Show that $Ha=Hb$ if and only if $ab^{-1} \in H$ where $a,b\in G$ and $H$ is a subgroup of $G$ Suppose $G$ is a finite group,  $a,b\in G$ and $H$ is a subgroup of $G$.

Show that $Ha=Hb$ if and only if $ab^{-1} \in H$. 
  (Notice that $Ha$ and $Hb$ are both right cosets of $H$ by $G$).

This is what I have so far:

(-->) Suppose $Ha=Hb$. We know that $H$ is a subgroup and therefore $1_G\in H$ and from this we can derive that $a\in H$ and $b\in H$. 

For a certain $h\in H$, we have:

$b\in Ha \implies b=ha \implies 1_G=h(ab^{-1})$

Is it enough to derive that $ab^{-1}\in H$ ?(because we know $1_G\in H$)

(&lt;--) Suppose $ab^{-1} \in H$. I'm not quite sure what direction should I take from here in order to show that $Ha=Hb$. Can you please give me an hint how to start this direction?

Thank you very much,
You have some mistakes. If $1_G\in H$ then we cannot derive that $a\in H$ and $b\in H$.

We have that either $Hg_1=Hg_2$ or $Hg_1\cap Hg_2=\emptyset$

$(-&gt;)$

If $Ha=Hb\Rightarrow a=hb$ for some $h\in H$.

Thus $ab^{-1}=h\in H$.

$(&lt;-)$ 

Let $ab^{-1}\in H$. Then $ab^{-1}=h\in H\Rightarrow a=hb=\rightarrow a\in Hb\Rightarrow Ha=Hb$.
For the converse, $ab^{-1}\in H$ implies:


$ab^{-1} = h_0$ for some $h_0\in H\implies a = h_0b\implies a\in Hb\implies ha \in Hb$ for every $h\in H$.


$\qquad\therefore x\in Ha\implies x\in HB$.


$(ab^{-1})^{-1} \in H$, by invertibility of group $H\iff ba^{-1} \in H$.


$\qquad\therefore x\in Hb\implies x\in Ha$, by applying similar argument in (1).

Hence $Ha=Hb$.

$\Box$