groups-rings-fields

[barren sierra]

but idk how to show that every map in Hom(A, B[m]) must take this form or even just constructing an inverse

[grand cliff]

Hey! I'm taking a second class in group theory. This second class has a specific focus on permutation groups.
What are some surprising/cool results or applications from this area?

It looks like the class culminates with the O'Nan Scott thm

[barren sierra]

ok I think I got something but idk how sound it is

[tacit hemlock]

what does it mean for a set to be generated by an element of that set?

I thought the <X> notation was reserved for generating sets of a subgroup

i.e. X is supposed to represent a set, not an element $$x \in X$$ of the set

[cloud walrus]

normalAtmosphericPa=101,325

[formal ermine]

<x> = <{x}>

like

<x> = { x^n | n in Z }

[tacit hemlock]

oh

so a set of only 1 element?

[formal ermine]

yes

it's the cyclic (sub)group generated by that element

[tender wharf]

well you could possibly also think about it as cosets, although that group is trivial

probably not what you want though

[static needle]

${a \in \mathbb{Z} \vert n < a < k}$

[cloud walrus]

messyinterval

[summer path]

i think you mean $(n,k) \in \mathbb{Z} \times \mathbb{Z}$ :)

[cloud walrus]

Tubular Cat

[timber oxide]

One application is the nonsolvability by radicals of polynomial equations of degree 5 or higher. Let $p\in k[X]$, $char(k)=0$, then $p(x)=0$ is solvable by radicals iff the Galois group of $p$ is solvable. A group $G$ is said to be \textit{solvable} if there exists a chain of subgroups $G=N_0\supset...\supset N_{k-1}\supset N_k={e}$ where $N_i$ is normal in $N_{i-1}$ and $N_{i-1}/N_i$ is abelian. For a general polynomial $p$ with $deg(p)=n$ ($n$ distinct zeros) we get that it's Galois group is $S_n$, which is not solvable for $n\geq 5$. As long as we don't work in finite characteristic, we can apply the theorem to prove that $p(x)=0$ is not solvable by radicals when $deg(p)\geq 5$.

[cloud walrus]

Timwestlund

[grand cliff]

What does this have to do with permutation groups?

[timber oxide]

The nonsolvability of the group $S_n, n\geq5$ is involved

[cloud walrus]

Timwestlund

[vagrant zinc]

One is generator for the subgroup and the other is generator of the group, I think in the books it is usually concentrated in subgroup and then in group, but it is the same tendency if "a" belongs to the group and it is generator then it means that it generates all to the group analogously for the subgroup, to identify that every element is generator of a group or subgroup we must take into account that gmd(k,n)=1 where k is the element "a" belongs to the group and n is the modulus of Z, that means that k and n must be coprime if they are then "a" is generator.

[south patrol]

My head hurts

[long nebula]

If K1/F and K2/F are both finite extensions and K1K2 is isomorphic to K1 \otimes_F K2 then why does that imply that a basis for K1 is also lin.ind. over K2? I've been reading about linearly disjoint extensions for so long and I'm so confused

[molten viper]

I realize, re: yesterday

I have no damned clue what a vector space actually is

I understand R as a vector space

But what’s the Abelian group you’re making a module over? And what’s the field?

[elder wave]

R and R

[molten viper]

Horrifying

We’re talking about field extensions in lecture and I’m sobbing

[elder wave]

field theory isn't a strong trait of mine either

[south patrol]

Does it lie outside your field of expertise?

[molten viper]

So like

The degree of a field extension

It’s the dimension of a vector space

[elder wave]

yes

[molten viper]

So if it’s F and R

[elder wave]

F/R?

[molten viper]

Like, F is the extension of R

[elder wave]

yes that's what F/R means

[molten viper]

Notation

Anyways

So

So we want the dimension of F as a K-vector space right?

