How to find identity and inverse for the group $(\mathbb{Z}, \ast)$, where $a \ast b = a+b-ab$ I'm trying to figure out how to find identity and inverse for the group defined as $(\mathbb{Z}, \ast)$, where $a \ast b = a+b-ab$?
I've found that closure and the associative property are both true, but I don't know where to go from here.
Any help in solving this would be appreciated? 
 A: At the outset, I need to warn you that this operation does not make $\Bbb Z$ into a group. We'll see why shortly.
Easy candidates for identity:
If $e$ were the identity for $\ast$, then you'd need $e\ast e=e$, or in other words, $2e-e^2=e$.
Can you see how this immediately narrows $e$ down to two possibilities? I'm sure you can do it: just rearrange the above equation and remember that $xy=0$ implies $x=0$ or $y=0$ in the integers. One of the two possibilities works as an identity for $\ast$, and one does not. 
The one that isn't the identity, call it $z$, satisfies $a\ast  z=z$ for all $a\in \Bbb Z$. If $(\Bbb Z,\ast)$ were a group, you'd then cancel $z$ from both sides to conclude that $a=e$ for all $a$, but this is not true. Thus $z$ can't have an $\ast$ inverse.
To correct the statement:
You can say that $(\Bbb Z,\ast)$ is a monoid, and in fact the element $z$ above is usually called a "zero element," so that this is a monoid with zero element.
A: For the identity, note that $$a*0=a+0-a\cdot 0 \\=a \\ = 0+a-0\cdot a \\ = 0*a $$ To find the inverse element, we need $a*b=0$. This means $$a+b-ab=0 \\ \implies a+b(1-a)=0 \\ \implies b(1-a)=-a \\ \implies b=-\frac{-a}{1-a} \\ \implies b = \frac{a}{a-1}$$ making $b=a^{-1}$. But there is a problem, because clearly $a$ cannot equal $1$, else you would be dividing by zero to get $a^{-1}$. Yet, $1$ is an element of your set. Are you sure this is a group?
