groups-rings-fields

[knotty badger]

i'm not defensive, i'm currently unclear on your position

diagrams are part of category theory

[elfin wraith]

My position is that category theory isn’t helpful for people studying linear algebra for the first time. It adds extra complexity and abstraction for no reason. I’ve been pretty clear about this lol

[karmic moat]

One can get far and get a lot out of using diagrams without actually defining categories imo

[knotty badger]

but i don't know what you mean by "category theory"

[knotty badger]

yes i agree with this

[elfin wraith]

And I don’t view just drawing a diagram as doing category theory, that’s just a way to represent maps. That can be category theory, but it isn’t inherently

[knotty badger]

i think my definition of category theory is significantly broader than yours

perhaps you misunderstand what i mean when i talk about "using cat theory at undergrad level"

e.g. i would say that venn diagrams are set theory

so to me this is the equivalent of saying "venn diagrams are not set theory"

[elfin wraith]

I would agree with that

Venn diagrams are a helpful tool to visualise sets, they are not set theory

[knotty badger]

ok then i think our disagreement is just a semantic one

[elfin wraith]

I don’t think it is, because as far as I’m aware you support more than just drawing diagrams in intro to linear algebra, which I agree is a perfectly sensible thing to do.

[knotty badger]

i feel like the idea you've constructed in your mind of what i want is far beyond what i actually want

[elfin wraith]

Or perhaps an informal discussion of universal products

[knotty badger]

(to echo an argument you made earlier, i believe the "majority opinion" would be that venn diagrams fall under set theory, for math educators)

[copper kestrel]

abstract algebra final tmr

[knotty badger]

good luck tabby!!

[formal laurel]

Good luck!

[copper kestrel]

ty yall i'm terrified LOL

[elfin wraith]

You’ve got it

Easy days

[noble nexus]

considering you can draw and understand diagrams without even knowing the definition of a category I think its safe to say that diagrams aren't category theory

[tulip otter]

any hints for part d?

[noble nexus]

conjugate tau by sigma and use the fact that n is prime

[quiet pelican]

Hint: consider the group generated by t and s^k for some suitable k

[tulip otter]

conjugating τ by σ gives [r+1 s+1], but i didnt see how that would really help me when i tried this earlier

[worldly canyon]

Do diagrammatic proofs become more or less common as you study more math in algebra/topology

[noble nexus]

more common

[tulip otter]

alright i will try this

[noble nexus]

its useful to think of some simple example, like say S_7 and let tau be idk, (36)

and just see what happens when you conjugate by sigma repeatedly

[worldly canyon]

Every time I see a diagrammatic proof I have an immediate reaction that “this is not rigorous enough” and I get caught up in technical details

[tulip otter]

alright

[noble nexus]

categorical diagrams are just a shortcut for a lot of equations so they're perfectly rigorous

[knotty badger]

Again this is “venn diagrams aren’t set theory” for me

[noble nexus]

algebraic topology diagrams are a shortcut for "this is obvious to anyone with a brain and writing out the explicit homotopy would be annoying as hell"

but its good to write it out a few times

[noble nexus]

even that I don't think is a very good comparison because people using venn diagrams know what sets are

at least on an intuitive level

[worldly canyon]

Yeah I’m starting to see once I do enough examples I don’t need to write some stuff out

[knotty badger]

Well it’s not like they’d know the formal def

[noble nexus]

saying diagrams are category theory is like saying if you write down the cartesian product of two sets you are doing category theory

[knotty badger]

And you can also have an “intuitive” idea for categories

[tulip otter]

well σ^kτσ^{-k}=[r+k s+k]

[noble nexus]

or if you compose functions its category theory

[knotty badger]

Just in terms of arrow composition

[quiet pelican]

Nitpick: everything there should be taken modulo n

[knotty badger]

I just feel like if you gatekeep what counts as category theory to only the hard stuff, then it’s a little silly to say “category theory can’t be taught early because it’s hard”

[noble nexus]

not really a nitpick its essential to the proof

[tulip otter]

right i was thinking about that, because we are dealing with the set {1,2,.., p}

[quiet pelican]

It’s more a nitpick because I assume you’re implicitly doing it

[noble nexus]

I mean if you count drawing a diagram to keep track of function composition as category theory then sure people should do category theory early in their career

because how wouldn't you

that feels kinda reductive to me though

[knotty badger]

If you use the universal property then sure

[noble nexus]

"learning category theory" to me is "learning category theory"

as in, learning what a category is

and what its structure is

[knotty badger]

Again it’s a semantic thing

It’s not like people learn the ZFC axioms when doing elementary set theory

But its still a useful language for describing what you do in maths

[noble nexus]

yeah but they learn what sets are and what you can do with them

[quiet pelican]

Learning category theory is (the process of) getting to the point you can understand the first 5 chapters of Riehl (or equivalent)

[knotty badger]

At an intuitive level sure

[noble nexus]

whereas you don't learn what categories are and what you can do wiith them until you... learn category theory

[knotty badger]

You can learn categories at an intuitive level too

[noble nexus]

I mean certainly you probably have an idea of what some of the concepts are, but you aren't actually using categorical concepts really at all in most cases

[knotty badger]

Again this

[noble nexus]

the closest I could see is universal properties, but im fairly sure universal properties predated category theory so they aren't really category theory

in that sense

[knotty badger]

I’m not really that interested in gatekeeping what counts and doesn’t count as category theory

I just use concepts that help my students

[noble nexus]

well sure but its not really learning category theory, maybe its learning concepts that appear in category theory but yeah I guess semantics

if someone doesn't even know the definition of a category I don't see how they can have learned category theory

in any reasonable sense of the definition

[knotty badger]

It’s strange to me that you’re so insistent on this

[tulip otter]

alright i see where this is going, choose k so that σ^kτσ^{-k}=[1 s+k] and then the result follows from part b. Tho i havent used the fact that n is prime yet.

[quiet pelican]

I feel saying “you’re not learning category theory until you’re making a genuine effort to learn about the full generality of natural transformations, limits, adjoints, monads etc” is similar to saying “you’re not learning set theory until you’re making a genuine effort to learn ZFC, ordinals, cardinals and proper first order logic”, and both statements are ones I’d probably stand by

Like if you say you’re learning set theory without further qualification, I’m not going to assume you’re learning naive set theory

[knotty badger]

Both statements are ones I’d disagree with

But then I’m not a mathematician so I guess I’m not as concerned about gatekeeping

[noble nexus]

I don't think those two are similar though really

because if you don't know what limits, natural transformations, adjoints are its really hard to actually use them outside of specific examples

its learning category theory that actually tells you how all those things are used

whereas you can do almost everything you need to do in set theory without knowing zfc

now of course this is mostly because we don't really teach "naive category theory", if we did then it would be more or less the same

but honestly I don't think there's really a purpose to naive category theory

well, that's a bit of a lie maybe I do think some of the concepts are good to know

but more towards mid to higher level undergrad

like im not against using the language of "this is a functor" and " these are adjoint functors" before actually doing category theory

without fully knowing the precise definition

[knotty badger]

From my perspective it feels like people have too many prior assumptions about “category theory”

It’s almost like there’s this axiomatic belief that category theory must be hard and abstract and useless

And when evidence to the contrary is presented people label it as not being category theory

It’s confusing to me as an outsider

[noble nexus]

idk I feel like almost nobody except you considers things like drawing a diagram to be category theory

of course in that definition then sure category theory is not hard and abstract because you've just defined something very simple to be category theory

[knotty badger]

Based on what I’ve seen category theory educators do, I’m not in the minority here

[noble nexus]

I also don't think cat theory is hard (at least, the basic stuff that most people will learn) if you have mathematically maturity

[knotty badger]

The people who actually teach this stuff don’t tend to have such rigid definitions for what does and doesn’t constitute cat theory

[noble nexus]

well if you are actually teaching category theory then its category theory of course

but like someone in a linear algebra class drawing V W and U on the board and three arrows between them and saying "oh its the composition" is not doing category theory

[knotty badger]

I’ve grown tired of this semantic argument

You can define useful cat theory out of existence if you wish

I’ll focus on what actually works pedagogically

[noble nexus]

lol "define it out of existence" is not even close to what I'm saying what

[knotty badger]

.

[rocky cloak]

I think if you're using a diagram in some type of argument. Like a universal property or a diagram chase or other cases(?) then you can safely say you're doing category theory.

But just everytime there are arrows between things seems a bit to broad.

Like saying "consider f: U -> V" or "the composition U -> V -> W" is category theory seems like a stretch

[formal laurel]

I feel like this is a tiny bit missleading. Pedagogically, one great easy use of categories is just definying products, without categories, every product definition feels the same but you can't tell exactly why without it.

[noble nexus]

sure and that's one of the great things about category theory, but also if someone has just come ito a first class on linear algebra then they have seen what, two examples of products?

