groups-rings-fields

[balmy python]

X ~ Y iff P(X = Y + c) = 1 for some c E R and P is a probability measure

how is the quotient space induced by this relation isomorphic to a subspace of random variables with mean 0?

[tribal moss]

That looks more like a job for #probability-statistics?

[balmy python]

oh damn okay 😭

[thorn jay]

equivalence relations aren't group theory

lol

[noble nexus]

or are they

fun fact equivalence relations on G that form a subgroup of G x G are in bijection with normal subgroups of G

[thorn jay]

yes, congruences

congruence relations

calling that group theory is like calling set theory topology lmao

[noble nexus]

ya I like the congruence perspective b/c it gives a bit more intuition on where normal subgroups come from

[thorn jay]

you'd like universal algebra

[noble nexus]

ya ik

[thorn jay]

it's basically the study of congruence relations

no one likes subalgebras

they're badly behaved

[vapid vale]

i do

[thorn jay]

do you?

name all of them

[noble nexus]

I like subalgebras

[thorn jay]

the only time I work with subalgebras is when they're the image of a homomorphism, so I can act as if they're a quotient

[vapid vale]

sub things are nice

elements are actual elements

[thorn jay]

and finiteness conditions

[thorn jay]

you can't quotient subthings

but you can quotient congruences

checkmate

[noble nexus]

true but really being able to quotient sub things is kinda a happy coincidence

not really as natural

Tbh I think the congruence relation perspective should be shown

b/c I think it's normal for ppl to get confused about normal subgroups and ideals

[thorn jay]

yeah totally

[noble nexus]

but it's actually all just congruence relations!

and really every quotient in math is a quotient by a suitably nice equivalence relation

[thorn jay]

congruences are just super nice objects, and it's crazy just how much more they say about an algebraic structure than their subalgebras

[balmy python]

why are the probability people all asleep at this time

[tough raven]

It's one extra concept for people to get used to, and I don't think people learning group theory (and also any kind of abstract algebra) for the first time are usually ready to handle that one more abstract thing.

[balmy python]

😭

[tough raven]

Moreover, normal subgroups are very useful in group theory - i.e., when doing group theory specifically, you really want to think that quotients are classified by normal subgroups and not congruence relations (e.g. because you might be inducting on size of a finite group and want to apply the inductive hypothesis to the normal subgroup).

[uneven vale]

you called?

[balmy python]

😭

[tough raven]

So it makes sense that in a group theory course, one would introduce normal subgroups whether or not the students know congruence relations already, and one would not introduce congruence relations at all, really.

[balmy python]

what does it mean for a variance of a random variable to be zero

[tough raven]

const

[balmy python]

but why

[tough raven]

Otherwise it would vary. If you want to be more technical, then Var(X) = E[(X - E[X])^2)] is the expectation of a non-negative random variable and can be 0 iff the random variable is (almost surely) 0, i.e., with probability 1, X = E[X], which is some constant.

Why this channel BTW

[uneven vale]

If you look at this definition of variance:
$$\text{Var}[X] = \mathbb E[(X - \mathbb E[X])^2]$$
you can intuitively imagine it as the average distance of the random variable X to its mean. If the average distance is zero, that means the random variable does not vary at all.

[uneven vale]

or you can do that

[uneven vale]

¯_(ツ)_/¯ reading is hard

[balmy python]

oh yeah

[cloud walrus]

clorox_1g

[balmy python]

well i started off asking here because i had a question on an isomorphism in some probability space but then they told me to ask in the probability section but they're all sleeping rn

so i came back here 😭

[balmy python]

yeahhh that intuitively makes sense

I was just looking at variance-covariance matrices and realised i needed a little bit of patchwork in my fundamentals

[uneven vale]

just play around with normal random variables and you're good. Everything's a normal anyway

[balmy python]

wdym everythings a normal 😭

[uneven vale]

Central Limit Theorem with n=1

[vapid vale]

“otherwise it would vary” is hilarious lmfao

[balmy python]

i'm getting onto that in a couple lectures time

[uneven vale]

fair enough, basically you'll learn that the normal distribution is the one true overlord of distributions and nothing else matters (approximately)

[noble nexus]

sure, but I would argue they are much easier to understand than normal subgroups

and the equivalence between the two is quite simple

(if H is any subgroup, then the equivalence relation it generates is a congruence relation iff H is normal, and if ~ is a congruence relation then the set of elements with x ~ 1 is a normal subgroup)

[cobalt heath]

Hello, is there a theory on approximate homomorphisms, like on modular rings?

Is it subsumed by lattice theory

[thorn jay]

what is a modular ring in this context?

[cobalt heath]

Rings like Z/qZ

More generally, Z[X]/(q, f) (ofc can go more general, but this one is what I am interested in)

I will elaborate later

[crystal vale]

We know that if R is a commutative ring with unity and I, J are ideals such that I + J = (1) then IJ = I \cap J.

Is the converse true if both I and J are non-zero ideals?

How crt helps me here?

[balmy gyro]

No idea where to put this, but does this proof work?

[rocky cloak]

Say you have finitely many maximal ideals, what happens then?

[crystal vale]

You have to use isomorphism here

[quiet pelican]

You haven’t proven every polynomial equation in F_2 can be written in the form x^n + f(a_n-1)x^n-1 + …

[rocky cloak]

It is not

[rapid cave]

Take an idempotent ideal I and J=I

[crystal vale]

Yes got it

[crystal vale]

Say M1,..,M_k are the only maximal ideals so I can say R \ ( \cup M_i) is finite.

And I am thinking different question in the same context, say if there is only one maximal ideal M, then R/M is field but R/M has only one element so can I say in this new question that it is not possible that R is local ring with finitely many units.

[rocky cloak]

It's not possible that R is an infinite local ring with finitely many units

Try to describe what the units of a local ring are

[crystal vale]

So that's an argument correct?

[crystal vale]

R-M

All elements outside of maximal ideals

[rocky cloak]

And why is that infinite?

[crystal vale]

No idea

I see

R is infinite ring

If R\M is finite then M has to be infinite

But in local ring, every 1 + x, where x in M is an unit element

Then we will get infinitely many units elements

[crystal vale]

Any hint here?

[rapid cave]

Jacobson radical

[rapid cave]

Iirc it has a characteristic similar to this

[rocky cloak]

If you have finitely many you can use CRT

[rapid cave]

@rocky cloak can this work?

[rocky cloak]

Yeah, if you can show the Jacobson radical is infinite you're done

[elfin wraith]

Yeah like I went to a group theory talk a couple weeks ago and it was kinda just this. A seemingly completely synthetic definition (I think it’s a vast generalisation of something that someone was kinda interested in before) and it was just working out what groups fit that definition

Fun, for sure, but like it just feels like a nice toy problem I’m not sure why I should really be all that bothered

[elfin wraith]

I was talking about this with some people the other day. I don’t know that I’d want to introduce quotient things via groups because normal subgroups are really opaque. At least with like rings you can clearly write down some elements and you’re saying these are the exact relations were introducing, but I think a lot of people get confused because normal subgroups are kinda weird

Like I don’t know how to motivate normal subgroups beyond “they’re the subgroups such that when you partition the group by them, the operation is still well defined” which is true and what you want, but I’m not sure that’s super motivating for your first introduction

[rocky cloak]

They are exactly the kernels of homomorphisms, just like ideals.

That also motivates the definition as gKg^-1 = K is clearly true for a kernel.

Or how are you motivating ideals?

[crystal vale]

Yes x in J iff 1 + xy is an unit for all y in R

[knotty badger]

There’s a few different ways you can motivate normal subgroups

They’re derived from congruences on groups for starters

They’re also highly “symmetric” subgroups in that they look the same from all perspectives

Kernels of homomorphisms is also a good motivator

[crystal vale]

So it gives me that R/(M_1...M_k) isomorphic to R/M1 ×..× R/M_k

[rocky cloak]

Indeed, so what happens if R/(M1...) is infinite? And what happens if it is finite?

[knotty badger]

I think the kernel pair perspective is good too

[crystal vale]

If it is infinite then one of them R/M_i needs to be infinite and if it is finite then all of them R/M_i needs to be finite

[thorn jay]

probably too categorical lol

this is first year UG

[knotty badger]

Well you don’t have to give the full abstract def

[thorn jay]

but then it's literally the same as a congruence relation

[knotty badger]

But I think it’s quite intuitive to think of quotients as “identify these things if they have the same property”

[rocky cloak]

Indeed, so maybe first consider if they're all finite, what does that say about
(M1...Mk)

[knotty badger]

And that’s all a kernel pair is

[thorn jay]

I don't know why you'd use kernel pairs rather than just the kernel tbh

why introduce fiber products

[thorn jay]

(kernel as equivalence relation i.e. congruence relation)

[knotty badger]

What actually is the issue, I genuinely don’t understand

All you need is

Given a homomorphism f : G -> H

You consider the subset of G x G given by {(g, h) | f(g) = f(h)}

[crystal vale]

(M1..M_k) is infinite

[thorn jay]

this is called the congruence relation perspective, they are the same

then

[knotty badger]

Not as far as I know

[rocky cloak]

And then you're able to construct infinitely many units, yes?

