# Prove abelian group

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

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

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

Then

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

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... –  sos440 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! –  sos440 Mar 20 '12 at 17:54
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Hint:

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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