Mathematics - 2016-10-24 (page 1 of 5)

@TedShifrin can I discuss with you something quick ?

Hi @PhysicsGuy

You just did, Karim :P

haha

So I just want to know if something is correct.

Suppose p is a prime and G is a group such that $|G| = p(p + 1)$. Prove that G has normal subgroup of order p or a normal subgroup of order p + 1.

so using counting argument I showed that there are p elements outside of all p-sylows.

@PhysicsGuy

i.e Let P be a sylow p-subgroup of G. Suppose it is not normal. Using that assumption I extracted that there must be p elements outside of all p-sylows.

12:07 AM

Um, assuming not just 1, then you account for $1+(p+1)(p-1) = p^2$ elements. Agreed.

yeah

call me crazy but ive never been interested in physics

I love physics, @user3502615 ... just don't know enough.

Physics Guy

@user3502615 I'm not calling you crazy

Some people don't like physics. that's just normal.

@Adeek by "counting argument" do you mean principle of induction?

12:08 AM

@user3502615 no

counting argument as @TedShifrin has shown above.

You need to use Sylow, though, @user3502615.

Then Let A be the set of elements that don't lie in any of the p-subgroup of G. We showed that $|A| = p$. Let q be any prime dividing p + 1. Then A must contain some element of g of order q ? Is that true ?

Agh, Karim. Don't make this complicated. Make it simple.

what would you suggest?

Physics Guy

Ok, I leave, bye ;)

12:10 AM

Bubye, @PhysicsGuy.

Hmm, I was being stooopid, Karim.

hm I mean if I show that A must contain some element g of order q. Then everything follows smoothly

But for any prime $q|(p+1)$, Sylow tells you there can be only $1$ subgroup.

but why is that the case (smoothly not in the case of infinitely differentiable haha)

Haha.

Oh, not if $p+1=q(q+1)$. Hmm.

@Adeek

sorry

12:12 AM

yeah by cauchy theorem we know that such an element exist but why can it be in A ?

Because $q\nmid p$.

ohh I see

I bet there's a good group actions argument that's less cumbersome.

yeah

is it true that $N_G(P) = P$ ?

What's $P$?

12:15 AM

P is the sylow p subgroup of G.

@BalarkaSen That's rather annoying. Is there a finite number for which that many points will correspond to a unique parabola? Thanks for giving specific counterexamples.

What do you know about the normalizer and the number of conjugate subgroups?

Hi @Axoren: Love it when you're annoyed :)

Most people do, it's a curse.

I'm sure more people try to annoy me than try to annoy you :P

You just know more people, likely.

12:16 AM

the number of the conjugates of the sylow P subgroup is given by the index of the normalizer.

Rather, more people know you.

Right, Karim.

So I win by default, @Axoren.

so we have $[G : N_G(P)] = p + 1$

Karim: Can you argue that the identity together with the other $p$ elements not of order $p$ must form a subgroup?

oh ok yeah I agree

12:19 AM

Yes, right. So what do you infer?

yeah $N_G(P) = P$

@TedShifrin At the very least, you're top billing.

"Most hated," you mean, Axoren?

so if we have g of order q it can't be inside of $N_G(P)$

First name on the sign.

12:19 AM

A lot of good that does.

let x be a generator of P. Then all powers $x^i$ are generators of P. Then none of the $x^i$ centralizes g.

Who's $g$?

otherwise we would get $g \in N_G(P)$.

g is an element of A of order q.

Oh, an, not the.

yeah

Therefore we have $\{g,xgx^{-1},...,x^{p - 1}gx^{-(p - 1)}\}$

are all distinict

all of them have order q.

Then, each of them must lie in A, so A must be that set above.

12:22 AM

Well, $A$ has way too many elements.

So you're arguing that anything in a $p$-Sylow subgroup commutes with the guys left out?

So every element of A has order q therefore q must be the only prime dividing p + 1, so p + 1 is a power of q, and so $A \cup \{e\}$ must be a unique sylow q-subgroup of G.

Nonsense, Karim

Take $|G|=5\cdot 6$.

A has p elements @TedShifrin

Actually, groups of order 30 are sorta interesting, as I recall. It's been 5 years or so since I last taught algebra.

wow

12:24 AM

$12$ is also interesting. Both are of the form $p(p+1)$.

So try actual examples to test out your reasoning.

hm what you don't like about my argument ?

oke

$6$ is not a prime power.

I proved that A which as a set has cardinality p. I proved that every element of A has order q.

That's clearly wrong, Karim.

Take $p=5$, $p+1=6$.

oh

12:27 AM

Uh huh. :D

yeah that works @TedShifrin

What works?

