rawr

:-O

@PedroTamaroff did you watch lost

Nope.

nobody has

@Mike Is it still mandatory to have two foreign languages to get a PhD in math?

@skullpatrol 2 presumably includes english :)

nope

@skullpatrol a university in my country demands the two languages to be english, german, russian or french

that may ahve been a his institution

requiremens are not universal

Laol

@IanMateus a serious computer programming language and/or Physics maybe a way out, no?

@skullpatrol I saw this in a site, so I can't say much. Highly dependent on your convincing skills.

@anon @seaturtles I don't know who to ping.

@PedroTamaroff ping me

doesn't

:14151196

really

mattress

WHAT THE FUCK

ho hum

@anon So, this is the question.

What are the possible orders of elements in SL(2,5)?

seven

:D

@anon ?

sigh

let's see here

the order is 5(5-1)/(5-1)=5

so...

WAT

120.

SL(2,5) is 2x2 matrices over F_5

oh right (25-1)(25-5)/(5-1)=120

So, I did some work.

I got elements of order 2,3,4,5.

that's, what, 120=2*2*2*3*5

so, elts of order 2,3,5 off the bat

Yes, sure.

One cannot have an element of order 8 as shown by Mariano, because $X^8-1$ is coprime to any polynomial of the form $X^2-tX+1$; $t\in\Bbb F_5$.

diag(2,3) has order 4

Yes, that's what I used.

This doesn't work with 6, because $X^6-1$ has $X^2+X+1$ and $X^2-X+1$ has factors.

so we're left checking 6,10,12,15,20,30,60

@anon Why not 40 or 30?

40 would yield element of order 8.

30 I forgot to write

Yes.

I just realized.

Good.

So $6,10,12,15,20,30,60$.

Wait.

$6$ is easy, I think.

let A be an elt of order three. then -A has order 6.

If we find diagonal matrices of order $2,3$.

@PedroTamaroff good luck with that route

3 don't divide 5-1

DERP.

=D

sure it does

$5-1 = 3\cdot \frac 43$

^ trying too hard

@anon Note you're in luck because the matrices are 2x2, so $(-1)^2=1$.

mmhhmm

So, 10,12,15,20,30,60.

do 10 the same way

Using 5.

heh heh

I'm pretty skeptical about finding an element of order 60.

@PedroTamaroff you should exclaim "o derp"

@anon ?

@robjohn Ordep.

@PedroTamaroff that works, too

@anon PSL(2,5) is isomorphic to A_5.

@anon it's a rational argument...

something is wrong with my understanding. the eigenvalues of a 2x2 matrix over F5 must be roots of unity in F25 so the order of a matrix should divide 24. but 5 doesn't divide 24.

@anon $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ has order $5$.

ah, acting nontrivially on the eigenspaces of dim>1

I don't understand one thing you said.

that's what I was missing

@PedroTamaroff F25 ain't Z/25Z

Why must it divide 24?

@anon I know.

@PedroTamaroff then what does phi(25) got to do with anything?

@anon I don't know.

@anon $\phi(25)=20$ not $24$

am I missing something?

@anon I don't know where the 24 came from.

@robjohn what does that have to do with anything?

Oh.

$F^\times$.

Sorry.

@PedroTamaroff 24 is the number of units in F25

Yes, yes.

stupid lag

my reasoning shows that any $n\times n$ matrix over $\Bbb F_q$ with distinct eigenvalues has order dividing $q^{{\rm lcm}(1,2,\cdots,n)}-1$.

