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 $a^{-1}bH \subset H$?
Best Answer
I'll prove one direction; hopefully seeing it will help you see how to do the other direction.
Let $h\in H$. Then $ah\in bH$ so there is another $h'$ such that $ah=bh'$ or $h=a^{-1}bh'\in a^{-1}bH$. Thus, $H\subset a^{-1}bH$.