groups-rings-fields

[plucky flicker]

How do I get a representation from G acting on itself via conjugation?

[robust pollen]

A representation of G on V is a group homomorphism G -> Gl(V), and conjugation with a fixed element is an (inner) automorphism of G.

[plucky flicker]

Uhh right

Thanks

But wait, what’s the vector space in that case?

[robust pollen]

what do you mean?

[plucky flicker]

Okay so a representation is a group homomorphism from G to GL(V). G acting on itself via conjugation is a group automorphism, but how does it induce a map from G to GL(V) for some V?

[robust pollen]

So, twisting "representation" by automorphisms is a general feature of the concept of "representation".
Take a group representation $\pi \colon G \to Gl(V)$. This is by definition a group homomorphism. Now take some automorphism $\phi \in \operatorname{Aut}(G)$ of $G$, which is bi definition an invertible group homomorphism from $G$ to itself. Surely then $\pi \circ \phi \colon G \to Gl(V)$ will be a group homomorphism, and thus again a representation.

[cloud walrus]

expectTheUnexpected

[robust pollen]

In general, if you have some algebraic object, then you can look at its automorphisms (i.e. "bijective" self-maps that preserve all the algebraic structure in question). This is a group, denote it Aut(X). If you denote your algebraic object by X, then we would call an action of a group G on X any group homomorphism G -> Aut(X). And all of these reps can be twisted by automorphisms of G - or even of X itself (although I never see this, now that I think about it)

[plucky flicker]

It’s clear now, thanks

[plucky flicker]

https://math.stackexchange.com/questions/1699593/sum-of-elements-in-row-of-character-table-is-positive-integer

In this proof what does that notation mean regarding the dimension in the second row, and why is that true that it equals to that inner prod?

[robust pollen]

We have [ U \cong \bigoplus_{X \text{ irreducible}} n(U)X X] and [V \cong \bigoplus{X \text{ irreducible}} n(V)X X], where the coefficients are natural numbers and e.g. $2 X$ means $X \oplus X$.
In particular, the characters are [\chi_U = \sum{X \text{ irreducible}} n(U)_X \ \chi_X] and similarly for $V$. Can you take it from here?

[cloud walrus]

expectTheUnexpected

[plucky flicker]

I guess we only need this for V, because U is irreducible, in fact we want to compute the sum of the row corresponding to U

or am I missing something?

in the char table only irred chars appear

[robust pollen]

In the answer, U is not assumed to be irreducible.

At least not at the beginning. Anyway, the dimension formula is true in particular for irreducible reps

[plucky flicker]

But what's that space? Space of class functions of G?

[robust pollen]

No, it's the space of homomorphisms of G-representations from U to V

Don't worry about the notation $\mathbb{C}[G]$. I will rxplain it, but don't worry if you don't understand it: This is known as the group algebra of $G$, it is basically a complex vector space which has a basis ${e_g | g \in G}$ and you define on it a multiplication $e_g \cdot e_{g'} =e_{g g'}$. It turns out that modules over this algebra are exactly the same thing as representations of $G$.

[cloud walrus]

expectTheUnexpected

[plucky flicker]

So basically we have linear maps between vector spaces preserving the G-action?

these maps form a vector space and the dimension of this vector space can be calculated from the inner product of characters

[robust pollen]

Yes

[plucky flicker]

Back to this: if pi is irred, then $\pi \circ \phi$ is irred too?

[cloud walrus]

Boti

[robust pollen]

What is an irreducible representation? How does phi act on elements in G?

[plucky flicker]

there are only two invariant subspaces

0 and V

phi is a faithful action

so $\pi \circ \phi$ is irred too

[cloud walrus]

Boti

[plucky flicker]

because pi send 0 and V to 0 and V for all g, but that composition do the same thing for another g

right?

[robust pollen]

that's what I would think too, yes. I mean, phi is just some permutation of the elements of G

[plucky flicker]

yes

[plucky flicker]

which book do you recommend to learn Galois theory?

[hidden haven]

milne's notes are good

[plucky flicker]

full title?

[hidden haven]

Fields and Galois theory

https://www.jmilne.org/math/CourseNotes/ft.html

[plucky flicker]

thanks

[eternal schooner]

Hey, I don't find any chat dedicated to group theory so I guessed this might be a good place to ask something about it, if I'm wrong just tell me. I was looking at the definition of the p-adic Heisenberg group I have to use (the one in the image). The operation defined over VxS^1 looks similar to a semidirect product but its not quite the same. Does anyone have a clue of what underlying structure there might be here?

the character \chi is just an additive character that goes from Qp to S1, it basically sends x to a exp(2\pi i x). (its not exactly that, but the idea is similar)

[viscid pewter]

this is where group theory usually goes, yes

[plucky flicker]

Can you help me to finish this proof? Or am I on a wrong track?

,rotate

[plucky flicker]

if |Z(G)| and d are coprime we are done I guess

but we don't know that

[robust pollen]

why "when g in Z(G), then |chi(g)| = d"?

[plucky flicker]

Shur's lemma

[robust pollen]

Schur's lemma doesn't say chi(g) = id though, it says chi(g) = x * id, where x is some number

[plucky flicker]

yes, and |chi(g)| = d iff phi(g) = x*id_V, where phi is the representation

[plucky flicker]

we have that g^k = 1 for some k, since G is finite

so x must be a root of unity so |x| = 1

[robust pollen]

right, I'm dumb lele

[plucky flicker]

any idea? I'm not sure this proof works :c why would d and |Z(G)| be coprime in general

[jagged field]

if I have a question about congruence classes would that go here?

[plucky flicker]

yes

[jagged field]

so I am a bit confused on the idea of congruence classes could someone explain whats the difference between this problem and solving x^2 = 2 (mod 119)

[chilly ocean]

no difference

[jagged field]

so what does Z_119 exactly mean? @chilly ocean

and what does the [] do

[chilly ocean]

Well, [] means the equivalence class

Because Z_119 is the group with equivalence relation that says that two numbers x,y are the same if x=y mod119

[x] means all numbers that are congruent to x mod 119

so eleements of Z_119 are [0],[1],...[118],

which are just numbers mod 119 pretty much

[jagged field]

so if the solutions I get for this x^2 = 2 (mod 119) are x = 11, 45, 74, 108

how would i write the anwsers using [x]^2 = 2 in Z_119

would it just be x = [11] or x = [45], or x = [78] or x = [108]?

[chilly ocean]

yes

[jagged field]

so im so confused why do they even need to write Z_119 or square brackets why not just say solve x^2 = 2 (mod 119)

[chilly ocean]

maybe to explain it better, Z_119 is basically all integers, but the ones that are equivalent mod 119 are treated the same. So for example 1=120, 2=121 etc This relation divides integers on 119 different sets

[robust pollen]

74 btw

[chilly ocean]

So Z_119 is short for Z/119Z if you know groups.

[jagged field]

so if I show x^2 = 2(mod 119) is that a proper way to represent [x]^2 = [2] in Z_119

[plucky flicker]

you can think of [x] as the box of all numbers whose division by 119 gives x as remainder

[chilly ocean]

yes, this is the same thing, just different notation pretty much

[jagged field]

wait but if x = [11] or x = [45], or x = [74] or x = [108] then whats x?

[chilly ocean]

exactly what you wrote, which means all integers congruent to lets say 11 mod 119

[jagged field]

hm im gonna need to google for a bit

im still confused

[plucky flicker]

[11] = {119*x + 11 where x is any integer}, but we prefer to work with small representative because all the work you can do with this stuff is independent from the representatives

so whatever you work with 11 or 119*1 + 11 you get the same answer

[robust pollen]

Side question, how did you prove that you have all answers?

[plucky flicker]

what do you mean?

ohh it's for Soap?

[jagged field]

oh so [11] is just representing the set of all solutions for the equality x = 11 (mod 119)

[plucky flicker]

that's right

[jagged field]

question how do we solve non linear congruences manually?

because the way i got the solutions for this is I ran through python

but im unsure how youre suppoused to get this by hand?

