groups-rings-fields

[rotund aurora]

very neat

[grizzled spindle]

Just bumping this cause another discussion took place

[south patrol]

Lol this came up in my masters thesis and has given me flashbacks

[delicate orchid]

This came up in my research lol

Such a nice combinatorial object

Absolutely horrendous in other ways but it’s nice enough for now

[south patrol]

Tbf you were the one who mentioned the trees iirc

[delicate orchid]

Yeah. This is the case even outside of S_p^k

For S_n with p-adic expansion n = \sum_i=1^k a_ip^i, The p-sylow subgroup is a direct product of the sylow subgroups of S_p^i, so it acts on a forest of p-ary trees

[rotund aurora]

can you consider infinite p-trees

[delicate orchid]

On a related note does anyone recall what the representation theoretic correspondence between GL_n and S_n is called? I think there’s an nlab page but I can’t find it

[delicate orchid]

Yeah but I’m not sure if it still works

[sly crescent]

Is it S_n being the Weyl group of GL_n?

[delicate orchid]

No. It’s the F_1 nonsense one

[sly crescent]

Weyl groups can supposedly be thought of as algebraic groups over F_1

[delicate orchid]

Although I don’t see why it shouldn’t

[sly crescent]

Is that it?

[delicate orchid]

Yeah that sounds like it lol

I don’t think I’ve heard of it like that but it makes sense

[sly crescent]

Oh

Did you know that quaternionic reflection groups are a thing and that they have been classified?

[delicate orchid]

https://tenor.com/view/locked-in-monkey-monkey-meme-monkey-locked-in-monkey-face-gif-940796321824170934

Googling rn

There’s a question on overflow about octonionic reflection groups this is crazy

[crystal vale]

They defined f: A->B and I want to show that if f is surjective then there exists g:B->A such that gf is identity function on B, yes they define composition in that way.

But I am not sure how they can say that g is well - defined? Does it fix that a ?

[crystal vale]

If it is a permutation group then it is always transitive and if X has more than 2 elements then that group will be non-abelian so X contains at most 2 elements. Thus it will be regular because there is no other element.

Is it correct?

Actually they defined the permutation group as a subset of Sym(X), now it makes sense

[inner harbor]

"prove that Q(isqrt2) has deg 4" tf ??

X²+2

Am i being dumb ?

[topaz solar]

is it Q(i, sqrt 2)?

with a silly little , in there

[inner harbor]

Nah

"prove that it's isomorphic to Z/2Z ²

U think they meant i,sqrt2 ?

[topaz solar]

yeah, since that has i <-> -i and sqrt 2 <-> -sqrt 2

which seems preeeeety C_2^2 looking

[inner harbor]

Kk

U have like
Q(i,sqrt2)
Q(sqrt2) Q(i)
Q

And
C_2 ²
C_2. C_2
0

?

[topaz solar]

yeah

[topaz solar]

theres Q(i sqrt 2) which is in there too though yeah?

theres another subgroup of C_2^2 that isnt the left or right copy

[inner harbor]

Oh yea

Can u write it

[topaz solar]

the one generated by (1,1)

which is isomorphic to C_2 yet again

[inner harbor]

How do I know this

Like Q( some combinations between alpha and beta)

Here P= x⁴+aX²+b for some rationnls

[rain grove]

When are two matrices of SL_2(R) similar?

signature? Im having a problem with some definitions... matrices $A,B \in M_n(\mathbb{R})$ are similar, if there exists an invertible matrix $P$, so that $A = PBP^{-1}$. If $A$ in $B$ invertible, we can equivalently say that $A$ and $B$ are conjugates of group $GL_n(\mathbb{R})$. Find similar matrices of $SL_2(\mathbb{R})$ that are not conjugates of group $SL_2(\mathbb{R}$)

[cloud walrus]

OHHELLNAH

[rain grove]

1. Why only if A and B invertible we can equivalently say that A and B are conjugates?

A,B in SL_2(R) are similar when there exists a invertible matrix P from M_n(R) so that A = PBP^{-1}

[rain grove]

is this right? cause im thinking maybe this invertible matrix that has to exists has to be from the same group

and M_n(R) is not even a group

[rain grove]

This is the problem im trying to solve

[mighty kiln]

https://en.m.wikipedia.org/wiki/Jordan_normal_form#Real_matrices

This may be of interest

[hidden cairn]

can i get some hints for the example part?

[rocky cloak]

If they're not invertible, they're not in GLn, so then it doesn't make sense to say they are conjugate in GLn

[wraith cargo]

Maybe consider like S_4 and a certain subgroup of index 2

[rocky cloak]

Yes, this is right. So the question is asking you to find A and B such that there is such a P, but no such P in SL2

[hidden cairn]

i got D_8 / <r>

that works right?

[wraith cargo]

Ye

[hidden cairn]

oki thanks

should i think of homomorphism or of normal subgroups when forming a quotient group?

i know N is normal iff N is a kernel of some homomorphism but when forming quotient groups, which pov is better?

[wraith cargo]

Just think about normals
Nobody really thinks about normals as the kernel of a homorphism

[hidden cairn]

hmmm

alrgiht thanks again

[rocky cloak]

I will have to disagree with Irony, it's usually much easier to think in terms of homomorphisms.

But it depends, if you want to understand the quotient group abstracly, then the homomorphisms for which N is the kernel is the quotient map G -> G/N. So that would be putting the cart before the horse.

But if you have some normal subgroup and want to find out what G/N is, it's usually best to try to find some homomorphism with kernel N and see what the image is.

[knotty badger]

Thinking about homomorphisms is more about considering what the quotient group “does”

[hidden cairn]

isnt it more like how the quitoent group is formed instead of what it does?

[knotty badger]

So what I mean is the following

[hidden cairn]

it does nothing right? it is formed as a result of the homomorphism

[glad osprey]

How would you think about GL_n(R)/SL_n(R) without using the determinant homomorphism?

[knotty badger]

It actually does do something, but I’ll explain

What it “does” is about it how it relates to other groups, and what you can use it for

Namely, there’s an alternative description for group homomorphisms out of G/N

Specifically, group homomorphisms G/N -> Z naturally correspond to group homomorphisms G -> Z that send N to the identity

You can interconvert between the two

In a bijective manner

Have you seen this?

[hidden cairn]

havent exactly read about this in dummit and foote so far but they briefly mention this at the end of the quotient group chapter

and it makes sense as G/N is a group iff N is a kernel of some homomorphism

[knotty badger]

So, I can be more explicit

Given a group hom G/N -> Z

We have the quotient map G -> G/N

If you compose these, you get a group hom G -> Z which sends N to the identity, right?

[hidden cairn]

yes

[knotty badger]

One important property of G/N is that this process is reversible

So if you have a group hom G -> Z which sends N to the identity

You can construct a corresponding group hom G/N -> Z

Do you see how?

[hidden cairn]

send 1N to 1 and mN to m?

[knotty badger]

So explicitly, you have $\varphi : G \to Z$ a group hom

[cloud walrus]

Pseudonium

[knotty badger]

And what I’m looking for is a group hom $\psi : G/N \to Z$ constructed from $\varphi$

[cloud walrus]

Pseudonium

[lone niche]

This looks like a combination of fundamental theorem of galois theory, and the constructive primitive element theorem proof.

[knotty badger]

I’m not entirely sure what this means

[hidden cairn]

what i wrote is just the natural proejction homomorphism from G to G/N

youre asking something different

[knotty badger]

That would be sending 1 to 1N, and m to mN

[hidden cairn]

ah yes

[knotty badger]

But you seem to be trying to do the reverse…?

[hidden cairn]

yes

[knotty badger]

And in general there’s not a reverse, cause you can have mN = rN

But m not being equal to r

[hidden cairn]

ah right

[knotty badger]

Despite this obstruction, what I said is still possible

[knotty badger]

.

[hidden cairn]

i was gonna say take the inverse of the natural projection and apply phi

but inverse of the natural projection is not how i thought it would be

[knotty badger]

Mhm, that’s what you’d like to do, but you can’t quite take the inverse as a function

It’s a relation - a single coset gN is related to many elements of G

[hidden cairn]

yes

so whats the correct approach

[knotty badger]

Would you like to see?

