Proving if the set is a group under the following operation So I’ve started learning about group theory this semester (actually just this week) and I’m a total newbie in the field. I’ve got the following set and I need to prove if it’s a group with respect to the operation stated below: 
$$  G := \mathbb R \setminus \{1\} $$
under the operation: 
$$a \circ b =  a +b-ab$$
for $$ a,b \in G $$
So far I've proved that It's associative and the identity is the number $0$, but I can't manage to prove the closure property and also can't find the inverse.
Any help in any form would be highly appreciated.
 A: Hint: For the closure property notice that $$a+b-ab-1=(1-b)(a-1)$$
For the opposite of an element $a$, you need to solve for $b$ the equation $a+b-ab=0$.
A: To prove closure, we need to show that if $a,b\in\mathbb{R}\setminus\{1\}$, then so is $a\circ b$.
Let $a,b\in\mathbb{R}\setminus\{1\}$. Our goal is to show that $a\circ b\in\mathbb{R}\setminus\{1\}$. Since $a,b\in\mathbb{R}$, $a\circ b=a+b-ab\in\mathbb{R}$. We'll be done if we can show that $a\circ b\ne1$. So suppose $a\circ b=1$. Then we have that $a+b-ab=1$. Hence $0=1-a-b+ab=(1-a)(1-b)$, which would imply that $a=1$ or $b=1$. So if $a\circ b=1$, then either $a=1$ or $b=1$. Since $a\ne1$ and $b\ne1$, we must have $a\circ b\ne1$. Hence if $a,b\in\mathbb{R}\setminus\{1\}$, then $a\circ b\in\mathbb{R}\setminus\{1\}$.
To prove that each element has an inverse, let $a\in\mathbb{R}\setminus\{1\}$. We want to show that there is $b\in\mathbb{R}\setminus\{1\}$ with $a\circ b=b\circ a=0$.
In order to do this, we will need to find out what $b$ is, so we'll first let $a\circ b=0$ and solve for $b$. Let $a\circ b=0$. Then $a+b-ab=0$. Solving for $b$ we get that $b=\frac{a}{1-a}$. Note: this exists since $a\ne1$.
Now we're ready to prove that each element has an inverse. Let $a\in\mathbb{R}\setminus\{1\}$. Note that $\frac{a}{1-a}\circ a=a\circ\frac{a}{1-a}=a+\frac{a}{1-a}-\frac{a^2}{1-a}=0$. Hence, for each $a\in\mathbb{R}\setminus\{1\}$ there is a $b\in\mathbb{R}\setminus\{1\}$ $\left(\text{namely }\frac{a}{1-a}\right)$, such that $a\circ b=b\circ a=0$.
A: I think an easy way to prove the closure property is to see that, $a\circ b \ne 1, \forall a, b \in \mathbb R \setminus \{1\}$. 
Using the identity $a+b-ab-1=(1-b)(a-1)$, we can see that $a+b-ab \ne 1 \Leftrightarrow  (1-b)(a-1) \ne 0, \forall a, b \in \mathbb R \setminus \{1\}$

The inverse of $a$ goes for resolving $a\circ a^{-1} = 0$, because $0$ is th identity element of the group $G$. So, $a\circ a^{-1} = 0 \Leftrightarrow a + a^{-1} - aa^{-1} = 0 \Leftrightarrow a + a^{-1} - aa^{-1} - 1 = -1 \Leftrightarrow (1-a^{-1})(a-1) = -1$.
Since $a, a^{-1} \ne 1$:
$(a^{-1}-1) = \frac{-1}{1-a} \Leftrightarrow a^{-1} = 1 + \frac{1}{a-1} \Leftrightarrow a^{-1} = \frac{a}{a-1}$