[knotty badger]

A congruence relation is an equivalence relation that respects the group operation

[crystal vale]

Yes

[thorn jay]

the congruence relation perspective in my eyes includes the correspondence of congruences <-> kernels of homomorphisms

[knotty badger]

A kernel pair arises from a homomorphism

You can look it up on nlab, they are not the same definition

[thorn jay]

I know they're not the same definition, but they are the same perspective

because congruences are kernels

[knotty badger]

I don’t agree with that either

Equivalence relations and partitions are different perspectives

Even if they turn out to be equivalent

[thorn jay]

a congruence is not a partition

it's an equivalence relation

[knotty badger]

I really don’t understand your point

A kernel pair arises from a homomorphism

A congruence does not

[thorn jay]

?? how does a congruence not arise from a homomorphism

[knotty badger]

When you specify a congruence, all you are doing is specifying an equivalence relation on G that respects the group operation

You don’t need to first define a homomorphism out of G to do this

[thorn jay]

I guess? but the fact that every congruence is a kernel and every kernel is a congruences makes the two equivalent in my book

[knotty badger]

Then I can go back to the equivalence relation and partition thing

Honestly just forget it

[thorn jay]

but both congruences and kernels are equivalence relations

[knotty badger]

I can’t understand you algebraists sometimes

[thorn jay]

lol

[knotty badger]

They arise from different things but I guess you can’t understand that

[crystal vale]

And if they are infinite, then one of them is infinite say R/M1 is infinite field, so how it helps me here?

[thorn jay]

maybe it's a concept with an attitude thing for me

[knotty badger]

I don’t care

[rocky cloak]

Think about the units in R/M1 x... x R/Mk

[thorn jay]

alright, alright, noted

[knotty badger]

uh huh

[crystal vale]

Infinitely many

[rocky cloak]

And what do they correspond to in R?

[crystal vale]

I am not sure about it

[elfin wraith]

None of this is a fully formed thought FWIW, just something I was thinking about the other day.

Ok yeah only way I know to motivate it is silly because I do know other ways, but I don’t know how much I love this as an introduction.

I also agree that it’s basically the way you want to introduce ideals, but I think my issue is more that I don’t know how, maybe “illustrative” that is, at least as to what they’re actually doing.

This kinda came off the back of a conversation with someone (another MSc student, mostly did logic but still knew a fair amount of algebra) mentioning that he still didn’t really “get” normal subgroups, and it made me think about how when they’re first introduced you typically only really know like S_n, D_n, maybe A_n and so when you see an example like S_4/A_4, I don’t know how clear it is what’s happening there?

At least when I compare it to quotients in rings, where you can very explicitly write down the relations you’re introducing, or topology where you have a very visual representation of what’s happening, so I wonder if the concept of normal subgroups either as a kernel or just being the condition such that your operation is still well defined is possibly more satisfying after some more experience with the other objects

[crystal vale]

I can say there are infinitely many x_i such that for each there are y_i such that 1-x_iy_i in (M1...M_k)

[knotty badger]

Are you saying ideals are easier to grok than normal subgroups

[thorn jay]

if he's into logic then quotients (rather, congruences) have a very nice logic-y meaning

[knotty badger]

Because this was not at all the case for me

[elfin wraith]

This was an idea I was considering, I was wondering if there’s some nice interpretation from how they act on the Cayley graph of your group or how they look in there, but I’m pretty bad at that kinda thinking and wasn’t sure if there’s something or not

[knotty badger]

You can do this for symmetry groups of shapes

Any subgroup can be interpreted as a stabiliser

[rocky cloak]

I'm not sure I follow what you mean by "very explicitly write down the relations". Are the relations for groups less explicit?

[knotty badger]

But it depends on whether the thing you’re stabilising looks the same from all perspectives or not

[elfin wraith]

For me, yes, I have found them to be. I think for quotients especially. This could just be a me thing though

[knotty badger]

I guess I’ll chalk that up to another thing I don’t understand about algebraists

[thorn jay]

once I got normal subgroups, intuition for ideals followed so it was not at all the case for me as for Nope

[delicate orchid]

Trying to understand the definition of an ideal only really clicked once I compared them to normal subgroups

[rocky cloak]

I can see rings and ideals being more intuitive, since you have a lot more examples of them than groups before going into an algebra course.

But just in terms of the definitions I don't see a big divide

[elfin wraith]

So like, comparing S_4/A_4 to for example, R[x]/<x^2-1>, or <a,b,c>/<a+b-c> or whatever, yes you can kinda unpack things for the groups example but in the typical presentation for these things, I find the ring examples far easier to understand

This could all just be a me thing, as I said this was just kinda my feelings, somewhat also felt by the guy I was talking to

[thorn jay]

maybe its weird that conjugation seemingly comes outta nowhere? where for ideals you've got the idea of "multiple of element"

[elfin wraith]

I wouldn’t generalise this to anyone past me lol

[rocky cloak]

And what do you know about elements with 1-z in the Jacobson radical?

[delicate orchid]

Ok I think I agree here in the sense it’s easier to get an intuition for quotients being “introducing relations” in the ring setting

[elfin wraith]

Yeah to me, at first anyway, the definition of a normal subgroup felt a bit odd (learning it was just because this is how we partition and still get a well defined operation satisfied me though, so maybe that’s enough) but ideals always felt pretty natural.

That being said I had already taken group theory at that point so maybe I just had better intuition

[knotty badger]

Idk, me not being able to understand algebra is a very general phenomenon

[rocky cloak]

So I guess you're saying that you usually consider rings that are free or have simple presentations, but you consider groups with complicated presentations. Hence when you think in terms of presentations, the ring perspective is easier?

[elfin wraith]

Which is kinda my point, I wonder if this is a nicer starting point? Because the situation is the same in topology I guess, I can show you how to make a torus as a quotient space, by introducing some relations, and this is the generic idea you “should” (at least this is how I think about it) have in mind for quotient objects

[delicate orchid]

Who up quotienting they fundamental polygon

[thorn jay]

well rings are in general better behaved wrt quotients than groups imo
like, it feels easier to work with ideals than with normal subgroups

[delicate orchid]

I think that’s because you’re CRAZY

[elfin wraith]

Yeah that could be case, I think it could be that

[quiet pelican]

Genus g surfaces my belovèd

[thorn jay]

esp comm rings but that's kinda cheating because normal subgroups of abelian groups are also well-behaved

[crystal vale]

z are units

but what if x1y1 = x2y2

[rocky cloak]

If x1y1 is a unit, what does that tell you about x1

[elfin wraith]

Well this could also be the point I guess haha, I did only give commutative examples there. But I think even if you get into the nitty gritty of left ideals you can still make the idea clear of relations

I mean this is kinda how I think about (some) noncom rings anyway, you take something like k<x,y>, and you say ok I don’t want total non commutativity, that’s not very interesting, so lets look at k<x,y>/<xy-1+yx> or something

Jagr may be right though that it’s just the case that the first rings you meet typically have much more simple presentations than the groups you tend to first meet

[crystal vale]

x1 is an unit, thank you @rocky cloak

You helped me a lot, nice exercise

Wait, I have to show more things

so x_i's are unit elements

[quiet pelican]

But but
$\langle x_1, …, x_n \mid (x_i x_{i+1})^3, x_i^2, x_ix_j = x_j x_i, |i - j| > 1 \rangle$
If it’s a Coxeter group it’s presentation is simple enough 🙃

[crystal vale]

Oh got it

x_i's are all different

[rocky cloak]

Lot of peoples first groups are Z/n, but I suppose these are too simple to be enlightening

[delicate orchid]

only real reflections
Cowardice

[elfin wraith]

Yeah I think quotients of Z are almost so nice they can be deceptive, but they’re certainly good examples to have in mind, usually the first thing I’ll go to for a sanity check

[cloud walrus]

mico the group birb

[elfin wraith]

I think the first groups I saw in my courses were like the symmetric, alternating and dihedral groups though, I don’t think my first course introduced anymore than those and like GL_n

[quiet pelican]

I’d assume some other matrix groups (SO, O, SL, U, SU, maybe Sp if you’re very lucky)

[rocky cloak]

I guess another thing is that normal subgroups are also groups.