[next obsidian]

,w factorize 119

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Chmonkey 2.0#9767

[next obsidian]

Rip

If it’s an integral domain then there’s at most 2 squareroots

So you can just find 2 of them and you’re done

Anyway I think the way you solve it is by doing number theory shit lol

Writing a script is a good way to handle it tho

Very realistic answer to the problem

[jagged field]

whats a integral domain?

[next obsidian]

Uh

It’s when multiplying two non-zero things is always non-zero

This fails cuz 119 isn’t prime

Like 7•17 breaks this

The point is if you know this then you’re looking for roots of x^2 - 2

Then if you have two roots alpha and beta

This turns into

(x-alpha)(x-beta)

So now if you have any root, call this one gamma

You know (gamma - alpha)(gamma - beta) = 0

[jagged field]

wait so is the fact of me having 4 solutions wrong?

[next obsidian]

But like, now if you had an integral domain

One of gamma - alpha or gamma - beta is 0 so gamma is either alpha or beta, so there’s at most two solutions

No

It’s cuz Z_119 isn’t an integral domain

Pls read what I wrote

[jagged field]

i havent taken number theory

[next obsidian]

Yeah me neither

[jagged field]

so how am i suppoused to solve this manually lol

[next obsidian]

¯_(ツ)_/¯

Enumerate all the possibilities or some shit idk

[jagged field]

bruh

lol

this is torture

lol

[robust pollen]

Soap, do you by any chance know chinese remainder theorem?

[jagged field]

ye

@robust pollen

[plucky flicker]

then you can compute idempotents with that

[jagged field]

whats idempotents

im in first year for reference so if its something advance idk it

[plucky flicker]

we call x an idempotent if x^2 = x

[jagged field]

isnt that only true for x = 1?

[plucky flicker]

nope

in an arbitrary ring nope

[robust pollen]

why idempotents though? we want x^2 =2

[plucky flicker]

uhh jeez

sorry xd

[robust pollen]

But, anyway. Z_119 = Z_7 x Z_17, so solve x^2 = 2 modulo 7 and modulo 17. If you have a solution x^2 = 2 mod 7 and y^2 = 2 mod 17, try to solve x = m*17 + y mod 7 for m, and you will get a solution mod 119

[jagged field]

im sorry the text is messing with me

so

do I solve the:

1. x^2 = 2 (mod 7)
2. x^2 = 2 (mod 17) I essentially solve this congruence set using CTR then what? @robust pollen

[robust pollen]

so these you can solve by hand, or even in your head, right?

admittedly, I didn't solve the mod 17 one in my head, too little time 😄

[jagged field]

x^2 = 2 (mod 7) means x= 3,4 then x^2 = 2 (mod 17) means x = 6, 11

now what?

[robust pollen]

So, if x^2 = 2 mod 119, then because 119 = 7*17, surely x must reduce to either 3 or 4 mod 7 and 6 or 11 mod 17

[jagged field]

hm okay?

but im still confused

on how that helps

[robust pollen]

so you know that (modulo 119), x is of the form [ (x = n\cdot 7 + 3 \text{ or } x = n\cdot 7 + 4) \text{ AND } (x = m\cdot 17 + 6 \text{ or } x = m\cdot 17 + 11 ]

[cloud walrus]

expectTheUnexpected

[robust pollen]

where m and n are natural numbers to be determined

[jagged field]

how do i determine them?

[robust pollen]

So if you "multiply out the AND", you get four possibilities. Let's look at $x = n \cdot 7 + 3 = m \cdot 17 + 6$. Reduce this again modulo 7, to obtain $3 = m \cdot 17 + 6 (mod 7) = m \cdot 3 + 6 (mod7)$, where the last equality is of course because $17 = 2\cdot 7 + 3$. Solving for $m$, you find $m \cdot 3 = -3$ mod 7, which is easily seen to have only the solution $m = 6$.
Therefore, $x = m \cdot 17 + 6 = 6 \cdot 18 = 108$

[cloud walrus]

expectTheUnexpected

[jagged field]

im so confused what you did

[robust pollen]

magic

But soap, try it yourself for, let's say $x = n \cdot 7 + 4 = m \cdot 17 + 11$. Just try to replicate my steps

[cloud walrus]

expectTheUnexpected

[jagged field]

how are you getting x = 7n + 4 = 17m + 11

[jagged field]

and im very confused on why youre doing this

[robust pollen]

What is the Chinese Remainder Theorem for you?

[jagged field]

hm

one sec let me try to understand this

because youre doing something similar to whats in my notes

but its not making sense

[robust pollen]

https://tenor.com/view/big-lebowski-the-dude-your-opinion-gif-4476328

[jagged field]

@robust pollenive been trying for 30 minutes trying to understand what you did but im lost

x is of the form $[ (x = n\cdot 7 + 3 \text{ or } x = n\cdot 7 + 4) \text{ AND } (x = m\cdot 17 + 6 \text{ or } x = m\cdot 17 + 11 ]$

what do we do from here

[cloud walrus]

Soap

[robust pollen]

Do you understand that this ("x is of the form...") follows from assuming that x^2 = 2 mod 119?

@jagged field

[jagged field]

not really

[robust pollen]

Ok. So, assume that x is a solution to x^2 = 2 mod 119. Since 119 = 7 * 17, we must also have that x satisfies (1) x^2 = 2 mod 7 and (2) x^2 = 2 mod 17. Do you agree?

[jagged field]

ye

[robust pollen]

Therefore, then, modulo 7 we must have x = 3 or 4, and module 17 : x = 6 or 11

[jagged field]

okay

[robust pollen]

So we know on the one hand that (everything modulo 119 now): x = 7 n + 3 or x = 7 n + 4, since modulo 7 these reduce to 3 and 4, but x also has to be of the form x = 17 m + 6 or x = 17 m + 11, since these reduce to 6 and 11 modulo 17

Here n and m are just some natural numbers that we want to determine

Alright?

[jagged field]

ye

[robust pollen]

In other words, x is of the form $[ (x = n\cdot 7 + 3 \text{ OR } x = n\cdot 7 + 4) \text{ AND } (x = m\cdot 17 + 6 \text{ OR } x = m\cdot 17 + 11 ]$

[cloud walrus]

expectTheUnexpected

[robust pollen]

This is equivalent to:
x is of the form
\begin{align}
& (x = n\cdot 7 + 3 \text{ AND } x = m\cdot 17 + 6) \
& \text{OR } (x = n\cdot 7 + 3 \text{ AND } x = m\cdot 17 + 11)\
& \text{OR } (x = n\cdot 7 + 4 \text{ AND } x = m\cdot 17 + 6) \
& \text{OR }( x = n\cdot 7 + 4 \text{ AND } x = m\cdot 17 + 11 )
\end{align}

[cloud walrus]

expectTheUnexpected

[robust pollen]

still agree?

[jagged field]

ye

[robust pollen]

so, in other words:
x is of the form
\begin{align}
& x = n\cdot 7 + 3 = m\cdot 17 + 6 \
&\text{OR } x = n\cdot 7 + 3 = m\cdot 17 + 11 \
& \text{OR } = n\cdot 7 + 4 = m\cdot 17 + 6 \
& \text{OR }x = n\cdot 7 + 4 = m\cdot 17 + 11
\end{align}

[cloud walrus]

expectTheUnexpected

[robust pollen]

And now try to figure out what n and m are in each row

still good?

[jagged field]

ye but how do i do this

[robust pollen]

Assume for example that $x$ is equal to
\begin{align}
n\cdot 7 + 3 = m\cdot 17 + 11,
\end{align}
this is the second line.
We can look at this equation modulo 7, right? Then what do we get?

