groups-rings-fields

[south patrol]

I thought everyone except Cambridge students preferred oxford as a city

Thanks

[delicate orchid]

whenever I visit london I feel like I'm going to be robbed constantly

I don't feel that way ANYWHERE else

[south patrol]

Lmao

[south patrol]

Accurate

[elfin wraith]

I mean it’s closer to London, so I guess it’s got that going for it

Cambridge is very pretty though

[south patrol]

Just don't hold a phone in your hand

[elfin wraith]

Simply a skill issue icl

[delicate orchid]

"just navigate the perplexing labyrinth without your trusty compass"

[karmic moat]

i need instagram reels... and youtube shorts... and tiktok...

[rapid cave]

https://tenor.com/view/akira-leave-me-alone-tiktok-meme-reaction-gif-9595462199354763206

[south patrol]

Dw this was a meme kinda

But also probably good advice

https://tenor.com/view/dog-stare-cute-look-at-me-math-dog-gif-11859458

[tulip otter]

if you want that feeling to become a reality come to brazil

[delicate orchid]

I'll pass

[elfin wraith]

Anyway, I’ve done literally no work today, I want to go home, I should probably email potential supervisors first though

[south patrol]

I feel like a band now

Being told to come to Brazil

[delicate orchid]

mood

[elfin wraith]

Rather than trying to convince you yokels that London is actually good

[tulip otter]

i mean i came to brazil approximately a year ago and everyone one warns me about being robbed but it never happened to me until now

[elfin wraith]

The set up for procrastinating is looking good these days though, got that desk set up now

[delicate orchid]

more than 1 person per square mile just makes me want to die

[elfin wraith]

Bonn isn’t even the nicest city you can see in Bonn

[karmic moat]

no way u got a mug telling u what to do

[delicate orchid]

I mog you with my "what's purple and commutes" mug but other than that ur smooth with it

[elfin wraith]

No one mogs snoopy

[south patrol]

An out of breath Londoner

[delicate orchid]

https://cdn.discordapp.com/attachments/728673494333915136/1418726223701479545/IMG_1081.png?ex=68e63d82&is=68e4ec02&hm=9b0e6f71f7d86fa32d63a3d5cbeb0c1bf3fb462c1054af7e1e01092d723e6d73& counter point

[south patrol]

Downside of marriage

[elfin wraith]

That is pretty incredible

[south patrol]

Have to share the Gromit mug

[tulip otter]

is #groups-rings-fields in fact #discussion-3?

[thorn jay]

only for the cool people

[delicate orchid]

it's #advanced-discussion sometimes

[karmic moat]

this is awesome as fuck

legit awesome as fuck

[elfin wraith]

I’ve made this point before, it’s a private server for like 10 of us that people occasionally ask bask group theory questions in

[thorn jay]

I vouch for photos of the mug being printed on other household objects

vouch is not the right word

[thorn jay]

fuck yes

[karmic moat]

how does he keep doing it

[elfin wraith]

Ok I will actually do work now, time to harass some random academics and beg for their advice and guidance

[delicate orchid]

I should probably do another hour of work before the wife gets home

[thorn jay]

I should also actually try and solve this problem

[karmic moat]

i should prob do my homework that i have split screened w discord

lol im on a mac

[rapid cave]

I should do AG

[karmic moat]

anyway mac has split screen

[thorn jay]

boooo

[karmic moat]

i can't tell a difference lol

it splits my screen and that's all i need

[rapid cave]

even my phone has split screen

[thorn jay]

yes but phone is too small for it to be any good

mine at least

[tulip otter]

i hate everything made by apple ngl

[thorn jay]

yes if you ignore the prices

[tulip otter]

especially iphone, imagine holding an iphone instead of samsung or something else

[thorn jay]

lmao

I sure love a site that doesn't give the information I want

[tulip otter]

nah, you are too restricted with iphone

[thorn jay]

apple is also so uptight about access to your device's files

that's so ass I hate it

[tulip otter]

yea i mean thats the main thing i require from a phone

like idc if its camera is good or no for example because i almost never use it

[elfin wraith]

This just isn’t really true anymore, especially not with the M series stuff. Their upgrades and stuff yes are still over priced but the base model MacBooks and Mac minis are incredibly fairly priced

The iPhone and whatever is silly expensive but not really any worse than other flagships

Theyre expensive, but thats not the same as overpriced, they just only really offer premium products

And fwiw I was a life long Samsung user and only recently stopped using a HP windows laptop (when it shit the bed and died)

[tulip otter]

i see, then it probably doesnt really matter much if you use either. Well personally i dont really use my phone much besides whatsapp, i use it also to get books (ofc from official websites ). But still, i like to have the liberty to download whatever third party app if i need to do that sometime

i see. I actually do the converse, i get the books on my phone and transfer them to the laptop

btw is this kind of conversation dangerous or is there no problem

anyway i will stop bc it may become dangerous if its not rn

[kind temple]

i like most of their stuff except their computers and laptops. i can’t stand macOS

[rocky cloak]

meh, better than windows at least

[kind temple]

true

i use ubuntu

windows for games

[rapid cave]

for personal use windows is good

but for dev windows sucks

[kind temple]

yea

[rocky cloak]

I feel like running python and grep from the terminal should qualify as personal use

[kind temple]

windows powershell

[rapid cave]

wsl!

[thorn jay]

alright I've done enough, I'm hungy

[tulip otter]

Ohhhh I see, I actually hate iphone because of what I think is an exaggerated security. Like you probably need a password/pin to breath

[kind temple]

nah lol. i don’t find myself bogged down by that on iphone

[tough raven]

Typically.

[karmic moat]

I’m pretty happy w my iphone and my mac

I have a windows desktop but I don’t use it much outside of video games (ie runescape)

[spark veldt]

hi this may sound a bit easy but i'm kinda struggling to show this idk. so I have an orbit $O={(x,y)g:(x,y)\in\Omega^2, x\neq y, g\in G}$. I want to show that $G$ is transitive here, so like given $(x_1,y_1)$ and $(x_2,y_2)$ with $x_i\neq y_i$, I want to guarantee that $g$ exists such that $(x_1,y_1)g=(x_2,y_2)$. How do I do this?

[cloud walrus]

bluepianist

[tribal moss]

What is G? What is Omega^2? What is the group action?

[spark veldt]

G is the group, Omega is the set. Group action is (x,y) = (xg, yg). The problem is concerned with proving doubly transitive stuff

[tribal moss]

Hmm, is this a counterexample to what you want to prove? If not, why not?
Omega and G are both the set of real numbers; the group action is addition.
But there's no group element that maps (1,2) to (1,3).

[spark veldt]

So this is the problem right and the solution says that in one of the orbits is {(x,y) ...}. So I'm just trying to see how they came to the conclusion in the last line that G is doubly transitive without showing that transitivity property required in the problem

[thorn jay]

is this Serre? lo

[spark veldt]

is Serre an author? this is from isaacs

[tribal moss]

You'll probably need to use the assumption about the average squared character, which I'm not entirely sure what means, but which you didn't say anything about in your initial post.

(Hmm, logically I suppose it must be the trace of the corresponding permutation matrix, i.e. the number of fixed elements, but I haven't any idea how to exploit this here -- sorry!)

[delicate orchid]

the character value of g is the trace of the corresponding permutation matrix

so you're not wrong

[spark veldt]

I understood the argument until the very last line. The last line it seems they drew the conclusion that G is transitive on the set {(x,y)...} after showing that the second orbit must equal to this same set

OH

wait i think it just clicked... because the orbit of {(x,y)} is equal to {(x,y)} then given two pairs i there is a g such that pair 1 * g = pair 2 ... so that's why they concluded that G is transitive on the second set

[delicate orchid]

we have <x(g), x(g)> = 2, meaning x decomposes as a sum of two irreducible C[G]-modules. If the action of G on either of these submodules was not transitive we would have further submodules, contradicting irreducibility. It is clear that (x,x) can never be mapped to (x,y) so these are our two orbit representatives

I think this works

[tribal moss]

(Ah, I see, the character thing is just Burnside's lemma in notation that was unfamiliar to me).

[delicate orchid]

you normally do <x, 1> not <x, x> which is throwing me

[delicate orchid]

because <x,1> exactly recovers burnside's lemma, as tropo has said :)

[tribal moss]

Yeah, they're just saying that if you have a != b and c != d, then (a,b) and (c,d) are in the same orbit -- because there's only that one orbit containing pairs of different elements -- so by the definition of orbit there must be a group element that takes (a,b) to (c,d).

[delicate orchid]

wait the question was just to show that a group acts transitively on its orbits?

[tribal moss]

The question was to show that the group acts transitively on the set of pairs-of-different-elements.

[delicate orchid]

ok then no worries

was just afraid I'd massively overkilled it lol

[tribal moss]

I think bringing in C[G] modules is overkill for that -- though speaking of characters certainly would steer one's thoughts in the direction of representation theory.

