groups-rings-fields

[dull marsh]

Probably a typo

[celest furnace]

Actually s stands for symmetry obviously

But u right that doesnt make much sense either

[lavish nexus]

Is infinite product defined in the ring of formal power series

Infinite sum is

[celest furnace]

It probably depends

[lavish nexus]

Yeah now that I think about it even infinite sum is a bit problematic

Like in a case such that the coefficient for a monomial term doesn’t converge

I’m grading some assignment and they need to prove every nonzero ideal in R[[x]] is of the form (x^n)
Instead of the easy way this guy’s doing infinite product
I feel very iffy

[hidden haven]

Yeah you can't do that

[hidden haven]

At least if this guy's using that, he needs to justify that it works in that specific case

I don't think it ever does though unless almost all factors are constants and then they should either be eventually all 1s or have a topology

Ah maybe not since degree isn't additive with power series

[steel pulsar]

Let G be a finite group of cardinal n. Show that G is isomorphic to a subgroup of Sn

I don't understand how to do that, even with the correction under my eyes...

[dull marsh]

Consider how elements of G act on G

[steel pulsar]

And why ? What in the question tells you that you need to do this?

[dull marsh]

I.e., think about what type of a mapping x -> ax with some a in G

[steel pulsar]

but I have the correction it's not the problem, I don't understand

[dull marsh]

Well it tells you to show that G is isomorphic to the group Sn and the latter consists of bijections from an n-element set to itself, so it could be useful to come up with a way of viewing the elements of G as such mappings

[steel pulsar]

the latter consists of bijections from an n-element set to itself

where do you see this?

[dull marsh]

What definition of Sn are you using?

[steel pulsar]

Sn has a particular definition? I thought it was just a notation like that

[dull marsh]

Sn is the group of permutations, i.e., bijections from {1, ..., n} to {1, ..., n}

[steel pulsar]

my teacher says that :

We consider the map τ: G → S_G given by the action of G on itself by
translations: g 7 → (τg : x 7 → g ∗ x). It is a morphism of groups: τg∗h = τg ◦ τh. It's a
injection, from inverse to left σ 7 → σ(e). The group G is isomorphic to the subgroup τ(G) ⊆ S_G.
By arbitrarily numbering the elements of G = {g1, . . . , gn}, we also have an isomorphism
S_n ≃ S_G given by σ 7 → (gi 7 → gσ(i)). In conclusion, G is isomorphic to a subgroup of S_n

[dull marsh]

Right, they considered how G acts on itself like I said

[steel pulsar]

but if

Sn is the group of permutations, i.e., bijections from {1, ..., n} to {1, ..., n}

what is Sg ?

[dull marsh]

$S_X$ is the group of permutations from $X$ to $X$, so $S_G$ and $S_n$ will be isomorphic given $G$ has the same cardinality as ${ 1, \dots, n }$, i.e., cardinality $n$

[cloud walrus]

A Lonely Bean

[dull marsh]

This makes sense because in a group of permutations we are considering the ways to rearrange a set of objects, what the objects themselves are is irrelevant

[steel pulsar]

Yes, well, it all seems very abstract to me, not intuitive

I understand nothing

[dull marsh]

Any particular source of confusion?

[steel pulsar]

All

Let G be a finite group of cardinal n. Show that G is isomorphic to a subgroup of Sn

I don't understand the question, I don't know what they want us to demonstrate, I don't understand is it necessary to use specific definitions from the course

typically you ask me that, it's total blankness in my head

[prisma ibex]

if you have a group G of size n you want to find a homomorphism G->S_n, a subgroup H of S_n, and an isomorphism G=H

if you can construct a homomorphism G->S_n which is injective, then this subgroup H will just be the image of this homomorphism

it's helpful to think about S_n as permutations of G, and then you can construct such a homomorphism by letting G act on itself in a natural way...

[steel pulsar]

Are you saying that if we find a homomorphism then we find an isomorphism?

or to find an isomorphism you must first find a homomorphism?

[prisma ibex]

if we find a homomorphism G->S_n with certain properties (namely that it's injective) then you can find the desired isomorphism easily from this

[steel pulsar]

And is this the course that says that? How do you know that ?

[prisma ibex]

I mean this is just a basic property of (injective) homomorphisms

if it's not explicitly mentioned in the course it's something you should be able to think through

like if you have a one-to-one map of sets then you can identify the domain isomorphically with its image

[steel pulsar]

and how do we found a homomorphisme G -> S_n?

[steel pulsar]

for me it doesn't seem intuitive

[prisma ibex]

think of S_n as Sym(G) (the set of permutations G->G of the underlying set of G)

a natural way you can let G act on itself by permutations is like

send an element g of G to the permutation l_g:G->G that acts by left multiplication by g, in other words l_g(h)=gh

you can (and should) check that this is an actual permutation of the underlying set of G

and then you can (and should) check that this construction gives you a homomorphism G->Sym(G)

and then you just have to check that this homomorphism is injective and you're done

[steel pulsar]

Ok I don't understand much. It's difficult for me.
I'll drop this question, thanks for trying to help me

[round hull]

this question is somewhat important for thinking about group actions

but there's probably a language barrier unfortunately

[cobalt heath]

Sad, G -> S_n is like one of the key connections of groups

Esp. as how it primarily interacts, i.e. by group action

Definitions are also hard to recall, meh

[steel pulsar]

I'm going to reread the course to see if it's clearer

on group actions

[cobalt heath]

Yea, that could be great!

Also it might help to work with concrete examples. Like cyclic groups, dihedral groups, and Q_8.

[chilly ocean]

how would I go about solving thios

[rocky cloak]

If k is in gH, that means k = gh for some h. Now, what is k^-1, does it live in some right coset?

[steel pulsar]

Let n be an integer. How many fixed points on {1, . . . , n} a does permutation belonging to Sn have on average? Check it for n = 3 and n = 4 by explicit calculations

for n = 4 I said that

4 = 4 (0 fixed point)

4 = 3 + 1 (1 fixed point)

4 = 2 +2 (0 fixed point)

4 = 2 + 1 +1 (2 fixed point)

4 = 1 +1 + 1 +1 (4 fixed point)

but in the correction they say

I don't understand how they give the cycle products for each partition

[rocky cloak]

If you have n elements, there are (n-1)! factorial ways to cycle them. Because if you fix one element as the "starting element", then each way of permuting the others gives a cycle.

So to find the number of permutations that break into a 3-cycle and a 1-cycle for example, you take the number of ways to split 4 elements into 3+1 (there are 4 such ways) and multiply by the number of ways to cycle each (3-1)! (1-1)! = 2. So there are 8 such permutations.

[steel pulsar]

Wait why do you say they are divided into 3 cycles and 1 cycles. Why 2 at the same time?

4 = 4 (0 fixed point)
4 = 3 + 1 (1 fixed point)
4 = 2 +2 (0 fixed point)
4 = 2 + 1 +1 (2 fixed point)
4 = 1 +1 + 1 +1 (4 fixed point)

One permutation = One cycle, no?

Wait, let me rephrase my question

4 = 4 here I have one element so (4-1)! ways to cycle

[rocky cloak]

A permutation can be written as a product of disjoint cycles. There can be more than one.

For example the permutation that send 1 -> 2, 2->1, 3->4 and 4->3 is the product of two 2-cycles

[steel pulsar]

4 = 3 +1 here I have two elements so (3-1)!(1-1)! way to cycle

[rocky cloak]

Yes, that's right

[steel pulsar]

and why in this question is it important to give the product of cycles which gives each partition? We do not care ?

Well, it's useless to answer the question.

[steel pulsar]

wait no it's not right

4 = 4 here I have one element so (4-1)! ways to cycle

the element = 1 so is (1 - 1)!

4 = 3 +1 here I have two elements so (3-1)!(1-1)! way to cycle

the element is =2 so (2-1)!

