Is $(\mathbb{R},*)$, where $a*b=ab+a+b$, a group? I have a problem with this question
is $(\mathbb{R},*)$, when $a*b=ab+a+b$ a group? If not, can you skip any element $a \in\mathbb{R} $ in that way, that $(\mathbb{R}$\{a}$,*)$ is a group?
I can prove, that $(\mathbb{R},*)$ is a binary operation, associative and it's neutral element is $0$. But I couldn't find an inverse element.
And I have no idea which element I can skip to get the other group. I tried to skip $a=0$ but it wasn't a good tip.
Thank you for your time.

you can start by looking at which elements lack an invert. Thus, we compute $a^{-1}$:
$aa^{-1}+a+a^{-1}=0\iff a^{-1}=\frac{-a}{a+1}$
So you should delete $-1$ out of $\mathbb{R}$.
However, you have to prove ($\mathbb{R}$\{-1},*) does make a group.
It is closed under multiplication (verify if $a,b\neq-1$ then ($a*b$)$\neq-1$), every element has an inverse (we got rid of $-1$), we have a unit $0$, and * is associative ($(a*b)*c=a*(b*c)=abc+ab+bc+ac+a+b+c$).
Hope I helped.
