groups-rings-fields

[white oxide]

ahhh

sweet

it's becoming semi-coherent 🙏

[void cosmos]

quasi-coherent sheaves over combinatorial representatinos of the suzuki group

[white oxide]

i agree

wait so is a ground field/ring literally just

some fixed field/ring lol

In mathematics, a ground field is a field K fixed at the beginning of the discussion.

bruh lmao

[night onyx]

if you're talking about a vector space over a field k, you call that field the "ground field of the vector space"

or I guess with a module over a ring you call R the "ground ring"

I prefer "ambient field" lol

sounds spookier

[white oxide]

oh damn i've never heard that terminology before

that's funny

aight

[white oxide]

ambient sounds p cool might have to go w that

[night onyx]

I remember seeing a proof that ended "whence the proposition"

I would use that all the time just to make prof's cringe lol!

[white oxide]

LMFAO i'll have to consider that

how does tau of pi_i give a direct sum composition?

[uncut girder]

What is the galois group of x^n - 2 over Q for arbitrary n?

[prisma ibex]

The splitting field of x^n-a over Q is the compositum K of Q(\zeta_n) and Q(\sqrt[n]{a}), and one has Gal(K/Q)=(Z/nZ)\rtimes(Z/nZ)* precisely if n is odd, or n is even and \sqrt{a} is not in Q(\zeta_n); otherwise it's some subgroup of this

[void cosmos]

wtf is this first line of the proof

[white oxide]

lang refers to anything "arrow-theoretic" as "abstract-nonsense"

in fact he even defines it in bold later on

😂

lang's so funny dude

[crystal turtle]

i hate the term whence

[night onyx]

lol I was a bit of a troll

[night onyx]

category theory is silly, the theory of putting arrows between symbols

although basically all of math is based on the theory of putting the "is contained in" symbol between symbols

and some axioms saying "hey, lets assume there's actually a thing you can do that with"

[open sluice]

from whence

[void cosmos]

i really wanna learn combinatorics

lmfao

i dont think there is cat theory in combinatorics

right

[night onyx]

https://tenor.com/view/do-it-get-to-work-gif-21630516

[void cosmos]

yea i will after im done with geometry

im set on it

i saw 2 problems

and i was just like yeah this is the shit

categorical combinatorics

lmfao

[night onyx]

I think in another life I could have been a set theorist

the most annoying type

don't accept that infinite sets exist

get nothing done ever

[crystal turtle]

ultrafinitists after accomplishing nothing

[open sluice]

wonder who that's a criticism of

[night onyx]

hey I don't really "think" infinite sets exist, I just like doing math built on the assumption that they do

it's more fun lol

ZFC is a fun set of rules to mess around with

[white oxide]

is the left expression the internal direct sum and the right expression the external direct sum?

it has to be right?

[delicate orchid]

categories are graphs

[void cosmos]

yo wait

thats correct as fuck

[delicate orchid]

functors are graph homomorphisms

but graph homomorphisms are not functors

[void cosmos]

you guys

should learn about morita theory

[abstract rock]

wouldnt they be directed multigraphs?

[hollow mica]

thin cats

[sly crescent]

But that still doesn’t tell you how the morphisms compose

[elfin prairie]

are all subrings of Z of the form bZ?

[torn warren]

Suppose r1, r2, ..., rn are roots for poly of degree n, then we can prove in general that Q(r1, r2, ..., rn) : Q(r1, r2, ..., r n-1)=1, right? because we have Vieta's formula.

[delicate bloom]

try to prove or disprove it

[primal beacon]

• what is your définition of ring

[hot goblet]

couple group questions regarding proving in all cases I'm stuck on

G has order 8

[hot goblet]

clearly you pick D5, and Z/10Z but showing these are the only ones is hard

[rocky cloak]

Do you know about semidirect products?

[hot goblet]

nope

[rocky cloak]

Alright, then you can go for a more direct approach. You're able to show that any group of order 10 has an element of order 5 and one of order 2, yes?

[hot goblet]

this is from an old past exam so its possible previous version of the course covered it and we didnt

since order divides 10

[rocky cloak]

Yeah, so Cauchy's theorem or the Sylow theorems

[hot goblet]

im clear that there can be one, does that imply there is one?

ie lagrange theorem

oh wait

yes cauchy theorem thats what to use

of course

yeah so that gets you to Z/10Z and Z/5Z*Z/2Z

but 5*2 is abelian and D5 isnt right?

[rocky cloak]

So then you have s of order 5 and t of order 2. Then it's pretty straight forward to show that s^m t^n are distinct for m=0, 1, 2, 3, 4 and n=0, 1

So that gives you all 10 elements

[hot goblet]

yep clear on that

[rocky cloak]

Then the question is what could ts equal?

Once you know that, you have completely described the group

[hot goblet]

hmm right

so currently our group has {e s s2 s3 s4 t} and it needs ts which is in the group

and st which needs to be ts4

then we also needs ts2 and ts3 and ts4

and to show that st2 is ts3 etc..

which is all dihedral group presentations

so the question from here is how do we close it and show this is the only other group

[rocky cloak]

So you have e, s, s^2, s^3, s^4, t, st, s^2t, s^3t, s^4t

And ts needs to be one of these. It can't be s^m, because then t would be s^m-1, so it must be s^n t.

Now it must be that t(ts) = s

What do you get if you compute t s^n t using that ts = s^n t?

[hot goblet]

s^-1 dihedral group presentation

srs=s^-1

so in this case youd get s since s^n*s=e

So yeah we can complete this group and I guess we can say this is the only other group since 10 has prime factors 2 and 5 so combinations are all that is possible?

[hot goblet]

Anyway thanks

[chilly ocean]

Use the generators are relations above to show that every element of $D_{2n}$ which is not a power of $r$ has order $2$. Deduce that $D_{2n}$ is generated by the two elements $s$ and $sr$, both of which have order $2$.

This is what I did:

Let $x = sr^k$ where $k \in [0, n -1]$. Consider $x^2 = (sr^k)^2 = sr^k sr^k$. Using the above relations ($sr = r^{n-1} s$) we must show that $sr^k sr^k = id$. So, I used induction with the following argument. In other words, $sr^k = r^{-k} s^{-1}$.

Base case: k = 0. Then this is trivially true since $s = s^{-1} \implies s^2 = 1$ which is true by the definition of the group $D_{2n}$. Inductive hypothesis: Let $k = m$. Assume $P_m$ true for some $m \in \mbb{N}$. Then we have:
$$sr^m = r^{-m} s^{-1}$$.

Inductive step. Consider $k = m + 1$, then multiply by $m$ on the right to obtain:
\begin{align}
sr^{m+1} &= r^{-m} s^{-1} r \
sr^{m+1} &= r^{-m} s^{-2} s r \
sr^{m+1} &= r^{-m} sr \
sr^{m+1} &= r^{-m} r^{-1} s \
sr^{m+1} &= r^{-(m+1)} s
\end{align}

Is this correct?

[cloud walrus]

zzzz

[elfin prairie]

Prove that the intersection of any collection of subrings of a ring R is a subring of R.

now how many subrings of a ring can possibly exist?

if it were countable I would have done an induction argument

my guess is it would differ from ring to ring

[coral spindle]

Really as many as you like. Choose any cardinal: there is a ring with at least that many distinct subrings.

And unital ones too.

[delicate orchid]

Induction is definitely overkill

[tribal moss]

The ring of real numbers has 2^(2^(aleph_0)) distinct subrings.

[rotund aurora]

is the finite group isomorphism problem solved in polynomial time?

[coral spindle]

No I don't believe so, but also iirc it's not known if it's np-complete.

[tribal moss]

It may also depend on how the group is specified -- a full multiplication table can be asymptotically larger than a set of generators written e.g. as permutations or matrices over a finite field, so "polyomial" can take different meanings depending on the input representation.

[elfin prairie]

then what should be the line of attack?

[rotund aurora]

given the multiplication table, doesn't it reduce to graph isomorphism somehow?

[tribal moss]

Here's a recent PhD thesis in the area that Google found for me; it may have some useful survey of the state of the art: https://math.colorado.edu/~tysc3584/Tyler_Schrock-PhD_Thesis.pdf

[coral spindle]

Use the definition of the intersection. It's really just that.

[coral spindle]

Yes this is true

I talked to an expert recently about this

[tribal moss]

The thesis above states that it is conjectured that isomorphism by multiplication table may be in P even if graph isomorphism isn't.

[coral spindle]

Yeah this lines up with what this person was saying. The converse reduction is being actively researched apparently.

