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, is a group under multiplication of complex numbers. b) Show that H={a∈G : a_1 ^2 + a_2 ^2 = 1}, where a_1 = Re a and a


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Question Number 193759 by Mastermind last updated on 19/Jun/23
$Ques. 5 Prove that if a,b are any elements of a group (G, ∗), then the equation y∗a=b has a unique solution in (G, ∗). Ques. 6 a) Show that the set G of all non-zero complex numbers, is a group under multiplication of complex numbers. b) Show that H={a∈G : a_1 ^2 + a_2 ^2 = 1}, where a_1 = Re a and a_2 = Im a is a subgroup of G.$
$Ques . 5$ $Prove that if a, b are any elements of a$ $group (G, ), then the equation y a = b$ $has a unique solution in (G, *) .$ $Ques . 6$ $a) Show that the set G of all non - zero$ $complex numbers, is a group under$ $multiplication of complex numbers .$ $b) Show that H = {a \in G : a_{1}^{2} + a_{2}^{2} = 1},$ $where a_{1} = Re a and a_{2} = Im a is a$ $subgroup of G .$
	Answered by aleks041103 last updated on 19/Jun/23

	$Q . 5$ $(G, ) i s g r o u p \Rightarrow \forall a \in G, \exists a^{- 1} \in G : {a a}^{- 1} = a^{- 1} a = e$ $\Rightarrow y a = b \Rightarrow y * a * a^{- 1} = b * a^{- 1} \Rightarrow y = b * a^{- 1}$ $w h i c h i s o b v i o u s l y u n i q u e .$ $y o u c a n p r o c e e d a l s o a s f o l l o w s :$ $s u p p o s e y_{1} \neq y_{2} \in G, s . t .$ $y_{1} * a = y_{2} * a = b$ $s i n c e \exists a^{- 1} \Rightarrow y_{1} * a * a^{- 1} = y_{2} * a * a^{- 1} = b * a^{- 1}$ $\Rightarrow y_{1} = y_{2} \Rightarrow c o n t r a d i c t i o n$ $\Rightarrow y i s u n i q u e$
	Commented by Mastermind last updated on 20/Jun/23

	$Thank you my boss$
	Answered by Rajpurohith last updated on 19/Jun/23
	$Solution:− Let a,b∈G (5)Suppose there are two solutions say y_1 & y_2 of the equation y∗a=b ⇒y_1 ∗a=b & y_2 ∗a=b ⇒y_1 ∗a=y_2 ∗a since a is in a group, a^(−1) exists. ⇒(y_1 ∗a)∗a^(−1) =(y_2 ∗a)∗a^(−1) ⇒By associativity y_1 =y_2 and we are done. ■ (6) (a) say (G, .)=(C−{0}, .) where ′.′ is multiplication of complex numbers which is clearly a binary operation on C. (i)For any z∈C−{0} , z_1 ∈C−{0} be such that z.z_1 =z_1 .z=z we get z_1 =1 which is the identity. (ii)It is obvious that associativity holds. (iii) for each z∈C−{0},z.((1/z))=((1/z)).z=1 hence the inverse of an element z equals ((1/z)) Hence (C−{0}, . ) forms a group. ■ (b) As per given H={z∈C : ∣z∣=1} clearly H⊂C−{0}. We shall use two step subgroup test let z,w∈H . Whether z.w^(−1) ∈H? ∣z.w^(−1) ∣=∣z∣.∣w^(−1) ∣=∣z∣.(1/(∣w∣))=1 (Why?) Hence H is a subgroup of (C−{0}, .} ■$
	$S o l u t i o n : - L e t a, b \in G$ $(5) S u p p o s e t h e r e a r e t w o s o l u t i o n s s a y y_{1} & y_{2}$ $o f t h e e q u a t i o n y * a = b$ $\Rightarrow y_{1} * a = b & y_{2} * a = b$ $\Rightarrow y_{1} * a = y_{2} * a$ $s i n c e a i s i n a g r o u p, a^{- 1} e x i s t s .$ $\Rightarrow (y_{1} * a) * a^{- 1} = (y_{2} * a) * a^{- 1}$ $\Rightarrow B y a s s o c i a t i v i t y y_{1} = y_{2} a n d w e a r e d o n e . ◼$ $(6) (a) s a y (G, .) = (C - {0}, .) w h e r e^{'} .^{'} i s m u l t i p l i c a t i o n$ $o f c o m p l e x n u m b e r s w h i c h i s c l e a r l y a b i n a r y o p e r a t i o n o n C .$ $(i) F o r a n y z \in C - {0}, z_{1} \in C - {0} b e s u c h t h a t z . z_{1} = z_{1} . z = z$ $w e g e t z_{1} = 1 w h i c h i s t h e i d e n t i t y .$ $(i i) I t i s o b v i o u s t h a t a s s o c i a t i v i t y h o l d s .$ $(i i i) f o r e a c h z \in C - {0}, z . (\frac{1}{z}) = (\frac{1}{z}) . z = 1$ $h e n c e t h e i n v e r s e o f a n e l e m e n t z e q u a l s (\frac{1}{z})$ $H e n c e (C - {0}, .) f o r m s a g r o u p . ◼$ $(b) A s p e r g i v e n H = {z \in C : ∣ z ∣= 1}$ $c l e a r l y H \subset C - {0} .$ $W e s h a l l u s e t w o s t e p s u b g r o u p t e s t$ $l e t z, w \in H . W h e t h e r z . w^{- 1} \in H ?$ $∣ z . w^{- 1} ∣=∣ z ∣ . ∣ w^{- 1} ∣=∣ z ∣ . \frac{1}{∣ w ∣} = 1 (W h y ?)$ $H e n c e H i s a s u b g r o u p o f (C - {0}, .} ◼$
	Commented by Mastermind last updated on 20/Jun/23

	$I do really appreciate it$
	Commented by Rajpurohith last updated on 21/Jun/23

	$P l e a s e d o p o s t G r o u p t h e o r y q u e s t i o n s,$ $I^{'} m t a k i n g a c o u r s e a g a i n i n A b s t r a c t a l g e b r a,$ $I t h e l p s m e t o o .$
	Commented by Mastermind last updated on 22/Jun/23

	$Wow!$ ${that}^{'} s good, i^{'} ll be posting it$
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