抽象代数/群论/同态/同态映射恒等元到恒等元

定理

设f是从群G到群K的同态。

设e_G和e_K是G和K的恒等元。

f(e_G) = e_K

证明

0.   $f({\color {Blue}e_{G}})\in K$	f映射到K
1.   ${\color {BrickRed}[}f({\color {Blue}e_{G}}){\color {BrickRed}]^{-1}}$	K中的逆
.
2.   $f({\color {Blue}e_{G}}\ast {\color {Blue}e_{G}})=f({\color {Blue}e_{G}})\circledast f({\color {Blue}e_{G}})$	f是同态
3.   $f({\color {Blue}e_{G}})=f({\color {Blue}e_{G}})\circledast f({\color {Blue}e_{G}})$	恒等元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.   ${\color {OliveGreen}e_{K}}={\color {OliveGreen}e_{K}}\circledast f({\color {Blue}e_{G}})$	恒等元 eK, 逆的定义
6.   ${\color {OliveGreen}e_{K}}=f({\color {Blue}e_{G}})$	单位元 eK