groups-rings-fields

[chilly ocean]

yes

are all finite groups cyclic?

na

[lavish nexus]

No

meaning g^k=e for some k

[chilly ocean]

think of Z/2Z x Z/4Z

[lavish nexus]

k!=0

[chilly ocean]

what is the single generator for this group

trick question

none?

[lavish nexus]

so like I have Z/10Z

[chilly ocean]

well think about (1,1)

does this generate the group

[lavish nexus]

Then {0,2,4,6,8} is a cyclic subgroup

generated by 2

and the order is 5

Now for any group G with prime order
consider the subgroup generated by any element of G

[chilly ocean]

if you work out examples of possible generators you will find there arent any that single handedly generate entire group

[lavish nexus]

it could be the identity element which generates the trivial subgroup

[chilly ocean]

so there are 0

yes

the group is finite and not cyclic

there are simplier examples though

[lavish nexus]

but for every other element what can the order be?

[chilly ocean]

fuck man

[lavish nexus]

|<g>| ||G|=p

[white jackal]

bruh just gimme the answer not a riddle

[lavish nexus]

what is |<g>|

[chilly ocean]

the order of one of the elements

in g

in G yea

[lavish nexus]

it divides a prime

[chilly ocean]

have you learned lagrange theorem @chilly ocean ?

[lavish nexus]

so it’s either 1

[chilly ocean]

i did bruh

[lavish nexus]

or the prime itself

[chilly ocean]

this is so confusing

what is confusing?

all lagrange says

is

order of h divides order of g

idk how you guys got all this it amazes me

im a junior in college

[white jackal]

"A group of order p prime has exactly one generator" is this true or false and why? can you guys help me on this problem?

[chilly ocean]

youll get there soon

yeah im trying

[chilly ocean]

can yoy think of a counter example?

[lavish nexus]

order of any subgroup H divides order of the whole group G

[chilly ocean]

yes i agree on this

[chilly ocean]

or better yet use lagrange theorem

[white jackal]

doesnt all the coprimes generate the group itself?

[chilly ocean]

thats what i thought

[chilly ocean]

what are coprimes?

[lavish nexus]

a cyclic subgroup is a subgroup

[chilly ocean]

[lavish nexus]

so the order of a cyclic subgroup divides the order of G

[chilly ocean]

yea

always

and what are divisors of prime numbers

boom you get your answer

[lavish nexus]

the order of G is prime, so the cyclic subgroup has the same order as the order of G

[white jackal]

1 and itsefl\

[lavish nexus]

then the cyclic subgroup is the whole group

[chilly ocean]

so for this statement are we assuming H is a cyclic subgroup

cause it says generators

?

no

[lavish nexus]

so the group G is cyclic and every non identity element is a generator

[chilly ocean]

generators are just elements of a group that generate a subgroup

okay

all elements of a group generate a subgroup

[lavish nexus]

you can always take any element of a group and generates a cyclic subgroup

[chilly ocean]

okay so A group of order p prime has exactly one generator

Yes that can be shown in three steps

where does lagranges theorem

take play in all of this

<g> is the subgroup of G generated by g

every subgroup H of G has order dividing order of G by lagrange theorem

so order of <g> must divide order of G

We know G is prime

so order of <g> must be 1 or the prime

if it is 1 then <g>=e

if it is the prime then it is G

[white jackal]

thank you u saved my ass tonight

[chilly ocean]

☺️

[lavish nexus]

we pick the subgroup generated by any element!= identity
then show its order is the same as the whole group
since order is the number of elements
if you think of <g> and G as just sets with <g> contained in G
and they have the same order hence same amount of elements
we must have <g>=G

then G is cyclic

[white jackal]

bro youre so smart that you dont even understand the question

[lavish nexus]

oh whatever you wanna take Zp go ahead

everything except 0 generates Zp are you satisfied?

his question was what is the number of generators of a group G of prime order
why don’t you show how they’re all isomorphic to Zp without proving G is cyclic

Oh right go talk about invariant factors

[chilly ocean]

?

wait

wtf is going on

[white jackal]

ookaayyy at least i touch grass

[chilly ocean]

It isnt much to show that finite cyclic groups of same order are isomorphic

Is a theorem

[lavish nexus]

yes but you have to prove prime order implies cyclic

to prove that well use Lagrange

[chilly ocean]

We already went over this though

[lavish nexus]

Indeed we were already done

[chilly ocean]

so if a group had order 11

only 1 and 11 would generate it?

[lavish nexus]

It’s prime order so it’s cyclic

[chilly ocean]

which means

it has

2 generators?

[lavish nexus]

which means every subgroup is either just the identity

or the whole group

because it has to divide the order of the whole group which is prime

[chilly ocean]

so 1 and 11

[lavish nexus]

every element except identity generates a nontrivial cyclic subgroup, which must be the whole group

so every element except the identity is a generator

now in Z11

[chilly ocean]

so just 11?

[lavish nexus]

the elements are {0,1,2,…,10}

identity would be 0

every other element generates the whole group

so the generators are 1,2,3…,10

They are also all coprime to 11

[chilly ocean]

ohhhh i think i get it it

thanks bro

so for 11

it would be

3,5,7,9?

@lavish nexus

[lavish nexus]

what does coprime mean

gcd is 1

[chilly ocean]

yes

[lavish nexus]

11 is prime

[chilly ocean]

yes

3 is prime

5 is prime

7 is prime

9 is prime

so those are coprime

[lavish nexus]

Yes those are but

what is gcd(4,11)

[chilly ocean]

1

woah

[lavish nexus]

Do you see the problem

[chilly ocean]

so its everything

yes

[lavish nexus]

yes

[chilly ocean]

so 11?

there are 11 generators?

