Menu Close

Ques-11-Let-H-be-a-family-of-subgroup-of-a-group-G-then-prove-that-H-is-also-a-subgroup-Ques-12-Using-GAP-find-the-elements-A-B-and-C-in-D-5-such-that-AB-BC-but




Question Number 193893 by Mastermind last updated on 22/Jun/23
Ques. 11       Let {H_α } ∈ Ω be a family of subgroup of  a group G then prove that ∩_(α=Ω) H_α  is also a  subgroup    Ques. 12        Using GAP, find the elements A, B and   C in D_5  such that AB = BC but A ≠ C.
$$\mathrm{Ques}.\:\mathrm{11} \\ $$$$\:\:\:\:\:\mathrm{Let}\:\left\{\mathrm{H}_{\alpha} \right\}\:\in\:\Omega\:\mathrm{be}\:\mathrm{a}\:\mathrm{family}\:\mathrm{of}\:\mathrm{subgroup}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{group}\:\mathrm{G}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\underset{\alpha=\Omega} {\cap}\mathrm{H}_{\alpha} \:\mathrm{is}\:\mathrm{also}\:\mathrm{a} \\ $$$$\mathrm{subgroup} \\ $$$$ \\ $$$$\mathrm{Ques}.\:\mathrm{12}\: \\ $$$$\:\:\:\:\:\mathrm{Using}\:\mathrm{GAP},\:\mathrm{find}\:\mathrm{the}\:\mathrm{elements}\:\mathrm{A},\:\mathrm{B}\:\mathrm{and}\: \\ $$$$\mathrm{C}\:\mathrm{in}\:\mathrm{D}_{\mathrm{5}} \:\mathrm{such}\:\mathrm{that}\:\mathrm{AB}\:=\:\mathrm{BC}\:\mathrm{but}\:\mathrm{A}\:\neq\:\mathrm{C}. \\ $$
Answered by Rajpurohith last updated on 22/Jun/23
(11) Let U=∩_α (H_α )  we need to prove that U is subgroup of G.  As e∈H_α    ∀α,  e∈∩_α  H_α   ⇒U≠∅.  let a,b ∈U   ⇒a,b∈H_α      ∀α  since H_α  is a subgroup for each α,  ab^(−1) ∈H_α     ∀α  ⇒ab^(−1) ∈(∩H_α )    ∀α  ⇒ab^(−1) ∈U   ∀a,b∈U  ⇒U=∩_α H_α   is a subgroup of G.
$$\left(\mathrm{11}\right)\:{Let}\:\boldsymbol{{U}}=\cap_{\alpha} \left({H}_{\alpha} \right) \\ $$$${we}\:{need}\:{to}\:{prove}\:{that}\:{U}\:{is}\:{subgroup}\:{of}\:{G}. \\ $$$${As}\:{e}\in{H}_{\alpha} \:\:\:\forall\alpha, \\ $$$${e}\in\cap_{\alpha} \:{H}_{\alpha} \\ $$$$\Rightarrow\boldsymbol{{U}}\neq\emptyset. \\ $$$${let}\:{a},{b}\:\in\boldsymbol{{U}}\: \\ $$$$\Rightarrow{a},{b}\in{H}_{\alpha} \:\:\:\:\:\forall\alpha \\ $$$${since}\:{H}_{\alpha} \:{is}\:{a}\:{subgroup}\:{for}\:{each}\:\alpha, \\ $$$${ab}^{−\mathrm{1}} \in{H}_{\alpha} \:\:\:\:\forall\alpha \\ $$$$\Rightarrow{ab}^{−\mathrm{1}} \in\left(\cap{H}_{\alpha} \right)\:\:\:\:\forall\alpha \\ $$$$\Rightarrow{ab}^{−\mathrm{1}} \in\boldsymbol{{U}}\:\:\:\forall{a},{b}\in\boldsymbol{{U}} \\ $$$$\Rightarrow{U}=\cap_{\alpha} {H}_{\alpha} \:\:{is}\:{a}\:{subgroup}\:{of}\:{G}. \\ $$$$ \\ $$$$ \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *