@Hippalectryon Can you skip that and look first at the whole proof? :-) I don't wanna spend time with trivial things now.

Sure

I'm doing that, actually :)

@Hippalectryon Thanks. :-)

@Chris'ssis Hmm what you write is not rigorous

@Hippalectryon Yeap, I know :-)

I know you handle the cases $n=k$, but the line above where you write the $n=k$ on the denominator feels bad :/

@Hippalectryon Yeap, sure, you're write. I should change that part. Well, you see? I didn't ask you in vain.

@Chris'ssis I don't fully get what you are trying to say by "the only values that count are for $n=k$" though

@Chris'ssis :-)

Patience, @UserX. This requires a little machinery, Probably.

@Chris'ssis Hey wait :O there's something weird there

@Chris'ssis $\sin((k-n)\pi)=\sin((k+n-1)\pi)=0,\forall k,n$

@MikeMiller Harro.

@UserX Have you progressed?

:18890708 Hello Daniel!

@MikeMiller got it!(almost). Using lagrange's theorem, the order of the elements of G can be either 1,2,p,2p. Let's ommit 1(id), and 2p(cyclic). Then it's 2 or p. Then let all elements be of order 2. Then we have a subgroup of order 4, then it doesn't divide |G|

@PedroTamaroff Tardes.

@PedroTamaroff working on it(if you're talking about my prob)

No idea for the element being of order p case though.

@Hippalectryon Well, that's clear. I'll change that approach a bit anyway.

Take a group of order $2p$. Pick an element of order $p$, call it $r$; and an element of order $2$, call it $s$. Let $H=\langle r\rangle$, so that $H\lhd G$ since $|G:H|=2$. This means that $srs^{-1}=r^i$ for some $i$. Can you take it from here, @UserX?

@Chris'ssis Well then your double sums should be zero

@Hippalectryon You also have $\sin((k-n)\pi)/(k-n)$ and here is the point where I consider that limit, when $k$ tends to $n$.

You should show that either $G$ is dihedral or cyclic.

Read my posts :P

@Chris'ssis I guess my problems comes from the lack of rigor in your notation :)

@UserX Nah, I'm too lazy. What I told you is the best way to go, at least to me.

Note that if $s$ and $r$ commute, then $sr$ has order $2p$, so you can assume they don't.

In that case, show that $G$ is dihedral.

@PedroTamaroff I'm not familiar with your way :P what does $\lhd$ mean

@Hippalectryon I think you got the idea behind my proof (or?). The idea is the most important thing of all.

@UserX "$H\lhd G$" means $H$ is normal in $G$.

@Chris'ssis Well that's true, but it's weird with how it's stated for now

@Hippalectryon I know, I'm aware of it. I try to improve that.

:-)

@PedroTamaroff I barely touched cosets lately, I'm not ready to use them this much. Can you give a simpler way?

@UserX You should know what a normal subgroup is.

If you don't, go learn it.

It is very important!

I do know what it is... I'm not familiar with it though(haven't solved any exercises)

@Chris'ssis Feel free to request more proof reading if you want !

Well, the same thing was told to me about lagrange's theorem and it turned out it was really important so maybe I'll do that

@Chris'ssis Btw at the end the notation $\chi_2$ might not be clear for everyone

@Hippalectryon $\chi_2(x)$ :D

@Chris'ssis I know $\chi$ as indicator functions

*remembering the days he told @Pedro to learn important things :) *

Used in cardinality problems

@Hippalectryon OK

hi @Hippa @Chris'ssis @UserX @DanielF and anyone else I've missed

@TedShifrin help me! Most elementary way to classify groups of order 2p?

I did everything but I'm stuck at the having elements of order p case.

"Most elementary"? Yikes.

@TedShifrin Heya.

You feeling better, @DanielF?

@TedShifrin what's the point in giving a solution I cannot understand?

Well, it's sort of the point, @UserX. I have no idea what you know and what you don't.

Heya @teadawg

Up to cosets. Not normal subgroups.

Whoa ... and you're trying to classify groups?

@TedShifrin Better. Not yet fully recovered, but close enough to cancel the appointment with the undertaker.

Hello @TedShifrin

@TedShifrin =(

@TedShifrin Well, you know that he seems not to know that you don't know what he knows

:D

I'm glad to hear that, @DanielF: I took your appointment ... had a basal cell carcinoma removed from my back today. :(

Oof...

Absolument @Hippa

@TedShifrin it's quite fun...

Hi @Pedro :)

@TedShifrin I hope it was completely removed.

@TedShifrin :c aren't that cancer things ?