[hidden cairn]

yeah

[knotty badger]

Here’s how you do it

You just define $\psi(gN) = \varphi(g)$, and you check that this is “well-defined”

[cloud walrus]

Pseudonium

[hidden cairn]

this looks so simple lol

[knotty badger]

Indeed, we wanted this to be the inverse to composing with the quotient map

So it better be the case that $\psi \circ q = \varphi$

[cloud walrus]

Pseudonium

[knotty badger]

Otherwise it’s not inverting the process

[knotty badger]

And if you plug an element of g into this

You get that $\psi(gN) = \varphi(g)$

[cloud walrus]

Pseudonium

[knotty badger]

So all you need to check is that if $gN = hN$, then $\varphi(g) = \varphi(h)$

[cloud walrus]

Pseudonium

[hidden cairn]

yes

[knotty badger]

That’s what it means for the map to be “well-defined”

And after that it’s not very hard to check this is a group homomorphism too

[hidden cairn]

then not all homomorphism from G to Z would lead to a homomorphism from G/N to Z right?

[knotty badger]

Yes

You need the homomorphism to send N to the identity

[hidden cairn]

cuz phi(g) = phi(h) is not always a must for arbitraty hmomorphism

[knotty badger]

Indeed

[hidden cairn]

i ll try to prove if this is the case then the homomorpishm from G/N to Z is well defined

but i have to go to class now

[knotty badger]

Ok

[hidden cairn]

thank you so much for the mini lecture

it is much appreciated

[knotty badger]

Yeah np

Feel free to ping me if you solve it

[marsh scaffold]

So I feel like this is not true and I have been trying for a counterexample

So they haven't specified G , H to be finite or infinite so I think it breaks in some infinite weird group

[delicate orchid]

this is a tricky one

if H is normal then this is true, I think

[marsh scaffold]

Yeah but H is not necessarily given normal

[delicate orchid]

that's the issue lol

[marsh scaffold]

If it helps this is in the group actions section

[rotund aurora]

If you consider g, g^2,...,g^n then one of the g^i will fall in H

[marsh scaffold]

Oh

[delicate orchid]

https://tenor.com/view/feel-that-friday-funny-two-gif-2768002361436862173

[marsh scaffold]

Wait okay so of group is finite then uh maybe we can do sth to get that g^mn is e

But what if the group is infinite

[sly frost]

if G is infinite then H has to be infinite too

[marsh scaffold]

Okay?

Infinite groups can have like an exponent that anhilates all elements tho right?

Actually @rotund aurora I think I might need one more hint

[delicate orchid]

croqueta's idea works in the infinite case as well. Just to convincee myself that his statement is correct:
If g, ..., g^n are all in different cosets of H then there is some g^i that's in H by the pigeon hole principle, if there are two g^i, g^j in the same coset g^iH, then g^ig^-j is in H?
Ok, now. Let g^i in H, then g^im = e... what I fail to see is why i needs to divide n

[rotund aurora]

note that H is of finite index, we don't care if G is finite or not. I did not use that G was infinite

[rotund aurora]

yeah so the problem is that i need not divide n, but if you take i minimal I think it does divide n

[rocky cloak]

This only proves that g^km = e for some k<=n though

[delicate orchid]

yeah I'm not seeing why i minimal would divide n, lemme think some more

[rotund aurora]

I believe it's an Euclid style argument, g,g^2,...,g^k are in different cosets, g^(k+1) is in the same coset as g, etc

[rocky cloak]

It won't

[delicate orchid]

this might also not be the intended solution. If we want to use group actions then perhaps the fact that G acts freely on G/H Gibberish, obviously false

[rocky cloak]

C2 in S3 is a counterexample, but it still satisfies the original question

[delicate orchid]

an infinite group acting on a finite set freely makes me raise an eyebrow personally

[marsh scaffold]

How tf did u think that up so fast 🔥🔥😭

Damn

[rocky cloak]

It's the smallest non-abelian group

[delicate orchid]

open problem

[rain grove]

Why is it like that?

[rocky cloak]

Don't know how to attack the original problem though

[rain grove]

Why not similar from SL_2{R}

[marsh scaffold]

Fr?

[delicate orchid]

yeah nobody knows what the smallest n0n-abelian group is

[marsh scaffold]

Huh 💀

[rocky cloak]

That's just not what the word similar means I guess

[delicate orchid]

jagr are you trying to find two matrices non conjugate in SL_2 but are in GL_2

like what's the deal here

[rocky cloak]

Yeah, or ohhelnah is anyway

[rain grove]

No I found them

Im a bit confused why this matrix that has to exists for the other two to be similar is from a different set of matrices

[rotund aurora]

oh yeah it doesn't work. I thought it would work, mb

[marsh scaffold]

Ok so yeah I guess I'll be skipping this problem

[rain grove]

Ohh nvm I think I got it its just from definiton it has to be a n*n matrix

[marsh scaffold]

Thanks for the efforts y'all 🙏

[shell pilot]

Do groups have to be of the same "type" to perform the external direct product of them? Would G and e_H have to be of similar structure in this question? e.g. e_H happens to be an identity element for a group of permutations and G is a subgroup of Z_n.

[delicate orchid]

there is no "type"

although writing it as direct summation implies that all groups involved are abelian

but that's probably just non-standard notation

[shell pilot]

So $G\bigoplus e_H$ I can write as $( g_1, e_H),(g_2, e_H), ...$ ? I'm wondering if $e_H$ is just a singleton or if I use the "elements" of $e_H$ in the product such as $G\bigoplus e_H= (g_1, e_H_1), (g_2, e_H_2), ...$

[delicate orchid]

there are no elements of e_H, it's not a set

ok I suppose it could be but that language is concerning

[cloud walrus]

Soap_Opera
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[delicate orchid]

but you're correct in that you can write $G \times {e_H} = {(g, e_H) \colon g \in G}$

[cloud walrus]

Wew Lads Tbh

[shell pilot]

That questions doesn't seem very interesting then...isn't that similar to saying prove that a group is isomorphic to the product of itself and {1}?

or any singleton for that matter

[delicate orchid]

yes that's exactly what they're asking you to prove

[rotund aurora]

what if you take the normalizer of H?

N(H)={ g in G : gHg^{-1} subseteq H }

[rocky cloak]

Alright, so I know you said you stopped working on this, but I had an idea.

||G acts transitively on G/H. Then for an element g consider the action of (g) on G/H. If the orbit of H has size k1, then we already argued g^k1m = e.||

||Now consider the orbit of some other coset xH. Then if it has size k2, then g^k2 x is in xH, hence x^-1 g^k2 x is in H. Since x^-1 g x has the same order as g we have g^k2m = e.||

||Now let ki be the sizes of the orbits. Then g^kim = e implies that g^gcd(kim) = g^gcd(ki)m = e.||

||Since sum ki = n, any common factor of them is also a factor of n, hence g^nm = e||

[rotund aurora]

wait nvm you could have N(H)=H

[marsh scaffold]

(g) is subgroup generated by g?

[rocky cloak]

Yes

[marsh scaffold]

let ki be the sizes of the orbits

These are the orbits of the action of (g) on G/H correct?

[rocky cloak]

Correct

[marsh scaffold]

What happened in g^(gcd(k_i m)) = g^(gcd(k_i))m

Oh wait nvm

Ok great

Thank you so much @rocky cloak

The solution actually dosent seem to have anything very non trivial

I guess the key is always to observe from the "right" group actions

But idk when I'll get that skill 😭

[rocky cloak]

No, nothing fancy. Just quite a few steps one would have to stumble into

[marsh scaffold]

I feel like i always have been weak with group actions 😔

[rocky cloak]

Basically my thought process came from looking at why it works for C2 in S3.

Like you have elements such that g^3 is not in C2, but only when g is conjugate to something in C2.

So okay, maybe in general you can find like a conjugate of g to make it work.

Then setting up the group action, looking at a conjugate of g is just the same as thinking of a different orbit of g. And from there the argument made itself I guess

[marsh scaffold]

Oh okay so looking at a concrete example

And then you found the right action

I never did that

Actually tbh all my examples are abelian 💀

[rocky cloak]

Well, you're given that G/H is finite, so that's really the only action to consider I guess

[marsh scaffold]

No but in all the actions I considered I was acting them up by G or by H

Or sth

I should have actually stepped back to find some non abelian example and see why it works

[delicate orchid]

very nice solution jagr

[marsh scaffold]

Anyways thanks for sharing your valuable insights also @rocky cloak

[storm axle]

Given a commutative ring R and a module M over R, I am asked to prove that if M is free then Tor(M) ={0}.
Do I have to assume that R is an integral domain?

My proof is as follows:
M is free so for all y in Tor(M) there is a unique representation: y = r_1 b_1 + ... +r_n b_n.
y is in Tor(M) so there is some nonzero scalar r in R such that ry=0 => r r_1 b_1 + ... r r_n b_n=0 => since M is free, r r_i =0 for all 1<=i<=n.

