groups-rings-fields

[solemn rain]

he means in VR

[tiny pagoda]

It's VR, of course they can.

I actually have Valve Index so I have individual finger tracking

[solemn rain]

well me and a friend just played once or twice visting servers and trolling in voice chat

[stone fulcrum]

Oh shit

[tiny pagoda]

I get to flex on people by doing F and Y signs natively

[stone fulcrum]

That's pretty cool actually

F

[tiny pagoda]

Yeah it's really fun!

[solemn rain]

yeaa

gotta get

the new halflife

tho

am i right guys haha

[tiny pagoda]

I have it

[solemn rain]

how is it

for u

[tiny pagoda]

Oh you can't play it yet

[solemn rain]

oh lmao

OKAY BUT YOU GOTTA BE EXICTED

FOR THE NEW DOOM ETERNAL

THO

RIGHT?

[tiny pagoda]

But I'm a bit of a HL nut, I speedrun the first game and I have like 400 hours in it

Hell yeah!

I'm the pink hair on the right

[solemn rain]

this is so getting out of off topic and u gotta sleep but fuck it

[tiny pagoda]

Sorry right

[solemn rain]

damn

[tiny pagoda]

Not person in middle with pink hair.

[solemn rain]

yellow

?

wearing?

[tiny pagoda]

Yes

Yellow vest white shirt

[solemn rain]

cool

u ordered doom/

[tiny pagoda]

👌

Not yet, just blew $350 dollars on fully body tracking for VR so I'm waiting a bit before buying more stuff.

[solemn rain]

yea

worth it tho

u can crack doom eternal later

Right?

[final gulch]

only 22 hours

[solemn rain]

yeaa

beware of spoilers tho on youtube

cuz UK ppl got the game

just a heads up

[final gulch]

yes I been avoiding that, there are so many videos about it

[solemn rain]

yea

cool abstreact algebra convos

[final gulch]

yes

what is the group of symmetries of a cacodemon?

[solemn rain]

D_666

[tiny pagoda]

I'm not cracking it I love Id too much

[solemn rain]

yea

and gordon

the doom music guy

OOOF

[tiny pagoda]

I need to play through Doom 2016 again anyways so I wouldn't be playing Doom Eternal immediately either

[woven delta]

Shitpost in a different channel please

Thanks

[solemn rain]

why tho

doom 2016 wasnt that even like

story driven

[woven delta]

#chill

[solemn rain]

@woven delta yea sure sorry

[tiny pagoda]

Anyways I should sleep but if you want to know more about the ASL stuff friend me on discord and I'd be happy to introduce you, the community is very friendly and actually way more wholesome than I'm used to online

[solemn rain]

[ripe crest]

you mean #chill

[ripe crest]

The difference between the direct product and direct sum of R-modules is just that we require finitely many coordinates to be non-zero in the sum?

I guess this isn't specific to modules

[stone fulcrum]

@ripe crest
Yes. The reason why we make a distinction is that there are different properties between the two

[cerulean wraith]

Anyone know what the difference between an field and a sigma - field is?

I don't understand the difference between finite intersection and countable intersection?

[mild laurel]

do you know the difference between finite sets and countable sets?

[cerulean wraith]

I guess not hahaha

[mild laurel]

countable means in bijection with a subset of N

[scarlet estuary]

a set is countable if it is finite or if there is a bijection between it and N

[mild laurel]

in other words, countable sets can be infinite

[scarlet estuary]

the notion of countable intersection turns out to be a very important one

[cerulean wraith]

So countable can be finite but finite can't be countable?

[stone fulcrum]

I can list every fraction that exists. The set of all fractions Q is countable.

But I can't list every real number. No matter what strategy I use, I'll always miss some. The real numbers R are uncountable.

All finite sets are countable

[cerulean wraith]

Oh I got that mixed up, so all finite sets can be countable, but not all countable sets are finite

[scarlet estuary]

precisely

and you'll hear "countably infinite" for an infinite countable set

an example of such a set is N itself

because well, x |-> x is trivally a bijection

a more interesting example is the aforementioned rationals, Q

or even the set of even numbers

or the set of prime numbers

in general, countability is the smallest "size" an infinite set can have

[cerulean wraith]

I see

[scarlet estuary]

for an example of an uncountable set

consider R, the real numbers

you cant construct a bijection between N and R

(this is cantor's diagonalization theorem)

so in a sense, there's "more" reals than naturals

and hence "more" reals than rationals and etc.

[cerulean wraith]

yes

srry caps

[mild laurel]

Also, fields and sigma fields are very different objects

[scarlet estuary]

also yeah theyre just completely different like definitionally

sorry, neglected what this was about

ignore the fact that they share a word

the definitions are completely different

[cerulean wraith]

So, going back to fields v.s. sigma - fields, fields has 'less' intersections than sigma - fields?

[mild laurel]

fields has no intersections in its definition at all

[scarlet estuary]

again theyre just very different things

im not sure what definition of "field"

youre looking at

[cerulean wraith]

Ohh right, I forgot to the define my field

Its from Probability Theory

Suppose you have a set A. Then F is a field if it is a collection of subsets of A, closed under the operations of finite unions, finite intersections and complementation, along with containing A,ϕ. F is a σ-field if it is a field but now allowing countable unions and countable intersections.

So, how would you try to explain the difference using the finite intersection v.s. countable intersection?

[scarlet estuary]

oh a literal

field of sets

lmao

sorry been a while

usually people just say "algebra over a set" in my experience butw/e

anyway

i mean it might be best to explain with an example here

consider an infinite set $S$ and some $s \subseteq S$ where $s$ is the collection of all subsets that are finite or of finite complement

[cloud walrus]

Namington:

[scarlet estuary]

then $s$ is a field, but not a $\sigma-$field

[cloud walrus]

Namington:

[scarlet estuary]

this is the canonical example

but theres a few more

we could also consider say

well lets think geometrically here

sorry this is off-the-cuff but i think it works

[cerulean wraith]

Like if you were to give me a sigma - field example?

