groups-rings-fields

[crystal vale]

Problem 1A.10

[delicate orchid]

that tells you about [G:N_G(P)] not [N_G(P):P] unless I'm missing someting obvious

[delicate orchid]

ok well that's not what you told us it said

[velvet hull]

lol

trolled

[crystal vale]

My bad

[delicate orchid]

it's ok boss

but yes now what you're saying does work

we just need that it's non-trivial

and that follows immediately from 1A10b

[crystal vale]

Thanks

[knotty badger]

wait what

is this true?

i know that $\text{Hom}(\oplus_{i=1}^\infty \mathbb{Z}, \mathbb{Z}) = \prod_{i=1}^\infty \mathbb{Z}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

it is surprising to me that the reverse holds as well

what do you mean by the indicator function for the support of Phi?

[velvet hull]

it just follows from the fact that ∏Hom is represetned by ∏ and ⨁Hom is represented by ⨁ lol

[knotty badger]

yeah yeah

your professor is correct in that divergence is not an algebraic concept

[velvet hull]

it's true, but I don't see how it is relevant to my idea

[knotty badger]

since you're not talking about any topology

[velvet hull]

we are just showing that it can't be equal to any element of Z

[knotty badger]

i don't understand your argument

[velvet hull]

So if phi is in the Hom, and e_i are the standard basis vectors, phi(e_i) has to be equal to something

So I construct a vector v, where v_i = 1 iff phi(e_i) is nonzero

Well, +-1 depending on the sign just to make it positive

[knotty badger]

ok

[velvet hull]

Then phi(v) = 1+1+1+1+…

[knotty badger]

that doesn't work

there's no notion of "..." for purely algebraic things

you need topology to talk about that

[velvet hull]

But I’m not doing it infinitely many times

I just show that it can’t be equal to any integer

[knotty badger]

why not

[velvet hull]

Because I can always pull out another +1

There’s no infinity there

[knotty badger]

i don't understand the logic

can you concretely show me why phi(v) can't be, say, 42?

[velvet hull]

Let’s say phi is nonzero on the first 43 entires WLOG

[knotty badger]

ok

is phi nonzero infinitely often on the basis vectors, or just on those 43

[velvet hull]

Then phi(v) = phi(v • e1) + phi(v • e2) …. + phi(v43 • e43) + phi(v - correctional terms)

[knotty badger]

yes

[velvet hull]

Then it’s 43+a positive term

[knotty badger]

this is where your argument breaks, unfortunately

i'm not convinced that phi(v - correctional terms) is positive

[velvet hull]

This is the first time I’ve ever seen you show up in this channel by the way but regardless

[knotty badger]

what does that have to do with anything?

[velvet hull]

Yeah it doesn’t, lol

[knotty badger]

#groups-rings-fields message

this is the first time i showed up in this channel

[knotty badger]

anyway i think this is the key point

you can't deduce algebraically that phi(v - correctional terms) is positive

you can deduce that phi(finite sum of basis vectors) is positive

but with algebra alone there's no way to extend this to phi(infinite sum of basis vectors) is positive

it could totally be the case that phi(infinite sum of basis vectors) was -69, and nothing would break algebraically

[velvet hull]

I’ll think about it

[knotty badger]

if you had some kind of continuity condition, then you can relate "finite approximations" to the infinite sum

[knotty badger]

maybe another way to say this is that if $e_i$ are the standard "basis vectors" in $\prod_{i=1}^\infty \mathbb{Z}$, then $\sum_{i \in \mathbb{Z} } e_i$ is an algebraically independent element of $\prod_{i=1}^\infty \mathbb{Z}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

there's no relation between the infinite sum and the basis vectors

or at least, none that the ring operations can detect

so you could totally have a map which was 1 on every basis vector, but -1 on the infinite sum

and nothing would break algebraically

heck you could consider $\text{span}\mathbb{Z} {e_1, e_2, \dots, \sum{i \in \mathbb{Z} } e_i}$ and define the map there

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[crystal vale]

Any hint for b part?

[tardy hedge]

Here comes Wew… he’s here 4 Yew.

[delicate orchid]

I'm gonna write f instead of theta. Does the following work?: It's clear that at least one Sylow p-subgroup of K has the form f(H). All Sylow p-subgroups of K are K-conjugate, so any Sylow p-subgroup S of K is conjugate to f(H), say by some k in K. Now because f is surjective, k = f(g) for some g in G, hence S = kf(H)k^-1 = f(g)f(H)f(g^-1) = f(gHg^-1), and gHg^-1 is another Sylow p-subgroup of G.

Alternateively, K = G/\text{Ker}(f) := G/Ker, then it's not hard to show that HKer/Ker is the Sylow p of the quotient

[crystal vale]

How is it clear that at least one Sylow p-subgroup of K has the form f(H).

Are you saying if we take H is the Sylow subgroup of G, then f(H) will be the Sylow subgroup of K?

[delicate orchid]

I've decided this is the better approach but I'll answer you anyway

[crystal vale]

You can tell me in terms of quotient of the group

[delicate orchid]

wdym

[crystal vale]

I think it is better that if we just treat the case that how Sylow subgroup of G/N looks like

[delicate orchid]

yeah I agree, ignore my first message

[crystal vale]

So if I take Sylow subgroup H/N in G/N, how do I show H is Sylow in G?

Cardinality argument?

[delicate orchid]

yeah you just mess around with the indexes of H, NH, and G

[crystal vale]

I think it is not true that H/N is Sylow subgroup then H is Sylow subgroup in G

Say |G| = 2^2 × 3 × 5, take abelian group now take subgroup N has order 2×3.

So G/N is 2×5, say H/N is 2-sylow subgroup then H has order 12

[delicate orchid]

Let S be a Sylow p-subgroup of G/N. So S = K/N for some K >= N. Let T be a Sylow p-subgroup of K, then because S is Sylow in G/N we have [G/N : S] = [G:K] are both coprime to p, so T is a Sylow p-subgroup of G. Hence TN/N is a subgroup of K/N = S. We know that TN/N must be a Sylow p-subgroup of S as [G/N : TN/N] = [G : TN] is coprime to p, so by maximality S = TN/N

[tardy hedge]

If R = Ann(M) then HomR(R,M) is 0 right

Also HomR(N,M) = 0 for N finitely generated ?

[rustic crown]

non-unital rings?

[tardy hedge]

Yea sorry ik i was kind of suspicious with that part

Lol

I dont know exactly whats wrong

The context was a proof i was reading it said I^t+Ann(M) = R so it annihilates Ext^i(R/I^t, M) so Ext^i(R/I^t, M) = 0…

And the context for that proof was showing how you can compute depth using local cohomology

I mean i guess if M an Rmodule is annihilated by R then M = 0 always

[tardy hedge]

I mean … this is true though, isnt it?

[tardy hedge]

I never understood what you meant by this

I am assuming unital rings

[thorn jay]

if R = Ann(M) then M = 0 in a unital setting

[tardy hedge]

Yea i just realized

[thorn jay]

(not in a nonunital setting: 2Z = Ann_2Z(Z/2Z))

[tardy hedge]

How about if R = Ann(N)+Ann(M) then HomR(N,M) = 0 if N is finitely generated

[thorn jay]

not even finitely generated: Let f : N → M be a hom. Then 1 can be written as a + b where a annihilates N and b annihilates M. For any n in N we see:
f(n) = f((a + b) • n) = f(b • n) = b • f(n) = 0
so HomR(N, M) = 0

[tardy hedge]

oh ok cool

[thorn jay]

this is intuited by the fact that there obviously cannot exist group homomorphisms from Z/nZ to Z/mZ where n and m are coprime

[south patrol]

Eh there is one

[thorn jay]

0 doesnt count as a homomorphism

[knotty badger]

mfw Grp is a pointed category

[south patrol]

Real

[thorn jay]

zero object

youve got pointed rings but still no zero object

[south patrol]

Augmented rings

[thorn jay]

i will augment your canonical simplicial object

[south patrol]

Aka nonunital rings lol

[knotty badger]

Anime rings

?

