groups-rings-fields

[pastel cliff]

[coral shale]

ok lemme stare at it again

[delicate orchid]

show me the money

[hidden haven]

Nice

[pastel cliff]

imma go watch dragon ball and avoid my problems

[coral shale]

I was told to read the proof in this

by algebraic nt lecturer (just stated the thm)

[delicate orchid]

warwick

[hidden haven]

c shouldn't be of the form (a - a')/(b - b') where a' is a conjugate of a and b' of b

Or maybe it was the reciprocal of this

Just make sure that it isn't either

[delicate orchid]

that's a lot more specific then I thought

does this generalise to longer sums in a nice way?

[hidden haven]

Induction

[delicate orchid]

I thought so

[hidden haven]

[delicate orchid]

so that just has to hold pairwise for all elements in the sum

[coral shale]

I don't think they state the condition in the proof

oh they kinda do

so we're looking at......

[hidden haven]

Ye they make that an implicit relation

[coral shale]

[hidden haven]

Just solve for x in this

You'll see the condition

[coral shale]

i see i see.

[plush wasp]

where does these come from ._.

experience? familiarity?

[coral shale]

what are 'these'

[delicate orchid]

nuts

[plush wasp]

"these" nuts answers

[coral shale]

shh

[hidden haven]

Good professors

[coral shale]

moldi, tell us how you do it

[delicate orchid]

moldi knows what's good

[coral shale]

are you an AI

[hidden haven]

[plush wasp]

an AI lmao,

I suppose just good brain :o

[coral shale]

lend me some

ill give them back to you in the same condition

[delicate orchid]

I imagine moldi's course on galois theory wasn't cancelled by their hack fraud university 3 months before I was due to start it

[hidden haven]

😼

[coral shale]

well tbh im not even halfway through my galois course

maybe i will be 10% as smart after it

[plush wasp]

:o meanwhile I'll just go and start with vector spaces

smh

[delicate orchid]

comparing yourselves to others

[plush wasp]

[coral shale]

$$K(a + cb) = K(a, b) \iff (\forall a', b')\quad c\neq\frac{a - a'}{b - b'}$$

Did I get that right moldi

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

im just being dumb and thinking c = 0

I think the implication might only be one way

[hidden haven]

c = 0 doesn't work because a = a'

[coral shale]

but if b is a conjugate of a

or something dumb

what are the conditions on a and b

K is a field

a, b algebraic

?

[hidden haven]

Ye

And separable

[coral shale]

Let $L:K$ be an extension. Let $a, b\in L$ algebraic.
$$K(a + cb) = K(a, b) \iff (\forall a', b')\quad c\neq\frac{a - a'}{b - b'}$$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

Separable?

i literally only just learnt the defn

[hidden haven]

oof

[coral shale]

oh right i see where it comes into the proof

i think

[hidden haven]

Yeah maybe you can tweak it somehow so it's not strictly necessary here

[coral shale]

I'm just thinking about the case when a, b are cube roots of unity say........

and then K(a) = K(a + 0b) = K(a, b)

[delicate orchid]

not again

[coral shale]

i think it is only a reverse implication

[hidden haven]

Ye it's just reverse

[coral shale]

or am i thinking dumb

D:

oki oki

Let $L:K$ be an extension. Let $a, b\in L$ algebraic and separable.
$$(\forall a', b')\quad c\neq\frac{a - a'}{b - b'}\implies K(a + cb) = K(a, b)$$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

So is this good 👀 (edit maybe I forgot a minus sign, am unsure for now^)

[hidden haven]

It doth look good

[coral shale]

waw

[hidden haven]

Ofc b ≠ b'

[coral shale]

(I like symbols, what are words again?)

[hidden haven]

Since in denominator

[delicate orchid]

slightly confused by the forall 😵‍💫 don't numbers only have one conjugate?

[hidden haven]

No

[coral shale]

conjugate in the polynomial sense...

[hidden haven]

This is not complex conjugate

[delicate orchid]

I wasn't thinking complex conjugate

[coral shale]

If you have a minimal polynomial f
All roots are considered to be 'conjugates' of each other

right?

yay

[hidden haven]

[delicate orchid]

oh right yes of course

[coral shale]

But that said, I don't think the conjugates partition do they.....

or maybe they do 🤔

aaa

[delicate orchid]

partition what

[coral shale]

uh so

[hidden haven]

[coral shale]

f = (x-a)g(x)
h = (x-a)k(x)

Is it possible for f and h to both be minimal polynomials

in some uh

field?

[coral shale]

ie. we have this in the big field

L : K

[hidden haven]

Minimal polynomial is unique

[coral shale]

Let f, h, g, k in L[x]

f = (x-a)g(x)
h = (x-a)k(x)

Can f and h both be minimal for something in K[x]

not necessarily a

[hidden haven]

Irreducible polynomials can't share a factor

[coral shale]

yh thats my Q

[hidden haven]

Because then their gcd is non trivial

And you can write gcd in terms of f and h

[coral shale]

So it comes from the fact F[x] is an ED

[hidden haven]

Ye

And gcd divides both

[coral shale]

hmm ima think about this

[hidden haven]

So you get factorisations of both

Basically you have to prove that gcd doesn't change if you go to a larger field

[coral shale]

yh that

thanks for the help tho

big thinks to do

[hidden haven]

[delicate orchid]

moldi... what's a polynomial...