[cloud walrus]

expectTheUnexpected

[jagged field]

wdym "look at this equation modulo 7,"

[robust pollen]

like, that's an equation mod 119. If it's true, then it must also be true if we reduce mod 7

[jagged field]

so you mean

$(n\cdot 7 + 3 = m\cdot 17 + 11) (mod 119)$

like that?

[cloud walrus]

Soap

[jagged field]

@robust pollen?

[robust pollen]

Well, that's the equation from above. I didn't write mod 119, because somewhere I wrote "(from now on mod 119)" or something like that, that was a bit sloppy, sorry 🙈

[jagged field]

okay so what do we do from here?

[robust pollen]

But now, you see, it's mod 119, so because 119 = 7 * 17, it must be true mod 7.

Write the euqation out modulo 7

[jagged field]

$(n\cdot 7 + 3 = m\cdot 17 + 11) \ (mod 7)$

[cloud walrus]

Soap

[robust pollen]

you can do better than that. What is 7 mod 7, 17 mod 7, and 11 mod 7?

[jagged field]

0 + 3 = 3m + 4 ?

[robust pollen]

Sure. Solve for m

[jagged field]

-1/3

but thats not what you got?

[robust pollen]

and -1/3 = 2

Remember, you're doing everything mod 7. While division is allowed, it is not the usual division.

2 * 3 = 6 = -1 mod 7, so -1/3 = 2 (modulo 7 of course)

[jagged field]

so do we solve for n now?

17n + 3 = (-1/3 * 17) + 11

in this?

[robust pollen]

No, you don't have to solve for n, because actually you only wanted to find out what x is, i.e. x solving x^2 = 2 modulo 119.

[jagged field]

so what do we do from here after we get m = -1/3?

[robust pollen]

And actually, you have only shown that m = 2 mod 7, so modulo 119 you would have something like $m = 7l + 2$, but that's ok:
[ 17 m + 11 = 17 \cdot 7 \cdot l + 17 \cdot 2 + 11 = 17 \cdot 2 + 11 = 45\text{ mod 119} ]

[robust pollen]

Don't write m = -1/3, write it as m = 2.

[cloud walrus]

expectTheUnexpected

[robust pollen]

So what this shows is: \begin{align*} x^2 = &2 \mod 119 \text{ and } x = 3 \mod 7 \text{ and } x = 11 \mod 17 \ &\implies x = 45 \mod 119 \end{align*}

[cloud walrus]

expectTheUnexpected

[jagged field]

but thats only one solution?

[robust pollen]

yesh

[robust pollen]

remember this

there are four equations, we just solved one of them. You will get 108 from the first line, 74 from the third, and 11 from the last

if you proceed as I have just tried to explain it

But, you also just hand in

[ x for x in range(120) if x*x % 119 == 2 ]

i guess

[jagged field]

thats what i did.. LOL

but i wanna learn how to do it manually

so for the first equation im getting m = - 1? @robust pollen

[robust pollen]

yes, m = -1 = 6 mod 7

use the 6, not the -1

[jagged field]

how are you getting this ?

[robust pollen]

What?

[jagged field]

where is the 6 coming from?

oh wait

nvm

for the fourth equation im getting m = 0? @robust pollen

cause i did 0 + 4 = 4 + 3m

m = 0

[robust pollen]

that's correct

what will be your x then?

[jagged field]

11

oh

OHHH

okok

now i get it

wait question

how would you solve x^2 = 2 (mod 17) @robust pollen

to get x = 6, x = 11

[robust pollen]

just do it. You only have to compute it for 5, ..., 16 at most, that's 12 small computations

[prisma thunder]

What’s the idea behind showing the Galois group of a finite compositum of fields over a base field F with pairwise intersection being the base field is the Cartesian product of the Galois group of each field in the compositum over the base field?

I thought First Isomorphism Theorem of groups would do it but I’m having a hard time arguing surjection—feel like a degree argument should do it but I’m trying to wrap the journey to that conclusion with a missing key detail 🤔

Gonna sit back and think about this a bit more

I think I got it now

Forgot isomorphisms imply equal dimension

[jagged field]

Could I have help with this proof

[rustic crown]

Follow their hint.

[cloud walrus]

det

[rustic crown]

how nice would it be if this was just 3 + 2 + 1 - 6

[cursive temple]

Does anyone have a hint for showing that if the blue, green and red sequences are exact, and the composition \eta \circ \pi = 0, then the yellow sequence is exact as well

these are just abelian groups with the arrows being homomorphims

[next obsidian]

Holy shit

I think this is like a general homological lemma

I’ve seen this diagram before

[cursive temple]

yeah its called the braid lemma

[next obsidian]

Have you been able to prove any amount of exactness?

[cursive temple]

unfortunately i cant find a proof anywhere

nope, i suck at diagram chasing

[next obsidian]

Let me try to see if I can show exactness at the first place

And if I can I’ll show you what I did

And you can maybe try to emulate it for the other spots

[cursive temple]

sure, thanks

[next obsidian]

Actually

Oh no you’d need to do it for arbitrary cats haha

[cursive temple]

also the braid lemma given in nlab is slightly different, and requires that the yellow chain is a chain complex

[next obsidian]

I was gonna try to use duality to reduce

Hmmm…

[hidden haven]

bruh there's the whole greek alphabet up there

[cursive temple]

pretty much yeah

[next obsidian]

My advice: if at some point you have to assume this, do it. Just try to assume more at first, then see if you can remove the extra assumption

Anyway I’m gonna try to figure this out

Oh also

This commutes right…

Everything?

[cursive temple]

yeahyeah

of course

for context this is part of an algtop course and we havent had any other diagram chasing exercises

so im not very familiar with those arguments

[next obsidian]

So first part

I can’t prove the exactness, I think maybe you have to start on the other side but

Commutativity implies the yellow is a chain complex

For beta•eta note this is he same as kappa•sigma•rho = 0•rho

For lamda•beta this is phi•tau•theta = phi•0

@cursive temple okay I got exactness at C

So take a c in ker lambda, then c in ker gamma (I am gonna say “so … a lot, it follows by commutativity or some amount of exactness”)

So pull back c to a g in G so that kappa(g) = c.

Note that I can fuck with g by anything in ker kappa, meaning if I replace g with g + a with a in ker kappa, then kappa(g + a) still maps to c

We will exploit this

[vestal snow]

Sorry

[next obsidian]

So the idea here is, if g is in ker tau we know g is in im theta so I can pullback g along theta, so find some b in B with theta(b) = g

Then, beta(b) = kappa(theta(b)) = kappa(g) = c

So we’ve shown c is in im beta like we want!

The thing is g might not be in ker theta, but I can fuck with g up to something in ker kappa so that it becomes inside ker theta

Let me explain, look at tau(g), then this is in ker phi

So we can grab a j in J with w(j) = tau(g)

Now sigma(j) in ker kappa, so I can fuck with g by sigma(j), namely I can replace g with g - sigma(j)

Now tau(g - sigma(j)) = tau(g) - tau(sigma(j)) = tau(g) - w(j) = tau(g) - tau(g) = 0

So g - sigma(j) is in ker tau like we wanted!

