TITLE: $\mathbb{R}$ isn't group with operation $\circ$ but $\mathbb{R}\setminus \{p\}$ is, find $p$
QUESTION [5 upvotes]: I have exercise for exam and I can't find solution for it

Set of real numbers with operation $\circ$ is not a group but set
  $\mathbb{R}\setminus \{p\}$ is. Find $p$ and check all axioms for group.
$$
a \circ b = a + b - \frac{5ab}{3}
$$

I tried to find solution by myself, and I begin with finding identity element:
$$
a \circ e  = a \\
a+e - \frac{5ae}{3} = a \\
e - \frac{5ae}{3} = 0 \\
3e = 5ae\\
a = \frac{3}{5}
$$
But this is value for what $(a = \frac{3}{5} \lor b = \frac{3}{5}) \Rightarrow a \circ b = \frac{3}{5}$, so it isn't identity element (it will be one if and only if $a = b = \frac{3}{5}$).

REPLY [7 votes]: The identity element is clearly $0$, the solution to $3e=5ae$ that you missed.
Now you need to find $a^{-1}$ for $a\ne 0$. You want
$$0=a\circ a^{-1}=a+a^{-1}-\frac{5aa^{-1}}3\;,$$
or $a^{-1}(5a-3)=3a$, i.e., $$a^{-1}=\frac{3a}{5a-3}\;.$$
This is fine unless $a=\frac35$, so you need to throw out $p=\frac35$ and show that $\left\langle\Bbb R\setminus\left\{\frac35\right\},\circ\right\rangle$ is a group.