Yes, @DanielF, I suppose so. It's on my back where I can't even bandage it :(

yup @Hippa ... but the best kind to have (I've already dealt with worse in my arm)

@TedShifrin did you at least get opiates? :P

@UserX: You know I'm not an algebraist ...

LOL, drinking gin ... will that do?

Not techincally an opiate.

@TedShifrin Oh. Were you feeling ill or something? How did you find it?

Hi

So there are two groups up to isomorphism.

Crush an oxy and pretend it's sugar

heya @Alizter

#420Blaiseit

Even after major cancer surgery, I hardly used opiates, @UserX

(Blaise Pascal)

So where are you stuck, @UserX? (And why isn't Pedro helping?)

@TedShifrin Pedro is being... well, Pedro.

He has his charms, @UserX ... you just have to know him :D

You're done with the case $p=2$, @UserX?

I make an ansatz that there are only 2; namely the cyclic(denote it $C$) and the dihedral_2p. Those are obviously not isomorphic. Then let $|G|=2p, p\in\Bbb P$. If G is cyclic it is isomorphic to $C$ via the isomorphism $f(a^i)=b^i$. Using lagrange's theorem, the order of the elements of G can be either 1,2,p,2p. Let's ommit 1(id), and 2p(cyclic). Then it's 2 or p. Then let all elements be of order 2. Then we have a subgroup of order 4, then it doesn't divide |G|

Gathered up my work

I'll brb, getting on my laptop

@UserX No, there always exist an element of order $p$.

Don't sit on it too hard, @teadawg :D

You cannot say "let all elements be of order $2$."

Without id

Hi ted

@UserX What?

Do you know Cauchy's theorem? @UserX (probably not)

hi @Mike

@MikeMiller Hello.

ansatz is a funny word

I love the word Ansatz :)

I got glasses today.

@DanielF can explain it to non-German speakers.

Mazltov @Alizter

@UserX The only finite groups in which all (nontrivial) elements have order $2$ are $(\Bbb Z_2)^n$.

Hi also to Pedro.

Everything is so much clearer

I know how you feel, @Alizter ... but I went through that at 30, not 15. :P

pfff

I am not that young

oh sorry, 16.

@Alizter Don't be too distressed. Sometimes, it is useful to see things clearer.

hi @Karl

hi @Ted

how are you?

Better than my students, @Karl :)

I can read music from 4m away which is new.

@TedShifrin $p| |G|\implies \exists x\in G\colon |x|=p$?

(I guess I won't get to say that next year.)

@Ted You can say "Better than Mike's students" instead.

That's Cauchy's Theorem, @UserX. Do you know that? (Are you doing a course or just meandering around on your own?)

Cauchy's theorem was meandering around to understand a proof. Define course though. I'm following a bunch of textbooks

Well, it turns out my average in my multivariable calc/analysis course was lower because I graded problem 6 out of 10 instead of out of 15. I only realized that grading an extra exam today. :( I asked them to give me back their papers.

The other TAs finally got around to inputting the grades... I'm out five bucks, one of them beat me by a point. :(

So you're doing an exercise in some book that assumes nothing at this stage, @UserX? Cauchy's theorem takes more stuff to prove than what you know.

@UserX Dat notation

@MikeMiller you bet your class would beat theirs?

@Hippalectryon it's curious you point out typos and mistakes, but no solution to fix the real issue (the one with the limit) in my proof :D

good, @Mike ... you need humility :P

Totally, @UserX.

@Alizter it helps me remember :D

Besides, you owed me lunch and I bought it :D

just space it

$p\ | \; |G|$

or something

use $\mid$ for the divides

or you could get away with $p | \operatorname{ord}(G)$

oops ... \mid

@Chris'ssis The problem is that it's that part that your notation makes tricky to fully get. Hence I can't say if a problem comes from your arguments, or from your way of writing things

$p\mid |G|$

I though \mid was for set builder?

Ask me if I care, @Alizter :) I use colons for that.

If I create a question I will. But on chat I barely get myself to latex. @TedShifrin I can't render latex so it's the same thing, with or without dollars

I remember reading not to use it for divisions

fair enough

Maybe drakmas would work better than dollars, @UserX :)

Hmm

@TedShifrin I'm not that old to remember them.

\ [

LOL @UserX ... I'm infinitely old.

I remember drakmas...

@TedShifrin Stop being a protective geometer.

@TedShifrin :/

@Hippalectryon :D

So there must be an element of order $p$. Either it commutes with the element of order $2$ or it doesn't.

