groups-rings-fields

[delicate orchid]

yeah by the character table it's simple, I believe it

[uncut girder]

Its simple

Thanks wew for saving me

[delicate orchid]

no problem boss

[uncut girder]

I still need a torsion free normal subgroup of PSL(2,Z) such that the quotient is isomorphic to Aff(F_7)

[delicate orchid]

very wacky request lol

[latent anvil]

What is Aff(F_7)? Polynomials ax+b for a, b in F7 under composition? @uncut girder

[delicate orchid]

they're usually written as matricies

[latent anvil]

sure

[uncut girder]

Affine transformations of the 1 dim vectorspace F_7

[delicate orchid]

they're the 1 dimensional affine transformations right

yeah ok then they're just those polynomials

[uncut girder]

Yeah basically.

[latent anvil]

I just thought this was the most compact way to write affine transformations lol

Okay thinking

[uncut girder]

Its order is 7*6. You have 7 translations and 6 rotations (6 elements of the multiplicative group), and compositions of them all

[delicate orchid]

hmmmm

scouring the wikipedia page on affine transformations it says that Aff_n(F) is the semidirect product of F^n and GL(n, F)

might be able to do some meme there

[latent anvil]

I don't think it's a hugely different perspective

wait isn't PSL(2, Z) the free product of C2 and C3 or something?

Yes

so defining a map from it into Aff should actually be easy

you just use the presentation <x, y | x^2, y^3>

[uncut girder]

It just so happens that PSL(2,3) is isomorphic to Aff(F_4). So letting Gamma(3) denote the kernel of PSL(2,Z) -> PSL(2,3) given by reduction mod 3, which is a surjective group homo, we see PSL(2,Z)/Gamma(3) is isomorphic to Aff(F_4)

[latent anvil]

so "Aff(F7) is a quotient of PSL(2,Z)" really means "Aff(F7) is generated by an element of order 2 and an element of order 3"

and this is crunchable

sound good?

[delicate orchid]

that's a super cool way of doing it

[uncut girder]

Yes PSL(2,Z) is a free product of Z/2Z and Z/3Z

[latent anvil]

so the elements of order 2 are f(x) = b - x for any b

a(a(ax + b) + b) + b = a^3 x + a^2 b + a b + b

so f(x) = ax + b satisfies f^3 = id iff a^3 = 1 and a^2 b + a b + b = 0

the third roots of unity mod 7 are 1, 2, 4

if a = 1 then b + b + b = 0, so b = 0, so we're actually f(x) = x

oh and a^2 b + ab + b = (a^2 + a + 1)b

so if a is a nontrivial 3rd root of unity then that's automatically zero

so the elements of order 3 are f(x) = 2x + b and g(x) = 4x + b for b in F7

so maybe f(x) = 1 - x and g(x) = 2x generate the group of affine transformations? just writing out relations we get 1 - 2x = 8 - 2 x = 2(4 - x)

[delicate orchid]

both 2 and 4 are also solutions to a^2+a+1... spooky

[latent anvil]

so the transformation h(x) = 4 - x is in there

[latent anvil]

yeah, that's bc they're the nontrivial 3rd roots of unity?

x^3 - 1 = (x-1)(x^2 + x + 1)

[delicate orchid]

veryyy spooky

anyway don't let me interrupt you

[latent anvil]

hmm i'm not sure that 3 x is in there

[uncut girder]

A set of generators of Aff(F_7) are x+1 and 3x (since 3 is a primitive root mod 7)

[latent anvil]

they have the wrong order though

3x is order 6 and x + 1 is order 7

[uncut girder]

Yeah I'm just saying if you can show 3x and x+1 are in the group you are generatign with an element of order 2 and order 3, then you have everything

[latent anvil]

ah yeah good point sorry

so if we can only start with a coefficient of 2 or 4

how could you get any other coefficients?

oh we start with coefficients of 2 and -1

1 - 2x = 1 + 5x

2(1 - 2x) = 2 + 3x

so if we can get h(x) = x - 2 we've got 3x

oh well any x + b for b ≠ 0 is basically the same

4(1 - 2x) = 4 - 8x = 4 - x
1 - (4 - x) = x + 4

Right?

so there we go, they generate the group

[uncut girder]

Yoooooo

[latent anvil]

f(x) = 2x, g(x) = 1 - x.
g(f(f(g(f(g(f(f(g(f(x)))))))))) = x + 1

lol

[uncut girder]

I'm just gonna write this down and double check everything

[uncut girder]

Yep this is right

[tall jay]

Is it true that the kernel of any f is a subgroup?

[latent anvil]

yes

f(e) = e, f(gh) = f(g) f(h) = ee = 1, f(g^-1) = f(g)^-1 = e^-1 = e

[uncut girder]

Yep, a normal subgroup in fact

Okay so to answer my original question

Consider the map PSL(2,Z) -> Aff(F_7) given by sending a to 1-x and b to 2x, where a,b generate SL(2,Z) = <a,b| a^2 =b^3 = 1>.
Let K be the kernel of this surjective group homo (we just showed 2x and 1-x generat Aff(F_7).
Then PSL(2,Z)/K is isomorphic to Aff(F_7)

Thanks so much shamrock and wewlads!

[latent anvil]

Np!

[delicate orchid]

no worries but I really didn't do anything

[latent anvil]

Wait what was the torsion free condition on the subgroup?

I don't know if it's obvious that K is torsionfree, but also this is basically the only way to do this problem right?

[tall jay]

Even if f is not a homomorphism?

[latent anvil]

Oh

In that case no

We don't usually use the term kernel unless f is a homomorphism

[uncut girder]

f needs to be a homomorphism

Hm

[uncut girder]

Can we explicitly describe the kernel of this map?

[delicate orchid]

I'm trying to come up with some presentation for K to try and see the torsion group

[tall jay]

For it to be just a subgroup? Not necessarily normal?

[delicate orchid]

the kernel of any homomorphism is always normal

[latent anvil]

you can't say anything really in this generality

When the map has no structure

it could be a normal subgroup or a subgroup which isn't normal or just an arbitrary subset

If S is a subset of G then we can define a function f : G -> Z/2Z by f(x) = 0 if x in S and f(x) = 1 if x is not in S. Then the "kernel" of f is S

[uncut girder]

One idea is to draw a binary tree, root node is identity, going left is applying 1-x, going right is applying 2x. Then see which left right paths give you identity

[delicate orchid]

good idea but I have no paper to hand

[latent anvil]

So we have a normal form for elements of PSL(2,Z)

an alternating product of a and either b or b^2

So if f(x) = 1 - x and g(x) = 2x, we want to understand when a product of f and either g or g^2 is the identity

[uncut girder]

I'm gonna work out my tree idea

[latent anvil]

wait what are the torsion elements of PSL?

Isn't it just like e, a, b, b^2?

[delicate orchid]

yeah every other product is non-trivial

due to the nature of it being the free product

[latent anvil]

well you need to worry about cancelation at the endpoints

[delicate orchid]

true

[latent anvil]

if x = aba then x^3 = ab^3a = e

So no it's not just those

Maybe it's better to look at matrices

[delicate orchid]

ha guess what I was about to post lol

I found that fg^3f(x) was the identity

turns out that's true even before you quotient (i.e. useless)

[latent anvil]

Right

That's not a problem though, right?

[delicate orchid]

no not at all

it's just not helpful

chatters I have located the character table

[latent anvil]

So if w^k = 0, can we say w = ava or w = bvb^2 or w = b^2vb?

