groups-rings-fields

[oblique river]

what is it youre trying to prove

are you missing a dollar sign at the end

oh

[cloud walrus]

creamy shits:

[oblique river]

ok

so are you just assuming that G is a set with some potentially-not-closed binary operation?

and for associativity is the axiom "for all x,y,z in G for which the following products are defined, (xy)z = x(yz)"?

k

[bleak abyss]

Define a group to be a subgroup of Sym(X) where X is some set

[solar wyvern]

@bleak abyss how does SymX differ from AutX

[bleak abyss]

Usually people use Aut when there's structure beyond just being a set

Automorphism group of a graph/group/etc

[sullen flint]

@chilly ocean is my proof different from what your prof gave?

[solar wyvern]

@bleak abyss what if it's an automorphism of isomorphisms?

[simple valley]

@solar wyvern Sym = Aut_Set

whereas Aut works in a general category C

[uncut girder]

Can someone explain to me what it means for a function on the space of n-by-n matricies to R is "linear in the rows of matricies"?

[earnest valley]

Assuming you have a function $f:V^n\to\mathbb{R}$ where $V=\mathbb{R}^n$. \
If $f$ is linear then for all $\alpha\in\mathbb{R}, v=(v_i){0\leq i<n}=(v_0, \dots, v{n-1})\in V^n$, you have $f(\alpha v_0, v_1, \dots, v_{n-1})=\alpha f(v)$.

[earnest valley]

Roughtly that what it means.

[simple valley]

can someone guide me through the proof of Goursat's lemma plase?

I keep getting lost

so we have groups $H \le G \times G'$, and we have projections $p_1 : H \to G, p_2 : H \to G'$

[cloud walrus]

mniip:

[simple valley]

by those projections do we mean $\pi_1 \rvert_H, \pi_2 \rvert_H$ ?

[cloud walrus]

mniip:

[simple valley]

we then define $N = \ker p_2 = \ker \pi_2 \cap H$ ?

[cloud walrus]

mniip:

[simple valley]

when we say we identify N as a normal subgroup of G do we mean the group $\pi_1(N)$?

[cloud walrus]

mniip:

[simple valley]

N is normal in H and by surjectiveity of $\pi_1$ we obtain $\pi_1(N)$ normal in G?

[cloud walrus]

mniip:

[thorny slate]

Yeah

[simple valley]

no wait

we need to use p_1

which is defined on a superset of N

[thorny slate]

I mean N normal in G implies N × id normal in G xH

[simple valley]

but N is not a subgroup of G

I mean what you said seems to have no relevance to this

[thorny slate]

Yeah I said something else

Wiki does the calculation

For why p1(N) is still normal

[simple valley]

right

if f is surjective then A normal in dom f implies f(A) normal in cod f

[solar wyvern]

@spark plank come

[spark plank]

explain

[solar wyvern]

okay

they don't define subgroup before that?

so it's a subgroup

certainly

example being the subgroup generated by the identity in every group

[spark plank]

subgroup = subset that forms a group with the operation of the "parent" group

[solar wyvern]

yes

exactly

[spark plank]

finite order => a^m = e for some m

[solar wyvern]

m finite yes

[spark plank]

I suppose I should prove these form a subgroup?

[solar wyvern]

yeah, shouldn't be hard for G commutative, even without assuming finiteness of G itself

in the meantime I'll whip up something for rings, ideals and modules (and why you only really need to consider rings and modules)

[spark plank]

mmm it basically comes down to the exponential rules?

[solar wyvern]

maybe

first I think you'd exclude anything with nonfinite order

[spark plank]

a is finite order => a^k is finite order for all integer k

[errant drum]

Closure is probably going to be the only difficult one

[solar wyvern]

actually might not be bad to consider what nonfinite order means, it's not that hard

[spark plank]

hm

[solar wyvern]

especially for a group with finite "generators"

i think you'd really like the part on generators and relations if you don't get scared by the few pages of formalism

[spark plank]

Actually isn't this pretty straightforward?

[solar wyvern]

(yeah should be)

[errant drum]

Associativity and identity is free

Inverses are basically there

[solar wyvern]

free

[spark plank]

[solar wyvern]

it's not bad to explicitly spell things out right now tbh

everything should be small enough to actually do that

[spark plank]

implicitly ig the hardest part is showing that $ab\in G_{\text{tors}}$

[cloud walrus]

Simple_Art:

[solar wyvern]

so assume a,b in G_tors

if you've shown e \in G_tors you can assume this

[errant drum]

That's like a few lines of working

[spark plank]

which follows pretty straightforwardly from exponential rules by raising it to the power of |<a>|*|<b>|

[errant drum]

Yeah

[spark plank]

\👌

define field again

[solar wyvern]

I think for commutative groups you might be able to do |a| |b|/gcd(a,b)

[spark plank]

eh whatevs :P

[solar wyvern]

commutative module over a field with a basis

[errant drum]

Fields = Rings + multiplicative inverses for non zero elements

[spark plank]

nice recursive definition

okay

[solar wyvern]

oh

field....

rip

yeah i thought you said vector space

but i can't read

[solar wyvern]

so let's do rings now I guess

[spark plank]

[solar wyvern]

do you really not know how matrices work

[spark plank]

I just don't wanna

[solar wyvern]

you won't have to deal with anything serious with them

[spark plank]

[solar wyvern]

maybe some 2x2 stuff

[spark plank]

kek

My thoughts were the matrices that appear when I try to solve quintics

[errant drum]

Define a ring in terms of a group ez

[spark plank]

[solar wyvern]

:GWnanamiCoconaSweat:

:sad

do I have to restart

[spark plank]

ring = abelian group over one operation and monoid over another operation + distribute rules

[errant drum]

Yup

[spark plank]

rip

already lost in this paragraph

oh wait so vector space = commutative module over a field with a basis...

define module (magnitude/absolute value/ |v| ?) and basis

||v||

||nani||

||­||

||||

since when did ||v|| appear as ||v||

@delicate chasm blame u fix Discord

[errant drum]

Yeah that's a vector space

Always weird seeing a vector space defined so concisely when you've been used to the 8 billion axioms that make up a vector space

[spark plank]

wuts a basis tho

[errant drum]

A set of linearly independent vectors that span the space

Also could be looked at as the generators

[spark plank]

ah okay, so unit vectors that are orthogonal for example?

