Mathematics - 2013-02-27 (page 5 of 5)

anon

21:00

@TobiasKildetoft ah, I am being silly. at any rate, applying $\varphi^{-1}$ to $\varphi(H)\subseteq H$ gives $H\subseteq\varphi^{-1}(H)$; how does that get us anywhere?

Tobias Kildetoft

@anon now $H\subseteq \varphi^{-1}(H)$ for all automorphisms $\varphi$

anon

ah, so $H\subseteq \alpha(H)$ for all $\alpha$, duh

user58512

I have this group H of order 660=11*10*6 generated by <x,A_5>, how could I say that a regular normal subgroup must be a 11-Sylow subgroup?

Tobias Kildetoft

@user58512 acting on what set?

user58512

it's kind of hard to explain what set it acts on

it's not just 5 symbols

Tobias Kildetoft

21:04

@user58512 the order of a group acting regularly and transitively on a set is uniquely determined

user58512

it's 10 unordered pairs plus one new symbol

hmm

Tobias Kildetoft

do you require your regular actions to be transitive?

user58512

that's perfect for what I need, ill try to prove it

yes, must be transitive

Tobias Kildetoft

so the order of the group must be the size of the set acted upon

user58512

thanks!

I don't see why it's an 11-Sylow group though. I mean is every group of order 11 an 11-Sylow group?

(in a group whose order is a multiple of 11 once)

I don't think I even need it to be 11-Sylow here..

Tobias Kildetoft

21:09

@user58512 yes, any subgroup of order 11 in a group whose order is divisible by 11 just once, is an 11-Sylow subgroup

just look at how p-Sylow is defined

user58512

oh... this makes the Sylow theorem even more amazing

Tobias Kildetoft

why?

user58512

I wasn't fully understanding it before

Tobias Kildetoft

ahh

it is indeed a very beautiful theorem(s)

user58512

I think I can say that this 11-Sylow subgroup is cyclic too, by Cauchys theorem

Tobias Kildetoft

21:12

The Hall theorems have a similar beauty

@user58512 it is a group of prime order, so it is cyclic by lagrange

user58512

the proof of Hall's theorem is very involved!

oh that's even better

Tobias Kildetoft

yes, Hall's theorems take a lot of work (especially one of the directions)

and the "weak" version of Hall's theorem is even nicer, though I am still not sure if all the details work out

7

Q: Is there a counterexample to this weakened converse of Hall's theorem?

Suppose that a finite group $G$ contains a Hall $\{p,q\}$-subgroup for every pair of prime divisors $p,q$ of $|G|$. Does it follow that $G$ is solvable?

user58512

@TobiasKildetoft, I'm should have taken representation theory

woah

Charlie

Hi

Tobias Kildetoft

21:29

hi

skullpatrol

hi

Argon

hi

skullpatrol

low

Argon

Hilow

skullpatrol

lowhi

user58512

21:43

what is the automorphism group of a code?

skullpatrol

Here

user58512

is there anything to knot theory other than knot invariants?

knots seem so beautiful but I don't care about the invariants....

I didn't realize noam elkies supervised john conway

does that even make sense

he is 30 years younger..

I got the name wrong, it was jeremy martin

skullpatrol

22:01

What does age have to do with it?

Kasper

22:20

I feel like this now, thank you all on MSE !
"Perhaps I can best describe my experience of doing mathematics in terms of a journey through a dark unexplored mansion. You enter the first room of the mansion and it’s completely dark. You stumble around bumping into the furniture, but gradually you learn where each piece of furniture is. Finally after six months or so, you find the light switch, you turn it on, and suddenly it’s all illuminated."
Andrew Wiles

2

Argon

@robjohn Are you around?

robjohn

@Argon yes

Argon

I was struggling with this integral: $$
\int_{0}^\infty \frac{\sin x}{e^{2 \pi x}-1}\, dx
$$

It was hinted that I use a rectangle contour that has vertices at $0, R, R+i$ and $i$ and let $R\to\infty$

robjohn

@Argon Can you compute $\int_0^\infty\sin(x)e^{-\alpha x}\,\mathrm{d}x$?

Argon

@robjohn Laplace transform, ya

robjohn

22:28

@Argon Contour integration is definitely one way to go. I wasn't sure if you could use that... if not, you can use the integral I mentioned and a geometric series.

Argon

@robjohn Oh, I was trying it with contour integration (it's a complex analysis book)

robjohn

@Argon Then that is the way they want.

Argon

Yes. My major problem is from $i$ to $0$

Oh, they also hinted to turn $\sin$ to $\exp(ix)$

So along the real axis part, the integral is the one we want. Along $i$, it is the integral times $e^{-1}$

robjohn

@Argon that is usually what you do for complex integration, $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$

Argon

As $R \to \infty$ the small jump from $0$ to $i$ seems to go

robjohn

22:31

@Argon yes, it will

Argon

But what is done with the contour on return to 0 from $i$?

