groups-rings-fields

[leaden finch]

oh okay, hmm so what should i do next D:

[thorn delta]

ac = 0 and bd = 0. a,b,c,d are real numbers, and R is a field (so its definitely an integral domain)

so what can you conclude?

[leaden finch]

hmm that its an integral domain and a field

so does that mean that we ac and bd can be any number?

[thorn delta]

im not sure on 3 to show if it has no zero divisors
wait a second, you should be showing that S does have zero divisors

instead of all of this, try to find two nonzero matrices in S which multiply to 0 (there are simple examples).

[leaden finch]

oh, we have to show that it doesnt have zero divisors

hmm

[thorn delta]

nono

it does have zero divisors

i.e. its not an integral domain

[leaden finch]

wait so how do we know it has a zero divisor

thats where i get lost

[thorn delta]

a zero divisor is a non-zero matrix A (so it has a non zero entry) such that there is some other matrix B such that AB = 0.

i claim that there are two simple-looking nonzero matrices in S such that they multiply to 0

[leaden finch]

oh okay

[leaden finch]

can someone check my number 3

part a

and part b i think its not an integral domain since it failed 1 commutuve ring

[latent anvil]

It's really not clear what you're doing in 1) of part a. Like you say you want to show it's nonempty but then you show it's commutative (which it isn't? I'm not sure how you got that)

also in 2) of that question at the bottom you say "since a, b, c, d, e, f in S" but that's not true, those are integers and S is a set of matrices

Also at the top you haven't specified what R is

Your computation in 3) of a is wrong, you computed the product correctly in 1)

[leaden finch]

ooo okie, i'll fix it

[latent anvil]

Your logic for b is correct. It's also true that S has zero divisors, can you find any?

also sunshine, do you mind if I post my question here while you try to fix your solution?

[leaden finch]

sure go ahead

[latent anvil]

Let $A = \prod_{i=0}^\infty \mathbb{F}_2$ and $X = \mathrm{Spec} A$. Is every closed subset of $X$ also open? Equivalently, for any ideal $I$ of $A$ is there an ideal $J$ of $A$ such that $I \cap J = 0$ and $I + J = (1)$. This is true for finitely generated ideals $I$, since those are all principal (as $I^2 = I$) and every element of $A$ is an idempotent.

[cloud walrus]

shamrock:

[latent anvil]

sorry @vestal snow, I'd try to help but I don't know what an algebraic function field is

[vestal snow]

Oh okay. Maybe I can rephrase it? Let $F,F'$ be fields with $F \subseteq F'$ and $K \subseteq K'$. Assume $K$ and $K'$ are algebraically closed in $F$ and $F'$ respectively. Assume that $F$ and $F'$ are finite extensions of $K(x)$ and $K'(x)$. Prove that if F'/F is separable and finite, then so is K'/K.

[cloud walrus]

Have a Banana Bitch:

[vestal snow]

I know that K'/K would be finite

But I have no idea how to show they are separable

Assume also that K is perfect

Oh wait

Doesn't K being perfect mean that all algebraic extensions of it are separable?

[latent anvil]

Yeah isn't that the defintion?

I'm gonna bump my post a little so it doesn't get buried

[cloud walrus]

shamrock:

[fierce perch]

Look at the ring k[x,y,u,v]/(xy+ux^2+vy^2), how does one see that D(x,y) is not principal?

[golden pasture]

@latent anvil lol i was also thinking of the same problem a few months ago and found https://qchu.wordpress.com/2010/11/22/boolean-rings-ultrafilters-and-stones-representation-theorem/

[fierce perch]

Looks like a stone cech compactification @latent anvil, not sure if it helps

[latent anvil]

@golden pasture ty

@fierce perch I think someone told me it's the stone cech compactification of N

[fierce perch]

Residue fields should be Z/2 and because of compactness I think it shouldn’t have great dimension and hence maybe your clain should be true

But it might not be true because infinite products should have weird dimensional behaviour so I dunno

[golden pasture]

yh it's the stone cech

[latent anvil]

residue fields are Z/2

I remember this scheme because A is an example of a ring which isn't noetherian but all of whose localizations at prime ideals are noetherian

coming up with such a ring was a homework problem in my algebra course last year

[oak grove]

Sucks to make my first question one i feel is dumb but I'm looking at $\langle u,v \mid u^4 = v^3 = 1 , uv = v^2 u^2 \rangle$

[cloud walrus]

jan Niku:

[oak grove]

The prompt is "Show v commutes with u^3 (show that v^2 u^3 v = u^3)"

im trying to understand how the parenthetical hint shows this

its given that it was previously proven that v^2 = v^-1 if that matters

heres the entire prompt

[latent anvil]

It's not a dumb question!

You're just getting familiar with this stuff

multiply both sides of v^2 u^3 v = u^3 by v on the left

in general, g and h commute iff g^-1 h g = h, and here we have g = v and h = u^3

[maiden ocean]

every question is dumb except "why do i not have a boyfriend" which is a very good question

[latent anvil]

no that's a dumb question

[maiden ocean]

: (

[carmine fossil]

What if you are gender neutral

[latent anvil]

what if

@oak grove does that make sense?

[oak grove]

so its just left multiplication of the inverse

i see

😄

idk why that took me a day lol

thank you

[latent anvil]

np we all get stuck on dumb stuff

[carmine fossil]

Sometimes because you don't read the text properly

[golden pasture]

sometimes you just miss somethinh completely trivial

[latent anvil]

sometimes because you're just learning the material and aren't familiar with it yet

[carmine fossil]

That's most times

[maiden ocean]

sometimes u have a mental breakdown and cannot do math good for a few hours

[latent anvil]

sometimes because you are possessed by dark spirits

[maiden ocean]

thats always happening to me

but usually it does not interfere with my math

unless i am doing analysis

[latent anvil]

hmm are you sure the dark spirits aren't just analysis vibes?

[oak grove]

i know DD

youre right

i didnt read the text properly

fwiw even with the information spoonfed to me i still dont comprehend

so i dont think your criticism holds

[maiden ocean]

sometimes it is helpful to reread from the beginning

or not the beginning beginning but like

the start of the chapter

[oak grove]

yea, im going to have to fix my mindset

[golden pasture]

sometimes useful to jus find a example

[vestal snow]

Is it possible to get an intuition for algebraic curves from Ch 4 of Hartshorne without knowing much scheme theory?

[fierce perch]

Read some easier text

Like Harris Geometry of Curves?

