In the paper The Simplest Protocol for Oblivious Transfer, in the scenario where the receiver chooses $c=1$, then $B$ is calculated as $Ag^b$.
Then $k_1$ is calculated as $H( (B/A)^a)$.
That means that the receiver can calculate the same encryption key, and decrypt message $M_1$ (because it's encrypted with $k_1$.
Does this imply that $H( (B/A)^a ) = H( Ag^b)$ ?
I don't understand how it works in this scenario (receiver choosing $c=1$)