Mathematics - 2018-12-10 (page 1 of 3)

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12:26 AM

I am trying to use the "method of shells" in calculus. I think I understand it when I understand where to get (y) the height of the "shell". But most questions I see phrase the question like: 0 < a < b. Revolve the disk (a -b)2 + y2 < a around the y axis. This doughnut shape is known as a torus. How do I get the equation for y from that inequality?

12:51 AM

How does one show that $S^2$ and the closed unit sphere $D^2$ in $\Bbb{R}^2$ are not homeomorphic? If one subtracts a point from $S^2$, then I believe the space is contractible. However, if one subtracts a point from $D^2$, do we get something homeomorphic to $S^2$, which is not contractible?

1:07 AM

@user193319 Your notation is very very unclear to me. What does $S^2$ mean, explicitly? What space does it live inside and what is the defining equation?

Secondly, you should call $D^2$ the closed unit ball or disc, not sphere. This is the set of points in $\Bbb R^2$ with coordinates satisfying $x^2+y^2 \leq 1$.

Whoops. Very sorry.

From my understanding, $S^2$ is the sphere in $\Bbb{R}^3$.

1:27 AM

whats the symbol used to refer to the set of all Gaussian integers?

@Startec What calc course are you in/completed of? Single/multi-variable? Are you comfortable with calc in polar coordinates, or only rectangular?

@nitsua60 single variable and right now I am only comfortable in rectangular coordinates.

@Adam $\Bbb Z[i]$

@user193319 Yes, that's correct. Your notation was fine, I had thought you meant your last occurrence of $S^2$ was supposed to be "$S^1$".

The punctured unif disc is definitely not homeomorphic to the 2-sphere. the latter is not compact!

However, it does deformation retract onto it's boundary, the unit circle $S^1$.

And you probably know the fundamental group of that.

1:43 AM

$S^2$ isn't contractible right

1:57 AM

No, but that is unnecessary hard for this argument.

3 hours later…

4:55 AM

Dummit and foote says 'In $D_8$ we may use the fact that the three subgroups of index 2 are abelian to quickly see that if $x\notin Z(D_8)$, then $|C_{D_8}(x)| =4$.'

Where do we use that subgroups of index 2 are abelian to derive $|C_{D_8}(x)| =4$?

Hi all

I see why $|C_{D_8}(x)| =4$, by noting that $<x>\le C_{D_8}(x)\le D_8$, and $|C_{D_8}(x)| =4$ or $|C_{D_8}(x)| =8$, but $|C_{D_8}(x)| =8$ would say $x\in Z(D_8)$.

I am not getting properly the concept of point at infinity for a elliptic curve. I am getting that to form a group it was considered, but we need to think that on top and bottom of every vertical line, we have that point $\mathcal{O}$.

The problem is, it is not a single point, though we consider that as a neutral element.

Can anyone explain it?

@taritgoswami you should learn about the projective plane

@loch Ooh, I never heard about it(Projective plane). Is it possible to explain that concept? So that it seems to me that taking the point in this way is fine..

5:46 AM

$C(x)$ means the centralizer? @Silent

3 hours later…

8:38 AM

@Daminark Sorry for responding so late! Yes, $C_G(x)$ is centralizer of x in G

Tobias Kildetoft

@Silent We use it to rule out the possibility of having the centralizer be of order $2$

How?

Tobias Kildetoft

any subgroup of order $2$ is contained in one of order $4$

(there are also other ways to see all of this of course)

So, i posted wrong reasoning earlier!

I wrongly assumed that it is given that x has order 4.

Tobias, thank you very much, I could not have thought this.

9:16 AM

Does anyone know how to simplify ((x+y)/sqrt(1-x^2-y^2))+2z? There was an example I did that had ((2xyz)+sqrt((1-x^2-y^2))+xy = 3xy.

10:02 AM

@Alessandro math.stackexchange.com/questions/3033601/…

I dunno how to disprove this

The first approach that comes to mind is there is a $\Bbb Z_2$-action on the set $S^2$ by switching the points in the fiber upstairs. But is there any reason this should be continuous?