[coral shale]

x

[hidden haven]

bro....

[delicate orchid]

bro like....

squaring things...? what if like... dude they're... like rectangles instead bro...

[hidden haven]

An element of the free commutative R algebra over a singleton is a polynomial in 1 variable over R

[delicate orchid]

you failed moldi I actually know exactly what that means

[hidden haven]

Alternatively the free pointed ring on R

[delicate orchid]

ok now you've done it

[hidden haven]

[coral shale]

back to before, so if L : K, if we take a ~ a' in K[x]

This relation partitions everything in L that is algebraic over K

right?

[hidden haven]

Yes

[coral shale]

is it an interesting partition 👀

[hidden haven]

And each equivalence class is finite

Yes

[coral shale]

A name?

[hidden haven]

Idk

But exercise

[coral shale]

D:

[hidden haven]

Prove that any K-homomorphism from an algebraic extension F/K to itself is an isomorphism

[coral shale]

WIll try, ty

I am used to seeing F : K

Me for a small while: F/K, F/K, what is this quotient

[hidden haven]

Oh F

Ye / is bad notation but is standard

[coral shale]

I think I done this exercise in a homework

[delicate orchid]

*Oh F/K

[coral shale]

you show think

did i write that wrong lel

[hidden haven]

As my prof said 5 times every lecture

Roots go to roots, and any root can go to any root

Important to remember when working with algebraic extensions 😌

[hidden haven]

Didn't you have it write

[coral shale]

f(a) = 0 <=> f(phi(a)) = 0

phi being the K-homomorphism

[hidden haven]

Yes

[coral shale]

I'm not sure I did

I'm sure now tho lul

[hidden haven]

[coral shale]

f(a) = 0 <=> phi(f(a)) = 0 <=> f(phi(a)) = 0

I think this is the steps to showing it

[hidden haven]

[coral shale]

right I was mulling over this and don't think I've fully grasped it yet

If L : K is an algebraic extension, if we take a ~ a' in K[x]

Then any K-homomorphism from L to L must leave these classes invariant

Like kinda

$\varphi:L/\sim \to L/\sim$ is the identity map

[delicate orchid]

use \sim

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

\simp

[delicate orchid]

_ _

[coral shale]

Mindblowing

I don't think I was really aware it worked for any f until recently

[hidden haven]

Ye

[coral shale]

Like I might think about K(a)

And the K-homomorphisms permute a, a', ...

but not about other elements

[hidden haven]

Yes

And given any pair (a, a') with a ~ a', you can find some endomorphism of the extension which maps a to a'

[coral shale]

I feel like there must be some kindof partition of K-homomorphisms... how to describe it.

[hidden haven]

Assuming algebraic extension

[coral shale]

If 3 of them map a1 to a2.
Then 3 of them must also map a1 to a3

Which means if the degree of the minimal polynomial of a is coprime with the degree of the extension

[hidden haven]

Yes

[coral shale]

phi(a) = a forall phi

is intuition speaking

[hidden haven]

Conjugates behave similar to each other

[hidden haven]

Because there's always an isomorphism of the field with this property

[coral shale]

yh been meaning to get my head around this

hmmmmmmm

L : K
We can consider K-homomorphisms from L -> L (elements of the galois group)
We can also consider K-homomorphisms from L -> L' (L' being the closure of L)

I've been trying to get my mind around the difference (for me, the top crops up in Galois, the bottom in Algebraic NT)

[hidden haven]

They're the same if your extension is normal

In particular if your extension is Galois

[coral shale]

I am familiar with normal, Galois haven't seen it yet

But I can see how it 'makes sense'

[hidden haven]

Ye since you have all the roots there anyway

Galois = normal + separable

You have all the roots and no repeated roots

[coral shale]

I have been thinking all my examples in Q, R, C so it's a bit of a struggle to internalise separable.

[hidden haven]

Char 0 everything is separable

[coral shale]

Right, the example I saw was F_p(t) : F_p(t^p)

[hidden haven]

Ye

[coral shale]

X^p - t^p

[hidden haven]

Because finite fields also everything is separable

[coral shale]

Is there some 'intuition' for the concept of separable?

[hidden haven]

And that's the simplest infinite char non zero field

[hidden haven]

'good'

[coral shale]

I think I am familiar with a linear algebra analogy kinda

[hidden haven]

I mean this

[lethal dune]

what's infinite char field?

[hidden haven]

Lol

[coral shale]

the additive order of 1

is the characteristic
if its not finite, its infinite 0

[lethal dune]

[hidden haven]

Infinite, char non 0

[coral shale]

waiiiiiit wait wait

[hidden haven]

Is what I meant

[coral shale]

😂

[lethal dune]

oh

[hidden haven]

Lol

[lethal dune]

[coral shale]

trolling

[hidden haven]

[coral shale]

I think an analogy for separable is Jordan Normal

Separable is like diagonalizable

non-separable you get 1's in jordan normal

idk if thats a good analogy (maybe not at all)

[lethal dune]

[hidden haven]

Oh I get what you mean

Similar but not exactly

[lethal dune]

isn't separable should imply thr Matrix is triangulable?

or am I wrong

[coral shale]

whats triangulable again ( 🤔 )

[hidden haven]

That's not the point

It's not a technically correct statement

[lethal dune]

what is

[hidden haven]

Shuri-Jordan lemma

[coral shale]

wat

[lethal dune]

🙈

[hidden haven]

[lethal dune]

nice

[lethal dune]

there is a basis wrt which the matrix is upper (lower) triangular

[coral shale]

I think my Linear Algebra is too weak (it literally stops at eigenvectors, diagonalisation, jordan-normal lel)

i thought you could do that for all matrices

[lethal dune]

it comes before jordan

[hidden haven]

I just intuit linear algebra

[coral shale]

You don't mean diagonalisable by this???

