groups-rings-fields

[formal ermine]

the derivative argument doesn't work here

because the polynomial is not irreducible

[tribal moss]

Yes, but it would probably be more straightforward that we know the polynomial has 13 different roots in C -- they're evenly spaced on the unit circle.

[formal ermine]

right

[tribal moss]

I mean, a derivative argument does work: The only point where the derivative of z^13-1 vanishes is z=0, but z=0 is not a root. But the derivative does vanish at a multiple root. So there can't be any multiple roots.

[formal ermine]

yeah

okay thanks

on that note

why do we know that if the minimal polynomials of the 'extended elements' in a field extension are separable, then the field extension itself is separable?

ah wait

it's simpler

we can use the fact that Q is char 0

every irreducible polynomial is separable in char 0

so every extension over a char 0 field is separable

got it

[elder wave]

Not obvious at all that you joined on an alt after getting muted, you could’ve at least phrased the question differently. I saw the unedited version

[tender bough]

here 𝔤𝔩(𝑛) is the general linear lie algebra, made of 𝑛×𝑛 matrices of complex number entries

𝔲(𝑛) doesn't seem to be a vector subspace, because it's not closed wrt scalar multiplication?
For 𝑋∈𝔲(𝑛), 𝑧∈ℂ,
(𝑧𝑋)ᴴ = 𝑧*𝑋ᴴ = -𝑧*𝑋

but we want (𝑧𝑋)ᴴ = -𝑧𝑋 somehow

[formal ermine]

I'd recommend you use latex here on discord, it's way easier to read

[tender bough]

$𝔲(𝑛)$ doesn't seem to be a vector subspace, because it's not closed wrt scalar multiplication?
For $𝑋∈𝔲(𝑛), 𝑧∈ℂ,$
$(𝑧𝑋)^\mathrm{H} = 𝑧^*𝑋 = -𝑧^*𝑋^\mathrm{H}$

[cloud walrus]

Mattuwu

[next obsidian]

It’s a real subspace

[chilly ocean]

this says -z*X = -zX, so you want z = z*...

[tender bough]

Ahh that makes much more sense

My lecture notes has the convention that everything is over ℂ, it might have missed a note here lol

[next obsidian]

Yeah so this is the Lie Algebra of the unitary group which is matrices with MM* = I

So it doesn’t form a complex vector space kind of heuristically because the conjugate isn’t holomorphic

Which well, showed up again in your computation

[tender bough]

@chilly ocean @next obsidian thanks

[solar glacier]

question on proving G is abelian if G/Z(G) is cyclic, for the trick showing elements of G can be written of the form x^mz where x \in G and z \in Z(G) is the following correct logic

as G/Z(G) is cyclic there exists an x \in G such that G/Z(G) = <xZ(G)>

then if g \in G we have gZ(G) = (xZ(G))^m =x^mZ(G) and as Z(G) \leq G is a subgroup this holds iff gx^-m \in Z(G)

i.e.., g = x^m z for some z \in Z(G) and some m \in Z^+

[warm wyvern]

are you TAing an algebra class?

[agile burrow]

No, these are from when I took an algebra course

[warm wyvern]

it says fall 2022 tho

[agile burrow]

Yes, I took an algebra course in fall 2022

[elder wave]

[solar glacier]

are you taking the algebra qualifying exam

[agile burrow]

No

[warm wyvern]

what year are we again?

[agile burrow]

My semester ended in early December

[warm urchin]

To be really precise , its x^-mg \in Z(G) , but in this case it doesn't really matter.

[plush wasp]

potato :o

G/Z(G) is cyclic, then G is abelian.
Why?

The proof was understandable but I'm probably slow.
Says: Assume G/Z(G) = <h>, G = union of h^kZ(G)

[solar glacier]

because every element of G can be written as x^mz where z is in the center

[chilly radish]

The point is if it's cyclic then it's already trivial

Since every two elements only differ by multiplication by an element they BOTH commute with

So like, for every g,h we have g=x^m h, and we are guaranteed that h commutes with x, so it commutes with g as well

[plush wasp]

how did you get g = x^m h? :(

that's what my doubt was

G/Z(G) is the group of left cosets right?

[chilly radish]

Yes

We have that for any element, it's of the form x^m z for some z in the center

So take any two elements, they're of the form g=x^m z, h=x^n z'

[solar glacier]

the z's are distinct as well ^

oh

lol

[chilly radish]

Yea I forgot a prime

[plush wasp]

hmm? of the form x^m z and not x^m ?

[solar glacier]

lol

[chilly radish]

You're looking at a coset

Or rather a member of a coset

[plush wasp]

right! 🤦‍♂️

understood!

[chilly radish]

So you get that g and h both differ by elements they both commute with

Since the x^m, x^n commute with each other and z,z' commute with everything

[plush wasp]

hmm

[chilly radish]

This wouldn't work if the quotient was only abelian, because while the elements would commute modulo the center, they wouldn't necessarily commute in G

When it's cyclic tho all the elements are just powers of x modulo the center so they have to commute in G as well

[plush wasp]

What is the group U(2) and SU(2)?

I was asked this:

Compute the center of U(2) and prove that U(2) = SU(2).Z.

[solar glacier]

unitary and special unitary 2 by 2 matrices

[plush wasp]

which are?

[solar glacier]

complex square matrices such that the conjgate transpose is the inverse of each matrix

SU i think is those with determinant 1

[plush wasp]

and conjugate transpose is?

transpose of conjugate matrix?

[solar glacier]

transposing the matrix and applying complex conjugate

yes

[plush wasp]

alr!

how to do 2?

[chilly radish]

What have you tried

[plush wasp]

Gimme a min.

[tender bough]

Is there a way to quickly show that 𝔱(3), the Lie algebra of upper triangular matrices of size 3x3, is not semisimple?
I currently am trying to compute its killing form using 𝐸ᵢⱼ's and show that the killing form is degenerate.

[plush wasp]

I'm trying to write the group with their properties but it doesn't look like working

I'm supposed to find a general X that satisfies AXA{-1} = X

so I wrote that down

[plush wasp]

ohhhh so I take samples of U(2) to check for characteristics of center

[tender bough]

Ah only need to spot this:
∀𝑌 ∈ 𝔱(3), 𝜅(𝟙,𝑌) = tr(0) = 0

[plush wasp]

Right.. so I used matrices [{1, 0}, {0, i}] and [{0, i}, {1, 0}] to figure that the center would be matrices aI where I is the identity matrix. Now, what's the second part of the question asking for?

I don't understand what's the SU(2).Z. mean ... is that a typo maybe?