Like A4 is a group people care about that has a name, etc. So you're sort of thinking about two groups when doing S4/A4.

Ideals don't do anything else besides being kernels. No one has given a name to or care anything about <xy-1+yx> by itself

[elfin wraith]

Yeah probably, if you introduce one the others just make sense , but my point is we mostly stuck to the first 3 and ran with those

[thorn jay]

ideals are R-modules I suppose lol

[quiet pelican]

Introduce surface and free groups and Coxeter groups and RAAGs and…

[rocky cloak]

That's more left ideals

[thorn jay]

yeah they're R-bimodules is more accurate

[elfin wraith]

Yeah I think this is true, i do have a slightly more complex idea of normal subgroups in my head than ideals, which I literally just think of as relations (perhaps as curves if someone’s forcing me to do AG…)

But yeah, idk, this was just a thought I was having the other day, pondering how I would introduce quotient objects if I was teaching intro algebra to someone

[rocky cloak]

But if you're thinking about that, then R is probably fixed and then the context is totally different

[rapid cave]

Absolute Galois Groups 🗿

[true bolt]

Yeah I was debating giving a talk that involved absolute galois groups for my uni's maths society

However I'd also like my talk to be understandable

woe

[south patrol]

What other type of Galois group is there

[sacred wharf]

Non absolute i imagine

[knotty badger]

uncommon for a math talk

[true bolt]

Potato when finite field extensions of non separably closed fields that aren’t real closed (or whatever the term is)

[true bolt]

Well the intended audience is ideally 2nd years and above. I’m considering doing something on monodromy

[rocky cloak]

It's good to pace your talk so that there is a part for everyone. You start out with some things everyone understands, then ramp up the difficulty slowly until only experts can follow. Then you should ramble for about 5 minutes about something only you understand and finish it of with about 5 minutes of rambling about something even you doesn't understand.

Recipe for a perfect talk

[tribal moss]

Any questions?

[knotty badger]

oh i asked this Q in #alg-top-geo-top

what is monodromy

[true bolt]

I currently think of it as essentially being representations of the fundamental group. I can yap more when I’m next at my computer

[sacred wharf]

2nd year undergrads? damn i need to start grinding

[south patrol]

Ye I meme

[true bolt]

The 2nd year undergrads who’d attend an optional talk here should know what a group/ring/module is, and if in the second semester they should also know some Galois theory and some point set

Oh and I guess some complex analysis

But there’s a difference between knowing some stuff and understanding it

[sacred wharf]

i am so behind damn

[sacred wharf]

i wanted to take galois theory in a few days but it's only offered every other year

[true bolt]

It’s probably some of my favourite maths

[rapid cave]

Galois theory is amazing

[sacred wharf]

Has someone got a good video on Smith Normal form ?

I can't understand jack shit

or maybe a pdf that is very pedagogical

[true bolt]

Specifically, I learn galois theory from Borcherd's lectures, and I remember when he drew up the subgroup diagram (labeled with indices of subgroups) and then the subfield diagram (labeled with degrees) and they were the same diagram, and that was probably the first time I've understood the idea of mathematics being beautiful as opposed to just like a fun puzzle

[sacred wharf]

you learn from lectures? I have never learnt any piece of math from a lecture

or anything meaningful at least

[rapid cave]

That is what you do at uni

Go to lectures

[true bolt]

Uhhh I mean I take the opinion that you use lectures to learn stuff for the first time and then if you want to learn it deeply you read it

[sacred wharf]

oh you're not meant to just skip all lectures and read the book

[knotty badger]

huh i see..

[sacred wharf]

is it more efficient than just reading the book more deeply

?

[rapid cave]

Yes

[sacred wharf]

https://tenor.com/view/side-eye-dog-suspicious-look-suspicious-doubt-dog-doubt-gif-23680990

[rapid cave]

In a recent lecture I went to, over 3h we covered some basic category theory and almost a whole section of Hartshorne, including some exercises

[crystal vale]

Oh man

[crystal vale]

Phd?

[elfin wraith]

MSc student interested in algebra and topology who doesn’t know Galois theory here!

[crystal vale]

Any hint for part b?

[true bolt]

I just sent you a small essay, my apologies

[knotty badger]

why apologies

[true bolt]

I view it as lectures are a more efficient way to see lots of things which is important to get an idea of the mathematical landscape

[true bolt]

Because now you have to read it :P

[true bolt]

Yeah the way my university sets things up is undergrads basically have to do Galois theory to do any further algebra past intro abstract

[knotty badger]

I don’t have to

idk it just seems like a weird thing to apologise for

[elfin wraith]

Yeah it was very much optional at my UG, and after going to the first lecture and realising the notes were some 150 pages I decided to drop it and take a quantum computing/monoidal category theory course lol

First time I heard about the Galois correspondence was in my AT class and I was somewhat lost

I am currently learning Galois theory though, I feel like I should probably know it

[sacred wharf]

mandatory galois in 2nd year can i ask where or what your institution is ?

[elfin wraith]

I don’t think that’s massively unreasonable, assuming you’re at a uni that just jumps straight into LA

[true bolt]

Australia

[elfin wraith]

It is quick but I think doable

[sacred wharf]

Maybe it is or isn't, but mandatory Galois in 2nd year I haven't seen in many places at least

[true bolt]

oh its not mandatory

its just the expected pathway if you're achieving high grades and want to go into algebra

[elfin wraith]

Half the reason I didn’t end up taking Galois in UG was because it seemed pretty straightforward and I was very confident that I could just do it myself whenever I needed to

[true bolt]

since you're locked out of algebra classes past the introduction if you haven't done the galois theory class

since it is "Algebra 2"

[elfin wraith]

Then my masters uni deleted half the algebra courses and so I’m now currently taking it lol

[sacred wharf]

interesting, i thought australia was generally more americanish

with calc 1- calc 4 sequence

before any pure math

[true bolt]

My institution is irregular

[sacred wharf]

ig it is a very good university

[sacred wharf]

is it true they have 26 weeks of college per year in aus

[true bolt]

For instance our first sem math class is Spivak calculus/rudin (and some more computaitonal linear algebra class) and then the next sem is axler linear algebra (and a collection of misc multivariable books)

The dedicated will also do an "intro to proof" class in first sem which does a lot of like discrete stuff and then in the second sem you typically take introductory abstract algebra

[true bolt]

my institution is 24 weeks of lecture per year

[supple ice]

pick a countable set of points in the interval that drifts toward the endpoint zero so it has no accumulation point inside the interval consider all continuous functions that are zero on that set this is a proper ideal so it sits inside some maximal ideal that maximal ideal is not an evaluation at a point ideal because you can build a continuous function that is zero on the whole set but equals one at any chosen point of the interval to get infinitely many such maximal ideals split the chosen set into infinitely many disjoint infinite subsequences and repeat

[elfin wraith]

It’s like this in the UK and it kinda annoys me, you could learn so much more if you weren’t off half the year. Granted it means academics get to do a bit more research and I got to work over summer so I could actually afford to live, but eh

[sacred wharf]

wha

over here it's 36-40

[elfin wraith]

At my UG it was 22 and I think at my masters it’s 20

[true bolt]

At least here its 24 weeks of lectures per year but then like 8 weeks of the year are the exam periods

[sacred wharf]

i can't lie having that time off to sort of study math at your own pace and do what you want seems nice but it's very low

[elfin wraith]

Yeah that was kinda how my UG was

[south patrol]

Ig mine was 16 lol

[elfin wraith]

Yeah that would be true if I ever really did any maths over the summers lol

[south patrol]

Cursed

[elfin wraith]

Oxbridge moment

[sacred wharf]

my uni doesn't have courses over the summer but they have retake exams for previous courses, so my plan is to just sign up for courses i'm not going to do and do the exam in summer

[true bolt]

Does oxford not do any lectures during the 3rd term where you half ass lecture attendance?

