# Abstract Algebra/Group Theory/Homomorphism/Homomorphism Maps Inverse to Inverse

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