[errant drum]

Yeah

[spark plank]

what's the group operation?

what operation for GL_n(F)?

matrix multiplication?

matrix addition?

[errant drum]

Multiplication

It's the set of invertible real matrices

[spark plank]

mmm, nonzero determinant => invertible matrices?

[errant drum]

Exactly

[spark plank]

well I'll take ur word for it

[errant drum]

I mean for matrices you know that det(AB) = det(A)det(B) so that's why you need to take the det 0 matrices out

[spark plank]

do I know this?

[covert vector]

wowie @spark plank is doing abstract algebra!!

[spark plank]

[covert vector]

I mean

[spark plank]

stupid matrices

[errant drum]

I'm pretty I've seen you talk about complex analysis here @spark plank

[spark plank]

dwai

[covert vector]

absolute value of determinant of matrix can be interpreted as the volume of the parallelopiped formed from the n columns of the matrix

if the volume is Nonzero

[spark plank]

[covert vector]

then no 3 are coplanar, no 2 collinear, no 4 co 3-whatever

etc

the vectors will be linearly independent

[spark plank]

kk

[covert vector]

if the n columns are linearly independent in Fⁿ then they form a basis of Fⁿ

so the matrix mapped a basis to a basis

invertible matrices hav exactly this property

if it didn't map to a basis then it wouldn't be surjective

or injective, for that matter

[errant drum]

That's actually much nicer than the adjugate matrix stuff ^^

[spark plank]

k

[covert vector]

I did not read the above context 🤔

I will assume it is not that crazy important

[errant drum]

I just meant the typical way of showing a matrix is invertible iff it's determinant is invertible isn't as nice as the way you explained it

[spark plank]

what's "bilinear form" mean

how do I get people who think I'm apparently a math god(des(slime)?) outta my dm's

[errant drum]

It's a function which is linear both ways basically

[spark plank]

ah okay

[errant drum]

I should mention it's a function from the vector space to the field of scalars

[spark plank]

field of scalars?

[errant drum]

So in a vector space you have scalars right

Those scalars need to form a field

[spark plank]

okay

[errant drum]

Yeah the definitions get very recursive

You basically have to start from group axioms and work your way up

[spark plank]

not really recursive, more like dependent on previous stuff

[errant drum]

Yeah although I think most people learn linear algebra separate to the rest of abstract algebra at the start

[solar wyvern]

sorry, back

was feeling feels

[spark plank]

T.T

[solar wyvern]

oh yay

[bitter mauve]

15+ problems solved

says not a math god

[solar wyvern]

um

[errant drum]

Exactly @bitter mauve

[solar wyvern]

@covert vector i don't think SA really needs to think through linear stuff too much do they?

[bitter mauve]

i also dm some ppl, im one of them 😦

[solar wyvern]

(right now)

[errant drum]

I swear SA knows like complex analysis and stuff?

[solar wyvern]

(it's tru

[spark plank]

[errant drum]

So what's with the abstract algebra then?

[covert vector]

complex analysis ≠ algebra

he isn't a math student

[errant drum]

Ohhh

[solar wyvern]

tbh what is SA even

[covert vector]

the learning does not necessarily follow the progression you would expect

[errant drum]

I see

[solar wyvern]

sa knows some stuff which no human should be subjected to

[spark plank]

SA = slime anime

[covert vector]

lol nice

[solar wyvern]

so where're you at right now, sa

[spark plank]

inb4 SA actually an intelligence skill-based Mana

[errant drum]

SA knows lots and lots of stuff

[solar wyvern]

did you get fields and stuff

[spark plank]

matrices

[solar wyvern]

knows ordinal collapsing functions, doesn't know what a field/matrix is

[covert vector]

I love group theory cuz u don't have to go far to have no idea wtf is going on lol

[solar wyvern]

yeah

[spark plank]

bbl8r

[solar wyvern]

although woog sorta fibbing

[bleak abyss]

Group theory best subject tbh

[errant drum]

Took me weeks to get my head around quotient groups

[covert vector]

fibbing! nuh uh

[solar wyvern]

you have to know about quotients and products and semiproducts and stuff

[covert vector]

who

for wat

[solar wyvern]

in some sense it's good to at least know what a ring, module, field is

[errant drum]

Like all of the isomorphism theorems work if you just treat the letters as fractions

[solar wyvern]

🤢

[spark plank]

also why wouldn't you wanna learn ordinal collapsing functions

[solar wyvern]

i am frail hooman

body cannot suffer such collapsing

[covert vector]

lol I love doing that @errant drum

[errant drum]

Look the H's cancel

[solar wyvern]

[errant drum]

Ez

[solar wyvern]

woog do you know how Ext and Tor work

at all

[covert vector]

not atm

anyone wanna figure out how a wreath product works

here's the definition on Wikipedia

I wanna be able to calculate a tiny wreath product

like Z2 by Z3

continued

[spark plank]

[bitter mauve]

when do we even learn this

grad?

[spark plank]

whenever u feel like

[covert vector]

yeah whenever u feel like it

I feel like it now, wanna learn it?