[scarlet estuary]

suppose we're thinking about subsets of $\bR^2$

[cloud walrus]

Namington:

[scarlet estuary]

say we consider the field containing just the open boxes and their unions

this is clearly a field of sets

since no matter what open boxes you smash together with union/intersection, you get an open box or union thereof

but is it a sigma field?

as it turns out, no

we can take an infinite union of open boxes

and construct, say, an open disc

or some other similar object

that definitely istn an open box

for an example of something that IS a sigma field

i mean, theres so many

obviously the power set of any set applies

but that doesnt really motivate the definition

[cerulean wraith]

Aaah ok, so fields is when you mash together an open box with another open box and it becomes an open box, but sigma - field is when you smash an open box with another open box but it becomes something else but can be still an open box if it wants to, just more shapes?

[scarlet estuary]

i mean thats just one example

and its because an infinite limit doesnt necessarily preserve geometry conventionally

like the specific reason that isnt a sigma algebra is

if you take a limit of a union of open boxes

that "approximate" an open disc

you end up with an open disc at the limit

[i.e. after a countable amount of unions]

but clearly an open disc is not an open box

thats just one particularly geometric example

[cerulean wraith]

Ooh okk

So, an example of a sigma - field is a field containing all elements of N?

Like a field would be a 2n, and a sigma - field would be n?

[solemn rain]

@chilly ocean

you can talk in voice tho but anwyays

so all i know is S_n and permutations

cycle notation

and i am given now the definition of a 2-cyle

called a transposition

or no the other way around but okay

now im given this definition

(triangle symbol whatever lmao ) = product (x_i-x_j) where 0<i<j<=n

a permutation acts on this product

by permutating the indexes

[chilly ocean]

are you talking about sign of a permutation?

[solemn rain]

yes

[chilly ocean]

that definition basically takes a poly

and then you have a ratio of polynomials with the same factors (but not necessarily same sign)

and this results in a number +1 or -1

imo it's a quite convoluted definition of a permutation's parity

but it's also pretty cool

[solemn rain]

okayy

so an even permutation

is when a permutation doesnt affect the triangle basicaly

and an oodd is hwen its -1

[chilly ocean]

Anyway uhh you said earlier you wanted clarification on even permutations and alternating group right

[solemn rain]

now what i dont know

is hwo to solve problems with alternating groups

how to find them

how to write them

like how do you find the set of even permutations on {1,2,3,4} forr exampl

A_4 ig

and more stuff

[chilly ocean]

ok so first of all

Do you the definition of even permutation right?

(Without using the sign definition)

[solemn rain]

an even permutation when

is when a peprmutaion

can be written as an even number of transpositions

thats one

[chilly ocean]

You can show that an even permutation is a permutation composed of an even number of transpositions correct

Ok cool

Can you see why even permutations would form a subgroup of S_n?

[solemn rain]

well

yea

yea yea

i do

[chilly ocean]

ok cool

alright now you want to find even permutations in S_4 right

[solemn rain]

yea

[chilly ocean]

Ok what's the order of A_4

[solemn rain]

24/4

/2

24/2

12 ?

[chilly ocean]

4!/2 yeah

Do you agree that |S_4 : A_4| = 2

[solemn rain]

yea

[chilly ocean]

well ok there's a naive way to do this and a more complicated way to do this

[solemn rain]

i just wanna understand stuff

its the next section

after simple and solvable groups

and i gotta prove soon that A_n is simple

for n>+5

and later on

im going to know that this is somehow related to polynomials lmao

like wtf XD

[chilly ocean]

well the naive way is to directly count the permutations

the more complicated way makes use of the fact |S_4 : A_4| = 2

[solemn rain]

the naive way is just

write all 24 permutations?

and see which of them are even?

[chilly ocean]

yup

and that's the difficult way too

[solemn rain]

im not that good with cycles lmao

[chilly ocean]

the simpler way (and technically more complicated)

observe the index right

[solemn rain]

okay

okay

it has to be cyclic ig

[chilly ocean]

Take any even permutation π

[solemn rain]

S_4/A_4 is ismoprhic to Z_2

[chilly ocean]

Let τ = (1 2)

Would you say πτ is odd?

[solemn rain]

yes

cuz even +1 is odd

right?

XD

[chilly ocean]

So for every even permutation we ahve a corresponding odd permutation right

[solemn rain]

yea

[chilly ocean]

ok cool

do you agree then that there are at least as many odd permutations as even permutations

[solemn rain]

yea

by waht you just said

[chilly ocean]

Now consider an odd permutation σ

We can similarly show στ is a corresponding even permutation right?

[solemn rain]

yea

[chilly ocean]

so there are at least as many even permutations as odd permutations

right

[solemn rain]

yea

[chilly ocean]

but wait

earlier we said there are at least as many odd permutations as even permutations!

this means there are an equal number of even and odd permutations

[solemn rain]

yea

okay

[chilly ocean]

So now you know the order of A_4 (and why it's that way)

[solemn rain]

okay

[chilly ocean]

and hence the number of even permutations right(?)

because all the even permutations in S_4 will form a subgroup of S_4

[solemn rain]

yea cuz A_4 iis set of even permuations

ig

opkay

yea

[chilly ocean]

any questions now?

