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Is $(R, *)$ with $* : R \times R \to R$ by $(a,b) \mapsto ab + a + b$ a group?

I've shown that it's associative and that $0$ is the neutral element, but couldn't find an inverse element.

Is there no inverse element? Or can I find an inverse element by omitting some elements?

JBL
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ChoRisk
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    Please your MathJax when you post here. Here is a tutorial: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Mark Apr 07 '23 at 00:22
  • Some elements indeed have no inverse. – Sassatelli Giulio Apr 07 '23 at 00:22
  • As for finding inverses, given some $a$ the equation $ab+a+b=0$ is a linear equation in the variable $b$ with parameter $a$, and this is high school stuff (assuming $R$ is $\Bbb R$). – Sassatelli Giulio Apr 07 '23 at 00:24
  • My solution ist $b = \frac{-a}{1+a}$ ,but if I then calculate $(a,b) = \frac{1}{a+1}$ – ChoRisk Apr 07 '23 at 00:31
  • Well your high school algebra might need some work. But also there is an obvious problem with that formula for $b$, sometimes. – JBL Apr 07 '23 at 00:33

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