groups-rings-fields

[south patrol]

if you take any element h of a group - say H - you can get a subgroup { e, h, h^2,h^3,...,...}

(if H is not finite / h not finite order you need h^-1 etc but yeah )

[oak grove]

oh, then you just include integers instead of nats

gotcha

[south patrol]

Yeah

[oak grove]

i see what you mean now

[south patrol]

Or define <g^7> as the intersection of all subgroups of G containing g^7,but i digress

Either way, we have a subgroup and it's order is k

We want to find k, so we must see what <g^7> is

Now k can be one of three numbers, agreed?

[oak grove]

im sorry im just staring at the problem

it can only be one of those numbers

n itself, 7, and 1 are the only divisors of n

[south patrol]

yes

so k is 1,7,or n

Can k be 1?

[oak grove]

no

[south patrol]

Correct

[oak grove]

then youd only have 2 subgroups

[south patrol]

Yeah.

Now try to finish it off

[oak grove]

oh

if n is 49

it has to be a square of a prime

that guarantees only 3 factors

[south patrol]

Yes.

[oak grove]

then you get 3 unique subgroups

[south patrol]

But also k can't be n, so k must be 7

And so g has order 49

:)

[oak grove]

thanks 🙇‍♂️ sorry im slow today

[south patrol]

np!

Now it may be interesting for you to know that a group G of order n is cyclic if and only if for each divisor d of n, it has a unique subgroup of order d

[oak grove]

this is uhh

[south patrol]

The upshot being that if G's order had factors other than n then it'd have subgroups of those orders

[oak grove]

fundamental theorem of finite abelian groups

[south patrol]

:)

[oak grove]

?

[south patrol]

Not quite

[oak grove]

its scribbled very roughly in my notes

[south patrol]

This is pretty special case, it was actually an exam question i did

[oak grove]

oh damn

[south patrol]

Whereas finite abelian groups is longer and requires more technology

[oak grove]

this is in the 'easy' set of exam review questions for us

[south patrol]

Sure

[oak grove]

hmm

[south patrol]

But yeah don't worry about the theorem other than that it's cool to know it exists, even just to guide intuition

[oak grove]

hmm ok 😄

[south patrol]

For example, suppose a cyclic group has one subgroup order 2, one order 3 and nothing else besides itself and the trivial subgroup - what order is G and how would you prove that

[oak grove]

i guess itd be 6

[south patrol]

yes

[oak grove]

you'd just cite this theorem i guess

[south patrol]

well without actually citing the theorem i mean

[oak grove]

its the only way to only pick up factors of 1 2 and 3

oh

[south patrol]

Use it for intuition, but use a similar method to what we just did to actually prove it

In particular, which order is <g^2> where g is a generator

[oak grove]

i guess you could just check brute force

[south patrol]

Yeah, sure

It's either 1,2,3,6 and do cases to show it has to be 3

But dw about it now aha i gave it just for an example

[oak grove]

yea my brain is mostly clogged lol

i appreciate the help

[south patrol]

Np

Mine is becoming clogged

[frank fiber]

what is a ground field?

[hidden haven]

Usually the smaller field in a field extension but depends on the context

[novel parrot]

last part

if a series has length of l(M) why is it a composition series

[hidden haven]

You can refine the series to a composition series and that would have the same length so must be the same series

the last line is saying this exact thing too lol

[novel parrot]

dont we need the converse now

[hot lake]

I think the last line is saying if k < l(M) then it's not a composition series

[novel parrot]

other direction

[hot lake]

so it's the converse

[hidden haven]

wait I didn't get it

[hot lake]

we know that all chains have length <= l(M)

the first sentence is saying if k = l(M) then it's acomposition series

[novel parrot]

agreed

[hot lake]

the second is saying if k < l(M) then it's not ?

[novel parrot]

wait no

[hot lake]

so k = l(M) <=> it is a composition series

ah no they want to show every chain can be extended

I should read the actual result they are proving

[novel parrot]

taking a composition series we see that it has length k

now take a chain of length k

im asking about showing its a composition series chain

[hot lake]

iii) says that all composition series have the same length l(M)

[novel parrot]

but we dont know all chains of lenght L(M) are composition series?

[hot lake]

iii) has like .. 3 parts

well if you have a chain of length l(M) that is not a composition series

you can extend it into a longer chain that is a composition series

so you would get l(M) > l(M)

which is a contradiction

[novel parrot]

how do we know that there cant be an infinite chain

so we cant extend to composition

[hot lake]

because we know M has a finite length composition series

and ii)

every chain has length <= l(M)

[novel parrot]

ok yes right

[novel parrot]

can this not be done by pigeon hole principle

and being noehterian and having homomorphic doesnt actually matter

[scarlet estuary]

are we just assuming every set is finite

[novel parrot]

yes

[scarlet estuary]

even then, isomorphic as sets ≠ isomorphic as modules

[novel parrot]

wait why

[hidden haven]

but u is a homomorphism

[novel parrot]

yeah we can keep u being a homomorphism mb

[hidden haven]

surjective homomorphism between finite modules is isomorphism

[scarlet estuary]

yes, i was disputing:

and being noehterian and having homomorphic doesnt actually matter

[hidden haven]

oh

right

[novel parrot]

yeah keep the homomorphism

[scarlet estuary]

if you know these modules are finite, then yes, thats enough

i dont think anyone cares about finite modules

but hey

it works

[hidden haven]

[novel parrot]

[golden pasture]

Hence F_{p^2} and F_p as F_p modules are isomorphic

[hidden haven]

Check the homomorphism conditions

[terse crystal]

You have a chain ker(f)<=ker(f^2)<=… since M is noetherian there exists n such that ker(g)=ker(g^2) where g=f^n now g is surjective therefore any x such that g(x)=0, there exists y from M such that x=g(y). g^2(y)=0 therefore x=g(y)=0 therefore g is injection, so f is injective

[uncut girder]

I think you're wrong

[golden pasture]

wat

[uncut girder]

how can they be isomorphic if they have different cardinalities

[upper pivot]

they have the same cardinality mod p

[terse crystal]

What? For a field k, k^m is isomorphic to k^n iff m=n

Otherwise you will have a matrix A with n rows m columns and a matrix B with m rows n columns where n is greater than m such that AB=I

[upper pivot]

no one who uses #groups-rings-fields can pick up sarcasm confirmed

[terse crystal]

😂😂

[novel parrot]

where should i start on this?

nothing , i got no idea

[upper pivot]

sorry one last meme b4 u guys get to the helping, but the context is modules and these are the same as F_p modules

[novel parrot]

well ive tried creating an exactness sequence

but i dont think that is the way here

[upper pivot]

(dim 2 vect space over field F_p)

[hot lake]

I would try to see if I can match ideals of A/a with submodules of M

[novel parrot]

wdym match

[hot lake]

I mean come up with a function

[novel parrot]

ok

[hot lake]

hopefully that preserves inclusion

then that would allow you to carry noetherianness from one to the other

[terse crystal]

I see,My bad

[novel parrot]

i dont think that can work tbh

like we wanna consider A/a as A module

and look for injective homomorphism

?

[hot lake]

???

[novel parrot]

maybe i misunderstood ya lol

[hot lake]

what does that have to do with showing A/a is noetherian

[novel parrot]

without injectivity it wont carry the inclusion?

[hot lake]

injective morphism from what to what

[novel parrot]

M -> A/a

[hot lake]

that's uuuuh

unlikely

[novel parrot]

but yeah idontthink we can find such a thing

[novel parrot]

what did u mean?