[south patrol]

What lol

[knotty badger]

Don’t you work with those

[south patrol]

Anime 😭

Animated rings

[knotty badger]

Yeah those

[south patrol]

But augmented just means like rings equipped with a map to Z

[knotty badger]

I see

[thorn jay]

how does that turn em into nonunital rings

[south patrol]

The assignment (A -> Z) |-> ker(A -> Z) promotes to an equivalence of cats

[thorn jay]

hm

awesome tbh

[south patrol]

Or like more explicitly like

You have provided an explicit choice of splitting of Z -> A so you can write A = Z (+) A' lol

[thorn jay]

right

I wonder

that must be generalizable

[karmic moat]

annihilator

[quiet pelican]

So there’s a copy of C_n inside G
How does conjugation affect order?

I don’t think there’s an easy way to do that without first proving the result tbh

[quiet pelican]

^this is how you prove the hint

after proving that n is prime, prove that it must be 2

Hint: ||consider an element h such that hgh^-1 = g^-1, where g has order n||

So h has order n (prove it!), so h^n \in C(g)
Also h^2 g h^-2 = hg^-1 h^-1 = g, so h^2 \in C(g)
If n =/= 2, then n is odd
So h \in C(g)

[cloud walrus]

mq in his algebra/stats arc

[quiet pelican]

Just head render /j

(Render the latex in you head /joke)

[quiet pelican]

Yes

How do I get that?

Errr
h^2 g h^-2 = h (hgh^-1) h^-1 = h g^-1 h^-1 = (hgh^-1)^-1 = (g^-1)^-1 = g

[cloud walrus]

mq in his algebra/stats arc

[quiet pelican]

Essentially, we want an element that acts on <g> with order 2, and with order n
Because if n = 2, this isn’t a problem, but for any other prime it is
Then the nicest order 2 (anti-)homomorphism is inversion, so we decide we want h to act like inversion

[covert cliff]

#linear-algebra message

I asked this question in the linear algebra chat

[quiet pelican]

Yes

[covert cliff]

Does anyone know why reducible representations have the name "reducible" if not all reducible representations can be reduced to direct sum representations

[thorn jay]

reducible has to do with direct sums

[quiet pelican]

No, indecomposable is “can’t be written as a direct sum of smaller reps”
Irreducible is “no non-trivial subreps”

[thorn jay]

oh right

haha

im used to simple because im based

so i read irreducible as indecomposable

[spice whale]

have you heard of the orbit-stabiliser theorem?
this problem can be solved very quickly with it

it says finite there

[thorn jay]

finite ordered element

[spice whale]

in the hypothesis

[quiet pelican]

Has element of finite order, not group is finite

[spice whale]

ohhh i misread

apologies

regardless it reduces the problem to proving such a group is finite

also hi mico

[quiet pelican]

Hallo!

[spice whale]

funny seeing you here

[quiet pelican]

Yeah I don’t think that’s easier than solving the problem directly

[spice whale]

yeah

[thorn jay]

are the two proofs homotopic though?

[quiet pelican]

I think proving the group is finite is most efficiently done by proving that it has order 2, by the exact proof above
So just do the proof above, and append the words “hence G is finite”

[thorn jay]

yeah

[delicate orchid]

yeah except the result is wrong if you allow infinite groups https://math.stackexchange.com/questions/88980/infinite-group-with-only-two-conjugacy-classes

so you should be able to conclude that G is finite directly from the fact that it has an element of finite order

[quiet pelican]

Doesn’t this just show there is an infinite group with 2 ccls
Like
We’ve only proven there is no infinite group with a finite order element with 2 ccls

[delicate orchid]

hence this comment

[quiet pelican]

How does that follow?

You can prove it’s finite from “finite order element + 2 ccls”, which is just the contrapositive ow what we’ve proven

[delicate orchid]

oh yeah it is just the contrapositive LOL

it's very interesting that it fails in the infinite case though

there has to be a result that says all exponent p groups have a non-trivial centre somewhere

[thorn jay]

using direct limits to construct an object where some property holds for all elements is goated

same technique is used for the proof that everg algebraic lattice is the congruence lattice of some (possibly horrible) algebraic structure

[delicate orchid]

I don't know what those words mean but I'm glad to hear it

[twilit wraith]

The word "used" is like the word "use" but done in the past

[thorn jay]

basically you construct a partial algebra with the desired properties, and iterate some construction that leaves the congruence lattice the same but "fills in the gaps" of the partial operations, of course though creating new gaps. then you iterate, take the direct limit and bingo bango proof complete

[delicate orchid]

thanks actually ts one was useful

[thorn jay]

"this shit one" 💔

[twilit wraith]

In times of darkness I am your light wew

[quiet pelican]

Do you know the direct product theorem?

[delicate orchid]

standard definition of the internal direct product

[quiet pelican]

(If there are subgroups K, H which commute with each other, such that KH = G, and such that they have trivial intersection, then G = K x H)

Then if K, H are normal with trivial intersection, if a \in K, b \in H, then aba^-1 b^-1 \in H \cap K, so it’s trivial, so a and b commute

[twilit wraith]

Probably any k in K commutes with any h in H

[quiet pelican]

If a \in K, b \in H then a and b commute

?

[delicate orchid]

we're adults now so we work up to isomorphism implicitly

yes, G is only isomorphic to K x H but that's as good as being equal

[twilit wraith]

Equal, isomorphic

[delicate orchid]

Equivalent. Homotopic.

[twilit wraith]

Equivalent, |a-b| < ɛ for all ɛ

[quiet pelican]

The = sign can mean any of “direct equality”, “canonical isomorphism”, “isomorphic by a non-canonical, but specific, isomorphism” or “isomorphic by a non-specific isomorphism” /hj

ab a^-1 b^-1 = e implies ab = ba by multiplying both sides on the right by ba

[twilit wraith]

I use = as an arbitrary group operation

[quiet pelican]

Because its in H \cap K, which is equal to {e}

[thorn jay]

analysis :frown:

[cloud walrus]

mq in his algebra/stats arc

[twilit wraith]

So true

[quiet pelican]

And it’s in H \cap K because both are normal, and aba^-1 b^-1 = (aba^-1)b^-1 = a(ba^-1 b^-1)

[sinful mortar]

hey guys

[thorn jay]

equivalent, there exists a (finite) sequence of Reidemeister moves from one to the other

[sinful mortar]

how would I go about showing a function mapping R2 to R is a homomorphism

like f(x, y) = x, is a homomorphism, how do we even prove this without knowing and inner operation

[thorn jay]

im assuming R is the real numbers? or just any ring

that comes equipped with operations

[sinful mortar]

R is the reals

my bad

[thorn jay]

and R^2 naturally comes equipped with the operations from R

i dont, its just hard to tell

[sinful mortar]

like it seems trivial,

[thorn jay]

it is dw

thats why the map is called natural

[twilit wraith]

Gotta prove that it preserves + and *

[sinful mortar]

but how does it type check

oh i see

[twilit wraith]

Just like for groups

In general a homomorphism is just an operation preserving function between two structures

[delicate orchid]

f((x,y)+(x',y')) = f(x+x', y+y') = x+x' = f(x, y)+f(x',y') and so on and so forth

[twilit wraith]

Two of the same structures

[sinful mortar]

thats what i thought

[knotty badger]

Getting flashbacks 💀

[thorn jay]

universal algebra!!

[quiet pelican]

A homomorphism of Xs is a morphism in the category of Xs /j

[delicate orchid]

if anybody mentions a god damn signature I'm gonna do something comedically disproportionate to the situation at hand

[thorn jay]

bbut

[twilit wraith]

As soon as anyone mentions something even remotely foundational this server goes crazy

Its always hilarious too

[knotty badger]

Well foundations are relatively low stakes

So

[thorn jay]

a homomorphism between algebras of signature F is a function h : A → B such that for every f ∈ F_n, we have h(f^A(a1, ..., an)) = f^B(h(a1), ..., h(an)) for all a1, ..., an ∈ A

[thorn jay]

i guess UA is foundations

but imo its not really

[twilit wraith]

I agree but theres a hint of model theory

And that alone is enough

[quiet pelican]

Foundations don’t exist, who needs to build a stable house?

[thorn jay]

equational logic my beloved

[thorn jay]

commutator theory:

[twilit wraith]

Ts is already past what I know

I already hardly know universal algebra besides the isomorphism theorems

[thorn jay]

its pretty doable in nice varieties like ones with commuting congruences

looking into some notion of homology for generalised congruences right now in the hope that it has some connections with commutators haha

[knotty badger]

homology @.@

i think it's quite neat

[thorn jay]

i love (co)homology

[knotty badger]

Oh to be clear, I am not currently taking charge for help requests in a topic I am not proficient in

[twilit wraith]

One day ill know what homology is

[knotty badger]

i have said before the intuition i like to have for homology

i could link it to you if you'd like?

