Coset proof: $aH=bH$ if and only if $H=a^{-1}bH$ In Gallian's Contemporary Abstract Algebra, 9TH edition, on page 140, he is trying to prove $aH=bH$ if and only if $a^{-1}b\in H$.
And he says to observe that:
$aH=bH$ if and only if $H=a^{-1}bH$.
How is this true?
"$\rightarrow$"
Assume $aH=bH$, so let $t\in aH$. Then $t=ah=bh$, so $a^{-1}t=h=a^{-1}bh$. 
How to go forward from this to prove $H \subset a^{-1}bH$ and $a^{-1}bH \subset H$?

Maybe this way of seeing it is clearer ? 
$\begin{eqnarray*}
aH = bH &\iff& \{ ah_1, \dots, ah_n \} = \{bh_1, \dots, bh_n\} \\
& \iff & \{a^{-1}ah_1, \dots, a^{-1}ah_n \} = \{a^{-1}bh_1, \dots, a^{-1}bh_n \} \\
& \iff & H = a^{-1}bH
\end{eqnarray*}$
