Name this abelian group.

[phukraut]

The set of real numbers (but with -1 removed) forms a group G under the operation *, where, for a,b in G, a*b = a + ab + b.

My question is, does this group have a special name or discoverer? I can't seem to find any more information on it than textbook exercises, which doesn't name it.

[vinaur]

Is that supposed to be a times b or a#b (or any other character for that matter)??

[somi]

* represents a binery operation: is it closed under *? are there any a,b element of the set of real numbers that would result in it being -1?

[Hain]

Nope. To get a -1, either a or b must be -1, thus the series is all real numbers except for -1.

[phukraut]

Another way to see that -1 can't be in the group is by looking at the inverse of any element. it is -x/(1+x). Thus if x=-1, then inverse(x) is not a real number. Therefore the group would not be closed, and hence, not a group.

[vinaur]

I'm not really following the inverse part. How do you get it?

[phukraut]

First step, let's find the identity element of G:
the identity "e" is the element such that, for any x in G,

x*e = e*x = x. Now, by our definition of G,

x*e = x+xe+e = x.

The only value for e that will make the statement true is e=0 since
x+x0+0 = x. Now, to find the inverse of any x (call it "y"), the following needs to be true:

x*y = y*x = e.

This statement should be familiar if you think of it this way: if you take an object (like a number), and operate with its inverse, you should get "nothing" left. Examples include

3 + (-3) = 0, and 2(1/2) = 1, and in functions, f^-1(f(x))=x.

In our case, since e=0, we have

x*y = x+xy+y = 0. Solving for y gives y(1+x)=-x, and thus y=-x/(1+x). In order for G to actually be a group, we must have that for any x in G, y=inverse(x) must be unique and be in G. Thus we exclude -1.