@anon sorry, it seemed as if someone was mixing up $24$ and $\phi(25)$. Sorry; I'll go back to sleep ;-)

@robjohn Peter was mixing up $\Bbb F_{5^2}$ and $\Bbb Z/5^2\Bbb Z$ :)

@anon I wasn't!

pout

whether he admits it or not

(:

@anon nods

hmm, I wonder what $\exp{\rm GL}_n({\Bbb F}_q)$ is as a function of $q$ and $n$

@anon I missed your reasoning.

decomposte $V={\Bbb F}_q^n$ into eigenspaces of a general linear matrix $A$. then $A$ acts on each eigenspace as multiplication by the corresponding eigenvalue. thus, $A$'s order is equal to the lcm of the orders of its eigenvalues (which are roots of unity in ${\Bbb F}_{q^t}^\times$ where $t$ ranges over degrees of irreducible factors of $A$'s characteristic polynomial).

crickets chirping

@anon I am reading the book.

The Book or the book?

Rotman. Kinda The Book for me at the moment.

@anon I think D&F is a bit too long winded.

Though it has cool exercises. In particular, a great deal of them, and some on classification of groups which are very instructive.

@anon Do you know about character groups?

what about them

Apparently they are to best solution to showing that if $G$ is finite abelian and $H\leqslant G$ then $G$ contains a subgroup iso to $G/H$.

"Our favorite PIDs are $\Bbb Z$ and $k[X]$."

ugh sage is taking forever to download

@PedroTamaroff heh heh

@PedroTamaroff hmmm

@Mike Ever tried sake?

It happens to be nice to drink it hot.

that is a very different beast than sage

ever beaten snake?

yes

daily

@anon never. saw a gif about it

9 minutes left

@Mike I have SAGE.

I also have Mathematica 9.

sage is better

WAT

@Mike I'm lissenin to Sarah Blasko.

I approve

I disapprove of that one comment

:(

Which one? Mine?

no

the other one

Oh sure,

It is... poof.

@Mike IZ? I need some music

@KarlKronenfeld What have you been reading lately?

@PedroTamaroff Lee intro to differentiable manifolds

@KarlKronenfeld Oh. I thought you were more into Algebra stuff.

That's how you responded last time you asked me this question.

@KarlKronenfeld I am sometimes forgetful.

@KarlKronenfeld This is the only question I enjoyed answering so far today.

@PedroTamaroff Manifolds seem to be everywhere, and I get the sense that they can geometrically motivate some otherwise abstract nonsense.

@KarlKronenfeld Cool. I look forward to studying diff. geo and whatnot.

@KarlKronenfeld Alex Grüber and I decided to use $K\blacktriangleleft G$ to say $K\,{\rm char}\, G$. What do you think?

@Pedro working

@Mike Working?

@PedroTamaroff mm, sure. :P

@Pedro @Mike: Taking a break from grading exams. @Skull: I use "trivial" only in the technical sense (like "the trivial solution"); I never say a proof or a problem is trivial -- perhaps easy or straightforward ...

@Karl: Every mathematician needs to understand manifolds. Varieties and schemes are generalizations, for example, and Lie groups are important even to algebraists, too.

@TedShifrin You must hate Lang to the bone!

@Mike: How can you say Sage is better than Mathematica?

LOL ... I actually like some of Lang, but only when I know what I'm doing :P

I do not like his algebra book, but his analysis is better ... probably because he's less an expert at it.

@TedShifrin I didn't know he had a book on analysis.

It's been so long since I did any analysis!

snif

Actually, I met Lang several times. He was brusque and egotistical, but I liked him anyhow. It was funny when he gave a lecture at Berkeley and announced in the middle of it something like, "Chern, I told you to learn this years ago!"

He has a graduate analysis book which is quite nice, @Pedro. That's actually where I first learned the right proof for the inverse function theorem (in Banach space).

@TedShifrin Ah!

How nice.

Why are you being starred for swearing, @Pedro?

@TedShifrin Look at what anon did before that.

@Pedro: Can't figure it out.

@Mike: I believe most serious Ph.D. programs still require 2 languages (reading knowledge) or 1 + programming.

@TedShifrin He simultaneously talked as seaturtles and anon.

Oh, anon has morphed into seaturtles?

I knew not that.

Well, he has those two accounts.

I dunno why people bother ...

Hi

@TedShifrin Bother with what?