[vestal snow]

Thanks. I'll check it out

[fierce perch]

Or fultons algebraic curves

[leaden finch]

can someone help me with this one on how to determine if its a subring

isnt just closed under addition and multiplication?

[golden pasture]

it must be a ring as well

There's a reason why the even integers do not form a ring

[leaden finch]

how can i explain my asnwer for this one

i created a table for it

[thorn delta]

you need closure under subtraction and multiplication to be a subring

also ariana, i don't think sunshine's rings have to have 1

[chilly ocean]

rngs

[golden pasture]

also ariana, i don't think sunshine's rings have to have 1
wtf?

wait u serious there isnt identity?

[leaden finch]

since we dont have closure under subtraction then what do we do?

[thorn delta]

idk, that's just sense i got when I helped earlier. Maybe u could confirm @leaden finch lol.
It aint that weird. Rings don't have identity in the book ive been using :p

[golden pasture]

[leaden finch]

loll i get confuse at this stuff

[chilly ocean]

ari u good?

[golden pasture]

wait sunshine what is your definition of the ring that you're using

[leaden finch]

seeing too many subrings make me dizzy ha

8 axioms?

[thorn delta]

send us the axioms

[maiden ocean]

rng

[thorn delta]

no 1 HA

[chilly ocean]

no 1

[leaden finch]

hmm i was thinking of saying this but i think its wrong

[thorn delta]

you also need to mention that its closed under taking additive inverses. i.e. for any n in S, -n is in S.
Either that or just show its closed under subtraction

[leaden finch]

what do i say for subtraction?

[golden pasture]

wait why isnt there a 1a=a1=a

ok uh just check everything lol

[4]-[2]=[2]

[leaden finch]

oo loll

[thorn delta]

you can just look at the table. Since every column has a 0 entry, the number corresponding to the column has an inverse in one of the rows.

[golden pasture]

or like you know how mod n arithematic works like
n+(-a)=-a mod n

so there you have your inverse already

[leaden finch]

oh okay

so do we show the addiitve inverse and addiitve identity

[thorn delta]

showing closure under subtraction is the same as showing closure under addition and taking additive inverses --like killing two birds with one stone. You don't need to show there is an additive identity, but you do need to show that R is nonempty (obvious in this case).

[leaden finch]

oh okay

[leaden finch]

can someone help me with this one

[somber jasper]

what do you need help with?

what did you try I mean @leaden finch

[leaden finch]

im probably wrong but i did this

[hot lake]

what's your definition of subring

[somber jasper]

the formula you used for f(a+bi) and f(c+di) isnt exactly right

cause you subbed in a+bi into the matrix where you were supposed to use the given homomorphism

you have the right idea though!

[leaden finch]

oh okay

ty

general defintion? @hot lake

[thorn delta]

you started with f(x) + f(y) = .... and you need to end with = f(x+y).
Then you do the same thing with multiplication f(xy) =...= f(x)f(y)

[hot lake]

are there many definitions of subrings ?

[leaden finch]

no, i know it has nonempty set, closed under subtraction, multiplication

[hot lake]

can't you copy paste it like the rest

[leaden finch]

wym copy paste it? lol

[hot lake]

like you did for the definition of ring

omg your definition of ring says the set of even integers is a ring

burn the heretics

[leaden finch]

ooo lol

[thorn delta]

its ok sunshine, rings don't have 1 for me either

[somber jasper]

same

[leaden finch]

oo yeah

[pallid ember]

rings have 1?

wtf

[next obsidian]

[golden pasture]

ya

rings should have 1

[coarse storm]

Undergrad courses may also assume commutativity. Mentioned once at the beginning and then completely taken for granted.

[deft plume]

True. Then things like matrices or symmetries make u realize how non-commutative sets can make our lives even slightly more complicated.

[marsh fractal]

question: $Let u \in \mathbb{Z}[\sqrt{d}]$ then N = 1 if and only if u is a unit

[cloud walrus]

ethane:

[coarse storm]

$\iff \text{is a unit}$

[marsh fractal]

ty

still learning tex lol

[cloud walrus]

Boni:

[marsh fractal]

but yes, am I going along the right lines by thinking: 1 and -1 is the only unit of those rings unless it is the ring where d = -1

also those are the rings where d is not a square

[coarse storm]

What's N?

[marsh fractal]

$N = | a^{2} - db^{2} | \text{, where } a + b\sqrt{d} \in \mathbb{Z}[\sqrt{d}]$

[cloud walrus]

ethane:

[marsh fractal]

its a norm

[deft plume]

Oh yah. It’s been a while since I remember norms, but that was also in a lemma if I remember correctly

[coarse storm]

but yes, am I going along the right lines by thinking: 1 and -1 is the only unit of those rings unless it is the ring where d = -1
I... think... that claim is false.

Regardless of its relevance to the proof.

[marsh fractal]

ok

[deft plume]

There’s a property saying that N(uv) = N(u)N(v)

[marsh fractal]

yeah

[chilly ocean]

yo boni long time no see

[deft plume]

For one direction if u is a unit, then uv=1

[coarse storm]

Hi! I don't check Maths servers much anymore!

[marsh fractal]

K cool, I guess I was checking if that a unit being 1 or -1 OR i or -i in the case where d = -1 was a short route

For one direction if u is a unit, then uv=1
@deft plume

[deft plume]

N(uv) = N(1)

for the other direction, i suppose choosing a product of conjugates in that ring

and probably applying that same property

Thats what i think off inspection

[marsh fractal]

k thanks

[deft plume]

its possible you might come across 1 or -1 in that direction after you take the product

[covert otter]

Theorem : If D is a division ring then any D-module V admits a basis.

So what do you guys think should be an approach for a proof of this, the one I've encountered (which I like to some extent) is a long proof using Zorn's Lemma

[golden pasture]

have always assumed ab=ba in my life but ig you can adapt the proof in the case of fields

[vital quail]

lol

[static temple]

it would necessary use Zorn's lemma @covert otter

I didn't check but isn't its proof quite similar to the one with vector spaces?

[covert otter]

umm hmm yeah ... I kinda thought one myself quite similar...I'd post the LaTeX after a while maybe

[leaden finch]

can someone help me with isomprhisms

[pallid ember]

[leaden finch]

part b this is what i did

[pallid ember]

looks good

[leaden finch]

okie, i did surjective

[marsh fractal]

Interjecting here, but what is the point of these Z[sqrt(d)] where d is square free rings?

like not practically or w/e , mathematically

[pearl bane]

this is what Artin says

[delicate bloom]

you can use them to solve some diophantine equations

you can factor things you usually couldn't and get some restrictions in this new setting that allow you to figure them out

[light tusk]

They're like the simplest possible way to extend the integers algebraically without dividing... If you're bored of linear, the next place to go is quadratic?