Now it seems to me that I have to assume that R is an integral domain to get that r_i = 0 (for all i) and then y=0.

[wraith cargo]

Idk how you defined Tor but I think r should be a non zero divisor

[rocky cloak]

It depends on how you define Tor(M) if R is not a domain

[storm axle]

So this claim is flase if R isn't a Domain?

I mean, could* be false.
Meaning, unprovable as long as algebra is consistent.

[rocky cloak]

By that definition it would be false yeah

[storm axle]

I see...

So they probably expected us to assume in our HW that R is an integral domain

Since when we defined Tor(M) is said "let R be an integral domain..."

[rain grove]

Let $G$ be a group, $H$ its subgroup, and $a$ an element from $G$. Suppose that $a^m = 1$ for some $m \in \mathbb{Z}$ and that $n$ is the smallest natural number with the property $a^n \in H$. Show that $n|m$

I started with proof with contradiction but idk what to do

[cloud walrus]

OHHELLNAH

[delicate orchid]

assume that n didn't divide m, then m = kn + r for some 0 < r < n and integer k. Now try thinking about a^ma^(-kn)

[tardy hedge]

Lets go Weewwwww

🙌

[glad osprey]

Are you using additive notation but with 1 as identity?

[tardy hedge]

Oh Hell Nah!!

[rain grove]

Idk that is the problem, im confusef about that too

[tardy hedge]

Oh Hell Nah Bruh!!

(Ok ill stop mods(

[delicate orchid]

don't use additive notation for non-abelian groups

[rain grove]

I think 1 is multiplicative unit

[rain grove]

a^m?

omg sorry guys

Let $G$ be a group, $H$ its subgroup, and $a$ an element from $G$. Suppose that $a^m = 1$ for some $m \in \mathbb{Z}$ and that $n$ is the smallest natural number with the property $a^n \in H$. Show that $n|m$

[cloud walrus]

OHHELLNAH

[rain grove]

damn i mistyped in my document and didnt check sry, i think i can solve now lol

[marsh scaffold]

i was able to get through a but am stuck on b

[cloud walrus]

[redacted]

[marsh scaffold]

i think theres a more group theoretic solution that dosent actually require looking at this equation

[rain grove]

U_n is by definition the set of n x n unitary matrices.. But in my problems we use U_n as that group connected with the unit circle, solutions of z^n = 1 I think... Is this the same thing or?

[delicate orchid]

U_1 is the unit circle

oh nvm I see what you're saying

they're absolutely not the same thing

rather obviously in fact, U_n is uncountably infinite, your group is finite

[rain grove]

Okk so in this problem:

Describe the sections of the group $U_{12}$ by the subgroup ${1, -1, i, -i}$

What is U_12?

[cloud walrus]

OHHELLNAH
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[coral spindle]

Well it looks like it's a subgroup of the group of roots of unity in C

So I imagine it's the 12th roots of unity in C.

[rain grove]

Ok then in abstract algebra the same notation is used for these groups

[coral spindle]

Sometimes people use overlapping notation

But notation for these groups is not totally agreed-upon

[delicate orchid]

in a first course people tend to use bespoke notation for some ungodly reason

I would just call this group the cyclic group of order 12 C_12

because that's what it is

[rain grove]

Yeah I like this ty

[delicate orchid]

I get that they're trying to draw on objects you're already familiar with but it ALWAYS leads to confusion later

[distant summit]

I might be having a bad brain day, but why is concatenation of {e, a} "automatically" associative?

We haven't been told that e is the identity element yet (i.e. in the first paragraph), so how do you know that (ae)a=a(ea)?

[coral spindle]

This is pretty much by definition of words

[delicate orchid]

(aa)a = aaa = a(aa)

[coral spindle]

Like this is just for free by the definition of concatenation. It's defined specifically so that it's associative.

[delicate orchid]

oh nvm haha I just read the 2nd line. Yeah it's what Boytjie said

[coral spindle]

You could spend time explicitly constructing this but it's not really helpful

[delicate orchid]

da associatorrr

[distant summit]

Hmmm

okie

[coral spindle]

Simply define words over an alphabet as being the free monoid

[delicate orchid]

circular...........

[coral spindle]

NOOOOO

[finite vine]

can somebody help me solve this question : Give an example for a group (G,* ) and a sunset H of G that (H,* ) satisfies all the axioms of a group except for the Identity element

[delicate orchid]

good luck having inverses without an identity element

[coral spindle]

We talked about this before...

Did you think about this hint I gave you at all? #discrete-math message

[delicate orchid]

oh wait yeah good point, I forgot about the ol \varnothing

[finite vine]

I tried it was nearly impossible

[coral spindle]

I disagree.

So ok let's re-copy my hint to make things easier.

Second hint: suppose x is an element of H. What do the closure and inverse properties together imply?

So what do you think of it. Tell me what comes to mind.

[finite vine]

sub group

[coral spindle]

Elaborate.

OK I will assume the silence means that you can't elaborate.

[finite vine]

I still can't figure

I might be stuck in some way of thinking

[coral spindle]

Let H be some subset of G that satisfies all the axioms except identity.

Suppose x is some element of H.

What does the inverses property tell us?

This is not a trick question and has an extremely simple answer.

[finite vine]

for every x in H there exists x**-1 in H such that x*x-1 is the id element in G right?

^

sorry I don't respond quick

[coral spindle]

So what does this property, that $x \cdot x^{-1}$ is the identity, tell us about $H$, since we have the closure property?

[cloud walrus]

Boytjie

[finite vine]

That the condition is false in the question?

[coral spindle]

No

Not quite

But why are you saying this? Say the reason explicitly.

Then we will talk about why it's not quite correct

[finite vine]

Can I tell what I tried to do since the beginning ?

[coral spindle]

Ahmed I am trying to bring you through this logic step-by-step, can you please just work with me?

[finite vine]

Damn I feel dump

[coral spindle]

We'll talk about the whole picture later

The logic so far is:

• Suppose x is some element of H
• Then x^-1 is in H by the inverse property
• Then x x^-1 is in H by the closure property

So what's the next step?

What is the next thing that tells us something very important?

[finite vine]

we have shown that the id element is in G then it must be in H so if H satisfies both the inverse and closure

[coral spindle]

we have shown that the id element is in G then it must be in H
This is faulty logic.

No, we have just shown the id element is in H.

So what does this tell us now. Remember, I made a key assumption in the logic:

suppose x is some element of H...

[finite vine]

for any group G, there is exactly one set H that might have this property

[coral spindle]

You just copied my message! That's really insulting.

[finite vine]

yeah I have short memory

[coral spindle]

If you're not going to use your own words and reasoning I'm not going to engage you.

We'll try one more time

[finite vine]

bruh

thanks

dont get tired with me

I will try to ask someone maybe understanding the question in english is not the best way was

I appreciate your help and I took your time

[coral spindle]

In my logic, I looked at some set H that satisfied the inverse and closure properties, but not the identity property.
I then assumed that H had some element x.
Then I showed that all together, this shows that H has the identity property.
What does this tell us about my assumption that H had some element x?

If you do not use your own words, we're done here.

That's right

[finite vine]

damn

[coral spindle]

Don't delete your messages

[finite vine]

assumption was flawed

[coral spindle]

It's better if you keep up your mistakes

yes

Indeed I assumed that H had some element x, and found a contradiction, so in fact H has no elements.

So what is H?

[finite vine]

empty set

[coral spindle]

That's right.

There you go, you've found the only set H with that property

It's the empty set.

[finite vine]

thank you man

I appreciate that you patience

[coral spindle]

Don't do this again, it's incredibly rude and completely defeats the point of this. I don't want to hear my messages repeated back to me, I want to hear your thoughts – even if they're wrong – otherwise I cannot teach you.

[finite vine]

I apologize for that I just didn't know how to translate it in my mind

in english

I forgot to add something in the question

H is not empty

[coral spindle]

Well good, you still have an answer.

[shell pilot]

Is there a shorter way to say "the quotient group of G by H" or "the factor group of G by H" for $G/H$?

[cloud walrus]

Soap_Opera

[coral spindle]

G over H

[shell pilot]

When I read G/H is there a shorter phrase for the symbol?

[coral spindle]

G mod H

[shell pilot]

Thanks!

[coral spindle]

or even "G quotient H"

[rain grove]

Describe cosets of $\mathbb{R}^3$ over subgroup ${(x, y, 0) | x \in \mathbb{R}}$

[cloud walrus]

OHHELLNAH

[rain grove]

In solution it says they are in form ${(x,y,z_0) | x,y \in R}$ where $z_0$ is a real number

[cloud walrus]

OHHELLNAH

[rain grove]

but why is there z_0, why not just (x,y,0)

mistake maybe?

