Question Number 191997 by Mastermind last updated on 05/May/23
$$\mathrm{Show}\:\mathrm{that}\:\mathbb{C}=\left\{−\mathrm{1},\mathrm{1},−\imath,\imath\right\}\:\mathrm{where} \\ $$$$\imath=\sqrt{−\mathrm{1}}\:\mathrm{with}\:\mathrm{addition}\:\mathrm{operation}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{group}. \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$
Commented by Rasheed.Sindhi last updated on 05/May/23
$${addition}\:{is}\:{not}\:{defined}\:{in}\:\left\{−\mathrm{1},\mathrm{1},−{i},{i}\right\}. \\ $$
Commented by Mastermind last updated on 07/May/23
$$\mathrm{Yes}! \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Answered by AST last updated on 05/May/23
$${For}\:{a}\:{group}\:{G}\:{under}\:{operation}\:\ast \\ $$$${By}\:{its}\:{properties},{if}\:{a},{b}\in{G},\:{then}\:{a}\ast{b}\in{G} \\ $$$${For}\:\mathbb{C}=\left\{−\mathrm{1},+\mathrm{1},−{i},{i}\right\}\:{under}\:+,{since}\:\mathrm{0}\:{is}\:{not}\:{in} \\ $$$${the}\:{set},\:\left(\mathbb{C},+\right)\:{cannot}\:{be}\:{a}\:{group}. \\ $$$$\left(\mathbb{C},×\right),{however},\:{is}\:{a}\:{group}. \\ $$
Commented by Mastermind last updated on 07/May/23
$$\mathrm{Wow}! \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much},\:\mathrm{i}\:\mathrm{really}\:\mathrm{appreciate} \\ $$