[thorn jay]

im pretty sure Serre's book has the same exercise

hence why i asked

[fallen ridge]

hello

i really dont understand how to start these proofs

they seem like memorizing but i need easy way to tackle these types of questions pls

[rocky cloak]

So I guess the place to start is to try to imagine what it would mean for this to be true.

Like say G was equal to HxK and G had order 8. What could the orders of H and K be? How is G being non-abelian related to whether H and K are?

[fallen ridge]

orders will be divisors of 8?

[rocky cloak]

Here it can be smart to think about the simplest cases first. What happens if Z(G) = G for example

[fallen ridge]

well since it is non abelian

identity ele?

im thinking of contradiction?

[tribal moss]

In 49 you're not assuming that the group is nonabelian.

[tribal moss]

Yes, so that allows you to say exactly what the orders of H and K have to be (up to which of them is which).

[rocky cloak]

Do you perhaps know what the groups of order 1, 2 or 4 are?

[fallen ridge]

no of elements in the group?

[fallen ridge]

yess. so they can be both of order 4

[rocky cloak]

You sure about that?

[tribal moss]

That would make their direct product have order 16.

[fallen ridge]

HUH

2,2,2 someone in help channel told me

would that work?

[rocky cloak]

Because A is an ideal

[tribal moss]

Looks like you need to flip back and check the definition of internal direct product.

[alpine plank]

What? Are you asking why BA is a subset of A?

[fallen ridge]

gotcha looked at it

[fallen ridge]

ah there are no subgroups that can be formed cuz the product cant have order 8??

no

nope

[rocky cloak]

Do then (B+A)(C+A) contains JKLV

[fallen ridge]

yess

[fallen ridge]

i need help with this too

[azure turtle]

i was thinking of the fact that if the order of a normal subgroup is n and then for any g in G then g^n is in the normal subgroup

[rocky cloak]

Say G is abelian, and x is of order 2 and y is of order 3. Do you know what the order of xy would be?

[fallen ridge]

6?

[rocky cloak]

Can you prove that?

[azure turtle]

Oh right, isn't that only if the normal subgroup is unique of that order

[fallen ridge]

well we can use the general formula of lcm?

[rocky cloak]

There are things involving the lcm you may or may not have proven yes

But maybe you can just prove this one case directly?

[rocky cloak]

I think what you're thinking of is that if the index of a normal subgroup is n, then g^n is in the subgroup.

I don't see it being particularly helpful for the exercise though...

[fallen ridge]

well the order is a^n = e

[rocky cloak]

It is the smallest such n yeah

[fallen ridge]

how is abelian relevant in that?

[rocky cloak]

It is very relevant, maybe you'll see if you try to prove it

[fallen ridge]

modulo whatever is given is identity

can we

take a group ourselves?

would it count as a proof

[rocky cloak]

I'm not sure what you mean

[azure turtle]

Yeah, i'm sorry, i was just confused

[fallen ridge]

if the order is 2

omly elements are 1 and -1

oh wait

6 is not a prime number

[azure turtle]

i'm just now confused over how the non-abelian case would be handled, like I can think of U(3, p) as an example of a group like this. Non-abelian, non-trivial center and exponent p.

[fallen ridge]

so it will prove that

[fallen ridge]

so different orders of elements

cant be there

unless we have p =q cuz otherwise lcm will be pq

damn it was easy asf

thank u so much for helping !! @rocky cloak

[rocky cloak]

So I'm not sure exactly which result you're thinking of when it comes to lcm.

But what is true (and is easy to prove) is that if x and y commute, and their orders are relatively prime, then the order of xy is the product of their orders.

In particular if x is in the center we can use this

[delicate orchid]

I cannot believe you would notate that group as U(3, p) as opposed to the obviously superior p^{1+2}_+

[azure turtle]

oh, that clears things up

[rocky cloak]

Never

(x) > (x^2) > (x^3) > ...

[fallen ridge]

gotcha. the topic is so much clearer now !! tysmmm

[azure turtle]

yeah I just started reading through Milne's notes last week so it was just the first example that came to my mind because proving that U(3, p) has exponent p whenever p odd, prime was a fun exercise

[delicate orchid]

it's a good example to have!

exercise: find the other non-abelian group of order p^3

[azure turtle]

yeah i'm not that deep in yet, to my chagrin

[rapid cave]

how would I go to prove the map
$\phi: k[x,y,z] \rightarrow k[u,v]$ defined by $$x \mapsto u^2-v^2 \quad y \mapsto 2uv \quad z \mapsto u^2+v^2$$
has kernel $(x^2+y^2 - z^2)$.

[cloud walrus]

ExpertEsquieESQUIE

[rapid cave]

This is a continuation of my question in #algebraic-geometry.
any approach using CA and/or basic AG is welcome.
Looking for a nice approach

[tardy hedge]

congrats on your Lemma man im really proud of you

[azure turtle]

it's just a random username, I know only the case for local rings

[tardy hedge]

Random?

[jovial oyster]

Kiand, you are aware that nakayama's lemma is a well known statement of commutative algebra and has been known at-least for 5 decades, right? /genq

[azure turtle]

is it not kosher around here to have usernames like that

[elfin wraith]

Grnq?

[rocky cloak]

I mean you can throw some theory on it to get it for free.

Like k[u, v] is an integral extension of R = k[u^2 - v^2, 2uv, u^2 + v^2] so they have the same dimension (i.e. 2).

So k[x, y, z]/kerphi is a 2d integral domain, so kerphi must be a minimal prime. Since k[x,y,z] is a UFD it must be principal.

[jovial oyster]

Fixed my tone tag

[tardy hedge]

yes i am aware

im a jokester ryan

[rocky cloak]

You can probably also prove it very explicitly, by writing down what a polynomial in the kernel could look like then do polynomial division. But that might be tedious

[rapid cave]

anyway the dimension of the image would be at most two (as a subring of k[u,v])

[rapid cave]

thought so but that is tedious

what about transcendence degrees?

[rocky cloak]

Like you want a proof that it's 2d without using integral extensions?

You can use transcendence degree then sure

[rapid cave]

I am not trying to avoid integral extensions. just thought it is unnecessary

[rapid cave]

is this right?

[rocky cloak]

I mean, it is true that any subalgebra of k[u, v] has dimension at most 2.

[rapid cave]

oh yeah I would need the other inequality

[rocky cloak]

I think that's true at least, it would be kinda specific to the polynomial ring

[rapid cave]

yeah but that wouldn't help me

[rocky cloak]

It wouldn't no

[rapid cave]

yes

why are minimal primes in UFD's principal?

[rocky cloak]

It's not hard to prove so you might try it yourself.

Think about an element in the ideal, then factor it into primes

[rapid cave]

take x not zero in a minimal prime ideal P, then x=p1....pn for irreducibles p1,...,pn. since P is prime one of p_i must be in P and so (p_i) is prime and contained in P and since P is minimal prime its equal to (p_i) and principal

yeah this is simple

[delicate orchid]

your hints are: it's a split extenstion and it's exponent p^2

[sly crescent]

Saying “it’s a split extension” without specifying the normal subgroup and/or quotient is very unhelpful, because every group is a split extension of the trivial group by itself and every group is a split extension of itself by the trivial group. Here specifically it’s even less helpful because the group in question is also a non-split extension.

[tardy hedge]

😲

[delicate orchid]

ok that's insanely nitpicky, and p_-^{1+2} is a split extenstion

I'm not talking about p_+^{1+2}

[sly crescent]

p_-^{1+2} is a non-split extension in two different ways

[delicate orchid]

and it's a split extension in at least one

[sly crescent]

If you exclude the trivial cases I mentioned it’s only a split extension in one way

[delicate orchid]

so it is a non-trivial split extension as I (albeit implicitly) stated

[sly crescent]

I guess

[fiery dirge]

Can anybody help me 🙏 I don't know what to do anymore I failed my abstract algebra exam 3 times now...

The tasks given are always so different from before I'm having a hard time thinking through them

If it's something I've seen before I will solve it no problem but every time I see something completely new...

[elfin wraith]

If you can only solve problems you’ve already seen, are you actually learning to solve the problems or just memorising the solutions? It sounds like you’re not actually engaging with the material and learning the methods and techniques the class is trying to teach you

[stone fulcrum]

Might help to have an example of a problem you're struggling with

Sadly "abstract algebra" can cover years of content

[elfin wraith]

Yeah, the main reason I said what I said is that they’ve failed 3 times and apparently aren’t quite sure why, which suggests to me a larger problem than just, “I’m bad at recognising when to use Sylow”

But it’s still worth knowing what’s exactly going wrong

[knotty badger]

This is a bad Q but… what’s the difference

[devout crow]

memorising sentence structures vs learning a language

[elfin wraith]

Between learning to solve a problem and learning to solve problems?

[knotty badger]

between learning to solve problems and memorising solutions

[elfin wraith]

I could teach a monkey to memorise the proof of a theorem, that’s a very different skill to being able to get the monkey to come up with their own, even with some prompting

[knotty badger]

you could teach a monkey to memorise the proof of a theorem?