I think using the universal property of products in an intro linear algebra class is like incredibly overly confusing

most linear algebra classes are already fundamentally built on a universal property (the fact that vector spaces are free) but you don't need to explicitly mention that

[knotty badger]

I think the latter might be a stretch, but things like the commutative diagrams I’ve seen for illustrating how to represent a linear map by a matrix feel very categorical

[quiet pelican]

I disagree that you can’t tell exactly why
Like, before category theory reason why is “the exact same proofs work the exact same way”
Category theory just allows you to prove them all at once*

* actually this isn’t as easy as it sounds, because you still have to construct the product
There is a systematic way to do this for sufficiently nice categories, but in general existence of (co)limits kinda needs to be proven more manually

[rocky cloak]

Yeah, I think that would qualify. Though that's kinda the edge I would pivot at I guess

[noble nexus]

the first situations where universal properties are more enlightening than they are confusing I would say are tensor products and free groups

but even then I think you don't need to actually use the language of category theory you can just call them universal properties

[quiet pelican]

Tensor products, that fun thing that every undergraduate course refuses to teach, and every postgrad course assumes you’ve seen /hj

[noble nexus]

if you count that as category theory then sure I guess teaching category theory is good but

[knotty badger]

I did use the universal property of direct sums/products in one of the classes I TA-ed this term

[noble nexus]

when most people hear "learning category theory" they don't think "doing some universal properties" they think "opening a textbook on category theory"

[knotty badger]

It’s funny to me how you went “sure learning category theory is good” and then instantly had to qualify the statement

Observing the knee-jerk reaction is fascinating

[noble nexus]

idk why you think im some category theory hater lol

I just don't think doing some universal properties and drawing diagrams is what people mean when they hear learning category theory, and I also think (and have seen first hand) that making things categorical is often a lot more confusing for people starting out

overly categorical I should say

[knotty badger]

I’ve also had a lot of first-hand experience with making things categorical in my explanations, and had a significant amount of positive feedback

[knotty badger]

Sure but doing anything “overly” is bad

[noble nexus]

hell I've seen third and fourth year students who cannot follow a proof of why a universal property characterizes an object up to isomorphism

[rocky cloak]

I mean I guess it's a context thing.

Is using a commutative diagram in a linalg class category theory? sure, whatever.

Is the statement "we're gonna use some category theory in this linalg class" particularly useful? Probably not

[knotty badger]

Sure, but that really is just a naming thing

I’ve had the strange experience where people seem to believe that “category theory” is some sort of magic spell

[noble nexus]

its certainly possible to inject enough category theory in explanations to make it more clarifying for early undergrads

[rocky cloak]

Isn't everything here a naming thing?

[knotty badger]

Where if I use some categorical concepts and mention on the side it comes from category theory, it suddenly makes the explanation significantly less understandable

[knotty badger]

Well I’ve said a few times that I’m more concerned with what works pedagogically, it’s just confusing to me how far some people are invested in ensuring “category theory” and “good pedagogy” have empty intersection

[rocky cloak]

If it's true as you say that students shiver at the mention of it, then I guess defining it away is a good solution.

But I don't know who these people are striving for this empty intersection

[topaz solar]

Based and good take

[knotty badger]

Oh to be clear, in my experience students don’t shiver at the mention of it

[knotty badger]

Because as it turns out the words are not a magic spell

[kind temple]

in my undergrad, there was a well-respected professor who advocated that one shouldn't begin to learn category theory until grad school.
i remember being influenced by this, in the sense that whenever there was mention of category theory, i tuned it out as something that i would understand later.
i also remember, a little later on, being somewhat confused by this, since my classmates and i were taking courses like lin alg, alg top, alg geo, group theory, comm alg, etc, where we found that category theory is useful, and not some scary thing that we had to learn at grad school.

[noble nexus]

tbh the philosophy behind delaying category theory is not because its hard or useless, but actually because its easy once you've already seen the examples of it being used

and you don't really need to have the full firepower of category theory to know what a universal property is or to talk about commutative diagrams

[topaz solar]

Also some undergrads are too afraid of computation

[knotty badger]

Well it wasn’t easy for me when I actually took the course so idk what you mean

[kind temple]

i also found there to be a pretty steep learning curve

[kind temple]

i get this philosphy, but it seems like you were arguing that any type of categorical thinking is just too much, no?

[noble nexus]

no? not at all

[knotty badger]

That was my impression as well

[thorn jay]

computation is scary tbf

but sadly necessary

[knotty badger]

Any time I presented something in the intersection of “cat theory” and “good pedagogy”, you seemed to argue that it didn’t belong to “cat theory”

As if, axiomatically, that intersection must be empty

[noble nexus]

that's also not my opinion at all

[kind temple]

do you mind clarifying?

[tulip otter]

maybe boring too

[wooden rain]

Do you mean theory of computation or actual computations? lol

[thorn jay]

computations

[quiet pelican]

I feel like this has a lot more to do with your opinions on algebra than you like to say it does

[topaz solar]

Actual computations

[knotty badger]

My opinions on algebra are what made the course I took difficult?

I don’t understand what you’re saying

[thorn jay]

that's why it's scary

[noble nexus]

two things (1) I think giving category theory explanations to people struggling with like, basic linear algebra or group theory or whatever is just really not good almost always, and in general I think it has very limited use for first/second year students and (2) I think self-studying category theory as a first/second year student is not really the most efficient (though of course, if people want to I guess they can)

[topaz solar]

Hard agree

[noble nexus]

however, once someone has a good amount of mathematical maturity under their belt category theory flips and becomes a super powerful tool for explaining why things are true

at least in a lot of fields I should say

[topaz solar]

If they’re struggling with matrices I don’t think universal properties are the ideal explanation

[noble nexus]

row operations are really just the yoneda lemma

[knotty badger]

I’ve had mixed experiences regarding (1) but I know the phenomenon you’re referring to

[topaz solar]

Explode

[knotty badger]

I did this literally today

[quiet pelican]

That you sometimes seem to treat algebra as this axiomatically difficult and scary thing, and not as one of the two primary inspirations of category theory

[thorn jay]

https://tenor.com/view/explosion-cat-gif-5858640239144030160

[knotty badger]

I don’t treat it as axiomatically difficult, I find it very difficult

[tulip otter]

#chill

[knotty badger]

That’s not an “opinion” of mine so much as my experience with it

[noble nexus]

universal properties in general are very confusing for people until they click, and its certainly amazing once they click but really I think you have to be very gentle when introducing them

[topaz solar]

Terminal algebra brain 🥀

[knotty badger]

Which I am, and have had success doing so

[thorn jay]

please don't banish me there have mercy 🙏

[noble nexus]

once universal properties click they're like, actually the best way of explaining things lol

[topaz solar]

I have seen you explain and have talked with multiple students who did not understand you

[quiet pelican]

?
you have an opinion (algebra is very difficult) informed by your experience (finding algebra hard)
Like I feel like this is an argument of semantics

[knotty badger]

And you’ll ignore the huge number that did understand me, because you seem to be interested in harassing me specifically

[tulip otter]

if you clicked on the linked message then you shouldve been redirected to this channel because this is #chill

[knotty badger]

“I find algebra difficult” is different from “algebra has to be difficult”

[topaz solar]

“Harassing” like cmon I just don’t think you can just say your pedagogy is flawless

[kind temple]

im not saying that category theory is the cure-all here tho.
if a student is struggling with row operations, lets address that issue first. but if you see students are somewhat comfortable with the material, why is it not a good strategy to mention a categorical result (perhaps even without naming it), for the sake of improving intuition, mental organization, etc. examples i have in mind are quotients, products, direct sums, images, and kernels.

[thorn jay]

groups-rings-fields-and-chill

[noble nexus]

if they're a strong student and are curious sure, and I also don't think there's necessarily a huge harm but I also think there are other approaches that are better

[kind temple]

such as?

[knotty badger]

I never did say it’s flawless, but you seem to be interested in continually harassing me and harping on about how bad it is

[wooden rain]

I'm always positively surprised how active these algebra channels are compared to like #advanced-probability

[noble nexus]

well products and direct sums I don't think really anyone has issues with, quotients I suppose using universal properties is a decent approach but I think there are better ways of building intuition

idk I could be convinced for quotients that using the universal property is good

[topaz solar]

Analysts aren’t so often talkative online

[kind temple]

what did you have in mind?

[topaz solar]

It doesn’t really help compute Z^3/<(1,1,1)> though

[tardy hedge]

Actually)

?

[topaz solar]

Yeah

[tardy hedge]

Why

[copper kestrel]

i looked at the solution to the practice final and oh my lord this is insane

[thorn jay]

else they'll face discrimination

[wooden rain]

I wonder why. If anything, I would've guessed it to be the opposite, considering how many people are only learning maths to apply it in a job

[kind temple]

too busy crashing out about some bound they can't get

[noble nexus]

well quotients are fundamentally about equivalence relations, they partition up an object into blocks that "look the same" and preserve algebraic structure

[topaz solar]

Empirically verified, but also a lot of analysis research is based on the right estimate trickery and they don’t wanna get scooped

[noble nexus]

I actually think the universal algebra approach to quotients is the most intuitive

[copper kestrel]

too busy analysising

[noble nexus]

though that has some crossover with category theory

[knotty badger]

If you’re referring to the example that happened recently, you’ll notice that I didn’t push the issue further once other people were helping

[wooden rain]

Or maybe analysis and other branches are simply not that philosophically deep to induce random conversations lol

[white oxide]

Ah thank you so much!

[fading acorn]

too busy compiling stuffs to do an experiment on numerical analys

[copper kestrel]

truth nyke

nuke

[kind temple]

sure, but have you ever tried to build a map out of a quotient space? checking every time that it is well-defined detracts from what you are trying to get done

im not the most familiar with UA

[thorn jay]

"look the same" 💔

you wish

[kind temple]

it has become my "scary" category theory

[thorn jay]

you wish your quotients were regular

[topaz solar]

No they’re philosophically deep, don’t fall down that “only abstraction is good” nonsense path, but a lot of analysts work to keep up publications are based on those tricks

[noble nexus]

from that perspective its actually quite easy to map out of a quotient since its just a map on the object that respects the equivalence relation

which is basically the universal property I guess

[knotty badger]

Yes exactly

[kind temple]

lol

[noble nexus]

but its not really categorical

[kind temple]

how so?