[cursive temple]

i like thinking about separable like you can separate double roots

in nonseparable fields they stay stuck

[lethal dune]

No

[coral shale]

Like you can keep extending the field and they'll always be stuck? Or wrong

[cursive temple]

i mean like if your polynomial has a double root then you can separate them :D

[lethal dune]

u can take algebraic closure

[hidden haven]

You can't tho

[delicate orchid]

Eigenvalues
Eigenvalues when the field is algebraically closed

[coral shale]

I can't find the term 'triangulable'

[hidden haven]

I've seen the term

Perhaps from wew lmao

[delicate orchid]

Isn’t this like, the exact condition required for non-separability lol

[cursive temple]

Yeah i brainfarted it

[delicate orchid]

Rest in peace

[coral shale]

oh yh so if we have non-separable

(x - a)^2 being minimal polynomial

And we consider K-homomorphisms

a can map to itself but not force the identity map

is that the idea?

Wrote that bad lel

===
K(a) : K

min_a(K) = (x-a)^2

K-hom f : K(a) -> K(a)

[hidden haven]

If the extension is generated by a, then it will force identity

[coral shale]

oh.

wat am I thinking then 🤔

[hidden haven]

Nothing to do with separability

Yeah idk any intuitive reason for the name

[coral shale]

Ok so K(a) === K[x]/((x-a)^2)

Normally when you quotient by a irreducible quadratic

You will have (x-a)(x-b)

And your galois group will have order divisible by 2

or just 2

[hidden haven]

K[x]

[coral shale]

I think I see but words fail me

[K(a) : K] = 2

[hidden haven]

Yes

[coral shale]

But |Gal| = 1

If separable |Gal| = 2

[hidden haven]

When non sep ye

Separated would be a better name I think

Roots are separated

[coral shale]

Ok I think I get the idea though

[hidden haven]

From each other

[coral shale]

If the roots cannot be separated

You have elements of the Galois group

'glued together'

kinda

a mapping to itself forces the identity

[hidden haven]

Ye ye

You done it

[coral shale]

And more generally (x-a)^n p(x)

yeah ok ok

woo

wee

[hidden haven]

[lapis trail]

As long as you are in a ufd you don't have to worry about the difference between primes and irreducibles? Ping when answering thanks

[delicate orchid]

yus

[coral shale]

https://math.stackexchange.com/a/2945602/243059
(yes)

[delicate orchid]

wow CHEATING

[coral shale]

😢

[rigid cave]

PID is sufficient right?

[lapis trail]

Thanks

[coral shale]

I was trying to look for sufficiency

UFD is a sufficient condition I think

PID => UFD

[delicate orchid]

^

I should get around to memorising that big ol list on wikipedia

[rigid cave]

aren't irreducibles primes in a PID?

[coral shale]

PID implies UFD

[delicate orchid]

they are but PID is a stronger condition than required

[coral shale]

I think UFD is as low as you can go?

[rigid cave]

right right

[delicate orchid]

GCD domains are slightly lower

[coral shale]

Does (prime <=> irreducible) imply the ring is UFD?

[delicate orchid]

cause then you can still do the funny bezout's (iirc)

[lapis trail]

PID is every ideal generated by an irreducible is maximal

No

[delicate orchid]

yeah

[lapis trail]

Oh ok cool

[coral shale]

Can someone fix my statement

[delicate orchid]

(2) is not maximal in Z[X] for instance

[lapis trail]

I got a quiz in a few hours

[delicate orchid]

(prime <=> irreducible) => ring is a GCD domain

[lethal dune]

there's a gcd domain?

[coral shale]

https://i.imgur.com/NyjW7U3.png

wait wut wut wut

oh nvm nvm nvm

[coral shale]

👌

[delicate orchid]

there are so many stupid ring domains

[lethal dune]

there's a factorization domain as well

not unique, but factorization domain

[hidden haven]

Call it stupid one more time and det will haunt you

[rigid cave]

domain domain when

[delicate orchid]

I imagine they do the proof that UFDs mean irreducibles are primes and then are like "oh we technically just need a notion of a gcd so lets define gcd domains and now this result is slightly more general"

[lapis trail]

Is gcdd>ufd>PID>Ed then

[delicate orchid]

yeah

[coral shale]

field

😄

[lethal dune]

like what if we have unique factorization for all but one

what should we call it

[delicate orchid]

with bezout domains in there somewhere

[coral shale]

someones going to prove this is impossible just for you, i bet

[hidden haven]

Exceptional domain

[lapis trail]

I hate not being in a ufd

[delicate orchid]

I was about to try

[coral shale]

I remember not long ago

we were fcking about in Z[x]

and had no clue

[lethal dune]

unique factorization but not commutative

[delicate orchid]

Z-algebras

[lethal dune]

[coral shale]