[bitter mauve]

yea but

[spark plank]

do I really need linear alg

[bitter mauve]

we'll need to learn its prereqs first

then prereqs of prereqs

[spark plank]

wut prereqs

@solar wyvern @covert vector maybe gimme shortcut to basic linear alg stuff

[covert vector]

shortcut?

@bitter mauve who cares about prereqs

let's just deal with that as it comes up

[spark plank]

[bitter mauve]

that sounds like me woog

[spark plank]

like basic linear alg stuff I should know to understand these things

[covert vector]

it's group theory, it's gonna look bizarre no matter how much you've done

[bitter mauve]

i spent 4 days on first exercise

of calc on manifolds

[spark plank]

well I kinda get 1.7 now

[bitter mauve]

since i knew little LA

[spark plank]

but 1.8 look like

[bitter mauve]

show full ques rimuru

[spark plank]

isomorphisms of V are simply functions mapping V to V

right?

[bitter mauve]

u just showed the let part, what to do after that

[spark plank]

these aren't questions

[bitter mauve]

oh

automorphism is bijection onto itself

isomorphism means bijection from one to another

can be itself

[spark plank]

I dunno, I semi-recall @solar wyvern or @covert vector saying iso's are f : V → V

[covert vector]

isomorphism is like equivalence of algebraic structures

[bitter mauve]

[covert vector]

automorphism is self isomorphism

[bitter mauve]

go on

[covert vector]

identity is not always the only way to have self equivalence

[errant drum]

Might be helpful to start with homomorphisms

[spark plank]

oh right

isomorphisms of a group G are functions f : G → G that basically preserve the group axioms

"equivalence of algebraic structures"

[covert vector]

codomain does not have to be G

[bitter mauve]

G to any

[covert vector]

well not necessarily any

if it's an isomorphism lol

if you have a map f:G→H that is an isomorphism

then G and H are equivalent algebraic structures

in whichever setting you have placed them

morphisms are structure preserving maps
adding on bijectivity means effectively a relabelling of elements x to f(x)

for example, suppose you have two rings, R and S

and a ring isomorphism φ:R→S

then, since rings are also abelian groups under its addition operation

this is also an isomorphism of the underlying abelian groups

as you just had an isomorphism preserving even more structure

(with the rings)

BUT

suppose you did not have a ring isomorphism

but instead just an isomorphism of the underlying addition groups

then it's not necessarily true that as rings this holds as a ring isomorphism

[solar wyvern]

woog what happening

[covert vector]

because you only know that the group structure is preserved

not sure I think I got distracted

[solar wyvern]

rip

@spark plank u there still

you already did the endomorphism ring, so automorphisms shouldn't be strange per se, they're just the (largest) subring where each endomorphism has an inverse.

[spark plank]

no

[solar wyvern]

isomorphisms you can just think of as bijective/invertible homomorphisms

but loosely you should also think that an isomorphism will preserve pretty much all the important structure between the domain and codomain

like woog said, it'll mean something specific for various structures

[covert vector]

Ya like if you have any new algebraic structure definition

if you wanna define a homomorphism

just look at ur list of axioms

[solar wyvern]

but usually it'll be something obvious like
if you have a binary op then f(ab) = f(a)f(b)

[covert vector]

take an arbitrary example

and write down the least amount of rules it needs to satisfy for the image under your map to also satisfy those axioms

and chances are that's the definition of a homomorphism

ya so f(a*b)=f(a)•f(b)
means
you have an operation in the domain, *
and operation in the codomain, •
the morphism will convert 1 operation to the other
a*b → f(a)•f(b)

[spark plank]

wait wuts an automorphism again

need like

a dictionary

[inner acorn]

isomorphism to itself

[solar wyvern]

maps into itself, invertible

i like how everyone has their own def

[spark plank]

[solar wyvern]

so um

[errant drum]

Homomorphism - function between groups.

Isomorphism - bijective function between groups

Automorphism - Isomorphism to itself

[inner acorn]

x-x soo many morphisms

[spark plank]

homo = one direction
iso = both directions
auto = domain and codomain are the same
?

[solar wyvern]

yes

[errant drum]

Yeah

[spark plank]

[inner acorn]

To be fair, it's better than ring theory xD where you remove letters from the term
i.e. a ring without an identity is a rng

[errant drum]

[inner acorn]

like.. who the hell decided that >.>

we have like... a million different names for morphisms

but we just remove the i here

[solar wyvern]

um

[spark plank]

homo = rape
iso = consented sex
auto = masturbation

[solar wyvern]

so no one except woog really point this out but

do you know what GL(n,k) and GL(V) are

[spark plank]

GL_n(V)?

[solar wyvern]

yeah

same thing

wait

change that V to an F or k

[spark plank]

GL(n, k) is (n x k) matrices??

[solar wyvern]

so I'll tell u something really cool

[covert vector]

k is a field

[spark plank]

[solar wyvern]

n x n

[covert vector]

some point along the progression in algebra

you go from F for a field

to k

[spark plank]

Vector spaces are fields right?

