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

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

https://arbital.com/p/group_homomorphism_image_of_inverse

by Patrick Stevens Jun 14 2016 updated Jun 14 2016

The operations of "taking inverses" and "applying a group homomorphism" commute: it does not matter in which order we do them.

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

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