groups-rings-fields

[chilly ocean]

or india

[hidden haven]

Do I need to be country to be the best

[chilly ocean]

yes

[pastel cliff]

djibouti

[hidden haven]

[delicate orchid]

idk I just think writing that and then saying "let a = Id" is waffly and overly verbose

[chilly ocean]

all number ones own number twos

[pastel cliff]

i have a tendency to be verbose...

[chilly ocean]

if you are great you own shit @hidden haven

that is the direct translation

[hidden haven]

I own a laptop

[pastel cliff]

i own deez nuts

[hidden haven]

Or do you mean shit literally

[pastel cliff]

some days im an adult

[hidden haven]

cringe

[pastel cliff]

some days im an adult mod 12

[chilly ocean]

this is getting off topic

[pastel cliff]

welcome to #groups-rings-fields

[hidden haven]

[chilly ocean]

i dont want to enter poop talk territory

[lavish nexus]

why is the center of SU(n) in the center of GLn(C)

[chilly ocean]

but i can if god wills it

[pastel cliff]

algebraic poop

[chilly ocean]

SU(n) is A+A^T=0?

and det 1?

[hidden haven]

multiplication not addition

[chilly ocean]

yeah mb

not ring

[hidden haven]

And conjugate transpose not transpose

[delicate orchid]

SU = SO?

[chilly ocean]

o

[lavish nexus]

AA*=I

[delicate orchid]

if this is the case it's because the centre of GL_n(C) is a subset of SU so restricting the centre to SU changes nothing about the centre

[lavish nexus]

det A = 1

[hidden haven]

Taking subgroups can make the center bigger

[pastel cliff]

SU = S

[delicate orchid]

oh yeah lol

something something abelianalisation of GL

[hidden haven]

something something what

[pastel cliff]

does showing that homomorphisms preserve inverses require first showing that the identity is preserved

i coincidentally did identity first and can only verify that inverses are preserved using that

[delicate orchid]

I think so?

[pastel cliff]

yeah yeah times of thing is thing of times

[hidden haven]

The standard way would be to do that, I do not know about requiring it

[pastel cliff]

but rigor

i just mean doing one before the other

but merci

[delicate orchid]

how would you define the identity in the codomain if not by the image of the homomorphism?

[hidden haven]

I don't get it

The requiring a lemma thing feels a lot like homotopy type theory lol

[delicate orchid]

like, you need to make sure that the inverse multiplies with whatever it's an inverse of to give you the identity

but how do you do that without knowing the identity

[pastel cliff]

is there any reason why i should care about the dihedral group beyond it being a pedagogical tool

[hidden haven]

Idk maybe you can prove that nothing else can give you identity lmao

[delicate orchid]

true

[pastel cliff]

did i do a dumb

[hidden haven]

D_n just shows up randomly is all I know

[pastel cliff]

rotations and flips of n-gons?

[hidden haven]

And if you see it you can just haha look this group is D_n

[hidden haven]

Not in that form

[delicate orchid]

I've encountered some D_n-esque groups appearing randomly

[hidden haven]

But isomorphic to that

[pastel cliff]

https://tenor.com/view/thats-it-yes-thats-it-that-right-there-omg-that-thats-what-i-mean-gif-17579879

[delicate orchid]

like, C_n semidirect C_4

and similar

[hidden haven]

It shows up a bit in galois theory

[coral shale]

👀 hi

[hidden haven]

Hello

[coral shale]

i see 'semi direct product' mentioned a few times from yesterday

never seen it

should i know it? 🤔

[hidden haven]

Semi direct product

[coral shale]

doing galois and algebraic nt rn

[hidden haven]

Might be useful to know it for Galois theory

[coral shale]

cus i saw you said yesterday for x^5 - 2

[hidden haven]

You can use it to break down larger groups into smaller ones

[coral shale]

and i had no idea what that group was

[hidden haven]

Right

[delicate orchid]

ok what is this x^5-2 meme

[coral shale]

shuri's polynomial

i skimmed wikipedia and didnt get it

wheres a good place 👀

[delicate orchid]

were you trying to find the splitting field/galois group?

[coral shale]

we were, and I gave up

well i found the splitting field

'found' lul.

[delicate orchid]

Q(2^1/5, fifth root of unity) :packwatch:

inb4 "wrong"

[coral shale]

it is.

[delicate orchid]

LOL

[hidden haven]

The idea is that when you have subgroups N and H of G such that every element of G is uniquely of the form nh, you would hope that maybe multiplication is also easy

Like (nh)(n'h') = nn'hh'?

If this happens, then G is isomorphic to N × H

But sadly this don't always happen

[cursive temple]

wat how is this wrong

[hidden haven]

And the multiplication can be weird

[hidden haven]

It's correct

[coral shale]

(I meant it is right)

Ok, sure.

[cursive temple]

lol

[hidden haven]

So in direct product N × H

Notice that both N and H are normal subgroups of this product

Automatically

[coral shale]

agreed.

[pastel cliff]

if isomorphisms are bijective homomorphisms that seems really powerful, can i get an abridged version of the thread

[hidden haven]

So we can try to generalise this slightly, by requiring that only one of N and H be normal, and here you can guess why I chose N and H

[coral shale]

let N be non-normal, let H be normal

[hidden haven]

In this case, we do get a nice multiplication law, unlike when both are not normal

Actually there are multiple choices of multiplication laws in such a case. Exactly one for each choice of homomorphism H → Aut(N)

And we call of these choices of multiplication the semi direct product

[coral shale]

This term 'multiplication law' is new to me... but I get the idea

[hidden haven]

I mean we are trying to define a multiplication on this set

[coral shale]

ah ok.

[hidden haven]

And we want N to be normal in the larger group that we get

[coral shale]

So we have N ? H = {(n, h)} and we are trying to make this into a group with N normal

[hidden haven]

So now I'm forgetting there's an ambient group G, and trying to reconstruct the multiplication operation on it from knowing that every element is uniquely nh, and N is normal

[hidden haven]

Yes

And G can't be uniquely constructed because there are multiple groups that satisfy these conditions

[coral shale]

oki, that makes a lot more sense to me now

from my skimming I saw inner and outer and was ?