[rotund aurora]

oh that's cool, I wanted to ask about that too

[tribal moss]

On the other hand it's stated to be a folklore result that graph isomorphism reduces polynomially to group-isomorphsm-by-generators.

[rotund aurora]

what kind of presentations could a group have?

[coral spindle]

Cayley table, matrix presentation, generators and relations, permutation generators, etc.

I say etc but that's really all that comes to mind

[rotund aurora]

Can you go from a reasonable presentation to another reasonable presentation in polynomial time? Sure, you could present a cyclic group as Z/f(m)Z where f is a very difficult weird non-injective function to compute, but that would be strange

[tribal moss]

Matrix presentation to Cayley table may need exponential space just for the output.

[rotund aurora]

ah okay I see

so it could well be the case that isomorphism problem for presentation A can be solved in polynomial time but isomorphism problem for presentation B not, right

[tribal moss]

Yes.

[rotund aurora]

I am confused by this notation. Shouldn't the R_i just be subsets of some A^k with k some natural number? In fact, there he denotes members of R_i as k-tuples for some unspecified k. As far as I understand, A^A is the set of A-tuples with entries in A

btw this notion of isomorphism is kinda wacky, because then <A, R, S> and <A, S, R> wouldn't be isomorphic in most cases

ig it doesnt matter lol

[cinder onyx]

Why not?

[rotund aurora]

R and S could have different sizes

[cinder onyx]

Well, yes, but that doesn't mean a different one-to-one mapping g' wouldn't exist

[rotund aurora]

and one could consist of k-tuples and the other of s-tuples with k and s distinct

[rotund aurora]

it does

(the map must also be onto, although this is redundant in the finite case)

[crystal turtle]

I hate the "one-to-one ... onto* terminology for this exact reason. Very easy to miss it

[tawny iris]

hello, there is a property that I'm not sure if it is correct in my math lesson (I was absent). Is the correct one "given a subset K of a subgroup H of a group G, K is a sub group of H if it is a subgroup of G" or "given a subset K of a subgroup H of a group G, K is a sub group of G if it is a subgroup of H"

[delicate orchid]

both of these are true

[cinder onyx]

both are true

[delicate orchid]

both are basically just "a subset is a subgroup if and only if it's a group"

[cinder onyx]

^

[tawny iris]

oh so it's equivalent

thanks a lot !

[cinder onyx]

they're not exactly equivalent, they're just both true but that's by the by :))

if K is a subset of H which is a subgroup of G and K is a group, then K is a subgroup of H and G

essentially what they're saying

[rotund aurora]

I don't think I understand this. The Cayley table of a finite group of size n has size n^2, I am also unsure what a Matrix presentation is. I guess what you mean is that a matrix presentation could have size like log(n) or something like that, where n is the size of the group?

[chilly ocean]

for my problem?

also could anyone can help do this in a non-combinatorial way (if there is one)?

I tried drawing them out

[delicate orchid]

a matrix presentation would just be a faithful representation

[rotund aurora]

I thought that the main point was to study the complexity of the isomorphism problem as a function of n where n is the size of the groups involved... But now I realize, the size is explicit in the Cayley table, but it is not explicit in a group presentation by generators and relations for example... Maybe it makes more sense to study how the complexity grows as a function of the amount of data of the presentation, idk

[delicate orchid]

except you don't have n matrices of dimension k, you have one matrix for each group generator

and it would be nk^2

[tribal moss]

In the matrix presentation you get just a set of generators as matrices, not all of the elements.

[delicate orchid]

the government cover up

[rotund aurora]

ah ok lol

So is then group matrix presentation (or ig just generators and relations group presentation) isomorphism problem polynomially reducible to graph isomorphism problem?

I guess what matters is the size of the input, so probably not

[stark helm]

If there is a three-side planar graph, and all three sides are same length, but I let all sides different by letting first side have a outward curve, and second side have two inward curve, while the third one is wavy lines. To find number of isometries, it definitely have no reflection. I am wondering if there still have three rotations here?

graph here

[delicate orchid]

are you literally just deforming the edges here

like, drawing them curved?

[stark helm]

Yes

[delicate orchid]

... that's the same as drawing them straight

[stark helm]

I want to find a n side graph that only has n elements of rotation for isometries, remove symmetry

[stark helm]

because I think this graph doesn't have symmetry

What I want is to find only rotation elements as isometries

[delicate orchid]

ok lets stop being wishy washy. This is the graph V = {0,1,2}, E = {01, 12, 20}

correct?

cause if so you're right in that there are reflection symmetries

[stark helm]

Yes

Then I want to confirm if isometries only has three rotations, to make sure if there are three rotations in this isometries

[stark helm]

You mean there still be a rotational symmetry?

[delicate orchid]

no, reflection symmetries

graphs don't care how you draw the edges that's not the point

[stark helm]

So in this graph, symmetry still exist?

[delicate orchid]

I'm struggling to come up with an example of a graph with a cyclic isometry group, I should really know this

every finite group can be realised as the automorphism group of some graph but I'm not sure about isometries

[stark helm]

my purpose is to keep the graph only has n elements of rotation, without symmetry. It seems like my graph mustn't have symmetry here because no reflection can be made

[delicate orchid]

ok this is a fundemental misunderstanding in what a graph is

a graph is not a subset of R^2

[delicate orchid]

it's two sets like this

it doesn't matter how you draw it as long as the connections between everything is the same

[stark helm]

I think I shouldn't say about a graph, Original question for any positive integer 𝑛
, there exists a planar figure whose group of isometries has exactly 𝑛
elements.

Therefore, I plan to figure out a kind of graph that only has n elements in isometries, so I aim to find out a kind of graph that does have rotation without symmetry

Only, r, r^2....r^(n-1), r^n=e

[crystal turtle]

trivial graph

[delicate orchid]

alright cadet

[crystal turtle]

Also what do you mean by isometry? Is this wrt some fixed embedding of it into R^2 as a planar graph? Or do you really mean a graph automorphism?

[delicate orchid]

a graph isometry is an automorphism that preserves the distances between vertices iirc

i.e. the length of the shortest path between two mfs

[stark helm]

In my graph, all distance between each vertice is the same, just want to make symmetry invalid

[crystal turtle]

Is that not automatically preserved by an automorphism?

[delicate orchid]

take a path of length 3 and the automorphism the swaps the left two nodes

I think that works?

[crystal turtle]

That's not an automorphism since it doesn'g preserve the graph relation though

[delicate orchid]

true cause then there's not an edge from geezer 1 to geezer 3

3-ary tree with 3 layers

automorphism that swaps the two left most branches

no that still preserves distances

hmmm

[crystal turtle]

I can see how it's not necessarily preserved if you embed a graph into a larger graph (since there might be a shorter path using vertices outside of the image). But this should be preserved by an isomorphism.

Since ya know

[delicate orchid]

I think you're right

[crystal turtle]

isomorphic = same up to relabelling

[delicate orchid]

then wtf is an isometry of a graph

[rotund aurora]

isomorphisms bring paths to paths and they preserve their length no?

[crystal turtle]

Right, that's why I am confused to what an isometry is here, since it sounds like we are just talking about the automorphism group.

[rotund aurora]

C_n has cyclic automorphism group

[crystal turtle]

No

It's D_n (symmetries on n-gon, also sometimes D_2n)

[rotund aurora]

yep, right

[delicate orchid]

I don't think taking the automorphism group of any cayley graph will work

if they're abelian and satisfy some other condition I can't quite recall then the automorphism group of the corresponding graph is something semidirected with C_2

aka not cyclic

[delicate orchid]

I've done some googling.... isometries don't have to be isomorphisms

https://math.stackexchange.com/questions/3504551/graph-isometry-versus-graph-isomorphism

[crystal turtle]

Hmm

I wager

[crystal turtle]

that for a finite (connected) graph it actually is

[delicate orchid]

they talk about inclusions being isometries

[crystal turtle]

We're talking about isometries of something on itself though

[delicate orchid]

yeah so they'd just be isomorphisms

the thing is I just don't really care about that problem anymore

I'm much more interested in finding out what groups can appear

[crystal turtle]

Isometries are injective, hence this must be a bijection. And moreover, two vertices are adjacent iff their distance is 1 using the path distance, so it must preserve the graph relation

[crystal turtle]

Based!!1!

[rotund aurora]

oh

[delicate orchid]

yeah so we can't use a cayley graph, the only boolean cyclic group is C_2

sad!

[rotund aurora]

I think I got it how to construct Z/nZ automorphism groups

[delicate orchid]

oh?

