# $\mathbb{R}$ isn't group with operation $\circ$ but $\mathbb{R}\setminus \{p\}$ is, find $p$

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 $a = b = \frac{3}{5}$).

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The correct conclusion from $3e=5ae$ is $a=3/5$ OR $e=0$. – Gerry Myerson Jan 6 '13 at 15:07
Great problem for me. Thanks for sharing it. – Babak S. Jan 6 '13 at 16:18

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.
Computations may be simplified by observing that $$(x+\frac35)\circ (y+\frac 35)=(-\frac53 xy)+\frac35.$$ – Hagen von Eitzen Jan 6 '13 at 15:28