[hidden haven]

The idea is that since N is normal, H acts on N by conjugation

Which gives a homomorphism H → Aut(N)

Now given such an ambient group G, try to derive the multiplication formula in terms of this conjugation

To be more precise

Suppose I give you a group G such that

• N, H are subgroups, N normal
• Every element of G can be written uniquely as nh where n ∈ N and h ∈ H
• You know what h^-1nh is for every n and h (and this product is forced to be in N by normality, so is an action of H on N), and how to multiply 2 elements of N or 2 elements of H
  Can you write the multiplication (nh)(n'h') in terms of this?

[coral shale]

I'll have a go after I eat. 🙇‍♂️ ty

[hidden haven]

So these 3 pieces of information are all that are needed to recover G. And then you ask, here we took a homomorphism H → Aut N (the conjugation action of H on N). Does every homomorphism give you a group structure with these conditions, if you use the formula you derive in the above situation? Turns out yes

[hidden haven]

This is an internal semi direct product, where you write a group as a semi direct product of its subgroups

[hidden haven]

This is external, you use the formula you got from the first case to abstractly construct a new group satisfying those conditions

[lavish nexus]

formula feels not right

is it D_(gx)a(g,x)

[pastel cliff]

is there any intuition for why GL_n is relevant

[next obsidian]

Aut(K^n)

[pastel cliff]

ive seen it has to do w lie stuff but i havent gotten there

well yeah, set of all invertible matrices means they all have nonzero det

[next obsidian]

It’s Aut(K^n)

[pastel cliff]

automorphism...?

what is K

ngl i dont know the relevance of either of those things yet

im admittedly groping around

[bright marsh]

quick question what do I actually have to show for this to be true?

I can break it up into it's basis vectors

x(1, 0) + y(0, 1), but then how do I make it convincing that it's the actual rotation of theta

like if i rotate (1, 0) by a theta,

i can't just say it's (sin theta, cos theta) or smthng like that

[lavish nexus]

it is (D_ (g,x)a)(0,v)
I read D_ (g,x)(a(0,v))

[lavish nexus]

before rotating you have {(1,0), (0,1)}

as the basis

after rotating the plane you have another {e_1, e_2} as basis

express your old basis as some combination of the new basis

[bright marsh]

okay, but how do I show that the expression of my old basis, wit a new basis is valid?

[thorn delta]

so, clearly the rotation matrix rotates the basis vectors (1,0) and (0,1) an angle of theta cc about the origin. If you can also show that rotation about the origin is linear, then, it follows that all other vectors in R2 are rotated cc by an angle theta under the action of that matrix.

Alternatively, you can use high school geometry to derive the rotated coordinates of a vector

i can't think of how to make the first option work, i.e. show that rotation about the origin is linear

[bright marsh]

like for all i could know

i could express my old basis as

(cos theta - sin theta, 0), (0, cos theta - sin theta)

[thorn delta]

huh?

[coral shale]

Thanks @hidden haven, think I clocked it.

Let $\sigma_h(n) = hnh^{-1}$.

$$(nh)(n'h') = (n\sigma_h(n'))(hh')$$

So we will write
$$G\cong N\rtimes H := {(n, h):n\in N, h\in H}$$
$$(n_1,h_1)(n_2, h_2) := (n_1\sigma_{h_1}(n_2),h_1h_2)$$

[cloud walrus]

Shuri2060

[coral shale]

Then for outer... we have to construct sigma

hmmmm

[bright marsh]

o i was going on Iteribus's idea

[cloud walrus]

Shuri2060

[coral shale]

I will look up the rest (edit, ah more than this, sigma_h needs to be an automorphism of N)

[thorn delta]

the matrix they wrote down sends (1,0) to (cos, sin) and (0,1) to (-sin, cos). Changing bases would just complicate things

[bright marsh]

ahhhh ok

[frail zealot]

showing that it rotates the standard basis is all that you need to show because you can just scale those vectors up or down

[bright marsh]

right i get that

but it's the showing that it rotates standard basis is the hard part

[coral shale]

We can get abelian normal subgroups that aren't a subgroup of the center? I think? Haven't got an example

An element is in the center if it commutes with everything, but the elements within the abelian normal subgroup only need to commute within themselves

[waxen hedge]

Try to look for a counter-example with a symmetric group

Since their center is trivial

[coral shale]

Oh I think C2

[delicate orchid]

is <t> in D_n normal

[coral shale]

👌

[waxen hedge]

S_3 would probably work

[coral shale]

Cn will always be an abelian subgroup

[delicate orchid]

it might not be normal

[coral shale]

just got to look for a normal example

Like in D2n, if you take just the rotations, that's normal

[delicate orchid]

there we go

[waxen hedge]

Take the subgroup of cardinal 3
It's index is 2 so it's a normal subgroup

[coral shale]

👌

[delicate orchid]

thinking about this now

I wonder how we'd characterise all groups that have an abelian normal subgroup

[coral shale]

rn I'm trying to internalise the semidirect product

thinking of random things

I think D2n is a good example

[delicate orchid]

ohhhh I see the connection you're making

yeah you're on a good route

[coral shale]

am I? can't remember how I got here lel

[delicate orchid]

did you get here by considering what normal subgroups of the semidirect product look like?

[coral shale]

Kinda yes

but kinda not lel

[delicate orchid]

well if we consider the outer semidirect product, $G = N \rtimes H$, and then use the fact that in the definition of the inner semidirect product N must be a normal subgroup of G...
you may see where this is going

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[coral shale]

👌 ill have a look

I'm thinking rn if we can spot a semidirect product from the group presentation

Should be able.... I think..................

[delicate orchid]

going back over this stuff is reminding me why the semidirect product makes sense too tbh

[coral shale]

Didn't know of it before, and I agree it makes a lot of sense

useless group theory course

[delicate orchid]

most of the time you'll get a aba^-1 = some function of b in the presentation

due to the conjugation in the definition of the semidirect group multiplication

[coral shale]

aba^-1 = b^5

will do for example?

[delicate orchid]

I don't know if that connection is rigorous though because you can have any number of presentations for the same group

[coral shale]

but you also need every element

able to be written in the form a'b' uniquely

kinda

[delicate orchid]

yeah, I think so

[coral shale]

ehhh ill have a think on this but maybe it wont be so easy to spot

If you have a few generators, then ok

more, not so

hmmmmmm

Then finally I have to get back to z^5 - 2

[delicate orchid]

need 2^1/5 add it in
need the complex roots add it in

[pastel cliff]

mow many wew lads make a wew lad

[delicate orchid]

this isn't an algebra question

I'll be right back in 14 minuites and 20-30 seconds

[pastel cliff]

set of wew lads

[coral shale]

https://i.imgur.com/606atCA.png

consistency 101

uhhh I feel like
xgx^-1
makes the most sense as the default conjugation action..... but idk is there any convention about this?