[rocky cloak]

I don't understand what you mean by "the element = 1"

[rocky cloak]

Yes, they do calculate some redundant information. No point in calculating stuff you then immediately multiply by 0 😛

[steel pulsar]

ok so here the element n is not 1 is 4 ?

last question

how do they calculate the average number of fixed points?

in the last image

[rocky cloak]

The partition 4=4 corresponds to a single 4-cycle. So you're cycling 4 elements

[rocky cloak]

There are eight (3+1)-cycles, having one fixed point each, six transpositions, having two fixed points each and one identity element having 4 fixed points.

8*1 + 6*2 + 1*4 = 24

Divide by 4! to get the average

[steel pulsar]

eight (3+1) cycles ?

there is only one

and there are 7 fixed points in total

[rocky cloak]

There are eight yes
(123), (132), (124), (142), (134), (143), (234) and (243)

[rocky cloak]

Not sure where you're getting 7 from

[steel pulsar]

I note them like this:

4 = 4 (0 fixed point)
4 = 3 + 1 (1 fixed point)
4 = 2 +2 (0 fixed point)
4 = 2 + 1 +1 (2 fixed point)
4 = 1 +1 + 1 +1 (4 fixed point)

and Why (3+1) cycles ant not (1+1+1+1) cycles ? or (2+2) cycles ?

[rocky cloak]

You have to count all the cycles

(or I guess you can skip the ones that have 0 fixed points)

[steel pulsar]

that's not what I said..

[rocky cloak]

You asked why not 1+1+1+1 cycles. I'm saying you have to count them as well

[steel pulsar]

but you say that all cycles are (3+1) cycles

[rocky cloak]

No, that's not what I'm saying at all

I'm saying there are eight permutations, that correspond to the 3+1 partition

[steel pulsar]

4 = 4 (0 fixed point)
4 = 3 + 1 (1 fixed point)
4 = 2 +2 (0 fixed point)
4 = 2 + 1 +1 (2 fixed point)
4 = 1 +1 + 1 +1 (4 fixed point)

and here you see that there are 7 fixed points and not 8

0 + 1 + 0 + 2 + 4 = 7

[rocky cloak]

So that gives you 8*1 fixed points

[rocky cloak]

Here you're only counting one of the eight (3,1)-cycles

[steel pulsar]

Ok I don't understand anything

[rocky cloak]

You know that there are 4! = 24 permutations in S4, right?

[steel pulsar]

yes

[rocky cloak]

And that each of them can break down as a product of cycles

[steel pulsar]

But what's the point of writing permutation if you don't use them?

It's all confusing for me: partition, permutation, cycles

[rocky cloak]

Now, one question you could ask is how many break down as the disjoint product of a 3-cycle and a 1-cycle

[rocky cloak]

Are you confused about what each of the three words mean?

[steel pulsar]

No I'm confused because we have an exercise, we write things down that we don't use

4 = 4 (0 fixed point) 
    4 = 3 + 1  (1 fixed point)
    4 = 2 +2 (0 fixed point)
    4 = 2 + 1 +1 (2 fixed point)
    4 = 1 +1 + 1 +1 (4 fixed point)

that for example we don't use it

[rocky cloak]

You most definitely use that

That's like the main thing you use

[steel pulsar]

When ?

[rocky cloak]

Throughout.

Like you identify how many fixed points a permutation has, depending on how it's partitioned into cycles (that's what's written).

Then you count how many permutations correspond to each partition.

Then you add it all up.

[steel pulsar]

I wrote the partitions of 4 and for each partition I wrote the products of cycles.

4 = 4 (0 fixed point) there is a 3! cycles product which gives this partition

4 = 3 + 1 (1 fixed point) there is a 2!0! cycles product which gives this partition

4 = 2 +2 (0 fixed point) there is a 1!1! cycles product which gives this partition

4 = 2 + 1 +1 (2 fixed point) there is a 1!0!0! cycles product which gives this partition

4 = 1 +1 + 1 +1 (4 fixed point) there is a 0!0!0!0! cycles product which gives this partition

After writing this normally I should be able to calculate the average fixed number of points

How?

[rocky cloak]

So your missing one step. The number 2!0! counts how many cycles you can make when you have picked the groupings of 3 and 1 element. But you also need to account for the 4 ways to pick such a grouping

[cloud walrus]

jagr2808

[rocky cloak]

Once you have that you just multiply
4*2!0!*1 + ... + (number of groupings) (number of ways to cycle) (number of fixed points)

and divide by the total number of permutations

[steel pulsar]

why 1 ?

2!0! is a cycles product

4 * 2!0! * 1 ?

Other example

[rocky cloak]

Each such permutation has 1 fixed point

[steel pulsar]

4 = 2 + 1 +1 (2 fixed point) there is a 1!0!0! cycles product which gives this partition

why 6 ?

[rocky cloak]

Right, so this gives you
6*1!0!0!*2

6 is the number of ways to pick out a group of 2 and two groups of 1, from 4 objects

[steel pulsar]

1 /2 is not 2

[steel pulsar]

do you have an example pls?

[rocky cloak]

(12, 3, 4), (13, 2, 4), (14, 2, 3), (23, 1, 4), (24, 1, 3), (34, 1, 2)
Are all six such groupings

[delicate orchid]

At this point just write out all 24 lol

[steel pulsar]

I think it might be the language barrier, I don't understand

I'll check on a discord in my native language...

thanks for the help, sorry again

[rocky cloak]

No worries.
A fun way to prove the general statement: ||if you pick a random permutation, the probability that any particular element is 1/n. By linearity of expectation, the average number of elements that are fixed are n*1/n = 1||

Less fun way: ||Burnside's lemma says the average number of fixed points equals the number of orbits, and there is 1 orbit||

[tribal moss]

Is there a name for non-commutative rings where the only two-sided ideals are {0} and the whole ring?

[coral spindle]

Simple rings

[tribal moss]

Thanks.

[coral spindle]

Famously under some mild conditions, all such things are matrix rings of some division ring, ofc

[tribal moss]

Interesting.

[delicate orchid]

Fartin wedderburn I presume

[coral spindle]

Yeah exactly

If your simple ring is artinian you get this

But iirc there are simple rings which are narsty... I cannot think of any examples off the top of my head

It is quite funny to think that not every simple ring is semisimple lol

[south patrol]

Lol

It came up in an exam for me to show simple rings are semisimple under some conditions and I felt very silly realising I'd never tried to prove that lol

[delicate orchid]

Do you recall the conditions

[south patrol]

No

Sadge

[delicate orchid]

WHAT

[coral spindle]

OK so apparently the Weyl algebra is an example of a non-Artinian simple algebra. And it's not semisimple. Tadah

[south patrol]

I think these were all k-algebras of finite dimension

so like artinian is probably enough

[delicate orchid]

Right

[coral spindle]

Artinian is definitely enough!

[delicate orchid]

Fartinian

[south patrol]

actually this is smth funny I found like I think thebcourse deliberately said f.d. k-algebra to avoid having to introduce the artinian condition etc

[coral spindle]

Valid tbh

[south patrol]

Yeah this was mostly a rep theory course anyway

[rocky cloak]

Let V be infinite dimensional vector space, let R be End(V) and let I be the ideal of maps with dimension of the image strictly less than the dimension of V. Then R/I is a simple non-artinian ring

[coral spindle]

Ew gross

I lie, this is in fact a fuckin sick example, very metal

[rocky cloak]

If anything non-artinian rings are a great source of counterexamples

I think both of these examples are non-Noetherian though. So it begs the question, does Noetherian+simple imply artinian?

Nvm, Weyl algebra is Noetherian

[icy bear]

that's quite cool

[rocky cloak]

It's pretty cool.

You have a group homomorphism GL(2, C) to functions on the Riemann sphere given
[a, b; c, d] |-> (ax + b)/(cx + d)
These are the möbiustransformations.

