Ques-5-Prove-that-if-a-b-are-any-elements-of-a-group-G-then-the-equation-y-a-b-has-a-unique-solution-in-G-Ques-6-a-Show-that-the-set-G-of-all-non-zero-complex-numbers-i

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Question Number 193759 by Mastermind last updated on 19/Jun/23

$$\mathrm{Ques}.5$$$$\mathrm{Prove}\mathrm{that}\mathrm{if}\mathrm{a},\mathrm{b}\mathrm{are}\mathrm{any}\mathrm{elements}\mathrm{of}\mathrm{a}$$$$\mathrm{group}(\mathrm{G},\ast ),\mathrm{then}\mathrm{the}\mathrm{equation}\mathrm{y}\ast \mathrm{a}=\mathrm{b}$$$$\mathrm{has}\mathrm{a}\mathrm{unique}\mathrm{solution}\mathrm{in}(\mathrm{G},\ast ).$$$$$$$$\mathrm{Ques}.6$$$$\mathrm{a})\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{set}\mathrm{G}\mathrm{of}\mathrm{all}\mathrm{non}-\mathrm{zero}$$$$\mathrm{complex}\mathrm{numbers},\mathrm{is}\mathrm{a}\mathrm{group}\mathrm{under}$$$$\mathrm{multiplication}\mathrm{of}\mathrm{complex}\mathrm{numbers}.$$$$$$$$\mathrm{b})\mathrm{Show}\mathrm{that}\mathrm{H}=\{\mathrm{a}\in \mathrm{G}:{\mathrm{a}}_1^2+{\mathrm{a}}_2^2=1\},$$$$\mathrm{where}{\mathrm{a}}_1=\mathrm{Re}\mathrm{a}\mathrm{and}{\mathrm{a}}_2=\mathrm{Im}\mathrm{a}\mathrm{is}\mathrm{a}$$$$\mathrm{subgroup}\mathrm{of}\mathrm{G}.$$

Answered by aleks041103 last updated on 19/Jun/23

$$$$$$Q.5$$$$(G,\ast )isgroup\Rightarrow \mathrm{\forall }a\in G,\mathrm{\exists }{a}^{-1}\in G:{aa}^{-1}=a^{-1}a=e$$$$\Rightarrow y\ast a=b\Rightarrow y\ast a\ast a^{-1}=b\ast a^{-1}\Rightarrow y=b\ast a^{-1}$$$$whichisobviouslyunique.$$$$youcanproceedalsoasfollows:$$$$suppose{y}_1\ne y_2\in G,s.t.$$$$y_1\ast a=y_2\ast a=b$$$$since\mathrm{\exists }{a}^{-1}\Rightarrow y_1\ast a\ast a^{-1}=y_2\ast a\ast a^{-1}=b\ast a^{-1}$$$$\Rightarrow y_1=y_2\Rightarrow contradiction$$$$\Rightarrow yisunique$$

Commented by Mastermind last updated on 20/Jun/23

$$\mathrm{Thank}\mathrm{you}\mathrm{my}\mathrm{boss}$$

Answered by Rajpurohith last updated on 19/Jun/23

$$\mathit{S}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}:-Leta,b\in G$$$$\left(5\right)Supposetherearetwosolutionssay{y}_1\mathrm{\&}{y}_2$$$$oftheequationy\ast a=b$$$$\Rightarrow y_1\ast a=b\mathrm{\&}{y}_2\ast a=b$$$$\Rightarrow y_1\ast a=y_2\ast a$$$$sinceaisinagroup,a^{-1}exists.$$$$\Rightarrow (y_1\ast a)\ast a^{-1}=(y_2\ast a)\ast a^{-1}$$$$\Rightarrow Byassociativity{y}_1=y_{2}andwearedone.\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fc.svg">}$$$$\left(6\right)\left(\mathbf{a}\right)say(G,.)=(\mathbb{C}-\left\{0\right\},.)where{}^{\prime }{.}^{\prime }ismultiplication$$$$ofcomplexnumberswhichisclearlyabinaryoperationon\mathbb{C}.$$$$\left(i\right)Foranyz\in \mathbb{C}-\left\{0\right\},z_1\in \mathbb{C}-\left\{0\right\}besuchthatz.z_1=z_1.z=z$$$$weget{z}_1=1whichistheidentity.$$$$\left(ii\right)Itisobviousthatassociativityholds.$$$$\left(iii\right)foreachz\in \mathbb{C}-\left\{0\right\},z.\left(\frac{1}{z}\right)=\left(\frac{1}{z}\right).z=1$$$$hencetheinverseofanelementzequals\left(\frac{1}{z}\right)$$$$Hence(\mathbb{C}-\left\{0\right\},.)formsagroup.\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fc.svg">}$$$$\left(\mathbf{b}\right)AspergivenH=\{z\in \mathbb{C}:\mid z\mid =1\}$$$$clearlyH\subset \mathbb{C}-\left\{0\right\}.$$$$Weshallusetwostepsubgrouptest$$$$letz,w\in H.Whetherz.w^{-1}\in H?$$$$\mid z.w^{-1}\mid =\mid z\mid .\mid w^{-1}\mid =\mid z\mid .\frac{1}{\mid w\mid }=1(Why?)$$$$HenceHisasubgroupof(\mathbb{C}-\left\{0\right\},.\}\mathrm{<img\; draggable="false"\; role="img"\; class="emoji"\; alt="◼"\; src="https://s.w.org/images/core/emoji/15.0.3/svg/25fc.svg">}$$$$$$

Commented by Mastermind last updated on 20/Jun/23

$$\mathrm{I}\mathrm{do}\mathrm{really}\mathrm{appreciate}\mathrm{it}$$

Commented by Rajpurohith last updated on 21/Jun/23

$$PleasedopostGrouptheoryquestions,$$$$I^{\prime }mtakingacourseagaininAbstractalgebra,$$$$Ithelpsmetoo.$$

Commented by Mastermind last updated on 22/Jun/23

$$$$$$\mathrm{Wow}!$$$${\mathrm{that}}^{\prime }\mathrm{s}\mathrm{good},{\mathrm{i}}^{\prime }\mathrm{ll}\mathrm{be}\mathrm{posting}\mathrm{it}$$

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