[elfin wraith]

You know what I mean, I could make a 5 year old recite a bunch of words, I could make it so that a 5 year old could spit out verbatim a proof of artin wedderburn but that doesn’t at all mean the kid understands ring theory, or what the methods being employed in the proof are, nor why those methods are being used

There’s a massive difference between learning how to solve problems and learning the solution to a problem

[fiery dirge]

Well it's not exactly that. I know how to test if an algebraic structure is a group but if I am given a weird set like (-1/2, 1/2) then it gets weird (especially if testing something additionally like number of elements of order 2,3,4 etc.)

Another example of unorthodox approaches would be in elementary number theory like n!m! as a base for modular arithmetic whereas I thoroughly learned how to apply Euler's function to examples that are n^m^p (mod k)

[knotty badger]

i don't actually know what you mean

could you actually teach a 5 year old to spit out verbatim a proof of artin wedderburn?

[elfin wraith]

Memorising != learning

[knotty badger]

hm...

[elfin wraith]

Yeah probably, but this is not the point, it’s an analogy not something to be taken literally

[knotty badger]

sorry i tend to take things quite literally 😭

hence why i'm having a hard time understanding your analogy

[vapid vale]

idk i think a simple indication of whether or not you are in the former vs the latter is if you can see an algebra problem and at least think to yourself or write down the tools you have available, what the answer could entail, etc., regardless of if you are able to make any headway on the problem

[elfin wraith]

I think in that case, assuming you do genuinely know how to test if something is a group (and I believe you do) then this is mostly psychological and you’re getting freaked out by unfamiliar groups. I think the only way to really beat that is to do a bunch of problems and see a bunch of different groups

[knotty badger]

but... hm, maybe i'm not understanding

isn't that just another form of memorisation?

knowing the tools you have available and stuff

[fiery dirge]

I'm "afraid" of proofs so that might be an issue. Why? Because I haven't ever dealt with proofs until I got into math major uni. And sure I can learn some common patterns like mathematical induction, direct proof, contradiction etc. but I can't see the wider picture and come up with them on my own in some new problems.

I need to learn to think

If that makes sense

[stone fulcrum]

Why is (-1/2, 1/2) a weird set? I think I'm missing context there

[knotty badger]

for the proofs used in lecture notes/the course, how confident are you that you can write them out if asked?

[elfin wraith]

Ok my point is this, if I solve a group theory problem and give the problem to a man on the street and tell him to memorise it, then a day later I give him a test with that question on it and he gets it right, would you say that he has learned group theory? He’s certainly learned the solution to that problem I’ll grant you, but he hasn’t learned the skills required to deal with group theory problems

It’s an issue of specific vs general

[knotty badger]

i guess he's learned some group theory, yeah

[elfin wraith]

A solution is pretty much useless in isolation, but a set of skills is not

[knotty badger]

in my case i don't really know how to distinguish between memorisation and learning

[vapid vale]

i mean at this level of granularity everything is memorization. there is a distinction between the units of things you are memorizing. if you are memorizing the solution to a specific problem as an entire unit, and the logical/algebraic atoms of the proof are not clear to you, the insights of that proof are not readily applicable to any other problem

[fiery dirge]

10% probably, majority of the proofs I approach with memorizing the steps rather than understanding them

[knotty badger]

ok this i don't quite get

[thorn jay]

well thats the problem then lol

[knotty badger]

if you're memorising the steps, why is your success rate only 10%?

[thorn jay]

if you dont understand why certain steps are made then you can't really understand the proof and know when to apply it

[fiery dirge]

Because memorizing can only get you so far, I will forget them easily since I don't have the logical foundation to back them up and make them make sense

[knotty badger]

hm...

i can speak to my experience, which is a bit of an unconventional one

but i often used memorisation as a tool for understanding proofs

and understanding generally

[elfin wraith]

This is true of 95% of students at uni, most people haven’t seen proofs before then. But your books and your lectures are full of proofs.

When you read those proofs you shouldn’t just be thinking word to word, you need to try to understand what are the methods being employed, why are they taking each step. In understanding that you’re starting to understand maths

This is why I’m generally against introduction to proof courses because the actual techniques like induction or whatever are easy, but knowing how and when to use them is both hard and subject specific and you can only learn it by engaging with the subject

[knotty badger]

it's not easy for me to just... stare at a proof for a long time and go "oh, now i get what the proof is doing"

[elfin wraith]

There an element of general skill but that just comes with general mathematical maturity

[knotty badger]

my prescription is actually like

more memorisation, but maybe in a different kind of way than you're used to

[elfin wraith]

I think you’re saying what we’re saying but you just understand it differently

[knotty badger]

one thing i did in my undergrad was, if i understood an aspect of a proof i didn't before, i'd explicitly make a little flashcard for that nugget of understanding

[fiery dirge]

That's true. If you asked me to prove H is a normal subgroup of G I'd know the exact steps I need to apply. But the problem is H and G are both very problematic notation wise on exam that I get completely confused.

[thorn jay]

i think that "different kind of memorisation" what youre referring to is "understanding"

(directed to pseudo)

[knotty badger]

right but i was literally using flashcards and stuff

[thorn jay]

yeah, sure

[knotty badger]

in a sense i was "memorising the understanding"

[elfin wraith]

People do that

This is what I would call learning

[knotty badger]

because i found that my understanding wasn't as durable as i thought

[thorn jay]

ive had periods where i understood a proof, next day understanding was gone

[elfin wraith]

We are saying the same thing in different words I think

[knotty badger]

yessssssss exactly this is what i mean

i had that a bunch, so i decided to memorise the understanding to make it more concrete

and that helped significantly for me

cause in a sense my understanding of a proof was now a monotonically increasing function of time

[elfin wraith]

I think quite simply you just need more experience at this point. You need to engage more actively with the material and you need to do more problems. I don’t think there’s another option but hard work

[thorn jay]

every misunderstanding is a semantics issue ahh lol

[knotty badger]

it seems to me that you might be suffering from some kind of "cognitive overload"

this is where memorisation actually did wonders for me

by making sure it was really really easy for me to recall definitions and theorem statements and proofs of stuff in the course

i had more "mental space" for the novel parts of the problem, so to speak

[fiery dirge]

I might go off topic slightly but the types of proofs I need to memorize are those that utilize a lot of different things simultaneously. For example, proving Euler's constant is irrational. Initial steps are assuming otherwise i.e. e = p/q where p is integer and q is natural. From there we would employ Maclaurin expansion for e, then use some Lagrange's property etc. etc. to end up with e being between 2 and 3 etc.

[knotty badger]

one thing that helped for me is breaking down a proof into a bunch of "subproofs"

to some extent i was forced to do this because of limitations of how much info i could actually put on a flashcard

[fiery dirge]

Yeah if there's more theorems involved I usually forget half of them in a task so if some "showing it's a normal subgroup" uses first isomorphism theorem for help I would genuinely not remember using that

[knotty badger]

then i might have a "skeleton" card which told me the high-level overview of a proof, and then cards which dealt with the subproofs if necessary

[knotty badger]

this is exactly the kind of thing i mean

some parts of a problem are "intrinsically hard"

but i think having unfamiliarity with the terms and common proof techniques is something that's "unnecessarily hard"

[vapid vale]

if your questions are big multi-part things like that, you may find it helpful to leave an intermediate step as conjecture and then continue with your proof. you cant control if your piece of confusion happens to be at the first step which prevents any progress – you can try to mitigate that

[knotty badger]

you can't avoid doing hard things, but you can try to make sure you're spending more time on things that are "intriniscally hard" rather than "unnecessarily hard"

in my experience, each of my courses had a finite "bag of tricks" which most proofs used variations of

[fiery dirge]

Holy shit bro is actually giving me the 1 AM motivation

[knotty badger]

but the trouble was, it's hard to isolate what "trick" is being used in a proof from the proof itself

one thing i did is try to memorise the standard proofs used in the course

[fiery dirge]

Exactly

[knotty badger]

and hope that somehow, my brain would be able to "infer" the trick from seeing lots of examples

even if i couldn't directly verbalise what the trick was

i found this to be pretty helpful for exams, cause i might be in the middle of a hard problem and my brain would be like "wait this kinda looks like [theorem X] which uses [proof technique Y], let's see if we can try something like that"

i guess i just had a very flexible notion of what was worth memorising

not just things like definitions and theorem statements, but also intuitions, or traps i kept falling into, or misconceptions i later resolved, or big-picture ideas about "why" something was important

the motivation was that i knew i made mistakes at least as often as other people, if not more

but that'd be alright so long as i only made that mistake once

[fiery dirge]

I don't even know what I'm thinking about while solving the exam 🙏

Probably about butterflies and grass outside

[fiery dirge]

Another issue is literature, we don't get much of that. So I usually go through foreign literature which isn't necessarily a bad thing but it follows a different doctrine teaching different things and relying on some different knowledge in the past from subjects I might not have

So even the tasks I go through are rather scarce

[tulip otter]

i recommend that you try proving the theorems that you learn before seeing the proof, this task may be hard/takes a bit of time but i personally think that its worth it. You may not be able to come up with the proof for the first few theorems you see, but eventually you will.