[solar wyvern]

because k on the home row and u dont need to shift

[covert vector]

no

[solar wyvern]

no

other way

[spark plank]

[errant drum]

No

[spark plank]

wait lemme thonk

[solar wyvern]

fields are vector spaces, over themselves

[spark plank]

what're vector spaces

[errant drum]

Vector spaces have vectors and scalars. The scalars are from a field

[spark plank]

kk

[covert vector]

like those thingies you saw in highschool, magnitude + direction

[spark plank]

oh right

[covert vector]

except not

[solar wyvern]

@spark plank gross oversimplicifcation, fields are where your coefficients are

[inner acorn]

SO(2) ≅ complex numbers with |_| = 1

^that's cool

[spark plank]

[solar wyvern]

like if you have a 3 x 3 matrix with stuff in it

that stuff comes from ur field

(or ring or whatever)

[spark plank]

anyways what were we saying

[solar wyvern]

oh, so GL(V) is actually same as saying Aut(V)

it's the automorphisms of a vector space V

but if you want to break it down a bit more, V is just a k-module with a basis

[inner acorn]

bleh

I'm lost now

[solar wyvern]

which means you can choose a subset of your module, then everything is just a linear combination of basis elements

[spark plank]

same

[inner acorn]

modules are vector spaces over rings?

[solar wyvern]

no

[inner acorn]

qwq

[solar wyvern]

vector spaces are modules over fields

modules are by definition over rings

[spark plank]

what're modules

[inner acorn]

that's what I meant ;~;

[solar wyvern]

so they're just commutative groups along with a compatible operation by a ring (your scalars/coefficients)

[inner acorn]

yeah, I think I have a course about Modules and Representation theory next year c:

[solar wyvern]

so we should probably go over rings again

[inner acorn]

will learn soon<3

[solar wyvern]

learn now

[inner acorn]

nuuuuu

[solar wyvern]

so the basic idea is that rings are the structure you want if you have a bunch of things and want to add and multiply

for pragmatic reasons, one of the operations is commutative

(usually called addition)

it's like the barebones boi for adding and multiplying stuff.

you'll typically want associativity

[errant drum]

Also it's called a commutative ring if the multiplication is also commutative

[solar wyvern]

so you can say things like a+b+c instead of a+(b+c)

[inner acorn]

associativity is lovely ~w~

octonions suck Dx

[solar wyvern]

not havine associativity is literally hell

[inner acorn]

I mean, if you dun have associativity, there's always power associtivity c: (x^2 * x) = (x * x^2)

[solar wyvern]

@spark plank did u die

[errant drum]

I mean there are weaker versions of associativity

[inner acorn]

mhmm

[solar wyvern]

tbh tho

[errant drum]

Jacobi identity and shit

[solar wyvern]

you will often want associativity, commutativity of both operations and a multiplicative identity

maybe even no zero divisors

gonna wait for sa to return

[spark plank]

no I was doing something

okay

go on

[solar wyvern]

so zero divisors will be important eventually, but all you need to know is that a (nontrivial) zero divisor is an element which, when multiplied with another element, is the 0 element in that ring

like you can multiply two nonzero matrices (even the same one) and get the zero matrix

also you can make up an example to show that there are 2x2 matrices such that AB \neq BA

[spark plank]

like watch-ya-ma-call-em's where ε² = 0

[solar wyvern]

yes, exactly

you can make a ring using those

[spark plank]

IIRC for matrix multiplication, AB is something like a rotation of BA

or wait no

maybe not

[solar wyvern]

so

in general not all matrices invertible

[spark plank]

yeah

[solar wyvern]

(if det=0 they aren't)

so you think you sorta get rings?

[spark plank]

ya

[solar wyvern]

okay

well I'll pick it up then

[spark plank]

:­nauseated_face: matrices tho

also not having definitions of literally everything

[solar wyvern]

what other things do u even know

[spark plank]

good question

[solar wyvern]

actually

[spark plank]

[solar wyvern]

you know l^p spaces right?

[spark plank]

sure

[solar wyvern]

so

do those form a ring?

[spark plank]

coughs something about knowing the basic definition

[solar wyvern]

do u know what a polynomial is

[spark plank]

no, I don't think so?

[solar wyvern]

rip

they're like eventually zero sequences

[spark plank]

wait no wasn't answering that question lol

lmao

[inner acorn]

lol

[solar wyvern]

so do you think there's an obvious way to add/multiply polynomials/sequences

and is there any other "nice" thing you can think of

[spark plank]

l^p is the set of functions/sequences f : N → (whatevs) such that Σ |f(n)|^p converges right?

[solar wyvern]

other than simply adding or multiplying entire sequences

[spark plank]

close enuf

oof

you mean other than doing like (f+g)(x) = f(x)+g(x)

?

[solar wyvern]

yes

[spark plank]

hm

well u could literally multiply the series with Cauchy products

[solar wyvern]

(this is tru)

[spark plank]

can't do function composition for sequences

[solar wyvern]

I was gonna say you can scale things

and that's where modules/algebras come in

[spark plank]

[solar wyvern]

so when you have square matrices, you actually have an algebra

since along with adding and multiplying matrices

you can scale a matrix by a constant

[spark plank]

ya

[solar wyvern]

you don't get an algebra if you just have a vector space though

since there's no "obvious" way to multiply vectors

not generally

(there's stuff like cross product tho)

[spark plank]

I wonder if there's a matrix dot product where u just multiply corresponding entries

[solar wyvern]

yeah

it's called uhhh

[spark plank]

creative name

but idk how I feel about calling it that

[solar wyvern]

https://en.wikipedia.org/wiki/Hadamard_product_(matrices)

[spark plank]

kk

[solar wyvern]

yeah, you literally just multiply componentwise lol

(this is usually not good tho)

[spark plank]

kek

[solar wyvern]

so in a nutshell that's the deal with modules/algebras, but there's small hitch you should be careful about

remember torsion subgroups?

[spark plank]

it basically turns it into a problem of the entries of the matrix and gets rid of the point of having the matrix in the first place ig

uh

[solar wyvern]

yeah

basically

[spark plank]

you never told me what torsion meant

[solar wyvern]

so now u will know

it's usually brought up wrt modules

[spark plank]

and pdf says something like the subgroup of a in G where |<a>| is finite.