(2, 3x+1) <-- 'is this maximallllllllllllll' or some sht

[delicate orchid]

oh I remember that

[coral shale]

me: lets try euclidean on it

[hidden haven]

It do be maximal tho do it not

[coral shale]

me: wtf how do we do that

[hidden haven]

Trivial from

[delicate orchid]

yeah I just gave up and showed the quotient was a field

[hidden haven]

How is that giving up

[delicate orchid]

it's literally cheating

[coral shale]

thats just cheating u know

u cant just

[hidden haven]

Bruh

[coral shale]

isomorphism

[hidden haven]

[lethal dune]

my book doesn't even require commutativity to be a field

[coral shale]

i-

[hidden haven]

No

[lapis trail]

Lol if this is wrong I don't want to be right

[delicate orchid]

R[Q_8] be like YES FUCKIN FINALLY LETS GOOOOO

[hidden haven]

[lethal dune]

like the author calls a skew-field module a vector space

[coral shale]

ok seriously, what?

[delicate orchid]

it just ISN'T

[coral shale]

commutativity of multiplication?

just checking

[lethal dune]

yes

[coral shale]

addition would kill lel

[delicate orchid]

addition is always an abelian group

[hidden haven]

That one I am fine with

[lethal dune]

die

[delicate orchid]

moldi when uhh

[hidden haven]

Modules over division rings are sometimes called vector spaces

[delicate orchid]

what fails

uhh

[coral shale]

wait wait whats an example of a non-commutative field

[hidden haven]

Their study is much more similar to vector spaces

[prisma ibex]

skew-fields are fine but it's very bad to call them fields

[lethal dune]

H

[coral shale]

these things are called skew-fields?

[delicate orchid]

(Av)^T != v^TA^T (^T being transpose chat chat that means transpose)

I think, if it's non-commutitive

[coral shale]

wut

we need matrices

u guys are telling me

[prisma ibex]

yeah skew fields are non commutative fields

[lethal dune]

skew vector spaces

[coral shale]

because matrix multiplication

ok ok

[delicate orchid]

yeah we're talking about skew vector spaces

[coral shale]

souns like crazy sheet

[lethal dune]

what about non commutative integral domains?

[south patrol]

[hidden haven]

[delicate orchid]

GL_2(Z)

I think

moldi are there zero divisors in that bithc

[prisma ibex]

yeah, these are just called domains

[rigid cave]

skewb domains

[hidden haven]

Not closed under add lmao

[lethal dune]

no i mean should u call a non commutative .... a integral domain

[delicate orchid]

@rigid cave domain domain just dropped

[prisma ibex]

the integer quaternions are an example of a domain which is not an integral domain

[rigid cave]

omg pog

[lapis trail]

Are zero divisors a pain or are they cool

[delicate orchid]

a pain

[prisma ibex]

they can be a pain

[delicate orchid]

a complete nightmare

[prisma ibex]

but they are important to consider sometimes

[hidden haven]

I've seen Lang call non commutative domains "entire" rings

[lapis trail]

I think they're cool just because it's different

[delicate orchid]

yeah they're incredibly important to be aware of

might end up dividing by 0 omegaLUL

[coral shale]

when i see the word quaternion I think physics. Is this just a big no. or quantum mechanics something

[lethal dune]

lang's LA book even says <1+i, 1+i>=0

[prisma ibex]

quaternions do show up in physics

[delicate orchid]

quaternonic representations

[prisma ibex]

so do octonions

[coral shale]

[hidden haven]

wat da physicicists doin

[prisma ibex]

octonions are kinda awful

[lapis trail]

That's algebra I want to learn probably after galois

[delicate orchid]

octonions aka who?

me when non-associativity resident sleeper bye bye

[lethal dune]

onion when

[prisma ibex]

they're non-associative but still close to associative in a precise way

[delicate orchid]

they're anti-associative

[hidden haven]

oh this is ryu

[delicate orchid]

right?

[hidden haven]

[delicate orchid]

cross product type memes

[prisma ibex]

no it's like

[hidden haven]

Pfp change is tragic

[lethal dune]

don't worry it'll get better

I'm just learning

[maiden ocean]

any subgroup generated by 2 elements is associative

[prisma ibex]

yeah

[maiden ocean]

Well the options are like R C and H so lol

[prisma ibex]

that's what you have to do for octonion computations, you always like, pick a pair of elements and then do computations inside this associative subalgebra

it's nutty

[lapis trail]

Doesn't it model behavior at a quantum level?

[prisma ibex]

so there's some like

nutty construction for the standard model particle generations in terms of octonions

although I'm not sure that this is like

necessarily the best way to think about it

[delicate orchid]

I thought it was spin group memes

[prisma ibex]

it's related yeah

[rigid cave]

lmao wew pfp change

[delicate orchid]

and I know there are quaternionic representations (of some) of the spin groups

[lapis trail]

That's just wild that our constituent particles are just algebra

[prisma ibex]

yeah

there's a very general principle that relates classification of particles with representation theory

[coral shale]

algebra > all

[lapis trail]

Symmetries and rotations

[prisma ibex]

but yeah octonions come up any time you deal with the exceptional Lie groups

that's the sense in which they're natural

[lapis trail]

Irreducible means that f=gh implies g or h is a unit, right?

