Finding an inverse. Group Theory

Question

I am given a set $G = \mathbb{R}$ and $x*y = x+y+x^3y^3$. I need to find out whether there exists an inverse. Please note that this is not a group since * is not associative. The identity element $e$ is $0$. So I did the following:

For the inverse to exists we must have the following: $x*y = x+y+x^3y^3 = e = y*x = 0$

So, $x+y+x^3y^3$ must equal to $0$. Imagine we fix $x$, then we get a cubic and we know that it has either 1 real solution or 2. For the inverse to exists, which is unique, we need 1 real root. Now, from here I am confused how to show if whether there are always 1 real root or not. I tried considering derivative, that is: $\frac{dy}{dx} = \frac{-1-3x^2y^3}{1+3x^3y^2}$

For $x>0$: $\frac{dy}{dx}$ is always negative and so the graph is decreasing and, thus will cross x-axis once and there will be 1 real root.

For $x = 0$, the inverse is $0$

As for $x<0$: I cannot say much since $\frac{dy}{dx}$ becomes $\frac{-1-3x^2y^3}{1-3|x^3|y^2}$ and it can be both negative and positive. So, would that mean that the inverse does not exist for all $x$? And so there is no inverse in $G$?

I hope my reasoning makes sense and I would appreciate any help! Thanks!

Answer 1

Fix $x<0$ and consider the polynomial $x^3y^3 + y + x$ in the variable $y$. Observe that $$\lim_{y \to -\infty} x^3y^3 + y + x = +\infty$$ and $$\lim_{y \to +\infty} x^3y^3 + y + x = -\infty,$$ so there is a solution $y$ to $x*y =0$ by continuity. Therefore every $x<0$ has a right inverse. By commutativity of the operation $*$, every such $x$ also has a left inverse. However, this inverse is not necessarily unique (so you might not want to call it an inverse at all). Indeed, consider $x=-0.1$ and observe (by plotting the function $-0.001y^3 + y -0.1$ that there exist three distinct $y$ such that $x*y = 0$.

Answer 2

I think you have already said the answer.

Since $x + y + x^3y^3 = 0$ is a cubic, it has at least one real root.

For any $x$ there exists a $y$ such that $x*y = y*x = 0$

However, we have not proven that for any $x$ there exists a unique $x^{-1}$

Rather than using implicit differentiation, continue with the assumption that $x$ is constant.

$\frac {\partial}{\partial y} (x+y+x^3y^3) = 0\\ 1+3y^2 x^3 = 0\\ y = \pm \frac 1 {\sqrt {3|x|^3}}$

Substitute this value of y back in, and suppose $x>0$.

If $x - \frac 4 {3x\sqrt {3x}}, x + \frac 4 {3x\sqrt {3x}}$ have oppose signs. then there are 3 real roots.

$x + \frac 4 {3x\sqrt {3x}} > 0$ for all positive $x$

$x - \frac 4 {3x\sqrt {3x}} < 0\\ 3x^2\sqrt {3x} < 4\\ 27x^5 < 16\\ x < (\frac {16}{27})^{\frac 15}$

$x\in (-\frac {16}{27}^{\frac 15}, \frac {16}{27}^{\frac 15}) \implies $ there are three $y's$ such $x + y + x^3y^3 = 0$