[lavish nexus]

everything except 0

gcd(0,11)=11

[chilly ocean]

ahhhh

okay

so 11 wouldnt be one

[lavish nexus]

No

[chilly ocean]

cause gcd(11,11) is 11

so its

1,2,3,4,5,6,7,8,9,10

[lavish nexus]

also 11=0 in Z11

Yeah

The number of elements coprime to some integer n is denoted φ(n)

[chilly ocean]

oh okay thanks bro

[lavish nexus]

so this is the answer you first got

[chilly ocean]

now i know what that symbol means

isnt it also similar to U(n)

[lavish nexus]

It would be in (Z/nZ)*

as the group of unit in Z/nZ under multiplication

[chilly ocean]

got it

thanks so much

[white jackal]

since |Imφ|·|Kerφ| = |G|, then |φ(a)| divides
|a|. Why can you not make this leap?

[chilly ocean]

oh this completely wrong bro

its pretty clear

[chilly ocean]

Every integer is coprime to a prime number

yeah

gcd(n,p)=1 for all n if p prime fo

idk how i forgot that bruh

[chilly ocean]

actually bro idek what this means im jk

[chilly ocean]

lol

bro

[lavish nexus]

assuming he knows 1st iso

[chilly ocean]

first iso theorem is more important than a boyfriend

this man drew a triangle

[hidden haven]

Lol

[chilly ocean]

im done bro this shit too abstract

no dont

i want you to stay

dude this class is killing me

im so lost

it gets way worse when you go higher, you will appreciate the struggle

i need a map from dora the explorer bro

or when you talk about sylow theorems ☺️

this my last math class

😣

[hidden haven]

F

[chilly ocean]

I wont blame you, if you dont enjoy it

i dont enjoy this class at all

I am a proponent for human rights

i loved combintorics

intro to proofs

[white jackal]

i love psych101

[chilly ocean]

You talked about symmetric group?

no

[lavish nexus]

well maybe you’re an analyst

[chilly ocean]

in this class?

[white jackal]

i know how to fit 5 choclolates into 4 boxes

[chilly ocean]

pigeonhole principle baby

[white jackal]

stars and bars baby

[chilly ocean]

breast first search

[hidden haven]

Then are you sure it's not just that your prof this time is bad

[chilly ocean]

he is bad

@hidden haven enablor

yall rocking with the bfs?

or yall dfs type guys

boyfriends!

[hidden haven]

I'm a tree exists kinda guy

[white jackal]

me in love with suffix trees

[chilly ocean]

any of yall CS majors?

imma make some abstract algebra nfts

[hidden haven]

Me

[rustic crown]

me too 🙈

[chilly ocean]

noice

same

Where do people typically go after finishing an algebra sequence in undergrad?

Finishing with galois theory for example

Id think they would jump to commutative algebra stuff

but when do they use the galois theory stuff?

[lavish nexus]

ag use it

not very sure about rep theory

[lethal dune]

nft deez zuts

[lavish nexus]

Just a taste of the horrors of cat theory diagrams

[fresh charm]

ECE here I'll appreciate your nfts

[lavish nexus]

I remember when I first saw it in intro to proofs

A month later it seemed completely natural

[chilly ocean]

ill make em soon

gotta finish this operating systems class first

[lavish nexus]

But also there’s algebra sequence in grad
at least the second semester would be basically all new stuff

[chilly ocean]

Yeah atm im taking second sem grad alg

and we learning comm alg and baby ag

but galois theory stuff on backburner

i also took an at course

and it felt like there was so many things i could build off of from there

[rustic crown]

i guess you can use lots of galois theory for doing alg nt

[maiden ocean]

Galois theory is geometry in disguise anyway

[rotund aurora]

How to interpret this kind of diagram ? The middle part is, I think, the usual way of putting the fundamental theorem. The rest I suppose is talking about the kernels of the different maps? But have no idea how to read it. Maybe this has to do with exact sequences or something likke that, but I have not gotten into that.

This is from the wikipedia article btw

thanks

[maiden ocean]

the long diagonal going from the bottom left to top right is an exact sequence yes

the only thing the diagram on its own says is that these maps commute

so like

f = iota circ pi

[wraith obsidian]

Not sure if this has an answer, but does the language of lattices help us in any way to formulate these statements in a more compact manner?
It's from Carter's book on lie algebras, and abstractly all of these statements are various compatibility conditions for the subspace / subalgebra / ideal lattices of a lie algebra L

like, (i) and (ii) tells me that the embeddings Id(L) → SubSp(L), SubAlg(L) → SubSp(L) are meet-semilattice-homomorphisms

Hm… does (iv) mean that Id(L) → SubSp(L) is also a join-semilattice-homomorphism?

[pastel cliff]

library's copy of artin is checked out

https://tenor.com/view/crying-emoji-dies-gif-21956120

what textbooks do yall use besides that

or judson which my library also doesnt have

[next obsidian]

I used D&F and Aluffi

[rotund aurora]

You can find the pdf in google

[pastel cliff]

the point is that i cant stand pdf's lol

[rotund aurora]

Check the pinned message in #book-recommendations of Several Sloths, he gives some recomendations there

[pastel cliff]

i... can't see book-recs, i have perma study

[rotund aurora]

I don't know what this means tbh

[pastel cliff]

i have a role that hides all the discussion channels

[rotund aurora]

pdfs or reading in any electronic device is much superior than physicial books, specially for maths. Because you don't have to hold the cover so that it doesn't close. And so on

but just my opinion

[pastel cliff]

ive heard a lot of people cite D&F, is there any benefit to one over the other

my library also has neither actually

https://tenor.com/view/kermit-falling-building-fall-woodz-gif-24715383

[final oasis]

Herstein

[pastel cliff]

this one?

we actually have it let's fucking go

[final oasis]

Topics in algebra

[pastel cliff]

oh

LETS FUCKING GO WE HAVE THAT TOO

[final oasis]

Topics in algebra is what I used

[next obsidian]

Bro what kinda library do you have at your school

One of them is at a higher level than the other

The lower level one doesn’t talk about eg group actions which is kinda monka

[pastel cliff]

ok correction we have em theyre just checked out

which means other people in my class probably checked em out which means they’ll be checked out the whole semester

[next obsidian]

What kind of school do you go to monka

[pastel cliff]

it’s like we all have stands and we’re gonna see who does best with what textbook

not actually i dont watch jojo or know most of my classmates lol

[next obsidian]

Okay other books which I know exist

Lang

Umm

[hidden haven]

Jacob son

[next obsidian]

True

Jacobson is insanely cheap too

Like <15 bucks I think

[hidden haven]

You bought it?

