hi all. is there an equivalent of LaTeX tabular in questions on this site?

George Lowther is completely insane...

@DavidWheeler I don't like too much choice when it comes to burgers.

@KannappanSampath Thanks!

Bleh I am having problems approximating integrals =/ I am supposed to show that doubling the number of points from $T_n$ to $T_{2n}$ decreases the actual error by $1/n^2$, where $\int_0^1 \sqrt{x},\mathrm{d}x$. But that is hard, buhuhu

@ymar I don't understand what you say when you say "two squares replaced with three squares"...

@tb But in a good way :-).

Where $T_n$ is using the Trapezoid method with $n$ steps. The problem is that $f''(x)$ is unbounded on the interval so I can not find a upper limit for the error.

but why burgers? i'm not judging you...it's just one of the more unusual aspirations i've come across.....

@MattN thanx

Welcome : )

@KannappanSampath A statement from Lam's Frist Course in Noncommutative Rings. For a field k we have that these two conditions are equivalent: (1) $(a,b,c,d)≠(0,0,0,0)\implies a^2+b^2+c^2+d^2\neq 0$ and (2) $−1$ is not a sum of two squares. I can see that (1) is equivalent to (3) $-1$ is not a sum of three squares.

@DavidWheeler It requires no skills, no thinking and no resposibility. And I like burgers.

@JonasTeuwen definitely. He's responsible for my own personal favorite answer on this site.

:-).

i believe you are underestimating the first 3 quantities and overestimating the last

@DavidWheeler Oh, I really do like burgers very, very much!

I worked as a shop assistant once for two months. It was the best two months of my life. I think.

@ymar I don't see any argument at all.

May be people into field theory should know...

@Mariano for instance can help us there, I guess.

@KannappanSampath Even for the equivalence between (1) and (3)?

@ymar I did that too. Except for that I couldn't stand it anymore after one day : )

@KannappanSampath (By the way this is about quaternions so you might be interested!)

@MattN Because it was physically tiring or boring? Or something else?

@ymar I am definitely interested in anything that smells algebraic. So, no worries.

@ymar Boring and because it involves dealing with people.

@ymar I see this.

Isn't it that $-1$ cannot be written as sum of two squares?

Yes: $- 1 = 0^2 + i^2$

@MattN Only for a bit in the shop I worked in. It was a small shop, and there were people in it only sometimes. But I agree, that was the worst part.

@ymar : D

@MattN We're in an arbitrary field.

Oh, ok : ) Sorry, didn't see this was part of an ongoing conversation.

@MattN It only just started, so you haven't lost anything. :)

@KannappanSampath What do you see?

@ymar $(1) \implies (3)$.

@KannappanSampath And the other direction?

Just a sec please.

What is $\lnot(1)$.

There is an Austrian village called Rottenegg.

Is it like: $a^2+b^2+c^2+d^2=0 \implies (a,b,c,d)=\underline{0}$

And one called Fucking, Oberfucking, Windpassing, Wankham,...

:-)

@KannappanSampath I'm not sure I understand. Perhaps we should move to the CA room?

^ Look what you've done, Jonas : )

Yeah, we can.

I 'll head there.

What I have done?

@JonasTeuwen I might flag that as offensive... : )

8-).

And I might force myself into bed soon.

@JonasTeuwen Legend has it that the first place is the only place on earth where the road signs are secured with video cameras :)

:D.

executioners needed

Hey what's the evolutionary benefit of wanting to be stroked?

Should there be one?

If there should be one, it might have to be with that it creates and improves bonds?

Sounds plausible : )

Hi there, is there any moderator around?

@Mariano's gravatar is around.

Good night ding ding ding : )

@MattN Yes!

@MattN Good night!

Good night, Matt.

@MattN good night!

to the-again-seeing

@dtldarek just flag the post (if it's that purple comet comment of yours that brings you here)

that's the one

it's just that I don't want any witch hunt around here

however this seems like something fishy

changing the user name like that is not normal

@MattN not as good as the silly walk sketch :D

ok, thanks

that's the Ministry of Silly Walks

wow. that video really killed the convo in here, huh? :D

Are you sure it was the video?

well, that was the last thing said

It was also one of the first things you've again said.

Just saying correlation <-> causation 8-).

(I don't mean to offend you, it was intended as a joke...)

Hi all. Would someone be willing to help me understand a small thing about groups?

I'm just wondering how the result linked in the comment of this question answers the question.

Just ask. I'm not sure if the group people are around, but you can try...

Hey I linked it.

Oh hah I didn't notice!

Hello Kannappan.

Hello Antonio.

I don't understand the way the question is worded. But, I assume that's what OP is talking about. Let me read the question properly.

Wonderful case of trolling

Yes, the same result. @Antonio. The OP is right.

@tb Is it?

It's not trolling if I don't understand what he's saying 8-).

(And the result I linked is precisely that.)

@JonasTeuwen the question is: Why is this foo a bar? the answer: this foo is a bar.

:-).

@KannappanSampath I see that since his permutations consist of disjoint cycles of odd length then the permutations must be even (members of $A_n$). For any such element, why can we find an odd permutation that commutes with it?

I wonder what Hendrik Lenstra is doing nowadays...

@KannappanSampath (Am I reading the groupprops article right? I think that's the splitting criterion.)