[marsh fractal]

ok!

They're like the simplest possible way to extend the integers algebraically without dividing... If you're bored of linear, the next place to go is quadratic?
@light tusk isnt square rooting in some sense dividing?

[light tusk]

With taking "just a square root", you're interested in monic polynomials such as x^2 - a. Throwing in a coefficient in front of the x^2, different things happen. A field you obtain by extending by all roots of linear polynomials like ax - 1, which is very different from solutions to higher degree polynomials.

With extensions of fields, monic isn't as big of a deal because you can divide the leading coefficient away. With rings, monic is important in order to preserve structure

[marsh fractal]

ok that makes sense!

somewhat 😅

is the idea something like: if you place a coefficient infront of the polynomial some restrictions are placed

so when square rooting you are considering only the monic polynomial

[leaden finch]

can someone help me with this one, how would prove this if it asked you to do well define?

[maiden ocean]

the map sending (a, b) to (b, a)

from Z2 x Z3 to Z3 x Z2

[chilly ocean]

[leaden finch]

i created my function

[maiden ocean]

hurb

why not just flip

[leaden finch]

i think i made a typo

i think its suppose to be like this

[maiden ocean]

ok good

[leaden finch]

okay sooo then i got stuck lol

[maiden ocean]

u must show that it is an isomorphism

[turbid pond]

Given F, G, H groups

Does (F x H) isom (G x H) => F isom G ?

[chilly ocean]

most likely

I think proj_1 (FxH) iso proj_1(G x H) which would prove it

[turbid pond]

proj_1 (FxH) refers to (f,h) -> f?

wait, what i said doesnt make sense

[chilly ocean]

yeah thats what I meant by proj1

but that doesnt help probably (unless you prove some statements about projections i guess, which is the same thing most likely)

I mean if theres isomoprhims f(a,b) = (c,d) then you can just define f(a) = c

oh wait it may not be correct nvm

[maiden ocean]

yeah

this is not as simple as it seems

[turbid pond]

but if b=/=d its not injective

[chilly ocean]

eyeyey

so moth what's the counterexample

[maiden ocean]

i am thinking

[turbid pond]

[maiden ocean]

maybe u can do something with Z4 and K4

idk

im sleepy

[turbid pond]

hmm, idk if relevant but im thinking about the finite abelian p-group case

[leaden finch]

is this right

@maiden ocean

[turbid pond]

@maiden ocean @chilly ocean @chilly ocean https://mathoverflow.net/questions/83395/cancellation-theorem-for-direct-and-other-kinds-of-products-between-groups
turns out it's true for H finite but it is non-trivial

[pallid ember]

{0}xR and RxR should be a counter

[turbid pond]

ah yes

[leaden finch]

im lost, can someone explain too me 😦

[golden pasture]

you have showed f is a homomorphism

still need the isomorphism step

there are a few equivalent ways to show it

• Find a inverse homomorphism
• Show it is a bijective function
• Show that it's kernel is trivial and it every element has a preimage

[jade crag]

could somebody help me with this?

im thinking i need to perform gaussian elimination on every row, and factoring out a term each time

how would i generalise it in this case for odd powers though

shd i be trying to eliminate by row or column

[maiden ocean]

Okay I'm really confused as to what "explain the structure of the field extension" is supposed to mean here

Can someone walk me through the case p1 = 2, p2 = 3 because im not really even sure what the problem is asking exactly

[stone fulcrum]

I'm not too sure haha, what was the case done in class?

Perhaps everything below is the "explain"part

[maiden ocean]

hnnnnn

ill try to find it

but im not even sure what day its from

[light tusk]

Those are each degree 2 extensions...

[oblique river]

maybe it means like

find a basis or something?

like a + bsqrt(p1) + csqrt(p2) + dsqrt(p1p2)

[stone fulcrum]

Simple enough and covers it pretty well I suppose

[maiden ocean]

i guess?

[light tusk]

You can do that, but I think it gets a lot harder the larger the extension

[maiden ocean]

maybe just writing it as a splitting field is enough

and then doing the rest of the problem

[light tusk]

It's easier, I think, to build it up one root at a time

[maiden ocean]

idek

so. is the basis over Q just (1, sqrt(p1), sqrt(p2), sqrt(p1p2))?

[golden pasture]

yea

[maiden ocean]

is

is that it ???

[golden pasture]

ig so lol

[stone fulcrum]

It describes the elements in that set fully

"Structure" makes me nervous though, as that usually means more than just the elements in the set

[maiden ocean]

[stone fulcrum]

But that's when you get into stuff like the Galois group ect

[maiden ocean]

i guess this is fine

sigh

[maiden ocean]

ok follow up

arent all 3 of the intermediary fields splitting fields?

cuz its Q(sqrt(p1)), Q(sqrt(p2)) and Q(sqrt(p1 p2))

[oblique river]

yes

[maiden ocean]

ok cool

i just have to tidy up the first problem then i guess

[maiden ocean]

i feel like an idiot lol but is there an easy way to verify that F64/F4 is normal or galois

can we write it as a splitting field somehow

[sturdy marsh]

F64 is the splitting field of x^64 - x over F2

[maiden ocean]

yeah

but is there a way we can translate that to over F4

[sturdy marsh]

you dont need that

[oblique river]

if you have E/K/F and E/F is galois

then E/K is automatically galois

[sturdy marsh]

it is galois over any intermediate extension

yew

[maiden ocean]

hnn

[oblique river]

proof: take the same polynomial

[sturdy marsh]

ye

[maiden ocean]

is there a proof

oh

ok

yea i was just overthinking it then lol

[maiden ocean]

ok last one haha

im not really sure what an obstruction would constitute here

some other condition that is equivalent to non-normality?

[sturdy marsh]

hmm

A definition of normal extension is: Let L/K be an extension, let K' be an algebraic closure of K containing L. Then L/K is normal if for every K-embedding $\sigma: L \rightarrow K'$, we have $\sigma(L) = L$.

