Prove that $H$ is a subgroup of $G$ if $H$ is non-empty and $gh^{-1} \in H$ whenever $g,h \in H$ The problem is what is stated in the title, if $G$ is a group and $H$ a subset of $G$. I have been trying to find a way to produce the identity element of $G$ by playing with the definition:

For $g \in H, gh^{-1} \in H$, $ghg^{-1} \in H$, $hg^{-1} \in H$,
  etc...

I also wonder what will happen if I modify the question to say that $gh^{-1} \in H$ whenever $g,h \in H$ AND $g \neq h$.
 A: If $H\subset G$ then $H$ is a subgroup iff it contains the identity, closed under multiplication and inverses. So in your case it is simple.
The identity is there because $H$ is not empty and contains at element $x$. then $e=xx^{-1}\in H$. 
If $x\in H$ then $x^{-1}=ex^{-1}\in H$, so it contains all inverses.
If $x,y\in H$ then as we already know $y^{-1}\in H$ and hence $xy=x(y^{-1})^{-1}\in H$. So it is closed under multiplication as well and is a subgroup. 
Now, if you assume that $gh^{-1}$ only when $g\ne h$ then it is not necessary a subgroup. Take $G=\mathbb{Z_3}$ and $H=\{2\}$ as a trivial counterexample. 
A: For your 2nd question take $G=(\mathbb{Z},+)$ and $H=(\mathbb{Z^{*}},+)$. Then for $x,y\in H$ and $x\not= y$ we have $x-y\in H$ but $H$ is not a subgroup of $G$
A: Since $H$ is not empty, choose some $h\in H$.  Then $1 = hh^{-1} \in H$ by assumption (put $g=h$), so you've got your identity.
As to your second question, in that case $H$ might not be a subgroup. Take the set of non-identity elements of the non-cyclic group of order $4$, for example.
