Last modified on 25 July 2010, at 02:01

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

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