[delicate orchid]

that's what I think as the default but thinking about it the other way produces identical results so

[coral shale]

Hmmm so for the semidirect product, I think the most intuitive way I see is

H is isomorphic to G/N with the isomorphism being the natural map H -> G/N

[next obsidian]

Semi direct product = split exact sequence

[coral shale]

not quite getting this though

[next obsidian]

It’s a special sort of isomorphism like that

[coral shale]

I have seen on wiki, but not come across those D:

[next obsidian]

In that you have a map H -> G/N

The point is just that you can find a group homomorphism one-sided inverse

Because you can just map H into N \semidirect H in the second coordinate

Like h -> (e,h)

This makes this different than the usual sorts of maps G -> G/N

Because those might not have a one-sided inverse

[coral shale]

👌 this I understand

[next obsidian]

The one-sided inverse

Is what makes it “split”

By definition

[coral shale]

@delicate orchid moar hint? D:

This was a suggestion towards hunting for subgroups?

[delicate orchid]

normal ones

[coral shale]

I've written down some algebra but uh

[delicate orchid]

chmonkey basically said it all though

[coral shale]

more think, i will do

[pastel cliff]

https://youtu.be/3aNeCWRjh8I

prof took a whole lecture to explain this with matrices

https://tenor.com/view/cat-death-lol-gif-20542436

[delicate orchid]

PERMUATION MATRICIES OMG

CAYLEY'S THEOREM.... WE GET GUARENTEED REPRESENTATIONS OF ALL FINITE GROUPS THROUGH PERMUTATION MATRICIES CALL THAT THE REGULAR REPRESENTATION

why the fuck you'd use permutations of basis set to teach S_n though

that's beyond brain dead

[pastel cliff]

havent done cayleys yet lol

wait

it's not on the syllabus

[delicate orchid]

every finite group is isomorphic to a subgroup of some S_n

not surprised it isn't

it's used like

twice

ever

[pastel cliff]

sounds neat at least

[delicate orchid]

that's a lie it's used constantly in combinatorics

[pastel cliff]

so like

twice

[delicate orchid]

thrice

[pastel cliff]

mice

[delicate orchid]

dice

[pastel cliff]

rice

set of -ice

this is algebra i promise

[delicate orchid]

just turn off permastudy

[coral shale]

Bah I'm still thinking about hunting for normal subgroups thing and don't have it

maybe I took what you said the wrong way

Are we looking for normal subgroups in any G

or a specific G that is the semidirect product of N and H

[delicate orchid]

idk I can't really remember

lol

[coral shale]

😂

I wasn't sure if you were referring to constructing new normal subgroups for N or H

or something else

[delicate orchid]

although if G is a semidirect product there has to be a normal subgroup so simple groups cannot be semidirect products

that's quite nifty imo imo

[pastel cliff]

NEVER

[coral shale]

yh sure

(looks up simple)

so many things

[pastel cliff]

.

[coral shale]

oh right no interesting normal, yes, i read that not 30 mins ago

[delicate orchid]

trying to think if we take normal subgroups $H_n \trianglelefteq H$ then is $(N, H_n) \trianglelefteq N \rtimes H$?

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[delicate orchid]

I'm gonna go with..... maybe

[coral shale]

phi : G -> G/N
phi : H -> G/N is iso

correspondence thm. Subgroups of G/N correspond to subgroups of G containing N

I think.

and then................

[delicate orchid]

correspondence thm
sheer agony

I can never remember that it's a thing (For groups)

[coral shale]

wait what

[coral shale]

why do u take normal subgroups of H

[delicate orchid]

because who cares about N we already know it's normal in the direct product

[coral shale]

oh nvm misread

[pastel cliff]

is S_n a subgroup of D_8

[coral shale]

So we're saying G = (N x Hn) x H/Hn

Probably some isomorphism thm spits this out

those x are semidirect (I hope)

[delicate orchid]

I'd just check the normal subgroup axioms

I have no clue what you're doing

[pastel cliff]

who me

[delicate orchid]

no not you boss

[pastel cliff]

le boss

[delicate orchid]

consider n >= 4

[pastel cliff]

yeah thats why it's a dumb

but how are they related then

[delicate orchid]

D_8 is isomorphic to a subgroup of S_4

[coral shale]

S_n is always a supergroup for anything finite Any finite group must be a subgroup of S_n for some n (down to isomorphism)

[delicate orchid]

*for some ns

[coral shale]

This case in particular D2n contains specific permutations of n-gons. Sn contains all permutations of n points
You can show D2n is isomorphic to some subgroup of Sn

[pastel cliff]

https://tenor.com/view/hmmm-thinking-batman-gif-14744673

[delicate orchid]

have... you not seen the elements of D_8 as permutations?

[pastel cliff]

maybe?? ngl ive been dozing off in lecture

not completely but to the point where im zombified

it's not my fault lecture is bland

looking through notes rn tho one sec

[stone fulcrum]

Start paying attention class gets real

[pastel cliff]

nah bro this class is sumn else

[delicate orchid]

stealing this from my presentation

[delicate orchid]

his syllabus and lecturer are crank frauds

I don't blame him for zoning out

[stone fulcrum]

Wait whot? Whom?

[pastel cliff]

mine

ive bitched about it in this channel lol

[stone fulcrum]

No who lecturer

Why so crank

[pastel cliff]

oh im not gonna doxx myself or him he's just wack and bland

[stone fulcrum]

That kinda sucks haha. This class should be a good time

[delicate orchid]

decided to introduce S_n through permutation matricies

[stone fulcrum]

That works I guess

Eh

[delicate orchid]

no, no it doesn't

[stone fulcrum]

You're better off having the action in your mind

[pastel cliff]

kek

[delicate orchid]

yes but you can just think about the action on the set {1, ..., n}

[stone fulcrum]

There's cycle notation in the top there

[delicate orchid]

and wow now permutation notation makes sense! the composition isn't N BY N MATRIX MULTIPLICATIONNN

[delicate orchid]

wow lying to me are we @pastel cliff

banned

[stone fulcrum]

You can very well be taught it but not know about it

[pastel cliff]

i wrote that afterwards lol

[stone fulcrum]

"This is cycle notation. Now for the important permutation matricies"

[pastel cliff]

he did teach cycle notation the lecture after introducing S_n

and also the hw is in permutation matrices

which is fine just yucky ig

[stone fulcrum]

Sad. Almost none of the theory is actually talked about in terms of matricies

[delicate orchid]

who tffff thinks about S_n as the matricies though that's so unwieldy

[delicate orchid]

oh yes lets teach intro rep theory at the same time as intro group theory (?!?!?!)