Like it's a conjugation

[delicate orchid]

uhhh

[latent anvil]

Like otherwise stuff won't cancel

[delicate orchid]

yes yes

yeah that's a result in the free group so it's definitely true in a free product

[latent anvil]

I think this is true

So then

v is also torsion

So the torsion elements are those of the form x a x^-1 or x b x^-1 or x b^2 x^-1

And these won't be trivial in the image

Bc they're conjugations of the nontrivial elements f, g

@uncut girder I think I have a proof sketch

[uncut girder]

I'm listening

[latent anvil]

So if a word is torsion, it has to either

1. End and start with a
2. End with b and start with b^2
3. End with b^2 and start with b

Do you agree?

Otherwise no cancelation can occur

This is if we write the word as an alternating product of a and b or b^2

(which is a unique representation)

[uncut girder]

Yep that's clear to me

[latent anvil]

So iterating this

[pastel cliff]

$P = {r+s\pi : r,s \in \mathbb{Q}}$ this isnt closed under multiplication is it (sorry to interject, just wanted super quick confirmation)

[latent anvil]

A torsion word is of the form x a x^-1, x b x^-1, or x b^2 x^-1

Just keep stripping off these matching endpoints

Right?

[cloud walrus]

μ₂ (46/47 🪲)

[uncut girder]

I'm not sure i understand

[latent anvil]

So you start with a torsion element w

[uncut girder]

No its not. Consider pi^2

[latent anvil]

You write it as w = a v a for a shorter word v

Yeah?

[pastel cliff]

thought so, merci

[latent anvil]

If w is torsion, so w^k = e, then v^k = (a w a)^k = a w^k a = a a = e

[uncut girder]

Are you claiming that v is necessarily torsion?

[latent anvil]

yes

it's a conjugate of a torsion element

[uncut girder]

I see

[delicate orchid]

yeah, say w^k = 0, then (ava)^k = av^ka = 0 <=> v^k = 0

[uncut girder]

Got it

[latent anvil]

So repeat this until v is too short to do it anymore

You get that w = x c x^-1 for c in {a,b,b^2}

Maybe you can show this with matrices too idk

Like look at similarity classes of integer matrices which have power I

But this works

[uncut girder]

Right

[latent anvil]

And no conjugate of f, g, g^2 in Aff is the identity

So no torsion element is in the kernel

[uncut girder]

No conjugate of a non identity element of any group is the identity. So does this show that the kernel of any map out of PSL(2,Z) sending a and b to nonidenity elements is torsion free?

[latent anvil]

Yep!

[uncut girder]

Awesome! Thanks again shamrock

[delicate orchid]

ok time to do it by lifting characters

[latent anvil]

What was the plan for that?

[uncut girder]

I want to get generators for this kernel.

[latent anvil]

Woah, if G is a free product of groups Ai then any finite subgroup of G is conjugate to a subgroup of some Ai

I was curious about how stuff generalized

[delicate orchid]

find the conjugacy classes 1-x and 2x get mapped to in the quotient, lift the characters back up and then look at the values that new character takes and maybe find a representative of the conjugacy class that's clearly torsion/not-torsion free good lifting practice ngl

[uncut girder]

I think doing the tree is my best bet. Or writing a sage program to compute it

[latent anvil]

the kernel does seem annoying to think about

I don't have any more tricks

[delicate orchid]

oohhh the group generated by 1-x should be normal isn't that cool

[pastel cliff]

There are infinitely many positive integers $m$ for which $\mathbb{I}_m$ is a domain. quick interjection again but i can prove this really quickly with just integers modded by a prime right

[cloud walrus]

μ₂ (46/47 🪲)

[delicate orchid]

wtf is I_m

[pastel cliff]

Z/pZ is a field, and fields are necessarily also integral domains (textbook uses domain interchangeably with integral domain)

stupid fucking textbook notation

integers mod m

[delicate orchid]

what the fuck

[pastel cliff]

thank you rotman very cool

[delicate orchid]

uh then yes that will work

[latent anvil]

Primes are the only ones that work

[south patrol]

Yes using p is even stronger as they are fields

[uncut girder]

Its true only for m prime

[latent anvil]

If n = ab then a, b will be zero divisors

[delicate orchid]

wait lemme do the proof

nooooooo

[south patrol]

Well also finite integral domains are fields lol

[latent anvil]

Sorry!

[uncut girder]

All finite integral domains are fields

[delicate orchid]

foiled once more

[latent anvil]

Yeah, try to prove that wewlad

[delicate orchid]

trivial ring

[latent anvil]

Not an integral domain!

[pastel cliff]

[delicate orchid]

there are no non-zero elements :troll: how can they multiply to give 0 :troll:

[latent anvil]

The 0 ideal is not prime bc it's not a proper ideal

[delicate orchid]

the ideal 0 is prime

[latent anvil]

Did yall know that finite division rings are cool?

i like this theorem

[delicate orchid]

I know zero non-commutative algebra :(
or algebra in general, I just divine properties from funny character tables - like a wizard reading a spell book
"mmm yess the first column is all 1s this group is abelian yes mmm"

[latent anvil]

Wair

Fuck

Finite division rings are fields

I just skipped ahead to "this is cool"

I should review character theory

[delicate orchid]

I need to learn more

[pastel cliff]

cringe rotman

[delicate orchid]

it's my specialty and I still know nothing

[latent anvil]

I'm taking quals in 4 months and I need to relearn complex analysis and group theory and galois theory and smooth manifolds stuff lol

[delicate orchid]

tfw never learnt complex analysis or galois theory

[pastel cliff]

mood

[delicate orchid]

and I barely even know what a manifold is

[latent anvil]

That is so fucked up μ2

What is rotman on

[delicate orchid]

"locally R^n" yeah cool I guess

I've seen \mathbb{I} to mean "the largest ideal that generates a given variety"

[latent anvil]

huh

I haven't seen that

[delicate orchid]

yeah so like

[latent anvil]

Just I(X)

[delicate orchid]

we'd write the weak nullstellensatz as $\mathbb{I}(\mathbb{V}(I)) = \sqrt{I}$

[latent anvil]

Wild

[cloud walrus]

Wew "Dreaming Discs" Tbh 💿

[latent anvil]

Actually that's not so bad

I didn't realize it was an operator

[delicate orchid]

I think it was so we can keep using I for ideals

[latent anvil]

Yeah okay I'm on board now lol

[delicate orchid]

yeah I like it too

[uncut girder]

Okay, these are the questions I care about. We have K, which is a torsion free normal subgroup of PSL(2,Z) with quotient Aff(F_7).
What i am interested in now is the action of K on the upper half plane via linear fractional transformations.
Question I have is: how many cusps does a this action have (i.e. how many equivalence classes of P^1(Q) are there modulo action of K?). What is the genus of the riemann surface given by quotienting out the upper half plane by the action of K? I want the answers to be 7 and 1 respectively

[latent anvil]

Yikes

[uncut girder]

Also, something that might be helpful is answering the following question: is K a congruence subgroup of PSL(2,Z)? Meaning does K contain Gamma(N) for some N? Where Gamma(N) is the kernel of the surjective homomorphism PSL(2,Z) -> PSL(2,Z/N) given by reducing mod N.

[uncut girder]

Yeah I have a big plan. These are all steps in my plan ◉◡◉

[delicate orchid]

horrifying

[latent anvil]

https://youtu.be/lxZpEFJhO6k

[latent anvil]

Okay so my first thought is N=7 lol

This would mean the map into Aff factors through PSL(2,Z/7Z)

Right?

hmm

[uncut girder]

I think N=7 doesn't work because of of the reason wew pointed out earlier. PSL(2,7) is supposedly a simple group, meaning it has no normal subgroups so it can't have Aff(F_7) as a quotient

[latent anvil]

Do the groups Γ(N) have easy generators?

ahh right ofc

[delicate orchid]

49 !

