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Ques-1-Let-G-be-a-group-and-b-a-fixed-element-of-G-Prove-that-the-map-G-into-G-given-by-x-bx-is-bijective-Ques-2-Let-G-be-a-group-and-g-be-an-element-of-G-Prove-that-a-g-1-1

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Question Number 193711 by Mastermind last updated on 18/Jun/23
Ques. 1 Let G be a group and b a fixed element of G. Prove that the map G into G given by x→bx is bijective Ques. 2 Let G be a group and g be an element of G. Prove that a) (g^(−1) )^(−1) =g b) g^m g^n = g^(m+n) Help!
$$\mathrm{Ques}.\:\mathrm{1} \\ $$$$\:\:\:\:\:\mathrm{Let}\:\mathrm{G}\:\mathrm{be}\:\mathrm{a}\:\mathrm{group}\:\mathrm{and}\:\mathrm{b}\:\mathrm{a}\:\mathrm{fixed}\:\mathrm{element} \\ $$$$\mathrm{of}\:\mathrm{G}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{map}\:\mathrm{G}\:\mathrm{into}\:\mathrm{G}\:\mathrm{given} \\ $$$$\mathrm{by}\:\mathrm{x}\rightarrow\mathrm{bx}\:\mathrm{is}\:\mathrm{bijective} \\ $$$$ \\ $$$$\mathrm{Ques}.\:\mathrm{2}\: \\ $$$$\:\:\:\:\:\mathrm{Let}\:\mathrm{G}\:\mathrm{be}\:\mathrm{a}\:\mathrm{group}\:\mathrm{and}\:\mathrm{g}\:\mathrm{be}\:\mathrm{an}\:\mathrm{element}\: \\ $$$$\mathrm{of}\:\mathrm{G}.\:\mathrm{Prove}\:\mathrm{that}\: \\ $$$$\left.\mathrm{a}\right)\:\left(\mathrm{g}^{−\mathrm{1}} \right)^{−\mathrm{1}} =\mathrm{g} \\ $$$$\left.\mathrm{b}\right)\:\mathrm{g}^{\mathrm{m}} \mathrm{g}^{\mathrm{n}} \:=\:\mathrm{g}^{\mathrm{m}+\mathrm{n}} \: \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$
Answered by Rajpurohith last updated on 19/Jun/23
(1) G be a group , b∈G be fixed. define f :G → G by f(x)=bx ∀x∈G clearly f is well defined. suppose f(c)=f(d) for c,d∈G ⇒bc=bd , by cancellation property c=d. so f is an injective function from G to itself. let y∈G so f(b^(−1) y)=bb^(−1) y=y Hence f is surjective. Thus f is bijective. ■ (2) (a)let G∈g say h=g^(−1) ⇒h^(−1) =(g^(−1) )^(−1) ⇒hg=e ⇒h^(−1) (hg)=h^(−1) ⇒(h^(−1) h)g=h^(−1) ⇒g=h^(−1) =(g^(−1) )^(−1) ∴ (g^(−1) )^(−1) =g ■ (b)g^m .g^n =(g.g.g...g)_(m times) (g.g.g...g)_(n times) =(g.g.g...g)_(m+n times) =g^(m+n) ■
$$\left(\mathrm{1}\right)\:{G}\:{be}\:{a}\:{group}\:,\:{b}\in{G}\:{be}\:{fixed}. \\ $$$${define}\:\boldsymbol{{f}}\::{G}\:\rightarrow\:{G}\:{by}\:{f}\left({x}\right)={bx}\:\:\forall{x}\in{G} \\ $$$${clearly}\:\boldsymbol{{f}}\:\:{is}\:{well}\:{defined}. \\ $$$${suppose}\:{f}\left({c}\right)={f}\left({d}\right)\:{for}\:{c},{d}\in{G} \\ $$$$\Rightarrow{bc}={bd}\:,\:{by}\:{cancellation}\:{property}\:{c}={d}. \\ $$$${so}\:{f}\:{is}\:{an}\:{injective}\:{function}\:{from}\:{G}\:{to}\:{itself}. \\ $$$${let}\:{y}\in{G}\:{so}\:{f}\left({b}^{−\mathrm{1}} {y}\right)={bb}^{−\mathrm{1}} {y}={y} \\ $$$${Hence}\:{f}\:{is}\:{surjective}. \\ $$$${Thus}\:{f}\:{is}\:{bijective}.\:\:\:\:\:\:\:\:\:\:\:\blacksquare \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\left(\boldsymbol{{a}}\right){let}\:{G}\in{g}\:{say}\:\:{h}={g}^{−\mathrm{1}} \:\:\Rightarrow{h}^{−\mathrm{1}} =\left({g}^{−\mathrm{1}} \right)^{−\mathrm{1}} \\ $$$$\Rightarrow{hg}={e}\: \\ $$$$\Rightarrow{h}^{−\mathrm{1}} \left({hg}\right)={h}^{−\mathrm{1}} \\ $$$$\Rightarrow\left({h}^{−\mathrm{1}} {h}\right){g}={h}^{−\mathrm{1}} \\ $$$$\Rightarrow{g}={h}^{−\mathrm{1}} =\left({g}^{−\mathrm{1}} \right)^{−\mathrm{1}} \:\:\:\therefore\:\left({g}^{−\mathrm{1}} \right)^{−\mathrm{1}} ={g}\:\:\:\:\:\blacksquare \\ $$$$\left(\boldsymbol{{b}}\right){g}^{{m}} .{g}^{{n}} =\left({g}.{g}.{g}…{g}\right)_{{m}\:{times}} \:\left({g}.{g}.{g}…{g}\right)_{{n}\:{times}} \\ $$$$=\left({g}.{g}.{g}…{g}\right)_{{m}+{n}\:{times}} ={g}^{{m}+{n}} \:\:\:\blacksquare \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Mastermind last updated on 20/Jun/23
Thank you so much
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

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