[next obsidian]

Cuz dover

No

[pastel cliff]

interesting

[hidden haven]

Nice\

[next obsidian]

But it’s a dover book so it’s insanely cheap

So if $$ is a concern

[pastel cliff]

thank u all for suggestions

[slate summit]

How do you solve x^5 - 5x^2 + 15 = 0 in radicals?

[chilly ocean]

Use the quintic formula

[slate summit]

like wikipedia says it's solvable

https://en.wikipedia.org/wiki/Quintic_function#Other_solvable_quintics

and it's galois group is D_5

[pastel cliff]

oh yeah, proof by desmos

for iv) isnt that just... by the way the elements are defined...?

[chilly ocean]

Pretty sure Wew can answer that

Wew buddy can you help, stop lurking

[pastel cliff]

he's muted

i talked to him tho lol

guys quick wew lads is muted post wew lads slander

https://tenor.com/view/epic-embed-fail-ryan-gosling-cereal-embed-failure-laugh-at-this-user-gif-20627924

[chilly radish]

Just to make sure i'm not being stupid: This exercise has me prove that the direct sum of quotient modules is isomorphic to quotient of the direct sums. Now, this is true for a direct product too right? (With an arbitrary index set ofc, otherwise both notiond are the same). I'm not seeing any reason why the obvious argument won't work, it just makes it a bit weird the exercise is specifically about direct sums

Direct sum = elements s.t. finitely many indices are nonzero, while direct product has no such restriction

[hidden haven]

What is the argument you are thinking of?

I don't have a counterexample but the proof I have in mind only works for direct sums

[chilly radish]

Just 1st iso

Where does it fail for direct products

Like define the obvious map from
\prod M_i -> \prod M_i/N_i
Ans apply first iso

[hidden haven]

Hmm what would be the kernel of that

is it product of the N_i

[chilly radish]

Yea would it not be

It's 0 exactly when all the elements are in Ni

Like I don't think the cardinality of the index set affects anything

[hidden haven]

Right

That seems like it works

[chilly radish]

Alright cool, just direct products with arbitrary index sets can be kinda funky sometimes so I wanted to make sure I wasn't missing anything

[hidden haven]

I might be missing something too

Like this argument seems fine but the statement feels somewhat sketchy and unnatural to me on sight

[chilly radish]

Same tbh

I don't see where it goes wrong if not finitely.many of your elements are nonzero

[hidden haven]

Yeah

Might be true then

Oh I guess this should just be a simple enough statement to google

[chilly radish]

I tried

Everything I found only deals with finite index sets yho

Tho

[hidden haven]

Same I don't see anything and that makes me more skeptical lmao

This is weird

[chilly radish]

Very

[weak oriole]

Moldi honourable now

OwO

[hidden haven]

[chilly radish]

Moldi has been honourable for long

[hidden haven]

omg when did that happen

[chilly radish]

Lmao

[weak oriole]

Oh color change nvm

[chilly radish]

Oh another thing that's semi-related that idr if I asked here. I know the contravariant hom-functor turns direct sums into direct products
(i.e. Hom(\oplus Ai,B)=\prod Hom(Ai,B))
But why does this fail w/ direct products

Loll

[hidden haven]

well while you are here

[weak oriole]

UwU?

[hidden haven]

do you know if direct products preserve quotients

it seems so weird because can't find anything by googling and it is not a left adjoint so why would it preserve coeqs

[hidden haven]

but this argument 😵‍💫

[hidden haven]

It works with direct products if the product is in the second slot

[chilly radish]

Yea ik, i'm asking specifically about the contravariant functor tho

Like, rotman gives the above as a theorem but doesn't talk about direct products in the 1st slot

[hidden haven]

Because if it worked with direct product, then direct product would satisfy the universal property of the direct sum

[weak oriole]

is rotman good

[chilly radish]

Is there like a more concrete reason?

I enjoy his books a lot

[hidden haven]

Well you could run the proof of uniqueness of universals

Use this isomorphism

to get maps going both ways

And you would get an isomorphism between the direct sum and direct product

There's a theorem by Yoneda (not the Yoneda lemma) which says that there is a universal morphism from x to G iff Hom(y, -) = Hom(x,G-) for some y

Where the y is in the domain category of G

[chilly radish]

I see

[pastel cliff]

hi sorry to interject but is this the same as saying G is "generated" by those two matrices

[chilly radish]

Do you happen to know an example where iso doesn't hold for the coproduct case

[chilly radish]

Yes, generated in the sense of algebras

[hidden haven]

Probably just take vector spaces since you can argue by dimension

You could maybe assume that that bijection holds and get a condition on the dimension

[chilly radish]

Hmm

[pastel cliff]

hi wew lads

[hidden haven]

and dimension of product and sum are different for infinite

[hidden haven]

yes

Assuming that your notion of generating is the correct one

[pastel cliff]

i ummm uhh

uhhh

i do not trust this professor so

[chilly radish]

But I mean in the case whwre both are products i.e.
Hom(\prod Ai,B) and \prod Hom(Ai,B)

[hidden haven]

Ye so for example take A_i = B = Q

As a Q vector space

[pastel cliff]

oh shit oh fuck he's back

[chilly radish]

Sure

[delicate orchid]

run

[hidden haven]

Then the right side is countable if the indexing set is countable

But countably infinite product of Q will have uncountable dimension

So the first thing should be uncountable

[pastel cliff]

what is the correct one then tho

[chilly radish]

Wait why would the right be countable then, isn't it also a countably infinite product of Q

[hidden haven]

Smallest subalgebraic structure containing those elements

[hidden haven]

Right is countable

left is not

[chilly radish]

I'm asking why

[hidden haven]

oh

Countable product of countable sets is countable

[pastel cliff]

i just imagine that G is the set of all matrices that can be reached by multiplying those two, and whatever they generate in turn, as long as it's closed under the operation

[hidden haven]

wait

[chilly radish]

This is definitely not true

[hidden haven]

did I just do something very stupid

ye

[chilly radish]

Also both sides will have Q^N

Hence why i'm confused

[hidden haven]

bruh moment

[chilly radish]

Lmao

[delicate orchid]

yeah that's right

[pastel cliff]

epicness

[chilly radish]

Multiplying and adding

[delicate orchid]

it's a group

[chilly radish]

It says subalgebra

[delicate orchid]

his prof is a crank

[hidden haven]

subalgebraic structure

[chilly radish]

Oh...