[uncut girder]

It does have Aff(F_7) as a subgroup though, which is misleading lol

[latent anvil]

If they're always finite simple groups...

Oh

For N prime

Mb

[uncut girder]

Good point

So it can't be a congruence subgroup of level p for any prime p

[uncut girder]

BTW this N is called the level

[delicate orchid]

the fact that you want K to contain gamma(N) rather than the other way around kinda makes me think that we're stuck

unless I'm thinking backways

[uncut girder]

Okay I've got to go for a while. Thanks for the discussion. This was really helpful. I might be back in a bit

[latent anvil]

So Γ(N) is stuff of the form
1 + a |b
c |1 + d
For a, b, c, d in NZ?

It would be nice if the map into Aff could be described on the matrix level instead of with generators

[uncut girder]

No, it's the matrices that reduce to the identity mod N.

[latent anvil]

The generators are z + 1 and - 1/z right?

Of the modular group?

[latent anvil]

Isn't that the same thing?

[delicate orchid]

yeah I agree with shamrock

[uncut girder]

Oh yeah ur right

[delicate orchid]

lemme tex it

[uncut girder]

Didn't read it correctly

[delicate orchid]

$\Gamma(N) = \left{\begin{pmatrix} 1+a&b \ c & 1+d \end{pmatrix} \colon a, b, c, d \in N\bZ\right}$

first try

[cloud walrus]

Wew "Devastation" Tbh 💀

[latent anvil]

Wait this isn't right z + 1 is infinite order...

Wait no that is the generator

I am confused now

[delicate orchid]

are we talking about generators of PSL(2, Z)?

[latent anvil]

Yes, sorry

The ones of order 2 and 3

[delicate orchid]

nah it's alright

lemme think about the matricies

(-1 1)
(0 1) is order 3 right

[latent anvil]

Ah if S(z) = - 1/z and T(z) = z + 1 then ST has order 3

[delicate orchid]

nah it's order 2

[delicate orchid]

nice

[latent anvil]

So as matrices S is {{0,-1},{1,0}} and T is {{1,1},{0,1}}

[delicate orchid]

ahh I was so close

[latent anvil]

So let R = ST, then T = S R

So if φ is the map into Aff(F7) we have φ(T) = f g

So φ(T)(x) = 1 - 2x

So if we represent everything by matrices

$$\varphi\left(\begin{bmatrix} 1 & 1 \ 0 & 1\end{bmatrix}\right) = \begin{bmatrix} -2 & 1 \ 0 & 1\end{bmatrix}$$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

$$\varphi\left(\begin{bmatrix} 0 & -1 \ 1 & 0\end{bmatrix}\right) = \begin{bmatrix} 2 & 0 \ 0 & 1\end{bmatrix}$$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

nothing is coming to mind lol

[delicate orchid]

fixed subspace detected

[delicate orchid]

ok so if we can decompose these into the products of those generators you gave that might give us something

yeah cause then if the image of those generators is zero we have that gamma(N) is in K right?

[uncut girder]

Let me just tell you where I'm going with all this

[latent anvil]

well the image of the generators is f and fg

which are not zeor

[latent anvil]

Pretty

[uncut girder]

This is a fundamental domain of Gamma(3)

[delicate orchid]

no no I mean, those matrices will be the product of some generators, and when that product is parsed through the homomorphism it might go to zero
it was just an idea

[uncut girder]

Also recall that PSL(2,Z)/Gamma(3) is isomorphic to Aff(F_4), which is the automorphism group of the tetrahedron

[uncut girder]

I've colored in parts of the fundamental domain of Gamma(3) so that the colors correspond to faces of a tetrahedron

There are 4 colors, which are the 4 triangular faces. All 4 faces are adjacent of each of the 3 others. This fundamental domain when you identify sides, becomes a sphere with 4 colors, ie a tentrahedron

You identity the left edge with the right edge, the green half circle with the other green half circle, and the blue half circle with the other blue half circle

Can you see how that's a sphere?

Well, with 4 cusps

Idl if any of this is making sense to you. But anyway I want to do this same thing for the heawood map, which is a seven color map on the torus

This one

[delicate orchid]

google chrome

[uncut girder]

The automorphism group of this heawood map is Aff(F_7), which is why I wanted to find a torsion free normal subgroup of PSL(2,Z) with this quotient

Thanks to you, I have such a subgroup, namely K, a certain kernel.

Now I want to plot the fundamental domain for K and actually verify that it has 7 cusps and the quotient surface is genus 1 (a torus)

Here's a question that will aid in plotting the fundamental domain of K in the upper half plane: whats a set of coset representatives for K inside PSL(2,Z)?

(If I have a list of coset representatives, then I can plot fundamnetal domain for K just by taking a union of fundamental domains for PSL(2,Z) transformed by the coset reps)

[delicate orchid]

wait I can actually do this

coset enumeration

wait NOOOOO I need a presentation of Aff(F_7) ok ok that can be arranged

[uncut girder]

I will dissappear now, but this is the question I leave you with

[delicate orchid]

so annoying, coset enumeration would produce exactly what we need but it's circular, we'd need to understand the kernel better for me to be able to do it

specifically I'd need to have a complete representation of Aff(F_7) in terms of the same generators we used for PSL(2, Z) and then I can get you a schreier transversal of the cosets

@uncut girder I know you're gone but if you can get back to me with the above ^ that would be great

I'm too tired to give it a crack myself

[uncut girder]

Hm

[uncut girder]

I see. This would be helpful to know

I'm just gonna keep making my tree

Once I get 7*6 distict elements in my tree, I have representatives for every coset

[uncut girder]

Here's a relation: g(f(g(f(g(f(g(f(g(f(g(f(x))))))))))))=x
where f(x)=1-x and g(x)=2x.
That is, (gf)^6=id in Aff(F_7)

[next obsidian]

:)

[latent anvil]

Is this the only relation you need?

[bleak abyss]

Sadly no, I had passed out for a while lol

I'll think some more though

[barren sierra]

Weird question

so we have certain finitely generated fields K/F such that

$F = K_0 \subseteq \ldots \subseteq K_n = K$ where $[K_{j}: K_{j - 1}] = 2$

[cloud walrus]

Spamakin🎷

[barren sierra]

My prof invented a term for these "2-radical"

Is there a standard term for these?

[latent anvil]

not that I know, but 2-radical makes sense to me

[barren sierra]

It does to me too

It's not that I dislike the name I was just curious if other names may exist since by his own admission the name is made up by him

[chilly ocean]

why is it called 2-radical

[barren sierra]

cause suppose $K / F$ such that $[K:F] = 2$ then $K = F(\sqrt{d})$ for some $d \in F \setminus F^2$

[cloud walrus]

Spamakin🎷

[barren sierra]

so then each of these subfields is generated by adding new square roots

[chilly ocean]

ih

i only see converse sadly

K:F=2=> |K/F|=2=>?

x in K such that x not in F

and x^2 in F

[barren sierra]

take some $\alpha \in K \setminus F$ so then $F(\alpha) = K$

[cloud walrus]

Spamakin🎷

[barren sierra]

call $f = x^2 + bx + c$ the minimal polynomial of $\alpha$ over $F$

[cloud walrus]

Spamakin🎷

[barren sierra]

Then $d = b^2 - 4c$

[cloud walrus]

Spamakin🎷

[barren sierra]

and $(2\alpha + b)^2 = b^2 - 4c = d$

[cloud walrus]

Spamakin🎷

[barren sierra]

and by construction $2\alpha + b \not\in F$

[cloud walrus]

Spamakin🎷

[chilly ocean]

why though?

