# Theorem

Let f be a homomorphism from group G to Group K.

Let g be any element of G.

f(g-1) = [f(g)]-1

# Proof

 0.   ${\displaystyle f({\color {Blue}g})\circledast f({\color {BrickRed}g^{-1}})=f({\color {Blue}g}\ast {\color {BrickRed}g^{-1}})}$ f is a homomorphism 1.   ${\displaystyle f({\color {Blue}g})\circledast f({\color {BrickRed}g^{-1}})=f({\color {Blue}e_{G}})}$ definition of inverse in G . 2.   ${\displaystyle f({\color {Blue}g})\circledast f({\color {BrickRed}g^{-1}})={\color {OliveGreen}e_{K}}}$ homomorphism f maps identity to identity 3.   ${\displaystyle {\color {BrickRed}[}f({\color {Blue}g}){\color {BrickRed}]^{-1}}\circledast f({\color {Blue}g})\circledast f({\color {BrickRed}g^{-1}})={\color {BrickRed}[}f({\color {Blue}g}){\color {BrickRed}]^{-1}}\circledast {\color {OliveGreen}e_{K}}}$ as f(g) is in K, so is its inverse [f(g)]-1 . 4.   ${\displaystyle {\color {OliveGreen}e_{K}}\circledast f({\color {BrickRed}g^{-1}})={\color {BrickRed}[}f({\color {Blue}g}){\color {BrickRed}]^{-1}}}$ inverse on K, eK is identity of K 5.   ${\displaystyle f({\color {BrickRed}g^{-1}})={\color {BrickRed}[}f({\color {Blue}g}){\color {BrickRed}]^{-1}}}$ eK is identity of K