Proving that the binary operator of a group is commutative if the relation on the group is a partial order

Question

What's probably worst is that this is a repost, as I was stuck on 3. before. 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.

You guys helped me through 3, so now I feel like a total jerk asking for more help, but there's not much more I can do.

If anyone could give me a hint that'd be great. Basically, never proven that anything is commutative before, so bear with me. The only real equation I have to work with is $s * x = y * s$, and it's inverse $s^-1 * y = x * s^-1 $, which, one of my numerous issues is that equation exists only if $y=x$ as the thing is a partial order, which means that if I go and show that $s^-1 * s = s * s^-1$, that's meaningless, as of course something times the inverse itself is equal both ways (as the thing is a group), and if I show $x*y = y*x$ this is equally as meaningless, as I'm not showing it for any $x$ or $y$, but for $x$ and $y$ which are equal, as per the whole partial order deal.

Basically here for guidance, any help is tremendously appreciated.

Answer 1

Let $x, y\in A$. Since $(A,*)$ is a group, $y$ has an (unique) inverse $y^{-1}\in A$, and letting $s = y^{-1}$ and $e$ denote the identity of $A$,

\begin{align} s * (y * x) &= (s * y) * x & (\text{since $*$ is associative})\\ &= e * x &(\text{because $s$ is the inverse of $y$}) \\ &= x & (\text{by definition of $e$})\\ &= x * e &(\text{by definition of $e$})\\ &= x * (y * s) &(\text{since $s$ is the inverse of $y$})\\ &= (x * y) * s &(\text{since $*$ is associative}) \end{align}

Hence, by definition of $\neg$, $$y * x\, \neg\, x * y.$$ The same argument with the roles of $x$ and $y$ switched shows that $$x * y \, \neg \, y * x.$$ Since $\neg$ is a partial order, which property does $\neg$ have that allows you to deduce $y * x = x * y$?

Answer 2

As you say, by part 3 and the assumption of 4, the equivalence classes of $\neg$ are all of size 1. In my last answer, I said that $\neg$ is the equivalence relation on $A$ with equivalence classes consisting of the conjugacy classes of $A$. If these are all singletons, then this means that $x^{-1}yx=y$ for all $x,y\in A$.

If $e\in A$ is the identity element (of the group) and $x,y\in A$ are arbitrary, then $$x^{-1}y^{-1}xy= (x^{-1}y^{-1}x)y= y^{-1}y= e$$ and so $xy=yx$, as required.