[coral spindle]

Not a mistake

The subgroup is not the whole group, so there is a coset other than the subgroup.

[rain grove]

but my understanding of cosets of group R^3 over that subgroup are sets (a,b,c)*(x,y,0) for every (a,b,c) in R^3

so then its always (ax,by,0) and because ax and by are again in R all cosets have the form (x,y,0) for some x,y in R

Idk where am I wrong

[coral spindle]

The group operation is not multiplication.

[rain grove]

Why?

[coral spindle]

Because it's not?

But anyway, if it were, what would be the identity, and what would be the inverse of (0, 0, 0)?

The group operation is vector addition

[rain grove]

Ohhh and if it said R^3 - (0,0,0) that would imply its multiplication

[coral spindle]

No, it wouldn't.

[wraith cargo]

Not necessarily

[coral spindle]

That still wouldn't work.

What's the inverse of (0, 1, 0) for example

[south patrol]

Not even multiplicatively closed

[rain grove]

I see ty

[sly frost]

you'd have to have (R-0)^3

or R+^3

[rotund aurora]

Let G be solvable. Is the minimal number of generators of G the same as that of G^ab?

wait nah

The abelianization of S3 fails

I still wonder if you can relate these two numbers somehow

[rotund aurora]

Btw, I know this is true for p-groups (Burnside basis theorem)

[south patrol]

Hadn't heard of this theorem lol thanks for sharing

[acoustic igloo]

is S_10 X S_2 likely to be the largest proper subgroup of S_12?

[delicate orchid]

,w 10!*2

,w 12!/2

no

A_12 is

[acoustic igloo]

ah yeah

besides that one?

[delicate orchid]

it's the largest parabolic ig uhh

actually no it isn't

S_11 is bigger lol

I'm kinda cooked 2nite...

[acoustic igloo]

oh right yes

sorry i totally forgot to mention S_11 and A_12 so i'm pretty cooked too

thanks! i'm gonna eat dinner

[delicate orchid]

https://tenor.com/view/its-peak-peak-yummers-its-yummers-peak-homelander-gif-4634796544937757943

[elfin summit]

I've come across some comments suggesting that there is some redundancy in the definition of a group, since in the characterizations of the neutral element and inverses, it is not necessary to state their respective conditions for both sides of the product, but rather it is sufficient to have it on one side only, and the other can be deduced.

To be more precise, what I mean is the following:
Given a non-empty set $G$ and a binary operator $\cdot : G \times G \to G$ on $G$, the pair $(G, \cdot)$ is a group if the following properties are satisfied:
\begin{enumerate}
\item The product $\cdot$ is associative.
\item There exists an element $e \in G$ such that for all $x \in G$, $x \cdot e = x$.
\item For each $x \in G$, there exists $y \in G$ such that $x \cdot y = e$.
\end{enumerate}

I have already verified that with these three conditions the remaining details can be derived to get the usual definition of a group. Now, what I am interested in knowing is whether these conditions are truly minimal and independent of each other, and this is where I need some help. I think that to prove this, I should find concrete examples of sets and operations that satisfy conditions (1) and (2), but not (3); another example where (1) and (3) are satisfied, but not (2); and another where (2) and (3) are satisfied but not (1). Is that correct? Can you think of any such examples? Or maybe there is a simpler way to justify it? Now I'm wondering if this should be posted on logic or foundations since this is really about axioms independence...

[cloud walrus]

アンドレー

[delicate orchid]

those objects are called, in order:

• monoids
• not possible, how are you defining 3 without the existence of e?
• loops

you also need xe = ex = x

[delicate orchid]

example of a monoid: the natural numbers (with 0) under addition

[elfin summit]

loops, like the ones for defining homotopies, right? 🤔

[delicate orchid]

absolutely not whatsoever

example of a loop? UHHHHH

natural numbers under subtraction?

[elfin summit]

thanks for the reference

[shell pilot]

Why are (1 2 3)A_3 and (1 3 2)A_3 not left cosets of A_3 in S_3?

[acoustic igloo]

aren't they?

[shell pilot]

ProofWiki and other sources say that only A_3 and (1 2)A_3 are left cosets of A_3 in S_3

[acoustic igloo]

oh maybe it's the notation

(1 2) swaps 1 and 2

[shell pilot]

Also, does [S_n: A_n] = 2?

(That's why I'm trying to work on the cosets of A_3 in S_3)

[acoustic igloo]

yeah

n! / (n! / 2)

[shell pilot]

Here is what I have written...
$S_3={(), (1 2), (1 3), (2 3), (1 2 3), (1 3 2)}$ $A_3={(), (1 2 3), (1 3 2)}$ $(1 2)A_3={(1 2), (2 3), (1 3)} = (1 3)A_3 = (2 3)A_3$ $(1 2 3)A_3={(1 2 3), (), ()}$ $(1 3 2)A_3={(1 3 2), (), ()}$

Let me fix the LaTeX

[cloud walrus]

Soap_Opera

[shell pilot]

Isn't each element in (1 3 2)A_3 its own inverse? And there is an identity. And the elements are associative.

(I'm asking, not telling...)

[acoustic igloo]

you got two identity elements in (132)A_3

there should only be one

(132)(132) is not ()

[shell pilot]

Oh, that is only for two cycles

So (132)(132)=(12)(13)(12)(13) and they are not commutative

Ok, that makes much more sense now, thank you!

[acoustic igloo]

you're welcome

[barren sierra]

I'm trying to understand this proof

this stuff in red brackets

how do they get this?

everything beforehand makes sense, standard Sylow stuff and whatnot

ok nvm you can get it via cardinality considerations

[south patrol]

K = Z/25 so H has order 11, so H = Z/11 and Aut(H) = (Z/11)* = Z/10, so we are studying maps Z/25 -> Z/10. These must send 1 to smth with order dividing 10 and 25, hence dividing 5. I.e. we must send 1 to +-1 or +-5. (And all these options work and determine the map). Lastly, flipping sign is an auto of Z/25

Oh lol

[barren sierra]

smth with order dividing 10 and 25
yea I came to this realization after sending

[south patrol]

Sorry I mean 0 or +-2

Lol

Oops

[rigid prairie]

is {0,1}^4 with xor is isomorphic to Z_16 with addition?

[topaz solar]

ok, well, what do you know about Z_16

[rigid prairie]

it the same size as {0,1}^4

so some homomorphism between them exist but it probably not the easy one

[topaz solar]

No

Or well yes it’s the same size, and there is a homomorphism

But like, sending everything to 0 isn’t good

What do you know about Z_16’s addition

Like, what’s the order of an element n in Z_16

[rigid prairie]

16 , I think I will sit on this on on my own later, though about it because of one-time-pad cryptography and who it can just use other groups

[topaz solar]

Well, consider

For a homomorphism, f(g^n) = f(g)^n yes?

So orders can’t be messed up too much

The over of 1 in Z_16 is 16

But if we tried to map any element of {0,1}^4 under xor there, we know x xor x = 0000

[rigid prairie]

ahhh {0,1}^4 is not cyclic group

[topaz solar]

So it’s order 2

[topaz solar]

Yeah that’s the most immediate killer, but the orders mean you can’t do much homomorphism stuff at all

Much less isomorphism

[rigid prairie]

ok thank you for the help

[distant summit]

If you see something like $ab^{-1}$, do you ever pronounce that $a$ divided by $b$, or is always $a$ multiplied by $b$-inverse?

[cloud walrus]

Douglas

[distant summit]

I'm not assuming that a and b are rational or anything, just elements of an arbitrary group

[rocky cloak]

I don't think I ever pronounce anything as "a divided by b". In special circumstances I might say "a over b", but then I'd probably write it a/b aswell

[distant summit]

Okie. Slightly different, but what about G/H where H is a normal subgroup of G and this denotes the set of left cosets?
I know you can call that the quotient group, so saying "G divided by H" seems somewhat natural, but idk if that actually happens

[rocky cloak]

"G mod H" would be the most common I think. I might also say "G over H".

[distant summit]

Makes sense

[prisma bear]

My group is a free group F2 = <a, b> quotient out the normal group generated by <ab^2, ba^2>, how do I show if it's an infinite or a finite group?

My intuition is that the normal group <ab^2, ba^2> is equal to the normal group <aba, bab>, and somehow I want to show a^n is not in the group for any n. The problem is that I need to consider the normal closure of the group generated by generators.

[rocky cloak]

Let N be the normal subgroup and let G = F2/N and f: F2 -> G the projection map.