[cloud walrus]

Brofibration:

[sturdy marsh]

This is the same thing as saying the map $i_*: Hom_K(L,L) \rightarrow Hom(L,K')$ is surjective

[cloud walrus]

Brofibration:

[sturdy marsh]

where i: L \rightarrow K' is the inclusion map

[maiden ocean]

hm

we didnt really cover anything like this in lecture unfortunately :c

ill think of something hnnn

[sturdy marsh]

You can define an obstruction as the cokernel of that map

The extension is normal iff cokernel is 0

but I have no clue on how you would compute that cokernel

[maiden ocean]

sigh

i guess i can come back t othis after

the only other thing left on this pset is uhh

whether or not F_p(x)/F_p is normal/galois

im not really sure about that one

[sturdy marsh]

What's your definition of Galois

[maiden ocean]

normal + separable

i think uh

[sturdy marsh]

what's your definition of normal

[maiden ocean]

E/F is normal if for any irreducible f in F[x], if f has one root in E then all of its roots are in E

my first thought is like if f(r) = 0 for r in F_p(x) then ur going to have r in F_p

but then f(x) = (x - r)g(x)

[sturdy marsh]

but doesnt the extension need to be algberaic

[maiden ocean]

and isnt irreducible

[sturdy marsh]

algebraic*

[maiden ocean]

yeah thats what made me go before too

it seems to say so in the textbook i have

[sturdy marsh]

well that's not an algebraic extension

so not galois

[maiden ocean]

yea

thonk

i guess probably im supposed to just be like "its transcendental"

[sturdy marsh]

ye

[maiden ocean]

¯_(ツ)_/¯

[sturdy marsh]

yeah my 'obstruction' seems to be pure bs lmao

it's just restating the definition

[maiden ocean]

i think thats what im supposed to do lol

so i just did that for my defn of normal

its kind of a silly problem

[sturdy marsh]

I was hoping there would be something more meaninful

meaningful*

[maiden ocean]

¯_(ツ)_/¯

[sturdy marsh]

Is there an obstruction that measures failure of normality of a subgroup of a group

the stupid answer would be G/normalizer

[maiden ocean]

[sturdy marsh]

which is hard to compute

[maiden ocean]

yeah idek what they want lol

[balmy dagger]

how would i prove that P(N) --> {0,1}^N is a bijection ?

[pallid ember]

u mean find a bijection between them?

[balmy dagger]

i am trying to show that this mapping is a bijection
A is a subset of M and
XA (x) = 1 when x ∈ A ,XA (x) = 0 when x ∉ A

I want to use this to prove that the power set of natural Numbers is uncountable but to do that i need to prove the bijection here. like i understand why it's a bijection but not sure how to write it formally 😅

[golden pasture]

not suitable for this channel but just show it is injective and surjective lol

[pallid ember]

''how do i solve this q'' ''just solve it lol''

[quaint ivy]

To show it's injective suppose two subsets have the same image, then more or less by definition they have the same elements. (Show this like you'd usually show set equality: suppose \chi_A=\chi_B and let x ∈ A, then... and conclude x ∈ B. Then show that x ∈ B => x ∈ A.)

[vestal snow]

Prove that if R is noetherian then R[[x]] is as well

I have proved that if if I is an ideal of R[[x]] and h in I is a multiple of x^k for a well-chosen k, then h is in a well-chosen finitely generated ideal

Here's how I chose the finitely gen ideal. Let A_i be the set of the dominant terms (first non-zero term) of all elements of I. Showed that this is an ideal of R and thus finitely gen'd by a_1,..., a_n with corresponding elements f_1,..., f_n of I

[woven obsidian]

Only the bottom of the image is relevant, but I included everything for context

I am having some trouble interpreting what a linear dependence relation for homogeneous coordinates really means

My guess would be that the expression at the bottom is short-hand for: "For every affine representative of point (x_i: y_i: z_i) there exists lambda_i such that the sum lambda_i (x_i,y_i,z_i) is 0

But then it seems trivial because we have five vectors in k^3

So I must be wrong

Really, what we want to conclude is that at least 3 of the point lie on the same line

[cinder bone]

um so <5> in this case would be.... 5, 10, 15, 20... etc?

or..

units of Z_26

yeah

okay so just 5 and 25 then?

12

including 5, 15, 25

9

right

yes

so the order has to be 1, 2, 3, 4, 6, or 12

i was thinking it was 3 at first

as in 5, 15, 25

[leaden finch]

has anyone done any research papers in latex

[cinder bone]

if im right, then it's {1, 3, 7, 9, 11, 17, 19, 21, 23}

yeah

oh, i thought it was saying something different

so

so U_26/<5> is a group under (<5>a)(<5>b)=<5>ab

or in otherwords its the set of right cosets of <5> in U_26?

and since 3 is prime

every group of order 3 is isomorphic to Z_3

sooo

i just have to show there are 3 right cosets of <5> in U_26?

[thorn delta]

this is where lagrange's theorem comes in

[cinder bone]

right so i have 1, 2, 3, 4, 6, or 12 as options

or wait no

actually idk

[thorn delta]

the order of G/H is the number of cosets of H in G, aka the index of H in G. So... what does lagrange's theorem tell us here?

err, by "lagrange's theorem," I am really referring to the equation |G| = [G:H]|H|.
is that familiar to you?

[cinder bone]

yeah.

so the index is the order of U_26 / the order of <5>

[thorn delta]

yep!

[cinder bone]

and the order of <5> must be 4 then

5,10,15,25

in Z_26

orrrr

[thorn delta]

<5> = {1, 5, 25, 125}

[cinder bone]

oh it has to be multiplicative

because 625 = 1?

in Z_26?

i mean, that's why there's not 625 in there

[thorn delta]

yea, 5^4 = (25)(25) = (-1)(-1) = 1 mod 26

i probably should have written
<5> = {1,5,21,25} to have everything fully reduced modulo 26, but that's the idea.

[cinder bone]

yeah i did that

thanks 🙂

[thorn delta]

np

[cinder bone]

i like the (-1) trick ive never seen that

that works for 125=(5)(25)=(5)(-1)=(-5)=21 too

[thorn delta]

yep. thats the beauty of working in quotients: choosing representatives that make your calculations simple 😎

[twilit pawn]

i’m trying to read chapter 1 of federer’s geometric measure theory and it’s pretty dense. i can probably handle it but does any recommend a reference for this content? here’s a pic

[cinder bone]

i have this solved for a finite group G but im pretty sure i cant do anything if G is infinite

since then i cant use lagrange

[thorn delta]

do u know what all of the groups of order 4 are?