I think my argument works as well

in this case

Can anyone in here help me show that the map provided in the answer to this question is surjective? math.stackexchange.com/questions/1972519/…

Karim: Seriously? $6$ is neither a power of $2$ nor a power of $3$, so you'd better fix your argument.

12:31 AM

oh I see

wait a sec

@Jessy: You're making life too difficult. If I take any nonzero rational number $a$, what is $\langle a\rangle$?

(Hint: The same is true in any field.)

The cyclic group generated by it.

Oh, I was doing rings. What are you doing?

groups

yes I see my issues @TedShifrin

12:33 AM

You need to clarify that in the question (or did I miss it?).

yeah

Yeah, I specifically mention groups in the question, @TedShifrin

OK, @Jessy, so let's look at a concrete example. (OK, sorry.) $a=1$ and $b=7/9$.

What's the isomorphism?

hm yeah

p + 1 is a power of q or it could be a multiple of q with something else.

Honestly @TedShifrin I don't know. Is that supposed to be somehting I'm supposed to be able to tell just by being given two numbers?

12:34 AM

Yup. :)

How??

but then why must it be unique ?

Hint: Whatever gets killed in the first quotient group has to get mapped to something that gets killed in the second quotient group.

hm

By killed, what do you mean?

12:35 AM

Become 0 in the quotient.

You mean 1

this is an additive group

Yes, and ...

what gets killed in the second?

Well, 1 in the first group is already dead

maybe I should ask it as a question

Karim: This is in all sorts of books. Dummit and Foote surely do this.

You were on the right track. I'm claiming that everything in $A$ together with the identity must be a subgroup that commutes with the $p$-group.

12:36 AM

In order to turn $b$ into 0, it needs to be added to its reciprocal

i mean it's negative

Not its reciprocal.

I see

No. Addition of a number is not a group homomorphism!!

$0$ has to map to $0$.

Usually

ALWAYS.

(FOR additive groups)

I know. sarcasm.

12:38 AM

At least always.

I never can read sarcasm in chat.

But how does that show me that f: x->x/a is surjective?

Unless it's mine.

Huh?

Poe's a jerk for inventing Poe's Law.

That's what I need to show in order to understand the answer someone posted to my question.

12:39 AM

Stick with my example. What homomorphism are you going to take $\Bbb Q/\langle 1\rangle \to \Bbb Q/\langle 7/9\rangle$?

I don't know

Review what I said.

something that maps 0 to 0

that's what I understood from what you said.

When I talked about killing things.

okay, f: x -> 9/7x?

12:40 AM

Yes, the equivalence class of 0 must map to the equivalence class of 0. Indeed.

Why 9/7?

x->-7/9x?

Stop and think, dammit.

Don't just guess.

I'm not just guessing. I am a very visual person and I have a very hard time understanding something unless I see it.

I don't really understand what you mean by killing things!

Well, you're wrapping an interval of rational numbers around and making a circle, identifying 0 and 1 in the first creature. Similarly, identifying 0 and 7/9 in the second creature.

When you look at $G/H$, you set things in $H$ equal to the identity element in the new world. That kills them all off.

What do you mean "identifying 0 and 1" and "identifying 0 and 7/9"

12:42 AM

I mean identifying. They become the same in the quotient.

Doesn't your teacher explain anything?

No, apparently not.

This isn't helping.

I'm sorry I asked.

OK ... I won't address any more of your questions.

Well, it would help if you didn't get so mean about it.

I'm not being mean. I am being frustrated, yes. But it's better that I just ignore your questions.

"think, dammit" and "doesn't your teacher explain anything" aren't helpful.

12:44 AM

Are you paying my salary here? :D

I wish I were.

I agree I'm not infinitely patient today. I will let you get help from others and I'll not deal with it.

Hi ted.

Jessy

Hi 0celo.

Just finished typing my review guide for analysis

@TedShifrin Also, I think $C^1$ and strongly differentiable are equivalent.

Strongly differentiable at $x$ implies $df(x)$ is continuous, and vice-versa.

So they're equivalent.

12:56 AM

Well, @0celo, I realized (and I think I commented to you) that there's a question of where.

Of where?

Strongly differentiable at $p$ is $C^1$ at $p$ plus knowing the function is differentiable near $p$.

If you work on an open set, then, yes, they're equivalent.

@TedShifrin Yeah.

The proof isn't trivial though (imo)

I still don't think I would ever teach this stuff, but it's a moot point now :)

No, it's not.

As with many things, one needs to cleverly use the mean value inequality

12:58 AM

If only a small segment of advanced researchers ever use this, I don't see the point of teaching it.

the $||f(x)-f(y)||\le M||x-y||$ thing

Oh, sure, mean value inequality gets used a hundred times in my multivariable math book.

Well, I haven't counted, so maybe that's an exaggeration.

