Question Number 204019 by mnjuly1970 last updated on 04/Feb/24
$$ \\ $$$$\:\:\:\:\:{G}\:{is}\:{a}\:{group}\:: \\ $$$$\:\:\:\:\:{prove}\:{that}\::\:\:\frac{{G}}{{Z}\:\left({G}\:\right)}\:\cong\:{Inn}\left({G}\:\right) \\ $$$$\:\:\:\:{Where}\:,\:{Inn}\left({G}\right)=\:\left\{{f}\:\mid\:{f}:\:{G}\:\overset{{f}\:{is}\:{an}\:{Automorphism}} {\rightarrow}\:{G}\right\} \\ $$$$ \\ $$
Commented by mokys last updated on 04/Feb/24
Commented by mnjuly1970 last updated on 05/Feb/24
$${thanks}\:{alot}\:{sir}..{so}\:{nice}\:{proof} \\ $$$$\:\cancel{ } \\ $$
Answered by witcher3 last updated on 04/Feb/24
$$\mathrm{G}\rightarrow\mathrm{Inn}\left(\mathrm{G}\right) \\ $$$$\mathrm{x}\overset{\mathrm{f}_{\mathrm{x}} } {\rightarrow}\mathrm{gxg}^{−} \\ $$$$\mathrm{x}\rightarrow\mathrm{g}\rightarrow\mathrm{gxg}^{−} \\ $$$$\mathrm{kerf}_{\mathrm{x}} =\left\{\mathrm{x}\in\mathrm{G}\mid\mathrm{gxg}^{−} =\mathrm{x}\right\}\Leftrightarrow\left\{\mathrm{x}\in\mathrm{G}\mid\mathrm{gx}=\mathrm{xg}\right\}=\mathrm{Z}\left(\mathrm{G}\right) \\ $$$$\mathrm{use}\:\mathrm{isomoprohism}\:\mathrm{Theorem}\Rightarrow\:\mathrm{f}\:\mathrm{surjective} \\ $$$$\mathrm{G}\overset{\mathrm{f}} {\rightarrow}\mathrm{G}'\Rightarrow\frac{\mathrm{G}}{\mathrm{kerf}\left(\mathrm{G}\right)}\approxeq\mathrm{G}' \\ $$$$\frac{\mathrm{G}}{\mathrm{ker}\left(\mathrm{f}\right)}\simeq\mathrm{Im}\left(\mathrm{f}\right)\Rightarrow\frac{\mathrm{G}}{\mathrm{Z}\left(\mathrm{G}\right)}\approxeq\mathrm{Inn}\left(\mathrm{G}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 04/Feb/24
$$\:{so}\:{nice}\:{solution}\:{sir}\:{wicher} \\ $$$${thx}\:{alot} \\ $$$$\:\: \\ $$
Commented by witcher3 last updated on 04/Feb/24
$$\mathrm{withe}\:\mathrm{pleasur}\:\mathrm{sir}\: \\ $$$$\mathrm{have}\:\mathrm{a}\:\mathrm{nice}\:\mathrm{day} \\ $$$$ \\ $$