[cinder bone]

no

so Z_4 and Z_2xZ_2?

i still can't deduce if G is finite or not from that right

[thorn delta]

its not asking about whether G is finite. Lagrange isn't going to help here

but, yes Z_4 and Z_2 x Z_2 are the only groups of order 4

[cinder bone]

i already proved that any group of order 4 is isomorphic to Z_2 x Z_2

i have that

ohhhh

my bad

so its isomorphic to either Z_4 or Z_2 x Z_2

makes ense

sense*

ok but still, the way my theorems are written from my teacher

I have to apply lagrange to get G and Z(G) from [Z:Z(G)]

and then a different theorem for |G/Z(G)|

like she doesnt equate the two

ok

i just have a theorem for each one is all

ok thats dumb then

so i should be able to just say

|G/Z(G)| = 4 without citing any theorem

the definitions i have only for finite tho

oh i see what u mean

ok

[thorn delta]

hint: || since you know what the groups of order 4 are, it suffices to show that G/Z(G) can't be Z_4, i.e. G/Z(G) can't be cyclic||

[leaden finch]

anyone know Latex?

[cinder bone]

ok i got

it i think

so since [G : Z(G)] = 4

they have different orders

so G is not equal to Z(G)

meaning G is not abelian

meaning G is not cyclic

?

it is not isometric to Z_4

isomorphic

sorry my b

I have a theorem that says it has to be isomorphic to either Z_4 or Z_2 X Z_2

wdym

doesn't that prove it's not isomorphic to Z_4

oh sorry

i meant to talk about G/Z(G)

oh i see

ok thanks

perhaps then i can show that G/Z(G) is cyclic implies G is abelian

which by contraposition means G is not abelian implies G/Z(G) is not cylic?

[thorn delta]

yep, that's an option.

i even thought about asking if you already knew that result

[cinder bone]

i didnt but i just realized

from some theorems

[barren sierra]

Anyone got any ideas?

need help

[thorn delta]

by ring congruence, you mean
x ~ y and a ~ b implies ax ~ by and x + a ~ y + b
right?

yea, that's right, just saw #proofs-and-logic
hint: ||its not, so think of examples where the conditions don't hold||

@barren sierra any progress?

[barren sierra]

I just saw these messages sorry

uh yes that's what I mean by ring congruence

hmmmm

ok ok

how did you know it wasn't

oh wait

ring on R

wait no

[thorn delta]

a quick counterexample i thought of + all ideals of a field are trivial or the whole thing

[barren sierra]

all ideals of a field are trivial or the whole thing
???

what are ideals

they're in my notes but idk really what they are

[thorn delta]

if you know what a normal subgroup is, they are like normal subgroups but for rings. Basically, if I is an ideal of a ring R, then the relation x ~ y iff x - y is in I is a ring congruence

[maiden ocean]

have you seen principle ideals in Z

[barren sierra]

I have no idea what anything to do for ideals is

and idk what a normal subgroup is

[maiden ocean]

theyre a certain kind of subset of a ring

they are important for quotienting reasons

dont worry about them for now

i dont think this is rly relevant to the problem lol

[thorn delta]

i.e. normal subgroup induce group congruences and ideals induce ring congruences. This is way overkill for this problem tho lol

[maiden ocean]

i mean like they kind of are but its overkill and its not helpful if we dump all these defn on him

@barren sierra ok lets say x = 3, y = 2

then x ~ y right?

can u find an a, b in R such that (3 + a) - (2 + b) is not in Z?

[barren sierra]

yes but that doesn't mean that a-b is in Z

[maiden ocean]

oh is that your defn of ring congruence

[thorn delta]

it fails for multiplication

[barren sierra]

x ~ y and a ~ b implies ax ~ by and x + a ~ y + b

[maiden ocean]

oh ok

[barren sierra]

that's a ring congruence

[maiden ocean]

then yeah 3a - 2b is the easier case

[barren sierra]

oh wait

oh duh that's easy

I was trying to use x + a ~ y + b

[maiden ocean]

mhm

[barren sierra]

ok that's not bad

ok time to figure out what ideals and normal subgroups and cosets are

[maiden ocean]

ideals r just subgroups that basically absorb the other elements by multiplication

[thorn delta]

u can think of all of these relations as partitions induced by what are called "cosets" of a ring/group.

[barren sierra]

@chilly ocean intro to proofs 💀

[cinder bone]

one last q friends

what’s a good way to show that for some H normal in G and K that’s a subgroup of G, H is also normal in HK?

I’ve seen some examples online that use subgroup generator [H,K]

but I’ve never dealt with generating groups from two elements before

[thorn delta]

H normal in G implies H is normal in every subgroup of G tho?

[cinder bone]

ah so just show HK is a subgroup of G?

actually proving the second isomorphism theorem but I did all the other work

just got this part left at the beginning

[thorn delta]

the key fact is that HK = KH whenever H is normal in G

[cinder bone]

im proving hk is a sugroup rn

oh i see thanks

[thorn delta]

np

[leaden finch]

can someone help me with latex format?

i need help inserting an imaged into my doc but its not displaying properly

[thorn delta]

wrong channel, but you can ping me in #bots if you have a specific question

[vital quail]

lol

[latent anvil]

Is this my first post?

lol I just did not get helped huh

[golden pasture]

what was my first post🤔

[latent anvil]

I think maybe someone told me to post on hopf lol. That might have been later tho

[golden pasture]

lol my first post was i use latex for notes

[coarse storm]

Latex notes. Less prone to water damage than paper notes!

[golden pasture]

[woven obsidian]

I want to find the resultant of two polynomials f=xu(x,y,z)+Y^rg(Y,Z) and l=X. With respect to X as a variable

I'm pretty sure the resultant is just Y^r g(Y,Z)

Is it true in general that the resultant is closed under "linear" combinations. R(f,l)=R(f+lh,l) if deg(h) < deg(f)

[golden pasture]

i dont think so? ill need to think of some examples but resultants are so painful to compute

[woven obsidian]

Yeah they really are

Well ok in this case the resultant was triangular so it wasn't that hard

Except remembering the definition

[golden pasture]

oh cool

[snow cliff]

I just started learning about euclidien domains and was just wondering about terminology. If a ring is a Euclidean domain and two elements (say a and b) of the ring have a gcd of 1, is it still valid to call them relatively prime?