[plush wasp]

oh nvm I got it!

ty

next how do I do 3

[rustic crown]

use that G/Z(G) is cyclic iff it is trivial iff G is abelian

this should help you show that it's not nilpotent

for solvable, use that for a group, if both G and G/N are solvable, then so is G

[hollow fjord]

Hi guys!, i've got a question, assume that $R$ is a PID (i'm not sure if this is needed, but in my case i'm working with a PID) first i have that $R[x] \cong R^{(\mathbb{N})} $ and $R[x_1,x_2] \cong (R[x_1])[x_2] $ and i need to do this recoursively (i do not know if that word exists) for $R[x_1,x_2,\dots , x_k], k \in \mathbb{N}$. Do yo have any ideas for this

[cloud walrus]

galois.theory

[rustic crown]

yee pid-ness is not needed

do you mean recursively?

idk what's the question lol

[hollow fjord]

sry

[rustic crown]

R[x1, x2, x3] = R[x1][x3][x2] or whatever you want

all of these are naturally isomorphic

[hollow fjord]

like generalize that for any number of variables

[rustic crown]

how do you define these objects in the first place. the proof is really dependent on the definitions

ig your definition of R[x] is just R^(N)

i.e. sequences in R where all but finitely many terms are 0

and you identify x with the sequence (0, 1, 0, 0, ...)

[next obsidian]

Isn’t this like

[rustic crown]

just the definition of normal

[next obsidian]

The degree of the extension is the same as the degree of that min poly

So it is the splitting field cuz it has all the roots

And the splitting field couldn’t possibly be any smaller

[rustic crown]

x is arbitrary i think

not the primitive element

[next obsidian]

Oh tru

Okay well then just use definition of normal

[rustic crown]

so the definition of normal means minimal poly of every element in the extension splits completely

[next obsidian]

It’s that if L contains a root

It contains all of them

[plush wasp]

can you explain please? I'm really new to this and it's hard trying to understand by myself

[rustic crown]

okie so since you wanna show that the group is not nilpotent, it's enough to show it's non-abelian and the center is 0

that way the sequence of subgroups you get will always stay constantly 0

[plush wasp]

Right, yeah

[rustic crown]

so non-abelian is given to us

and if Z(G) was non-trivial, then G/Z(G) will be trivial or will have prime order

which means it's cyclic

but this is a standard exercise that if G/Z(G) is cyclic, then G is actually abelian and this G/Z(G) is trivial

the proof is pretty straightforward check

say that cyclic group is generated by the class of g

so any element in G can be written as g^n * z where z lives in the center

now simply check it's abelian :3

[south patrol]

i do wonder if this G/Z(G) cyclic => trivial thing generalises to anything else

[rustic crown]

(g^n * z) * (g^m * z') = .... = (g^m * z') * (g^n * z)

[plush wasp]

Guys can I ask something non acadmic :c

[rustic crown]

(here everything commutes with everything else, which is why we can do this

[south patrol]

#❓how-to-get-help

jk

[rustic crown]

yee sure

[plush wasp]

what happens if I fail the first semester in a Bmath course

[rustic crown]

[plush wasp]

Am I allowed to retake the course in the following sem or what-

or right away removed from the course

[barren sierra]

this is more a question related to your university

than something anyone here can help you with

but uh usually universities let you retake

[plush wasp]

what's the general procedure though ... hmmm I see

[barren sierra]

but otherwise I would talk to some academic advisor

I could not say anything more, cause I probably do not attend your university

[plush wasp]

ty I'll see

Q) The additive group of rationals cannot be written as a semi direct product of two proper subgroups.

Why?

[rustic crown]

wait, so a non-trivial semi-direct product would give you something non-abelian

so you're essentially asking why we can't write Q as a direct sum of two proper subgroups

[coral shale]

mhm mhm

[rustic crown]

and this is then easy because any two non-trivial subgroups of Q will intersect non-trivially

if a/b is in one and p/q is in the other, then both subgroups contain ap

so one of subgroups would be 0 which would force that the other subgroup is the whole thing, so isn't proper

[coral shale]

hmm

uh

[rustic crown]

[plush wasp]

not wishing I could exchange brains with det for a day

[coral shale]

Q is a field not a group, this is too confusing

[rustic crown]

lol

.<

[coral shale]

All non-trivial subgroups of Q must contain a non-trivial subgroup of Z, and the same result can be said of non-trivial subgroups of Z. Interesting.

[rustic crown]

this took me some time to think lol, i was initially hoping that A+B = Q would imply that either A or B is Q... but you could take A = rationals with denominator a power of 2, and B = rational with odd denominators

but then realized A n B = {0} was the problem :p

[coral shale]

you have subgroups of R which have {0} intersection. So its false for R?

[rustic crown]

ah like Z and sqrt(2)Z

[coral shale]

wait no not necessarily

[rustic crown]

for R, ig you can just use some linear algebra

it's a Q-vector space

pick a basis and then write it as a direct sum

[coral shale]

wait in which case u can

[rustic crown]

yee

[coral shale]

R = A + B

ok

ic ic

[rustic crown]

i wonder if there is a nice example which doesn't require you to choose basis and stuff

[coral shale]

this aoc right

no explicit choice?

[rustic crown]

yea

[coral shale]

😬

If its false with not aoc, then there shouldnt be a way

Well ig u can like

R/piZ justify this is a group

then done

or wait no

Show R/piZ is isomorphic to some subgroup

back to beginnning

[rustic crown]

yea :p

[plush wasp]

[rustic crown]

[plush wasp]

det what are the general ways to show if a group is:

1. nilpotent
2. not nilpotent
3. solvable
4. not solvable (⁠ヘ⁠･⁠_⁠･⁠)⁠ヘ⁠┳⁠━⁠┳

[rustic crown]

i only know the definitions lol

[upper pivot]

in general this is very hard

[plush wasp]

perfect. what are the definitions

[upper pivot]

for a first year course though, its often noticing right subgroups for the definitions

basically

[rustic crown]

i still don't know why one cares about nilpotent groups. ig one motivation comes from lie groups and lie algebras, but i have hardly looked at them, so dunno

[plush wasp]

it's really helpful if, at this point, I can grasp the definition perfectly

if it's not too hard, could someone explain this on VC for a bit?

[upper pivot]

well the definitions you can look up and they are pretty straight forward

try looking at them and if you have confusion you can ask us about it

[rustic crown]

ig there is a good answer for checking solvability

[plush wasp]

this is what I'm looking at!

[coral shale]

wait wth nilpotent groups and nilpotent element of ring looks so different

[plush wasp]

,rotate

[coral shale]

nilpotent element of ring is simple -...-

[rustic crown]

lol

[plush wasp]

-,- Shuri, I have an exam in 2 hrs aaaaaaaaa

[rustic crown]

F

[plush wasp]

F

[rustic crown]

good luck bro

[plush wasp]

Yeah I'd definitely do my best begging the teacher to let me retake the course

[coral shale]

I would drop these 2 defns if u dont think they will be worth many marks

not so worthy of a time investment to me.

