Let $M$ be a set, not null, and $*$ an operation that is associative. If there exist $m,n\geq2$ such that $x^{m}y^{n}=yx$, for any $x,y\in M$, then prove that the operation is commutative.
Now, I obtained a result by $x{\rightarrow}yxy^{-1}$ such that $yx=(yxy^{-1})^my^n=yx^my^{n-1}=yx^my^ny^{-1}=yyxy^{-1}$ and from here $yx=yyxy^{-1}{\Rightarrow}xy=yx$.
However, this works only if my set is a group. How can I adapt this to the form of the set?
In the four Lemmas that follow, $z$ is an arbitrary element of $M$.
Proof: $z^2=zz=z^{m}z^{n}=z^{m+n}.$
Proof: $z^3=zz^2=z^{2m}z^{n}=z^{2m+n}.$
Proof: \begin{align} z^2 &\stackrel1=z^{m+n} \\&=z^3z^{m+n-3} \\&\stackrel2=z^{2m+n}z^{m+n-3} \\&=z^{m+n}z^{2m+n-3} \\&\stackrel1=z^{2}z^{2m+n-3} \\&=z^{m+n}z^{m-1} \\&\stackrel1=z^{2}z^{m-1}. \end{align}
Proof: Should be obvious by the symmetry of the assumptions in $m,n$.
With these Lemmas under our belt, we can solve the problem at hand.
\begin{align} xy &=y^mx^n \\&=x^{mn}y^{mn} \\&=x^{m-2}x^{2}(x^{n-1})^my^{n-2}y^2(y^{m-1})^n \\&\stackrel{3}=x^{m-2}x^{2}(x^{n-1})^my^{n-2}y^2 \\&\stackrel{4}=x^{m-2}x^{2}y^{n-2}y^2 \\&=x^my^n \\&=yx \end{align} In step $\stackrel{3}=$, we apply Lemma $3$ a total of $n$ times with $z=y$, each time eliminating one copy of $y^{m-1}$ next to the $y^2$. In $\stackrel{4}=$, we apply Lemma $4$ $m$ times with $z=x$.
Note $x^0$ refers to the empty product, so $x^{m-2}$ is valid (here is where we need $m,n\ge 2$). This is not an element of $M$, but the absence of a product.