Ah, there's an answer which cleverly uses compactness

Basically the diameter of the preimage is a continuous function on $S^2$ which is bounded below by $0$, hence is bounded below by some positive constant. Very cool

Alessandro Codenotti

10:18 AM

@BalarkaSen the preimage of what?

Oh, it's written in the answer

Let me read it

Akiva Weinberger

11:08 AM

@BalarkaSen Why is $d(x,y)$ continuous in $p$?

11:27 AM

@Akiva You should be able to prove that if $p_n \to p$ then $x_n \to x$ and $y_n \to y$ (or, well, $x_n \to y$ and $y_n \to x$).

By continuity of $f$

Ya that's not hard

Akiva Weinberger

@BalarkaSen Hm… first we need to decide, given $x$ and $y$, which point in the preimages of $p_n$ is $x_n$ and which is $y_n$

I guess one first step is to check that, for every $p$, there's an open set containing $p$ whose preimage is disconnected and with $x$ and $y$ in separate components

Is that true in general? If a point has two preimages in a continuous function, there's a neighborhood whose preimages separates them?

Well, you need Hausdorffiness, clearly

Akiva Weinberger

12:05 PM

Ah, this is why I had a hard time proving it — it's not true! (The weaker version, I mean, where only $p$ has two preimages and all the other points can have as many preimages as they want)

It's definitely true if the domain is compact.

Akiva Weinberger

Yeah, my counterexample isn't compact

Take the plane, minus the $y$-axis, plus two points on that axis added back in

projected onto the $x$-axis

Suppose $f^{-1}(p) = \{x, y\}$ and $B_n$ balls of radius $1/n$ around $p$. Suppose $f^{-1}(B_n)$ all contain $x$ and $y$. Then the connected components of $x$ in $f^{-1}(B_n)$, let's call it $C_n$, are a decreasing sequence of closed subsets.

Akiva Weinberger

$\Bbb R^2\setminus\big((\langle-\infty,a\rangle\cup\langle a,b\rangle\cup\langle b,\infty\rangle)\times\{0\}\big)$

Domain is compact so they are also compact.

Akiva Weinberger

12:09 PM

Oh, I was thinking in general, not in a metric space

Is it true if $C_n$ decreasing sequence of connected compact subsets of a metric space then $\bigcap C_n$ is connected? Should be.

Akiva Weinberger

(Still Hausdorff though)

@AKiva I don't care about bullshit spaces bro

:P

Akiva Weinberger

Hausdorff implies not bullshit

All spaces are CGWH

Akiva Weinberger

12:10 PM

(usually)

I think there's a theorem, actually, that says "Hausdorff implies not bullshit". Probably from some paper in the 20s

(/s)

lmao

is there a statistical measure which estimates oscillatory behavior of a sample? I have a stream of numbers coming in and I need to check how much it oscillates.

Oh yeah this is Cantor's intersection theorem. So $\bigcap C_n = \{x, y\}$. Take separating closed ball $B$ around $x$ disjoint from $y$. $\partial B \cap C_n$ are all nonempty compact sets and it's a decreasing sequence of such things, so is nonempty.

But that gives one more point in the intersection, which is bollocks

@AkivaWeinberger I suppose I don't technically need a metric. Which topological spaces have the property that for any point $x \in X$ there is a decreasing sequence of closed sets $C_n$ containing $x$ such that $\bigcap C_n = \{x\}$?

Akiva Weinberger

If I had to guess, all Hausdorff

Well, you want closed neighborhoods, right?

Er, yes. Closure of open sets containing $x$.

Thanks.

Akiva Weinberger

12:20 PM

Does the sequence need to be countable

or just any decreasing chain

I'd like that

But if it's a decreasing chain I can pass to a countable subchain, no?

Akiva Weinberger

I think the (compact) long line would be a counterexample

$[0,\omega_1]$

shudders

Akiva Weinberger

but it's pretty pathological

Yeah, I don't like that name because that's not a manifold with boundary

12:23 PM

The "boundary" points have neighborhood homeomorphic to long lines, right?

