groups-rings-fields

[rocky cloak]

That's right, notice

1 = x^2(x^2 + 4xy + 6y^2) + y^2(y^2 + 4xy + 6x^2)

Then multiply by it on both sides

[rotund aurora]

yeah I see, cool argument

[rocky cloak]

Yeah, so
phi(sqrt(2))^2 = phi(sqrt(2)^2) = phi(2) = 2.

So if you can show that no element squares to 2, you're good.

[stark helm]

I am confused about why phi(sqrt(2)^2)=phi(sqrt(2))^2=2?

[rocky cloak]

Well, by definition of a homomorphism
phi(xy) = phi(x)phi(y)
so in particular phi(x)^2 = phi(x^2)

We also have phi(1) = 1, thus phi(1+1) = 1+1 = 2

[stark helm]

so we are firstly assuming that Q(sqrt(2))=a+b*sqrt(3) as isomorphism, then use contradiction to show it is impossible to be homomorphism?

[rocky cloak]

That's right

[stark helm]

I got it, really appreciate that

[rustic crown]

that's the type of problem that would occur if you did with closed sets

you'll just have radical ideals I and J such that I + J = (1) and I n J = nil(A), and from this its not immediately clear how to manufacture an idempotent without doing some messy newton-ish work

you'll have i+j=1 and ij nilpotent. so i(1-i) = nilpotent

which is back to the original thing

sowwy det's internet died

or just on the open cover U u V

yee, but you would be dividing by a unit element.

lemme give the details, so you have a ring A and a nilpotent ideal I, and an element e such that e^2 = e mod I. (if the ideal is locally nilpotent, just replace I with (e^2 - e))

for newton's method you look at f(x) = x^2 - x and f'(x) = 2x - 1. Notice that f'(e)^2 = (2e-1)^2 = 4e^2 - 4e +1 = 1 mod I. hence f'(e) is a unit as I was nilpotent. so it makes sense to look at
e' = e - f(e)/f'(e) and blah blah

https://stacks.math.columbia.edu/tag/00J9 this has a few more details

ah they do it by hand >.<

lemme just finish it as well

so for for h in I you know f(e + h) = f(e) + f'(e)h mod I^2.
putting h = -f(e)/f'(e), you simply get
f(e') = 0 mod I^2. and I^2 is again a nilpotent ideal (with smaller nilpotency)

how does jagr keep such stuff in head

[rocky cloak]

It helps that I'm teaching this semester

Gave an exercise sheet with this as an exercise not long ago

[stark helm]

how to justify f(x)=x^5+ 2x^2 + 5 can not factor the quadratic factor? I am considering f(x)=(x^2+ax+b)(x^3-ax^2+cx+d), is it correct or is there any better way to show it can not be factored into quadratic factor times cubic factor?

[south patrol]

Hm good question

[celest furnace]

Turns out it’s irreducible mod 23

[south patrol]

I think basically you can go yeah

mod different primes

Probably easiest thing here is to go mod 5 and 2 to try to get to a contradiction

But you can always just brute force it as you are doing, which is probbly fastest in many cases lol

[errant shadow]

Okay so the quotient ring here is just the set of equivalence classes ${f(x) + I | f(x) \in F(X, R)}$ ? Is that really it?

[cloud walrus]

CHTRGPT

[errant shadow]

I keep on feeling like there's more here

If I take evaluation homomorphism is this isomorphic to just R by first theorem of isomorphism?

[white oxide]

what exactly would this embedding be given by? I know that $H_i \leq G_i$, but i'm not sure if $x/H_{i + 1} \mapsto x/G_{i + 1}$ makes sense

[cloud walrus]

okeyokay

[white oxide]

agh wait okay nvm

[errant shadow]

What difference does it make if I take continuous function here?

[crystal vale]

Question 3, I conclude that the kernel will be {1}.

And fix an element of A, if I do mapping f:G->A such that f(g)= g(a).

Then it will be bijective mapping by using the given condition and transitive property.

But I do not know where abelian property required.

[dull ginkgo]

Is A and G finite

[crystal vale]

No

[stark helm]

if mod 5, then f(x)=x^5+2x^2 and its root is x=1. If mod 2, x^5+1 implies root x=1. it seems that it can not turns out that f(x) is irreducible I think?

[dull ginkgo]

What’s the preceding exercise

[celest furnace]

Mod 23 it’s irreducible

Smallest prime that works too

[stark helm]

you mean mod 2？

[stark helm]

for mod 23, all roots x will generate nonzero right?

from x=0 to x=22, f(x) not zer

[celest furnace]

Like if f(x) is irreducible mod F_p for some prime p then it’s irreducible over Q too

So u just have to find a prime that works

[stark helm]

OK, it seems that 23 is the smallest prime that works

actually this questoin gives the hint: to let us show that it does not have linear factor and quadratic factor. And I also tried p=2, 3, 5......these don't work

[south patrol]

I didn't mean to imply it is irreducible over those. I meant you can use this info to narrow down possible factorisations

But brute forcing should be fairly OK. Like write it as (x^2 + ax + b)(x^3 + cx^2 + dx + e) with everything an integer. Then 0 = a + c, 0 = b + a and 2 = ad + bc = ad + a^2 which means a = 0 or a = -d.

That's only a couple of lines and nearly solves it

[crystal vale]

This one

[crystal vale]

G is a group of order p^2 and wants to prove that it has a normal subgroup order p.

Since G is a group of order p^2, it has an element of order p, say a , so G/(a) is cyclic so G is an abelian group.
So take subgroup generated by a, (a) so it is normal subgroup.

Is it correct?

And I want to show that the Quaternion group is not isomorphic to a subgroup of S_n for n<=7.

Let A be any set of order <=7.
So in this case, I want to show that stabilizer of any point a belong to A must contain subgroup (-1).

Hint?

[sonic coral]

this is fine.

every group of order p^2 is abelian, so every subgroup is normal. in particular, the one generated by the element of order p that is guaranteed by cauchys theorem.

[crystal vale]

Thank you, any hint for the last one?

[sonic coral]

actually, how do you know G/(a) has group structure

you are implicitly assuming that (a) is normal in G

[crystal vale]

Oh

[crystal vale]

How every group of order p^2 is abelian?

If Z(G) has only identity element then?
If Z(G) is non-trival then G is abelian group.

[sonic coral]

well G is a p-group so it has a nontrivial center

if the center is G then it is trivially abelian

if the center is of order p then G/Z(G) is cyclic so G is abelian

we know that the center is normal so the quotient has group structure

[crystal vale]

How can I show that?

[crystal vale]

But I have one result that G is a finite group of order n and p is the smallest prime dividing |G|, then any subgroup of index p is normal.

So if I have an element of order p, say a then (a) has index p in G so (a) is a normal subgroup.

[rocky cloak]

G acts on itself by conjugation. The size of an orbit divides the order of G, so is either 1 or a multiple of p.

An element has orbit of size 1 iff it's in the center.

So |G| = |Z(G)| (mod p). So the center has at least p elements.

[crystal vale]

Yes, the last equation comes from the class equation?

Got it, thank you.

[long obsidian]

Why are the eigenvalues at the bottom roots of unity?

This is about the standard representation of S3 permutations group

[languid trellis]

#1182793396776091708

[rocky cloak]

tau has finite order, so tau^n is the identity, hence eigenvalues^n = 1

[long obsidian]

Oooh thank you

[tawdry plover]

https://media.discordapp.net/attachments/811392071117701120/1214472798009954326/image.png?ex=65f93cee&is=65e6c7ee&hm=5d9ee0daa2a9042e8f9376a559c3e1a286a5be6e0de073807a7f82c2c8c03682&

All little bit help for the first bit pls

All my attempts have been aimed at finding an explicit map

But have failed to do so

have tried (a,b) |-> ((ab)^d,a^l/m b^l/n)

[languid trellis]

Hint: mn = lcm(m,n) * gcd(m,n)

[tawdry plover]

Okay?

I was aware of that I couldn't proceed tho

Actually I have tried a bunch of natural maps uptil now

Nothing is working 😭

[delicate orchid]

If n = p_1^a_1…p_k^a_k for some primes p_i you can write C_n as C_{p_1^a_1} \times … \times C_{p_k^a_k}. Result then follows by exploring commutivity of the direct product of groups

[tawdry plover]

I see but the book hasn't stated the abelian groups split into cyclic groups yet

[delicate orchid]

We’re not using that

[tawdry plover]

Oh okay

Actually not

It's chinese

[delicate orchid]

I’m not citing the fundamental structure theorem this is a consequence of the Chinese remainder theorem

Yurrrr

[tawdry plover]

Yeah

[tawdry plover]

Actually this is cool

It didn't think of this

Holy shit

Thanks a million

[minor wraith]

How many homomorphisms are there between $\varphi: \mathbb (Z/20\mathbb Z,+) \mapsto D_{12}$, where $D_{12}$ denotes a dihedral group

[cloud walrus]

Kakaka

[minor wraith]

any tips for this Q?