[hidden haven]

Ye

And I think you want f non 0

[lapis trail]

Oh ok

[coral shale]

'Irreducible' only applies to non-units, and non-0 for my notes

Think that makes sense. You want prime <=> irreducible in GCD domain and you don't want to think of units as primes in general

https://i.imgur.com/vjnN7LJ.png

being dumb, how does the below follow D:

ab = ac => b = c

[lapis trail]

Hmmm oh

[coral shale]

oh i got it

b =/= c => ab =/= ac
Suppose ab = ac.
ab - ac = 0
a(b-c) = 0

[lapis trail]

a(b-c)=0

[coral shale]

so a must be 0

oki oki oki

[lapis trail]

Or b is c

[coral shale]

'prove cancellation'

😂

[lethal dune]

see this definition doesn't require it to be commutative

[coral shale]

ic ic

[lethal dune]

ya but most (all) authors do require commutativity

[coral shale]

so we have rng and r1ng

we also have ring and gnir

[lethal dune]

[hidden haven]

gnir

[coral shale]

$\vbox{\ialign{\hfil$#$\hfil\crcr r_i\crcr\noalign{\kern-.6ex}g!{\font~cmr12 at15pt\lower.4ex\hbox{~n}}!\crcr}}$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[lethal dune]

[ornate arch]

i know theres no algorithm to find automorphisms, but is there a mthod i can use to identify inner automorphisms from automorphisms

[coral shale]

Given an automorphism (somehow), identify whether/not it is an inner automorphism?

or did I misunderstand

[delicate orchid]

check if it maps conjugacy classes to themselves

if it does for every conjugacy class it's an inner automorphism I do believe yes quite

[ornate arch]

yep

[ornate arch]

could u give me an example, im a little confused by that statement

[coral shale]

call the automorphism f

I think you are looking at what repeatedly applying f on an element does

f(x), ff(x), fff(x), ...

That would be the orbit of f

[delicate orchid]

oohh group action memes I like it

[coral shale]

Now why is this helpful... god knows

https://i.imgur.com/ClxWYDV.png

I think this is the relevant bits for thought 🤔

[delicate orchid]

we established last time we talked about this that outer automorphisms act on the conjugacy classes right?

[coral shale]

I wasnt too awake

[delicate orchid]

lol

[coral shale]

they indeed do though

ah and the class of inner automorphisms

[delicate orchid]

ok yeah so Inn(G) is the identity element in Aut(G)/Inn(G)

[coral shale]

is the id map

[delicate orchid]

yes

exactly

[coral shale]

yh yh yh

boom

[delicate orchid]

so if an automorphism maps conjugacy classes to themselves it must be inner

(as all automorphisms preserve conjugacy relations (x ~ y => y = gxg^-1 => f(y) = f(gxg^-1) = f(g)f(x)f(g)^-1 so f(x) ~ f(y))

[coral shale]

which classes do we need to check though

[delicate orchid]

all of them

well ok

[coral shale]

i think you would need to check the classes of the generators of G only?

[delicate orchid]

not quite

since automorphisms are bijections we should only need to check permutation of conjugacy classes that are the same size? right?

[delicate orchid]

wait yeah I think so, that's clevar

oh ho ho ho ho shuri you've done it again

[coral shale]

im only guessing around

(who proves stuff smh, just intuit and get things wrong half the time)

[delicate orchid]

uhh waitttt ttt

what if the generators are conjugates

eh whatever

it'll work out I'm sure of it

[coral shale]

doesnt that imply......... something

[hidden haven]

deep

[delicate orchid]

$\Rightarrow$ something

[cloud walrus]

Wew Lads Tbh (200 🍇) ✓

[delicate orchid]

idk!

[coral shale]

i knew it

[delicate orchid]

all I know is bro, check the conjugacy clas- :realisation:

hold on a moment

[coral shale]

maybe generators is no good now i think about it

Given $x^G$, what do we know about $(x^2)^G$?

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[delicate orchid]

yes linear characters yes... constant on conjugacy classes just like inner automorphisms yess yes we must map the inner automorphism group of abelian groups to the irreducible characters yes yes bijection
wait fuck this is just a proof that Inn(G) is trivial when G is abelian holy hell I'm an idiot

y = gxg^{-1} => y^2 = gx^2g^{-1}

spoookkyyy

so y ~ x => y^2 ~ x^2

[coral shale]

$x^Gy^G = (xy)^G$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

So I gather we have this.

$(x^G)^{-1} = (x^{-1})^G$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

quite cool

and then wut

[delicate orchid]

isomorphism type beat

y = gxg^{-1} => y^-1 = g^{-1}xg

yeah ok I'll believe you but I don't trust you

[coral shale]

huh what

[ornate arch]

wait

[coral shale]

does it not follow from the algebra

[coral shale]

u typoed

[ornate arch]

do we care about this

[coral shale]

x^-1

we kinda do

[delicate orchid]

I care

it basically means that if the automorphism swaps any two conjugacy classes it cannot be inner

[coral shale]

doesnt this show conjugacy classes form a group or am I being silly

[delicate orchid]

there's no binary operation between conjugacy classes there

[ornate arch]

seems like you two are saying diff things

[coral shale]

I just made one

[delicate orchid]

literally everything there is within one conjugacy class

[coral shale]

Is this no good for u

[delicate orchid]

nope

I ain't buying it

[coral shale]

uh oh

[ornate arch]

wth

[coral shale]

what went wrong there

[ornate arch]

isnt there smth we can do like

[delicate orchid]

y = x^-1

[ornate arch]

finding isomorphisms to Aut(g) and then its related to its normal group

[coral shale]

so when x = y it works

but else not

what is going on

[delicate orchid]

ok shhhh

shhh

[delicate orchid]

this ain't how it is

it ain't how it be

[coral shale]

ok ok i wrote it wrong completely lul

[delicate orchid]

a = gxg^-1
b = hyh^-1
ab = gxg^-1hyh^-1

it ain't so

[coral shale]

yh yh yh

[ornate arch]

😕

[coral shale]

a ~ x
b ~ y
Then
ab ~ xy

silly me

[ornate arch]

what yall talking about man

[delicate orchid]

no?

lol

[coral shale]

huh?

