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.
Best Answer
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 $-1 + H$.