[barren sierra]

cause we know $[K : F] = 2$

[cloud walrus]

Spamakin🎷

[latent anvil]

F(alpha) is an intermediate field, right?

and it's not F, because alpha isn't in F

so [F(alpha) : F] > 1

right?

[barren sierra]

yea

[latent anvil]

I'm talking to icup sorry

[barren sierra]

oh right 🤡

[latent anvil]

but 2 = [K : F] = [K : F(alpha)][F(alpha) : F]

so @chilly ocean what can we say about [K : F(alpha)] from this?

[chilly ocean]

oh

it is 1 or 2

[latent anvil]

but it's not 2

[chilly ocean]

because the other one is greater than 1

[latent anvil]

if it were 2 then [F(alpha) : F] would be 1

yep, exactly

[chilly ocean]

so it has to be 1

so alpha is in F

[latent anvil]

I don't get the last conclusion

[chilly ocean]

why is it minimal polynomial of K coefficients though?

[latent anvil]

so we get [K : F(alpha)] = 1 right?

[chilly ocean]

yes

[latent anvil]

this tells us K = F(alpha), right?

[chilly ocean]

yes

i meant alpha has to be in K

because index 1

yea

i intuitively understand why there should be minimal equation of alpha in base field but dont have any exact reasonings

or maybe i do

i definitely do nvm

because adjoining elements means exactly that

they are a root to a polynomial in F coefficients

why does the polynomial have degree less than or equal to index?

i think i know why again

because if [K:F]=n then K/F has n elements. Supposing alpha is not in F then it can have all factors of n besides 1 as possible orders of alpha

and in the case of n=2 it means the only order is 2

So this means that alpha in K/F has order 2

[strong yacht]

The most standard term I've seen is 'field extension of degree 2'

[strong yacht]

Because of this fact here however, they are also known as quadratic extensions. But some may use quadratic extension in special reference to the base field being the rationals.

[barren sierra]

hm these are extensions of degree 2^r

not just 2

[strong yacht]

Sorry, just woke up and did not read what you said correctly at all

[tough raven]

Thanks! This is interesting.
Why does the basis have to consist of irreducibles?

Not sure if these has been answered (or your deadline passed), but
(4) if Hi is the stabiliser of Li in an action τi on a space Vi, H should be the stabiliser of L1 × L2 in the action τ(M)(v1,v2) = ( τ1(M)(v1), τ2(M)(v2) ) on V1 × V2.

I think this does work to turn any set action on a set S into a vector space action, and if a subgroup H is the stabiliser of a subset S', it is also the stabiliser of span(S') in the vector space action.

[bleak abyss]

The thing that basically worries me is that you're in principle only allowed to take finite linear combinations

I haven't thought about whether this causes us problems here

[tough raven]

I don't think it does. What issue are you worried about?

Not getting well-defined linear maps for individual group elements?

Or that it's not a group homomorphism?

[bleak abyss]

Nah it was more the stabilizer side but let me think this out in detail

Ah I guess I was worried about something different but this should work out

What

What's wrong?

[tall jay]

This is a follow-up from a question I have before, but if we don't know that f is a homomorphism, we can't say that ker(f) is a subgroup, right?

[fickle brook]

@bleak abyss @prisma ibex apparently stabilizer doesnt mean pointwise stabilizer!

it means stabilization as a set!

[bleak abyss]

Oh

[fickle brook]

ie the Stab(W) group of all g st W is a g-invariant subspace

also the conj rep isn't faithful

[bleak abyss]

That probably makes life so much easier lmfaooooo

[fickle brook]

[bleak abyss]

Oh tru

lambda I

God wait so this changes the whole game

[fickle brook]

part 5 worked at least!

the one with diag(a,1)

as did 4

[bleak abyss]

So for permutation matrices I think you take the space of vectors the sum of whose entries is 1

Maybe?

[hidden haven]

Ye

[tall jay]

So if we want to prove the theorem

If f:G->H is a homomorphism, then Ker(f) is a normal subgroup of G,

can we just say that Ker(f) is a subgroup of G, because f is a homomorphism (or do I need to spell tht out)? and then from there, just prove that Ker(f) is normal?

[hidden haven]

If you have proven that the kernel of a homomorphism is a subgroup then sure

[fickle brook]

is that even a subspace

cause idt it is

[bleak abyss]

Err, 0

The additive 1

[hidden haven]

Bruh

[tough raven]

The space of fixed points of a subgroup might end up bigger than you expect (e.g. for finite S_n, fixedpoint space of whole group is non-trivial) but stabilisers of spaces generated by a subset of the original basis are the same.

[tough raven]

Oh

[coral shale]

I'm sleepy, when we have group actions, the codomain is a symmetric group right

f : G -> Sym(S)

What's so special about
f : G -> Aut(H)
for some group H

or even f : G -> Aut(G)

[coral shale]

For example, left/right multiplication wouldn't qualify this because e . x neq e for non-identity x

[next obsidian]

This lets you form semi direct products

[coral shale]

I'm tryna figure out this on my own but I'm not getting too far

A point to a useful resource would be appreciated

[barren sierra]

Somehow got an 89 on my algebra midterm

This channel has been mad helpful

Glad I'm taking this grad course but fuck it's hard

[lavish nexus]

doesn't seem to use anything from Lie groups

[pastel cliff]

the center of a group will always be a normal subgroup right?

[chilly ocean]

yes

[untold basin]

Hello, can someone help me to find the center of the dihedral group ?

[pastel cliff]

you know what the center is right

[untold basin]

Yep

[pastel cliff]

do you wanna chance a guess based on that

[untold basin]

I actually wrote by equivalence what does an [element in Z(D_n)] mean

?

[pastel cliff]

idk what you meant there

but D_8 has few enough elements that you can just manually find the center

[limber vale]

How do i prove the second part of proposition 14? For the first part I just used corollary 12.1and split up the sum. For the second part i tried doing the same but i am left with a double sum (If this is at all the correct approach)

\sum_i\sum_j phi(g_i^-1g_j^-1 g g_jg_i)

I think i might have to now sum over g_jg_i instead of the double sum since phi(g_i^-1g_j^-1 g g_jg_i)=0 if the inside is in H

[coral shale]

@next obsidian if u see this. Still lost

[untold basin]

The definition I learnt is that D_n = <sigma, p_n> where sigma is the conjugation application and for all complex z, p_n(z) = e^{i * 2pi/n} * z

So I wrote [x in Z(D_n)] <=> x in D_n and for all d in D_n, dx = xd
x in D_n <=> there exists an m natural >= 1 such that x = a finite product of elements in {sigma, p_n, p_n^{-1}, sigma^{-1}}

[pastel cliff]

huh

center is the subset of all elements that commute

oh but you're doing D_n not D_8 ig

[untold basin]

yes

[next obsidian]

Idk, in any textbook it should probably talk about semi direct products

[kind jungle]

are all Euclidean domain with valuation-> Integral domains?

[lethal dune]

definition of ED required ID

[uncut girder]

I think a presentation of Aff(F_7) is
<p,q | p^7 = q^6 = 1, qp = p^3q>

p(x)=x+1, q(x)=3x

[upper cape]

Could anyone help me understand what the question is asking?

I assume it is just going to be showing that some diagram is commutative

but I can't figure out which diagram

[chilly ocean]

you didn't post the full question

[uncut girder]

Writing f(x) = 1-x and g(x) = 2x, we have
p = gfggf and q = fgfggfgg

[upper cape]

oops here's all of it

[chilly ocean]

Idk what you are supposed to do

all id know is that you probably use universal property of tensor and then product

so you are trying to show there is isomorphism between W and V* such that V* -> W -> V tensor W -> k = W -> V* -> V tensor V* -> k ?

maybe?