Oh waairjwaiojdsiadjaio

[delicate orchid]

does "gxg^-1hyh^-1" look like a conjugation to you?

[coral shale]

being dumbbbbbbb

[ornate arch]

alrighty im out

[hidden haven]

[delicate orchid]

I speak the truth and nothin but the truth

[coral shale]

x ~ y
x^n ~ y^n

[ornate arch]

yall gonna confuse me before my exams

[delicate orchid]

except for polynomial and integer multiplication I don't cover that

[ornate arch]

hahaha

[coral shale]

sry lel

What we do know is Out(G) := Aut(G)/Inn(G) acts on conjugacy classes

[ornate arch]

do u guy sknow about smth like this

[delicate orchid]

yeah it permutes those mfs

[delicate orchid]

what do you mean "its normal group"? is inn(g) the only normal group chat chat is it the only normal one?

I have concluded that it is

[coral shale]

Inn(G) is the identity element in Out(G)

[delicate orchid]

and isomorphisms from what to Aut(G)?

brbbb

[coral shale]

All of the automorphisms in Aut(G) can be classified using Out(G)

Out(G) acts on {x^G : x in G}, the conjugacy classes of G

Now suddenly I'm unsure

@delicate orchid when u back help D:

[delicate orchid]

no

[coral shale]

nvm nvm nvm nvm nvm

Let $f\in \textnormal{Aut}G$
$$f(x^G) = f({gxg^{-1} : g\in G})$$
$$= {f(gxg^{-1}) : g\in G}$$
$$= {f(g)f(x)f(g^{-1}) : g\in G}$$
$$= {gf(x)g^{-1} : g\in f^{-1}(G)}$$
$$= f(x)^G$$

[chilly ocean]

checks out

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[chilly ocean]

idk what x^G means tho

[coral shale]

The definition is written in the first line

[chilly ocean]

unless thats ur notation for all conjugations of x

[coral shale]

yh

[chilly ocean]

never saw it until today

[coral shale]

my lecturer

What notation is common?

[chilly ocean]

why does f(g)=g?

[coral shale]

i substitute

f(g) -> h

[chilly ocean]

oh

[coral shale]

actually i dont

[chilly ocean]

i see tho

[coral shale]

But this holds cus f is an automorphism

[chilly ocean]

i guess it isnt that f(g)=g

its more of a substitution

but it seems slightly misleading to have same variables ig

[coral shale]

I'm like viewing g in f^-1(G)

[chilly ocean]

otherwise it checks out

yeah

[coral shale]

its no sub actually

[chilly ocean]

call em g’ or something

[coral shale]

no no, its the same g

[chilly ocean]

it isnt tho

[coral shale]

it is tho

[chilly ocean]

o

it is

[coral shale]

or is it not

[chilly ocean]

lol

[coral shale]

[chilly ocean]

i can see why my bad lol

[chilly ocean]

yeah

[coral shale]

staring staring

[chilly ocean]

your second line is the g such that g in f^-1(G)

[coral shale]

basically if f wasn't an automorphism this would fail in one direction or the other

f(g) in G
equiv
g in f^-1(G)

[chilly ocean]

yeah

u were just rewriting it

oh

ur showing this ^G action of conjugation is a homomorphism?

[coral shale]

im just thinking if this kinda writing works for preimages rn 🤔

[chilly ocean]

what are u trying to prove?

[coral shale]

I was convincing myself all automorphisms act on conjugation classes

Hence Out G = Aut G / Inn G also does

In particular I want to show f in Inn G iff f is an identity action on the conjugation classes

[chilly ocean]

oh ok

[chilly ocean]

yeah ok

i guess i forget

[chilly ocean]

this is ur next step?

[coral shale]

yh

[chilly ocean]

well one direction is obvious

[coral shale]

Let $f\in \textnormal{Aut}G$
$$f(x^G) = f({gxg^{-1} : g\in G})$$
$$= {f(gxg^{-1}) : g\in G}$$
$$= {f(g)f(x)f(g^{-1}) : g\in G}$$
$$= {gf(x)g^{-1} : g\in f(G)}$$
$$= f(x)^G$$

I think I wrote the above wrong lel

tripping myself up by sticking to set notation

[chilly ocean]

yeah its cringe

just pick elements

it doesnt matter since arbitrary

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

im very used to writing proofs like this

when I can

$f(x^G) = f(x)^G$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[chilly ocean]

what precisely is x^g?