[solemn rain]

no no

so we showed

for every even permutation there is odd permuation

and the otherwise

so they are equal

now thas why |A_n| = n!/2

[chilly ocean]

Yup

[solemn rain]

yea i totally get it

by that i getr that the index of A_4 in S_4 is 2

now how can we use that ig

[chilly ocean]

well what do you want to use the index for

I basically just showed you why A_n has the order it does (and consequently how to count the total number of even permutations in S_n)

[solemn rain]

yea

now

is this how i find A_n

[chilly ocean]

Well, I showed you how to find the order of A_n and why its that way :p

[solemn rain]

yea

cool lmfao

lmfao

okay

so can you help me with these problems

[chilly ocean]

uh I might be able to help

post them

[solemn rain]

The permutation u is odd if and only if the number of cycles of even length in its cycle decomposition is odd.

now i get so fucking confused

[chilly ocean]

oh shit this was one of my homework problems 2~3 weeks ago lmao

[solemn rain]

an m-cycle is an odd permutation if and only if m is even.

am i supposed to be able tro prove those?

if ur oging to ask

idk group actions yet

thats next chater and its a big one

[chilly ocean]

lol you dont need group actions dw

[solemn rain]

okay XD

[chilly ocean]

you only need to know about the symmetric group

[solemn rain]

yea i do know about it abit

[chilly ocean]

cool

then you can totally do these questions

[solemn rain]

Prove that pi^2 is an even permutation for any permutation pi

Show that Sn = ((1 2), (1 2 3 .. . n) } for all n :::: 2

[chilly ocean]

these are a lot of questions

[solemn rain]

ugh srry

[chilly ocean]

and I can answer exactly one

It;s kinda late here so I'll help with one and recommend asking the rest in #❓how-to-get-help

choose whichever one and I'll try my best to help rn

[solemn rain]

wow ur so good

umm

can you give me a hint for 2?

i was going to exhaust

everything

let pi be odd

then pi^2 is odd(odd) then its even

[chilly ocean]

Question #2 is "an m-cycle is an odd permutation if and only if m is even." correct?

[solemn rain]

yea

yea

ooh no

[chilly ocean]

Ok so let's do some computations first

[solemn rain]

i didnt mean this

but yea

[chilly ocean]

oh ok nvm lol

[solemn rain]

lets do this 1q

no no

lets do it

[chilly ocean]

sure why not lol

[solemn rain]

Question #2 is "an m-cycle is an odd permutation if and only if m is even." correct?

now let m be even

[chilly ocean]

Ok so let's write out an m-cycle in S_n

[solemn rain]

okaay

[chilly ocean]

Before we even begin the proof

Let's try to get some idea of why it's true

[solemn rain]

(a_1a_2a_3...a_m)

[chilly ocean]

$\sigma = (s_1 ; \cdots ; s_m) = (s_1 ; s_m) \cdots (s_1 ; s_2)$

[cloud walrus]

kakuhen:

[chilly ocean]

in general we can write the cycle as a product of 2-cycles right

how many 2-cycles do we end up with?

[solemn rain]

in general we can write the cycle as a product of 2-cycles right'

why

i am fuzzy with

the mechancis here

of cycles

tahts why im bad

[chilly ocean]

has your professor shown you that result yet

that you can write permutations as product of transpositions

[solemn rain]

oh yeah

nvm i read it

[chilly ocean]

if you dont believe this then just do the computation

[solemn rain]

and its later

proved

[chilly ocean]

multiply the individual 2-cycles in my equation above

and you should get the m-cycle back

[solemn rain]

okay just an example cuz im fuzzy with mechancis

[chilly ocean]

e.g. (1234) = (14)(13)(12)

[solemn rain]

(1234)

okay so thats how you write a cycle

as product of transpositions

and this is unique right?

[chilly ocean]

Not necessarily

[solemn rain]

what

okay

okay okay so

back to our problem

(a1a2a3....am)

=(a1am)(a1a2)

...

yea

okay

[chilly ocean]

Given an m-cycle, how many 2-cycles do we end up with?

count the individual 2-cycles

[solemn rain]

let me think cuz im stupidi now

idk ugh

wait

XD

okay inm stupid let me do an example fuck it

oh okayh

m-1?

[chilly ocean]

Correct!

[solemn rain]

is it bad

that i couldnt do this

on my own

and saw ur example

and counted?

lmao

[chilly ocean]

That's not being bad

That's literally what I wanted you to do.

[solemn rain]

i used thise.g. (1234) = (14)(13)(12)

my brain was too weak for the m case

lmao

yea

[chilly ocean]

yeah that's fine that you needed some guidance

[solemn rain]

so if m is even

[chilly ocean]

you got the right answer

[solemn rain]

m-1 is odd

hence number of transpositions is odd

lmlao

cool

cool af lmao

u up for another 1?

lmao

jk

is lcm(2,2) = 2?

[chilly ocean]

Hold on

One last thing

[solemn rain]

oh

[chilly ocean]

So you basically got the idea of how to prove that statement

[solemn rain]

its and iff

yea yea

[chilly ocean]

Now let's go back to your "triangle thingy"

[solemn rain]

okay

[chilly ocean]

usually we write that as sgn(σ)

i.e. the "sign of the permutation σ"

[solemn rain]

so sgn(phi) = 1 iff phi(triangle ) = triangle

[chilly ocean]

Notice that what you just did implies sgn(σ) = (-1)^m-1

[solemn rain]

sgn(phi) =-1 iff phi(triangle) = -triangle

[chilly ocean]

so that's another way to see the problem

[solemn rain]

where m is length of phi

yea cool

[chilly ocean]

glad I was able to help you :p

anyway I'm gonna shower now lol

[solemn rain]

yea cool

im so bad tho

lmao

[chilly ocean]

nah you're just learning

and the fact you feel bad is evidence

[solemn rain]

i dont think im able to do most problems

[chilly ocean]

It wasnt uncommon for me to spend hours on some exercise and then ask a professor for a hint

and after that hint everything suddenly seemed so obvious and I felt stupid

lol

[solemn rain]

harder exercises tho lmao

not this

but anyways

tysm

[chilly ocean]

uh no I did literally everythingyou did

so yeah dont put yourself down so much

[solemn rain]

ur using df?