[hot lake]

I said a function from the set of ideals of A/a to the set of submodules of M

because to show that A/a is noetherian you got to talk about ideals of A/a

[novel parrot]

yea

[hot lake]

and you know that M is noetherian, that is, you know stuff about submodules of M

[novel parrot]

how will that function help

to do anything

[hot lake]

you will say

[golden pasture]

maybe i need to explicitly /s

[hot lake]

"suppose that A/a is not noetherian, Take an infinite ascending chain of ideals. Use the function to get an infinite chain of submodules of M, contradiction"

[novel parrot]

im kinda iffy about this method tbh

we also know that a = ann(M)

so surely that is nudging us to the direction of finding atleast a homomorphism

if we just had a function between ideals, we dont really know that inclusion is respected

[hot lake]

well the point would be to define a function in a way that it has good properties

where inclusion-preserving is basically trivial and uh injectiveness maybe has some actual work to do

[novel parrot]

hmmm

[terse crystal]

M=ΣAx_i a=ann(M) equals the intersection of ann(x_i) therefore A/a is isomorphic to a submodule of ΠA/ann(x_i) for any i A/ann(x_i) is a noetherian A/a module.

[novel parrot]

oh damn i completly forgot that M was finitely generated

but that last part u said

A/ann(xi) why is that Noetherian A/a-Module?

@terse crystal

[terse crystal]

A/annx_i is isomorphic to a M submodule and any submodule of M is annihilated by a

[novel parrot]

tysm!!

[wraith obsidian]

Does anyone here have examples of transitive antisymmetric relations which are not everywhere reflexive?

reason: I've noticed that A ~ B iff cl(A)≤B for any closure op cl(-) is transitive and antisymm, but reflexive only for closed elements

and I wanted to know how „nonstandard“ examples of Tr/AS relations look or whether every such relation arises from a closure operator

(please ping if responding)

[plush quartz]

The inverse doesn't exist in H because it's not an integer.

[chilly ocean]

under addition

[plush quartz]

I thought multiplication is the default operation?

[chilly ocean]

it says in your screenshot that we're thinking of G as a group under addition

the integers with multiplication aren't even a group anyways (for the reason you state)

[plush quartz]

I only see multiplication here

[hidden haven]

We use multiplication notation

But that doesn't mean the operation is multiplication

We need to use some notation

[plush quartz]

ahhh crap ty thats why

I misunderstood the notation

[wraith obsidian]

welcome to the world of abstraction

where you can use symbols but have no clue what they might mean

[hidden haven]

I don't think it has to be in the centre of R (anyway that doesn't make sense since a is in R')

It should commute with everyone in the image of R

Reason is that R[x] is a commutative ring and the image of a commutative ring is commutative

For all x in R yes

[next obsidian]

Moldi

I’m pretty sure the map is surjevtive (this is false but w/e)

I mean I don’t know what the map is from and to

Or well… no

[hidden haven]

Oh wait

[next obsidian]

But like

[hidden haven]

Evaluation into the ring itself

[next obsidian]

Yeah

[hidden haven]

I see

[next obsidian]

I mean it depends on how you defined R[X]

I don’t like it when R isn’t commutative

[hidden haven]

Right

[next obsidian]

But if you let cX = Xc

Then you just apply it

And you get ca = ac

And then c is arbitrary

I mean yes

But when R isn’t commutative

It gets dicey as to whether you’ve defined cX = Xc

So does X commute with everything?

If so then you get what you described

Cuz when you do f(cX) = ca

Then this has to be equal to f(Xc) = ac

Idk

I don’t see how the problem is correct unless X commutes with everything

So this is a shitty way to justify X commuting with everything

But it makes me think it’s been defined like that

I just forget if you actually get a ring doing that

[hot lake]

usually when you write R[X], X commutes with R

[next obsidian]

I’ve lived in commutative land for too long

Ok

Then just do what I suggested

[plucky flicker]

Hi! If i have a free group F(X) of rank = r, how can i express the rank of F(X)^2 = <w^2 : w \in F(X)>?

Maybe Nielsen-Schreier theorem? Idk if F(X)^2 is a subgroup of F(X) or not, but if it is, and i would know the index of it in F(X) then i would have the answer

[mild laurel]

It's not a subgroup

Unless r=1 I guess

@plucky flicker

[plucky flicker]

Maybe you're right

it was just an idea

[mild laurel]

Then what are you even asking

It doesn't make sense to ask for the rank of something that's not a group

[plucky flicker]

I have this question: Express the rank of F(X)^2 = <w^2 : w \in F(X)> using that the rank of F(X) is r.

[hidden haven]

Yeah but unless X is a singleton, the set of squares of F(X) is not a subgroup

And you can only talk about the rank of a group, not a set

[plucky flicker]

i know that

but i do not have the set of squares, but the group generated by the squares

[hidden haven]

Ahh

[mild laurel]

Oh my bad

[plucky flicker]

no problem

so any idea would be appreciated 😄

[hidden haven]

I'm guessing you can go case by case to some extent. For example when r = 1 then it is just 1 again. If r = 2 then the group is countable, so rank ≤ |N|, and it's non abelian, so rank > 1, which means rank = 2 immediately

So I think rank will be r itself

Wait I think I just assumed that F(2) surjects onto F(N) when it's that F(N) injects into F(2)

So the finite rank case should be the interesting one

[mild laurel]

I want to say that we can use the Nielsen Schreier index formula. F(X) modulo that subgroup is a Z/2 vector space and it should be of dimension r I think

[plucky flicker]

i'm guessing that so

[mild laurel]

Yeah this works

[plucky flicker]

but what would be the homomorphism whose kernel is F(X)^2?

[mild laurel]

It should just be the map that sends x_1 to (1,0,0,...,0) of Z/2Z^r

And so on

[plucky flicker]

Z/2Z^r? not Z^r/2*Z^r?

[mild laurel]

Aren't these the same thing

[plucky flicker]

i mean how can you factor out 2Z^r from Z?

[mild laurel]

I mean

(Z/2Z)^r

[plucky flicker]

ooooh, okay

sorry mate

[mild laurel]

I'm on mobile so I was too lazy sorry

[plucky flicker]

its fine

thanks 😄

[hot lake]

abelianization ?

[upper inlet]

yeah that's called the commutator subgroup

[hot lake]

I don't know if there is a word for rings

[upper inlet]

oh I didn't see ring

[plush quartz]

This is not true because <-1> has two outputs where <1> has only one

[hot lake]

<x> means the subgroup of (Z,+) generated by x, I am not sure what you are calling outputs

you are making the same mistake as earlier today

the operation that makes Z a group is addition

not multiplication

[terse crystal]

Consider positive integers and R={(m,n): m divides n and m doesn’t equal n}

[next obsidian]

Empty relation I think

But this is incredibly hurbed

[terse crystal]

[F(X):F(X)^2] is 2 when |X|>1? Since x_j = x_1 x_1 ^-1 x_j therefore x_j is contained in x_1 F(X)^2

[mild laurel]

I'm not sure how you're getting that conclusion

How do you know that x_1^-1 x_j is in F(X)^2? @terse crystal

[terse crystal]

I confused that subgroup with the subgroup whose elements have even length.

[wild sapphire]

Hey guys, I'm trying to prove that there is no homomorphism between Z8xZ2 and Z4xZ4. With some work I've found that I think the issue should lie in the fact that there are elements of order 8 in Z8xZ2 and Z4xZ4 only has elements of order up to 4. From this and some more work I can show that the kernel of a hypothetical homomorphism can't be trivial.

Is the fact that both groups have the same order, but all mappings aren't surjective enough to show that there can't be a homomorphism?

I feel like I'm a missing a small piece, is there any theorem that says something like "homomorphisms between groups of identical order are always surjective" or something?