[uncut girder]

@delicate orchid I think this is a presentation of Aff(F_7)

[delicate orchid]

What was the other group again? PSL(2, Z)?

[delicate orchid]

sweet jesus

here we go

[barren sierra]

💀 is there any fast way to find this polynomial

[delicate orchid]

ok doing this manually will not work

GAP time

[barren sierra]

yea I wolfram alpha'ed this

but I was just wondering if there was some trick

[delicate orchid]

I was not referring to your galois theory problem

[barren sierra]

to do this nicely by hand

oh

[delicate orchid]

I was referring to that absolute MONSTER of a coset enumeration

[barren sierra]

this also looks awful

yea

[delicate orchid]

oh no no no wait a second pepeLaugh oh no no no the cosets they are not finite oh no no no no

[barren sierra]

[strong yacht]

I think there is a very small saving you can make, by reducing the polynomial mod 2 ?

@barren sierra

So you don't have to expand out terms that are a multiple of 2 e.g. 2x^2

And irreducible over a finite field iff irreducible over Z (if the coef of the largest power is not a multiple of char of finite field)

Having said that i don't think the direct way is not so bad either (Eisenstein is immediate)

[barren sierra]

I meant like finding the polynomial

once you find the polynomial then yea rational root or eisenstein is immediate

[barren sierra]

hm ya I guess it isn't that much

[strong yacht]

Btw wdym by rational root exactly

[barren sierra]

rational root theorem

[strong yacht]

Right but this checks if you have a root or not

What if there was a factorisation into irreducible polys that weren't linear?

[barren sierra]

oh

ok then yea Eisenstein ig

[strong yacht]

Although rational root is a good if you need to check a cubic since one of the factors would have to be linear if it were not irreducible

Good to keep in mind

[barren sierra]

yea

[pastel cliff]

if i want to show that Z/9Z is not isomorphic to Z/3Z x Z/3Z should i a) show that there are elements of different order or b) make multiplication tables for them

ig the second is a longer version of the first but im really asking like

in general, what's a good way to disprove the existence of an isomorphism

[strong yacht]

Isomorphisms preserve all the algebraic content of the groups, i.e. they behave identically as groups up to re-naming the elements. So finding an element of an order in one but not the other is good way to disprove isomorphism, or perhaps one of the groups is cyclic and the other one isn't, or perhaps one of them is abelian and the other isn't etc.

[uncut girder]

Wdym

There should be finitely many cosets

[delicate orchid]

good

lemme try and get GAP working

[uncut girder]

I could just do this in GAP or Sage probably myself, but I was trying to do it by hand

[uncut girder]

Whats that

[delicate orchid]

the beginning of my aborted manual coset enumeration attempt

[delicate orchid]

I'm not sure the number of cosets is finite you know

[pastel cliff]

seems like it's at least 7

[delicate orchid]

PSL(2, Z) is infinite and Aff(F_7) is finite, implying whatever you're quotienting by must also be infinite as their index is finite (|Aff(F_7)|)

[pastel cliff]

whats psl

[delicate orchid]

the projective special linears

major blunders

[pastel cliff]

latex segfault

join the dark side

study CS

[delicate orchid]

I spent 6 hours today debugging matlab code

no thanks

[coral shale]

Let Sn. Consider the subgroup isomorphic to Sm which fixes n-m elements.

Sm is a normal subgroup of Sn, correct?

[delicate orchid]

no, completely wrong

for n >= 5 the only normal subgroup of Sn is An

[coral shale]

where my thinking going wrong... hmm

OH nvm yes nvm i see

[delicate orchid]

just look at the character table 🤪

[coral shale]

I feel like that Sm has something to do with a quotient? But given what you said about Normal subgroups...

[delicate orchid]

you're wanting to restrict the group action of Sn on the set {1, ..., n} to the set {1, ..., m} to get Sm, correct?

like this is where your intuition is leading you

[coral shale]

im wanting to glue the fixed points together maybe?

An equivalence relation that leads from Sn to Sm

But that is no Group quotient then

[coral shale]

more than just glue them together

like pin them down in place ig..

[delicate orchid]

write them as permutation matrices, a projective map should map the representation of Sn to the representation of Sm

if I'm visualising it right

[coral shale]

That sounds correct

Then fixed points correspond to fixed axes

so project onto origin

[delicate orchid]

and then you get a split exact sequence if you take those permutation matrices to be over C I think

so you get a nice direct sum meme

[coral shale]

What objects are in that split exact

GL subgroup?

[delicate orchid]

yeah if phi is the projection, then a projective map from the representation "V" of Sn to the representation "U" of Sm gives you that V = ker(phi) \oplus img(phi) = ker(phi) \oplus U
kinda cool

so the representations (which are faithful) are kinda little wholesome direct sums of each other

this might be where your "feels like a quotient" meme comes in I guess?

[coral shale]

,,V = \ker(\phi) \oplus img(\phi) = \ker(\phi) \oplus U

[delicate orchid]

anyway that's just an aside it's not that important

[coral shale]

yh well

[delicate orchid]

I've forgotten what the original question was

[coral shale]

i wonder why i keep thinking its a quotient

maybe it is what you said

[delicate orchid]

it's definitely not a group quotient

[coral shale]

I am simultaneously thinking examples to do with Rubiks cube.

The Rubiks Cube group is as usual.
Take the subgroup which fixes Corners, say. This is Normal (pretty sure).

Maybe my confusion comes from not seeing that this is a different kind of fixing

[delicate orchid]

Rubiks cube

yeah I mean

hmmm

[coral shale]

Scramble. A move that changes edges only. Unscramble.
Corners will remain the same as they don't interact with edges, hence normal

[delicate orchid]

thinkin bout the fixed point theorem

there's something relating stabilisers to conjugacy classes I just can't remember it

[coral shale]

hmm ok i think i at least cleared up the confusion

[delicate orchid]

ah if y is in the orbit of x then their stabiliser groups are conjugates, ok that's not as useful as I thought it was

[coral shale]

If you arent fixing an orbit... you won't get a normal subgroup? Is that what I'm thinking?

And if you fix an orbit, you will get a normal subgroup?
Guessing.

[delicate orchid]

well a normal subgroup is just a union of conjugacy classes

that's where I was thinking

[coral shale]

Doesnt have to be 'an' orbit oops

[delicate orchid]

so what's the difference between you're two examples

[coral shale]

In the 2nd examples

the corners are an orbit

[delicate orchid]

on a rubicks cube corners can't get mapped to non-corners right?

[coral shale]

i cant remember the proper defn, but i think im using the term correctly

yes

[delicate orchid]

ok ok good

[coral shale]

whereas in the first example, that is not so

[delicate orchid]

so yes

yes yes yes

they're all in the same orbit so their stabiliser subgroups are all conjugates

[coral shale]

So fixing unions of orbits -> Normal subgroup (to check)

[delicate orchid]

I'm not sure about unions

but fixing an orbit might just get you there

[coral shale]

No paper, i might have a proper think tomorrow

[delicate orchid]

fix an orbit, then all their stabiliser subgroups are conjugates, meaning that gG_xg^-1 = G_x' for all g in G x in X for some x' in X

it smells normal

[coral shale]

When I said orbit, I mean some set that is invariant under action by G

Maybe wrong term

yh prolly wrong term, no notes around me

[delicate orchid]

oh my god what's that thing I've just thought of something and now I can't find it

[coral shale]

I really mean Fix(G)

no wait

[delicate orchid]

the class equation that's it

[coral shale]

gah i give. Will put on hold until I have my notes with me and paper

[delicate orchid]

rip

[coral shale]

Symptoms of not taking group actions seriously the first time round appearing

[delicate orchid]

same from me

and all I do is fuckin group actions

[uncut girder]

Welp thanks anyway wew

[delicate orchid]

sorry I couldn't do it :despaireline:

[coral shale]

I have all I need to work with tomorrow, should be fine

[uncut girder]

Consider <a,b|a^2 = b^3 =1>.
Consider the subgroup generated by
R^6, T^7 and R^5T^4RT where
R := ba, T := abab^-1

Question: is this subgroup finite index, normal, torsion free and have quotient isomorphic to Aff(F_7)?

