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Ques-6-Let-G-be-a-group-and-let-C-c-G-c-a-a-c-a-G-Prove-that-C-is-subgroup-of-G-hence-or-otherwise-show-that-C-is-Abelian-Note-C-is-called-the-center-of-group-G-Ques-7-




Question Number 193804 by Mastermind last updated on 20/Jun/23
Ques. 6        Let (G, ∗) be a group. and let  C={c∈G : c∗a = a∗c ∀a∈G}. Prove  that C is subgroup of G. hence or   otherwise show that C is Abelian.    [Note C is called the center of group G]    Ques. 7       If (G, ∗) is a group such that (a∗b)^2    = a^2 ∗b^2  (multiplicatively) for all   a,b∈G. Show that G must be Abelian
$$\mathrm{Ques}.\:\mathrm{6}\: \\ $$$$\:\:\:\:\:\mathrm{Let}\:\left(\mathrm{G},\:\ast\right)\:\mathrm{be}\:\mathrm{a}\:\mathrm{group}.\:\mathrm{and}\:\mathrm{let} \\ $$$$\mathrm{C}=\left\{\mathrm{c}\in\mathrm{G}\::\:\mathrm{c}\ast\mathrm{a}\:=\:\mathrm{a}\ast\mathrm{c}\:\forall\mathrm{a}\in\mathrm{G}\right\}.\:\mathrm{Prove} \\ $$$$\mathrm{that}\:\mathrm{C}\:\mathrm{is}\:\mathrm{subgroup}\:\mathrm{of}\:\mathrm{G}.\:\mathrm{hence}\:\mathrm{or}\: \\ $$$$\mathrm{otherwise}\:\mathrm{show}\:\mathrm{that}\:\mathrm{C}\:\mathrm{is}\:\mathrm{Abelian}. \\ $$$$ \\ $$$$\left[\mathrm{Note}\:\mathrm{C}\:\mathrm{is}\:\mathrm{called}\:\mathrm{the}\:\mathrm{center}\:\mathrm{of}\:\mathrm{group}\:\mathrm{G}\right] \\ $$$$ \\ $$$$\mathrm{Ques}.\:\mathrm{7} \\ $$$$\:\:\:\:\:\mathrm{If}\:\left(\mathrm{G},\:\ast\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{group}\:\mathrm{such}\:\mathrm{that}\:\left(\mathrm{a}\ast\mathrm{b}\right)^{\mathrm{2}} \: \\ $$$$=\:\mathrm{a}^{\mathrm{2}} \ast\mathrm{b}^{\mathrm{2}} \:\left(\mathrm{multiplicatively}\right)\:\mathrm{for}\:\mathrm{all}\: \\ $$$$\mathrm{a},\mathrm{b}\in\mathrm{G}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{G}\:\mathrm{must}\:\mathrm{be}\:\mathrm{Abelian} \\ $$
Answered by Rajpurohith last updated on 20/Jun/23
  (7)I shall omit the notation ∗.  i.e, By ab I mean a∗b.  Given that (ab)^2 =a^2 b^2       ∀a,b∈G  ⇒(ab)(ab)=aabb                  ∀a,b∈G  ⇒abab=aabb                         ∀a,b∈G  ⇒a^(−1) (abab)b^(−1) =a^(−1) (aabb)b^(−1)    ∀a,b∈G  ⇒ba=ab    ∀ a,b∈G  ⇒G is Abelian.      ■
$$ \\ $$$$\left(\mathrm{7}\right){I}\:{shall}\:{omit}\:{the}\:{notation}\:\ast. \\ $$$${i}.{e},\:{By}\:\boldsymbol{{ab}}\:{I}\:{mean}\:\boldsymbol{{a}}\ast\boldsymbol{{b}}. \\ $$$${Given}\:{that}\:\left({ab}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} \:\:\:\:\:\:\forall{a},{b}\in{G} \\ $$$$\Rightarrow\left({ab}\right)\left({ab}\right)={aabb}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\forall{a},{b}\in{G} \\ $$$$\Rightarrow{abab}={aabb}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\forall{a},{b}\in{G} \\ $$$$\Rightarrow{a}^{−\mathrm{1}} \left({abab}\right){b}^{−\mathrm{1}} ={a}^{−\mathrm{1}} \left({aabb}\right){b}^{−\mathrm{1}} \:\:\:\forall{a},{b}\in{G} \\ $$$$\Rightarrow{ba}={ab}\:\:\:\:\forall\:{a},{b}\in{G} \\ $$$$\Rightarrow{G}\:{is}\:{Abelian}.\:\:\:\:\:\:\blacksquare \\ $$$$ \\ $$$$ \\ $$