wait nvm

i tood me

[pastel cliff]

is there any intuition for the definition of cosets

can i just think of it like

[chilly ocean]

yuh

for integers

it comes down to partitioning by a multiple of some number

so think of a number n

we knows nZ is a subset of Z

[pastel cliff]

and a subgroup

[chilly ocean]

have yoy learned about quotients b4 i continue

[pastel cliff]

yesnt

[chilly ocean]

because this is where they come in handy

ok

well

say we have 3 in Z

and we look at the set 3+nZ

what does that look like you might ask

well nZ is the set of multiples of n

and +3 is adding each element by 3

now what happens if we look at 4+nZ

or n+nZ

well 4 is like last time and is just adding each element by 4

but n is adding each element by n

[pastel cliff]

n+nZ is in nZ

[chilly ocean]

but if we do that we are still in nZ

yea

it turns out that if you look at the set of all cosets

its not only a set, it has a multiplication also

it gives the set of cosets of a subgroup a group structure

[pastel cliff]

is that a thm

[chilly ocean]

well it depends on the subgroup you are looking at

[pastel cliff]

we learned cosets for the sake of lagrange's thm

hi shuri

[chilly ocean]

they have a condition called normal which means it stays fixed under conjugation, but it isnt important

it is a theorem

infact this group structture makes something called a quotient group

[coral shale]

I like to think of quotienting as gluing

https://i.imgur.com/mYQhDJS.png

[chilly ocean]

yeah its grouping

[coral shale]

You do the same for abstract structures as you do for numbers

[pastel cliff]

oh whoa

damn i just checked syllabus

quotient groups are two weeks from now :/

[chilly ocean]

quotienting only works on a special type of subset called a normal subset though

[chilly ocean]

yeah you probably need to cover conjugation and normality to talk about it next

[chilly ocean]

what else in ur slyabus screenshot?

[coral shale]

@chilly ocean oh https://math.stackexchange.com/questions/494042/distinguishing-inner-automorphisms

I'm a bit confused now

[chilly ocean]

what in the speed run is this

[pastel cliff]

@delicate orchid would you like to keep shitting on my prof

[delicate orchid]

not right now sorry I'm busy making a mockery of latex

[chilly ocean]

i remember spending a lot of time on symmetric groups/group actions/sylow subgroups

what are you supposed to take away from ur class?

[pastel cliff]

oh this isnt the full thing

[chilly ocean]

the straight jump to ring theory to a cocktease of galois theory at the end

[pastel cliff]

that's starting from week 6

[chilly ocean]

yeah

[pastel cliff]

though 1-5 weren't the best either tbh

[chilly ocean]

but im sure u didnt talk about group actions at all

[pastel cliff]

wdym

maybe

[chilly ocean]

a group action on a set is a function F:G x X -> X

[pastel cliff]

[chilly ocean]

have u seen it b4

[pastel cliff]

in this channel yeah

[chilly ocean]

bruh

educational system

[delicate orchid]

we didn't do group actions in our first group theory course

[chilly ocean]

i applied for that school too

[pastel cliff]

dont.

[chilly ocean]

wht the fuck

isnt group actions main way people think about groups

because symmetries on objects

[delicate orchid]

NO?

ohh ok ok

[pastel cliff]

ok coset though, can i just think of it for now as like multiples of a subgroup

[delicate orchid]

I see

[chilly ocean]

it also is very helpful to study group actions

[delicate orchid]

cosets are just a set of mofos multiplied by some mofo

[chilly ocean]

like its only way to talk about symmetries of surfaces without sounding handwavy

[delicate orchid]

yes but in a first course?

[pastel cliff]

times of the thing is the thing of times

[delicate orchid]

I do think they should be in there but towards the end

[chilly ocean]

i did that in a first course

[delicate orchid]

yes but this time that's literally all there is to it, a coset of a subgroup is just some element times that subgroup

[chilly ocean]

bruh tbh

idk what is covered before cosets,quotients

[delicate orchid]

group axiomz

[chilly ocean]

like subgroups product groups?

homomorphisms ig

[delicate orchid]

group axioms, homo+isomorphisms

uhh

[chilly ocean]

thats all

[delicate orchid]

cyclic groups

abelian groups

[chilly ocean]

bruh wasted 6 weeks

[delicate orchid]

order of element nonsense

uhh

LIE ALGEBRAS

[pastel cliff]

i iwish

[delicate orchid]

no, no you don't

[lethal dune]

truth algebra

[delicate orchid]

boolean "algebras" or as I like to call them, fields, are a fake construction

[chilly ocean]

structure theorem of fin abelian groups was last topic i did in my first course

[delicate orchid]

wtf lol

[chilly ocean]

we were based i guess

[delicate orchid]

I learnt about that through the structure theorem of Z modules

[chilly ocean]

yeah

structure theorem of modules over pids

[delicate orchid]

horrifying and yet ZASED

[chilly ocean]

useful af

it was used casually in algebraic topology

[pastel cliff]

or plus right?