Well, long line with one end

Akiva Weinberger

Whatever you wanna call it

Yeah @BalarkaSen

Well, depends on if it's long at both ends (like $[0,\omega_1]$ versus $[-\omega_1,\omega_1]$, to abuse notation 'cause I dunno what they're actually called)

"Long stick"

"Long compact ray"

"A pathological topologist speaks softly, but carries the long stick"

2

I feel like there's a "that's what she said" joke somewhere in here but I can't be bothered to --- there you go

Great minds think alike

That's a Teddy Roosevelt joke

12:32 PM

Oh ok I just interpreted it otherwise

Akiva Weinberger

There's a mountain range in Antarctica called the Executive Committee Range

and I can't help but assume the name was voted on

12:57 PM

Image and cokernel seem to be endofunctors on the arrow category on an abelian category

Akiva Weinberger

1:22 PM

The forest moon of endofunctor

A category is an endofunctor on the category of categories

Deep within a great chasm on the forest moon of endofunctor one finds the malevolent wizard Spectrator

"I will not abut before I chase my enemies to infinity"

@BalarkaSen By Cat you don't mean the 2-category with 0-cells being categories, 1-cells being functors and 2-cells being natural transformations do you? Unless you are identifying an identity endofunctor with the category or something?

I just mean the 1-skeleton of the 2-category you are describing

And by "category is an endofunctor" I do mean the identity endofunctor

I was just meme-ing tho

Oh I see precisely what you mean now

Excellent

:P

1:33 PM

Your memes are of an excellent quality, mathematically and otherwise

Finally, recognition!

born in the right age

Is there a name for functors preserving monos, or a functor that preserves epis?

(rather than reflecting them)

@MatheinBoulomenos or @loch or @LeakyNun might know

(and isn't necessarily stronger, like preserving limits or colimits)

Yeah I had right and left-exact functors in mind but they are stronger

Maybe not even equivalent. Monos and epis are different ideas than trivial kernel and cokernel. Maybe they are the same in an abelian category?

Where kernel and cokernel makes sense

1:37 PM

@BalarkaSen Please stuff the gaps in my knowledge, and lock the gaps shut so they don't leak

Lol

Good puns

I had to translate Boulomenos

@BalarkaSen I think they are the same in a concretizable abelian category (monos=injective and epis=surjective there I mean)

Gotcha

Akiva Weinberger

All of the user#####s are like the character from Night Vale who no one can remember his face

@AkivaWeinberger You will remember me from the feelings my words will draw out of you

1:45 PM

@user616128 Wait, don't all abelian categories have a faithful functor to R-Mod?

No

The standard statement is small abelian categories, but iirc you can change that to locally small

That's freyd mitchell

There are non-concretizable abelian categories

Ah right

Thanks

4

A: Counter-example for abelian category that is not concrete

The term you want is concretizable (meaning admits a faithful functor to $\text{Set}$). The category of representations of a quiver does not work: if $Q$ is such a quiver with vertex set $Q_0$ and $V$ is a representation, then $$V \mapsto \prod_{q \in Q_0} V(q)$$ is a concretization. More gene...

I found this super helpful a bit ago

Yikes

2:43 PM

what do I call this type of functional equation?
$$f(g(x,z),g(y,z))=g(f(x,y),z)$$

Abhas Kumar Sinha

associative functions

@MikeMiller isn't the adjective pathological redundant there? I mean all topologists are psychopaths it's a given

LOL

Abhas Kumar Sinha

I want to know more about the Rienmann's Zeta function, what's it?

Riemann zeta function

In mathematics, the Riemann zeta function is a prominent function of great significance in number theory. It is so important because of its relation to the distribution of prime numbers. It also has uses in other areas such as physics, probability theory, and applied statistics. It is named after the German mathematician Bernhard Riemann, who wrote about it in the memoir "On the Number of Primes Less Than a Given Quantity", published in 1859. The Riemann hypothesis is a conjecture about the distribution of the zeros of the Riemann zeta function. Many mathematicians consider the Riemann hypothesis...