I think next you want to prove exactness at B, then at F

I think exactness at F will be the hardest, maybe it uses exactness at B

The idea is you want to start moving elements until you can find a way to pull it back along a map by noting it’s in the kernel of some map = image of a different map

This lets you “go backwards” through a map and is well-defined up to an element in the kernel of the map you went backwards through

This is good! Because as we saw in the proof I outlined you can sort of mess with the element you got by pulling back up to something in the kernel, and this is good, in our case it let us basically assume g was in ker tau, which we needed to pull it back along theta

Exactness at B isn’t too bad

Take b in ker beta

Theta(b) is in ker kappa, so we can write theta(b) = sigma(j)

Note that w(j) = tau(theta(b)) = 0 so j in ker w, so we can write j = rho(f)

Now, theta(eta(f)) = sigma(rho(f)) = sigma(j) = theta(b)

So eta(f) = b up to something in ker theta, cuz they map to the same thing under theta

b = eta(f) + x for x in ker theta

Now write x = alpha(a), then look at f + epsilon(a)

Note that eta(f + epsilon(a)) = eta(f) + eta(epsilon(a)) = eta(f) + alpha(a) = eta(f) + x = b

So b is in im eta

I don’t know how to do exactness at F

And I have to do other stuff (like sleep lol)

[cursive temple]

Wow

Yeah ill have to read this

[next obsidian]

I can get an x in ker eta as the image of an i in I up to something in ker rho

I however haven’t used the full strength of x being in ker eta tho

And I don’t really see how to

[robust pollen]

I got exactness at F, I think 🥳

[cloud walrus]

expectTheUnexpected

expectTheUnexpected

expectTheUnexpected

expectTheUnexpected

[cyan raft]

urH
sorry if this question is too simple
i'm still kinda confused by the proof of lagrange's theorem

in the proof, they said that they would have a subgroup H of a group G. Then, the cosets of H all have the same size (which i do understand), and those cosets form a partition of G (why? i don't understand why they would form a partition?)

so can anybody help me?

[chilly ocean]

Cosets are the equivalence classes on G

So they partition G

[cyan raft]

sorry?
i didn't get that
(i just got started on abs algebra a few days ago, so forgive me)

[chilly ocean]

Well have you learned about the equivalence relations?

[cyan raft]

uh
no

[chilly ocean]

How do you define two cosets are the same?

[cyan raft]

like

a coset is basically multiplying (or doing whatever other operation) to all members of a set with a value not in that set

i would just prove that there wouldn't be any duplicates

[robust pollen]

Can an element be in two different cosets?

[chilly ocean]

Ok so like the crucial observation is every element in G is in one of the cosets

[cyan raft]

don't think so

[cyan raft]

that's the question i was asking
(i didn't know why that was true)

(i still don't)

[hidden haven]

Each g is in a coset

[cyan raft]

right

[hidden haven]

Now suppose g is in 2 cosets aH and bH

[cyan raft]

i mean

i'm not trying to prove that 2 cosets can't overlap

(can't have anything in their intersection)

i'm trying to prove that there are no elements that are in G but aren't in any cosets

[hidden haven]

Yeah so you take this

And prove that aH = bH

[robust pollen]

if g is in G, then it's in the coset gH

[hidden haven]

Oh

mb

[cyan raft]

so then
after that
do i just prove that cosets do not intersect?
(and after that i'm done?)

[robust pollen]

You said you're not trying to prove it's a partition??

[cyan raft]

i mean
after that?

[hidden haven]

You can do it in any order

[cyan raft]

uh

ok

thanks for your help!

[robust pollen]

Yeah, I mean, Set^op = Set, so, they might as well be called sets

Wait. Is that even true?

Bope. Forget it, that was a brainart.

https://ncatlab.org/nlab/show/opposite+category#opposite_of__and_

sure, I was just making a joke which requires the highest of intellects!

(Recall also my dumb mistake)

" COSET was used in 1910 by G. A. Miller in Quarterly Journal of Mathematics. [OED] "

as per this (very good) page: https://mathshistory.st-andrews.ac.uk/Miller/mathword/

[plucky flicker]

Do you have any clue how to prove this statement?

If $g \in G$ has prime order p and $\chi(g) = 0$ for an irreducible character $\chi$, then p divides $\chi(1)$.

[cloud walrus]

Boti

[hot lake]

look at the eigenvalues of rho(g) ?

I'm not sure if irreducibility really matters

[wooden ember]

In French we say left and right classes

[plucky flicker]

what am I supposed to do with the eigenvalues?

those are p-th roots of unity, right?

[hot lake]

yes

[plucky flicker]

and the sum of the eigenvalues are 0 since $\chi(g) = 0$

[cloud walrus]

Boti

[hot lake]

yes

[plucky flicker]

it's some linear algebra shit, that the order of a matrix divides the dimension?

I don't really see why would p divide chi(1)

[hot lake]

have you heard of field theory and minimal polynomials

or algebraic number theory

[plucky flicker]

I know some Galois theory

[hot lake]

then think about Q(exp(2ipi/p))

[plucky flicker]

alg number theory is next semester

[hot lake]

you want to figure out that the minimal polynomial of exp(2ipi/p) has degree p-1

and that it is 1+x+x²+...+x^(p-1)

so that if an integer combination of pth roots of unity is 0

then that integer combination must me a multiple of 1+x+x+...+x^(p-1)

[plucky flicker]

why?

[hot lake]

because if you have an integer combination of pth roots that is 0

it means you have a polynomial of degree <= p-1 with integer coefficient that vanishes at exp(2ipi/p)

and if you know the minimal polynomial has degree p-1

then it must be an integer multiple of it

[plucky flicker]

So we have $\epsilon = e^{\frac {2i\pi} {p}}$ and we know that $\sum_{i=0}^{p-1} \epsilon^i$ = 0, right?

[hot lake]

from 0 to p-1 yes

usually it's called zeta_p

1+x+x²+...+x^(p-1) is the minimal polynomial of exp(2ipi/p) only when p is prime

use brackets { }

[cloud walrus]

Boti

[plucky flicker]

because we can write $1 + x + x^2 + ... + x^{p - 1}$ as $(x - \epsilon^1)...(x - \epsilon^{p-1})$

[hot lake]

the thing on the left has degree p-1 and the thing on the right has degree p

[cloud walrus]

Boti

[hot lake]

that equality is also true for nonprime p

so I'm not sure what the "because" is referring to

[plucky flicker]

yes, this is the p-th cyclotomic polynomial, which is indeed the minimal polynomial of the p-th roots of unity

[hot lake]

yes

[plucky flicker]

and why do I get from here that p divides chi(1)?

[hot lake]

because chi(1) would then be P(1)

[plucky flicker]

brrr

right

[plucky flicker]

but wait, if this is true then the dimension of the representation is exactly p

[hot lake]

no

P could be 2 times the minimal polynomial

or 3 times

etc

[plucky flicker]

hmm then I'm not sure why is P(1) exactly the dimension

P is an integer*the minimal polynomial of exp(2ipi/p), right?

[hot lake]

P is the polynomial whose coefficients of x^k tell you how many times rho(g) has an eigenvalue exp(2ikpi/p)

if p=5 and rho(g) has the eigenvalue 1 two times and exp(4pi/5) 3 times, P would be 2 + 3x²

the trace of rho(g) is P(exp(2ipi/p))

and the dimension of the space is P(1)

because that just counts the number of eigenvalues

if the trace is 0 then P must be an integer multiple of the minimal polynomial, and so P(1) must be a multiple of p

[plucky flicker]

Now I see it. Thanks for your help!

[covert storm]

i am having a problem with this definition, what does Φ|R = ϕ mean?

in the last line

it is bosch "algebra from the viewpoint of Galois theory" and this is the universal property of polynomial rings

[robust pollen]

How is R[M] defined here? I'm asking because it's not clear what X is.

But probably there is a canonical injection R -> R[M], so you can view R as a subring of R[M], and Phi|R simply means restriction of Phi to that subring

[covert storm]

that actually solved my problem

thanks

[chilly ocean]

I’m confused about the terminology of the going up and going down theorems

A includes into B

If q is an ideal

We can contact this to an ideal of A

Is this called

Going up

Or going down

[chilly ocean]

wdym by that and q is ideal of which ring

Maybe check out the wikipedia page it should clear it out I guess

[cursive temple]

@next obsidian @robust pollen thanks a lot for the help, i just went through the proofs, and probably wouldnt have made any progress without your help

[wooden ember]

Idk any character theory but I found this pretty interesting https://m.youtube.com/watch?v=wSWBk0LFvPc

[next obsidian]

Neither

Going up is about saying if you have a prime p lying over q, and then a prime q’ contained in / containing q if you can find a prime p’ lying over q’

It’s called “going up” or “going down” because you’re like going up a chain of primes or going down a chain, and then being able to find a prime lying other still

[chilly ocean]

Yes but

Which direction is lying over

A \subset B

p in A, q in B

Does p possibly lie over q, or does q possibly lie over p

I just don’t know where A is “above” or “below” B

[small bison]

B is bigger so it’s above

Like with field extensions you usually draw the bigger one above

[dense mulch]

Hello!

