@BalarkaSen

I have a question for you that I just solved

Let $g_1,g_2,...,g_r$ be representatives of the conjugacy classes of finite group G and assume pairwise of the representatives pairwise commute. Prove G is abelian

the way I did it is by contradiction

We have $g_i \ in C_G(g_j) \forall i,j \in \{1,....,r\} \implies |C_G(g_i)| \geq r$

Now suppose G is not abelian $ |G| > r$

Suppose $Z(G) = \{1,...,n\} s.t $1 \leq n < r$ from class equation and lagrange we get $|G| \leq r$ I am pretty sure you can fill in those missing steps.

but do you have another way to prove it ?

Could you take a look at the edit part of my question: math.stackexchange.com/questions/1461063/… ? @robjohn

@MaryStar Take a look at my answer.

Are you "smarter" than a football player?

:-)

@KarimMansour You got to use the class equation there, yes.

hmm, probably there is a different way. your representatives $g_i$'s generate a subgroup $H$ of $Z(G)$. if $G = H$, done. otherwise, partition $G$ into conjugates of $H$ (given an arbitrary elt $x$, some $g_i$ represents it's conjugacy class, so you can do this). there are no more than $[G : H]$ distinct conjugates. so you have $[G : H]|H|$ elts. however, you are overcounting identity, so $[G : H]|H| - [G : H] + 1$ for that. this is clearly strictly smaller than $|G| = [G : H] |H|$.

that breaks everything, resulting $G = H$.

good

@Karim Can you prove that a group of order $300$ is not simple?

I don't think |G| = 300 is simple

That's not a proof. :P

1 sec let me factor 300

$300 = 2^2 * 3^2 * 17$ $n_5 = 1$

or it is 6

If it's $1$, you're done.

if it is 1 then we don't have anything to prove as it will be normal

yeah

What if $n_5 = 6$?

um, by the way, your factorization is wrong

5^2

not 17

it's $2^2 \cdot 3 \cdot 5^2$.

2^2 * 3 * 5^2

I am sleepy haha

so if it is 6

if $n_5 = 6$

we will have transitive subgroup of $S_6$ 6! = 6 * 5 * 24 = 6 * (100 + 4 * 5) = 6 * (100 + 20) = 6 (120) = 6 * 20 + 6 100 = 720

300 doesn't divide 720

erm, huh?

why would we have a subgroup of $S_6$? you should elaborate on that.

what do you mean by a "transitive subgroup" anyway?

suppose we act with G on the set S of sylow p-subgroups of G for that specific prime 5

so

we will get the permutation representation

how does $G$ act on those?

as a homomorphism

?

I mean, how do you make $G$ act on the set of 5-sylows?

by conjugation

ok. and then?

I gotta sleep though have alot to do tomorrow lets carry the rest tomomorow

sure.

cya tomomorow @BalarkaSen

hello!

Hola

how are you?

Good good

excellent!

Working a bit with factorials and learning how fast they overflow a 64-bit variable :P

Needed the answer to 64!

sounds pretty cool!

I just wrote another answer

Good on you :D

lol - no upvotes, BUT no downvotes, so all is good

but I put a pretty picture ...lol

nah, got rid of my answer

Hi can anyone please explain the purpose of tagged partitions in a Riemann integral?

?

may u explain more what should we explain ?

I've just started Riemann integrals and came across the concept of tags and tagged partitions. While I think I understand why we need partitions for the integral, why do we need to introduce the concept of tags?

oh those duplicated tags..

cant see what s the point, i see many symbolic tag-names derived from one subject like geometry-arithmetics,etc, and they keep proliferating

hope it wont get any worse if ever someone commit whichever of them

i dont think there is so much difference between riemann-iintegration and riemann-summation, or is it ?

I don't know yet. I just started it so my knowledge is very limited.

@Paradox101 its so much easy to make a tag

@Agawa001 why?

I mean aren't partitions enough? The concept of tags is confusing

yes thats my point

i see the faar light-spot coming nearby !

keep coming on dear, keep doing

@robjohn, how can I show that Bolzano-Wierstrass theorem implies least upper bound property (i.e. completeness axiom) in $\mathbb R$? I can't think how to make a sequence who has subsequence that converges to the sup of the given set and nowhere ele. I first thought to take a sequence that converges to the sup itself but somehow it seems 'unethical'

@Silent If $a_n\ne s$ where $s$ is the sup, then there must be some element $a_{n+1}\ge\frac{a_n+s}2$. Is that what you were thinking felt like cheating?