you also have to do many problems

you dont have to restrict yourself to say the textbook that the prof is using

you can pick another textbook on abstract algebra if you want, it wont be much different than the one the prof is using (if different at all) in terms of the content

but the thing is that you should avoid just memorizing the proofs, you can memorize (and should) memorize the statements of theorems/propositions etc.. of course along with knowing what the statements are saying. You can also memorize the key points of the proofs that you see if you want (the very big ideas of the proof) but only after understanding the proofs

[karmic moat]

In case no one said it: talk to your professor and ask them for advice too. Be open abt ur struggles but show that u wanna improve

[tulip otter]

the ideas will more or less automatically be memorized once you understand them well

[earnest hatch]

Hey all, Im struggling with a question in Dummit and Foote: Prove that A_n does not have a proper subgroup of index < n for all n >= 5

Truthfully, I have ZERO idea about how to begin. The chapter is just a proof of the Theorem that A_n is simple for all n >= 5

[vapid vale]

are you able to use the fact that A_n is simple

[velvet hull]

I’ll instead give you this different, but related question:

prove that if a group (finite or infinite) has a subgroup of finite index, then it must have a normal subgroup of finite index

[earnest hatch]

yes, I can use the fact that A_n is simple, its proven and the only thing given to us in the chapter

[velvet hull]

If you’re able to prove this question then the proof idea in that should be able to solve your problem

[earnest hatch]

Alright I'll give that a shot, thanks

[earnest hatch]

Would the easiest way to prove this to define a group homomorphism from G to some group such that the kernel is normal in G?

[velvet hull]

That’s the idea!

[earnest hatch]

I am just confused as to where I should map G to?

I want G to act on the G/H

[velvet hull]

You’re very close

[earnest hatch]

So would I have a mapping G x G/H -> G/H?

[velvet hull]

So what is the action?

Don’t worry about the mapping for now

Just try and figure out what this action could be

Everything follows from that

[earnest hatch]

So in G/H I would have xH = {xh | h \in H}

So I define a left coset right

[velvet hull]

Sure

[earnest hatch]

Then for some g \in G I can have g * xH

Can I act on it by left multiplication

[velvet hull]

Try and see

[earnest hatch]

So g * xH = gxH?

[velvet hull]

Try and see

[earnest hatch]

Ah could I have then x(gH), so it would be some permutation of xH?

[velvet hull]

So what do you conclude

[earnest hatch]

That I would be mapping G to a subgroup of S_n? Some symmetric group ?

[velvet hull]

Is this enough to solve my problem?

[earnest hatch]

Ah no not yet we would have to find the Kernel

[velvet hull]

It is enough

The kernel, whatever it is has to have finite index

Because the image is in a finite group

[earnest hatch]

Ohhh

[velvet hull]

So now, looking at this proof try and adapt it for your original question

[earnest hatch]

So I want to define some mapping from A_5 to its cosets?

Sorry A_n

Ah then I can see that I have a mapping from A_n to a subset of some symmetric group which is of finite index

[velvet hull]

That’s not good enough for your problem unfortunately

A_n clearly embeds into S_n and that’s no problem

[earnest hatch]

lol yeah that makes sense

but I am assuming the notion of using cosets is vital here?

[velvet hull]

It’s the same action

[velvet hull]

You have yet to use the fact that the index is < n

[earnest hatch]

Ohh can I use that becuase A_n is normal in S_n, I can act by conjugation on A_n to permute it?

so something like sA_ns^-1 for all s \in S_n

[velvet hull]

What would you get from that?

[earnest hatch]

I would get some A_m

[velvet hull]

Why would we want to conjugate A_n in S_n when we’re worried about what’s inside A_n itself

[earnest hatch]

So would I act on A_n by an element of itself?

[velvet hull]

Try and think about what you did for my question

Go back to that, there are good ideas to be found there

[earnest hatch]

So if I say G is a subgroup of index m < n

Then I would have a mapping from A_n to A_n/G?

[velvet hull]

Okay, what next

[velvet hull]

It’s not a mapping, it’s an action

[earnest hatch]

So I have x \in A_n so xG

Then I act on it by a \in A_n

so I have a(xG) = x(aG)?

Oh wait, I would have (ax)G

Which would be well-defined?

Ah and then because this is well-defined, we can define a homomorphism from A_n to the permutations of A_n/H

[velvet hull]

So what do you conclude

[earnest hatch]

That the Kernel is normal in A_n

Which shows that any subgroup of finite index has a normal subgroup of finite index?

[velvet hull]

But that’s not a strong enough result for your original question

How are you sure that the kernel isn’t trivial, for instance?

That’s perfectly good and possible

[earnest hatch]

OH WAIT

Because A_n is simple, it has a trivial kernel?

[velvet hull]

Why can’t it be the entire group?

[earnest hatch]

And by action by cosets, we can make a map from A_n embedding into S_k for k < n?

[velvet hull]

Why is this action not trivial

[earnest hatch]

If the kernel were the entire group, then wouldn't G = A_n

Which would mean the subgroup isnt proper?

[velvet hull]

Okay, good

[velvet hull]

So what happens from here

[earnest hatch]

Well, I suppose we can see the orders of the mapping? A_n has order n!/2 and S_k has order k! ?

[velvet hull]

Okay, and why is this bad?

[earnest hatch]

Well we know its an injection yes?

So the order of A_n must be less than or equal to that of S_k?

Wiat it would have to be EQUAL

So |A_n| = |S_k| because its an injective homomorphism

[velvet hull]

Injective does not imply equal

Once again A_n injects into S_n and there’s no problem there

But we are almost done

[earnest hatch]

Well

So we have that n!/2 > k!?

By the orders of the sets?

[velvet hull]

Which is only true when n>2

But yes!

[earnest hatch]

Ahh

But because A_n is being mapped into a smaller group S_k

that contradicts the injectivity?

Thus no proper subgroup of index k < n can exist?

[velvet hull]

Yes

Yep!

[earnest hatch]

So, I have a question

[velvet hull]

Mhm

[earnest hatch]

When faced like a question like this on lets say an exam

Do you have any advice for approaching these?

[velvet hull]

Work with what you’re given

All you know is that there is a subgroup of index less than n

So you should immediately think of what you can do with that

And the only meaningful action you can do of a group on a subgroup is multiplication on coasts

[velvet hull]

Part of it is also experience with similar problems yes, but this is what I would say right now

[earnest hatch]

So, honestly, it seems like just having a "library" of definitions and seeing what we're given?

[velvet hull]

Oh also, if G acts on a set X do not think of it as a function from G x X to X

Either think of it as an action or as a group homomorphism from G to S_X

[earnest hatch]

Alright, so if its an action its always a group homomorphism from G to S_X?

[velvet hull]

Sometimes it is useful to think about it as an action

[earnest hatch]

What do you mean by an action?

[velvet hull]

But it’s never just “some function”, that description is forgetting too much structure

For instance left multiplication

[earnest hatch]

Ah

So think of it as action by left multiplication of conjugation?

[velvet hull]

Something of the like, yes

[earnest hatch]

So think of it by its set structure almost?

[velvet hull]

Not sure what you mean by that

[earnest hatch]

Or what it looks like element wise?

[velvet hull]

Also not sure what you mean by that, lol

[earnest hatch]

Breaking it down into elements of the group and the set

If that makes sense?

[velvet hull]

No not really but just finish the problem first lol

[earnest hatch]

lol alright

Well thanks !

[limber sequoia]

@fiery dirge With a costudent of mine something that worked very well in a galois theory course was asking him to prove exam questions he thought he knew and then asking "why" repeatedly at every point. Sooner or later it would reveal the weakness of his understanding of the material.

It doesn't sure like your exact problem but it's an option to try... This abstract courses are very different and require a readjustion and a different approach to material than before.

[spark veldt]

Hi kinda missing something here why do z1 and xa2 commute here?

Is xa2 in the center implied somewhere?

[quiet pelican]

Because z1 is in the centre, so commutes with everything

[spark veldt]

OH bruh lmao right my bad

thanks

[thorn jay]

I indeed have

for any algebra A you've got your algebraic sets X, which you can assign a coordinate algebra A[X]. Then it turns out that a subset X of A is a neighborhood (central object of study in TCT) if and only if X is an algebraic set and there is a splitting A[X] -> A[A] -> A[X]

and two neighborhoods are isomorphic (in the way of TCT) if their coordinate algebras are isomorphic

[ivory grotto]

hello, we are doing set theory in class and we were introduced to equivalence classes, the set Z_m, and the operations of addition and multiplication. I have a very vague question about what it means for an operation to be well defined in a set; is a well-defined operation in a set determined by whether or not it partitions the set exactly one time?

[thorn jay]

a well-defined operation is more just something you say; an operation cannot be not well-defined, because then it would not be a function

it's usually said when you, like in this case, already have an operation on some set and then pass to a quotient by some equivalence relation

in this case the operation on the quotient set is well-defined (i.e. is an actual function) if and only if the equivalence relation "respects" the operation on your original set

[ivory grotto]

so we go from the quotient to the entire set and prove that the operation holds there too?