[mild laurel]

No, you always have the identity morphism. In other words, for any groups G, H, there's a homomorphism from G to H that sends every element of G to the identity

[wild sapphire]

ah, yeah. I didn't think of that trivial case. So I guess I'm just wondering how to prove there is no homomorphism. I'm not sure where to go from the mismatch of element orders, to showing that no homomorphism can preserve the structure. It makes perfect sense intuitively but something is missing

[mild laurel]

Well, I'm saying that I'm not sure if your question is correct. It's not true that there's no homomorphism because you have the identity homomorphism

[wild sapphire]

oh... yeah 🤔

[mild laurel]

is this a question in your book or something?

Maybe the question meant to ask if there are any surjective homomorphisms?

[wild sapphire]

a question in my course's extra problems. Yeah I'm pretty sure the surjective part was just left out

[mild laurel]

yeah, there are a lot of other homomorphisms other than just the trivial one too

[wild sapphire]

really? I feel like the structure of the two groups isn't that similar

[mild laurel]

You can take (1,0) in the first group and map it to (1,0) in the second, and (0,1) in the first group and map it to (0,2) in the second

[wild sapphire]

oh yeah, that makes sense. It's not a nice mapping but I can see how you picked out the structures that were needed. Could you instead map (2,0) to (1,0) to avoid the double mapping? it feels better at least.

[mild laurel]

If you map (2,0) to (1,0), then where does (1,0) go to?

[wild sapphire]

oh yeah, we can't throw away items from the domain in the same way we forgo elements in the codomain, oops 😄

[mild laurel]

Yeah, this has to do with what you said earlier

about (1,0) having order 8

so that the kernel can't be trivial

[wild sapphire]

yeah, that makes sense. So with your mapping we're still kinda keeping structure in an overlapping way, it doesn't work to just poke holes in the structure to establish a mapping

[mild laurel]

Right yeah

[plush quartz]

Is the operation addition or multiplication?

[chilly ocean]

it could be either or neither. it's just an abstract group operation, and those are commonly denoted like multiplication

#groups-rings-fields message

[plush quartz]

I need to know what the operation is because I'm about to show that n1 & n2 are closed under something to prove the first axiom

I'll try multiplication first I think

[mild laurel]

It doesn't really make sense to ask if an operation is addition or multiplication

It's written "multiplicatively" here by just writing elements next to each other but

[plush quartz]

When we say the integers are a group, they are a group under addition. We include the operation name but in this problem we are not given that information.

[chilly ocean]

the operation name in this problem is just "the group operation of the group G"

G is an arbitrary group, its operation could be anything

all you need to know is that it's a group

[plush quartz]

For the problem I posted, what exactly are we intersecting over? I understand $S_1\cap S_2$ but am having difficulties seeing the sets in this problem

[cloud walrus]

bunkermush

[chilly ocean]

it's the intersection of all of the sets of the form xHx^{-1}, where x is in G. such sets are those consisting of all elements of the form xhx^{-1}, for some h in H

[wraith obsidian]

Thanks! Thinking about it when falling asleep I noticed that there's not too many conclusions one could draw – adding or removing „reflexivities“ x ~ x still produces a transitive, antisymmetric relation.
So we can classify them by giving a partial order + specifying which reflexivities are omitted.

[plush quartz]

Either x or h can be in N, correct?

[hidden haven]

x is in G, h is in H. N hasn't been defined yet

[novel parrot]

how F-module

any element of E roots a polynomial over F

but we got powers

oh

it means E is a vector space over F

with a finite dimension

[next obsidian]

THE KOSZUL COMPLEX STAYS WINNING

[novel parrot]

any hints on this?

i think we should use this

so i just need to show that B is a finitely generated B^G module

are the xi fixed by G?

that would be enougn i think

[novel parrot]

and another question

A field of characteristic 0

a finitely generated one

why finitely generated over Z?

[hidden haven]

Finitely generated as a ring?

I don't think Q is finitely generated as a Z algebra

[novel parrot]

"K is finitely generated over Z"

do they mean as rings

or

[hidden haven]

They are assuming that to get a contradiction

[novel parrot]

i thought we assuming Char 0 for contradiction

[hidden haven]

oh wait yeah

[novel parrot]

let me show the whole thing

[hidden haven]

Yeah they mean finitely generated as a ring

That implies finitely generated over Z

After you embed Z through the characteristic map

[novel parrot]

dont get it

how

K is finitely generated by say (x1,x2 , xn)

has char 0

we can embed Z

then how?

[hidden haven]

Then it's a Z algebra which is generated by those same elements

[novel parrot]

hmmmmmm

how does it become Z algbera?

[hidden haven]

A Z algebra is just a ring with a ring homomorphism from Z to it right?

Every ring is a Z algebra

[novel parrot]

oh

so not K = Z[x1,x2,.xn] ?

having it finitely generated over Z could miss out some elements no?

[hidden haven]

That equality is correct

[novel parrot]

ok so in any ring A

A = Z[some elements of A]

thats true?

[hidden haven]

As long as that's a generating set for A as a ring

[novel parrot]

how we know for the field K that f(Z) is all our coefficients

[hidden haven]

I didn't get it. Generated as a ring by some collection {x_a} means that every element is a sum of products of these x_a

You can view any such expression as a polynomial in the x_a's with all coefficients 1

[novel parrot]

but

isnt it a sum of Kx_a

Where we take any combination of K and x_a

[hidden haven]

You're thinking of ideals

In subrings you don't take coefficients from the whole ring

[novel parrot]

ok

alright i think i see it

generated as a ring means:

hm

in the generating set

we also need a 0 and 1 yeah?

[hidden haven]

You get 0 from empty sum, 1 from empty product

If you wanna set that convention

[novel parrot]

what

[hidden haven]

But like they will always be there in any subring

You don't need to add them to generating set to get those

Because subring generated by something is smallest subring containing it

[novel parrot]

oka

okay

[hidden haven]

And all subrings contain 0 and 1 implies the smallest does too

[novel parrot]

ive never generated a ring before

but yeah the rest makes sense now

[hidden haven]

Kids these days relying on pre processed stuff

[novel parrot]

are char p fields automatically finite?

[hidden haven]

nope

[novel parrot]

all elements can be broken into sums of 1?

[hidden haven]

Thats only F_p

[novel parrot]

okay

[hidden haven]

alg closure of F_p is infinite

[novel parrot]

sure

ok

[chilly ocean]

I want to show that Z[sqrt(3)] is a pid

it seems the best way to do thsi would be to find euclidean function

but im not sure how to find euclidean functions

opps this should go in number theory

[novel parrot]

just use Norm

for euclidean function

a + bsqrt3 I-> a^2 - 3b^2 i think

[chilly ocean]

We need |a^2-3b^2| instead

It's still multiplicative but outputs non-negative integers

[novel parrot]

yh that

why does that even matter

[chilly ocean]

Sorry wrong

[novel parrot]

how we appy nakayama lemma

we saying that m^n is finitely generated A module

and m is in jacob radical of m^n?

[thorn delta]

m^n = m^(n+1) = m(m^n). m is equal to the jacobson radical since A is a local ring

[novel parrot]

oh

[bronze jay]

any ways of writing this cleaner?

[latent anvil]

Notably this can fail even if the field is finite

Take Fq for q a prime power which is not prime

For instance F4 = F2[x]/(x^2 + x + 1)

[mild laurel]

Is there a reason you're asking this?

I guess one reason is that if you have a group homomorphism from f: G to H, then you consider the group homomorphism from G to im(f) and this is surjective. So you can kinda reduce the study of homomorphisms to surjective homomorphisms

[south patrol]

Is the hypothesis that x is in the normaliser of A? Then by definition xA = Ax and from that you can get the solution p quickly

A = x^-1(xA) = x^-1(Ax) = x^-1 Ax

[frank fiber]

if i have a presentation of a group G , how can in know if G is trival?

