groups-rings-fields

[topaz solar]

I.e. taking an object in the kernel gives you a 0 in the image

So this gives you a way to describe quotients/extensions

Don’t worry about this shit atm, just thought it relevant

[solemn rain]

i dont understand this notation

ker(f) ---> G ?

you mean there is a function between them

?

[topaz solar]

yes

[solemn rain]

im slow wXD

[topaz solar]

dwai, it’ll make sense more later

[solemn rain]

okay

can you like give me any exercises or anyshit

so ic an see if i understand?

so i did this one on my own

which like cool af

G is a cyclic group with generator a

G is isomorphic to Z/kZ

[topaz solar]

Have you shown image is a subgroup?

[solemn rain]

G is of order k

yea

and shown ker is normal too

ker is normal to G

[topaz solar]

Also try R/Z with R as reals, Z as integers

[solemn rain]

and img is subgroup to H

wait let me do this again

[topaz solar]

Good

[solemn rain]

the img one

img(H) = { phi(a) | a is in G }

[topaz solar]

It’s fairly obvious, just making sure you’ve done it

[solemn rain]

oh okay

i thoguht it was important

yea its easy a bit idk

Also try R/Z with R as reals, Z as integers

[topaz solar]

$\mathbb R/\mathbb Z$

[solemn rain]

Z is normal to R?

[cloud walrus]

Darkrifts:

[topaz solar]

it’s abelian so yes

[solemn rain]

obv under addition right

yea

what about R/Z?

[topaz solar]

I want you to look at it

[solemn rain]

hows

how

R/Z = {a+Z | a is in R}

[topaz solar]

Better exercise: Show what (G x H)/H is

Yes, just gave R/Z as a basic thing to look at

[solemn rain]

okay i didnt read this def yet

but i assume

G x H = { (g,h) | g in G h in H }

[topaz solar]

product of groups is Cartesian product

[solemn rain]

yea

[topaz solar]

With pointwise operatiobs

(ab, xd)

[solemn rain]

how can i multiply

an element in GxH times a coset

[topaz solar]

well,

1. show there is a subgroup 0 x H iso to H

2. show it’s normal

3. find the quotient

[solemn rain]

0 x H = { (0,h) | h is in H }?

[topaz solar]

Ye

(g,h)(g’, h’)=(gg’, hh’)

Also note: there are canonical maps from G x H to G and to H

Where you throw away one of the terms

Use this knowledge to your advantage

[solemn rain]

cant i just say

phi : 0 x H ---> H

phi(0,h) = h

why cant i just say that

xd im so stupid

[maiden ocean]

I'm a little bit confused about the argument used here to prove this lemma:

[cloud walrus]

Sloth:

Sloth:

[mild laurel]

is N(P) the normalizer of P

[maiden ocean]

yes

also change ur pfp degen

[cloud walrus]

Sloth:

[mild laurel]

look at the def of sylow p-subgroup

hm

Yeah hm

[cloud walrus]

Sloth:

[maiden ocean]

i dont see how this implies that k must = 0

[mild laurel]

HP is a subgroup of N(P) which is a subgroup of G

[maiden ocean]

right

[mild laurel]

so

[maiden ocean]

um, p^k divides |G| i guess?

[mild laurel]

more than th

that*

[maiden ocean]

hm

[mild laurel]

HP is a subgroup of G

[maiden ocean]

right

[mild laurel]

its order divides G

[maiden ocean]

im kinda lost 😦

sorry

[mild laurel]

u got this

[maiden ocean]

hm

lemme think abt it more for a bit

oh my goddddd

@mild laurel how did i not get it

im so dumb

lmao thanks

[mild laurel]

tru

well you got it faster than i'm figuring out this question

and im pretty sure this should be obvious but

[maiden ocean]

its probably an actually hard question tho

jesus i cant believe i took so long to get that

[mild laurel]

eh it might just be tedious

yea that was kinda dumb

[maiden ocean]

i havent done any math in like 2 weeks b/c of school and ive already reverted to absolute brainlet

[mild laurel]

it happens

[ornate verge]

Not to me

[mild laurel]

Sorry, I should have said that it happens to those of us that aren't geniuses like you

[chilly ocean]

lmfao

[solemn hollow]

Is that the real Thomas Bayes :OOOOOOOOO

[barren plank]

I think not, Thomas Bayes has died in 1761

[maiden ocean]

why am i being so god damn dumb agh

say we have the dihedral group of order 2n

generated by the relations $\langle x, y \mid x^n, y^2, (xy)^2 \rangle$

[cloud walrus]

Sloth:

[maiden ocean]

how do we show that x and y are conjugate in D_2n?

[mild laurel]

xyxy=e

[maiden ocean]

holy shit im fucking stupid

galaxy brain

fucking expanidng

of the absolute largest size

[mild laurel]

crazy idea

[maiden ocean]

reeeeeeee

why is this tied to odd n tho

im trying to show that if u have two involutions u and v in G and the product uv is of odd order

then u is conjugate to v

so my strategy was liek, consider <x, y>

but i dont see how this argument is limited to odd

maybe im just smoothbrained to the highest degree

oh wow im going to kill my self

i had a basically complete argument

and then i fucked up because i couldnt complete the obvious last step

gdi

ugh wow

[mild laurel]

happens

[maiden ocean]

blegh

[mild laurel]

I wrote that \binom{n-d}{n} = \binom{d}{n} earlier today

[maiden ocean]

lol

man today is not my day math wise

[mild laurel]

happens

you can take breaks

[polar hazel]

what algebra is that @mild laurel

[mild laurel]

I have to finish this hw lmao

part of an alg geo assignment

[polar hazel]

Oh ok

[maiden ocean]

aw man

i just wanna finish these last 2 problems

so that i can understand the characterization of S_5 he gives

[mild laurel]

not even midnight yet

[maiden ocean]

true

[mild laurel]

u got it

[maiden ocean]

thanks

[mild laurel]

I have like 4 more problems left to do on this assignment lmao

[maiden ocean]

lol

is it due tmrw?

[mild laurel]

yeah

[topaz solar]

Gotta hit that grind my dude

[maiden ocean]

oof

im sure ull get it

[mild laurel]

its not really a grind cause idk how to do these problems lmao

[topaz solar]

Grind that reading to answer them

[solemn rain]

darkrifts remembeer that group you told me to look at

i gave up , googled it

turns out its isomorphic to the unit circle

i g

can some1 tell me how to 'look' at R/Z

[mild laurel]

It's the unit circle

[solemn rain]

yea, how did you get this

[mild laurel]

Think of an isomorphism

[solemn rain]

i dont even know how R/Z looks like

like

R/Z = {a+Z | a is in R}

does this describe the set of cosets of Z in R?

should i try to find a surjective homomorphism easier?

where the kernel is Z?

or would that be too specific and hard to find?