[stone fulcrum]

Maybe students were struggling to get that Sn is a set of "things you do"? It is kind of abstract

[delicate orchid]

swapping things is incredibly intuitive

a toddler can do it

[stone fulcrum]

You know a toddler that does group theory?

[delicate orchid]

now visualising swapping basis vectors in high dimensional space is a lot harder

[stone fulcrum]

Get them over here

[delicate orchid]

you're talkin to em buster

[stone fulcrum]

Prodigy

[pastel cliff]

he hasn't even expressed this idea tho, at least not in lecture

maybe he's assuming it's implied??

which ig it is but idk

[coral shale]

we should teach group theory to toddlers

[stone fulcrum]

Worth knowing this version of Sn, especially when group theory gets more "action-y"

[delicate orchid]

residentSleeper

[stone fulcrum]

If I swap block 1 with block 3, what I've done is an element of S_3

[delicate orchid]

yes, it's worth knowing but it is an awful teaching method for an introductory class

[stone fulcrum]

No I mean it is worth knowing the "swap elements" version

[delicate orchid]

I'm speaking from experience btw, my 3rd year group theory course was all matrices

and it was awful

[coral shale]

How is Sn done with matrices?

[delicate orchid]

permutation matricies

[coral shale]

permuting the standard basis vectors?

or

[delicate orchid]

yeah

[coral shale]

oh ok

that sounds like a lot of 0's and 1's

[delicate orchid]

see how easy it is to go the other way around, kaynex?

[pastel cliff]

so many

[delicate orchid]

from abstract Sn -> permutation matricies?

rather than permutation -> abstract?

[stone fulcrum]

I wouldn't even bring it there.
[1 0]
[0 1]

And
[0 1]
[1 0]
Are the two elements of S2. Multiplying them together gives the operation

[coral shale]

whats so good about seeing it this way?

[delicate orchid]

not even an irreducible representation residentSleeper

[stone fulcrum]

I'm not disagreeing haha. It's clearly the better way

[delicate orchid]

oh ok

mb

[pastel cliff]

wew you said sumn about D_8 as permutations

like yeah it's obvious

[delicate orchid]

kaynex is right about it being a very useful way to see Sn

[coral shale]

do u know what dihedral group is

[pastel cliff]

but is there any like subtle relevance

[coral shale]

i dont see how for now hmm

[delicate orchid]

they've seen it as matricies

and only matricies

[pastel cliff]

matrices then in general

[coral shale]

lets start from the top shall we...

[delicate orchid]

it follows from the action of D_8 on the corners (or edges) of a square

[pastel cliff]

like fuckin 2 lectures later

[delicate orchid]

mspaint time

[stone fulcrum]

It's a very computational way. Determinant gives even/odd which is sick. But It's stupid to think about.

[coral shale]

dihedral group 2n is the symmetry group of a regular n-gon

[delicate orchid]

sign character my beloved

[coral shale]

so d8 is symmetries of a square

[pastel cliff]

i know what d8 is lol

im just tryna reteach myself some stuff

[coral shale]

o ok lel

run from matrices

[delicate orchid]

see this

[coral shale]

oh i thought you would use a nxn matrix for Sn

[delicate orchid]

the whole point is to avoid matricies

and you do

this is a representation specific to D_8

[coral shale]

oki

if i were to write a program, this looks good ig

[delicate orchid]

if you wrote a program you'd use the matrix form

[pastel cliff]

ok im gonna be dumb for a second

actually

[delicate orchid]

1. Label the corners of the n-gon
2. Consider how elements of D_2n permute these labels
3. write these permuations in cycle form

boom

done

[pastel cliff]

that's what im trying to do lol

but i wanna make sure im not missing something silly before asking

[delicate orchid]

it's like a baby version of the proof cayley's theorem (that I've got in my head atm) 🥺

[pastel cliff]

ok fine i'll just ask - set of corners of a square is {a (top left),b (top right),c (bottom left), d(bottom right)} - a permutation might be {b, a, c, d} - what elements of D_8 do this

[delicate orchid]

there isn't an element of D_8 that does that

[coral shale]

The only things in D2n are rotations and reflections

You have to keep the edge relations

[pastel cliff]

so im not dumb

[delicate orchid]

D_8 is isomorphic to a subgroup of S_4 not S_4 itself

[pastel cliff]

[delicate orchid]

the only cases when D_n = S_n are n = 3 (or n = 1 and n = 2 if you want to be weird, don't be weird)

[pastel cliff]

i am sick in the head

[coral shale]

D0

[pastel cliff]

ok im gonna go be dumb in private i'll be back in a min

[delicate orchid]

here's D_3 as permutations

in this case it just so happens to be S_3 but ignore that

[coral shale]

hmmm I'm thinking about

(F[x]/(f))[y]

This doesnt simplify does it

[pastel cliff]

unrelated but should i spend much time on alternating groups

[delicate orchid]

it's S_n but they're even permutations

not too much time

[pastel cliff]

he defined them and then completely forgor about them

kk

[coral shale]

pretty much

if it comes up in a Q, just know what it is

[delicate orchid]

pop quiz what's the group of rotations of this bad boy

[coral shale]

s4?

[delicate orchid]

A_4

[coral shale]

oh u said rotations

why u gotta trick

[delicate orchid]

no trick

the group of rotations is all the symmetries anyway

[coral shale]

uhhhh

[pastel cliff]

"A_n is the set of nxn permutation matrices with det = 1"

[coral shale]

why are reflections no good?

[pastel cliff]

is this analogous

[coral shale]

yes, even permutations

[delicate orchid]

yeah, but fuck me holy shit why

[delicate orchid]

what reflections

[pastel cliff]

im not even gonna bother digging into that

[coral shale]

how do i even describe it

[delicate orchid]

WHY define A_n as the kernel of the sign character

[stone fulcrum]

An is the group of actions that can be done with an even number of two-object swaps

[coral shale]

you just switch 2 of the points

and leave the other 2 alone

this can be done by reflection cant it?

use a knife to chop it in half

[stone fulcrum]

Or, sorry, an EVEN number, right?

