@LeakyNun Do you have any hint as to how I could think about finding an embedding from $f:T^2 \to T^3$ and a closed 2-fomr on $T^3$ such that $\int_{T^2} f^* \omega \neq 0$? I dont want it given away, but Im not even sure how to start thinking about this

@KevinDriscoll you're asking the wrong guy

Cool

I am stumped on a curve sketching problem.

if anyone feels like looking at it

@MatheinBoulomenos can we have G acting on the 3 sylow subgroups by conjugation ?

Geeze thats a lot of conditions

yes ;P

@KevinDriscoll lol

@KasmirKhaan sure, $G$ acts on the $3$ Sylows by conjugation

my idea is to find a stablizer of some subgroup

that is non trivial

any stablizer is a normal subgroup

so making G non simple

But I dont see how to make that step

not every stabilizer is a normal subgroup

oups i meant centralier

or wait wait

that is wrong too

what was it called

Zenter :D

You have to use that fact that the 3-Sylows intersect non-trivially

Hmm hint? :(

I dont know how many of those will intersect non trivially

we have 10 , 9 elemtn subgroups

Just take two P and Q that intersect non-trivially

they can intersect in the identity and 2 elements

okay ._. i allways try to find the whole picture

im not sure if there's even a solution

we define H = P intersect Q

Now look at the normalizer of H in G

Call the normalizer N

@Kasmir Can you see why P and Q are both subgroups of N?

hmm

i drew apicture

two leaves with 9 elemnts

and a H intersecting 1 elennt of Q and 1 of P

and the identity

well, H needs to have 3 elements

it does that way

1 from P 1 from Q and identity no?

okay

if i take an element from P

all the elements in H are both in P and in Q

pHp'=H

qHq'=H

because?

because all p subgroups are contained in psylow subgroup no?

no

well all 3 sylow subgroups are conjugate

yes

if i take an element p that is not in the intersection

not in H, when we conjugate by it, we dont leave P

that's correct

but elements in H are defined to be the elements which are both in P and in Q

how do you know that pHp' is again contained in Q?

Hint: you know something about groups of order 9

This is related to an exercise you did earlier

well Groups of order 9 are abelian

but dont see how that is helpfull

what do you know about subgroups of abelian groups?

ahaaaaaaaaaa

:D

they are normal :D

so we found our non trivial normal subgroup

not so fast

oh okay ._.

H is a normal subgroup of P

and of Q

but it is not necessarily normal in G

hmm true true

@MatheinBoulomenos what is an algebra? why are various concepts called algebra?

what is the commonality?

hmm, good question

i was wondering about that too

division algebra

whats that

@KasmirKhaan but do you see now why P and Q are contained in N?

division algebra is just a ring in which every element is invertible

like a field but without commutativity

exept 0 :D

@MatheinBoulomenos is it?

right

ehm I dont see that yet but hmm let me see

wiki says it's an algebra over a field

@LeakyNun the center of a divison algebra is always a field

he was answering me leaky

@MatheinBoulomenos alright

@MatheinBoulomenos well by definiton of H being normal in Q and P

means that they both normalize it

@KasmirKhaan correct

now you have to do some casework

how large can N be?

I mean what are the possible orders of N

well it is at least 15

P+Q-H

thats what I did

well i need to add 1

but you also know that N is a subgroup of G

hmm thinking

well i find 17 elements at least

but that cant be it because 17 dont divide 90

right

P is a subgroup of N

aha

i Think i can see it now just one second :D

must be mutiple of 9

Okay that was a full 2 seconds Kasmir

Fricc I got sniped

the only possible choiuce for order of N to make sense is 45

thus we have a subgroup of index 2

making it normal

but I need to find that with testing numbers

hmm

I think there are a few more possibilities

this number must be divisible by 90 and multiple of 9

hmm

thats it mathein ._. i find it to be only 45 tha tmake sense

@Daminark hello dami

How's it going?