When we put $z=2$, we get $\frac{\pi^2}{6}$

That's just bait not falling for that I got banned when i started a lol-a-thon on the RH last click bait campaign

Abhas Kumar Sinha

2:48 PM

Though the function looks converging but I dun know to what it converges

@Adam what?

nothing i haven't taken my antihistamines the autisms were just playing up

Abhas Kumar Sinha

oh okay

do you know what is the cool thing about that function?

This is valid for $s>0$
$$\zeta(s)=\lim\limits_{k\to \infty}\left(\sum _{n=1}^k \frac{1}{n^s}+\frac{k^{1-s}}{s-1}-\frac{k^{-s}}{2}\right), \tag{1}$$
$$1/a^{b+i c}=1/a^b (\cos (c \log (1/a))+i \sin (c \log (1/a)))$$

Abhas Kumar Sinha

I didn't get that @MatsGranvik Do I need to prove that?

that conjucture?

Mathematics knows it. It is the Euler Maclaurin formula without the Bernoulli numbers.

Abhas Kumar Sinha

2:52 PM

oh okay understood what you wanted to write

hmmm.....

what's cool in that zeta func? It's kinda something very very simple :P

what?

what we know?

what we know now?

we know that we know that we know

wow

hello @Semiclassical

hi chat

Abhas Kumar Sinha

You know I proved $e^{\pi i} = -1$

What is the most coolest eqs in mathematics you ever saw?

@AbhasKumarSinha Do you want to solve the Riemann Hypothesis with me?

Abhas Kumar Sinha

Yes

@user616128 Let's talk here - chat.stackexchange.com/rooms/86876/…

Anyone who wants to talk about RH - come and join ^ room

3:45 PM

Hmm, since closed $3$-manifolds with finite fundamental group $G$ are $S^3/G$ where $G$ acts freely, you can give $S^3$ the standard Riemannian metric, make it $G$-invariant by averaging over the translates of the metric by $G$ and push it down. Since the standard metric was everywhere positively curved, the averaged metric is also everywhere positive curved, right?

Isn't that an argument for saying every closed $3$-manifold with finite fundamental group admits a positively curved metric?

Oh lol I am using Poincare conjecture implicitly

Universal cover of closed $3$-manifolds is a homotopy $3$-sphere. By Poincare conjecture it's $S^3$.

I have not just proved elliptization by fluke

Akiva Weinberger

1

Q: are there meaningful binary operations on the set of Catalan objects?

Is it possible to come up with some kind of meaningful closed binary operation $\star$ on sets $C_n$ of Catalan objects? By Catalan objects I mean objects that correspond to a given Catalan number. I'll use well-nested parentheses since they're easy to write here, but feel free to choose your fav...

soft-question catalan-numbers binary-operations

4:08 PM

trying to figure out how to extract the following from sagemath: Given a polyhedron and a point in such, what are the lowest-dimensional faces containing that point?

I can easily enough deduce which facets contain said points. Seems like that should be enough info...

Akiva Weinberger

How are the facets encoded? @Semiclassical

If it's a list of vertices, then just intersect them, no?

Incidentally, interesting quote from a PDF I just found

user image

(Specifically the first two paragraphs)

I'm trying to understand why a continuous harmonic function on an open set minus a closed interval can actually extend over the interval. Does anyone have any ideas? I can't quite see how to use the mean value property to my advantage here.

@AkivaWeinberger by default, they're encoded as equations

$u:\mathbb{\Omega}\to\mathbb{R}$ is continuous, $u:\mathbb{\Omega}\setminus I\to\mathbb{R}$ is harmonic - why is $u$ therefore harmonic on $\mathbb{\Omega}$?

not sure how to get sagemath to say which vertices generate a particular facet

I want to do this more cleverly if possible: If I know how many facets a given point lies on, should I be able to figure out the dimension for the intersection of these facets?