[delicate orchid]

even number yeah

[delicate orchid]

you'd have to reflect about a plane

3 points define a plane

[coral shale]

whats wrong with this

[delicate orchid]

[coral shale]

eh?

[delicate orchid]

it doesn't preserve order

[coral shale]

wat

[delicate orchid]

I dunno though, dropping the trolling act

[coral shale]

take one of the edges

[delicate orchid]

maybe if you reflect in a plane orthogonal to one of the edges it can work

[coral shale]

and the midpoint of the edge opposite

that gives you where to chop

or put your plane

[delicate orchid]

you'd have to pick a corner and the midpoint of the edge opposite, no?

[coral shale]

no no, no corner

You pick an entire edge

and the midpoint of the edge opposite

[delicate orchid]

fixing the midpoint of two edges results in a 180 degree rotation

a corner and the midpoint though? I can definitely see that being a reflection

[coral shale]

almost this

is what im referring to

[delicate orchid]

time to use my rep theory power point again

[coral shale]

you are able to permute 2 points via reflection

[delicate orchid]

here's the rotations

ah now that I see it both our planes work

everybody wins

[pastel cliff]

what are these slides 👀

[coral shale]

[delicate orchid]

_ _

[delicate orchid]

no deal

it's not... funny or anything

[pastel cliff]

i can offer you seven wulongs

[delicate orchid]

tempting

[coral shale]

Is the subset of Sn which fixes no points anything special

🤔

[delicate orchid]

n-cycles

uhh

probably

[coral shale]

it is in combinatorics

[delicate orchid]

fixed point memes come up sometimes in c-

[coral shale]

idk about group theory

[pastel cliff]

isnt that just Sn

[delicate orchid]

NO definitely not

[coral shale]

huh what

[delicate orchid]

fr though fixed points come up in some group action theorems

[delicate orchid]

(12) in S_3 fixes 3

[coral shale]

lul ok

e fixes all

[delicate orchid]

haha fermat's little theorem is a corollary of the fixed point theorem

that's funny

number theorists btfo'd

[pastel cliff]

https://tenor.com/view/thats-it-yes-thats-it-that-right-there-omg-that-thats-what-i-mean-gif-17579879

i actually knew that

yeet

[delicate orchid]

nice

[pastel cliff]

only time ive said that today

[delicate orchid]

I think the fixed point theorem follows from orb-stab which is a funny name

I scrolled too far down on the wikipedia page

[pastel cliff]

is D_8 a subgroup of S_4 or isomorphic to one

i feel like it's a subgroup

[azure seal]

think so

[delicate orchid]

it's isomorphic to one

[pastel cliff]

but

but

[delicate orchid]

D_8 is defined using the funny presentation <a, b : a^4=b^2=1, bab = a^-1>

plus

ok

an actual reason is that the permutations you get depend on the labelling of the square

you pick different labels? different permutations

[pastel cliff]

ok i was thinking in the sense that it's generated by a flip and a rotation

[coral shale]

We rarely refer to equality in group theory

[azure seal]

ye, take a = (1 2 3 4) and b = (1 3)

[pastel cliff]

and those are definitely in S_4

[coral shale]

Things are the 'same' if they are isomorphic

[delicate orchid]

turns out all of the "different" labellings each give you a coset of D_8 in S_4 :troll:

iirc

[pastel cliff]

wew lads i might ask you to hop in vc tm if that's aight

i am mental

cosets

[delicate orchid]

you can think about this as first applying any permutation a from S_4 to the labels and then applying D_8 to them, you get aD_8

of course this still is coset representative dependant but I can't be bothered you know how it is

[delicate orchid]

sure

[coral shale]

https://imgur.com/RexWqu7

lemme see D:

oh wait this is crap

i need S4 if i wanna check

[delicate orchid]

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

[pastel cliff]

,av wew lads

[cloud walrus]

Wew Lads Tbh#0864's Avatar

Click here to view the image.

[frail zealot]

symmetry group of cube?

[delicate orchid]

with reflections?

[coral shale]

ahhh i getchu

[delicate orchid]

S_4 without reflections S_4xC_2 with reflections

got bored of waiting so there's both!

[delicate orchid]

this guy gets it

[pastel cliff]

was this the wrong way to think

like

[delicate orchid]

no that's correct

like that's objectively true

[frail zealot]

I have a question, how would you go about showing that O(n) is the symmetry group of a n-1 sphere

[pastel cliff]

then why cant we say D_8 is a subgroup

[azure seal]

it is

[coral shale]

isnt

[azure seal]

bruh

[coral shale]

its just technically wrong

[delicate orchid]

it's isomorphic to a subgroup by cayley's

[coral shale]

theyre starting out in group theory, get things right

[delicate orchid]

brutual

[coral shale]

wat

[pastel cliff]

i am sick in the head

[azure seal]

the "right" way is the up to isomorphism way

[delicate orchid]

NUMBER 1! show that any reflections are generated by just a fixed reflection and some combinations of rotations
NUMBER 2! uhhhhhhhhhhhhhhhhhhhhhhHHHHHHHHHHHHH

[pastel cliff]

this video is omnipresent in my mind

[frail zealot]

[coral shale]

When we have to groups that look the same, we dont say they are equal

we say they are isomorphic

[delicate orchid]

I mean... SO(n) is just.... rotations they all.. rotate

[pastel cliff]

i get that

[delicate orchid]

tbf I usually do

[coral shale]

Thats why D8 is not a subgroup of S4. It is isomorphic to one of its subgroups

[delicate orchid]

I'm just being incredibly pedantic

[coral shale]

well yh, we say that casually

theyre 'equal'

[pastel cliff]

ok are the elements of D_8 in S_4 as well

[azure seal]

well for me at least D8 is only defined up to isomorphism so "D8 is a subgroup of S4" is fine

[coral shale]

S4 can be the permutations of any 4 things

4 points, 4 chairs, 4 cats

[pastel cliff]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

in the case where S_4 holds the corners of a square it's a subgroup

not in general?