[mild laurel]

impossible in general https://en.wikipedia.org/wiki/Word_problem_for_groups

[next obsidian]

Kekw

[south patrol]

That's when you know you asked an interesting question tbf

Undecidable moment

[scarlet estuary]

"mathematics is the study of the impossible" is the kind of phrase that sounds a lot cooler than what it actually means

[next rain]

there's also a quote like "physics is the study of what man can't do, mathematics is the study of what God can't do"

[chilly ocean]

hey guys dumb q

but vector space axioms how do you show 0x = 0?

for x vector

[novel parrot]

(0 + 0)x = 0x + 0x = 0x

[strong valve]

[chilly ocean]

okay nice

thanks

and you are allowed to assert additive inverse okaaaay

[novel parrot]

yup everything has inverse

[chilly ocean]

i gotta say this made me break a sweat HAHA

im not suited for this

[wooden ember]

What zoph said. You can find examples of really convoluted presentations with only a slight difference (a power changing from a 3 to a 4 or smth) and one will be trivial while the other will be infinite lol

It’s sad

I wish we could just work with presentations and get everything easily

[novel parrot]

i was also stuck on this before lol

[chilly ocean]

the additive inverse is crucial

[novel parrot]

yes

[sullen island]

https://math.stackexchange.com/questions/134234/transcendence-degree-of-a-field-extension

so i understand the accepted answer for this post, but idk why it proves the desired result - doesnt it just show that B_1 U B_2 is algebraically independent over F? it doesnt yet show that it is a maximal algebraically independent set

dont we need to show that its a maximal algebraically independent set over F to conclude the equality?

<@&286206848099549185>

[terse crystal]

Given a connected graph and a base point,why the rank of its fundamental group equals the first Betti number of the graph? And why this first Betti number equals the number of edges minus the number of vertices plus the number of connected components (which is 1 in this case)?

[hidden haven]

Connected graphs are homotopy equivalent to a wedge of circles, by contracting a maximal spanning subtree

@terse crystal

The number of circles is the number of edges not in this tree

[terse crystal]

Thanks so I understand you saying that the rank equals |edges|-|vertices|+1. Now why this number equals the first Betti number?

[hidden haven]

Betti number of a wedge of circles is the number of circles

Because the 1st homology is Z^(no of circles)

[terse crystal]

Did you change the underlying space? I mean given a connected graph X and a base point p, surely π(X,p) is a free group with rank n. Then I can define Y= wedge product of n circles and according to you the first Betti number of Y equals n but X is not necessarily a wedge product of circles so how do I know that the first Betti number of X is also n?

[hidden haven]

Homotopy equivalence preserves homology groups, and Y is h equiv to X

So we can work with Y

Like once you establish the h equiv, everything is in Y

[terse crystal]

How do I know that the connected graph X with e edges and v vertices is homototy equivalent to Y=wedge product of e-v+1 circles?

Wait wait I kind of get it. By contracting a maximum tree to a point?

[hidden haven]

Contracting the maximal subtree is an h equivalence. This is proved in Hatcher I think (definitely in May)

Yep

But what toki is pretty good too

[terse crystal]

Thanks a lot

I tend to ask algebraic topology questions in both channels 😂😂

[hidden haven]

Ask in top only lol

Alg top is supposed to go there

[terse crystal]

I see got it

[median pawn]

hello friends in algebra

i learned about free groups in algebraic topology (apparently they will be useful there)

[hidden haven]

free groups are useful anywhere there are groups

[cloud walrus]

Hausdorff

[median pawn]

S is a set (the alphabet), F(S) is the free group generated by it

Of course it is trivial but how do I do such proofs

[hidden haven]

isn't that the definition of the elements of F(S)

[median pawn]

For me F(S) is the set of equivalence classes of words

[cloud walrus]

Hausdorff

[median pawn]

I think we have to use induction somewhere

Perhaps induction on the number of elementary contractions or something like that

[hidden haven]

elementary contraction?

[median pawn]

lemme define

[hidden haven]

And that definition seems like just "equivalence closure of elementary contraction as a relation"

just nice terminology you can use to make things shorter

not too hard to see what it would mean

[hidden haven]

cancelling one thing with its inverse I am guessing?

[cloud walrus]

Hausdorff

[median pawn]

yes

[hidden haven]

nice

This seems very Church Rosser like let me have one think

[median pawn]

x in S union S inverse but yeah

[hidden haven]

Try this: if a word w can be contracted to both y and y' then y and y' have a common contraction z

This is what one of the steps in the proof of CR looks like too

[cloud walrus]

Hausdorff

[hidden haven]

Yes, I am saying that the above lemma will help

it should let you argue, by induction, that w1 and w3 must also have a common contraction

but they are reduced so their only contractions would be themselves

[median pawn]

Interesting, this is true indeed

[median pawn]

This is just painful case work right

[hidden haven]

Case work yeah but I don't think it's painful

Not too many cases

[cloud walrus]

Hausdorff

[hidden haven]

😵‍💫

Suppose in x you contacted some pair a'a and in y you contacted a pair b'b (' denotes inverse) then you can write w as
pa'aqb'br

Where p,q,r are arbitrary strings

But one more case

Where a = b or b'

Then w = the above, or you can overlap the 2 pairs in different ways

like a(a'a) = (b'b)b'

I guess it is painful

Not I didn't understand what you were doing lol

[median pawn]

It is painful yeah lol

That's why I prefer Analysis

Anyway, let's assume this lemma lmao

[hidden haven]

Wow analysis less painful than algebra

Who ever checks this stuff anyway smh

[median pawn]

this also is not easy

[median pawn]

because aa^{-1}b ---> b is a contraction, but aa^{-1}b ---> aa^{-1} bb^{-1} b ---> bb^{-1} b----> b is also a valid way of showing that aa^{-1}b and b are related lol

basically you can also have elementary expansions if any two are related

but we can probably relax this by strictly considering only contractions from words to reduced forms

[hidden haven]

How are you even typing all that Looks like pain

[median pawn]

yeah lol

[hidden haven]

I don't see how that's a counterexample though

Like pick any chain that exists between the 2 words and induct on that

It's fine if there are multiple, that won't matter

[median pawn]

pick a chain consisting of contractions only

this can always be done

[hidden haven]

I don't think so, aa'b ~ bcc'

You can't go one way

[median pawn]

here you can

because w_1, w_3 are reduced

https://www2.unb.ca/~ntouikan/MATH6022/IntroCGGT/html_output/reductions.html use diamond lemma to prove theorem 1.2.2 lol

[hidden haven]

I think that's equivalent to there being a reduced form

Oh diamond lemma

That's what my claim is called I think

oh neat

Yeah from this it's just induction

[median pawn]

yup!

[hidden haven]

🎉

[urban acorn]

you just need to explicitly construct free groups once and show they have the universal property, and then you don't need to worry about it anymore

it's like constructing the integers as equivalence classes of pairs of natural numbers (where (a, b) is taken as a - b) and the rationals as equivalence classes of (a subset of) pairs of integers (where b is nonzero and (a, b) is taken as a / b)

sure, someone at some point should formally construct them, but once you did, you don't need to think about it anymore

[wild sapphire]

Are there certain criteria for when subgroups of a group ovelap? like do two cyclic subgroups of different order overlap? Do two subgroups of prime order overlap?

I feel like the cyclic subgroups of different order don't overlap, that one makes sense to me

I'm just wondering if I'm missing some more general fact

[south patrol]

By overlap I assume you mean have a non-trivial intersection

[wild sapphire]

well correction. The cyclic subgroups of prime order don't overlap

[south patrol]

Often Lagrange's theorem is useful for finite groups in doing this

[wild sapphire]

yes, non-trivial intersection

[south patrol]

that seems more like what you're after?