[twilit pawn]

yeah, and this coincides with notion of coprime ideals in a ring: https://en.m.wikipedia.org/wiki/Coprime_integers#Coprimality_in_ring_ideals

in any euclidean domain you get bezout’s identity, so a, b in R are coprime (meaning their gcd is 1) if and only if (a) and (b) are coprime ideals

actually it’s true in PIDs too just clicking through some wiki articles here

[snow cliff]

Ty

[marsh fractal]

the only ideals of a finite field are {0} and the entire finite field correct?

[maiden ocean]

of any field

[marsh fractal]

(the same as the infinite field)

[maiden ocean]

yes

[marsh fractal]

ok cool great thanks

another mini question

are the units of $Z[\sqrt{-2}]$ just 1 and -1?

[cloud walrus]

ethane:

[maiden ocean]

uhhh lets find out

let a + bsqrt(2) be invertible so we have some a', b' with (a + b sqrt(2))(a' + b'sqrt(2)) = aa' + (b + b')sqrt(2) + 2 b b' = 1

[marsh fractal]

yeah thats the line im on now

so you gotta be able to cancel out the sqrt{-2} term

[maiden ocean]

then b = -b' so 2b b' = -4b and we have a a' - 4b = 1

[marsh fractal]

so you need (a + b sqrt(2)) (a -b sqrt(2))

[maiden ocean]

wait lol i multiplied that wrong

im silly

[marsh fractal]

nps haha

so then i end up with a^2 - 2b^2 = 1

a solution is a = +-1 and b = 0

but im not sure of any others

[maiden ocean]

(a + b sqrt(2))(a' + b'sqrt(2)) = aa' + (a'b + ab')sqrt(2) + 2 b b' = 1

[marsh fractal]

ok thats not a bad one

[maiden ocean]

so we have the system a a' + b b' = 1 and a'b + ab' = 0

[marsh fractal]

so a'b = ab'

yea

[maiden ocean]

negative ab' but yea

[marsh fractal]

yes

[maiden ocean]

i think from herei ts just algebra

like elementary algebra

[marsh fractal]

ok

just subbing around now that we have this fact

[next obsidian]

The units are just 1 and -1

And this is actually easier to do

Use the normal norm on C

Like sqrt(a^2 + b^2)

Where your point is z = a + bi

[marsh fractal]

ahhh

[next obsidian]

Then show this is multiplicative

Then show anything which isn’t 1 or -1 has norm > 1

So it’s inverse has norm < 1, which doesn’t exist

If you want

You can just go to C

And do stuff there

And you probably don’t need to justify

Then show this is multiplicative

So basically the trick here is

If you know what a field of fractions is

You basically want to pass to that, this isn’t quite true since Z[sqrt(-2)]’s field of fractions is smaller than C

But if you have an integral domain you can just pretend fractions exist to work in a field

And if you’re careful you can work there, then basically clear denominators to go back to your integral domain

And the results still@hold

[marsh fractal]

ok im not aware of field of fractions

but if i can hop in here

[next obsidian]

It’s like

You know how Q is defined from Z?

Just fractions

[marsh fractal]

yeah

[next obsidian]

You can do that in an integral domain

Always

Same rules for addition and multiplication

Then you get a field which your ring “lives inside of”

So you can divide at will

Don’t worry too much about it right now I think

[marsh fractal]

no i would like to understand it

[next obsidian]

You can just do what I described, or follow through on what Moth was saying

[maiden ocean]

[next obsidian]

You can probably jump forward in your textbook and read about a topic called “localization”

[marsh fractal]

ok i have heard of that yes

[next obsidian]

This is like where you add only some fractions

Basically you symbolically add inverses to some stuff

By just adding in 1/x for some x’s

The field of fractions is where you do this for all x which aren’t 0

[maiden ocean]

simple problem about units in Z[sqrt(2)]
impulse is to start explaining localization
this is your brain on AG

[marsh fractal]

ok so how would that explain how the units of Z[sqrt(-1)] are 1,-1,i,-i ?

[next obsidian]

Same idea

Use the norm idea

[marsh fractal]

does the field of fractions come into that too

[next obsidian]

I mean essentially

The idea here is

Let’s say ab = 1

The norm of 1 is 1

So the norm of a is 1/norm of b

This really is just intuition from working in C

[marsh fractal]

yes and it works with i in the same way

[next obsidian]

Yeah pretty much

[marsh fractal]

my complex numbers intuition is pretty bad

[next obsidian]

I mean you can even pretend it’s R

The idea is just

[marsh fractal]

ive done like a decent bit of maths but almost nothing with complex numbers lol

[next obsidian]

If a is bigger than 1

Then it’s inverse is smaller than 1

That’s literally all you’re saying

And in both of these extensions of Z

It’s easy to show if your norm is < 1 it must be 0

[marsh fractal]

ah

[next obsidian]

Since your elements look like a + bi where a,b >= 1

[marsh fractal]

thats actually pretty nice

[next obsidian]

This won’t always work tho

Like imagine Z[1/2i]

Then you can have something with norm 1/4

[marsh fractal]

yeah

[next obsidian]

And in fact 1/2i is invertible!

[marsh fractal]

ok that makes the sense

ty

to both !

[next obsidian]

👍

[next obsidian]

Okay but listen. Yes

[marsh fractal]

what is AG?

your brain on AG

[next obsidian]

Algebraic geometry haha

[marsh fractal]

ah ok haha

[next obsidian]

They’re just poking fun at me

[marsh fractal]

yeah gotchu

i guess localisation is a big concept in AG

or often used or whatnot

[next obsidian]

Absolutely

It’s also important outside of it but it’s kind of a meme that everything to me is AG and I do weird things sometimes because of it

[bronze trench]

noncommutative localization is cooler

[maiden ocean]

oh jesus fuck another one

[bronze trench]

but see, noncommutative means no AG

[maiden ocean]

this is not better

[bronze trench]

AG can die, algebra is cool just for being algebra 😂

[marsh fractal]

maths holy wars

[bronze trench]

as a professor of mine once said, commutativity is too artificial. Try and put your socks on after your shoes

life is noncommutative 😂

[maiden ocean]

if i enjoyed life i wouldnt be doing math

[next obsidian]

Okay but look

If you can’t put your socks on after your shoes how can you possibly deal with schemes and memes

[bronze trench]

also the biggest counter to commutativity is matrices 😂

[next obsidian]

Matrices are dumb anyway

[maiden ocean]

working with matrices is lame

[next obsidian]

Uhhh, fix a basis oooooo

[chilly ocean]

matrices commute additively

[maiden ocean]

who's like "ahh yes i really wish i was working in SO(n)" or whatever

[bronze trench]

try composing linear transformations additively lol

life if matrices commuted
insert that cool post scarcity society pic meme

too lazy to go find that sorry guys

[next obsidian]

I gotchu

[bronze trench]

as with everything in math, just imagine it

yeah ok that works thanks xD

anyway popped in just to say I'm in knee deep trouble as I have another GT homework sheet and haven't been paying attention to class lol

fml

[maiden ocean]

is GT geometric topology

[bronze trench]

group theory 😂

[maiden ocean]

sad!

idk why i didnt think of that

[bronze trench]

what even is geometric topology? isn't that like... topology? xD

[maiden ocean]

manifolds

basically

[bronze trench]

soooo... just geometry?