[wooden rain]

I'm not sure if I understand the last part

[noble nexus]

since when you work with equivalence relations you aren't really working in category theory

[topaz solar]

Calling it a universal property makes it sound more special than it is 🥀

[noble nexus]

I mean you can make it categorical but its kinda jank

[knotty badger]

.

[kind temple]

lol wot

[topaz solar]

Like you don’t want others to figure out the “trick” you use to build your results before you can use it, not a very healthy ecosystem they run through

[rocky cloak]

Maybe analysists actually have friends in real life 😉

[topaz solar]

Also this, they can afford to do things

They actually get money

[kind temple]

idk, im on pseudo's side here. your alternatives were, 1. UA, 2. the "janky" category theory is not worth mentioning

[noble nexus]

II mean I just don't think every universal property is category theory lol

[topaz solar]

I don’t think calling it a universal property “helps”

[noble nexus]

in fact I think quotients are not really best understood through category theory unless your category is abelian really

[wooden rain]

Before you use it = before you publish your paper? Sorry for keeping asking, but I'm genuinely confused 🥲

[noble nexus]

or "nice"

[knotty badger]

topological spaces??

[kind temple]

sure, especially given the "stigma" that cat theory has around it

[noble nexus]

topological quotients using category theory imo is really not helpful

[topaz solar]

Yeah, or the secret machinery you use to figure out what’s true quickly before making a “cleaner” proof to publish

[noble nexus]

like how do you set that up in the category of topological spaces

[knotty badger]

i…

[wooden rain]

I see. That definitely souns a bit toxic

[noble nexus]

you need to I geuss build your relation as a topological space, then have the left and right projections mapping to X

[knotty badger]

universal property of the quotient is literally what saved me trying to do algebraic topology

[noble nexus]

yes but how is that category theory

[topaz solar]

Which this happens sometimes, so you like generate results by a different means than how you prove them

[knotty badger]

it’s hard for me to engage with you when you keep playing this redefinition game

[wooden rain]

Though on the other hand, I guess there must be a reason behind it... But shouldn't it be the same for other branches as well? Like are all algebra folks chill just because they do algebra? lol

[kind temple]

do you think that universal properties are not category theory?

[noble nexus]

not necessarily no

[topaz solar]

This is more in specific sub-fields, but they’re active ones and they have to keep up the pace to get tenure/funds, so…

[noble nexus]

considering they predated category theory I'm fairly sure

also like, explain to me how you would set up a topological quotient in the language of category theory

by an equivalence relation

[topaz solar]

Well, there’s less “tiny improvements in regularity” grinding, or “inequality tricks” that can go on in, say, PI rings

[knotty badger]

If R is a relation on a space X (not necessarily an equivalence), the universal property of X/R is that continuous maps X/R -> Z naturally correspond to continuous maps X -> Z which respect R

[wooden rain]

Fair enough. Are you a scholar/PhD student, Sharp?

[noble nexus]

the issue is you have to realize your equiivalence relation as an object in some category, and that is really not that intuitive (how do you make your relation into a topological space?)

[topaz solar]

Now, there’s algebraists who do that thing too, but there’s a lot broader question types too

[knotty badger]

No, you don’t?

[noble nexus]

like you can do it, but I don't think its best understood using category theory

[knotty badger]

At least not for stating the universal property

[noble nexus]

sure but how is it category theory if the universal property doesn't even use any categories

I guess in a sense it is but I really don't consider it

[topaz solar]

Whereas for PDEs there’s, like, not as broad of “question types,” and there’s instead a lot of tiny fine structure e.g.

[knotty badger]

do you see what i mean here c squared

[noble nexus]

idk why you want to claim like all of mathematics as being in the domain of category theory lol

[topaz solar]

Obviously there’s a lot of PDEs but there’s only a few classes of big interest

[knotty badger]

I’ve never once said that

[topaz solar]

Things like that

[knotty badger]

Nothing even close to that

[noble nexus]

to me category theory approach would be setting it up as a relation R mapping onto X via the left and right projection, then taking some sort of limit

and you can do this, but its also very much not intuitive imo

[topaz solar]

You do kinda shoehorn it in every time

[kind temple]

yes lol

[knotty badger]

Stop harassing me

[topaz solar]

Yes

[wooden rain]

That's great! what do you work in?

[noble nexus]

yes but to phrase "respecting R" in the language of category theory you need to make the relation and object and phrase respecting R in terms of maps

[topaz solar]

Analysis, groups, and logic

[noble nexus]

which is not intuitive in my opinion

[topaz solar]

Wrong reply but wtv

[knotty badger]

No, you don’t

[noble nexus]

??

then how is it category theory

[knotty badger]

You can phrase it as representing a copresheaf

[wooden rain]

I think I get the point, but it's still surprising. It's not like people ask research-level questions on discord. Even here I can usually understand what a discussion is about

[knotty badger]

Which does not require R to be an object in your category

[noble nexus]

oh yes that's way more intuitive

[topaz solar]

It’s a universal thing in a category of maps that already respect R

[wooden rain]

But maybe algebra is just interesting even when dealing with basic stuff, whereas analysis less so

[knotty badger]

Again, you can phrase it without explicitly using the word “copresheaf”

[noble nexus]

I mean you can I just don't think you are actually using any of the machinery of category theory so I don't get it

[topaz solar]

Like, the category of maps X -> B respecting R for some B \equiv category of maps X/R -> B

[topaz solar]

This is way cleaner than saying copresheaf

[knotty badger]

This is more a problem with your understanding, then

[topaz solar]

Right, but in analysis, research level doesn’t necessarily mean a lot of fancy adjectives

Can just be a lot of technical bounding

[topaz solar]

Can ask this to be an isomorphism even

[noble nexus]

yeah I'm not saying it can't be explained using category theory I should say

but I would not say "maps out of X/R are maps out of X that respect R" is category theory because you have to explain what "respect R" means

and the most obvious way to explain it is to go inside the objects and talk about elements of sets

[knotty badger]

Which is fine

[noble nexus]

which is in my opinion very intuitive (and this is basically the universal algebra approach)

[knotty badger]

That’s the approach i use too

[topaz solar]

Now, if you actually wanted to explain this instead of just throwing out terminology, you could specify how this X/R object that makes this work is exactly the kind of thing that represents the appropriate copresheaf by that functor

Because that gives a functor to specify maps respecting R

[noble nexus]

and I don't think that just because you are talking about maps it means you are doing category theory

[knotty badger]

That’s not what i said, but you’ve already shown you’re willing to misrepresent my position multiple times

[noble nexus]

I mean, you were talking as if the universal property of the quotient for topological spaces was some big example of category theory no?

[topaz solar]

More the other way around imo , compare how you need to introduce a lot of machinery to define some problems, whereas major problems in analysis can be as explicit as “There is a smooth solution to this pde”

[knotty badger]

I’m done with this conversation

[topaz solar]

No abstraction there at all, somehow Navier Stokes is still a big deal because the details are horrendous

Which you also can’t say “so much abstraction, garbage” to things either ofc

[white oxide]

Does this just follow from the fact that (a/p) = \pm 1 and the set {m_1, m_2, \dots} coincides with {1, 2, \dots}?

Or do we need to show that the +-1 agrees with (a/p)

I'm confused

[rocky cloak]

What is f?

[thorn jay]

I think it may feel like this because the "simple stuff" is still far away from what you're used to so it feels more new

[white oxide]

f(z) = exp(2 \pi i z) - exp(- 2 \pi i z)

[tulip otter]

sotrue

[fading acorn]

Theorem. groups-rings-fields is chill

so there is nothing to do

[rocky cloak]

Then I guess it's just

f(la/p) = f((a/p)(l/p)) and because (a/p)=±1 this is (a/p)f(l/p)

So the products are just equal factorwise

Hmm, no up to a (a/p)^(p-1)/2 thing I guess...

[thorn jay]

proof: Omitted

[wooden rain]

That's a good point. Honestly, it must be quite sad that research boils down to very small iterations

[topaz solar]

Well, it usually doesn’t

But some research directions do

Which, to some extent, that’s just the Ph.D. student/early career researcher paper mill for you

And those minor improvements can be significant at times

[thorn jay]

I was abt to say

might be the one part missing in someone working on a proof

[topaz solar]

Since there’s also a lot of proofs/papers of the form “We know something \leq A, but if something \leq B (and A-B is small), then [big result]”

There’s a few equivalences of RH stuff in that direction

[thorn jay]

I wouldn't count that as a "small iteration" then

[topaz solar]

Or see bootstrapping for elliptic pde regularity stuff

[thorn jay]

[topaz solar]

Well, on its own the result is a small iteration

And those “this gap improvement is critical” papers are not always well known

So sometimes things aren’t known to be sufficient, or that minor improvement was known to be sufficient by some old researcher but he never wrote it down because he couldn’t get that jump, so it appears minor

Etc

So there’s some small “pass/fail gaps” these minor improvements in estimates (minor as in being only a small improvement) can overcome

But yeah it’s kinda stinky lmao

[thorn jay]

right, I see, that makes sense

[thorn jay]

ts is what happens when you don't operate as a hivemind 😔

[topaz solar]

“This proof failed because I was this small bit off” “Oh hey see this minor improvement”

[thorn jay]

this minor improve your what? 🤨

[topaz solar]

NOOOO

[wooden rain]

😭

[thorn jay]

I shall now use this for slendering purposes

[thorn jay]

literature research is important people

[topaz solar]

The comparable algebra slop results are like proving some specific (class of) objects you never come across has a property you never use

[thorn jay]

ts results kind of annoy me cuz I'm always like "cmon there's so much more you can do these objects are so cool"

maybe they just don't do it for the love of the game

[topaz solar]

Sorry they just needed another paper

[thorn jay]

the sad reality

[topaz solar]

“We peeled off this minor assumption from this other paper”

[thorn jay]

Mohamed Elhamdadi moment

[topaz solar]

What did they do

[wooden rain]

On the other hand, I guess a good portion of research has some uses even if boils down to small improvements since grants aren't just money thrown in random directions... right?

