# TheoremEdit

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

Let eG and eK be identities of G and K.

f(eG) = eK

# ProofEdit

 0.   $f({\color{Blue}e_{G}}) \in K$ f maps to K 1.   ${\color{BrickRed} [} f({\color{Blue}e_{G}}) {\color{BrickRed}]^{-1}}$ inverse in K . 2.   $f({\color{Blue}e_{G}} \ast {\color{Blue}e_{G}}) = f({\color{Blue}e_{G}}) \circledast f({\color{Blue}e_{G}})$ f is a homomorphism 3.   $f({\color{Blue}e_{G}}) = f({\color{Blue}e_{G}}) \circledast f({\color{Blue}e_{G}})$ identity eG . 4.   ${\color{BrickRed} [} f({\color{Blue}e_{G}}) {\color{BrickRed}]^{-1}} \circledast f({\color{Blue}e_{G}}) = {\color{BrickRed} [} f({\color{Blue}e_{G}}) {\color{BrickRed}]^{-1}} \circledast f({\color{Blue}e_{G}}) \circledast f({\color{Blue}e_{G}})$ 1. . 5.   ${\color{OliveGreen}e_K} = {\color{OliveGreen}e_K} \circledast f({\color{Blue}e_{G}})$ identity eK, definition of inverse 6.   ${\color{OliveGreen}e_K} = f({\color{Blue}e_{G}})$ identity eK