[viscid pewter]

pain

[maiden ocean]

dan

[bronze trench]

I'm getting a bit tired of this GT class to be honest, it's just a bunch of nonsense. I'm definitely an algebra guy, but more like rings and algebras and shit

GT is pain for me rn

[next obsidian]

Rip

What are you covering?

[bronze trench]

ok this second exercise sheet covers Hall subgroups and \pi-groups, nilpotency, all the fancy series you can do (derived, composition, upper and lower central), frattini subgroup

and for some weird reason, a little linear algebra (lol)

[next obsidian]

Oh nice

Those are interesting

I mean if you like group theory lol

[bronze trench]

well I realised I don't particularly like this lol

[sturdy marsh]

The bit of linear algebra might be to study lie algebras

and to do Lie's and Engel's theorems

on solvable and nilpotent lie algebras

[bronze trench]

yeah because I recognize it from my lie algebra course

yet we're finished with the linear algebra thing in our GT class and we won't study lie algebras here

it's a mystery as to why we did that detour in linear algebra lol

[snow flint]

im standing in fromt of the elevator rn

[bronze trench]

but one of the hw problems is basically prove the jordan-chevalley decomposition for matrices over an algebraic closed field

[sturdy marsh]

Spending time on linear algebra is normally a good investment I think

[bronze trench]

and I can't remember that from our lie algebra course and I feel that going back on the notes and checking is cheating :/

well yeah in undergrad. I feel the undergrad courses in my uni are shit so they are teaching us linear algebra we should have learnt then, now in the masters

[sturdy marsh]

the proof isnt bad at all

you should be fine

I miss doing algebra lol

[sturdy marsh]

Huh, just looked up the Jordan-Chevalley decomposition

Apparently it can be done over a perfect field

[bronze trench]

yep my prof mentioned that in the lie algebra course

just in passing tho, we always worked over the complex numbers I think

so algebraically closed, which is cooler 😂

[sturdy marsh]

the proof for the perfect field case is pretty cool

and surprisingly simple

modulo stuff about algebras over fields

[barren sierra]

what is a kernel?

I don't quite get it

just 0?

ok so the answer is integer multiples of pi

[stone fulcrum]

Yeah yeah the integer multiples of π

Is the kernel

[barren sierra]

this is the definition given

it doesn't mention anything about the identity value

[stone fulcrum]

Oh

[barren sierra]

the kernel is an equivalence relation

soo is it just

for all r in R, r ker(sin) r + 2pi?

[chilly ocean]

[stone fulcrum]

Any two points are equivalent if they are equal after sin is applied

[barren sierra]

ok so

that second answer

got it

[stone fulcrum]

That's better stated as a fiber lol

[barren sierra]

man this midterm is gonna murder me

we stan inconsistent naming 🥳

[scarlet estuary]

category theorists fuming rn

[stone fulcrum]

Wait, it's not so simple

There's two points on the unit circle with the same sin. These will be equivalent

[barren sierra]

it isn't because I know you have symmetry around other points but that's way too hard to figure out

[stone fulcrum]

That is, x Ker(sin) (π - x)

[barren sierra]

wait

thats just it?

[fair shard]

I'm learning algebra from an introductory level textbook and watching michael penn's youtube series just to reinforce some of it, how does that compare to a real first semester course in algebra?

[stone fulcrum]

Than any point is equivalent to any coterminal point ofc

Yeah that's it

[barren sierra]

hmmm

I asked this on class forum but no one responded

[scarlet estuary]

@fair shard its hard to say since "real first semester course in algebra" as a description varies heavily

many first semester algebra courses never get past groups, others go all the way to galois stuff

[fair shard]

oh yeah

[scarlet estuary]

in general though, as long as youre actually doing the problems from the textbook

[fair shard]

this book does groups, rings, vector spaces, modules, fields, galois theory

[scarlet estuary]

you should be getting a good education

[fair shard]

yes this textbook is awesome it's like 30% lesson 70% proof exercisss

[stone fulcrum]

What is it?

It's K I did too

[fair shard]

"a book of abstract algebra" charles c pinter and I'm also looking at one by dummit and someone else

the dummit one is far more advanced and I'm gonna study it I think next year when I finish this book

[stone fulcrum]

Everyone talks about dummit, everyone forgets foote

[fair shard]

is that the co author?

hahaha

[stone fulcrum]

Dummit and whoever is a famous book is all

But yeah, advanced and dry. See whatever else you can sink into

[fair shard]

what topics are usually involved in a course called "commutative algebra"

[stone fulcrum]

It's good to have multiple sources though, always

Comm alg is the study of commutative rings.

But, Jesus does this study go deep

Rings are an interesting object

[fair shard]

that's interesting

[solemn rain]

@fair shard try reading atiyah macdonald

do all exercises

hard

fun

gl hf

[fair shard]

I haven't grown an apprentice rings yet

[solemn rain]

but dont do my mistake

[stone fulcrum]

Gotta get through your abs alg class first lol

[solemn rain]

and make sure to know your topology

point-set topology

[fair shard]

they seem like the awkward middle sibling between fields and groups

[solemn rain]

oh u dont know algebra?

[stone fulcrum]

They are and that's a good argument for why they are difficult to pin down

[fair shard]

yeah currently my algebra sucks and my topology too

[stone fulcrum]

Not having multiplicative inverses but having the structure of the additive group makes a very weird thing.

[solemn rain]

learn algebra or topology first

the topology you need is like basic like

munkres first 4-3 chaps

but know good algebra

but why do u want to study CA

in the first place?

[fair shard]

I have learned abstract linear algebra that's at least a start

[solemn rain]

yea thats a start

try dummit and foote for algebra or jacobson

[fair shard]

any good references on where to learn topology?