I have only seen variations of this type of Q for between cyclic groups, i.e., $\varphi: \mathbb Z/20\mathbb Z \mapsto \mathbb Z/ 6\mathbb Z$

[cloud walrus]

Kakaka

[minor wraith]

then I can just reason by looking at the restrictions on where the generator 1 can be sent, i.e. possible values of $\varphi(1)$ in modular artihmetic

[cloud walrus]

Kakaka

[minor wraith]

would it make sense to construct an isomorphism of D_n to some cyclic group so I can use this strat?

[lusty marlin]

Any homomorphism whose domain is a cyclic group is determined by the image of a generator under it

[minor wraith]

or is there supposed to be some more general way that I am missing to do these "counting homomorphisms" Qs

[minor wraith]

what if it were teh other way around

domain dihedral

codomain cyclic

[lusty marlin]

Any group homomorphism is determined by the images of a generating set of the domain

[minor wraith]

makes sense

but how do I choose images if the codomain isn't cyclic

I can't use modular arithmetic anymore

do I go by a case-by-case

depending on the type of group the codomain is?

i.e. for instance

1 is a generator of any cyclic domain

suppose I sent that to the Identity map in D_n

[delicate orchid]

If by identity map you mean identity element then yes that would work

Actually that gives a group homomorphism between any two groups

[minor wraith]

omg i recognise this username

u helped me in like a number theory q last yr or soemthing

and it made me so happy

[delicate orchid]

If u say so boss lol

[minor wraith]

wait how do I see that

because for cyclic domain + codomain with the same group operation

I can jsut do modular arithmetic

so using homo properties

[delicate orchid]

I don’t mean to be rude but it’s very obvious

[minor wraith]

I have f(k) = f(1)+...f(1) k times

[minor wraith]

i am like 1 week into group theory soz

[delicate orchid]

f(x) = 1 for all x in G implies f(xy) = 1 = f(x)f(y) immediately

[delicate orchid]

I can tell because you still care what the actual operation is

[minor wraith]

what do u mean hahaha

i need to be careful of the operation on groups don't i

[delicate orchid]

Groups don’t care what the operation is just what structure that operation gives

[minor wraith]

because it needn't be the same between groups after homo

[delicate orchid]

Anyway

As previously mentioned homomorphisms out of cyclic groups have cyclic image, so this is just classifying cyclic subgroups of D_12 with orders dividing 20

[minor wraith]

i mean like

elements in D_{12} are flips and rotations right

symmetries of an undirected n-gon graph

so supposse x = a flip, y = a rotatoin by 2π/n (for n vertices of the n-gon)

how can I see that if f(1)=Id

then f(xy)=Id = f(x)f(y)

wait, but identites needs to map to identities

oh

i did that

oh that's the trivial homomorphism

which always exists

[delicate orchid]

I’ve already given a full proof that the trivial map between any two groups is a homomorphism

Yes

And it’s image is the trivial group, which is a subgroup of D_12

[minor wraith]

so how do i find the nontrvial maps

[minor wraith]

is Pinter's book good

[delicate orchid]

I told you

[delicate orchid]

I have never read an introductory algebra textbook so idk.

Anyway the seminar is starting chat

[coral spindle]

Have fun wew

[crystal vale]

If we have |G: Z(G) | = n then what can we say about |G: C(a) | ?

[delicate orchid]

Z(G) <= C_G(a) so [G:C_G(a)] <= [G:Z(G)]

[delicate orchid]

I sort of did ig lol

[coral spindle]

Good

[crystal vale]

Yeah, but how can I show that?

[delicate orchid]

It’s obvious?

[crystal vale]

I want to prove that if Z(G) has n index then every congruency class has at most n elements.

So I think it will work.

[crystal vale]

Not for me

[delicate orchid]

Like it follows from definitions in one line:
Z(G) <= C(a) => |Z(G)| <= |C(a)|

And [G:H] = |G|/|H|

[crystal vale]

Yeah but it is not given that G is finite

[delicate orchid]

This is true with cardinal arithmetic so I don’t care about finiteness

[crystal vale]

Okay thank you

[delicate orchid]

An actual proof though? Each gZ(G) is contained in gC(a)

[crystal vale]

Oh

[crystal vale]

Is it correct?

[delicate orchid]

yeah that implication holds

[crystal vale]

Thank you

(12)(34) has a length of 4 in S_n, right?

[dull ginkgo]

Hello chat what’s up

[delicate orchid]

good question uhhh

oh wait no, luckily (12) and (34) are both of the form (i, i+1) so this should have length 2

[crystal vale]

So what about (13)(24)?

[delicate orchid]

now that's a bit more interesting

[dull ginkgo]

by length do you mean the order of the element here

[delicate orchid]

they're conjugate by a transposition in S_4 - so I will cite the result that multiplying by a transposition either increases or decreases the length by 1
we're conjugating - aka multiplying by two transpositions, so this is either length 0 or length 4 (and I don't think it's length 0...)

[delicate orchid]

nah it's a coexter group thing. For S_n it's the number of transposititions of the form (i, i+1) you need to bash together to make your element

[dull ginkgo]

[delicate orchid]

ah wait, I didn't consider it decreasing and then increasing... hmm

[dull ginkgo]

Bubbleshawty

[crystal vale]

Oh I didn't know about it, thank you

[delicate orchid]

huh?

ok maybe I'm using the wrong notion of length

[dull ginkgo]

I.e the power n such that x^n = e or the cock texter group

[crystal vale]

No, I think your definition makes sense

[delicate orchid]

that's good, what definition do you have?

[south patrol]

Group theory

[delicate orchid]

(23)(12)(34)(23) = (13)(24), so yeah it should be length 4

[crystal vale]

They didn't define length just say (123......n) has n length

[delicate orchid]

weird

[south patrol]

Never heard of length of elements in S_n

Hm

[dull ginkgo]

goop theory

[delicate orchid]

I'll give u the exposition just for real reflection groups cause I cba with coexter systems

[hardy scaffold]

length as in in terms of a generating set, with specifically the adjacent transpositions as the generating set of S_n

[dull ginkgo]

bubble sort moment

[crystal vale]

They defined length of a cycle is the number of integers which appear in it.

[dull ginkgo]

root systems?

[delicate orchid]

yes, actually

[dull ginkgo]

Root systems be nice and simple but somehow churn out bullshit like this

[delicate orchid]

let <a_1, ..., a_n | a_i^2 = 1, (a_ia_j)^(m_ij) = 1> be a real reflection group (set m_ij = 3 for S_{n+1}) then you can decompose an element into this basis, and the number of generators is invariant based on decomposition and is called the length

[dull ginkgo]

I did a little root system shit when I was interested in why E8 existed for a time

Turns out it’s a rather basic module exercise why it does

[delicate orchid]

it's just one of those things....

[dull ginkgo]

it is one of those things yes

[delicate orchid]

your dinky diagrams are small enough where the thingy just so happens to be randomly positive definite

[crystal vale]

Where can I find this ?

[dull ginkgo]

Don’t make fun of my dinky diagram

[delicate orchid]

I don't know of a source, I know it from Humphreys book on reflection groups

although, this maybe?
https://en.wikipedia.org/wiki/Word_metric

no not quite

[dull ginkgo]

Wikipedia links are just chat zipbombs

[delicate orchid]

https://en.wikipedia.org/wiki/Length_of_a_Weyl_group_element this is closer but is more technical and YUCKY!!

[hardy scaffold]

it is a word metric

[delicate orchid]

yeah but my mans just wants to know about lengths of permutations for pete's sake!

[hardy scaffold]

fair

[delicate orchid]

he don't need to know about the got dang word metric

[hardy scaffold]

where the expression is just a string in the generators of the group, with the product being w

i.e for S_n, write it as a string in transpositions (i i+1)

[crystal vale]

Okay thank you

Then what can I say about if I have two elements g and a in S_n , where length of g is s and length of a is t. Then what will be the length of gag^(-1)?

[hardy scaffold]

anything between l(a)-2l(g) and l(a)+2l(g)

[crystal vale]

I don't know how but okay

[hardy scaffold]

congruent to l(a) mod 2

you can show that l(ab) <= l(a) + l(b)

[crystal vale]

I want to prove that Z(S_n)= (1) for all n>=3 . I thought I could use the C(a), but I think it does not work.

