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


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Question Number 193804 by Mastermind last updated on 20/Jun/23
$Ques. 6 Let (G, ∗) be a group. and let C={c∈G : c∗a = a∗c ∀a∈G}. Prove that C is subgroup of G. hence or otherwise show that C is Abelian. [Note C is called the center of group G] Ques. 7 If (G, ∗) is a group such that (a∗b)^2 = a^2 ∗b^2 (multiplicatively) for all a,b∈G. Show that G must be Abelian$
$Ques . 6$ $Let (G, ) be a group . and let$ $C = {c \in G : c a = a * c \forall a \in G} . Prove$ $that C is subgroup of G . hence or$ $otherwise show that C is Abelian .$ $[Note C is called the center of group G]$ $Ques . 7$ $If (G, ) is a group such that {(a b)}^{2}$ $= a^{2} * b^{2} (multiplicatively) for all$ $a, b \in G . Show that G must be Abelian$
	Answered by Rajpurohith last updated on 20/Jun/23

	$(7) I s h a l l o m i t t h e n o t a t i o n * .$ $i . e, B y a b I m e a n a * b .$ $G i v e n t h a t {(a b)}^{2} = a^{2} b^{2} \forall a, b \in G$ $\Rightarrow (a b) (a b) = a a b b \forall a, b \in G$ $\Rightarrow a b a b = a a b b \forall a, b \in G$ $\Rightarrow a^{- 1} (a b a b) b^{- 1} = a^{- 1} (a a b b) b^{- 1} \forall a, b \in G$ $\Rightarrow b a = a b \forall a, b \in G$ $\Rightarrow G i s A b e l i a n . ◼$
	Commented by Mastermind last updated on 20/Jun/23

	$Thank you so much, God will bless you Inshallah$
	Answered by gatocomcirrose last updated on 20/Jun/23

	$Note that e \in C since e * a = a * e, \forall a \in G \Rightarrow C \neq \emptyset$ $Let x, y \in C,$ $\Rightarrow x * a = a * x \Rightarrow a = x^{- 1} * a * x \Rightarrow a * x^{- 1} = x^{- 1} * a, \forall a \in G$ $\Rightarrow x^{- 1} \in G$ $(x * y) * a = x * (y * a) = x * (a * y) = (x * a) * y$ $= (a * x) * y = a * (x * y), \forall a \in G \Rightarrow x * y \in G$ $Then C < G (subgroup) .$ $Let x, y \in C, since C < G ∴ x, y \in G$ $(x \in C) \Rightarrow x * a = a * x, \forall a \in G$ $for a = y \in G \Rightarrow x * y = y * x \Rightarrow C is abelian$
	Commented by Mastermind last updated on 20/Jun/23

	$Thank you my man$
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