[elder wave]

yes

[south patrol]

R-vector space

in dis case

[elder wave]

[molten viper]

What

[elder wave]

you said R earlier

[molten viper]

Ope

My bad

[south patrol]

Also F/K rather than K/F is lol

[elder wave]

but i get what you mean

Do people use K/F

i've only seen bigger field / smaller field

[south patrol]

lol yes same

I meant like traditionally K should be bigger than F lol

[elder wave]

ahh lmao

for me it's L/K

[south patrol]

Yes sure

M/L/K/F/E

though M is on thin ice

[elder wave]

L/M/F/E/K for me

[south patrol]

Bruh

[elder wave]

M, F and E in no particular order

[south patrol]

alphabetische

[molten viper]

So

[elder wave]

important problems being discussed

[south patrol]

Lol

[molten viper]

It’s a R-module

With the multiplication from R also

[elder wave]

yes

[south patrol]

That's what it means to be an R-module

[elder wave]

We can stick to vector spaces tho since everything is a field

[molten viper]

Well I’d like to understand the “simpler” structure

[elder wave]

"A vector space is a special case of a module "

[elder wave]

what structure

[molten viper]

Modules

[elder wave]

i mean

the problems you face when moving from Vector spaces to Modules don't exist here

so just sticking to a lin alg perspective is more helpful imo

[molten viper]

I’ll think about this more later

[elder wave]

i'm not sure what your question is

tbh

[molten viper]

Me neither

[molten viper]

Who’d have though linear algebra is related to algebra

Thought

[topaz solar]

Modules are by no means simpler, vector spaces are definitely way nicer

[long nebula]

Does anyone know anything about linearly disjoint extensions? I'm trying to figure out why K1K2 being isomorphic to K1 \otimes_F K2 is equivalent to a basis for K1 being lin. ind. over K2

[remote nymph]

I'm trying to compose two permutations: (1 2 3)(2 3 4). Just by writing it out by hand I get (1 3)(2 4). However, I'm trying to calculate it without writing out each step, but I keep getting the wrong answer:

Say a=(1 2 3) and b=(2 3 4)
b keeps 1, and a maps 1 to 2, so we get (1 2)
b maps 2 to 3, and a maps 3 to 1, so now we've finished a cycle
b maps 3 to 4, and a keeps 4, so now (1 2)(3 4)
b maps 4 to 2, and a maps 2 to 3, so the second cycle is complete and the answer is (1 2)(3 4)

This is obviously wrong, but I don't understand where I made a mistake

[barren sierra]

how are you writing it out by hand

[remote nymph]

(2 3 4) maps 1, 2, 3, 4 to 1, 4, 2, 3
(1 2 3) maps 1, 4, 2, 3 to 3, 4, 1, 2

[delicate orchid]

you seem to be doing left-to-right multiplication but then right-to-left composition

[barren sierra]

^^^

[delicate orchid]

those two permutations don't commute with one another so you'd get different results

[barren sierra]

ab = a o b

where o is composition

cause really permutations are functions

so you apply b, then a

if you are getting 1 -> 3

[remote nymph]

alright, could you give me an example of how to carry out the first step of computing where 1 maps to?

[delicate orchid]

if you worked it out "by hand" right to left you'd get (12)(34) as you stated

[barren sierra]

this is right

what you have written as "obviously wrong" is actually right

[remote nymph]

ooh

[delicate orchid]

a maps 1 to 2, b maps 2 to 3

a maps 3 to 1, b fixes 1
(13)

so on

[barren sierra]

wait am I doing it backwards

[delicate orchid]

it's just convention which way round you do them

every (finite, at least) group is isomorphic to it's opposite group

[remote nymph]

is this left-to-right or right-to-left?

[delicate orchid]

left-to-right

a is on the left and we're applying it first

[barren sierra]

((1 2 3) o (2 3 4))(1) = (1 2 3)(1) = 2

is this not the way everyone does the symmetric group operation?

you apply the right most one first like composition

[delicate orchid]

no I do it backwards to that

[barren sierra]

[delicate orchid]

I also do function composition backwards to everyone else

[barren sierra]

I mean arguably you do it the correct way if looking at diagrams but no one does it that way

I'm 99% sure @remote nymph's teacher will want (1 2 3)(2 3 4) = (1 2)(3 4)

not (1 3)(2 4)

[topaz solar]

The incorrect way

[delicate orchid]

never claimed it was correct

[remote nymph]

so (1 2)(3 4) is the result of multiplying right to left

[delicate orchid]

I might start writing f(x) as xf if ur not careful

[delicate orchid]

yus

[topaz solar]

I can respect that one

[barren sierra]

ye

because the group operation respects function composition

[remote nymph]

but this also feeels like its right to left, but i get the left to right result?

because im applying the right permutation first

[long nebula]

what book is that 👀

[next obsidian]

Matsumura

[barren sierra]

(1 2 3) maps 1, 4, 2, 3 to 3, 4, 1, 2

[long nebula]

ty!! I'll check it out

[barren sierra]

what

1 -> 2

not 1 -> 3

[next obsidian]

This is the entirety of the proof

[long nebula]

Oh lol

I'll check it out anyways might be nice to look through for the future

[barren sierra]

petition for #groups-rings-fields and #adv-algebra plz

[delicate orchid]

this tbqh

[topaz solar]

There seems to be a missing b->c bit

[next obsidian]

…

[elder wave]

...