[crystal turtle]

I(G) is the automorphism group of its cayley graph?

[delicate orchid]

yes

[rotund aurora]

Choose some orientation, say clockwise orientation. Then in the n-gon, for every edge AB going in clockwise orientation substitute it for the following

this works right?

[crystal turtle]

I want to say it would? I think this should kill off the reflections

[rotund aurora]

yeah, that is the idea

btw this uses 8n vertices, I wonder what is the minimal number of vertices required to produce Z/nZ automorphism group

I think I dont need to use like 3 and 4 "towers"

[crystal turtle]

You should be able to justify that any automorphism of this guy restricts to an automorphism of C_n. So we can consider the automorphism group of this guy as a subgroup of D_n. Except it cannot have any reflections. Which makes it Z/n

[rotund aurora]

this also works I think

uhh

maybe not

yeah no

[delicate orchid]

this is a really sick solution

[crystal turtle]

Don't think that one works. But you can just do one and two length branches in your original, instead of two and three length branches

[rotund aurora]

its like you are substituting the vertices for triangles, so its essentially the same as C_n

[rotund aurora]

yup

[rotund aurora]

I learned the trick before I have to admit

[delicate orchid]

could you get away with 0 and 1?

[rotund aurora]

https://www.cs.cmu.edu/~glmiller/Publications/Papers/Mi79.pdf

[delicate orchid]

the 0 concerns me

[crystal turtle]

I don't think so?

[rotund aurora]

it's essentially theorem 1 (page 3 of the pdf)

I realized that if C_n was directed then we would not be able to reflect, so that's it

[crystal turtle]

Like you should be able to reflect and then shift by one, which wouldn't be a rotation

Or uhh, essentially you should have D_n again, since if you consider the branched points as your vertices, it looks a lot like the n-gon again

[delicate orchid]

yeah, shifting the nodes around from the corners to the 0 or vice versa was my concern

ahhhhhh good point

[rotund aurora]

Given G and H two graphs, form another graph X by joining every vertex of G to every vertex of H. Then Aut(X)=Aut(G) x Aut(H), right?

[crystal turtle]

No

Consider two one point graphs

[rotund aurora]

oh right

I can fix

[crystal turtle]

The box product (as opposed to categorical product) will also have similar issues

[rotund aurora]

Substitute every vertex of H by a star of some fixed large degree (you are only supposed to join every vertex of G to the center of every star in H)

[rotund aurora]

I'm only considering finite things

[crystal turtle]

Right. These are just two different graph products you can consider

(The two closed, symmetric, monoidal products on the category of graphs)

[rotund aurora]

Ah, not aware about products of graphs

I am just trying to give every finite group as the automorphism group of a graph

[stark helm]

For my question, I think windmill-like stuff should work here

[crystal turtle]

Yeah if you were looking for a graph with just rotations as it's automorphism group, then this works

[rotund aurora]

wait this doesn't work, because stars have automorphisms. But just surround vertices of H by many paths of increasing length, all of large length

anyway

[untold basalt]

Let $G$ be a $p$-group and suppose there exists $H \leq G$ not normal in $G$. Show that there exists $g \in G$ such that $H^g \neq H$ and $H^g \leq N_G(H)$.

\vspace{1ex}

Let $X := {H^g \mid H^g \neq H}$. From the $G$-action $\rho: G \to \Sigma({\text{subgroups of } G})$ such that $\rho(g) = \gamma_g$ (where $\gamma_g$ denotes the permutation by conjugation) we have $|O_g(H)|=|G|/|N_G(H)|$, so $X=p^k-1$ for some integer $k>1$. Since $|X| \not\equiv 0 \pmod p$ the action $\tau: N_G(H) \to \Sigma(X)$ with $\tau(g)=\gamma_g$ has a fixed point $H^g$, which means that there exists $g \in G$ such that $(H^g)^h=H^g$ for all $h \in N_G(H)$, so for such $g$ we have $H^g \triangleleft N_G(H)$ and $H^g \neq H$. $\square$

[cloud walrus]

Seagull

[untold basalt]

I found that such a group has to be normal too in N_g(H), is this correct?

[south patrol]

What does your notation sigma means

[untold basalt]

$\Sigma(X)={\text{bijective functions $\phi:X \to X$}}$

[south patrol]

Sure

As a topologist/htpy person I approve :))

[cloud walrus]

Seagull

[untold basalt]

Do you approve the notation or the poof?

[south patrol]

Anyway, this looks great to me

[south patrol]

I meant the notation, but the proof seems great!

[south patrol]

I knew the reaction would be you wew

Tbh you can afford to just use words rather than writing out notation for the action and then also explaining it in words

I think that'd make it easier to read too

Just my two pennies worth tho

[next obsidian]

That’s worth two cents

[untold basalt]

hmm ok thanks, I'll ask my professor about it because he didn't actually prove that the subgroup has to be normal too

[south patrol]

So less valuable

[delicate orchid]

if a group is fixed under the action of conjugation then it's normal by definition

or is that not what you mean

[south patrol]

Wew now give a character-theoretic proof

[onyx trout]

It should be suspension

[south patrol]

True ig I mean maybe I'd write sigma subscript

But Sigma_n is standard for permutations ig

I'd just write Sym(-)

[delicate orchid]

no, no I don't think I will

[onyx trout]

I think especially with the notation for this object being X it is easily misconstrued

[south patrol]

It was by me lol

[onyx trout]

I did not realize at first glance it was a group

[delicate orchid]

who out suspending they groups

put a lil \sigma BG on that thang

[south patrol]

Yooo

Now we're talking

Me and my homies are weakly homotopy equivalent to G

[delicate orchid]

I don't know much about suspension topologically, just as a functor

and the way u draw the little cone on ur circle

[south patrol]

Good

I mean I only really think about suspension as a functor or when it comes up lol

Well

[delicate orchid]

specifically a functor of simplical complexes not like

CW complexes or anything. They have to be general or I won't get it

[south patrol]

By topologically do you mean like point-set-wise?

Ah ok

[delicate orchid]

I don't know what a "gluing" is and I never will

[south patrol]

Good

[delicate orchid]

WTF

[south patrol]

How do you suspend a simplicial complex

[delicate orchid]

same way you would a CW complex I think

[south patrol]

Well like what makes them special

[delicate orchid]

like you just copy everything one dimension up and then collapse via an appropriate quotient

[delicate orchid]

I've learnt about them

that's the unique feature!

[south patrol]

Nice

[south patrol]

oh sure lol

[delicate orchid]

yeah, looking at the topological version if you just take geometric realisations it's exactly the same in both cases

which makes sense

[south patrol]

hm by simplicial complexes are you working with like

abstract ones

i am interested what they are used for tbh hm

[delicate orchid]

yeah that's the thing right there's like 4 different levels of abstraction on these things

I'm not working with the most general "set of subsets closed under inclusion" or whatever

they're usually nerves of finite categories for me

[south patrol]

hm so simplicial sets?

or nah

simplicial complexes scare me slightly lol

[delicate orchid]

yeah they're just sets for me

but I think it would work the same regardless of what you're mapping into

I can't say for certain though

[torn warren]

could you help me on my yesterday's question?

[crystal turtle]

Don't randomly ping people for help. You can just ask here and see if someone answers.

[torn warren]

Suppose r1, r2, ..., rn are roots for poly of degree n, then we can prove in general that Q(r1, r2, ..., rn) : Q(r1, r2, ..., r n-1)=1, right? because we have Vieta's formula.

[torn warren]

anyone here?

[glossy crag]

Depends on the polynomial, I suppose.

If you take something like x^{n-1}(x-i) and r_1=...=r_n-1=0, then no, it doesn't work. If it's a polynomial over Q, then it does.

But your essential idea is of course correct, if it's a polynomial over a field K, then the product of the roots is the equal (up to sign) to the constant coefficient and is hence in K, so adjoining n-1 many of the roots gets you the n-th for free.

[torn warren]

yes, it is in Q, thank you

[torn warren]

Here how can it guarantee those are distinct? if a and b are distinct zeros of poly p(x), it is possible that $a^m=b^m$, where $m=p^t$ in this proof.

[cloud walrus]

WT

[white oxide]

A fellow Fraleigh enjoyer i see

Fraleighs fantastic for Galois theory

[torn warren]

I feel Fraleigh skip a lot of details for the proofs in the field extention chapters.

[white oxide]

yeah i mean a little bit

but certainly not to the extent that other textbooks do

besides i feel like textbook authors should leave some parts out to the reader

[torn warren]

Tao?