@KasmirKhaan 18,45,90 ...

these are the only three possibilities

technically 9 is also a multiple of 9 and divides 90, but you ruled that out already by seeing that it has at least 17 elements

hmm true ._.

Oh are we counting stuff? I'm excited

we can exclude 90 because we want it to be nromal

how exactly can we exclude 90?

what happens if N has 90 elements?

Hi Dami

Yo Akiva, how's it going?

Good¨

Oh god how'd you even do that?

Oh whoops

@MatheinBoulomenos hmm sorry about that ._.

I like that ö looks like someone screaming

90 is accutly also good for our cause

you don't have to apologize

ÓoÒ

@KasmirKhaan right, because if N=90 .... ?

N(H) =90 means that H is normal in G

that is accually good for us :D

correct!

:D

also 45 is good for us

two cases to go

because of index 2

right

now we just have to exclude that 18 busness

ermm let me Think a sec :D

18 cant happen because if we have 1 elemnt that normalize H , means its inverse normlaize H as well

because if gHg' =H

so does g'Hg =H

no ?

in fact we might have 3 mor eelemnts or 9

yes, it is true that N is closed under taking inverses, but I don't see how that implies that 18 is not possible

so we have our original 17 elements that come from P+Q

if we want to add just 1 element from Another 3 sylow subgroup

we have to take with it its inverse at least no?

why do we take another element from a 3 sylow subgroup?

we could add an element of order 2

well that works too :)

even better :D

no, it doesn't, an element of order 2 is its own inverse

><

true

Bamboozled again

You know that both P and Q are subgroups of N

they are even Sylowsubgroups of N

> division algebra is just a ring in which every element is invertible

Every element slash additive identity :P

aha

@Secret right

by that logic 18 wont work

18 =2*3^2

this is saying that we have one 3sylow subgroup

One will probably need to blow up at least power associativity in order to algebraically divide by zero

but in fact we have 2

in a nonassociative algebra, there can be no zero divisors and every element has left and right multiplicative inverses, but they may not be equal

n_3 congruent to 1 mod 3 and divisible by 2

making it n_3 =1

@anon sup anon :D long time no see

@KasmirKhaan that finishes the argument. Good job!

sup

@MatheinBoulomenos thanks :D and thanks for being patient with me :D

What do we call for a nonassociative algebra that is neither alternative nor power associative nor lie?

horrible to work with

jordan

wait, those are power associative

joke failed

part c) is to prove that a Group of order 90 is not simple

should one use what we did?

or what kind of question is that :D

@KasmirKhaan that's a bit strange

you basically already proved it

yes ._. but i Think he wanted to give us 1 Point for free

the other question were 3 -2 points

=p

well i Think the argument is , when 3 sylow had no intersection G was proven to be not simple

and when it had intersection was also not simple

but that is not very creative

I think you mean "not simple", but yeah

oups yes =p

does that argument work in general?

i mean alot of info is missing

what do you mean by "that argument"?

if we had the biggest p sylow subgroup

by biggest i mean cardinality

oh never mind ><

bad question =p

Well, you can always try to make different cases depending on how the p-Sylows intersect, sometimes that's helpful and sometimes not

it is very Clean the subject of algebra ><

also very hard

but the arguments once one sees them , they make sense

unlike analysis =p

I feel similarly

that's why I like algebra

:D

I Think ill chose more algebra Courses =P

allready gonna take rep theory for finite Groups on january

and shall take more!

Suppose R is non necessarly commutative divison algebra , prove that the only two sided ideals of R are 0 and R

Oh rep theory sounds like it'll be a fun time

Rep theory is really beautiful

yes but i need while dember moth to prepare for it

iam still struggling with abstract algebra

and i assume am gonna need to be fluent in AA to do well there

=P

hi everyone

Hey

@Daminark I am trying to understand few things about hermitian metric on a manifold

Is that just like, the complexified version of a Riemannian manifold?

yeah

I see

hello

If $G/Z(G)$ is abelian, how can I show that $[x,y]\in Z(G)$?