Then G is generated by f(a) and f(b). Since f(a) = f(b)^-2, G is generated by just f(b), hence G is cyclic.

f(b) = f(a)^-2 = (f(b)^-2)^-2 = f(b)^4 so f(b)^3 is the identity

[prisma bear]

thanks a lot. I'm really rusty at doing this.

[languid trellis]

I did end up thinking about this map but I'm still a little lost. In particular, I'm not convinced this map is bilinear, hell, I'm not even sure what f(x, ei + ej) would be.

[rocky cloak]

Well, you want the map to be bilinear, hence it should be
f(x, ei) + f(x, ej)

[languid trellis]

I would like to say (x, ei + ej) = (x, ei) + (x, ej) but I can't justify this jump, because we are just in the cartesian product

oh, we just define it to be bilinear

[rocky cloak]

Like the task is to cook up a bilinear map

[languid trellis]

and I am playing god so I can just declare it bilinear

[rocky cloak]

Yeah, you can define it however you like. Just as long as it's well defined

[languid trellis]

right

[rocky cloak]

You know a general element of MxN looks like
(x, sum_j rj ej)
so you define f of that to be x*ri

[languid trellis]

in some sense then, a general element of M \otimes N is entirely defined by the x \in M and the scalar in front of ei

phi = f \circ i (the inclusion MxN -> M \otimes N) is an isomorphism between M and M \otimes N I feel

[rocky cloak]

Yeah, so M(x)N is in fact isomorphic to the direct sum of copies of M (same amount as rank of N)

[languid trellis]

oh that would make more sense actually

So I should define an n-multilinear map in a similar way, Claim there is an isomorphism to M + ... + M, and show it to be so by jsut writing an inverse

Let me write up what I'm thinking more fully

[rocky cloak]

You might be able to do something like that yeah. But it's a little clunky for my taste.

At that point I would rather throw some machinery like left adjoints preserve colimits onto it

[languid trellis]

Unfortunately i have no such machinery at my disposal

[rocky cloak]

Constructing maps to a tensor product is always pain

[languid trellis]

I'll give it a go and if it's too much pedantry then I'll just move on haha

I appreciate the help jagr

[languid trellis]

Does this just follow from ( mr ⊗n) = (m ⊗rn) in the tensor product? So, (a mod I ⊗ b mod J) = (1 mod I(a) ⊗ b mod J) = (1 mod I ⊗ ab mod J) where the first equality follows because R/I is an (R/I,R)-bimodule and the second follows from the relations imposed on the tensor product?

[delicate orchid]

I'd do it like (a+I) (x) (b+J) = a((1+I) (x) (b+J)) = (1+I) (x) (ab+J) so u don't need to worry about bimodule nonsense but either is fine

the "commutative" is doing a lot of heavy lifting here

[languid trellis]

wait

[delicate orchid]

waiting

[languid trellis]

i was going to ask why we can left multiply by "a" but here the left action on R/J is defined by first mapping a : R -> R/J by the canonical surj. then multiplying as per usual (in the ring R/J)

okie dokie

thanks wew

[delicate orchid]

np boss

left/right structure doesn't really matter cause R is commutative btw. Everything is a (R,R)-bimodule

[languid trellis]

mhm

[delicate orchid]

but not every (R,R)-bimodule is a two sided R-module

[languid trellis]

R not commutative i guess

[delicate orchid]

no even for R commutative

[languid trellis]

huh

give example

[delicate orchid]

the example I know is to let the subring of diagonal matrices in M_whatever(R) with R commutative (which is itself commutative) act on M_whatever(R)

the left/right actions are different

[languid trellis]

wati waht do you mean by left/right actions are differnet

[delicate orchid]

so like, given a diagonal M and some matrix N, it need not be the case that MN = NM

[languid trellis]

oh right

[delicate orchid]

maybe "two sided module" is the wrong terminology for what I mean

[languid trellis]

so what we're saying is that M_n(R) isn't commutative

[delicate orchid]

I'm saying that the centre of M_n(R) is strictly smaller than some commutative subring

[languid trellis]

oh we're acting on the whole matrix ring

[delicate orchid]

yurrr

[languid trellis]

i'll let that sit in the noggin

[rocky cloak]

Another example: let R = k[x] be the polynomial ring, and let M = k be the module where f(x)m = f(1)m and mf(x) = f(0)m

[languid trellis]

This example feels very constructed but I still quite like it

[south patrol]

I mean it's just there's no reason they should be the same

And this is just an easy couple of homs

[rocky cloak]

It is in some sense a general example. If R = k[x] then any (k-linear) bimodule is given by a vector space M and two commuting linear endomorphisms A and B such that
f(x)m = f(A)m and mf(x) = f(B)m

In general a bimodule is exactly the same as a module over R(x)R^op. So for R=k[x] a bimodule is just a k[x, y]-module

[molten viper]

What exaclty does this problem from Atiyah mean? https://media.discordapp.net/attachments/1260434282644504646/1265198559611846676/image.png?ex=66a0a37a&is=669f51fa&hm=4c5359f43c04a5c0319827b1100299731766422461ce8bcd4c78ff43104fc514&

[mighty kiln]

Find the points and "draw" the topological spaces

[molten viper]

What would that look like for spec(z), assuming thats the simplest

[languid trellis]

I'll come back to this message once I know what the words mean haha. Thank you

[mighty kiln]

The maximal ideals are (p) and there is a generic point (0)

So it would be infinitely many closed points indexed by primes plus a generic point over them

[rocky cloak]

And I guess it's nice to think about dimension/codimension. Like the maximal ideals are points, and then (0) is curve passing through all those points.

[molten viper]

“Over” in the sense that (0) is a subset of (p)?

[mighty kiln]

Yea

Also in that the closure of {(0)} is the whole space

[molten viper]

Idk much topology yet so some of this goes over my head haha

[mighty kiln]

Oh

Then you can just draw the poset of prime ideals

[hidden cairn]

just proved the homomorphism from G/N to Z is well defined if ker phi = N. now i see that quotient groups actually do something (they bring another homomorphism to the table)

[knotty badger]

Yay! Do you want a challenge?

[hidden cairn]

sure

[knotty badger]

Oh wait

You want N to be contained in ker phi

It need not be equal to ker phi

[hidden cairn]

oh

lemme check my proof again

yeah

okay just inclusion

not necessarily equality between kernel phi and N

but G/N is a group iff ker phi = N

phi : G ---> Z

psi : G/N ---> Z

psi(gN) = phi(g) iff N \subseteq ker phi

but G/N is a group iff ker phi = N

[tribal moss]

That definitely belongs in #algebraic-geometry rather than here, by the way.

[hidden cairn]

so already we have ker phi = N no?

[distant summit]

I'm guessing for this you can just say that 2n≠n! for n>3, so it is impossible to have a map from a dihedral group to the symmetric group that is surjective, let alone bijective

[knotty badger]

You only need N to be a normal subgroup of G for this

[hidden cairn]

but we have this proposition:

hmmm

so phi is just any homomorphism

N is a normal subgroup, because there exists some homomorphism f ( f != phi) with kernel f = N. but if we take any other homomorphism phi: G ---> Z, then we get a homomorphism psi: G/N ----> Z iff N \subseteq ker phi

okay i think i get it now

[molten viper]

Soz

[delicate orchid]

one of the strangest exercises I've ever seen

[sonic coral]

literally

[distant summit]

Tbf it was from lecture notes rather than a textbook

So it's prolly intended more as a "is your brain working?" type exercise

[knotty badger]

Neat!

Do you still want the challenge?

[hidden cairn]

yeah

[knotty badger]

Ok so, I talked about the “universal property” of the quotient, right? How it relates to all other groups in the universe?

Specifically, a group hom G/N -> Z naturally corresponds to a group hom G -> Z sending N to the identity

[hidden cairn]

yes

[knotty badger]

What about if we just have an arbitrary subset S of G? Is there some group “G/S” such that group homs G/S -> Z naturally correspond to group homs G -> Z sending S to the identity?

That is, S is contained in the kernel (but not necessarily equal)

[hidden cairn]

"G/S" becomes a group only when S is a normal subgroup. what is the definition of G/S for an arbitrary subset S?

cuz if G/S is a group, then S is necessarily a normal subgroup and G/S is a quotient group and then we have the same question as before. so i dont properly understand what youre asking

[delicate orchid]

they want you to say G/<<S>>, the quotient of G by the normal closure of S. Calling the group that satisfies the universal property stated "G/S" is misleading

[south patrol]

Nice

[shell pilot]

In this video, they call 1 + 5Z a coset, but in the book I'm reading, they say a coset is a subgroup. Which is correct?