[icy bear]

damn

I've heard of mobius transformations some time

[tribal moss]

Notably, a Möbius transformation is uniquely determined by its value at any three different points.

[steel pulsar]

hey

to come back to this question

Let G be a finite group of cardinal n. Show that G is isomorphic to a subgroup of Sn

[celest furnace]

Hint: look at the left regular representation of G

[steel pulsar]

Can someone try to explain to me again by details? please ?

corrections are at the top

[celest furnace]

Can you reply to what you’re responding to

[tardy hedge]

I remember doing some of that in complex analysis

For conformal mappings or something

I forgot a lot of that course tho tbh

[delicate orchid]

use the fact that multiplication by some g in G is a permutation of G

[celest furnace]

I had a final project on this for my topology class!

[placid ferry]

hi im unsure if this is the right place to ask, but im looking for clarification on a definition: A solvable group has a normal chain with abelian factors; does this chain have to be finite, or can it be infinite?

[delicate orchid]

it has to be finite

infinite groups can strangely have a solvable series which never stabilises but in a rigorous sense stabilises at infinity

there was a name for them I forgot

[placid ferry]

thank youu that makes this problem im trying to solve much easier

[barren sierra]

Depending on how to define finite and infinite in this context

Sometimes what you really mean is the chain "stabilizes"

So there exists some N > 0 where for every group G_n in the chain with n > N, G_n = G_N

So it's infinite but stabilizes

What I'm saying is nothing more important than just semantics

[placid ferry]

you could equivalently require that the inclusion is proper, but then you lose nice chain properties

[delicate orchid]

and there he goes

[white oxide]

??? what are you talking about ???

[coral spindle]

Too late, I burned everything into my retinas

[placid ferry]

that is extremely confusing, congradulations

[celest furnace]

What properties?

[placid ferry]

idk

[chilly ocean]

I have abstract algebra final exam tomorrow aaaaah

[next obsidian]

Real

[chilly ocean]

literally looked up the definition of a principal ideal a few minutes ago

for the first time

[next obsidian]

Umm

[chilly ocean]

Alright so

a field is

an integral domain

where uh

I forgot

[next obsidian]

Brother

I will pray for you

[chilly ocean]

thanks brother

so anyways a field is

where division makes sense

But not really division cause we don't have that operation

[delicate orchid]

every non-zero element is a unit

[chilly ocean]

Alright can someone explain why we need commutativity to hold in a field?

[delicate orchid]

by definition

[chilly ocean]

fancy way of saying every element that isn't 0 is invertible

[delicate orchid]

?

[chilly ocean]

ik but what motivates it

[delicate orchid]

linear algebra doesn't work the same over division rings (non-commutative fields)

and we'd like vector spaces to be over fields

[chilly ocean]

cool cool

We didnt do much with fields

[round hull]

you want the space of linear maps between modules to be a module

that only happens with commutative rings

[chilly ocean]

all ik is

[delicate orchid]

that feels far too advanced for someone who struggled to remember the definition of an integral domain

no offense

[chilly ocean]

if you quotient out a ring with a maximal ideal

you get a field

[chilly ocean]

*field

a field is just a commutative integral domain

[delicate orchid]

that is absolutely not true by any metric

I can name thousands of commutative integral domains that are not fields

[round hull]

you shouldn't worry about this, and instead do computational exercises

[delicate orchid]

^

[chilly ocean]

wait what's a field extension

[white oxide]

integral domains are already commutative btw

[next obsidian]

I can name infinitely many

[chilly ocean]

so a field is just an integral domain where every element is unit except 0

[delicate orchid]

there aren't infinitely many. The Z[x_1,...]s stop after Z[x_1, ... x_563]

[next obsidian]

Yeah but then u start adjoining y

[delicate orchid]

you made that up

[chilly ocean]

what are the most popular fields?

[delicate orchid]

Q, R, C

[round hull]

data science

[delicate orchid]

F_p for p prime

[round hull]

machine learning

[delicate orchid]

lol

[chilly ocean]

F_n gets no love

[delicate orchid]

probably because it doesn't exist

F_1, F_4

[chilly ocean]

wait what's F_p again

[delicate orchid]

Z/pZ

[chilly ocean]

oh so just Zp with division

[delicate orchid]

? what

[chilly ocean]

Z mod p

[delicate orchid]

no it's just Z/pZ

[chilly ocean]

that's like Z/pz

it's all just labels

[delicate orchid]

there's no "division" operation that magically gets added to fields?

[chilly ocean]

well multiplicative inverses

semantics

[delicate orchid]

no

still this is worrying me

Z/pZ has multiplicative inverses just as is

you don't need to add them in

[chilly ocean]

time to use fields to grow some grass

[chilly ocean]

?

[delicate orchid]

feel that? that's true

[white oxide]

my guy just wanted to share a proposition

and i wanted to spread the word

[chilly ocean]

billion dollar question how is this useful

If M is finitely generated, then isn't it noetherian?

[round hull]

OH WAIT were you the same person who remarked that aa was just a collection of random topics

omg

[chilly ocean]

you know im right

[delicate orchid]

Noetherian modules are "finite" in flavour

[barren sierra]

Imagine dealing with non-Noetherian rings

[delicate orchid]

it really isn't. It's all triangles

[round hull]

we literally gave a bunch of applications below and you never responded

[delicate orchid]

not if R isn't notherian

[chilly ocean]

Yeh cause im not on discord all day

[white oxide]

also bro you're not gonna get that much enjoyment out of this stuff if you can't recall the basic definitions

it's just gonna be a pain in the ass

[chilly ocean]

I gotta touch grass in actual fields

[delicate orchid]

for example, R is finitely generated as a module over itself but R isn't noetherian lol

[chilly ocean]

not grass in Z/pZ

[white oxide]

LMFAO

[chilly ocean]

grass doesn't grow on rings

[white oxide]

that's actually pretty funny

[round hull]

it is

[delicate orchid]

it isn't really

[rustic crown]

char p grass

[delicate orchid]

I got a good joke

what's purple and commutes

[chilly ocean]

a field?

[rustic crown]

purple diagrams?

[coral steeple]

Good luck tomorrow

[delicate orchid]

an abelian grape

[chilly ocean]

you know I think it's not bad

like we barely did anything with fields

except try to construct F4 or F25

I don't think we did anything interesting besides making fields of certain sizes

[rustic crown]

hewwo tubu

[next obsidian]

Hello det

[summer path]

How do abelian grapes taste compare to non abelian grapes

[rustic crown]

chmuwu

[summer path]

Hewwo det

[next obsidian]

They taste better, they have a more well rounded flavor profile cuz the flavors drive around your various taste buds in a car

(They are commuting)

[celest furnace]

W

[rustic crown]

i thought they tasted less interesting since they're always stuck in the center

[coral steeple]

🎵 when you eat your non-abelian smarties do you eat the red ones last? 🎵

(sorry I had to)

[karmic moat]

dont eat nonabelian grapes and drive

[white oxide]

it's not really clear to me why this is true - isn't this equivalent to the claim that R is finitely generated?

[delicate orchid]

I really don't know how to make this clearer to you

it's just the definition of each term

[white oxide]

maybe i misread the original sentence wrong

[delicate orchid]

R is sums of the form ab

linear combinations of a basis are sums of the form ab

like

[white oxide]

wait

wait give me one sec

i probably misread

[delicate orchid]

so it's an L_2-vector space, so if we have a basis B, then every element in R can be written as a sum of terms that look like l_2b

if this basis also generates L_1, then each b is in L_1, so this is a subspace of R. But since the basis generates all of L_1 this subspace must just be all of R

[white oxide]

hm ok

[round hull]

let $a_i\in L_1$, $b_i\in L_2$. let $c_j$ be a basis of $L_1/K$. then there exist $\gamma_{ij}$ such that $a_i = \sum_j\gamma_{ij}c_j$, and
[ \sum_i a_ib_i = \sum_i \sum_j\gamma_{ij}c_jb_i ]

[cloud walrus]

quasi_semi_group

[delicate orchid]

yeah thank you

I did not want to resort to latex lol

[white oxide]

bravo

thank you

[round hull]

the $\gamma_{ij}$ are in $K$, and so in particular for each $j$, $\sum_i\gamma_{ij}b_i$ is in $L_2$, and the sum is
[ \sum_j\left(\sum_i\gamma_{ij}b_i\right)c_j ]

[cloud walrus]

quasi_semi_group

[white oxide]

no yea i got it lol

thank you tho

[delicate orchid]

this idea of a "double decomposition" comes up occasionally it's a good thing to keep in your mind

[white oxide]

yeah some tower theorem shenanigans

[cobalt heath]

If F/k is an arbitrary field extension, is Vect_F categorically equivalent to Vect_k? My assignment seems to assume so.