@robjohn, yes! how can i use that $s$ if i have not proved yet?

@Silent Well, you could take $m$ which is bigger than any element in your set and then choose the smallest $k$ so that there is an element in your set so that $a_{n+1}\ge\frac{ka_n+m}{k+1}$.

oh my gawd sometimes when you fiddle with decimal numbers, some wierd remarks and assumptions cross your mind, that this numerical structure isnt just a human invertion, everything is cyclic and pefectly-knitted as it goes though a regular loop.

Hello

@UserX hey

invention*

This may be the wrong place but it's the only math chat room I know. If anyone studies math in Ioannina, please message me, I have a bunch of questions about our confusing website's pdfs. Sorry for disturbing the rest of you.

i mad the first matlab request , sec, third, then i guessed, i f i make the fourth, i would recieve this set of numbers, waaw extrapolation is marvellous

Oh and @robjohn I addressed this to you about a year ago. Did you figure a workaround to make your script for tex in chat run on android?

@UserX I believe that there are people who use it on Android. Are you able to create the bookmark on your phone?

@UserX Look at the bottom of the installation page

Oh that wasn't there before

Nice

troubleshooting part wasnt there indeed

I can confirm that it works for me on Android, @UserX.

@Hippalectryon

this is the first time i figure out that google+ is associated to youtube :S

Google is associated to youtube. It bought it about 2-3 yers ago I think.

"bought" ?

i was wondering all thetime why are my youtube activities passing thru my g-mail, until now.

google seems to offer great services, such a gigantic server!

lol

someone's been living behind the moon

(it was bought in 2006)

lol

For 1.65 billion

apparently i live where there is no gravity , so i cudnt memorise a new once happened, i couldnt catch this information so it couldnt stabilize there inside my brain

This is nice.

Hello@Balarka

hi

this site is not quite what I was expecting

@Ghost why ?

you can ask and get answered whenever you want

not really

well, just to remind, the main is more reponsive than chat

because not all se users are in one room

that's what I am refering to

meh, its all good

you just said "this site", where it is more inclusive statement

semantics

Hi!!! Could you take a look at my question?

^ see this example, the question is put in main, besides in chatroom, and you accentuate your chance of responsiveness

already self answered my question a while ago

also wrote a separate answer, and another that I deleted

@TedS !

@Balarka \o

hi

How's it going?

alright

@Studentmath!! :)

hi @Balarka

hi

you're quite taciturn for a change

i have exams at 5th. will end at 13th.

ah ... you studying hard?

kind of.

rolls six of eight eyes

wonder what the remaining two eyes are doing.

keeping you guessing

maybe you are investing those two, to glare at the guy who parked the car incorrectly when he appears.

oh oh, @MikeM broke his vow of disappearance

@Ted: I haven't said anything wrong here, have I?

Standing room only on the bus means I'm not going to be reading.

@TedShifrin he has done that before already.

You did say something wrong at the beginning, @MikeM.

It acts by $d$ on $T^*M$, not by the identity!!

Thanks, ofc

I'm not skew-symmetrizing, am I?

$d_\nabla$ should still have the vector bundle in it, shouldn't it?

No, you don't skew-symmetrize, but you do use wedge, not tensor, on the form stuff when you do $d_\nabla$.

I'm aware. I mean when defining the higher derivatives up above

No, there you map to full tensor algebra.

Where's my map $T^*M \to \otimes^2 T^*M$?

OK, that's what I thougth

I weas mostly worried about his complaint abt the gradient. My comment there is fine, yes? (Made edits to fix the errors you mentioned.)

@TedShifrin: OK, so when we say $d$ here, locally we mean $f_i dx^i \mapsto \frac{\partial f_i}{\partial x_j} dx^j \otimes dx^i$? So it's still symmetric, we just haven't 'forced it' to be by passing to $\Lambda^2$

I think he probably is right, that he wants to talk about the gradient. The important thing is that higher-order derivatives of a function do require a connection to define. I need to leave to be fingerprinted (again!), but I'll think about this later.

@MikeM: I don't ever write $d$ for that. But what you wrote down is not well-defined.

I think he's saying the exact same thing I am. I am adamant that in gauge theory literature you do what I say you do, and I don't see what the difference is, or why you would care about $k$-fields instead. Seems like a worthless switch.

No, the hessian is not a well-defined 2-tensor except at a critical point. You actually need a connection to make sense of this.

