Should my book have considered $\mathbb{R}-\{1/3\}$ instead of $\mathbb{R}$ or am I doing something wrong? In my book, I have a question saying
"Prove that $\mathbb{R}$ with the following operation
$$x\times y=x+y-3xy$$
is an (abelian) group."
Just like this. My point is that in order to compute the identity element $e$, I wrote
$$x\times e =x \Rightarrow x+y-3xy=x \Rightarrow e=\frac{0}{1-3x}.$$
Should my book have considered $\mathbb{R}-\{1/3\}$ instead of $\mathbb{R}$ or am I doing something wrong?
 A: Your objection is somewhat correct:
For $x,e\in\mathbb R$ you have $x\times e=x\iff e(1-3x)=0\iff e=0\lor x=\frac 13$.
From this you can see that $(\mathbb R,\times)$ cannot be a group:
$\frac 13\times 0=\frac 1 3\times\frac 1 3$, so applying the inverse of $\frac 13 $ from the left would yield $0=\frac 13$.
A: Your equation should be
$x\times e =x \Rightarrow x+e-3xe=x \Rightarrow e=0.$
Then the inverse of an element $x$ is given by $\frac {x}{3x-1}$ which works for all $x\ne \frac{1}{3}$.
A: Lemma: The only idempotent of a group $G$ is the identity $e$.
Proof: Let $x^2=x\in G$. Then $xx=x=xe$ so, multiplying by $x^{-1}$ on the left, we get $x=e$. $\square$
Suppose $r\in \Bbb R$ such that $r\times r=r$. Then $r=r+r-3r^2$ implies
$$\begin{align}
0&=r-3r^2\\
&=r(1-3r).
\end{align}$$
Thus either $r=0$ or $r=\frac{1}{3}$. We cannot have both, by the Lemma, if we are to have $(\Bbb R,\times)$ be a group. Indeed $0$ is an identity (as one can check easily).
But what about $r=\frac{1}{3}$? Well, for arbitrary $x\in\Bbb R$, if $x\times \frac{1}{3}=x$, then $x+\frac{1}{3}-x=x$ implies $x=\frac{1}{3}$, a contradiction (by letting,
say, $x=1$); thus we must indeed remove $\frac{1}{3}$.