Do we know that if G is not cyclic that G/H is not cyclic?

or is that not necessarily true

(G being a group, and H being a subgroup generated by a subset)

[next obsidian]

no

Z/2Z x Z/2Z

quotient by {(0,1), (0,0)}

[robust pollen]

What's a nice example of two non-isomorphic finite groups with the same representation theory?

By this I mean that the group algebras are Morita equivalent

Or in other words, the categories of representations are equivalent. Whatever terminology you like better

[hidden haven]

Would you say that you want a simple intuitive example of this?

[robust pollen]

That would be preferred, but I'd like to just see an example tbh

also

wtf

is up with your name

or your picture

wat

[hidden haven]

wat

[robust pollen]

moldicats

[hidden haven]

[fossil shuttle]

i laughed out loud

[robust pollen]

wait what

idgi

😭

ok, i know now that a sufficient condition is "equal number of conjugacy classes". But moldi plox, what did you mean by simple intuitive? Pls assplain

[next obsidian]

Hello. Can you give a simple intuitive example of a subspace whose (covering) topological dimension is greater than that of the space containing it?

[hidden haven]

its just a stupid joke lol

says a lot about clerk's sense of humour

[next obsidian]

Can someone does is give me exampul?

[robust pollen]

Oh, lmao

I remember that question lel

now that is indeed very funny huehuehuehue

But, anyway. Canonical example is complex reps of Z_2 x Z_2 vs Z_4.

[vocal wadi]

Is a product of 2 polynomial rings a polynomial ring too?

[next obsidian]

Sure, it’s a polynomial ring in 0 variables

Maybe try to be a bit more specific?

Do you mean to ask if like
R[x] x S[y] = (R x S)[x] or something?

Or is there some ring A abstractly such that the product is isomorphic to A[x]?

Either way, I don’t know the answer

[hidden haven]

If you are taking polynomial rings over the same ring, then no

[next obsidian]

Why’s that?

I was hoping to cheese it with some dimension theory type crap, but it didn’t work

[hidden haven]

Product not integral domain lol

So there's a counterexample, idk if that will be a theorem

[chilly ocean]

those are questions

do you need help with any?

[lethal dune]

those are options

which one is true and why?

[hidden haven]

which ones have you worked out so far

[lethal dune]

ik 3 is false

[hidden haven]

ok so for 1, just realise that finite implies finitely many ideals

[lethal dune]

so 1 is false

[hidden haven]

yes

2 implies 4

so if you know that exactly one of these is supposed to be true

you know which one it has to be

[lethal dune]

lol that's one way of solving it

hmm so like if $R$ is not finite then for each element $r \in R$ we can get the ideal $\langle r \rangle_R$

[cloud walrus]

Ryu?

[hidden haven]

These need not be distinct

[lethal dune]

yeah that's what I was thinking

[hidden haven]

2 is false by taking a field lul

[rustic minnow]

Im trying to solve 33. I was just wondering if I assume equality as the relation when checking the properties of the equivalence relation or is it any relation R where n,m have the same number of digits base ten?

[hidden haven]

They are defining R by saying that 2 numbers are related exactly when they have the same number of digits

[lethal dune]

yeah lol

[rustic minnow]

well I guess im confused about the transitive property since if I given number representations for n,m in Z+ clearly reflexivity, symmetry will hold, but how do I compare with a third element from the set?

[hidden haven]

if m and n have the same number of digits, n and k have the same number of digits, do m and k have the same number of digits?

I think you are misinterpreting transitivity

[hidden haven]

All you have to check is this

[rustic minnow]

oh, right. Now I get it. Guess I was just accustomed to seeing some mathematical operand associated with the relation

And for the partition this will be a subset of Z^+ where elements have the same number of digits base ten?

[hidden haven]

ye

The partition will be into subsets of that kind

[rustic minnow]

cool, thanks

[weak oriole]

where are these from

[coarse storm]

Is it supposed to be H-stable, rather than H-invariant?

[hidden haven]

Aren't those the same thing?

[coarse storm]

Oh right. They are defined to be the same in some areas.

For me, invariance requires say hw=w, whereas stable needs only hw in W.

[hidden haven]

makes sense

[upper pivot]

what do you need help with here, and what did you try

[chilly ocean]

It is always so suss when you get weirdly cropped documents

[chilly ocean]

lol

[molten silo]

Im trying tocome up with a homomorphism mapping sym(x) to sym(y)

ls help

[chilly ocean]

what have you tried

[molten silo]

nothing really

[chilly ocean]

a place to start is to consider a bijection f: X -> Y and to try and come up with a nice bijection Sym X -> Sym Y

now you can try things

[molten silo]

im having a hard time coming up with any function that maps Sym X -> Sym Y

[chilly ocean]

if g: X -> X then how can you use f to get a new function Y -> Y from g?

try composing with f and/or its inverse

[molten silo]

f composed with g maps X to Y, so maybe (k composed with f) such that k is in Sym(Y)

that maps Y to Y

[frank fiber]

Let g:X -> X and element of Sym(X), then fgf^-1: Y->Y is an element of Sym(Y)

[chilly ocean]

or you can just give the answer. that works

[molten silo]

isnt that x-> x though

i think fgf^-1 maps Y to Y

[chilly ocean]

there you go

so you've mapped g in Sym X to fgf^{-1} in Sym Y. all you need to do is check whether this is a group homomorphism

[molten silo]

Yeah i knew what to do, it was just defining the homomorphism

[chilly radish]

I did this exercise where a unital ring of order p^2 is commutative. I used the fundamental theorem of finitely generated abelian groups to show that it's isomorphic to either Z/p^2 or (Z/p)^2 as groups where one of the cyclic subgroups is generated by 1, and then from there show that the multiplicative structure passed (More or less). I fail to see where this proof fails if R is not unital, although i've seej examples of noncummitative rngs of order p^2

p prime.of course

[next obsidian]

I am really suspicious of how you can associate the multiplicative structure to the additive one

Granted this was years ago but I tried to do something like this and it went terribly

I feel like that’s where something breaks

[rustic crown]

i haven't seen rngs at all, but for rings, i remember you really really need the ring to be unital. that's like the first thing,
we get the characteristic map Z --> R
if the kernel is p^2Z then done! else the kernel is pZ and R is a Fp algebra
we'll get Fp[x] --> R surjective and this would show R is either Fp[x]/(x^2) or Fp * Fp or F_{p^2}

(depending on whether the quadratic is a square, factors or is irred)

[chilly radish]

Writing out my process again I realised where I need the unit

[next obsidian]

Is it in that step?

Or is it somewhere else?

[chilly radish]

I did it in a bit more of a basic way

Maybe I went wrong somewhere there, but I basically took the subgroup generated by 1, either it's R and we're done or it's Z/p, quotient out and find an element a not in it, it generates a subgroup of order P, so every element in R is of the form na+k1 for n,k natural

[next obsidian]

Oh sure

[chilly radish]

It's in the step of multiplying that you need 1 and a to commute

Otherwise it's not necessarily true

So yea in associating the multiplicative structure

Cuz otherwise the cross terms won't commute obviously

[chilly radish]

This pset is technically before we've learned ideals and isomorphism thms, but that's p slick

How is the map Fp[x]->R given?

Mapping 1 to 1 and x to the other generator and the rest to 0 and extending by linearity?