[twilit wraith]

Sure

If its a topological intuition I think i know it

[thorn jay]

every adjunction has a naturally associated cohomology for every abelian group object in the codomain of the left adjoint

[twilit wraith]

But itll still be good to see again

[knotty badger]

before i do that, would you say i am being authoritative and/or monopolising the discussion in this channel?

[twilit wraith]

Uhh

No, why

[knotty badger]

ok, just checking :)

[twilit wraith]

Alright

[thorn jay]

youre good dw

[knotty badger]

#alg-top-geo-top message

[knotty badger]

for the time being i need to double check every time

[knotty badger]

the summary is that exact sequences are like "indicator functions for algebraic structures"

and homology is about having imperfect tests/multi-stage tests

often this takes the form of passing a "local" test but failing a "global" test

[thorn jay]

local global stuff sounds like you get something to do with sheaves lol

[knotty badger]

yeah, sheaf cohomology is a good example

[knotty badger]

but this is homology generally

it just turns out that "passing a local test but failing a global test" is a common example of a multistage test

[twilit wraith]

So homology is built upon what is essentially more sophisticated indicators for normal subgroups

[delicate orchid]

right that's it

[knotty badger]

yeah categorically most categories of algebraic structures do not form a topos

[thorn jay]

intuitive no?

[knotty badger]

there's no analog of {0, 1} that lets you classify substructures

but this is essentially because asking for a {0, 1} is kind of the wrong question

as i lay out in the link, the answer to an algebraic test is "yes" or "no, but here's why"

it's a richer notion of indicator function than a purely set-theoretic one

[thorn jay]

measuring how much something fails rather than just that it does

[knotty badger]

exactly!

and that's a lot of what homology is about

measuring obstructions, right?

[thorn jay]

and abelian groups/categories are the right setting for that because theyre simply that damn nice to work with

[knotty badger]

essentially because you get a correspondence between substructures and quotient structures

via kernel and cokernel

this lets you set up an equivalence between tests and collections

[delicate orchid]

ermm but the triangulated categories!!! the trianglelated cateogires tho

[thorn jay]

hell, this is fucking reflected in universal algebra where any structure that is ""nice"" MUST be a fucking abelian group in some way

[knotty badger]

analogous to the equivalence between predicates and subsets in set theory

[thorn jay]

literally one the topics in the introduction mathematics course 🔥

[knotty badger]

what do you mean?

[thorn jay]

somewhere along the line they introduce functions and have a section about that equivalence

[knotty badger]

oh yes yes

[south patrol]

Which course lol

[knotty badger]

this is actually part of a much bigger story in mathematics

i would go into it so long as it doesn't come across as me being authoritative and monopolising this channel

or taking charge of a help request in a topic i am not proficient in

[thorn jay]

Stone duality and its a big part of topos theory right

subobject classifiers

[thorn jay]

"introduction mathematics"

I honestly just make the homework as a formality

[knotty badger]

i mean more "indexed-fibred duality", i.e. grothendieck's relative point of view

[thorn jay]

i am too logic pilled 💔

[south patrol]

Idk how much I would equate that with his relative pov but ye

[knotty badger]

afaik the relative pov is about considering objects "parametrised by" another object, via slice categories and stuff?

[south patrol]

I mean sure but also just emphasis on considering maps between stuff and things, which is important in the treatment/development of ag. like the idea of thinking of a map a-> b as b parametrising stuff feels kinda older / more basic

[knotty badger]

oh, interesting

is this like focusing on properties of morphisms rather than properties of objects?

[south patrol]

I would say so yes, with the knowledge that you can always take the target terminal if you want something truly absolute

[knotty badger]

mhm mhm

i suppose i'm familiar with the more old-fashioned notion of the relative pov, which is the "fibred" half of indexed-fibred duality

[south patrol]

Like whenever you say a k-scheme you can think of that as the object X -> Spec k and things are often defined in terms of that whole structure

Ye ig there is also a separate thing that like idk

(Un)straightening is very cool and deep imo

[knotty badger]

oh i've heard that's kinda like indexed-fibred duality in an infinity-categorical context?

[thorn jay]

unstraightening? why so queerious

the internet is doing terrible things for me

[south patrol]

Yeah sort of but I am being ahistorical with terminology - point being it goes back to something older (using fibrations (of 1-cats) to encode pseudofunctors

[knotty badger]

right right, grothendieck construction is about going from indexed pov to fibred pov

[south patrol]

And one can soup this up to infinity and beyond

Yes this is what I mean, exactly

[knotty badger]

phew, i was worried i was taking charge with a help request on a topic i was not proficient in

[glad osprey]

lol, why are you taking passive aggressive stabs at the mods? If you feel you have been mistreated, this is not a very sensible response

[knotty badger]

what would you suggest i do?

[glad osprey]

I dunno, I don't know the specifics well enough

[knotty badger]

well, until i get clarification, i'm gonna have to be extra cautious

[glad osprey]

but you're asking to be banned if you continue like this

[knotty badger]

oh?

[glad osprey]

I'm not saying the warnings were warranted, but you're clearly being antagonistic. I know you have had bad experiences with people here, who have been elitist and dismissive. But starting to call other people's opinions "algebra brainrot" and stuff isn't a solution. I don't wanna take sides, because I think both you and people you interacted with have said things you shouldn't have. It's like every interaction you have is colored by your previous interaction, so each interaction gets worse and worse

[knotty badger]

well, the way i see it

either i've been mistreated, or my calibration for detecting the complaints is off

if it's the latter, then it's actually worth me double-checking so i can fix my calibration

if it's the former, i'll get banned anyway

[tardy hedge]

unless stuff was deleted idk who was being rude in the previous messages

[south patrol]

I mean yeah I am not really sure of what you are trying to achieve with constantly saying you hope you aren't monopolising since it just seems you are trying to complain at the other mods until someone is stricter or something

[tardy hedge]

but im just a bystander so who cares

[knotty badger]

well, it could help fix my calibration

[south patrol]

What do you mean by that

[knotty badger]

it may be the case that i am monopolising but unable to recognise that

hence, by explicitly asking i can check for those scenarios

[south patrol]

It comes across as you just memeing on it rather than actually checking

[glad osprey]

the context isn't really in this channel, it goes way back AFAIK

[knotty badger]

well, it kinda serves a dual purpose

if my calibration is off then it'll help

but if i'm being mistreated i'll get banned anyway

might as well make a little fun of it

[knotty badger]

not like there’s much else I can do if the mods refuse to communicate

if they wanna ban me they’ll ban me, nothing I can do about that

[wraith cargo]

I think it might be good for you to consciously try to tone it down internally a bit and if the mods still aren't happy they'll be glad to let you know trust me
With the stuff you're saying it just feels like you're being actively salty about the mods complaining

[south patrol]

I thought other mods had communicated

(I don't really want to say much esp as I wasnt there for earlier stuff and don't know what has been said)

[knotty badger]

I’ve DMed modmail and haven’t received a reply

I suppose I’ll wait

But yeah no need to drag yourself into this potato

[knotty badger]

If I’m being mistreated they will ban me no matter what

Acting compliant won’t solve anything

[wraith cargo]

Genuinely take a second to be at least a little bit self aware about the shit you're saying
Because there is no Grand conspiracy against you

[elfin wraith]

I also don’t have the full context for this, but just as a general point, people aren’t out to get you pseudo, and yet you often come across as very defensive and antagonising. I know that you’ve spoken about why before, I’m sorry you’ve had some shitty interactions with mathematicians in the past, but if you constantly act like you’re being persecuted and be kinda defensive I’m not sure you’ll have many better experiences

[knotty badger]

It doesn’t have to be a grand conspiracy

There have been people who’ve wanted me gone ever since I joined the server, is all

[knotty badger]

I was bullied and belittled a fair amount when I joined the server

It’s lessened recently thankfully

[elfin wraith]

I can’t speak to that, but it doesn’t appear to be the case anymore and continually being so defensive around other people will just negatively affect any other interactions you may have here

I really don’t care to get into anything, but FWIW, I think you have some really interesting perspectives on stuff, and I really haven’t seen any of this persecution, but I have seen a lot of defensiveness on your part which can be a bit much, so idk do with that what you will

[knotty badger]

I do try gradually to open up and be vulnerable

And then stuff like this happens

It’s ok, you don’t have to believe me or take sides or whatever

I just get tired of being gaslit into thinking that I alone am the problem

[topaz solar]

Can the pitying and complaining and such be done in a non-topic channel

[knotty badger]

Right, it seems that I monopolised the discussion here

[delicate orchid]

I'm gonna side with pseudo here. If this was an important channel like #alg-top-geo-top you might have a point but just let her vent

where else is this discussion realistically gonna happen? #serious-discussion ?