[thorn jay]

90% of his papers are just introducing some new cohomology theory for some new variant of quandles

[topaz solar]

Whatever helps you sleep at night

[thorn jay]

the other 10% are great though

[topaz solar]

Now this is peak slop

Maybe they’re useful but these kinds of “turn the dial and print” papers are what I was talking about

[thorn jay]

the first couple are cool but then it's just kind of annoying

whatever gets you money though

you go girl

I read a blog post on knot theory once where the author was complaining that not enough people doing knot theory were looking for invariants for things they actually cared about in knots

i.e. people were just inventing invariants with whatever they could find rather than proving relationships with fundamental properties of knots lol

[tardy hedge]

Lol

[topaz solar]

Right

Not that that’s easy though

[thorn jay]

it's not! but they were frustrated that no one was even making an effort

[noble nexus]

knot that that's easy

[thorn jay]

which is kinda how I feel when looking at all the algeslop

[topaz solar]

Could just be effort that never made it far enough for papers too

[thorn jay]

argh I suppose

[wooden rain]

Is this how you denote an action of a group on a set?

[topaz solar]

I mean, “hey there’s this logical connection between these PDEs and this algebra” doesn’t always pan out enough for a publication

[thorn jay]

left action yes

[noble nexus]

its pretty common yes

[wooden rain]

Great, thank you

[topaz solar]

The only kind fr

[thorn jay]

ts

[quiet pelican]

I would say that but I also assume like
If they’re making that comment, they know the relevant people well enough personally to know that their statement is correct, even in terms of failed efforts

[thorn jay]

I mean it doesn't have to be a publication

[quiet pelican]

angry group theorist noises

[topaz solar]

Right probably true, at least to some degree

[topaz solar]

Yeah but I meant enough to write down, tell people about, etc

[thorn jay]

I see where you're coming from

and I suppose the same is true for much algebra

[topaz solar]

Variations on “it dies on the whiteboard”

And analysis

Math in general

[thorn jay]

(probably because there isn't much reason to care about many of these objects in the first place)

[topaz solar]

People have found all manner of things in unpublished notes once they get ahold of them

Sometimes only once the author dies

[thorn jay]

hmm

noted, to get noticed you should kys

[topaz solar]

NOOOO

[thorn jay]

I can imagine

sometimes that random thing you wrote down ages ago becomes useful

[topaz solar]

Or there were very useful lemmas you just never followed up on

But nobody ever knew (or thought) about

[thorn jay]

right

[topaz solar]

Can’t say I’m the biggest fan of the state of some literatures

Folklore, errors, citation issues, and problems between mixing of conventions

[thorn jay]

citations issues? how

like how do you mess that up lol

[topaz solar]

1 cites 2 cites 3 cites 1

None have the proof they claim is in the next

Also I’ve seen misspelled names

Like vro copy and paste atp

[thorn jay]

this has to be coordinated

[topaz solar]

Anyway, citation loops, the cited thing being irrelevant/wrong, citing unavailable private notes/communications etc

[thorn jay]

doesn't arXiv literally have a citation button

[topaz solar]

Yes

Though not all are recent, not all are arXiv, etc

Also things to be said about the quality of the claimed peer reviewing as well

Some old ones have name typos because of transliteration things for non-Latin characters, or publishers who don’t get the name conventions

Like differences in using i, j, and y for Й

[wooden rain]

Let the order of G be equal to pq where p and q are prime numbers (not necessarily distinct). Prove that either G is abelian, or the centre of G is trivial.
If G is abelian, then Z(G) = G != 1, and we're done. If G is not abelian, then Z(G) is a proper subgroup of G, so its order must be an element of {1, p, q}.
If |Z(G)| = p, then Z(G) is cyclic and so is generated by a non-trivial element of G. However, that would mean that G is abelian. From symmetry, |Z(G)| != q. Hence Z(G) is trivial.
Does this work?

[thorn jay]

If |Z(G)| = p, then Z(G) is cyclic and so is generated by a non-trivial element of G. However, that would mean that G is abelian
needs more justification

also I'm not quite sure it follows from that

hint: look at G/Z(G) too

[rocky cloak]

I guess everything works if you just change it to "... G/Z(G) is cyclic..."

[worldly canyon]

Should this say p^n?

[rapid cave]

Yes

[formal laurel]

Yes

[wooden rain]

Let H = Z(G). Let H = <x> and |H| = p. Then |G / H| = q, so G / H is cyclic and G / H = <yH>. In other words, G = <x, y>. However, if x in Z(G), then xy = yx, so y in Z(G) as well. Contradiction

Like that?

[thorn jay]

exactly

this can be extended and you get the neat theorem that G/Z(G) can never be nontrivial and cyclic

because everything in Z(G) commutes with a representing element of the generator of G/Z(G), but that of course commutes with itself, and G is generated by Z(G) plus that generator, hence G is abelian

[wooden rain]

Makes sense!

Thanks for the help 🙏

[thorn jay]

of course!

[wooden rain]

And also for elaborating on that. It sounds pretty handy

[thorn jay]

are there any uses for it besides this theorem?

[wooden rain]

Oh, I meant the theorem haha

[thorn jay]

I was kinda asking all the people here lol are there any uses for the fact that G/Z(G) can't be cyclic besides this particular case?

maybe as like one of the cases of the order of Z(G)

[topaz solar]

Was it used in the “finite p-group has nontrivial center”?

[quiet pelican]

No, that’s just “consider the class equation”

[thorn jay]

that seems backwards

lmao

[candid patrol]

You can show that |Aut(G)| is never prime

[topaz solar]

Right right, just meant if it could be

[thorn jay]

oh, yeah I see

[south patrol]

I am a p-group.

[thorn jay]

prismatic p-group

[quiet pelican]

I don’t think so
Like, you can’t use it to rule out Z(G) = e very easily

[topaz solar]

Oh yeah that would work, there’s some tomfoolery around this for infinite G tho

[south patrol]

I forgot this was still my nick

Lol

[topaz solar]

Unfortunate, can’t think of any other use of this then

[topaz solar]

As in, there’s choice bs involved

Since if you lose choice you can make infinite F_2 vector spaces with like order 7 automorphism groups or some such stupid bs

I think

[thorn jay]

yes Im an infinite F_2 vector space

yes I have an ORDER SEVEN AUTOMORPHISM GROUP?????

WHAT THE FUCK

consequences of having basis

[topaz solar]

Well, not having one

[delicate orchid]

I wonder what happens if G/Z_2(G) is cyclic. I think G would have to be solvable but can we say more

[topaz solar]

Does that imply nilpotent?

[thorn jay]

eeoooe higher centres

scary

[quiet pelican]

If you have a basis, you always have an order 2 element (swap two basis vectors)

[thorn jay]

yeah I meant the negation lol

[delicate orchid]

That’s what I’m thinking

If G is a p group this is a very restrictive condition

[topaz solar]

It’s a quotient but can yoy, like, get Z3(G) = G?

[delicate orchid]

Obviously G would be nilootent but still

[topaz solar]

Infinite groups are a hoax

[delicate orchid]

Maybe. That’s what I’m thinking. Z_3(G) = {g in G | [g,h] in Z_2(G) for all h in G}

U figure it out from there im going to sleep

[thorn jay]

have fun or sum

[topaz solar]

I don’t touch higher centers ever it’s just what I was thinking (something something are you chill and get this for Zn generally too?)

[topaz solar]

Good night freaky lad

[worldly canyon]

Should these d be n/d?

like fixed field of C_d

would be F_{p^n/d}

[formal laurel]

no

[worldly canyon]

but like

p=2

n=6, d = 3

C3 = <phi^2_2> right

and everything fixed by that is the alpha thta tsatisfy alpha^2^2=alpha

not alpha^2^3=alpha

[formal laurel]

What's the fixed field of C_6?

[worldly canyon]

F_2 right

Since C_6 is the whole galois grouup

[formal laurel]

Are you sure?

[worldly canyon]

F_p^n/F_p has galois group C_n

[formal laurel]

Gamma is the generator of the Galois Group right?

[worldly canyon]

oh yeah phi(a)= a^p here

[formal laurel]

I see what you mean. I think he should have written C_d=gamma^d
Or equivalently what you wrote

[worldly canyon]

but if C_d = gamma^d

like then C_3 = gamma^3, but this isnt true

gamma^3 only has order 2?

so how could it be C_3

[formal laurel]

Wait.

But Gamma^3(x)=x^p^3 so the fixed field is F_p^3

I'm really sleepy so somebody else should help you

[worldly canyon]

yeah this is correcct

but the subgroup generated by <gamma^3> is C_ 2 right

not C_3

[formal laurel]

ah yes, you are refering with C_d the cyclic group

So yep, you are right then

[wooden rain]

Let H <= K <= G. I want to show that [G : H] = [G : K] * [K : H]. Is there a way to prove this by constructing an isomorphism between G/H and (G/H) x (K/H)?