[thorn jay]

no, other way around

we start with on operation on X, and want to prove that trying to induce an operation on X/~ (~ being some equivalence relation) is well-defined and yields an actual operation

[ivory grotto]

ohhh okeoke

thank you for that

[tardy hedge]

Random ahh question, is there a name for elements who can add to 1 in a ring?

[wraith cargo]

there are in certain situations

[tardy hedge]

Characteristic

[wraith cargo]

if they're orthogonal idempotents

then they have this property

but they have some additional properties than just this

[thorn jay]

they generate coprime ideals, for example

[rocky cloak]

I mean, it's called complements in probability.

Any context where you're thinking about this?

[tardy hedge]

Yeah i was just reviewing in my head why Z can’t be a Q-module

[rocky cloak]

Ohh, you mean elements like 1/n

[tardy hedge]

Yeah i was thinking the problem was u have elements in Q that can add to 1

[rocky cloak]

They're usually called unit fractions

[tardy hedge]

Making an equation in Z that cant be solved

Yeah i guess any localization has elements with that property

Or

[south patrol]

a common way thi sis phrased btw is that like "Z is not a divisible abelian group", i.e. n: Z -> Z is not surjective. and your argument is showing it doesn't hit 1 lol

i just think yeah divisible is the terminology you probably want even if it isn't what you asked for aha

this is then very useful for proving things relating Q and Z. e.g. showing that Q is not a free Z-module

[rocky cloak]

It funnily enough, showing that Q is an injective Z-module

[karmic moat]

by assumption that your ring has no non-zero nilpotent ideals

[crystal vale]

I don't know much about context but (RmR)^2 = 0, and R has no non zero nilpotent ideal so it implies RmR = 0

[thorn jay]

I don't get how they go from R(mRRm)R to 0 lol

[karmic moat]

x \in R is arbitrary and they claim that mxm = 0. assuming this claim, wouldn't you just have mRRm = 0?

[thorn jay]

no that would be if mxm = 0 for all x

[karmic moat]

doesn't that just follow because x is arbitrary

i.e. it holds for all x

[thorn jay]

no?

ah well I guess

that's terribly written though

just add "for all x in R"

[karmic moat]

yeah it was a bit confusing for me too lol

[tribal moss]

It reads confusingly because the arbitrariness of x is "discharged" (by universal generalization) behind the scenes in the middle of a sentence, somewhere between "we claim that mxm=0" and "R(mRRm)R=0".

[white oxide]

Could I have a hint for showing that (4, t) is not a power of m? My idea was to show that if it was a power of m, then 4 could be written as a product of n elements from (2, t). I could then use the binomial expansion theorem to derive a contradiction

If this were the case, then $4 = \bigl( 2 f(t) + t g(t) \bigl)^n$ for some $n > 0$, $f, g \in \mathbb{Z}[t]$. If $g(t) \neq 0$, then $\deg 4 \geq 1$, which is absurd

[cloud walrus]

okeyokay

[white oxide]

Wait no, if n = 2 and f(t) = 1, g(t) = 0 this is fine

Nvm then

[rocky cloak]

I guess you can start by finding the smallest n such that m^n is contained in (4, t) then show they're not equal

[tardy hedge]

Isnt t in (4,t) but t is not in (2,t)^n unless n = 1

[rocky cloak]

Indeed

[tardy hedge]

So thats enough isnt it

[white oxide]

you too quick with it

[thorn jay]

yeah

[fallen ridge]

how to start with these types of questions?

[tardy hedge]

probably start playing around with easy examples like G = Z

[fallen ridge]

wouldnt that take time?

[tardy hedge]

well how else are you gonna solve the question?

[fallen ridge]

i actually dont have time rn :(

i have its exam tom

[tardy hedge]

the question is asking to find examples of groups with such and such property so of course u need to try around examples

[fallen ridge]

i want to learn the approach

[fallen ridge]

there can be many non isomorphic groups but how can we be sure of automorphism

[tulip otter]

am i missing something or does this not use the fact that G is abelian?

if so then this counts as a proof to cauchy's theorem right?

[fallen ridge]

omg im doing the same thing

[fallen ridge]

i think abelian is used in showing that the elements commute

so we can take a cyclic subgroup

[tulip otter]

ohhh

[fallen ridge]

that H was taken cuz it is abelian

[tulip otter]

sure but that H can be generated in any group

[rocky cloak]

It's using that H is normal, which is not true in general for subgroups of non-abelian groups

[tulip otter]

i suspected that but then where does it use this

i mean (G:1)=(G:H)(H:1) for any subgroup H of G right?

[rapid cave]

yes

[tulip otter]

then where exactly is normality used

[fallen ridge]

can anyone pls help

[rapid cave]

G/H wouldnt be a group otherwise

[rocky cloak]

They are using that G/H is a group with exponent dividing n

[tulip otter]

ohhh yes the induction implicitly uses that G/H is a group i see

tysm everyone

[karmic moat]

for me the first thing i think of is the bare minimum like having aut(G) and aut(H) be trivial

[kind temple]

think about the automorphism group of Z

[crystal vale]

I think if I take appropriate finite cyclic subgroup G and H then it will be good

[rocky cloak]

If you just classify Aut(G) for all cyclic groups G you'll find an example

Which I guess is the general take away: start by checking easy groups and see if you figure it out

[crystal vale]

|| Z/12Z and Z/10Z, maybe work ||

[glad osprey]

If you just classify all finite simple groups, you'll find an example

[fallen ridge]

U()?

[kind temple]

huh?

[fallen ridge]

they can be written as U(n) right?

[fallen ridge]

i dont know how to take the cycles

thats the problem im facing

[kind temple]

think about where you have to send 1 in order for the homomorphism to be invertible

[crystal vale]

You know a group of form Z/nZ?

[rocky cloak]

Like do you know what the automorphism group of the trivial group is? What about Z/2, what about Z/3?

[fallen ridge]

yess

[fallen ridge]

im cooked for toms exam holy

lemme think

[fallen ridge]

send? i dont understand the send part

[kind temple]

maps out of Z are uniquely determined by where 1 is mapped to

[fallen ridge]

im not that dumb i swear . im just starting this chapter 💔 💔

[kind temple]

i don’t think anybody thinks that ha!

[fallen ridge]

it will map to the kernel right

[kind temple]

a homomorphism Z -> Z has to have trivial kernel if it is going to be invertible. so you can’t send 1 to 0

[fallen ridge]

well then

1 maps to 1?

but if it is multiplication

[kind temple]

what homomorphism does that determine? its Z with addition as the group operation

[fallen ridge]

we only have homomorphism till kernel

[kind temple]

i don’t know what that means

[fallen ridge]

identity

[kind temple]

yes

nice

that is an automorphism of Z

are there any others?

[fallen ridge]

ohhh

wait

if we prove this then aut(g) ~ aut(h)

due to identity or its like step 1

[kind temple]

just focus on finding Aut(Z) first

[fallen ridge]

well x -> x?

[kind temple]

that is the identity

you already got that one

[fallen ridge]

well its defined by

axa-1

would that help anywhere

[kind temple]

i don’t think so

the homomorphisms Z -> Z, what are they? can you classify them?

[fallen ridge]

are u asking like the properties?

[fallen ridge]

well its like simple mapping right

2 maps to 2?

[fallen ridge]

auto, inner? its the f(ab) = f(a).f(b) funcn

[kind temple]

im just trying to get you to realize that a homomorphism Z -> Z is just multiplication.
so for each integer m, you get the “multiplication by m” homomorphism n |-> mn

and if this is going to be invertible, then it should be injective and surjective

there are only two choices of m that achieve this

[fallen ridge]

oh ok wait i understand now

[fallen ridge]

1 and 0?

[tardy hedge]

keep in mind that a homomorphism just needs to satisfy f(ab)=f(a)f(b). there can be many such maps that accomplish this. Since you are looking at automorphisms though, u need it to be injective and surjective which restricts your choices more

[kind temple]

not zero, remember? multiplication by 0 can’t be invertible

[fallen ridge]

should i just pray that this question doesnt come in exam 😭

[kind temple]

its just m = 1 and m = -1

[fallen ridge]

generators

[kind temple]

yes

[fallen ridge]

yess

[rapid cave]

you should pray it does come up

[kind temple]

lol

[rapid cave]

because a new question might give you more trouble

[fallen ridge]

im crying lmao

[tardy hedge]

you should review what it means to be a homomorphism, isomorphism etc all that stuff @fallen ridge

at the very least make sure u really know what the definitions are

[fallen ridge]

nah actually im really good in other questons. that doesnt need proof

[tardy hedge]

then when u get a question on the exam u at least know where to start

[fallen ridge]

i know that 😭

[rapid cave]

most interesting questions are proof questions

[fallen ridge]

i memorized them im gonna be honest

[tardy hedge]

do u understand why they are defined that way?

