Here is the problem I am attempting to solve: Let $S$ be a set with a composition law $\cdot$ possessing the following properties:
$$(i)~x\cdot y = y\cdot x~\forall~x,y\in S$$ $$(ii)~x\cdot (x\cdot y) = y~\forall~x,y\in S.$$
Now define a new composition law by fixing $e\in S$ to be $$x+y = e\cdot(x\cdot y).$$
Prove that $+$ is commutative, that $e$ is the unique identity, that $a+x=y$ has a unique solution for all $x,y\in S$. Then find a condition on $\cdot$ so that $+$ is associative if and only if your condition holds. Prove your claim.
I have completed everything except that there is a unique solution to $a+x=y$. I have one solution: If I choose $a= x\cdot(y\cdot e)$, then $a+x = y$, as is verifiable. My question is this, how do I show that this is unique? If we were working over a group, I could use associativity easily. However, I am effectively told this is not an associative structure. I did discover (and prove!) that $+$ will be associative, hence $(S,+)$ will be a group, iff $z\cdot (e\cdot(x\cdot y)) = x\cdot (e\cdot (y\cdot z)).$
Hint: If $a+x=y$, then $$y=e\cdot(a\cdot x)=e\cdot(x\cdot a).$$ What operations can you perform to this equation to end up with $a$ alone on the right-hand side?
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