# Group Theory, Binary Operation

$a$ is a non-zero real number.

Define a binary operation on the set of real numbers by:

$$x*y = x + y + axy$$

1) Show that $*$ is associative.
2) Show that $(G,*)$ is a group, where $G$ is the set of all real numbers except for one number which you should identify.
3) Find a $2$ element subgroup of $(G,*)$

I have a very hard time proving $(x*y)*z = x*(y*z)$ because I always get either $3z$ or $3x$ which does not appear on the other side...

I can't think of the one number that wouldn't be a member... is it $0$ because $0$ does not have an inverse?

The $2$ element subgroup is $\{0,a\}$?

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Where do you get $3x$ or $3z$ from? –  Tobias Kildetoft Feb 13 '13 at 19:54

We can view $*$ as a perturbed multiplication, i.e. there is a bijection $f\colon G\to\mathbb R$ such that $f(x*y)=f(x)f(y)$. This transports the group laws (thus solving the first two parts of th eproblem) from $(\mathbb R^\times,\cdot)$ to $(G,*)$ so that $f$ becomes in fact a group isomorphism.

If we make the attempt $f(x)=ux+v$ with $u,v\in\mathbb R$, then we need $$u(x+y+axy)+v = (ux+v)(uy+v),$$ i.e. $$au\cdot xy + u\cdot (x+y)+v = u^2\cdot xy+uv\cdot (x+y)+v^2,$$ Which works out if we let $v=1$, $u=a$. We find that $f(x)=0$ implies $x=-\frac1a$, hence $G=\mathbb R\setminus\{-\frac1a\}$.

Since we know the only two-element subgroup of $\mathbb R^\times$ is $\{1,-1\}$, there is exactly one two-element subgroup of $G$, namely $\{f^{-1}(-1),f^{-1}(1)\}=\{-\frac{2}a,0\}$.

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Suppose $\,\alpha\,$ is the group's neutral element, then:

$$\forall\,x\in G\;\;,\;\;x=x*\alpha:=x+\alpha+ax\alpha\Longrightarrow \alpha(1+ax)=0\Longleftrightarrow$$

$$\alpha=0\,\,\,\vee\,\,\,1+ax=0$$

But the second option above cannot be since then $\,a\,$ would depend on $\,x\,$ , which is impossible since the above must be true for all real numbers $\,x\,$.

Thus, $\,0\,$ is the neutral element. The operation's clearly commutative, and we must have that

$$\forall\,x\in G\,\,\exists!x'\in G\,\,\;\,s.t.\;\; 0=x*x'=x+x'+axx'\Longrightarrow$$

$$x'(1+ax)=-x$$

But for this equation to have a solution it must be that

$$1+ax\neq0\Longleftrightarrow x\neq-\frac{1}{a}\ldots$$

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That is why I wrote specifically and on purpose that the elements are in the group, without remarking what's the real element that must be taken out of the game... –  DonAntonio Feb 13 '13 at 20:18
+1 Okay..I get it..., I missed the fact that to get $x*\alpha = x\;$ implies...$~~$;-) –  amWhy Feb 13 '13 at 20:23