This seems straightforward, but I'm stuck

Any hints?

I have: $[x,y] = xyx^-1y^-1$, so by the abelieness of $G/Z(G)$, we have $[x,y]Z=[y,x]Z$. I don't know see how that helps though.

@Daminark do you want to discuss it ?

I want to see if I am understanding this correctly ?

$[x,y]=xyx^{-1}y^{-1}$

I could try? I guess? Though I don't know any Riemannian Geo so I'm worthless

So A hertmian metric is a positive definite inner product on the tangent space

poisitive definite hermitian inner product

Aight

Then we have an induced real inner product

and the reason we have an alternating induced bilinear form

is because we are looking at the holomorphic tangent space

so in particular elements of the conjugate are killed

right ?

@orbit-stabilizer So, I think it's a general fact that $G/K$ is abelian iff $[G,G] \le K$

But I've never proven it yet, so we can try

What is $K$?

I think their is a mistake in GF page 27

@Adeek Do you mean like a $\frac{d}{d\overline{z}} f(z) = 0$ for $f$ holomorphic type of thing? I could see that

I think we are looking only at Tz(M) \otimes Tz(M)

not bar

@orbit $K$ is any normal subgroup

yeah

@Adeek that's how I interpreted "elements of the conjugate are killed"

Though again I don't know anything about that stuff, like I only know very basic differential topology and complex analysis, haven't even touched Riemann surfaces yet

@Daminark hmm okay. I think I need to spend a bit more time with nilpotent groups before I proceed. I found that lemma in Rotman, I'm going to try to understand it

@orbit Okay so if $G/K$ is abelian, then for any $k\in K$, you have $xyk = yxj$ for some $k,j\in K$

Nvm, didn't find it, it was something else

Right, following

This means $x^{-1}y^{-1}xy = jk^{-1}\in K$

@Daminark I am not sure

@Daminark I will ask my advisor tomorrow

Ted hasn't been here lately.

@Daminark ah.

So that's just $[x^{-1},y^{-1}]$. And this must hold for all $x,y\in G$. So this implies that $[G,G] \le K$

@Adeek :(

@Semiclassical ?

@Adeek probably a good idea. And yeah now that I think about it, it's been a while

@Daminark got it. How's your group theory going?

thanks

re: Ted not being here lately

@Semiclassical Yeah I missed him

It's going pretty well. We finished Sylow a while back so we've just been doing a bunch of things

@Semiclassical You want to hear something funny ?

Same. We did solvability and nilpotence recently

it is about physicists

Showed that $A_5$ and $\mathbb{Z}/p\mathbb{Z}$ are the only simple groups of order under 100

Then talked a tiny bit about composition series and solvability

sure

Oh, interesting. Didn't know that

Today we did a bit of Mobius transformations

And next class we're doing finite subgroups of $SO(3)$, apparently it's got something to do with Platonic solids

@Semiclassical I was talking to a theortical physics prof today about Principles of algebraic geometry by GH. He was mentioning to me the difference between a geometer and a physicist is that geometer finished all of GH and physicists only read chapter 0.

lol

lol

Huh, that's outside of standard group theory class

Yeah, our prof wants to do geometry and the like

Like this is just supplementary material that isn't really coming up on the psets, aside from the simple groups of small order business, it's just, here's some stuff

(Given that we've already had our last pset of the class...)