[kind temple]

would say to go over a few examples then. especially homomorphisms out of Z

[fallen ridge]

i suck at proofs. im so good at discrete maths but this subject is making me cry

[rapid cave]

you will need to get better

group theory is one of the better places to do that

[fallen ridge]

i wouldnt have to read about this subject after december

so im sure its just like basic things not too deep

[tardy hedge]

yea tbh always just go to Z and Z/nZ when ur trying to prove things here

for examples and stuff to play with

[fallen ridge]

i will keep that in mind

THANK U SO MUCH !!!!

[kind temple]

good advice in general: start with the simplest examples

[tardy hedge]

so true

[kind temple]

anyways, if you still need help on this question,
finish proving that Aut(Z) = Z/2Z.

then, find another group, not isomorphic to Z, that only has 2 automorphism

[kind temple]

-# how ironic of me to not use the simplest example here 😭 🤦

[fallen ridge]

cyclic?

[kind temple]

yes its cyclic

[fallen ridge]

U(2)?

[tardy hedge]

Gulp! GPT!

[thorn jay]

huh

[fallen ridge]

is this correct

i will just memorize

[tardy hedge]

Be careful with that

If ur not sufficiently prepared to critique ai responses

[tough raven]

chole kulche

[thorn jay]

"is also of order 168. It is known that it is [group of order 16], but a more detailed analyis shows that its GL_3(F_2)?????

[fallen ridge]

is this not correct

i was so confused

[thorn jay]

vro ...

[tough raven]

What would be the point of that?

[fallen ridge]

passing in my exam

its tom. i dont have any other option

[crystal vale]

But it does not help you

[thorn jay]

bro is so close to figuring out that you shouldnt use ai for math 💔 💔

[tough raven]

Is the exam going to ask you for the automorphism group of Z_4 x Z_2?

[fallen ridge]

i didnt trust it 💔 thats why i came here

[thorn jay]

then dont use it in the first place 💔 save some water and brain cells 💔

[fallen ridge]

waittt

[tough raven]

and electricity

[fallen ridge]

lemme show why im using google lens

[tardy hedge]

Well hey im sorry to be that guy but ai has helped me parse some stuff before

U just need to be careful with it

[thorn jay]

sure but dont ask it to solve questions for you

[tardy hedge]

Yea

[thorn jay]

personally i despise ai chatbots and anything similar (like generative ai) but thats honestly a personal preference

[rapid cave]

chat gcd

[fallen ridge]

these are the examples of the book. its so cramped up i can barely understand them

[tardy hedge]

chat gpt find the gcd

[karmic moat]

the other day i was listening to frank zappa and heard some part that sounded a lot like a famous piano piece but i couldn't remember which one

so i asked chatgpt and it didn't know either that stupid robot

[kind temple]

you seem really anxious about this.
it feels like you aren’t taking your time and thinking through your answers while you’re studying.
maybe take a brain break and come back to it in a few. if the exam is tmrw, its best to be well-rested and be comfortable with the basics

[karmic moat]

kept giving me wrong answers

eventually i found the answer on some random music blog

[rapid cave]

Its mainly because of the people using the AI and not the ai itself

[thorn jay]

nono its also the AI itself e.g. the tone and style of how it writes

[karmic moat]

Great question! Great question! Great question!

[tardy hedge]

You are correct! You are amazing! You are brilliant!

[thorn jay]

aargrgghhh

[fallen ridge]

i didnt listen anything in class and didn't revise (BIG mistake i get it now) 😭 cannot take a brain break since its like 11pm here

[rapid cave]

you can get it to write however you like with some work

[tardy hedge]

Omg i am? 🥹

[fallen ridge]

thank u so much thoo

[tardy hedge]

Gpt thinks i am very smart and i have great questions

[kind temple]

in these situations, i would just take a two-four hour nap, then keep studying

[crystal vale]

[tardy hedge]

I think gpt is in love with me

[thorn jay]

"this product that you dont need wont annoy the fuck out of you with some work"

[rapid cave]

yeah I am not selling you anything

just saying that ai can be used in good ways, and in bad ways

[thorn jay]

hehe i know i get your point but still

[rapid cave]

and most people use it badly

[fallen ridge]

yeah i will prolly sleep 😭 btw can u tell the solution of the question

one is z/z2

so other will be cyclic group of order 2?

[kind temple]

yea, so the one i was leading you down was Aut(Z) = Aut(Z/4Z), but there is a minimal example with Aut(1) = Aut(Z/2Z)

[karmic moat]

i'm so prophetic

[thorn jay]

you should become a seer fr

[rapid cave]

you should become a musician

[karmic moat]

literally like i should ditch math

[thorn jay]

nah everything i make would probably suck ass

[karmic moat]

don't listen to what mq says

your music could be awesome who knows

[kind temple]

become a meteorologist

[rapid cave]

the only instrument you know is universal algebra

[thorn jay]

fr fr

[thorn jay]

in this economy?

[karmic moat]

no, weather is too unpredictable these days even prophecy can't do that

[kind temple]

pick your poison between math and weather, in this economy

[karmic moat]

well one is really applicable to the real world (especially universal algebra), and the other is weather, so....

[thorn jay]

exactlyyy

[kind temple]

dawg 💀

[thorn jay]

i see the vision

[fallen ridge]

since factor groups arent always iso to cyclic groups. we just need to find an example such that their orders are same but structure arent?

practice is really needed in this subject dawg

[rapid cave]

in every math subject

[somber sleet]

If I have two subgroups H and K of G, where K is normal in G and H \isom K. Must then H be normal in G as well?

[kind temple]

idk what you are trying to say here, sorry

[fallen ridge]

literallyy

[thorn jay]

no

[fallen ridge]

if we take one cyclic and other as factor then their auto might be equal but they can never be iso cuz of different structure

we just need to find the Z

cuz we can tell the cyclic one from there right

[thorn jay]

let G be any nonabelian, then consider GxG, along with the normal subgroup 1xG, and the diagonal (not normal) subgroup { <x, x> | x ∈ G }

normality can be seen as a property of a homomorphism f : H → G (f is normal if im f is a normal subgroup of G), and not of groups themselves, so you wouldnt expect it to be an invariant under isomorphism in that way

[kind temple]

you can just analyze cyclic groups, if that is what you are asking

[fallen ridge]

https://tenor.com/view/silly-funny-reaction-meme-stan-twt-horse-gif-2791469803182064015

[fiery dirge]

Can anybody help me understand what are group actions?

In my book, action of group $G$ onto set $S$ is defined as mapping $f: G \times S \rightarrow S$ but what does this mean? Does this mean we take the operation that's in $G$ and apply it to set $S$ and then output set $S$ with that operation applied to its elements or how does it work?

[cloud walrus]

HMD

[vapid vale]

yes

an element of G x S is an element of your group and an element of your set, and it gets sent to another element of the set. you should think of it as g "acts on" s by sending it to another element

this definition isn't precise however, its not just an arbitrary mapping

denote Sym(S) the group of permutations of the set S. then a group action is a homomorphism from G to Sym(S). if you understand the sort of structure which should be preserved by homomorphisms then this should be a clear picture

[kind temple]

group actions are like a way to take a set S and define “multiplication by elements of G”

[vapid vale]

a lot of groups that you (might) know are already defined in terms of group actions. S_n acts on the set {1, 2, ..., n}. D_n acts on the vertices ofthe n-gon

[kind temple]

if you look at vector spaces, for a familiar example, scalar multiplication is a group action F x V —> V.
the field F (really, its multiplicative group) acts on the vector space V because 1 * v = v and a * (b * v) = (ab) * v

the intuition is the same for group actions

[fiery dirge]

im completely lost here. A homomorphism is a mapping of groups that keeps the group operation f(a * b) = f(a) . f(b), I don't understand the sort of structure which should be preserved by homomorphisms this part

[glad osprey]

Which Zappa song? I wanna see if I'm better than chatgpt

[vapid vale]

right

[karmic moat]

Little Umbrellas

[vapid vale]

so firstly if we have a homomorphism G to Sym(S), then for g, h in G, we should want them to correspond to permutations of the group

the way we get a permutation is from g "multiplying" with s

the elements of the set

so the statement is now that for f: G -> Sym(S), f(g)*f(h) = f(gh)

[fiery dirge]

I can see this yeah but cant really connect it with actions

[kind temple]

where is the disconnect?

[vapid vale]

meaning, we shouldnt get discrepancy if we perform two separate permutations consecutively vs the permutation done by multiplying the group elements

let the group be S_n, and S = {1, 2, ..., n}

then we consider when homomorphism f is just the identity – S_n acts on the set in the way you already know it does

[fiery dirge]

probably because I look at a group as a set with some additional properties like having an operation, inverses, identity...

Whereas scalar and vector are completely different

[vapid vale]

S_n was defined so that multiplication of group elements is precisely composition of permutations

so this is a group action

many groups you know are defined, or can be defined, as G acting on something. and then the fact that composition of actions and multiplication of elements respect each other should feel tautological

[kind temple]

well, just forget that the vector space and the field had that additional structure!
for example, the multiplicative group of R is R - 0, and R - 0 acts on R^n by scalar multiplication in each coordinate, not even mentioning the vector space properties.

[vapid vale]

this definition now allows you to use that notion more broadly

[kind temple]

hk is also going over a standard example

[glad osprey]

Is it ||Chopin's funeral march||?