[delicate orchid]

idea! concept! plan! @frail zealot if you can show that all SO(n) matricies are some rotation of the sphere and combine it with the "any reflections are generated by just a fixed reflection and some combinations of rotations" meme then you're done, right? Well consider that all linear transformations can be decomposed into rotation and scaling via the eigenvalue decomposition thingy (M = PDP^-1), but we know that the determinant is 1 cause it's in SO(n), so it must be a pure rotation (equivalenently, it scales space by it's determinant which is just 1 - so it's an isometry)

[coral shale]

thats the idea more or less, ig.

[pastel cliff]

epic

[coral shale]

D8 also doesnt have to be the 4 corners of a square as well

[azure seal]

still true in general, but a d8 subgroup wont be literal rotations and reflections any more, just an abstract d8 group

[coral shale]

this can be viewed as matrices... like your lecturer did...

[delicate orchid]

could be every other corner of an octahedron :troll:

[pastel cliff]

that's what i mean

[frail zealot]

[coral shale]

Well as long as you understand this

then yh sure D8 is a subgroup of S4, you can casually say this

[pastel cliff]

i get the point you were making tho

and i appreciate it

[frail zealot]

So SO(n) are the rotations and then you can represent the rest of the elements of O(n) just as pure reflections

[delicate orchid]

you can think of O(n) as being SO(n)+reflection*SO(n) (indeed these are the cosets in O(n)/SO(n))

so you actually only need ONE "pure reflection"

[frail zealot]

(-1 0
0 I )?

where I is nxn idenity

[delicate orchid]

sounds good to me

I should now inform the class that this is, unfortunately, not the group of symmetries of the sphere it's the group of isometries of the sphere

[frail zealot]

[delicate orchid]

which for anyone who isn't MENTAL

is the same thing

[pastel cliff]

I AM MENTAL

[delicate orchid]

but technically I could map any point on the sphere to any other and still get a sphere

[pastel cliff]

I AM A NUTTER

[delicate orchid]

something something axiom of choice, btw

[coral shale]

you could look at the isometries of a sphere and suddenly find 2 of them

[frail zealot]

[delicate orchid]

however, by the axiom of choice, I shall simply choose not to do so

[pastel cliff]

ok so when trying to describe the relations that generate S_n

do you just refer to each element as s_i...?

i dont wanna do it with matrices lol

[delicate orchid]

do I tell em chat

do I spoil the fun

[coral shale]

you name them what u want

[pastel cliff]

"deez" and "nuts"

[coral shale]

exactly

[delicate orchid]

consider the cycle decomposition of representations

[coral shale]

no matrices

[delicate orchid]

that's all I'm gonna say

[coral shale]

lul ok wasnt sure what u were referring to with naming

[delicate orchid]

the standard symbol for a permutation is

and I kid you not

$\sigma$

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[pastel cliff]

let sugma be an element of S_n

[frail zealot]

that reminds me of these lectures on topology i was watching and the prof was like there's no Letter for Real projective space so you can name it what you want and i call it butthole space because they way he draws it is as a sphere with a mobius strip glued on but it makes it look like a butthole

[delicate orchid]

do you know about this? btw?

[coral shale]

Usually you consider Sn to act on {1, 2, ..., n}

[pastel cliff]

most likely not

this feels like so much man

[delicate orchid]

like, (a_1a_2a_3...a_n) = (a_1a_2)(a_1a_3)...(a_1a_n)

[coral shale]

bruh

[delicate orchid]

we're jumping between like 5 topics

[coral shale]

just call them 1...n lel

[delicate orchid]

no

because they might not be

in numerical

ordrrrr

[coral shale]

o thats what u were writing

[delicate orchid]

yeah $a_i \in {1, \dots, n}$

[cloud walrus]

Wew Lads Tbh (200 🍓) ✓

[pastel cliff]

i feel slightly less bad lol

but goddamn justaskjfhaklsbflakshbdflkhasbdlhksabdfbas

[coral shale]

(1234) means
1 -> 2
2 -> 3
3 -> 4
4 -> 1

[delicate orchid]

considering the generators for S_n is kinda non-trivial if you don't think to use cycle decomp

although now my head is thinkin.... me noggin... is joggin

[pastel cliff]

i mean since he used matrices as the generators of S_n

the first thing he did was state the relations

[delicate orchid]

oh yeah you've literally got the permutations written under

[pastel cliff]

which actually make sense with matrices

[frail zealot]

i mean were they matrices or just the notation for permuations

[delicate orchid]

matricies

permutation matricies

n by n

[pastel cliff]

matrices

[delicate orchid]

the whollleee schbang

[pastel cliff]

color

neighbor

[delicate orchid]

anyway my noggin is joggin because I am convinced there's a theorem somewhere that says every finite group can be written with 2 generators at most

uhhh

[pastel cliff]

oh wow

[delicate orchid]

permutation with

uhh

p-cycle q-cycle, gcd(p, q) = 1

repeat

whole group

there we go

ok ramble over

wow I actually had to think there

very scary

[coral shale]

??? wat

[delicate orchid]

oh lord apparently it relies on the classification of finite simple groups

[frail zealot]

just do the groups with prime order and then it's easy

[coral shale]

I think its non-abelian finite simple

which have that property

more???

[delicate orchid]

trying to think of a counter example to the non-simple case

[pastel cliff]

im done

[delicate orchid]

can't be bothered

that's enough for tonight

[pastel cliff]

mr wew lads could you summarize relevant S_n shit

[delicate orchid]

time to relax by going through my presentation again

ok

uhh

the S stands for "Swag"

because it is a very cool group

idk what to say other than it's the group of all permutations of n objects

[frail zealot]

symmetric group because of cayleys theorem. the group is all possible (finite) symmetries

[pastel cliff]

so im overthinking this

oh what fun

idk uhhhhh

[delicate orchid]

ok wise guy

it's the group of all bijections of a set of size n to itself

[pastel cliff]

dont do this to me

[delicate orchid]

ok wise guy, it's the group of all SET AUTOMORPHIS- ok I couldn't keep a straight face for that one

[pastel cliff]

https://tenor.com/view/dies-cringe-math-dies-from-math-death-gif-22822700

[delicate orchid]

what are you confused about

how you compose permutations?

