Question Number 171314 by Mastermind last updated on 12/Jun/22
$${If}\:{C}_{{r}} ,\:{C}_{{s}} \:{are}\:{cyclic}\:{groups}\:{such}\:{that} \\ $$$${g}.{c}.{d}\left({r},{s}\right)=\mathrm{1},\:{then}\:{show}\:{that}\:{C}_{{r}} ×{C}_{{s}} \:{is} \\ $$$${a}\:{cyclic}\:{group}. \\ $$$$ \\ $$$${Mastermind} \\ $$
Answered by mindispower last updated on 12/Jun/22
$${C}_{{r}} \:{cyclic}\:\Rightarrow\exists{a}\in{C}_{{r}} ...{such}\:\forall{g}\in{C}_{{r}} \:\exists{n}\in\left\{\mathrm{0},....{r}−\mathrm{1}\right\} \\ $$$${a}^{{n}} ={g} \\ $$$${C}_{{s}} {cyclic}\:\Rightarrow\exists{b}\in{C}_{{s}} ..{such}\:\forall{g}'\in{C}_{{s}} .\exists{m}\in\left\{\mathrm{0},....{s}−\mathrm{1}\right\} \\ $$$${b}^{{m}} ={g}' \\ $$$${we}\:{use}\:{standar}\:{definition}\:{of}\:{product}\:{of}/{Groups} \\ $$$${card}\:\left({C}_{{r}} \ast{C}_{{s}} \right)\leqslant{rs} \\ $$$$\left({a},{b}\right)\ast.....\left({a},{b}\right)...\left({rs}\right)\:=\left(\left({a}^{{r}} \right)^{{s}} ,\left({b}^{{s}} \right)^{{r}} \right)=\left({e},{e}'\right){Times} \\ $$$$\left({a}^{{x}} ,{b}^{{x}} \right)=\left({e},{e}'\right)\Rightarrow{r}\mid{x},{s}\mid{x}\Rightarrow{rs}\mid{x}...{gcd}\left({r},{s}\right)=\mathrm{1} \\ $$$$\Rightarrow{rs}\mid{card}\:\left({C}_{{r}} \ast{C}_{{s}} \right)\Rightarrow{card}\:\left({C}_{{r}} \ast{C}_{{s}} \right)={rs} \\ $$$${and}\:{C}_{{r}} \ast{C}_{{s}} =<\left({a},{b}\right)\:> \\ $$$$ \\ $$$$ \\ $$
Commented by Mastermind last updated on 14/Jun/22
$${Thanks} \\ $$
Commented by mindispower last updated on 19/Jun/22
$${withe}\:{pleasur} \\ $$