We're both too old.

smacks Alizter for sub-par pun

@Chris'ssis Try rewriting your proof with proper "math grammar" :)

I like lagranges theorem

Or a light definition of normal subgroups?

@TedShifrin Would you recommend approaching lie groups from abstract algebra or manifolds?

What do you mean by that, @Alizter?

Can I be oblivious to the existence of manifolds and still learn about lie groups?

You need both, @Alizter. You can do just Lie algebras more algebraically.

@MikeMiller didn't you lose the bet?

I am not at anything fancy yet, @UserX, am I?

Look up "Naive lie theory". Great book whose idea is to treat these things without knowing much manifold theory, if any.

Normality may well be implicit in what comes next, though, @UserX, although I think you can do it ad hoc.

One can study matrix groups - which are your main examples - without knowing much manifold theory at all. I don't know how the author deals with the Lie algebra, but I'm sure he does it deftly.

So if I say 'Naive' I get bashed for not using the double dotted i.

Really cool elementary book: Curtis's Matrix Groups. Get it @Alizter.

I'm on my phone so I have an excuse.

Ditto on the Curtis rec

First time @Mike's ever agreed with one of my recommendations :D

@TedShifrin $50 :O

I usually agree with your recommendations...

What is the conformal mapping that sends $\{z \in \Bbb C : |\Im z |<1\}$ to the right half space?

You both agreed on Artin I think

What have you tried, @Twink?

@TedShifrin Everything

Well, in that case, you would have found the answer.

@UserX Yeah, I did.

Everything... that I know

You could, for example, translate it up or down.

How much linear algebra should I have studied before lie?

Perhaps everything subexponential, @TedShifrin.

Indeed, @DanielF.

How much linear algebra do you know?

I'm feeling good about lie groups today since I finished my project for that class...

@DanielF: I've gotten basically no reaction to that probability write-up. I guess it sucked :D

@TedShifrin I know, the problem is to expand it conformally

@Twink: Put my hint together with @DanielF's: Translate it and then exponentiate.

@TedShifrin I think you got a couple of upvotes.

Oh, wait, you don't even need to translate. You just need to expand by a constant.

I don't know how to quantify my knowledge

Probably from you and @Studentmath @DanielF :)

@TedShifrin I didn't really get anything from your hint yet... I also have to go to sleep cause tommorow I have a ton of classes(10h). So I'll think about it and ttyl

I failed the exam anyway

Night, @UserX ... Sleep well.

Maybe I should study linear algebra first.

Sigh @Twink :(

@TedShifrin Voting is anonymous, we'll never know.

@Twink What exam?

An exam I just did

getting a cracker with blue cheese

Blue cheese :/

bashes blue cheese

I hate exams

I study all the theorey and exercises

and then I don't come up with any ideas for the problems of the exam

@Alizter If you want to study it, why worry about the prerequisites (assuming you know a basic course worth's of linear algebra)? Just read what you want and go back if it's not enough.

@MikeMiller OK.

@Twink What topic, I mean.

@MikeMiller How much does lie rely on knowledge built from finite groups?

@TedShifrin AGH. Fungi.

Not very much, @Alizter.

@Alizter Lie groups aren't usually finite.

We spend so much time on finite groups when most of the theory in that case doesn't carry forwards...

fungi? mold, yes :P no mushrooms

@MikeMiller Ah ok.

@Twink: My best advice is lots and lots of practice — like learning to do antiderivatives.

That's not true, @Pedro. There are lots of finite Lie groups.

Infinitely many, even.

@PedroTamaroff conformal mappings

Anyhow, @Twink, think polar coordinates on the right half-plane.

Volume used to cause me nightmares, until I realized I wrote down the formula wrong when I learned about it last year

Not interesting as Lie groups, thanks, @Mike.

@MikeMiller Really? But finite topological groups seem uninteresting to me.

@Twink Remember any question?

volume of what, @teadawg?

@PedroTamaroff why?

You never demanded that they be interesting, @Pedro.

So do I get Matrix Groups or Naive Lie Theory?

Sometimes @Pedro prefers boring, but not always

@MikeMiller Every object of study should be interesting, useful, or both. =D

@Twink Just out of curiosity. Maybe I can help you get them right now.

I don't know the latter, @Alizter.

@Hippalectryon I improved my previous proof, but just to a certain extent (I'm also very tired now).

@Ted I like it a lot.

Show me :-)

@Alizter Aren't you at college? Libraries, man.

I think @Alizter is another sub-college one, @Mike.

My library does not have those books.

Seriously, our mean age is younger than you and Pedro.