I know camb at least used to run classes during a 3rd term but I don't really follow camb too closely now that most people I knew there have graduated

[south patrol]

Yeah 3rd term was basically just exam prep and exams

[knotty badger]

Same at cam

[south patrol]

There are also sometimes misc. courses but like 1/2 a course rather than multiple lol

[true bolt]

yeah that's what I was thinking of

[thorn jay]

wtf do you think lectures are for vro

[sacred wharf]

an excuse to pay professors

[south patrol]

So that you can skip them and feel superior

[sacred wharf]

superior until you're in the exam hall

then you feel inferior

[thorn jay]

skill issue git gud

[south patrol]

Tbh I did not go to lectures much through uni but they had very thorough lecture notes for (essentially) everything

[sacred wharf]

lecture notes > book

for some reason

idrk why

[south patrol]

Though it was bad when profs deviated a lot from the notes and then like I found out too close to the exam

[sacred wharf]

me in probability fr

[thorn jay]

because the lecture notes go more in depth with more examples or diagrams, a textbook usually has wayy more to cover

[south patrol]

E.g. my pointset topology exam was not great

[thorn jay]

why tf even have notes then

[sacred wharf]

books i've read had more examples than lecture notes

[thorn jay]

the Zariski incident™

[delicate orchid]

Same at Notts RAAAAAAAHHHH Nottingham mentioned RAAAAAHHHH 🦁 🦁 🦁 🐺 🐺

[south patrol]

Big up Notts

I like it

[delicate orchid]

Deleting your post…. A craven manoeuvre

[thorn jay]

mods ban mq

reason: cowardice

[sacred wharf]

soundcloud ban enpeace 'music'

[delicate orchid]

Nottingham is the only place south of Durham I will defend as not being a shithole

[sacred wharf]

scandinavia is the only place north of germany that i will defend as not being a shithole in europe

[quiet pelican]

Places south of Durham don’t exist

[delicate orchid]

Anyway 40 weeks of lectures is crazy. Are your undergrads like 1 year

[delicate orchid]

This is fair

[sacred wharf]

it's not 40 more like 38

[delicate orchid]

“It’s not 69 it’s more like 67” thats still basically double what it is here

[sacred wharf]

but no it's just that our exam studying periods include lectures wheras in other unis they typically just study a similar amount since it's before exams

[delicate orchid]

Ah I see

So you still get summers off?

[sacred wharf]

fs

[delicate orchid]

are we not counting august as summer

Based

[sacred wharf]

i am counting august

and i miscalculated

it's not 38 weeks

but 37

[delicate orchid]

AI use mods mods

[fading acorn]

real

[thorn jay]

you didnt calculate shit

[sacred wharf]

nah I did, skill issue

[crystal vale]

Nice one, thanks

Is there any ring where no factorization exists, means there is an non unit element which cannot be written as product of irreducible elements

And why we do factorization as a product of irreducible elements, why not as a product of prime elements?

[elfin wraith]

Manchester’s nice

[elfin wraith]

Literally why use AI for this lmao

[thorn jay]

wahhh i cant add numbers wahhhh i cant look stuff up wahhh

[tardy hedge]

Why was it poppin at Groups Rings Fields wee early in the mornin

[elfin wraith]

I’m pretty sure Google has like a built in function for exactly that anyway

[rocky cloak]

For example
k[[x]][x^-2^i : i in N]

[sacred wharf]

wahh i get emotionally triggered when i see someone use ai

[elfin wraith]

Why use the often wrong, more expensive less efficient version of something that there are already better ways to do though? I’m not triggered I just don’t understand

[thorn jay]

im not triggered lol

[rocky cloak]

Primes are (in an integral domain anyway) irreducible, so a product of primes is in particular a product of irreducibles.

Typically we do care about things being products of primes, but it's easier to ensure something is a product of irreducibles

[noble nexus]

Yeah the ring of continuous functions in that question is an example

say on the closed interval [0,1]

then the function f(x)=x you can keep splitting it again and again forever into products of smaller power roots

I believe you cannot write it in any way as a product of irreducibles

in general rings that are very "large" you may fail to have factorization

I think you need (or at least it's enough) to have the ascending chain condition for principal ideals

[noble nexus]

problem with this is that (and I still don't have an entirely satisfying answer) not every normal subgroup is a characteristic subgroup (which are fixed under all automorphisms rather than just inner ones)

I suppose you could say that for a quotient you only need to to respect internal symmetries but

not sure if that's entirely satisfying

[noble nexus]

This is why congruences are nice, because they are the same for groups, rings, algebras, and basically every structure in math

[rocky cloak]

Your can have some weirdness like if x^2 = x, then x is not irreducible even though it's principal ideal might be

[crystal vale]

But what if we are in non integral domain?

[crystal vale]

But what if we are in non integral domain?

[crystal vale]

Oh man

How can I think about such a counterexample?

[noble nexus]

notice that both of our counterexamples are actually quite similar (involving an element that can be forever broken up into increasing powers in some sense)

[rocky cloak]

I mean, it's just you have an irreducible element x. Then you make it not irreducible by adjoining sqrt(x). Now sqrt(x) is irreducible, so you adjoin sqrt(sqrt(x)), etc

[crystal vale]

I see

[noble nexus]

there are a lot of standard counterexample rings that are good to know about

[mighty spade]

$Z[-\sqrt5]$ is the certified hood classic, but may i suggest the coordinate ring of the circle:
$k[x,y]/(x^2+y^2-1)$?

[cloud walrus]

sergeEmbedding

[noble nexus]

p sure z root -5 still has factorization it's just not unique

Still a good example to have though

[mighty spade]

oh shit, i misread him

thought he was looking for non-ufds

[noble nexus]

actually if you relax this condition to say, measurable functions (or all functions) on [0,1] then you can easily prove that there is no nonzero elements that can be factored

p sure that's true for cts functions as well but maybe a bit harder to show

[crystal vale]

Yes

But in Z[√-5] every element has factorization, right?

[noble nexus]

if f is a function, you can write it as $e^{i\theta}|f|$ where theta is some phase function. Then write f = $(e^{i\theta}|f|^\frac{1}{2})|f|^\frac{1}{2}$ and if f is not a unit, then neither of these two factors will be a unit either

[mighty spade]

try a field

[noble nexus]

rings are ufds

[crystal vale]

If I want to show x+3 is irreducible in Z/8Z [X], how do I show?

[noble nexus]

(vacuously)

[cloud walrus]

Blake

[rapid cave]

Its a polynomial of degree 1

[crystal vale]

In the field, we can get vacuously

[rapid cave]

And the leading coefficient is invertible

[crystal vale]

Yes but the ring is not an integral domain

[mighty spade]

fields dont work

[elfin wraith]

We should start teaching UA before intro abs

[noble nexus]

true and based

[rapid cave]

The deg 0 element in any factorization is invertible because the leading coefficient of x+3 is invertible

[noble nexus]

this also proves every matrix ring over C has this property I believe

using the polar decomposition and taking matrix roots

(which ultimately just means there are no irreducible elements in either of these rings)

those I suppose these are not integral domains so I guess it's not surprising

[crystal vale]

Let Z/6Z[X], then 5x + 1 = (2x+1)(3x+1)

And i think 2x +1 and 3x+1 are not unit

[noble nexus]

Tbh I think the definition of irreducible is just not really used much for non domains because its poorly behaved like this

[crystal vale]

Yeah i think so

[noble nexus]

However as for this question I'm guessing it has something to do with the specific numbers 1 and 3 and their relation to 8 so maybe you can rule out certain factorizations

[crystal vale]

So your example of ring of continuous function, it is non integral, right?

[noble nexus]

Yeah

because you can have a nonzero function that is zero on [0,0.5] and another that's zero on [0.5,1]

[crystal vale]

How?

[noble nexus]

I put it below

it's just easier to prove for all functions because you can do a polar decomposition and then use square roots

for arbitrary elements

[spark veldt]

Can anyone like explicitly explain what are the <g>-orbits here? I know what orbits are, but what exactly are these <g>-orbits? This Isaacs book give 0 explanation

[noble nexus]

orbit of the subgroup generated by g?

[spark veldt]

what does it mean precisely? Like why is <g>-orbits suddenly popping up in a discussion about permutation groups

[tribal moss]

<g> is a particular permutation group.

[noble nexus]

because it focuses on a specific element

which is often useful

[tribal moss]

In general, "why is this lemma here?" is best answered by reading on and finding out which theorem the lemma helps prove.

[crystal vale]

I have to show iff ACC holds on all principal ideals of A, integral domain and any two elements of A has a GCD, then A is ufd.

ACC implies existence of factorization, how do I show the uniqueness of factorization by GCD?

[wraith cargo]

assume you have two different factorizations and calculate their gcd

[spark veldt]

Where is the first line from?

This is the lemma.

God I hate this book it skips so many explanations. Uses words like "trivial" or gives zero elaboration on many of its steps. Honestly looks like its goal is to make each chapter as short and as frustrating to learners as possible.

[noble nexus]

well it says lemma 4.2 so

oh wait I can't read

[spark veldt]

Well yes I mean where does it get the theta not in Alt Omega from

[noble nexus]

by assumption g induces an odd permutation

if a book doesn't elaborate on something and you don't understand, it can be frustrating but it's a sign that there's a "standard" argument you haven't yet internalized, and working through the details is very important

[spark veldt]

yeah that's probably right i still have a long way to go

Can you remind me again what does it mean for g to 'induce' an odd permutation? Like the map of g?