[knotty badger]

You could use third iso theorem? But that only works if the subgroups are normal

I’d recommend a more elementary combinatorial argument

[noble nexus]

You can prove it very quickly using group actions and orbit stabilizer if you have done those

If I recall

[quiet pelican]

Yes there is, but even that argument requires you to prove its well-defined
Choose coset reps for G/K, label them g_1, …, g_n
Then define G/H -> G/K x K/H by gH -> gK x gg_i^{-1} H where g_i K = gK
But that needs a lot of checking to show that it actually works

[noble nexus]

hmm actually I might've been thinking of a different proof, though I do think there's probably an action for which this will work

ok I guess if you allow the group to be finite you can do this p easily by considering the action of G x K on G/K x K/H

[wooden rain]

I see. So all of these seem to follow a similar path to the proof I found

[noble nexus]

actually I think the action of G on G/H x G/K will also work and doesn't require the group to be finite

[wooden rain]

Where we first get rid of the infinite order cases, and then try to manipulate the cosets

[noble nexus]

that's the cleanest one, and its always what I was remembering since I needed to prove [G:H\cap K] = [G:H][H:H\cap K] for finite index subgroups of infiinite groups on an assignment recently

[wooden rain]

Oh. What's the idea?

[noble nexus]

the orbit should be everything and the stabilizer is H\cap K = H

and orbit stabilizer says that the index of the stabilizer is equal to the size of the orbit

actually wait the orbit isn't everything, but it should be G/H x H/K

let me think ab this for a second

[wooden rain]

Sure, take your time

[wooden rain]

Yes, I don't think I see why the function is well-defined 😵‍💫

Wait. Shouldn't gg_i^{-1} H be g_i^{-1}g H?

Since we want the representative be an element of K

[quiet pelican]

I think either should work

[noble nexus]

ok at the least this one is quite simple, sinice the orbit of any point is the entirety of G/K x K/H which has size [G:K][K:H] and the stabilizer (of say, the pair (K,H)) is equal to K\times H

[wooden rain]

So we need to ensure that the representative is an element of K. How do we know that gg_i^{-1} is in K?

[noble nexus]

so then you just need to compute the index of K x H in G x K, which is just |G|/|H| = [G:H]

you might be able to show this abstractly it doesn't seem too hard

[quiet pelican]

Oops I miswrote

Actually no

[quiet pelican]

actually yeah flip it

[wooden rain]

How would it act on the product?

Sorry for not following. Group actions are one of the topics I don't understand well yet, but I thought this could be a good way to see how it works

Actually, maybe I'm trying to go too fast. I'll first go through some more problems on group actions and then come back to this

[noble nexus]

you would have (g,k)(xK,yH) = (gxK,kyH), but yeah probably goood to cover more on actions

pretty much any combinatorial problem involving group theory is going to involve the orbit stabilizer theorem or some group action stuff

its really overpowered

[wooden rain]

I'll do my best!

And thank you guys for the help. I really appreciate it

[tall igloo]

lol what is the group

(I assume you mean subgroup?)

[thorn jay]

well there arent many groups of order 7

[thorn jay]

@tall igloo

[tall igloo]

sure but I meant how is it generated 💀

but cool I'll take a look

[topaz solar]

In a choiceless set theory

[tall igloo]

Actually there isn't even a set that can contain all of these

of course theyre all isomorphic 💀

[copper kestrel]

set theory ruh roh

[tulip otter]

axiom of choiceless

[twilit wraith]

The fuck

Wait genuinely how do you do that

Does it have something to do with the existence of a basis-less vector space

[worldly canyon]

HOw do they conclude the sattement highlighted/

oh nevermind

[topaz solar]

Yes

[south patrol]

I feel this makes sense ye like

To me a better question is how do you even prove the existence of any nontrivial auto of an infinite F2 vector space without having a basis lol

Like I wonder if it is consistent with ZF that there is an infinite F2 vector space with trivial automorphism group and would not be surprised if it is

Maybe I am being silly though

[worldly canyon]

Should this box circled in red be |H|? or |G/H|? It seems like we have |G/H| distinct roots, and |H| copies of each root

[topaz solar]

probably is?

I think it is but I hate these set theory things

[thorn jay]

the set theory things hate you too i think

nothing personal of course

[topaz solar]

very personal, actually

[thorn jay]

...nothing personnell kid

[south patrol]

Ye this was my guess just idk

And I was paranoid I had missed something obvious LOL

[wooden rain]

Just to make sure, if I want to create a quotient group, the subgroup I divide by must be normal, right?

Oh, I think I know where my confusion comes from. I can consider the index of any subgroup, but if I wanted to consider the quotient, it wouldn't be "properly structured" if the subgroup isn't normal

[knotty badger]

Yes

[knotty badger]

You still get a quotient set, it just doesn’t have a natural group structure

[wooden rain]

That makes a lot of sense now. I finally get why normal subgroups are "special"

[knotty badger]

Yeah there’s lots of reasons for this but this is a good one

[wooden rain]

It seems I somehow solved this right by accident then. It only works because the centre of a group is always normal

The more you know... :')

[knotty badger]

mhm

[wooden rain]

Thanks! I feel like my eyes have just been opened lol

[knotty badger]

yayy

im a little surprised because "normal subgroups are what make quotient groups possible" is how i was introduced to them

[wooden rain]

If I'm being honest, I sometimes come late to class. I'm sure it was covered, but I probably just happened to miss it...

💔

[knotty badger]

ahh ic

have you come across the "congruence" perspective on normal subgroups?

[wooden rain]

Not really. I recall the term appeared at some point in my linear algebra class, but that's pretty much it

[knotty badger]

the idea is that congruences are the natural generalisation of equivalence relations to algebraic structures

an equivalence relation on a set $X$ is a subset $R \subseteq X \times X$ which satisfies reflexivity, symmetry, and transitivity

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

so, a congruence on a group $G$ is a subgroup $R \leq G \times G$ which satisfies reflexivity, symmetry and transitivity

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

you take the equivalence relation definition, and just replace "subset" by "subgroup"

you can also think of this as an equivalence relation which "respects the group operation" - if $g_1 \sim h_1$ and $g_2 \sim h_2$ then $g_1 g_2 \sim h_1 h_2$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

in linear algebra, a congruence on a vector space $V$ is a subspace $W \leq V \times V$ which satisfies reflexivity, symmetry and transitivity

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

in ring theory, a congruence on a ring $R$ is a subring $S \leq R \times R$ which satisfies reflexivity, symmetry and transitivity

[wooden rain]

I see. So that's the background behind constructing e.g. exterior algebras

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

mhm - in general, the way you construct quotients of algebraic things is with congruences

[knotty badger]

in the case of group theory, any congruence determines a normal subgroup and vice-versa

if $N$ is a normal subgroup, you can define a congruence on $G$ by $a \sim b \iff a^{-1} b \in N$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

conversely, if $R$ is a congruence on $G$, then the equivalence class of the identity element is a normal subgroup

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

and these operations are inverses of each other

so every normal subgroup corresponds to a unique congruence, and vice-versa

does that make sense?

[wooden rain]

I'm still processing why the equivalence class of the neutral element is a normal subgroup

But the rest does, yes

[knotty badger]

ah so we derive it as follows

you can think of it as the set of $x \in G$ such that $(e, x) \in R$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

since R is reflexive, $(e, e) \in R$, so you get $e \in N$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

also if $(e, x)$ and $(e, y) \in R$, then since $R$ is a subgroup we have $(e, x)(e, y) = (e, xy) \in R$, so $xy \in N$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

so it's closed under multiplication

also if $(e, x) \in R$ then since $R$ is a subgroup we have $(e, x)^{-1} = (e, x^{-1}) \in R$, so $x^{-1} \in N$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

that means N is at least a subgroup, right?

[wooden rain]

Yes

[knotty badger]

to get normality, you use reflexivity

if $g \in G$ and $(e, x) \in R$ then $(g, g) (e, x) (g^{-1}, g^{-1}) \in R$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

meaning that $(g e g^{-1}, g x g^{-1}) \in R$, so $(e, g x g^{-1}) \in R$, so $g x g^{-1} \in N$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

thus the equivalence class of the identity is a normal subgroup

[wooden rain]

And we know that it's in R because R is reflexive, so for every g, (g, g) is an element of R, right?

[knotty badger]

yep!

[wooden rain]

Then I understand that, yes!

Very nice proof

[knotty badger]

in fact the congruence is determined by the equivalence class of the identity

$(a, b) \in R \iff (a^{-1}, a^{-1})(a, b) \in R \iff (e, a^{-1} b) \in R \iff a^{-1} b \in N$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

that's why normal subgroups produce a unique congruence

[wooden rain]

What about other equivalence classes in the quotient set of R? Do they have some relevant properties?

[knotty badger]

they're the cosets of N!

note that $a^{-1} b \in N \iff b \in aN$, i.e. $b$ and $a$ are in the same coset of $N$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[wooden rain]

Right, that makes a lot of sense

That's really wholesome

[knotty badger]

:)

if you do this for rings, then you get that congruences correspond to ideals (as the equivalence class of 0)

[wooden rain]

I see. So that's why we divide by ideals

[knotty badger]

mhm mhm

[wooden rain]

Because there's no other choice

I also understand why normal subgroups pop up in so many exercises

[knotty badger]

right - in a sense, the "congruence" is the fundamental concept for quotients, and then it turns out that, for algebraic structures, they're always determined by the equivalence class of the neutral element

(well at least most of the time - for monoids you don't have a similar result and actually need to work with congruences)

[wooden rain]

I'll try to think how this relates to abelianisation

Or more like what abelianisation looks like in these terms

[knotty badger]

ah that's when you quotient the group by the commutator subgroup, right?

