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

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