[thorn jay]

basically no one's active here at this time

[topaz solar]

Fair

[elfin wraith]

Groups rings and fields is just alg-chill 2

[knotty badger]

Damn, I wasn’t expecting that from you wew

[delicate orchid]

it's why it's my favourite advanced channel

[thorn jay]

you get it 🤝

though i suppose a name change would be rather hard to get through

[elfin wraith]

It’s essentially a private server for like 10 of us that people occasionally ask group theory problems in

[tardy hedge]

Lol

[thorn jay]

wouldnt have it any other way tbh

builds community or smt

[knotty badger]

i do like the community here, my grievances aside

i think this was where i learned why the first iso theorem was cool

it was a rare event of me actually enjoying algebra

[thorn jay]

i mean if you enjoy monads youre technically alr enjoying algebra

[elfin wraith]

Category theory is algebra no matter what those weird foundations folks will tell you

[merry harness]

algebra: substitution math

[south patrol]

Controversial lol

[knotty badger]

I think it took me a while before i understood this

[tardy hedge]

Ya

This is this and that is also that

[knotty badger]

But I don’t feel as put off by it now

[thorn jay]

no fr thats basically what a deductively closed set of identities is

skull

[karmic moat]

The only categories that matter are Ab CRing R-mod and Sch and all it’s derviatives

Everything else is just a hoax to publish papers no one understands

[knotty badger]

What’s a derviative?

[thorn jay]

insane take

[south patrol]

Bro also made two grammatical mistakes in one word

[thorn jay]

derviative

[karmic moat]

I just mean stuff like chain complexes of modules and stuff

[south patrol]

Hopefully anamono is native english speaker or I feel bad

[knotty badger]

interesting, didn’t realise

[karmic moat]

English is my 10th language you evil person

[wraith cargo]

me when Bun_G

[proud vigil]

sorry discord latex is too frustrating i hope this is readable :(

[knotty badger]

I had some friends that went bungee jumping

[proud vigil]

would love any ideas!! but its ok if this is just a bad approach

[delicate orchid]

you said R-mod three times

[knotty badger]

Oh maths rime

[south patrol]

Sch comes from CRing and set anyway

[karmic moat]

TRUE

[south patrol]

Etc

[thorn jay]

CRING????????????????????????

[delicate orchid]

think about it. But not too much

[wraith cargo]

wait I thought this was discussy

[elfin wraith]

Hilb is controversial but I am going to allow it

[thorn jay]

i should add a few more question marks to make the joke more obvious

guys what about the category of congruences of an algebra 🥲 😢

[delicate orchid]

category of E_n rings. call that Cringe

[knotty badger]

This seems promising

The group you’re considering is the group of affine maps mod m, I believe

[thorn jay]

yee, translation + scaling on Z/mZ

[knotty badger]

And then you’re considering the subgroup generated by a particular pair (a, c), and letting that act on Z/mZ

Mhm

[thorn jay]

I KEEP WRITING Z_m

[knotty badger]

Orbit-stabiliser sounds like exactly what you’d want to use here

Whoops same

But yeah I do see your point about computing the order…

[thorn jay]

I can appreciate the Top erasure though

[south patrol]

Why

[knotty badger]

Ok actually here’s one way to think about it

The group action is faithful

[delicate orchid]

CRingE

[knotty badger]

I.e. if an affine map acts trivially, the map itself must be (1, 0)

So now, suppose (a, c) goes through all of Z/mZ

So that the orbit size is m

Then it must act via a cyclic permutation on Z/mZ, right? An m-cycle specifically

And m-cycles have order m

[karmic moat]

Tbf this is Potato’s preferred notation, esp when m is a prime

[knotty badger]

But since the group action is faithful, this implies the original group element has order m

So, if (a, c) works, then the stabiliser must have order 1

[proud vigil]

oo that makes sense

[knotty badger]

Trying to see if the reverse is true

[thorn jay]

ive been doing too much local property stuff lately to see Z_m as anything but some Z localised at the maximal ideal m

[knotty badger]

Ok so, any affine map acts via a permutation

Hm…

The issue is I think it’s possible to have stabiliser order 1 without having an m-cycle

E.g. something like (12)(34)

So it’s a necessary but not sufficient condition

[karmic moat]

No erasure here just take the forgetful functor Sch -> Top

[thorn jay]

darn

[knotty badger]

I’m a little too eepy to continue but hopefully someone else can help further

[thorn jay]

use m to denote a prime number

its a shame that radboud doesnt provide an elementary number theory course this year, i wouldve totally used m for prime numbers in the exam

[elfin wraith]

Literally the first person to ever complain that they can’t take ENT

[thorn jay]

no its shame only because of this reason

yk whats weird theyve got a shelf with course textbooks which i assume are for bachelor/masters courses with an algebraic K-theory book there?

they dont have that

inconsistent ahh

[south patrol]

Fair

[south patrol]

Naur

[south patrol]

Ragebaited

[karmic moat]

Oh i just realized i said it’s instead of its

Oops

Didn’t realize tbh

Lemme think of an excuse

[thorn jay]

bro is still thinking

[karmic moat]

Nope i got a burger and im eating it

I’m onto burger and better things

And I’m watching family guy clips

[thorn jay]

bro better be talking about hypercommutators

[tardy hedge]

Yaaas i love burger yaaas

I also love saying yaaasss

[karmic moat]

No I’m talking about the delicious burger I just ate

With like 9 toppings on it

And a bunch of cajun fries

Omfg 🤤

[tardy hedge]

Omg incredible

Let me have some 🤪

[karmic moat]

❌

[proud vigil]

is there any special property that governs when an element x of a (probably finite?) ring has some integer exponent n such that x^n = x?

specifically i'm trying to figure out when x^m = x modulo n, and I can't tell if an m generally exists / if there's any theorems about rings that describe this (this could be extremely basic, I just don't know anything about rings)

[south patrol]

also just like should be "their"

But that's ok

[karmic moat]

Oh true

[proud vigil]

ugh is this not a basic thing

for some reason i cant find things online

in my head i feel like u could maybe create some sort of "subgroup" structure? but im thinking about everything in terms of Z/nZ and that probably isn't generalizable 😭

[covert cliff]

Hello

Im reading this book and they first mention that SU(2) has a unique irreducible representation for every dimension and they call it the spin representation. They define the representation on the space H_j which is the space of degree 2j homogenous polynomials in two complex variables and is defiend as: let g in SU(2) and let f in H_j then p(g)(f) = f \circ g^{-1}

now why did they chose the space of 2j homogenous polynomials?

[crystal vale]

if TN/N is sylow p-subgroup of S, and S itself p-subgroup of G/N, isn't directly comes that TN/N = S?

[astral ivy]

How do I prove closure of a group in non-trivial examples?

[rapid cave]

It depends a lot on what are you working with

[astral ivy]

for example: Let $G = \mathbb{Q}\setminus\left{0\right}$ and $p\ast q = \left{\begin{matrix}
p\cdot q & p > 0 \
\frac{p}{q} & p < 0 \
\end{matrix}\right.$

[cloud walrus]

danilojonic

[rapid cave]

you want to show closure? this is easy in this case since a product of rationals is rational

[astral ivy]

so its trivial?

[rapid cave]

for you it may not be trivial, for me it is

[astral ivy]

my prof said that theres no need to show closure if its trivial whatever that may mean

like if G = Q\{0, 1, 2} then it wouldn be trivial

in both cases 0 isnt taken into account so it doesnt matter

but what about 1 and 2

for example

how would I show it?

[rapid cave]

you mean 1/2 * 4

[astral ivy]

it wouldnt be in group?

but 1/2 wasnt in a group to begin with

because elements p = 1 and q = 2 arent

[rapid cave]

its fine to use 1/2

[astral ivy]

okay so the main point is to find a counterexample, is that enough?

[rapid cave]

yes

[astral ivy]

what if we had an opposite scenario? Example:

$G = \mathbb{R}\setminus\left{-1\right}$ and $x\ast y = x+y+xy$. How would I demonstrate closure here? Would I go straight for $x = -1$ or $y = -1$ and try to find that it doesnt hold?

