Mathematics - 2016-10-13 (page 1 of 5)

Forever Mozart

12:05 AM

feel like I'm about to have a stroke

ran 2 miles to get my blood flowing

Adeek

wtf @ForeverMozart how old are you

Balarka Sen

I sympathize, @Forever. Drink some water.

Forever Mozart

mid 20s

i'm not overweight I just don't run that much

Balarka Sen

I am underweight and I don't run at all

dsillman2000

I am average weight and do not exercise

Adeek

12:08 AM

Yeah I need to also start excerise

Forever Mozart

if you run then stand on your head, you get blood flow to the brain

dsillman2000

Lol xD

Forever Mozart

i don't know if that is recommended ;)

PVAL-inactive

@MikeMiller Isn't the actual (the one not involving Skein relations) definition already very geometric.

Mike Miller

I have more or less never ever thought about these things.

PVAL-inactive

12:21 AM

I mean you build the infinite cyclic cover, and evaluate the action of the induced "translation" transformation on H_1 of this thing. To me that that is quite geometric.

Hola qué pasa

Hi, Akiva

Hey

@MikeMiller Actually I don't know what's being asked.

Now I think he asking if theres some relation between numerical hyberbolic invariants and the alexander polynomial. No idea.

Balarka Sen

@AkivaWeinberger What are you doing these days?

I hardly see you here.

Akiva Weinberger

12:30 AM

School stuff mostly

Balarka Sen

Me too. I got a largish holiday though

Mike Miller

@PVAL Is the new SG paper interesting?

Adeek

@BalarkaSen Suppose K and H are groups such that $\psi : K \rightarrow Aut(H)$ is a homomorphism. Suppose
$p \in Aut(K)$ and set $\psi_2 = \psi \circ p$. Prove that $H \rtimes_{\psi} K$ ~ $H \rtimes_{\psi_2}K$. Is this problem even correct ?

Balarka Sen

@Adeek Sorry, I think I am done with group theory for today.

Adeek

ok

Balarka Sen

12:33 AM

It does sound correct at a glance

PVAL-inactive

@MikeMiller Maybe. It's certainly not real progress towards the question in the title. I'm very inclined to not take it that seriously given what the title is and what they actually prove.

Potentially you'd like to know these aren't symplectically fillable, and they don't even talk about that.

Balarka Sen

Maybe I'll stay awake today and reboot my sleep cycle

Or maybe I won't. Shrug.

Forever Mozart

1:00 AM

got my new dvd in the mail

McCabe and Mrs. Miller

arctic tern

@Adeek yes. apply p^-1 to the K component.

Adeek

@artic tern to K component ? what is K component ? the problem here is that p lives in Aut(K) right ?

essentially what I want here is to show that $\psi$ and $\psi_2$ are conjugates

arctic tern

@Adeek send (h,k) to (h,p^-1(k))

Balarka Sen

@Adeek why would you?

Adeek

oh

I thought maybe to use the first part which I proved

Balarka Sen

1:13 AM

that's only for cyclic $K$.

Adeek

maybe not yeah I guess $(h,k) \mapsto (h,p^{-1}(k))$ would maybe define a isomorphism

I am just in the car now I will check the details when I go home.

oh I see @BalarkaSen

I didn't notice that

arctic tern

Let's use $\phi=\psi\circ p$ instead. Then $H\rtimes_\psi K$ is $H\ast K$ mod the relations $khk^{-1}=\psi_k(h)$, and $H\rtimes_\phi K$ is $H*K$ mod the relations $khk^{-1}=\phi_k(h)$. since $\phi_k(h)=\psi_{p(k)}(h)$, replace $k$ with $p^{-1}(k)$ and these relations are $p^{-1}(k)hp^{-1}(k)^{-1}=\psi_k(h)$. to go from the relations $khk^{-1}=\psi_k(h)$ to $p^{-1}(k)hp^{-1}(k)^{-1}=\psi_k(h)$ inside $H\ast K$, apply $p^{-1}$ to $K$ and $\rm id$ to $H$

Forever Mozart

going to get some coffee. back shortly

arctic tern

(Since $H\rtimes_\psi K$ and $H\rtimes_\phi K$ are both quotients of $H\ast K$, to exhibit an isomorphism between them it suffices to exhibit an automorphism of $H\ast K$ which sends one kernel to the other.)