[pastel cliff]

very this

[weak oriole]

So do direct products preserve quotients
Or have you moved on

[hidden haven]

The argument seems legit

[pastel cliff]

algebra doesnt imply algebra over a field in this class

[delicate orchid]

he's 2 weeks into group theory Shin

[pastel cliff]

just an algebraic structure

[chilly radish]

I.c.

[weak oriole]

Bruh

[chilly radish]

Universal algebra moment

Well i'm still thinking about it but I want a counterexample to contra hom functor preserving products

[pastel cliff]

still couldn't define most of these damn terms

[hidden haven]

Ye I think too

[pastel cliff]

monoid? sounds like a single-balled individual

wew lads you fucking monoid

[chilly radish]

Why would u not provide a counterexample rotmans. Does it require choice or smth

[delicate orchid]

the way I think about generating groups is G is generated by a, b then G consists of all elements of the form a^nb^m for n, m integers (subject to some relations )
this isn't the "formal" way but you know what I do not care

[hidden haven]

Non constructive proof of counterexample by Yoneda

[chilly radish]

And b^na^m too for nonabelian groups

[delicate orchid]

very true

[chilly radish]

U don't need explicit relations yoy just assume everything is in simplest form

Generated groups = all finite products of elements and inverses in generating set

[chilly radish]

I very much hope a concrete counterexample exists

If there is one, idt one of the modules can be free

[pastel cliff]

in this example then G is just four elements right

[hidden haven]

why is that?

It shouldn't work for free modules either

By the yoneda argument again lol

[chilly radish]

Oh chrew

But I feel like it would be much harder to find a concrete counterexample

[hidden haven]

perhaps

[chilly radish]

Maybe uhh, if ur ring doesn't have ibn?

[hidden haven]

[chilly radish]

Idk any rings like that

I saw a paper.on arxiv that said that if contra hom functor preserves all products then the base ring is 0

In zfc

[hidden haven]

That makes sense yeah

[delicate orchid]

no, it's 8 I think
the matrix with just one -1 is order 4 and the other is order 2
and both their subgroups are disjoint

[hidden haven]

By Yoneda

[chilly radish]

What scares me is the need to specify

In zfc

[hidden haven]

Bruh

[chilly radish]

[hidden haven]

Wait

In the example I gave

The left side is bigger than the right side still

[chilly radish]

Idts?

Isn't left side still just Q^N as a Q vector space

Or uhh

Wait

[hidden haven]

It's N^(N^N)

[chilly radish]

Wait why

Wouldn't
Hom(Q^N,Q)
Just cardinality N^N

[hidden haven]

Because the direct product of Q with itself |N| times is already Q^N^N

Wait

What the fuck am I saying

[chilly radish]

Both sides have a direct product of Q with itself N times

[pastel cliff]

idk what that last bit means but im gonna assume i should check order of each element from now on for shit like this

[hidden haven]

No direct product of Q definitely isn't countable dimension though

[chilly radish]

Oh yea I agree

But both sides have that factor

But if you're like, doing Hom with the base field idt that increases the cardinality

[hidden haven]

Ah true

[delicate orchid]

checking their order is a good place to start, the only problem is when you get something like a^3 = b or something like that, that could interfere with the order
or like ab = ba, weird nonsense

[chilly radish]

I think that holds for both contra and covariant hom

[delicate orchid]

alternatively just compute all 8 combinations and show they're all distinct explicity

[hidden haven]

It does

Dual is larger

[chilly radish]

Wait seriously

Fuck this

[hidden haven]

Ye for infinite dimension

[chilly radish]

Contravariant hom functor can suck my dick

[hidden haven]

Dual is power set dimensional

[chilly radish]

I suspected as such moldi

But I didn't know how to show it

Ok so that's a counterexample

Albeit a very unsatiafying one

[hidden haven]

Damn this is so dumb I can't believe I'm struggling to count dimensions

[chilly radish]

It's not even dimension it's cardinality

[hidden haven]

Am I going senile already

Lmao ye

[chilly radish]

You're good moldi, just go to sleep

And think about product preserving quotient

[hidden haven]

[sturdy marsh]

what's the problem?

[hidden haven]

It's done lol

[sturdy marsh]

O

[hidden haven]

He wanted an explicit counterexample for contravariant hom preserve product

[chilly radish]

Ye

[hidden haven]

And we were struggling to count the dimensions in the counterexample I have for like 15 minutes

[chilly radish]

Cuz rotman said it turns sum into product but no further discussion

I still have no intuition for why this fails but ig I'll come back to what u said about universality a bit later

[hidden haven]

Just learn Yoneda smh

[chilly radish]

I need to get more category brained

I know yoneda I just don't have a feel for it

[hidden haven]

Oh actually this can also be proved using yoneda lemma

Rather than the small theorem he proved that I was using so far

Well you gotta use both

But with both it's actually just an equation and you are done

[chilly radish]

Does what u said about it saying that product will have universal property of coproduct require those

[hidden haven]

Well the equation is
Product of Hom(A_i, B) ≈ Hom(Product of A_i, B) ≈ Hom(Sum of A_i, B)
And yoneda lemma says that Hom(X, -) ≈ Hom(Y, -) implies X ≈ Y

so easy

[hidden haven]

Require what

[hidden haven]

Oh ye I straight used yoneda lemma here nothing else actually

Along with the assumption that this holds for both

Of course I'm assuming that the isomorphism you have is natural in B

There's no way you can describe an isomorphism here that works for everything but isn't natural lol