[topaz solar]

[elder wave]

wait which matsumura is that

[next obsidian]

Commutative ring theory

[elder wave]

wtf

why is mine so much worse

print wise

[barren sierra]

skill issue

[next obsidian]

This is an old hardcover copy

[remote nymph]

yes so this is kind of confusing me. if (2 3 4)(1,2,3,4) = 1, 4, 2, 3
and (1 2 3)(1, 4, 2, 3) = 3, 4, 1, 2
was that not basically the same as multiplying from right to left?

[next obsidian]

I bought on eBay

[elder wave]

ah

i have a softcover from cambridge thingy

[next obsidian]

But I feel like my original paperback was not really that much worse

In print quality

[elder wave]

okay i just rechecked

[barren sierra]

this notation is confusing me. By saying (2 3 4)(1,2,3,4) = 1, 4, 2, 3 are you saying that 1 gets mapped to 1, 2 gets mapped to 4, 3 gets mapped to 2, etc?

[elder wave]

it isn't that much worse, idk why i thought that

[remote nymph]

sorry the comma separated list of numbers is the list we're applying the permutation (2 3 4) to. So we map 2 to 3 in the list, 3 to 4, etc

[barren sierra]

you say you mapped 2 to 3 but the 2 turned into a 4

like

(1, 2, 3, 4)
(1, 4, 2, 3)
just looking at it vertically

this looks like 2 turned into 4

3 turned into 2, etc

[remote nymph]

oh my god

ive been doing permutations wrong

[barren sierra]

happens to the best of us

[remote nymph]

i was thinking positionally, and not in terms of the actual values

thanks for your help

[fervent agate]

Didn't want to interrupt the conversation that was going on, so I posted a question in #help-26

[fast stratus]

i only saw cyclic groups defined with integers… so is there an extended version of it where cardinality can be more than Z?

[north sand]

all cyclic groups are countable

[nocturne bone]

Okay, I am trying to solve this by finding some way to extract the minimal distance from the generator polynomial. I know that the generator polynomial is the smallest polynomial that generates the code so the number of coefficients of the polynomial that aren't 0 are the minimal distance. But how do I express that fact mathematically?

[fast stratus]

i see why they must be countable… but can’t you extend the definition somehow that’s what i’ve been wondering

[north sand]

well the question is how you would want to extend it. what properties do you want to keep and which ones do you want to give up

[honest edge]

I’m just starting abstract algebra and got this problem

I suck at proofs and that kinda stuff and I don’t really know how to do this

[molten viper]

Why must they be countable?

[fast stratus]

seems like you either have to giveup unique generator or powerlaws if you use R

[molten viper]

How do you prove a biconditional statement?

[honest edge]

I don’t know

[molten viper]

Also you might be better in #foundations for this particular problem

[honest edge]

Alright thanks

[north sand]

dont go into #foundations with that. try one of the early university channels

[honest edge]

Oop ok

[molten viper]

Oh ok my baaaaaad

I have early uni muted lol

[glass vine]

I have a question related to notion of abelianess in Universal Algebra and was recommended to look at it here. I was reading a paper by D. Stanovsky titled Commutator Theory for loops. At one point he mentions that it's easy to see that a loop is abelian if and only if the multiplication is commutative, I decided to try and . One direction is easy, but I am strugling to find a term that would give me the => direction.

[honest edge]

Uh

Why’d you reply to my message for that

[glass vine]

Mistake

[honest edge]

Oh

[glass vine]

Sorry, discord isn't my cup of tea, I was referred by a friend who also couldn't think of an idea.

[topaz solar]

I hate the wacky magma, loop, semigroup, etc structure names,

what was a loop and what's the condition to be abelian?

[glass vine]

Quasigroup is a set with an operation , for each element x and each element y there exist a and b such that xa = b*x= y, loop is a quasigroup with an identity element. .

