0

I am trying to solve a problem since a while, adding the remaining constraints along the process of learning the concepts. It's been a long journey.

The best approach (for the weak version of the problem) was nicely given in this answer here.

The remaining question seems to be how to ensure, if possible, no commutativities and no self-inverse elements - except the identity element, obviously.

The unitriangular approach is good to fit exactly the order needed for the group, but has many self-inverses due to symmetries along the secondary diagonal and probably due to other causes also.

The composition of permutations seems more resilient to such problems, but the group order does not fit exactly into $2^{128}$. It is either $34!$ or $35!$.

So, I am still in doubt if I am missing something obvious that a seasoned mathematician would spot right on. Is such problem trivial? Is it impossible?

dawid
  • 347
  • 2
    By Cauchy's theorem, in a group of even order there must be an element with order $2$, i.e. self-inverse. If further it doesn't commute with other elements, then it has more conjugates which are all self-inverses. – Berci Dec 12 '20 at 12:55
  • 1
    Also, what exactly do you mean by "no commutativities"? – Derek Holt Dec 12 '20 at 13:24
  • What others said. No can do. You will need to tolerate many elements commuting. If you can describe a precise measure, then we can work on the resulting optimization problem. Though it may be most efficient to describe the problem, and then wait for Derek Holt to give the answer. That was the algorithm followed in the newsgroup sci.math in the 90s on pretty much all the questions dealing with finding a group with specific properties :-) – Jyrki Lahtonen Dec 12 '20 at 13:31
  • @DerekHolt No (trivial) commutativities would mean $A . B \neq B . A, A \neq e \neq B$, $e$ being the identity. But as Jyrki already told us, that's impossible. So I suppose I can overcome that by detecting when the result of the operation is from a commutative pair, and apply a second operation for correction that leaves some kind of mark for me to be able to know that the correction would need to be unapplied - in case I want to reverse the operation in the future. I feel like a dog trying to bite its own tail. – dawid Dec 12 '20 at 14:20
  • @JyrkiLahtonen That seems like a property of the universe. I cannot have associativity and only trivial commutativity. Perhaps having associativity and only the trivial self-inverse is a possible bet (?). Not surprisingly, the application I intend for all of this would be really useful... if at all possible. – dawid Dec 12 '20 at 14:26
  • I've read the proof about using product of dihedral groups to achieve arbitrary small commutativity $P_r$: https://www.maa.org/sites/default/files/pdf/cms_upload/Clifton57596.pdf . However, that is far from something I can use in a computer program for a practical application processing numbers from the real world. – dawid Dec 12 '20 at 15:08

1 Answers1

3

If $S$ is a group with $128$ elements, then it has an element of order $2$, by Cauchy theorem (as Berci already noticed), that is a non trivial self-inverse element.

Worse, it is known that the center $Z(S)$ is non trivial. Hence $S$ has a non trivial element which commutes with every element of $S$. Applying Cauchy theorem to $Z(S)$ shows furthermore that you have a nontrivial self-inverse elements which commutes with all elements of $S$.

GreginGre
  • 16,641