[solar wyvern]

and it's exactly how it is with groups, since modules are commutative groups with a ring

[inner acorn]

Let G be a group, then the torsion of G is T(G) = { all g in G : |g| is finite } ?

[solar wyvern]

that is, you can add a nonzero vector to itself some finite number of times and get the zero vector

note that this is the big distinction from modules and vector spaces

besides vector spaces always being over fields

incidentally even if you have a module over a field, you can still have torsion.

i think..

[raw moth]

just need a field of non-zero characteristic I guess?

[bitter mauve]

how to say that <x,z> = <y,z> for all z implies x=y? or is this false?

[solar wyvern]

@raw moth im sorta ded can u help me out here

[raw moth]

idk what a module is :^)

[spark plank]

but what are modules

[inner acorn]

xD

[spark plank]

[solar wyvern]

okay

[bitter mauve]

rimuru has best emotes

anyway, my ques how to say that <x,z> = <y,z> for all z implies x=y? or is this false?

in #help-2

[spark plank]

don't post the question in multiple channels or channels that are being used by other people already!

[bitter mauve]

do people even look at question channels

[solar wyvern]

So you remember what endomorphisms are right?

[spark plank]

are you any special to give yourself the privilege of disrupting other conversations and ignoring the rules?

[bitter mauve]

...

im asking dude, not saying i posted in multiple channels because of that

[spark plank]

fock too many morphisms @solar wyvern

[solar wyvern]

[spark plank]

wait

not necessarily preserving structure?

[solar wyvern]

nah, usually preserves dat

[spark plank]

auto = iso + endo?

[solar wyvern]

yes

this.

[spark plank]

but endo doesn't necessarily preserve structure?

iso = preserve structure

­

[solar wyvern]

thank u yes

[inner acorn]

wow

that's actually *writes this down*

[spark plank]

wait but which ones are necessarily preserving the group axioms and stuff?

[raw moth]

I think for groups morphism is another word for homomorphism but idk#

or like any morphism on a group is a homomorphism or something

[covert vector]

every morphism is assumed to preserve structure

otherwise its useless

[spark plank]

kk

kek

[solar wyvern]

heres the first wiki def of a module
Suppose that $R$ is a ring and $1_R$ is its multiplicative identity. A left $R$-module $M$ consists of an abelian group $(M, +)$ and an operation $⋅ : R × M → M$ such that for all $r, s \in R$ and $x, y \in M$, we have:

$$r\cdot (x+y)=r\cdot x+r\cdot y$$
$$(r+s)\cdot x=r\cdot x+s\cdot x$$
$$(rs)\cdot x=r\cdot (s\cdot x)$$
$$1_R\cdot x=x$$

[cloud walrus]

flimflam:

[spark plank]

r\cdot s on the third line down there?

[solar wyvern]

yeah

that's your "scalar multiplication"

[spark plank]

ic

Vector space = module + basis?

[solar wyvern]

+ring=field

[spark plank]

field = ring + module?

[solar wyvern]

field = ring with commutattive everything and inverses

(necessarily this means fields have multiplicative identities)

[spark plank]

what're u saying "+ring" to

[stone fulcrum]

Field = two groups

And distributes

[spark plank]

yaya

[solar wyvern]

there's the "nice" definition that a module is a commutative group with a ring action

[stone fulcrum]

Oh cool I never knew that one

[solar wyvern]

u should

[spark plank]

[stone fulcrum]

I just know it as a vector space except the scalars are from a ring not a field

[solar wyvern]

we could actually spell it out now and say a module is a abelian group G, a ring R and a map
$$R \to Endo(G)$$ which satisfies certain props

[cloud walrus]

flimflam:

[solar wyvern]

but this will probably seem weird for now

[bitter mauve]

@spark plank what book was that screenshot u sent a while ago from?

[solar wyvern]

also the way I thought was really intuitive is that formal sums typically get you a module

like with l^p spaces

[spark plank]

abelian = commutative
monoid = group - inverse
ring = abelian group + group + distribution
field = abelian group + abelian group + distribution
      = ring + commutative and inverses of both operations
module of M = ring with distribution over M(?)
vector space = ???

[solar wyvern]

or even non l^p sequences

you can just brainlessly add two sequences together componentwise or scale them componentwise right?

[spark plank]

ya

[solar wyvern]

and if you scale by 1 it's the same, and if you scale by 0 it's the zero sequence

so then if the elements of your sequences are in $\mathbf C$ for instance, which is a field, you write
$$\mathbf C[[t]]$$ for the formal power _series

but there are just sequences with the X's there to keep track of stuff

[cloud walrus]

flimflam:
Compile Error! Click the reaction for details. (You may edit your message)

[spark plank]

ew u use _ ... _ for italics lol

[covert vector]

lol

[solar wyvern]

or if the sequences are eventually zero

[spark plank]

$smh not using both$

[solar wyvern]

$\mathbf C [t]$

[cloud walrus]

Simple_Art:

flimflam:

[solar wyvern]

ie polynomials but you're not really considering multiplication

(if you did, you'd have an algebra, not a module)

[spark plank]

wait how're polynomials defined here?

[solar wyvern]

polynomials are just sequences which are 0 for all but finite entries

this is actually the better way to consider polynomials imo

[spark plank]

uh

so like in R, the sequence [3, 5, 7, 9, 0, 0, 0, ...] is a polynomial?