[chilly radish]

No, the frobenius is injective

Since it's a field homomorphism

i.e. the map x->x^p is injective

[white oxide]

huh

[tardy hedge]

does a little bit of weed help anyone else study sometimes?

[torn warren]

oh, right, my bad I forget, thank ;you!

[open sluice]

math is a drug

[tardy hedge]

I've noticed when Im just a bit high, since i find everything more fascinating I tend to pause for longer on a "simple" point , leading me to think about it at a more fundamental level for longer than normal

instead of having the faster urge to move on

[tardy hedge]

i definitely agree

[torn warren]

Tao is not leaving some parts to readers, but leaving entire book to readers

[open sluice]

that can be fine depending on the skill level

for intro it's funny

[white oxide]

terence tao analysis?

[tardy hedge]

bruhhh im trying to understand first isomorphism thm

the quotient group with kernel or whatever chunks G into bigger chunks that map to the unique things of H

Right??

something like that

am i right or just high guys

Bruh no response

pls respond

[lusty marlin]

This isn't coherent

[tardy hedge]

chunks the elements of G into bigger chunks

ok, partitions G

damn

Good?

[abstract rock]

start with a concrete example

you're not gonna get anywhere trying to explain something you are having trouble understanding

[tardy hedge]

ok

can u help me out here with it pls

with a concrete example

can we do like a back and forth

[abstract rock]

take G to be Z_4, what are its normal subgroups

[tardy hedge]

i think i do understand the thm intuitively thouhg

why do we have to do normal subgroups

ohh is it cause

[tardy hedge]

ok how about ill get back to u once i do more reading

I didnt read much about normal subgrps but im slowly seeing now why they are relevant and important

bruhh quotient groups are so cool too

bro im learning so much rn

[abstract rock]

hell yeah

[tardy hedge]

quotient groups are like bruhhh

so cool

right??

Did you guys also find this stuff so cool when first learning it

I feel like im the weird one here

[abstract rock]

No they feel plenty interesting and not very obvious as a condition

atleast to me

[tardy hedge]

Yeah its like in my actual class i feel like im the only one who finds it so cool

In class i didnt pre read and my prof is not good so i dont learn

[open sluice]

quotient groups were hard for me in the beginning
I still don’t entirely understand why they’re called or written as quotients tbh

[tardy hedge]

I was like hm this looks kinda cool but ill learn it on my own later

[tardy hedge]

Cause ur "dividing"

based on the equivalence relation

its like quotient of numbers is like "regrouping" the numbers

its like the same regrouping idea but based on the equivalence relation defined by the "dividing" group

@abstract rock im on the right track?

with what im saying?

[open sluice]

I guess it makes sense if you look at it that way

I think of it as identifying a subset of elements as one element

[tardy hedge]

yeah its like that too tho

[open sluice]

so in a way you are just dividing a group into subsets of elements with roughly the same properties

[tardy hedge]

yea

[abstract rock]

That's a good analogy if im reading you right

[tardy hedge]

LETS GOOOOO

[white oxide]

damn bro

i can relate hardcore

the first time i was learning about quotient groups i was so fucking confused lmao

i had to read the section 4 times to understand it

but man are they important LOL

soon the first iso theorem will be bread and butter to ya

[tardy hedge]

Yeah the first iso theorem makes a lot of intuitive sense too

Its awesome

[uncut girder]

if R is a free Z module of rank n then R \otimes_Z Q is an n-dimensional Q algebra, right?

[chilly radish]

What's the Algebra structure

[uncut girder]

i forgot to say R is a ring

[chilly radish]

Ah

Then yea that should be right I think

As a module it's definitely an n dim Q-vs

[cobalt heath]

Oh, this might be viewed as a product by localization, right?

Where the multiplicative set is entire Z \ 0

[tardy hedge]

Lol Q15 is so cute

So like whats really the difference between 2b and 2c

Besides now the homomorphism is onto

[coral shale]

none.

[crystal turtle]

That's the difference.

Just noting the range lies in {1, -1}

[coral shale]

The working is precisely the same

[tardy hedge]

Yeah i guess ur also like one step closer towards making it an isomorphism

[uncut girder]

i have a homogenous polynomial f(x,y) = ax^3 + bx^2y + cxy^2 + dy^3 with integer coefficients that is irreducible over Q. Is it true that the specialization f(x,-a) is also irreducible?

We know that a is not zero, as if a were zero then f(x,y) would be reducible

So f(x,-a) = ax^3 - abx^2 +a^2cx - a^3d = a(x^3 - bx^2 + acx - a^2d)

[tardy hedge]

So all homomorphic images on cyclic groups will be cyclic right

[uncut girder]

I believe, if f(x,-a) were reducible over Q, since its a degree three polynomial, there would be a rational root alpha such that f(alpha,-a) = 0.
I THINK this implies that (ax + alpha y) divides f(x,y). And since, a is not zero, this is a nonconstant factor of f(x,y), and thuss f(x,y) is reducible

I dont know how to justify the 'I THINK' step

[uncut girder]

yes

Question: Let f(x,y) be a homogenous polynomial in two variables with coefficients in a field k. Is it true that the linear polynomial (ax+by) divides f(x,y) if and only if f(a,-b) = 0?

[tardy hedge]

G1 to G2 onto homomorphism with G1 cyclic means that G2 is cyclic, but if instead G2 was cyclic it doesnt imply G1 is cyclic right

[uncut girder]

no

it doesnt

[tardy hedge]

I was gonna say it does mean G1 has aome sort of subgroup thats cyclic, but then of course all groups do anyway

[uncut girder]

yes

[tardy hedge]

Thx

So in that second case the homoemophism has to be mapping some cyclic subgroup of G1 to G2

[tender wharf]

the idea is something gets mapped to the generator of G_2

*a generator

not the, sorry

[tardy hedge]

Yeah

Which also means its mapping a cyclic subgroup of G1

Onto G2

Righ

Its like pairing up some cyclic subgroup of G1 with G2

[rocky cloak]

Yeah, since f(x, -a) still has the same degree as f, if you can factor f(x, -a) then just homogenizing the factors should give a factorization of f

[uncut girder]

its a wierd kind of homogenization tho

its like sending (x-alpha) to (ax + alpha y)

[rocky cloak]

Right, that minus sign might make things weird

Okay no, it's a little more complicated than I first thought, but maybe it works out the same anyway

[tribal moss]

Is the minus sign important? It would seem to disappear if you negate y (and also the coefficients b and d, but there doesn't seem to be any sign assumptions about those anyway).

[rocky cloak]

So f(x, y) should be irreducible iff f(x, y/(-a)) is. So you should be able to just some -a = 1

[white oxide]

here, does lang mean show that these two, altho sets, are naturally isomorphic if we can consider them as functors in a way? maybe by replacing E with - and considering it as a hom functor?

or does he mean just construct a canonical isomorphism between the two (or maybe demonstrate that there are two homomorphisms which are inverses to each other

because here Lang described these isomorphisms as "natural" but they weren't considered categorical

anyone?

[delicate orchid]

This is the tensor-hom adjunction if I’m reading it right. So yeah these are natural

[white oxide]

so like we consider them functors?

[delicate orchid]

I don’t know what the L means but yes these are functors

[white oxide]

sorry how exactly would we consider them functors? i know that for the right one we would just consider it as Hom_A(- F_A) right, a functor from the category of A-modules to sets?

and then for the left one would it be Hom_B(B \tensor_A _, F), also from the category of A-modules to Set?

sorry im just tryna get my bearings here

[delicate orchid]

I don’t know what F_A means either I just went on pure pattern recognition

[white oxide]

i'll probably just look up a proof online of this given that i don't even know what an adjoint is

damn

[coral spindle]

It means the functors specified are naturally isomorphic

e.g. L(E, L(F, G)) is describing a functor with three arguments

L(-, L(-, -))

So this would be a functor, at least, R-mod^op x R-mod^op x R-mod → Set

[white oxide]

ye so for example how would the functor on the left side be described

[coral spindle]

See above.

[white oxide]

aren't you talking about this tho

i'm asking about $\text{Hom}_B(B \otimes_A E, F)$

[cloud walrus]

okeyokay

[coral spindle]

OK.

[white oxide]

would we just replace everything with -

[coral spindle]

No

[delicate orchid]

Does L mean linear maps??

[coral spindle]

Note that B is determined by one of the categories.

