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 =
 * {| Style = "width:75%;boarder:1px"

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 * 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
 * }