[tribal moss]

What is the role of RT^2RT in that description?

[uncut girder]

We know what R and T are

[tribal moss]

So R and T are not themselves generators of the subgroup in question?

[uncut girder]

oops

Fixed

Fixed*

[tribal moss]

OK, then it makes more sense.

[uncut girder]

F i messed up

Let me fix the relations

The last relation should be R^5T^4RT

YES it has the right index

@delicate orchid @latent anvil I got the generators for a normal subgroup of <a,b|a^2=b^3=1> with quotient isomorphic to Aff(F_7)
The subgroup is generated by
R^6, T^7, R^5T^4RT
where R=ba and T=abab^-1

[tribal moss]

One observation is that all three generators have an even number of a's, so you could rephrase the whole thing in terms of b and c=aba.

[latent anvil]

nice!

[uncut girder]

Thats a good point

[latent anvil]

let $R = \Z[2^{1/2}, 3^{1/2}, \ldots, n^{1/n},\ldots]$

[cloud walrus]

tr.deg.Shamroc/k

[latent anvil]

is there an easy proof that this ring is not noetherian?

qual problem

[latent anvil]

@next obsidian found a proof!!!

let $p$ be any prime. I claim the radical of $pR$ cannot be finitely generated. If it were then there would be some natural $N$ such that $f^N \in pR$ for any $f \in r(pR)$. Since $\lim_{k \to \infty} \frac{kN}{p^k} = 0$, there is some $k$ such that $kN < p^k$. Clearly $p^{k/p^k} \in r(pR)$ (since $(p^{k/p^k})^{p^k} = p^k \in pR$) so $p^{kN/p^k} \in pR$. Then we can write $p^{kN/p^k} = a p$ for $a \in R$, so $p^{kN} = a^{p^k} p^{p^k}$, and hence $1 = a^{p^k} p^{p^k - kN}$. Then $p^{p^k - kN}$ is a unit, and since $p^k > kN$ this implies $p$ is also a unit, so $1/p \in R$. But every element of $R$ is an algebraic integer!

[cloud walrus]

tr.deg.Shamroc/k

[minor badger]

I would like some help to compute the radical series of this representation

I get something else compared to the book and Im new to this

[cloud walrus]

Hausdorff

[median pawn]

Just confirming, flat abelian group means flat Z-module, right?

[median pawn]

Okay yeah

[coral shale]

What's so imporatant about the Normaliser of a subgroup?

$H\trianglelefteq N_G(H)$

[coral shale]

This is the maximal subgroup of G such that H is a normal subgroup of it, right?

[south patrol]

Yeah, or the stabiliser of the action of G on H by conjugation

[coral shale]

Is it uh... 'obviously' useful?

and the same for centraliser ig

[cloud walrus]

Hausdorff

[coral shale]

group action things

[hidden haven]

Localisations of the underlying ring are flat

Equivalently localisation is an exact functor

π_1 is linear by assumption

ℤ-linear map is the same as abelian group homomorphism

[median pawn]

Yes, or Z-module homomorphism

[median pawn]

I haven't seen this before, could you refer me to somewhere I can find a statement or something?

[hidden haven]

You might have seen this

S⁻¹R ⊗ -
and
S⁻¹-
are naturally isomorphic functors

So one is exact iff the other one is

[iron vessel]

Hey guys, I'm trying to figure out the subgroup lattice of the Galois Group of the polynomial $x^4 + 2x^2 + 2$ over $\mathbb{Q}$.

[cloud walrus]

Évariste Galois

[iron vessel]

I arrived to the conclusion that the splitting field is $L = \mathbb{Q}(\sqrt{-1+i},\sqrt{-1-i})$ and that there is 8 elements in $Gal(L/\mathbb{Q}) \cong D_8$.

However im struggling to draw the subgroup lattice of it

[cloud walrus]

Évariste Galois

[minor badger]

anyone here worked with the GAP program? would like some guadince

[prisma shuttle]

yea i've worked with it before but very limitedly

its really useful for counting the number of homorphisms from a group to another group

and it can also do so up to conjugacy which is also really nice

is there anything in particular u want help on?

[minor badger]

So i was to use the qpa package, but i cant even define a quiver

[prisma shuttle]

oops i am not too familar with qpa

but i found thi sonline

https://www.gap-system.org/Manuals/pkg/qpa/doc/chap1.html

https://folk.ntnu.no/oyvinso/QPA/

hopefully they're helfpul

[coral shale]

bump actions/orbits stuff #963391851367374868

[minor badger]

its not group action things

[prisma shuttle]

i think shuri2060 was refering to a different question

[minor badger]

oh

okey so when i type LoadPackage("qpa"); i get return fail

[pastel cliff]

is it just me or are the socratica videos on algebra really damn good

[strong yacht]

Some comments (hope this helps): sqrt(-1+i) * sqrt(-1-i) = 2 so you're really dealing with Q(sqrt(-1+i)), sqrt(-1+i) can be written as a+bi so I think you can deal with Q(a, bi)? Then remember the dihedral group is generated by two elements, so think of two automorphisms to write all your automorphisms in terms of, and then write a lattice of subgroups. Then (Galois correspondence) these subgroups correspond to subfields, to figure out which correspond to which find the fixed fields of your automorphism subgroups

Finally replace the groups in your subgroup lattice with the corresponding subfields to get the subfield lattice

[civic tapir]

With no additional context, if G is an abelian group, what does the notation G/pG, for p prime, represent?

[coral shale]

pg := g + ... + g p times

pG := {pg : g in G}

Verify it's a subgroup

[civic tapir]

Makes sense, thanks

[coral shale]

There's a reason why G^p might not be a subgroup for non-abelian groups

[delicate orchid]

zwed mwoduwle

[iron vessel]

Alright, Ill try this, thanks 🙂

[iron vessel]

Also, do you mean (-1+i)*(-1-i) = 2?

[strong yacht]

Sorry i think what i wrote was complete nonsense at the start (thinking without paper)

@iron vessel check out Kaplansky's theorem on page 65 it does a full treatment on X^4+aX^2+b

https://www.maths.ed.ac.uk/~tl/galois/lastyear/Galois.pdf

[iron vessel]

Amazing! Thanks!

[broken stirrup]

Hi. Why don't we define the action of R by (rf)(x)=f(rx) but f(xr) in order to show that Hom_Z(R,J) is a unitary left R-module where J is an abelian group?

[coral shale]

https://math.stackexchange.com/questions/4426296/pointwise-stabiliser-of-a-union-of-orbits-is-a-normal-subgroup
I figured out what I wanted to ask earlier, if anyone has any ideas on it

[coral shale]

Looking for a source/reference that mentions this and related results.

[south patrol]

the action of G on S should restrict naturally to one of G on G.X

which then reduces it to the usual case

unless ofc you want a proof of the usual case as well, with the 'usual' stabiliser?