[kind temple]

another example is that a group always acts on itself by left multiplication, or conjugation

[karmic moat]

Yea

[fiery dirge]

what's the difference between a Fix set and a stabilizer? Is stabilizer just a single element whereas a Fix has to have multiple elements therefore being kinda like a set of stabilizers?

[glad osprey]

(I had to google to confirm it was Chopin, but me + google > chatgpt at least )

[azure hull]

all of them sux

[vapid vale]

the stabilizer is a subgroup which fixes a particular element in S; the fixed set is a subset which is invariant to a particular element in G

[azure hull]

i almost everytime find my queries on stackexchange or anywhere else

i prefer discussing with my friends cuz everytime anyone comes up with something new or has something to add up

[thorn jay]

so real

[vapid vale]

greggary peccary

[thorn jay]

gpt wouldnt know what im talking about anyways

[glad osprey]

I didn't want to derail the convo into Chopin and LLMs btw, maybe continue in #serious-discussion or #advanced-lounge ? So that HMD can learn about group actions in peace 🙏

[azure hull]

most of gpts arguments r circular lol

[tribal moss]

More succinctly: a fixed set is a subset of S; a stabilizer is a subset of G. So they don't even contain the same kind of elements.

[vapid vale]

:D

one has set in it

[tribal moss]

(Yes, this is already part of what Hk said, but I thought this point deserved to be singled out).

[vapid vale]

ya

it did

[glad osprey]

There's a stabilizer subgroup for each element x in S, but there's only one fixed set per group action, right?

[tribal moss]

That's one notation for it. In other words, given some g in G, we're looking at { x in X | gx=x }.

(X or S, whatever ...)

[tribal moss]

In the sense I think we're talking of, each group element has it own fixed set.

[glad osprey]

Ah, I see 👍 basically just the fixed points of the automorphism induced by g?

[tribal moss]

Yes.

There's something here that connects to both the orbit-stabilizer theorem and to Burnside's lemma, since each of them can be thought of as counting pairs (x,g) with gx=x sorted into buckets by either g or by x.

[fiery dirge]

the thing is I cant actually distinguish them

sure G is a group and S is a set that is acted upon but I just cant think of an example to explain it to myself

[tribal moss]

Do you mean you have trouble remembering which word is which of them?

[fiery dirge]

no, I have troubles with finding an example that's clear enough

[tribal moss]

I looked it up on Wikipedia a few days ago, because someone mentioned an exercise that used it :-p

[glad osprey]

So the stabilizer subgroup of x in A is the fixed points of g |-> gx, while the fixed set of g in G is the fixed points of x |-> gx

[fiery dirge]

A mapping G x S -> S where G is the group that acts on set S

so I know the words but dont understand them well

[glad osprey]

There are some more properties required in the definition - it's not just any function from G x S to S

[fiery dirge]

what does that mean?

a from G and b from S so a * b where * is operation of G?

[fiery dirge]

true, e*x = x for x in S and associativity also needs to hold

[glad osprey]

It's also nice to know this equivalent definition: a group action of G on A is a homomorphism from G to Aut(A)

[crystal vale]

Here A is a group?

[thorn jay]

Aut as in a general category lol

[crystal vale]

Oh

[glad osprey]

Usually A is just a set, in which case Aut(A) is just the group of bijections on A, but you can consider group actions where A is an object in any other category too

[thorn jay]

but Sym(A) is probably clearer

[tribal moss]

How about this:
Let the group (Z,+) act on subsets of the plane R² in the following way: The number n acts on a subset by rotating the entire plane n degrees clockwise around the origin.
Then (forgive my hastily mouse-drawn shapes):

[thorn jay]

the shapes look good

[tardy hedge]

Yayyy cute shapes and stars and circles 🥹

We love cutesy math 😝

[spark veldt]

I’m really struggling to find some idea constructing T here

Like I can picture these transversals by themselves, but S and T together I don’t really know how to connect both of them, tried drawing diagrams but to no avail.

[quiet pelican]

I’d be tempted to use Zorn’s lemma here
I don’t think it’s overkill
Consider the set of partial transversals of H which contain S
||First show that S is a partial transversal
Then it’s a standard zorn argument||

[tribal moss]

Yeah, you do need some kind of choice. Otherwise, by setting G=R, H=Q, K={0}, you could set S={0} (as you must) and then produce T, which would be a Vitali set.
So Zorn is fair game.

[thorn jay]

maybe groups are taken to be finite?

[elfin wraith]

All groups are finite, don’t buy into the Lie

[alpine plank]

Finite groups are Lie

[quiet pelican]

If the groups are finite I’d still argue the same way, you just don’t need to appeal to zorn for the existence of a maximal partial transversal

[thorn jay]

yes yes

it was in response to the use of Zorn / choice

[quiet pelican]

Why are you doing algebra without Zorn 🙃

[thorn jay]

I mean idk what level this is at cuz depending on that it's weird to have to use zorn

[fiery dirge]

When I am given 2 permutations g and f from S_n, and I rewrite them as disjoint cycles, how do I calculate g (or f) to some large power?

Like for example g^3007. I didn't give a specific example because I want to try to generalize. I've seen 2 different ways of doing this, first way is to take those disjoint cycles and see their orders (for example for g let that be 5 and 2 so (1 2 3 4 5)(6 7) or whatever). So we take 3007 and figure out what it is congruent to mod 2 and mod 5 seperately. Then since 3007 == 1 (mod 2), that means the (6 7) goes to itself (if 3007 were congruent to 0, then it would go to identity and we wouldnt write it all right?) and we don't write it at all. Next, 3007 is congruent to 2 (mod 5) so that means our elements all shift for 2 positions (1 2 3 4 5) = (1 3 5 2 4) so the final result is g^3007 = (1 3 5 2 4)(6 7)

Another way I saw was to take lcm(2, 5) = 10. So 3007 is congruent to 7 (mod 10) so g^3007 = g^7 and for the first one that means we shift by 2 (because shifting by 5 in (1, 2, 3, 4, 5) we get the identity) and similarly we shift by 1 in the (6, 7). So both ways seem to be correct.

What would happen if it were g^3006 instead? Would we write (6 7) in the result at all or we wouldnt since 3006 is congruent to 0 (mod 2)?

Also, I know g and f are conjugates iff number and order of their respective disjoint cycles is equal. However, I can't quite understand tasks where it's asked to find h such that f = hgh^-1

[tough raven]

In each cycle you "jump through" n steps at a time.

[fiery dirge]

in this particular case I just gave as a quick example we are working in S_7

so what youre saying is that I shouldve divided by 7 instead?

[tough raven]

More precisely, if g has a cycle (a_1 ... a_k) then g^n has a cycle (a_1 a_{n+1} a_{2n+1} ... a_{k-n+1}). Here you continue the cycle until the next term would be a_1 again, wrapping around (that is, taking a_{k+1} = a_1, a_{k+2} = a_2, etc.) if necessary until that happens. This cycle may or may not cover all of a1, ..., ak; if it does not then you should make more cycles starting at a_2, at a_3, etc. until every one of a_1, ..., a_k has been covered exactly once.

[spark veldt]

wow... okay haha I'll try this. thanks mico

[prisma thunder]

Any Dedekind domain that’s finitely generated as a Z-algebra does not mean it’s a free Z-module, right?

[south patrol]

You can do stuff like Fp or Fp[x]

[prisma thunder]

Noetherian integrally closed domain of Krull dimension 1

That’s the definition I’m using

[prisma thunder]

Kinda like Z[1/p]?

[rocky cloak]

So then Fp[x] would work

Z[1/p] would also work

[prisma thunder]

I guess the above allows for torsion while being free mean no torsion

[south patrol]

Exactly

And Z[1/p] is "p-divisible" unlike Z

[rocky cloak]

The finitely generated free Dedekind domains should exactly be rings of integers of number fields

[prisma thunder]

Yes

[rocky cloak]

Because being fg free is the same as fg torsion free and being fg as a module is the same as being integral extension

[south patrol]

Indeed their feaction fields are then finite dim over Q and they're already integrally closed

[quiet pelican]

A is in F
If A is prime, then A contains the the product of the single prime A

So A is not in F if A is prime

[tardy hedge]

You need Zorn’s lemma to prove that every ring with 1 has a maximal ideal

And that fact is very important for ring theory in general

[quiet pelican]

You don’t need it
Product of primes includes product of a single prime

It’s a product over the singleton

This is kinda the definition of a product over the singleton

[tardy hedge]

Also … the proof does need that R has 1 right

[quiet pelican]

Yes
Consider Q with the zero multiplication

(Ie, product of any two elements is 0)

[tardy hedge]

{a}

[quiet pelican]

A set with a single element

[tardy hedge]

U just learn things as u go

Math be like that

[quiet pelican]

I think it’s something you usually first learn by seeing it in a proof usually lol

[thorn jay]

books

you still do except the books are now on the internet

[tardy hedge]

Well i know personally my learning would have been slower if this discord didnt exist

[thorn jay]

I mean, none probably explicitly define it but plenty use it

[thorn jay]

like this

I mean idk? It tells you visually? or something?

just "product" in general

yeah products over singletons, whatever your notion of product may be

[tough raven]

One thousand KinGs

[tardy hedge]

Kind of a weird question but, in a ring after localization like R_x1x2, technically something like 1/x1x^2 as written wouldn’t be in the ring because x1x2^2 is not in the multiplicatively closed subset, but we can still treat it as being in there because its the same as x1/(x1x2)^2?

