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I'm aware that posting exam questions is probably frowned upon, but this isn't homework, I think I'm genuinely misunderstanding some part of the algebra. The question is this:

Throughout this question, we shall denote by $\neg$ the relation on a semigroup $(A, * )$ defined so that elements x and y of the semigroup satisfy the relation $x \neg y$ if and only if there exists some element $s$ of the semigroup $A$ such that $s * x = y * s$

  1. Prove the relation $\neg$ is a transitive relation on $A$ for all semigroups $(A, *)$.

  2. Prove that the relation $\neg$ is a reflexive relation on $A$ for all semigroups $(A, *)$.

  3. Prove that if the semigroup $(A, *)$ is a group, then the relation $\neg$ on $A$ is an equivalence relation.

  4. Prove that if $(A, *)$ is a group, and if the relation $\neg$ is a partial order on $A$, then the binary operation $*$ of the group $A$ is commutative.

I've proven it's transitive and reflexive without too much hassle, but can't get past 3.

My approach was such that I'd start with some $s$ such that $s * x = y * s$, and then play around with this equation until I'd get some new term, in terms of $x^{-1}$ and $y^{-1}$ such that (some term) * y = x * (some term), but I keep running around in circles. Then I thought $s * x = y * s$ doesn't capture enough information about the problem in order to be molded into the solution, but I can't see what else I can add into it.

Thanks in advance, sorry about posting an exam question, but I figure this should expose something I'm missing in all similar problems.

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1 Answer 1

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For 3, in a group the relation $\neg$ is called conjugation. If $x\neg y$ for some $x,y\in S$, then there exists $s\in S$ such that $s*x=y*s$ and so $s^{-1}*y=x*s^{-1}$, which implies that $y\neg x$.

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  • $\begingroup$ Yep. Someone posted an exact solution and deleted it moments earlier, which is curious. But yeah, cheers, should have asked hours ago instead of just spelunking around the equation. $\endgroup$ Commented Dec 21, 2014 at 11:16

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