Mathematical Reflections 2011, Issue 6 - Problem U211

Problem:
On the set $M=\mathbb{R}-\{3\}$ the following binary law is defined:
$$x \ast y = 3(xy-3x-3y)+m,$$ where $m \in \mathbb{R}$. Find all possible values of $m$ such that $(M, \ast)$ is a group.

Proposed by Bogdan Enescu.

Solution:

First we observe that $x \ast y = 3(x-3)(y-3)+m-27$.
We want that $\ast$ satisfies the group axioms.

(i) Closure: For every $x,y \in M$, $x \ast y \in M$. It must be $3(x-3)(y-3)+m-27 \neq 3$, i.e. $3(x-3)(y-3)+m \neq 30$ and this is clearly satisfied for $m=30$.

(ii) Associativity: For every $x,y,z \in M, (x \ast y) \ast z = x \ast (y \ast z)$. It must be
$$\begin{array}{lll} [3(x-3)(y-3)+m-27] \ast z = 3[(3(x-3)(y-3)+m-27)-3][z-3]+m-27 \\
= 9xyz-27(xy+yz+zx)+3(27x+27y+(m-3)z)-8m \end{array}$$
and
$$\begin{array}{lll} x \ast [3(y-3)(z-3)+m-27] = 3(x-3)[3(y-3)(z-3)+m-27-3]+m-27 \\
= 9xyz-27(xy+yz+zx)+3((m-3)x+27y+27z)-8m, \end{array}$$
so the associativity holds if and only if $m=30$.

(iii) Identity element: There exists $e \in M$ such that for every $x \in M$, $x \ast e = e \ast x = x$. It must be
$$3(x-3)(e-3)+3=x \iff 3e(x-3)=10(x-3) \iff e=\dfrac{10}{3}.$$

(iv) Inverse element: For each $x \in M$, there exists $y \in M$ such that $x \ast y = y \ast x = e$. It must be
$$3(x-3)(y-3)+3=\dfrac{10}{3} \iff (x-3)(y-3)=\dfrac{1}{9} \iff y=\dfrac{1}{9(x-3)}+3.$$
So $(M,\ast)$ is a group if and only if $m=30$.