Question Number 193893 by Mastermind last updated on 22/Jun/23
$$\mathrm{Ques}.\:\mathrm{11} \\ $$$$\:\:\:\:\:\mathrm{Let}\:\left\{\mathrm{H}_{\alpha} \right\}\:\in\:\Omega\:\mathrm{be}\:\mathrm{a}\:\mathrm{family}\:\mathrm{of}\:\mathrm{subgroup}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{group}\:\mathrm{G}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\underset{\alpha=\Omega} {\cap}\mathrm{H}_{\alpha} \:\mathrm{is}\:\mathrm{also}\:\mathrm{a} \\ $$$$\mathrm{subgroup} \\ $$$$ \\ $$$$\mathrm{Ques}.\:\mathrm{12}\: \\ $$$$\:\:\:\:\:\mathrm{Using}\:\mathrm{GAP},\:\mathrm{find}\:\mathrm{the}\:\mathrm{elements}\:\mathrm{A},\:\mathrm{B}\:\mathrm{and}\: \\ $$$$\mathrm{C}\:\mathrm{in}\:\mathrm{D}_{\mathrm{5}} \:\mathrm{such}\:\mathrm{that}\:\mathrm{AB}\:=\:\mathrm{BC}\:\mathrm{but}\:\mathrm{A}\:\neq\:\mathrm{C}. \\ $$
Answered by Rajpurohith last updated on 22/Jun/23
$$\left(\mathrm{11}\right)\:{Let}\:\boldsymbol{{U}}=\cap_{\alpha} \left({H}_{\alpha} \right) \\ $$$${we}\:{need}\:{to}\:{prove}\:{that}\:{U}\:{is}\:{subgroup}\:{of}\:{G}. \\ $$$${As}\:{e}\in{H}_{\alpha} \:\:\:\forall\alpha, \\ $$$${e}\in\cap_{\alpha} \:{H}_{\alpha} \\ $$$$\Rightarrow\boldsymbol{{U}}\neq\emptyset. \\ $$$${let}\:{a},{b}\:\in\boldsymbol{{U}}\: \\ $$$$\Rightarrow{a},{b}\in{H}_{\alpha} \:\:\:\:\:\forall\alpha \\ $$$${since}\:{H}_{\alpha} \:{is}\:{a}\:{subgroup}\:{for}\:{each}\:\alpha, \\ $$$${ab}^{−\mathrm{1}} \in{H}_{\alpha} \:\:\:\:\forall\alpha \\ $$$$\Rightarrow{ab}^{−\mathrm{1}} \in\left(\cap{H}_{\alpha} \right)\:\:\:\:\forall\alpha \\ $$$$\Rightarrow{ab}^{−\mathrm{1}} \in\boldsymbol{{U}}\:\:\:\forall{a},{b}\in\boldsymbol{{U}} \\ $$$$\Rightarrow{U}=\cap_{\alpha} {H}_{\alpha} \:\:{is}\:{a}\:{subgroup}\:{of}\:{G}. \\ $$$$ \\ $$$$ \\ $$