Let $G$ be a group. Prove the equivalence relation: If $H$ is a subgroup of $G$, let $a \sim b$ iff $ab^{-1} \in H$ Let $G$ be a group. Prove the equivalence relation: 

If $H$ is a subgroup of $G$, let $a \sim b$ iff $ab^{-1} \in H$

To prove an equivalence relation my guess is to show that reflexivity, symmetry, and transitivity hold, but I am not completely sure if I am doing this correctly. So, here goes.
First we must show that if $a \sim a$ then $aa^{-1} \in H$. But $aa^{-1}=e$ and since H is a subgroup, $e$ exists in $H$.
(not sure if I am doing this correctly) Next, we must show that if $a \sim b$ then $b \sim a$, and $ba^{-1} \in H$. If $a \sim b$ then $ab^{-1} \in H$, so $a, a^{-1}, b, b^{-1} \in H$. Thus, $ba^{-1} \in H$.
(not sure if I am doing this correctly) Next, we must show that if $a \sim b$ and $b \sim c$, then $a \sim c$, and $ac^{-1} \in H$. But since $b \sim c$, $bc^{-1} \in H$, and so $$a, a^{-1}, b, b^{-1}, c, c^{-1} \in H$$ So, $ac^{-1} \in H$.
Is this even remotely correct? Is my approach correct? Since this is a biconditional statement, I am assuming I must prove the converse?
 A: Your proof of reflexivity is fine.
For symmetry, you have an error. A correct answer uses $(ab^{-1})^{-1}=ba^{-1}$. See if you can complete the proof.
You also have an error in transitivity. Use the fact that $(ab^{-1})(bc^{-1})=ac^{-1}$.
A: It's not necessarily true that $a,b\in H$. Note that since $H$ is a group, it is closed under inverses. What is the inverse of $ab^{-1}$? Similarly, since $H$ is a group, it is closed under products. If $ab^{-1},bc^{-1}\in H$, what do you know about $(ab^{-1})(bc^{-1})$?
A: Hint:
$$e = b\ b^{-1} = b(a^{-1}a)b^{-1} = (ba^{-1})\underbrace{(ab^{-1})}_{\in\  H} $$
You may use the same idea for $ac^{-1}=(ab^{-1})(bc^{-1})$.
A: The easiest way to prove this is to note that $ab^{-1} \in H$ iff $aH = bH$.
Then the equivalence classes are just the fibers of the function $f(a)=aH$.
Every relation in a set $X$ defined by $x \sim y \iff f(x) = f(y)$, where $f:X \to Z$, is an equivalence relation because equality is an equivalence relation. There is nothing really left to prove after you make this observation.
In your case, $X=G$ and $Z=\{ aH : a \in G \}$.