[chilly ocean]

Nope

[solemn rain]

conincidence ig

these are fro mdf

[chilly ocean]

my professor made his own textbook

[solemn rain]

from*

[chilly ocean]

but like these are standard exercises

[solemn rain]

ye aayea

[chilly ocean]

everybody ends up doing them in algebra classes

[solemn rain]

yea

so

these problems are supposed to be taking hours from me?

[chilly ocean]

Yes

Unironically

[solemn rain]

give or take?

[chilly ocean]

Yeah

That problem I just helped you on

When I was doing it myself for the first time ever

I think I spent at least 30 minutes

and that was the quickest exercise I've done

[solemn rain]

yea man cool

okay this is going to waste your itme

sorry but

is this triangle thingy

related to determinants?

thats outside the book but just wanna know more

[chilly ocean]

no idea tbh

[solemn rain]

cool

[chilly ocean]

anyway can you snapshot the triangle thingy

im curious on the notation lol

[solemn rain]

its just a triangle

discrimant

triangle

[chilly ocean]

oh

[solemn rain]

reversed nabla

[chilly ocean]

huh

[solemn rain]

ig

[chilly ocean]

interesting

[solemn rain]

wait

and yea its written up how wto write

something as transpositions

yea i suck

.

[chilly ocean]

ohhh

now that I tihnk about it my prof did this too lol

[solemn rain]

yea then its probably related to some higher shit

that we are ggoing to use later

ig

[chilly ocean]

nah

trust me i've only seen that cursed polynomial like once

and never again

in my algebra class

lmao

[solemn rain]

lmao does it

get worse?

[chilly ocean]

No

[mild laurel]

yes

[solemn rain]

fuck

[chilly ocean]

oh it depends on "it"

[mild laurel]

uh

wrong thing I'm replying to

[chilly ocean]

I thought you meant like "the sign of permutations"

[solemn rain]

i wanna do verify for my7 self

[chilly ocean]

at least in an intro to group theory class

[solemn rain]

that A_n is kernel of the homomoprhism

[chilly ocean]

you're only gonna see that cursed polynomial once

[solemn rain]

S_n ---> sgn

yea

[chilly ocean]

maybe in more advanced courses it'll reappaer

[solemn rain]

cool

[chilly ocean]

when you let the idea sit in your head for a week or two

it's actually really cool

[solemn rain]

what idea

[chilly ocean]

you'll see it 2 weeks from now

[solemn rain]

the kernel?

[chilly ocean]

uh that definition of a permutation's parity

when I first saw it I was like "wtf is this non-sense"

and understood 0%

and like 1~2 weeks later it suddenly clicked

[solemn rain]

okayyy

what was the next section

for you after this?

groupa ctions also?

[chilly ocean]

this was the entire syllabus basically

[solemn rain]

thats exactly same as df

but df has 1 more chapter ' furthur topics in gt'

[chilly ocean]

lmao my professor straight up made a shittier copy of D&F then

[mild laurel]

you should confirm that the discriminant corresponds to the normal idea of a discriminant for quadratics

[solemn rain]

what determinant

[mild laurel]

i.e., if you let the x_i be the roots of a polynomial, then that equation gives you the determinant of your polynomial

b^2 - 4ac

[solemn rain]

discirinat?

discrimnant u mean?

[mild laurel]

or sqrt(b^2 - 4ac) i guess

discriminant yeah

[solemn rain]

yea thats the name i know it bye

by*

[mild laurel]

its the right name

[solemn rain]

oh

okay

cool question :

determiannts

are realted to this?

related*

[mild laurel]

idk probably

[solemn rain]

liek if we write the permutation usingg array or matrix notation

and find the determinant

if 1 even

if -1 odd?

i think finding determiannt would be really hard ig

a problem that i attempted just wanna check if profo is right

pi^2 is an even permutation for any permutation pi

proof: pi is an m-cycle

if m is even then pi has odd number of transpositions

then pi^2 would have (odd number) times odd numebr of transpositions hence even transpositions

if m is odd then same

ig

is that irhgt argument>?

[lament pilot]

I'll follow @mild laurel 's advice and post my question here.
I'm trying to find the left cosets of $\mathcal{S}{n}$ in $H = \mathcal{S}{-1}$.
I've thought about it some more and this is my current reasoning: ($\mathcal{S}{n}$ : $H$) = $\frac{|\mathcal{S}{n}|}{|H|} = \frac{n!}{(n-1)!} = n$.
So there are $n$ left cosets. Now, by definition, $\sigma \sim \sigma'$ if $\exists \rho$ such that $\sigma = \sigma' \cdot \rho$. Since $\rho \in H,$ $n \notin Supp(\rho)$. Now here is where I struggle to rigorously write this: $\sigma$ is not equivalent to $\sigma'$ iff $\sigma(n) \ne \sigma'(n)$. Is this correct?

[cloud walrus]

Smor Brother:

[lament pilot]

There are n choices for sigma(n) and hence n left cosets

[mild laurel]

Okay, wait what does Supp(p) mean?

[lament pilot]

Oh, is the notation different in english? My bad.
$Supp(\rho) = { i \in {1, ... ,n}$ such that $\rho(i) \ne i}$

[cloud walrus]

Smor Brother:

[lament pilot]

Basically all the {1,..,n} that are changed by the permutation

[mild laurel]

Ah no, that's standard, I had just forgotten about that

[lament pilot]

The $n$ from $\mathcal{S}_{n}$

[cloud walrus]

Smor Brother:

[mild laurel]

Yeah, that statement isn't true

[lament pilot]

[mild laurel]

like if you take n = 3

then you can take the permutations f,g such that f(1) = 1, f(2) = 3, f(3) =2 and g(1) = 3, g(2) = 2, g(3) = 1

but these two permutations are equivalent since you can take the permutation h such that h(1) = 2, h(2) = 1, h(3) = 3

[lament pilot]

They're not equivalent though. There is no way to swap 1 and 2 in any of those to get the other

[mild laurel]

and if you compoe this with g, you get f

[lament pilot]

Oh nevermind

Indeed

[mild laurel]

so its not that easy

[lament pilot]

[mild laurel]

you're close though

[lament pilot]

Let me think for a bit

I'll ping you if I don't find another way to arrange it

[mild laurel]

actually, this example might be wrong

Yeah this example isn't right, your statement is correct

And its important you're looking at left cosets here and not right, for right cosets, my example would be a counterexample

[lament pilot]

Wait so I'm right afterall ?