In the book it is given that if g is length of m then | C(g) | = m•(n-m)!.

[hardy scaffold]

this can be done with cycle notation

[delicate orchid]

assume $z \neq 1$ is in the centre, then z acts non-trivially on the set ${1, ..., n}$ then use the fact that V5_ is about to type out for me

[crystal vale]

Like for all elements there exists one element which does not belong to the Centraliser set of it.

[hardy scaffold]

essentially you show that conjugating by an element of S_n corresponds to applying a permutation ot the numbers in the cycles

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

only stated here for cycles but it holds in general (decompose your permutation into cycles, then add \tau^-1\tau between the cycles)

[crystal vale]

Yes

[delicate orchid]

see if you can use that to show that if z is in the centre, z must be the identity

[crystal vale]

Okay

[south patrol]

Chat is this real

[delicate orchid]

it might just be...

[south patrol]

Oh centre of S_n

[delicate orchid]

change of basis of a F_1-vector space

[south patrol]

Lol

Centre is a normal subgroup and not A_n or the quotient would be cyclic, that solves for n >= 5

Then you have a finite problem

Overkill solution lmao

But yes I'd do what you did

[delicate orchid]

I'm not following - well I am but what do you mean

[hardy scaffold]

also showing A_n is simple is much longer and I think still uses the fact

[south patrol]

I mean like centre is normal and can't be An or Sn or Sn would be abelian

[delicate orchid]

using A_n is simple is crazy

[south patrol]

I was joking though

Note though that our approach doesn't work for S2 so I'm confused why

[delicate orchid]

"centre can't be non-trivial because there's only one non-faithful character of S_n"
Thanks buddy

[south patrol]

Yeah lol

[delicate orchid]

boss that's abelian

[hardy scaffold]

(12)=(21)

[south patrol]

That's what I mean

But yes that is the issue

[crystal vale]

So if I let z be an element that belongs to Z(G), then zgz^(-1) implies that for all i, z(i)=i, right ?

[delicate orchid]

indeed!

[south patrol]

That you can rotate stuff in cycles

[crystal vale]

Thank you

[south patrol]

I guess the key thing is that n! has 2 prime divisors iff n >= 3

Lol

Like commuting with a p cycle means you are 1 or a p cycle

And then obvs cannot be p and q cycle

[hardy scaffold]

no? (123)(45)=(45)(123)

[delicate orchid]

this is deranged

[south patrol]

Oh wait no

[delicate orchid]

I was thinking something with a (p+p)-decomposition but this is even better

[south patrol]

I mean when you restrict to that bit but yes

Sure

I should stop memeing

[delicate orchid]

the key thing is the thing we used imo

I get that ur memeing though

[south patrol]

Lol ye

[dull ginkgo]

Chat is going bonkers

[tardy hedge]

Its always pretty lit here

[crystal vale]

0 Matrix is diagonal matrix, right?

[mighty kiln]

Yes

[crystal vale]

Okay thank you

[delicate orchid]

Huh. I guess it is.

[coral spindle]

It has an eigenbasis

[rocky cloak]

More than one even

[crystal vale]

Okay thank you

[delicate orchid]

I just never thought of asking that question!

good one to ask

[crystal vale]

Sarcasm 😂?

[delicate orchid]

nope

[crystal vale]

Okay 😂

[dull ginkgo]

Just going to type random thought shit out here becaused bored

Assume M is a free Z module with symmetric bilinear form B.

Then it has associated quadratic form Q(x) = B(x,x)

M/2M is a vector space over Z/2Z = F_2

B(x + 2a, y + 2b) = B(x,y) + 2(B(x,b) + B(x,a)) + 4B(a,b), so mod 2 it is B(x + 2a, y + 2b) = B_2(x,y), so it's well defined over M/2M

If M is self dual and admits an isomorphism B(c,x) = L(x) in M*, then likewise there should be an isomorphism of B_2(c,x) in M*/2M* = (M/2M)*

But it's also important to note that Q_2(x + y) = B_2(x + y, x + y) = B_2(x,x) + B_2(y,y) = Q_2(x) + Q_2(y) so Q_2 is actually a linear form, thus Q_2(x) = B_2(x,x) = B_2(b,x) for some b in M_2

or b - x and x are orthogonal for every x

Need to do the big think here

[delicate orchid]

where we going with this boss

[dull ginkgo]

The dimensions of even unimodular lattices must be divisible by 8

[delicate orchid]

holy guacamole...

[dull ginkgo]

Well, depending on the form lol

If the form is like, just the standard dot product then yeah it must be 8

If you have negative terms like space time the dimension would be congruent to the number of negative terms mod 8

It boils down to if the norm of x is even then the norm of 2x must be divisible by 8

Then you just kind of look at the dimension of your free module

Going to see if you can determine the minimum dimension such that there is no < 2k norm elements in these lattices

For 1 I know it’s 24

[delicate orchid]

huh

[dull ginkgo]

I think Serre has a similar approach but I don’t remember specifically what approach he used

[delicate orchid]

wonder if that's related to sphere packing somehow

[dull ginkgo]

Probably Sylvester law of inertia

[delicate orchid]

probably a conincidence

[dull ginkgo]

Pontryagin duality
locally compact abelian group
is group but also is topology? Which channel does it reside in?

hello chat

Idk about sphere packing but it sounds like some Pontryagin duality shit revolving around modular forms

[delicate orchid]

I literally heard "24" and "lattice" and made the connection to sphere packing

nothing concrete lol

[dull ginkgo]

Leech lattice and E8 respectively I think

[delicate orchid]

yeah

although we don't use the Leech lattice anymore due to the theory of four humours being superseded by germ theory

[dull ginkgo]

The what

OH I got the joke

There’s probably a way fucking easier way to do this

[delicate orchid]

there always is

[stark helm]

I think it should be a+c=0, b+ac+d=0, and bc+ad+e=2?

[dull ginkgo]

I am also typing on phone lol

Waiting for the pasta wuder to boil

Lunch of champions

[delicate orchid]

I just ate like

fifteen raw mushrooms for lunch

[dull ginkgo]

whimsical

[delicate orchid]

positively

[dull ginkgo]

Are you sure you aren’t Icelandic at heart

[delicate orchid]

I ain't no huldufolk I'm more of a Sedihkonur

[dull ginkgo]

true

[delicate orchid]

did have to look up the spelling of that second word I will admit

anyway we're a tad off topic lol

[dull ginkgo]

I mean if L is a unimodular lattice of dim N then it basically is self dual and has the symmetric form (inner product) by default lol

Looking at the F_2 vector space L/2L, the inner product is antisymmetric if you can call it that mod 2 idgaf

[delicate orchid]

u can't

there must be another way....

[dull ginkgo]

Or you could just keep shifting x forward or behind in increments of 1 and check eisenstein's criterion

I don't think it's factorizable

[delicate orchid]

Gauss' lemma lets us reduce this down to checking ur poly in Z[x]

can we eisenstein it or something?

[dull ginkgo]

My lazy way is eisenstining the fuck out of it by shifting x or by either checking a fuckton of modular congruences

[delicate orchid]

no we can't (in reference to eisenstein)

[dull ginkgo]

f(x) is factorizable iff f(x + n) is

[delicate orchid]

yeah shifting is a good idea

[stark helm]

Can someone help me check x^5+2x^2+5 in Q[x] and use brute force to factorize it into (x^2+ax+b)*(x^3+cx^2+dx+e) and show why it is irreducible? I write a+c=0, b+ac+d=0, and bc+ad+e=2, bd+ae=0, and be=5, but don't know how to proceed it

[delicate orchid]

the shadow government trying to silence my homedawg out here

[dull ginkgo]

if you're mega lazy you can use decarte law of signs to guess if there are strictly positive factors and see if it's even worth it lo

[stark helm]

Also, i factorize it into (x^2+ax+b)*(x^3-ax^2+cx+d), but still not work?

[dull ginkgo]

what do you mean that I can't?

[delicate orchid]

I don't allow it

[dull ginkgo]

being gaslit by staff smh my head

[celest furnace]

Maybe try polynomial long division of x^5+2x^2+5 by a quadratic x^2+ax+b and notice that you get a factorization iff the remainder is 0

This should reduce to just 2 variables I believe

[dull ginkgo]

i tried for a lot of values and I could not get a consistent fucking prime for the non-leading coeffs for the life of me

which i guess that method is only really useful if you know a priori that it's irreducible or it's low degree

[stark helm]

We will have (b^2+a^2*b-2a+a^4)x+(5-2b+a^3b) as remainder, but question is how to show that these coefficients must be non-zero?