Adeek

oh i see

@arctictern that is nice way to see it

i.e seing it as mod relations

seeing *

arctic tern

1:19 AM

mmhmm

Adeek

brb few min just going to apartment

dsillman2000

Really quickly fact-checking here; for the first order d-e $\frac{dy}{dx}=\frac{1}{\sec ^2y}$, the solution is $y=\tan ^-1x$, right?

shai horowitz

does the banarch tarski paradox work on a discrete space?

dsillman2000

I meant $\tan ^{-1}x$, sorry

shai horowitz

i would imagine so because i can go from all even numbers to all integers by division by 2 which is kind of the same idea

arctic tern

1:30 AM

@dsillman2000 that's a solution

there are many solutions

the others are $\tan^{-1}(x+C)$

dsillman2000

Ok cool thx

arctic tern

@shaihorowitz the fact that an uncountable set can be bijected with any countable number of copies of itself is not very surprising. Banach-Tarski applies specifically with rigid motions of space (rotations and translations), which is why it's so surprising. Although really BT is a fact about group actions, and indeed it can be generalized to other group actions. (I'm not really familiar with the details, even though I love group actions.)

Forever Mozart

yumm

Adeek

back

shai horowitz

thanks, that suffices to answer what i was curious about @arctictern

enthdegree

1:37 AM

Hello I have a small question, the weak open unit ball in a hilbert space $\mathfrak{H}$ looks like: $B_\varepsilon(0) = \{x\in\mathfrak{H}|\,|(x|y)|<\varepsilon \forall y \in \mathfrak{H}\},$ right?

Balarka Sen

I am out of ink. Does that mean I have to go to sleep?

Forever Mozart

it looks like your cat is typing

0celo7

No, get another pen

Balarka Sen

Maybe it's an ill omen. What if I don't listen to it? Will I die?

0celo7

You're gonna die regardless

shai horowitz

1:41 AM

not according to wuantum immortality

quantum*

Forever Mozart

what is the best halloween movie

shai horowitz

the first

the sequel halloween movies sucked

Forever Mozart

i mean in general

Balarka Sen

@ForeverMozart The Nightmare before Christmas

Forever Mozart

doesn't have to be a "haloween"

shai horowitz

1:42 AM

the rocky horror picture show

Forever Mozart

0celo7

@BalarkaSen I have a question for you

Forever Mozart

that is a cartoon

0celo7

Hatcher or video games?

Balarka Sen

Hatcher

@ForeverMozart it is indeed an animation.

shai horowitz

1:43 AM

its an animation not a cartoon

Balarka Sen

and quite a good one, too!

0celo7

@BalarkaSen yes but Fallout New Vegas

Balarka Sen

never played fallout.

shai horowitz

real still frame pictures 30 pictures for one second of film thats dedication to art

Balarka Sen

yup

0celo7

1:45 AM

I haven't played a video game in months

Balarka Sen

I also like his animation "Frankenweenie".

0celo7

I think I'll take a break from math/physics tonight

Balarka Sen

definitely

I thought that question was for me.

0celo7

are you talking to me?

I didn't ask a question

shai horowitz

whats a question?

Balarka Sen

1:46 AM

"Hatcher or video games?" I was referring to that

shai horowitz

im bored sorry....

0celo7

oh

yes that was @BalarkaSen

Hatcher is too long, it's intimidating

Balarka Sen

I would read Hatcher. But you need not, taking breaks is perfectly fine

Forever Mozart

i cannot prove :(

Balarka Sen

I would probably be tempted if you had one of those Empire Earth-type strategy games. I used to like those, would have played if I had them

0celo7

1:48 AM

@BalarkaSen Ever heard of "strong differentiability"?

Forever Mozart

anyone who solves this I give 100 dollars:

Balarka Sen

Not by that name

0celo7

@BalarkaSen Me neither.

The next lecture in analysis is on "strong differentiability"

Balarka Sen

shrug

0celo7

My prof has an obsession with really old books so maybe it was standard back then

Forever Mozart

1:49 AM

Is there a metric continuum $X$ and a point $x\in X$ such that every two points of $X\setminus \{x\}$ are contained in a nowhere dense subcontinuum of $X$, and every non-degenerate subcontinuum containing $x$ has nonempty interior.

Adeek

@0celo7 it is strong

0celo7

The proof is trivial! Just view the problem as a
combinatorial
poset
whose elements are
greedy
bijections

Forever Mozart

solve that problem and be famous

0celo7

@BalarkaSen I do have a question

Balarka Sen

go on

Forever Mozart

1:51 AM

$|X|>1$

0celo7

Is the "invariance of domain" theorem on page 172 of Hatcher the same one as on wiki?

