Showing Two Groups are Isomorphic

[Oxymoron]

So if my set S was the set of all real numbers excluding -1. And (S, *) was a group where x*y = x + y + xy. How would I start proving that * is a binary operation?

 

[matt grime]

you would show it satisfied all the axioms defining a binary operation, that's all. It's a "just do it" proof. Clearly given two inputs there is a unique output, how about closure? Note, since you've called (S,*) a group, then * must be a binary operation, or it isn't a group.

 

[mathwonk]

The hints here have been outstanding. I wanted to add something though, and I can only think of this summary of what has been said:

1. to show two groups are isomorphic, you must find an isomorphism between them,

2. to show two groups are not isomorphic, you must show there cannot be any isomorphism, not just that one particular attempt fails. this is harder, as your argument has to apply to all potential isomorphisms. hence you need to find a property of groups that would be preserved by all isomorphisms, and yet which your two groups do not share, such as being commutative (which is called "abelian" for groups, in honor of Niels Abel).

In general the search for properties that would be preserved by all isomorphisms is a deep and fundamental one in every area, sometimes called the search for "invariants".

for example in algebraic curve theory, to show the projective plane curve x^3 + y^3 = z^3, is not rationally isomorphic to the line, can be done by outright cleverness, but is most efficiently done by producing the invariant called the genus. I.e. topoloogically the cubic is a doughnut and the "line" is a sphere.

the proof of the fundamental theorem of algebra in topology, is done by finding some way of discerning the difference between the punctured plane and the plane itself, which eventually becomes the first homology group. i.e. you have to show why the unit circle cannot be pulled away from the origin without passing through the origin. this is usually done by computing the integral of dtheta, and applying greens theorem from calculus.

in number theory one uses reduction "mod n" which says that any solution of an equation in integers would also yield a solution mod every n. Hence, since after division by 4, the equation x^2 = 2 has no solution (the left side always has remainder 0 or 1 after division by 4,) hence the equation x^2 = 204,840,962 also has no solution.