[coral spindle]

Surely

[delicate orchid]

Yeah

[white oxide]

yea

[delicate orchid]

Strange notation

[coral spindle]

$\operatorname{Hom}_B(B \otimes_A -, -) \cong \operatorname{Hom}_A(-, {-}_A)$

[coral spindle]

Lang moment

[cloud walrus]

Ramify it (extensions) down

[coral spindle]

These should be naturally isomorphic as functions A-mod^op x A-mod → Set

[delicate orchid]

Yurrrrrr

[white oxide]

ok cool, thanks

will attempt

can i ask how you were able to know which arguments to replace with -?

[coral spindle]

Intuition

[white oxide]

damn

[delicate orchid]

Pattern recognition wins again

Ah it’s not quite tensor Hom directly is it

It’s restriction-extension

That’s what _A means

Got it now

[coral spindle]

No it goes through Hom_B(B, -) being the identity functor

right

As in I agree

It's tensor-hom but composed with this

[delicate orchid]

Not sure what Hom_B(B, -) has to do with anything

[coral spindle]

$\operatorname{Hom}_B(B \otimes_A -, -) \cong \operatorname{Hom}_B(B, \operatorname{Hom}_A(-, {-}_A)) \cong \operatorname{Hom}_A(-, {-}_A)$, no?

[cloud walrus]

Ramify it (extensions) down

[delicate orchid]

\text is right there

[coral spindle]

But it's not right

[white oxide]

tfw when it takes 1 hour to understand the problem

[delicate orchid]

Oh ok now I buy it

That’s just how you show restriction-extension are adjoint right

[coral spindle]

Yeah but ofc you can just write down the isomorphism directly idk

[delicate orchid]

Yur

Kinda obvs if u draw the triangles

[coral spindle]

true and also facts

[delicate orchid]

Not the triangle identities but like

The tensor triangle

[white oxide]

ok so what's the easiest way to do this problem 😭

should i read about adjuncts or whatever

[delicate orchid]

Show that for each map out of the tensor product there’s a unique one into the restriction

[white oxide]

what's the restriction?

[delicate orchid]

F_A

[white oxide]

oh okay, thanks

[white oxide]

can somebody explain to me what he means by writing down the kernel of the map E' --> E? how am i supposed to know what it is? i know that the image is equal to the kernel of E --> E'', does he mean just append a zero at the beginning?

[strong jasper]

ñ

[delicate orchid]

Use the fact that the first sequence is exact

Also what’s F here

[white oxide]

a flat module

"which should be called tensor exact"

[delicate orchid]

Yeah, that’s the definition of flat then ok

[white oxide]

i mean i know that $0 \to E' \to E \to E'' \to 0$ exact implies $F \otimes E' \to F \otimes E \to F \otimes E'' \to 0$ is exact

[cloud walrus]

okeyokay

[delicate orchid]

Because the 0 is on the other side here

So flatness is important

It basically follows from just tacking the kernel of the map onto the end and arguing by flatness that the entire thing has to be exact

[white oxide]

if $F \otimes E' \xrightarrow{f} F \otimes E \xrightarrow{g} F \otimes E''$ is exact that doesn't necessarily imply that $f$ is injective right

[delicate orchid]

Seems really weird to not write it down it’s a one liner

[cloud walrus]

okeyokay

[delicate orchid]

Not unless there’s 0s at the end

[white oxide]

ah ok

[delicate orchid]

Then it does

[white oxide]

algebra is confusing

[white oxide]

sorry what do you mean precisely by this

i'm lost

[delicate orchid]

Just Google the proof. I’m in the pub.

[white oxide]

british moment

[summer path]

alcohol

[white oxide]

alcohol

[coral shale]

init

[karmic moat]

incredibly real. last night at a party i had a beer in one hand and rotman pdf in the other

[rotund aurora]

what differs Isaacs "Character theory" from other representation theory books like Kowalski's ?

[white oxide]

aight time to chat gpt this hsit

[delicate orchid]

Isaac’s character theory is, unsurprisingly, very focused on character/modular character theory with not much attention paid to the module perspective

[white oxide]

\documentclass{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}

\begin{document}

\textbf{Proof:}

Given an exact sequence:

[
E' \xrightarrow{f} E \xrightarrow{g} E''
]

we want to show that the sequence:

[
F \otimes E' \xrightarrow{F \otimes f} F \otimes E \xrightarrow{F \otimes g} F \otimes E''
]

is exact.

1. First, observe that the sequence $0 \to E' \xrightarrow{f} E \xrightarrow{g} E''$ is exact. Since we're given that for every injection $0 \to E' \to E$, the sequence $0 \to F \otimes E' \to F \otimes E$ is exact, we can apply this to the injection $0 \to E' \to E$ to conclude that:

[
0 \to F \otimes E' \to F \otimes E
]

is exact.

2. To show that $F \otimes E \xrightarrow{F \otimes g} F \otimes E''$ is exact, we need to prove two things:

   a. $\text{im}(F \otimes f) \subseteq \text{ker}(F \otimes g)$

   b. $\text{ker}(F \otimes g) \subseteq \text{im}(F \otimes f)$

a. To prove $\text{im}(F \otimes f) \subseteq \text{ker}(F \otimes g)$, let $x \in F \otimes E$, and let $y = (F \otimes f)(x)$. We want to show that $y$ maps to the zero element under $F \otimes g$, i.e., $(F \otimes g)(y) = 0$. Using the properties of tensor products and the fact that $y$ is in the image of $F \otimes f$, we can write $y$ as $y = (F \otimes f)(x) = F \otimes (f(x))$. Therefore, $(F \otimes g)(y) = F \otimes (g(f(x)))$. Since the original sequence $E' \xrightarrow{f} E \xrightarrow{g} E''$

[delicate orchid]

Oh my golly gosh

[white oxide]

that wasn't a one liner @delicate orchid

cmon now

[rotund aurora]

what does the adjective "modular" entail? Also, characters are 1d representations, right? It's only 1d representations in Isaacs book? 🤔

[delicate orchid]

Characters are not 1D representations

And modular entails working over a ring with characteristic dividing the order of the group, so the group algebra isn’t semisimple

[delicate orchid]

As a character theorist I am offended

[rocky cloak]

A character is the trace of a representation. A 1D representation just equals its character, but other representations also have characters

[rotund aurora]

ah wait its the trace

xD

[rocky cloak]

And if you're doing modular representation theory, then I guess characters are defined in a little more complicated way, to avoid them being 0 all the time

[sonic coral]

How can i show that a normal subgroup of a p-group intersected with the center of the p-group is nontrivial

i think i should be using the class equation and the fact that p-groups have nontrivial center but i’m not sure where to go

[rotund aurora]

I think you can do it like this. Let P be the p-group and N the normal p-subgroup. Then let ||P act by conjugation on the elements of N||

and ||look at the class equation|| 🙂

[sonic coral]

hm okay let me think about this some

so |P| = |Z(P)| + the sum of the index of P with the stabilizer of N right

since the action is conjugation

[chilly ocean]

what is the order of an n-cycle?

permutation

is it just n

[sonic coral]

yes

[chilly ocean]

oh okay

so (1 2 3) has order 3

and any (a1 a2 a3 . . . an) has order n

[sonic coral]

yup

[chilly ocean]

thank you

any hints?

[rotund aurora]

it's a bit unclear to me what you mean by "the sum of the index of P with the stabilizer of N", but to finish the proof just recall the last step to prove that p-groups have non-trivial center, it's the same

[chilly ocean]

can I try to say that this is in a bijection with N

i.e. $S_\Omega \to \mbb{N}$ is a bijection

[cloud walrus]

zzzz

[abstract rock]

try to find the order of an element that acts as a derangement of the set omega

[chilly ocean]

but i'd need to prove its surjective and injective

[abstract rock]

ie an element where every element in omega goes to a different element

[chilly ocean]

can you explain this

[abstract rock]

how would you usually find the order of an arbitrary permutation in S_n

[rotund aurora]

this is far from true

[chilly ocean]

lcm of the disjoint cycles' lengths

[chilly ocean]

wdym

im asking if i can construct a bijection to show its countably infinite

[rotund aurora]

I am saying you can't

[chilly ocean]

or is it uncountably infinite?

[abstract rock]

that's a shortcut, what is that process supposed to represent on the permutation

[chilly ocean]

how come

[chilly ocean]

i'm not sure

[abstract rock]

do a few examples

[chilly ocean]

like $(1 2 3)(4 5)$ for example? in $S_5$

[rotund aurora]

do you know about cantor diagonalisation?

