@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

hello, is this means that R is bounded from upper ? $\forall x\in\mathbb{R},\forall \varepsilon>0, x<\varepsilon$

?

@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

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/…

@MatheinBoulomenos all right , I think I see the idea, can I write what I understood and correct me if am wrong?

@Jacksoja go ahead

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

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!

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

Okay if am stuck I will ping you for hint , but not strong hint :)

@MatheinBoulomenos Okay hint please :D

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

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

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

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