# 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.

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 $$a+b-ab=0$$.

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 $$a\circ b=b\circ a=0$$.

• Thank you very much for your elaborate and clear answer. Not to dismiss the fact that the other answers hinted at the same solution, I chose yours as the most simplest one. Jan 19, 2020 at 14:44
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 $$a, a^{-1} \ne 1$$:
$$(a^{-1}-1) = \frac{-1}{1-a} \Leftrightarrow a^{-1} = 1 + \frac{1}{a-1} \Leftrightarrow a^{-1} = \frac{a}{a-1}$$