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?
Best Answer
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 $(ab^{-1})(bc^{-1})=ac^{-1}$.