[plush wasp]

That's the point. I can only scrap whatever marks I can if I focus on the topics - Sylow theorems, Nilpotent groups, Solvable groups

is what the prof said

[coral shale]

bruh

[rustic crown]

oops

[coral shale]

the last part of my gt course i didnt bother with

[rustic crown]

our gt didn't do nilpotent and solvable at all uwu

[plush wasp]

Bruh you had this in the last part of your gt...

wha-

[coral shale]

group theory

yes

[plush wasp]

yha I think we're living in different worlds for sure

[rustic crown]

i just think there's no point in doing solvable groups until you study some galois theory

[plush wasp]

yes yes... I'm dropping.. Name your uni and I'll maybe try online courses

[rustic crown]

you have 2 more hours!!!

[plush wasp]

oof sad life

tf do I do with 2 more hours -,-

[rustic crown]

i'll cheer for you

[coral shale]

im not sure which of these 3 things is the most approachable

possibly sylow

[rustic crown]

https://tenor.com/view/plusle-minun-cheer-dp161-gif-25271349

[coral shale]

just knowing how to apply it

theres no need to regurgitate proof for sylow probably

so just learn the thms and how to apply

[rustic crown]

i shoudl sleep >.<

it'll be 4am soon

[plush wasp]

night night Eevee

[rustic crown]

gn uwu

[plush wasp]

Kiwi

[rustic crown]

i'll probably be up for 12 more minutes for fun

[agile burrow]

This probably isn't helpful for Arya right now, but I'll just throw it out there. One thing I only realized recently is that nilpotent groups and solvable groups can be realized as certain kinds of extensions

For example, nilpotent groups are built by repeatedly taking central extensions

[coral shale]

category thing yh?

[rustic crown]

btw do you know why 1 --> N --> G --> G/N --> 1 is called extension of G/N by N?

[agile burrow]

Not really, group extensions are kind of natural to study just in the context of group theory

[coral shale]

i think it is category though

[rustic crown]

i always confuse the directions

[coral shale]

youll find at least

[agile burrow]

I've asked that before det but I never get a straightforward answer. I think it's just terminology that stuck

I'll see if I can find something on it or at least try to come up with a satisfying answer

[rustic crown]

(btw really cute pfp )

[agile burrow]

Thanks It got cold so I added a scarf

I guess one way to justify it is that if N is abelian, then it becomes a G/N-module via conjugation inside G. In this sense, you can start with G/N, specify a G/N-module N, and then consider extensions which induce this action. In that sense, the extension group "extends" G/N by N

[rustic crown]

the first one follows because G by definition fixes L^G

wait, not the best way to phrase that

like if there way no K, then also what i said is true

say G is a subgroup of Aut(L)

[plush wasp]

Quick question

[rustic crown]

then by definition of L^G, G will fix that

so G is actually a subset of Aut(L/L^G)

[plush wasp]

what are the conditions to write G = H x K?

I remember their intersection has to be {e}

[rustic crown]

H, K normal subgroups
H n K = {e}
HK = G

[plush wasp]

is it fine if HK = G is not given but we know |H| |K| = G?

[rustic crown]

the other one is more like the argument you were giving above

[rustic crown]

yea if everything is finite, because |HK| = |H| * |K| / |H n K| = |H| * |K|

[plush wasp]

hmm .. this might just be all that I need to understand

so, how do you conclude if a sylow subgroup is normal or not? I read arguments about "there exists only one such 11-sylow subgroup. now, since all 11-sylow subgroups must be conjugates, it must be normal."

I didn't understand what it meant though

[rustic crown]

so sylow-1 says that sylow p-subgroups exist

sylow-2 says that all sylow p-subgroups are conjugates

and sylow-3 says that number of sylow p-subgroups is congruent to 1 mod p, and it divides the size of the group

[plush wasp]

what does sylow-2 mean?

[rustic crown]

it means if P and Q are two sylow p-subgroups then Q = gPg^-1 for some g in your group

[plush wasp]

if there's only one sylow p-subgroup, it must be normal, why?

[rustic crown]

because gPg^-1 is also a sylow p-subgroup

so it's forced to be P

[plush wasp]

ohh,

right ✓

[rustic crown]

uwu

det didn't sleep

[plush wasp]

ahahaha, totally helped though.. Will pray you get a nice rest

[rustic crown]

arigatou

[plush wasp]

Burnside theorem is: ord(G) = p^a q^b, then G is solvable right?

[agile burrow]

that's right

[steep mesa]

is an "affine Z-module" a real thing? or, at least, will people know what I mean if I say it?

[agile burrow]

I'm not familiar with the word "affine" being used to describe modules

[steep mesa]

then maybe this is the wrong term

[agile burrow]

what are you trying to describe

[steep mesa]

I mean an affine space where instead of the vectors being elements of a vector space, they're elements of a Z-module

[agile burrow]

I see

[steep mesa]

which now that I'm thinking about it

kinda ruins the whole point of an affine space because you can't take convex combinations of the points

but it's still well defined, I guess

you still get integer affine combinations of points

[ebon gyro]

decent reach but you may be thinking of an (integer) lattice? in which you consider the free Z-module on some number of generators e.g. basis vectors in Euclidean space

[cosmic oriole]

In the Lawvere theory for groups, how are the group laws encoded?

[shy turtle]

Hello! Is there some writing like «real analysis for an algebraist» out there? Given that I understand Abstract Algebra (more or less), what can help me learn the «analysis» stuff?

It seems that every introductory book that talks about limits and integrals was written as if algebra did not exist (unless it is an algebra of functions over ℝ, sad laugh).

[rustic crown]

tell me if you find one :p

[lethal dune]

hmm...

[tender wharf]

Wouldn't you have sufficient mathematical maturity to take on analysis though assuming you have already done algebra

[formal ermine]

analysis just feels like remembering a bunch of random unrelated facts to me

real analysis I mean

complex analysis in one variable is gucci

[tender wharf]

[field in math] is just knowing the definitions /s

[formal ermine]

[rustic crown]

ig the question is whether you could do analysis much more efficiently once you know enough algebra

[tender wharf]

im guessing the mathematical maturity is helpful

[rustic crown]

for example, like if you're constructing R from Q, you can say consider the Q-algebra of cauchy sequences and the ideal of this of cauchy sequences that converge to 0, this is maximal so the quotient is a field something something

[chilly ocean]

Quick ques, do I need abstract algebra for CS ?

[rustic crown]

instead of having to verify all the algebra by hand

do you need math beyond linear algebra to do cs?

[tender wharf]

cs is just discrete math tbh

I mean you don't need it?

It's nice to have

[chilly ocean]

Oh ok

[rustic crown]

ig knowing some group theory and galois theory woudl be useful

[formal ermine]

is there any efficient way of finding a primitive element of a splitting field over a char 0 field

I mean by the primitive element theorem we know that such an element exists

[rustic crown]

if you're like doing some cryptography

[formal ermine]

but how do we actually compute it

e.g. in my case x^3 - 2 over Q

[rustic crown]

find finitely many generators

and then combine them to get a single generator

[formal ermine]

splitting field is Q(cbrt(2), zeta_3)

[formal ermine]

how?

[rustic crown]

just like in the proof of primitive element theorem

it says cbrt(2) + c*zeta_3 is a generator for all but finitely many c

usually c = 1 should work lol

[cosmic oriole]

For CS, semigroups and monoids are important. Oh and categories.