Like informally speaking

Maybe if they were vector spaces because choosing bases breaks naturality

But you can even do that for modules

[chilly radish]

Yea I think it would have to be natural

Ok I see, that's really nice

Yoneda lemma is incredible

[hidden haven]

Doing my topology hw and spamming yoneda

For simple stuff

Hope they don't mind 😌

[chilly radish]

Lmao

[pastel cliff]

you ever just forget that matrix multiplication isnt commutative

[hidden haven]

Oddly enough this was never proved in our classes

[chilly radish]

Oh if you don't.mind expanding a bit on why yoneda implies that. Like from the formal statement

Huh

[hidden haven]

no because that's like the most prominent example of a non commutative ring lol like you can construct a very large class of nice non commutative rings from it

[hidden haven]

It comes from the fact that X mapping to Hom(X, -) is a fully faithful functor

[delicate orchid]

rudeness...

[hidden haven]

All fully faithful functors reflect isomorphisms (if the image is an isomorphism so is that preimage)

[next obsidian]

Uhhhhhh

I'm chmonkey mode

[hidden haven]

[chilly radish]

Wait so where does yoneda come in

[hidden haven]

And to prove that it is fully faithful you need to identify what maps from Hom(X, -) to Hom(Y, -) look like

Yoneda does it slightly more generally

Classifies maps from Hom(X, -) to any F

Those are exactly the elements of FX

[chilly radish]

Ahh I see

[hidden haven]

So take Hom(Y, -) = F you get Hom(Y, X)

[pastel cliff]

[hidden haven]

And if you retrace the bijection everything works out

[chilly radish]

I c

[pastel cliff]

when you said this did you consider both left and right multiplying matrices...

[delicate orchid]

[prisma shuttle]

does anyone get this

[pastel cliff]

does the suffering ever stop

im at 8 elements of G already only using the first matrix

[delicate orchid]

ok something's gone awry

it's definitely order 4

and 2

yeah

should be 16 unless the matrices so happen to commute

[pastel cliff]

i found one pair that commutes so far

im just going through and doing every combination

[delicate orchid]

is it a^4 and b^2 perchance

[hidden haven]

Bruh what are you trying to do

[pastel cliff]

find all the elements in G

[pastel cliff]

but also idk if this is specific enough bc a*a^2 is not the same as a^2 *a

unless im overthinking it

[delicate orchid]

actually the second matrix squared is in the centre of Z(GL_2(R)) so yeah it makes sense some should commute

[delicate orchid]

no

this is never the case

because groups are associative

[hidden haven]

Think about it geometrically

[pastel cliff]

yeah i was trying to see the transformations

but then i lost track

[hidden haven]

One matrix is reflection of the plane along x=-y

[pastel cliff]

and the other is a rotation ccwise yeah

[hidden haven]

The other is clockwise rotation by 90

[pastel cliff]

clockwise

[hidden haven]

Is it not

It is one of the 2 who cares

[delicate orchid]

it's counter clockwise but yeah same thing

[hidden haven]

I would claim that you have reflections along the 2 axes, the 2 diagonals, and rotations by multiples of 90°

That should be all

Isn't this exactly D_4

Yes

[pastel cliff]

probably yes

[hidden haven]

This is it

[pastel cliff]

bc there's a question about D_4 on this homework as well

[hidden haven]

Nice

[pastel cliff]

i was trying to do it numerically and i got more matrices tho that's what im confused

[hidden haven]

For D_4 here's a cheat code

[pastel cliff]

like i saw the connection early on but ignored it

[hidden haven]

Take any reflection call it s

[pastel cliff]

up up down down left right left right b a

did i do it

[hidden haven]

Take a 90° rotation call it r

Then your set will be exactly {r, r², r³, 1, sr, sr², sr³, s}

Because you will be able to prove that

[pastel cliff]

r^4 being the identity tho

[hidden haven]

Ye

There

[hidden haven]

r⁴ is 1

s² is 1

And rsrs is 1

[delicate orchid]

<#groups-rings-fields message>
🚎

[pastel cliff]

i mean yeah i can see all this but then multiplying those two matrices from before in different ways gives more shit

unless actually wait

[hidden haven]

It can't check your multi

[pastel cliff]

no yeah it def does

[pastel cliff]

wdym

[hidden haven]

D_4 is a subalgebra of this

And D_4 contains both of those matrices

[delicate orchid]

it definitely doesn't

[hidden haven]

So the subalgebra generated by them must be contained in D_4

By closure properties of D_4 itself

[delicate orchid]

you're fucking up your matrix multiplication I hate to say it think about the transformations, you cannot rotate by 90 more than 4 times before ending back up where you started, and reflecting twice gives you fuck all (the identity)

[pastel cliff]

[delicate orchid]

I fucking hate these exercises btw can I just interject for a second

why force students to do 16 matrix multiplications (8 if they notice r^2 is in the centre of GL_2(R))

[pastel cliff]

yeah i dont like this professor at this point lol

maybe ignorance wouldve been more bliss

[hidden haven]

It might help to just think of how these transformations behave in the x and y axes

[pastel cliff]

yeah doing it like that makes perfect sense

[hidden haven]

Ignore all else and it becomes a counting problem

[pastel cliff]

im just gonna redo my multiplications i guess

[hidden haven]

All that these 2 things can do is permute the half axes

In a way that once you know what happens to the positive x axis you know what happens to the negative x axis

So they just map positive x axis to 1 of 4 choices and positive y to 1 of remaining 2

They and none of their compositions can do anything wacky other than these 8 things

Now just check that you can get all 8 however you want

[delicate orchid]

yeah and then map them back into matricies

[hidden haven]

[subtle ivy]

this is why dummit and foote is the way.

some of the best exercises i have seen in any math text

[pastel cliff]

ok if a is $\begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$

[cloud walrus]

nitezba

[delicate orchid]

calculate a^2 real smooth like

[pastel cliff]

it's just -1's on the diagonal

[delicate orchid]

yup

no it isn't

infact

[pastel cliff]