[chilly radish]

Wait this wouldn't work

[rustic crown]

if the kernel is pZ, then we get the quotient map Fp = Z/pZ --> R
now pick any alpha outside the image of Fp, so for a, b in Fp; a + b*alpha are different things, this shows how all of R looks like. In particular alpha^2 = a + b*alpha for some a, b, which shows minimal polynomial of alpha is quadratic.
the map Fp[x] --> R is defined by extending Fp --> R by sending x --> alpha

[chilly radish]

I see

[modest wedge]

Ayo, is this channel the right one to talk about Geometric Group Theory ?

[rustic crown]

yep, that's one of the first ways you show irreducibility. Just need to make sure that the leading coefficient doesn't lie in that prime.

it's like a very basic result, so not sure if it has a name.

maybe try to prove it yourself, it's a nice exercise

[modest wedge]

perhaps no one here is familiarized with the most intricate things of the subject, but my question is not that difficult, I just want to know an overall description of the Reidemeister-Schreier process for rewriting group presentations, just a synopsis of the thing.

[rustic crown]

polynomials probably behave very badly over non-integral domains, so better to only look at integral domains.

[cloud walrus]

det

[rustic crown]

@ember field this is the precise statement, which i just made up right now. hopefully it's correct.

the proof should follow from the very first definitions

i sleep now, bai bai

[wind locust]

hey, I've been struggling on this for a while now

The proof is trivial when R is commutative, but I don't even know if this holds true when it is noncommutative

I tried to come up with a counterexample in M_n(R) but had no luck

[wind locust]

I found a solution online, if anyone's curious. Just take 1=i+j, then i and j must commute. But then, raise (i+j) to a sufficient power so it will be in I^m+J^n

[tropic spade]

I'm trying to make sense of the proof for this corollary from Fraleigh and I'm not sure I understand the reasoning. Here is the corollary along with somebody else's proof:

So, in Fraleigh the proof is like this:

Take tau_i an isomorphism of E onto a subfield of the closure of F (there are {E:F} such tau_i's). By thm 49.7 there are {K:E} extensions of tau_i to an isomorphism of K onto a subfield of the closure of F.

(So the result follows by counting rules basically)

Here is thm 49.7

So for Fraleigh's proof, I don't understand why it does anything but give us an upper bound on the number of extensions of tau_i that he's talking about.

In that other guys proof, I'm not exactly sure why the u he gives actually works.

[gleaming quarry]

can I use Langrage Interpolation to find the exact solution here?

[tropic spade]

Doesn't Lagrange interpolation work in any field?

I thought it did.

I'm not sure what ur P_3 notation stands for tho.

But it doesn't seem like it would hurt to try Lagrange interpolation?

[gleaming quarry]

neither do I tbh

prob some weird notation of my prof

or it has to do with Horner's method

[tropic spade]

Seems like something you'd need to know. 🤔

To do the problem I mean.

[gleaming quarry]

Substration in a field e.g. F7 works like
0-5= 5?

[tropic spade]

Isn't F_7 just isomorphic to Z_7?

[gleaming quarry]

yes

[tropic spade]

Then I don't think 0-5=5

Since 2 is not 5 in Z_7

[gleaming quarry]

ah
was it 0-5 = 7-5 = 2?

[tropic spade]

Maybe I am misunderstanding?

Yah that looks right to me.

[gleaming quarry]

been a while

thanks

[tropic spade]

No problem!

[tropic spade]

For this proof here, I think my main confusion is that I don't see why u is an isomorphism. I see that it is 1-1 and fixes F.

But when I start grabbing up elts of K and working thru cases to show u(a+b)=u(a)+u(b) it doesn't seem to work out for me.

[balmy compass]

Describe all groups G that contain no proper subgroup. For this I had the idea that if G had more than two elements then we could pick some x that isn't the identity and then <x> would be a proper subgroup so we only have to look at the case when |G|=1,2, but this doesn't seem right. It should be something similar though since this is in the cyclic groups chapter. Could i have a hint?

[tropic spade]

Take any group G with more than 1 elt, can u think of a proper subgroup G contains?

You don't really need cyclic stuff, every group by definition has a specific single elt you can use to construct a certain subgroup of that group.

[balmy compass]

just to confirm, you're not talking about {1} right? i don't think this book considers {1} a proper subgroup

[tropic spade]

Hmm

That was exactly what I was thinking of. Why would that not be a proper subgroup?

[balmy compass]

yeah artin doesn't allow it

[tropic spade]

Weird

[viscid pewter]

bruuuh

just say non-trivial subgroup like a normal person

[tropic spade]

That's what my textbook called those things

Rather than proper

[balmy compass]

lol

[tropic spade]

I guess with picking random elts and considering <x> you might need to worry about when x generates G right?

For ex Z_2?

[balmy compass]

yeah exactly

should i just exclude cyclic groups then

[viscid pewter]

no

[balmy compass]

and then i'm good with my argument?

[viscid pewter]

consider Z4

it has Z2 as a proper subgroup

[tropic spade]

Maybe Artin is being dumb and saying "proper" when he means like a proper subset rather than proper subgroup lol?

[viscid pewter]

nah

the answer is still quite elegant

for non-trivial subgroups rather than proper subgroups

[ionic quest]

I feel like you're most of the way there. You need to worry about when an element x generates G, like you said.

[balmy compass]

wait don't Z4 and Z2 have the same cardinality

[viscid pewter]

?

[ionic quest]

No, Z4 has 4 elements, Z2 has 2

[balmy compass]

oh artin calls Zn the subgroup of (Z, +) generated by n bruh

[viscid pewter]

yeah

wait what

no

oh come on

[ionic quest]

Wait, what?

[balmy compass]

lol

[tropic spade]

Lmao

[ionic quest]

Oh, I definitely assumed you meant Z_2 and Z_4

With the numbers as subscripts

[viscid pewter]

oh that makes sense

no one cares about Za because Za is always basically just Z

it's all about Z_n

[balmy compass]

is that {0, 1, ..., n-1}?

[viscid pewter]

yes

[balmy compass]

haven't seen it yet

ok

[viscid pewter]

??

they're just the cyclic groups

Z_n is the cyclic group of order n

[balmy compass]

yeah i see

mb

[viscid pewter]

ok so anyway yeah

Z4 is cyclic

yet it has subgroup Z2

[balmy compass]

ok so just playing around it seems like any integer <n which is coprime to n also generates Zn. so we just pick x^k with gcd(k,n) not 1, and then < x^k > is a proper subgroup? and for groups that aren't cyclic we can pick any element?

[viscid pewter]

that's correct

so

when is there no x coprime to n

wait no

when is there no x not coprime to n

[balmy compass]

when are all the elements coprime basically yeah?

so Zp

[viscid pewter]

very nice

so yeah groups of prime order have no non-trivial subgroups

and only those

[balmy compass]

awesome

thanks

that's quite cool

[ionic quest]

No. For instance, the reals

[frank fiber]

No, but the converse is true

[balmy compass]

for the non abelian case, i assume i'm supposed to come up with some matrices as examples but i don't see how

[next obsidian]

I think you can grab an example by nothing that
1 1
0 1

Has infinite order

I’m pretty sure you’re able to write this as the product of two finite order matrices

I think…

Or something along these lines

[tropic spade]

I just realized for this question I was asking, I should probably mention, the notation {K:F} refers to the number of isomorphisms from K to a subfield of the closure of F that fix F.

[tropic spade]

Oof I see it now. I was missing the fact that if F is a subfield of E that any isomorphism on F has the same amount of extensions to an isomorphism on E as any other isomorphism on F, which was the whole point of that theorem.

I didn't know this was called the separability degree

Then I stumbled on the proof of it in Lang.

[hidden haven]

Take free group on 2 letters x and y, quotient by relations x² and y², then x and y have finite order but xy doesn't

This isn't a complete proof and you'll have to realise this quotient group properly, because it's not immediate that this quotient isn't trivial for example

But this gives you a starting point, and now you know you to create the group you are looking for

This should be free product of Z/2Z with itself, if you are familiar with free products

[kind temple]

every algebraically closed field is infinite?