I can't really picture it

[noble nexus]

so a group action on a set basically means for each element of the group you get a permutation of the set

given by the action of g

the axioms of a group action mean that this defines a homomorphism from G into Sym(Omega) as mentioned in the lemma

but the important thing is that an action is just a way of viewing any element of your group as a permutation on the set

[spark veldt]

ohh i see

thanks blake

[crystal vale]

Sorry, i don't see any progress here

Can I show every irreducible element is prime?

Say, q is irreducible element and q divides ab

[wraith cargo]

this isn't enough to show it's a UFD tho

[crystal vale]

I think it is

[rocky cloak]

Every element product of irreducibles and every irreducible prime is enough for ufd

[crystal vale]

Because ACC gives us that existence of factorization

[wraith cargo]

ah okay I forgor about the first part

sorry I thought you meant like totally forgetting the factorization thing lol

[crystal vale]

No problem

Do I need to consider GCD(a,b)?

[wraith cargo]

yeah this is the way to show that every irreducible element is prime
Basically you wanna assume that some irreducible q|ab, and then calculate what GCD(q,a) and GCD(q,b) can be

this basically solves it ig lol

[crystal vale]

but it doesn't gives gcd(q,a) = q or gcd(q,b) = q, on contradiction we can assume they are 1.

[wraith cargo]

well I mean you wanna show that gcd(q,a) = q or gcd(q,b) = q that proves your claim

oh wait I see your point

oh no I mean it doesn't change anything

still works

[rocky cloak]

So I think a good approach is say
q divides ab. Then consider the gcd of ab and qa. both q and a divide both, so there must be an element that is a multiple of both.
If qa divides ab, then q divides b and you can do induction on the number of irreducibles in a decomposition of b

[crystal vale]

So here are two things gcd(ab, qa) = d is a multiple of qa.

If qa divides ab we are done.

If qa doesn't divide ab

Isn't d divides qa and qa divides d?

Oh my bad

d is not a multiple of qa

q and a divides d.

I don't understand the case, qa doesn't divide ab

[rocky cloak]

I'm saying d is a multiple of both a and q, and it divides ab.

If d = qa, then q divides b.

If d = ab, then b divides q, so they are associate.

If d is a proper divisor of ab, then you can write d as ab' where b' is a proper divisor of b, and continue by induction

I guess you can probably assume b irreducible from the onset to avoid induction

[sacred wharf]

How do I show that if an abelian group is divisble by p then it contains an element of p

ik i can use FTAG

but i'm not sure howt o

I don't fully understand ftag

[rocky cloak]

Well, from FTAG you just need to note what the order of a product of cyclic groups is, to reduce it to that case.

You can also prove this by induction if you don't won't to involve as complicated machinery as FTAG

[sacred wharf]

So FTAG states that every abelian group is isomorphic to cyclic groups with order of a power of a prime

I'm not sure if i'm understanding it correctly

[rocky cloak]

It is the product of those yeah.

And do you know what the order of a product of groups is?

[sacred wharf]

The order of each group multiplied

i don't see how to use it

So hold on

[rocky cloak]

So then if the order is divisible by p, what does that say about the order of the cyclic groups appearing in the product

[sacred wharf]

wait maybe i do

So I know that there must be a subgroup of order of power of p right

[rocky cloak]

Yup

[sacred wharf]

So now I just have to show that if a group contains order p^n it contains element of order p

I reckon it's a contradiction proof then

[rocky cloak]

Well even more specifically you have to show that the cyclic group Z/p^k has an element of order p

[sacred wharf]

oh

yea

then that is simpler i guess

[sacred wharf]

So of course it has an element of order p^k, say A^{p^n} = 1

Would it be correct to say

(A^{p^{n-1}}) has order p ?

Since

Yeah exponenet laws

[sacred wharf]

I'm not sure if this correct

[rocky cloak]

What part?

[sacred wharf]

A^{p^{n-1}} is our element of order p

[crystal vale]

if b is irreducible then we can say, since d is a multiple of a then d = at, and d divides ab implies ab = ats implies b = ts implies d associate with a or d associate with ab.

[sacred wharf]

$A^{p^{n-1}}$

[cloud walrus]

mq

[rocky cloak]

Indeed, if d associate with a, then q divides a.

If d associate with ab, that means ab divides qa

[rocky cloak]

Well, try to prove what the order of A^p^n-1 is. Start by unwrapping the definition

[sacred wharf]

i'm 98% sure it's correct since if it's anything less than that would contradict A's order

If it's more well it doesn't matter because exponenet laws already give 1 before

[crystal vale]

Yes

[crystal vale]

b' is a proper divisor of b, is it necessary that b' is the some reduced form of factorization of b?

[rocky cloak]

I'm not entirely sure what you mean, but probably not.

It's just that if you have a chain of proper divisors
b -> b' -> b'' -> ...
it must terminate

[crystal vale]

I see

I have one question for you guys, how do you maintain yourself to touch within all mostly areas of mathematics?

[tribal moss]

I don't think that is humanly possible for the last 100 years or so.

[crystal vale]

But as you, you know undergraduate and postgraduate concepts very well

[noble belfry]

how can i show that there's a subgroup of order p^n-1 inside a group of order p^n?

[thorn jay]

thats far from all areas of mathematics

[thorn jay]

atom vs universe type shit

[crystal vale]

My bad

[crystal vale]

You can show in a group of order p^n there is subgroup of order p^i, for all i≤n.

Use induction and use the fact centre of p group is non-trivial

[crystal vale]

Okay ignore that, but if you get what I mean you can answer

[thorn jay]

do lots of math

also, once youve learn something usually you can reread the definitions to refresh your brain

[sacred wharf]

reminds me of the cool properties of p^n groups
Center is non trivial
Every normal group has non trivial intersection with the center
Every subgroup is properly contained in its normalizer,
what did I miss ?

[thorn jay]

almost every group is a 2-group

[knotty badger]

I remember looking at the table of number of groups of a given size

It jumps up so much for powers of 2

[delicate orchid]

my dumbass thought you meant a 2-group instead of a 2-group there - that would've made be look silly!

[noble belfry]

there must be an element of order p by cauchy's theorem. So let H = <p> and let H act on Z(G). We have that H is contained in Z(G), so H is normal. So consider G/H which has order p^{n-1} and therefore has a normal subgroup Q of order p by the same reasoning. I want to then send Q back to G... idk if this is possible

[noble nexus]

99% of the first 50 billion groups have order 1024 I think I saw that stat somewhere

[thorn jay]

right? and statistically it makes sense but its still crazy

[thorn jay]

2-group instead of 2-group uhuh right

[delicate orchid]

I genuinely thought you meant "monoidal groupoid with all objects invertible" for a solid 2 minutes and I tried to think of a group that couldn't be viewed like that

[thorn jay]

my god

you are so far gone

[chilly ocean]

It’s weird that a group object in Grpd and a group of order 2 have the same name

[elfin wraith]

Brother lay down the nlab

[delicate orchid]

the answer is any non-abelian group if you do the standard delooping thing btw. It has to be abelian by eckman-hilton

[quiet pelican]

It makes sense for me because like
There’s a lot less restrictions (from Sylow, Cauchy, etc) on what your group can look like if you only have one prime
And powers of 2 are exponentially smaller than the same powers of higher primes

[thorn jay]

yes that's what I meant by "statistically"

[quiet pelican]

And finite group theory is basically Combi /hj

[elfin wraith]

One day I will learn how to use eckman Hilton. May be the only result I’ve ever proven without understanding literally anything about it

[thorn jay]

its basically when the thingy and thing and then abelian

[elfin wraith]

I mean this is the extend of my understanding

[quiet pelican]

The funniest thing I remember about Eckmann Hilton is when one of my friends called it would be on our algtop exam despite it not being taught in the course (although it was of the form “topological group has abelian fundamental group”)

[elfin wraith]

The proof I did was draw this diagram, play with it and prove these things commute

[elfin wraith]

Yeah I know it’s got something to do with homotopy groups but I’ve no idea what

[thorn jay]

it's for proving higher homotopy groups (n > 1) are abelian

[elfin wraith]

But hey, I’ve proven it’s true in any monoidal category with sufficiently many adjectives

[thorn jay]

cuz you've got two+ compatible operations and that satisfies the hypothesis

[elfin wraith]

Ah I see

[quiet pelican]

I like the cute visual moving squares arguement

[thorn jay]

yeah

[noble belfry]

how can you construct a homomorphism from Q to G?

