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I have a problem with this question

is $(\mathbb{R},*)$, when $a*b=ab+a+b$ a group? If not, can you skip any element $a \in\mathbb{R} $ in that way, that $(\mathbb{R}$\{a}$,*)$ is a group?

I can prove, that $(\mathbb{R},*)$ is a binary operation, associative and it's neutral element is $0$. But I couldn't find an inverse element.

And I have no idea which element I can skip to get the other group. I tried to skip $a=0$ but it wasn't a good tip.

Thank you for your time.

Daniel R
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martina
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2 Answers2

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Given any $a \in \mathbb R$, we want to find some $b \in \mathbb R$ such that $a * b = 0$. We need to solve for $b$ as a function of $a$. Indeed, observe that: $$ ab + a + b = 0 \iff b(a + 1) = -a \iff b = \frac{-a}{a + 1} $$ So now we know how to get the inverse of any group element. Except...I implicitly made an assumption that excludes one element from being invertible. Which is it?

Adriano
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you can start by looking at which elements lack an invert. Thus, we compute $a^{-1}$:

$aa^{-1}+a+a^{-1}=0\iff a^{-1}=\frac{-a}{a+1}$

So you should delete $-1$ out of $\mathbb{R}$.

However, you have to prove ($\mathbb{R}$\{-1},*) does make a group.

It is closed under multiplication (verify if $a,b\neq-1$ then ($a*b$)$\neq-1$), every element has an inverse (we got rid of $-1$), we have a unit $0$, and * is associative ($(a*b)*c=a*(b*c)=abc+ab+bc+ac+a+b+c$).

Hope I helped.

Junkai Dong
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