[rustic crown]

there is a exact list of c which you need to avoid

[formal ermine]

so I just compute the powers of x := cbrt(2) + zeta_3 and solve for zeta_3 and cbrt(2) in terms of x?

[rustic crown]

yea that shoudl work, but i suppose there are better ways

[rustic crown]

if F = k(alpha, beta) and alpha_i and beta_j are respectively the conjugates of alpha and beta, then alpha + c beta would be a generator if you avoid c = - (alpha - alpha_i)/(beta - beta_j)

[formal ermine]

conjugate?

[rustic crown]

roots of their minimal poly

[formal ermine]

ah

[rustic crown]

so like in this case it's better to do
(x - cbrt(2)) = zeta_3

use that zeta_3 satisfies t^2+t+1=0 to kill it off

then solve for cbrt(2) in terms of x

similarly do that for zeta_3

[formal ermine]

ok

[rustic crown]

and here if you square and kill sqrt(2), then it should give you the minimal poly for x

assuming it has small enough degree

like 6 in this case

which it should since we cubed once and squared later

[formal ermine]

cbrt(2) not sqrt(2)

[rustic crown]

oh my bad :p

okie no

so i think then it's much better to do the other way

you wanna keep the quadratic guy with x

so that the left side is f(x) + g(x) * zeta_3

this way you won't have to solve some quadratic equation

x - zeta_3 = cbrt(2)

if you cube this up, you should get zeta_3 in terms of x

and then use cbrt(2) = x - zeta_3 to write cbrt(2) in terms of x

yee this is nice

[formal ermine]

okay that's gucci

[shy turtle]

You will gain a lot in Software Engineering from understanding Category Theory, which is kind of like a branch of Abstract Algebra…

[formal ermine]

thanks

det can you give me some random polynomials

so I can practice this more

[rustic crown]

(never use random polynomials)

(it's like learning integration and then trying to integrate a random function)

(it will just make you sad >.<)

[formal ermine]

uh

that's what I did back then

[rustic crown]

[shy turtle]

What is the density of symbolically integrable functions among all closed form expressions?…

Sketch:

1. Only a tiny amount of products are symbolically integrable among all product expressions.
2. As size grows, the proportion of expressions that contain at least one symbolically non-integrable product grows.

Therefore, the density of symbolically integrable expressions among all expressions is zero.

However, when we speak of random stuff, we need to account for probability density. You cannot have uniform density for a random generator of finite expressions, can you?

You will tend to generate smaller expressions, so your chances of generating a symbolically integrable expression improve.

[formal ermine]

I get a quadratic in terms of zeta_3

3xzeta_3^2 - 3x^2zeta_3 - 3 + x^3 = 0

[rustic crown]

right and zeta_3 satisfies the quadratic t^2 + t + 1 = 0

so you can write this whole thing as f(x) + g(x) * zeta_3 = 0

[formal ermine]

hmm

[rustic crown]

which gives zeta_3 as a rational function of x

[formal ermine]

why

[rustic crown]

um

because it satisfies t^3-1 = (t-1)(t^2+t+1) = 0

and it's definitely not 1

have you not computed minimal polynomial of zeta_3 till now lol?

and in general minimal polynomial of z_n is the nth cyclotomic polynomial

if n = p is prime, this is just t^(p-1) + ... + 1 = 0

(everything over Q)

[formal ermine]

ah

[vocal patrol]

How does commutativity imply the existence of inverses?

By definition the difference of rings and fields is that for a field also the multiplicative property is commutative

This implies inverses for every non zero element right?

But I don’t understand why

[formal ermine]

no?

[rustic crown]

Z is a commutative ring which isn't a field

[vocal patrol]

Oh so it’s two commutative groups for a field right

And the multiplicative group has inverses for every non zero element?

[formal ermine]

a field is a commutative ring with multiplicative inverses for every element except 0

if it would be two groups then 0 would have an inverse

[rustic crown]

(R, 0, +) and (R\{0}, 1, *) are abelian groups

[formal ermine]

I'm not sure how I would do that

[delicate bloom]

I think I found a tricky way of doing it, since zeta_3 = (-1+sqrt(-3))/2 you could instead think of the field Q(cbrt(2), sqrt(-3)). Then you could pick alpha = cbrt(2)*sqrt(-3).

Now you have $\frac{\alpha^3}{-6}=\sqrt{-3}$ and $\frac{\alpha^4}{18}=\sqrt[3]{2}$.

[cloud walrus]

mOwOsity

[delicate bloom]

I think this should work every time you have two or more elements where their roots (the denominator of the exponent) are relatively prime

[formal ermine]

did you find these equations by just trying out?

[delicate bloom]

well I knew 2 would kill a square root and not a cube root, and 3 would kill a cube root and not a square roote

I needed to put 4 since I needed the 2 to "roll around" back to 2 again, since 2^2=4

kind of thinking group-theory-ily

like in my mind I was thinking of an order 3 and order 2 element multiplied together

that makes an order 6 element, so squaring it makes it order 3, while cubing it makes it order 2

maybe that illuminates my intuition here

I guess I could say I'm literally thinking about the group $\bQ^\times (\sqrt[3]{2},\sqrt{-3})/\bQ^\times$?

[cloud walrus]

mOwOsity

[delicate bloom]

because I'm thinking of it being the identity when it's a rational number again

I feel like me saying this is making it sound scarier than it actually is :x

[delicate bloom]

like let's start over,
if you look at sqrt(-3), its first 6 powers look like: sqrt(-3), -3, -3sqrt(-3), 9, 9sqrt(-3), 27, ...
if you look at cbrt(2), its first 6 powers look like: cbrt(2), cbrt(4), 2, 2cbrt(2), 2cbrt(4), 4, ...

[formal ermine]

yes

[delicate bloom]

that's how I think about them, then look at their products

you can kind of think one has period 2 and the other has period 3, so you can find a place where they don't line up, since we can remove rational number multipliers

[formal ermine]

don't line up?

[delicate bloom]

so make a new list with the products like this:

oh I should have 2 not 8 in the last line

ugh

it should say -6sqrt(-3) if you were to multiply the 3rd entries together for instance

[formal ermine]

yes

[delicate bloom]

by not lining up I mean we have only one of the original roots appearing

I could make up more examples like this if you want, maybe that'd help

$\bQ(\sqrt[5]{17},\sqrt[7]{23})$

[cloud walrus]

mOwOsity

[delicate bloom]

now $\alpha = \sqrt[5]{17}\sqrt[7]{23}$ will work, so we have $\bQ(\sqrt[5]{17},\sqrt[7]{23})=\bQ(\alpha)$

[cloud walrus]

mOwOsity

[formal ermine]

ah

got it

thank you!

