xy+x+y=0 What is the inverse Element?

I have a task to show that $G:= \mathbb Q\setminus \{-1\}$ is a group to some binary operation *

(x is the "times" symbol) with $A*B:=A\times B + A + B$

I could show that it's associative, that the neutral Element is $0$, but I'm stuck finding the inverse Element...

Maybe I started in the wrong place and don't see the wood for the trees, but I just found a Element for $1$ with $A=1$ and $B=-1/2$

so that $A*B=[1 \times (-1/2)]+1 - 1/2 = 0$

but because of the one you can't derive a general rule from that

but is there a general inverse Element??

• Hint: $G\to\mathbb Q\setminus\{0\}$ $a\mapsto x+1$ is a group homomorphism. Commented Oct 19, 2014 at 19:46
• You have to define some things here: what is that Q, what then does $\;Q\{-1\}\;$ mean, what is a "linkage operation" ...? Commented Oct 19, 2014 at 19:49
• ok i see....it escaped the \ ....sorry Commented Oct 19, 2014 at 19:53
• The English term you're looking for to mean "linkage operation" is "binary operation." I don't know if I can help with the rest of your question, though. Commented Oct 19, 2014 at 20:01
• thank you...ill remember that :-) Commented Oct 19, 2014 at 20:12

If the neutral element is $0$ and you want to find the inverse of $a \in G$, that means you want to find $B$ such that $0 = a*b$. This implies:

$0 = a*_{\scriptsize G} b = a\times_{\scriptsize \mathbb Q} b+_{\scriptsize \mathbb Q}a+_{\scriptsize \mathbb Q}b = (a+_{\scriptsize \mathbb Q}1)\times (b+_{\scriptsize \mathbb Q}1) -_{\scriptsize \mathbb Q} 1$

$\iff 1 = (a+_{\scriptsize \mathbb Q}1)\times_{\scriptsize \mathbb Q}(b+_{\scriptsize\mathbb Q}1)$

$\iff 1/_{\scriptsize \mathbb Q}(a+_{\scriptsize \mathbb Q}1) -_{\scriptsize \mathbb Q} 1 = b$

$\qquad \qquad \qquad \qquad \qquad \:\:= -_{\scriptsize \mathbb Q}a/_{\scriptsize \mathbb Q}(a+_{\scriptsize\mathbb Q}1)$

• Corrected it, I made a mistake. Commented Oct 19, 2014 at 20:07
• It seems to me you are multiplying by the inverse of $(a+1)$, but isn't part of the problem to determine the inverse of an arbitrary element? Commented Oct 19, 2014 at 20:14
• @Clayton Your confusion arises from mixing up the two multiplication operations. Commented Oct 19, 2014 at 20:15
• thank you very much flawr Commented Oct 19, 2014 at 20:19
• @Clayton I now denoted each operation with the corresponding algebraic structure. Commented Oct 19, 2014 at 20:26

Hint: What happens when you expand $(a+1)(b+1)$?

• thanks...afterwards it's so simple Commented Oct 19, 2014 at 20:23