Commented by Mastermind last updated on 20/Jun/23
Thank you so much, God will bless you Inshallah
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much},\:\mathrm{God}\:\mathrm{will}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{Inshallah} \\ $$
Answered by gatocomcirrose last updated on 20/Jun/23
Note that e∈C since e∗a=a∗e, ∀a∈G⇒C≠∅  Let x, y∈C,  ⇒x∗a=a∗x⇒a=x^(−1) ∗a∗x⇒a∗x^(−1) =x^(−1) ∗a, ∀a∈G  ⇒x^(−1) ∈G  (x∗y)∗a=x∗(y∗a)=x∗(a∗y)=(x∗a)∗y  =(a∗x)∗y=a∗(x∗y), ∀a∈G⇒x∗y ∈G    Then C<G (subgroup).    Let x,y∈C, since C<G∴x,y∈G  (x∈C)⇒x∗a=a∗x, ∀a∈G  for a=y∈G⇒x∗y=y∗x⇒C is abelian
$$\mathrm{Note}\:\mathrm{that}\:\mathrm{e}\in\mathrm{C}\:\mathrm{since}\:\mathrm{e}\ast\mathrm{a}=\mathrm{a}\ast\mathrm{e},\:\forall\mathrm{a}\in\mathrm{G}\Rightarrow\mathrm{C}\neq\varnothing \\ $$$$\mathrm{Let}\:\mathrm{x},\:\mathrm{y}\in\mathrm{C}, \\ $$$$\Rightarrow\mathrm{x}\ast\mathrm{a}=\mathrm{a}\ast\mathrm{x}\Rightarrow\mathrm{a}=\mathrm{x}^{−\mathrm{1}} \ast\mathrm{a}\ast\mathrm{x}\Rightarrow\mathrm{a}\ast\mathrm{x}^{−\mathrm{1}} =\mathrm{x}^{−\mathrm{1}} \ast\mathrm{a},\:\forall\mathrm{a}\in\mathrm{G} \\ $$$$\Rightarrow\mathrm{x}^{−\mathrm{1}} \in\mathrm{G} \\ $$$$\left(\mathrm{x}\ast\mathrm{y}\right)\ast\mathrm{a}=\mathrm{x}\ast\left(\mathrm{y}\ast\mathrm{a}\right)=\mathrm{x}\ast\left(\mathrm{a}\ast\mathrm{y}\right)=\left(\mathrm{x}\ast\mathrm{a}\right)\ast\mathrm{y} \\ $$$$=\left(\mathrm{a}\ast\mathrm{x}\right)\ast\mathrm{y}=\mathrm{a}\ast\left(\mathrm{x}\ast\mathrm{y}\right),\:\forall\mathrm{a}\in\mathrm{G}\Rightarrow\mathrm{x}\ast\mathrm{y}\:\in\mathrm{G} \\ $$$$ \\ $$$$\mathrm{Then}\:\mathrm{C}<\mathrm{G}\:\left(\mathrm{subgroup}\right). \\ $$$$ \\ $$$$\mathrm{Let}\:\mathrm{x},\mathrm{y}\in\mathrm{C},\:\mathrm{since}\:\mathrm{C}<\mathrm{G}\therefore\mathrm{x},\mathrm{y}\in\mathrm{G} \\ $$$$\left(\mathrm{x}\in\mathrm{C}\right)\Rightarrow\mathrm{x}\ast\mathrm{a}=\mathrm{a}\ast\mathrm{x},\:\forall\mathrm{a}\in\mathrm{G} \\ $$$$\mathrm{for}\:\mathrm{a}=\mathrm{y}\in\mathrm{G}\Rightarrow\mathrm{x}\ast\mathrm{y}=\mathrm{y}\ast\mathrm{x}\Rightarrow\mathrm{C}\:\mathrm{is}\:\mathrm{abelian} \\ $$
Commented by Mastermind last updated on 20/Jun/23
Thank you my man
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{man} \\ $$

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