[cloud walrus]

zzzz

[chilly ocean]

yes i do

[abstract rock]

try to go for a derangement like i mentioned

[rotund aurora]

so it's pretty much the same

[chilly ocean]

not sure what derangement means sorry

[rotund aurora]

at least to show the cardinality is >= continuum

[abstract rock]

i explained already

[chilly ocean]

oh

[abstract rock]

but in other words

a permutation with no fixed points

[sonic coral]

the last step used the fact p divides the index of P and the centralizers of g_i in P as well as divides P to conclude that it must divide the order of the center

i don’t think i’m seeing it

like how it follows from that same argument

[rotund aurora]

you have $\sum_{\mathcal O} |\mathcal O|\equiv 0\mod p$ right? and we know that $|\mathcal O|=p^k$ for some k for every $\mathcal O$

[cloud walrus]

Croqueta

[rotund aurora]

but note that the orbit of 1 (the identity in the group which is also in N) has size 1, therefore there must be at least one other orbit of size 1

[chilly ocean]

im really still suck on this

[sonic coral]

there being another orbit of size 1 implies that there is more than one element in N that is also in the center?

[rocky cloak]

What's the simplest type of nontrivial permutation? How many of those are there?

[rotund aurora]

if that is the orbit of x with x different from 1, it implies that gxg^{-1}=x for all g in P, which is what you wanted

[chilly ocean]

$\begin{pmatrix} 1 & n \end{pmatrix}$?

[cloud walrus]

zzzz

[chilly ocean]

where $n \in \mbb{N}$?

[cloud walrus]

zzzz

[chilly ocean]

in other words, a transposition?

[rocky cloak]

Yeah, there are infinitely many transpositions, so there you go

[chilly ocean]

so I define a function $f : \mbb{N} \to S_\mbb{N}$ by $f(n) = \begin{pmatrix} 1 & n \end{pmatrix}$?

[cloud walrus]

zzzz

[chilly ocean]

don't I have to prove injectivity to say that $S_{\mbb{N}} \geq \mbb{N}$?

[cloud walrus]

zzzz

[rocky cloak]

Sure, but different transpositions are different, so that shouldn't be a problem

[chilly ocean]

yeah ig

[chilly ocean]

doing exercises a) and b)

I realized

a pattern

but I'm not sure if it's true

I came to the conclusion that for an $n$-cycle $\tau$, $\tau^i$ has order $n$ (or is also an $n$-cycle) if and only if $\gcd(i, n) = 1$

[cloud walrus]

zzzz

[chilly ocean]

is that correct?

[daring nova]

Yes

[daring nova]

For general groups, g^n has order ord(g) / gcd(n, ord(g))

In this case, you'd need to argue that such powers are still cycles

[safe widget]

Fairly trivial question -

Doom was discussing with a friend on Ring Theory proof for this question -
Prove that for a Field, F and a Ring R - a ring isomorphism, if g:F->R is not the zero homomorphism, then it is injective.

Doom's friend proved it using this manner. Here's the question -

Given that we only utilized the non-injectivity and \phi being a homomorphism, can we weaken the given condition to work for any two abelian groups and a homomorphism, assuming the proof is indeed true. (which Doom says it is because homomorphisms preserve identity elements and the inverse operation is well defined)

[delicate orchid]

Kernels of ring homomorphisms are ideals, the only ideals of fields are 0 and the field itself, therefore if the map isn’t the zero homomorphism it must have a trivial kernel and therefore is injective

It looks like your proof is essentially proving “zero kernel => injective”

[next obsidian]

This proof is incorrect, why can’t g(c1) and g(c2) not both equal 0_R?

It ends up being that yes, this is true (because the map is injective if it’s not 0), but nothing’s been done to justify it.

You have to use the multiplicative structure in order to prove this fact, which is never once used.

As-is, this proof would apply to maps of abelian groups as you noted, but clearly the statement that nonzero maps of abelian groups are injective is false so it tells you that this proof is not correct

[delicate orchid]

They can’t both equal 0 because we’re assuming they’re different towards a contradiction

I agree that this doesn’t hold as is, because I could use this proof to show that any group homomorphism is injective or zero

[safe widget]

why can’t g(c1) and g(c2) not both equal 0_R?

Like Ghost mentioned, yea. The assumption is non-injectivity to go towards a contradiction.

You are correct about the non-zero maps of abelian groups are injective being false because we were able to come up with counterexamples. So, what is actually incorrect then?

[tribal moss]

The trouble with the fake proof is that it has unfolded the assumption "f is not injective" wrongly. It says

Suppose there are c1 and c2 such that c1=c2 but f(c1) != f(c2)
which is absurd for any function, injective or not and homomorphism or not. The assumption should have been
Suppose there are c1 and c2 such that f(c1)=f(c2) but c1 != c2
and then the entire rest of the reasoning breaks down.

[safe widget]

AH! That makes sense!

Thanks!

[tribal moss]

(Or rather, it becomes an argument that if f is not injective, then there is something nonzero that maps to zero -- which is in itself true, but not enough to reach a contradiction).

[sonic coral]

i really feel like i should understand, but i don’t think i do

[tribal moss]

They are saying that elements of the center are exactly those whose orbits (under conjugation) have size 1.

[next obsidian]

Ah okay, so you already were taking that to be true

[next obsidian]

Oh wait

Me bad at math

[torn warren]

Why does it simply say it is common zero and reach the conclusion?

Let $p(x)=irr(\gamma, F(\alpha) )$ be the irr-poly in $F(\alpha)$, then $p(x) | h(x)$ in $F(\alpha)$. Since $\gamma$ is the only zero for $h(x)$, it implies $\gamma$ is the only zero for $p(x)$, hence, $p(x)=(x-\gamma)^k$ in $F(\alpha)$.
\
Case(1), if the characteristic of F is 0, then $k\cdot 1\neq 0$, so $p(x)=(x-\gamma)^k$ in $F(\alpha)$ implies $(x-\gamma)$ in $F(\alpha)$ and we are done for this case.

Case(2), if if the characteristic of F is p, and I stuck here.

[cloud walrus]

WT

[torn warren]

anyone here?

[chilly radish]

I don't understand your reasoning at all. h may have other irred components shared with p, why would p contain just the root, or even the root at all

Also if p.does equal (x-gamma)^k then k=1 since the extension is separable

Wait my bad, I misread

But note

h,g only share 1 zero by definition. p as the min poly of gamma divides both h and g, so it divides their gcd

Their gcd is just (x-gamma), since if it wasn't then they would share another root necessarily (if you want you can pass to the algebraic closure to see this, but this is not necessary, namely if their gcd had degree >1 it would have as a root one of the other gamma_i, since its roots are a subset of the roots of g)

The reason it has mult 1 is because both polynomials are irreducible over F

So they are separable, so have distinct roots

Really you only need that g has distinct roots to conclude this

[torn warren]

I miss the condition that E is separable in my work, now I think I understand it, due to E is separable, g(x) can be factorized into the product of linear factors with mul=1. Also p(x) divides g(x), so p(x) is the product of some of those linear factors of g(x). But actually p(x) can only have one factor from those linear factors of g(x), due to gamma is the only root for p(x).

[chilly radish]

Yea exactly

Separability is very important, it's what makes PET work

[torn warren]

I stuck at $(x-\gamma)^k$ due to I didn't use the fact E is separable, hence k=1

[cloud walrus]

WT

[chilly radish]

Yea

I understand the confusion now

[torn warren]

thank you very much!

what is PET stands for?

[chilly radish]

Np

Primitive element theorem

[teal vessel]

I am incredibly stupid: when it just says "map" here, I assume it is talking about a homomorphism, and not simply an arbitrary mapping?

[daring nova]

doesn't matter

surjectivity for homomorphisms is the same as surjectivity for functions

[teal vessel]

yes, but this is, strictly speaking, not describing a single function

[daring nova]

there's only one function from G to G that maps x to x^k for all x

[teal vessel]

except the notation isn't describing that according to the definitions layed out in the book

[open sluice]

consider the function f : G -> G defined by f(x) = x^k
this is a proper, good old function, you can show that it makes sense

now show that f is surjective

[teal vessel]

now that ^ is unambiguous

[open sluice]

what other function did you have in mind

[daring nova]

also it's only an homomorphism because G is cyclic. It wouldn't be a homomorphism in any finite group

[teal vessel]

that I'm aware of.

[teal vessel]

the rooted arrow is describing mapping a particular element to another particular element. It could be the case that the domain of definition isn't the entirety of G, or a constant function. They've been using "x" as a distinct element of some of these groups too often and it got me paranoid.

[open sluice]

usually if people want to define a function on a particular proper subset they will make it clear

[teal vessel]

so my initial read was "there exists some function f which maps the element x to the element x^k" without reading x as an arbitrary element.