[mild laurel]

They're equivalent

[solemn rain]

okayy

did i write the group right?

nvm boys

[magic owl]

Mo2

Take R

Start with [0,1] what happens here

Well, we identify 0~1

And that’s obviously a circle

Then what happens on [0,2]?

Well we glue 1~0

And then we glue like, 1+\epsilon to \epsilon

And we keep doing that

Wrapping around again and again

For all of R

@solemn rain

Does that make sense

[bleak abyss]

Formally, use the exponential map to get that [0,1] with endpoints identified is the circle

[magic owl]

formally you can just take the quotient but formalism wasn’t the point

[bleak abyss]

I mean showing the quotient is topologically S^1

I don't believe in Hatcher-style topology, I want my proofs lol

[barren ice]

Hey guys! my exam is coming up soon and my teacher mentioned something about problems that combine both arithmetic and geometric sequences and series. However I don't understand what he meant by that as I have never encountered a question like this. have any of you encountered questions like this, and if you have can you please @ me or private message me as I do have a few questions. Thanks 🙂

[bleak abyss]

That topic is not suitable for this channel, look at #prealg-and-algebra

[south goblet]

Hey there! recently I'm reading GTM150 to give my first try to commutative algebra, but the book seems somewhat difficult to me (the first chapter), anyone could give me some advice pls?

[mild laurel]

That's the eisenbud one right?

[south goblet]

yap

[mild laurel]

Do you have any idea why you're finding it difficult?

Have you seen these topics before?

[south goblet]

No, it's my first time attach this area

I can read the proof

but sometimes I just don't know the purpose

[mild laurel]

Can you give an example of that?

And yeah, usually commutative algebra textbooks assume you've seen the basic rings stuff before and so they go very quickly through that

Commutative algebra is definitely something you study after a first algebra course, which contains some stuff on rings

[south goblet]

like some invariant theory

yeah, I have already read GTM73, Algebra by Hungerford

[mild laurel]

what do you mean by invariant theory/

[bleak abyss]

So the point of invariant theory geometrically is that you have groups acting on varieties, and you're wondering whether the quotient is a variety or not

[south goblet]

maybe that sort of thing will become clear when I go further😂

[bleak abyss]

That's the entirety of my knowledge about it tbh

But like for instance Z/2 acts on S^n by the antipodal map, right?

The quotient is RP^n, which is a manifold

[mild laurel]

Yeah, sometimes things in commutative algebra aren't fully motivated until you learn more alg geo, which is what Eisenbud tries to do but

[bleak abyss]

Yeah, I'm not familiar with Eisenbud in particular really

But yeah I mean, sometimes you're just gonna have to hold tight and take things at face value a bit. Or you can try to see if you can ask around in case there's a nice elevator pitch out there

[mild laurel]

I've read bits and pieces of it but

Oh you're talking about the first chapter

Yeah I wouldn't too much about realizing why everything in that first chapter is important

It's supposed to be more of a motivation chapter, in that it tells you why commutative algebra is important for all these other things

If you've read all of Hungerford, you're definitely ready for commutative algebra

[south goblet]

@mild laurel @bleak abyss thanks so much! I will keeping going on!

[mild laurel]

good luck

[cerulean siren]

can I ask for homework help?

I'm learning rings and I'm having trouble with one of the proofs

The book says: define Z[i] to be { r = a + bi | a, b are integers }. Then, norm(r) = a^2 + b^2. Show that r is a unit in Z[i] <=> norm(r) = 1

I got as far as I want rs = 1, which means (a+bi)(c+di) = 1, but I'm not sure where to go from there

[mild laurel]

try something

what's something you could try

[cerulean siren]

I tried expanding the multiplication to ac -bd + bci + adi = 1

but I don't see a way to get only a and b on one side

[mild laurel]

you don't need to

when are two numbers in Z[i] equal?

[cerulean siren]

when a=c and b=d

[mild laurel]

Use that here

[cerulean siren]

like set r = s?

[mild laurel]

Hm there are nicer proofs

Have you shown that the norm is multiplicative?

norm(rs) =norm(r)norm(s)?

[cerulean siren]

yes, I have N(rs) = N(r)N(s)

[mild laurel]

okay, use this

rs = 1

[cerulean siren]

hmm I could apply norm to both sides

and so I have norm(rs) = norm(1) = 1!

hold on I think I might have it from here 🙂

[mild laurel]

good luck!

[cerulean siren]

ooh and it works out because we're working with Z[i], not C

so (a^2 + b^2) only has a multiplicative inverse if it's 1, right?

I know Z only has 1 and -1 as units

yeah and a^2 + b^2 is in Z, not in Z[i]!

and since it's squared it can't be -1 🙂

[mild laurel]

yeah

the norm for this is always a positive integer

[cerulean siren]

I am not, we're using Abstract Algebra by Saracino

[uncut girder]

@chilly ocean I am!

Just gonna start the journey

[chilly ocean]

So I'm stuck with computing quotient thing. The answer should be in a form a+bX + cX^2 (X=theta) right?

the closest I got was (x^3-2)/(2x+4)

[mild laurel]

Find the inverse of the denominator

[chilly ocean]

ohh ok yeah that solves it, thx. Too much calculations for that tho

[primal mantle]

i hope this is the right channel lol

for the proof of lagrange's theorem im finding it difficult to understand one part

The bit where u show each coset has the same amount of elements in it

i dont get

[topaz solar]

the coset $aH = {ah\vert h\in H}$

[cloud walrus]

Darkrifts:

[topaz solar]

if ah=bh, what can you say about a and b @primal mantle

[primal mantle]

a = b @topaz solar

[topaz solar]

Think about how this means $\vert aH\vert = \vert H\vert$

[cloud walrus]

Darkrifts:

[primal mantle]

a=b implies H -> aH is a unique mapping (sorry if im being wack with terminology lol)

1 to 1 mapping sorry

@topaz solar

[topaz solar]

ye

So this holds for any coset a

[primal mantle]

oh of course lol

wait hmm

ah ok yh i get it

how do we know it'll be an integer amount of cosets in the group?