[delicate bloom]

cool cool

[formal ermine]

this only works when they're different nth roots

so it wouldn't work for sqrt(2), sqrt(3)

would it

[delicate bloom]

yeah

[formal ermine]

yeah so my strategy would be try this thing => try a + b => try a + cb

[delicate bloom]

that's why I switched over cause cbrt(2) and zeta_3 are both cube roots

[formal ermine]

ye

[delicate bloom]

I guess so, yeah

unless we can come up with some other trick

[rustic crown]

oopsie, i went for some lunch >.<

3xzeta_3^2 - 3x^2zeta_3 - 3 + x^3 = 0
plug zeta_3^2 = -zeta_3 - 1 here

and collect terms with zeta_3 and without zeta_3

[rustic crown]

for same square roots, you can use some highschool algebra
(sqrt(3)-sqrt(2))(sqrt(3)+sqrt(2)) = 3 - 2 = 1

so something like (x + 1/x)/2 and (x - 1/x)/2 should give you what you want

[delicate bloom]

oh nice, I'm gonna play with that

[cloud walrus]

mOwOsity

[formal ermine]

ah ok

thanks

the galois group of Q(cbrt(2))/Q is trivial because its the splitting field of the real roots of x^3 - 2 and autos send roots to roots?

(x^3 - 2 has only one real solution)

is that correct?

[rustic crown]

yee

but i wouldn't call it "galois" group as the extension isn't galois :p

[formal ermine]

right

the splitting field part doesn't sound right to me

cuz like for a splitting field we always add all roots over the algebraic closure, no?

[rustic crown]

oh lol my eyes skipped that

what does "splitting field of the real roots" mean?

[tribal moss]

A "splitting field" always has to contain all the roots.

(That is, enough roots that the polynomial factors into linear factors).

[formal ermine]

yeah

[tribal moss]

So "splitting field of the real roots" doesn't make sense.

What you want to say is "the field generated by the real roots".

[formal ermine]

right

but then my auto sends roots to roots argument still holds?

[tribal moss]

Why would using correct terminology to state it cause it to break?

[formal ermine]

I wasn't sure whether the correct terminology changes anything

like if my "idea" was correct

or if it was totally wrong and the terminology was off too

anyway

yeah looking at the proof of that thing it seems to work too

thanks

[solar glacier]

is this proof ok for normal cyclic subgroups have all normal subgroups

[tribal moss]

You should probably state explicitly earlier on that what you're proving is that a subgroup of H is normal in G (in contrast to K normal in H, which is trivial).

[solar glacier]

true, I didnt indicate WHERE its normal

i should add that in

other than that is it ok ?

[tribal moss]

I didn't see anything else screaming to me, but I didn't bother to read it very closely.

[shy turtle]

Found two:

http://www.tac.mta.ca/tac/volumes/20/10/20-10.pdf — Algebraic Real Analysis by Peter Freud.
https://www.maths.ed.ac.uk/~tl/glasgowpssl/banach.pdf — A universal Banach space by Tom Leinster.

(Somewhat advanced but can be managed.)

[south patrol]

find it weird you use the notation K := <h^d> when K has already been introduced

But other than that sure this is fine

[knotty frigate]

well if H is cyclic then its abelian. so its subgroups are necessarily normal

[formal ermine]

they want to show that any subgroup of H is normal in G

[next obsidian]

This seems not true

Oh, we’ve assumed H is also normal in G

[knotty frigate]

oh

oh

okay

[formal ermine]

yeah only really works when we assume that it's cyclic I think

because C2 normal V4 normal S4 but C2 not normal S4

[rustic crown]

ig another way to see that is by using that any subgroup of a cyclic group is automatically characteristic

any group theory book >.<

i used aluffi's book. you could check the pins in #book-recommendations

aluffi tries to do algebra more categorically, so won't be nice for everyone

[ebon gyro]

artin, dummit & foote were what i used

i then used aluffi more for a second pass which i loved as well

[tender bough]

fun: I was computing the killing form of 𝔤𝔩(2) manually. I'm supposed to show 𝜅(𝐴,𝐵) = 4tr(𝐴𝐵) - 2tr(𝐴)tr(𝐵) and I must have made a mistake somewhere 💀

[ebon gyro]

oh no 😭 messing up these commutators can be lethal

let me look through and see if i can find anything but the number of these involved lie algebra calculations i've messed up as well is astronomical

[tender bough]

@ebon gyro pls don't toture yourself with my "latex", I have decided to embrace technology...

machine power 💪

[ebon gyro]

machine power !!!

[tender bough]

lol turns out I messed up the signs three or four times, and I multiplied the wrong column and row together, missed several opportunities to cancel out terms 💀
Maybe shouldn

shouldn't have listened to music while doing this

[solar glacier]

I’m showing normality in G not in H

question for proof of LaGranges Theorem. Are we using the fact that left (or right) cosets partition any finite group into a disjoint union of left cosets of a given subgroup

taking some g_1 \in G we note that the sizes of g_1H and H are the same and G = U_{g \in G} gH

so its some n times the order of H

thus order of H divides order of G?

how can I make this argument more rigorous

[coral spindle]

His name is Lagrange, not La Grange. I'm not sure what problem you're having with the rigour.

We are indeed using the fact that the cosets partition the set, along with the fact that they are all the same size.

[solar glacier]

so thats pretty much it then?

[coral spindle]

Yes

[lime badge]

I'm curious if there's a simple example of an R module homomorphism g: F ---> F, where R is an integral domain but not a field and F is a finite rank R module, where im g^(n+1) is strictly contained in im g^n for all non-negative n

For a vector space homomorphism g, the kernels and images of the powers g^n stabilize but that's apparently not true for the image in the general case above

[coral spindle]

R = Z, F = Z, g(n) = 2n.

[lime badge]

Lol yep, just came up with that rn. Thanks!

[rustic crown]

uwu

[chilly radish]

The reason the stabilisation fails is that that, even though Z is a PID so that every submodule is free, it's not true that a proper submodule has strictly lesser rank

[coral spindle]

If you were a cool kid you'd say it's because it's non-Artinian

[rustic crown]

me wanna learn some commie algebruh

[coral spindle]

Me too, comrade

[chilly radish]

Artinianity is fake. Too strong

[coral spindle]

Hey it happens!

[chilly radish]

This orbit isn't necessarily inside F, but the min poly splits in L, and we know that in a Galois extension the galois group acts transitively on roots of irreducible polynomials. To me it seems like they're proving 1 iff 3 first (every min poly already splits in F iff every element of the galois group sends F to itself)

Then they procceed to show 3 iff 4 by relating it back to groups

[rotund aurora]

So in general, even though an ideal is contained in a maximal ideal, this maximal ideal need not be unique, correct? What are some examples?

[coral spindle]

Choose the ring Z. If p is a prime, pZ is a maximal ideal.

