# Consider a set $G\subseteq \Bbb R$ and a binary operation * defined on $\Bbb R$ as $a*b=a+b+ab$, such that $(G,*)$ is an Abelian Group. Determine $G$.

My question differs from this one(also, the binary operation differs slightly). Here, I'm supposed to determine $$G$$, as opposed to proving a given $$G$$ to be an Abelian group. I know that for $$(G,*)$$ to be a group, it has to satisfy the following postulates:

1) Closure, i.e., $$a*b\in G$$, $$\forall a,b\in G$$

2) Associativity, i.e., $$(a*b)*c=a*(b*c)$$, $$\forall a,b,c\in G$$

3) Existence of a unique identity element $$e\in G$$ such that $$a*e=e*a=a$$, $$\forall a\in G$$

4) Existence of inverse elements, i.e., $$\forall a\in G$$, $$\exists a^{-1}\in G$$ such that $$a*a^{-1}=a^{-1}*a=e$$

I begin by assuming $$G$$ to be identical to $$\Bbb R$$, because I have no idea how to proceed with an arbitrary subset of $$\Bbb R$$ whose elements and properties are unknown. Since addition and multiplication are both closed on $$\Bbb R$$, the closure property is satisfied by $$(G,*)$$. $$(G,*)$$ is also associative(as can be shown by trivial calculations). Calculations prove that $$0$$ is the identity element. So far so good. Now to evaluate the existence of inverse elements, I use postulate 4 $$a*a^{-1}=e$$ This means $$a+a^{-1}+aa^{-1}=0$$ This equation proves that $$a^{-1}$$ does not exist when $$a=-1$$. Intuitively, it seems reasonable that $$G=\Bbb R\backslash \{-1\}$$. However, this requires to check the consistency of the first 3 postulates again. I'm not sure if I can rely on addition and multiplication being closed operations on $$\Bbb R$$ anymore, now that one of the elements has been excluded. I'm confused as to how to proceed forward and seek guidance for the same.

Edit: Apparently I overlooked the postulate of commutativity for Abelian Groups, but as it stands I think proving $$(G,*)$$ to be a group is the primary challenge for me.

Define a map $$T:\mathbb R\to \mathbb R$$ by $$x\mapsto x+1$$. This turns your binary operation into ordinary multiplication, i.e. $$T(a)T(b) = T(a\ast b)$$. It follows that the multiplicative subgroups $$H$$ of $$\mathbb R$$ give you the possibilities for $$G$$, namely as $$-1 + H$$.

You first need to determine what is $$e$$: $$e*e=e$$, so $$e+e+e^2=e$$, so $$e^2=-e$$. Hence $$e=0$$ or $$e=-1$$. Then $$e*x=x$$, so $$e+x+ex=x$$ or $$e+ex=0$$ for every $$x\in G$$. If $$e=-1$$, then the only $$x$$ is $$x=-1$$ and so $$G=\{-1\}$$. So we can assume $$e=0$$. Then for every $$x\in G$$ there should be $$y$$: $$x*y=0$$, that is $$x+y+xy=0$$.So $$y=1/(x+1)-1=x/(x+1)$$. So with every $$x\in G$$ this $$y$$ should be in $$G$$ too. Finally $$G$$ must be associative which gives some more restrictions on $$G$$: $$(a*b)*c= (a+b+ab)*c=a+b+ab+c+ac+bc+abc=a*(b*c)=a*(b+c+bc)=a+b+c+bc+ab+ac+abc,$$ so for every $$a,b,c\in G$$: $$0=0$$ - no new restrictions. Hence the only restrictions on $$G$$ are

1. Either $$G=\{-1\}$$ or $$0\in G, -1\not\in G$$;

2. For every $$a,b\in G$$, $$ab+a+b\in G$$;

3. If $$0\in G$$ then for every $$x\in G$$, $$x/(x+1)\in G$$.

For example $$G$$ can be one of the following sets

a) $$\mathbb{Q}\setminus \{-1\}$$;

b) $$\mathbb{R}\setminus\{-1\}$$;

c) $$\mathbb{Q}_{\ge 0}$$

but also many other subsets of $$\mathbb{R}$$ (continuously many).

• I'm sorry, but this does not seem to resolve my dilemma. I've already determined that $e=0$ and $-1\notin G$. What I'm precisely concerned about are these restrictions on $G$ that you have mentioned. How do I find the values permitted in $G$? Commented Jun 12, 2020 at 0:48
• It is not true that $-1\not\in G$. Either $\{-1\}=G$ or $G$ contains $0$ and satisfies some other conditions. Commented Jun 12, 2020 at 1:38
• The fact that the operation $*$ is commutative is obvious. Commented Jun 12, 2020 at 2:20