In universal algebra notion of abelianess is a little messed up. Algebra A is abelian if whenever t(x,u) = t(x,v) then t(y,u) = t(y,v) here x,y,u,v can be tuples of elements of A

And t is a term in A, so basically a sequence of operations from A on elements of A

For example terms in rings are polynomials, in groups they're things of the form x1x2x3... where xi's are not necessarily distinct

[topaz solar]

So you need abelian->commutative, or commutative->abelian

[glass vine]

For this problem it's important to note that we consider the loop L to be a universal algebra (L,*,1,/,) where * is multiplication, / and \ are "division" operations and 1 is the identity operation

[glass vine]

abelian => commutative

[topaz solar]

Have you tried throwing the classic commutator at it

[glass vine]

That would be t(x,y) = x'*y'xy?

[topaz solar]

yeah something like that

this is obviously not something I know the answer to offhand

[glass vine]

Yes yes I don't expect you to hahah

I don't think it works but it might depending on what you plug in as x and y here

Give me a few to try some things with it because it might actually work

[topaz solar]

well if you can get t(x, e) = t(x, u) = e, then you'd get t(y, u) = e

but uhhh, that's not quite helpful yet hm

[glass vine]

Yeah I am pretty sure that the term ought to involve some usage of the loop properties

[topaz solar]

I mean strictly speaking we'd need some bracketing because associativity

[glass vine]

Yes, strictly speaking I meant (x'y')(xy) here

[topaz solar]

ye ye

We just kinda need xy = yx, so xy(yx)' = e right

[glass vine]

yes

[topaz solar]

It's not associative so it's hard to play the usual commutator game

x = (yx)y' might be more amenable?

[glass vine]

if you manage that it's just as good

[barren sierra]

thing in brackets, I don't get it

[next obsidian]

You just showed that one of those is a submodule of B strictlyncontaining C

With b not in it

You took C to be maximal for this property

[barren sierra]

Ah

[next obsidian]

So true

[novel parrot]

If M is torsion free then why are R/(a) forced to be 0 ?

I know that R/(a) are all torsion modules

for any a

except 0

[next obsidian]

(1,0,…,0) in M is killed by a1

So it has torsion

[novel parrot]

oh

got it

so M = R^n-m iff M is torsion free

free module

[next obsidian]

Yeah

This is true more generally even if M isn’t finitely generated that free <==> torsion free over a PID

Although the proof is different

[novel parrot]

how u know im in a PID

[elder wave]

[next obsidian]

Because you wrote the form of a finitely generated module over a PID

If it wasn’t over a PID this would be weird

[novel parrot]

oh

my professor prooved this only using matricies

[next obsidian]

Yeah it follows from some normal form

[novel parrot]

and the whole thing was only like half a page long

but i dont understand lol

so i need to look elsewhere

[topaz solar]

So the trick I’d think would be that we need t(x, e)=t(x, u)=e like with the commutator, and this implies t(y, e) = t(y, u) for any y?

Can we just plug t(e, e) = t(e, u) into the commutator (xy)(yx)’

[elder wave]

smith normal form

had to compute those in lin alg

[novel parrot]

rip

[glass vine]

I will say that t(x,y)=(e/xy)*xy will always work for this but I'm not sure if it's enough

[topaz solar]

Well idk exact premises for the substitution

but t(e, e)=e obviously

And t(e,u) = uu’ = e

So can we sub in y for e?

Yeah idk @glass vine, not my wheelhouse (i don’t even know the Abelian condition ) but they did say it was obvious

[glass vine]

They indeed did, which doesn't necessarily make it true hahah

Thank you either way

[chilly ocean]

I know some universal algebra

[chilly ocean]

okay so first I found this and it's a presentation, and he says that loop is abelian iff its an abelian group

[main needle]

Cayley-Hamilton thm for fields implies the thm for integral domains directly just by considering the matrix as a matrix with entries in the field of fractions ?

[next obsidian]

You have to have a tiny bit of care because you have to make sure your polynomial is defined over the integral domain and some other stuff, but this will work

[glass vine]

https://scholar.google.cz/citations?view_op=view_citation&hl=cs&user=HLqQjH8AAAAJ&citation_for_view=HLqQjH8AAAAJ:MXK_kJrjxJIC It's a paper and you're right, I need more than just commutativity, I need associativity as well

[main needle]

isn't the characteristic polynomial is already defined over the entries of the matrix which will be inside the integral domain?