[solar wyvern]

yepppp

[spark plank]

ew

lmao

[solar wyvern]

the one you'd normally write $9x^3+7x^2+5x+3$

[cloud walrus]

flimflam:

[spark plank]

oh right okay

[solar wyvern]

but with a ring you assume there's some algebraic structure

[spark plank]

that's what u meant

[solar wyvern]

namely that you can add sequences

[spark plank]

oh right formal power series

[solar wyvern]

like if it's just a sequence of points idk if you can add points generally, maybe probably not

so yeah, algebra just means module plus you can multiply things (any two things)

[spark plank]

pretty confused tbh but okay

[solar wyvern]

okay

so like

[spark plank]

I feel like

I need absorb the definitions more

[solar wyvern]

sorry if this is getting tedious, and I'd even go so far to say that if it's a bit boring, might just skim over definitions

you don't actually need to know some of the formalism at this point

that's what I did tbh

[spark plank]

[solar wyvern]

you'll probably want to go back later tho

[spark plank]

a lot of not understanding what ur really saying cuz I don't fully know the words in ur sentences

[solar wyvern]

this is a fair complaint.

there are many words.

[spark plank]

[solar wyvern]

words are actually a technical term in group theory too

so you want to know what a group is definitely

[spark plank]

stupid strings of letters

[solar wyvern]

yeah

this is actually it

letters with concatenation.

so cat*dog=catdog

[spark plank]

[solar wyvern]

so you know what a group and subgroup are right?

what sort of groups do you know if any

[spark plank]

ya ik groups/subgroups

[solar wyvern]

also you know the order of a group/element of a group

[spark plank]

I can figure out whether or not a set with an operation is a group e.g. N is not a group with normal + or *, Z is a group with normal + but not *, Q, R, and C is with both (so a field without multiplicative inverse of 0), and so on

[solar wyvern]

you need inverses

[spark plank]

order of group = cardinality of group?

[solar wyvern]

yes

[spark plank]

order of element = least natural satisfying a^m = e?

wait can't you define exponentiation as a completely different thing, given a field M you can define a^m for m in M?

and then apply some ordering over M?

[solar wyvern]

tbh I prefer the (positive, by convention) integer such that a^{order a} = e, and for a^m = e, order(a) divides m

[spark plank]

trivially equivalent

[solar wyvern]

only with Z-modules

[spark plank]

oh hey Z-modules

it finally makes sense lmfao

progress

[solar wyvern]

when I think Z, I don't think ordering

because I'm not a heretic.

[spark plank]

[solar wyvern]

I think unique prime factorization

[spark plank]

oic

that's where it comes in

[solar wyvern]

yeah

[spark plank]

fock

[solar wyvern]

it'll be useful to think about it in terms of factorization later

like

[spark plank]

rip ordering

[solar wyvern]

in a week

[spark plank]

guess that means I should start thinking more algebra less ordinal

[solar wyvern]

it begins

[spark plank]

[solar wyvern]

tbh if it makes you feel better Z>N

[spark plank]

N > Z

well-ordering ftw!

[solar wyvern]

well, you still use N and well ordering a lot, tis tru

like for induction 🤷 :

[spark plank]

[solar wyvern]

so there's an important definition you definitely should know

the centre/center of a group

is a subset such that everything commutes

[spark plank]

[solar wyvern]

ie the center of a commutative group is itself always

[oblique river]

you should be more precise

[solar wyvern]

yeah ok

[oblique river]

the center of a group is the set of all elements that commute with everything in G

[spark plank]

ab = ba for a,b in centre?

[oblique river]

which could be smaller than, say, "a large abelian subgroup of G" which is kind of what "a subset such that everything commutes" kinda sounds like

[spark plank]

kk

[oblique river]

no. Z(G) = {a in G | ab = ba for all b in G}

[solar wyvern]

yeah, so what buncho saying important here

it's not simply that you have a,b such that ab=ba

[spark plank]

oh okay

[solar wyvern]

they have to commute with everything

if you don't want to do matrices, there's a silly sorta example i think

[spark plank]

[solar wyvern]

oh yeah

i should give it

[covert vector]

the center of an abelian group G is G

the center of a non abelian group will be a proper subset(group) of G

[spark plank]

ya trivially

[solar wyvern]

okay

there's one with rotations that you can think about

has to be in 3D though

oh

[spark plank]

[covert vector]

have you seen the prototypical finite groups

@spark plank

well ill just list them

[spark plank]

too many concatenated letters

[covert vector]

integers mod n

[oblique river]

Z isnt finite lol

[solar wyvern]

shhh

[covert vector]

o yea

[spark plank]

{0}

[covert vector]

integers mod n

thats integers mod n=1

dihedral groups D_n

these are the symmetries of a regular n-gon

Q_8

[solar wyvern]

dihedral before symmetric always seemed weird to me

oh yeah

[covert vector]

{+1,-1,i,-i,j,-j,k,-k}

[solar wyvern]

quaternions

[covert vector]

thats Q_8

[spark plank]

nani

[covert vector]

note that theres no Q_n for n not equal to 8

[solar wyvern]

nani

[covert vector]

its just for 8

lol

[solar wyvern]

whats Q woog

Q_n

[covert vector]

rationals

nothing

[solar wyvern]

p-adic rationals

[covert vector]

o tru

[spark plank]

{}?

[covert vector]

thats something else tho

need identity

groups are then necessarily nonempty

[spark plank]

[covert vector]

so ya the quaternions satisfying i^2=j^2=k^2=ijk=-1

just those 8 dudes form a non abelian group with multiplication

[solar wyvern]

@spark plank first exercise is basically checking quaternions are elementwise order 2 but as a group there's 8, and some other stuff

[covert vector]

and S_n = permutations of {1,2,3,...,n}

order 2? wat

[spark plank]

many confused

[solar wyvern]

wait...