(Or is M_n(F) fails to become a k-algebra)

[delicate orchid]

Uhhh I think so?

Definitely for finVect lol, just take skeletons

There’s almost definitely a theorem about morita equivalence of field extenstions that I just don’t know

[cobalt heath]

How do I construct functor finVect_R -> finVect_Q, especially the map between morphisms?

[delicate orchid]

That sounds very non-constructable if I’m being honest

Or at least I’m not going to figure it out at 3am

[cobalt heath]

Ouch

Have a good night!

[delicate orchid]

They’re equivalent because their skeletons are both isomorphic to the poset category (N, <=)

[prisma ibex]

I don't think these can possibly be equivalent, how are you getting a fully faithful functor if the hom sets are so different

[delicate orchid]

Am I getting confused again

[barren sierra]

FinVect is finite dimensional vector spaces right?

[cobalt heath]

Yes

[delicate orchid]

Did the original question just say non-transcendental field extension or something

[barren sierra]

R is infinite dimensional extension over Q

that's ringing alarm bells already for me

[prisma ibex]

I feel like things already go wrong if you have a nontrivial extension

[barren sierra]

so how could these be equivalent

[cobalt heath]

Nah, it is just what I came up, because it seems like it satisfies the condition of assigment

[delicate orchid]

I don’t know what that is

[cobalt heath]

Oh yea, no specification of non-transcendental

So a non-algebraic(transcendental) extension is permitted.

[delicate orchid]

Ok then I just don’t believe this result anymore lol

[barren sierra]

yea I'm sure you can get a counterexample here

[delicate orchid]

@prisma ibex do you know anything about morita equivalence of field extensions

Wait are your fields finite? @cobalt heath

[cobalt heath]

No such restrictions.

[delicate orchid]

Hmm

[prisma ibex]

uhhh two commutative rings are isomorphic iff they are Morita equivalent

[delicate orchid]

Well there we go then

[prisma ibex]

so this cannot possibly be true for fields

[cobalt heath]

Hmmm

Yea I guess the assignment is missing a condition

[faint fractal]

Given a cyclotomic polynomial $\Phi_p(x)$ where $p$ is a prime, show that the the only fields $\mathbb{F}_k$ in which $\Phi_p(x)$ is irreducible are those in which $k$ has multiplicative order $p-1$ mod $p$

How do I prove this?

[cloud walrus]

xpoes

[faint fractal]

(Or is this even true?)

like this

[rustic crown]

it doesn't make sense when p divides k. but otherwise it's true.

show that that the splitting field of x^n - 1 over F_q with n, q coprime is exactly F_q^d where d is the order of q (mod n).

and so if beta is a root of phi_p(x) then splitting field of x^p-1 is exactly F_k(beta), which means the degree of this exactly the order of k (mod p). phi_p is irreducible if and only if this degree is p-1 iff order of k (mod p) is p-1.

the statement follows from standard theory of finite fields, F_q^d is the splitting field of x^(q^d) - x. So n | q^d - 1 means x^n - 1 divides x^q^d - x. so q^d = 1 (mod n) implies that x^n - 1 splits over F_q^d.

conversely, if x^n-1 splits over F_q^d, then the roots of x^n-1 form a subgroup of order n inside (F_q^d)* which has size q^d - 1, so q^d = 1 (mod n)

[chilly ocean]

What does degree of elements in a ring mean

[untold basalt]

What is this notation?

[south patrol]

fr ye

this is a question i thought about recently lol

[icy bear]

is there any slightly fast way to verify this?

like, without multiplying general matrices and their inverses

actually I'll see the determinant

cus it has to match

at least

[south patrol]

Nah the char polys are the same

hence the determinant in particular

[icy bear]

aaa the determinant is the same

I was wishing so much it wasn't

[south patrol]

They are likely conjugate ig

[icy bear]

I think I'll just skip this exercise

[south patrol]

Well, actually they are lol

[south patrol]

because of this

B is diagonalisable as A

as B has eigenvalues 3, 2

[icy bear]

oh

[south patrol]

(though in general, same char poly doesn't mean conjugate)

[icy bear]

the book didn't yet talk about this subject which makes me think of what they were expecting me to do

[mighty kiln]

Are there similar results for n×n
How do you uniquely determine conjugacy class

[south patrol]

just in this case the char poly splits into distinct factors so the matrices are diagonalisable

oh lol

[south patrol]

e.g. jordan block vs diagonal matrix

[mighty kiln]

For this the matrices are small enough to be hand-solvable

[south patrol]

you can do jordan normal form which is unique up to reordering

[cobalt heath]

Quick question, when was two fields isomorphic?

[mighty kiln]

Oh right

[icy bear]

searching in the internet I only found people saying that they have necessary but no sufficient conditions

[south patrol]

Uh

Not sure what kind of answer you want lol

[mighty kiln]

When an isomorphism exists

[south patrol]

When there exists an isomorphism

[cobalt heath]

Well I just realized

[icy bear]

yeah but
it's not even linear 😭

[cobalt heath]

That was too unspecific

So let's qualify with F/k a field extension.

[icy bear]

cus some repeated stuff multiply

[cobalt heath]

When is F isomorphic to k?

[mighty kiln]

Degree 1

k/F

[icy bear]

guess I'll at least try solving directly this one

since these are very nice matrices

[cobalt heath]

Oh, then any finite degree means they are not isomorphic?

[south patrol]

well 1 is finite

[chilly ocean]

What does 'degree of elements of ring' mean

[cobalt heath]

I mean yea

[south patrol]

But anyway no you could have a field extension where both are isomorphic

e.g. algebraic closure of C(X) is isomorphic to C i'm pretty sure

[cobalt heath]

Yea, I was asking about that; if it is possible for a nontrivial extension to be isomorphic

[south patrol]

Ye

in fact there is this theorem uh uncountable algebraically closed fields are isomorphic iff they have the same characteristic and cardinality

[mighty kiln]

[cobalt heath]

Wait, so countable fields might not be as isomorphic?

[mighty kiln]

Wait why does this get more the more I stare at it

[south patrol]

"be as isomorphic"

In what way

Well I'm not sure if I can think of an example where K and F are iso but K is also a non-trivial extension of F and both are countable

[cobalt heath]

Like, can countable algebraically closed fields fail to be isomorphic under similar conditions

Ahh

[south patrol]

Model theory

[mighty kiln]

Cool stuff

[icy bear]

yeah this one's pretty easy

[mighty kiln]

Do you have pog model recommendations

[icy bear]

maybe the other one also was and I was just being too lazy

[south patrol]

this is also a nice fact uh

If you know about Jordan normal form then uh turns out every Jordan block is conjugate to its transpose

[south patrol]

Idk much model theory sorry - i learnt it from a uni course lol

[icy bear]

huh

I'll actually try to compute the first one cus I'm ashamed of not even trying

[cobalt heath]

So this one is the assignment that I found issue with, it says k-algebra but not finitely generated condition. Did I make some logic error?

[rapid junco]

What do we mean by a subgroup will be centralized by another subgroup?