[next obsidian]

yes, but a priori couldn't your polynomial get smaller over the field? In the end it doesn't matter, but I'm just saying that a priori you are dealing with different polynomials

I mean for the characteristic polynomial it's actually insanely clear they're the same but

I guess this is more directed towards the minimal polynomial

[main needle]

Oh yeah I think that makes sense. Thanks.

[long nebula]

might be a dumb question, but what does K[K'] mean

[next obsidian]

KK'

[long nebula]

ah okay

ty

What's the motivation for the notation K[K']?

[solar shore]

i intuitively know that $\forall a \in G, |a| = |\langle a \rangle|$, but how does one go about proving this?

[cloud walrus]

blanket

[solar shore]

should i be using division algorithm here?

its a given corollary in my book but i feel awkward when there's no proof

[next obsidian]

Taking polynomials in elements of K’ with coefficients in K gives you the smallest subring containing both

[chilly ocean]

write out elements in <a>

[solar shore]

right, but especially for finite groups, how do i go about showing that a seemingly infinite set $\langle a \rangle = {e, a, a^2, \ldots}$ is indeed finite?

[cloud walrus]

blanket

[solar shore]

(and is surely then equal to |a|)

or am i overcomplicating this?

[chilly ocean]

If you had n = |a| = 3

what would <a> be

[solar shore]

e a, a^2

[chilly ocean]

why

[solar shore]

since its 3 distinct elements and 3 is the minimum positive integer

such that a^n is e

[chilly ocean]

[solar shore]

.

cough

[chilly ocean]

using result you want to prove

this is illegal

you're going to jail

I want a justification

[solar shore]

.

oh

do i use division algorithm here?

[chilly ocean]

yes

[solar shore]

oh

wait

so

[chilly ocean]

anyway

I have another point to make before

[solar shore]

[chilly ocean]

<a> = {e, a, a^2, ...} won't give you a group if a has infinite order

<a> = {a^m : m is integer} is really the group generated by a

[solar shore]

oh so it's necessarily finite then

[chilly ocean]

I mean that there's no equality between those sets if a has infinite order

[solar shore]

ah

[chilly ocean]

you need to write this

[solar shore]

gotcha okay

[chilly ocean]

and I guess you don't know a priori that you can write this either

[solar shore]

as in we can just state that?

[chilly ocean]

what?

[long nebula]

oh hm okay

got it, ty!

[solar shore]

as in, we can say that <a> = {e, a, a^2, ...} is not a group if |a| is infinite

[chilly ocean]

the equality you're putting here is not true

<a> = {e, a, a^2, ...} is not true

[solar shore]

oh i see

[chilly ocean]

how is <a> defined

its defined as the intersection of all groups containing a

subgroup*

then, proposition: we can describe <a> algebraically

how do we do it?

<a> = {a^m : m is an integer}

[solar shore]

$\langle a \rangle = {a^n|n \in \mathbb Z}$

right yeah

[cloud walrus]

blanket

[chilly ocean]

this is the representation that you want, not just natural powers

[solar shore]

ah

[chilly ocean]

well I don't like you used n because I wanted to use n as shorthand for |a|

but lets name k = |a|

k is by definition non-zero and natural

so when we have a^n, we can divide n by k

okay divide

[solar shore]

n = kq + r, 0 =< r < k?

[chilly ocean]

yes

[solar shore]

oh

.

a^r is left

thus its in the set?

well for some a^n at least

[chilly ocean]

a^r was in the set

froim the beginning

[solar shore]

oh

[chilly ocean]

what you showed is that a^n is a^r

so technically

[solar shore]

ohh

[chilly ocean]

any element of <a> is also element of {e, a, ..., a^(k-1)}

[solar shore]

right okay

[chilly ocean]

so they're equal

done

that doesn't show |<a>| = |a| yet

the mapping r -> a^r from {0, ..., k-1} to {e, a, ..., a^(k-1)} is a bijection

now it shows it

why its a bijection? Fromd efinition of order of a

its a minimum

simple

[woeful sage]

representation

[long nebula]

If K_1/F, K_2/F are field extensions, when is K_1 \otimes_F K_2 not a field? Is it when 1 \otimes 1 = 0?