[dull ginkgo]

There is a kind of logic-y interpretation of this in a way

[delicate orchid]

oh brother this GUY STINKS!

[dull ginkgo]

boolean logicy

alg e bruh

Self-dual vector spaces with a symmetric bilinear form over F_2 are themselves kinda fucky

if they have a basis e_1...e_n, then basically the only options you got are e_m e_n being orthogonal or not

Which actually is interesting because that restricts the amount of possible symmetric bilinear forms

[dull ginkgo]

Might try this with E8 actually

[dull ginkgo]

bruh it's nice and even except for a single fucking basis vector that just tags along like a tumor

[rotund aurora]

how

[delicate orchid]

this is kinda craazzyyy

[rotund aurora]

isnt it one or two lines

[delicate orchid]

maybe but I can't find them

first thought is that multiplication by anything in G is a homemorphism from G to itself

[rotund aurora]

yeah

[delicate orchid]

don't know how to use compactness here. It's too pointsetty for my tastes

[rustic crown]

#groups-rings-fields message

[delicate orchid]

or hell - why does it have to be closed?!?! who knows!

je ne ce pas!!!!

[rotund aurora]

let me think for a second

first of all

why do you need compactness, what is an example where K,F closed and KF not closed

[dull ginkgo]

Are topological groups hausdorff in this case

[rotund aurora]

the book I'm following assumes they are Hausdorff, yeah

[rustic crown]

consider G = R and subgroups Z and αZ for irrational α

Z + αZ is dense

hence not closed

[rotund aurora]

hot

[dull ginkgo]

I haven't done much topology here at all lol

if any

[rustic crown]

more or less infinite union of closed is not closed, compactness makes it "finitey" union

[dull ginkgo]

But FK is the union of the (compact) images xK of K by left multiplication of elements in x

[rotund aurora]

yeah, I noticed you can write it as $\bigcup_{k\in K} kF$

[cloud walrus]

croqueta3385

[rotund aurora]

so it's like a compact union of closed stuff

but idk how to make this intuition precise

[delicate orchid]

oh yeah

check that out

[dull ginkgo]

Also literally xK is bijective to K because of x^-1K lol

it's a goop

goopy

[rotund aurora]

btw does this generalize

[dull ginkgo]

Parsing

[rotund aurora]

like if I have a continuous function $f\c X\times X\to X$ such that for all $x\in X$ both $f(-, x)\c X\to X$ and $f(x, -)\c X\to X$ are homeomorphisms, does it follow that the image of $f(K, F)$ is closed, where $K$ is compact and $F$ closed?

[cloud walrus]

croqueta3385

[warm urchin]

you can prolly do this directly too with proving compliment open, or even with nets if you add Hausdorffness

[rustic crown]

this uses the inverse map as well, so at least this solution doesn't generalize ><

[dull ginkgo]

seems like the proof of proving in-Hausdorff & Compact -> Closed

if you're a real major dingus you can use the product topology probably lmao

[rocky cloak]

Does this work?

Consider a sequence in f(K, F) that converges. Lift this by preimage to a sequence in (kn, xn) in KxF. Since K is compact kn has a subsequence converging to k. So wlog say kn converges to k.

Then I believe we can say f(k, xn) has the same limit as f(kn, xn). But since f(k, F) is closed, the limit is in f(k, F).

So I guess I'm using the assumption that f(x, F) is closed, not that F is closed.

[rotund aurora]

I'll have to read that but I think det's proof works verbatim

[rotund aurora]

the left multiplication by g^(-1) would be the inverse of f(g, -), which you are allowed to take

unless there's some subtletly I'm not seeing

[rustic crown]

(det doesn't like to think about generalizing general topology ><)

[rotund aurora]

that's neat

thanks guys

[rocky cloak]

Im still a bit unsure about the step with the convergence of f(k, xn), but meh.

[delicate orchid]

it's topology dear you don't need to be rigorous

[dull ginkgo]

Rigor mortis

[rotund aurora]

you are doing this $\lim_{n\to\infty} f(a_n, b_n)=\lim_{m\to\infty}\lim_{n\to\infty} f(a_n,b_m)$, assuming that $\lim_{n\to\infty} f(a_n,b_m)$ exists for each $m$

[cloud walrus]

croqueta3385

[rotund aurora]

i think you can just do that because it's continuous in the two variables

[rotund aurora]

Chat, I didn't make much progress

ah second

mmh connected meant that you can't have clopens besides the whole space or the empty set

so I have to show that the subgroup is clopen

I know open subgroups are closed

[delicate orchid]

oh yeah I remember this one

[rustic crown]

yep and it's pretty quick to show that the subgroup is open

[rotund aurora]

yeah so if U is open and S is some subset then US is also open

and the subgroup is like U*U^{-1}, and you keep doing this

also U^{-1} is open

so QED

[delicate orchid]

swagg

[rustic crown]

nu

you need more

[rotund aurora]

yeah I meant you keep going

you don't need to finish in finitely many steps

ok let me be more careful

at limit stages you take unions

union of open is open

so you can keep climbing

lmao

This technically uses transfinite induction

[rustic crown]

no

[rotund aurora]

how I did it yes

[rustic crown]

[rotund aurora]

because I constructed the subgroup by an inductive process

and at each step everything is open

idk how you would show the subgroup is open otherwise

[rustic crown]

a slightly easier way is to look at V = U u U^-1 and then consider union of V^n for n >= 1

[rotund aurora]

how do you know that's the subgroup?

[rustic crown]

because it's closed under multiplication and inverses

[rotund aurora]

right

I mean that's what I did

I just didn't realize the construction stopped at omega

induction is a special case of transfinite induction :p

[rustic crown]

<

[rotund aurora]

thanks

[dull ginkgo]

horror

[tough raven]

Transfinite induction is recreational mathematics.

[rocky cloak]

You don't really need any inductive construction though.

Like let U be the neighborhood of identity, and let H be the subgroup generated by U. Then for any h in H, H also contains hU, hence H is open, thus clopen. Thus equal to G.

[delicate orchid]

I may be stupid but how are we concluding H is closed here

oh

[rocky cloak]

Open subgroups are closed

[delicate orchid]

wait no

yeah

[rocky cloak]

Because G\H is the union of cosets

[delicate orchid]

yus

I knew it had something to do with cosets but couldn't remember the argument properly

[toxic zephyr]

if $R$ is a ring and $A\in R^{n\times n}$, then can we consider $R^n$ an $R[x]$ module via
$$p(x)\cdot v=p(A)v$$
or must R be a field for that to be well-defined?

[cloud walrus]

eigentaylor

[coral spindle]

Sure you can do that

The ring doesn't need to be a field, why would it?

[toxic zephyr]

and more generally if T is a module endomorphism on a module M over a ring R, can we consider a R[x] module via the similar definition
p(x)*m=p(T)m?

[rocky cloak]

I guess R would need to be commutative

[pastel pollen]

You need to give R^n a R^nxn module structure, but i assume its just the obvious one implcitly

[coral spindle]

Yes

[toxic zephyr]

okay cool

[coral spindle]

This is in fact all R[x]-modules...

[toxic zephyr]

just when this sort of idea was proposed in a class i took, it was in the context of a vector space V over F and a linear operator T such that it was an F[x] module
p(x)*v=p(T)v

wanted to know how general we can get

[toxic zephyr]

oh really? as in, if the scalars are elements of R[x], the scalar multiplication action must be p(x)->p(T) for some module endomorphism T?

[coral spindle]

Every R[x] module is an R module in the obvious way, and all the extra data you need is how x acts

[toxic zephyr]

and the way it must act is necessarily some module endomorphism?

that sounds very reasonable

[pastel pollen]

thats more or less part of the definition of an action

[toxic zephyr]

ive just never thought about it that way before. thats really cool

[chilly radish]

The map r->xr is almost by definition a module endomorphism

[pastel pollen]

giving M and A module structure is just a morphism A -> End(M)

[toxic zephyr]

i suppose scalar multiplication in vector spaces is similar. cv is just (cI)v

that's kind of trippy but also in some ways kind of obvious

[rocky cloak]

Actually, I don't think you need R to be commutative if you're just a little careful about what p(A) means.

[coral spindle]

If it's an R-module endomorphism there's no problem I think

But the matrix alone will be an issue, surely

In any case I think we're in commutative land still

[pastel pollen]

yeah the rest is specified by how R^nxn acts on R^n, that has to be part of the data

[rocky cloak]

Yeah, so if you think of R^n as a left module. Then endomorphisms look like right multiplication by matrices (viewing R^n as row vectors).

