Let $G = \mathbb R$ together with the operation $x*y = x + y + xy,\; \forall x,y\in \mathbb R$ Let  $G = \mathbb R$ together with the operation $x*y = x + y + xy,\; \forall x,y\in \mathbb R$. Prove that $G$ is a group.
I know I have to verify the group axioms:
Closure
associative
identity
inverse
I don't know how to go about proving these, please help
 A: One of the axioms is very quick to check: closure – since $\mathbb{R}$ is a group under addition and multiplication, we know that $x+y,xy\in\mathbb{R}$ for all $x,y\in\mathbb{R}$, and thus (applying the former again) $x+y+xy\in\mathbb{R}$ for all $x,y\in\mathbb{R}$.
For associativity, just write out by hand what $(x*y)*z$ and $x*(y*z)$ are, and you'll see that they agree. For the identity, we want $e\in\mathbb{R}$ such that $x*e=e*x=x$. But this, after writing it out, is asking that $x+e+xe=e+x+ex=x$ which rearranges to $e+ex=0$ (noting that $*$ is commutative, i.e. $x*y=y*x$) and thus $(1+x)\,e=0$. But we want one $e$ to be the identity for all $x\in\mathbb{R}$, and so we must have that $e=0$.
Finally then, knowing the identity is $0$, we want to see if we can find an inverse for each $x\in\mathbb{R}$. That is, we want $y\in\mathbb{R}$ such that $x*y=0$. Writing this out gives us $x+y+xy=0$ which rearranges to $y\,(1+x)=-x$ and thus $y=\frac{-x}{1+x}$. But this is only defined if $x\neq-1$, and $-1\in\mathbb{R}$, so we have a problem. In fact, we see that $(-1)*x=-1+x-x=-1\neq0$. Thus $G$ is not a group.
(As a note for these sort of problems, it is usually the inverse axiom that fails, and if it is one of the others that fails instead it is usually quite a lot easier to spot.)
A: $\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\renewcommand{\phi}{\varphi}$$\newcommand{\R}{\mathbb{R}}$
It is a typical example of transport of structure.
Consider the map $\phi : \R \to \R$ given by $z \mapsto z - 1$. Note that 
$$
\phi(x \cdot y) 
=
x y - 1 
= 
(x - 1) + (y - 1) + (x - 1) (y - 1)
=
\phi(x) * \phi(y).
$$
So $\phi$ is an isomorphims between the sets with a binary operation $(\R,\cdot)$ and $(\R, *)$. 
It follows immediately that all properties of $(\R, \cdot)$ transport to $(\R, *)$. So $*$ is associative because $\cdot$ is, the neutral element for $*$ is $\phi(1) = 1 - 1 = 0$, the image of the neutral element $1$ for $\cdot$.
And since $0$ has no inverse in $(\R,\cdot)$, then $\phi(0) = -1$ has no inverse in $(\R, *)$. But take $-1$ away, and $(\R \setminus \Set{-1}, *)$ becomes a group, exactly like $(\R \setminus \Set{0}, \cdot)$ does.