[thorn jay]

yum

this is the actual homotopy

[thorn jay]

[delicate orchid]

we are in an algebra channel. MY algebra channel. You will write down SYMBOLS in your proof and not COLOUR IN your PRETTY PICTURES as a SUBSTITUTE

[noble belfry]

maybe this is obvious but i can't think of any nice maps from Q to G that are well defined

[thorn jay]

Q is normal so you can go to G/Q, use the induction hypothesis, and then take the preimage via the surjection mapping p : G -> G/Q

[noble belfry]

Q is a subgroup of G/H

[quiet pelican]

Suppose we have a path connected covering p: G -> H where H is a topological group
Prove that there is a unique group structure on G (if we fix a preimage of e_H) where p is a group homomorphism, with this structure, ker p is abelian

[noble belfry]

so G/(Q) isnt defiend is it

[thorn jay]

ah well then use the induction hypothesis on G/H

this gives a subgroup of order p^n-2

then you take the preimage of that subgroup via the projection p : G -> G/H to get a subgroup of order p^n-1, as desired

[thorn jay]

totalitarian dictator

[noble belfry]

how do you guarantee the preimage is a subgroup though 💀

[thorn jay]

that's always true..?

[noble belfry]

is it

[thorn jay]

like not even just for groups but for general algebraic structures

[noble belfry]

oh wait

[elfin wraith]

How do you feel about this as a proof

Or, well, a statement of the theorem

[thorn jay]

are these string diagrams?

or wtv

[elfin wraith]

Yuh

[delicate orchid]

I've forgotten how to read these icl

[thorn jay]

can these strings be braided

and like

braid groups and stuff

[elfin wraith]

Yes, you see the braiding in the second diagram

[noble belfry]

Ok yeah so the preimage is a subgroup (since (q1 * q2) -> q1q2H implies q1q2 are in the preimage, and same for inverses).

[elfin wraith]

Though, the graphical calculus is only known to be sound and complete for braided categories in like up to 3 dimensions IIRC

[thorn jay]

I love it when stuff can get braided

quandle-pilled

[delicate orchid]

I remember reading some insaneo paper where they attempted string diagram esque constructions for 4-categories

[elfin wraith]

I love drawing things which cannot be pictured

[true bolt]

This looks like my first intro to abstract algebra

[thorn jay]

operad-pilled first year UG course

[elfin wraith]

If you started out with "Categories for Quantum Theory" then fair play lol

[manic cairn]

eckman-hilton? i heard that in one of the g++ videos

[thorn jay]

but gee, you might say

[balmy python]

How do I work out Inn(Sn)

[tidal schooner]

Hint: For any group G, we have an isomorphism G/Z(G) = Inn(G) (why?). So, it suffices to find the center of Sn

[noble belfry]

Let G be a finite group and let N be a normal subgroup, with |N|=3 and N is not contained in the center of G. Show that G contains a subgroup with |G:H|=2.

no idea how to start on this

i'm thinking about N acting on G by conjugation

if u know how to do don't tell me yet just give hint if u can 😵

[vapid vale]

you should probably do it the other way around

G acting on N

then that should give you some information about G

[noble belfry]

oh

we know that the orbit size has to be 2

so we have that

|G:Gx| = 2

done

[twilit wraith]

exactly

[noble belfry]

wait

oh yeah that works

[noble belfry]

every group G of order 99 is abelian.
n_3 = 1 mod 3
n_3 | 11
so n_3 = 1
n_11 = 1 mod 11
n_11 | 9
n_11 = 1

so P_3 x P_11 = G
= C_11 x P_3
P_3 is abelian since it has order p^2.
C_11 is abelian since ithas order p.
so the product is abelian

[twilit wraith]

yep

[sacred wharf]

sylow computations are so fun fr

[noble belfry]

Let |G| = 168=7*2^3*3 and G is simple.
a) Then n_7 = 1 mod 7
and n_7 | 24
so n_7 = 8
So there are 8(6) elements of order 7.

b) G has a subgroup of order 21 since:
|G: N_G(P_7)| = 8 -- so N_G = 21.

c) Suppose G has a subgroup of order 14, H.
n_7 = 1 mod 7
n_7 | 2
so n_7 = 1 in H
We know that H is in the normalizer of P_7 so 14 | N_G(P_7) which is a contradiction

Let |G| = 7 x 43 x 17 and suppose the center contains an element of order 7. Show G is abelian.
n_7 = 1 mod 7
7 | 43 * 17
7 could equal 1 or 43
n_43 = 1 mod 43
n_43 | 7*17 => n_43 = 1
n_17 = 1 mod 17
n_17 = 1.
The center contains an element s of order 7. so <s> is a p-sylow subgroup and s is in the center which implies <s> is normal. So G is isomorphic to P_7 x P_17 x P_43 which we know is cyclic since they have relatively prime orders.

[twilit wraith]

an easier conclusion is that G is abelian because its the product of abelian groups

[noble belfry]

external direct product u mean right

like we still have to show normality

but yeah i agree thats nicer

[twilit wraith]

i mean technically this ones internal

but i havent seen someone care about the difference between internal and external direct product in a while

[noble belfry]

yeah i just mean direct product 😵

[twilit wraith]

well we dont have to show normality

[noble belfry]

why

[twilit wraith]

unique sylow subgroups are normal

[noble belfry]

yeah thats how we showed normality

but if they weren't normal then it wouldnt work

[twilit wraith]

im confused

[noble belfry]

like

[twilit wraith]

whats the problem

[noble belfry]

no problem xd

[twilit wraith]

oh

i mean yes you are right

if they werent normal we couldnt do a direct product

[noble belfry]

Let G be a group of order 5x11x17.
n_5 = 1 mod 5
n_5 | 11*17 so n_5 = 1 or 11

n_11 = 1 mod 11
n_11 | 5 * 17
(5*17)mod 11 = 5 * 6 mod 11 = (30 mod 11) != 1
so n_11 is 1

n_17 = 1 mod 17
n_17 | 5 * 11
n_17 is 1

G must contain an element of order 187 = 11*17 since there is a subgroup of order 187 (the normalizer of P_5) and we have that any group of order 187 is the direct product of C_11 x C_17 so its cyclic so theres an element of order 187.

I want to show that if G has an element of order 55 G is cyclic

not rlly sure why

somehow i want to show that n_5 is normal

if s has order 55 <s> = C_5 semidirect C_11 (C_11 is the normal one)

[twilit wraith]

pretty sure some slight number theory can be done here but idk

actually wait no nvm

[twilit wraith]

normal one goes first

at least how i learned it

[noble belfry]

it depends which way u point the triangle 💀

[twilit wraith]

fair enough

i thought $\rtimes$ was the standard

[cloud walrus]

hiidostuff

[noble belfry]

in class i see both a lot

wait

oh this is obvious

wait no this isn't obvious 🙁

im wrong

I was thinking we could take the element of order 55 and element of order 187

and say G = C_55 x C_187

but we don't know that <s> where |s| = 55 is a normal subgroup

wait

we can semidirect product shit tbh

[twilit wraith]

then G would have order 5 11^2 17

[noble belfry]

wait...

ur right 💀

[twilit wraith]

C17 semi C55 could work

[noble belfry]

we can do G = C_55 x C_17.
We know 17 is normal

yeah

and then since 17 doesnt divide 54

[twilit wraith]

automorphisms from C55 into C17

theres only one

[noble belfry]

yeah

we get that G = C_(55*17)

damn i have to relearn all of the semi direct product stuff

[noble belfry]

Why is every cyclic subgroup of Q_8 (the quaternion group) x C_n where n is odd normal?

oh

every subgroup of Q_8 is normal since they all have index 2

and then the rest follows

[noble nexus]

fun fact Q8 is one of the only nonabelian groups with all normal subgroups

https://en.wikipedia.org/wiki/Dedekind_group

[noble belfry]

That’s pretty cool

[noble nexus]

and in a sense its the only one since all others are Q8 times some abelian stuff

(not anything abelian, they have a specific form)

[tidal schooner]

One of them has index 4, but that one's the center, so it's also normal

[noble belfry]

Yeah <-1>

So far I haven’t been convinced the typical groups that we work with in group theory are very interesting

I like doing stuff in cyclic groups and arbitrary finite groups and Sn and S_\infty

but i dont like the dihedral group very much

[noble nexus]

the infinite dihedral group is p cool

matrix groups over finite fields are also quite cool and imo underrepresented in courses on group theory

although I believe that saying a lot about them is more advanced which maybe explains why

[noble belfry]

i dread doing stuff in matrix groups

[vocal pebble]

Matrix groups over finite fields are also cool, they give you a proof of the sylow theorems for example

[noble nexus]

yeah that's my favorite proof of the sylow theorems

[noble belfry]

whats the proof?