[delicate orchid]

matricies that are scalar multiples of the identity commute with fucking everything

literally EVERYTHING

this violates associativity on top of that

[pastel cliff]

so uhhh

i did my multiplication wrong

[delicate orchid]

a^3 is $\begin{pmatrix} 0 & -1 \ 1 & 0 \end{pmatrix}$ and I can tell you that for sure without even doing matrix multiplication

[cloud walrus]

Wew Lads Tbh
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[pastel cliff]

cringe pmatrix

[delicate orchid]

lmfao

[pastel cliff]

i need a break

gonna go hide in a cave for a little while, might fuck around and fall in love with some shadows idk

[delicate orchid]

alternatively note that (a^2)^2 is the identity

kinda funnY

[pastel cliff]

i promise im not always this fucking braindead

[delicate orchid]

yeah sure, and I'm not braindead either

we're all braindead other than moldi (and det)

[hidden haven]

[spice whale]

I've started Knapp and the NT at the start has completely lost me

[pastel cliff]

modulo stuff?

[spice whale]

no

[pastel cliff]

euclidean algo?

[spice whale]

like weird divisor stuff

yeah

euclidean algo is fine

just

all the proofs are losing me

[pastel cliff]

yeah proofs are funky

i understood the at one point

that point has since passed

[spice whale]

i just skipped the last few theorems in the section

[pastel cliff]

mood

[spice whale]

I'm sure i won't need them...

[pastel cliff]

famous last words

[chilly ocean]

Guys where does this highlighted implication come from?

whats confusing me is that the rationals under addition arent finite

[gritty sparrow]

If [Q:H]=n then the group Q/H has n elements, so if x is any element of the group you can see that the subgroup generated by x must have order dividing n, in other words nx=0 in this group. so picking any element r of Q we see that rn belongs to H

[chilly ocean]

okay gimme a sec

lemme read this lol

[lavish nexus]

lol this is a cute formulation

[chilly ocean]

what i wrote? LOl

[lavish nexus]

no the problem

it is just a field has no nontrivial ideal

but said in terms of groups

[chilly ocean]

can you help me understand this

i understand up to the highligted part now

[lavish nexus]

because N is normal Q/N is a group

the elements are cosets

[Q:N]=n means there are n cosets

so Q/N is a group of order n

[chilly ocean]

i already knew the index of a subgroup is a group

why did we have to show it was normal first

[lavish nexus]

otherwise Q/N isn't a group

[chilly ocean]

but didnt we assume that [Q:N] = n?

[lavish nexus]

that's the number of cosets

[chilly ocean]

and the def of index in this case would be Q/N ?

[lavish nexus]

cosets don't require N to be normal

[chilly ocean]

so the first highlighted portion

is basically saying

wait

i honestly dont know why they do Q/N

like how did that even come up and why does that matter

[delicate orchid]

_ _

something wacky happens because this is finite

[chilly ocean]

is this a formula?

i dont get why we make the jump from being normal to that statement

[south patrol]

Q/N has as its underlying set the cosets of N in Q, so (provided it exists) it has cardinality [Q:N]

[delicate orchid]

it's a standard result of quotient groups

[chilly ocean]

we havent even went over quotient groups smh

[lavish nexus]

so now that Q/N is a group of order n
the elements of Q/N are qN
but then qN+qN+...+qN n times must be N

hence nq is in N

then nQ is in N

but nQ is Q

the end

[delicate orchid]

such a wholesome ending

[delicate orchid]

this is true because q is an arbitrary rational number btw

[lavish nexus]

yes

[chilly ocean]

fuck man

[delicate orchid]

if you haven't done quotient groups yet idk why you're looking at this result

[chilly ocean]

im missing a piece of the puzzle mentally

this homework was assigned last friday

[delicate orchid]

maybe there's an obtuse way to formalise it without quotient groups?

[delicate orchid]

I could go through the steps of iteribus' proof in very high detail if you want

[chilly ocean]

okay so let me ask a few questions lol

if H is normal to G then |G/N| = [N:H] ?

[delicate orchid]

H normal in G then |G/H| = [G:H]

[chilly ocean]

got it

how did you know the elements look like qN

[delicate orchid]

definition of a quotient group

[lavish nexus]

actually should be q+N

[delicate orchid]

it should

[chilly ocean]

so quotient groups are just left cosets>

?

[delicate orchid]

they're cosets

because H is normal the left cosets are equal to the right cosets

that's why normal subgroups are defined like they are

[chilly ocean]

how is qN + qN .... = N

[delicate orchid]

I'll let you take this one iteribus

[lavish nexus]

everything raised to the order of the group is identity

[chilly ocean]

yes thats in my notes lol

i see it

[delicate orchid]

it follows from "the order of the element divides the order of the group"

[lavish nexus]

identity in Q/N is N = (q+N)+...+(q+N) n times = nq+N

[chilly ocean]

thats the identity?

[delicate orchid]

(the group operation on G/N is (g+N)+(g'+N) = (g+g')+N btw, and e+N = N so N is the identity)

[rotund aurora]

Has someone got Dummit and Foote pdf with bookmarks?

[chilly ocean]

how did he get nq + n

and then say nQ is in N

and noticed that nQ is in Q

[delicate orchid]

nq is in N because nq+N = N

and nQ = Q

[chilly ocean]

how are those equal bruh

man im so done with this

i literally read the cosets and lagrange chapter

i took notes

this is nothing like what the book has

[delicate orchid]

nQ is equal to Q because if you multiply every fraction by n you still just get every fraction

[lavish nexus]

you know 1 is in Q

is 1 in nQ?

[chilly ocean]

yeah

[lavish nexus]

ok then everything that is in Q is in nQ

[delicate orchid]

alternatively, show that Q -> nQ, a/b -> na/b is a group isomorphism (since we don't care about the field structure here)

[lavish nexus]

infinite group isos are problematic

[delicate orchid]

nah just check da axiomz

actually you can just show it's a straight up bijection from Q to itself, not even isomorphic it's just equal

[lavish nexus]

Z is isomorphic to nZ

[chilly ocean]

bro

i cant pass this class

im not made for it

literally none of this makes sense to me besides definitions

[delicate orchid]

true, you'll have to consider the identity map Q -> nQ, a/b -> na/nb ig

[proud bear]

wait what's wrong with a/b->na/b?

