抽象代數/群論/同態/同態對映恆等元到恆等元

定理

設f是從群G到群K的同態。

設e_G和e_K是G和K的恆等元。

f(e_G) = e_K

證明

0.   $f({\color {Blue}e_{G}})\in K$	f對映到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是同態
3.   $f({\color {Blue}e_{G}})=f({\color {Blue}e_{G}})\circledast f({\color {Blue}e_{G}})$	恆等元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.   ${\color {OliveGreen}e_{K}}={\color {OliveGreen}e_{K}}\circledast f({\color {Blue}e_{G}})$	恆等元 eK, 逆的定義
6.   ${\color {OliveGreen}e_{K}}=f({\color {Blue}e_{G}})$	單位元 eK