[covert vector]

ok so these are basically all the groups you care about

for all intents and purposes

[solar wyvern]

i might be wrong about things

[covert vector]

like not really

[oblique river]

some people care about GL_n(F_p)

[covert vector]

but these are the first ones you learn

and actually work with

[solar wyvern]

@oblique river those are automorphism groups

@spark plank already knows those

they just don't know it yet

[oblique river]

lol what

[spark plank]

nani

[covert vector]

i mean sure but u can look at them as groups on their own

[spark plank]

:­eyesb:

[covert vector]

anyway so the groups i threw at you

those will be your main sources of examples and counterexamples in group theory

if u ever wanna test a certain property just test one of those groups on it

integers mod n are abelian for all n

all the others i listed are not abelian

[solar wyvern]

@spark plank check out p15

neat table

[spark plank]

still veri confused but okay

[oblique river]

what was the original question @spark plank

[covert vector]

so like, obv you know non abelian groups exist, but these are concrete examples

[solar wyvern]

@oblique river rimuru just learning gt

[covert vector]

that u can explicitly write down without being too massive

[spark plank]

even more confused

[covert vector]

ok well ill use them for examples

[spark plank]

wut

hol up

[covert vector]

lets see D_3

ok

[solar wyvern]

i can't believe woog killed SA

[spark plank]

I can easily make my own non-abelian groups

[solar wyvern]

oh ya

try finding the lowest order nonabelian group

[covert vector]

but are they smol

[solar wyvern]

lowest order noncyclic group

[spark plank]

can easily make a non-abelian group with 2 or 3 elements?

I think

[solar wyvern]

do it then

:>

[spark plank]

[oblique river]

let's start iwth something a little more basic

@spark plank how many groups of order 2 are there?

[solar wyvern]

[spark plank]

wdym

[oblique river]

how many groups of order 2 are there

[solar wyvern]

what did buncho mean

[oblique river]

can you write down two different groups of order 2?

let me help get you started

[covert vector]

good question buncho !

[oblique river]

our group of order 2

must have 2 elements in it. one of them has to be the identity element (because that's what groups must have). let's call it e

our group has one more element in it, let's just call it a

[solar wyvern]

@spark plank how many nonisomorphic groups are there of order 2

[spark plank]

too many words

[oblique river]

write out the possible multiplication table for this group

you know what e*a and a*e have to be, as well as e*e. but what about a*a?

[spark plank]

by groups of order 2 you mean a*a = e for all a?

[oblique river]

no

[spark plank]

[oblique river]

a group of order 2 is a group with 2 elements in it

it's that simple. the order of a group is just it's size.

[spark plank]

okay then

then you need a*a = e

[solar wyvern]

why?

[oblique river]

great, why can't you have a*a = a?

[spark plank]

because then you don't have an inverse of a

[oblique river]

great

[solar wyvern]

we're onto something

[spark plank]

[oblique river]

so you just showed that there is a unique group of order 2

there's only one!

[spark plank]

k

[oblique river]

there aren't any other possibilities for our group because we were "forced" at every step when we tried to write down its multiplication table

so going back to a previous question, there can't be any nonabelian groups of order 2

cuz there is a unique group of order 2 (the cyclic one) and it is abelian

sshhhh! no spoilers!

[inner acorn]

sowwy!

[oblique river]

haha :)

[inner acorn]

I will go back to the corner and watch

[oblique river]

but yes, it's true that there is also a unique group of order 3. you should try to prove it by hand @spark plank

[solar wyvern]

also probably just just start doing this now but you can write a group by its generators and relations

so like

for the one you just did

<a|a^2=1>

[oblique river]

no, katzen... I think that's a bit much

[covert vector]

YA TOO COMPLICATED

[solar wyvern]

r u kidding

[covert vector]

no U

[oblique river]

absolutely not

[solar wyvern]

incidentally <a> is just the infinite cyclic group

with no relations

well

if that's confusing then don't

[oblique river]

i'm serious, writing things in terms of generators and relations and being able to understand it requires a little bit more group theory

[solar wyvern]

literally next chapter

[oblique river]

I mean you should know how to manipulate groups first

[solar wyvern]

fair

[oblique river]

sorry for getting distracted @spark plank

so @spark plank how did you feel about that last example

[spark plank]

{e, a, b}
e*e = e
e*a = a*e = a

(a*a = b*b = e) or (a*b = b*a = e)

(a*a = b*b = e)
=> a*(a*b) = (a*a)*b = e*b = b = a*x => x can't equal e or a, so, x = b, so a*b = b = e*b => a = e which is a contradiction

(a*b = b*a = e)
(a*a)*b = a*(a*b) = a*e = a = x*b => x = b => a*a = b
similarly b*b = a

Can I not assume b*x = b*y <=> x = y?

pretty sure that's a thing

[oblique river]

yes you can, because you can multiply on the hleft by b^(-1)

[spark plank]

okay...

[oblique river]

you dont seem convinced

[spark plank]

then I messed up somewhere?

cuz a*b does not equal b*a

[oblique river]

this group will be abelian

it seems like you just showed that a*a is not equal to e

but what's wrong with ab = ba = e?

[spark plank]

nani

I didn't get to that yet

what's wrong with aa = bb = e?

[oblique river]

but you hit a contradiction when you assumed aa = e

you concluded that ab = b

but that's impossible

multiply ont he right by b^(-1)

[spark plank]

oh tru

[oblique river]

you get a = e

this lemma might be helpful: in the multiplication table of a group, every row and column must contain every element of the group exactly once

(so you can use this to kind of "sudoku" your way to a solution)

[spark plank]

So the unique group of order 3 is:
G = {e, a, b}
ee = e
ex = x = xe
aa = b
bb = a
ab = ba = e

[oblique river]

"group of order 3" not "third order"

but yes that is correct

now you can try groups of order 4. this is the last one that's reasonably brute-force-by-hand-able

[spark plank]

kk will thonk in bed

[oblique river]

but this time there will be two groups of order 4

at some point, you will have to make a choice, and both options will lead to a valid group

[spark plank]

I assume this is interesting?