@PedroTamaroff I'm ready for the zone

@Mike What are you working on?

Bother with multiple personalities.

OK, I have to go back to grading. It'll be a late night.

@TedShifrin Cheers.

Tanx.

@PedroTamaroff Tutoring.

@TedShifrin Darn, I missed you.

@Mike Cool. What did they ask you?

@TedShifrin Neither Berkeley nor UCLA require 2.

What if every entry of $A$ is zero?

$A$ is nonzero

Still there are cases, I'm too lazy to type out the matrices.

The case where $A$ is the zero matrix is an easy case for the proof, so I am neglecting this choice of $A$.

Ok

@PedroTamaroff Lots of stuff

Chain rule is the most recent pain

"Have you tried thinking about it?"

I have always wanted to ask an OP that

@anon That question won my "Sillyest question of the day" award.

@PedroTamaroff There's still time left. Let's see if I can beat that.

@anon I want to do that to people who deserve being mean to

@KarlKronenfeld Do you want to co-author one?

@Mike I'm having a bit of trouble, so sure.

My first question on MSE.

Perhaps we should make a joint account like we talked about, this is our second co-authorship.

What subject do you think we should write about?

elementary number theory

hm

what about using CRT to find a mutual solution to $x \equiv 1 \mod 2, y \equiv 2 \mod 4$

lol

too dumb?

I like the idea, maybe too dumb though

here's another ida

not dumb enough

using wilson's theorem to test primality in linear time

check if $(n-1)! \equiv -1$ using stirling's approximation

why is $2\in2\Bbb Z$? I tried $2=2$ but then I can't.

@anon you're missing a factor of $\Bbb Z$ in there.

$2=2\Bbb Z$ but then I can't

@anon I love those. "but the I can't"

Wrong side

as you've written it it's untrue

Ah, with a fake account the comments would be hilarious.

in reality $2 \Bbb Z = 2$

@anon But $2=2\Bbb Z$ gives $1=\Bbb Z$.

can't divide by 2 because it's a zero divisor (0 = 2 times 0)

2 is a unit

but it divides 0

oh, since it's a unit and a zero divisor it must be the trivial ring

therefore $2\Bbb Z$ is the trivial ring :)

beat me to it

how show gcd(ab,c)=1 implies gcd(a,c)=1??? I though maybe if p divides both ab and c, then ab=qp,c=rp so abc=qrp^2. HUH??

@KarlKronenfeld Set up the account

I have to logout, right @anon?

Oh, maybe just different browser.

Email me the 'deetz

I want to too.

imitation is the sincerest form of flattery

@seaturtles I agree. Partially.

@seaturtles I just found the magnetic thingy on the book I borrowed.

So tempted.

why do people care?

@KarlKronenfeld suggest a name

@Mike "Iluvmaths"

I made up a pseudonym, we can change the name later

only 1 name change per month

(except by special request from the gods)

(which I do about once every April 2nd)

that's weird

i made a second account which I will try something with first

it should be obviously me by the name

Oh, I only created a login name. The username is still the original one. math.stackexchange.com/users/133670/user133670

Anyone owns a flamethrower?

@FernandoMartin

Hello.

@Mike amazingmathguy

ok, I am about to post

tell me how you feel

it is posted

it is not showing on the front page, curiously

so it's not the laplace transform question? (which sounds a little like a joke)

@Mike Just link to it, I guess.

I'm not seeing it.

Ah, but that would link me to a crime

Heh, through email.

@Mike HAHAHHA

I just saw it.

I really lolled.

hates the self-learning tag So good choice there.

Everything is effing self learning for crying out loud.

It still won't show for me.

Yeah, it's good @Mike

Oh, it's showing now.

"Perhaps school geometry should be taught from Euclid's Elements in the original language with a translation for those who are beginners – Henry 4 hours ago"

WTF

"your question body does not meet our quality standards." damn

@Mike david bowie is scary as fuck

haha what was in the question body

@PedroTamaroff hawt as hell

I'm getting frustrated during tutoring

Why?