[wild sapphire]

yeah I was trying to apply it to just two general subgroups of different prime order, but I'm struggling to see what the intersection could be if it's there

[south patrol]

Ah, sure

Well if H,K are subgroups of a finite group G, then H cap K is a subgroup of both H and K

[wild sapphire]

ah yeah, i remember that

[south patrol]

hence its order must divide both H and K's orders, so if H and K are of prime order...

[wild sapphire]

oh..

[coarse stag]

Is there any closed form way to compute the order of an automorphism group for a given group G? I’ve been trying to find some solution with Burnsides that takes advantage of orbits but I can’t find any.

[wild sapphire]

well turns out its a lot easier than I thought, thanks @south patrol !

[south patrol]

np :)

[viscid pewter]

i've looked into this a bit, although i'm bad

afaict it's one of those types of things that's sorta manageable for abelian groups and extremely non-trivial and an open question otherwise

that's not even touching infinite groups, which can get really weird i think

part of the difficulty is like

how would you actually define any given group such that you can input it into a hypothetical automorphism-group-order-determining-formula

just plugging in the order of the original group is clearly insufficient

i was thinking you'd have to work with a presentation and try and go from there

maybe that works, i haven't tried that much, but my instincts say it's hard

can you specify it all sufficiently with burnside's lemma? like the data about the fixed points? that's sorta interesting

but anyway yeah probs you can only get anything like what you want for abelian groups

that's actually feasible, although there's some fun corner cases

[topaz leaf]

yeah i suspect that calculating automorphism groups is in general a very difficult question, unless you're looking at particular families of groups (e.g. cyclic groups, finite abelian groups, dihedral groups, maybe symmetric or general linear groups, etc.)

[viscid pewter]

yeah it's almost trivial for cyclic, dihedral, symmetric groups, finite abelian generally is describable, general linear i haven't looked into

[sullen island]

hi, so i posted this question here yesterday but it got ignored and i still can't figure it out, could anyone help?

[mild laurel]

@sullen island I think you're right in saying that you still need to prove maximality

[sullen island]

so suppose that B_1 = {x_i} and B_2 = {y_j} are transcendence bases for L over K and K over F respectively (where L/K/F) and that there exists an element, say z, such that B_1 U B_2 U {z} is algebraically independent.

then B_1 U {z} is not algebraically independent over K, so there exists a nonzero polynomial f in K[X_1, ... , X_m+1] such that f(x_1, ... , x_m, z) = 0

[sullen island]

im not sure how to proceed to derive a contradiction to prove maximality, this is what i have so far

[full panther]

is this a valid proof?

i understand the idea, im not sure i put it in words correctly though

because it is the intersection of the sets

so the elements in that set must be equal

do i have to specify the index of the x and ys in the set

to make the statement

the intersection just means the elements are in both sets

if the element is in the intersection, then it is in both X and Y

im trying to say that element in X is equal to the one in Y

Ok

thats what i meant with the index

if A = {a,bc} and B = {b,c,d} and alpha \in A and beta \in B, then im looking for the pairs of alpha and beta that match up

Ok so the heuristic picture in my mind for this is that X \cap Y, includes all the pairs of elements and inverses in both X and Y, so they are in X \cap Y, then i am trying to show that that set preserves closure under operation

is that at least correct?

if X is a subgroup, then it includes an element x and x^-1

pairs of elements that are inverses of each other

Ok thanks

[median pawn]

I have proved this lemma

Could someone help me understand this hint better?

[full panther]

So i tried saying that if x \in X \cap Y, then x \in X and x \in Y, but X and Y are groups, so then x^-1 \in X and x^-1 \in Y, hence x^-1 is in X \ cap Y

is this better

Is this enough to conclude that X \cap Y is a subgroup of G though?

Ok

[hidden haven]

Hmm I don't see why they are taking the minimal length chain. I think you can just induct like this:
For any chain of length n, the 2 end points of the chain can be reduced to the same word.
Induct on n.

[median pawn]

I think minimal length makes sure that almost every step is a contraction

There are no unnecessary insertions, I mean

[hidden haven]

Yeah but I don't see why that matters

In the induction hypothesis you'd want to assume that the 2 end points can be multi-step-contracted to the same thing, not directly contracted ofc

And no contractions also counts as a multi step contraction

[vestal snow]

Let $k$ be an algebraically closed field and consider the divisors on $\mathbb{P}_k^1$. Let $D$ be a divisor on $\mathbb{P}_k^1$ such that deg(D)=0. Then show that $D=div(f)$ for some rational function $f$.

Proof: Let $[(x-a_i)]$ be all (finite) places where the valuation of $D$ is not zero. Then define $f$ as $\frac{(x-a_i)^{v_{a_i}(D)}}{x^{\text{deg}(D)-\sum_iv_{a_i}(D)}}$. Then $f$ is the required element.

[cloud walrus]

Finitely Many Bananas

[vestal snow]

Can someone verify if this proof is correct?

There is also this problem:

If $k$ is a perfect field and $\overline{k}$ is its algebraic closure, then prove that the set of divisors on $\mathbb{P}k^1$ is the set ${D\text{ a divisor on } \mathbb{P}{\overline{k}}^1:\sigma(D)=D \text{ for all } \sigma\in \text{ Gal}(\overline{k}/k)}$

[cloud walrus]

Finitely Many Bananas

[vestal snow]

I don't think that the second problem makes sense

Because places in $\overline{k}$ will, in general, not be places in $k$.

[cloud walrus]

Finitely Many Bananas

[vestal snow]

Ping me if/when someone answers

[terse crystal]

I am thinking taking f to be just simply Π(x-a_i)^v_a_i(D) since k is algebraically closed degree being zero implies Σv_a_i(D)=0.

[terse crystal]

σ(D) is the pull back of D in terms of morphism induced by σ? I think it’s correct if you can prove that k(x)={r from k bar (x): any σ from Gal(k bar/k), σ(r)=r} .I don’t know how to prove it

[cursive temple]

does this have any nontrivial structure

it isnt a ring because stuff doesnt distribute

[maiden ocean]

Its called a near ring sometimes

You need it to be abelian to say much more

[cursive temple]

yeah jacobson does consider that case in the text itself

im just a bit lost on what to really write down

[maiden ocean]

Id just prove every ring property you can

[cursive temple]

ill go with that, thanks!

[wooden ember]

im asked to formulate the notion of a free nilpotent group of class c, not having seen this before. Is this inductive definition correct? We let $N_1(S)=A(S)$ the free abelian group on $S$. Suppose $N_{c'}(S)$ has been defined for $c'<c$. Then we define $N_c(S)$ to be the group with presentation $(S,R_c)$ such that $R_c={[s,t] : s\in N_{c-1}(S), t\in S}$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

hmm no that doesnt seem right

wait I don't have to define it inductively

i can just use $F(S)^i$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

i have no idea how to go about this

[silent osprey]

sorry for the ping but what book is that?

[terse crystal]

An introduction to combinatorial and geometric group theory by Nicholas Touikan

[terse crystal]

We know that given a finitely generated free group G of rank n, G has finitely many subgroups of index m. But how many?

[next obsidian]

P^1_F should be dimension 1 for any field F so divisors are just cycles of closed points. The points of affine space over k are like in bijection with points of k-bar where you glue Galois conjugates

So I don’t have a proof but this seems really related, I think this might not be a literal equality as sets

As much as it just is like, for every divisor on P^1_k-bar fixed by the Galois group there’s a very very obvious divisor on P^1_k associated to it and the correspondence is bijective or something

Like I’m gonna work on affine space rn, and work with C and R, but the points z,z-bar correspond to (x^2 - z(z-bar)) [note here the divisor we are associating to this latter point is the formal sum z + z-bar]

Or for the real points of R of the form (x - a) the corresponding complex number is just already fixed by the Galois group

[vestal snow]

What about the valuation at infinity?