So it's not really right to say p(A). Since if p is for example rx, then the resulting action should be left multiplication by r and right multiplication by A.

This is what makes r and x commute even though r and A don't

[toxic zephyr]

so how about the simple case of R=F a field. would this necessarily have Torsion based on the minimal polynomial of A? i.e. if P(x) is the minimal polynomial of A, would P(x) generate Tor(M) (forgive me, i'm super rusty on the notation, idk if Tor(M) is right)

[rocky cloak]

The entire module would be torsion yes, and p(x) would generate the annihilator.

[toxic zephyr]

oh yeah because p(x) would annihilate everything

so could we instead make it an F[x]/(P(x)) module via
(q(x)+(P(x))v=q(A)v?

i think that should work actually...

it would be well defined since if p(x)+I=q(x)+I, where I=<P(x)>, then (p(x)-q(x))*v=0 for all v so
(p(x)+I)v=(q(x)+I)v

unless i'm makin some crazy mistake. i am rusty

[rocky cloak]

Yes that's right. If M is an R-module, and IM = 0, then M is naturally an R/I module, for any ring R and ideal I.

[stark helm]

A quick question: if we have a form of a+b2^(1/3)+c2^(2/3)=2^(1/3), can we get the conclusion that a=0, c=0 and b=1?

[rocky cloak]

Yes, that's true

[placid heart]

Let $A$ be a ring, $I$ be an ideal of $A$, and $B$ a set in $A$. Is it true that if for every $a \in A$ there are $b \in B$ and $i \in I$ such that $a=b+i$ then $B$ is a set of representatives of $A/I$?

[cloud walrus]

Casiel368

[rocky cloak]

I guess you probably want there to be a unique b for each a. Or I guess "set of representatives" can have several representatives for each class?

But yeah, that's right.

[placid heart]

Yeah I think asking for a unique b is the way to go.

How can I prove that?

[rocky cloak]

Well, you have to prove that
a + I = b + I

Then for the uniqueness thing, think about what happens if
b + I = b' + I
and you pick a to be b.

[placid heart]

Got it. Thank you

[narrow marsh]

Here R[f; F] where R is of char p, is the free left R-module generated by 1,f,f^2,.... such that fr=r^p f

Why does this V=U+Kf generate the algebra? Specifically, i dont see how to get the f^2 from this say. Does Kf mean the smallest subalgebra containing f instead?

[dull ginkgo]

[narrow marsh]

[rustic crown]

kaori

[cobalt heath]

Seems like R[f; F] should be the smallest K-algebra containing V.

[narrow marsh]

😭

[dull ginkgo]

this sucks when I actually started to try it

First I tried tackling abelian-ness. If x, y, and xy are all in I, then x commutes with y:
a(xy) = (xy)^-1 = y^-1 x^-1
a(xy) = a(x)a(y) = x^-1 y^-1

Which implies that $I \cap xI \in C_G({x})$

[cloud walrus]

Dyfunction Executive

[tawdry plover]

Isn't I closed under inverses ? As a is automorphism if g -> g^-1 then g^-1 -> g

[dull ginkgo]

yes it is

it's closed under that automorphism, inverses, and xy for two elements in I is also in I iff they commute

[tawdry plover]

Right

[hoary nexus]

What do yall think of this derivation of derrnagements

For example, this can be used to count the number of Permutations of $N$ elements that fix $k$ elements. First, let's lay down notation, let $P_{k}(N)$ be the number of permutations of $N$ elements that fix $k$ elements.

This sum runs over all the permutations of $N$ elements so it will clearly be equal to $N!$.
[
\sum_{k=0}^{N} P_{k}(N) = N!
]
The number of ways to fix $k$ elements is equal to the number of ways to NOT fix $N-k$ elements multiplied by the number of ways to fix $k$ elements.
[P_{k}(N) = P_{0}(N-k) \cdot \binom{N}{k}
]
Using the symmetry of binomial coefficients, we make a cosmetic change to the sum.
[\sum_{k=0}^{N} P_{0}(N-k) \cdot \binom{N}{k} = N!
]
Substituting this into the sum.
[\sum_{k=0}^{N} P_{0}(k) \cdot \binom{N}{k} = N!
]
Now I will use a sleight of hand and claim that we've already solved this problem. This is because we saw earlier that if you sum the differences of a sequence against combinatorial coefficients, you get the sequence back. Thus, I claim that the number of "Derangements" of $N$-elements is equal to the $N$-th difference of the factorial. Let's put this to the test.
[
\begin{array}{c|ccccccc}
n & 0 & 1 & 2 & 3 & 4 & 5 & 6 \
\hline
0 & 1 & 1 & 2 & 6 & 24 & 120 & 720 \
1 & & 0 & 1 & 4 & 18 & 96 & 600 \
2 & & & 1 & 3 & 14 & 78 & 504 \
3 & & & & 2 & 11 & 64 & 426 \
4 & & & & & 9 & 53 & 362 \
5 & & & & & & 44 & 309 \
6 & & & & & & & 265 \
\end{array}
]
This table says that the number of derangements of 2 elements is 1, and the number of derangements of 6 elements is 265. This table lines up with the known values and leads to the known formula
[
\sum_{i=0}^{N} (-1)^{N+i} \binom{N}{i} i!
]
Simplifies to
[
\sum_{i=0}^{N} (-1)^{N+i} \frac{N!}{(N-i)!}
]

[cloud walrus]

Hundred Humans In a Rat Suit

[hoary nexus]

It's my own

[dull ginkgo]

$|I \cap xI| = |I| + |xI| - |I \cup I| = 2|I| - |I \cup I| > \frac{3|G|}{2} - |G| = \frac{|G|}{2}$

[cloud walrus]

Dyfunction Executive

[dull ginkgo]

Jacobson is a cheeky mf

Can't have a subgroup index less than 2 unless it's the whole group

that is a fucking absurd exercise my god

[tawdry plover]

Hmm?

What group are we talking about

[dull ginkgo]

G

[tawdry plover]

One with index 2

[dull ginkgo]

So here's the whole thing, but not using that fucking I because it sucks

[tawdry plover]

Oh you meant C_g{x}

Has index less than 2?

[dull ginkgo]

Let group $G$ be finite with automorphism $\alpha$ such. Let $A = { x \in G | \alpha(x) = x^{-1} }$ and assume $\frac{3|A|}{4} > |G| \ \$

Assume $x, y \in A$ commute, then $\alpha(xy) = \alpha(x)\alpha(y) = x^{-1}y^{-1} = (yx)^{-1} = (xy)^{-1}$ thus $xy \in A$. It's not hard to see that it's necessary and sufficient. Therefore, for a fixed $x$, $A \cup xA \in C_G({x^{-1}}) = C_G({x})$. Now using inclusion-exclusion: $|C_G({x})| \geq |A \cap xA| = |A| + |xA| - |A \cup xA| = 2|A| + |A \cup xA| > \frac{3|G|}{2} - |G| = \frac{|G|}{2}$. Because the centralizer's order must divide the group's order but is also greater than half of the group's order, the only possibility is that the whole group commutes with $x$. Since $x$ was arbitrary, the whole group is abelian

[cloud walrus]

Dyfunction Executive

[dull ginkgo]

Fuck the second part is hard too

If |A| = 3|G|/4 then |A| can't be a subgroup because |A| must divide |G|. If A lies in it's own centralizer than A must be a subgroup so clearly there must elements that don't commute with another in A

like assume a is not in C_G(A), then A \cap a^-1A must be an abelian subgroup of order 2

[cobalt heath]

Wow, you managed it

Man of skill

[dull ginkgo]

It was the inclusion exclusion that fucked me over

and I realized it and felt dumb

lol

[cobalt heath]

Well, at least you are not mathematically dumb

[dull ginkgo]

This is the first jacobson problem that took me an hour

[cobalt heath]

Was everything else so easy for you

[dull ginkgo]

no,

just somewhat difficult

much harder than D&F's group theory exercises

but usually they took me around 5-10 minutes

or less if they are immediate

[cobalt heath]

Wow, seems you are very good at this

[dull ginkgo]

I do sometimes get caught up in being overly rigorous

which isn't always necessary especially for exercises

the sylow stuff scares me

though admittedly the proof of sylow's theorems are really cool applications of orbit stabilizer in itself

[cobalt heath]

Btw no offense, but there seems a few typos here

Like 2 |A| + |A \cup xA|

[dull ginkgo]

fair enough

I was kind of typing out my thoughts here to make it concrete

not really caring about typos or whatnot as long as I got the idea

[cobalt heath]

Also A \cup xA \in C_G({x}) >.>

[cobalt heath]

Ahhh

Just that I was smooth brained to follow it

[dull ginkgo]

Trying to prove A \cap xA is a subgroup if x \in G\Z(G)

[cobalt heath]

Isn't A \cap xA always a subgroup

[dull ginkgo]

Maybe?

xy is in A iff x and y commute for x and y in A, but I can't guarantee they commute

[cobalt heath]

From the previous exercise, you can see operation is closed over commuting elements in A

[dull ginkgo]

but i can't gauruntee x and y commute in A \cap xA

Not x and y

crap, I was doing it on paper

I used an element b in A \ Z(G)

I tried writing it out on scratch paper

For fixed b in A \ Z(G), B = A \cap b^{-1}A is what I had down but I was considering a fixed x here and reused it lol

If I can show B is a subgroup then commutativity comes with it

[cobalt heath]

Ow notation

But you can do the same with b, no?