[noble nexus]

so basically because every group can be embedded into some general linear group over Fp for any prime p

you can do some reductions and its enough to prove the sylow theorems for the prime p in GL(Fp)

this is probably the least intuitive step tbh and requires a bit of work

but once you have GL(Fp) the proof is really nice (if you are familiar with some linear algebra)

suppose H is a p-subgroup of GLn(Fp)

then H acts on the nonzero elements of Fp^n (since invertible matrices don't map anything other than zero to zero)

but since there are p^n-1 nonzero elements and the order of H is a multiple of p, there must be a fixed point by orbit stabilizer

i.e. a vector v fixed by the entire action of H

then you quotient out by v and basically repeat this argument

and what you get is a basis v1, v2, v3, ... , vn for Fp^n

such that v1 is fixed by H, v2 is fixed by H up to a multiple of v1, v3 is fixed by H up to a linear combinatino of v1 and v2, etc...

and if you think about it, that just means the with respect to this basis, the group H is upper triangular

so in particular choosing some change of basis matrix P

you find that PHP^{-1} is a subgroup of the upper triangular matrices

if you just compute orders, its easy to see that the upper triangular matrices form a sylow p-subgroup

[noble belfry]

thats pretty crazy actually

[noble nexus]

and so you automatically get that every p-subgroup is a subgroup of a sylow p-subgroup and that every sylow p-subgroup is conjugate

I believe you should be able to get the rest of sylow but you might have to just use the same argument as in the usual proof

[noble belfry]

math is wild

wait can you explain to me the importance of O(n) and SO(n)

[noble nexus]

SO(n) is the group of rotations in n dimensional space

and O(n) is rotations + reflections

[noble belfry]

Yeah

[noble nexus]

which are nice because they preserve length

and SO(n) also preserves orientation

so they show up a lot in physics just because rotations are useful

they're also nice because they are compact groups, which makes them a lot easier to deal with mathematically (this is in the theory of topological groups)

[noble belfry]

What do the finite subgroups of SO(n) represent

[noble nexus]

finite rotation groups of the sphere, of which there aren't actually that many. I believe they correspond to nice shapes in 3D

platonic solids have a group of their rotations and there are some more

also fun fact: the finite subgroups of O(n) are exactly the finite groups that admit a planar cayley graph

which is a really awesome theorem proved by Maschke

I did a preserntation on that for a 4th year group theory class

although in hindsight I put wayyyy more effort into it than I'm pretty sure the prof was ever intending

these are all of them

S4 is the group of rotations of a cube or octohedron and S4 x C2 is the group of reflections + rotations

A5 and A5 x C2 are the groups of rotations and reflections + rotations respectively of the dodecahedron and icosahedron

or sorry

[noble belfry]

understanding stuff like this is probably where i'm the weakest out of the group theory content i've covered so far

i always want to actually imagine rotating the cube 💀

but im bad at doing that in my head

i need a physical cube in real life or something lmfao

[noble nexus]

and then A4 and A4 x C2 come from the tetrahedron

ok that's correct now

[noble nexus]

its alright its kinda hard for everyone I think to orient shapes in their head

[vapid vale]

you can see A4 as the symmetry of the tetrahedron by imagining the symmetries of the vertices - you can't rotate a tetrahedron so that you just swap two vertices

[twilit wraith]

i just envisioned a 3 dimensional cow with a helmet on doing a sick ass kickflip in the middle of the street i live on in my head

[noble belfry]

https://tenor.com/bDxZJ.gif

[noble belfry]

i can imagine cows rotating but i cant keep track of the vertices lol

[vapid vale]

S4 you can maybe imagine via the space diagonals, but thats less clear to see imo

i dont know of a nice way to see symmetries of the icosahedron as "even permutations of 5 elements"

[noble nexus]

I think there's probably some way

even-ness corresponds to the fact that orientation is preserved

[twilit wraith]

isnt this only clear after knowing that the icosahedron can be given as an embedding of 5 congruent cubes in a certain way

[vapid vale]

i have seen a way but its not nice

like you could color the faces in a certain way with 5 colors

[vapid vale]

this is probably that

[noble nexus]

if you want groups that are really impossible to visualize, you can take each of the group of rotations of the sphere and take their double cover in SU(2)

[twilit wraith]

rip i havent done covering spaces yet

learned about the fundamental group on wednesday tho so were getting there

[noble nexus]

you don't entirely need to know covering spaces, just the fact that SU(2) maps onto SO(3) with kernel {+-1}

[vapid vale]

there is a pretty elementary way to "see" that SU(2) double covers SO(3)

[karmic moat]

i have some homework assignment on the universal covering space of SO(2)

[twilit wraith]

i mean i dont know what SU(2) or double covering means

SU(2) im guessing is the unital subgroup of SO(2)

[vapid vale]

its a subgroup of 2x2 complex matrices

[noble nexus]

its 2 by 2 unitary matrices

[noble belfry]

unitary matrices

[noble nexus]

with det 1

[noble belfry]

^

[karmic moat]

oh wait sorry

[twilit wraith]

oh ok

[karmic moat]

universal covering space of SL(2)

SL_2(R)

not SO(2)

[noble nexus]

oof thats a nasty space if I recall

[karmic moat]

yeah the homework is to show it's not a linear lie group

[noble nexus]

p sure the universal cover of SL2(R) isn't even finite dim as a lie group

yah

[karmic moat]

but no spoilers please

[noble belfry]

damn u guys know so much abt math

[noble nexus]

idk how to prove it dw

[karmic moat]

lfg

[noble nexus]

I just heard about it in a borcherds video

[karmic moat]

the hint our prof gave us is half a page long lol

[noble nexus]

by the way, his group theory lecture series is quite good (although lacks a lot of details)

[waxen tinsel]

is this the symmetries of a circle

[noble nexus]

its the symmetries of the integers

[velvet hull]

almost but not quite

it's the symmetry of a countably infinite n-gon

[noble nexus]

if you think about it, if you take a polygon with infinitely many sides you basically get a big line

[velvet hull]

but on a circle you can have rotations at irrational angles

[vapid vale]

you can essentialy see that SU(2) is isomorphic to a 3-sphere, and SO(3) is a ball with antipodal points identified. this gives you a double cover

[velvet hull]

so the symmetry group of the circle is at least uncountable

[noble nexus]

where you include both shifts (rotations) and reflection (multiplication by -1)

so its actually isomorphic to the group of functions from Z to Z of the form x -> ax+b where b is an integer and a is plus or minus 1

[twilit wraith]

its a subgroup iirc

[noble nexus]

it is not, since the symmetries of the circle are abelian

[twilit wraith]

fr?

[noble nexus]

even if you include reflections I believe it isn't

[twilit wraith]

no yeah ur right

[vapid vale]

wat

you still have the same reflect then rotate =/= rotate then reflect issue

[noble nexus]

yeah I meant just the rotations

but I don't think you get a copy of the infinite dihedral group even with reflections

[twilit wraith]

wait so isnt the shifts of Z also abelian

[noble nexus]

yeah

but you also have a reflection

[twilit wraith]

but like

[vapid vale]

oh

[noble nexus]

Z is a subgroup of the circle

[twilit wraith]

your point about abelianity doesnt make sense anymore at least to me

[noble nexus]

the infinite dihedral group is nonabelian, so cannot be a subgroup of rotations of the circle

[twilit wraith]

but it can be of the symmetries of a circle right

or if it isnt then its for a different reason

[noble nexus]

I don't believe so

to have a copy of the integers you would need an irrational rotation

lets call this irrational rotation r

then lets let s be a reflection

to obtain a copy of the infinite dihedral group we would need srs = r^{-1}

actually maybe you can do this

[twilit wraith]

and we do have this correct

[vapid vale]

i think you do

this just works

[twilit wraith]

i remember accidentally creating this subgroup on accident on my first ever algebra hw iirc

ahh look at me screw up

where has the time gone

(this was 5 months ago)

[vapid vale]

noooo

[twilit wraith]

hey it was a good screw up at least

[noble nexus]

yeah actually it does give a copy of the infinite dihedral group that's nice

although you are going to have uncountably many subgroups isomorphic to it so

perhaps less nice

[twilit wraith]

tbf this is expected

[noble nexus]

yeah

[twilit wraith]

not expected by me 5 months ago but it is expected now

[noble nexus]

anyway probably my favorite finite groups are the symmetry groups of mobiles

[karmic moat]

mobiles like baby mobiles?