[rotund aurora]

👍

[south patrol]

So yeah I guess a fairly general set of examples is any ring A with multiple maximal ideals, since they all contain the ideal 0

(i say fairly general cause if a ring A has multiple maximal ideals containing I, then A/I has multiple maximal ideals containing 0)

[solar glacier]

and since pZ is maximal, Z modulo pZ i.e., Z/pZ is a field 🙂

just practicing my theorems 🙂

[rotund aurora]

uhhh facts

[solar glacier]

or proposition lol

[formal ermine]

...or definition

[solar glacier]

actually this is a fact that requires some proof so its not a defn lol

a field is by definition a commutative division ring

[coral spindle]

I think illum knows

[south patrol]

... or corollary

[agile burrow]

It's more of a remark, really

[rustic crown]

observation uwu?

hewwo walter

[south patrol]

OwO

[agile burrow]

Hi det

[formal ermine]

walter!

[agile burrow]

Hi rectangle cube

[formal ermine]

call me illu or illum

[agile burrow]

Sure thing illu

[rustic crown]

hewwo illu or illum

[south patrol]

Sure thing rectangle cube

Jk hewwo illum

[formal ermine]

at least you don't call me Iluminator03

[rustic crown]

illuminator3#0001

[formal ermine]

that is what this one guy calls me

[rustic crown]

[formal ermine]

the origin of illum is this btw lol

[rustic crown]

sick and shy >.<

[formal ermine]

shy in the wrong situations

I also don't like to call myself annoying but rather say that I have a ✨ special ✨ personality

[inner acorn]

[rustic crown]

Hewwo Sasha

[inner acorn]

Hi det, have the vees overthrown the cats yet?

[rustic crown]

in abstract alg yee :3

[inner acorn]

nice

[rotund aurora]

🐊

[rustic crown]

what exactly do we need to prove

i'm confused with a,b,c and i, ii, iii

[rustic crown]

oh i didn't hear the ping for some reason >.<

okie so
for (i) => (iii),
you need to show that sigma(F) lies inside F for every sigma. (this is enough, as one can argue via degree, or another way is to say it will also hold if you replace sigma with sigma^-1)
to do this, pick x in F, and show that sigma(x) also lies in F. since F/K is galois, sigma(x) is another root of the minimal polynomial m_{x, K} of x, and thus it must also be in F (by normalness)

and
for (iii) => (i)
we need to show that F/K is normal, so pick any x in F and show that the minimal polynomial m_{x, K} splits completely over F. since L/K is galois, you know m_{x, K} does split over L, and so if y in L is another root of m_{x, K} we can find an automorphism sigma in Gal(L/K) which sends x to y. Since sigma(F) is contained in F, we must have sigma(x) in F, which says y lies in F. since y was an arbitrary root, it shows m_{x, K} splits completely over F.

@willow geyser

[tender wharf]

if G/H is a group then H is normal in G.

Is the right-multiplication by a valid here?
Proof: H = eH = aa^-1H = aHa^-1H => Ha = aHa^-1Ha = aHa^-1 a = aH

[next obsidian]

Huh

Thinking

I think no

Uhhhhhhh

This is confusing actually like

I’m not sure if G/H is defined as left cosets or right cosets

When H is normal they’re the same and it’s fine but like

[tender wharf]

Yeah I was thinking right multiplying felt a bit

iffy

G/H is left cosets iirc

[next obsidian]

Is saying “G/H is a group” fixing what side of coset you’re using?

Hmm then I don’t see how it is

[tender wharf]

I'll have to find another way to do it then

The hypothesis is that for all a, b in G aHbH = abH

[next obsidian]

Yeah

It might be easier to go by contradiction

[tender wharf]

Actually I may have just missed the obvious one since

Oh I probably didn't

[next obsidian]

Yeah so

[tender wharf]

Yeah I'll play around a bit more ig

[next obsidian]

I think because you only can ever deal with left cosets

I think you have to like do an element by element thing at some point

I’m not sure tho

[tender wharf]

I'm gonna look at how G/H being a group was proved

Thanks for the help

[hidden haven]

G/H is left cosets

Unless otherwise specified

[knotty frigate]

its proved by first proving H being normal entails that aH=Ha and then proving that entails that left coset multiplication is well defined and then the rest of the properties follow form the properties of G and H

[next obsidian]

And H\G is right cossets

[hidden haven]

H\G is sometimes used for right cosets

[next obsidian]

[hidden haven]

Lmao

There's also double cosets

[tender wharf]

Yeah my book provides the proof

[hidden haven]

I think that is G/H\G

[knotty frigate]

[next obsidian]

Yeah except with different G

[tender wharf]

How is this defined just curious

hGk?

[hidden haven]

The set of all double cosets is denoted by H\G/K

[next obsidian]

Yeah

[hidden haven]

Ok so this is H-K-double coset

[tender wharf]

{hGk | h in H k in K} im guessing

[hidden haven]

HxK for each fixed x

[tender wharf]

ah

[hidden haven]

is the double coset corresponding to x

[tender wharf]

interesting

[next obsidian]

Oh wait

Isaacs finite group theory from like 1 week of Junior year

It returns to meeeeee

[hidden haven]

I am supposed to know this stuff since I am working with Mackey functors

[fossil shuttle]

how u doin moldi

[hidden haven]

Good

[fossil shuttle]

why u crying

[hidden haven]

oh no wrong emoji

[fossil shuttle]

Good

[hidden haven]

[knotty frigate]

oh hey

[delicate orchid]

Wholesome Mackey’s theorem

[hidden haven]

[rustic crown]

because the map sigma : F --> sigma(F) is an isomorphism of K-vectorspaces

injectivity follows as everything is a field
surjectivity is clear
K-linearity because sigma fixes K

okie uwu?

the other idea was to say "as we proved sigma(F) is inside F for every sigma", it's also true for sigma^-1, and so sigma^-1(F) is inside F which is the other inclusion

uwu

[smoky ivy]

is there a way to define a multiplication on R^2 so that R^2 and R are isomorphic as rings?

we know that R and R^2 are isomorphic as Q-vector spaces and thus as additive groups, but i was thinking of a multiplication maybe…

is it possible somehow?

[rustic crown]

use that isomorphism of Q-vector spaces to define the multiplication :p

say f : R^2 --> R is the isomorphism, then define a * b := f^-1(f(a) * f(b))

[smoky ivy]

i don’t quite see how they would be isomorphic as rings then

[rustic crown]

because this definition literally makes f also a ring homomorphism

f(a*b) = f(a)*f(b)

[smoky ivy]

ohh, i see

yeah, right

thank you…got it

[rustic crown]

[formal ermine]

R and R^2 are isomorphic as Q-vector spaces because they have the same dimension or?

[rustic crown]

yee

[formal ermine]

so V iso W iff dim(V) = dim(W) also holds for non finite vector spaces?

[rustic crown]

yep

[formal ermine]

cool

I started reading la :p

[smoky ivy]

Yeah

[coral shale]

means there's a bijection between basis vectors

and then

[formal ermine]

yes I knew it holds for finite spaces

[coral shale]

each vector

can be considered as a function from your basis set

to the field

[rustic crown]

basically you also have to prove the "inifnite" version of replacement theorem, which will then also require AoC

[formal ermine]

replacement theorem?