[formal ermine]

this may be of interest to you

https://mathoverflow.net/q/82083

[long nebula]

Ngl I tried to read that earlier and could not

[novel parrot]

how this be done ?

i can get B = R/p^a + ... + R/p'^b

for primes p

but dunno where to go from there?

For some b, pb = 0 so take any x from ANN(B), xb = pb for that b

so xb is in (p)

so suppose b in (p)

[delicate orchid]

that's what I'd say from there

[novel parrot]

am i on the right track ?

using the fundamental theorem

or ..

[delicate orchid]

definitely imo

[novel parrot]

oh

good

[delicate orchid]

the slight thing is you can only guarantee that p^ab = 0 for some b in B here, so you'll have to think about that

unless I'm mistaken

[novel parrot]

wait what do you mean

[delicate orchid]

nevermind I'm being silly

[novel parrot]

i dont really know what to do after using the fundamental theorem

lol

so uh, taking x from ANN(B), we want to show that p divides x, as ring elements

i dont think it leads anywhere

maybe its as simple as saying that ANN(b) > (p)

since pb = 0

but (p) is maximal

so ANN(b) = (p)

and ANN(B) < ANN(b) = (p)

I think this works

but why require B be a torsion module ?

[sonic coral]

Is this argument solid?

G would be simple by Sylows second theorem*

[pastel cliff]

im having a little trouble seeing this bijection

this eventually arrives at the isomorphism between n x m matrices and Hom(R^n, R^m)

okay i see the first part a bit better

we can identify each basis element with the map in Hom(R^n, R^m) sending e_i -> x_i in some collection of n-many elements

this feels almost like shifting(?) the basis?

[novel parrot]

its the same as linear maps between vector spaces and how their correspond to matricies

[pastel cliff]

im trying to read through this step by step

first it shows that given a collection of n elements in R^n, we can identify that whole collection with the map sending basis elements to corresponding elements of the collection

on the other hand, given a map in Hom(R^n, R^m), since these are free modules the map is determined by where each basis element is sent - but how does this identity a map with a collection of n eleements

oh it literally just does

i just had to read it enough times ig

in particular, we're identifying the map with the collection of x_i's right

[karmic moat]

I’m trying to construct the quotient field of an integral domain R. Let Q be the set of pairs (a,b) for a, b in R. (a,b) ~ (c,d) iff ad=bc. I already proved this is an equivalence relation. We also defined addition of equivalence classes as [(a,b)] + [(c,d)] = [(ad+bc, bd)]. the additive identity is (0,1). I’m stuck trying to find its additive inverse; my best attempt is (a,b) + (-a,b) = (0,b^2), but this isnt the identity

[coral spindle]

Remember you're working with equivalence classes of pairs, not pairs themselves

[karmic moat]

oh

[coral spindle]

So you have [(a,b)] + [(-a,b)] = [(0, b^2)]. Now what is (0,b^2) equivalent to?

[karmic moat]

(0,1) ~ (0,b^2) because 0b^2 = 1*0

[coral spindle]

Yup

[karmic moat]

gotcha thanks

[pastel cliff]

ok and in this matrix representation that we arrive at

we have an n x m matrix

a column corresponds to a basis of R^n...?

3b1b linear algebra videos gonna come in clutch for the third year in a row

ok maybe it's more accurate to say that the columns of an m x n matrix represent each basis element of R^n as a linear combination of m elements of R^M...?

[elder wave]

are you familiar with the way this works in lin alg

or not

[pastel cliff]

probably a bit rusty but a quick reminder would make it click fast

i remember a 2x1 matrix is essentially a map from R^2 -> R^1

[elder wave]

you choose a basis, get the matrix representing your linear map wrt. to that basis by taking the images of the basis vectors to be the columns

and given a matrix that obviously defines a linear map so you get a one to one correspondence

free modules are basically just vector spaces

[coral spindle]

Other way around

[elder wave]

oh i didn't even spot that lmao

[south patrol]

Depends on whether multiplication on the left or right is the ting 🤓

[coral spindle]

Phooey

drive on the side of the road that's normal wherever you are

[south patrol]

no

less fun

[elder wave]

say you choose {(1,0), (0,1)} as basis of R^2 then ( phi((1,0)) phi((0,1)) ) is the matrix representing phi

[coral spindle]

