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 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