[topaz solar]

There’s an integer number of elements

@primal mantle

[stone fulcrum]

I personally like to show that "belonging to the same coset" is an equivalence relation, thus forming a partition on the group. Then show that every coset is the same size as the subgroup. @primal mantle

[bitter mauve]

please help with 3.1.2

[mild laurel]

what did you try

[bitter mauve]

well, i looked at examples

[mild laurel]

okay and?

Did you try showing either direction?

[bitter mauve]

yea i was thinking of showing that a chain with non finitely generated I1 won't stabilise but how'd I even show that. is the trick being like infinite sets could go on and on? like <G> is a subset of <G - blah> and on and on?

$$\langle G\rangle \subseteq \langle G - F_1 \rangle \subseteq \ldots \ldots$$ for $F_i\subseteq F_{i+1} \subset G, |F_i|<\infty$

??

idk if im correct with the math notation tho

[cloud walrus]

nnutannep:

[mild laurel]

"non finitely generated I1"?

[bitter mauve]

i mean order of G being infinite

sorry if wrong words

infinitely generated ideal I_1

oh wait

feels like this proves it

in my head at least

[outer estuary]

Is the concept of index in rings analagous to the one in groups?

[mild laurel]

I don't see how this proves anything

[bitter mauve]

😦

[mild laurel]

As in, what are these F_i's

and how do you know that this chain doesn't stabilize

[bitter mauve]

finite subsets of an infinite set G

[mild laurel]

@outer estuary are you talking about index of subgroups?

[outer estuary]

Yes

Index of ring was referenced with context to ideals and modules

[bitter mauve]

i thought like process of subtracting finite sets from an infinite would never stop

[mild laurel]

@bitter mauve Are you sure that <G> is contained in <G - F_1>?

[bitter mauve]

hmm....

its not?

[mild laurel]

@outer estuary I'm still not quite sure what you're asking but no, people don't really talk about the index of an ideal in a ring

why not?

[bitter mauve]

im asking lol

i felt like it is

contained in a smaller set

[mild laurel]

Well let's say that f is in F_1

then since f is also in G, then f is in <G>

but is f in <G - F_1>?

[bitter mauve]

yea ok

no

yea im wrong

proves nth

nvm

its so wrong lmao

just realised

[marble bolt]

Seems like this channel is busy, so disregard this question until it is free

Im trying to find all ring homomorphisms from Z to Z/30Z.
I got {0,1,6,10,15,16,21,25}
Are those right?

[mild laurel]

This isn't super clear

0 is not a ring homomorphism

You probably mean sending 1 to 0

[marble bolt]

Are the ring homomorphisms supposed to be integers?

[mild laurel]

Uh no?

[bitter mauve]

maps

[mild laurel]

Ring homomorphisms are functions

in this case from Z to Z/30Z

[bitter mauve]

$\langle F_1\rangle \subseteq \langle F_2 \rangle \subseteq \ldots \ldots$ for $F_i\subseteq F_{i+1} \subset G, |F_i|<\infty$

hmm.... bot?

[mild laurel]

I can read it its fine

[bitter mauve]

yea actually i was supposed to say it the other direction

[mild laurel]

So what are the F_i here

[marble bolt]

Ok thanks

[bitter mauve]

they are finite subset of the infinite set G that generates some ideal (infnitely many generators) in the non noetherian ring. feels like this is the contrapositive version

i know that <F1> is contained in <F2> lol

[mild laurel]

And why doesn't this chain stabilize?

[bitter mauve]

cuz there doesnt exist a number i_0 else we'd have order of G finite

thanks for your help man, my shift ended gotta run

[mild laurel]

Not sure what you mean by that

Technically the way you have it written

We can have F_1 = F_2 = F_3 = ...

[bitter mauve]

hmm... if proper containment then it works no?

or like we do union

[mild laurel]

yes but you need to give a better argument for why the chain won't stabilze

[bitter mauve]

oh, thanks. ill try

[mild laurel]

@marble bolt Do you understand what we were saying?

[bitter mauve]

what about each F being disjoint and I talk about <F1>, <F1>+<F2>, <F1>+<F2>+<F3>

etc

[mild laurel]

it'll work, you just have to be careful

i.e., elements in G don't need to be linearly independent

[solemn rain]

idk how you use group action

group actions to prove cayley's theorem

can some1 help

[willow garden]

D&F chapter 4 (pp 119-120) has what you're looking for

[solemn rain]

still lmfao

can some1 prove iit for me

[mild laurel]

just read it

ask questions about what you don't understand

[solemn rain]

okay i understand the details , its just unfolding definitions

but whati odnt understand is the motivation behind defining some certain maps

or like using a certain lemma

so

Theorem: Every finite group G is isomorphic to a subgroup of S_n

where n = |G|

so now this from nowwhere lemma

let G act on an enmtpy set X

define m : G---> S(X) as g--> m_g

where m_g is a map : X--->X as x-->gx

i have shown m_g is in S(X)

now the claim of the lemma is that m is a homomorphism

how would you come up with this

if i were asked to prove this for the first time

how would i think

of this

now i think the proof was then

showing the kernel of this homomorphism is e_g

hence injective hence isomorphism

[mild laurel]

Idk I mean the further you read into math, its going to be hard to ask for that sort of motivation on every single proof you read

[solemn rain]

well ik im not getting any further but

proofs are getting hard af

is it normal to not be able to do this on your own

for the first time?

i may understand the technicality of the proof

but wtf

[stone fulcrum]

Generally the proofs are not the original proofs. These are clever remanufactured proofs made to quickly attain the result

[solemn rain]

ohhh

so basically these are the polished clean stuff

lmao

[stone fulcrum]

You aren't expected to come up with these yourself, but are expected to understand why they work

[solemn rain]

i dont think i do that tbh

like i dont think i know the 'why'

like really the 'why'

how can i improved

improve*

[stone fulcrum]

I don't mean "why they used the proof" but "why the proof they used works"

[solemn rain]

lmao

thats easy ig?

like they did nothing wrong

idk tbh XD

im sorry

[stone fulcrum]

Because I agree with you, this stuff can be very out of nowhere

[solemn rain]

wait

really

i thought it was just me im stupid

im like mega confused

like 70% of the time

the 30% is where i read the defintions and or examples lmao

[stone fulcrum]

Especially group actions, this shit is magic. They're like "this complicated result is now obvious if you use the correct action lol"

[solemn rain]

okayy

so i should just

moveo n?

move on?

without taking too much time?