[chilly ocean]

since finite fields can't be algebraically closed

yes

[hidden haven]

goob proof strategy

[rustic crown]

you can give a very simple proof of that btw. Repeat Euclid's proof to get that k[x] has infinitely many monic irreducibles. if k was algebraically closed, each of these have roots in k, but no two can be same because of irreducibility. this shows k is infinite

[kind temple]

idk what an irreducible or Euclid's proof is, my aa is lacking lol. but i think arguing along the lines of TTerra, if you have a finite field, you just want to produce a polynomial without a root in the field

so it seems like multiplying (x - r) for all r in the field, then maybe adding 1 works

because the polynomial always evaluates to 1 no matter what the input, and has no root in the field

[rustic crown]

ah that's pretty neat

[kind temple]

thanks

[rustic crown]

whenever i see finite fields, i automatically assume we're using the classification theorem xD

[kind temple]

that joke went over my head lmao i need to get out of the aa channel

[rustic crown]

you can say a lot more about a finite fields lol, and you know exactly how many irreducibles of each degree there will be

also irreducible = cannot be reduced further 😶

[lethal dune]

[kind temple]

uhh

so x^2 - 2 is irreducible over Q

[rustic crown]

and euclid's proof is the same thing lol, multiply finitely many primes and add 1 to get a contradiction

yep

[kind temple]

assuming prime is just, only factors of the polynomial are itself and 1

[lethal dune]

why are vector fields called fields

[rustic crown]

yea, upto units, we don't care about trivial factorizations... like 1/2 * (2f)

[kind temple]

cool

[kind temple]

pretty pictures look like grass fields and i also ignorantly blame physicists

[rustic crown]

vector fields make a little more sense than algebraic fields

[lethal dune]

lol

[rustic crown]

if you look a piece of ground with fencing, then the fence forms a ring and the ground inside is a field

that's the closest i ever got with the names rings and fields

[lethal dune]

it's nice

so what are groups here? the fence woods ?

[chilly ocean]

rings are called that cuz of distrivutive proprty is like a ring marrying + and * together 👰‍♀️ 👰

[lethal dune]

what about fields then?

[robust pollen]

Hm, in Danish, Dutch, French, German, Spanish, Turkish ... fields are actually called (the equivalent of) "bodies". weird

[lethal dune]

[robust pollen]

a nice read about that: https://mathshistory.st-andrews.ac.uk/Miller/mathword/f/

[modest wedge]

and in Portuguese too

[modest wedge]

but we still call vector fields, well, (the portuguese equivalent of) "vector fields".

[chilly ocean]

true, im portuguese and confirm that anøk is not lying

[winter thorn]

hey I've done 80 pages of Gallian but still feel insecure in my Group theory knowledge

how do I know whether I got it or not?

[robust pollen]

When you can prove Feit-Thompson with a hangover

[winter thorn]

I'm sure I'd be terrified if I knew what you were talking about too, but really, how do I know I've got the theory?

More practice?

I've already done 60+ questions...

[coarse storm]

When you can answer random questions about groups on a certain Mathematics Discord server.

[robust pollen]

Well, look at exercises, and if you immediately have an idea how to go about it, that's good. If not, train.

[viscid pewter]

what's the normaliser of a normal subgroup

[winter thorn]

I've done 70 pages of Gallian, at least have the facade of mercy.

[viscid pewter]

ok so what exactly have you done

[winter thorn]

Prereqs, definition of a group, uniqueness of inverses and identity, and center of a group and centralizers of elements, cyclic groups, fundamental theorem of cyclic groups and some other things in between.

I can't remember all of them right now.

[viscid pewter]

weirdaf order ngl

have you done any subgroups

[robust pollen]

quotients?

[winter thorn]

oh yeah, those too.

[viscid pewter]

i don't think they've done normal subgroups

so probs no quotients

[robust pollen]

oof

[winter thorn]

proper subgroups, subgroup tests

[coarse storm]

Anything you have learnt so far that made you go, "wow, this is neat. They fit together so well"?

[robust pollen]

so you definitely want at least to get a grasp on normal subgroups and quotients and stuff like that. Since imo they're one of the most fundamental things in abstract algebra

[winter thorn]

I guess... how cyclic groups generate the underlying group, like how some elements of Z_n generate Z_n.

[hidden haven]

I feel like if you feel insecure in your knowledge then you haven't completely got it

[viscid pewter]

name three distinct groups where the center is:
the whole group
half the group
only one element in the group

[hidden haven]

I am interpreting the question as not whether your knowledge is sufficient but whether you understood what you learned

[viscid pewter]

well obviously

[hidden haven]

By feeling insecure in your knowledge do you mean that you wouldn't be able to answer foundational questions?

[viscid pewter]

wait actually does a group where the center is half the group even exist

depression

ok but you should be able to give the answer to the first and third instantly

[winter thorn]

The Whole Group - Z_n?
Half the group - Well, rotations are half of D_n, right? And rotations always commute.
Only one element - no idea

[hidden haven]

rotations don't commute with reflections

[viscid pewter]

rotations are half of D_n, and rotations commute with each other, but they don't commute with everything

[rustic crown]

prolly not. G/Z(G) cyclic implies Z(G) = G

[viscid pewter]

yeah ok

[winter thorn]

oof, I guess I do need to read more.

[coarse storm]

How did this become a quiz? xD

[viscid pewter]

??

[viscid pewter]

.

if you can answer questions fast then you know you've got the theory

[winter thorn]

Wait, is e an answer to the third one?

[viscid pewter]

tf is e

[rustic crown]

also one thing you can try is classifying groups of small order

[winter thorn]

identity

[viscid pewter]

oh ok

fiiiine

but distinct groups

[rustic crown]

it can be a fun exercise, see how far you can go

[hidden haven]

do you mean trivial group

[viscid pewter]

technically Z1 is cyclic

[hidden haven]

identity is an element of a group

[viscid pewter]

so you can't use it for the first one then

you should just look up more groups

[hidden haven]

if you want to talk about a group you shouldn't say e

[viscid pewter]

groups are cool

[robust pollen]

meh.

{e}

[hidden haven]

better

[robust pollen]

1

[hidden haven]

[viscid pewter]

Z1!

[winter thorn]

They are, but what I'm really interested in is Geometric Constructions, the very idea of it makes me want to cry tears of joy. exaggerating a bit

[viscid pewter]

hmmmmm

also what exactly is wrong with Z1 as a notation

[robust pollen]

I guess instead of e or 1, one should write it as 0?

[rustic crown]

just use 1

[viscid pewter]

no

[viscid pewter]

??

[hidden haven]

0 for abelian case is better because it is the 0 object

[viscid pewter]

no i hate this

[rustic crown]

well in exact sequences you do write
1 --> N --> G --> H --> 1

[hidden haven]

but for non abelian idk I have seen 1 in a couple places

[robust pollen]

for all groups, right?

[hidden haven]

oh ye that's true

[viscid pewter]

i hate this

[hidden haven]

I am all for 0 now

[viscid pewter]

no

[robust pollen]

cats = love = life

[viscid pewter]

i refuse

[hidden haven]

[robust pollen]

0 --> N --> G --> H --> 0

[winter thorn]

Do abstract algebra and number theory unite into some other esoteric branch of math?

[robust pollen]

https://en.wikipedia.org/wiki/Algebraic_number_theory

[viscid pewter]

do you know lagrange's theorem?

[rustic crown]

no, additive should be more
0 --> A --> B --> C --> 0
lol

G,N,H feel too multiplicative

[hidden haven]

why need additive

[winter thorn]

That's chapter 7, I think, I just finished reading chap 4.

[viscid pewter]

hoooow

[robust pollen]

But {e} is the zero object in Grp, so I will just write it 0 -> N -> G ... because it's more confusing

[hidden haven]

0 is because 0 object of category

[viscid pewter]

so slow!