The hypotheses seem slightly different

Forever Mozart

Who can solve that???

Balarka Sen

I don't beremember Hatcher's version. I think he parses it as "R^m is not homeom to R^n if m \neq n"?

0celo7

probably Poincare

@BalarkaSen No

I actually do not understand hatcher's formulation

Forever Mozart

$100 I promise

0celo7

1:53 AM

he says "if a subspace $X$ of $\Bbb R^n$ is homeo to an open set in $\Bbb R^n$, $X$ is open in $\Bbb R^n$"

Maybe I'm being dumb, but isn't that trivial?

Forever Mozart

I am interested in very simple problems like that. Easy to understand but unsolved.

Balarka Sen

@0celo7 How would you prove it?

0celo7

@BalarkaSen Homeomorphisms are open maps.

Oh wait

Balarka Sen

So what?

0celo7

subspace

So with the subspace topology?

Balarka Sen

1:55 AM

of course, what else?

0celo7

Well wiki says something completely different!

Balarka Sen

No it doesn't

0celo7

ok that's what I'm asking about

it seems completely different

In Hatcher, he starts with a homeo, then concludes it is open

Wiki starts with an open set, concludes the map is a homeo

Forever Mozart

$h:X\hookrightarrow \mathbb R ^n$ homeomorphic embedding and $h[X]$ open in $\mathbb R ^n$.

Balarka Sen

Hatcher says if $U$ is open in $\Bbb R^n$ and $f : U \to \Bbb R^n$ is an embedding , then $f(U)$ is open. Wiki says the same but with "injective continuous". Replace that by embedding (because that bit is obvious once you can show for injective continuous)

0celo7

1:59 AM

is your $U$ Hatcher's $X$?

Forever Mozart

now let $x\in X$.

Balarka Sen

I don't see an $X$ in page 172. I have the online version open in front of me.

0celo7

wtf

that's completely different than my version

in my version, Hatcher says exactly what I quoted above.

Balarka Sen

Sounds like the same thing to me, with $X = f(U)$

$f$ is the homeo with the open set $U \subset \Bbb R^n$.

0celo7

@BalarkaSen I still don't get it. Hatcher starts with a homeo as a hypothesis

is Wiki saying the converse?

enthdegree

2:06 AM

This can't be right

Balarka Sen

@0celo7 No, just replace wiki's hypothesis "injective continuous" with "homeomorphism".

Then compare.

0celo7

Unless I'm missing something, those are not the same hypotheses...

Balarka Sen

Sure, it's a bit more general in that sense. But part of the statement of the theorem (the version in wiki) says that it doesn't matter, @0celo7.

@enthdegree What are you referring to?

0celo7

@BalarkaSen Well, yeah

that's all I was wondering about

enthdegree

Can anyone explain how on earth you get your hands on the weak closure of a subset of a space?

I have a normed vector space X and its dual X*, and X* induces a 'weak topology' on X (the weakest topology where all the functionals in the dual are continuous, i.e. a sequence (xn) --> x weakly in X iff (psi(xn))--> psi(x) for any psi in X*)

Adeek

2:31 AM

hey @arctictern I am classifying all groups of order pq where p and q are distinct prime. THe way I am solving this is as follows we can assume that q < p. Let H be a sylow p subgroup and K be a sylow q subgroup. We can prove H is normal in G and HK = G.

then this means that $G = H \rtimes K$

If K is also normal in G then we get $G = H\times K$

why is it true that we will have a non-trivial homomorphism iff q divides p - 1 ?

Forever Mozart

is an image of hitler stuffed with straw

Balarka Sen

@Adeek One direction is clear from Sylow's 3rd theorem.

Adeek

what about <--

ok let us say I know that the theorem above is true

hm yeah I guess I would be done

Balarka Sen

You need to explicitly work out a homomorphisms K --> Aut(H). H is a cyclic group so that shouldn't be hard.

Adeek

I see

Balarka Sen

2:39 AM

I was messing up a matrix multiplication for the last few minutes. Maybe staying awake wasn't a great idea.

Adeek

yeah ok I proved it @BalarkaSen

what are you doing anyway @BalarkaSen

Balarka Sen

Differential geometry.

As in, surfaces in $\Bbb R^3$.