[noble nexus]

yeah

[vapid vale]

my math ice breaker is usually favorite group and favorite soup

[karmic moat]

hmmm

i can't think of a favorite group

wow

[noble nexus]

heres a mobile and its symmetry group

[karmic moat]

PGL_n i guess

[noble nexus]

its actually a really good conceptual exercise to convince yourself that this is actually the correct group

and also to figure out what the action is

[karmic moat]

that's a complex mobile lol

[noble nexus]

we gotta get the babies started on semidirect products early

[karmic moat]

ive never seen a mobile with mobiles

genius

[vapid vale]

is this also C3 wreath C3

[noble nexus]

yeah its an iterated wreath product

[vapid vale]

oh the bottom things are also triangles

[noble nexus]

yeah its (C3 wreath C3) wreath C3

[vapid vale]

kewl

isnt that the semidirect symbol

[noble nexus]

and of course you can continue

it is yeah

[twilit wraith]

i know my least favorite group

[noble nexus]

but wreath products are just semidirect products

[thorn jay]

i love K4 the klein 4 group
delightfully many coincidences because its so small

[karmic moat]

what is it

[vapid vale]

soup

[karmic moat]

my least favorite group is SL_3

[twilit wraith]

frobenius group of order 20

[noble nexus]

at least for mobiles you couuld continue infinitely (in theory) but idk how to actually write down the resulting group

[twilit wraith]

reason being i had an exam on it last week and am 90% sure i got a C on that jawn

[noble nexus]

the group would be ((((C3 W C3) W C3) W C3) W C3) ...

[vapid vale]

exam on one group

[noble nexus]

where W is wreath

[twilit wraith]

i should say

exam where that group was used to test knowledge on different topics

[vapid vale]

this looks like an owl face

[karmic moat]

i hate SL_3 because i've been learning algebraic groups and all the examples i've done are GL_3 and SL_3

and i'm sick of seeing those damn 3x3 matrices

[twilit wraith]

like semidirects, groups acting on themselves by conjugation, and nilpotency/solvability

[noble nexus]

the frobenius group is quite nice

[karmic moat]

in my sleep im probably mumbling some bullshit like "let G be SL3 and B its borel of upper triangular matrices and T the maximal torus and P a parabolic"

[mental lake]

how do u show that {f : f(0)=0} is a maximum ideal for R^R

[quiet pelican]

Show that the quotient by it is a field

[mental lake]

and why can u do that?

[quiet pelican]

Because the quotient by an ideal is a field if and only if the ideal is maximal (if you haven’t seen this, try and prove it from the correspondence theorem)

[mental lake]

ok ty

[thorn jay]

stalks o.O

[tribal moss]

But if you want to prove it from first principles, it's straightforward to show that an ideal that contains {f:f(0)=0} and at least one g with g(0)!=0 must contain everything.

[lean sail]

Hey all, I plan on reading about rings, fields, and modules in D&F. I already did most of the chapters on groups, probably around 60 exercises so far in total. For rings, fields, modules, in D&F, that would include chapters 7, 8, 9, 10. I want to do some exercises from each chapter to test my knowledge. A minimum of 1 exercise per section. I’m having trouble figuring out which problems to do. Anyone here familiar with D&F have any recommendations or old homeworks from those chapters that I can go over?

[wraith cargo]

I have some psets I can send you when I was taking algebra

we were following D&F

I'll send the links here

[lean sail]

That would be great!

[wraith cargo]

https://mathweb.ucsd.edu/~asalehig/math200a-22-f.html
https://mathweb.ucsd.edu/~asalehig/math200b-23-w.html
https://mathweb.ucsd.edu/~asalehig/math200c-23-s.html

[lean sail]

Awesome, thank you

[white oxide]

How do we know that this equation is minimal

[rocky cloak]

If s had a smaller minimal polynomial, then you would also get a smaller minimal polynomial for y

(because y = sx)

[merry summit]

are you guys familiar with "mind maps" they're like VERY short notes connecting the dots in a subject. does someone have one for introductory group theory? because i get soo confused with the theorems and results.

[noble nexus]

the whole point of mind maps is to do them yourself

[tribal moss]

Yeah, it's not having the resulting diagram that helps any; it's the mental effort of deciding how it should go.

[sacred wharf]

I would argue both are useful

But i agree trying to find the connections yourself and building the mind map is the key benefit of mind maps

[elfin wraith]

Yeah having it definitely still helps, but it’s kinda just like less effective flash cards

Which I’ve decided to start experimenting with this year, I’ve never done it before but I’m enjoying it so far. It makes reviewing what ive been up to recently much easier. I just start every day by quickly flipping through them

We’ll see if it translates to tangible exam benefits, but I’m hoping yes. One of my biggest issues in maths is recall, I lose things quite quickly if I’m not actively using them

[sacred wharf]

nahh i'd say they have quite different purposes, flashcards are primarily for recall, wheras ind maps are trying to see how different ideas connect to each other, of course you do both in either, but i'd say the key purpose of each one is different

[elfin wraith]

Yeah that’s fair

[sacred wharf]

I started anki for ring theory, it is amazing

[knotty badger]

I used Anki throughout my undergrad

[tardy hedge]

Anki is good

[elfin wraith]

Anki is like £20 on Apple, fuck that

I’ve just got some classic little flip books, I’m liking it so far though

[sacred wharf]

I use it free on computer instead

Cant believe it costs money on ios

[elfin wraith]

Yeah and it’s a kinda silly amount as well

[thorn jay]

quizlet on sheaf theory concepts 🔥

[thorn jay]

this is so funny because I always associate flash cards with highschool stuff

[tardy hedge]

igneous sedimentary and metamorphic rocks type Shi

the water cycle type shi

[elfin wraith]

I’ve never used them for anything before, but I’ve decided to give them a go haha

It usually takes me a while to warm up to doing actual work in the morning so this is a nice way to drill definitions and proof outlines while I had my coffee but yeah like spaces with polynomial cohomology is a crazy flash card

[thorn jay]

yurrr

geology was genuinely one of the more fun parts of geograph

y

[tardy hedge]

Aw Hell Nah

[thorn jay]

lmao yeah

[tardy hedge]

there was a guy here who was named OH HELL NAH!!

[karmic moat]

i do wordle to warm up my brain

first thing in the morning is wordle

[elfin wraith]

I do the linked in games when I first wake up

But it still takes me a while to get to maths readiness

[tardy hedge]

i am most motivated in the morning by far

[karmic moat]

first person ive ever talked to that actually plays linkedin games

[elfin wraith]

They’re so good

Well half of them are

[karmic moat]

yeah but it's linkedin

[tardy hedge]

ya f dat shi

[elfin wraith]

I’ve got beef with the word games and the sudoku is whatever

[tardy hedge]

we hate jobs we hate the real world

[thorn jay]

when I wake up I drag myself out of bed, die, resurrect, eat cornflakes (straight out of the box), then die and rebirth

that's my morning routine

[tardy hedge]

good routine

[thorn jay]

you guys have linked in?

[karmic moat]

i do but i dont use it

[thorn jay]

I should get my head in the game

[tardy hedge]

i finally did make one this year

[elfin wraith]

I genuinely just got it for the games

[karmic moat]

i used to be an engineering physics major so i had a linkedin

then i changed to math before my first semester lol

[vapid vale]

linkedin games

[tough raven]

Wait but I think the desktop app is (was?) free? How do they manage to charge for a mobile app with that?

OMG I'm yellow!!!?

Why?

[thorn jay]

congrats!

or something

piss colour

[sacred wharf]

People want both apps

[knotty badger]

Afaik the mobile app is their only source of income

Also

The web version kind of sucks on iOS

Also congrats on pissing

[tough raven]

Have you considered dihydrogen monoxide supplements?

[thorn jay]

scary words that must be bad element

[elfin wraith]

I mentioned you ask really cool questions and give helpful answers a couple weeks back when they were discussing the updates to honourable, well deserved!

[knotty badger]

-# noted, I ask bad Qs and give unhelpful answers

[thorn jay]

the dichotomy fallacy

or something

THE DICHOTOMY CONJECTURE HAS BEEN PROVEN RAHHH ANOTHER WIN FOR UNIVERSAL ALGEBRA

[knotty badger]

the what

[sacred wharf]

How u do that weird subtext?

[knotty badger]

-# that’s for me to know, and for you to find out~

[thorn jay]

dichotomy is when only two options

so you (jokingly) implied that the only options were good questions and good answers or bad questions and bad answers

[rocky cloak]

Honour has been bestowed

[vapid vale]

holy crap

[knotty badger]

well they are! /s

in all seriousness i know i'm not the honourable type

[rocky cloak]

I'm not even sure who decides the honour role or on what grounds

[vapid vale]

me neither

im the honorable type

🦅

[knotty badger]

gojo be like

[thorn jay]

purely vibes

I'm sure

[thorn jay]

I alone am the honored one-

[knotty badger]

unfortunately you're not pissy enough for that enpeace

[tardy hedge]

Can i be dishonourable

[rocky cloak]

I think that's just called banned

[thorn jay]

I do not have white hair, nor the rizz of gojo mcgojoguy

[elfin wraith]

There is a slight in between with the minus role

Annoying enough to be banned from all the actual channels but not enough to be from the server