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.