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@TedShifrin you are right about the 8 figure it gonna always burn In the time of the slowest burning loop of the 8 and you can always find a solution to get X/3
but this is a bit of a hack , I guess my true question is : can you create 3 front of flame who are gonna die simultaneously without calculating but only using the shape of the rope
Of course the rope is non-homogeneous thus the speed of the front of flame is variable 8:33 PM
hello, is this means that R is bounded from upper ? $\forall x\in\mathbb{R},\forall \varepsilon>0, x<\varepsilon$ 8:43 PM
? @WilliamSun huh, interesting @MatheinBoulomenos Hi
I have a question about the proof of Cauchy theorem @Jacksoja hi we have G is finite and p prime dividing order of G
We want to consider this set S of p tuples such that the product of all the coordinates is 1
are you familiar with this one? yes 8:52 PM
very good
by product being 1
they mean identity right? yeah because the notation 1 never been used on this book
they used "e"
I was thinking could be product of indices being 1 but that makes no sense no 1 is used in group theory sometimes for the neutral element okay thank you
I will try on my own to solve it, that part was not clear so I skipped the problem What is the strategy for computing the determinant of a Hankel matrix? One could add all the rows to the first one and get a row of equal elements, but where would one go from there on? math.stackexchange.com/questions/3065461/… 9:11 PM
@MatheinBoulomenos all right , I think I see the idea, can I write what I understood and correct me if am wrong?  first I showed the order of S
is |G|^(p-1)
since we can freely chose the first p-1 entries
but the last one has to be the inverse
after that we can show that given any element of S
we can consider an equiv relation on S
given by a R b if b is a permutation of a
cyclic perumtation of a in S , is also in S, since we do not change order of product ( i showed this) an arbitrary permutation? Are you sure you don't want cyclic permutation? yes ! sorry np 9:17 PM
i meant, moving the product but keep them in the order that gives 1
abc = bca
for example ok after that ,we can see that some equi classes will contain either p elements
or 1 element
the identity is one such
since p is prime
we can then consider elemnt of this form
( a,a,a,a...,a) p tuple
this is also in its own equi class
now I think, showing that there exist such element would finish the proof yeah
so far so good this seems like the class equation
but I do no see what acts on what haha group action is a very good idea! 9:22 PM
my conjecture is, the action is conjugation no and the center would be elements of size one
okay let me think a min or two :) no, class equation is the wrong way to go here okay , i try other strategy if you want a hint, let me know, but I won't spoil it 9:23 PM
Okay if am stuck I will ping you for hint , but not strong hint :) the word "cyclic" in "cyclic permutation" points in the right direction for what group action to consider now am getting comfused
in the cycle notation
(12345) = (23451) right?
but what is the element that take us from one to the other?
or am not supposed to think in that way?
aha elements of S are ordered tuples so they are different elements of S but equivalent right I think in that case you can define a group action of the cyclic group of order p in S such that orbits are precisely the equivalence classes you described
I think cycle notation is more confusing than helpful here 9:37 PM
indeed haha, but I see the idea I think I think you can work out the rest we have |S| = m + pq
we have orbits of order p and 1
if I can show that m ,does not count only for neutral element I should be done
since all divisors of p
but yeah that is it
m cannot be 1 you got it! haha :D it's a cool proof 9:41 PM
yes but finding the action has been a challenge
cycle notation as you said hurt more than help here
@MatheinBoulomenos thanks so much ! the thing to remember is that if a p-group acts on a finite set X, then |X| is congruent to the number of fix points mod p
that comes up a lot in proofs I would defintely write that down in particular, if |X| is divisible by p and we can find one fix point, there must be at least another one but do you have a paper or book on group actions ? I really want to master this topic uh not really sorry 9:43 PM
since it will come up on rings and modules as they say in the book
oh no worries, thanks anyway , Ill keep looking at examples / questions from the book
I assume you mean by fix points element that are alone in orbit yes like conjugation for abelian group ,each in its own orbit
ok now it is super clear, it uses a bit of action and a bit of number theory

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