\sim_R}$. Thus \begin{align*} [x]_{\sim_R} = Hx. \end{align*} Since $x \in G$ was arbitrary, every equivalence class is the corresponding right coset. [guided] Fix $x \in G$. The equivalence class of $x$ under $\sim_R$ is the set of all elements of $G$ related to $x$: \begin{align*} [x]_{\sim_R} := \{y \in G : x \sim_R y\}. \end{align*} Using the definition of the relation, this becomes \begin{align*} [x]_{\sim_R} &= \{y \in G : yx^{-1} \in H\}. \end{align*} We want to prove that this set is exactly the right coset \begin{align*} Hx := \{hx : h \in H\}. \end{align*} The natural method is double inclusion. First suppose $y \in [x]_{\sim_R}$. Then, by the displayed description of the class, \begin{align*} yx^{-1} \in H. \end{align*} Define $h := yx^{-1}$. This definition gives $h \in H$. Multiplying the equation $h=yx^{-1}$ on the right by $x$, and using associativity and the inverse axiom, gives \begin{align*} hx = (yx^{-1})x = y(x^{-1}x) = ye = y. \end{align*} Therefore $y=hx$ for some $h \in H$, which means $y \in Hx$. Hence \begin{align*} [x]_{\sim_R} \subset Hx. \end{align*} Conversely, suppose $y \in Hx$. By the definition of the right coset, there exists an element $h \in H$ such that \begin{align*} y = hx. \end{align*} Multiplying this equality on the right by $x^{-1}$ gives \begin{align*} yx^{-1} = (hx)x^{-1} = h(xx^{-1}) = he = h. \end{align*} Since $h \in H$, this proves \begin{align*} yx^{-1} \in H. \end{align*} By the definition of $\sim_R$, this is exactly $x \sim_R y$, so $y \in [x]_{\sim_R}$. Hence \begin{align*} Hx \subset [x]_{\sim_R}. \end{align*} The two inclusions give \begin{align*} [x]_{\sim_R} = Hx. \end{align*} Since the element $x \in G$ was arbitrary, the equivalence class of each element $x$ is precisely the right coset $Hx$. [/guided] [/step]