[robust pollen]

makes me look smarter

[hidden haven]

I have heard Gallian is not great

[rustic crown]

cuz Z-mod?

[hidden haven]

we are talking about non abelian groups

[winter thorn]

For what it's worth, I understood it better than Artin.

[rustic crown]

oh, i don't think anyone uses + for non-abelian stuff

[hidden haven]

I have also heard Artin is not great

lmao

I have used Artin

Some chapters are bad

[rustic crown]

artin bad

[hidden haven]

did anyone do that here?

[robust pollen]

yeh but 0 because it's the zero object in the category, not because of addition or multiplicatoin. It'sstupid dumb fun

[viscid pewter]

god i can't think of any interesting questions to ask with the basics you know about

[winter thorn]

Artin compressed all of intro group theory into some 20 pages, I live in fear of that man.

[hidden haven]

[viscid pewter]

read lang

[robust pollen]

dummit foote is also goote

[viscid pewter]

i use d&f's

[winter thorn]

Monoids made me gtfo.

I'm only in high school, have merce.

[hidden haven]

classify all simple groups of order < 10 billion

[robust pollen]

monoids are just decategorifications of monads

[viscid pewter]

monoids are literally more boring than groups

[robust pollen]

wat

[viscid pewter]

they don't know what simple groups are

[hidden haven]

It was a joke on d&f

[robust pollen]

oh, I didn't even know what a set was back then lel

[rustic crown]

reminds me of little narwhal

[winter thorn]

hmm, simple groups are easy to understand definition wise.

[hidden haven]

exactly lel

[viscid pewter]

??

so you know what normal subgroups are?

[winter thorn]

bruh, literally skipped that word, and am searching it now

sometimes i hate me

[viscid pewter]

depression

okay here's one

show QxQ is not cyclic

[winter thorn]

... QxQ, what notation is that?

[viscid pewter]

screaming

do you know what Q is

[winter thorn]

The group of all the rationals?

You didn't specify the operation

[viscid pewter]

addition

[robust pollen]

QxQ is a man with funny glasses, don't mind these jokesters

[viscid pewter]

ok fine

show Q is not cyclic under addition

show Q\{0} is not cyclic under multiplication

should be able to do this fast

[winter thorn]

hahaha, I think this is enough discord for today.

I'm going back to studying group theory now, don't mind me.

[viscid pewter]

bruuuh

[coarse storm]

Stay on Discord! Learn the theory of being in a group!

[winter thorn]

I should've heeded the aphorism of Socrates when said, "The only thing I know is that I know nothing."

[robust pollen]

yes yes, heeded indeeded. I concur. 🥂

[winter thorn]

I've seened I'm incompetened.

[robust pollen]

But yeah, if you feel you know everything or hjave a good understanding, that's probably not true. At least for me lel ... as soon as you feel confident, something is going to come up which you can't solve. Probably

[winter thorn]

Those gallian exercises didn't prepare me for shit.

[robust pollen]

Look at an exercise 20 pages later in your book and you probably won't understand anything

[winter thorn]

Might as well have given clown makeup for free with the book smh

[robust pollen]

So, make it your goal just to understand what's being asked.

[rustic crown]

it took me a 3-4 reads until i could internalize sylow's theorem

before that the proof was magic

[waxen hedge]

Arithmetic geometry

Actually, algebraic number theory is not an algebraic version of number theory, it's the study of algebraic numbers

There is a lot of analysis as well in algebraic number theory

[modest wedge]

I think of arithmetic geometry as something closer to algebraic geometry instead of algebraic number theory, but I mean, you're not wrong tho.

[waxen hedge]

(Commutative) Algebra = algebraic geometry

[winter thorn]

Numerical algebra

[modest wedge]

truuu 😆 , Im taking classic algebraic geometry this semester and its like... spicy ring theory 😂

[rustic crown]

i'm taking algebraic geo too, but i'm scared lol. our prof started with sheaf cohomology

[winter thorn]

in which uni year are you all in?

[rustic crown]

i still have little to no intuition for that course

i'm in my third year of ug

[modest wedge]

based

[modest wedge]

Im finishing high school rn but taking some doctorate classes

[robust pollen]

tf is wrong with you people

[winter thorn]

bruh you really had to do a drive-by and kill my sense of security, didn't you?

[modest wedge]

what ? it doesnt make me smarter of something like that tho

[winter thorn]

not about smarts

[coarse storm]

I'm not sure what a doctorate class is.

I assume no frogs are dissected in those.

[modest wedge]

isnt it called doctorate in english ?

[coarse storm]

Doctorate typically refers to a PhD, I think.

[modest wedge]

Yes

[robust pollen]

yeh. and there are usually no classes

[winter thorn]

I found some practice questions online. I don't understand shit so here you all go

[modest wedge]

well, Im referring to classes you take in a Phd; rn Im taking classes on algebraic geometry and geometric group theory, with phd students, in an phd program (they also do research tho, but in my case Im just taking the classes)

anyways, I think this is kinda off-topic, we should be talking about modules or something

[robust pollen]

but after a masters? Or is it like in the US where people just go from bachelor program to phd program

[robust pollen]

that's quite a list

[modest wedge]

they usually take this after the masters, at least all of my colleagues have masters degrees

(btw my english is kinda broken, so pls be patient with some things)

[robust pollen]

I know how to show the inversion using column orthogonality. But I want to do it differently, since there was a hint for exercise, namely: Write the right hand side as the trace of an endomorphism of CG.

so, what one can do is $\dim V tr_V(\rho(g^{-1}) \hat{\phi}(\rho)) = \sum_h \phi(h) \dim V tr_V(\rho(g^{-1} h)) = \sum_h \phi(h) \tr_{\mathbb{C}G}( g^{-1} h)$, but then I don't see how to proceed

[cloud walrus]

expectTheUnexpected

[robust pollen]

I guess this is then (tr_CG * phi)(g), so that the trace is the unit for the convolution? But I do not see.

[robust pollen]

oops, I ofc forgot the sum over irreducibles in the first two expressions, but other than that it's correct.

[fossil shuttle]

same

that's a hard proof

[hidden haven]

I have never seen a proof of sylow

[modest wedge]

😭 Im still sad bc idk where to talk about GGT ;-;

[next obsidian]

Monkey

[obsidian sleet]

Sylow good. I need to review it (aka read it 4 more times)

[viscid pewter]

stupid q, if H, K and HK are subgroups of G, must H be a subgroup of the normaliser of K?

i think it should?

HK = KH so HKH^-1 = K?

is that how it works?

god i'm tired

[ancient night]

I have a subset of integers $S$ and a function $f: \mathbb{Z} \to \mathbb{Z}$. I want to consider the quotient set $S/ker(f)$ where $ker$ denotes the equivalence kernel of $f$. Every subset $A$ of $S/ker(f)$ can be assigned the value of image $f[A]$ of $A$ under $f$. I want to define it for the purpose of taking the $k$th order statistics of $S/ker(f)$. What kind of space or algebraic structure is $S/ker(f)$ in this case? I want to assign an integer value to every set-element of the underlying set.

[viscid pewter]

that's probably not enough really

[cloud walrus]

JohnDark

[ancient night]

$$
ker = f \mapsto \lbrace (x,x') \in dom^2(f) : f(x) = f(x') \rbrace
$$

[cloud walrus]

JohnDark

[ancient night]

And $ker(f)$ in $S/ker(f)$ is interpreted as a graph of an equivalence relation (up to $f$)

[cloud walrus]

JohnDark

[obsidian sleet]

yes one of the two normalizes the other

[viscid pewter]

oh sick

[obsidian sleet]

wait actually reviewing now the implication was only proved in one direction 🤡

[viscid pewter]

d&f's?

[obsidian sleet]

yes

[viscid pewter]

same here lol

[obsidian sleet]

[viscid pewter]

that's why i was thinking about it

[obsidian sleet]

HK = KH iff HK leq G in the first part

then hmm

yeah sounds about right lol

oogers