[stone fulcrum]

I mean get why the proof works

[solemn rain]

on a specific theorem or 'topic'

yea i think i do

i dont have any objections ig

[stone fulcrum]

If you're really stuck with actions, review actions

[solemn rain]

this is my first time with actions for me

[stone fulcrum]

Because they have some very nice properties that generalize a lot of group stuff

[solemn rain]

yea i haveeent gotten into the cool stuff

havent defined orbits or stabilizers

yet ig

would it be stupid to be asked to prove such a theorem

for the first time

in like a test?

as cayleys theorem

or like lagranges theorem even

like i wouldnt have ever thought of

like defining a relation --> classes are cosets ---> cosets partition G etc

is this normal lmao

even fucking using the theorems in some problems can be a bit hard

[tiny pagoda]

Is there a good example of a simply defined field where 2 + 2 would equal 5?

I always love the 2 + 2 = 1 mod 3 joke

[delicate bloom]

if by "good" you mean "finite" then the answer is no

I don't think there's a way to magically do this, but what do I know

[tiny pagoda]

Hmm

[delicate bloom]

to rule out finite fields just look at F_p, after p>=5 it's clear you can't make it

then check by brute force for less than that

[tiny pagoda]

I mean there's a weird one with

{2, 5} is isomorphic to Z_2, and 5 is identity element

IDK the exact syntax for that but it could be a dumb t shirt idea

[delicate bloom]

how?

5+5=2 not 2+2=5 for your field

[tiny pagoda]

2 + 2 through the isomorphism is just 1+1 in Z_2

[delicate bloom]

what's the isomorphism, you're saying f(2)=1 and f(5)=0?

[tiny pagoda]

Yes

0 is additive identity, and so is five

Of course someone already asked this on r/math lol

[delicate bloom]

that's a group, not a field

[mild laurel]

F_2 is a fine field

[delicate bloom]

but it's not a field where 2+2=5

[sharp sonnet]

i mean, first you must be clear what you mean by 2 and 5

[delicate bloom]

"Is there a good example of a simply defined field where 2 + 2 would equal 5?"

[sharp sonnet]

is it just symbols you give to random elements

or is it actually 2 = 1+1 and 5 = 1+1+1+1+1

where 1 is the multiplicative identity

[delicate bloom]

yeah, we could just redefine 2 and 5 to be whatever but that'd be cheating imo

[tiny pagoda]

Yeah, true

The Z_2 thing is still kinda cheating

[mild laurel]

just localize Z at the multiplicative set {0,1}

[tiny pagoda]

Just slightly less IMO.

Also wait why is Z_2 not a field?

[sharp sonnet]

there is no multiplication defined on it

[tiny pagoda]

Can't you define

1 * 1 = 1

0 * 1 = 0

1 * 0 = 0

[sharp sonnet]

yes

[delicate bloom]

write your addition and multiplication tables with 2 and 5

to show how you get 2+2=5

[sharp sonnet]

and then you get the field GF(2) or F_2 or wtv you want to denote it

[tiny pagoda]

I mean basically that's what you're doing with the {2, 5} onto Z_2 thing but with a bit of abstraction

As in just rewriting multiplication

[delicate bloom]

show me

[tiny pagoda]

Hmm?

[sharp sonnet]

the answer is that every field has prime or 0 characteristic

[delicate bloom]

power of a prime*

[sharp sonnet]

so 2+2 = 1+1+1+1 is never 5 = 1+1+1+1+1

im talking about characteristic, not order

[tiny pagoda]

If you map 2 -> 1 and 5 -> 0

Then do addition in Z_2, you're basically just doing a fancy way of redefining the numbers (at the added bonus of murdering every number other than 2 and 5)

[delicate bloom]

oh my bad

[sharp sonnet]

any field of prime power order has prime characteristic

so, unless you randomly denote an element that is not (1+1+1+1+1) with the symbol "5"

or an element that is not (1+1) with the symbol 2

then you will not get this result

if you are willing to call random elements "2" or "5", this result is trivial

[tiny pagoda]

How would you write that in LaTeX? (Not specifically just this, I didn't ever learn most of the notation for it)

[delicate bloom]

0 = 0+0 = f(5 + 5) = f(2+2+2+2+2) = 1+1+1+1+1 = 1

@tiny pagoda

[tiny pagoda]

Ah

[sharp sonnet]

this is just looking at an isomorphism of F_2, where you renamed the identities for no apparent reason

i mean, using 0 and 1 for the identities is pretty standard, so don't change that

[delicate bloom]

I was showing that f(5)=0 and f(2)=1 is not a field isomorphism because you can use it to prove 0=1

[tiny pagoda]

I mean the idea is not to use this rigorously

I'm trying to find good examples of how abstraction can break notions about math

2+2=5 is a particularly notable one

It's used in 1984 and the like a ton as a metaphor

[delicate bloom]

you can't really make 2+2=5 unless you just make the symbols meaningless

[sharp sonnet]

ofc you can define a field isomorphism as f(0) = 5 and f(1)=2

[tiny pagoda]

^

[sharp sonnet]

but 2 and 5 won't behave like the numbers 5 and 2

they will just be symbols

[delicate bloom]

that's what I'm saying

[tiny pagoda]

That's what I was thinking would be the best option

[delicate bloom]

like sure, you can just write any gibberish and just redefine it to the core

[tiny pagoda]

It also illustrates that symbols don't always mean the same thing

[sharp sonnet]

it's like defining f(0) = 😸 and f(1) =

[delicate bloom]

but that's not really what I'd consider an honest misconception like you're trying to achieve here

[tiny pagoda]

Well sure but honest misconceptions are usually more advanced

I think the best one of those I have currently is just how * and + get redefined in Boolean algebra

[delicate bloom]

it's like saying "I like your shoes" but then redefining "like" to mean "hate" and "shoes" to mean "hat"

[tiny pagoda]

LOL

[delicate bloom]

sure, fine, whatever

[tiny pagoda]

If you guys have other good examples that would be more applicable to the thing I'd love some

[delicate bloom]

the rationals inject themselves a handful of places

so it's not really a wise idea to start rewriting the definition of integers for a t-shirt imo lol

[tiny pagoda]

I mean, I wouldn't actually put this on a t-shirt lol

All my ideas for those are just bad puns

[sharp sonnet]

(a+b)^p = a^p + b^p in characteristic p

[tiny pagoda]

"I like the strong inductive type" with the Hindley-Milner rules

Hmmm

How would you recommend dumbing down characteristic?