Just a sec.

Oh, I am reading it wrong. We need that any permutation which commutes with such an element is even.

Wait, I think I am lost. We want to show that if there exists an odd permutation that commutes with a cycle, it must be even, no?

Yes. I agree.

So we have a permutation like (123)(45678), and we want to show that any permutation which commutes with it must be even.

I think.

So, we have an even permutation there. (123)(45678)

Yes. The requirements were that it is composed of disjoint cycles of odd length, and that the lengths of the cycles are different.

So (123)(456) isn't allowed.

Yes. OK. So, we are in case 1 now.

I was thinking of the case where it splits and was wondering why this example. :)

Case 1?

But, we know those cycle that commutes with this guy explicitly, right?

@AntonioVargas splitting happens in the conjugacy class.

(It's getting late here. Sorry about that.)

@KannappanSampath Ah, ok. Yes.

So this (123)(45678) is an element of Aspirin's set C(h).

Those who commute with this will be powers of this element or cycles disjoint from this one.

I agree.

@AntonioVargas Yes.

BRB in a few minutes.

Ok.

Back.

Welcome back.

So here is where my confusion is coming from:

Say we are in $S_{12}$ with our example (123)(45678). Then (9 10 11 12) commutes with this but is an odd permutation.

Wouldn't that mean that the conjugacy class does not split?

let's close this sad thing.

@tb It just got two more downvotes! Put it out of its misery!

@tb I just voted..

thanks.

In $A_{12}$, it won't split then? Looks weird to me.

Well looks ike 4 people on chat voted to close it!!

Sick shit!

@Antonio Distinct odd length! So, there are 5 fixed points of length 1 here.

@dtldarek thank you for spotting this! Did you see this and this?

@tb Why not people with powers that be delete this question altogether?

@KannappanSampath I voted but the others aren't around.

how much rep do you need to vote to delete?

@KannappanSampath What do you mean by 5 fixed points?

@BenjaminLim 20k if it is immediately after closure, 10k a few days afterwards.

ok

@AntonioVargas I should have said 4 fixed points. Sorry for adding to the confusion.

the 9, 10, 11, 12 are fixed by your example and they are cycles of length one after all.

@KannappanSampath Zev was there a few minutes ago and moved some of the "answers" there to the comments. So the mods are informed and we can all take a deep breath :)

@tb Right.

Yes, that "answer" was an answer in this case.

@Antonio Did I make sense?

@KannappanSampath I don't understand. To show that the conjugacy class splits we want to show that $C_{S_{12}}((1\,2\,3)(4\,5\,6\,7\,8)) \subseteq A_{12}$. But $(9\,10\,11\,12) \in S_{12}$ commutes with $(1\,2\,3)(4\,5\,6\,7\,8)$ and $(9\,10\,11\,12) \not\in A_{12}$, so $C_{S_{12}}((1\,2\,3)(4\,5\,6\,7\,8)) \not\subseteq A_{12}$.

So, you have that the conjugacy class does not split, right?

Right. So Aspirin is wrong?

No, he is right. He should demand cycles to be of distinct odd length.

But the cycles of (123)(45678) are of distinct odd length.

(123)(45678)(9)(10)(11)(12)....

Hmmmmmm. I seeeeeee....

That is why, I said, there are four fixed points. :)

Indeed. Now I understand. Thank you!

I missed this point too. I did it a few months back. I did not have an occassion to recall if not this question of yours now. So, thank you @Antonio.

Can I take this as an example of well-written question and point this to new users?

I like it. May be a few more upvotes if you feel like it -- simple but well-posed.

@KannappanSampath So if I understand correctly, the last step follows simply from the fact that one of Aspirin's permutations has at most one fixed point? Anything that commutes would have to consist of some of its cycles.

which last step? I would actually say, Aspirin uses independent cycles without definition.

(Sorry, I was not following fully.)

So I think the full proof would be...

Any permutation $\sigma$ which consists of disjoint (independent?) cycles of odd length, none of whom have the same length, must have at most one fixed point. Then anything which commutes with $\sigma$ must be composed of some subset of $\sigma$'s permutations, and would thus be even, giving $C_{S_n}(\sigma) \subseteq A_n$.

Yes, correct. The argument is true. :)

Great. I'm currently studying for an algebra exam and this has been good practice. Thanks again.

some subset--"some cycle or its power" may be a better substitute. what do you think?

Why does time go so faaaast? 8-).

@AntonioVargas As I said, I like the feeling. In this room, I take a lot of help and am capable of doing very little like this. :)

@KannappanSampath Ah, yes, it should be "some subset of $\sigma$'s cycles".

@AntonioVargas Yeah. True dat. :)

@JonasTeuwen Because, I told something correctly off late today. :)

8-).

I had bloopers after bloopers in a row in a few days and I felt like I was fit for nothing. Then, I did not understand Jordan forms and cursed myself. I hate when I cannot be precise. I know, I am not good in anything at all. This is becoming clear from my activities in the few weeks. </rant>

@tb only one of them.

Good night guys!

good night ding ding ding. @Jonas

:-).

@KannappanSampath That is how I feel whenever I do algebra.

Or try to do algebra.

Me too. but, that I like it keeps me going.

If not for that feeling, I would already be asked to leave my institute.

@Jonas: sleep well

Cya @Jonas.