[coral shale]

usual case?

[south patrol]

If G acts on a set S then the stabiliser is a normal subgroup of G, I mean

[coral shale]

[south patrol]

I assumed you knew that case and were trying to generalise it, but if you've not seen that then I could explain that too

[coral shale]

I'm aware it is always a subgroup

Not that it is always a normal subgroup

[south patrol]

Ah

[coral shale]

It is?

[south patrol]

Well have you seen representations?

[coral shale]

no

[south patrol]

And yes it is

Well so

[coral shale]

What's wrong with my notes

[south patrol]

Oh no I mean so

You can prove this just by brute force but this might be a nice way to view it

Say a group G acts on a set S by the left. This in fact induces a homomorphism ρ: G -> Aut(S), where ρ(g) is the action by g (i.e. ρ(g)(s) = g.s)

[coral shale]

wait could you elaborate. The stabiliser of what is a normal subgroup

https://i.imgur.com/89jqGZA.png

[south patrol]

Yeah I meant sorry

The case where it's the stabiliser of a point lol

mb

[coral shale]

Is G_x not a stabiliser of a point?

[south patrol]

ye it is

[coral shale]

This isn't always Normal?

np

[south patrol]

Uh what am I on about

[coral shale]

but like, take the symmetries of n points. If you fix 1 point, that gets you S_{n-1}

S_{n-1} is not a normal subgroup of S_{N}

[south patrol]

Wait no what I'm talking about is correct aha

I'll keep going lol

[coral shale]

yh sure

[south patrol]

Ok so

To repeat

Say a group G acts on a set S by the left. This in fact induces a homomorphism ρ: G -> Aut(S), where ρ(g) is the action by g (i.e. ρ(g)(s) = g.s)

You can check that's well-defined

[coral shale]

thats the definition of action I am used to 👌

[south patrol]

oh epic

Ok, so what is the kernel of ρ?

[coral shale]

uhhh it's a normal subgroup of G. Trivial if the action is faithful

uhhh

[south patrol]

Well yes, but it's also the set of g such that g.s = s for all g

so in fact in your notation, it's the pointwise stabiliser of S

[coral shale]

oh right

[south patrol]

And so that proves what you want

(after restriction to an action on G.X and noting you get the same thing)

[coral shale]

Right that makes a lot of sense

aha thanks - makes total sense now

[south patrol]

Np

[coral shale]

So what about a converse hmmm

If N is a normal subgroup of G...

[south patrol]

I'm not sure what sort of converse you had in mind?

[coral shale]

uhhhh

If N is a normal subgroup of G and G acts faithfully on S

idk what comes after

N must be some something perhaps

Uhhh perhaps G/N acts faithfully on some subset ???

no wait

G/N acts faithfully on some quotient set of S

I think is the kindof idea but idk

[coral shale]

Well initially I kindof wanted to reverse this statement. Saying all normal subgroups must somehow be of this form (probably not true without certain conditions - think I need a faithful action for starters)

[south patrol]

Well you can let G act on G/N in the natural way and then N ought to be the pointwise stabiliser right

[coral shale]

yh, N leaves each coset fixed (and is the largest set that does so)

[south patrol]

gaN = aN for all a in G iff (gN)(aN) = aN for all a in G iff gN = N

[coral shale]

But more like I want to still let G act on an arbritrary set

The initial statement tells me how I can find Normal subgroups given an action

But it doesn't tell me how to find all of them

[south patrol]

I'm not really sure what you'd be able to do though

you only have a limited number of actions on any given (finite) set

[coral shale]

Well a faithful action

like fully represents the group

[south patrol]

Don't you want the action to not be faithful if you're to find normal subgroups?

[coral shale]

So if our action is faithful, I feel like we should be able to find all normal subgroups

Well more like - I'm taking a faithful action and then restricting that action onto certain subsets

to find normal subgroups (hopefully all)

[south patrol]

Ok that makes more sense

[coral shale]

Given a faithful action G on S,

N is a normal subgroup of G iff N is the pointwise stabiliser of some union of orbits

is what I want to claim

alternatively,

N is a normal subgroup of G iff N is the kernel of the restriction of the action onto some subset of S

I think these are equivalent

[coral shale]

I think

If we restrict the action to act on Fix_N(S)

We get there... thinking it through . . .

[south patrol]

Isn't that the opposite of what you want?

Like G should act trivially on that

[coral shale]

Fix_N(S)

N acts trivially on it

but the rest of G might not

[south patrol]

Noic

[coral shale]

Let G act faithfully on S.
Let N be a normal subgroup of G.
Restrict the action of G to Fix_N(S).
Clearly N is a subgroup of the kernel of this action. Remains to show the other inclusion (if true)

[tribal moss]

How about a group acting on itself by its group operation? There's only one orbit, no matter how many normal subgroups there are.

[coral shale]

hmmm

That's true.

uh

Does it look like I'm unable to conclude any sort of reverse statement

I thought faithful would be enough but maybe not

[tribal moss]

On the other hand given G and a normal subgroup N, we can always find an S with a faithful action such that N is a pointwise stabilizer.
(Namely take S to be the disjoint union of G with G/N).

[coral shale]

hmm a bit baffled (didn't have much intuition for faithful beyond the definition)

I kindof assumed there wouldn't be any point looking at different faithful actions - assumed they all 'look' the same

[tribal moss]

Faithful just means that different group elements can be told apart from their actions.

In my proposal above, I'm really just interested in the action on G/N, but I get faithfulness artificially by adding in a separate copy of G that it can act on by multiplication.

[coral shale]

I'm not interested in faithful at all if it isn't a restriction I need to impose

so I suppose I just want the action on G/N

[tribal moss]

Yeah.

[coral shale]

Ah rip, thanks

I suppose all groups act on themselves faithfully (one orbit)

[tribal moss]

Yes.

[coral shale]

But there are some faithful actions with multiple orbits, should've spotted that

not sure what to make of that

[tribal moss]

The group really acts separately on each orbit. Once the action is faithful, it cannot stop being so just by adding additional orbits to S.

[coral shale]

right.

[coral shale]

I'm not sure those 2 earlier statements were equivalent, then

N is a normal subgroup of G iff N is the kernel of the restriction of the action onto some subset of S

This may be true still ... ? (not disproven by your counterexample)

[tribal moss]

I'm not sure how that restriction of the action is something that has a "kernel" unless the subset is a union of orbits.

A bit more, perhaps: We can create a "universal" S by taking the disjoint union of all the quotients of G. Then by construction, the normal subgroups of G are exactly the pointwise stabilizers of unions of orbits in that particular S.

[coral shale]

Ah true, I just had very specific concrete examples in mind when writing these and failed to spot.

[coral shale]

Will have a think - thanks

[tribal moss]

Even easier to specify is just to let S be the power set of G, with the "obvious" action.

The elements we're really interested are still just the cosets of normal subgroups, but we avoid needing to speak about disjoint unions.

[coral shale]

Has this something to do with transitivity

We want n-transitive actions for n the size of the group?

nvm no

[lavish nexus]

G finite group, L a simple G-module

then there is a copy of L in CG?

[mint seal]

what's CG?

[lavish nexus]

group ring with complex numbers

[pastel cliff]

can someone explain chinese remainder thm to a 3 year old pls

[upper pivot]

not to a 3 yr old but i could try explaning it to u if u wanted

lol

[coral shale]

make them play zpordle

[pastel cliff]

goo goo ga ga

[upper pivot]

lol, so you want an explanation?

