Prove: $aH = bH \iff Ha^{-1} = Hb^{-1}$

Question

I have to prove this exercise for my math-study:

Let $G$ be a group and $H \subset G$ a subgroup. Prove that for every $a,b \in G$ holds:

$$aH = bH \iff Ha^{-1} = Hb^{-1}$$

I tried this, but I'm not sure if it is correct or if I'm making too big steps at once:

\begin{align*} & aH = bH \\ \iff & \exists h \in H \text{ such that } b = ah \\ \iff & \exists h \in H \text{ such that } a^{-1}b = a^{-1}ah = h \\ \iff & \exists h \in H \text{ such that } a^{-1}bb^{-1} = a^{-1} = hb^{-1} \\ \iff & a^{-1} \in Hb^{-1} \\ \iff & Ha^{-1} = Hb^{-1} \end{align*}

Is this prove correct? and if not, how can I make it correct? Thanks in advance!

Answer 1

You are on the right way. In the last step you should arrive at: $Ha^{-1} \subset Hb^{-1}$ (not an equality). In a similar way you Show $Hb^{-1} \subset Ha^{-1}$; you start with "it exists $h \in H$ such that $a = bh$" and then take the inverse of $b$ from the left...

Answer 2

I'm also new with this topic, so I might be wrong. But I would explain the first iff and the last one. $aH=bH\iff \exists h_1,h_2 \in H : ah_1=bh_2 \iff \exists h_1,h_2 \in H : b=bh_2h_2^{-1}=ah_1h_2^{-1}$ Similarly at the end. I'd like to receive a comment whether I have mistakes, I don't have ones or there is no need for such explanation.