# 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

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