[pastel cliff]

yes please

i kinda get it but i could use a different wording/motivation i think

[upper pivot]

ok so let me first state its origin, in number theory right

so its like, if i had a system of congruence eqn (e.g i told u what a number is mod 2 and 3), how much can i recover about it

right

and the answer is this, if m,n are coprime

and i know what a number is mod m and mod n

then i know what the number is mod mn

that intuitively makes sense right

[pastel cliff]

yes

[upper pivot]

right, so now let me state this in a ring theoritic language

if m,n are coprime, Z/(mn) iso Z/m \times Z/n

do u see why this is equivalent to what i said before

[pastel cliff]

one moment pls

im writing stuff out

[upper pivot]

nice

[pastel cliff]

sorry, backtracking a bit

[pastel cliff]

so it's of interest to consider what information we get about x when we're given x = a mod m and x = b mod n

i know that's probably just a rephrasing but double checking nonetheless

[upper pivot]

mhm

[pastel cliff]

oooo and cyclic groups of relatively prime order will have no elements that have the same order

[coral shale]

0

[pastel cliff]

thank you shuri

[upper pivot]

lol

[upper pivot]

btw to be clear when i said this i meant it as rings

not just groups

[pastel cliff]

ahhhh

my prof introduced it with groups

[upper pivot]

oh i see, well in modern math chinese remainder theorem is a statement about certain quotients of rings in general

[pastel cliff]

he's a bit of a crank

[upper pivot]

what i was describing is the origins

where ppl wanted to combine congruences

[pastel cliff]

i can see how this might be true but i cant put it in words

ik that's not real understanding but like

[upper pivot]

oh so how much about rings do you know

[pastel cliff]

well i know what they are

i know what an integral domain is but idk why i care about it

[upper pivot]

oof, what about ideals, or the isomorphism theorems

[pastel cliff]

i know a lil about ideals but only from very basic lie algebra stuff im doing outside of class

im grappling with first iso alongside this

[coral shale]

lie algebra

[upper pivot]

i mean its technically less structure so lol

well let me try my best to explain this ig

[pastel cliff]

my advisor likes it so by the transitive property i like it

[upper pivot]

so take the map $f:\bZ \longrightarrow \bZ/m \times \bZ/n $ defined by sending $a$ to the class of $[a]$ mod m in the first coordinate, and to the class $[a]$ mod n in the second

[cloud walrus]

JohnDS

[upper pivot]

i.e, $a\mapsto ( a\pmod{m}, a\pmod{n})$

[cloud walrus]

JohnDS

[pastel cliff]

gcd(m,n) = 1?

[upper pivot]

yeah assume coprime

so what chinese remainder really tells us from like, the original thousand years ago thing

is that, if m,n coprime, then $x=a\pmod{m}$ and $x=b\pmod{n}$ has a solution for any choice of $a,b$

right

[pastel cliff]

ok quick interruption

you keep specifying the "thousand years ago thing"

[cloud walrus]

JohnDS

[pastel cliff]

should i think of it as something separate from CRT in the context of algebra

[upper pivot]

well its a specail case

when we say it today it means a more general statement

just want to emphasize that this is a specail case and not the general one

ok, so far so good?

[upper pivot]

right so what does this tell us?

this tells us the f we defined above is surjective right

because for all pairs (a,b) there is some integer that solves this

so $f:\bZ \longrightarrow \bZ/m \times \bZ/n$ is a surjective homomorphism right

[cloud walrus]

JohnDS

[upper pivot]

what theorem do we like using when we have surjective homomorphisms?

[upper pivot]

indeed

and what is the kernel of this homomorphism?

[pastel cliff]

https://tenor.com/view/drake-computer-notebook-explaning-gif-21152267

im kinda seeing it

wait dumb question

first iso works on rings?

we've barely studied rings lol

[upper pivot]

yeah its the same theorem basically

[next obsidian]

Except prove it instead of ask

But yeah you can use the first iso for groups to get you the map

[coral shale]

the isomorphism theorems extend to a lot of Algebraic structures

[next obsidian]

And all you need then is to show the map is also a ring hom

[coral shale]

check wikipedia

its cus its a category theory kinda statement I think

[next obsidian]

Eh

I don’t really think so

[coral shale]

idk then

[next obsidian]

It’s a “just do it” kinda thing if you remember what the map is supposed to be you can just compute it and oh wowza it’s a ring map and I already know it’s bijective!

[coral shale]

universal algebra is the generalisation

[upper pivot]

btw ping me when ur ready to continue @pastel cliff

[pastel cliff]

typing a long thing

[upper pivot]

i c

[pastel cliff]

ok im gonna be stupidly verbose for a moment for my own silly brain:

Z/mZ is the set of all cosets of mZ (i.e. mZ + a for all a < m) so it like partitions Z ; the map Z -> Z/mZ where a \mapsto [a] basically just places the integers into the correct element (coset of mZ) in Z/mZ so that we end up with an injective map, and since it's already surjective it's a bijection and it's a homomorphism bc i said so, so first iso applies. same goes for a map Z/nZ and if m and n are coprime then the only elements that would get mapped to the identity in the map Z -> Z/mZ x Z/nZ are elements that are multiples of mn, hence the kernel is Z/mnZ (why doesnt this apply if m,n are not coprime? - if m and n are not relatively prime, then the kernel will be Z/mnZ, as well as Z/kZ, where k is divisibly by both m and n but not a multiple of mn (like 12 = 0 both mod 4 and 6))

sorry that's super rambly

but i also feel like ive now digressed from CRT

[upper pivot]

yeah excellent

mnZ is exactly the kernel

so to conclude, $\tilde{f}:\bZ/mn \longrightarrow \bZ/m \times \bZ/n$ is an iso due to first iso

[cloud walrus]

JohnDS

[upper pivot]

or $\bZ/mn \cong \bZ/m \times \bZ/n$

[cloud walrus]

JohnDS

[upper pivot]

if you will, this is the ring theoretic version

of the original CRT

[pastel cliff]

so without realizing it i just explained CRT to myself here...?

[upper pivot]

yeah lol

[pastel cliff]

[upper pivot]

right so ready for the general version over arbritary rings?

[pastel cliff]

mmmm one minute i need to now write that in my notebook lol

[upper pivot]

yeah sure lol

[pastel cliff]

ok this might be a bit lame but for the time being can i "ignore" the thousand year old version of CRT

[pastel cliff]

and just think of it as this statement

[upper pivot]

yeah this is def the right way to think of it these days

i was just historically motivating it

as long as you understand like, what this tells u ig

[pastel cliff]

bc i semi understand that yes, the fact that we can solve systems of eqns modulo coprimes leads us to that statement about isomorphisms mod coprimes

but the bridge between the two things doesnt fully exist in my head yet

[pastel cliff]

even though i do actually get this

as long as that's aight then yes pls continue

[upper pivot]

yeah i think thats fine

the general CRT is used a lot so ull get used to it by practice lolz

ok so for a second we r going to go back to the thousand yr old statement, that basically gives us surjectivity of the map right

and basically you can try working out a proof for it, its not too hard iirc, but the main ingrediant is bezouts theorem

i.e if $m,n$ are coprime, then there is $a,b$ such that $am+bn=1$

[cloud walrus]

JohnDS

[pastel cliff]

damn u beat me to it

[upper pivot]

in a ring theoretic language, you would say

$(m)+(n)=(1)$

[pastel cliff]

but i actually knew that i feel smart

bezout i mean

[cloud walrus]

JohnDS

[upper pivot]

right nice, yeah just wanted to write it out incase u werent like, nt

do u understand what the statement i wrote means

[pastel cliff]

( ) <=> [ ] ?