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8:07 PM
@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?

8:52 PM
very good
by product being 1
they mean identity right?

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

9:17 PM
i meant, moving the product but keep them in the order that gives 1
abc = bca
for example

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

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 :)
@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

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

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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