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Given $\mathbb{R} \setminus \{-1\}$ with the operation $a*b=a+b+ab$, check whether is it or not a group.

My solution:

  • Associative $$(a*b)*c=(a+b+ab)*c=a+b+ab+c+(a+b+ab)c=\\ =a+b+ab+c+ac+bc+abc;$$ $$a*(b*c)=a*(b+c+bc)=a+b+c+bc+a(b+c+bc)=\\ =a+b+ab+c+ac+bc+abc$$

  • Existence of identity el.

If $e$ is the identity, then, for each a in the set, $a*e=e*a=a \rightarrow a*e=a+e+ae=a$. Here, it seems that the identity is $0$, but if $a=1$, the equation has no solution. So there is no identity element and therefore the set is not a group.

Is that right?

Vincenzo Tibullo
  • 10,955
Allonsy
  • 659
  • But $01=0+1+0\cdot1=1$ so $e=0$ works for $a=1$ too. Indeed it is obvious that for any $a$ we have $0a=a*0=a$. – almagest Jun 12 '16 at 15:23
  • If you mean that the equation $$a+e+ae=a$$ cannot be solved for $e$ when $a=1$, that is mistaken. If $1+e+1\cdot e=1$ then $1+2e=1$ so $2e=0$ and so $e=0$. $\qquad$ – Michael Hardy Jun 12 '16 at 15:26
  • BTW, notice the proper way to typeset $\mathbb R\setminus{-1}$ in MathJax. I edited the question accordingly. $\qquad$ – Michael Hardy Jun 12 '16 at 15:28
  • Hi Michael. However $1+2e=1$ means subtracting $-1$ to each side, which is impossible since $-1$ is not in the set. – Allonsy Jun 12 '16 at 15:29
  • @Allonsy No, just work direct from the definition: $0*1=0+1+0\cdot1=1$. – almagest Jun 12 '16 at 15:31
  • @Allonsy The inverse is slightly trickier: the inverse of $a$ is $-\frac{a}{1+a}$ which explains why $-1$ was excluded. – almagest Jun 12 '16 at 15:36

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