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Prove that a semigroup $S$ in which $x^3=x$ and $x^2y^2=y^2x^2$ for every $x,y \in S$ is commutative.

My attempt: From $x^3=x$, I can deduct that $x^n=x$ if $n$ is odd and $x^n=x^2$ if $n$ is even for every positive integer $n$. The problem that I can't solve is the fact that no matter what I do, I can't get rid of that $2$ that is present in the exponents of the second equation. I have also tried a lot of other combinations such as taking $x=ab$ and $y=ba$ in the second equation, but nothing seems to work. I don't see a point in writing here all of my attempts of this kind because it would take a lot of time, but if you ask m for them, I will try to write some of them. I also don't have that much experience working with semigroups. I solved dozens of problems regarding monoids, groups, rings and fields, but this is within the first problems that involves semigroups.

The semigroup's definition is very weak, and I can't even imagine how a solution to this problem would look like. Every time I plug something instead of $x$ and $y$ the problem only seems to get more and more complicated.

Shaun
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alien2003
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  • This is interesting. What's the source of this question? – Shaun Feb 09 '23 at 20:09
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    I don't know exactly. My professor gave it to me along with other problems that were selected from undergraduate contests. I will ask him in the following days if he can give me the exact source. – alien2003 Feb 09 '23 at 20:18
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    I believe I saw a very similar question not too long ago. Here's one: https://math.stackexchange.com/questions/4634054/how-to-prove-a-semigroup-with-properties-x3-x-and-x2y2-y2x2-is-commutat/4634381#4634381 – student91 Feb 09 '23 at 20:26
  • I meant this question: https://math.stackexchange.com/questions/4634097/proving-a-semigroup-is-abelian It has a link to an answer in the comments but it is a different question – student91 Feb 09 '23 at 20:30

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