[cloud walrus]

danilojonic

[astral ivy]

this is a commutative group so yes

but I have to show the closure

everything else is easy for me

just closure can be difficult

[rapid cave]

if x+y+xy = -1 then 1+x+y+xy=0

factor the left side as (1+x)(1+y)

one of x,y must be -1

so x+y+xy=-1 is impossible proving closure

[astral ivy]

ahh I see

[dim wagon]

would anyone be able to help me with the forward direction?, i’m quite stuck

i know a maximal ideal exist by zorn’s lemma and that if I R/I is a field if I is maximal

the proof probably will involve showing that (R/I)^n is isomorphic to (R/I)^m as vector space iff n=m

but i’m not sure how to get there

R/IR would just be trivial right? because IR=R

[tardy hedge]

what question, 24?

[dim wagon]

yes

[tardy hedge]

Why is IR = R?

[dim wagon]

i put 21 and 7 there because they might help give context

ah okay, i got confused i thought that ideals must contain identity but clearly that’s not true

[tardy hedge]

yeah if an ideal does contain identity (or any unit) then its the whole ring

[dim wagon]

oh instead I=IR right?

[tardy hedge]

Yeah

[dim wagon]

ah okay, i think i have it thank you

[crystal vale]

do i have to solve this type of problems?

[elfin wraith]

No one has to do group theory

[twilit wraith]

https://tenor.com/view/family-feud-survey-says-strike-buzzer-wrong-answer-gif-16312959

[elfin wraith]

My point is like, how could anyone possibly answer the question of “do I have to do this exercise”

No you don’t have to, but it will probably help your understanding

[twilit wraith]

i mean im not sure thats what they were asking

i think they were wondering if, were they to take a course involving group theory, would they have to answer questions like that, assuming they wanted a good grade?

and to that the answer is yes

[crystal vale]

i mean they seems dry questions, leave it

[twilit wraith]

the questions on the picture you posted seem dry?

i guess they can come off that way if you havent taken group theory yet, but i bet theyre more interesting than it seems

[crystal vale]

how?

[twilit wraith]

Have you done some group theory? Whether or not im able to explain why the problems could be interesting kinda depends on how much I can actually explain

The part b question has a cool result bc of the universal property you find

[crystal vale]

yes i done undergraduate group theory

[twilit wraith]

Ah then good

I guess if you dont really care about algebra then the questions would be dry, but if you do, you get some cool results like seemingly finding the minimal normal subgroup of P

Which you do in part b

The nontrivial automorphism problem is also cool

I guess just the fact that you're finding so many different properties of the quotient group that the problem cares about is what's cool to me at least

[crystal vale]

oh

[velvet hull]

oh, C is actually an interesting result

you are showing that any minimal generating set for a finite p group has the same size

which is not true in general for groups or finite groups

[white oxide]

Could I have a hint for the reverse inclusion of ii? I know that if p is in Supp(A/a), then S^{-1}A/S^{-1}a is nonzero, that is S^{-1}a is a proper ideal of S^{-1}A. I want to show that S^{-1}a is prime however. This happens if and only if S^{-1}A/S^{-1}a is an integral domain, but I don't know if this is the right track

Nvm I got it

a^e = S^{-1}a \neq (1), so that a n (A \ p) = \varnothing, so a is contained in p

[white oxide]

Does anybody know if this sum is a direct sum

[velvet hull]

likely not since they wrote it that way

[white oxide]

Over any arbitrary indexing set?

[south patrol]

Ye

[white oxide]

Also I'm confused when authors write \bigoplus_i M_i I often don't know whether to interpret it as a direct sum (all but finitely many summands are zero) or as coordinates (x_1, \dots, x_n, \dots)

I guess they're isomorphic in that case

[velvet hull]

same thing

with the caveat that whenever oplus shows up, you need to implicitly understand its the set of all vectors with finite support

[white oxide]

So how do I interpret \sum M_i? Could it be possibly the module of all infinite sums?

I'm assuming that localization does not commute with arbitrary sums in general

[karmic moat]

it does

[next obsidian]

It does not

[karmic moat]

localization can be written as a tensor product, the latter commutes with arbitrary sums

no?

[next obsidian]

Sum ≠ direct sum

[karmic moat]

oh

oops

[next obsidian]

Or well

Wait

No I agree sorry

Product is not a sum

[karmic moat]

🫂

[next obsidian]

I think for this exercise you want a direct sum

But I think the result might still be true with direct products

Because any direct product is a colimit of finite ones

[white oxide]

I'm sorry I'm lost. If as in q \subseteq r(q) = p, then s is not in q, for otherwise it would be in p. Therefore, a^n in q for some n, but how does this show that a in q?

Oh I see, s^n is not in q for all n (because otherwise s would be in p), so we must have a in q

[south patrol]

Pretty sure this is false, I mean even in sets or Ab

Maybe I am misinterpreting what you mean

But like you want it to be the limit of the finite ones rather than colim

[thorn jay]

you get an inverse system indexed by the finite subsets of your indexing set

[south patrol]

Indeed

[thorn jay]

yes

[next obsidian]

Wait yeah

I’m being real dumb

[thorn jay]

or in words: every zero divisor of R/q must be nilpotent, so as s is not nilpotent (it is not in the radical p) a must be zero in R/q, i.e. a ∈ q

[south patrol]

Naur

[velvet hull]

let V and W be f-d vector spaces over a field F, is there a nice way of showing dim(V ox W) = dim(V) dim(W) without having to explicitly choose a basis?

[next obsidian]

Follows by universal property I think

But this is kind of like picking a basis without “picking” a basis

[velvet hull]

so your idea is to use the fact that dim(V) = dim(V*)?

[next obsidian]

Not really

If V is of dimension n its a free module on n things

Similar for W

[velvet hull]

oh, right

[next obsidian]

And this is a mapping out thing

[velvet hull]

yeah apply adjoint twice

yeah that's good

thanks

[next obsidian]

Yurr

[hidden wind]

i’m having a moment

i feel that seeing m = (x1-a1,…,xn-an) is maximal in k[x1,…,xn] by noting that clearly k[x1,…,xn]/m = k, is too simple

for a1,…,an in k

[velvet hull]

it works

[hidden wind]

could someone challenge me on that “clearly” ?

[quiet pelican]

You justify it by the first isomorphism theorem on “x_i -> a_i”

[velvet hull]

well the xi's gets sent to constants in the quotient map

and it contains k already

so it's k

[quiet pelican]

(And that is the kernel because reducing modulo that ideal, you can turn f(x) into f(a))

[hidden wind]

ok yeah it works but it still feels too simple, this comes with the consequence that a polynomial vanishes at (a1,…, an) iff it’s a linear combination in the xi-ai and that’s not at all clear to me 😭

[quiet pelican]

@hidden wind try doing this manually for a couple polynomials to get a feel

[hidden wind]

i did already

maybe i should do some more, thanks

[quiet pelican]

More formally it should look something like err
f(x) - f(x_1, …, x_n-1, a_n) = sum g_i(x_1, …, x_n-1)(x_n^i - a_n^i)
And that latter term is divisible by x_n - a_n
So then done by induction

[thorn jay]

what you could also do is write V and W as direct sums of the vector space k, and then use the distributivity property and the fact that k ⊗ k = k

[velvet hull]

ok yea that also works

that feels kinda hacky though

[south patrol]

Nah it is clear

[quiet pelican]

That’s essentially choosing a basis without saying you’re choosing a basis

[thorn jay]

i am aware

[south patrol]

Ye lol

This is the nice algebraic way

Imo

[velvet hull]

yeah but it hides the choice of a basis well

whatever Im typing out the adjoint proof as we speak

[thorn jay]

but this really is a choose a basis question

:P

[south patrol]

But also like I mean you need to show that tensor product of basis vectors give a basis

Etc

[hidden wind]

wait why the ^i here

[south patrol]

Idk I think enpeace's is best lel

[quiet pelican]

i is what we’re summing over (it ranges from 1 to the x_n degree of f)

[thorn jay]

tyty

[topaz solar]

Factors work too, not just sums

In the n=1 case, it says you can factor out (x1-a1), then you get sums of those since they indeed vanish

[hidden wind]

the n=1 case is clear to me by polynomial division

[topaz solar]

Yeyeye

But yeah I mean, if you evaluate it at the point, along any line you have a 1d thing so can’t be too different

Vaguely

[hidden wind]

hmm

i feel maybe there is an argument by partially evaluating k[x1,…,xn] -> k[x1] but i dont see it

[quiet pelican]

You can induct with the n = 1 case, using k[x1, …, x_{n+1}] -> k[x1, …, xn]

[topaz solar]

There’s an actual argument by noether normalization too

[hidden wind]

i forget, how is xn^i - an^i divisible by xn - an again?

