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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?
 
@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
 
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
 
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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