Akiva Weinberger

4:25 PM

Each intersection takes you down a dimension, usually

I know that, in 4D, two planes can intersect in a point, but that shouldn't happen in polyhedra, right?

no idea tbh

Akiva Weinberger

If they're homeomorphic to spherical polyhedra, anyway

An example of the counting I'm running into: My polyhedron is some full-dimensional convex hull in 6D. For a particular point, I find that it lies on 12 facets (each of dimension 6-1=5). The lowest-dimensional face to which this point belongs is an edge (a 1-dim face).

For a different point, I instead find that it lies on 4 facets and belongs to a 2-dim face.

Trying to figure out if I can infer the dimension of the face from the number of facets (i.e. 1 from 12 and 2 from 4)

Akiva Weinberger

Oh, I see… if it's codimension 2 then it can be in any number of facets

Well, any number${}\ge3$

@Semiclassical You'd need to find some way to tell if the co-1 face $A\cap B$ is different from the co-1 face $A\cap C$, where $A$, $B$ and $C$ are facets

Or are facets codimension 1 and these intersections codimension 2? Depends on what you consider your ambient space, I guess

Yeah, let's say the facets are co-1 and $A\cap B$ and $A\cap C$ are co-2

Probably the simplest way to do this, in the end, is to find some way to represent a facet by its vertices. There's got to be fancy computer science-y ways to do that

especially if this is given as the convex hull of some points

4:44 PM

Yeah. This just revolves around being careful with what sage gives you

The main thing I’m worried about is whether the face I get is simplicial

For instance, consider a point on the surface of a square pyramid

The only non-simplicial facet is the square on the bottom

So if it lies there, then there’s multiple ways I can represent said point as a convex combination

But if it lies anywhere else on the surface, then there’s only one convex combination

Is there a group G where, $Z(G)$ is not subset of [G,G] and [G,G] is not subset of Z(G)?

where Z(G) is the center of the group and [G,G] is the commutator subgroup

@Eran $GL_3(\Bbb R)$ should work. The center is the scalar matrices, the commutator subgroup is $SL_3(\Bbb R)$.

Akiva Weinberger

Hm

What about $A\times H$ where $A$ is abelian and $H$ has trivial center

The center is $A\times\{e\}$ and the commutator subgroup is $\{e\}\times[H,H]$

Right?

wonderful, thank you.

Akiva Weinberger

4:59 PM

In general, $Z(A\times B)=Z(A)\times Z(B)$ and $[A\times B,A\times B]=[A,A]\times[B,B]$ (correct me if I'm wrong)

so I was combining both extremes

That's completely correct

5:15 PM

Hi @Ted!

Hi a @Balarka!

well, at least this polytope stuff has made me work on understanding sage

(which seems like good practice for python)

I know nothing

Hey everyone!

Hey @TedShifrin :)

[G, G] being a subgroup of Z(G) is an interesting condition. It means that commutators commute with everything. More aptly, [G, [G, G]] = 0

5:22 PM

Hi, Perturb.

That means it's nilpotent of depth 2 I think

Nilpotent groups always split as direct product of it's Sylow subgroups, if I recall correctly

Right, because normalizers always shrink in nilpotent groups, but that if I have a Sylow subgroup of my group, it's normalizer will shrink - but intersection of a Sylow with a normalizer of another Sylow is precisely intersection of the two Sylows

So Sylow subgroups of nilpotent groups are normal

The splitting thing is just an induction argument then, I think

Just like the proof that finite abelian groups split into direct product of it's Sylow subgroups (that's how the Sylow proof of fundamental theorem of f.g. abgrps go)

Makes sense, 'cuz nilpotency is like abelian "after some finitely many iterations"

Ah, a Balarka is back to being an algebraist :P

I'm reading a little bit of combinatorial group theory from Lyndon & Schupp

As a break from stratified thingys

Not that that has anything to do with finite group theory :)

Fun problem from my rep theory final which I didn't get

If the probability that two random elements in a group commute is strictly greater than 5/8, the group is abelian

Ya that's true

5:31 PM

oh, that's sorta cool

howdy, Demonark

How's it going?