But yeah I dunno, I'm a bit less into this part of the class. I was the one who asked him to do solvable groups and all, hoping it'd replace Mobius transformations and whatnot. I'm just not really into the geometric stuff. But he ended up doing this instead of group presentations, which he originally planned to

alright I am gonna back to work

@Daminark Why geometry is awesome

It's just not really my thing somehow, every time in the past I've engaged with geometry it's been eh

@Daminark whaat geometry is great

you're in the minority lol

Like when we did diffgeo in the summer I was just like "Yeah... I dunno... I guess this stuff is neat and all but like... meh"

I love diff geometry I learned about Riemannian metrics and stuff for my things

it is very nice

interesting, I really liked the very little bit of diff geo i did

Also I'm kinda bad at thinking geometrically, which has made problems where you want to be thinking geometrically really frustrating/discouraging

I think we're all a bit geometrically challenged haha. I can't visualize stuff well

It's nice when it works out though

Like in topology, algebra, analysis, I can sorta pretend that I know what I'm doing

But the diffgeo was a train wreck for me, and when we were doing the sort of geometric stuff in group theory I was also having a bad time

It took forever for me to realize how the dihedral group worked

regarding D_2n, I watched the visual group theory yotube videos, they really helped me

regarding symmetry groups in general

and how they're different from symmetric groups

I only was completely convinced that the dihedral group had $2n$ elements when I thought of it as the automorphism group of the cycle graph. Once I realized that the point was you had to preserve adjacency, everything else made total sense

In any event, I may give other parts of geometry a try, in particular I've heard that algebraic geometry has a place for those who aren't geometrically inclined so that might be better suited

Yeah, symmetry groups of geometric objects are (almost always?) the automorphism groups

I wish I could do algebraic geometry haha. Way beyond me atm

I mean same, just like, in the long run it may be something worth looking at

In the long run, we're all gonna be dead :(

Okay here's another group theory question

We have different definitions of long run, I'm thinking more, the next few years

I've shown $[x,y] = [xz,yz']$ when $z,z'\in Z(G)$.

I need to show this induces a map $G/Z \times G/Z \rightarrow Z$.

I define it as $f(xZ,yZ) = [x,y]$.

$\to Z$? What do you know about $G$?

Yes, to G

Huh?

$\rightarrow G$

oh. you wrote $\to Z$, not $\to G$.

I know that $G/Z(G)$ is abelian. Oops - just noticed.

okay, then yes, $\to Z$

a priori $G/Z\times G/Z\to [G,G]$

I want to show it is well-defined. So, if $xZ = x'Z$ and $yZ = y'Z$, then $[x,y] = [x',y']$

yes

you've already done that

$xZ=x'Z$ iff $\exists z$ s.t. $x'=xz$, and similarly $y'=yz'$

so verifying $[x,y]=[x',y']$ amounts to verifying $[x,y]=[xz,yz']$, as you did

****

sometimes I wonder why I'm in math

thank you

Hey chat

hello

Yo

next, you know $G/Z\times G/Z\to [G,G]$, so now you want to observe that if $G/Z$ is abelian then $[G,G]\subseteq Z$

(really, we have $G/N$ abelian iff $[G,G]\subseteq N$ for all normal subgroups $N$)

Why do I need that part?

If I just want to show the map is well-defined, I'm done before that step, right?

sure, but you said you wanted to show it induces a map $G/Z\times G/Z\to Z$

showing it's well-defined doesn't establish the range is in $Z$

brainfart. jeez

you're totally right

Ah, but I've already shown [x,y] is in Z(G)

previously

anyone here into topology?

Sorta? But I'm a noob so like... Not sure if I'll be helpful

algebraists >:(

is this fun-pictures topology or point-set topology?

Why all the hate for algebra? It's like, the child of number theory, which is the queen of math

Elementary Number Theory is scary

But it's so much fun!

@anon, i'd say fun pictures

a little algebra as well

hi @anon

hi

@gian what kind of thing are you doing right now?

Hey, here's a fun problem that was posted in this chat a while ago for number theory

a little research project of mine

oh you mean in school?

$2^{n}(2^{n!}-1) \equiv 0$ (mod $n!$)

Prove or disprove

@gian in life, and also whatever you were asking about just now

@orbit hmm, I'll try it out and see

in life i just study math. right now im working on some ways of computationally analyzing stratified spaces

Here's a tough question