Under a group homomorphism, the image of the inverse is the inverse of the image

For any group homomorphism $f:G\to H$, we have $f\left(g^{-1}\right)=f(g{)}^{-1}$.

Indeed, $f\left(g^{-1}\right)f\left(g\right)=f\left(g^{-1}g\right)=f\left(e_G\right)=e_H$, and similarly for multiplication on the left.