[mild laurel]

i.e., its important that rho acts before sigma'

yeah

[lament pilot]

[mild laurel]

i had the order of composition backwards

[lament pilot]

Damn I'm glad, cosets are weird man

[mild laurel]

permutation groups are weird

[lament pilot]

So is my formulation correct?

Or is it just intuitive?

[mild laurel]

you should be a bit more rigorous, but the right idea is there

[lament pilot]

Like I don't think it's rigorous but I don't see how I can write it more rigorously though

[mild laurel]

your statement really should be rewritten as

$\sigma$ is equivalent to $\sigma'$ iff $\sigma(n) = \sigma'(n)$

[cloud walrus]

Zopherus:

[mild laurel]

but you can do it more rigorously by showing that both directions are true

[lament pilot]

Now the statement is of course a lot better

[mild laurel]

double negatives were hard to wrap my head around

[lament pilot]

However I'm still trying to think how I'd rigorously write it.
Say I have $\sigma \sim \sigma'$. By definition, $\exists \rho$ such that $\sigma = \sigma' \cdot \rho$.
However $\rho \in \mathcal{S}_{n-1}$ so $n \notin Supp(\rho)$.
$\Rightarrow \sigma (n) = (\sigma' \cdot \rho) (n) = \sigma'(n)$.

[cloud walrus]

Smor Brother:

[lament pilot]

That's the implication done rigorously right?

Now the other way around

[mild laurel]

yeah

[lament pilot]

Can we reason via equivalence in my previous message there?

Say I have $\sigma \sim \sigma'$. By definition, $\exists \rho$ such that $\sigma = \sigma' \cdot \rho$.
However $\rho \in \mathcal{S}_{n-1}$ so $n \notin Supp(\rho)$.
$\Leftrightarrow \sigma (n) = (\sigma' \cdot \rho) (n) = \sigma'(n)$.

[cloud walrus]

Smor Brother:

[lament pilot]

Is that still correct?

[mild laurel]

Hm, now its a bit more awkward what the second half of the statement means

[lament pilot]

Let me rewrite it then

Say I have $\sigma \sim \sigma'$. By definition, $\exists \rho$ such that $\sigma = \sigma' \cdot \rho$.
However $\rho \in \mathcal{S}_{n-1}$ so $n \notin Supp(\rho)$. Thus
$\sigma (n) = (\sigma' \cdot \rho) (n) = \sigma'(n)$.
Hence $\sigma \sim \sigma' \Leftrightarrow \sigma (n) = \sigma'(n)$.

[cloud walrus]

Smor Brother:

[lament pilot]

Not sure if I can already conclude an equivalence there is my question

Or is it merely a one-way proof

[mild laurel]

its really only one way

[lament pilot]

Damnit

[mild laurel]

starting from \sigma(n) = \sigma'(n)

you have to show that some rho exists so that \sigma = \sigma' \cdot \rho

[lament pilot]

Roughly speaking, I'd say I can just change the rest to my liking with a certain permutation rho.

[mild laurel]

yeah

[lament pilot]

However I have no idea how it's done rigorously there. Do you mind helping me a bit here?

[mild laurel]

if I have f,g so that f(1) = 2, f(2) = 3, f(3) = 1, and g(1) = 3, g(2) = 2, g(3) = 1

what rho are you going to pick so that f = g \rho

[lament pilot]

Let me write it down

I'll pick p = (1 2) ?

[mild laurel]

yeah, and how did you get that?

[lament pilot]

[mild laurel]

okay, maybe this is a dumb example because there's only one thing you could pick lmao

[lament pilot]

By looking I guess lol

[mild laurel]

since obviously the identity doesn't work

hm

maybe an idea is that

if f(1) = m for some m, then you have that g(k) = m for some k, but this means that you want rho(1) = k

so that g(rho(1)) = m

[lament pilot]

🤔

[mild laurel]

and so this is how you can construct a rho

so that it has the properties you want

[lament pilot]

Seems like this part is the hardest

So, given any two $\sigma, \sigma'$ such that $\sigma(n) = \sigma'(n)$.
Let $\sigma(1) = m \in {1,..,n-1}$.
Because $\sigma'$ is a bijection, $\exists k \ in {1,..,n-1}$ such that $\sigma'(k) = m.
Let $\rho \in \mathcal{S}_{n-1}$, if $\rho(1) = k, (\sigma' \cdot \rho)(1) = \sigma'(\rho(1)) = \sigma'(k) = m = \sigma(1)$.

Do I try induction here?

[mild laurel]

nah

uh, what's r

[lament pilot]

Oh whoops

[mild laurel]

also it should be \sigma(1) at the end

but anyways to finish, you did this for 1

[cloud walrus]

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

[mild laurel]

so now the functions match up at 1

but they have to match up everywhere

you know they already match at n, so you don't have to do anything about that but

basically do the same thing you just did for 2,..., n - 1

[lament pilot]

So then we do induction for i in {1,.., n-1}

but you said nah

[mild laurel]

its not induction lmao

you don't have any base cases or anything

or inductive steps

[lament pilot]

So I just take i in {2, .., n-1} and prove they match up to i then?

since I don't specify i, then they match up for any i

[mild laurel]

you prove that you can construct rho so that it matches up

[lament pilot]

Alright I'll give it a shot and come back with a written proof for you to look at.