[daring nova]

that is the standard meaning that is shortened to x -> x^k

[coral spindle]

This isn't correct. For some fixed k, the map x |-> x^k is a homomorphism in any Abelian group, not just cyclic ones. Furthermore for specific values of k, it is a homomorphism in non-Abelian groups too. For example if G is any finite group, the map with k = |G| is a homomorphism, trivially.

[teal vessel]

that moment when you get paranoid about x -> x^k vs x |-> x^k ;-;

[open sluice]

i don't think this is really something to be concerned about, it's as i said

and if it's not apparent from context then that's the author's fault

[teal vessel]

in retrospect, my concern would probably be justified only really if it said that x ∈ G, but meh.

[daring nova]

yes abelian ofc

[open sluice]

right, so when people want to say that, they will say "let x \in G be fixed ... define y \mapsto f(x, y)"

[daring nova]

therefore not any group

[open sluice]

it's kinda improper to say "let x \in G, define a map x \mapsto f(x)"

but if someone says that you can probably recognize they mean "define a mapping from G to somewhere by x \mapsto f(x)"

or "define y by y = f(x) for each x \in G"

or something weirder idk

[teal vessel]

anywho, got it fixed in my own notation now so it's clear (don't ask, it's a mess)

[velvet steeple]

What does a ring denote? Is it like a field?

[karmic moat]

thing of a ring as generalizing the integers

you have addition and multiplication, additive inverses (a and -a), but you dont have multiplicative inverse

additive identity 0, multiplicative identity 1

[open sluice]

including 1 👌

[karmic moat]

fields are a special kind of ring

[karmic moat]

blah blah ring vs rng blah blah

[untold basalt]

I've written a piece of python code that is supposed to give me a list with all the elements of $GL(n, \mathbb{Z}_3)$ which is supposed to have 48 elements, but the code produces 50. I've checked, the determinants are all non-zero and the matrices are actually 50, it's not a problem with the counter. Can anyone help me out here please?

\begin{verbatim}

group = []

def det(x):

return x[0][0]*x[1][1]-x[0][1]*x[1][0]

def diff(A, B):

return [[A[0][0]-B[0][0], A[0][1]-B[0][1]], [A[1][0]-B[1][0], A[1][1]-B[1][1]]]

K=(0,1,2)

for a in K:
for b in K:
for c in K:
for d in K:
if det([[a, b], [c, d]]) !=0:

                A=[[a, b], [c, d]]

                group.append(A)

for i in range(len(group)):

for j in range(len(group)):

    if j!=i and diff(group[i], group[j])==[[0, 0],[0, 0]]:

        print(i,j)

for el in group:

print(el[0])
print(el[1], det(el))
print("")

s=0

for el in group:
s+=1

print(s)

\end{verbatim}

[cloud walrus]

Seagull

[karmic moat]

texit is gonna have a stroke with that one man

[daring nova]

use backticks please

[untold basalt]

what's that?

[daring nova]

python code

wrap it in
``` python
{your code}
```

[delicate orchid]

holy moly

[open sluice]

you have addition
you have subtraction
you have multiplication, albeit it isn't necessarily commutative
you don't have division in general

even though multiplication doesn't work exactly the way it does with integers, it's still reasonable to call it multiplication, because you have the distributive property

[daring nova]

it's fine, |K| = 3

[delicate orchid]

you can get rid of one of the for loops and the conditional and just set d = bc/a

[karmic moat]

yandere dev is that you?

[open sluice]

do the same thing but put ` three times at the start and end of your message
it's the same key you'd use for ~

[daring nova]

rip a = 0 ?

[delicate orchid]

yeah fuck a = 0

[daring nova]

also -
Fuck F2, this isn't F2, sign matters

wait what

why would you want d = bc/a

or d = -bc/a

there's 3 possible determinants

[delicate orchid]

I don't care

[karmic moat]

i care.

[daring nova]

I care too

[delicate orchid]

is each matrix in the output actually different

[karmic moat]

just kidding i dont actually care

[daring nova]

honestly, I got things to do too

[karmic moat]

yeah im reading a book

[daring nova]

yeah no I'm debugging my command prompt

[delicate orchid]

I've got to stare at 6-tuples of numbers for a few hours

[daring nova]

explain yourself

[delicate orchid]

you first

[karmic moat]

damn i was just looking at 5-tuples of numbers for a few hours yesterday

[delicate orchid]

why you got your own command prompt u weirdo

[untold basalt]

yes?

the 4 cycles ensure that don't they?

[delicate orchid]

considering I'm not seeing a single mod 3 in here

[daring nova]

trying to run ocaml programs compiled under a linux system to then execute it in a linux system, on a windows machine. Except the program calls for the execution of another file, and does it in the windows cmd because it doesn't expect anything but linux. And the windows cmd doesn't know how to run that file

[daring nova]

I have 4, actually

[delicate orchid]

my guesses are you've accidentally got some det 3 matrices in there or two matrices that are equal mod 3

[untold basalt]

determinant is from K to R isn't it?

[daring nova]

windows cmd, cygwin cmd, the other cygwin cmd, the linux VM terminal

[delicate orchid]

no? it's to the base field

[untold basalt]

crap

[delicate orchid]

it's from M_n(K) to K

determinants are an evaluation of a polynomial expression in K[x] why would it be in R?

[daring nova]

now you

[delicate orchid]

I'm having to manually decompose these characters because the representation ring I'm working over isn't freely generated

so I can't just linear algebra it cause the matrices involved are A) not unique B) not square

[karmic moat]

just let macaulay 2 do it fo ru or something duh

[daring nova]

such an #groups-rings-fields moment

[untold basalt]

yes yes you are right, brain fart of mine duh

[untold basalt]

but $\begin{pmatrix} 2 & 1 \ 1 & 2 \end{pmatrix}$ is invertible

[cloud walrus]

Seagull

[delicate orchid]

not over F_3 it isn't

2^2-1 = 0

[velvet steeple]

Oh

[untold basalt]

ok yes thanks

[karmic moat]

fun thought for you

[untold basalt]

got it now

[karmic moat]

can you think of some conditions you could put on a ring to make it a field?

think about Z and Q

[velvet steeple]

Wait, what are the conditions we need for a field

[karmic moat]

brainstorm

what are all the things you can do in Q and all the things you can do in Z

and see if you can find any differences

[teal vessel]

A field is a spicy ring, which is a salty group, which is a sweet set.

[velvet steeple]

a + b = b + a
a * b = b * a
a * (b + c) = ab + ac
a * (b * c) = (a * b) * c
(a + b) + c = a + (b + c)

[velvet steeple]

those are associativity, distributivity, commutativity

[daring nova]

rip distributivity

[karmic moat]

for a in Q, does there exist a b such that a*b = 1?

[velvet steeple]

?

[velvet steeple]

Yes

[daring nova]

at least write it correctly

[karmic moat]

what about for Z?

for a in Z, does there exist b in Z such that a*b = 1?

[velvet steeple]

ac ofc, sry

[velvet steeple]

No

[karmic moat]

boom theres your answer

a field is a ring in which every element has a multiplicative inverse

wait is the requirement also that it's an integral domain

i forgot

[teal vessel]

0•C=1

[delicate orchid]

that follows from every non-zero element having a multiplicative inverse

[karmic moat]

ah yeah

[velvet steeple]

So Z is a ring and Q a field?

[delicate orchid]

rather trivially, as you can't be an unit and a zero-divisor

[teal vessel]

Yes

[karmic moat]

yep

[delicate orchid]

you're probably thinking of the fact that fields have to be commutative

[karmic moat]

yeah i should've mentioned that

i just habitually assume we're working over commutative rings

[teal vessel]

Sounds like a dangerous assumption

[daring nova]

fun fact: Z and Q are the smallest rings and fields containing 1

[delicate orchid]

F_2:

[daring nova]

the real 1

[delicate orchid]

they're the smallest char 0 rings and fields

[teal vessel]

all other 1? fake

[daring nova]

I said the real 1

[delicate orchid]

Z is even better than that, it's inital in the category of rings

and every ring of characteristic n contains Z/nZ

[teal vessel]

yeh, all other 1s are posers

[daring nova]

guys the goal was to not overwhelm him with characteristics

[daring nova]

perfection

[delicate orchid]

an improperly phrased statement would be more confusing

[teal vessel]

when you're dumb and you have to go all the way back to the euclidean algorithm to remember that (k,n)=1 implies that there exists an inverse of k (mod n)

ok, so if we define a function f(x)=x^k for some k coprime with the order n<∞ of our cyclic group G, because (k,n)=1 that implies there exists some a such that ak ≡ 1 (mod n), therefore we can define an inverse function g(x)=x^a. Thus, because f(x) has an inverse, it is a bijection, which must be surjective by the definition of bijection.