[wooden rain]

Yep

[knotty badger]

you can think of that as defining a relation on a group $G$ by saying $a \sim b$ iff $a^{-1} b$ is a commutator $g h g^{-1} h^{-1}$ for some $g, h$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

and then taking the congruence generated by that relation

[wooden rain]

Ok, maybe that's too much for now. I understand the equivalence relation, but I wanted to see how it works in the terms of congruence. But let's ignore it for now

I'll try to do some more exercises

[knotty badger]

that's alright

i guess abelianisation also satisfies a convenient universal property

[wooden rain]

For example this one: Let G be a finite group, H its subgroup and N its normal subgroup. If gcd(|H|, |G/N|) = 1, then H is a subgroup of N.
Now that I know that I can only consider quotients when dividing by normal subgroups, I see this way: consider G/N. Let A = {hN : h in H}. A is a subgroup of G/N:

• Since e in H, then eN in A.
• If xN, yN in A, then we can assume that x and y belong to H, so xy does too. So xyN is in A.
• If xN in A, then again, we can assume x in H, so x^-1 in H, so x^{-1}N in A.
  Since |H| and |G/N| are relatively prime, then by Lagrange's theorem, A must be a singleton. Since eN in A, then all of elements of H must be in N, so H <= N.

Does that make sense?

[quiet pelican]

Yes

[knotty badger]

yeah that tracks

[wooden rain]

Awesome! And we don't really need the finiteness of G, right? We only need |H| and |G : N| being finite

[quiet pelican]

Yeah

[wooden rain]

Yay. That was the thing I was concerned I somehow skipped

[knotty badger]

can you spell out the lagrange's theorem argument again

i didn't quite understand it

[wooden rain]

Oh, wait. That is a problem

[knotty badger]

like A is a subgroup of G/N, so the size of A divides G/N

i don't quite get why the size of A should divide |H| though

[wooden rain]

Yes, that's wrong. I somehow used the size of H in a place where the size of A mattered

Let me think how to fix that

A is definitely smaller or equal to H (size-wise)

But also |A| must divide |G/N|

[quiet pelican]

Hint: ||first isomorphism theorem||

[wooden rain]

And A is the image of a projection of H onto G/N

So it's isomorphic to G/Ker projection

[knotty badger]

what if you take $G = A_4$, $H = \langle (123) \rangle$, and $N = \langle (12)(34), (13)(24), (14)(23) \rangle$?

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

hm no that doesn't work

[quiet pelican]

gcd(|H|, |G/N|) = 3

[quiet pelican]

The theorem is correct

[knotty badger]

it is??

[wooden rain]

So if pi = the projection, then |H| / |Ker pi| = |H/Ker pi| = |Im pi| = |A| => |A| divides |H|

[quiet pelican]

Why do you think it isn’t?

[wooden rain]

But if |A| divides |G/N| and |A| > 1, then gcd(|H|, |G/N|) > 1. Contradiction

[knotty badger]

i don't understand the argument for why |A| divides |H|

[quiet pelican]

A = H/ker(H -> G/N)

[knotty badger]

hm

[quiet pelican]

It’s pretty much exactly the second isomorphism theorem

A = HN/N \cong H/(H \cap N)

[knotty badger]

i see..

[wooden rain]

That part was tricky... I completely disregarded it

But I'm glad we figured it out!

Good catch, Pseudo

[knotty badger]

ah i think mishu was more helpful 🙃

[wooden rain]

You both were

🙏

[tardy hedge]

Uh why am I uncomfortable that M/(N1 cap N2) injects into M/N1 (+) M/N2

[rocky cloak]

Idk, but this is basically Chinese remainder theorem

[tardy hedge]

Maybe thats why. I never learned that theorem properly.

Never took the time to understand it

[rocky cloak]

Well the injectivity part is just if m+N1 = N1 and m+N2=N2, then m is in both N1 and N2.

I guess the tricky part of CRT is that you get surjectivity when M = N1+N2

Which is just one of the isomorphism theorems really:
(N1+N2)/N1 = N2/(N1capN2)

[wooden rain]

Let A be an abelian and normal subgroup of G, and B a subgroup of G. I want to prove that A \cap B is a normal subgroup of AB.
I understand that x(A \cap B)x^{-1} \in A for all x in AB (because A is normal), but I don't know how to prove that it's also an element of B

[quiet pelican]

Now you need to use that x = ab for some a \in A, b \in B

[wooden rain]

Hm, all I see is that I end up with aya^{-1} where y is in A \cap B. Oh, so since A is abelian, then it's equal to y

That makes sense

Thank you!

[wooden rain]

Let M and N be normal subgroups of G such that G = MN. Prove that G/(M \cap N) iso with (G/M) x (G/N).
I define a phi : G -> (G/M) x (G/N), phi(g) = (gM, gN). Then Ker(phi) = M \cap N, so by the first isomorphism theorem, I get what I wanted. But I don't really use the fact that G = MN, which is quite concerning. Where did I make a mistake?

[kind temple]

phi is an isomorphism onto the image of phi in (G/M) x (G/N), but you haven’t proved yet that phi is surjective

[wooden rain]

Right

[kind temple]

well

phi isn’t an iso

i had a brain fart, my b

phi descends to an injective map out of the quotient

[wooden rain]

So you're saying it's not an isomorphism because it's not surjective?

[rocky cloak]

You have to prove surjectivity and that uses G = MN

[kind temple]

i misspoke. let me restate.

phi : G —> (G/M) x (G/N) has kernel M \cap N. phi descends through the quotient to a homomorphism f : G/(M \cap N) —> (G/M) x (G/N). the map f is an isomorphism onto the image of phi, by the first isomorphism theorem.

in order to show that to show that f is an isomorphism G/(M \cap N) —> (G/M) x (G/N), it suffices to prove that phi is surjective

[wooden rain]

Since G = NM = MN, then I can factorise each element of G as g = nm = m'n'. Since N is normal, I can assume that m' = m. Then
phi(g) = (gM, gN) = (nmM, mn'N) = (nM, mN)
The elements n and m were arbitrary, which proves that phi is surjective

Thank you!

[marble hinge]

I like how different algebra books motivate introduction/interplay between normal groups, quotient groups and homomorphism differently. For example in Hertstein's "Topics in Algebra" normal groups naturally followed from his desire to make cosets into groups, this required the operation (Nx)(Ny)=N(xy) to be well-defined, so it necessitated putting some additional constraints on subgroup N (that it must be normal), and it all worked amazingly and looked elegant and motivating, I was happy 🙂 Then I looked at Dummit and Foote, and they motivate all quite differently and also in an eye-opening fashion: they didn't even bother much with that coset business and started from homomorphisms and show quotient groups as groups of "fibers" of the homomorphism.

this image stays with me now 🙂 On the top the vertical strands are those "fibers" that are subsets of elements of G taken to the same element of H

and also multiplication of fibers using the "representatives" and that it is independent of the choice of representatives:

And the third insight - that Pinter's book gave me - is that quotient groups (or "factor groups") can "factor out" certain properties to form groups without them, which is quite cool. I.e. if you want to factor out "non-abeliannness" in your group G, you can form a quotient G/H where H is the subgroup containing all "non-abelianness", i.e. all commutators (elements of form xyx^{-1}y^{-1} for x, y from G, if they are distinct from identity elements this means that xy != yx). And if you do that, then G/H is abelian. The same about factoring out elements with finite order (subgroup H of elements with finite order) and forming group G/H where all elements are of infinite order

[tribal moss]

Yeah, that's probably a better way than to pull normal subgroups and cosets out of a hat.

[marble hinge]

I mean, this is probably not all that different and are just several sides of the same coin (can a "coin" have more than two sides in larger dimensions?) - but still, accents are different, point of views are somewhat different, I like contrasting different approaches

[tulip otter]

lang starts with normal subgroups by wishing to characterize kernels of homomorphisms

[marble hinge]

to characterize in which sense? I.e. to find some properties of kernels?

and then shows that they must have normality? and hence are normal groups by definition?

[tulip otter]

yea

[twilit wraith]

thats for sure the best way to do it

immediately give your audience the reason why we care

[tribal moss]

And this generalizes better to other algebraic structures:
"Which subsets of a group can be the kernel of a homomorphism?" gives you normal subgroups.
"Which subsets of a (commutative) ring can be the kernel of a homomorphism?" gives you ideals.
... even though at first sight the conditions for being a normal subgroup vs an ideal don't look like they have much in common.

[tulip otter]

so it turns out that there more than 2 sides of a coin

[tulip otter]

yea lang doesnt yap

[twilit wraith]

i mean im a fan of yapping

[tulip otter]

thats why i decided to go through his book even tho its hard/time consuming

[twilit wraith]

usually when im reading a textbook the overall goal is to think like a person who studies the content in the book

and yapping exposes you to that more in my opinion

[tulip otter]

ohhh right thats a nice remark

[tulip otter]

i mean one should think about the content regardless if there is much yapping or no

[twilit wraith]

well its more than just knowing the content

i want to know how a topologist or an analyst or an algebraist actually approaches these things

thats much different than just being able to state theorems

and even different than applying the theorems to homework problems (though both of these things are super important steps to knowing the approach)

[tulip otter]

well i dont consider this as thinking about the content, this is more like memorizing the content imo

[twilit wraith]

thats why i added that last bit

i realized that same thing as i was writing it out lol

thats my bad

[tulip otter]

by thinking about the content i mean to actually think about how to prove these stated theorems, and this is independent of whether the author yaps or no

[tulip otter]

dw all good

[thorn jay]

to UA too; we wish to characterise the equivalence relations that come out as kernels of homomorphisms

[wooden rain]

UA?

