Let $G$ be a group, and $H$ a subgroup of $G$. Let $a, b \in G$.
Prove $Ha=Hb$ iff $ab^{-1} \in H$.
$\rightarrow$ If $Ha=Hb$, then $h_1a=h_2b$ for some $h_1, h_2 \in H$.
So, $ab^{-1} = h_1^{-1}h_2$.
Therefore, $ab^{-1} \in H$.
$\leftarrow$ If $ab^{-1} \in H$, then $ab^{-1} = h_3$ for some $h_3 \in H$.
So, $a=h_3b$, and thus $a \in Hb$.
Therefore, $a \in Hb$ implies $Ha=Hb$.
Is this proof correct? I am unsure about the first step (If $Ha=Hb$, then $h_1a=h_2b$ for some $h_1, h_2 \in H$.).
Best Answer
Your argument is correct (including the first step), but I think your last assertion requires a bit more justification. That is, you should explain why $a \in Hb$ implies $Ha = Hb$; just to be clear, it is true, but I don't think you've clearly explained why it is true. As $a = h_3b$, $ha = hh_3b \in Hb$, so the fact that $Ha \subseteq Hb$ follows fairly easily (but may still be worth pointing out). The reverse inclusion is not as trivial.
Added Later: As Sayantan Kolgy mentions in his answer, you may be using the fact that distinct cosets are disjoint so as $a \in Ha$ and $a \in Hb$, we must have $Ha = Hb$. While this is a much more direct way at arriving at the conclusion, it isn't immediately clear that this is the fact you are using (as my answer demonstrates).