[topaz solar]

Difference of whatevers

There’s like a specific expression but uhhhh run the division algorithm

[quiet pelican]

It’s like x^i - a^i = (x - a)(x^{i-1} + ax^{i-2} + … + a^{i-1})

[topaz solar]

I can never remember the statement but its vibes are like, you can’t send the variables to nowhereseville, gotta stay in k

[hidden wind]

ah right the binomial theorem ig

[topaz solar]

Because integral

[topaz solar]

Almost

[hidden wind]

guh

[topaz solar]

Ultimately it’s like

[twilit wraith]

You can also use the factor theorem

[topaz solar]

Run the synthetic division trick

[hidden wind]

seems to be known simply as “difference of powers”

[hidden wind]

right

[twilit wraith]

Since a is a root of x^i- a^i

[topaz solar]

You can also prove it model theoretically easily in the uncountable case by like drawing the diagram and throwing an ultrafilter iirc

[languid trellis]

just what i was thinking

[topaz solar]

I don’t remember the exact shape of it but it be like that sometimes

I really should know it offhand but just kinda push in 1/(x-s) for each s or what have you

[hidden wind]

i love 1/(1-x), idk about throwing an s in there

[topaz solar]

Well I mean, if you could for every s in k

As in, there was no contradiction with x-s -> 0

[hidden wind]

ok i think i got it thanks so much 💖

actually it turns out one can be very explicit here, which is just what i wanted mwahaha

[astral ivy]

my literature defined normal subgroup as:

"A subgroup $\mathbb{H}$ of a group $\mathbb{G}$ is normal if it is invariant under the action of the group $Inn\mathbb{G}(\mathbb{H}) = \mathbb{H}$, i.e. $\sigma_a(\mathbb{H}) = \mathbb{H}$ for all $\sigma_a\in Inn\mathbb{G}$. The fact that a subgroup $\mathbb{H}$ is normal in a group $\mathbb{G}$ is denoted by $\mathbb{H} \triangleleft \mathbb{G}$."

I don't understand this definition at all. What does being "invariant" mean? What exactly is the action of group $Inn{\mathbb{G}}$? I also didn't understand what is inner automorphism.

[cloud walrus]

danilojonic

[thorn jay]

I guess it's equivalent, but if the book is supposed to be introductory then this is an ass way to introduce it lmao

[astral ivy]

what is $\sigma_a(\mathbb{H}) = \mathbb{H}$ for all $\sigma_a\in Inn\mathbb{G}$ tho? Is that some oddly denoted element?

[cloud walrus]

danilojonic

[thorn jay]

what book is this?

[astral ivy]

it's a script my prof wrote

[thorn jay]

has he introduced group actions and automorphisms?

[velvet hull]

what a terrible way to teach normal subgroups lmao

[thorn jay]

im saying 😭

[astral ivy]

he did but it was very brief. I don't have anything to say about automorphism other than it being a bijective endomorphism

[thorn jay]

is this introductory group theory?

or are you supposed to be familiar with some algebr

a

[astral ivy]

introductory abstract algebra

[thorn jay]

wtf

[astral ivy]

done on 2nd year, right after linear algebra

[astral ivy]

I guess that's why average completition time for my faculty is 8 years...

[thorn jay]

okay, a better definition of normal subgroup is a subgroup H < G such that, for all g in G, we have gH = Hg

[astral ivy]

I agree, I know for that definition as well as the conjugate one

[thorn jay]

well the definitions are equivalent as s_a(H) = aHa^-1

but I just think it's strange

[velvet hull]

teaching normal subgroups using the action of Inn(G) on Sub(G) looks terrible however you cut it

[astral ivy]

but, I heard somewhere that $aHa^{-1} \neq aha^{-1}$ and same goes for the seperate left/right cosets, can anybody explain why? I know one is a set and the other one is element but why both don't hold true always?

[cloud walrus]

danilojonic

[thorn jay]

it may have context but in general this is not the right way to see normal subgroups

[velvet hull]

aHa-1 is a coset/ subgroup, while aha-1 is an element of the group

they're just incomparable

[astral ivy]

I know that

[thorn jay]

what do you mean by "why dont both hold true"

[astral ivy]

Here's the entire page where that's introduced

[astral ivy]

I need to find a youtube tutorial where I found that, will be back

[velvet hull]

it's because H = gHg-1 implies that g only needs to permute the elements of H

while h = ghg-1 implies that g needs to fix every element of H exactly

which is a much, much stronger condition

[thorn jay]

N_G(H) vs C_G(H) ahh

[proud vigil]

omg i just realized why theyre called centralizers

oops

[velvet hull]

lol

[astral ivy]

"Whole cosets (left and right) are equal but individual elements from them may not correspond"

[thorn jay]

yeah okay

that's because the set gHg^-1, while maybe being H, the map h -> ghg^-1 may permute the elements of H

[astral ivy]

if a group is abelian, then this will hold true for individual elements as well right?

[thorn jay]

yeah that's the definition

G is abelian iff the center of G is G

[astral ivy]

I like how you described it but I simply can't visualize it. Can you give me a simple example?

[velvet hull]

how many different types of groups do you know of right now?

[astral ivy]

integers, dihedrals, symmetric

[velvet hull]

ok, good

[astral ivy]

I knew quarternions but I don't know them well enough

[velvet hull]

in S3, the subgroup given by H = {e, (123), (132)} is a normal subgroup

but (12) (123) (12)-1 = (132)

so although (12)H(12)-1 = H, (12)h(12)-1 does not have to be equal to h

[astral ivy]

what is the inverse of (1, 2)? its the one we multiply with to get the identity (1, 2, 3) but idk how to 'reverse engineer' it?

[velvet hull]

the inverse of (12) is (12)

[astral ivy]

ahh yeah

it just swaps them

[astral ivy]

i got it! that conjugate can give us anything in H but not necessarily the element from H we used in the conjugate. thank you

[sage sand]

could i check that this question probably assumes that R is commutative?

otherwise the thing we are trying to might not even be R-modules correct?

[velvet hull]

the question is specifying that M and N are left and right R-modules, so probably not

[sage sand]

hmm but generally the tensor product might not be an R-module if R is not commutative right

[velvet hull]

it can still be well-defined, you just have to keep track of the parity

[glad osprey]

Wouldn't M and N need to be bimodules for the tensor prod to be a module?

[thorn jay]

why can't you do
r * (n (x) m) = n (x) r * m
and
(n (x) m) * r = n * r (x) m

ah, this forces sr to be the same action as rs I think

lol

[south patrol]

The tensor product gives you an abelian group though

And you can ask for those to be iso

[thorn jay]

is there any reason not to treat it as a bimodule like this?

I suppose it doesn't really help with any universal property stuff

eh

[south patrol]

Wdym

[velvet hull]

hint - ||syl_p(N_G(P)) = {P}||

you don't need to do the last part

it is a p subgroup of N_G(P)

so by sylow 2 we are done

huh

let me pull up dummit and foote

no, I need to see the proof

because if you use the right proof for sylow 2, you should be able to get this as a corollary

hold on im getting the book

wait, this is in D&F

why can't you use it

so what do you have

[formal summit]

Not sure why I'm stuck on this problem but I need to show that the order of the p-torsion subgroup of an (additive) abelian group is p. Any hints?

[thorn jay]

that's not true though

the p-torsion subgroup of Z/p^2Z is of order p^2

[formal summit]

Ah true. The original problem was showing that if p divides |G|, then the number of pth roots of each x ∈ G is either 0 or p. Which I thought reduced to the same problem as above?

[thorn jay]

hmm

[formal summit]

(Via considering the kernel of the homomorphism f: G-> G that sends x to px)

[thorn jay]

well, that's not the p-torsion subgroup, because that looks at prime powers of p

[formal summit]

Is p-torsion subgroup of G not just g \in G such that pg = 0?