[noble nexus]

even better: normal subgroups correspond to congruence relations

oh yeah enpeace just said that oops

[twilit wraith]

universal algebra

[thorn jay]

:>

[tribal moss]

So perhaps a possible conclusion is that the generic concept of "quotient" should first be taught for abelian groups (which seem to be the most general setting where we can just look at the preimage of {0} rather than an entire equivalence relation) and only later extended to rings, general groups, vector spaces.
(Modular arithmetic can still be a main example, even though it's a bit awkward that respecting the ring structure is only in the "later" part).

[thorn jay]

before vector spaces even?

[tribal moss]

This is just thinking aloud, mind you. :-)

[thorn jay]

considering how ubiquitous the concept is id imagine it being taught in an introductory LA course

[thorn jay]

ah alright

[tribal moss]

But yeah, you make a good point there.

[marble hinge]

I somehow missed the quotient set construction in any set theory I was using/doing, so for me this whole business of "quotient something" was new when seen in group theory. Only later I realised that there was a quotient set construction based on equivalence relations

there is a survey article on GroupProps about this: constrasting quotient set vs quotient group

https://groupprops.subwiki.org/wiki/Understanding_the_quotient_map

[thorn jay]

the big thing here is two things really:

1. congruences of groups are 1-regular, i.e. if they agree on their 1-coset they are the same
2. 1 is an idempotent; the 1-coset always forms a subgroup

[tribal moss]

On the other hand, a radical answer to that might also be that the introductory (proof-based) LA course really ought to treat some basics of abelian groups on the way to abstract vector spaces, instead of dumping the entire abstract definition of vector space on the students in a single package.

[thorn jay]

im pretty sure i could cough up a proof using Malcev conditions that if these two are satisfied in an algebra then the "normal substructures" can be defined equationally lol

[thorn jay]

hmm, i see some points here

but that could also be turned around to use vector spaces are examples of abelian groups

[marble hinge]

mm, so this is probably a terser way of saying this?

[tribal moss]

Perhaps first treat R^n, then "let's look at the facets of R^n and R itself and Z and Q that all work the same way, namely addition and subtraction of elements".

[thorn jay]

well it does say more but the reason kernels can be defined equationally as "special subgroups" is because of the two i mentioned

[marble hinge]

which to me can be summarised as "groups have more structure, so congruences on them have additional properties" 🙂

[thorn jay]

congruence uniformity doesnt have to do with that but it is the reason we can make certain counting arguments

[thorn jay]

lmao that is true in a way, but the better way of phrasing this, is that it has the right additional structure for these properties to appear

this can be phrased very specifically but that requires a background in UA so i will not bother you with it lol

[tribal moss]

That seems to break down a bit for rings, since ideals are not substructures.

[thorn jay]

yeah i phrased it wrong, better would be that the 1 generates a trivial substructure

or you could take idempotent to mean additionally that c^A = 1 for any constant operation c, as that still fits into the pattern of f(1, ..., 1) = 1

:P

wait i have my terminology wrong such an element is called a unit or identity

[thorn jay]

but it doesnt if you remove your 1!

rngs are congruence representable

[tribal moss]

So there are some useful algebraic structures where kernels are substructures and others where they aren't.

[thorn jay]

yes

like lattices for example super useful but not congruence regular

quandles/racks too, if you want to argue them to be useful

rings are super super special

because their congruences arent substructures, but they are exactly the sub-bimodules

[elfin wraith]

I’ve been saying this

[wooden rain]

Especially wedding rings

[worldly canyon]

If a polynomial is solvable by radicals, does its splitting field necessarily sit atop a tower of simple radical extensions?

[copper kestrel]

I'M FREE I TOOK MY FINAL

[quiet pelican]

How’d it go, if you wanna talk about it?

[copper kestrel]

i feel okay about 9 out of 10 questions, the 10th one i really couldnt figure out, the sun was BEAMING down upon me and i was sweating so bad so i ended up just saying smth about generators being mapped to generators

i also realized that denoting a subgroup of order 2 as N = {n_1,n_2} is a bad idea bc one of them needs to be the identity

[copper kestrel]

i was stuck on a question for at least 20 minutes until i realized i needed the identity and i got it within the next 3 minutes 😭

overall very fun exam but knowing how i feel good i probably did not do great :')

hopefully i get above a 46 though

[wooden rain]

Fingers crossed for the results!!

[copper kestrel]

if i get above a 46 then i pass the class, if i get above a 77 i can get a B- in the class

i think for an A i have to get a 95

lord knows that aint happening

hopefully (if i pass) i'll do better next semester bc i wont be taking a notoriously difficult topology class

AND my midterms wont be on the same days like they were this term

[wooden rain]

I've proven that if G is a permutation group acting on a set A, then if I take a σ in G and a in A, then σG_aσ^{-1} = G_{σ(a)}. Also, if G acts transitively on A, then \bigcap_{σ in G} σG_aσ^{-1} = 1.
Now I want to prove that if G is an abelian and transitive subgroup of S_A, then σ(a) != a for all a in A and all non-trivial σ's

Actually, let me rewrite it in tex because it's a mess...

What I've proven: Let $G \leq S_A$ act on $A$. If $\sigma in G$ and $a \in A$, then $\sigma G_a \sigma^{-1} = G_{\sigma(a)}$. Furthermore, if $G$ acts transitively on $A$, then
[
\bigcap_{\sigma \in G} \sigma G_a \sigma^{-1} = 1.
]
What I want to prove: If $G$ is an abelian and transitive subgroup of $S_A$, then
[
\forall a \in A , \forall \sigma \in G \setminus { 1 }. ; \sigma(a) \neq a.
]

[cloud walrus]

thingoln

[wooden rain]

I think I've got to a contradiction. Fix $a \in A$ and $\sigma \in G_a \setminus { 1 }$. Let $\psi \in G_a$. Since $\sigma G_a \sigma^{-1} = G_{\sigma(a)}$, then $\sigma \psi \sigma^{-1} \in G_{\sigma(a)}$. However, since $G$ is abelian, then $\sigma \psi \sigma^{-1} = \psi \in G_a$. Hence, $G_{\sigma(a)} \subseteq G_a$, so in particular $\sigma(a) = a$. Since $a$ was an arbitrary point, then $\sigma = 1$

[quiet pelican]

I don’t think that intersection is always trivial without further assumptions

[cloud walrus]

thingoln

[quiet pelican]

Namely I think you need to assume the action is faithful

[wooden rain]

What I've proven is the statement of an exercise

It comes right before the one I'm tackling now

The hint said to use it, but it seems I'm doing something wrong :')

[quiet pelican]

Wait yeah what you texed and what you originally wrote aren’t the same

Namely, what you texed is for a subgroup of the permutation group on A, not any old permutation group

I agree with everything you wrote then

[wooden rain]

Ah, I see

In that case, why did I arrive at a contradiction?

Like, I take a non-trivial sigma and prove that it's actually trivial. That would mean that G is a singleton

Oh, wait

[quiet pelican]

How does this follow?

In particular the second part

[wooden rain]

The sigma's coming from G_a

Which is good because it's supposed to be trivial

[quiet pelican]

You prove that something that stabilises a, stabilises a?
I’d hope so

Sigma depends on a, so you can’t let a vary and keep sigma fixed in the last line

[vapid vale]

i think its great that you enjoy algebra and are taking more of it, and improvement within a semester is a great indicator for future algebra since starting out the notions are so foreign

[wooden rain]

Yes. Sigma should be coming from G. Sorry, I have at least that part right in my notebook, but I miswrote it here

[wooden rain]

This is something I did mess up, though

Since we're in an Abelian group, I can simply say that sigma G_a sigma^{-1} = G_sigma(a) implies that G_a = G_sigma(a). I don't know where it leads yet, but let's write it down

[quiet pelican]

Yes

[tribal moss]

Just because a and sigma(a) have the same stabilizer doesn't mean they're the same element of A.

[wooden rain]

That's true

[tribal moss]

But because G acts transitively on A, sigma(a) can be anything, so you have proved that all the stabilizers are the same subgroup.

[wooden rain]

Okay, I think I see it. We've just proven that a permutation from $G$ stabilises a iff the same permutation stabilises $\sigma(a)$. And that works for all non-trivial $\sigma$'s. So let $a \in A$ and $\psi \in G_a$. Then, for all $\sigma \in G \setminus { 1 }$ we have $\psi(\sigma(a)) = \sigma(a)$. Let $b \in A \setminus { a }$ and choose a $\sigma$ such that $\sigma(a) = b$. Then $\psi(b) = b$, so $\psi = 1$

[cloud walrus]

thingoln

[wooden rain]

In other words, $G_a = G_b$ for all $a, b \in A$

[cloud walrus]

thingoln

[wooden rain]

So there is no non-trivial stabiliser in G, which concludes the proof!

