TITLE: Prove that $a * b = a + b - ab$ defines a group operation on $\Bbb R \setminus \{1\}$
QUESTION [6 upvotes]: So, basically I'm taking an intro into proofs class, and we're given homework to prove something in abstract algebra. Being one that hasn't yet taken an abstract algebra course I really don't know if what I'm doing is correct here.

Prove: The set $\mathbb{R} \backslash \left\{ 1 \right\}$ is a group under the operation $*$, where:
  $$a * b = a + b - ab, \quad \forall  \,\, a,b \in \mathbb{R} \backslash \left\{ 1 \right\} .$$

My proof structure:
After reading about abstract algebra for a while, it looks like what I need to show is that if this set is a group, it has to satisfy associativity with the operation, and the existence of an identity and inverse element.
So what I did was that I assumed that there exists an element in the set $\mathbb{R} \backslash \left\{ 1 \right\}$ such that it satisfies the identity property for the set and another element that satisfies the inverse property for all the elements in the set. However I'm having trouble trying to show that the operation is indeed associative through algebra since
$$\begin{align}
a(b * c) & = a(b+c) - abc \\
& \ne (a+b)c - abc = (a*b)c \\
\end{align}$$
So in short, I want to ask if it's correct to assume that an element for the set exists that would satisfy the identity and inverse property for the group. Also, is this even a group at all since the operation doesn't seem to satisfy the associativity requirements.

REPLY [2 votes]: Something that you always need to check is if the set is closed under the operation $*$. Let's assume that it is not closed and there is $a$ and $b$ in the set, such that
$a*b=1$
Then, we can deduce that
$a*b=a+b-ab=1 \implies a(1-b)=1-b \implies a=1 \hspace{0.1cm} or \hspace{0.1cm}b=1$
which is not possible, as both $a$ and $b$ are taken from $\mathbb{R} \backslash \left\{ 1 \right\}$.
In a group, you only have one operation defined, i.e. the operation $*$. So, for associativity, you have to show
$a*(b*c)=(a*b)*c$
rather than
$a(b*c)=(a*b)c$
where you have assumed the regular multiplication as an extra operation.
For the identity element, you may make a guess first. By definition, an element $x$ is the identity element of a group, if
$a*x=x*a=a \hspace{0.2cm} \forall a\in \mathbb{R} \backslash \left\{ 1 \right\}$
Then
$a*x=a+x-ax=a \implies x=0$
So, $x=0$ seems to be a valid identity element and then you need to proof, using the definitions, that it is actually the identity element.
Note that you need to find the identity element before you deal with the inverse of each element.
Finally, the inverse of an element $a$ is $b$ if and only if
$a*b=0$
Note that $0$ is used as the identity element.
Again make a guess
$a*b=0 \implies a+b-ab=0 \implies a(1-b)=-b \implies a=\frac{b}{b-1}$
and finish the job.