[dull ginkgo]

idk how to show closure under product

not just product with b

let me try it out real quick

[cobalt heath]

Ah, A \cap b^{-1}A might be different, since b^{-1} may not be in A

..wait it should be in it

[dull ginkgo]

A \ Z(G)

though I wonder if C(b) must be A \cap b^{-1}A

[cobalt heath]

Then, similar logic still follows from
Fo y \in (A \cap b^{-1}A),
y = b^{-1} x for some x \in A (and y \in A)

[dull ginkgo]

trying to find aforementioned similar logic

[cobalt heath]

Idk, y x^{-1} = b^{-1} \in A

[dull ginkgo]

holy shit I'm a moron

[cobalt heath]

No

[dull ginkgo]

Same logic as the first part

wait nvm

assume c lies in C({b})/B

it's 12:30 am I'm gonna go to bed

[cobalt heath]

Yeah, suspected that

Good night+!

[stable thistle]

is it possible to come up with a functioning algorithm that is the sort of inverse of hyperoperational iteration

Like, an operation that operates on an operation, and returns an operation which shares the same relationship to it as addition to multiplication

It seems intuitive that you can't just always produce such a thing but is there a strategy to construct one if possible

Also interested in how unique they would be as solutions

[toxic zephyr]

what is the name of the theorem about what direct sum of rings a given module is isomorphic to?

[mighty kiln]

Classification of finitely generated modules over a PID?

[minor wraith]

https://www.youtube.com/watch?v=HUT-hkmemMk

dont ban me pls

[delicate orchid]

Banned

[coral spindle]

Channel name

[delicate orchid]

Really makes you think huh

[languid trellis]

#1182793396776091708

[chilly ocean]

Hey if you don't mind asking which book by Jacobson are you currently venturing through

[languid trellis]

Basic Algebra I

[chilly ocean]

Ah okay

Thank you for answering

[delicate orchid]

no problem!

[dull ginkgo]

Gaming

[languid trellis]

Np

[languid trellis]

I did one exercise in like 15 minutes then spent an hour dying to the next one

[dull ginkgo]

I was up for like 2 hours on this exercise

I gave up because it was eventually like 12:30 am when I stopped and I was tired as fuck

Back to this: If $|G|$ is a finite Abelian group with automorphism $\alpha$ and subset $A = { x \in G | \alpha(x) = x^{-1}}$ such that $\frac{3|G|}{4} = |A|$, then $G$ has an abelian subgroup $K$ of index $2$.

[cloud walrus]

Dyfunction Executive

[delicate orchid]

let me find my solution to this

[dull ginkgo]

A has the property that $\forall x, y \in A \left[ xy = yx \Leftrightarrow xy \in A \right]$

[cloud walrus]

Dyfunction Executive

[delicate orchid]

if this is true then set y = 1

[dull ginkgo]

So if the whole damn group is abelian, then A is a subgroup which can't happen because |G|/|A| = 4/3

[delicate orchid]

ah nvm, x in A

[south patrol]

Hi friends

[dull ginkgo]

Therefore, there must be a noncommuting element b in A

[south patrol]

I'm interested in this now

Oh wait finite abelian means you can kill it right

[delicate orchid]

here's how I did it yesterday.

[languid trellis]

I am not reading this conversation for when I inevitably try this problem in like 2 weeks

[delicate orchid]

it took me like 40 minutes and I think I got lucky

[dull ginkgo]

don't answer immediately noooo

[delicate orchid]

topologist make an actual fuckin rigorous statement challenge

[tough phoenix]

Must be a typo, G should be a finite group

[dull ginkgo]

rigor mortis

[delicate orchid]

I'm not

[south patrol]

I mean if 3 |G|/4 = |A|, doesn't that imply 2 | |G|, which gives you a subgroup of index 2

if G is finite abelian

Or am I smoking

[dull ginkgo]

no, not necessarily, because 4 divides |G|

[tough phoenix]

.

[dull ginkgo]

no wait

We haven't gotten to sylow stuff

[delicate orchid]

personally I don't know where you've found an abelian subgroup from lol

[south patrol]

Well okay I was overkilling it

[dull ginkgo]

"anyway back to this"

Anyway

[south patrol]

Like from the classification it follows, I wasn't sure what tools we were meant to have aha

[delicate orchid]

G isn't abelian

[south patrol]

Was that a typo thenlol

sure

[dull ginkgo]

if 3|G|/4 = |A|, then |A| can't be a subgroup of G, which happens if A lies in C_G(A) i.e all of it's elements commute. So therefore there are noncommuting elements in A

who's product therefore are also not in A

consider b

[delicate orchid]

or at least it better not be abelian or else this is really boring

[dull ginkgo]

it can't be

[south patrol]

Okay lol

[tough phoenix]

Yeah

[dull ginkgo]

Let b be in A \ C_G(G)

Now consider A \cap b^{-1}A

Which I'm pretty sure is equal to C_A({b})

If x in A commutes with b then x is in b^-1A too, so yeah

I'm wondering if it's actually C_G({b}) itself so I don't have to prove it's a subgroup which grinded my progress to a halt last night

[delicate orchid]

why do authors decide to make every exercise the most annoying technical bullshit possible btw

[dull ginkgo]

but also probably because it was 12:30 am and I was tired as fuck

[delicate orchid]

it is, I used this fact in my proof for the first part of my question

[dull ginkgo]

D & F exercises: calculate the order of A5
Jacobson Exercise:

[dull ginkgo]

hmm, gotta consider c \in G \ C_G({b})

[delicate orchid]

A \cap b^-1A = {x in G | x in A, b^-1x in A} = {x in G | a(b^-1)a(x)b^-1x = 1} = {x in G | bx^-1b^-1x = 1} = C_G(b)

[dull ginkgo]

damn

way easier than me trying to prove that it's a subgroup lmao

[delicate orchid]

wait hold on I'm reading the wrong line

lemme do it properly

[dull ginkgo]

oh my fucking god I should've just considered the commutator

[delicate orchid]

\begin{align*}b^{-1}A &= {x \in G | b^{-1}x \in A} = {x \in G | \alpha(b^{-1}x)b^{-1}x = 1} \ &= {x \in G | \alpha(b^{-1})\alpha(x)b^{-1}x = 1}\end{align*}

[\Rightarrow A \cap b^{-1}A = {x \in G | bx^{-1}b^{-1}x = 1} = C_G(b)]

[dull ginkgo]

[b,x] = b x b ^{-1} x^{-1} = b x a(b) a(x) = b x a(bx) = b x (b x)^-1 = 1

lol

[dull ginkgo]

wow that was easy

[delicate orchid]

oh yeah that's way cleaner

I imagine this is what Jacobson wanted you to spot

and the idea of intersecting your set with some convolution of that set (which is the part I got lucky in spotting lol)

sorry it was annoying me lol

[cloud walrus]

Wew Lads Tbh

[dull ginkgo]

however, that's assuming x is already in A

the way I did it

[delicate orchid]

well you're showing that elements of A commute

[dull ginkgo]

I mean that's what I based the whole thing on

elements in a A commute <=> their product in A

Arguably this proof should work for any antiautomorphism

instead of (xy)^-1

so where a automorphism and an antiautomorphism overlap

[delicate orchid]

also I don't think you need to take the compliment of Z(G) when you choose b, it's still true just a lot more obvious

[dull ginkgo]

I still don't get how we can assert that A \cap b^-1A is C_G(b)

like itself if we don't know a priori that that C_G(b) is in A

Which is basically equivalent to if x commutes with b, then x is in A

I know the one way

but not the other way

NAH WAIT

nvm

all I can assert is that a(x) commutes with b

but not that it's necessarily in A nor b^-1A

[dull ginkgo]