[coral spindle]

Nowhere do they say a coset is a subgroup

There is only one coset of H which is a subgroup, namely H

[shell pilot]

Yeah

[coral spindle]

Are you forgetting that the group operation for Z is +?

So cosets of a subgroup H of a group with additive notation look like g+H, not gH

[shell pilot]

Ok, you're right. I misread the paragraph

[coral spindle]

You should elaborate on your confusion if this doesn't answer your question

[shell pilot]

All good! Just need to read the paragraph closer haha

Thanks @coral spindle

[knotty badger]

Yeah G/S wouldn’t be the right notation, but giving the correct notation would also be a spoiler…

[shell pilot]

Here is hopefully an easy question....but why is Z/Z_5 not a subgroup of Z if it contains all the elements of Z?
{Z_5, 1+Z_5, 2+Z_5, 3+Z_5, 4+Z_5}

[coral spindle]

What is Z/Z_5

What does this notation mean

Z_5 is not a subgroup of Z so I don't know what this means

[rotund aurora]

Is the abelianization of SL_n or GL_n known?

[coral spindle]

I can't remember for sure but I think SL_n is perfect over big enough fields, and GL_n's derived subgroup is SL_n with quotient the field. But there will be some conditions on the field.

Algebraic closure should do the trick

[shell pilot]

Z/Z_5 is the factor group of Z by Z_5. So it is the set {Z_5, 1+Z_5, 2+Z_5, 3+Z_5, 4+Z_5} where Z_5={...,-10,-5,0,5,10,...}

[coral spindle]

No this is meaningless

Z_5 is not a subgroup of Z

Do you mean 5Z?

Z_5 is defined as Z/5Z

[shell pilot]

Yes lol

Sorry

[coral spindle]

Furthermore, your claim that Z/5Z contains all elements of Z is not true

Cosets of 5Z are not elements of Z

They are sets of elements of Z

Z contains infinitely many elements. Z/5Z contains 5 elements.

[shell pilot]

Ah, ok, so it's not a "combination" of all the cosets

[coral spindle]

No.

[rotund aurora]

btw, @coral spindle it is actually true that every p-group embeds in a p-group generated by 2 elements (all groups here finite). There is this statement about countable groups that say that you can embed them in a group generated by 2 elements, if you watch out the construction for a p-group you can make the resulting group a p-group. This is done in this paper, it uses wreath product twice it is pretty nice https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/jlms/s1-34.4.465

I thought maybe you would find it interesting.

[shell pilot]

Thank you very much again

[coral spindle]

Neumann and Neumann paper spotted

Wow this is powerful

[rotund aurora]

oh that's interesting. But what about SL_n(Z)? (I forgot to include the ring fo coefficients, mb). iirc, the abelianization of SL_2(Z) is Z/12Z

[coral spindle]

I have no idea, the classical groups over non-fields is beyond my ken

[sly crescent]

I’m pretty sure the abelianization of SL_2(Z) is Z/4Z

[coral spindle]

No, the result that it's Z/12Z is well-known

bringing up a reference now

[sly crescent]

Oh

[coral spindle]

See theorem 3.8 here: https://kconrad.math.uconn.edu/blurbs/grouptheory/SL(2,Z).pdf

[delicate orchid]

all groups are 2-generated this is true

[coral spindle]

@rotund aurora he doesn't have anything on SL_n for n>2 :(

[sly crescent]

I was about to ask that 😦

[rotund aurora]

yeah, what a nerd

[delicate orchid]

what's the character table

oh wait Z hahahah

what's the character table.

[coral spindle]

NO.

[sly crescent]

What about Sp_2n(Z)

[delicate orchid]

do you think I have any idea what the fuck a "sympletic" is

[sly crescent]

Do you know what a unitary group is?

[delicate orchid]

ok don't patronise me

[mighty kiln]

I'm a symplectic for you

[rotund aurora]

wait SL_n(Z) is a quotient of SL_2(Z) for n>=5

[delicate orchid]

WHAT

@coral spindle confirm/deny?

[rotund aurora]

actually this is kconrads pdf

[sly crescent]

A symplectic group is a special case of that

[delicate orchid]

yeah I googled it on limewire

I really do not want to think about Sp_2n(Z)

[coral spindle]

What the fuck

What if you changed it to Sp_2n(Z/pZ) haha what if

[mighty kiln]

This just means SLn(Z) is generated by an order 2 element and an order 3 element right

[coral spindle]

What if you looked at schur indices of this group

haha

imagine

[delicate orchid]

what's the character table

[coral spindle]

that would be so wild

[delicate orchid]

,w character table of Sp_2(2)

[coral spindle]

idk ask me about normalisers

[cloud walrus]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by wewladstbh

[coral spindle]

Sp_2(q) is SL_2(q)

you need to look at least at Sp_4(q)

[delicate orchid]

https://tenor.com/view/kangerscream-cat-funny-sound-kanger-gif-10800162690937707549

[sly crescent]

Sp_2 \cong SL_2 and Sp_4 \cong Spin(2,3)

[delicate orchid]

ok I actually know the character table for that one

[coral spindle]

If you ask me to do Deligma-Lusty theory one more time I will beat ur ass

[rotund aurora]

you would have to pay attention to the relations too tho

[delicate orchid]

yo my supervisor has complained to me about t hat before. No idea what it is

[mighty kiln]

Isn't this like universal property of coproduct

[delicate orchid]

huh?

[south patrol]

Universal property

[coral spindle]

Vile algebraic geometry (rep theory)

[mighty kiln]

Since SL2(Z) = Z/2 ⨿ Z/3

[south patrol]

Étale

[coral spindle]

Blocked and reported

[rotund aurora]

idk what the cup means, but I think you should factor by +-I first? then I think PSL_2(Z) has presentation {x^2=y^3=1}

(also on kconrad)

[south patrol]

kconrad

[coral spindle]

In group theory we write this as * and people call it the free product

[sly crescent]

PSL_2(Z) is also D(2,3,∞)

[coral spindle]

But yeah this is true but like

Not easy

lmao

"Simply prove this harder fact"

[mighty kiln]

No but like the statement feels more reasonable when phrased this way

[rotund aurora]

yeah

You can write the surjections explicitly

[rotund aurora]

but also, you know what x and y are. You just have to make sure that there is no relation between them. The proof is just a page in kconrad

[knotty badger]

Oh that’s so cool

And then it makes perfect sense why (2, 3) generated groups are quotients of SL_2(Z)

[mighty kiln]

Oop it's PSL2(Z)

[knotty badger]

Bruh

I guess that’s a quotient of SL_2(Z) already though

And a quotient of a quotient is a quotient

[marsh scaffold]

Is it true that if the permution groups of two sets are isomorphic then they are g isomorphic

But I don't think looking at the map G-sym(X) would be of help here

[delicate orchid]

use the mark homomorphism :troll_face:

[marsh scaffold]

What's mark homomorphism

[delicate orchid]

it would be circular to use marks here

[marsh scaffold]

[delicate orchid]

what I'd do is decompose each G-set into a disjoint union of it's orbits, then use the fact that each orbit is isomorphic as a G-set to G/H for some H <= G, then prove the result for those transitive G-sets. Then it's simply a matter of showing that if two G-sets have the same orbit sizes, then they're isomorphic

oh and you'll need to show that Fix plays nicely with disjoint unions of G-sets but that's pretty easy

[marsh scaffold]

Oh I did try decomposing into disjoint union of G/C(x)

Oh wait

👍

Yeah

Thanks ❤️

[marsh scaffold]

Also sorry to bother again but after we've reduced the case to transitive G sets the wouldn't just considering the action of {e} on both would give that they have the same cardinality

So they should be G iso ?

[delicate orchid]

I'm gonna be real I have forgotten how the argument goes

but yeah I believe that would work

wait

is G finite

[marsh scaffold]

Oh

Lemme check

[delicate orchid]

ohhh lorddd I've done it againnnn ohhhh heavens

[marsh scaffold]

Yeah it is

No worries 💀

[delicate orchid]

ok then yes that definitely works

[marsh scaffold]

Theyve mentioned it before the question

[delicate orchid]

I don't even want to think about marks for infinite groups @coral spindle u know anything about this? I forget if ur burnside ring pilled or not

[mighty kiln]

Does this fail for infinite G

You just need to check all the finite index subgroups right

And you get the decomposition

[delicate orchid]

there could be an infinite orbit

oh wait

it says they're also finite G-sets

[mighty kiln]

finite G-sets

[delicate orchid]

yeah it should work

[coral spindle]

I'm sorry I don't know

[rotund aurora]

what is the absolute galois group of C((T)) (Laurent series in T with coefficients in the complex numbers)?

the algebraic closure are the Puiseux series

[marsh scaffold]

Okay @delicate orchid sorry for the ping but apparently I'm stuck on why if I have two disjoint unions of G sets for the for G/C(x) and they satisfy the property in question then they should be g-iso, I was able to show it for transitive ones but I am really lost on why given two decompositions we can find a way to like match their components ??