Adeek

a non-trivial homomorphism gives a non-abelian group right ?

we can prove that

I guess I am trying to show that only two groups exist with order pq where they are distinict and show that they are not isomorphic.

@BalarkaSen can we actually show what kinda group is the semi-direct product ?

or it is too general to answer.

in the case where it is non-trivial homomorphism

Balarka Sen

@Adeek That's right. If it was abelian, H and K both would have been normal.

Adeek

yeah

Balarka Sen

2:47 AM

I don't know what you mean by "what kinda group". You can write down a presentation for it: is that sufficient?

Adeek

yeah maybe

yeah

Balarka Sen

Then the presentation comes from the very definition of semidirect products.

Adeek

i see

@BalarkaSen I attended this seminar today I thought it might interest you

Title : Calculus of functors and the De Rham complex

Abstract : The calculus of functors, pioneered by Goodwillie in the 1980’s, refers to a way of examining functors which topologists and algebraists often use to distinguish and classify algebraic or topological objects like algebras. The De Rham complex of a commutative algebra over a field, the algebraic analogue of De Rham cohomology of a manifold, is precisely one of these functors. The De Rham complex is closely related to cohomology theories for non-commutative algebras, such as cyclic (co)homology and crystalline cohomology. …

was very interesting

Balarka Sen

I know nothing of Goodwillie's calculus, but I have heard of it.

enthdegree

Does anyone have any idea how to complete this? imgur.com/Zpxdwjo.png

Adeek

2:54 AM

yeah it was kinda of interesting but too advanced

for me

enthdegree

I can't figure out how to show that the epsilon ball intersects \mathfrak{I}, even given the hint!

I guess i should add, I^{-w} denotes the closure of I in the weak topology

dsillman2000

3:08 AM

Hey guys

Is $\int 3x^2y\ dx = x^3(yx) = x^4y$ because y would be treated as a constant, as in partial differentiation?

Forever Mozart

no

dsillman2000

Or by that I meant yx^3

Why not?

Forever Mozart

yes

dsillman2000

So it would be yx^3?

Because I think someone made a fatal error in their answer to someone's question but I don't have enough rep to comment xD

Forever Mozart

you can edit your answer

enthdegree

3:13 AM

alright, it looks like i have some fundamental misunderstanding of what the weak* topology is

does anyone online right now know functional analysis?

Adeek

3:24 AM

@BalarkaSen this question is really cool

classify all groups of order 8

very fun

Balarka Sen

I don't see why that's interesting. There's just one.

Adeek

order 8

I dunno why I said order 5

Balarka Sen

Hi @TedShifrin

Adeek

fixed it

hi @TedShifrin

I was actually wanted to talk to you

Balarka Sen

Sylow hard.

Adeek

3:26 AM

no @BalarkaSen

by structure theorem

Ted Shifrin

Hi, all. Order 8 is still pretty easy ... For the non-abelian, anyhow.

Adeek

you can classify 3 abelian group of order 8

Balarka Sen

@Adeek Structure theorem? Huh?

Adeek

then Suppose now that G is not abelian

Balarka Sen

That's for abelian.

Adeek

3:26 AM

then there is an element of order 4

Forever Mozart

i have killed so many trees

Adeek

because otherwise it have an element of order 8 which would make it cyclic

or all elements will have order 2

Balarka Sen

Sure.

Forever Mozart

i just write and write until I finally discover something

Adeek

which mean it would be isomorphic to $Z_2 \times Z_2 \times Z_2$

Forever Mozart

3:27 AM

and there is a pile of paper

Adeek

denote this element by a so <a> is subgroup of G of index 2 so it is normal

Ted Shifrin

No, Karim, that's not clear

Adeek

?

Ted Shifrin

Why must it be a direct product?

Adeek

what is not clear ?

Balarka Sen

3:28 AM

@TedShifrin If it has no element of order 4, there are only two choices for the orders.

Adeek

yeah

Ted Shifrin

Balarka, hush.

Balarka Sen

Oops.

I'll scurry, as usual.

Ted Shifrin

Why if every non-1 element has order 2 must it be .... .

Go do interesting diff geo, Balarka.

Adeek

okay so by lagrange we can only have elements of order 2 or 4 or 8. If we have all elements of order 2 then it would be isomorphic to $Z_2 \times Z_2 \times Z_2$

Balarka Sen

3:30 AM

I need interesting stuff. I am going to keep myself awake today and reboot my sleep cycle.

Ted Shifrin

You keep saying that. That doesn't suffice.