Size of {(g, h) \in G x G : gh = hg} is less than 5/8|Z(G)| if G is nonabelian

Akiva Weinberger

Strange fraction

It's a combinatorics argument. Basically if G is nonabelian Z(G) has index at least 4 in G.

hi DogAteMy

Akiva Weinberger

5:32 PM

Hi

2 finals down, one to go. And one short writeup

Hey everyone

It seems that I joined just when all the nerds did. Truly cursed...

So say I have a function $f : \mathbb{R}^m \to \mathbb{R}^n$, to show that $f$ is of class $C^{1}$, all I have to do is look at the Jacobian of $f$, see if all the entries are continuous and then I get $f$ to be $C^1$. To check if $f$ is of class $C^{\infty}$ then all I have to do is proceed by induction and differentiate each of those entries in the Jacobian with respect to the canonical basis correct?

5:55 PM

why can't I see all the mathematical formulas you guys are putting up do I need to download a plugin?

@Rick There's a link to a thing with instructions in the top right (if you're on a computer).

Basically you set a script as one of your bookmarks, and then click that bookmark when you're in the room.

cool!

but it sucks that i have to click it each time to see it render when I visit the chat.

How do I show that every closed path $\gamma$ in a $\mathbb{C} \setminus \{0\}$ is contained in a subannulus?

6:11 PM

ok a Wikipedia article has used the notation $\sum_{n \operatorname{mod} c}$ now I have guessed this must mean the same thing as $\sum_{n | c}$, but this is just really dumb notation because it isn't defining a congruence relation as you should, and it is using $\operatorname{mod}$ in the manner as standard for the modulo operator, ie integer remainder, but this is completely meaningless unless you specify a sum over specific values of the integer remainder when $n$ is divided by $c$,

I've been told not to post questions complaining about notation several times therefore this is where it needs to be queried QED

@BalarkaSen Here I meant the normalizer of proper subgroups always expands. That is, $H < N_G(H)$. The argument for normality of Sylow is that for any Sylow subgroup $P$, $N_G(N_G(P)) = N_G(P)$, which forces $N_G(P) = G$.

Closely related to $N_G(P) \cap Q = P \cap Q$ but not quite.

Alexander Gruber

@Adam the convention in programming is for $n \operatorname{mod} c$ to denote the representative in $[0,c)$.

Hi @AlexanderGruber!

Speak of finite group theory and the devil arrives

Alexander Gruber

@BalarkaSen i have been summoned

What shall I permute?

Can you give me an example of a group where every subgroup is self-normalizing?

I don't know too many simple groups

MatheinBoulomenos

6:55 PM

the only such finite groups are cyclic of prime order.
Proof: Let $G$ be a finite group such that every nontrivial subgroup is self-normalizing. Let $p$ be a prime and $P$ be a $p$-Sylow subgroup, let $x \in Z(P)$ of order $p$, then $P$ normalizes (even centralizes) $x$, so $P \supset \langle x\rangle = N_G(\langle x \rangle) \supset P$, thus $P=\langle x \rangle$. Thus every $p$-Sylow subgroup of $G$ is cyclic of prime order. This implies that the order of $G$ is square-free, which implies that $G$ is supersolvable. But the condition also implies that $G$ is simple, so solvable+simple gets…

Anyone here like complex analysis?

@MatheinBoulomenos The center has to be trivial if every subgroup is self-normalizing.

Oh, center of P

MatheinBoulomenos

yeah, center of a p-group is always nontrivial

Right

This is cool!

So, I was talking to someone the other day about infinite cardinalities, and they were saying that $\aleph_1$ is simply defined as the next largest infinite cardinality after $\aleph_0$, which conflicts with what I learned a little bit. Isn't it true that we don't know if there exists an infinite cardinality between $|\mathbb{N}|$ and $|\mathbb{R}|$? (On mobile at the moment, so apologies if there's any broken syntax in there.)