[marsh scaffold]

If youve forgotten the problem here it is

[delicate orchid]

maybe this solution isn't the intended one. But the idea is that if you define a function taking in a H \leq G (up to G-conjugacy) and a transitive G set G/K (with K up to G-conjugacy) and gives you Fix_H(G/K), then these functions are linearly independent allowing you to "match their components"

In hindsight this approach is a lot more effort and you can probably do it via something dumb like Burnside's fixed point lemma

if you actually write out the values those functions take you get a lower triangular matrix after reordering, fun fact

but perhaps I should try and think of a better way of doing this

[marsh scaffold]

No tell me more about it pls

[delicate orchid]

no problem lol

the function I described is most often denoted m(H, K) (which is just #Fix_H(G/K)) and is called "the mark of H on K" for some reason

these marks are ring homomorphisms in the sense that m(H x H', K) = m(H, K)m(H', K) and M(H u H', K) = m(H,K)+m(H',K)
they're like characters but for G-sets rather than G-reps

[marsh scaffold]

Oh

Wait im a bit unclear on what the function is exactly

f takes what was input

[delicate orchid]

it sends H and K to #Fix_H(G/K)

[marsh scaffold]

Ohkay

these functions are linearly independent

But we only have one such function?

[delicate orchid]

here's this matrix for D_8

[delicate orchid]

as K varies

[marsh scaffold]

Oh okay

[delicate orchid]

really I should write m(-, K)

[marsh scaffold]

Okay so let me try proving their linear independence

[delicate orchid]

it suffices to show that the matrix M_{H,K} := m(H,K) is lower triangular, which it is

[marsh scaffold]

Thanks for your patience but i might ping you again if i fail 😞😭

[delicate orchid]

be warned ⚠️ two G-conjugate subgroups will give you the same mark

but that's ok because G/K \cong G/K' as G-sets for G-conjugate K, K' so it's all fine

[knotty badger]

Woah you use matrices for this?

[delicate orchid]

there's almost certainly an easier way to prove it but I take any opertunity to talk about marks and the burnside ring

and it requires me to think less

[sly crescent]

What is a mark

[delicate orchid]

I should really know the proof

https://en.wikipedia.org/wiki/Burnside_ring#Marks

but like, can anyone remember why characters classify G-reps up to iso? fuck no!

actually yes I can but that's besides the poin...t...

wait

[sly crescent]

Huh

[delicate orchid]

yeah nvm it's the same argument, conjugate subgroups have equal marks and G quotiented out by them are isomorphic as G-sets

identical argument just over F_1

I'm going full babble mode now chat 🐺 🌕 🌙

[sly crescent]

I had a thought

Homogeneous spaces can be thought of as elements of Burnside rings of Lie groups

[delicate orchid]

are they just a space with a lie group action

[sly crescent]

They’re a space with a transitive Lie group action

[delicate orchid]

then not only are they elements, they form the basis of the ring as a Z-module

[marsh scaffold]

Wait okay so why is this true

[delicate orchid]

ok we'll need a finiteness condition in order for that to be true

[delicate orchid]

m(H,K) is only non-zero when H is conjugate to a subgroup of K

m(K,H) = #Fix_H(G/K) = #{gK in G/K : for all h \in H, hgK = gK} = #{gK in G/K : HgK = gK}
HgK = gK <=> g^-1HgK = K <=> g^-1Hg \leq K

in hindsight why the fuck did I choose this route to try and prove it there's way more machinery here than I thought

[marsh scaffold]

Yeah I guess this will be really useful later on 🔥🔥

[delicate orchid]

so now we just order our conjugacy classes of subgroups so if H_i is conjugate to a subgroup of H_j, we put H_i before H_j in the table

[marsh scaffold]

Really cool stuff 🔥

[delicate orchid]

boom upper triangular, and it's clear that m(H,H) = |G/H| so the diagonal is non-zero

[delicate orchid]

have you encountered characters in rep theory before?

[marsh scaffold]

No 😭

I'm just undergrad

[delicate orchid]

that's ok, when you do you'll see a LOT of parallels between them and these marks

[delicate orchid]

and now finally....... it is no longer circular....

[marsh scaffold]

Okay so now that we've established that these functions are LI

Ow what do we do

Now

[sly crescent]

I don’t think they do

[delicate orchid]

we need some finiteness condition on the sets the lie group can act on

like, there needs to be a finite number of orbits

[sly crescent]

Closedness*

[delicate orchid]

I have no idea what the difference is

I haven't seen a topological group in 2 years

[sly crescent]

That gets rid of stuff like SO(3,R)/SO(3,Q)

[delicate orchid]

write your two G-sets as $X \cong \bigsqcup_{i=1}^{|X/G|} G/K_i$ and $Y \cong \bigsqcup_{i=1}^{|Y/G|} G/K'_i$ where the $K_i, K'i$s are some representatives of conjugacy classes of subgroups of $G$. Now letting $H \leq G$, $#\text{Fix}H(X) = \sum{i=1}^{|X/G|} m(H,K_i) = \sum{i=1}^{|X/G|} m(H,K'_i) = #\text{Fix}_H(X)$. But by linear independence of marks, we therefore have that $K'i = K_i$ for all $i$, and so we have $X \cong \bigsqcup{i=1}^{|X/G|} G/K_i \cong Y$

it's crying at me but that's literally what I wanted

[normal oasis]

Hi, I am just wondering about a compact way of writing the identity element axiom of groups. $\exists e \in G \implies \forall a \in G \implies a \ast e = a$. I want it to not use any words like such that. So I thought first-order logic style was a good bet, but is my expression ambigous? Like do I need ∃e∈G⟹(∀a∈G⟹(a∗e=a)) nested parantheses or the parenthetical scope thing ∃e∈G(∀a∈G(a∗e=a))?

[cloud walrus]

Optima
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

Wew Lads Tbh

[sly crescent]

Apparently I wasn’t the first person to consider this https://www.heldermann-verlag.de/jlt/jlt18/fauskla2e.pdf

[delicate orchid]

I've thought of a way to do it but it's gross

[marsh scaffold]

Wait so we have for all subgroups of H the sum of those marks are same , then how by the linear independence part comes in

[delicate orchid]

if they weren't linearly independant that decompistion might not be unique so we can't conclude that K'_i = K_i for all i

really what I should write is instead of K'_i = K_i is "G/K_i \cong G/K_i' as G-sets"

slight blunder on my part

[delicate orchid]

actually no I haven't and I'm not sure it's possible. I remember @topaz solar chatting about this a few months ago where you needed to include the identity element as a special thing in the F.O.L for groups but I don't remember the rigorous formulation of any of it lol

[marsh scaffold]

All good just that I am still stuck on

The sums to be equal in each subgroup of G and m(_,K) are LI so why are the marks of each respective component same

[delicate orchid]

by linear independence? If they were different then we'd immediately have a linear depencence of marks

we're saying that the sum of two "basis vectors" is the same, so each "basis vector" must appear the same number of times in the sum

[topaz solar]

$\exists e.\forall a.(ae = a \land ea=a)$ works, but note that this lets you have substructures where things are no longer groups (since you’re now looking at subsemigroups), and if you don’t specify e of G maps to e of H in a mapping, you can have groups mapping to subgroups of semigroups

[cloud walrus]

Sharp for President 2036

[delicate orchid]

oh yeah you just use a .

also sharp u really need to learn to tailor ur responses to an audience dawg

[topaz solar]

Yeah probably

[delicate orchid]

but that's ok, it'll make you sound smart during talks

[marsh scaffold]

Okay thanks @delicate orchid i got to learn a lot of new things , as for the marks part maybe I'll understand this problem better aftee maybe like spending some more time with marks

[delicate orchid]

I doubt they'll come up again for you but sure

lemme find a good source

https://ncatlab.org/nlab/show/table+of+marks

just kidding!

[marsh scaffold]

Thank god

[topaz solar]

To make this more clear, @normal oasis, there’s a subtle difference between “there is an identity e” and “we pick out an explicit identity e”, but if the question is just notation, you do not use the => arrow between quantifiers no

[topaz solar]

I definitely go into silly digressions too much

[delicate orchid]

I can't believe this there are NO good sources.

https://en.wikipedia.org/wiki/Burnside_ring is unironically the best one I've found so far

[delicate orchid]

section 1 of this is our new winner...

this is DIRE.