@TedShifrin: I'm aware, which is why my answer is covered in connections. There are 2 issues here: I'm a little bit unclear what you mean by $d: T^*M \to T^*M \otimes T^*M$ when you say we act by $d$ on that part (this is my fault; I should know this). Secondly there's the difference between doing everything w/ the gradient and w/ the connection. I think these are probably trivially equivalent.

Ah, I see. So I was wrong about $d$ on $T^*M$. You need to use the Riemannian connection on $T^*M$ to differentiate that part.

I was thinking about $d_\nabla$ when I said that. Apologies.

Right... I just got that. Oops fromme too.

OK, I need to head off. Time to get fingerprinted and sent to the FBI for the second time.

It's my stop too.

If @Studentmath ever shows up, tell him I'll be back in a few hours.

Ta ta.

later

regex is a hell, i spent this evening just vainly

hi @TedShifrin @BalarkaSen

why do people get fingerprinted and sent to the FBI

o_O

@KarimMansour hello

have you heard of frattini subgroup ?

yes

I am supposed to do project on it this week

I don't know anything about it yet

so today I will use all day to solve my algebra assignment

and do it

if sent is refered to fingerprints, because they have an itouch, dont touch the screen of your ipod with your bare finger

databases are continuously updated, wait we have just one fingerprint that follows us all our life :S im screwed

have someone already spent half a day debugging 10 letters instruction ?

Hello

I was wondering if anybody would be able to help me with a big theta question?

I would really appreciate any guidance!

oops read the info on the sidebar, I'll just ask

@DanielFischer @Huy Could you take a look at my question? math.stackexchange.com/questions/1462595/…

I've done big theta on basic problems but how would I do big theta on a ln question. For example, big theta of $n^2$ ln$(n) + n$ ln $(n^2+1)$

n^2ln(n) + nln(n^2+1)

*

@Paradox what's your question?

I mean, since n*ln(n^2+1)~n*ln(n^2)=2n*ln(n) is dominated by n^2*ln(n) [because 2n is dominated by n^2], we can say that is ~n^2*ln(n) [which is even stronger than big-theta)

it's a bit ambiguous to say "the big theta of [blah]," because there will be infinitely many things that are big-theta of blah

So then one of the big thetas of n^2ln(n) + nln(n^2+1) could be n^2ln(n)?

yes

Thank you!

just writing up a more complete proof for it rn

but how did you go from nln(n^2) to (2n)ln(n)?

do you know what a logarithm is? ln(n^2) is 2ln(n)

lol so dumb

yea my bad x)

But since its n^2ln(n) + nln(n^2+1) then would big Oh be 2n^2ln(n)?

no

read what I wrote

for instance, x^2+2x is ~x^2, not ~2x^2

the 2x term is inconsequential

just as nln(n^2+1) is a drop in the ocean compared to n^2ln(n)

I've seen a notation: {x | x + S2 \subset S1}

what does x + S2 means here?

hi @anon

@FloydChen presumably x+S={x+s : s in S}

hi Ted

Hi guys, suppose to have a CTMC, if i have the stationary distribution and the distribution at time t how can I check how fast does it converge?

when I dealt with those stuff in high-school I always memorized the formula

The way I see permutation formula and why it works is as follows suppose we have 5 objects as A B C D and suppose we want to get 4 ways to arrange A B C D, so for a specific fixed choice for the first one the second one will have 4 choices, but for each of those 4 choices we can have 5 different ways of getting first choice, so we get 5 * 4. One can easily now use this following logic inductively to see that permutation of K object in n system is $\frac{n!}{(n - k)!}$.

now I am trying to get somewhat similiar intuition for the combination formula

@Karim: At the end you divide by $k!$ because you don't care about the order of those $k$ things chosen.

^ +1

I see so you remove that subset of k things because those arrangements

are equal

in a sense that we don't care about them

No, you don't remove. We're dividing, not subtracting.

How many ways are there to list the numbers 1,2,...,k?

k!

So each subset gets listed $k!$ times.

Therefore we divide by $k!$.

I see

You can be fancy and figure out a way to do this with group actions and stabilizers, too :P

there is actually a very nice proof about sylow theorem that rely on combination

combinations

I don't like the proof in DF

the proof of the first part of sylow theorem isn't bad in it

I don't remember what they have. I learned this from Mike Artin and I teach it the way he taught it to me.

but I didn't like the 2nd or 3rd one

this reminds me of an answer about combinatorial handshakes i could never understand

I like group actions everywhere, whenever possible.

I like group actions too, but you know DF sometimes what they do in their proofs they present the most neatest argument that is out of intuition in order for it to look nice

I like proofs that have intuition in it not something they pulled from their ass to make things work out

Herstein is worse.