Do you need a list of exercises, B?

Adeek

@TedShifrin I don't understand what is the issue ?

Balarka Sen

@Adeek He's asking you why all elts having order 2 imply the group is Z/2 x Z/2 x Z/2

Ted Shifrin

What justifies your claim, Karim?

Adeek

oh

Balarka Sen

3:32 AM

@TedShifrin Yeah, definitely.

Adeek

oh I see

yeah I can't answer that hmm

Ted Shifrin

-30

Alan

pokes on "I don't suppose anyone knows of any decent textbooks on Banach Lattices? Having trouble getting into the one my professor gave me."

Ted Shifrin

No clue, Alan.

enthdegree

alright i cant believe i was being this dense before, i was basically not interpreting my definition of the weak* topology

Mike Miller

3:34 AM

Are we doing anything fun?

Adeek

oh I see

I got it @TedShifrin

Alan

Pretty sure the classify all groups of "x" order when "x" has a simiple factoroization is easiest done via semidirect products, but it's been a few years since I did algebra qual prep

Balarka Sen

Karim's classifying groups. I'm trying not to fall asleep.

none of these are fun though.

Alan

I'm trying to find a text on Banach Lattices that I can understand :(.

enthdegree

with the weak topology we assert that sets of the form {x: |phi(x-z)|<\varepsilon, z in X} (for phi in the dual, varepsilon positive)

are open

Mike Miller

3:35 AM

I have no idea what a banach lattice is

enthdegree

what is a banach lattice? is that just a banach algebra with more structure

Adeek

if we have every element has order 2 then take the group generated by the elements a,b,c then <a> <b> <c> all intersect trivially and they all generate the whole group.

Ted Shifrin

@Balarka: Make sure you know how to do #2 (linear alg, of course), 5 (note that problems with * have hints/answers at the back), 10, 12, 13, 14 (important), 15, 17, 18, 19 (sorta cool), 21 (might interest you because there's a blowup in there).

Adeek

so this means that $G = <a>\times<b>\times<c>$

Ted Shifrin

Total handwave, Karim.

Balarka Sen

3:36 AM

I worked out #2 a couple hours ago.

Alan

it's a banach space (Not necessariolly an algebra) that has a lattice order that is compatible with the vector operations. A lattice is a partial order in which every two elements has an inf and a sup (though not necessarily every subset, so think of it as an lcm and gcd)

Ted Shifrin

Why is it abelian? You keep missing the main point.

OK, cool, @Balarka.

Balarka Sen

Thanks for the list, by the way!

Ted Shifrin

23 is actually interesting, Balarka, but it might take you down a rabbit hole.

Alan

the compatibility requirements being if $x<y$ then $x+z<y+z$ and if $a\in mathbb R , a>0$, $x<y$ means $ax<ay$

Mike Miller

3:39 AM

@Alan Why is this interesting?

Alan

@MikeMiller I've yet to get to that part. I wanted to do something in Functional Analysis, and the main professor at my school who does that turns out to work in this area.

Balarka Sen

I bookmarked that too for kicks.

Mike Miller

:P I think step one in studying anything is to find out why you care.

Ted Shifrin

@BalarkaSen What?

Mike Miller

If you can't find that, I suggest trying something else.

Balarka Sen

3:40 AM

@TedShifrin Problem 23. I bookmarked it along the others.

Ted Shifrin

Students always complain when professors don't show them why things are interesting, @MikeM, and professors shrug.

Alan

Yeah :). I really enjoyed Kreyzig's Functional Analysis. I'm year 4 of grad school, I may switch thesis advisors at the end of this semester if I can't seem to get a hold of tghis

Ted Shifrin

Oh, ok.

Alan

Maybe I should see if anyone's working in topological data analysis....I found that fascinating when we had a talker come in about it, and I enjoyed both general and algebraic toplogy

err, speaker. Yeah. I can think right now :)

Mike Miller

@Alan Your professor would be a better person to ask than us. This sounds like pretty specialized stuff.

Ted Shifrin

3:42 AM

It's almost late, @Alan.

Balarka Sen

Top. data analysis is fantastic.

Mike Miller

@Ted: When a student asks me why something is interesting, I certainly don't shrug. I think that's one of the best questions you can ask.

Ted Shifrin

@MikeM: I don't disagree. I say that in general teachers do a terrible job of making math motivated/interesting.

Alan

Yeah. I...think I'm coming to that point where I should switch tracks before I sink too much time/energy into this.

