Ytukyg

Let$ G = {x ∈ R | x neq -1} $and * a link to ¨G with $x * y: = x + y + xy.$
Show that $(G, *)$ is a group.

To determine if $ G$ is a group I have to make the associative proof:

Associative?

$(x * y) * z = (x + y +xy)+z -(x + y +xy)*z$

$= x + y + z + xy - xz - yz - xzy$

$= x + y + z + xy -z (x + y + xy)$

$= x + (y + z + xy) -z (x + y + xy)$

I kinda stop here, because I think I made a mistake

I don't know from where minus sign from came in your solution .
If you're a beginner one easy way is to make seperate equation for LHS and RHS and show they are equal.

$(x*y)*z=(x+y+xy)*z=x+y+xy+z+xz+yz+xyz$ ....(1)

Now consider ,

$x*(y*z)=x*(y+z+yz)= x+y+z+yz +xy+xz+xyz$ ...(2)

from (1) and (2) and using the fact real numbers commute, you can prove associative

It's best not try to force $(x*y)*z = .... = something = .... x*(y*z)$, but to instead do $(x*y)*z = .... = something$ and $x*(y*z) = ... = the same something$.

$(x*y)*z = (x*y) + z + (x*y)z = x + y + xy + z + (x+ y + xy)z = color{blue}{x + y + z + xy + xz + yz + xyz}$.

And $x*(y*z) = x + (y*z) + x(y*z) = x + y + z+ yz + x(y + z + yz) = color{blue}{x + y + z + xy + xz + yz + xyz}$.

So $(x*y)*z = color{blue}{x + y + z + xy + xz + yz + xyz}= x*(y*z)$

Now if you wanted to do it in one line you could with some factoring:

$(x*y)*z = (x*y) + z + (x*y)z = x + y + xy + z + (x+ y + xy)z = x + y + z + xy + xz + yz + xyz= x + y + z+ yz + x(y + z + yz)=x + (y*z) + x(y*z)= x*(y*z)$.

But why kill our brain cells and give us a headache? (Okay, I know. A single equation is more idealist but... let's get practical.)

The fact that $Bbb Rbackslash{-1}$ is a group under $(x+1)(y+1)-1$ is equivalent to the fact that $Bbb Rbackslash{0}$ is a group under $xy$; the two groups are isomorphic, since the invertible function $sigma (x):=x+1$ of inverse $sigma^{-1}(x):=x-1$ satisfies $xast y:=sigma^{-1}(sigma(x)sigma(y))$. In particular, you can prove the associativity in terms of this isomorphism without any messy arithmetic: $$(xast y)ast z=sigma^{-1}(sigma(xast y)sigma(z))=sigma^{-1}(sigma(x)sigma(y)sigma(z))=xast (yast z).$$