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Question Number 179955 by Shrinava last updated on 04/Nov/22

	Answered by aleks041103 last updated on 04/Nov/22
	$(K,+)≇(K/{O_2 }, ∙)⇒ ∄s:K→K/{O_2 }:s(x+y)=s(x)∙s(y)∧s−bijective denote ((a,b),((−b),a) ) :=(a,b) then (a,b)+(c,d)=(a+c,b+d) (a,b)∙(c,d)= ((a,b),((−b),a) ) ∙ ((c,d),((−d),c) ) = = (((ac−bd),(ad+bc)),((−(ad+bc)),(ac−bd)) ) = (ac−bd,ad+bc) But for complex numbers we have (a+bi)+(c+di)=(a+c)+(b+d)i and (a+bi)(c+di)=(ac−bd)+i(ad+bc) ⇒obviously(if this is not obvious, I′ll clarify − just ask) (K,+)≅(C,+) and (K/{O_2 },∙)≅(C/{0},∙) Therefore to prove that (K,+)≇(K/{O_2 }, ∙) is equivalent to proving that (C,+)≇(C/{0}, ∙). suppose (C,+)≅(C/{0}, ∙). then ∃s:C→C/{0}:s(z+w)=s(z)s(w) and ∃s^(−1) :C/{0}→C:s(s^(−1) (z))=z and s^(−1) (s(z))=z but if s(z+w)=s(z)s(w)⇒s(z)=v^z , for some v∈C/{0} ⇒v=re^(2πif) , r∈R^+ and f∈[0,1) ⇒s(a+bi)=r^a e^(−2πbf) (r^(ib) e^(2πifa) )=e^(a ln(r)−2πbf) e^(i(2πfa+b ln(r))) now we need to find (a,b) for given s(a+bi)=e^u e^(i(p+2kπ)) , where u∈R^+ and p∈[0,2π). ⇒ { ((aln(r)−2πbf=u)),((bln(r)+2πaf=p+2kπ)) :} i.e. (((ln(r)),(−2πf)),((2πf),(ln(r))) ) ((a),(b) ) = ((u),((p+2kπ)) ) but det( (((ln(r)),(−2πf)),((2πf),(ln(r))) ))=(ln(r))^2 +4π^2 f^2 >0 ⇒we can always invert the matrix and find (a,b) for ∀k∈Z. ⇒if s:C→C/{0} : s(z+w)=s(z)s(w) then for ∀w∈C/{0}, ∃z_1 ≠z_2 (z_(1,2) ∈C):s(z_1 )=s(z_2 ) ⇒s is not bijective ⇒ every homomorphism from (C,+) to (C/{0},∙) is not bijective. ⇒(K,+)≅(C,+) ≇(C/{0},∙)≅(K/{O_2 },∙) ⇒(K,+)≇(K/{O_2 },∙) Q.E.D$
	$(K, +) ≇ (K / {O_{2}}, \cdot) \Rightarrow$ $∄ s : K \to K / {O_{2}} : s (x + y) = s (x) \cdot s (y) \land s - b i j e c t i v e$ $d e n o t e (\begin{matrix} a & b \\ - b & a \end{matrix}) := (a, b)$ $t h e n (a, b) + (c, d) = (a + c, b + d)$ $(a, b) \cdot (c, d) = (\begin{matrix} a & b \\ - b & a \end{matrix}) \cdot (\begin{matrix} c & d \\ - d & c \end{matrix}) =$ $= (\begin{matrix} a c - b d & a d + b c \\ - (a d + b c) & a c - b d \end{matrix}) = (a c - b d, a d + b c)$ $B u t f o r c o m p l e x n u m b e r s w e h a v e$ $(a + b i) + (c + d i) = (a + c) + (b + d) i$ $a n d$ $(a + b i) (c + d i) = (a c - b d) + i (a d + b c)$ $\Rightarrow o b v i o u s l y (i f t h i s i s n o t o b v i o u s, I^{'} l l c l a r i f y - j u s t a s k)$ $(K, +) ≅ (C, +) a n d (K / {O_{2}}, \cdot) ≅ (C / {0}, \cdot)$ $T h e r e f o r e t o p r o v e t h a t (K, +) ≇ (K / {O_{2}}, \cdot)$ $i s e q u i v a l e n t t o p r o v i n g t h a t$ $(C, +) ≇ (C / {0}, \cdot) .$ $s u p p o s e (C, +) ≅ (C / {0}, \cdot) .$ $t h e n \exists s : C \to C / {0} : s (z + w) = s (z) s (w)$ $a n d \exists s^{- 1} : C / {0} \to C : s (s^{- 1} (z)) = z a n d s^{- 1} (s (z)) = z$ $b u t i f s (z + w) = s (z) s (w) \Rightarrow s (z) = v^{z}, f o r s o m e v \in C / {0}$ $\Rightarrow v = {r e}^{2 π i f}, r \in R^{+} a n d f \in [0, 1)$ $\Rightarrow s (a + b i) = r^{a} e^{- 2 π b f} (r^{i b} e^{2 π i f a}) = e^{a l n (r) - 2 π b f} e^{i (2 π f a + b l n (r))}$ $n o w w e n e e d t o f i n d (a, b) f o r g i v e n s (a + b i) = e^{u} e^{i (p + 2 k π)},$ $w h e r e u \in R^{+} a n d p \in [0, 2 π) .$ $\Rightarrow$ ${\begin{cases} a l n (r) - 2 π b f = u \\ b l n (r) + 2 π a f = p + 2 k π \end{cases}$ $i . e .$ $(\begin{matrix} l n (r) & - 2 π f \\ 2 π f & l n (r) \end{matrix}) (\begin{matrix} a \\ b \end{matrix}) = (\begin{matrix} u \\ p + 2 k π \end{matrix})$ $b u t d e t ((\begin{matrix} l n (r) & - 2 π f \\ 2 π f & l n (r) \end{matrix})) = {(l n (r))}^{2} + 4 π^{2} f^{2} > 0$ $\Rightarrow w e c a n a l w a y s i n v e r t t h e m a t r i x a n d f i n d$ $(a, b) f o r \forall k \in Z .$ $\Rightarrow i f s : C \to C / {0} : s (z + w) = s (z) s (w)$ $t h e n f o r \forall w \in C / {0}, \exists z_{1} \neq z_{2} (z_{1, 2} \in C) : s (z_{1}) = s (z_{2})$ $\Rightarrow s i s n o t b i j e c t i v e$ $\Rightarrow e v e r y h o m o m o r p h i s m f r o m (C, +) t o$ $(C / {0}, \cdot) i s n o t b i j e c t i v e .$ $\Rightarrow (K, +) ≅ (C, +) ≇ (C / {0}, \cdot) ≅ (K / {O_{2}}, \cdot)$ $\Rightarrow (K, +) ≇ (K / {O_{2}}, \cdot)$ $Q . E . D$
	Commented by Shrinava last updated on 05/Nov/22

	$cool dear professor thank you so much$
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