[lavish nexus]

an infinite group can be isomorphic to one of its own proper subgroups

[delicate orchid]

shows the sets are isomorphic but not equal as sets

very important distinction between the two

[chilly ocean]

What you think

[delicate orchid]

drop the "and the elements are" and just write "consider"

[lavish nexus]

get rid of |Q|/|H|

they're infinite you can't divide their order

[chilly ocean]

done and done

[lavish nexus]

also change every N to H lol

[chilly ocean]

oh yeah LOL

Better?

why did we do the addition chain

[lavish nexus]

they add up to identity which is H

[delicate orchid]

exploiting the fact that every element raised to the power of the group is the identity

[lavish nexus]

but they also add up to nq+H

so nq is in H

[chilly ocean]

so those qs

would be diff qs

[lavish nexus]

the same q

[delicate orchid]

no

if they were different we would've written them q_1, q_2, etc

it's literally x+x+x+x+... = nx just with a +H after it

[chilly ocean]

i get how you got nq +H

the thing is

q is just and arbritary element

[lavish nexus]

that's the point

they are arbitrary

[delicate orchid]

that's why we can go from nq in H to nQ in H

[chilly ocean]

so any q that i would choose from Q

would have this behavior

[lavish nexus]

yes

[delicate orchid]

yes

[lavish nexus]

so for any q in Q, nq is in N
so nQ is in N

[delicate orchid]

"if every element of a set is contained in another, that set is a subset" basically

[chilly ocean]

yeah i think i get it

kinda of a bummer i didnt know how quotient groups worked

but the hw is due fri

[delicate orchid]

if you're expected to do this without quotient groups I will eat my own shoe

[chilly ocean]

lol

dude my prof is a troll

did i tell u the class median for the midterm was 10/30

so everyone basically failed

[delicate orchid]

maybe that should've been a wakeup call for him

[chilly ocean]

our foundation is so weak

im still lost in this class

like im playing catch up

the whole quarter

[delicate orchid]

shame

[pastel cliff]

just wait till analysis

[delicate orchid]

anyway fun fact this generalises to any infinite abelian group

[pastel cliff]

friend of mine got a 22 on the final

[chilly ocean]

this my last math class lol

[pastel cliff]

crimge

[delicate orchid]

crigne

[chilly ocean]

this was an optional cs elective bro

major mistake

i wake up at 6am and still cant get all my work done

[pastel cliff]

that's not very sigma of you

[chilly ocean]

my other classes are killer lol

[delicate orchid]

literally just groups just the funny groups lolooolol

[lavish nexus]

just like my organic chemistry once

I never took organic chemistry again

[chilly ocean]

lol

i took gen chem and called it

[lavish nexus]

also you might consider just going through exercises on dummit

they're usually not very difficult

[chilly ocean]

dummit?

[lavish nexus]

what book you using

[chilly ocean]

im using the gallian book rn

it makes more sense to me

contemporary abstract algebra

[subtle ivy]

last semester the averages on my analysis exams were never above a 60%

[subtle ivy]

this exercise is lifted directly from dummit and foote's text. now, there is no real context for this problem in the book outside of the statement, but it still might be worth looking over.

[dire summit]

ay me too lol :)

i've got a bit of a silly question: my homework problem is "characterize those n such that the only idempotents of Z_n are 0 and 1"

without my putting any thought into it, am i crazy for coming to the answer that n=p^k works for all primes p?

i tried running a few examples and the first ring with nontrivial idempotents was Z_6 (a=3), followed by Z_10 (a=5) and Z_12 (a=4), which makes me think i have the right idea

[lavish nexus]

you want a^2=a

a^2-a=0

this has two roots

wait

[dire summit]

no wait i think i got it

[lavish nexus]

nvm

[dire summit]

i want a^2 congruent to a mod n

[pastel cliff]

oh my god do i need to do the subset both ways bullshit for 2a

[lavish nexus]

I'm just concerned about the zero divisors

[dire summit]

so n | a^2-a -> n | a(a-1)

[pastel cliff]

@delicate orchid you're awake hi do you have any bleach

[dire summit]

since a and a-1 are relatively prime the only way a can exist is when n has two relatively prime divisors

[delicate orchid]

https://cdn.discordapp.com/attachments/359052581022203914/940656085893799936/unknown.png

[delicate orchid]

Exactly the same as for the matrices

[pastel cliff]

i know it's isomorphic to D_8 yeah

[delicate orchid]

In fact it’s the same group

Oh

Well excuseeee me

[pastel cliff]

princess

i realized that much already i mean like specifically the "smallest subalgebra" part

[delicate orchid]

I knew you’d get the zased ref

[pastel cliff]

like im assuming that needs proof right

[delicate orchid]

By definition the subgroup generated by a set is the smallest subgroup that contains that set

No proof should be required

[pastel cliff]

yselkhfasdljfh my prof is a horsecock man idk

he took a point off a proof in my last hw bc i didnt do a "claim + proof structure"

[delicate orchid]

Just write this and cite “Wew lads tbh” he’ll know what’s up

[pastel cliff]

submitting this jpeg instead

[delicate orchid]

Alternatively you could consider a larger algebra containing those elements and then show it has extra bonus ones you don’t need

Because by definition, the subgroup generated by elements is a SUBGROUP

[ember field]

What do it mean for a group to act on another by automorphisms?

[pastel cliff]

do i do the wew lads and be lazy or do i do both ways to be safe

t-minus two hours remain

[delicate orchid]

“The wew lads way”
Underestimating me again are we

Say there exists a subgroup H <= G that contains all elements of a set X and H is a proper subset of <X>. Then there exists some element h in <X> such that h is not in H, but H must be closed under multiplication for arbitrary products of elements in X, and because h is in <X> h is equal to such a product so it must be in H - contradiction, so H = <X>

I think this is correct but I can hardly be bothered to check

[delicate orchid]

[pastel cliff]

noooooo

who's gonna tell me how much of a doodoo prof i have

,ti moldi

[cloud walrus]

This user hasn't set their timezone! Ask them to set it using ,ti --set.