[sharp sonnet]

it's a simple enough concept

also a pretty common joke

i heard it like at least half a dozen times just last year

[tiny pagoda]

Yeah

Wouldn't GF(2) work as an example for that?

[sharp sonnet]

it has characteristic 2

[tiny pagoda]

Yeah

[sharp sonnet]

like any finite field has order p^n

and characteristic p

and all infinite field are extensions of the rational numbers

[mild laurel]

I really don't know who uses GF(2) outside of programming languages

[sharp sonnet]

it's the simplest example of a finite field

i guess

[mild laurel]

I mean like the notation

Over $\mathbb{F}_2$

[cloud walrus]

Zopherus:

[sharp sonnet]

oh

🤷

[tiny pagoda]

True but this is a "class" that's much more likely to have programmed (actually a mandatory class for some of the students) than have touched AbstractSingletonProxyFactoryBean algebra

What what

Google

Oh no what have I done...

That's how an autocorrect option for when I capitalize Abstract

Well it isn't anymore

God how old is that, it's like from a HackerNews thread called "Everything wrong with Java in a single class" or something

Anyways, these students are not gonna understand F_2 or GF_2 or anything like that as is

They would get me saying "This is like your bool in C "

God I'm stupid

I just spent the last 2 minutes wondering why GF(6) doesn't have a prime field characteristic

[delicate bloom]

just relabel 6 to be a prime number

6 := 1+1+1

[tiny pagoda]

LOL

Seriously though it's like 2:30 am and I should probably get to bed if I'm sitting here potatoing so hard I didn't remember that Z6 is not a field and is in fact the example used to teach that not all quotient rings are fields

Either that or I should just drop out

And it's the freshman's dream problem to boot

[solemn rain]

can some1 explain to me orbit-stablizier theorem

class equation?

i tried but i cant

at all

all i know is the def of a group action

[sharp sonnet]

i liked this blog entry by timothy gowers: https://gowers.wordpress.com/2011/11/09/group-actions-ii-the-orbit-stabilizer-theorem/

[solemn rain]

this was not how the orbit satblizier thoerem was stated for me

|X| = sum(|G(x)|) over x in R

= sum([G:G_x]) over x in R

what does G(x) or G_x mean

maybe i should redo the whole topic

since im stupid

[delicate bloom]

don't be hard on yourself, it will only make it harder for no reason

just sleep on it and come back to it fresh tomorrow

[solemn rain]

im being extremely hard now im so frustruate

d

problem is even id ont think im good enough with the formeer sections lmfaaaaop

[delicate bloom]

that's fine, the further you go, the more context you have for what/why things earlier on were important to go back to, it's just how it goes

[chilly ocean]

Can confirm

[solemn rain]

yea i shouldnt have not payed attention when

i was reading about system of represenatations if i remember correctly

[chilly ocean]

@chilly ocean sup

hello there

[golden pasture]

lol orbit stab stuff took me like a few rereads to internalize it just sleep reread and repeat if it doesnt work

[solemn rain]

so fucking hard lmai

and practically i cant do any problems

[topaz violet]

Can anyone tell me why this is true? It's a step in the proof of Sylow's Third Theorem so it's supposed to be obvious, but I can't figure it out.

[mild laurel]

Hm, can you show the whole proof?

[topaz violet]

Does this help?

[mild laurel]

Well okay, one containment is clear

You can see why the right set is contained in the left one righ

[topaz violet]

Yes, P<N(P) by definition, so 1*N(P) is a member of the left set.

[mild laurel]

Hm, what's lemma 6.3?

[topaz violet]

Apologies that I didn't provide this earlier, I didn't think it was being used for that particular equation.

[mild laurel]

I mean yeah, this lemma exactly does what you want

The first thing in the lemma shows why X^P = the set

Then you can also see that X^P = {N_G(P)} since every element of P is trivial in X

[topaz violet]

I see that the lemma gives the first equation. Can you expand on that last part a bit more?

[mild laurel]

Hm I think I misunderstood, the action of P on X here is the left multiplication action, not the conjugation action

[hollow quarry]

is there anyone here willing to quench some knowledge thirst i have for near-rings and near-ring modules? I have some basic questions about it

[tawny pine]

just post your q and whoever's available to help you will help if it's been at least 15min since you asked a q, feel free to use the Helpers ping

[hollow quarry]

its not exactly a question from a book, its mostly a bunch of possibly bad questions, but i'll give it a try

how would one go about making a non-abelian group be an F_2-module (F_2 meaning Z/Z_2)? obviously if it is non-abelian it can't be an F_2-module, but I've start reading into near-ring modules, and there it seems you can stabilish an R-module over a non-abelian group G, but I don't think I have the mathematical pre-requisites to know much more about it

so where could I go to learn more about modules over non-abelian groups in a way that is accessible to an undergrad?

[fathom whale]

I need help <@&286206848099549185> idk what topic this is

[chilly ocean]

are you fucking kidding me

[charred dew]

you didn't even ask anything

[chilly ocean]

there is so much wrong with what you're doing here

[sharp sonnet]

lmao

[fathom whale]

?

[chilly ocean]

this is abstract algebra

channel

[fathom whale]

Idfk whay thay is

[charred dew]

read #❓how-to-get-help

[chilly ocean]

jesus fucking christ

[fathom whale]

I’m in algebra 1 honors in school and this said algebra

[chilly ocean]

just go to #precalculus

or pre algebra i dunno

[solemn rain]

chill boys

that was just a ping

i never understood why ppl hate being so pinged

@solemn rain

[charred dew]

I don't mind informed pings

if you're patient and genuinely need help then sure

[solemn rain]

unperceived rules do not exist

[charred dew]

but if you just @ helpers without asking anything and demanding to be helped, then we're going to be angry with you

[spark bear]

@solemn rain Agreeing

[solemn rain]

did you just ping me lmfao

ur done

[spark bear]

OH NO

[solemn rain]

wdym agreeing

😠

[solemn rain]

oh my god i cant do anyproblems in group theory

like 0 problems

wtf is wrong with me how can i improve

[chilly ocean]

draw out the operation table

[solemn rain]

whatt

lmao

Let G be a group with 27 elements. Given that G has exactly 2 elements of order 3, exactly 6 elements of
order 9. Prove that G is an abelian group

help

[mild laurel]

What have you thought about

[solemn rain]

nothing , i dont know how to approach this at all

i dont see how having certain orders of elements can help lmao

[mild laurel]

Play around with it

See if you can figure anything more out at all

[solemn rain]

like keep playing with the elements?

holy fuck i dont know at all

can u give any more hints

[mild laurel]

Think of anything that could be relevant here

[chilly ocean]

Damn I should be able to do this

More hints pls

Is that some sylov stuff

[willow garden]

well, if |G| = 27, then yeah, it's a p-group for p=3

there's only so many groups (up to isomorphism) that have 27 elements. using sylow's theorems and the information given in the question, can you figure out which one it is?