Thank you so much for the help you've already given me, this is not my strong suit

[mild laurel]

i'll probably be asleep by then

but hopefully someone else can help

[lament pilot]

I managed, thanks

[ripe crest]

@stone fulcrum cool - thanks

[mild laurel]

If it were reducible, what would be the dimensions of the irreducible parts have to be?

yes

Ah, I misread what you said

You don't need to use an arbitrary vector, you only need to give an example of some vector that when you apply the matrices, don't give a vector thats a scalar multiple of the original one

If your representation is indeed the direct sum of two 1-dimensional representations, then the matrices applied to every vector will have to be just scaled

So if you can find just one example where its not scaled then

[chilly ocean]

Let $G$ be a finite group with an even number of elements. Prove that $G$ contains $a \neq 1$ such that $a^2 = 1$.

[cloud walrus]

downtown:

[chilly ocean]

How would you approach this problem?

Imagine a group with n elements, n is even.

Try to pair elements with their inverses

The identity pairs with itself, and now do the rest

Oh right and since $1$ maps to $1$ we will get $a$ maps to $a^{-1}$

[cloud walrus]

downtown:

[chilly ocean]

This should give you an idea of why the statement is true

for some a

play around with my suggestion and try to make a proof out of it

if u still cant then ask us about what youre stuck with

I see what you mean, the elements will get overloaded if you try to map them to something other than themselves

since 1 goes to 1

oh not quite, let me think

I understand it intuitively but I am having trouble formalising it

since a*a^-1 = a-1 * a you have some odd number of pairings

I can draw it graphically, and it makes me think there is a combinatorial argument

oh!

@chilly ocean if the identity is the only element that maps to itself then there is an odd number of elements right?

and we can get a contradiction this way

you dont need contradiction

but youre definitely have the right idea

elaborate more on that counting argument

then youre done

@chilly ocean it would mean that you can pair up elements that aren't 1, so there are an even number of elements not including 1

hence an odd number including 1

how would you say that formally?

Have you learned about the order of an element yet?

yeah

ok

so lets try to do this step by step

Your idea makes use of the fact that the order of an element equals the order of its inverse, right

uhh

does it?

Yes. Try to think where you used that fact when arguing that the group has an odd number of non-identity elements

Anyway for now try to write the proof since you have the idea in your head

hint: youre gonna want to talk about the orders of elements and the order of rhe group

I came up with a proof that doesn't use a contradiction argument

If we just take all the elements of $H = {g \in G: g \neq g^{-1}}$ we can show that it is even

[cloud walrus]

downtown:

[chilly ocean]

and then $G - H$ is even, and $1 \in G - H$

[cloud walrus]

downtown:

[chilly ocean]

I think that works, as long as you show why H must be even.

a little more convoluted than the idea i had in my head, but try writing a proof with that idea

[gritty fractal]

if $\frac{G}{H_1}\simeq\frac{G}{H_2} $, does that imply $ H_1\simeq H_2 $

[cloud walrus]

StevenDoesStuffs:

[mild laurel]

No

[gritty fractal]

counterexample?

[mild laurel]

The first example that came to my mind was to take an infinite direct product of Z

This quotiented by one copy of Z is still isomorphic to the original

And this quotiented by two copies of Z is also isomorphic to the original

But obviously Z is not isomorphic to Z x Z

One's cyclic and the other isn't

if G is finite I think its still false but I haven't thought about a counterexample yet

[golden pasture]

https://math.stackexchange.com/questions/40881/isomorphic-quotients-by-isomorphic-normal-subgroups
found this

[chilly ocean]

if we're allowed to use trivial subgroups then uhh

D_3 ≅ S_3 ≅ S_4 / V but {e} isn't isomorphic to V.

where V is klein 4 group

[mild laurel]

Uh, wait I don't see how that example works, is G = S_4?

[chilly ocean]

mine?

oh shit I'm stupid lol I misread the question

[mild laurel]

@golden pasture that question goes in the other direction

[golden pasture]

yea

[mild laurel]

i.e., does isomorphic subgroups imply isomorphic quotients

[golden pasture]

but gives a counterexample too

[mild laurel]

oh yeah

should've read past the first paragraph

okay that's what I was thinking

[mild laurel]

@soft elm what do you mean by constructed?

[soft elm]

suppose i have a set

and i want to make a group with operation *

[woven delta]

Do you know what a symmetry group is?

[soft elm]

and the set elements are {a1,a2,a3...}

uh permutations something?

i havent studied it yet

[woven delta]

So groups arise naturally as groups of symmetries of some object

[soft elm]

hm

[woven delta]

That's why they're interesting

[soft elm]

like symmetries of an equilateral triangle

[woven delta]

Yeah

[soft elm]

so i was seeing the integers

[woven delta]

Yeah that's a bad intuition

[soft elm]

and then i saw that they can be constructed from 1

itself

and the identity element

then i saw that this is a cyclic group

[woven delta]

Yeah

[soft elm]

how would i get a group which is not cyclic

what is the basis of construction of the elements

[woven delta]

So one example is the symmetry group of 3 elements

Ie the group of all permutations of 3 elements

[soft elm]

hm

[woven delta]

There are 3! = 6 elements in this group

[soft elm]

yeah

[woven delta]

And you can write them all out

Some good notation to do that is to write it in terms of cycles

[soft elm]

suppose my set is {a1,a2,a3......}

hm

how would i construct a group out of these

with an operation *

[woven delta]

Lol you can always make some dumb shit

[soft elm]

lol

[woven delta]

You don't want to make groups randomly

You want them to arise from something else you studied

[soft elm]

oh

so my group can be a1a2,a2a3

and anything i want right

[woven delta]

Yeah given any set there are lots of group with that underlying set

[soft elm]

hm

ngl group theory nice

thanks bye

[chilly ocean]

symmetric groups are hot

ngl

[scarlet estuary]

in general we dont really care about the set of a group

like at all

we care about the group's structure

not what the individual elements represent

[chilly ocean]

^

[solemn rain]

every group is just S_n ig

[scarlet estuary]

theres plenty of groups that arent S_n

[mild laurel]

Every group is just F_n

[chilly ocean]

I think he means that one specific corollary of cayley's theorem

the one that's like every group is isomorphic to some subgroup of S_N

[solemn rain]

isnt any group ismorphic to a subgroup of S-N?