[coral shale]

space V = {f : B -> F}, B the basis set, F the field

yes the above does need aoc

[formal ermine]

dim(V) = dim(W) for non finite spaces compares cardinality?

[rustic crown]

yea something like if B is a basis and S is a spanning set, then there is an injection B --> S

[coral shale]

cardinality of bases

[formal ermine]

yeah

oke

[coral shale]

not just for non finite ofc

[formal ermine]

I like using size for finite and cardinality for non finite

makes things clearer imo

[rustic crown]

the idea here to replace elements of S one by one with elements of B and such that at each point S remains a spanning set

[formal ermine]

ah

[rustic crown]

when you're done, B would be literally a subset of S, and so you have your injection

[formal ermine]

yeah

[coral shale]

vector spaces my beloved

[formal ermine]

vectors spaces are eh

[coral shale]

idk if u were there yesterday when i briefly talked about galois

[formal ermine]

I don't like rings nor modules nor vector spaces

[formal ermine]

no I wasn't

[coral shale]

L : K a field extension, L is a vector space over K

[rustic crown]

rings are uwu

[coral shale]

is a nice idea

[formal ermine]

yes

my alg course is doing galois theory right now

we did fundamental theorem last lecture

[coral shale]

ur studying bigger and better things i see

[rustic crown]

do infinite galois theory now

[formal ermine]

[rustic crown]

[formal ermine]

I can't wait till I'm through this la book

then I can do real anal

[rustic crown]

.<

[formal ermine]

shush

[barren sierra]

sounds like you don't like multiplication

[formal ermine]

I like fields

and algebras

only associative ones

[agile burrow]

Z-algebras

[rustic crown]

Rings very uwu

walter you also very uwu

[agile burrow]

det is uwu too

[river nebula]

Can anyone help me know how to find the maximal torus of some classical groups?

[lavish spoke]

Tbf the linear algebra used in e.g. Functional anal is basic to begin with

[next obsidian]

What about me

[coral shale]

chuwu

[next obsidian]

Chmuwu

[warm wyvern]

the "picture" here can be a torus

but it can also be just a circle, right?

[elder wave]

what do you mean by this

[warm wyvern]

tbh, I'm not sure

I'm assuming that's what the exercise is asking, no?

coz R/Z iso S^1

[chilly ocean]

R/Z is the circle, but this is R^2 / Z^2

[warm wyvern]

well yeah

but RxR/ZxZ iso R/Z x R/Z

[chilly ocean]

circle x circle is not circle

[warm wyvern]

oh woops

[chilly ocean]

it's like a circle of circles (torus)

[warm wyvern]

I meant sphere, sry

[next obsidian]

No

It’s not a sphere

[chilly ocean]

nope

not sphere

[warm wyvern]

[next obsidian]

Take a unit square

To get the sphere you identify all the boundary points with one point

Like you scrunch it up, and pack the entire boundary as the top point or whatever

When you do R^2/Z^2 you’re only identifying the top and bottom edge, and the left and right edge

Basically

But also idk how to put a group structure on S^2

Lmao

[chilly ocean]

it won't be a nice one

[chilly radish]

I think you can't if you want it to be compatible with the topology

[chilly ocean]

S^n a top group implies n odd

[chilly radish]

Adams' Theorem

[warm wyvern]

why not just S^1 x S^1?

[next obsidian]

Because

Those aren’t the same thing

A torus and a sphere are not homeomorphic

They’re not even homotopic

S^2 has trivial fundamental group and the torus has fundamental group Z^2

[warm wyvern]

why are you guys talking about algebraic topology? lmao

this is just group theory

I was thinking just "give the points in x shape a group structure"

[next obsidian]

Because it isn’t that

[chilly ocean]

because it's a precise way to explain how a torus and a circle aren't the same

[next obsidian]

Like your picture is wrong

[rustic crown]

chmuwu

[formal ermine]

why is this?

[chilly radish]

Prove it!

[formal ermine]

is it because R^2/Z^2 = (S^1)^2 and S^1 = R/Z?

[chilly ocean]

S^1, not C

no one uses C for the circle

[formal ermine]

okk

so S^2 is the torus?

[chilly ocean]

no

S^2 means the sphere

[formal ermine]

what's S^3

[warm wyvern]

the 3-sphere

lel

we don't have special names for high dimensional sphere, I don't think

they're just called n-spheres

[formal ermine]

S^0 is a point?

[warm wyvern]

there is a cute proof using universal properties

are you familiar with those?

[formal ermine]

from cat theory?

[warm wyvern]

ye

[tribal moss]

The principled thing would be to make S^0 mean { x in R^1 : |x| = 1 }, which is two points.

"Two points" is also the space whose suspension is (homeomorphic to) S^1.

[warm wyvern]

tf is a suspension?

[tribal moss]

https://en.wikipedia.org/wiki/Suspension_(topology)

[warm wyvern]

yeah

lol

I should've went there before asking haha

that's cool

[formal ermine]

yes

[prisma ibex]

the (-1)-sphere is the empty set

[solar glacier]

Question, an let me go all the way through with it before commenting, it may be elementary but the whole set notation throws me off so

let $\phi: R \rightarrow S$ be a ring homomorphism and let $J \subset S$ be an ideal show that $\phi^{-1}(J):={r \in R: \phi(r) \in J}$ is an ideal of $R$

[cloud walrus]

MyMathYourMath

[formal ermine]

I remember doing that one at 2 am once

[warm wyvern]

ok so using the universal pwopewty for products we have two canonical projections pi_1, pi_2: RxR->R

[formal ermine]

pwopewty

[warm wyvern]

combine those with the canonical projection R->R/Z

[cloud walrus]

MyMathYourMath

[warm wyvern]

now you have two morphisms going from RxR to R/Z

[cloud walrus]

MyMathYourMath

[solar glacier]

thus x-y is in phi inverse of J

[warm wyvern]

now according to the universal property for products that means there is a morphism going from RxR to R/Z x R/Z

[formal ermine]

because J is a subring, yes

[solar glacier]

sorry yes

[warm wyvern]

and the kernal of that morphism is ZxZ

[solar glacier]

subgring property

[formal ermine]

hm

[warm wyvern]

RxR/ZxZ iso R/Z x R/Z (first iso)

[formal ermine]

ok nvm I don't know enough cat theory for this

[solar glacier]

then for absorbtion by R

[formal ermine]

I'm still trying to comprehend adjunct functors

[solar glacier]

let a \in R and x \in phi^{-1}(J)

then ax \in R

and phi(ax)=\phi(a)\phi(x) \in J and THIS is by property of ideals since phi(a) \in S

[formal ermine]

yes

btw both of these hold for both + and * so you can just use \circ imo

(and you need to show both)

[solar glacier]

yeah i use a star or circ

for groups though usually for rings i put the actual sign

and to show somethings and ideal its enough to show for x,y \in I x-y \in I and for all r \in R that rx \in I for every x \in I

[formal ermine]