Now it's my turn to be 🤓 and say you have to choose an ordering on the basis too

[elder wave]

too many nerds here

[south patrol]

Bases should be tuples

not sets

that's my pnion on lin alg

*onion

[elder wave]

🧅

[pastel cliff]

https://tenor.com/view/timi-gif-20112884

[pastel cliff]

kek

i think im overthinking it

[elder wave]

it's really just linear algebra

[south patrol]

zwiebel homologische(r?) algebra

[sonic coral]

When I am working with these Sylow theorems, when can I do this simple counting argument (group of order 30) versus this more involved argument (group of order 96)? Also, is the argument for 96 sufficient?

is it because for a group like 30, the prime factorization is just some primes to the first power, and so the intersection of the potentially more than one sylow-p subgroup is always 1?

[delicate orchid]

groups of order p_1p_2...p_n tend to be a lot easier to deal with sylow wise than ones with funny exponents

mainly cause of what you said yes

maximum power of p_i that divides |G| is just p_i => Syl_p_i(G) are just cyclic subgroups of order p_i => the intersection between them either has to be the whole sylow-subgroup or the trivial group as prime order cyclic groups have no proper subgroups

[sonic coral]

yes okay, makes sense. So I cant do the simple element counting contradiction for any group of order p^a * q^b * r^c ect? Im going through all groups of order 1-101, so im just trying to figure out which ones I have to consider the intersection of the Sylow-p subgroups on because its not at all obvious to me

[delicate orchid]

I'm going through all groups of order 1-101
love how u stuck 101 on the end cause it's free

[sonic coral]

love my professor for that

[delicate orchid]

right uhhh

[sonic coral]

I thought it was funny too

[delicate orchid]

this looks like something I've seen before but I can't for the life of me remember

[sonic coral]

what does?

[delicate orchid]

N_G(P_1 \cap P_2) specifically is triggering something in my brain

[sonic coral]

I know burnsides theorem would make this very easy but I obviously dont have the tools to prove and therefore use it.

[delicate orchid]

I can't remember, but I think it applies whenever one of the prime divisors isn't just p, it's p^something

[sonic coral]

My professor told me to look specifically at Sylow-2 subgroups on groups of order 24, 48, 56, and 96 and then Sylow-3 subgroup for order 36.

I think I can do the longer argument for the groups of that order, but I just dont understand why the simple counting argument works on some of the other ones I havnt done yet which are also some primes to some power > 1.

I guess maybe I have to consider the intersections of multiple Sylow-p subgroups if |G| = p^n * q * r

[prime sundial]

trying to understand notation regarding some homology stuff.
i have a set {a,b} that spans some F vector space. is the notation F{a,b} the vector space itself?

im trying to write what the kernel of a boundary map is, e.g. finding Z_1(L; F) for some simplicial complex L. if Z_1 is spanned by {a,b} can i write ker(d_1) = F{a,b}?

if not, could someone let me know what im confusing the notation for?

[white oxide]

does transitivity apply to isomorphisms?

That is, if $X, Y, Z$ are groups, then if $X \simeq Y$ and $Y \simeq Z$, is $X \simeq Z$?

[cloud walrus]

okeyokay

[south patrol]

Yes

[barren sierra]

Can you prove it @white oxide

[south patrol]

Try this

[barren sierra]

isomorphism forms an equivalence relation

[white oxide]

yes i'll try to prove it soon

oh really?

that's cool

[barren sierra]

Yes it should be easy to prove

[white oxide]

ok thanks guys!

[tender wharf]

it pops out super fast

let f be an isomorphism from X to Y

g be from Y to Z

all you gotta do is check gf is an isomorphism

[white oxide]

ah shucks i was gonna do the proof later and try to figure that out on my own

but regardless ty

[prime sundial]

well nothing has been spoiled lol

that's just the template for transitivity which is kind of a given

[tender wharf]

this isnt a complete proof at all lmao

[white oxide]

nah

i meant like a start

but ur a good person for that

[tender wharf]

how could I spoil a proof

that would be so sad

[sonic coral]

The group of order 1 is simple right, super basic question but just trying to make sure 😅

[prime sundial]

what are it's (normal) subgroups?

[sonic coral]

just itself

[tender wharf]

[sonic coral]

like it seems so basic since eG=Ge

[tender wharf]

it is literally the trivial group lmao

but a simple group cant be trivial no?