Does it mean that first subgroup is contained in the centralizer of the other one?

[delicate orchid]

the 2nd is contained in the centraliser of the first

[cobalt heath]

Maybe should have added "possibly" infinite extension

Anyway I think if F/k is a field extension, an F-algebra is also a k-algebra. Maybe this reasoning is faulty?

[cobalt heath]

Did I word it confusingly again

[rocky cloak]

Q(x) and Q(x^2) for example

[rocky cloak]

Yes, if k is a subfield of F, then you can make any F algebra (or any F-vector space) also naturally become a k-algebra (or k-vector space)

[cobalt heath]

Thank you for confirming!

I guess assignment can have small omissions, perhaps I am too sensitive

[rocky cloak]

And I don't if you figured this out yet or what, but
Q(x) / Q(x^2) is a degree 2 extension of countable fields where the ground field and extension are isomorphic.

One thing you can say is that if one of the fields are algebraic over it's prime field, then isomorphism happens iff the degree is 1.

[delicate orchid]

that's just a dimension argument right

[rocky cloak]

More like Galois theory argument I guess

[delicate orchid]

lameeeee

[cobalt heath]

Ah, so algebraic extensions would be not isomorphic

[rocky cloak]

Any homomorphisms must fix the prime field and permute roots. So if a field is algebraic over it's prime field then any endomorphism is surjective.

[rocky cloak]

I'm not entirely sure what you're asking.

[cobalt heath]

Sorry, I just misread there

So by the prime field, you mean F_p where p = char k?

[rocky cloak]

Yes, Fp or Q

[cobalt heath]

Yep, I see.

[south patrol]

Lmao yeah

[rapid junco]

How does one prove for all natural numbers there is a group of size n?

Just define the p groups first and then use products?

[south patrol]

Cyclic group of order n

Though unfortunately there is no group of size 0

[sand geyser]

Hey, could u give me a hint, why this group is unsolvable?

[next obsidian]

Is that even a group?

I feel like it isn’t closed under inverses or multiplication or something

[sand geyser]

It is a group

[rocky cloak]

Seems like a group to me

[next obsidian]

It feels weird cuz of the asymmetry

But I might just be on one

[sand geyser]

I've proved that G'=this, but idk what's next

[rocky cloak]

Find (G')' I guess

[wraith cargo]

this is the way

||you'll see you get the same thing back||

[sand geyser]

This looks like a way, but I thought maybe there's an easier one

[delicate orchid]

it's a parabolic subgroup dear

[rustic crown]

was i summoned

[next obsidian]

[rocky cloak]

Show that it has GL2 as a quotient group, then assuming your field has at least 4 elements PSL2 is simple nonabelian.

Proving that PSL2 is simple is probably harder, but assuming you already know that this is pretty pain free.

[sand geyser]

What is PSL2?

[wraith cargo]

SL2 but quotient out by {+- I}

[rocky cloak]

Projective special linear group

So you have a normal series
{±1} < SL2 < GL2
Which has PSL2 as a composition factor. So if PSL2 isn't solvable then neither is GL2, and thus neither is your group.

[rocky cloak]

I think you have a mistake here actually, since shouldn't any commutator have determinant 1.

[sand geyser]

So why is it a contradiction

[rocky cloak]

Because what you found includes matrices that don't have determinant 1

I think if you just add the condition det=1 it should be correct though

[winter shore]

If a group G has order $p^am$, does it follow that for any $x<a$ there's a subgroup with order $p^x$?

this doesn't seem to follow from Sylow because the sylow p-subgroup may not be normal

[cloud walrus]

Shiranai

[rapid junco]

When is A x B isomorphic to AB for A B subgroups.

[south patrol]

I don't see what normality has to do with it?

You have a subgroup of order p^a, and then all you need to do is show that every p-group has subgroups of all possible orders

You can do this by induction, say

I'll try to remember how lol

[winter shore]

yeah you're right I was thinking about some deeply flawed argument lol

[rose prism]

when do you think?

[south patrol]

Oh sure, it has non-trivial centre Z(G).

[rapid junco]

Okay so I did some googling and A int B is trivial with A, B both normal and AB = G implies this.

[south patrol]

Pick a non-identity element g in the centre, then <g> is a normal subgroup of G and you can consider G/<g> which is a smaller p-group

[rapid junco]

But nessecarily do both A and B have to be normal?

I dont know

[south patrol]

By induction, G/<g> has a subgroup of index p, so G has a subgroup of index p as well

The only exception is if G = <g>, in which case it's easy since G is cyclic

[south patrol]

Yes, if you want the "canonical" map A x B -> AB to be an iso

Because A,B are normal subgroup of A x B

[rapid junco]

What condition do we have that the kernel is trivial.

I.e. What do we impose on A and B such that ab = 1 implies a = 1 and b = 1

Is this where we use A cap B = 1

almost like a direct sum proof for v-spaces

[rocky cloak]

ab = 1 means a = b^-1, so a is in both A and B.

[winter shore]

this stumpted me for a while, but the proof looks good, thanks!

[delicate orchid]

it's basically the same idea yeah. Normal subgroups are the "subspaces" of groups

[south patrol]

Yes basically the main thing is

Well Wew is much better at this than me lol

[delicate orchid]

Society if Group was an abelian category:

no potato ur proof was perfect lol

[rapid junco]

God

[south patrol]

But a lot of things with p-groups boil down to using non-trivial centre to reduce size

[rapid junco]

Why did I not see this

[south patrol]

Like quotient out by smth small, reduce size and then pull that back to your original group

How do you show non-trivial centre lol it's orbit stabiliser right

Yeah

[delicate orchid]

indeed

[south patrol]

G is union of orbits under conjugation action, each orbit has size a power of p, so |G| = Z(G) mod p

nice

[rapid junco]

What if we know something special about our groups. Say they are both cyclic subgroups. Then neasexariky only 1 must be normal right?

Or still.

[south patrol]

Well I guess you mean "only 1 must necessarily be normal"

in which case that's right

You may want to look at semi-direct products

[rapid junco]

Are semi direct products the natural language like direct sums for vector spaces?

[delicate orchid]

kind of

[south patrol]

Hm the more natural thing is products

Semi-direct products are sort of inbetween

[delicate orchid]

in the sense that every "vector space extenstion" splits, and every split group extenstion is a semidirect product

[rapid junco]

And this follows because if both groups have trivial intersection we guarentee normality under that map?

[south patrol]

No, normality is smth you have to hypothesise for semidirect products

[rapid junco]

Why just 1 normal works though.

[south patrol]

Essentially if you have a group $G$ and subgroups $A,B \subseteq G$ where $A \cap B = 1$ and $AB = G$ and $A$ is a normal subgroup, then normality of $A$ implies that $B$ acts on $A$ by conjugation i.e. for all $b \in B, a \in A, bab^{-1} \in A$ already

[cloud walrus]

potato

[rapid junco]

Why is this?

[south patrol]

So this means that $AB$ is kind of like $A \times B$, except the elements of $A$ needn't commute with elements of $B$

[cloud walrus]

potato

[south patrol]

Oh I mean I just gave a proof in what I just said at the end

[rapid junco]

Oh a is normal

So then primallity comes in where?

Oh cyclic implies abelian and so then they do commute

[south patrol]

I'm not sure where primality was mentioned?

[south patrol]

Hm well A and B could be abelian, but not commute with each other as part of G

[rapid junco]

Right

[south patrol]

For example, the dihedral groups are semidirect products of like

C_n and C_2

[delicate orchid]

as an e- wow.

[south patrol]

if i'm not mistaken

[delicate orchid]

I see how it is

we were both about to give the same example

[south patrol]

Lol

I think it's probably the most natural one (and the only thing I could come up with lmao)

[rapid junco]

The main reason I ask is because most sylow proofs its useful to construct two sylow groups as a direct product.