(or x + y in I)

and yes

wait

I think you mean

for every r in R and x in I you have that r+x in I

[solar glacier]

no its rx \in I

[formal ermine]

o wait

yes

you're right

[solar glacier]

its absorbtion by r in R

[warm wyvern]

you prolly do, I'm just too sleep deprived rn to write an answer half assed lel

sry

[formal ermine]

lol dw

[warm wyvern]

if you're still interested you can ping me tomorrow

[formal ermine]

on the 30th?

or later today

[solar glacier]

30th could be either 36 or less than 24 hours away depending on where you stay on this planet lol

[warm wyvern]

lmfao

later today ig

[formal ermine]

we are in the same timezone

[solar glacier]

ahh

[formal ermine]

how do I show that the galois group of $x^{13} - 1$ is $(\bZ/13\bZ)^\times$? I mean it obviously has 13 elements

[cloud walrus]

$$\rotatebox{90}{▢🧊▢🧊▢🧊▢🧊}$$

[formal ermine]

not sure how I would continue though

[solar glacier]

think you could walk me through a group theory problem lol

[formal ermine]

I probably only have as much knowledge as you but sure

[solar glacier]

if not its chill

ok and heres my attempt

let N be a normal subgroup of index n in G show g^n \in N for all g \in G

so i know if N has index n then G/N has size n

thus elements of the form gN have order n

unsure if those emojos are good or bad lol

[rustic crown]

lol, your idea is correct

[solar glacier]

oh ok lol

[rustic crown]

but saying "order" isn't the word

you only know that (gN)^n = eN

[solar glacier]

yes!

[rustic crown]

and not that n is literally the order

[solar glacier]

that

oh thats right order is min value correct?

[rustic crown]

yea

[solar glacier]

and eN = N

since every group absorbs e

[rustic crown]

yee

[solar glacier]

so g^nN \subset N

[tribal moss]

I think the only thing to object to is that you used the phrasing "order of".

[solar glacier]

yeah i got that cleared up lol

i havent used normality of N yet

[tribal moss]

Ah, I thought you were trying to defend against that yet.

Sorry.

[solar glacier]

oh no lol ur good

[rustic crown]

you did that at the start

when you claim G/N is a group

or did you not do that?

[solar glacier]

oh thats right

[rustic crown]

.<

[solar glacier]

u can only quotient out by normal subgroups

[rustic crown]

you could only use lagranges theorem once you know it's a group

[solar glacier]

true

am i basically finished after the g^nN=eN step

or i feel i have more to go

i need to show g^n \in N

[tribal moss]

Hmm, but even if N is not normal, the whole group subgroup generated by g still acts on the (say) left cosets of N, so we can appeal to the orbit-stabilizer theorem, can't we?

[formal ermine]

you can try showing a more general statement: gH = H iff g is in H for any subgroup H and any g in G

[solar glacier]

thats true

ive shown this before i feel like

ibe certainly used it

ive *

[chilly ocean]

one liner from the definition...

[solar glacier]

its by the closure of subgroup

[formal ermine]

why is (gN)^n = g^nN

[rustic crown]

okie i'm not sure i fully follow the argument >.<

[formal ermine]

why sully

[chilly ocean]

are you prompting them with that or is it a real question

[formal ermine]

real question

[chilly ocean]

think about it

just like.... what is (gN)^n? lol

[formal ermine]

eN

[chilly ocean]

no

forget that for a second

what IS it

how is it defined

[solar glacier]

think cosets

[formal ermine]

oh my god

yes

ok got it

[chilly ocean]

lol

[formal ermine]

I was thinking euler

but in the wrong moment

[chilly ocean]

(gN)(hN) = (gh)N so...

[solar glacier]

it has to do with the cosets

and mult by cosets

[formal ermine]

yeah lmao

[solar glacier]

^^^

[formal ermine]

I was thinking (gN)^order = neutral element = eN

[rustic crown]

det is sleepy >.<

[formal ermine]

same

[formal ermine]

bump

[chilly ocean]

it happens to us all

[tribal moss]

Given G, N and g, with |G|=n. The subgroup generated by g has an order the divides n. It acts on the left cosets of N by left multiplication. By the orbit-stabilizer theorem, the orbit of N itself has a size that divides the order of g, which divides n. But everything in that orbit has the form g^kN for some k, so multiplying by g must simply cycle through the orbit. Multiplying by g n times cycles through the orbit a whole number of times, so we end up at N again.

[chilly ocean]

sully is best used as a "ur overthinking it buddy"

[formal ermine]

I use sully as wtf are you saying that shit makes no sense

[rustic crown]

ah, i think we're only given the index [G:N]=n

[tribal moss]

Ah, whoops.

[rustic crown]

i was also thinking how you could relate action of <g> with the size of the new orbit >.<

nvm, me go sleep now

good night

[tribal moss]

An orbit under the action of a cyclic group is always a cycle.

[formal ermine]

gn!

[tribal moss]

Ah! Counterexample for the non-normal case:
G is S_3 and N is the (non-normal) subgroup {e, (2 3)}, of index 3. If now g is the transposition (1 2), then g^3·e is not in N.

[solar glacier]

NICE!

[solar glacier]

but is S_3/N a group since N is not even normal so you cant talk about its index? or that just means you cant consider S_3/N as a group

but |S_3:N| still makes sense?

[ebon gyro]

Index isn't exclusive to normal subgroups

[solar glacier]

ok thats what i was wondering ^

but in assuming my problem is normal, the normality gets used from forming G/N into a group of order index of N in G?

[formal ermine]

yes

[solar glacier]

ok sweet makes sense

i had a prof who once mentioned

when you hear normal subgroup

it should scream out to quotient by it

for the most part

all but a set of measure zero lol

[solar glacier]

means "for the most part" lol

btw im glad i joined this community, ive learned tons even when not asking and just reading the chats about questions! everyones super friendly too

[formal ermine]

same

[solar glacier]

Merry Late christmas everyone and happy holidays!!!

[ebon gyro]

this is good advice LMAO

[solar glacier]

lol right?!

[ebon gyro]

I remember seeing this exact phrase except like "if you ever encounter a normal subgroup in the wild. you should want to take a quotient"

[solar glacier]

loooool

yeahi had a problem on my midterm that was super tough

but i remember seeing it was a normal subgroup

and needing to use to first iso theorem so i just put down a map from G to G /N and got partial points lol

i have the problem if you wanna try it lol

[ebon gyro]

ooh I'm curious to see if you have it

[ebon gyro]

this is a solid strategy always

[solar glacier]

here it is

I also have the solution if you get stuck lol

[cloud walrus]

MyMathYourMath

[solar glacier]

you really just need to show its a subset of N

since its already a subgroup it has the subgroup property of closure and inverse

but this problem had the entire class

i meant , prove if GCD is 1

not is coprime that doesnt make sense lmao

@ebon gyro

or if anyone wants to attack that exercise feel free

[tribal moss]

||The size of the image of H in G/N must divide both |H| and |G/N|, and it contains e||

[solar glacier]

unless youve seen it before