[dreamy jewel]

Bruh commutative algebra is driving me crazy

And it’s only like the first week…

[pastel cliff]

welcome to the club

[dreamy jewel]

Thanks Nitezba

Appreciate your vote of optimism

Maybe I am not made for this cutthroat industry

[pastel cliff]

YES YOU ARE

KEEP FIGHTING

WAX ON, WAX OFF

[pastel cliff]

https://www.youtube.com/watch?v=KxGRhd_iWuE

,ti

[cloud walrus]

The current time for nitezba is 11:09 PM (EST) on Wed, 09/02/2022.

[pastel cliff]

my hw is due midnight and im still pushing you can do it

[dreamy fiber]

Nothing is impossible

Yesterday you said tomorrow

So JUST DO IT!

[pastel cliff]

MAKE YOUR DREAMS COME TRUE

JUST

DO ITTTTTTTTTTTTT

[dreamy fiber]

Yes you can!

[pastel cliff]

YES YOU CAN

[dreamy jewel]

Bruh 😂

Thank you

[pastel cliff]

you knonw you're in college when that video stops being funny and just becomes actually helpful

[dreamy jewel]

Good luck on your homeworks too

[dire summit]

"prove that the field of quotients of a field F is isomporphic to F" what

[dreamy jewel]

Mine too is due in less 50 minutes

[dire summit]

imma be real with you guys i don't know where to begin

[dire summit]

good luck lol

[dreamy fiber]

Lol mine’s due in an hour and 1 day

[dreamy jewel]

Thanks

[dire summit]

mine's due in ~10 hours

but ideally i'd like to sleep before then

[dreamy jewel]

Best of luck to everyone here

[dreamy fiber]

Well if you are talking about field of fractions of a field F then just consider the map from f: F to Frac(F) where f(a) = a/1

[dire summit]

oh shit that totally works yeah

let me get back to you once i finish the proof

[dreamy fiber]

You have got this 👍

[dire summit]

ok surjectivity's proving a little tricky to word out but i think this isn't too bad

thanks for the tip :)

[prisma shuttle]

can someone explain why this mapping is a ring isomorphism over multiplication

because ima getting that we need that $(r_1r_2)bn+(s_1s_2)am = r_1r_2(bn)^2+s_1s_2(am)^2$ but idk how to show that

[cloud walrus]

JustKeepRunning

[prisma shuttle]

also what does it mean by showing that this map is "well-defined"

[tribal moss]

"this map is well-defined" is a common sloppy way of saying "the definition actually manages to define a map".

[prisma shuttle]

you mean like every output is unique?

but isn't that obvious

[tribal moss]

Not really. The definition pretends that (r,s) is an element of R/(m) × R/(n), but actually the elements of R/(m) × R/(n) are pairs of congruence classes.

And by writing (r,s) and then using r and s in an expression where they need to be elements of the ring, the definition effectively says "whatever, just pick a member from each congruence class at random".

In order for that to work, we need to be sure that the result points at the same congruence class in R/(mn) no matter which random picks you make.

[prisma shuttle]

wait but isn't it just like

because its (integer * m + r) * bn + (integer * n + r) * am = r * bn + s * am regardless of wut the "integers" are

[tribal moss]

Yes, and that's the "one can check" being alluded to.

[prisma shuttle]

ok thx so much

do you get my other question about the isomorphism

idk like why its not working out

[pastel cliff]

in this definition, rho composed with tau is its own inverse right?

[tribal moss]

Hmm, good question that I don't have an answer for right away...

[pastel cliff]

i know it's the dihedral group but i realized that ive been doing transposition wrong this whole time so now im overthinking things

[prisma shuttle]

what is "it" referring to here?

[pastel cliff]

rho tau reverses rho tau

[prisma shuttle]

no

[pastel cliff]

[tribal moss]

Actually anything with a tau in it is its own inverse.

[prisma shuttle]

wait

[pastel cliff]

yea so it is right

ok cool

[prisma shuttle]

oh yea i think u are right

[pastel cliff]

it's late and im overthinking everything, just needed some other eyes on it

[prisma shuttle]

yea for these types of things u can just draw it out with like a square

and label the vertices 1,2,3,4

and just try the two operations

[pastel cliff]

yeah that's what ive been doing

[prisma shuttle]

idk if there's a faster way (maybe using rs = sr^{-1}) but this usually how i do it

[tribal moss]

It's generally true about dihedral groups that all the reflections have order 2.

[pastel cliff]

tropo can u confirm that the only element here whose inverse isnt itself is rho, since its inverse is rho^3

[prisma shuttle]

by order you just mean that if i do the same reflection twice i just get back to itself right

[tribal moss]

Yes.

[pastel cliff]

was that @ both things

[pastel cliff]

ik this much is true

[tribal moss]

Yes was to JKR.

Nitezba appears to be right too, though, for this particular group.

[pastel cliff]

thank you tropo

hw was due 30 min ago but we out here grinding

[subtle ivy]

this is a very good explaination.

[subtle ivy]

it may help to notice that you can formulate the identity for the multiplicative equality as an equality of inner products...

[pastel cliff]

i just realized how dumb my homework is

[subtle ivy]

then you can minimize notational clutter, and the result follows rather nicely.

[pastel cliff]

arent ii)and iii) pretty much asking for almost the same thing

since 5 of the 8 elements are of order 2

showing their order to be 2 should be just the same as showing they are their own inverse

[tribal moss]

Ah, I got it. We have am + bm = 1 = (am+bn)² = (am)²+(bn)² (the cross terms vanish), and therefore am - (am)² = (bn)² - bn. Since this common difference is a multiple of both m and n, and since they have no common factor it must be a multiple of mn. Thus am = (am)² and bn = (bn)² modulo mn.