[delicate orchid]

And for a non-example, Q_8

[south patrol]

or A5

lol

I do wonder if there are any super streamlined proofs of sylow theorems

[delicate orchid]

I tried to pick a metacyclic group

[south patrol]

I always forget the relevant actions

But I guess the proofs I've seen are pretty enough and there is just some combinatorial trickery necessary to get everything so fair enough

[delicate orchid]

F_S(G) is a saturated fusion system with S \in Syl_p(G) it's pretty quick!

[south patrol]

Lol

How do you prove that

[delicate orchid]

it's circular if you're using it to show the sylow theorems

[rapid junco]

And it's always true that two P Q groups have trivial intersection right?

Because of orders

[delicate orchid]

yus

that comes up occasionally

[rapid junco]

Word

[dim widget]

I think if you already know the statements it’s easy to construct the actions but i don’t really know if I would’ve come up with them before the proliferation of group theory

[south patrol]

Yes

[dim widget]

It does feel like there’s essentially only one proof

tho

[open latch]

Let G be a group with order 75. Suppose it has a subgroup of order 25 that is cyclic. How can I prove that G is Abelian?

I know that the subgroup of order 25 is normal by the Sylow's theorem.
If I could prove that G has a single 3-sylow subgroup, then G is the direct product of both subgroups and I'm done.

[delicate orchid]

there's a theorem that says if every sylow subgroup is cyclic then the group itself is metacyclic

lemme see if I can find the proof of that and then plug in some numbers

oh wait there's no need for that, just use the results you know about the number of sylow subgroups

the number of Sylow 3-subgroups is congurent to 1 mod 3 and has to divide 75. This gives us two options - there is only one sylow subgroup, or there are 25 of them

I'll leave it up to you to see why it can't be 25

[timid latch]

what makes a group cyclic

[open latch]

this is exactly where I am stuck haha

Can't prove it

[open latch]

being generated by an element

[timid latch]

(i like ur pfp btw, karoo) so like are the integers a cyclic group under additioj

because they can be generated by adding and minusing q

1

[open latch]

this much I know. I just don't know how to prove that there can't be 25 3-sylow subgroups

if there were 25 3-sylow subgroups, then it would have 25*(3-1) = 50 distinct nonzero elements

[timid latch]

no i dont know much group theory

[open latch]

and a 5-sylow subgroup has order 25, which means it has 25-1 distinct nonzero elements
so 50 + 24 + 1 with the identity = 75
So it matches the order of the group

[delicate orchid]

yeah you make a good point

the theorem who's proof I was emulating had a condition excluding the p^2 case

[timid latch]

whats a sylow subgroup

[open latch]

a p-sylow subgroup is a subgroup which has order p^n

[rocky cloak]

Since the group of order 25 is normal, G is the semidirect product of C25 and C3. |Aut(C25)| = 20, so there is no non-trivial action of C3 on C25

[open latch]

and p divides #G

[delicate orchid]

yeah there we go

[delicate orchid]

it's the maximal p-subgroup, that's important

[timid latch]

how is that definition useful

[open latch]

sure. although there is a theorem that states that there is a subgroup of order p^n if p^n divides the order of G

[delicate orchid]

that follows from the existence of Sylow subgroups

[timid latch]

ahhh

[open latch]

wow i've never heard of semidirect product 😭

[delicate orchid]

and then an inductive argument using the fact that p-groups have non-trivial centre to show that each group of order p^n has a subgroup of order p^k for each k < n

[delicate orchid]

annoyed at myself. I literally started this conversation by quoting "sylow subgroups cyclic => metacyclic" but didn't make the connection that the 5-Sylow was normal hence it was a semidirect.

[dire siren]

can't we just use the lemma "If G/Z(G) is cyclic, then G is abelian"?

[delicate orchid]

lol

hmm

We know the C_25 is normalised but not centralised

so showing this would be equivalent to Jagr's proof

[rocky cloak]

Since C3 is quite small you can prove it directly:

Let x be your element of order 25 and y your element of order 3. Then
yxy^- = x^i
for some i.
y y x y^- y^- = x^(i^2)
x = y y y x y^- y^- y^- = x^(i^3)

So i^3 = 1 modulo 25.

You can check that i=1 is the only solution so x and y commute, and your group is abelian

[delicate orchid]

but even if we have a C_5 or a C_3xC_5 that's centralised anywhere then we know that groups of order pq are cyclic so it's slightly easier, maybe

could also use burnside's fusion theorem and one line it

[open latch]

Oh this i've done before.
You are checking the commutativity between those elements.
Since the only solution is yxy^-1 = x
then yx = xy and therefore they commute, so we have the direct product G = <x> X <y>

because they are coprimes

[rocky cloak]

This is really the semidirect product argument written out

[open latch]

lol true but then it is cheating xD

[delicate orchid]

which is what you have to do when you don't have semidirect products

[open latch]

I see! Thank you for your help!!

My book doesn't have semidirect product

[delicate orchid]

I suggest learning about them, they come up a lot

[open latch]

I will. Decomposing groups into other isomorphic products seems really important and relevant

[chilly ocean]

I fucked up and said A4 was a group of size 4

I should've just picked Z2 x Z2

[chilly ocean]

Why is the semidirect product defined as G=NU, U is a subgroup and N is a normal subgroup. Why must be N a normal subgroup?

[coral spindle]

Because that's the definition

What are you really wanting to know about the semidirect product?

Maybe you can try and crystallise the question more clearly

[chilly ocean]

Wait nvm, i think i know now why

If N is a normal subgroup G=NU=UN, right?

[coral spindle]

Sure

[chilly ocean]

Ok, i was just very dumb

Thanks

[coral spindle]

Btw, you've also missed out the requirement that U and N have trivial intersection

[chilly ocean]

Yeah

[rocky cloak]

This can also happen without N and U being normal. In which case, it's called a Zappa-Szep product
https://en.m.wikipedia.org/wiki/Zappa–Szép_product

[chilly ocean]

Thanks!

[rapid junco]

Thank you to everyone who has answered my questions this semester.

[blissful zinc]

Hi, it might be super trivial for you but not for me:
Let G group, H and K subgroups of G such that |H| = m and |K|=n, gcd(m, n) = 1
Prove that H intersection K = {e}
(e = identity of G and sry for the non-latex)

From left to right subset: it is simple but Idk how/what to start for the other way.

[rapid junco]

Look at the orders of your elements in H and those in K

[coral spindle]

Good hint above

[rapid junco]

Which are in the intersection

Another hint ord(g) | G

[blissful zinc]

My first idea was something like that:
Let x = e,
o(x) = 1 = gcd ( m, n )
...

But it is like I "re-write" the same thing as the left to right inclusion.

[coral spindle]

Let's make this more explicit

H n K is a subgroup of H and of K

So what do you know about its order?

[blissful zinc]

the order divides m and n

[coral spindle]

One might even call it a common divisor of m and n.

You should be able to do this now.

[blissful zinc]

Okay I got the proof in mind now.
Ty guys

[faint fractal]

Why is the splitting field of x^n-1 over F_q F_q^d? Neither would contain the roots of unity right? Like take n = 4, q = 5, then we have x^4-1 over F_5. Our roots would be like zeta_4, but those aren't contained in F_q^d? Or am I misinterpreting this?

oh is this the same quesiton?

[rustic crown]

wdym by zeta_4?

all non-zero elements of F_5 satisfy a^4 = 1

that's lagrange's theorem

[faint fractal]

oh ok

[rustic crown]

yep

although i used that F_q^d is the splitting field of x^(q^d) - x to prove it.

the statement follows from standard theory of finite fields, F_q^d is the splitting field of x^(q^d) - x. So n | q^d - 1 means x^n - 1 divides x^q^d - x. so q^d = 1 (mod n) implies that x^n - 1 splits over F_q^d.
if you knew (F_q^d)* is cyclic you can do this instead: say q^d-1 is divisible by n, then that cyclic subgroup has an element of order n, call it beta. now all roots of x^n-1 are exactly beta^0, beta^1, ..., beta^(n-1) so x^n-1 splits over F_q^d.

