Showing that if $Hg^{−1}_1 = Hg^{−1}_2$, then $g_1H = g_2H.$ $Hg^{−1}_1 = Hg^{−1}_2 \implies g_1H = g_2H $ 
I have a really quick question. I was doing some Group theory exercises, for this particular question I couldn't understand the proof: 

"Since $Hg^{-1}_1=Hg^{-1}_2$ there is an $h \in H$ with $g^{-1}_1=hg^{-1}_2$ and $g_2=g_1h$."

How do I know that? I couldn't understand the equality and how $g_2=g_1h$. Thanks a lot in advance!
 A: If 
$Hg_1^{-1} = Hg_2^{-1}, \tag 1$
we choose 
$e \in H, \tag 2$
where $e$ is the identity element of $G$; we know (2) binds since $H$ is a subgroup of $G$; then
$g_1^{-1} = eg_1^{-1} \in Hg_2^{-1} \tag 3$
by virtue of (1); thus 
$\exists h \in H, \; g_1^{-1} = hg_2^{-1}; \tag 4$
this yields
$e = g_1g_1^{-1} = g_1hg_2^{-1}, \tag 5$
and this further yields
$g_2 = eg_2 = (g_1hg_2^{-1})g_2 = g_1h(g_2^{-1}g_2) = g_1he = g_1h, \tag 6$
the desired relation 'twixt $g_1$, $g_2$, and $h$.
A: Since $Hg^{-1}_1=Hg^{-1}_2$, there exist $h_1, h_2\in H$ such that $h_1g^{-1}_1=h_2g^{-1}_2$. Cancel $h_1$ to get $g^{-1}_1=(h_1^{-1}h_2)g^{-1}_2$ and note that $h:=h_1^{-1}h_2\in H$. Cancel $g^{-1}_1$ on the left and multiply by $g_2$ on the right. Then $g_2=g_1h$.
A: $g_1^{-1}$ belongs to $Hg_1^{-1}$ since it is expressible as $eg_1^{-1}$ (as $e \in H$).  But $Hg_1^{-1}=Hg_2^{-1}$ so $g_1^{-1}$ must be expressible as $g_1^{-1}=hg_2^{-1}$ for some $h \in H$.  Solve for $g_2$ (right multiply by $g_2$, left multiply by $g_1$) to get $g_2=g_1h$, as claimed.
A: Let $X^{-1} = \{ x^{-1} : x \in X \}$. Then
$$
Hg^{−1}_1 = Hg^{−1}_2
\implies 
(Hg^{−1}_1)^{−1} = (Hg^{−1}_2)^{−1}
\implies
g_1 H^{−1} = g_2 H^{−1}
\implies
g_1 H = g_2 H
$$
A: Hint: if $k=h_1g_1^{-1}=h_2g_2^{-1}$ (so $k\in Hg_1^{-1}=Hg_2^{-1}$) then what form does $k^{-1}$ have?