@Axoren 5

Oh oh ... Balarka didn't sleep again.

any conic is determined uniquely by 5 points

@TedShifrin I did actually, this time!

12:59 AM

Indeed ...

Any cubic is determined by 9. :)

what time is it in your part of the world?

Yup

Too late.

about 6:30

8:59, and I want to go to bed somet4ime tonight.

12:59 AM

I can never get used to time zones being off by a half hour (mod 1 hour).

Daylight savings something something?

@BalarkaSen Thanks. Is there a succinct argument for why that's true?

In the US daylight savings shifts by one hour.

@Balarka: Do you want me to answer so you can get ready for school?

Yes @Axoren :P

@Axoren Totally! Think of a generic conic as a polynomial of 6 terms, with 6 coefficients each term.

Huh?

Six terms, actually.

1:02 AM

Fixed.

6 coefficients TOTAL, yes :)

Oh, that makes sense.

And passing through a point is a linear condition on those coefficients, @Axoren.

I meant one coefficient for each term.

But the equation can be multiplied through by a nonzero constant, so that takes away 1 degree of freedom.

1:03 AM

Yah. So you're going to boil down to intersecting 5 hyperplanes in a 5-space

That's a unique point, generically.

Only if you projectivize.

Five hyperplanes in 6-dimensional vector space.

You should have let me take over :P

I'm shoving that compactification under the rug.

I should just leave. :0

wait, no

Nah, go on. I'd leave soon anyway

1:05 AM

I have a question about the definition of $C^2$

LOL. Have a good day at school, @Balarka.

When you start talking about projective spaces, that's where I know very little about how the geometries work.

And I start thinking less geometrically than I normally do.

Hence the shoving under the rug. But Ted's mad about that, so...

That's why I left it out. You get 5 hyperplanes in $\Bbb R^6$, @Axoren, all independent, so they intersect in a $1$-dimensional subspace.

That $1$-dimensional subspace corresponds to $c(a_0,\dots,a_5)\in\Bbb R^6$.

Hardly "mad," @Balarka.

When you say subspace. Which type of space?

Linear or projective?

1:07 AM

Linear.

Huh. Okay.

I said $\Bbb R^6$ :)

A line in $\Bbb R^6$.

I don't immediately see why it's a line in $\mathbb R^6$

So your conic has the equation $a_0x^2+a_1xy+a_2y^2+a_3x+a_4y+a_5=0$, or any constant multiple thereof.

Independent equations each reduce the dimension by $1$.

i wish i were dead.

1:08 AM

Oh, the coefficients.

That's clearer, actually.

@TedShifrin A function $f:U\to\Bbb R^n$ is differentiable at $x\in U$ iff each $f_i:U\to\Bbb R$ is differentiable, right?

Yuppers @0celo.

You want $\to \Bbb R$.

@TedShifrin Just joking. Also, it's not school today, just have to go to a physics tuition I signed up for, but now regret for it.

@MikeMiller thanks.

1:09 AM

tuition ≠ tutorial, @Balarka :P

@BalarkaSen Tuition?

grr

You ok with it @Axoren? Lots of pretty arguments like this in baby algebraic geometry.

I've taught Balarka both to smack and to grrr. My life is complete.

It makes sense. It's just weird to think about it like that at first.

It's pretty weird, yes

1:11 AM

Well, you're learning about projective duality without even knowing it :)

But you just learnt about the moduli space of conics

See, @Axoren, now you know soooo much :P

@JessyCat If you mean this, there are plenty of people and organizations you can talk to, and should. If you don't, you shouldn't joke or insinuate.

Is this somehow a group action? Because $\{a_i\}$ is from $R^6$, but then you have this other bit $x^2, y^2, xy, x, y, 1$. This seems like a linear transformation with two different fields.

OK, gotta go.

1:14 AM

Bye, @Balarka.

No, @Axoren. We're just looking at the vector space of (non-homogeneous) quadratic functions. It's a 6-dimensional vector space.

But when we set one = 0, constant multiples don't matter. That's why God invented projective space.

Right, there's an equivalence between anything on lines intersecting the origin.

There is a group action of $\Bbb R^\times$ (nonzero reals under multiplication) on the vector space.

That's what you just said, actually.

Yeah, I was wondering if that's what it was.

I might have said it backwards

But it's either one of those groups acting on the other.

I would just say the points of each $1$-dimensional subspace are all identified.

Huh? Only one group so far. :)

Alright, that's where I'm missing it.

1:17 AM

Group acting on vector space.

I was under the assumption that all vector spaces were groups.

@TedShifrin Ok, it basically amounts to showing $df(x)[h]=(df_1(x)[h],\dotsc, df_n(x)[h])$

But you don't want that. As an additive group, a vector space needs a bazillion generators.

