$a$ 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 $\{0,a\}$?
Best Answer
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 $\{f^{-1}(-1),f^{-1}(1)\}=\{-\frac{2}a,0\}$.