[daring nova]

euclidian is impressive. But imo Bezout is fairly common

[teal vessel]

probably is, but I don't know that one atm

technically I only went to the EEA, not the strict EA

[daring nova]

Bezout just states (a, b) = 1 iff there exists u,v such that au + bv = 1

which surely you saw somewhere between EA and Z/nZ

[teal vessel]

probably, just didn't commit it to memory because it follows directly from the EEA

or, rather, I remember, it, just not under the name of Bezout's theorem

now to prove that this statement on cyclic groups is actually true for all finite-order groups.... using Lagrange's theorem?

this statement: for finite cyclic groups, x to x^k is surjective

OH wait, since this mapping implies a cyclic structure, you're observing all cyclic subgroups of G, and because |H| divides |G| that means that anything coprime with |G|=n will also be coprime with |H|, therefore you have split G up into a collection of cyclic subgroups all of which have orders coprime to K, therefore the above surjective argument applies individually to each cyclic subgroup, and since the domain of this mapping is G itself, it forces all elements of G to be put into these cyclic structures, thus every element is accounted for, and it must be surjective.

(these cycles may have common elements from distinct generators, so it's not a partition)

[daring nova]

doesn't require Lagrange's ?

[teal vessel]

the question told me to use it, and tbh once I thought about it, that makes pretty good sense to use

[daring nova]

(k, n) = 1 so we have ka = 1 [n] for some a
Then f^a(x) = x^ka = x^1
hence f^a is a bijection. This is only possible if f itself is a bijection

which you may say uses a corrolary of lagrange

[teal vessel]

yeah, pretty much, just stating that for all subgroups H of G, (k,n)=1 implies that k is coprime to the order of H as well.

[daring nova]

seems so much simpler than this for not even more theory
Yours I struggle to follow

[teal vessel]

¯_(ツ)_/¯

order of H divides order of G for all subgroups H
the union of all cyclic subgroups of (the finite group) G is equal to G
f(x) is bijective on cyclic groups.
[in this particular case, at least] this bijection applied to the union of these cyclic groups is still a bijection, since it's technically a bunch of automorphisms

[daring nova]

but what is H ?

[teal vessel]

any arbitrary subgroup of G, a subset of those subgroups are cyclic.

it was a statement of Legrange

that moment when it tells you that if (k,n)=/=1 it must fail, right before it asks you to prove it

[daring nova]

it's honestly fine for cyclic groups

cyclic groups are simple enough that most things about them are easy

[teal vessel]

yeah, it's pretty simple, I just thought it was funny

especially since I basically proved half the logeq in order to do 25

[untold basalt]

How do we know what big groups are like? Conjugacy classes, subgroups etc?

Like GL(n, Z_p) and S_n when the order is big

[delicate orchid]

very vague question

[untold basalt]

Or do we even know them that well?

[delicate orchid]

we know those groups very, very well

S_n especially

I'd argue those two groups are the most well known

[untold basalt]

But do we know them form the theory or from computation?

[delicate orchid]

I don't really see the divide between those two?

again, vague question

[rocky cloak]

Since there are infinitely many such groups we clearly can't compute something for every single one individually. So you have to use some theory somehow, I guess it comes down to where you draw the line between theory and computation

[cloud walrus]

zxxz21

[rocky cloak]

I guess consider a polynomial h in the kernel, and consider h a polynomial in Y1. Do polynomial division to write h as something times (Y1 - f1) plus something that doesn't depend on Y1. Repeat until you eliminante all Y, then h is something in the ideal (Yi - fi) plus a polynomial in the Xs. Deduce that the Xs part must be 0, because h is in the kernel.

[wet zodiac]

wait what

is n^2=1 for all n in G not a counterexample?

[delicate orchid]

I'm spooked solid

[wet zodiac]

(p is an odd prime)

[delicate orchid]

this would imply 2 divides the order of G, infact it implies that G is C_2^k for some k

wait

odd prime

[wet zodiac]

G is a group of order 2p

[delicate orchid]

oh well that would have been nice to know lil nerd

[wet zodiac]

because it seems like it is

G is indeed a group and all elements are either order 1 or 2

[delicate orchid]

if that's the case then it cannot be of size 2p with p odd

[delicate orchid]

because of this

which seems like something they really should have mentioned before hand because not knowing this really does make it look like a counter example

[tribal moss]

That shouldn't be possible, due to Sylow.

[wet zodiac]

we did not cover sylow

[delicate orchid]

ok then, due to lagrange+what I've said above

[tribal moss]

Lagrange would be happy enough with that, I think.

[wet zodiac]

damn what should i do then

[tribal moss]

It n²=1 for all n, then the group is abelian because abab = 1 = aabb, and you can cancel the leftmost and rightmost letters on each side.

If you have the structure theorem for f.g. abelian groups, that ought to finish it.

[wet zodiac]

i do not

[tribal moss]

Hrmmm.

[delicate orchid]

no need for the structure theorem

say this n^2 = 1 group is generated by a_1, ..., a_n

then argue inductively that G = <a_1>G_1 = <a_1><a_2>G_2 = ...

these all commute with one another (and are obviously disjoint) as you've just shown so the entire group is a direct product of C_2s

I feel like we've gotten a tad off topic though

[wet zodiac]

wait a minute

this is a torsion group

[tribal moss]

Finite groups tend to be that.

[swift raptor]

Hi, I would like to show that for distincts f_1, f_2...,f_n in Hom(G,K*) and a_1,a_2,...a_n in K, not all equal to 0, there always exists g in G such that sum(a_if_i(g))=0, does someone have an idea? I havent got any clue (G is a group and K a field)

[tardy hedge]

Hey guys

So part of my homework this week is like this

Prof wants to do “coloured tables” like that on the board

Showing that group of cosets idea

Quotient/factor group thing

But, one question was do a coloured table for {e, (2,3)} , that subgroup of S3

But that subgroup isnt normal

And the multiplication of cosets idea was shown to be well-defined and it was shown that that operation with the cosets formed a group if the subgrp was normal

Anyone?

Yeah like this doesnt make sense right

U can see he said “do coloured tables for S3, cosets of {e,u1}” (u1 is (2,3))

The other two examples make sense cuz both of those subgroups are normal

[amber plank]

yeah that's very suspicious

[tardy hedge]

Right?

[amber plank]

just the fact that left and right cosets don't coincide will make that an absolute mess

[chilly ocean]

could someone help with this problem

[tribal moss]

Hmm, what even is R[Z/nZ]? The group ring of the cyclic group of order n?

[chilly ocean]

I'm not really sure, I'm not very good at finding representations

or more like I don't even know how to do it

I need to find like Z/nZ -> Gl(R) right?

[tribal moss]

I asked because I don't even understand what the question is asking (so you might be ahead of me ...)

[chilly ocean]

Wouldn't it just be like, R[Z/n]-module are the elements r + rg + rg^2 where g^3 = 1?

i have this

[delicate orchid]

yeah, but the module structure is given by a specific scalar multiplication by elements in R[Z/nZ] determined by the corrisponding representation

[delicate orchid]

there's a canonical way of representing C as a 2-dimensional matrix algebra over R, use this to find the 2 dimensional irreducible R-reps of Z/nZ

or, as the hint suggests, take the matrices corrisponding to a rotation by 2pi/n

[chilly ocean]

so I can use the rotation matricies right? and then I rotate them by like m2pi/n degrees to get the various 2d representations?

[delicate orchid]

yur

[chilly ocean]

what about 1d representations

would that just be the identity?

and how do endomorphism algebras work? What does that even mean

[delicate orchid]

it would be the identity unless n is even

because you do have 2nd roots of unity in R

[chilly ocean]

wait why?

oh

wait

what are 2nd roots of unity?

[delicate orchid]

-1...

[chilly ocean]

what does roots of unity mean in R?

[delicate orchid]

the same thing it means in any ring?

[chilly ocean]

just roots of unity with only real partS?

o

I see

so if n is even we have 2 1-d representations?

[delicate orchid]

yeah

[chilly ocean]

either the identity or like reflection

[delicate orchid]

so if n is even we have a subgroup Z/(n/2)Z in it, then the one dimensional would be the lift of the non-trivial rep of (Z/nZ)/(Z/(n/2)Z)

[chilly ocean]

what does lift mean?

[delicate orchid]

doesn't matter then