[oblique river]

you already know one group of order 4: the cyclic one {e, a, a^2, a^3}. so what is the other one?

I mean I think so

that "order 4" is the first time you hit a noncyclic group

relevant to the previous discussion: is that group abelian?

(you'll find out as you solve the problem)

[inner acorn]

Somewhat relevant
https://ih1.redbubble.net/image.562569675.4448/flat,550x550,075,f.u2.jpg

[oblique river]

clever!

[inner acorn]

and a kitty! o3o

[spark plank]

tfw ur message don't send

aa = e => aab = b => ax = b
ae = a and aa = e
ab = b => a = e => bad
so ab = c.
Likewise ac = b, ba = c, bc = a, ca = b, and cb = a.

I thonk

[oblique river]

i'm not sure what x in your first line is

[spark plank]

oh ya and bb = cc = e

[stone fulcrum]

Cayley Diagrams are useful af for small groups

[oblique river]

but yes this is one of the groups of order 4: {e, a, b, ab}

where ab = ba (so the group is abelian)

and a^2 = b^2 = e

[spark plank]

My first line is saying that if aa = e, then aab = b, so ax = b where x = ab

I then work out possible values of x

[oblique river]

i'm not sure if you need the first line then

[spark plank]

🤷

[oblique river]

in any case it looks right

[spark plank]

Hm

Kk

time to brute force order 5?

[stone fulcrum]

Order 5 is interesting cuz theorems you'll learn soon

Well, it's interesting because it's not interesting

[spark plank]

[stone fulcrum]

There is only one group for every prime order

So the only group of order 5 is the cyclic group of 5 elements

[spark plank]

inb4 n groups of order k, where n is the numbre of prime factors of k with multiplicity

[oblique river]

that fails for k = 1 ;)

[spark plank]

shh

[stone fulcrum]

Sadly it isn't that easy. Number of groups tends to be messy

[spark plank]

as they say, proof by ignoring the cases this fails

[oblique river]

although it is true for k between 2 and 7

[covert vector]

did u know theres 51 groups of order 32

[oblique river]

but sadly there are 5 groups of order 8

[spark plank]

[covert vector]

or something crazy like that

[oblique river]

3 abelian and 2 nonabelian

[covert vector]

267 of order 64

[spark plank]

how fast does it grow tho

[oblique river]

that's the thing, it varies like crazy

cuz for primes it's 1

and for powers of 2 it's HUGE

[stone fulcrum]

Inb4 groups of order 1024

[covert vector]

well i think at 2^11 its not known how many there are

[oblique river]

99% of groups of order less than 2000 have order a power of 2

[covert vector]

[stone fulcrum]

Half a billion groups of order 1024

[covert vector]

wat @oblique river

oh

[oblique river]

wat wat

[spark plank]

f(n) = sup{k : k groups of (m < n)}

[covert vector]

yea that sounds about right lol

[stone fulcrum]

No wait 50 billion

[spark plank]

[stone fulcrum]

I got the wrong number lel

[spark plank]

sounds exponential-ish

[covert vector]

yea luckily its just 2 thats crazy

[spark plank]

[covert vector]

other prime powers arent as nasty

[stone fulcrum]

But then only one group for every prime order

[spark plank]

[covert vector]

well actually

im not too sure on that

[spark plank]

need to invent faster growing algebra property stuff

[oblique river]

that's not really true woog

there are more groups of order 3^k than 2^k

[spark plank]

ya but 3^k > 2^k

[oblique river]

the thing is that 3^k grows much faster than 2^k

exactly

so if you're just looking at groups of order less than n, it's dominated by powers of 2

[spark plank]

Compare 3^m to 2^n where 3^m < 2^n but order 3^m got more groups than 2^n

[covert vector]

hmm idk with respect to what metric i was using, but i was thinking more like the growth of groups order 2^n relative to like

how big 2 is

[raw moth]

just take rational approximations of log_2(3)

[covert vector]

ok doesnt make sense lol

[oblique river]

https://groupprops.subwiki.org/wiki/Groups_of_order_p^n

[covert vector]

woa tha formula at bottom

[stone fulcrum]

All groups of p² order are abelian. I learned today.

[oblique river]

yeah it's kinda crazy that there are formulas for it

but what really blows my mind is that they're only stable formulas

[spark plank]

[oblique river]

meaning they're only valid for p large enough

(relative to n, the power of p dividing G)

[covert vector]

yaaa

turns out smol primes are spooky

[oblique river]

I dont really know anything about that, though. for example I'm not sure if it's known or conjectured that there should be a formula for groups of order p^n for large enough p (n fixed)

modern group theory research is pretty specialized and not really close to much other math

[covert vector]

theres a bunch of finite field theory results for like

[oblique river]

basically the rest of math already has "what it needs" from group theory haha

[covert vector]

prime p but p must be over 9000

or something like that

[spark plank]

idk, imo it seens normal to know information about large values but not small values or vice versa

[oblique river]

I mean it happens all the time

I just wouldn't have expected it here

like, I would have expected an asymptotic or something

but to have an exact formula is kinda crazy

imo

[spark plank]

Ah tru

[stone fulcrum]

We know (and use most often) lots about the small groups. We also know a lot about things that relate to primes cuz sylow was smart

[spark plank]

Something probably happens at large values

🤷

[oblique river]

I mean, something must happen at large values in order for it to be true

so yes you are correct because we know the theorem lol

[spark plank]