If $\displaystyle G$ is a group, $\displaystyle K \lhd G$ is proper and $\displaystyle \forall x \in G \backslash K, x^2 = e$, show that $\displaystyle \forall t \in G \backslash K, \forall a \in K, t^{-1} at = a^{-1}$.

I'm completely confused on how to solve this. I just seem to be going round in circles. (Headbang)

Can anyone help?