7:03 PM

@MatheinBoulomenos If the order of the group is a product of distinct prime, it's also very easily not simple, I think.

You can take product of the Sylow subgroups of order not equal to the least prime

Hm, why is that a subgroup, though.

Oh, because the Sylow of the highest order. There is only one such thing, no?

If $|G| = p q r$, $p < q < r$, $n_r = 1$.

MatheinBoulomenos

I'm sure there are easier ways to go from squarefree order to not simple

Well, if $|G| = p_1 \cdots p_k$, then $n_{p_k} = 1$, that's all. Automatically implies the $p_k$-Sylow is a normal subgroup.

Nice idea to use the center of the Sylow

@Rithaniel and where is the conflict?

$\aleph_0 = |\Bbb N|$, $\aleph_1 \le |\Bbb R|$

MatheinBoulomenos

@BalarkaSen so if for example $|G|=2 \cdot 3 \cdot 5$, how do you conclude that $n_5 \neq 6$?

@BalarkaSen made me realize that I don't know jack about central / lower / upper series and nilpotent groups and all that bull

7:10 PM

@MatheinBoulomenos Either $n_3 = 1$ or $n_5 = 1$

Take their product, that's cyclic of order $15$

But then every subgroup is characteristic

Characteristic in normal implies normal so actually both $n_3 = n_5 = 1$

Maybe I am not sure this generalizes to multiple primes

So I'd always thought $\aleph_1$ was defined as $|\mathbb{R}|$.

Ya it shouldn't. Oh well.

Alessandro Codenotti

@Rithaniel nah, just the smallest cardinal above $\aleph_0$

Okay, I need to remember that, then.

Actually $n_r = 1$ is also not true. Take $2 \cdot 3 \cdot 7$.

7:14 PM

And "bethe one" is the cardinality of the reals?

2*3*5 crucially uses group of order 15 is cyclic.

Alessandro Codenotti

Yes

And $\aleph_1=\beth_1$ is independent of ZFC

Meaning it can't he proved by ZFC?

Alessandro Codenotti

And it cannot be disproved either

@MatheinBoulomenos Need a group of order $p_1 \cdots p_k$ have a normal Sylow $p_i$-subgroup, actually?

Tobias Kildetoft

7:17 PM

@BalarkaSen No

Only for the smallest prime divisor

But then by induction, it has a chain of normal subgroups of orders $p_1$, $p_1p_2$,...

I don't mean for every $i$, I just mean if for at least one $i$.

Tobias Kildetoft

Right, it will have one for the $i$ giving the smallest prime (assuming the primes are distinct of course)

? $D_{30}$ has no normal Sylow $2$-subgroup

Tobias Kildetoft

Woops, for the largest prime I mean

The chain goes from the other side

OK, that's what I meant. How do you prove that?

Tobias Kildetoft

7:19 PM

with difficulty :)

MatheinBoulomenos

probably some p-complement theorem

Hah

Tobias Kildetoft

It follows from Burnside's transfer theorem that if $p$ is the smallest prime divisor and the Sylow $p$-subgroup is cyclic, then the group has a normal $p$-complement

@MatheinBoulomenos I still don't know what I want to do

Tobias Kildetoft

Not sure if there is some way to avoid that theorem here (except when $p=2$)

7:21 PM

@TobiasKildetoft Thanks, very cool!

Tobias Kildetoft

The theorem also applies if that Sylow subgroup has order $p^2$ and $p\neq 2$ (but is still smallest), or in the same situation with $p=2$ if $3$ does not divide the order

Oh, wait. Say I assume it's true by induction. Suppose $p_1 \cdots p_k$ is the order with $p_1 < \cdots < p_k$. I think one of $n_{p_i} = 1$ for $i = 2, \cdots, k$ by counting argument. Let $H_{p_i}$ be the $p_i$-Sylow subgroup. Then $H_{p_1} \cdots H_{p_k}$ is a subgroup, and a normal subgroup, in fact, because index is the smallest prime $p_1$. By induction the Sylow $p_k$-subgroup $H_{p_k}$ is normal in $H_{p_1} \cdots H_{p_k}$, but normal Sylow implies characteristic, right?