@KarlKronenfeld Oh, you!

huh? ;)

Someone was trying to find the zeroes of $\cos(2x)$

They figured out (with a lot of work) the zeroes of $\cos(x)$

@Mike Aha.

but cos2x is completely different

damn you karl

If we're sharing that account, we can't upvote it.

You're right

I deupvoted it

I absolutely love "$a$ goes into $b$".

lol I lost my password

err

for the other acct

email

@PedroTamaroff Try fixing the formatting. ;P

@Mike I think you emailed it.

@Mike hm?

@KarlKronenfeld The foundations account

oh lol

isn't the answer here actually too complicated? math.stackexchange.com/questions/702586/help-me-with-gcd/…

yes

@KarlKronenfeld [probablility-theory]

Look mortals.

@Mike Do you like George Lopez? @KarlKronenfeld

@PedroTamaroff hey, should I bother you with an "analysis" question?

@IanMateus Definitely.

Not.

>=)

Just ask.

Don't ask to ask.

@PedroTamaroff no

@Mike Why not?

@KarlKronenfeld you should fix it

Suppose $\displaystyle \sum_{k\geqslant 1}f(k) /k^s$ converges for $\Re(s)\gt a$. What's the radius of convergence of $\displaystyle \sum_{k\geqslant 1}f(k) x^k$?

Awesome, got an answer from bill

I think this one is hard unless it is very easy.

"So the derivative of $\frac{\pi x}{2}$ is technically $\pi$ times one half $x$, right?"

> latexing each individual letter

@IanMateus Where are you getting that from?

@PedroTamaroff I came up with this one alone. No reason to expect a nice proof

@KarlKronenfeld omg I love arrowthingy

Interesting, @Mike. Berkeley sure did when I was there. So did MIT, and we still do here.

Ugh ... so tired of students' doing poorly on exams. Sigh.

@TedShifrin Yeah, at least the UCs don't anymore.

sup @PedroTamaroff

heya @Fernando

MIT's is now "Doctoral candidates in the Department of Mathematics must pass a language exam in one of the following languages"

Hi @Ted

well, more and more undergrad degrees in math don't require proofs ... so why should I be surprised by the degradation of graduate requirements? :D

I strongly disagree that this is a degradation

one still has to read old stuff that is not in English ... that hasn't changed

I don't object to a strong programming skill replacing a foreign language, since so much of math--even pure--now involved serious computation

I do support the programming requirement

but there's still a necessity to read stuff NOT in English ...

Right, hence the single language requirement

even though no one knows what a library is anymore

But much of that has been ranslated

a lot hasn't

sometimes one needs to read stuff not from the last twenty years :)

@anon

much of Grothendieck's work isn't translated, iirc

Pedro ... you should try my test. I bet you'd beat out most of my students! Sigh.

@TedShifrin You can totally email it.

@FernandoMartin I heard he doesn't want his work to be published.

Okey dokey ...

@FernandoMartin no need

I recall your last test being really long!

Well, this wasn't. Besides, they have 75-85 minutes.

Sent.

@TedShifrin I like that test. =D

Seems totally doable.

Well, wait 'til you do it and then see what you say :)

grr

@TedShifrin Well, I scanned it.

Well, all my tests are quite reasonable, it turns out ... Homework is far, far harder.

I understand that it might not be obvious what substitution to do for $\int \frac{x}{1+\sqrt{x}} dx$

but when you identify the substitution I don't know how you can say "I'm stuck"

You plug in...

@Mike: Students have very disappointing algebra skills.

@Mike: I'm really fond of the $u=x$ substitution

@Fernando: That explains why you've got so far in math!

All this time I thought that was unexplainable...

@FernandoMartin Is there a more obvious way to go than $x=u^2$?

I know ...

No @Mike

Well, better might be $1+\sqrt x = u$ ?