[south patrol]

it is fine, because 1/a should standard for the multiplicative inverse of a, which exists here

[tardy hedge]

I was surprised when i found out that when localizing at the set S, its not only those in S that become invertible but also the factors of anything in S become invertible. Is there any other cases that could happen?

[noble nexus]

that shouldn't be too surprising, as if s and t are both invertible then their product st has inverse t^{-1}s^{-1}

the other thing to keep in mind is just that the localization map is not in general injective

[balmy wraith]

Sometimes I think about how far mathematically I would have gotten if I was living in primal cave man ooga booga times without any of the knowledge I have now. Like would I have been able to count, to generalize it to addition and perhaps even multiplication/division?

[balmy wraith]

Wicked cool username btw

[somber sleet]

What must X satisfy such that the map G \times X \mapsto X, (g,x) mapsto gx is a left-action?

[velvet hull]

what is gx?

[somber sleet]

that's also my problem

there is no assumption

Is it multiplication? I don't know

Can I show you a piece of skript in german?

[velvet hull]

you can, im just not gonna understand it

lol

[mint seal]

(gh)x = g(h(x))

and 1_G x = x for all x

X is just any set, it doesn’t have to satisfy anything

gx is the action of g in G on x in X. It’s an element of X

[somber sleet]

Okay yeah, but there is no way you can show that (g,x) \mapsto gxg^-1 is a left action without knowing what X is right=

[mint seal]

I think if gx is already defined by some action, then conjugation gxg^-1 is also an action

mm, I guess it would have to be a two sided action for that to make sense

if X is itself G or a subgroup of G, then that conjugation action makes sense

[somber sleet]

yeah I did the same reasoning, but otherwise you need other assumptions right?

[mint seal]

yeah, I think the right generalization is X is a set, and there’s a left action and right action of G on X, satisfying (gx)h = g(xh) for all x, g, h

[thorn jay]

in general im pretty sure this is captured by the poset of distinguished open subsets D(f)

[somber sleet]

okay that's good, thank you`!

[mint seal]

I should have said normal subgroup of G here

[rocky cloak]

The statement that (g, x) maps to gx is more in defining notation.

It is saying that gx is the notation for the image of (g, x).

Saying gxg^-1 doesn't make any sense unless you're in a situation where that notation means something.

[somber sleet]

that's exactly what I was aiming at: what are the assumption needed to X for that map to be a well defined action

[rocky cloak]

I mean, I'm not sure what kind of answer your looking for.

The notation would need to be define, and it would need to satisfy the definition of being an action

[twilit wraith]

hey yall, i have to prove that if |G| = 203 and H is a normal subgroup of G of order 7 then H is a subgroup of Z(G)

just wanting to check my logic

i know that N_G(H) / C_G(H) is isomorphic to a subgroup of Aut(H)

but Aut(H) is order 6

and N_G(H) = H is order 7

so the subgroup its isomorphic to has to be the trivial group and so N_G(H) = C_G(H)

but that means C_G(H) = H and thus H is central

is that right?

[rocky cloak]

Central would be CG(H) = G

You should also use that H is normal

In particular you can't have NG(H) = H

[twilit wraith]

oh right

still essentially the same argument tho fortunately

N_G(H) = G so divides both 6 and 203

only number that does that is 1

and then C_G(H) = N_G(H) = G

i guess i just screwed up the equalities a bit

[glad osprey]

you say "that map", but (g, x) |-> gx isn't a concrete map, it's just notation. It's like asking, what must X satisfy for f(x) to be continuous

[crystal vale]

why?

but then how f_i = 0 ?

why do you think so?

but you have to multiply x^{deg F_N - deg f_i } with f_i so isn't give you polynomial of degree { deg F_N } ?

i think we always prefer polynomial ring where coefficient comes from commutative ring

yes thats how we define multiplication in polynomial ring

what do you mean f_i switched spots with x?

[cloud walrus]

mq

[crystal vale]

yes they commute

they are same thing

[tardy hedge]

in a graded module, whats the reason that 0 cannot be written as a sum of nonzero homogeneous elements? is it because if like n1+n2 = 0 with n1 and n2 nonzero then you could add that to any other decomposition of an element and then the uniqueness of the graded decompositions would not be satisfied?

[glad osprey]

you can write 0 = n1 + n2 where n1 and n2 are nonzero homogenous, but not if n1 and n2 have different gradings

like in R[x], 0 = x - x, but you can't write 0 = f(x) + g(x) where f and g have different degrees

[tardy hedge]

yeah ok thanks

[rapid cave]

this is probably some universal property of direct sums

[knotty badger]

Looks like it

[south patrol]

Though I would strongly discourage using direct sums for k-algebras

Not your fault ofc lol

[thorn jay]

huh, this would imply that direct sum of k-algebras is the coproduct in k-alg no?

wait no im stupid

[knotty badger]

It’d be product

[thorn jay]

yeah its the product of k-algebras grghrgr i hate that notation of direct sum

[south patrol]

Ye lol

[empty kernel]

how to prove that a group doesnt have a subgroup of a certain order?

[elfin wraith]

Lagranges theorem would be my first thought, but in general I’d need a little more information

Lagranges theorem tells you that the order of a subgroup must divide the order of the group so that rules out a lot of options straight away

It of course doesn’t guarantee the existence of a subgroup though, just that if it were to exist it must satisfy that

[empty kernel]

its abt the group A_4 (even permutations) and it not having a subgroup with 6 elements

so lagrange wouldnt work cuz 6 in fact divides 12 but idk how to actually make a proof i was thinking abt something with the orders of the elements but idk im kinda lost

[thorn jay]

think about the orders of elements in a group order 6

[tough raven]

When you think 4⨯2 = 6 🤡

[thorn jay]

or, notice that a subgroup of order 6 of A4 must be normal and derive a contradiction from that

yes, by the structure theorem

[elfin wraith]

Yeah it’s true for some other classes of groups too, which I can’t remember off the top of my head

Soluble groups possibly? Something like that idk

[karmic moat]

A4 is solvable with no subgroup of order 6

[vapid vale]

A4 is not soluble – A4 paper is made of cellulose

[karmic moat]

Oh…

It was just yesterday I was mixing some A4 paper in water and wasnt sure why it wasnt dissolving

[elfin wraith]

Then it’s certainly not soluble groups I’m thinking of lol

[karmic moat]

https://bearworks.missouristate.edu/cgi/viewcontent.cgi?article=4484&context=theses

[elfin wraith]

It’s been a long day, no more maths can happen

[karmic moat]

This says supersolvable

[elfin wraith]

Yay at least I’m not completely dumb

Dumb enough to not realise that the canonical example when you start looking at soluble groups (at least in my class) was soluble but still

On the right track

[karmic moat]

Ive never even heard of supersolvable

[elfin wraith]

There’s so many random ass conditions for groups

[thorn jay]

ive heard of them once

[thorn jay]

people seem to be interested in them for some reason

[karmic moat]

https://oasis.library.unlv.edu/cgi/viewcontent.cgi?article=1462&context=rtds
this says supersolvable is true but not polycyclic

[elfin wraith]

Me when I look at polycyclic byfinite complete intersection groups

[karmic moat]

Me when I go to club bed featuring DJ pillow and MC blanky

[elfin wraith]

That seems quite cool, i wonder if that’s a complete classification or if there’s anything in between/outside

[thorn jay]

arent finite abelian groups.. nilpotent

[thorn jay]

its strictly between supersolvable and polycyclic, it seems

[elfin wraith]

It’s not complete intersection groups I’m thinking of is it? That’s for rings, but I can’t remember the thing in group theory, they’re the counter example to the unit conjecture

[elfin wraith]

Interesting, cool result

Is the machinery in the proof pretty funky or can you do that relatively elementarily?

[thorn jay]

no clue havent read the paper

hmm

[tulip otter]

a partial converse holds when the divisor is prime (cauchy's theorem)

but idk if there is a more general case

[thorn jay]

maybe you start with a supersolvable group G and an invariant series
1 = A0 < A1 < ... < An = G
then consider the factor group G/A1 and use some inductive argument? maybe

the fact that A1 is cyclic is probably helpful

[fiery dirge]

How do I find all Abelian groups of order 1800 in which there exist element of order 18 but doesn't exist element of order 25? How do I think through these tasks? I genuinely have no idea how to solve this

[rocky cloak]

Do you know the classification of finite abelian groups?

Do you know what the order of elements in product groups are?

[fiery dirge]

Umm the amount of times we use group operation on that element to get it to identity

So we need our groups of order 1800 to be divisible by 18 and not by 25 right?