Tobias Kildetoft

right

More generally normal and with coprime order and index implies characteristic

I think Schur-Zassenhaus says the stronger thing that it also splits if that happens?

Anyway @Mathein so I think $n_{p_k} = 1$ was elementary after all

Tobias Kildetoft

right

7:30 PM

Hmm, maybe it takes more work to show $H_{p_1} \cdots H_{p_k}$ is a subgroup. Just knowing one of the subgroups in the interior direct product to be normal is not enough, is it?

Ugh too annoying

Tobias Kildetoft

@BalarkaSen What are you trying to show?

I was trying to show that the Sylow $p$-subroup of a group of square-free order with largest prime divisor $p$ is normal. For $|G| = pqr$ it's really simple; $n_q$ or $n_r = 1$ by counting. Then the interior product of the Sylow $p$- and the Sylow $q$-subgroups is normal in $G$. But the Sylow $r$-subgroup is characteristic in the product as it's normal, so it's normal in $G$.

I was trying to analogize the argument

(Here $p < q < r$)

Tobias Kildetoft

Yeah, I am not aware of a proof that does not use transfer as I mentioned (and thus applies equally well to all $Z$-groups, i.e. groups with all Sylows cyclic)

Very strange.

It felt like that argument would generalize for a moment

7:45 PM

@TobiasKildetoft Do you ever get more than one referee report on your papers?

I often see people discuss say Reviewer 2

But I think that might only be in non math fields

Hm, if $|G|$ is coprime to a prime $p$, $H^n(G; \Bbb Z_p) = 0$, right?

Tobias Kildetoft

@MikeMiller Not "really" (I have had one paper that had a couple of informal reviewers who gave a few comments but declined to review, but never more than one actual reviewer)

MatheinBoulomenos

@BalarkaSen @BalarkaSen yes

Tobias Kildetoft

As far as I know most math journals make do with one reviewer. It is just all other fields that want more

But it also seems like there is a lot more work put into the review by most math reviewers than what I hear about from other fields

@MatheinBoulomenos So I don't need Schur-Zassenhaus to see that if I have a normal Sylow $p$-subgroup then the group splits over it, I guess

7:48 PM

@BalarkaSen Here is a natural question (which you do not have the tools to answer)

Because splitting of a group extensions is determined by vanishing of $H^1$ and $H^2$

MatheinBoulomenos

@BalarkaSen group cohomology is the standard proof for Schur-Zassenhaus (at least the part where one group is abelian)

How do you relate the cohomology of $G$ with $\Bbb F_p$ coefficients to that of a p-Sylow?

@TobiasKildetoft Interesting. I guess in some cases that can be quite good

I would not want more than one person to suffer reading what I write

@MikeMiller Oh interesting, no clue really. If the p-Sylow is normal the group splits over it by Schur-Zassenhaus, and then I can run a spectral sequence.

The cohomology of the fiber vanishes because that has order coprime to $p$

So it should be a simple enough computation

Tobias Kildetoft

@MikeMiller No, that is all wrong. You need to wish for everyone to suffer reading your stuff.

2

7:52 PM

Hah!

That tells you that when the p-Sylow is normal the cohomology is the same as the p-Sylow

Yeah, good point, it does. Because only the bottom row of the spectral sequence survives

@TobiasKildetoft Poor souls ...

(Alternatively, cohomology of fiber vanishes just means I can floop any cocycle down to the base by sliding fiberwise)

Good proof

7:55 PM

So maybe I should compute $H^*(\Bbb Z_p \rtimes \Bbb Z_q; \Bbb Z_q)$ and get back to you

To see an example when it's not normal

How about $S_3$

@TobiasKildetoft suffering leads to enlightenment

I'll do the calculation right now with my pen and paper. Give me a few minutes (I'll make sure not to make it an hour)

with $S_3$

Tobias Kildetoft

@Rick You seem to have high expectations of my (or I suppose Mike's) writing.

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