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

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Theorem[edit]

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[edit]

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