-1
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ What did you try? $\endgroup$ – user261263 Oct 16 '16 at 16:28
  • $\begingroup$ For example: try to find an identity element. $\endgroup$ – lulu Oct 16 '16 at 16:29
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan Oct 16 '16 at 16:30
2
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ Much appreciated i'll take your note on board next time for this sort of question $\endgroup$ – Bradley Oct 16 '16 at 17:18
  • 1
    $\begingroup$ 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. $\endgroup$ – MJD Oct 16 '16 at 19:01
4
$\begingroup$

$\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.

$\endgroup$
  • $\begingroup$ +1, really nice. Is there a good heuristic to "guess" the isomorphism? $\endgroup$ – Andres Mejia Oct 16 '16 at 17:02
  • $\begingroup$ 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. $\endgroup$ – Andreas Caranti Oct 16 '16 at 17:04
  • $\begingroup$ @AndreasCaranti very interesting note about the Jacobson radical! $\endgroup$ – Tim Oct 16 '16 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.