[pastel cliff]

idek i think ive been staring at things for too long

im just gonna finish up some easy notes and call it a night

[delicate orchid]

fairs

[pastel cliff]

i appreciate the help and patience tho

https://tenor.com/view/exterminador-do-futuro-gif-20118743

[delicate orchid]

no patience required to chat shite

[pastel cliff]

oh one quick thing

how should i think about the relations that exist in S_n

like sigma^2 = id

[delicate orchid]

now this is fun

[pastel cliff]

actually do those hold for all n

[delicate orchid]

say you have a permutation like

(123)(4657)

no wait

(12)(354)

something smaller first

the order of this element is lcm(2, 3) = 6

isn't that funny

[pastel cliff]

the fuck

[delicate orchid]

if a permutation is the product of a n_1 cycle with a n_2 cycle, the order is lcm(n_1, n_2)

and then you can use induction (with slight modification)

so if a permutation is the product of a n_1 cycle, n_2 cycle, n_3 cycle, n_4 cycle... then the order is lcm(n_1, n_2, ..., )

[pastel cliff]

what exactly do you mean by an n_1 cycle

i know what a cycle is

[delicate orchid]

in this (4657) is a 4 cycle, (123) is a 3 cycle

[pastel cliff]

ok easy enough

ya never know

[delicate orchid]

n-cycles have order n

this is kinda true for regular group elements as well, btw

[pastel cliff]

wait yeah that makes sense then

[delicate orchid]

if a is order n and b is order m then the order of ab is lcm(n, m)

you can prove this using permutations quite easily and then CAYLEY'S THEOREM

[pastel cliff]

how does this tie back to sigma^2 = id tho

[delicate orchid]

consider that 2 is prime

so it has to be a product of 2 cycles

or else the lcm won't be 2

for example, say you have an n-cycle and a 2 cycle, then the order of this element is lcm(n, 2) >= n > 2

[pastel cliff]

ok rq

[delicate orchid]

my favourite visual example of this is in... ||sheet music ||

[pastel cliff]

does this all look fine

[delicate orchid]

it's not simple :troll:

[pastel cliff]

you're right that y looks ugly

[delicate orchid]

yeah that looks fine

[delicate orchid]

anyway I've launched sibelius so I might as well explain nvm don't want to shit up the channel

[barren sierra]

How is this different to G being solvable?

so then all solvable groups are nilpotent?

and all nilpotent groups are solvable?

[thorn delta]

wait n

o

[viscid pewter]

considering the group generated by i and j, this is just Z/3Z, right?

[barren sierra]

sounds right?

[next obsidian]

I don’t think solvable implies nilpotent

If it did the two are equivalent

[barren sierra]

but then?

[viscid pewter]

because, assuming group properties, the top two things give that i^3 = j^3 = 1

[thorn delta]

maybe it means G_i is normal in G for all 0 <= i <= r

[viscid pewter]

but then the bottom thing gives that i and j are inverses

[barren sierra]

oh you know what that might be it

[thorn delta]

this is a bit stronger than being solvable

[barren sierra]

ok so nilpotent => solvable

got it got it

[thorn delta]

right

[viscid pewter]

yeah so this is absolutely trivial if you even have the four group criteria

idk how you stretch this into a 38 minute video

[lavish nexus]

do a group ring

[barren sierra]

they might be appending this to the integers or something

or reals

[viscid pewter]

maybe

[barren sierra]

similarly to the complex numbers

so you consider a + bi + cj

[viscid pewter]

oh, right

[barren sierra]

however I am pulling this out of my ass I have not watch the video

[lavish nexus]

which is a group ring

[viscid pewter]

yeah i'll have a look

[barren sierra]

yea

[viscid pewter]

idk what a group ring is, so

[barren sierra]

so take a ring R

and a group G

[viscid pewter]

i just saw 'hey this looks like a group presentation'

[barren sierra]

R[G] is all elements of the form r_1 g_1 + r_2 g_2 + ... + r_n g_n

with coordinate wise addition

[viscid pewter]

that makes sense

[barren sierra]

and distribution to give you multiplication

[viscid pewter]

so G is C3, but R[G] might be interesting?

[barren sierra]

yea

[viscid pewter]

right

[barren sierra]

cause since G is C_3

everything can be combined

and you get a + bi + cj

[viscid pewter]

no, yeah

[barren sierra]

just learned about group rings today in class, how fun

[delicate orchid]

is G C_3 or C_3xC_3

nvm yeah it's C_3

[barren sierra]

ye

[delicate orchid]

they are swag

[barren sierra]

they're not too much at the front of my mind

cause uh

midterm Wednesday

and we have done
Free Groups
Automorphisms
Group Actions
Sylow's Theorems and Classification
Composition Series, Solvable, Nilpotent Groups

[delicate orchid]

wait and you're on group rings?

[barren sierra]

yea we started ring theory on Friday

[delicate orchid]

have you taken an advanced lin alg course before hand or something?

[barren sierra]

this is a graduate level algebra course lol

and I am dying

[lavish nexus]

Note that H is not a group ring of Q8

[delicate orchid]

ok... then why are group actions and sylow theorems in there

sounds very based though

[barren sierra]

well they weren't introduced in my undergrad level lmfao

[delicate orchid]

devastating

[barren sierra]

it is very based tho

https://faculty.math.illinois.edu/~rezk/500-sp22/lecture-notes-500-sp22-1.pdf

[lavish nexus]

lol rezk

I had him for my 416H

[delicate orchid]

universal properties

[lavish nexus]

A long long time ago

[barren sierra]

everytime I talk to him its like meeting him for the first time

it's so strange

I talk with him after class, go to his OH

and it's just as awkward every time

also the man writes so fast and not fast at the same time lmao

I'm happy I'm taking the course with him it's just I hope the course has a curve

[lavish nexus]

Actually since when did they start offering 500 in spring

[barren sierra]

idk lol

Happy they did tho

[lavish nexus]

Not when I was there

They’ll curve bigly it’s a grad class

You do things you get B

half of the people get A

[chilly ocean]

um

for proving that each coset gK where G is a group and K a kernel of a homomorphism have the same number of elements

would you argue that

K has a fixed number of elements which you're multiplying each g by

[bleak abyss]

Pretty much, show that's a bijection then gg

And now I'm hijacking this channel to do some Weil rep stuff

Iteribus stay it'll be fun 🙂

[chilly ocean]

Simone Weil?

[lavish nexus]

😂

[bleak abyss]

Close, Andre Weil 😛

So we have the Heisenberg group

[prisma ibex]

you going through Gan's notes tonight haha

[bleak abyss]

Prasad's

[prisma ibex]

oh cool

[bleak abyss]

They're better written according to Simon

[prisma ibex]

yeah these are better

[cloud walrus]

Several Sloths

Several Sloths

Several Sloths

[bleak abyss]

This is the Schrodinger rep

Nice theorem here is by Stone and von Neumann

Heisenberg group has up to iso one irrep acting by a given central character

[neat valley]

If F is a field of characteristic zero and A is an abelian group, then

F ⊗_Z A = F^rank(A)

why is this true?