@TedShifrin Groups, like men, are known by their action.

lool

I could never make out what DF did all over his Sylow theory chapter when I was first reading it.

It all made sense when I studied Artin and learnt what a short exact sequence is.

DF's exercises are nice, though.

yeah that fucken part when he said since Q was arbitrarily we may take $Q = P_1$ is very bad argument

like it works but it is very intuitive

unintuitive

@BalarkaSen : Btw I have a beautiful question for you in group theory when I solved it I had to even use some number theory argument in particular solutions to linear diophantine equation, altough it doesn't exist as a group. Assume G is non-abelian group of order 15. Prove there is only one possible class equation.

there is no nonabelian group of order 15, so there is no such class group. I deny to even try that problem.

its in DF

I don't care.

:P

@BalarkaSen is artin good I am considering reading it along with DF this semester

yes, it is good.

Sylow theorem, I recall I always went to combinatorics in group theory as long as I could

My safe place

@Studentmath Classify groups of order $1031$

Just kidding, it's a prime :P

@TedS I am back - I actually wanted to ask your advice, I am going to start my M.Sc soon.. I won't have many options regarding the choice of university as I won't be a regular student and I will need universities that accept that (there are few here), but I don't know how to choose the subject to focus on, and you quite need to before you begin the M.Sc

@Balarka should hand a test with "classify groups of order X" question, put some huge prime as X and see the faces of the students

hehe.

I didn't know that $1031$ was a prime, but I checked and was amused to see it was after I posted the message.

Haha

@TedS also catch up with how is it in the new place and etc..

@Balarka what are you working on now?

THERE's @Studentmath :)

@Balarka: You'll be amused to know I answered an "algebraic number theory" question:P

Not much, preparing for my exams. Last time, it was calculus, having done with singular cohomology.

@Teds !!!!

You've been hiding, @Studentmath. You doing ok in the service?

@Balarka good luck, if you need it

Extremely busy, barely have time to do anything sides work eat and sleep

Barely at home too

But yeah, it's fine actually

@TedShifrin hah. I'm sure you know more algebraic number theory than that.

Well, you're still young and energetic :)

@Studentmath I need it, thanks.

The first part is true, hopefully the second is too :P

How is it in the new place?

I'm still proud of you for how much math you assimilated so quickly ...

Enjoying the more relaxed life?

Yeah, for the most part I am. I am trying to line up volunteer math tutoring gigs, and having to do all sorts of fingerprint/background checks. They don't want a criminal cavorting with minors. :)

Indeed, @Studentmath knows much more math than I do now. I don't know any probability/analysis for one.

No, @Balarka, you guys know different things. You are a snob about applied things, and he loves them.

I wouldn't have managed without the internet and specifically this forum... Plus, it was either that or not assimilating any Math :P

@Balarka nah that's completely wrong

What Prof. Ted said is true, at least the second part

BTW, @Studentmath, did I ever tell you to look at Babai/Frankl's wonderful book/notes on combinatorics and linear algebra? You'll find all sorts of cool stuff in there.

@TedShifrin I just know a bit more topology. And yes, I agree with the applied thing you said. I don't understand any thing about applied stuff.

@TedShifrin They always do that in the US?

In the state of California, they sure do. They should care half as much about people buying guns!

@TedShifrin Not yet, I am actually looking for a book like that to keep me sharp until the Masters.. will look it up

You are living in an odd country, and I say it from here..

You can find it on the internet (probably not legally, but ...) ... All sorts of tremendous things and you might want to do something like that in your masters work.

Oh, I'm 200% ashamed of what's going on in this country.

Maybe I should come hide out with you :D

Thanks, will look it up! Will try to buy it if I manage to afford it..

It's going crazy here again, too

Yeah, Syria and the migrants everywhere is a total mess all over Europe, too.

I guess it's Canada or Australia now.. Australia has freaky animals, so..

Australia is pretty backwards on gay rights, so I'll opt for Canada :D

Most important advice for your MSc: Pick a subject you have a passion for. You don't need to have a thesis topic when you apply!

They have less cool accent, but that's much less important

LOL ... Well, I can work on my Québecois if I go to Québec :D

That's what I am trying to figure out.. I know I have a passion to all these combinatorics question and 'tricks', probability.. But I am not sure if it's enough

@Karim the idea is common in combinatorics: overcount by a factor of $m$, then divide by $m$. which is also similar to what's going on with orbit-stabilizer, as @Ted mentioned: the symmetric group $S_n$ acts transitively on the collection $\Lambda$ of $k$-subsets of $\{1,...,n\}$ (i.e. there is a single orbit) and $\{1,..,k\}\in\Lambda$ has stabilizer $S_k \times S_{n-k}\le S_n$, so the orbit has size $n!$ divided by $k!\cdot(n-k)!$.

