Can we say that h_1g_2=g_2h_1? Unfortunately not. But let’s play around a little. We know that H is closed under conjugation, so we might try to find a conjugation. And we can! Rearranging the equation we were hoping for gives us that g_2^{-1}h_1g_2=h_1. There is no reason to suppose that that is true, but we do at least know that the right hand side belongs to H…

Should be left hand side.

]]>Do you evaluate the level of jornal ‘Discrete applied mathaematics’ as high as ‘Discrete mathematics’?

]]>(Note, without connection with the previous post, that I have a conjecture that is stronger than weaker FUNC (remplacing 1/2 by any c smaller then 1) here :

https://mathoverflow.net/questions/290906/abundance-of-full-couples

and that there is a few nice things to discuss futher about it…

]]>Let be set of classrooms. Each classroom is a set that contain a set of teachers ,. For any we denote . is a University if

and if it satisfy the following conditions :for any latex (C-t)\in \mathcal{U}$

2)

3) if then latex C^*=K^*$ then

4) say that a classroom is well defined by the set of its teachers.(note that then the only classroom without teacher is the empty classroom, and that any student in the university is the teacher of the classroom where he’s the only member)

1) and 3) tells that when somebody leave a classeroom, the people still there form a classroom of the Univesity, if and only if the one who left is a teacher of the classroom that he left. And 2) say that when a teacher leave, no teacher become a student in the new classroom….

…but some student can get promoted to be a teacher when a teacher leave, and we will say that these students are disciples of order 1 of the leaving teacher. More precisely :

If , then we define $latext(C):=(C-t)^*\setminus C^*$ to be the disciples of order 1 of $t$ in $C$. If then is then the set of all disciples of order 1 that can have in some classroom, and we can define recursively to be the set . It is not hard to see that the process has to stop for some integer $ latex j$ where is empty. We’ll state that is still empty, and note $t^U=\bigcap_{k\in \mathbb{N}}t^k$ the set of all disciples of $t$.

Conjecture of classrooms :

For any , Univesity, there exists such that .

If is the set of all union closed family with a given ground set , and the teachers of a family $C$, is the set of its generators (members that are not the union of 2 other members) then we get the FUNC in the lattice version, considering $C$ as a lattice. Indeed, iff $matex a\subset b$)

But, what is fun, is that we can consider plenty different university, The classroom conjecture restricted to downset would then be :

For any down set there exists a maximal element in such that there at least as much element that are not included in then element that are included in it.

I did not find a proof or a counterexample of this last statement, but I think even if the conjecture does’t hold for some University, classifying the universities where it holds can be a nice thing to do

]]>An alternative, if anyone is interested, would be for me to start a new post with a serious discussion of some of these follow-up problems. I would be particularly interested to disprove rigorously the strong conjecture that the “beats” tournament is quasirandom. I think it’s a doable problem.

Perhaps the best option would be to do both — put a draft on the arXiv and try to get a discussion going of the follow-up problem.

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