Prove a group is abelian Let $S = \mathbb{R} \space \backslash$ $\{-1\}$ and define a binary operation on $S$ by $a*b = a + b+ ab$. Prove that $(S,*)$ is an abelian group.
This is problem 7 in chapter 3 of Judson's Abstract Algebra book. I was trying to work on it and can't figure it out.
 A: The idea is that the set of all non-zero real numbers forms an abelian group under multiplication. The group in question is the same group except every number has been moved to the right by one unit on the real number line. So $-1$ is the new zero.
In other words, if you take non-zero $x$ and $y$ , and define the operation as $z = x\cdot y$. Substitute $x=1+a$, $y = 1+b$ and $z = 1+c$, then
$$1+c = (1+a)(1+b) = 1+a+b+ab \implies c = a+b+ab =: a*b.$$
Armed with this idea, I hope you will find it easy to prove $S$ is a group.

Edit: (answering the comments below)

*

*$a*0 = a$. So $0$ is the identity element of the group.


*Show that $\left(\frac1{a+1} - 1\right)$ is the inverse of $a$. Now you see why $a=-1$ is not allowed.
Now consider the map $\phi: S \to \mathbb{R}\backslash \{0\}$  given by $\phi(a) = a+1$. My answer above claims $\phi$ is a group isomorphism.
A: First show $S$ is a group. The fact that $-1 \notin S$ is necessary for this to be true. Once you have shown it is a group, here is how you show it is abelian. Let $a,b \in S$. Then $$a \star b=a+b(1+a)\\ =a+b+ab \\=b+a(1+b) \\ =b\star a$$ So $S$ is abelian. 
A: Let $a,b \in S$. Then $$a \star b=a+b+ab\\ =b+a+ba \\ =b\star a$$ So $S$ is abelian.
Look actually the map ($x \mapsto (x+1) )$  from  $\mathbb{R} \space \backslash$ $\{-1\}$ to $\mathbb{R} \space \backslash$ $\{0\}$ is a automorphism. So as $\mathbb{R} \space \backslash$ $\{0\}$ is abelian, so is $\mathbb{R} \space \backslash$ $\{-1\}$.