[thorn jay]

no, not as it stands on wikipedia but I suppose conventions may differ

[formal summit]

Ah I see

[thorn jay]

anyhow, consider the abelian group Z/pZ x ... x Z/pZ, p+1 many times

as in there are p+1 copies of Z/pZ

[formal summit]

Yes

[thorn jay]

then any element (0, ..., 1, ..., 0) has order p, so there are at least p+1 pth roots of (0, ..., 0)

[formal summit]

Oh hm

[thorn jay]

what you can show, however, this set must be a subgroup of the p-sylow subgroup of your abelian group, so it must have order p^k for some k

[formal summit]

Right I see, that's what I thought originally. Thanks for the help

[thorn jay]

(i dont even know if p-sylow is defined for arbitrary abelian groups)

[formal summit]

They should be for any finite abelian group, I think?

[thorn jay]

yes

and profinite too

[formal summit]

Ah yes

[thorn jay]

but I think you can just take the p-sylow subgroup of an abelian group just to be the p-power torsion subgroup

[thorn jay]

with like generalized integers or something

profinite integers is the name

[chilly ocean]

Just for curiosity’s sake, is there any paper or book that defines groups as fundamental groups of spaces? And then homomorphisms as certain kinds of functions from one fundamental group to another(afaik, not every homomorphism of groups is in the image of the fundamental group functor, so they can’t be defined as the image of certain continuous functions under that)? It might be a little awkward to define some stuff but it seems like a potentially interesting read. And I don’t just mean defines it as such, then treats it like usual, but actually uses that for intuition and proving some stuff about groups?

[earnest delta]

82 so here my thoughts are

|Z|=1 means X^2+Y^2=+-1

elements would be 1,-1,i,-i which forms cyclic group

[quiet pelican]

What about the element (1+i)/sqrt(2)

Or, actually better example, e^(2 pi i sqrt(2))?

[earnest delta]

ahha

it would be B

[quiet pelican]

Do you mean 2? If so, yes

[earnest delta]

yes

thanks

[wraith cargo]

??????????

were you thinking of Z/pZ x Z/pZ?

[south patrol]

This is correct

I would say "p-power torsion" or similar for the other concept (I.e. p^n g = 0 for some n). The corresponding subgroup of an abelian group A is often written A[p^oo]

[thorn jay]

wikipedia you fucking ass 😭

the only time double checking myself made me wrong

[south patrol]

Ah lol

Though they do say p power ayy

Gotem.

I might edit this lol

[thorn jay]

pls do

[delicate orchid]

if someone says "p-torsion subgroup" I think of the stuff that vanishes when you tensor with F_p

[crystal vale]

i want an counterexample of group of order 56, such that its 8 sylow subgroup is not unique

I think Z/4Z × D7 works

[white crescent]

hey, I assume this question falls in here.

currently started studying abstract algebra literally the first session. I read on that lie algebras are defined on finite dimensions and wondered, is it because differentability breaks down at infinite dimensions?

[sonic coral]

does this have a normal sylow 7

it can’t have both

[white crescent]

also secondary question, if the conditions for something to be lie algebra is to be differentiable and for its inverse to be as well(i.e 1/x), does that mean what is considered in lie algebra can never be 0? afterall otherwise it will be divided by 0

[south patrol]

You can have Lie algebras in infinite dimensions just fine

[white crescent]

I see, possibly a misreading on my part then

[south patrol]

This doesn't really make sense to me. Differentiabikity doesn't enter the definition of a Lie algebra - it enters the definition of the Lie algebra of a Lie group or similar

[glad osprey]

Yeah, it sounds like you're talking about smooth maps or something when you say it has to be differentiable and its inverse has to be too. But this is only tangentially related to Lie algebras

[white crescent]

I see, thank you, its my first encounter with this mateiral so sorry for the waste of time

[glad osprey]

If you learned something then it's not a waste of time

[tidal schooner]

Pun intended?

[versed dagger]

dihedral group of order 56

[glad osprey]

oh, uh... yes

[tardy hedge]

😍🥰

[astral ivy]

Hello, can anyone tell me the difference between elementary and normal form of abelian group?

In example I had, $\mathbb{G} \cong \mathbb{Z}_8 \times \mathbb{Z}_12 \times \mathbb{Z}_30 \times \mathbb{Z}_75 \times$

So $\mathbb{Z}_8 = \mathbb{Z}_2^3$, $\mathbb{Z}_12 = \mathbb{Z}_2^3 \times \mathbb{Z}_3$, $\mathbb{Z}_30 = \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_5$ and $\mathbb{Z}_75 = \mathbb{Z}_3 \times \mathbb{Z}_5^2$.

This then gives me $p_2 = 3, 2, 1, 0$, $p_3 = 0, 1, 1, 1$ and $p_5 = 0, 0, 1, 2$.
From here it's pretty straight forward to determine the elementary form (rewrite the above formula and just regroup orders with each other). But how does normal form differ? In this case I think it'll be the same, just put $p_2 = 3, 2, 1$, $p_3 = 1, 1, 1$ and $p_5 = 2, 1$ in descending order. So in this particular case it's no different from elementary form (except the order will slightly be different), but when will it actually differ?

[cloud walrus]

danilojonic

[quiet pelican]

C_8 is not equal to C_2^3

G = Z_8 x (Z_4 x Z_3) x (Z_2 x Z_3 x Z_5) x (Z_3 x Z_25) = (Z_2 x Z_4 x Z_8) x Z_3^3 x (Z_5 x Z_25)

[astral ivy]

but I need all orders to be primes. Z_8, Z_4, Z_25 aren't primes...

[quiet pelican]

And the normal form is just group everything up nicely
So it’s err Z_6 x Z_60 x Z_600 I think

[quiet pelican]

You can’t do that
You need them to be prime powers, not primes

[astral ivy]

I mean they are prime powers no?

2 and 3 are prime

[rapid cave]

also 8, 4, 25 are prime powers

so no need to do anything

[astral ivy]

is my elementary form good or will it look different?

and normal

[rapid cave]

I don't understand what you did

what are p1,p2,..

[astral ivy]

Primary components. Subgroups of G whose orders are powers of the given prime number. So p_2 references all powers of 2, p_3 all powers of 3 etc. It's not necessary but I guess it's a bit easier to see what's going on when making the elementary or normal form.

[glad osprey]

Maybe you meant $\mathbb{Z}8 = \mathbb{Z}{2^3}$? Which is very different from $\mathbb{Z} = \mathbb{Z}_2^3$

[cloud walrus]

sheddow

[astral ivy]

Wow yes!

Latex abuse of notation

I didn't even notice it came out like that

[rapid cave]

@glad osprey you can use \bZ

its shorter then \mathbb{Z}

[rapid cave]

Notice the difference \bZ_{1}{2} and \bZ_12, the {} are important

[glad osprey]

I was so confused to begin with, because I thought $\bZ_12$ meant two copies of $\bZ_1$ or something

[cloud walrus]

sheddow

[astral ivy]

$\mathbb{G} \cong \mathbb{Z}8 \times \mathbb{Z}{12} \times \mathbb{Z}{30} \times \mathbb{Z}{75}$

So $\mathbb{Z}8 = \mathbb{Z}{2^3}$, $\mathbb{Z}{12} = \mathbb{Z}{2^2} \times \mathbb{Z}3$, $\mathbb{Z}{30} = \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}5$ and $\mathbb{Z}{75} = \mathbb{Z}3 \times \mathbb{Z}{5^2}$.

Rewritten

[cloud walrus]

danilojonic

[rapid cave]

for you too @astral ivy

[astral ivy]

huh? For latex editor I usually use that's invalid syntax

weird

[rapid cave]

its a macro defined in this server

[astral ivy]

nice

[merry harness]

You can add it to preamble if you want to use it

[rapid cave]

anyway, elementary form will be grouping all the same groups together

[merry harness]

|| or typst... It's just ZZ 🙃 ||

[astral ivy]

also, regarding this task (and in general)

How do I find number of elements of certain order? Like here specifically it's asked for number of elements of order 120.

