We define a new operation
$$x*y= x+y+xy,$$
on the set of real numbers with the usual addition and multiplicaton. Has this operation got an identity element?
It seems clear for me that there is one such an identity element since for all reals $x$, we have that
$$x*0 = 0*x = x.$$
But I'm confused because the book where I found the exercise claims the opposite. That such an identity element doesn't exist because from the definition of the identity element we arrive to
$$e= 0/(1+x),$$
which isn't well defined for all $x$.
I need someone confident enough about this elementary abstract algebra topic to confirm me that the book is wrong or explains to me what I'm missing!
Best Answer
You are right. A simple check indeed verifies that for all $x\in\Bbb{R}$ we have $$x\ast0=x+0+x\cdot0=x\qquad\text{ and }\qquad0\ast x=0+x+0\cdot x=x,$$ so $0\in\Bbb{R}$ is an identity element with respect to $\ast$. It is not hard to show that $0$ is the unique identity element of $\Bbb{R}$ with respect to $\ast$: If $e\in\Bbb{R}$ is an identity element with respect to $\ast$, then $$1=1\ast e=1+e+1\cdot e=1+2\cdot e,$$ which clearly implies that $e=0$.
But do note that $(\Bbb{R},\ast)$ is not a group, because $-1\in\Bbb{R}$ does not have an inverse with respect to $\ast$. To see this, suppose toward a contradiction that $y\in\Bbb{R}$ is inverse to $-1$. Then $$0=-1\ast y=-1+y+-1\cdot y=-1+y-y=-1,$$ a contradiction. So $(\Bbb{R},\ast)$ is not a group.
On a more abstract note, the operation $\ast$ is 'essentially the same' as the usual multiplication on $\Bbb{R}$. To see this, let $(\Bbb{R}-\{0\},\cdot)$ denote the non-zero reals under the usual multiplication, which is a group. Consider the map $$\varphi:\ (\Bbb{R}-\{-1\},\ast)\ \longrightarrow\ (\Bbb{R}-\{0\},\cdot):\ x\ \longmapsto\ x+1,$$ which is clearly a bijection, and note that $$\varphi(x\ast y)=x+y+x\cdot y+1=(x+1)\cdot(y+1)=\varphi(x)\cdot\varphi(y),$$ which shows that $\varphi$ is an isomorphism.
It follows immediately that $\varphi^{-1}(1)=0$ is the identity element of $(\Bbb{R}-\{-1\},\ast)$, and that $(\Bbb{R},\ast)$ is not a group because $\varphi^{-1}(0)=-1$ does not have an inverse with resepct to $\ast$, as $0$ does not have an inverse with respect to $\cdot$.