[vestal snow]

I see. That makes a lot more sense

@next obsidian when k is not algebraically closed, what are the places of k? Are they just places corresponding to irreducible polynomials + the place at infinity?

[next obsidian]

Idk what your definition of a divisor is

I’m working with the definition that it’s a formal Z-sum of codim 1 closed irreducible subsets

But in the dim 1 case this literally just means closed points

And in the affine case I was working with the closed points are maximal ideals of R[x] or C[x] which are either irreducible polynomial over R or are just points of C

[vestal snow]

My definition is the same as yours

[next obsidian]

Okay

Idk what valuations have to do with anything then haha

Anyway you can cover P^1 with two A^1’s

So I think you might wanna expand that thought or something

Or maybe just take the k[x0,x1]/~ definition or something I dunno

[vestal snow]

You know how if k is alg closed, then the codim 1 irred closed subsets are [(x-a)] and the point at infinity [1/x]

[next obsidian]

Uh

Okay

:)

This makes sense

[vestal snow]

For the projective line

[next obsidian]

I haven’t really explicitly thought about it but it makes sense

Yeah okay

[vestal snow]

I was wondering if there was a good description of those when k is perfect instead of algebraically closed

[next obsidian]

I dunno

I think the two A^1 things

Is what you want to shoot for

That description of P^1 when algebraically closed

[vestal snow]

Got it

[next obsidian]

Follows by covering with like

P^1 minus 0 and P^1 minus infinity

And those are the two A^1’s

And when you expand that out you should get the description you gave

I bet you can do the same thing when k isn’t algebraically closed or sometbinf

It just comes down to evaluating the intersection I think

[vestal snow]

Alright. I better get to work on it

I need to give a presentation on it in about 3 hours

[next obsidian]

Bruh moment

[hidden haven]

Try Galois correspondence maybe? You need to classify all connected m sheeted covers of wedge of n circles upto base point preserving isomorphism

Idk if that's easy or not, but seems easier than doing it algebraically at least

[chilly ocean]

Let A be differential graded algebra

Let M be differential graded module over A

Suppose A the only non zero degree term of A is A_0

then M a differential graded module over A

should be the same as a complex of A modules

oh yeah

i was thinking that it doesn't make sense because d will kill any a in A immediately

but it does work because of the liebnitz rule on A

[terse crystal]

Thanks. Yeah definitely easier than pure algebraic methods.

[dense pelican]

I am not sure about what rank-one convexity is. Does it geometrically mean that we are checking convexity for some cross section along some one dimension plane of our space?

A definition I found reads that rank-one convexity, for $f:R^{m\times n} \rightarrow R$ implies:

$$f((1-\theta)\mathbf{X}_1+\theta\mathbf{X}_2) \leq (1-\theta)f(\mathbf{X}_1)+ \theta f(\mathbf{X}_2) \qquad \qquad(1)$$

where $0<\theta<1$ and $\mathbf{X}_2=\mathbf{X}_1 + \mathbf{ab}^T$ which means rank($\mathbf{X}_2-\mathbf{X}_1)\leq1$

[cloud walrus]

Allan_M

[weary terrace]

My brain is kinda lagging today so apologies for the dumb question:
Let V be n-dimensional complex vector space with basis v1,...,vn. Consider the complex vector space V \otimes V. How does a homomorphism f: V\otimes V -> \mathbb C works? e.g. how can I prove that f(v1\otimes v2 + v3\otimes v4) = f(v1\otimes v2) + f(v3\otimes v4) without additional info?
I mean, we can't add v1\otimes v2 + v3\otimes v4 to become a pure tensor, so how does f operate on this sum?

[terse crystal]

This is required in the definition of that morphism but most of time it’s omitted

Like elements having the form x \otimes y are generators so you just need to define the image of those elements

Then f(Σ)=Σf is simply omitted

[weary terrace]

So it requires a kind of a linear extension?

[terse crystal]

They just don’t say the image of sum is defined to be the sum of image.

They assume that you are clear about it so they don’t bother add that sentence.

[weary terrace]

Ok great, thanks 🙂

[wooden ember]

just reiterating on this if anyone has any ideas, it's frustrating me

[mild laurel]

@wooden ember maybe its true that the homomorphic image of a nilpotent group of class n has at most class n?

[wooden ember]

oh right and hence nilpotent groups with greater class cannot be a homomorphic image 🤦‍♂️

thanks lol

[mild laurel]

I guess the only question is if the class of nilpotent groups with two generators is unbounded

[wooden ember]

fair

wait so if i have a unique surjective homomorphism from $N_c({a,b}) \to M$ does this mean $M$ has nilpotence class at most $c$?

[cloud walrus]

𝓛ittle ℕarwhal ✓

[mild laurel]

yes this is true

[wooden ember]

how would i show this 🤔 im struggling a bit

i really dont like nilpotent groups lol im so edgy with them

[mild laurel]

I mean, you can show more generally that for any surjective map G -> H, the nilpotence class of H is at most that of G

[wooden ember]

fair enough

that might be easier to think about

[mild laurel]

Uh, you do basically the only thing you can do tbh

[wooden ember]

i may have already shown it

gimme a sec

yeah lmao ive already shown this thought so

easy proof too

big brain fart moment

the surjective map doesnt even have to be unique smg

smh*

but why are two generators important then 🤔

doesnt this apply to just N any nilpotent group?

because if M is a homomorphic image of N then we have a surjective homomorphism from N to M

what's the point of focusing on two generator

[mild laurel]

maybe because of the fact that we have to find nilpotent groups with two generators with unbounded class

[wooden ember]

oh right that part

[mild laurel]

which is easy if we don't look at two generators

but I haven't thought of a two generator example yet

[wooden ember]

You cant just take free nilpotent groups on 2 generators of any class size?

[mild laurel]

oh i didnt know this was a thing

[wooden ember]

so we have to show the normal closure of the set of iterated commutators on two elements is never the free group on two elements 🤔

so that we can take nilpotency class of free nilpotent group arbitrarily big

[wooden ember]

oh right i realize i never mentioned this was in the context of free groups lol

[median pawn]

okay so i am kinda struggling with this easy exercise

[cloud walrus]

Hausdorff

[median pawn]

I have provided some definitions here: https://math.stackexchange.com/questions/4260001/prove-that-x-v-is-an-affine-subspace-of-x-where-v-subset-t-is-a-subspac

[hidden haven]

doxxed

Your interpretation for what you need to do to prove the first condition seems wrong

You say that you need to show that (assuming some stuff), the sum of 2 differences is in T

But any difference is in T and so is any sum

[median pawn]

Lol true

[wooden ember]

wtf lang calls group actions group operations

based

[hidden haven]

why is that based

[wooden ember]

does anyone actually call them operations?

or is this just historical cause lang's algebra is getting a bit old

what am i saying that's so stupid lmao

[chilly ocean]

[median pawn]

Okay, edited

Can you see?

[cloud walrus]

Hausdorff

[hidden haven]

Yes

Both for the first condition alone

[median pawn]

That's right

but how do you go about proving these

Looks nasty

If d was known explicitly I could do something, but this is too abstract

[hidden haven]

Do you have a picture in mind for what affine spaces look like?