I think this is incorrect because this shows an inclusion not a two way

a(b^-1)a(x)b^-1x = 1 AND a(x) = x => [b,x] = 1 is a one way, not a two way

I cannot think of a single fucking way to do this

That would imply over the whole group that it is necessary and sufficient for x to commute with b that a(x) = x and a(b^-1x) = (b^-1 x)^-1

@delicate orchid is it specifically because of any assumptions we made?

a([x,b]) = [a(x),b^-1] = e which just shows a(x) commutes with b

fuck this bullshit

[delicate orchid]

I'm googling this

[south patrol]

I am being dumb on group theory again lol

[dull ginkgo]

a(x)x = a(x)b^-1 bx = a(xb) bx = a(xb)bx = a(xb) xb

[delicate orchid]

https://math.stackexchange.com/questions/619594/an-automorphism-of-a-finite-group-which-sends-more-than-three-quarters-of-elemen

oh wait this is only for the first part

[dull ginkgo]

y e p

[delicate orchid]

ok at this point you know what I'd do?

anything else

anything else is more productive than this

[dull ginkgo]

whatever fuck this problem

[delicate orchid]

questions that rely on some bullshit "trick" rather than testing fundemental understanding of the concepts have no place in textbooks

save that shit to fuel the olympiad sweats' egos

[dull ginkgo]

This fuckin answer key is like

yeah lol it's trivial

[delicate orchid]

this is what we done

it's just the last part

oh right you

ok

|A| = 1/2|G|

[dull ginkgo]

It's the whole

fuckin $C(b) \subseteq A \cap b^{-1}A$ direction which keeps fucking me over

[cloud walrus]

Dyfunction Executive

[delicate orchid]

I'm convinced what I did works

[dull ginkgo]

I'm not because it's a one way not a two way from what I can see

[delicate orchid]

=> |A U h^-1A| = |A|+|h^-1A| - |A \cap h^-1A| = 3/2|G|-|A \cap h^-1A| < |G| => |A \cap h^-1A| >= 1/2|G|

not quite getting equality there either

[dull ginkgo]

If we can show A \cap h^{-1}A is a subgroup then it's immediate

[delicate orchid]

equality comes immediate from proving that it's equal to C(h) though

yeah

[dull ginkgo]

But I can't prove that it is

might try proving it's closed under multiplication

[delicate orchid]

I remain convinced that what I did worked

maybe I should write out every step

[dull ginkgo]

x, b^{-1}x, y, b^{-1}y in A
a(xy) = a(x)a(y) = y^-1 x^-1 = y^-1 b b^-1 x^-1 = (b^-1 y)^-1 (xb)^-1 = a(b^-1y)a(bx) = a(yx)

so at least they commute

WAIT

so their product, xy, must be in A

and thus so must b^-1 xy

so it's a subgroup

[delicate orchid]

alright whatever u say boss

[dull ginkgo]

nah wait I fucked it up lol

I can't find a goddamn resource for this problem either

that isn't this one that just skips over the proof that C(h) = A \cap x^-1 A

If b is in A \ Z(G), then b doesn't commute with an element c. Clearly bc = d is NOT in A, so c is in A \ b^-1 A

[delicate orchid]

@rocky cloak HELPPPPPPPPPPPPPPPPP HELPPPP HEEELLLPPPP

[dull ginkgo]

Gotta love exercises that make you feel like a moron and a complete failure lol

[delicate orchid]

yeah, I got my PhD to do that

don't need this nonsense on top of it

[dull ginkgo]

jagr is still typing

[languid trellis]

Jagr about to spit some supa hot fire I can feel it

[rocky cloak]

So b is in A\Z(G) then you want to prove that
C(b) = A cap b^- A?

So certainly C(b) is a proper subgroup. And you've shown that A cap b^- A is contained in C(b), so since
|A cap b^- A| >= |G|/2 it must be everything

[delicate orchid]

it's so unbelievably over

[dull ginkgo]

omfg i forgot about the greater than or equal

[rocky cloak]

So yeah I guess it's just a counting argument

Don't know if you could generalize it to a setting where G isn't finite...

[delicate orchid]

considering we're ordering things by cardinality and using lagrange everywhere, I'd be surprised

[tough raven]

What was the question?

[rocky cloak]

.

[dull ginkgo]

A \cap b^{-1}A is obviously abelian because of the commutativity <=> product-in rule

Next one I need to try is showing $\frac{\bigcap_{i \in I}K_i}{H} \cong \bigcap_{i \in I}{\frac{K_i}{H}}$

[delicate orchid]

WHY

[dull ginkgo]

what

[delicate orchid]

NOBODY UNIONS GROUPS

NOBODY DOES THIS

[dull ginkgo]

oh, I copied and pasted

put bigcup

instead of cap

because lazy to type it out

[delicate orchid]

oh ok

yeah then that's a good problem

[dull ginkgo]

how rigorous do you recommend doing this problem lo

[delicate orchid]

write down "obvious" then continue on with ur day

[dull ginkgo]

hm

[cloud walrus]

Dyfunction Executive

[delicate orchid]

well the thing I notice is that on the left you have things that look like kH with k in the intersection, and things on the right that look like kH with k in the intersection (xH is a coset of all of these dudes iff xH = kH for some k in the intersection)

[rocky cloak]

Step 1: ||correspondence theorem says there's a lattice isomorphism between subgroups containing H and subgroups of G/H||
Step 2: ||lattice isomorphisms preserve meets and joins||

[delicate orchid]

the FOURTH iso theorem makes an appearence

I am SCARED and ALONE in this DARK SCARY WORLD

something something Zassenhaus

[dull ginkgo]

isn't that the "butterfly" one

[delicate orchid]

think so

you don't need it though I'm babbling

[tough raven]

This is the best way though TBH.

[dull ginkgo]

isn't this proving the lattice morphism preserves meets

[tough raven]

No.

If P and Q are posets and f: P -> Q is monotone, f need not preserve meets or joins.

But if f is an order isomorphism, it does have to preserve meets and joins.

Very convenient.

[delicate orchid]

Dyfunction means this is showing that the morphism in the corrispondence theorem is infact a lattice isomorphism

I think

[dull ginkgo]

yes

[tough raven]

This asks to show that the correspondence preserves (arbitrary) meets, yes.

[dull ginkgo]

that's what I was

[tough raven]

My point (and @ jagr2808#0's IG) though is that the easiest way to show that is to show that it is monotone (in both directions).

[dull ginkgo]

Epimorphism of the projection map $G \twoheadrightarrow{} G/H$

shit I forgor

[delicate orchid]

\leftrightarrow

[cloud walrus]

Dyfunction Executive

[dull ginkgo]

RECOVERED

[delicate orchid]

oh right you wanted the surjection

[dull ginkgo]

Epimorphisms of groups are always surjective when forg'd I think

[delicate orchid]

... yeah? where are you going with this

[dull ginkgo]

ring morphisms seem foreign and scary given Z -> Q is fucking bimorphic

i was just making sure

[tough raven]

That's not so bad honestly; after a while it seems natural for localisations to be epi.

[delicate orchid]

epi <=> surjective is just true in Set so I really don't know what you mean here, presuming you're also "forg'd"-ing these too

if you're not, then yeah it's GROSS

[dull ginkgo]

lol

[rocky cloak]

How does the proof that epimorphisms of groups are surjective go again?

[dull ginkgo]

i forgor the category theory version

i just know quotient groups projections are surjective because of the congruence relation

[delicate orchid]

something something all SES of groups are split something something

I don't actually know

good question, I shall google

[rocky cloak]

I remember it being somewhat complicated

[delicate orchid]

the proof on nlab does actually use a split SES so I wasn't far off

https://ncatlab.org/nlab/show/epimorphisms+of+groups+are+surjective

[tough raven]

But I'm sure there are epis which are not combinations of quotients and localisations .

[tough raven]

I think there's a proof that shows all subgroups are equalisers that is more complicated, but for showing that epi injection => isomorphism, there's a more elementary argument.

[tough raven]

Given a subgroup H, let G act on G/H + a new point infty the usual way (fixing infty), and define another action by conjugation by swapping H and the point.
This defines two group homomorphisms to Sym(G/H U {infinity})

Any g in G fixes infinity, so it commutes with the transposition iff it fixes H iff it lies in H.
So equaliser of the two group homomorphisms into Sym(G/H U {infinity}) is H.

Oh, IG you do get subgroups are equalisers.

