Group of rational numbers

What is the inverse of $1$ under this group operation? There isn't one: $1 * b = 1+b-b=1 $ for any $b$.

Well. Now I figured it. For any two element a & its inverse b in Q1 we will get $a*b=0$ as 0 is its identity element. This will give $a+b-ab=0$ and $b=a/(a-1)$. So if the set consists 1 then inverse will not exist.