$120=2^3\cdot 3\cdot 5$ is the way I'd start it to make it more simple

[cloud walrus]

danilojonic

[rapid cave]

you use the fact is that if $g=(g_1,...,g_k)$ is an element of $\bZ_{n_1} \times \bZ_{n_2} \times \dots \bZ_{n_k}$ then the order of $g$ is the lcm of the orders of $g_i$

[cloud walrus]

ExpertEsquieESQUIE

[rapid cave]

$(\bZ_{2} \times \bZ_{4} \times \bZ_{8}) \times \bZ_{3}^{3} \times (\bZ_{5} \times \bZ_{25})$

[cloud walrus]

ExpertEsquieESQUIE

[rapid cave]

you sort of need to create the prime power order

so if you look at $g = (g_1,g_2,g_3)$ then $g_2$ must be of order 3 in $\bZ_{3}^{3}$

[cloud walrus]

ExpertEsquieESQUIE

[astral ivy]

so $ord(g) = lcm(ord(g_2), ord(g_3), ord(g_5))$ and since $120 = 8\cdot 3\cdot 5$ that means $ord(g) = lcm(8, 3, 5)$?

[cloud walrus]

danilojonic

[rapid cave]

of course the last thing is true

you assumed ord(g)=120

[cloud walrus]

mq

[rapid cave]

seems fine

same operations

A,B,C,M have the same module operations

[quiet pelican]

that works

[velvet hull]

yeah it's kind of trivial

the isomorphism is an R-module isomorphism

no, the isomorphism is an isomorphism of R-modules

[quiet pelican]

It also means that the R-action commutes with the isomorphism

[velvet hull]

for instance the map Z -> Z given by x -> 2x

is a group homomorphism

but is not a ring homomorphism

so when we talk about isomorphism we need to specify what structure is being preserved

in this case we specify that the R-module structure is being preserved

[astral ivy]

Show that $\b{H} = \left{\pi \in \b{S}_8: \pi(2)\in \left{2, 3\right}, \pi(3)\in \left{2, 3\right}, \pi(7) = 7 \right}$ subgroup of group $\b{S}_8$ and find order of $\b{H}$ as well as $a$ and $b$ from $\b{H}$ such that $\omega(a) = \omega(b) = 2$ and $\omega(ab) = 10$

[cloud walrus]

danilojonic

[quiet pelican]

What have you tried?
And where are you stuck?

[astral ivy]

Okay so I tried with $a, b\in \b{H}$ then $ab^{-1}\in \b{H}$ definition to check if it's a subgroup

[cloud walrus]

danilojonic

[astral ivy]

but $b \neq a$ and in first case $\left(2, 3\right)$, 2 and 3 are inverses to each other

[cloud walrus]

danilojonic

[astral ivy]

so my logic is kinda all around the place I'm confused

not to mention the single element of $7$, it's its own inverse but wouldn't that also get me the identity element

[cloud walrus]

danilojonic

[quiet pelican]

So H is the set of elements which map {2, 3} to {2, 3} and {7} to {7}
Can you see why that would (a) contain the identity (b) be closed under multiplication and (c) be closed under inverses?

[quiet pelican]

Tell me what S_8 is

[astral ivy]

symmetric group of permutations {1, 2,..., 8}

[quiet pelican]

So what is an element of S_8

[astral ivy]

everything $\pi$ describes really

[cloud walrus]

danilojonic

[quiet pelican]

I’m going to be irritating here and ask for something more explicit - I feel like this may be a problem here

[astral ivy]

mapping 7 to 7 is part of S_8 as well as mapping (2, 3)

is this what youre asking me?

[quiet pelican]

I think you’ve gone a step further than I was asking, but I also feel you understand enough that what I was asking is irrelevant so errr

[dull ginkgo]

Automorphism group moment

[quiet pelican]

Why are such elements closed under multiplication?

[astral ivy]

im not sure how to answer properly on this

[quiet pelican]

Suppose σ, τ \in H
What is σ(τ(7))?
What is the set {σ(τ(2)), σ(τ(3))}?

[astral ivy]

they all map into their original source

[quiet pelican]

?

[astral ivy]

σ(τ(7)) = 7 and {σ(τ(2)), σ(τ(3))} = {2, 3}

[quiet pelican]

Yup

Therefore στ \in H

Now do the same for inverses

[astral ivy]

that's interesting

what's bugging me is

sigma and tau are functions right

[quiet pelican]

Yes

Bijections from {1, 2, 3, 4, 5, 6, 7, 8} to itself

[astral ivy]

and $ab \in H$ are 2 elements from H whose product is in H

[cloud walrus]

danilojonic

[astral ivy]

so that's where I don't get it

how did you use functions instead of elements

[quiet pelican]

The elements of H are functions from {1, …, 8} to itself

[astral ivy]

ohhh

[dull ginkgo]

A lot of algebraic constructs are either families of functions or isomorphic to such

[quiet pelican]

Well they’re all isomorphic to families of functions for groups /pedantry

[dull ginkgo]

Many of the algebraic operations are actually compositions of those functions. It’s a perfectly well defined algebraic operation (it’s binary, associative, and even has an identity, the trivial map)

[astral ivy]

so for inverses it will just be

Suppose σ, σ^{-1} \in H

if σ(7) = 7 \in H, then σ(7)^{-1} = 7 \in H
?

actually no

I mixed up 2 definitions

[dull ginkgo]

Many of the functions on finite sets do nothing to most of the set

Which is why the simplest automorphisms (invertible maps) are transpositions, they only swap two elements

And don’t touch the other ones

[quiet pelican]

Continue

[astral ivy]

the one single we can check would be:

Suppose σ, τ^{-1} \in H
Then σ(τ(7)^{-1}) = σ(7) \in H

[quiet pelican]

How do you know t^-1(7) = 7?

[dull ginkgo]

What are you trying to check homie

[astral ivy]

similarly, Suppose σ, τ^{-1} \in H

Then {σ(τ(2)^{-1}), σ(τ(3)^{-1})} = {σ(2), σ(3)} \in H

[quiet pelican]

^

[astral ivy]

wouldn't group property of S_8 be that every element is its own inverse?

[quiet pelican]

7 doesn’t have an inverse
7 isn’t an element of S_8

[astral ivy]

how?

[quiet pelican]

Elements of S_8 are permutations of {1, …, 8}
Not elements of {1, …, 8}

[astral ivy]

okay so permutation σ(7) will be an element

[quiet pelican]

No, σ(7) is another element of {1, …, 8} (which is forced to be 7, as σ \in H)

[astral ivy]

and why wouldn't σ(7)^{-1} be its inverse? Or how do I show it has an inverse? Because for this to be a subgroup I need to show inverse

[quiet pelican]

σ(7)^-1 doesn’t make sense
Because σ(7) is not an element of a group

[astral ivy]

uhhh idk how to make the notation work sorry

[glad osprey]

I think you need to review what the elements of S_8 actually are, like mico asked you here

[astral ivy]

Okay so I revisited the definition of symmetric group:

S_8 = {f: X \rightarrow X | f being a bijective funciton and X being non-empty set}

[quiet pelican]

Where X = {1, …, 8}

[astral ivy]

yes!

[quiet pelican]

So try this

Step 1: what is the inverse of σ? Like, as a function, what does it do?

[astral ivy]

it just 'undoes' what function σ did

|X| = |Y| and σ is a bijective function.

σ: X -> Y.
σ^{-1}: Y -> X.

[quiet pelican]

Yup
Now what conditions does σ^-1 need to satisfy to be in H?
Can you prove it satisfies them?

[astral ivy]

Now what conditions does σ^-1 need to satisfy to be in H?
I have no idea

[quiet pelican]

So we need σ^-1(7) = 7 and similarly σ^-1({2, 3}) = {2, 3}

Can you prove that?

[astral ivy]

no, I have no idea

[astral ivy]

is it showing this but for inverse function this time

[quiet pelican]

Yes

[astral ivy]

but I don't need 2 inverse functions (?), I only need that if σ(7) in H, then σ^{-1}(7) and this last part is the one I show.

because σ() in H was already shown

[dull ginkgo]

@astral ivy what exactly are you trying to do homie

[glad osprey]

.

[astral ivy]

This is the task

[glad osprey]

You don't want to show that sigma(7) is in H, because sigma(7) is an integer, but H is a set of functions

[dull ginkgo]

Oooooh neat problem

[glad osprey]

Do you see the difference? We want to show that H is a subgroup of S_8. So we use the subgroup teat: assume f, g is in H, then show that fg is in H. Then assume that f is in H, then show that f^-1 is in H

[astral ivy]

Well by definition for any element (or function) from H I need to show closure as well as inverse.

for it to be a subgroup

[dull ginkgo]

You can also do this a cheeky way, in that each of the conditions pi(7) = 7 and transposing 2,3 is a subgroup

[glad osprey]

Yes, but you're mixing up functions and elements

[dull ginkgo]

The group is the permutations