@caveman

what

@caveman Let $G$ be a group, $g\in G$. I wanna show something Tobias said some days ago:

ok

$g$ can be uniquely written as $g=hk$ where $|h|=p^m$, $(|k|,p)=1$ and $hk=kh$.

Now, let me tell you what I tried.

Let $|g|=p^m r$, where $(p,r)=1$.

@κρανίοπεριπολία Guess... What does the following code produce?

If we take $h=g^r$ then $|h|=p^m$

\documentclass[pstricks]{standalone}
\SpecialCoor
\begin{document}
\begin{pspicture}[showgrid=false](4,3)
	\psset{linecolor=blue}
	\psframe*(0,2.25)(4,2.75)
	\psframe*(0,0.25)(4,0.75)
	\psset{linewidth=3pt}
	\rput(!1.5 3 sqrt 6 div neg 1.5 add){\pspolygon(1;60)(0,0)(1;0)}
	\rput(!1.5 3 sqrt 6 div 1.5 add){\psscalebox{1 -1}{\pspolygon(1;60)(0,0)(1;0)}}
\end{pspicture}
\end{document}

@Karl'sstudents AH! Please! Was that really necessary? Use PASTEBIN

@PeterTamaroff, ah yes, every element commutes with its own powers

@PeterTamaroff OK. Sorry!

@caveman OK, yes. But I haven't finished.

I need to find $k$. It seems it should be $g^{p^m+1}$

But I'm not sure if this idea is working.

$|g| = p^m r$, $|g^r| = p^m$ and $|g^{p^m}|=r$

@caveman Yeah.

Let $ar + bp^{m} = 1$ then what are the orders of $(g^r)^a$ and $(g^{p^m})^b$?

@caveman Let me see.

Idont know the answer

@PeterTamaroff Hey, what's up

$G$ is a group, $g\in G$.

@AlexYoucis Tobias said that a few days ago.

I want to prove it,.

I think you started with a good idea, but I am not sure if bezout completes the proof or not

Now, suppose $g=p^mr$, where $(p,r)=1$:

Then I chose $h=g^r$.

oh wait, $a$ is coprime $p^m$, and $b$ is coprime to $r$ so we are done.

So that $|h|=p^m$.

@caveman Oh, I see now.

@PeterTamaroff What the hell is that? What are $g$ etc.

that's a nice litle problem

@AlexYoucis, g is short for goat

@AlexYoucis $g$ is an element of a group $G$.

$|g|$ denotes the order of $g$

$(a,b)$ is the $\gcd$ of $a,b$.

So you want to prove that every element of a group can have the "p-part" of its order factored off?

@AlexYoucis Mmmm, well. Tobias said that given any $g\in G$ and $p$ a prime, we could write $g=hk$ where $|h|=$ a power of $p$; and $|k|$ is coprime to $p$; and $hk=kh$. And this factorization is unique.

that's quite interesting we can iterate it to write and element g as a product of elements which have orders all powers of a prime, and they all commute. Does that imply the Sylow groups for various primes always meet?

no

can you add more than a single picture to an answer?

@DominicMichaelis Yeah.

@PeterTamaroff, what does this theorem tell us about the structure of groups?

maybe it's a weak sort of version of Sylow basis

@caveman No idea, but I think I should use it to prove that the subgroup generated by elements of odd order has index a power of two.

I dont remember that

Well, it was 2 days ago I think.

Tobias said "here is a hint that may be helpful: for any $g\in G$ there is some power of $2$ such that raising $g$ to that power gives an element of odd order"

hmm

That is for $|G:H|=2^k$

if a and b have odd order, does ab have odd order?

@caveman Sounds too good to be true.

@AlexYoucis Do you have any input on this?

well the group of elements generated by odd order elements is a subgroup of G and the only things it doesn't have are some of the even elements of G

every element of G may be written as $e o$ for some commuting elements, $e$ of even order $o$ of odd

oh that doesn't help

because $eoe'o'$ doesn't necessarily equal $ee'oo'$

@caveman OK, so the $g=hk=kh$ theorem is proven with $h=(g^r)^a$

And $k=g^{p^m b}$

Where $ar+bp^m=1$.

yeah

We're taking $a,b$ to be the "minimal" solutions in Bezout yes?

it doesnt matter what a and b are

Tobias said it was unique.

can you upload .avi too ?

@DominicMichaelis I'd say you can just upload the video to YouTUBE or similar (Vimeo, Dailymotion) and link to it,.

@Karl'sstudents I give up, what does the code produce?

@κρανίοπεριπολία You have to compile it first!

@Karl'sstudents How?

pdftexify for example

call the group of elements generated by those of odd order $H$

how do we show $|G:H|$ is a power of 2

@PeterTamaroff, I think I got it

@caveman OK. Small hints please.

I don't want to give hints since I'm not sure if I'm actually right and don't want to mislead you

In fact: let P (e.g. P={2}) be a set of primes, and H the subgroup generated by elements whose orders are coprime with everything in P. Show that H is normal (hint: conjugation preserves orders of elements) and then suppose p|[G:H] for some p in P and use Cauchy's theorem to obtain a contradiction.

@PeterTamaroff yes, let $n=\prod p^{v_p}$ and suppose $g$ has order $n$. by bezout's, there are $a_p$s indexed by $p\mid n$ s.t. $p\nmid a_p$ and $\sum a_p(np^{-v_p})=1$. hence $g=\prod g^{a_pnp^{-v_p}}$, where the $h_p:=g^{a_pnv^{-v_p}}$ all commute and $h_p$ has order $p^{v_p}$. dunno about uniqueness.

errata: I mean suppose q|[G:H] for some q not in P.

@anon Yes, I was about to tell you that =)

But I knew you knew!