YEA xd

[scarlet estuary]

yeah but

[chilly ocean]

isomorphism isnt equality

[scarlet estuary]

"a subgroup"

[chilly ocean]

we say it jokingly

[scarlet estuary]

is a very different matter

[chilly ocean]

but dont take it seriously

[solemn rain]

okayy

[scarlet estuary]

you lose a lot of information from the "sub-" prefix

[chilly ocean]

I guess you could probably say something like "Every finite group is equal to some subgroup of S_n (up to isomorphism)"

[scarlet estuary]

like Q is a subfield of R but theyre not really that similar

their cardinalities are totally different for one

[solemn rain]

yea

got it

[scarlet estuary]

also yeah this only applies to finite groups

and finite group theory is solved anyway

[chilly ocean]

RUDE

[solemn rain]

oh really

i thougth there was this one thing

wait let me get it

[scarlet estuary]

you might be thinking of cayley's theorem

like the actual statement

that applies to all groups

[solemn rain]

yea

[scarlet estuary]

but the S_n corollary

is only finite groups

[solemn rain]

?

im given that

the actual satement

[scarlet estuary]

im not sure how youd make an infinite group isomorphic to S_n?

[solemn rain]

is the corollarly

[scarlet estuary]

oh huh

usually i see it as

every group is a subgroup of the symmetric group acting on that group

[mild laurel]

Wait, infinite groups should all be subgroups of a symmetric group with large enough cardinality too right?

[chilly ocean]

im not sure how youd make an infinite group isomorphic to S_n?
The closest to an example I can think of is like the infinite dihedral group

but that's still not an example

[solemn rain]

well caayleys theorem is still like 3rd section

[scarlet estuary]

zoph i dont think thats true

[solemn rain]

so untioll i get tthere

[scarlet estuary]

but maybe im a dumbass

the statement of cayley's theorem is a statement on group actions

it just so happens that by considering a trivial group action

you get the corollary

[solemn rain]

i will get there

[scarlet estuary]

but only where it makes sense (i.e. in the finite case)

[solemn rain]

thanks

ohh

when i saw it the first time

i was usin gaillan

[chilly ocean]

oh wait

[solemn rain]

gaillan didnt cover group actions

[chilly ocean]

I think Zoph is correct

[solemn rain]

so thats why

i get coroallyh

[scarlet estuary]

how so kaku

doesnt the symmetric group of N only have like

4 normal subgroups

[chilly ocean]

what's stopping us from applying a group action that's a permutation on an infinite group

[scarlet estuary]

or some shit

[chilly ocean]

oh wait I see your point now

or maybe I do

the proof of cayley's theorem I recall works on infinite groups if you want to establish G being isomorphic to a subgroup of Sym(G)

but not necessarily S_n

like if G is finite then certainly Sym(G) is isomorphic to some S_n

but if infinite then idk

[scarlet estuary]

yes thats the point

its a statement on group actions

[chilly ocean]

welp this is where I bother my grad student friends with random questions at midnight

[golden pasture]

lol

[chilly ocean]

they're used to it

we often argue about random shit in math while drunk at pizza places near midnight

well, not anymore thanks to covid-19 but yeah

[chilly ocean]

ok after thinking about it

I missed the forest for the trees or whatever

isomorphisms are bijections so you can't possibly have one between an infinite group and a finite group

so yeah the whole S_n thing should only ever apply to finite groups

[solemn rain]

i think you can play with cardinality

[mild laurel]

You can have infinite symmetric groups

[chilly ocean]

right that's what I said earlier

but you wont have the case where like Sym(G) ≅ S_n for some S_n if G is infinite

and mo2men's question was isnt any group ismorphic to a subgroup of S-N?

[chilly ocean]

ok actually I think I see your point Zoph

if the action is faithful then even if G is infinite it could be isomorphic to some subgroup of S_|G|, maybe

im taking way too much time to think about this than I should :s

[lime skiff]

Asking here because this feels a bit too low level:

I have $P\in R[X]$ where $R$ is a ring. I need to show that $P$ can always be written as a sum of the form $\sum_{i=0}^n c_i (x-b)^i$ for any $b\in R$.

How do I approach this? Induction?

Do I just slap on the binomial theorem and work from here?

[mild laurel]

Uh, I assume your sum should be c_i (x-b)^i

[lime skiff]

Ah my bad

Let me rewrite this

[cloud walrus]

kuri:

[mild laurel]

But you can see that the degree d term in your sum only depends on the indices of the sum that are >= d. In other words, the degree n term only depends on c_n, the degree n-1 term only depends on c_n and c_{n-1} etc

[golden pasture]

map like x to x+b

[lime skiff]

Oh hey Ariana you're here as well

[golden pasture]

yee

[chilly ocean]

wow small world

[mild laurel]

So you can start at the top, if P(x) = a_n x^n + ... + a_0, then you can see that you have to take c_n = a_n, but then yeah, you use binomial theorem stuff. Comparing degree n-1 terms gives you that c_n * n + c_{n-1} = a_{n-1}, and you can just repeat this all the way down

[lime skiff]

I get that I need c_n = a_n but I wasn't really sure what to do from here

But in the end it's really just manipulating the binomial coefficients right

Thanks!

