Mathematics - 2013-03-24 (page 4 of 4)

Peter Tamaroff

21:09

@caveman

caveman

what

Peter Tamaroff

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

caveman

ok

Peter Tamaroff

$g$ 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 $(p,r)=1$.

Karl's students

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

Peter Tamaroff

21:11

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

Karl's students

\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}

Peter Tamaroff

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

caveman

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

Karl's students

@PeterTamaroff OK. Sorry!

Peter Tamaroff

@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.

caveman

21:15

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

Peter Tamaroff

@caveman Yeah.

caveman

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

Peter Tamaroff

@caveman Let me see.

caveman

Idont know the answer

Alex Youcis

@PeterTamaroff Hey, what's up

Peter Tamaroff

21:20

10 mins ago, by Peter Tamaroff

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

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

@AlexYoucis Tobias said that a few days ago.

I want to prove it,.

caveman

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

Peter Tamaroff

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

Then I chose $h=g^r$.

caveman

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

Peter Tamaroff

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

@caveman Oh, I see now.

Alex Youcis

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

caveman

21:22

that's a nice litle problem

@AlexYoucis, g is short for goat

Peter Tamaroff

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

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

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

Alex Youcis

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

Peter Tamaroff

@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.

caveman

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

Dominic Michaelis

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

Peter Tamaroff

21:27

@DominicMichaelis Yeah.

caveman

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

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

Peter Tamaroff

@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.

Remember I tried to prove that but failed the other day?
Tobias told me that then.

caveman

I dont remember that

Peter Tamaroff

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"

caveman

hmm

Peter Tamaroff

21:39

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

caveman

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

Peter Tamaroff

@caveman Sounds too good to be true.

@AlexYoucis Do you have any input on this?

caveman

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 $ee'oo'$

Peter Tamaroff

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

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

Where $ar+bp^m=1$.

caveman

yeah

Peter Tamaroff

21:49

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

caveman

it doesnt matter what a and b are

Peter Tamaroff

Tobias said it was unique.

Dominic Michaelis

can you upload .avi too ?

Peter Tamaroff

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

κρανίοπεριπολία

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

Karl's students

21:59

@κρανίοπεριπολία You have to compile it first!

κρανίοπεριπολία

@Karl'sstudents How?

Dominic Michaelis

pdftexify for example

caveman

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

Peter Tamaroff

@caveman OK. Small hints please.

caveman

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

anon

22:11

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.

Peter Tamaroff

@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.

Gustavo Bandeira

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

caveman

I forgot that I have it! thanks for reminding me

Gustavo Bandeira

=)

κρανίοπεριπολία

7 hours ago, by Haytham

Where's the greek boy and the greek boy's girl?

wtf?

Peter Tamaroff

22:24

@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 $|G/H|=|G:H|$.

We may write $G=\bigcup_{i=1}^n x_i H$ where $x_1=e$ and $x_i\notin H$ for $i\geq 2$
In particular, $|x_i|$ is even for each $i\geq 1$.
By Cauchy, for some $x_i\; ; i\geq 2$; $(x_i H)^p=H$

But then I fucked up, @anon.

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

anon

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

Peter Tamaroff

@anon Contra-friggin-diction.

@anon Thank you man. Made my day.

κρανίοπεριπολία

Chucky & Skullpatrol's victims

@Haytham

19 mins ago, by κρανίοπεριπολία

7 hours ago, by Haytham

Where's the greek boy and the greek boy's girl?

caveman

22:46

I needhelp with stop being lazy

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

Karl's students

@κρανίοπεριπολία Install TeX Live 2012

Tessa Danger Bamkin

hello

Peter Tamaroff

@MarianoSuárez-Alvarez Hola!

TheAlchemist

Hello :)

caveman

hi

Gustavo Bandeira

23:00

hi

@caveman DO IT NOW!

Tessa Danger Bamkin

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

oh wai

i got it

caveman

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

Gustavo Bandeira

@caveman I have the same problem. Procrastination.

Tessa Danger Bamkin

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

caveman

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

which is what I should be doing

Gustavo Bandeira

23:09

@TessaDangerBamkin What do you study in philosophy?

Tessa Danger Bamkin

lots of things

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

Gustavo Bandeira

Nice.

I have interest in epistemology, metaphysics and perception.

But I'm still a noob.

Tessa Danger Bamkin

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

Gustavo Bandeira

I just read a few pages from some books.

Time travel?

Tessa Danger Bamkin

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

Gustavo Bandeira

23:11

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

Tessa Danger Bamkin

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

Gustavo Bandeira

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

caveman

but time travel isn't real!

Gustavo Bandeira

@TessaDangerBamkin I don't know about that.

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

caveman

time travel cant work though

Tessa Danger Bamkin

23:15

thats what it talks about

whether its possible and if it were the paradoxes that arise

Gustavo Bandeira

Yep.

caveman

but why

its obvious that it cantbe done

Gustavo Bandeira

Is it?

anon

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.

Gustavo Bandeira

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

@anon We knew?

caveman

23:17

I heard that nothing Wittenstein wrote makes any sense

Peter Tamaroff

Tessa Danger Bamkin

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.

Gustavo Bandeira

@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.

Tessa Danger Bamkin

not really

Gustavo Bandeira

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

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

Tessa Danger Bamkin

23:21

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

Gustavo Bandeira

You're better than me.

Tessa Danger Bamkin

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

Gustavo Bandeira

Oh, let me find something.

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

Tessa Danger Bamkin

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 >:

caveman

oh that sucks

Gustavo Bandeira

23:25

Haha

caveman

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

Tessa Danger Bamkin

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

Gustavo Bandeira

I discovered this today:

►

Karl's students

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

3 decades later...

Gustavo Bandeira

@GitGud Hello!

How are you?

Git Gud

23:34

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

Gustavo Bandeira

Me too.

What ya doing?

Git Gud

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

Gustavo Bandeira

How was your day?

Git Gud

Same as any other.

Except I crapped two pizzas that I ate yesterday.

Charlie

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

do you guys knew This shit?

Git Gud

23:44

Sad people.

Charlie

@Karl'sstudents Hi

Karl's students

@Charlie HI

Charlie

@Karl'sstudents :D

Karl's students

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

Charlie

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

Karl's students

23:46

@Charlie Hmmm...

Charlie

@Karl'sstudents because of this

Karl's students

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

@Charlie Oh my ghost!

Gustavo Bandeira

@Charlie Tatuagem no olho! Que útil!

@GitGud XD

Charlie

@Karl'sstudents I love figure skating

Git Gud

xDD

Charlie

23:50

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

Karl's students

@Charlie He is crazy!

Charlie

@Karl'sstudents yes!!!

Karl's students

@Charlie Are you an English native speaker?

Charlie

@Karl'sstudents nope, why?

Karl's students

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

Charlie

23:58

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

Karl's students

But the manuscript has not been completed yet.

:-)

Charlie

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

Karl's students

@Charlie Good idea.

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

caveman

I don't understand this question

Karl's students

@caveman Which one?

caveman

23:59

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

Charlie

@Karl'sstudents okay :)

caveman

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

Transcript for

$Mathematics$ Mathematics