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

Theorem

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

Let e_G and e_K be identities of G and K.

f(e_G) = e_K

Proof

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 e_G
.
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 e_K, definition of inverse
6. ${\color{OliveGreen}e_K} = f({\color{Blue}e_{G}})$	identity e_K

Last modified on 25 July 2010, at 02:01