I think I should check the last few thesis the university published in few subjects and see what interests me the most there, too

You might look for some interesting combinations of ideas, @Studentmath. That's why I mentioned that book. Fabulous questions in there.

@Ted you know french, right?

@Studentmath Bien sûr.

Have you done much with computer projects, @Studentmath?

nice

Yeah, I won't see I really know how to work things out in that area (I have seen people that really do...), but I am constantly working and learning

Specially logical programming

No, I just meant to find some projects that require computational work (that you can't do by hand) or exploration.

quebecan accent is wierd

I still want you to learn some differential geometry, too. All sorts of cool physics-y applications there.

Oh, in that

@anon: You just did the exercise I assigned to Karim! GRR :)

I still have your book on that subject on my to-read list

The Seminar I wrote was based on few articles that exploited some CS techniques and programs

You should also look into things like the discrete Laplacian and graphs, @Studentmath.

come to canada @Studentmath its nice here

no shootings 24/7 haha

Well, when Trump builds the wall to keep Americans out of Canada, that'll change :D

@Ted the book is Linear Algebra Methods in Combinatorics, right?

Yes, @Studentmath.

You really think he has a shot at the presidency..?

@Karim that sounds relaxing

No, but there are so many insane angry dumb white guys in this country ...

And a lot of liberals like me get disenchanted with the disappointing Democratic candidates and don't vote ... :(

Funny, Liberals here means the opposite of what it means in the US

Well, the politics in Israel are scary right-wing these days.

There's this bernie something who's a rather socialist, isn't it so?

Not all that much, but that's the press, yes.

@KarimMansour no thanks i rather prefer to die by heat-wave than frozen with ice

Hey could somebody help me to solve the inverse of y = (1-e^-x)/(1+e^-x)? I tried myself but got the inverse to be y = -ln((1-x)/(1+x)) which is different from the answer on wolfram alpha.

Did you solve for $e^x$, @Paradox?

Anyhow, I am afraid I have to go now - Thanks for the book and recommendations @Ted, it really helps :) Enjoy California! Bye @Balarka and hi-bye @Pedro

I swapped x with y and solved for y

so yes

bye @Studentmath

@Studentmath, I got to lne^-y = ln(1-x/1+x) then of course -y = ...

then my answer

@Paradox: Simplify your answer one step and I bet it matches.

stilll wrestling with my regex stuff, no exit ! i wont spend another 7 hours with this uggh

so -ln(1-x/1+x) = ln(1+x/1-x) ?

Yes. Why?

@TedShifrin

because on wolfram alpha it showed the inverse to be log((-x-1)/(x-1))

What is $\ln (1/x)$?

-ln(x) ?

I did not see mr @Pedro sneak in!

@Paradox: Right. So use that with your answer and simplify.

@TedShifrin so wouldn't that just simplify to -ln(1-x/1+x) = ln(1+x/1-x) ?

and x<1

With parentheses in there ...

ln(1/(1+x/1-x)) ?

If you're going to be that pedantic, @Agawa, you'd best say $|x|<1$. :)

:24494704

@Paradox, No: $\ln((1+x)/(1-x))$.

ln((1+x)/(1-x))

Right.

how did the answer on wolfram alpha of log((-x-1)/(x-1)) come to be though?

That looks wrong.

Oh, no, it's right.

Multiply inside by $(-1)/(-1)$.

hey no negative amount inside

If you're going to play with Wolfram Alpha, you have to do algebra to make sure your answer isn't the same.

wont you explode the system ?

Look at it carefully, @Agawa.

ah 1-x flipped = x-1

why did the program multiply inside by that though?

It did the algebra slightly differently from you.

It's odd that LaTeX isn't in chat for mathematics

When you solve an equation, you can move stuff to the left side or move stuff to the right. And the algebra will look different.

so my answer would still be correct even if I don't multiply by that?

ohh ic

It is ... See the top entry on the right column.

I prefer the simplest possible answer, @Paradox, which here is what I said.

@TedShifrin top entry in the right column?

Where it says Chat guidelines | $\LaTeX$ in chat ...

ohh okay!

start ChatJax

$\in$

$\in$

good evening