"hint: conjugation preserves orders of elements" Yes, that is what I used to show it is normal.

@caveman Have you read the paper I've found to you?

I forgot that I have it! thanks for reminding me

=)

wtf?

@anon Now, in my proof I did something very similar to that, man! I said "Assume now without loss of generality that $p$ an odd prime (i.e. $\mathbb P\setminus \{2\}$) divides $|G/H|=|G:H|$.

But then I fucked up, @anon.

I am glad my idea was the special case of your idea =)

right, there is some x such that (xH)^p=H, so x^p is in H, so x^p has odd order, so x has odd order, so x is in H

@anon Contra-friggin-diction.

@anon Thank you man. Made my day.

Chucky & Skullpatrol's victims

@Haytham

I needhelp with stop being lazy

I have to work hard and revise, but I don't do it

@κρανίοπεριπολία Install TeX Live 2012

hello

@MarianoSuárez-Alvarez Hola!

Hello :)

hi

hi

@caveman DO IT NOW!

how do i go about proving the residual sum of square formula of a regression line

oh wai

i got it

@GustavoBandeira - I was too busy answering a question on the site :.. but maybe I will try now

@caveman I have the same problem. Procrastination.

now how about that the slope of a regression line has a normal distribution of N(B,o^2/sxx)

this one i have no idea on

math stack exchange help, because at least Im wasting time doing math

which is what I should be doing

@TessaDangerBamkin What do you study in philosophy?

lots of things

at the moment im doing metaphysics about supertasks and whether they are logicaly coherent

Nice.

I have interest in epistemology, metaphysics and perception.

But I'm still a noob.

but next year when i get to specialise im doing time, time travel, semantics and formal methods (cos its easy maths)

I just read a few pages from some books.

Time travel?

I recommend Duncan Pritchard for epistemology intro, he writes well and happens to be from my uni

I didn't know there was philosophy on time travel.

yep we have a specialist in time travel

coursera.org/course/introphil you can do a bit of it there

anyway now you can give me my answer about normal distributions! :D I have no idea about how to prove that something follows one

I was using this one on epistemology: books.google.com.br/books/about/…

but time travel isn't real!

@TessaDangerBamkin I don't know about that.

@caveman Whatever technology we have today wasn't real 1000 years ago.

time travel cant work though

thats what it talks about

whether its possible and if it were the paradoxes that arise

Yep.

but why

its obvious that it cantbe done

Is it?

the laws of physics are obvious? yes, I suppose that is why we knew about curved space, black holes and entanglement from our caveman days.

@TessaDangerBamkin I also kinda have some interest in the limits of languange, I guess Wittenstein wrote about that.

@anon We knew?

I heard that nothing Wittenstein wrote makes any sense

from what ive heard from secondary places wittgensteins philosophy is quite coherent and I like it a lot more in some respects otherwise we all get a bit silly. he goes on about how language is a game and we just need to know the rules of each word and each play.

@anon's Transcendental nature tricked me.

@TessaDangerBamkin Do you know something about languange extension? I mean - I've heard a premise that states that we are able to know what our languange allow us to know.

not really

I supose we may be able to know more with a "bigger languange".

But I'm not really sure. I'm a noob.

most likely. we often have to define a new language to deal with new concepts. i mean maths is full of them. but yeah im a relative noob too

I have a surface level understanding of most areas of philosophy now

You're better than me.

but Im not an expert in anything apart from laughing at religion

Oh, let me find something.

@TessaDangerBamkin 3.bp.blogspot.com/-Uo7D8b5VgnM/UU756FnR60I/AAAAAAAAEcs/…

argh dont you hate it when you hought you bought something and you didnt

i have to rebuy my train tickets and now they may be more expensive >:

oh that sucks

Haha

if only you could travel back in time and buy them then

and now im scared that I did buy them

and I'd love for someone to create a better algorithm for buying cheap tickets in the UK

the internet said i could get it for £34 but I could get it for £24 after using pen paper and trial and error.

youd get a lot of money

I discovered this today:

@PeterTamaroff Can you paste it to PASTEBIN? :-)

3 decades later...

@GitGud Hello!

How are you?

@GustavoBandeira Yo. I'm fine. What about you?

Me too.

What ya doing?

Waiting for my body to dry. Just got out of the shower.

How was your day?

Same as any other.

Except I crapped two pizzas that I ate yesterday.

Oh, guys...oh guys....

do you guys knew This shit?

Sad people.

@Karl'sstudents Hi

@Charlie HI

@Karl'sstudents :D

@Charlie whilte(true){I am fine! and you?}

@Karl'sstudents I'm fine....a bit disgusted

@Charlie Hmmm...

@Karl'sstudents because of this

@Charlie I am watching this youtube.com/watch?v=COp3tq_hD6U now. OK I will visit your link now.

@Charlie Oh my ghost!

@Charlie Tatuagem no olho! Que útil!

@GitGud XD

@Karl'sstudents I love figure skating

xDD

@Karl'sstudents there are no words for that, Karly.....

@Charlie He is crazy!

@Karl'sstudents yes!!!

@Charlie Are you an English native speaker?

@Karl'sstudents nope, why?

@Charlie I am looking for some English native speaker to proofread my free tutorial on PSTricks voluntarily. :D

@Karl'sstudents Ah, hmm, try Robjohn :)

But the manuscript has not been completed yet.

:-)

@Karl'sstudents okay, well, I can read :)

@Charlie Good idea.

@Charlie OK. I will let you know when it is ready to proofread.

I don't understand this question

@caveman Which one?

here math.stackexchange.com/questions/340170/…

@Karl'sstudents okay :)

does he mean ability to compute discrete logarithm mod every prime? mod n (tbut there is no discrete log mod n)?.. it doesn't make sense