Yay! That was a tough one lol

I really messed up at first, but thanks to your help, guys, we've done it

[tribal moss]

In fancier words:

1. We know that the stabilizers of elements in the same orbit are conjugates.
2. Since G is assumed to act transitively, there's only one orbit.
3. So all stabilizers are conjugates.
4. Since G is abelian, conjugate subgroups are equal.
5. Thus all stabilizers are equal, and an element of G that fixes anything fixes everything.

[wooden rain]

nvm, I misread your message. I understand it now

[white oxide]

Does anybody know why x^{-1} mapsto zero? Why is it contained in B? I'm assuming they are saying that it is in the kernel of the extension B -> \Omega

I understand that the extension factors through the projection, and since x^{-1} is contained in m', it is killed, but I don't understand why it's in the domain

I also understand that x^{-1} is contained in A', but I don't see why A' should be contained in B

Because B only extends A, not necessarily A'

[thorn jay]

dont crosspost, man

[wooden rain]

Let's say that G acts on a set X. What is meant by X^G?

[quiet pelican]

The elements of X fixed by every element G

[wooden rain]

Thank you!

[chilly ocean]

Equivalently, the set of objects of the action groupoid whose connected components are singletons

[tough raven]

Let A = (a_ij) the generic m⨯n matrix over the polynomial ring R = ℤ[a_ij] in mn variables. What is a minimal spanning set for the kernel of A as an R-linear map from R^n to R^m?

[tough raven]

TBH maybe this is an algebraic geometry question. I will shift it.

[crystal vale]

So can I say additive group of group ring G over R is isomorphic to direct sum of |G| copies of additive group R ?

[quiet pelican]

Yes

[crystal vale]

And if r^2g = r^2h for all r, is it true that g = h ?

[quiet pelican]

Yes, you only need r = 1

[crystal vale]

Here they are not assuming R has to be a unitial ring

[quiet pelican]

Then it fails if we have the 0 multiplication

[crystal vale]

I don't get it

[quiet pelican]

Consider the ring R with underlying additive group Z, and multiplication given by ab = 0
Then r^2 = 0 for all r, so r^2 g = 0 = r^2 h for any g, h \in G

[crystal vale]

I see

[quiet pelican]

It holds if and only if there is some r with r^2 =/= 0

By definition of a direct sum

[crystal vale]

How do I show if R(G) is commutative then R and G are commutative.

For R, I did in the following way,

(r e) (s e) = (s e ) (r e) => rs e = sr e => rs = sr

[quiet pelican]

It’s false
If R has 0 multiplication, then we don’t need G to be commutative

[crystal vale]

So I need some r and s in R such that rs ≠ 0.

[crystal vale]

But it is given in this paragraph

[quiet pelican]

R(G) is commutative iff the multiplication on R is 0, or both G and R are commutative

Yes

[crystal vale]

Okay thank you

[crystal vale]

I showed that in the finite ring, if there exists either a left zero divisor or a right zero divisor it will imply there is non-zero, zero divisor.

So, any hint for an infinite ring, where it will not hold

Also I want a ring such that ab = 0 but ba ≠ 0 but ca = 0 for some c≠0.

[noble nexus]

you should be able to find an example by looking at matrix rings I think

for the last thing you mentioned

[quiet pelican]

For the infinite ring here, consider left shift and right shift operations on the natural numbers

[crystal vale]

No, I want a ring where there exists left zero divisor but no non-zero zero divisor

[crystal vale]

Yes you are right, I got it

[tulip otter]

doesnt this follow immediately from the fact that a ring homomorphism is a monoid homomorphism (by viewing the ring as a monoid under multiplication) along with the fact that the units of a ring form a group under multiplication?

[south patrol]

Yes

Well you also want to check that the units of a product are the product of the units

but that's straightforward

[tulip otter]

ah true

tysm

have a great day/night

[crystal vale]

How to show R has an identity? Any hint?

I am trying to show ba is an identity

[formal laurel]

Use the fact that R is a domain.

this basically tells you that you can cancel stuff. So you have to make an equation that tells you when ba is the identity and find a way to tell is zero.

[south patrol]

No u

[tulip otter]

🫵

[solemn topaz]

in a ring without zero divisors a nonzero idempotent is a unit

[tribal moss]

How do you prove (a) there?

[thorn jay]

let y be such that xy = 0 and x≠0, and b such that xbx = x
=> x(b + y)x = (xb + xy)x = xbx = x
but b is unique so b + y = b, and so y has to be 0

[tardy hedge]

hi

[thorn jay]

ih

[tardy hedge]

HELLO MY MATHLETES

[copper kestrel]

ih

[formal laurel]

ih

[white oxide]

Does anybody know how this implies that v^{-1} is algebraic over A?

I know that v is algebraic over A since the algebraic closure is a ring

Does it have to do with the result that if A \subset B are integral domains, and B is integral over A, then A is a field <=> B is a field?

Is it just that x is in the integral closure of Frac(A)

So that v is in the integral closure of Frac(A) since it's a polynomial in x with coefficients in Frac(A)

and since Frac(A) is a field, v^{-1} is in the integral closure of Frac(A)?

Nvm I got it thanks!

Wait what nvm

Well I guess if an element is integral over Frac(A) then it's integral over A

Since we can just multiply the polynomial by the product of the denominators I think

Since $x \in \overline{\text{Frac}(A)}$, $v \in \overline{\text{Frac}(A)}$ (it is a polynomial in $x$ with coefficeints in $A \subseteq \overline{\text{Frac}(A)}$). Now, $\overline{\text{Frac}(A)}$ is integral over $\text{Frac}(A)$; thus Proposition 5.7 implies that $\overline{\text{Frac}(A)}$ is a field, so that $v^{-1} \in \overline{\text{Frac}(A)}$. We have an equation of the form
[\frac{a_n}{a_n'} v^{-n} + \frac{a_{n - 1}}{a_{n - 1}'} v^{1 - n} + \dots + \frac{a_0}{a_0'} = 0] Let $a = a_n' a_{n - 1}' \dots a_0'$. Then
[(a_n a_{n - 1}' \dots a_0') (v^{-1})^n + a_{n - 1} a_n' a_{n - 2}' \dots a_0' (v^{-1})^{n - 1} + \dots + a_0 a_{n - 1}' \dots a_1' = 0] so that $v^{-1} \in \overline{A}$.

[cloud walrus]

okeyokay

[merry harness]

How to show that $\bQ(\sqrt{2}) \ncong \bQ(i)$ as rings

[cloud walrus]

qiu

[quiet pelican]

Bleh

Easier way: does Q(sqrt(2)) has a square root of -1?

[merry harness]

no, i see

[quiet pelican]

(Manually compute it, if you’re not willing to take “||it embeds into the reals||” for an answer)

[merry harness]

Okay thanks

yeah of course the intuition with embedding into reals vs only being able to represent Gaussian rationals(?) makes sense

[rapid cave]

If $u$ is a unit in $O_K$ for some imaginary quadratic field $K$ such that $$u = \pm 1 \bmod fO_k$$ where $0 \neq f \in \bZ$. How can I show that $u = \pm 1$.

[merry harness]

Oh nice they are actually called gaussian rationals

[cloud walrus]

ExpertEsquieESQUIE

[quiet pelican]

Manually compute it for the two imaginary quadratic fields with non-trivial units?

[rapid cave]

right

[iron arrow]

This is may be pretty silly but there's an exercise that says to classify G = <x,y|y^3, (x^2)yxy>, and I've gotten to x=y^3 and therefore order of x is 9 but not further (due to x^-3 = x^-2 * x^-1 = yxyxyxy=y(x^-1)xy=yy=y^2, and obviously y^-2 = y) (mainly because I don't think I have to do such large combinations and see which ones are the same as others now to check up which order should it be)

As for classify ig they just mean "isomorphic to a known group"

[quiet pelican]

How do you get x = y^3?

Because that’s not right

Namely we have a surjection G -> C_3 by y -> 0, x -> 1
And in that group, 1 =/= 3*0

[iron arrow]

Wait sorry

x^3 = y

It's 2 am I'm becoming dyslexic

[iron arrow]

I also have absolutely no idea how you define those surjections but whatever

[quiet pelican]

So then we can rewrite the presentation as <x | x^9> so we’re done

[iron arrow]

Is it just the cyclic group of order 9?

But I'm not getting how that holds true, like, couldn't it be of order a multiple of 9?

Oh wait right

Because any construction via x and y will be with x and x^3

Damn 2 am really fucking me up

[iron arrow]

Is there a more standard way than doing this btw

[quiet pelican]

Not really no
You can prove there’s no general algorithm that works

[velvet hull]

in general figuring out what a group looks like from a presentation is not possible

this is known as the word problem, and is equivalent to decidability in CS

[iron arrow]

I would imagine that, but I rather meant for this specific case or "similar" to this

[iron arrow]

Has sense

[quiet pelican]

Specific cases don’t tend to be any easier than the general case, unless you have really nice relations

[iron arrow]

I guess that's true

[crystal vale]

yes got it, thank you

but when i am saying ring has no zero divisors, does it imply it has neither left divisor nor right?

[crystal vale]

Basically i have to show if R has no zero divisor then right and left cancellation holds

I got it

[velvet cave]

What does faithful representation mean in this context?

and what is the degree of a representation?

[tall igloo]

faithful means that the group action is only trivial at the identity element, so the homomorphism G --> Aut(V) is injective. i think by degree they mean the dimension of the underlying vector space

[crystal vale]

I can show f(n)f(1/n) = f(n)g(1/n) for all n, how can I show f(1/n) = g(1/n) for all n?

[quiet pelican]

Prove that f(n) is a unit

Actually are your rings unital or not?

[crystal vale]

Not