[median pawn]

If T = X then yes

Otherwise I don't

[hidden haven]

That is what all affine spaces on T look like

You don't need to prove that for this problem

You can just sketch this situation in T, then try to generalize the solution you get

[median pawn]

on X*

[hidden haven]

No I mean if T is the vector space over which you are taking affine spaces

[median pawn]

Hmm yes

[hidden haven]

Then all those affine spaces look like T does with its usual affine structure

[median pawn]

Okay, let me think

[frank fiber]

Let $A$ be an abelian group with a filtration $0=F^0A \subset ... \subset F^{n-1}A \subset F^n A=A $, then the graded group is $gr_k A:=F^k A/F^{k-1 } A$. Suppose there are numbers $a,b,c$ such that $gr_aA=Z,gr_bA=Z/2Z,gr_cA=Z/4Z$ and all the others $gr_k A$ are zero. Then how can i know the possible values of $A$?

[cloud walrus]

Or x1

[tame grove]

This is vague, but how do I become good at abstract algebra? Ive spent probably 15 hours in the past week working on pretty basic proofs and I cant seem to get them with any consistency.

Just doing modular arithmetic and divisors. I feel like I understand the content, but the proofs are just really hard for me to pin down

[frank fiber]

i think that the only way is practice

the cosets partition the group https://math.stackexchange.com/questions/1357195/show-that-left-cosets-partition-the-group

[tame grove]

Gotcha. I am still searching for that revelation that makes abstract algebra make complete sense. It has happened in every math class to me thus far so I was hoping someone would be able to point me to it. thanks

[frank fiber]

it may be helpful to do lots of examples

[next obsidian]

I thought I was really bad at algebra for like a quarter, and thought I would just go do analysis instead and now I literally only do algebra

You never know what will make it click, it just took doing problems for me

And also, group theory is only a small subset of algebra. Maybe rings will make more sense for you?

[opal osprey]

Let $V$ be a finite dimensional vector space over a field $\mathbb{K}$, I want to define $\forall k > 0$ the map:
$$
d^{k} : \bigwedge^{k}(V) \rightarrow \bigwedge^{k+1}(V)
$$
Such that $\forall \alpha \in \bigwedge^{0}(V) = \mathbb{K}$ we have $d^{0}(\alpha) = 0 \in \bigwedge^{1} (V)$ and $\forall k \geq 1, \omega \in \bigwedge^{k} (V)$ and $u_{0}, \cdots, u_{k} \in V$ we have:
$$
(d^{k}(\omega))(u_{0}, \cdots, u_{k}) = \sum\limits_{i=1}^{k} (-1)^{i} \omega(u_{0}, \cdots, \widehat{u_{i}}, \cdots, u_{k})
$$
What I want to verify is if the map
\begin{align*}
d : \bigwedge& (V) \rightarrow \bigwedge (V) \
\omega& \mapsto d^{k}(\omega)
\end{align*}
Is a degree 1 antiderivation with respect to the wedge product of forms, i.e $\forall \omega \in \bigwedge^{k} (V)$ and $\forall \tau \in \bigwedge^{l} (V)$ we have:
$$
d(\omega \wedge \tau) = d(\omega) \wedge \tau + (-1)^{k} \omega \wedge d(\tau)
$$

[cloud walrus]

MisterSystem

[opal osprey]

We can suppose nice things about the field K if it is necessary

like having finite characteristic or characteristic different from 2

[lavish nexus]

If G has p+1 sylow p-subgroups, they must contain a normal subgroup of index p

Can anyone give a hint

I know I need to prove their intersection has index p

[stone fulcrum]

@chilly ocean
Post it here for more attention. See #get-advanced-access if you can't post here

[terse crystal]

Using structure of finitely generated abelian groups I think $Z \oplus Z/2Z \oplus Z/4Z$ and $Z \oplus Z/8Z$ are all possible A s

[cloud walrus]

Cogwheels of the mind

[steady axle]

Can I define Q as Z_(p) module?

[gritty sparrow]

Are you talking about Z/pZ or the p adic numbers?

[next obsidian]

It could even just be Z localized at (p)

[gritty sparrow]

Oh yeah lol

[steady axle]

that's what I mean

[gritty sparrow]

Yes Q is a Z_(p) module in a natural way

Because Z_(p) is a subring of Q

Hence you can define the multiplication of an element of Z_(p) and an element of Q by just multiplication as elements of Q

[chilly ocean]

We know that for the self-adjoint matrices there's an order A ≥ B iff A-B is a positive semi-definite

is there any characteristic property of orders which arise in such way?

(more generally, self-adjoint operators)

Apparently such order, in general, isn't a lattice, hence my question

This is order theory I guess so I'm not sure if it fits here, but I'm not really sure where it fits

[median pawn]

How do we get the stuff in blue?

[cloud walrus]

Hausdorff

[median pawn]

definitions if you need them

[regal carbon]

Please recommend me an abstract algebra book which has questions with Solved solutions because I'm facing difficulty in solving problems and proofs and exams are not too far.​

[wooden ember]

i thought the classification of finite simple groups was completed in 1981 but lang here says that as of 2001 the proofs of classification are still incomplete

[scarlet estuary]

classification of finite simple groups was completed in 1981
it wasnt

it was completed in 2004 with a slight error

that error was resolved in 2008

(it didnt take 4 years to fix, it just wasnt noticed for 4 years)

(dumb gap in one of the programs used)

(easily patched up)

a lot of parts of the proof have now been computer-verified so its "probably" correct

but its so massive that an error mightve slipped by unnoticed still

hopefully not.

[wooden ember]

that would really suck lmao

[hidden haven]

Yeah imagine the real world impact if they find out that there is actually a simple group of order 10^13 + 1 when they thought there are none

[median pawn]

Any idea what's going on here?

Also is there a better place to learn about affine spaces than this book that I am reading

It seems convoluted

[wooden ember]

still it would be pretty sad : proofs of this length are only going to become more common in the future so hopefully we can trust them

[hidden haven]

what are parallel subspaces?

[hidden haven]

proof checking using lean, coq etc are also becoming more common so I doubt it will be a big issue

[median pawn]

is the notation clear?

i realize it seems non-standard

[hidden haven]

Sorry I was busy for a while. So you have that d(x', y") = s+t, so x'+s+t = y, and x'+s is in X'. Converse should be similar

[median pawn]

makes sense. should be y''

this question (and next) is from an ongoing exam - don't answer it

[hidden haven]

which exam? tag mods

[median pawn]

it's a take home exam at my uni <@&268886789983436800>

[paper flint]

Thanks for notifying.

[cloud walrus]

Hausdorff

[hidden haven]

yep, nice

[wooden ember]

If p iterates over all primes then isn’t m undefined?

The orders of each x_p is a power of p so the least common multiple should be unbounded

unless we can show there exists p so that A(q) is trivial for all q>p

which is obvious when A is finite

[mild laurel]

x is an element of the direct sum so x_p is not 0 only for finitely many p

[wooden ember]

right of course

bruh i really do be making stupid comments again lately

[wraith obsidian]

if L: V→ℝ is ℝ-linear, then we know it factors as the composition V→ℂ→ℝ through L': V→ℂ and the real part Re: ℂ→ℝ.
We can see this explicitly by setting L'(x) := L(x)-iL(ix).
Is there any abstract reason for this? I feel like there must be something involving the extension of scalars

[mild laurel]

what does ix mean?

[wraith obsidian]

i (in ℂ) multiplied with x (in V)

Perhaps I should've said that V is a complex vector space

[cloud walrus]

Or x1
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[median pawn]

For the second part of the proof, I have been able to show that $d_y^{-1}(t) = d_x^{-1}(t + d(x',y'))$ but we need $d_y^{-1}(t) = d_x^{-1}(t - d(x',y'))$ for the proof to go through. Any ideas how to show that?