[mild laurel]

no lmao

There are way easier ways to do this

[chilly ocean]

Ye thats what I thought but cant remembrr

[mild laurel]

No sylow required

[willow garden]

(I'm also not very good at AA, so better you should listen to zoph than me)

[chilly ocean]

I remember doing Such problems some time ago but I have a feeling I didnt fully understand it

Especially since I cant do it in my head right now

[mild laurel]

Took me a little bit to see it too

[solemn rain]

order of element divides order of group

rest of elements must have order divide 27

27 has 1 , 3 , 9 , 27

so there must be an element of order 27??

is that true?

[mild laurel]

Why is that true

[solemn rain]

umm

we are given that there are 2 elements of order 3 and 6 of 9

1 elememy of order 1

identity

the rest must divide order tho

so like

exhaustation ? lmfao

the rest must be of ord 27?

correct?

[mild laurel]

Yes

[solemn rain]

lmai

hence cylic hence abelian?

correct?

[mild laurel]

Yes

[solemn rain]

lmao thars like rhe first problem i like sorta did ever in gt

tysm for hint

thats*

[mild laurel]

Congrats

@willow garden @chilly ocean

If you want to see the solution

Also sylows theorem can never tell you anything when your group is of prime power order

[chilly ocean]

Yee got it actually, thx tho

What I didnt realize at first was that cyclic implies abelian

[chilly ocean]

Any books for abstract algebra?

[mild laurel]

Lots

which one you should read depends on a lot of things

[chilly ocean]

But there’s probably like the best one out of lots

[mild laurel]

How do you define best

[chilly ocean]

Good explanation, exercise problems and solutions

And beginners

[mild laurel]

All good books have that

Some good books could be geared towards more advanced people

It's hard to say a book is worse just because it doesn't cater to absolute beginners

[chilly ocean]

A beginner to advance would be good too

[mild laurel]

Anyways the idea of best book is nonsense

Some algebra books assume knowledge of linear algebra, some don't

Some take a more formal approach to the subject, some take a more intuitive approach

Some are more terse and more faster, some are wordier and move slower

It's hard to say that any one of these things makes a book "better" than another

[chilly ocean]

Right

Perhaps grabbing free PDFs is my plan

[mild laurel]

So which book is best for you will depend on all these things

How much math you know, how comfortable you are with mathematical arguments, why you're learning algebra etc

[chilly ocean]

So I’m gonna have to find the correct book by searching?

[mild laurel]

Or you can answer these things and I can give my best recommendation

[chilly ocean]

That one

[mild laurel]

well answer them

"How much math you know, how comfortable you are with mathematical arguments, why you're learning algebra etc"

[chilly ocean]

I know calculus and I’m good at algebra. My linear algebra is basic

And I’m learning this just for hobby

[mild laurel]

Okay, probably Pinter's book is best

On the easier side, a bit less formal

[chilly ocean]

Cool

I’ll search him up

[fading wagon]

Infinitely Large Napkin is catered to people with experience in mathematics olympiad

[upper pivot]

artin is nice if you dont have strong background in LA

but it assumes you know about proofs

or are familiar with proofs i should say

[golden pasture]

jacobson does assume some la but isnt too much

[outer estuary]

Does cokernal intuitively measure the difference between the image of f and the object f maps to?

Essentially measuring surjectivity of f?

[woven delta]

Distance is a bad word

But yeah

If cokernel is trivial then the map is surjective

The way I usually think about coker f where $f:A \to$ B is in terms of the exact sequence
$0\to \ker f \to A \to B \to coker f \to 0$

[cloud walrus]

Liquid:

[magic owl]

I think if you think about it

The formal defn of@coker as codomain/im

Gives the intuituition

It measures distance from surjectivity in some@sense

[bleak abyss]

Artin assumes practically nothing but the English language lol

[chilly ocean]

@chilly ocean one single book: Topics in Algebra by Herstein

If you study this book, you will learn it the proper way

or at least have solid base

[golden pasture]

ngl i didnt rlly like herstein

[chilly ocean]

Dummit foote Best one imo

[fickle brook]

🅱️est?

[simple agate]

aluffi's algebra: chapter 0 😳

[worthy kindle]

you could learn algebra from Bourbaki for the meme

[simple agate]

if I,J are ideals of R isn't this just equivalent to I union J?

[worthy kindle]

mhmm {0}R seems to be equal to {0}, and not R 🤔

[simple agate]

oh right 🤦

how would I show that IJ is closed under addition? assuming I approach it with $x, y \in IJ, x = \sum_i a_ib_i$ and $y = \sum_i a'_ib'_i \implies x + y = \sum_i a_ib_i + a'_ib'_i$ is that good enough? it doesn't feel complete

[cloud walrus]

alex:

[worthy kindle]

There could be more $a_ib_i$'s than $a_i' b_i'$'s, you can' t fuse the sums like than without saying anything

[cloud walrus]

Tuong:

[simple agate]

well you could take the smaller one to have the same number of i's but either a_i or b_i = 0 couldn't you?

[worthy kindle]

you could

actually, what if someone wasn't convinced that fusing the sums like this gives the result?