[somber sleet]

Hey guys, how do you rigorously calculate the signum of following permutation:

$\sigma= (4 5 1 3 2) \in S_5$

[cloud walrus]

damn_guuurl

[somber thorn]

a cycle is even iff it has an odd number of elements

or like i mean its length is odd

[somber sleet]

I know that, but is that rigorous enough? I wanted to split the above in transposition, but I don't remember how to do it

(I have to explain it to my class)

[somber thorn]

i mean it's completely rigorous

one way u could do it

is counting the number of inversions

and then finding -1^that number

[somber sleet]

Yes exactly, but the head TA of the class I'm teaching to, asked that we have to explain it by splitting sigma in a concrete product of transposition. I don't know if the above you said is enough anyway

[somber thorn]

u can split it like $(45)(51)(13)(32)$

[cloud walrus]

sean

[somber thorn]

that's 4 so it's an even permutation

[rocky cloak]

A general method to split a cycle into transpositions:
Take your cycle (a b c d ...)
Then you want a to map to b so include (a b), now you want b to map to c, but it currently maps to a, so do
(a c)(a b)
Now a and b map correctly, but you want c to map to d, so
(a d)(a c)(a b)
etc.

There are many other general methods you could use as well.

[untold basalt]

Proof check, basic group theory

[somber sleet]

yes I got it too, now I'm able to explain it, thank you!

[somber sleet]

I see it now, thank you as well!

[languid trellis]

another less general method of seeing this is by considering the cyclic subgroup generated by a permutation and the sign homomorphism \varphi : S_n -> {-1, 1}. If the cyclic subgroup generated by a permutation has odd order, then the restriction of the sign hom to said subgroup must lie within the kernel of the sign hom. (if it didn't then the counting formula would imply that the subgroup has even order), thus it is an even permutation.

this is a way of seeing that a permutation is even iff it has an odd number of elements, as sean mentioned earlier, andI think it's quite cool

[icy bear]

does every finite group who's order factors into exactly two primes have at least one element of order of each factor?

[mighty kiln]

Yes by Cauchy

[icy bear]

just realised the argument I used for this was wrong

what dym by cauchy

like

is it a theorem

by cauchy

[mighty kiln]

Cauchy's theorem: If p divides |G| then there is an order p element.

For prime p

[icy bear]

oh

I was thinking about this problem and building a group with all elements of order 5 except the identity

and this lead me to a wrong argument of this

[languid trellis]

I skipped this exercise when I was reading this chapter but I can try help. If a subgroup of order 5 or 7 exists, then what can we say about it?

[icy bear]

I'm thinking about it

I'm guessing for no real reason that two cyclic subgroups of the same order are either equal or have intersection e

which is probably not true but it'd be nice if it was

[languid trellis]

they're isomorphic

[rocky cloak]

Not true for cyclic, but true for prime order

[icy bear]

oh yeah

I'm stupid

then it's proven ig

I mean it lacks the argument maybe I'm still wrong

well

[languid trellis]

consider the factorisation of 35

maybe

[naive whale]

So what do you guys think about my ring theory wall

[rocky cloak]

No love for noncommutative rings 😢

[naive whale]

There's still plenty of room on the adjacent wall

This is just the beginning

[static glen]

Dedekind domains missing 😦

[naive whale]

I was also thinking about whether a map of functors could be cool

[languid trellis]

get a tattoo of an insane diagram instead

[static glen]

Multiplier algebras of reproducing kernel Hilbert spaces are cool as well

[icy bear]

wait the amount of subgroups should also divide |G| right?

or not?

[languid trellis]

I'm not sure

[icy bear]

noo

[languid trellis]

what I was trying to get to was: the only integers that divide 35 are 1, 5, 7 and 35

[icy bear]

I know

[languid trellis]

so if no subgroups of order 5 or 7 existed, then G would have no subgroups, and each element would have order 1 or 35

[icy bear]

I came to the conclusion that each element having order 5 would make me have 12 cyclic soubgroups which's union gives G

[languid trellis]

only the identity has order 1, so each non-1 element in G would have order 35

[static glen]

i think Z/4Z has just 3 subgroups

[icy bear]

ok but I need 5 AND 7 to exist

or not

wait no

I don't

I'm stupid

oh yeas

I need

es

yes

[icy bear]

that's why I'm supposing all the elements except the identity have order 5

I want to arrive at some contradiction from this

[rocky cloak]

Think about how different subgroups of order 5 overlap

[icy bear]

preferably I want this 12 to be impossible but idk how

[icy bear]

yeah only 1 in common

I mean, the identity

[rocky cloak]

Exactly, so there's 1 identity and then each subgroup provides 4 other elements each

[icy bear]

if all the elements except the identity have order 5, we have separated the group into subgroups of order 5
which means we need to have 4*something disctinct elements and the identity to fill up the group
that is, 4*k - (k-1) = 35
since I only can have one identity and each subgroup would have one identity
3k - 1 = 35
3k = 36
k = 12

[rocky cloak]

So how many can that be in total?

[icy bear]

that was my counting of the amount of subgroups

[rocky cloak]

I'm not sure what's happening in your counting, but of the 35 elements 1 is the identity, so that's 34 elements you have to divide by 4

[icy bear]

oh

ohhhh

yeah my counting was wrong

I think

no no

the algebra was wrong

[rocky cloak]

I'm not sure where -(k-1) came from

4k + 1 = 35 would make more sense

[icy bear]

oh yeah

I did a stupid

it was supposed to be 5*k - (k - 1)

not 4*k

[rocky cloak]

Then it makes sense

[icy bear]

;-;

at least I got the right idea

I guessed in the beggining that there was going to be some counting error

[rocky cloak]

I guess this extends to a nice and simple argument for why a group of order pq has both an element of order p and q, at least whenever p-1 doesn't divide q-1.

[icy bear]

yey

[rocky cloak]

No group actions or Cauchy's theorem needed

[icy bear]

:))

pk - (k-1) = pq
(p-1)k + 1 = pq
k = (pq-1)/(p-1)
now idk if p-1 can divide pq - 1

[rocky cloak]

Think you have the same off by 1 error again

[icy bear]

hkgkghgmhkmgjkmhhh

AHHHH

YEH

why do I do that

well for p and q = 2 that's false but that case would be the 4 elements thing that's very well classified

for p = 2 that's false actually

in general

for q = 2 while p =/= 2, that's obviously true

(that you can't divide)

[rocky cloak]

So I guess I'm implicitly assuming p<q

Or at least that the condition is symmetric.

But yeah, still lots of cases that aren't covered

[icy bear]

oh well

[rocky cloak]

Have to wip out some stronger machinery at some point

[icy bear]

ye

[icy bear]

found more counter-examples

[rocky cloak]

5 and 13 for example

By Dirichlet's theorem I guess about 1/phi(p-1) of primes will have p-1 dividing q-1

[tardy hedge]

For q8, Q(D) is just Q?

[static glen]

it's $\mathbb{Q}(i)={a+bi:a,b\in\mathbb{Q}}$ if i guess the definition right

[cloud walrus]

Max

[tardy hedge]

Quotient field of D is not Q?

[static glen]

You have for instance $i$ in the quotient field

[cloud walrus]

Max

[rocky cloak]

Think you're reading question 9 there

[static glen]

Ah you're right nvm

i answered question 9

[tardy hedge]

Oh ok lol sorry yea i circled 9

[rocky cloak]

But yeah it's Q

[tardy hedge]

Ok thanks im like a little shaky on it but

I used that theorem i sent the other day

Extending the map idea

[static glen]

I think 8 is a pretty forward computation

[tranquil fern]

What is 5x5

[tardy hedge]

I used this theorem to show in Q8 that Q(D) is Q

[static glen]

i think you can argue more elementary

[tardy hedge]

Probably

How would u?

