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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?

the_fox
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    Your notation is inconsistent. You say you're using an operation $\ast$, yet you compute using concatenation; this is fine, I suppose, but it is a bit sloppy. – Shaun Mar 06 '19 at 19:30
  • Just a bit, not a lot. But fair point, @MikeEarnest. – Shaun Mar 06 '19 at 19:45
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    Even assuming there were inverses for $y$, from $(yxy^{-1})^my^n$ you get $y(yxy^{-1})=y^2xy^{-1}$, not $yx$. – Arturo Magidin Mar 06 '19 at 20:59
  • That's consistent with the exercise, @ArturoMagidin. (EDIT: I think I see what you mean now though . . . ) – Shaun Mar 06 '19 at 21:49
  • @Shaun: Sorry, I'm confused. What is "consistent with the exercise"? The OP starts with $yx$, and claims this is equal to $(yxy^{-1})^my^n$, and I don't see how this is justified. – Arturo Magidin Mar 06 '19 at 22:05
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    No, I'm the one that's confused, @ArturoMagidin. Sorry. You're quite right. What I thought you meant was that $(\color{red}{yxy^{-1}})^m\color{blue}{y}^n \neq \color{blue}{y}\color{red}{x}$ was the issue (without the OP's reasoning, just yours if you see what I mean . . . ). (Please excuse the handwaving colours.) – Shaun Mar 06 '19 at 22:13

1 Answers1

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In the four Lemmas that follow, $z$ is an arbitrary element of $M$.

Lemma $1$: $z^2=z^{m+n}$.

Proof: $z^2=zz=z^{m}z^{n}=z^{m+n}.$

Lemma $2$: $z^3=z^{2m+n}.$

Proof: $z^3=zz^2=z^{2m}z^{n}=z^{2m+n}.$

Lemma $3$: $z^2=z^2\cdot z^{m-1}$.

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}

Lemma $4$: $ z^2=z^2\cdot z^{n-1}. $

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.

Mike Earnest
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