Expert: Sherry Wallin - 11/8/2010

Question
Dear Sherry Wallin,

I am trying to show that the below operation is a group. I have showed that it is closure and assocative but how can I find the identity and the inverse? Please kindly advise. Thanks

Let G={a in R: a^2<1} where a binary operation * on G is a*b=(a+b)/(1+ab)

Can I also know is a^2<1 has any restriction to the operation?  

Answer
Raymond~

First you must have an identity in order to have an inverse. So let x be in G so that x*e = x = e*x.

By the definition of G, x*e = (x + e)/(1 + xe)
= (x + e)/ (1 + x) but x*e = x so
(x + e)/(1 + x) = x =>
x + e = x + x^2 =>
e = x^2

and e*x = (e + x)/(1 + ex) = x =>
e + x = x + ex^2 =>
e + x = x + x^2 =>
e = x^2

If e = x^2 then the following is true:

thus (x + x^2)/(1 + xxx) = [x(x+1)]/(1+x^3)
= [x(x+1)]/[(1+x)(1 -x + x^2)]
= x/(1 -x + x^2)
but remember this is equal to x:
x/(1 -x + x^2) = x =>
x^2 -x + 1 = 1 =>
x^2 - x = 0 =>
x(x-1) = 0 =>
x = 0 or x-1 = 0 => x = 1 BUT x~=1 because x^2 < 1 so x must be 0.

x^2*x = (x^2 + x)/(1 + x^2x) = x and using the same trick above we have:
[x(x+1)]/(1+x^3) = [x(x+1)]/[(1+x)(1 -x + x^2)]
= x/(1 -x + x^2) = x => x^2 -x + 1 = 1 so
x^2 - x = 0 =>
x(x-1) = 0 =>
x = 0 or x-1 = 0 => x = 1 BUT x~=1 because x^2 < 1 so x must be 0.

This shows the identity is 0.


As for the inverse:

Now that you have an identity 0 you can find the inverse.
Let b be in G and suppose b = a^-1 in G
a*b = (a+b)/(1+ab) = 0. The only way for (a+b)/(1+ab) to be 0 is for the numerator to be 0, thus a + b = 0 => b = -a

and b*a = (b+a)/(1+ba) 0 => b + a = 0 => b = -a

this shows that -a is the inverse of a. You should always be concerned about the denominator that it doesn't become 0. Your denominator is 1 + a^-1(a) = 1-a^2 > 0 since all a^2 < 1. It is only important that the denominator isn't 0 and it is not.

Please feel free to ask if you have any more questions.

Math Prof