Finding an operation that makes a group

[learningphysics]

Hi. I'm working through the book "Elements of Abstract Algebra" by Allan Clark. This question has stumped me... find an operation on (0,1) (set of reals x such that 0<x<1 ) that makes (0,1) a group and makes the inverse of x, 1-x.

I'd appreciate any help. Thanks.

 

[AKG]

Do you know of an operation on R that makes it into a group with -x the inverse of x? Think of a way to make (0,1) correspond to R such that you can define your operation on (0,1) in terms of the very familiar operation on R.

 

[NateTG]

Another thing you might want to think about is what the identity of the group will be.

 

[learningphysics]

Thanks AKG and Nate. I believe I've found the answer but I'm not sure why the method I used "works"...

I looked for a way to make (0,1) correspond to R as AKG said. I used the function: $$\frac{1}{2}tanh(u) + \frac{1}{2}$$

From here I decided to multiply out $$\frac{1}{2}tanh(u + v) + \frac{1}{2}$$

Then after setting $$x = \frac{1}{2}tanh(u) + \frac{1}{2}$$ and $$y = \frac{1}{2}tanh(v) + \frac{1}{2}$$, I rewrote the above as:

$$\frac{xy}{1+2xy-x-y}$$ which is the product.

(I got this idea because the previous problem had the function $$\frac{x+y}{1+xy}$$ which you can get from substiting $$x=tanh(u)$$ and $$y=tanh(v)$$ into $$tanh(x+y)$$ but I really don't understand why it works)

Why did this method work? Thanks again for all your help.

 

[Hurkyl]

Why did this method work? Thanks again for all your help.

Because the target of a group isomorphism is a group. You just constructed the isomorphism before you figured out what the group was.

 

[mathwonk]

that is a wonderful book. hang in there and you will learn a lot.

 

[learningphysics]

Thanks Hurkyl and mathwonk. The book hasn't covered group isomorphisms yet, but I looked it up and everything makes perfect sense.