[simple agate]

tbh I wanted to prove it by saying that if a_i is in I and b_i in J which is a subset of R then obviously a_ib_i is in I

then do separate cases for zero ring and when 1_R =/= 0_R

but it seemed overkill

rings are commutative btw

[worthy kindle]

I think anyone can see that IJ is stable by +, but a priori, if you want to show an element is in IJ, you have to show the existence of k in N, of a k-tuple of elements of I and of a k-tuple of elements of J such that blablabla

[simple agate]

I did take a look at this too

https://math.stackexchange.com/questions/329079/let-r-be-a-commutative-ring-and-let-i-and-j-be-ideals-of-r-show-ij-is

which is why I started to believe what I had written was sufficient

although they distinguish between n and n' for their sums

[worthy kindle]

basically, what they are doing here is "hey look it's obvious"

if you're satisfied with this then it's alright

[simple agate]

I don't really care about it too much since I know I could do it better if I needed to but this is only a subsection of a subsection of a question lol

thanks

[worthy kindle]

You're welcome

[chilly ocean]

@simple agate aluffi is worst choice imo you need to first get used to concrete examples then proceed to abstraction

[simple agate]

yeah lol I didn't mean it seriously. I'd like to read aluffi in the near future but I don't plan on it until I finish undergrad

same with atiyah-macdonald

[woven delta]

That's not why aluffi is a bad choice

It's just a bad book

[solemn rain]

bad problems

[bleak abyss]

Yeah problems are bad and it seems like you have to dig through too much chit chat to get to the content

Whether category theory is a merit or not depends on you. If you get fixated on it, it's bad

But if you go in with a mindset of, this is for reorganizing information, but the content is in the actual group theory and ring theory, then it's useful to learn

[magic owl]

Tbh category theorys best asset is how much it allows you to compress definitions

Like “groups actions are just set valued functors”

And then you can peel it apart but you always remember the slogan

Presheaves are a better example of this imo

[topaz solar]

it’s also good because commutative diagrams let you make things nice

[chilly ocean]

are there multiple meanings for the word kernel, or is it the same concept in different fields like harmonic analysis, linear alg

[fading wagon]

Well for group homomorphisms it's what goes to the identity. For linear maps, it's the same idea (goes to 0). For convolutions, it's something else entirely.

Convolutions/harmonic analysis looks like the same idea

[earnest valley]

👀 .

[primal mantle]

Is this channel being used

i need some help

[chilly ocean]

[primal mantle]

Cool, is there an efficient way to find all the subgroups of a finite group?

in my textbook they used cayley tables to find subgroups but i have no idea how they used it

[mild laurel]

how is the group given to you

[primal mantle]

I need to find all subgroups of {0,...,7} under addition mod 8

so the sizes of the subgroups have to be 1,2,4 and 8

but idk where to go from there

thats all the info i have

[mild laurel]

Well just do each case

what are the subgroups of size 1

[primal mantle]

e

0

Hm yh u could do this case by case

but in general is there a way to get every subgroup of a finite group

and be certain u have all of them

[chilly ocean]

not in general I think

[mild laurel]

Again this depends on how the group is given to you

[primal mantle]

O dam rip

[chilly ocean]

yeah if a group has some properties its easy to eliminate some possibilites

[mild laurel]

@chilly ocean There are definitely ways

[primal mantle]

for example?

[mild laurel]

you could just try every subset

[chilly ocean]

yeah but not in 'general'

[mild laurel]

That's a terminating algorithm that will always work

[primal mantle]

Yeah and most subsets would be eliminated by lagrange's

but thats still a ton probably

in most cases

But yeah i get u it just depends on context

[mild laurel]

Sure, but he was saying that there's no general algorithm, which is wrong

[primal mantle]

yh

fairs

cool

So given a group of small order your best bet is to just go through different possibilities?

[chilly ocean]

Ok so whats the algorithm

Without being checking all elements

I mean subgroups generated by them

No information about the G itself

[primal mantle]

I dont think he claimed that

he just said you have to check every element

[chilly ocean]

yeah but you asked for a different than that one

[primal mantle]

lol yh tru

[mild laurel]

For a group of small order yes

in this case, you could use the fact that subgroups of a cyclic group are always cyclic but

[primal mantle]

oh yeah i forgot about that

[ivory dust]

I'm confused

This question is asking me to prove that if G/Z(G) is cyclic then G is abelian. But wouldn't that mean that Z(G) = G and thus G/Z(G) is the trivial group?

I mean that still works because the trivial group is trivially cyclic

(Don't give me the proof. I'm still working on it)

[velvet delta]

You can think of Z(G) as the "largest abelian subgroup" of its parent group

[oblique river]

that seems like it could lead you down a bad path though

[velvet delta]

If you cut that out and everything else is cyclic then that means G is abelian

[oblique river]

there are groups that have trivial center but nontrivial abelian subgroups

[ivory dust]

No I have the proof

But the question itself seems weird

[oblique river]

I mean, a consequence of the proof is that it's impossible for G/Z(G) to be cyclic unless it's trivial

[ivory dust]

Yeah it seems weird that it wouldn't mention that detail

[oblique river]

the question could be rephrased as "prove that if G/Z(G) is cyclic then it must be trivial"

[ivory dust]

Yeah

It seems weird that it wasn't phrased that way

[oblique river]

but that's not a great way to phrase it because it might be misleading in terms of the proof

[ivory dust]

I guess yeah

[oblique river]

because you don't actually end up proving that G/Z(G) is trivial directly, you prove that Z(G) = G

of cousre those are equivalent

but the entire proof is doing stuff in the group G, not in the group G/Z(G)

[ivory dust]

You could say something like "prove G is abelian and thus G/Z(G) is trivial"

That seems like the best phrasing

[oblique river]

do you mean trivial in that lsat line?

[ivory dust]

Trivial yeah

Typo

[oblique river]

yeah I agree that it could be phrased that way. I guess it's just about economy of words and stuff

[ivory dust]

Maybe?

Still seems weird

[mild laurel]

It is weird

But it's just that showing its cyclic is a lot easier than showing its trivial

[velvet delta]

@mild laurel Are you sure it belongs in here

This is literally intro alg

[mild laurel]

Yeah it's fine, I mean, so was the last question

[velvet delta]

oh im not talking about the guy above me

i was just here waiting for them to finish

Am i asking: I'm looking to disprove that A4 (group of even permutations over 4 things) has a subgroup H of order 6 -- and I confess I'm searching for a time-efficient method. It seems to me that the two popular ways to do it are...

🅰️ ..by looking for a contradiction through an isomorphism to S3
This method 👇 seems to depend on inspecting the order of each subgroup in S3.

(from source i lost) says: Find the 3 elements of S3 of order 2, argue they are in H, and shown the contradiction is that these three form a subgroup of order 4, which cannot be in H which is of order 6

🅱️ ...through cosets -- but this depends on finding all the elements of A4. I'm not keen on this one because I'm not sure how to systemically find all the elements

[fading wagon]

@velvet delta A4 has only 12 elements, the even permutations
e (123) (132) (124) (142) (134) (143) (234) (243) (12)(34) (13)(24) (14)(23)

[chilly ocean]

If G is a finite group of even order, and half the elements of G are of odd order, then G has no subgroup of index 2, right

@fading wagon

[fading wagon]

ah that hmm...

actually I can't really show that

well, if it's a subgroup of index 2, then it's a normal subgroup

[velvet delta]

oh hello

ya i tried that already

i wanted to shave off more time avoiding having to calculate

e (123) (132) (124) (142) (134) (143) (234) (243) (12)(34) (13)(24) (14)(23)

[fading wagon]

Okay, we know that a group of order 6 must have an element of order 3

and the automorphism group of this contains the S4

WLOG we can consider (123) being inside the subgroup.