[static glen]

your integral domain already sits inside Q and Q is a field

[rocky cloak]

I mean I guess that's the argument they're making. Just comes down to exactly how you're defining Q(D), and how formal you want to be.

[static glen]

Yes from a higher point of view it is pretty obvious using some identifications

[rocky cloak]

D -> F being a 1-1 ring homomorphism is just saying D sits inside a field after all

[static glen]

i guess you mean the injection D -> Q(D)

[rocky cloak]

No, I was talking about the map D -> F.

[static glen]

does one-to-one mean bijective here?

[rocky cloak]

Injective

[static glen]

well the existence of those is not guaranteed by the theorem just if smt like this exists

the embedding into a field is D->Q(D)

[static glen]

aah okay, never used that terminology

[rocky cloak]

I'm a little lost in what you're trying to say.

We have a map D -> Q, so the theorem gives that Q(D) is a subfield of Q

Which means Q(D) = Q

[static glen]

i think we talk about different stuff, so whatever

[rocky cloak]

One-to-one = injective, and
Onto = surjective
Are the usual English translations for people who hate Latin

[static glen]

what is bijective then?

[rose prism]

both

wym latin

they are french

introduced by bourbaki i think

[static glen]

nvm i will just stick to saying injective, surjective and bijective

[rose prism]

see https://jeff560.tripod.com/i.html

[rocky cloak]

Correspondence is used sometimes

[tardy hedge]

Why does that mean that

You cant have a proper subfield of Q

[rocky cloak]

Not very many subfields of Q no

[delicate orchid]

lol

[tardy hedge]

Like yea i know

Ok yeah subfield of Q well its a subset of Q thats also a field. So if u just try working that out then its obvious you cant have a proper subset thats a field

Ok cool yea lol.

[delicate orchid]

any characteristic zero field has to contain Q

[tardy hedge]

Yas

[tardy hedge]

Is that like another explanation/viewpoint for it?

[chilly ocean]

If r is a non-zero non-unit and not irreducible element in a domain D, then how is it shown that r has a factorization r=xy?

[delicate orchid]

do you know about prime fields? I.e. the fact that every field of characteristic p has to contain F_p/what I just said about containing Q

it's quite simple to see, since it's characteristic 0 your field has to contain a copy of Z (just keep adding the multiplicative identity to itself), but then it has to contain multiplcative inverses to everything inside that copy of Z (and their products) - which is Q

[south patrol]

only 2

:)

[void knot]

Let $H\le G$. How does the partition [G=\bigcup_{gH\in G/H}gH] gives a bijection $G\to G/H\times H$?

[delicate orchid]

F_1 RAAAARRRGHHHHHHHHHHHH

[cloud walrus]

RaD0N

[south patrol]

i mean

[tardy hedge]

Ik any integral domain has characteristic 0 or p for prime p

[south patrol]

Well

[delicate orchid]

for each x in G, we can write x = gh for g in the transversal of H in G and h in H

cosets are disjoint, so if we fix our transversal this is unique

[void knot]

Uhh, I see. So it's straightforward

[chilly ocean]

If D is finite, then you can get a factorization from left-translation by some non-zero element s in D.

But what about when D is infinite?

[void knot]

Thx

[void knot]

Yeah.. I was a lil bit skeptical that for a given g in G, there is a unique h in H.

[delicate orchid]

specifically h = g^-1x (for x in gH)

[void knot]

Intuitively that h is the projection of x onto H, right?

[delicate orchid]

I suppose that's a good way of thinking about it

it's the "h-part"

[rocky cloak]

Is F1 "morally" a subfield of Q?

[prisma ibex]

[dim widget]

it's the absolute base no?

[rocky cloak]

Idk

[delicate orchid]

hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

[mighty kiln]

What is F1

[delicate orchid]

No. But it is morally a "subfield" of C

[mighty kiln]

Wikipedia be not helping

[delicate orchid]

The way I'm justifying this crankery is that the burnside ring over a group is essentially the representation ring over F_1, and it decomposes like the representation ring over C does

[rocky cloak]

The only encounters with F1 I've had is linear algebra on F1 beeing this funny pointed set thing. Which does sort of imply F1 is a subfield of every field, but idk...

[rocky cloak]

Field with one element

[delicate orchid]

oh wait yeah that's a good point

[mighty kiln]

But what is it actually

[rocky cloak]

Now, that's a good question

[delicate orchid]

to me it's a strange corrispondence between facts about GL_n over finite fields and the symmetric group

I know there's an actual categorical way to make it some object but that's not relevant to me

[mighty kiln]

Is it like "category of fields would've been nice with this extra object"

[rocky cloak]

The explanation I've heard is polynomial rings are PIDs, Z is a PID, thus Z is the polynomial ring over some field. Let's call this field F1 blablabla algebraic geometry or something

[delicate orchid]

from the nlab page

but to me the pointed set thing comes from the symmetric group statement

for example, the p-local theory of S_n is exactly the same as GL(n, q) if we set q = 1

[rotund aurora]

I remember a prof mentioned a connection with F1 and the Riemann hypothesis, I wasn't sure if he was trolling then but it is true apparently

[mighty kiln]

Oh right nLab exists

[rocky cloak]

I don't think anything can make the category of fields nice

[delicate orchid]

adjoin objects until it turns into CRing

[void knot]

And when H is normal subgrup of G, does that bijection become isomorphism, or we need more?

[prisma ibex]

Yeah this is like, one of the main motivations for F1

[delicate orchid]

you need very specific conditions for that to become an isomorphism

[void knot]

There is a name for these conditions?

[delicate orchid]

mainly you need A) A subgroup X of G isomorphic to G/H and B) for H and X to centralise each other

yeah it's called "an inner direct product"

[delicate orchid]

the main motiviation is that it's REALLY funny

[void knot]

Thx

[mighty kiln]

Alternatively H and X are normal and intersect trivially

[chilly ocean]

Are factorizations in a domain always unique up to units and ordering?

[mighty kiln]

You need UFD

Literally unique factorization domain

[rocky cloak]

A very appropriate name

[chilly ocean]

Is there an extra simple condition to put on a domain so that factorizations are unique?

[slim kayak]

What's even the intuition behind F1-like object?

[rocky cloak]

Z[sqrt(-5)] is an example of a domain that isn't UFD.

[slim kayak]

Like some analogue to how algebraic groups are group objects F1 might be some weird categorical construct that "behaves like a field with 1 element"?

[delicate orchid]

yeah that's pretty much it

nG can tell you what the corrispondence is actually called

[rocky cloak]

PIDs are UFDs, otherwise I don't know of any simple restriction

[mighty kiln]

Poly rings over UFDs are UFDs

[prisma ibex]

Yeah you want things like rings of integers in number fields to be like curves over F_1

You want to be able to make sense of something like “Spec(Z)x_Spec(F_1)Spec(Z)” and do intersection theory on it

We somehow know what this looks like around the point “(p,p)” and it’s complicated

[cobalt heath]

Is F_1 legal

[tardy hedge]

Can i have an example of integral domain D and a 1-1 map into a field F such that Q(D) is not isomorphic to F?

[slim kayak]

Damn the french have a better name for F_1. F_un

Groethendieck might have been onto something

[rocky cloak]

Z -> R

[delicate orchid]

for some reason my mind went to Z -> Q(sqrt(2)) before R lol

[tardy hedge]

Thx

Damn so weird how i just dont feel satisfied with the quotient field stuff yet

I know its not complicated but some of it just doesnt feel like its clicking

In the way i want it to

I suppose i just need to move on eventually

[tardy hedge]

I think this example does help me a bit though

[delicate orchid]

fractionszzzzzzzzz

[tardy hedge]

Yeah

A fraction is just blah times inverse of blah

[delicate orchid]

how many examples do you know

you've seen the isomorphism between a field and it's "field of fractions" I know that much