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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.

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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 $y * x = x * y$?

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  • $\begingroup$ You pretty much spelt it out, but that was brilliant, as answers go. Thanks a ton! It's easy to see how it works once written out. Are there any tips or tricks you could give me, common stuff to try when trying to arrive at this sort of proof? $\endgroup$ – intoTheMandelbrot Dec 23 '14 at 17:01
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    $\begingroup$ @intoTheMandelbrot I need more information on the kinds of problems you're dealing with. There weren't really any "tricks" in the explanation above, but when dealing with a problem of partial order where you have to show equality between two elements $a$ and $b$, you want to get $a \le b$ and $b \le a$, so that by antisymmetry you can claim $a = b$. $\endgroup$ – kobe Dec 23 '14 at 17:23
  • $\begingroup$ I'm preparing for a scholarship exam, and whilst the basic course covered the basics of groups, we went only as far as proving something is a group (as in, verify the identity, the inverses, associativity etc.) and very briefly touched on homo and isomorphisms. So I started doing past exam papers, and this was one of the many which cropped up. The questions are, I believe, supposed to test one's knowledge of groups, and how well one understands the concept, so I figured if I did all the exam papers and understand the solutions I should have a good feel for the thing. $\endgroup$ – intoTheMandelbrot Dec 23 '14 at 17:28
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    $\begingroup$ Then my advice is to find an abstract algebra text for reference, keep doing more exam problems, and ask around here if you have any questions - there are several people here who can break things down for you so you can grasp the concepts. $\endgroup$ – kobe Dec 23 '14 at 17:58
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    $\begingroup$ You're right, I've totally neglected the library. You've been wonderful, thanks a whole lot! $\endgroup$ – intoTheMandelbrot Dec 23 '14 at 18:01
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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.

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  • $\begingroup$ Thanks again, I'm terribly sorry to be a nuisance. I've got you up until conjugate classes, which I'm trying to wrap my mind around. So entities in the set are partitioned by whether or not there exists an $s$ such that $s * a * s^-1 = b$, which is the definition of the relation just played around with some (stop me there if I'm wrong, I get the feeling I'm wrong). Then as per partial order, these classes can only be of size one, as only things equal to each other may be within the class. How come does the fact that these are singletons imply $x^-1 * y * x = y$? $\endgroup$ – intoTheMandelbrot Dec 23 '14 at 17:22
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    $\begingroup$ For every $y$ in $A$ there exists an $x\in A$ such that $x^{-1}*y*x=y$, the identity of the group $A$. Alternatively in a group $x\neg y$ if and only if $x$ and $y$ are conjugate, and so you already showed that $x\neg x$ for all $x\in A$ in part 1. $\endgroup$ – James Mitchell Dec 23 '14 at 19:22
  • $\begingroup$ Hm, am I misunderstanding? So $x$ in $x^-1 * y * x = y$ is, some $y$ multiplied by the identity and it's inverse, analoguous to writing $e * y * e^-1 = y$? But then you wrote for all $x$ and $y$ members of $A$, so I think I'm seriously misunderstanding. Again, really sorry to be taking your time up with what seems to be really basic to you. $\endgroup$ – intoTheMandelbrot Dec 23 '14 at 19:42

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