Hmm, also there's only 6 elements in this subgroups, so one of the (abc) cycle structure is not in the subgroup. Hmm...

[chilly ocean]

Hold on my foot is asleep but usinf the conjuntandy classes seems like a lot of work

[fading wagon]

So, we consider eH, (123)H and (132)H. Each of these have 4 elements [wait, this is skechy]

[velvet delta]

I think this sounds nice but I don't think I'm going to be allowed to talk about automorphisms on my exam

😂

[fading wagon]

and the cosets can only be these cosets combined

so there is no such subgroup

Ah got it

[chilly ocean]

Er

Element did u see that

[fading wagon]

1. Elements are e (123) (132) (124) (142) (134) (143) (234) (243) (12)(34) (13)(24) (14)(23), in particular there are 8 elements of order 3.
2. At least one element of order 3 is outside the subgroup. WLOG (123) else do a relabelling automorphism changing that to (123)
3. Hence, H and (123)H are the 2 cosets. Consider which coset (132) is in. If it's in H, then contradiction, if it's in (123)H, contradiction too.

Hey I fixed it

[chilly ocean]

If g^2 in H, then all elements of order 3 in A_4

Which isnt possible

And we can prove this because H is normal

@fading wagon some guy on SO mentioned the theorem on of odd order, then no subgroup of index 2

But I can’t prove that

[fading wagon]

If an element has odd order, it's either the identity, or you can pair it with the inverse, which means, in any group, there is an odd number of elements with odd order.
Suppose G is a group with H subgroup of index 2. If |G| is divisible by 4, then there must be at least |G|/2+1 elements with odd order (due to parity), hence let the element with odd order not in H be g, then the two cosets are H and gH. Consider g^-1, it cannot be in H, so it has to be gH, so g^-2 and hence g^2 is in H. However, g has odd order k, so (g^2)^((k+1)/2)=g is in H, contradiction.

@chilly ocean I am only able to show it if |G| is divisible by 4.

Same idea as what I did above.

If G has H subgroup of index 2, then consider the orders of each element in gH. If g has odd order, g^-1 is in there, so g^-2 and hence g^2 is in H. However, g has odd order k, so (g^2)^((k+1)/2)=g is in H, contradiction.

So, it follows that |G| is 2 mod 4

but consider D10

so, the theorem is false

[harsh tundra]

4y(2x + Ay) - 2y(2y + Bx) = 0
A = ?
B = ?
Plz help

can someone please help me with this question?

[sharp sonnet]

read #❓how-to-get-help

[south coral]

Unless I'm mistaken, this is the Cayley table for the group of integer addition mod 4. Am I correct in believing that "1" and "3" are indistinguishable here?

If we had any order-4 cyclic group we could identify 0 as the identity and 2 as the only element whose square is the identity, but it seems there would be two options for 1 and 3

[topaz solar]

Think on this for a moment

is 3 a valid generator for Z/Z4?

Also, they’re not indistinguishable, 1=/=3, 2+3=/= 2+1, but they can be swapped for an isomorphism

And isomorphisms are usually the finest level of “uniqueness” you consider for groups

[stone fulcrum]

@south coral

[south coral]

The multiples of 3 become 3, 2, 1, 0, so I think 3 is a generator, as well as 1. Isomorphism seems to be the notion of "indistinguishable" I was looking for, thanks. We haven't covered it yet.

[hot lake]

yes, there is an automorphism swapping 1 with 3

[topaz solar]

Ah ye, automorphism’s the word

[waxen iron]

So the kernel of an homomorphism is a a normal subgroup. Proving it is pretty easy but what's the intuition behind it? Is there any or is just a thing that's true because we can prove it is and it's useful?

[chilly ocean]

Well, a subgroup is normal if it is invariant by conjugation @waxen iron

That’s as much intuition as you can get, I think

[mild laurel]

a priori, I'm not sure why it'd be clear that the kernel of a homomorphism is invariant under conjugation

So maybe the better answer is that a subgroup is normal if the natural multiplication is a group structure on the set of cosets

But the natural multiplication exactly corresponds to the multiplication in the image subgroup

So it's clear that the natural multiplication will form a group structure on the set of cosets

This is essentially what the first isomorphism theorem says

[ivory dust]

"Prove that every finite abelian group G has a subgroup of prime index"

🤔

Um... trivial group?

[mild laurel]

??????

[ivory dust]

I mean I clearly holds in all other cases

It's a problem from my book (not Jacobson)

[mild laurel]

?????????????????????

[ivory dust]

Prove that every finite abelian group G has a subgroup of prime index

Is a problem from my book

But it doesn't say non-trivial

[mild laurel]

Oh I thought you meant the trivial subgroup

[solemn rain]

maybe try showing G/H is abelian

?

oh no

nvm

can you use cauchys theorem

[ivory dust]

Me?

I mean the problem isn't that hard

It just follows from the classification of finitely generated abelian groups

[solemn rain]

fuck i dont know this yet

okay sorry 😂

[chilly ocean]

Can u write the proof

[ivory dust]

Yes

I did

I can paste it here

It's just take the subgroup of G genrated by all the generators of G but with the first one raised to the power k/p

Where k is its order and p is a prime dividing k

[chilly ocean]

Can u like write it out fully

And what you use etc

Cause I dont get it

But I should

[ivory dust]

Sure

Gimme a sec

(It has index p because it has order |G|/p)

[chilly ocean]

Oh ok thx

[ivory dust]

np

[woven delta]

Gamma wtf are you saying

Why are you name dropping theorems to prove things that are proven much earlier in a course

[ivory dust]

The structure of finitely genrated abelian groups is really simple though

[woven delta]

It's not

[ivory dust]

Wait no

[woven delta]

I mean it is if you know smith normal form

[ivory dust]

I confused two theorems

[woven delta]

Lmao

[ivory dust]

At least I think I did

[woven delta]

What theorem did you confuse it with

[ivory dust]

The one that states that every finitely generated abelian group is the direct sum of fintie cyclic groups

Or is that the right one