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I am given $((0,1),*)$ Where $x,y\in (0,1)$ and $*$ is defined as $x*y=\frac{xy}{1-x-y+2xy}$ How should I go about finding the inverse and identity elements?

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The title does not reflect the question. – lhf Mar 20 '12 at 18:15
Once you prove that this operation defines a group structure, then it'll be clearly abelian because the expression is symmetric in $x$ and $y$. – lhf Mar 20 '12 at 18:16
up vote 3 down vote accepted

Well, to find the identity, you need to find an element $y$ such that for every $x$, $$x*y = \frac{xy}{1-x-y+2xy} = x.$$ This leads to $$xy = x - x^2 - xy + 2x^2y$$ or $$x^2-x = (2x^2-2x)y$$ hence to $$y = \frac{x^2-x}{2x^2-2x} = \frac{1}{2}.$$ Now verify that $\frac{1}{2}$ is actually the identity of this operation.

Once you know the identity, you can try to find inverses. Given $x\in (0,1)$, you are looking for a $z$ such that $x*z = \frac{1}{2}$ (assuming the above was correct). Solve for $z$ in terms of $x$ and verify that it lies in your set.

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Thanks for your help @Arturo. – Mike Mar 20 '12 at 17:02
@Mike: Of course, you have to verify $*$ is an associative operation on $(0,1)$ (if $x,y\in (0,1)$, then $x*y\in (0,1)$; and if $x,y,z\in (0,1)$, then $(x*y)*z = x*(y*z)$) if you want to prove you have a group. – Arturo Magidin Mar 20 '12 at 17:11

If the problem is to verify that $G = ((0, 1), \ast)$ is a group, in fact what we have to show the first is whether the associativity holds or not. Let

$$ f(z) = \frac{1-z}{z} $$

with the inverse

$$ f^{-1}(w) = \frac{1}{1+w}.$$


$$\frac{xy}{1-x-y+2xy} = \frac{xy}{(1-x)(1-y) + xy} = \frac{1}{\left(\frac{1-x}{x} \right)\left( \frac{1-y}{y} \right) + 1} = f^{-1}(f(x)f(y)),$$

thus we obtain

$$ \begin{align*} x \ast (y \ast z) & = x \ast f^{-1}(f(y)f(z)) = f^{-1}(f(x)f(y)f(z)) \\ (x \ast y) \ast z & = f^{-1}(f(x)f(y)) \ast z = f^{-1}(f(x)f(y)f(z)), \end{align*}$$

so that they coincide. Furthermore, this shows that $f$ is an isomorphism from $G$ to the group $(\mathbb{R}^{+}, \cdot)$ of positive real numbers equipped with usual multiplication. Thus both identity and the inverse can be trace back from this isomorphism as follows:

$$ \begin{align*} e &= f^{-1}(1) = \frac{1}{2}. \\ x^{-1} &= f^{-1}\left( \tfrac{1}{f(x)} \right) = 1 - x. \end{align*}$$

In fact, most of artificial operations $\ast$ in the problems are given in this way. That is, they are a disguise of some familiar operations $\ast'$ driven by some bijection $f$, so that the operation takes the form

$$ x \ast y = f^{-1}(f(x) \ast' f(y)).$$

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Good answer. :-) +1 – user21436 Mar 20 '12 at 17:10
Thank you! But I'm just afraid if it is hard to understand... – Sangchul Lee Mar 20 '12 at 17:25
OP will relish some aspects of your answer better at later point of time, but I thoroughly enjoyed reading it. :-) – user21436 Mar 20 '12 at 17:30
A $y$ went missing in the denominator in the third display equation. – user21436 Mar 20 '12 at 17:42
@KannappanSampath, I brought my runaway $y$ back to the equation. Thanks! – Sangchul Lee Mar 20 '12 at 17:54


Let $e$ be a identity in $((0,1), \ast)$ and $x^{-1}$ denote a inverse of $x$.

  • $x \ast e=x$ for all $x$.

  • $x \ast x^{-1}=e$

From your answer, prove that they are the identity and the inverse respectively.

That the answer has been spelt out, I'll leave you with another exercise (Sorry!).

Try this exercise:

(Dec. 2011, Math. Reflections) On the set $M~=~\mathbb R-\{3\}$, the following binary law is defined: $$ x \ast y = 3(xy-3x-3y)+m$$ where $m \in \Bbb R$. Find the values of $m$ such that $(M, \ast)$ forms a group.

(Proposed to Math. Reflections by Bogdan Enescu, ``B. P. Hasdeu" National College, Buzau, Romania)

Ping me here in case you want hints, and if you want an answer google, Math. Reflections and I have given you the reference for the problem here. I had written a solution to Math. Reflections, which if you're particular, I'll add it here later.

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