# 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

• What did you try? – user261263 Oct 16 '16 at 16:28
• For example: try to find an identity element. – lulu Oct 16 '16 at 16:29
• First, try to find an identity. Let $x*e=e*x=x$ and solve for $e$. To prove associativity, just go to the definition. Compute $(x*y)*z$ and $x*(y*z)$ and observe they are the same. Closure comes from the closure of the reals under addition and multiplication. – Ross Millikan Oct 16 '16 at 16:30

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.)

• Much appreciated i'll take your note on board next time for this sort of question – Bradley Oct 16 '16 at 17:18
• Might be worth pointing out that it is a group if you just leave out $-1$; that is, $\langle \Bbb R \setminus \{ -1\}, \ast\rangle$ is a group. – MJD Oct 16 '16 at 19:01


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.

• +1, really nice. Is there a good heuristic to "guess" the isomorphism? – Andres Mejia Oct 16 '16 at 17:02
• Thanks! Well, this particular operation (possibly with a variation, i.e. a minus sign in front of $x y$) is critical in the theory of the Jacobson radical for rings without identity, so it's easy to recognise it. In general, I am not sure. – Andreas Caranti Oct 16 '16 at 17:04
• @AndreasCaranti very interesting note about the Jacobson radical! – Tim Oct 16 '16 at 20:46