Ted Shifrin

One of my colleagues when I first went to UGA did lattice ordered groups. It left me cold.

But the math I spent my life on doesn't interest plenty of people ... so it's a matter of individual taste and strengths.

Alan

3:44 AM

I know I always have real world examples of interesting stuff for when I teach. Sometimems little things, like the hash code for Florida's driver's license.

Yeah, I think this was a case of me thinking I'd like something then trying it and finding out I don't.

Ted Shifrin

Well, yeah, talk to other people, @Alan.

Adeek

@ted I don't understand how to make it more rigorous the way I see it is that if we have all orders of order 2, then take a particular element of G and denote that by a. Then $<a> = Z_2$. suppose b not in <a> and go like that and we can then construct 3 groups of order 2, which I call <a>,<b>,<c>

Alan

Time to talk to my advisor/the graduate studies head. Thanks for the feedback.

Ted Shifrin

There's lots of other things in functional analysis, but if no one there can advise you on other things, talk to other people.

Adeek

Since they are all p groups so they intersect trivially

Ted Shifrin

3:45 AM

OK, Karim. But you're far from a direct product.

Mike Miller

@Ted: Part of why I don't judge people's interests. I couldn't spent my time writing down estimates for gradient flows of analytic functionals on Banach manifolds, but I'm glad someone can, since I like to use them.

Adeek

We know the following theorem

Ted Shifrin

There's no denying taste, @MikeM.

Mike Miller

@Alan Good luck.

Ted Shifrin

Yes, @Alan, we're rooting for you.

Forever Mozart

3:46 AM

Is there a metric continuum XX and a point x∈Xx∈X such that every two points of X∖{x}X∖{x} are contained in a nowhere dense subcontinuum of XX, and every non-degenerate subcontinuum containing xx has nonempty interior.

Alan

De de gustibus non est disputandum

Adeek

if (1)$H \cap K = \{1\}$ (2) H and K are both normal (3)HK = G then $G = H\times K$

Forever Mozart

i might have a way to solve this problem

Ted Shifrin

You're far from having hypotheses for that, Karim.

Adeek

yeah

yeah I see what the trouble I am facing now

Alan

3:48 AM

Have any of you worked with extremely disconnected spaces? I ran into them trying to understand tghis stuff.

Balarka Sen

@ForeverMozart, perhaps?

Adeek

is there a more general version of this theorem

I guess by induction

Alan

definition is the closure of every open set is open.

Ted Shifrin

You haven't justified normality, Karim.

Forever Mozart

yes it is deep

usually induction doesnt work in topology

Adeek

3:49 AM

normality comes from having index 2

Forever Mozart

in continuum theory anyway

Adeek

@TedShifrin

Ted Shifrin

You have order 2, not index 2, Karim.

Alan

Do you have semidirect products to work with?

Ted Shifrin

The induction comment wasn't aimed at you, @Forever.

Adeek

3:49 AM

yeah

Alan

(whoever's working on this classification problem, I lost track :))

Balarka Sen

Am I sleepy or is this conversation becoming confuzzling

Adeek

yeah @Alan

Ted Shifrin

Yes @Balarka

Alan

@adeek I'm about 95% sure you can recognize any $p^3$ group as a semidirect product, and work from there.

Ted Shifrin

3:51 AM

Stop trying to kill a fly with a cannon, guys.

Alan

grin But.....cannons are fun! :)

Ted Shifrin

I won't even go there ...

Adeek

yeah I thought about it @Alan but then the problem should have been more simpler

hm

Ted Shifrin

There's an exercise that appears early in any undergraduate group theory course.

Adeek

why is subgroups of order 2 in group of order 8 are normal.

hm maybe by sylow theory ? let us see

Ted Shifrin

3:52 AM

That's false.

$D_4$ has plenty of subgroups of order 2 that are not normal.

Adeek

yes

Alan

Don't $p$ groups have subgroups of every power of $p$ up to the group?

Ted Shifrin

Stop it with all this generality. We're talking $p=2$.

Mike Miller

Normal subgroups even

Ted Shifrin

How the hell do I get dragged into undergraduate algebra? I'm leaving ...

Balarka Sen

3:54 AM

And now, the whole room is talking about group theory.

Alan

laugh Sorry, reminds me of one of my professors who always started with the most general case.

snorts Just saw the bordism thing on the sidebar, had to email a link to my topology professor

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$Mathematics$ Mathematics