How to define a non-abelian group over the set of integers $S=\{0, \cdots, 2^{128}-1\}$ with no commutativities nor self-inverse elements?

Question

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?

Answer 1

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 $S$.