@PedroTamaroff thanks but the exam is over, I can't answer anymore

@Hippalectryon (did you take it?)

@TedShifrin Surface of a revolution, mainly involving areas between curves (I didn't take very advanced math in high school, so I'm making up for it now)

@Chris'ssis Yep

good @Twink

still, @Twink, you came in and asked a question, and we're trying to help. If you aren't interested in thinking, don't ask us the questions :P

@Ted I don't think it's entirely true that the finite case is uninteresting in general. I'm particularly thinking of the relation between group Cohomology and the Cohomology of $BG$; that they're identical motivates similar studies for Lie groups...

@Twink Don't you want to know how to solve them? You can learn from your mistakes

@TedShifrin you already told me the answer

@teadawg: I wish students would wait for college to get to the good math, and then do it right. The sad fact is that most colleges don't do it right, either :(

I didn't pass my first complex analysis exam either.

I think of Lie groups as a generalization of finite groups :)

I didn't pass the course

@Mike: I most definitely do not.

I'm gonna take it again

Hmm... maybe that's not quite right. I think of compact Lie groups as a generalization of finite groups.

@Chris'ssis It's indeed clearer now

OK, I'm lying. But it's fun to say.

@Mike: I still vehemently disagree.

growls and smacks @Mike

@Hippalectryon Thank you.

In one sense you're totally on the money, @Mike.

You can average over a compact group and make things invariant, just as you can in the finite case.

The typesetting of matrix groups is not the best.

I skipped two years of math between 6th and 7th grade, but my work ethic in 7th grade threw off my mathematical development. I should've finished high school taking Calc BC, I ended up finishing with Calc Honors

@Ted Fun fact: Stasheff says that for a Lie group $G$ and discrete coefficient group $A$, $H^*(BG;A)\cong H^*_{Borel}(G;A)$, where the latter is group Cohomology with Borel measurable cochains.

Is this the right generalization of the finite result? ...I dunno.

@teadawg: I find that most people don't learn enough in BC to skip Calc II, but I suspect you've learned more than your share of it. You're way beyond most mathematicians on integrals. What about Taylor and series?

I did hardly any work before I was 14.

@Mike: I dunno what you're talking about.

@BumSkeeter The number of the fastest way to move all the pieces is the part "necessary" or "sufficient"??

@Ted I don't believe you...

I think you just want me to shut up.

LOL ... think what you want.

@Ted The calc book I breezed through didn't go into much detail on Taylor Series (Stewart Calc: ET 7E). I'm fairly decent with series, but I'm not very good with using them with integrals

@TedShifrin I think matrix groups covers everything in naive lie theory. However it may be too much. I will start with naive lie theory as it is cheaper. And less dense.

@Hippalectryon well, things should be written in a simpler form though, and then I make them completely clear. Wait a bit. No need for the ugly part with the limit.

@teadawg: If you want to email me, I can send you my exams from my calc theory class (I've sent them to others on here). Not as good as taking the course, but it'll give you something to work on.

@Alizter: I can't comment, as I don't know both.

Wow hardcover is cheaper than paperback

@Chris'ssis Ok

@TedShifrin What's your email ?

it can easily be discerned from my profile, @Hippa

@Hippa hey, that's my line >:(

@teadawg1337 ?

I was joking, nvm...

Oh indeed @TedShifrin

@PedroTamaroff

I wish y'all would stop changing your pings

@Twink Yes?

this is the exam

No habla Espanol

wow, I had no idea @Twink was in the Spanish-speaking portion of the hemisphere

lol

@Twink: Don't make it disappear if you want us to look at it!

But I've spoken in spanish here a lot of times

Making a lot of stupid mistakes ... (damn...being terribly tired)

Not that I've seen.

@TedShifrin do you speak spanish?

You and Pedro can commune. No, I speak French, German, and a bit of English.

save it

OK

GRRR

So, could you do point 1? I actually solved that same exercise some days ago.

@TedShifrin Here.

no

Well, they all look standard, but you need to have done your homeworks. @Pedro had a few problems similar in recent times.

@TedShifrin Sure, I'd enjoy that

I know I should had used Rouché's theorem

but I didn't find the other function

Yes.

Well, you're asked to look at $|z|=1$ and $|z|=2$.

#2 is the conformal invariant of an annulus ...

I had never seen a problem like that

just in the unitary disk

for |z|=1

If you look at the individual terms, note that the term that leads when $|z|=1$ is $4z$.

So you could take $f(z)=$ your function and $g(z)=-4z$. You should be able to prove that $|f(z)-g(z)|<|g(z)|$.