Let $S$ be a non empty subset of a group $G$ and define a relation on $G$ by $a\sim b$ if and only if $ab^{-1}\in S$. Show that $\sim$ is an equivalence relation if and only if $S$ is a subgroup of $G$.
My attempt: Let $\emptyset \neq S\subseteq G$ and $\sim =\{(a,b)\in G\times G\mid ab^{-1}\in S\}$. $(\Rightarrow)$ Supppose $\sim$ is an equivalence relation on $G$. Let $a,b\in S$. Since $ae^{-1}=a\in S$ and $be^{-1}=b\in S$, we have $(a,e),(b,e)\in \sim$. By symmetry, we have $(e,b)\in \sim$. By transitivity, $(a,b)\in \sim$. That is $ab^{-1}\in S$. Thus $S\neq \emptyset$ and $ab^{-1}\in S$ for all $a,b\in S$. Hence $S\leq G$.
$(\Leftarrow)$ Suppose $S\leq G$. Since $gg^{-1}=e\in S$, we have $(g,g)\in \sim$, $\forall g\in G$. So $\sim$ is reflexive. Let $(a,b)\in \sim$. Then $ab^{-1}\in S$. So $(ab^{-1})^{-1}=(b^{-1})^{-1}a^{-1}=ba^{-1}\in S$. Thus $(b,a)\in \sim$ and $\sim$ is symmetric. Let $(a,b)\in \sim$ and $(b,c)\in \sim$. Then $ab^{-1}\in S$ and $bc^{-1}\in S$. So $(ab^{-1})(bc^{-1})=ac^{-1}\in S$. Thus $(a,c)\in \sim$ and $\sim$ is transitive. Hence $\sim$ is an equivalence relation. Is my proof correct?