[golden pasture]

why not like
Let φ be a homeomorphism from R[x] to R[x] mapping R to itself as the identity and x to x-b
then just show it has a inverse

[lime skiff]

Wait does that work?

That's what I had in mind at first

An obvious inverse would be x+b right

[golden pasture]

well the inverse is trivial and you can get the coefficients from it explicitly if need should work

yup

[lime skiff]

Well this shows they're isomorphic

And I can just define the polynomial in x-b to be my polynomial in x composed with the iso right

[golden pasture]

yea

[lime skiff]

And obviously this works for any b in R since R is also an additive group so +/-b is always defined

[golden pasture]

ye

[lime skiff]

Thanks Ari

And since you're here, thanks again for commutative algebra!

It's a big help having you give solutions to the project

[golden pasture]

yw<3
lol bored highschooler here xd

[steel owl]

jesus you geniuses xD

[blissful ice]

anyone knows the basics of maclauray2?

so basically

I know what I have to do and how. I just don't know how to do it efficiently (for my time, not processor speed)

I know that to solve that, you compute the grobner basis for the ideal wrt. lex order

then you get one equation which is singlevariate

you find the roots, plug into 2 variable equation. Find roots, into 3 variable equation

and that's all fine and dandy but how do I do that in maclauray2 without wasting my time
I know how to compute the gb

and solve solves an univariate equation

is there anything to idk, solve the matrix that gens(gb(ideal... puts out? That matrix and searching for when is every term zero

i13 : S = CC[x,y,z, MonomialOrder => Lex]

o13 = S

o13 : PolynomialRing

i14 : gens(gb(ideal(x^2+y^2-z, x^2+y+z^2, -x+y^2+z^2)))

o14 = | z6+1.5z5+1.25z4+.125z3-.125z2-.25z yz2+.5yz+.5y+z4+z3+.75z2+.25z y2-y-z2-z x-y-2z2-z |

this
ping me if you know

or even matrix->list function would be a nice start

[blissful ice]

flatten entries works if anyone needs it in the future

[zenith heath]

A relation cannot be simultaneously symmetric and antisymmetric

Waht you guys think about it

[somber bramble]

what do you think about it?

||I can come with at least two examples of relations which are both||

[golden pasture]

@blissful ice idk but is sage avil lol if so just do like print(solve([x^2+y^2-z,x^2+y+z^2,-x+y^2+z^2],[x,y,z])) or if want to have nice latex code print('\n'.join(map(lambda x:latex(x),solve([x^2+y^2-z,x^2+y+z^2,-x+y^2+z^2],[x,y,z])))) and gives you closd form rightaway

gb=groebner basis? if so doesnt rlly work here its still quadratic

sage: x,y,z = AA['x,y,z'].gens() 
....: Ideal(x^2+y^2-z,x^2+y+z^2,-x+y^2+z^2).groebner_basis()  
[x^2 + 1/2*x + 1/2*y - 1/2*z,
 y^2 - 1/2*x - 1/2*y - 1/2*z,
 z^2 - 1/2*x + 1/2*y + 1/2*z]

[blissful ice]

gb is groebner basis yeah

I mean technically the exercise just asks for the points that solve it, but I wanted to get familiar with Macaulay2
I ended up throwing it in mathematica to solve it 😦

[golden pasture]

rip

use sage for algebra

sage also does use macaulay2 below i believe

[blissful ice]

you can call m2 from sage yeah

it has a nice interface to it

[golden pasture]

yeeeee

sage is pretty

[blissful ice]

but you still kinda have to know how to use some of M2, I just started using it today

and wanted to learn more. It's good for AG and CA

at least my prof says it's really good, suggested us to learn it

a bit lost on some regards, not gonna lie

[golden pasture]

honestly yea the docs is kinda confusing to navigate

thats why i only know how to sage cuz its docs are so pretty

[blissful ice]

haha yeah I gotchu
that's the only reason I still use mathematica from time to time too

although I want to switch to sage. Much more free

[golden pasture]

mathematica is like nice for analysis and numerical stuff only tbh

[ripe crest]

Let a/b \in Frac(R). Is it true that a \in Frac(R) and b \in Frac(R)?

R \subset Frac(R), so that must be the case, right?

Where R is an integral domain

[zenith heath]

What are with these symbols

[ripe crest]

Do you mean \subset and \in?

$a/b \in Frac(R)$

[golden pasture]

frac R like the fraction field of R?

[ripe crest]

Yeah

[cloud walrus]

silent flower:

[golden pasture]

then yea, normally we just define it as a/b=(a,b) in frac(R) where a,b are in R

[ripe crest]

so, a and b are in Frac(R) as well, right?

at least, a/1, b/1

[golden pasture]

(btw need commutativity iirc if it isnt commutative then you need certain conditions and fancy constructions)

yeaa

[ripe crest]

R injects into Frac(R)

yeah, that's why I specified that it's an integral domain xD

thanks!

[golden pasture]

oops missed that ll

[clear obsidian]

Hopefully this is right

[mild laurel]

Yeah

[clear obsidian]

Ty

[ripe crest]

k(X) is generated by k, X, and 1/X as a k-algebra, right?

k(X) = Frac(k[X])

[mild laurel]

just X and 1/X

[ripe crest]

ah yes - we can get any a \in k via (X + ... + X)/X, where there are a Xs in the numerator,

thanks

[mild laurel]

Eh, that doesn't quite work, this only gives you the image of the map from Q to k

[ripe crest]

ah right whoopsies

hmmm

oh actually k(X) is not generated by X and 1/X

prof just responded

Consider 1/(X+1)

[mild laurel]

Ah hm you're right

[uncut girder]

Generated by x and 1/p(x) where p(x) is irreducible