[cloud walrus]

Hausdorff

[frank fiber]

i think i find some presentations for the posibilities of $A$. $\langle a,b,c|b^2,c^4\rangle$,$\langle a,b,c|b^4,c^2\rangle$, $\langle a,b,c|a^2,c^4\rangle$, $\langle a,b,c|a^2,b^4\rangle$, $\langle a,b,c|a^4,b^4\rangle$ and $\langle a,b,c|a^4,c^2\rangle$

[cloud walrus]

Or x1

[frank fiber]

but i find this supposing that n=3, how can i find the cases when n is not 3?

[wooden ember]

the classification of finitely generated abelian groups has a dope proof

it's long but the outline is nice (ugly details though)

[wooden ember]

Could someone describe a use for inverse limits less complicated than something like the Tate group?

Nvm I just got one: apparently the inverse limit of Z/p^nZ with the canonical homomorphisms is the p adic integers

Which I suppose makes some sense actually 🤔

[upper pivot]

yep, and theres something more general called profinite groups which are also inverse limits, if you wanna look into them

[wooden ember]

What are those?

[upper pivot]

its kind of similar to this, you take the inverse limit of some directed system of finite groups

[wooden ember]

Is it in the case where you have an infinite normal tower and you take G/H_n with the canonical homomorphisms?

[upper pivot]

yeah thats also a thing, called the profinite completion

it reads as making a group where you only care about the finite quotients

[wooden ember]

Oh actually I’m being a bigot

[upper pivot]

(you want the G/H_n to be finite)

[wooden ember]

I see that the word profinite appears on the next page

[upper pivot]

the nice thing is if you give these groups the discrete topology then the inverse limit will be compact, haussdorf and completely disconnected

oh nice lmao

the p-adic integers is a profinite group right

and so it has these topological properties too

[wooden ember]

I see

[upper pivot]

another application is infinite galois theory

you can view the galois group of an infinite extension as the inverse limit of the finite extensions galois groups

oh are you familiar with galois theory?

[wooden ember]

I’m afraid I’m not haha

[upper pivot]

ah i see thats fine

[wooden ember]

I’ll give it another think once I get properly acquainted with galois groups though

[upper pivot]

yeah its one of the cooler applications

for reference pierre guillot's local class field theory, ch3 and 4 are a good read on the topic (with p much no prereqs except basic top and galois required) if you are interested in the future

[wooden ember]

Taking note

Thanks!

Yup the text mentions it

[upper pivot]

oh nice

what book are you reading out of

[wooden ember]

I don’t understand most of it but it does clarify the use of the word limit

[wooden ember]

Lang

[upper pivot]

ah i see

[wooden ember]

I’ll probably work simultaneously with it and DF from now on: I was just reading through the group theory section to check if DF didn’t miss anything and it was a good idea as there was a decent amount of stuff I hadn’t seen

[upper pivot]

to be fair like, you'd probably learn things like profinite groups when learning about their application as opposed to your algebra 1 class right

so i wouldnt be too scared of missing out on stuff like this from DF

[wooden ember]

Oh no of course

But like for example there was an actual proof of the fundamental theore of finitely generated abelian groups

Which DF left out for later

[upper pivot]

i see just blackbox that

[wooden ember]

And the main upshot is Lang didn’t gloss over cat theory

So I got to rethink about some of the stuff in simple cat theoretic terms

[wooden ember]

I did but it’s nice to see the proof

[novel parrot]

he proves it later

and easier proof

[chilly radish]

The proof of the fundamental theorem of finitely generated abelian groups sucks anyways

[terse crystal]

Whether n equals 3 doesn’t matter at all

[terse crystal]

Lots of them are isomorphic, you can keep only the 4th and the 5th. And you forgot to add relations such as aba^-1b^-1. The 4th one is the direct sum of Z, Z/2Z and Z/4Z. And I don’t think the 5th one is correct, it supposed to be isomorphic to direct sum of Z and Z/8Z

Like <a,b,c: commutators, a^4,b^2a^-1> is isomorphic to direct sum of Z and Z/8Z

[terse crystal]

In order to extend L to L’ from V to C . Write L’=L+iH where H is a map from V to R. In order to make L’ to be a C-linear map it should satisfy that L(ix)+iH(ix)= L’(ix)=iL’(x)=-H(x)+iL(x). Therefore H(x)=-L(ix)

[frank fiber]

this is because gr_kA=0 for k diferent from a,b and c?

[terse crystal]

Yeah

[frank fiber]

thanks

[ripe night]

I’m working in Dummit and Foote and there’s a question about using the subgroup lattice diagram to find the centralizers of the elements of a group, and I’m not completely following their example. The book says that to find the centralizer of s in D_8, for example, we see that <s,r^2> is a subset of the centralizer because r^2 is an element of the centralizer (which we’ve apparently already calculated), and there’s no other group except the entirety of D_8 for which <s,r^2> is a subgroup, so <s,r^2> must be the centralizer. I’m fine with all of those steps, but for some larger group that we’re not necessarily as familiar with, how would we even know to have checked if r^2 in particular is in the centralizer? Was r^2 chosen because <s,r^2> is the only subgroup other than D_8 containing <s>, or in practice, would I just have to guess and check?

[hollow topaz]

I have a quick question, for a group such a^3=e (e identity), then so does a^-3=e?

[ripe night]

Yep @hollow topaz , consider
a^3 = e
(a^3)(a^-3)=e(a^-3)
e=a^-3

[hollow topaz]

Oh thank you

[opal osprey]

@fossil shuttle Sorry for the ping, but do you happen to know anything about quantum groups and their representation theory at all? I am looking for some references.

Also, if possible, could you give a brief description of what people mean when they say a certain algebra is a ''quantization/deformation'' of another algebra? For instance, what does it mean when people say a Weyl Algebra is a ''quantization'' of the Symmetric Algebra or that a Clifford Algebra is a ''quantization'' of the Exterior Algebra?

[next obsidian]

https://math.stackexchange.com/questions/4260440/theorem-26-10-the-cartier-equality-in-matsumuras-commutative-ring-theory?noredirect=1#comment8861293_4260440

If any of you can answer this I’ll give you a big Chmonkey kiss

[void cosmos]

king cartier

[novel parrot]

how to do this

so nilradical is prime

then kinda lost what to do

[golden pasture]

how would you define the nilradical

one of the definition gives you the answer immediately

[novel parrot]

i figured it out!!

but my nilradical is intersection of all primes

[novel parrot]

which definition?

[golden pasture]

yup

[novel parrot]

what does closure of topology mean?

[hidden haven]

closure of a set in topology means something

never heard of closure of topology

[novel parrot]

i actually mean set

😅

closure of Y meaning, the smallest closed set containing Y in X?

[hidden haven]

Yes

which exists because intersection of closed sets is closed

[novel parrot]

Y also has a topology yeah?

[hidden haven]

Yeah the subspace topology

but that is not relevant here

[golden pasture]

would suggest learning some basic pointset before doing AM

but yes closure of A is intersection of all closed sets that contains A

[novel parrot]

ive already finished AM

[golden pasture]

oh wat

[novel parrot]

but just doing the problems i skipped

🙂

[golden pasture]

wait how

[novel parrot]

well i didnt do the section of completions

other than that

[novel parrot]

been studying this for maybe 2 months now lol

[golden pasture]

isnt most of AM in the exercises anyways

2 months is q short

[novel parrot]

but i also studied section 15 of dummit and foote

lol

before

[novel parrot]

yeah i like it

short proofs to everything

only the section on completions needs topology

[novel parrot]

here again, the closure has a topology the subspace topology?

[hidden haven]

Every subset of a topological space can be given a subspace topology

Unlike algebraic structures 😌

[novel parrot]

but can topology do anything cool like algebra can

[hidden haven]

Yes

[light sluice]

Hmm

[novel parrot]

like what