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Let $G=(-1,1)$ and the binary operation such that $x*y=\frac{x\sqrt{1-y^2}+y\sqrt{1-x^2}}{\sqrt{1-(xy)^2+2xy\sqrt{(1-x^2)(1-y^2)}}}$. Prove that (G,*) it is a isomorph group with the group (R,+). I tried to find the bijective function $f:G\to R$ such that $f(x*y)=f(x)+f(y)$ but I could not.

Wang Kah Lun
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  • Parenthesis were not so clear for the square roots, I tried to fix them, could you check if that is what you meant ? If not you can use {} together with sqrt to do square roots. – P. Quinton Dec 22 '20 at 12:55
  • Yes it is what I meant . Thank you! –  Dec 22 '20 at 12:59
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    I'm not sure, but think that this relation may be useful: https://math.stackexchange.com/questions/672575/proof-for-the-formula-of-sum-of-arcsine-functions-arcsin-x-arcsin-y – Luca.b Dec 22 '20 at 13:41
  • It could be if it would not be for the denominator. –  Dec 22 '20 at 14:44

1 Answers1

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Let $x=\sin\alpha$, $y=\sin\beta$, so $\alpha,\beta\in(-\pi/2,\pi/2)$. (always do such substitution when there is a $\sqrt{1-x^2}$)

Then $x*y = \frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\sqrt{1 - \sin^2\alpha\sin^2\beta + 2\sin\alpha\sin\beta\cos\alpha\cos\beta}} = \frac{\sin(\alpha+\beta)}{\sqrt{\cos^2\alpha\cos^2\beta + \sin^2(\alpha+\beta)}} = \frac{t}{\sqrt{1+t^2}} = \tanh\mathrm{asinh}\,t = g(t) $ where $t = \frac{\sin(\alpha+\beta)}{\cos\alpha\cos\beta} = \tan\alpha+\tan\beta = \frac{x}{\sqrt{1-x^2}} + \frac{y}{\sqrt{1-y^2}} = f(x)+f(y) $.

(I have replaced $1$ in the denominator by $(\sin^2\alpha+\cos^2\alpha)(\sin^2\beta+\cos^2\beta)$ because we need something symmetric with respect to $\alpha\leftrightarrow\beta$, and we need to get rid of that complicated term with the factor $2$; we did it by including it into $\sin^2(\alpha+\beta)$.)

So $x*y=g(f(x)+f(y))$, and since in fact $g(f(x))=x$, $f$ and $g$ indeed give the desired isomorphism.

colt_browning
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  • I am sorry @colt_browning but I do not understand who is g(t)? –  Dec 22 '20 at 16:05
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    @iuli23 Consider the equalities "$...=g(t)$" and "$...=f(x)+f(y)$" as the definitions of $g(t)$ and $f(x)$. Sorry for not being clear about it in the answer. – colt_browning Dec 22 '20 at 16:10
  • $g(t) = t/\sqrt{1+t^2}$. $f$ is the mapping from $G$ to $\mathbb{R}$, $g$ is the inverse mapping. – colt_browning Dec 22 '20 at 16:17
  • Alright so f is the isomorphism I am looking for,right? @colt_browning. And nice again I am really sorry –  Dec 22 '20 at 16:20
  • @iuli23 Well, to show that two groups $(G,)$ and $(\mathbb{R},+)$ are isomorphic, any of the group-operation-preserving bijections $f:G\to\mathbb{R}$ and $g:\mathbb{R}\to G$ suffices because $(G,)\cong (\mathbb{R},+)\Leftrightarrow(\mathbb{R},+)\cong(G,*)$. – colt_browning Dec 23 '20 at 08:50