[solemn rain]

munkres

[fair shard]

oh ok

[solemn rain]

im using introduction to topology and modern analysis by simons

and im having extreme fun

check it out

0 prereqs

other thazn like how to prove stuff

[fair shard]

wait r u serious

[solemn rain]

yes

and it teaches you alot of math

like alot

[fair shard]

niceeee

[solemn rain]

it starts with set theory

[fair shard]

how fast of a read is it?

[solemn rain]

for me not at all

[fair shard]

I usually rip through textbooks

[solemn rain]

as it has more prose than the usual textbooks i read

[stone fulcrum]

I liked "topology without tears" throw that on the pile

[solemn rain]

but idk maybe im slow

[fair shard]

well I think I have a problem with thoroughness

[stone fulcrum]

I am very slow with textbooks. That's both good and bad

[solemn rain]

why is that bad

kaaynex

[stone fulcrum]

I can take a long time to get any math in! Also it can be boring.

[solemn rain]

yea ig it can be boring but i mean

not bad tho ig

[stone fulcrum]

Intro courses can have that problem

[solemn rain]

yea

i get u

[fair shard]

I have lots of 300 page textbooks, like not too huge

[solemn rain]

anyways ethan i recommend ( im bad dont takee my advice haha ) reading simons or munkres ( simons just part 1 the rest is funtional analysis ) and then reading algebra

then ur set to read AM ig

[fair shard]

what's am

[solemn rain]

atiyah macdonald

golden text in CA

[fair shard]

oh ywah

[barren sierra]

quick question for ring homomorphisms do we have to check that distribution holds?

[solemn rain]

yes

[barren sierra]

ok

[solemn rain]

wait whats distribution

u have to check that it preserves both adddition and multiplicative structure

thats iit

if thats what you meant

[barren sierra]

so you check the function is a group homomorphism for addition, monoid homomorphism for multiplication, and then also check distribution holds

uh

[solemn rain]

f(a+b) = f(a)+f(b) and f(ab) = f(a)f(b)

check those

[solemn rain]

okaay

[barren sierra]

so I don't need to check distribution?

[solemn rain]

no

[barren sierra]

ok

and then how do I check this descent holds

is it just this?

[fair shard]

when studying rings is there ever such thing as line a = b (mod c) where they are all polynomials

I feel like that's obvious sorry

[delicate bloom]

not sure what you're asking

[barren sierra]

<@&286206848099549185> for my question. I put the picture of the question and then my answer for why the f~ descends to f but I want to know if that reasoning is sound

[stone fulcrum]

Yes, mod is really just a quotient in disguise and all rings have them

[barren sierra]

so my solution works as to why it descends?

all rings have them
@stone fulcrum
All rings have what? Quotients? what does that tell us though

[stone fulcrum]

@barren sierra
No your thing isn't a mod

The mapping is f(x) = (x,x)

Which is weird, if the domain and codomain aren't equal

[barren sierra]

oh you were answering Ethan

my bad

[stone fulcrum]

So honestly I don't know what to say about your question

[barren sierra]

damn :/

this midterm bout to murder me ☠️

this is what i get for taking honors section

yea so you want to show that

does my answer show this tho

cause I assume left hand side and then wrote why the right hand side is true I think

ok

[next obsidian]

Uhhh

...

If that’s a problem on your midterm

You shouldn’t be asking for help

Id you’re saying your upcoming midterm is gonna murder you

Then that’s okay

[fair shard]

trying to prove that there is a bijection from cosets of the centralizer of a to the set of conjugates of a

a proof is also not all I'm looking for, I want intuition for this

[thorn delta]

im guessing u haven't seen group actions?

[fair shard]

I have

but I'm confused by them

there should be a pretty simple proof

[chilly ocean]

maybe something like "each conjugate gag^{-1} of a corresponds to a coset"

kxrider knows more than me tho so im not gonna butt in

[maiden ocean]

TTerra moment

[chilly ocean]

no i just think they'll be able to provide a better intuition

[maiden ocean]

TTerra moment

[thorn delta]

in general, if you have a group action, the cardinality of the orbit of an element is the same as the order of the corresponding stabilizer subgroup

im guessing u haven't seen that?

[fair shard]

yes I have

just gotta think about that for a sec

[thorn delta]

when G acts on itself by conjugation the stabilizer subgroups are the centralizers and the orbits are the conjugacy classes

[fair shard]

why should that be the case?

the first thing?

[thorn delta]

if you have a group action, the cardinality of the orbit of an element is the same as the order of the corresponding stabilizer subgroup
this?

[fair shard]

yes

I'm trying to think about that

it isn't obvious to me

trying to prove that first statement rnt

that will give me intuition

maybe I'm slow and stupid

but elements of the orbit seem like completely opposite of elements of the stabilizer subgroup

so idk why they should have the same #

[carmine fossil]

Note that this map is well defined (If g.a=h.a, h^-1g is in C(a),implying g C(a)=h C(a) ) and surjective as well as injective

[fair shard]

unless it's cosets of the stabilizer subgroup

then it makes sense!

[thorn delta]

oh, umm yea i said something wrong my bad.

if you have a group action, the cardinality of the orbit of an element is the same as the order of the corresponding stabilizer subgroup
the index of the stabilizer group in G has the same cardinality of the corresponding orbit

[fair shard]

yes totally

[thorn delta]

that should make more sense haha

[fair shard]

yes

I was so confused

but I get it :)

so the proof is ✓

but now

I'm thinking more

forget actions

the cosets of c(a) are just c(a)x

so is it the case that xax^-1 = yay^-1 for all y in c(a)x?

sorry I bet this stuff is so elementary, I studied this stuff over the summer so stuff is familiar to me but I forgot most of it and I'm trying to relearn for real this time

[thorn delta]

this is true. xax^{-1} = a = yay^{-1}

[chilly ocean]

xanax = yay?

[thorn delta]

ik u said forget about group actions, but the elements of centralizer fix a by the action of conjugation. thats legitimately how i prefer to think about lol

[fair shard]

yeah that makes sense

so it's like

the set of all conjugates of a

are the left and right cosets of c(a) the same?

it seems to be that right cosets make more sense to me

for c(a)

[thorn delta]

why is that?

[fair shard]

here is why

[thorn delta]

I don't think it really matters, but if you think of conjugation as g.x = gxg^{-1} then it seems like left cosets would make more sense (because you have a left action).

[fair shard]

I meant left

I was thinking more like

if xax^-1=y, then you can find other group elements who conjugate a to get y by just multiplying x on the left with any element of the centralizer of a