Coset proof: $aH=bH$ if and only if $H=a^{-1}bH$

Question

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 $a^{-1}bH \subset H$?

Answer 1

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, $H\subset a^{-1}bH$.

Answer 2

Maybe this way of seeing it is clearer ?

$\begin{eqnarray*} aH = bH &\iff& \{ ah_1, \dots, ah_n \} = \{bh_1, \dots, bh_n\} \\ & \iff & \{a^{-1}ah_1, \dots, a^{-1}ah_n \} = \{a^{-1}bh_1, \dots, a^{-1}bh_n \} \\ & \iff & H = a^{-1}bH \end{eqnarray*}$