Supericy

This probably isn't the right place to ask this, but simple question: What syntax is used is everyone using for their math equations? Like "$e^{-1}$", what language is that?

robjohn

@Argon I get that it is $$\frac12\int_0^1\frac{\sin((1-\pi)t)}{\sin(\pi t)}\,\mathrm{d}t$$

@Supericy You mean LaTeX?

Argon

Is that little bit from $i$ to $0$ $$\int_0^1 \frac{e^{i(ix)}i}{e^{2 \pi i}-1} \, dx$$ ?

robjohn

@Argon yes, but mine is in the opposite direction so it should be the negative

Supericy

22:41

@robjohn Yeah, I think thats it. Thanks.

Argon

@robjohn Yes, yes. But this goes over a pole, no?

robjohn

@Argon Note that this blows up at $i$

Argon

Exactly. Should I loop around it?

robjohn

@Argon yes. and include the quarter circle, which is $\frac\pi2i$ times the residue at $i$

Argon

@robjohn Hmmm, the book didn't tell me that in the hint!

Will the integral not blow up then? How do I change this integral, it just now goes to $i-\epsilon$

robjohn

22:44

@Argon Do You know the Cauchy Integral Theorem?

Argon

Yep

Ok

But what do I do with the integral? I cannot find it's value, nor can I seem to be able to make it in terms of the integral that I am trying to find.

robjohn

@Argon Hang on for a bit

Argon

Thanks. I will get that residue

When multiplied by $i\frac{\pi}{2}$, it is $\frac{i}{4e}$

BenjaLim

23:02

@robjohn Hi.

@Sanchez Around?

robjohn

@BenjaLim Hey there

BenjaLim

@robjohn They're still at topology though not yet at the measure theory

robjohn

Punishment will ensue ;-)

BenjaLim

@robjohn They don't take attendance

@robjohn But the differential geometry class is better

I have been to every lecture

robjohn

@BenjaLim I wouldn't miss either

BenjaLim

23:15

@robjohn So sad Jonas is not coming here. He was supposed to but due to some private reasons could not.

@robjohn I understand you met him/ @Ilya?

robjohn

@BenjaLim Today, or ever?

BenjaLim

@robjohn I mean he was supposed to come to visit someone here. But that was cancelled.

robjohn

@BenjaLim Yes. Ilya had a stopover on the way to Hawaii

BenjaLim

@robjohn You and Ilya met on the internet yea? That was the first time you met him in person?

robjohn

@BenjaLim Ah. Good. I was afraid we had lost another

@BenjaLim Yes. We had only known each other from MSE

BenjaLim

23:17

@robjohn Two people I have met on math.se in real life. We are now close friends :)

@robjohn Out of curiousity, do you find a lot of people named Liam in where you are (san francisco)?

zero

Hi @GitGud

Git Gud

@zero Hello.

zero

@GitGud I heard you and Charlie had an argument?

Git Gud

@zero She might have had one with me. I didn't notice, to be honest. Not until it was too late, I mean.

zero

@GitGud Wasn't there a suspension involved?

robjohn

23:23

@BenjaLim I am not in SF and I know noone named Liam

Git Gud

@zero What? She was suspensed because of that little thing? I didn't even know. I guess I can see why she'd be suspended because of that, zero tolerance. But it was up to me she wouldn't have been suspended.

2

zero

@GitGud That is why I'm here zero tolerance :)

Git Gud

So I guess people here hate me now. Should be fun.

zero

Come on don't be that way... I don't hate you

robjohn

@Argon: I made an error, that the real part of that integral from $i$ to $0$ comes to $\frac12\left(\frac1e-1\right)$. The imaginary part might blow up, but not the real part

zero

23:29

@GitGud I just wanted to get your side of the story pal

Git Gud

@zero Can't you access the chat history?

zero

@GitGud Yes, but some added context would help, no?

POV

Git Gud

Everything is in there, there's not much else to it. It was the first time I ever spoke to her.

I like POV, great revolution.

zero

Oh... I didn't know about it being your first time?

...so just getting off on the wrong foot, yeah

Git Gud

@zero Not my first time on the chat, but it was the first time I spoke to her. I've been on the chat a half a dozen times maybe, that episode with Charlie was the only second time I had a conversation here. This is the third time.

zero

23:37

@GitGud Is it now?

Git Gud

@zero Yeah, as far as I remember.

Peter Tamaroff

@GitGud Dafuq happened?

Git Gud

@PeterTamaroff Not much. She wanted me to tell her where I was from because I knew she's from Brazil. But I didn't say anything. It was amusing.

3

I'm off now. See you all.

Argon

@robjohn But it is the imaginary part we want, isn't it?

zero

@GitGud later

robjohn

23:40

@Argon the answer is real, so the imaginary part should be $0$

skullpatrol

@zero Why don't you mind your own business?

Argon

@robjohn But we changed $\sin$ to an exponential so we take the complex part. Bleh, I am missing something here.

zero

@skullpatrol Was I talking to you?

skullpatrol

@zero You're a good for nothing divisor ;-)

zero

lol

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$Mathematics$ Mathematics