[bleak abyss]

Is A finitely generated?

If so then it boils down to showing tensor product commutes with direct sums

Any ring tensor with Z is itself

And then you'd show char 0 field tensor torsion abelian group is just 0. Which is just oh take a pure tensor scale it by a number which is invertible in the field (thanks to char 0) but murders the group (thanks to torsion)

[cloud walrus]

Several Sloths

[bleak abyss]

Let's denote the intertwining equation by *

So $\rho_{\psi}(gw,t) \cdot \omega_{\psi}(g) = \omega_{\psi}(g) \cdot \rho_{\psi}(w,t)$

[cloud walrus]

Several Sloths

Several Sloths

[bleak abyss]

So we have an exact sequence

[cloud walrus]

Several Sloths

Several Sloths

[bleak abyss]

Now let's do some computations

[cloud walrus]

Several Sloths

Several Sloths

[bleak abyss]

Focus is struggling rn

[uneven folio]

😣

Idk why thats persevere but

[chilly ocean]

Is this proof correct?

[next obsidian]

The idea is, but you should write more clearly

For example, you introduce a and b but don’t specify what that means

Granted, I know you mean that the element (a,b) has order lcm(|a|,|b|), but you should write that

Also maybe you want to justify why the largest possible order is lcm(m,n)

[chilly ocean]

I will adjust my proofs accordingly. Thanks

[bleak abyss]

Let's try again

[bleak abyss]

So question is why that's the case

[cloud walrus]

Several Sloths

[bleak abyss]

Or better yet

So we have our symplectic basis

So we just verify commutation relation

[cloud walrus]

Several Sloths

Several Sloths

[bleak abyss]

Umm

Uh oh

I'll ignore that part for now

Anyway

The result is

[cloud walrus]

Several Sloths

[bleak abyss]

I'm guessing absolute value means psi here

Now on the other hand

[hidden haven]

That's it nice

For external, you don't need to define sigma, it's a homomorphism H → Aut N and the claim is that the same formula always defines a group structure

[barren sierra]

what is the quick way of doing 6ii?

C(x) is the centralizer of the element x and Cl(x) is the orbit of x

6i was easy. So if you want C( (1 2) ) then that's the powers of (1 2) of which there are 2 or the elements that are disjoint from (1 2) which there are I believe 5?

[terse crystal]

You just calculate it by definition…

For example |C(x)| for a x in S_n =Π(λ_i !)i^λ_i

|C(x)| in A_n need to be divided by 2 if there exists an odd b commutative with x

Where x has λ_i many i-cycles

C_n trivial

|Cl(x)| = 1/1/2/2/2 in Q_8 if my calculation didn’t go wrong

[barren sierra]

I'm talking about identification

so like S_n has 120 elements, the number of 2 cycles is 5 choose 2 = 10 so by orbit stabilizer C( (1, 2) ) = 120 / 10 = 12

but then what group is C( (1 2) )

like idk what more I can say than < (1 2) > x S_3

[terse crystal]

You just … by definition

A permutation b satisfying ab=ba must map any cycle of a to another cycle of a

Order preserving also

Like say you want to determine C(x) in S_7 where x=(146)(25)(37) , then y mapping 1,4,6 to 4,6,1, mapping 2,5 to 7,3 mapping 3,7 to 2,5 is from C(x)

[barren sierra]

sure

so for that cycle type 3 + 2 + 2 there are (3^1 * 1!) * (2^2 * 2!) = 24 elements of that cycle type

[terse crystal]

Yeah

[barren sierra]

so then C(x) = 24

[terse crystal]

||=24, yeah

[barren sierra]

but

C(x) is isomorphic to ????

that's what I was asking lol

like C(x) is a group and I want to classify it

that's what that problem is asking

is it just S_2 x S_2 x S_3?

[terse crystal]

Oh I see not hard

It is isomorphic to

Direct product of S_λ_i and λ_i copies of Z/iZ

And direct product in terms of i again

So Π(S_λ_i Π(Z/iZ)^λ_i)

Since permutation of λ_i i-cycles, and i ways mapping a i- cycle to another i-cycle

[barren sierra]

Hmmmmm that is annoying lol

[terse crystal]

But it satisfies the condition, only symmetric groups and cyclic groups used

So it was my English reading problem… I thought you were going to calculate |C(x)| for those S_n,A_n,…😂

[barren sierra]

ah yea I already did that lol

it's the classification which is >_>

[terse crystal]

Oh I forgot to mention, direct product of (S_λ_i Π(Z/iZ)^λ_i) in terms of i of non-zero λ_i , I forgot non-zero

So
Z/5Z
Z/4Z
Z/3Z times Z/2Z
S_2 times Z/3Z
S_2 times Z/2Z times Z/2Z
S_3 times Z/2Z
S_5

[coral shale]

$$(F[x]/(x^5-2))[y]/(1+y+y^2+y^3+y^4)$$
$$\cong F[x,y]/(x^5-2, 1+y+y^2+y^3+y^4)$$

[cloud walrus]

Shuri2060

[coral shale]

This should be true right?

[next obsidian]

Yes

This follows because if I is an ideal of A, A[x]/IA[x] = (A/I)[x]

Take A = F[x], I = (x^5 - 2)

And then you’re using the 3rd isomorphism theorem

Or maybe the 2nd

But I think the 3rd

[coral shale]

Thanks

[coral shale]

Why IA[x] as opposed to I[x]

[next obsidian]

same thing

[coral shale]

Am I right in thinking left and right Cayley graphs should be isomorphic?

Left, right multiplication, same generators

[delicate orchid]

graph theory monkaS

[coral shale]

Ah I think I got it. G^-1 is isomorphic to G. Map the generators to their inverses (ab)^-1 = b^-1a^-1

[coral shale]

Hmmm I can't seem to find this anywhere which makes me unsure 🤔

[stoic rose]

Yeah it's true that a group is always isomorphic to its opposite via g -> g^-1

[coral shale]

I'm thinking about isomorphism of graphs and I think the above shows it

[next obsidian]

no

this isn't a homomorphism unless G is abelian

oh wait

its opposite