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

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

Let eG and eK be identities of G and K.


 * f( eG ) = eK

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

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 * 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
 * }