[chilly ocean]

and i was confused because i didnt know it at time

[chilly ocean]

yeah

operations arw arbitrary

plus notationally implies you are in abelian group

times implies not necessarily abelin

[delicate orchid]

same thing

literally same thing

could be fuckin

uhhh

[pastel cliff]

ACKCHUALLY

[delicate orchid]

g♫H for all I care

g╝H

[chilly ocean]

there is just linguistic context i guess

[chilly ocean]

i saw notation like this

in differential topology

[rigid cave]

Just use :checkmark:

[pastel cliff]

plague

[delicate orchid]

let <a, a♫12 = e> be the group of chromatic notes under the operation of transposition up by a semitone

[pastel cliff]

https://tenor.com/view/flight-flights-flightreacts-scream-screaming-gif-23350989

[delicate orchid]

I wrote a little paper about group actions in music actually

for latex practice

turns out all symmetries of the chromatic scale is $C_12 \times C_2$

[cloud walrus]

Wew Lads Tbh (200 🍇) ✓

[delicate orchid]

wow looks like that latex practice was fuckin worthless

[chilly ocean]

,w unmanly

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Toysem Teans#6295

[delicate orchid]

should be semidirect maybe

idk I forgor

yeah it was D_24 epic winnn

[pastel cliff]

wait random thought

[coral shale]

_{..}

[pastel cliff]

by definition if H is a subgroup of G then order of H is less than G right

[delicate orchid]

no :troll:

do you mean a proper subgroup :troll:?

[chilly ocean]

no

think about integers

[coral shale]

if A is a subset of B, then |A| \leq |B| (assuming finite)
infinite, a similar statement about cardinality can be said

[delicate orchid]

finite groups boss finite groups

infinte groups residentSleeper

[chilly ocean]

yeah

[pastel cliff]

okay nvm

[chilly ocean]

no

[delicate orchid]

were you going to construct a chain perhaps
going to apply zorn's lemon were we?

[chilly ocean]

for infinite groups it gets weird

[pastel cliff]

i was gonna make a sea shanty

[chilly ocean]

they are all necessarily infinite

but the cardinalities can either be equal or less than

[coral shale]

no

{e} is a subgroup of any infinite group

[delicate orchid]

I'm not sure about this

[chilly ocean]

for abelian groups

equal or less than

[delicate orchid]

even simpler, I was thinking $\bZ/2 \leq \bZ/2 \bigoplus \bZ$

[chilly ocean]

e is trivial

[coral shale]

$\abs{A}\leq\abs{B}$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

I mean, this notation is still well defined for infinite sets

it just means something slightly different

[delicate orchid]

wait I mean

bigoplus

jesus

[coral shale]

why the big one lel

[cloud walrus]

Wew Lads Tbh (200 🍇) ✓

[delicate orchid]

big big chungus big chungus big chunugs

[chilly ocean]

Z/2 x Z

the order two subgroup ig

[coral shale]

wasn't paying attention to chat

you are saying for infinite abelian groups

[chilly ocean]

its still not true

[coral shale]

there are no non-trivial finite subgroups?

[chilly ocean]

i need more conditions

[coral shale]

oh

=...=

If the order of all the generators are infinite

then maybe we can say something

[chilly ocean]

yeah

[coral shale]

im not totally sure though

$$\langle a, b \mid (ab)^2\rangle$$

[chilly ocean]

there cant be finite order generators but then thats trivially saying that the group is infinite and has no finite subgroups

[coral shale]

I think this has finite subgroup

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[chilly ocean]

like free group and ab=e?

[coral shale]

yh yh

[chilly ocean]

abab=e?

[coral shale]

yh

[chilly ocean]

bruh lol

[coral shale]

am i writing dumb

[chilly ocean]

yeah

[coral shale]

$$\langle a, b \mid ab=ba\rangle$$

Ok, what happens here again lel

[chilly ocean]

so if it isnt necessarily true

[coral shale]

This is the lattice in Z^2 right

[chilly ocean]

yeah

[coral shale]

wait wait no no

wait

yes

=..=

[chilly ocean]

its abelian free group

ab=ba i get u

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[coral shale]

so we have a finite subgroup rip

right

[chilly ocean]

which one?

[coral shale]

idk

[chilly ocean]

we do not

[coral shale]

man, whats the example im tryna make up

Generators have infinite order

[chilly ocean]

free groups work

[coral shale]

but there is a finite non-trivial subgroup

oh wait nvm nvm its impossible

[chilly ocean]

idk if its possible yea

[coral shale]

i think its impossible

cus

[chilly ocean]

any finite relation you put on generators implies the group has finite generators

after the abelian part

[coral shale]

$$\langle a, b \mid R\rangle = \langle a, b\rangle/\langle\langle R\rangle\rangle$$

[cloud walrus]

𝐒𝐡𝐮𝐫𝐢𝟐𝟎𝟔𝟎

[delicate orchid]

langle my rangle

[coral shale]

Must any finite subgroup of the left

be a finite subgroup of <a, b>

?

🤔

maybe not again.....

[chilly ocean]

shuri

i can prove that any abelian group with finite condition on generators is not abelian

or maybe i can go one step above

[chilly ocean]

this is gibberish

ignore me ill be back

[coral shale]

im chucking the idea of abelian, I don't know if we need to talk about those

[coral shale]

.

[chilly ocean]

well like

[coral shale]

i was tryna think in general

[chilly ocean]

counter examples are easy

D infinity

[delicate orchid]

me when R = {a^2, b}

[coral shale]

🙏

[chilly ocean]

lots of non abelian groups are counter examples

[coral shale]

yeah so we can't have restrictions that kill the order of a , b

[chilly ocean]

yeah

[delicate orchid]

yeah cause then you just take the subgroup generated by that element boom finite

[chilly ocean]

and we know because of structure theorem thet any additional conditions besides commuatative puts a finite condition

[coral shale]

Really?

what about {ab}

[chilly ocean]

yeah