[delicate orchid]

anyway I'm gonna get distracted reading this now

[topaz solar]

Make one

[delicate orchid]

I should at some point but there's probably some paper by tom Dieck or whomever from the 1820s that covers it all

[delicate orchid]

found it:
https://www.lamfa.u-picardie.fr/bouc/burnamscorr.pdf @marsh scaffold Bouc is cracked at this sort of thing so this will probably be great

[topaz solar]

Make a new one that someone could actually find

[delicate orchid]

omg his proof is basically the same as mine

am I... swag?

nah now THIS is the swag way of doing it

[topaz solar]

Ok this kind of homotopy trickery going into finite group combinatoric-y things I’ve recently seen some comments on wrt logic

No I don’t have cursed theorems stashed, it’s more like someone asking if homotopical combinatorics could be used/could arise from so and so logic nonsense

(Which a lot of logic nonsense is related to silly finite combinatorics locally)

[delicate orchid]

well it definitely exists as far as group actions go

[topaz solar]

[round jay]

Any tips for this guys? feels very finicky. So far my approach consists of showing that the size of H is less than the size of the subgroup of A_n which fixes 1.

[normal oasis]

So those before 5 are trivial😂🙂🙃🫠

Oh and above 2

Btw for 2 elements is the only possible operation a cycle? Like a 2 cycle bidirectionally?

[acoustic igloo]

that and the identity

[normal oasis]

Oh

Forgot

Too trivial 😂

[main flower]

So part (a) is easy and G is abelian implies H is normal is trivial, but proving H is normal implies G is abelian has been difficult for me. I spent a long time doing scratch work and got nowhere. The best I could do was create a homomorphism between GxG and G where H was the kernel only if G was abelian but I do not think that is sufficient.

[normal oasis]

Isomorphic is the best word in math no?

So you need to prove a biconditional. Then it has to be something that is scalefree. Because it needs to be an extension of the normal subgroup (if it isn't the other way around) maybe at least to 4 times that

If I am trippin' just tell me btw

[topaz solar]

Well, is this an exam

[topaz solar]

What is bro on about

[main flower]

No it’s old practice qualifying exams. I can send a link to the webpage I got them from if it’s important.

[topaz solar]

Very well then

Abelian -> normal is trivial yeah

But here’s a q, whats the definition of normal

[normal oasis]

By scalefree I mean something that is true for the normal subgroup and for a transformed version of it by a scalar-like thing to make it into abelian group

[topaz solar]

What

[topaz solar]

I mean this as a leading question of course

[main flower]

yH=Hy for all elements y in GxG

https://math.utdallas.edu/graduate/qualifying-exam-archive/

here's the list of exams

[topaz solar]

Ok so, let’s say yHy^-1 = H is equivalent yes?

[main flower]

yes

[topaz solar]

Suppose G is not abelian

What does that mean

As in, how do you know a group is not abelian

[main flower]

if xy does not equal yx for some x,y

[topaz solar]

Ok xy \neq yx

So x \neq yxy^-1

[main flower]

ahh ok I get it.

[topaz solar]

Indeed

[main flower]

thank you for the help. It's been a while since I have done abstract algebra so my skills are rusty

[normal oasis]

If we make truth table it would be:
...

jk, I am too lazy

Anyway just the complement of equals in the permutation space

[main flower]

Wow I was way overthinking it.

[normal oasis]

0,1 for all variables ya know

[normal oasis]

Nah, proof by negation are annoying and doesn't ever give you insights, jk

[topaz solar]

Yeah it’s one of those

[main flower]

Wait I don't think this is enough

[topaz solar]

Ok

What is conjugation by (g1, g2) on (x, y)

@main flower

[main flower]

(g1x(g1)^-1, g2y(g2)^-1)

@topaz solar

[topaz solar]

Indeed

Well, care about the diagonal, so we’re conjugating (x, x)

But g1, g2 can be different right?

[main flower]

yes in my scratch work,since we haven't shown abelian yet, g1 and g2 can be different

[topaz solar]

Let’s pick some pair of elements where xy and yx are different

what can we do about conjugating (x, x) with this

[main flower]

conjugating (x,x) with (y,y)?

[topaz solar]

No

Why don’t we conjugate (x, x) by (y, e)

[main flower]

(yx(y)^-1, ex(e)^-1)

[topaz solar]

Yeah

Well, e is the identity

So

Is that in the diagonal?

[main flower]

what do you mean by the diagonal?

oh H

it is not in the diagonal

(e, e) is in the diagonal but (yx(y)^-1, ex(e)^-1) is not

[vapid vale]

done

[main flower]

I get the idea that I am supposed to show x1x2=x2x1 but every time I try to manipulate the original formulas I come up with nothing

[topaz solar]

Why not

Why is (yxy^-1, x) not in the diagonal

[main flower]

because yxy^-1 \neq x unless y=e

oh wait

or yx=xy

[topaz solar]

That’s not true, but yes we chose x and y such that they didn’t commute

So (yxy^-1, x) is not (x, x)

So the diagonal is not normal

[main flower]

I think I understand.

Yes I have it thank you so much @topaz solar

[glad osprey]

You could prove this directly without contradiction, right? Since (yxy^-1, x) is in H, we have yxy^-1 = x?

[topaz solar]

Yes

[main flower]

Ahhh

[acoustic igloo]

there can be more than one isomorphism between 2 groups right?

[topaz solar]

Yes

Consider the cyclic group of order 3

{0,1,2} basically right?

[acoustic igloo]

yes

[topaz solar]

Well, C_3 iso to C_3 by 0 -> 0, 1 -> 2, 2->1

That’s not the identity

[acoustic igloo]

neat

thanks

[topaz solar]

Having multiple isomorphisms is less “how same-y are they”

It’s more like, “how homogenous is one of them”

Since it’s really just like, Aut(M)

[acoustic igloo]

because an isomorphism composed with an automorphism is an isomorphism

[topaz solar]

Yep

And two isomorphisms gives an automorphism

f^-1 g

[sly crescent]

Is (Z_p,*) isomorphic to (Z_p,+)*Z/(p-1)Z?

[glad osprey]

Also, for cyclic groups, there is one automorphism for each generator, which in this case are 1 and 5

[acoustic igloo]

thanks

[lone niche]

Can you elaborate more on the homogeneous part please? In the sense of how is "homogeneous" intuitively measured, and would more heterogeneous groups have lower aut groups?

[topaz solar]

Yea

Sorta

I mean, if I have more automorphism that move a point x, there’s more points in x which look the same

[lone niche]

Ahhhh

[topaz solar]

You can like, get more explicit about this with logic stuff, but similarly for pairs of objects etc etc

[lone niche]

I understand thanks!

[topaz solar]

I mean, contrast with R, where every point is distinct

Because you can pin em down by exactly what rationals are above it, below it, because you have to preserve being positive, etc

[acoustic igloo]

you can't negate them?

[topaz solar]

On the other hand, there’s the Rado graph, where any isomorphism between finite subgraphs extends to an automorphism

[topaz solar]

No, try and prove that there’s no field automorphism of R except the identity

[acoustic igloo]

oh fields have more operations

[topaz solar]

Ye

[acoustic igloo]

i was thinking of a group

[topaz solar]

Yeah fair, there’s a lot of automorphisms as a group

(It’s a very large Q vector space!)

[barren sierra]

I've proven that $|Z_k(G) | \geq p^k$. Basically what I used is that $Z(G)$ is a $p$-group and so $|Z(G)| \geq p$. Then each $Z(G/Z_i(G))$ is a $p$-group. Thus $|Z(G/Z_i(G))| \geq p$ which gives that $|Z_{i + 1}| \geq p \cdot |Z_i(G)|$.

[cloud walrus]

Spamakin🎷

[barren sierra]

But this only shows that the nilpotency class is $\leq a$ not $\leq a - 1$.

[cloud walrus]

Spamakin🎷

[barren sierra]

My only other intuition for this fact is that if $|G| = p^2$ then $Z(G) = G$ and the nilpotency class in that case is $1 = 2 - 1 = a - 1$.

[cloud walrus]

Spamakin🎷

[topaz solar]

Ok so, you know |Z_i(G)| \geq p^i

Yes?

Once you reach p^{a-2}, you have p^2, so you jump up 2 at once

Or you already did, in which case you’re already on pace, because you add at least 1 to the exponent at each step

Yeah?

iirc, being nilpotent or solvable of (ofc finite) length n is first order?

[barren sierra]

oh I can do this?

ah yea I can ok