I see. We avoid calling it a group to avoid justifying it as a group. The fact that it's a vector space is enough for group action. The fact that it's a set is enough for a group action.

Sure, @0celo, but that is clear. Just like you can always break down a linear map $T\colon V\to\Bbb R^m$ to $m$ linear functionals.

Not justifying, @Axoren. That's too little structure.

1:19 AM

@TedShifrin yeah but you have to show that each component is $df_i(x)$

Not difficult, 0celo.

I didn't say it was

Just checking

When defining a group action on a vector space, is it generally easier? I haven't really done it enough to know if the structure has some streamlining benefit.

Easier than what?

I've only really approached the concept in the barebones setting of $G \times A \to A$ where $A$ is just a set.

1:20 AM

That's fine.

I was just objecting to your thinking of the set as a group. :)

What does the additional structure of a vector space $A$ give in that context?

It means multiplication by nonzero scalars makes sense :)

Wait.

It means that the map $A \to A$, given by $v \mapsto gv$, is linear for all $g$.

Do we really need that, @MikeM?

1:22 AM

Oh jeez, that's huge, actually.

@TedShifrin Huh? Yeah?

I just need to know that multiplication by a scalar sends a line to itself.

I mean, ultimately, it matters, of course, when I want to define mappings on projective space, etc.

A group action on an object should preserve the structure of that object...

Which means it preserves linearity.

That's funky.

Well, ok, but Axoren was just talking about the vector space as a set. I don't disagree with your wanting more structure.

1:23 AM

@TedShifrin He asked what the addition structure of a vector space gives... It gives two things.

Did he ask that? All I care about is scalar multiplication. :P

I was objecting to his thinking of the vector space as an additive group.

OK, presumably I didn't pay enough attention to the conversation.

Sorry for butting in.

I do that a lot. Nah. Butt in entirely.

Additional structure, not addition structure.

But your chiming in was still helpful.

@Axoren: Agreed. But originally you wanted to think of the vector space as an additive group, remember?

1:25 AM

Typo.

That's when I objected.

Yeah.

Anyhow ... I think you understand.

I need to learn function spaces better.

It's one of the topics of my Monday Wednesday classes

Supposedly, we're going to be talking about Hilbert spaces, and might touch upon space-filling curves.

Nice topic.

I don't see space-filling curves connected to Hilbert spaces, though. Maybe the Hilbert cube, though.

1:29 AM

@TedShifrin what's gonna be on a first semester analysis midterm?

Don't ask me, 0celo. Ask your professor.

@0celo7 Analysis

I've already said I'd never teach such a course the same way.

Good answer, @Axoren.

@Axoren well, or topology

Also, spelling.

If you get your name wrong, the wrong person gets the grade and you end up with a 0.

1:31 AM

He probably knows my handwriting

That's good. One less subject to study.

:P

LOL

Has he told you whether it's all proofs, some examples/counterexamples, etc.? Such courses often ask for examples or counterexamples.

Surely you know all this crap anyway.

@MikeMiller If he asks me to prove $\ell^p$ is Banach I'll be in trouble

Knowing things and knowing the proofs are different

That's not an unreasonable proof if there's time.

1:33 AM

although I did work that one out for extra credit on a homework

@TedShifrin not if you need to prove the Minkowski inequality too

But if it was only extra credit homework and not done in class, that's not a reasonable test question. ... No, I would accept basic facts.

But something that appeared only on extra credit should not show up on an exam.

Minkowski and Holder.

I don't think Minkowski is absurd to prove.

There's only an hour and a lot of stuff has been covered ...

> Included: all the handouts posted, including HW solutions (be sure to check you have the latest posted version)
Sections in Fleming's text: 2.4 to 2.10, 3.1, 3.3, 4.3, 4.4.
You may be asked for definitions and statments of theorems (in addition to short proofs.)

Not that it's important to be able to do tedious manipulations.

1:36 AM

dropbox.com/s/wxfvpiz9383bj22/m447%20study%20guide.pdf?dl=0

this is basically what we've done so far

I typed this up while reviewing

Fleming. So he's teaching differential forms and Lebesgue integration?

yes, in the second semester

That course was my one B as an undergraduate. I've always found that ironic :P

idk why he uses Fleming

all the Banach space stuff is not in there

nor is Arzela-Ascoli, Stone-Weierstrass

I like the book, but it's very idiosyncratic. Well, he's off the deep end. Fleming is plenty hard enough. ... That's all single-variable, Fleming is multivariable analysis.

1:39 AM

He's off the deep end?

I've already said that's my opinion several times.

Single variable? Check page 3 of the notes

We did all of those in great generality

I mean they standardly appear (e.g., in Rudin) in a single-variable analysis course. They can be put in more general settings, yes.