[rocky cloak]

Nice

[south patrol]

Woah very cool

[tough raven]

IG the idea is that
(i) in S_{n+1}, the elements of S_n (embedded by fixing n+1) which commute with (n n+1) are precisely the things fixing n. (Let n be infinite if necessary .)
(ii) Stabiliser of any point in a group action is an equaliser (using (i)).
(iii) Any subgroup is a stabiliser in left-coset space, hence an equaliser.

Is there a version of linear independence of characters evaluated at finitely many points?

That is, for some characters χ1, …, χm: G -> F^× and some elements x1, …, xn in G with suitable hypotheses, the vectors (χi(x1), …, χi(xn)) in F^n (i = 1, …, m) are linearly independent.

[dull ginkgo]

[tough raven]

OK, so a standard proof can be used to show that if χ1(x), …, χm(x) are distinct, then for any y, linear independence holds after evaluation at y, xy, …, x^{m-1}y.

Is there a version of invertibility of Vandermonde matrices where instead of x, we have a character on a group, and instead of 1, x, …, x^{n-1}, we evaluate the character on a subset of the group?

Actually, I guess that amounts to double-posting, since it's practically equivalent to my previous question…

[chilly ocean]

The mapping to order which was a natural choice for me didn't work

Cause we loose injectivity

Is such a map possible

If not how does one generally try to prove that such a map can't exist

[delicate orchid]

you'd have to find two non-isomorphic groups with a similar enough subgroup structure that you can't distinguish between them using this map

gah if it was just solvable finite groups we might be able to do something funny here

[chilly ocean]

Well I don't grasp the full potential of what u said

But it sounds game to me

[delicate orchid]

thinking about subgroup lattices, given a H < G by the correspondence theorem we have the subgroup lattice of H is the lower set from H, and the subgroup lattice of G/H is isomorphic to the upper set from H

nah this doesn't work, there's non-isomorphic groups with isomorphic lattices

[rocky cloak]

Some hints:
The composition factors of a group is independent of composition series.
There are infinitely many primes.

[delicate orchid]

wait you can still use the composition series idea even if G isn't solvable?

oh yes of course

[chilly ocean]

I am not very familiar with composition series...I know the definition and what composition factors mean

[delicate orchid]

jesus christ jagr you're smart

haha

[rocky cloak]

Actually these are bad hints cause they overcomplicate it.

A better hint is that C2xC2 and C4 are not isomorphic.

[chilly ocean]

Hmmmm

I'm dumb

I still don't see how

[delicate orchid]

ok maybe I'm over complicating it then

[rocky cloak]

Well say f(C2) = n

What is f(C2xC2)?

[chilly ocean]

nk?

[rocky cloak]

What is k?

[chilly ocean]

some no.

[rocky cloak]

Okay, but which number?

[chilly ocean]

Huh

[delicate orchid]

2

[chilly ocean]

Why

[delicate orchid]

f(C2xC2) = f(C2)f(C2xC2/C2) = f(C2)(C2)

[chilly ocean]

Oh right C2*C2/C2 is iso to C2

[rocky cloak]

Now, the next question would be, what is f(C4)?

[chilly ocean]

f(C2)*f(C4/C2)

🤯

[delicate orchid]

ok yeah wow I was absolutely overcomplicating it

I was associating each prime to one of the simple groups and then using their exponents to count their appearances in compositions series - which I think would satisfy the required property? and just trusted that this would somehow be injective

[rocky cloak]

So the main point is that the value is completely determined by the composition factors of the group.

[chilly ocean]

Hmm

I need to learn more about composition factors

Could one do this without this example

I mean this specific objects

Anyways this was cool @rocky cloak thank you

[tough raven]

Enumerate all finite simple groups as S_1, S_2, …
Enumerate all prime numbers as p_1, p_2, …
Given a finite group G with composition factors S_i1, …, S_in (possibly with repetition), let f(G) := p_i1 … p_in.

[delicate orchid]

wait my idea WORKED???

this is unprecedented chat

[chilly ocean]

But f does not exist?

[delicate orchid]

yur

[chilly ocean]

What's yur

[coral spindle]

yes

[chilly ocean]

Okay

[dull ginkgo]

The british can be strange and unusual sometimes

[next obsidian]

https://tenor.com/view/fnaf-freddy-fazbear-fnaf-sb-golden-freddy-fredbear-gif-24705482

[delicate orchid]

it's just me

[dull ginkgo]

Everyone with an honorable role in Discussy past 7:00 est

[coral spindle]

it's just Wew

[dull ginkgo]

Hello chat what’s a good use of Sylow stuff

[coral spindle]

You can use it to show that a bunch of orders can't house simple groups, that's the classic example.

[delicate orchid]

^

[coral spindle]

But they're more like

[delicate orchid]

they tell you global data about the group from local data

[coral spindle]

You just need them to understand the p-structure of a group. It's just super important for later stuff.

[delicate orchid]

that's the gist

all of the examples that come to my mind are way too high level lol

shout out to 95% of proofs involving Brauer's characterisation of characters

[dull ginkgo]

They are interesting in general, a bit confusing

The proof is neat though by basically just considering the group action on prime-size subsets

[coral spindle]

Yes and it also tells you a bit about the structure of p-groups

[delicate orchid]

a bit

does he know chat??

[coral spindle]

No, who's chat?

Jk

[dull ginkgo]

hi chat

[coral spindle]

yes I know

[dull ginkgo]

P groups are groups with elements all of prime power order right

The group section of Jacobson is like 70 pages total lmao

[chilly ocean]

You're also doing basic algebra by Jacobson?

What's the difference between his basic algebra and lectures in algebra?

[dull ginkgo]

why else would a single exercise that sticks out like a sore thumb from the others is awful

[dull ginkgo]

idk I didn't see his lectures

Might do that eventually, but I do much better doing problems on my own then watching videos or lectures

[chilly ocean]

Youre thinking lectures in algebra is videos?

[dull ginkgo]

no idea homie

[chilly ocean]

It's a book

[dull ginkgo]

Oh

[chilly ocean]

I am confused which one I should start with

Rate the one you're doing out of 10?

[dull ginkgo]

uhhhh

8/10 but it's also the second textbook I've ever gone off of so i can't give a good relative score

I got to the Galois stuff in D&F and didn't go further, but D&F didn't make a lot of sense comparatively speaking to Jacobson

[chilly ocean]

Okay then I'll also start Jacobson tomorrow help me if I get stuck okay..?

Thank you for the suggestions

Have a nice day

[rocky cloak]

Surely the main use of Sylow subgroups is that the representation type of a group in characteristic p is equal to the representation type of its sylow subgroup.

I mean what else could you possibly use them for.

[dull ginkgo]

.... U H

I'm not very good at algebra

There's much better people to ask, I am just grinding for no good reason and as a motivation to not just lay in bed all day feeling miserable

[delicate orchid]

calling that the "main use" shows your bias

I never use that fact really

p-solvability is a lot more important for characteristic p reps in my opinion

[chilly ocean]

Is there any simple tool, compass-like, that is actually able to draw cubic roots?

[rocky cloak]

Imagine caring about anything more than just the representation type of your algebra 🤓

[dull ginkgo]

Nuesis construction I think?

[south patrol]

Source?

[south patrol]

Okay this is kinda cool

[dull ginkgo]

does Jacobson do group representations or is that like it’s own dedicated course

[delicate orchid]

it better be

I recommend Benson II for an introductory representation theory course joy cat joy cat

[dull ginkgo]

It’s in the second book nvm

The second book is apparently like the fucking gauntlet from what I’ve heard

[delicate orchid]

oh and the first one isn't?

[dull ginkgo]

Idfk I have only done D&F before this which is also apparently just as bad

Relative to what idk

I’m going to go work on Abbott lmao

[languid trellis]

bro u havent even read it what r u on abt

[delicate orchid]

oh I've seen the exercises

and to answer your question I'm on about 5 tabs and 2 litres of pure ethanol

[languid trellis]

you've only seen the exercises that we struggle with enough to post here

selection bias moment

[south patrol]

Hot

[rotund aurora]

Does this generalize for finitely generated k algebras

so quotients of k[x1,....,xn]

mmh I guess it's still true

because maximal ideals of k[x1,...,xn]/I where I is some ideal correspond with maximal ideals in k[x1,....,xn] that contain I

[coral spindle]

Yes that's completely correct

[rotund aurora]

this led me wonder if there is a criterion in general that tells you when, given a map A-->B the induced map Spec B-->Spec A respects maximal ideals

ok I guess surjectivity is enough in general

[south patrol]

Calling this the Nullstellensatz is a bit funny to me though lol

[south patrol]

Well an important thing to mention is Jacobson rings