Defining an equivalence relation via $x^{-1}y\in H$ w.r.t a subgroup $H$ of a group $G$.

Question

Let $G$ be a group and $H$ a subgroup of $G$. Does the formula $$x^{-1}y\in H$$ define an equivalence relation on $G$? It is clear that $x^{-1}y\in H$ is both transitive and symmetric. But is it reflexive on $G$? That is to say, does the formula $$(\forall x)(x^{-1}x\in H\Rightarrow x\in G)$$ hold? If not, then does that mean that one instead has to use the formula $$x^{-1}y\in H\ \land\ x\in G\ \land\ y\in G$$ in order to get the desired equivalence relation on $G$?

Edit 1:

Obviously $x^{-1}x=e\in H$ for all $x\in G$. I want to know whether $x^{-1}x\in H$ implies $x\in G$ without assuming that $x\in G$.

Edit 2:

The question is which of the following formulas defines the desired equivalence relation on $G$:

1. $x^{-1}y\in H$
2. $x^{-1}y\in H\ \land\ x\in G\ \land\ y\in G$

Do you wish to say that these are logically the same? The usual way of checking reflexivity implicitly intersects the graph defined by (1) with $G\times G$.

Edit 3:

Ok it seems clear that I am not able to get my point across. Forget the part about the equivalence relation. Can one prove the formula $$(\forall x)(x^{-1}x\in H\Rightarrow x\in G)$$ given the assumption that $G$ is a group and $H$ is a subgroup of $G$?--Is it the case that one cannot suppose $x^{-1}x\in H$ without making any other assumptions about x? i.e. is it an ungrammatical string of symbols in set theory?

Answer 1

For the reflexive property you have to show that for every $g\in G$ we have $g^{-1}g\in H$

Since $H$ is a subgroup we have $g^{-1}g=e\in H$ for every $g\in G$

Answer 2

It is indeed an equivalence relation. The conditions are essy to check using the fact that groups (subgroups here) are closed under inverses and contain the identity element.

Answer 3

The proof that the relation is reflexive is short and sweet. Since $H\le G$ is a subgroup, $e\in H$. Thus for any $x\in G$, we have $xx^{-1}=e\in H$. Thus $x\sim x$.