[sonic coral]

algebra makes me lack confidence

[tender wharf]

its like how

[prime sundial]

oh didn't know this was the convention

[tender wharf]

1 isnt prime

[sonic coral]

a group is simple if it’s only normal subgroups normal subgroups are trivial is the definition i have been using

[tender wharf]

oh then that definition says it is

[sonic coral]

but maybe you mean something different

[tender wharf]

I guess the trivial group counts then

[prime sundial]

it's actually funny how often this convention shows up

[sonic coral]

i never understood why 1 wasn’t prime until my professor asked me to consider the fact that prime factorizations had to be unique

[white oxide]

Up to isomorphism, there are two abelian groups of order p^2 where p is a prime, correct?

namely Zp^2 and Zp x Zp

nobody saw that

is it possible to find a formula of the number of abelian groups isomorphic to p^n where p is a prime?

[tender wharf]

classification of finite abelian groups

[white oxide]

good man that derp

[tawny pine]

note the meat of the proof isnt algebraic, its just checking that a composition of bijections is bijective

[white oxide]

cool!

[white oxide]

right

[formal ermine]

when is the radical of an ideal prime?

[rotund aurora]

the radical of an ideal I is the intersection of the prime ideals that contain I

And intersection of prime ideals is radical

So you are asking when is the intersection of prime ideals prime

You can have prime ideals in a chain, and intersections of those ideals will be prime obviously

And I think that's the only case when an intersection of prime ideals can be prime. Because if $A$ and $B$ are prime ideals with $a\in A$, $a\not \in B$, $b\in B$, $b\not \in A$, then $ab\in A\cap B$ but neither $a$ or $b$ is in $A\cap B$

[cloud walrus]

Croqueta

[rotund aurora]

so you would want only one minimal prime above your ideal

and idk if there's a nice characterization for that or something, just reformulated the question lol

[chilly ocean]

Yeah. Intersection of ideals not containing each other is not prime

[simple valley]

The standard meaning of the tensor product of two representations of $G$ is a representation with action $g \cdot (u \otimes v) = (g \cdot u) \otimes (g \cdot v)$, how is this expressible, if at all, using $\bC[G]$-modules?

[cloud walrus]

mniip

[simple valley]

This is neither $\otimes_\bC$ nor $\otimes_{\bC[G]}$ it seems?

[cloud walrus]

mniip

[coral spindle]

Oh this is tensoring over the field, so the former

It would certainly not be tensoring over the group ring, that would be very weird to do in this case

You can see this as akin to the diagonal action, in group actions

so if you have a G-set X, then X x X is a G-set with the diagonal action

This is really the same thing. In fact since k[X x Y] is isomorphic to k[X] \otimes k[Y], it is literally the same thing in the case of permutation representations

[simple valley]

Diagonal in the sense of the diagonal functor?

I don't think I've encountered this term before

[coral spindle]

I'm trying to think about how it would be described in terms of modules yeah

[simple valley]

This gives us a monoidal functor but I don't think taking the grothendieck ring of it (to get the rep ring) is much related to what the product is/was

[coral spindle]

Yeah I was wrong about that, sorry

It ought to have a universal property but I'm struggling to formulate what it is

[south patrol]

Hm I remember asking about this and they said either tensor as C and then define rep separately or tensor C[G] mods

But the latter sounds like it would be a different set right

I do wonder if there is a more intrinsic way yeah

Hm I guess you can view this as being from the ljke map GL(V) tensor GL(W ) -> GL(V tensor W) right

Wonder if you can use that to get a nice interpetatiin

[solid olive]

in order to use this definition do you have to express p(x) and q(x) in summation notation?

and if so how do you experss -5+x in summation notation?

[formal ermine]

what is confusing you?

the p(x) = sum a_m x^m?

[solid olive]

yeah and trying to find it for p(x)=-5+x

[formal ermine]

a_0 = -5

a_1 = 1

and M = 1

[solid olive]

I got that we're summing from 0 to 1

[south patrol]

Well by definition each polynomial can be uniquely written as a sum a_i x^n with inky finitely many nonzero coeffs

So it is fine to do this

And it matches with just regular nultiplication you'd expect

If x were an element of your field/ring

[simple valley]

Tensor as C[G] mods doesn't make sense for G noncommutative

[solid olive]

the problem is 5x doesn't work

[south patrol]

Why?

[solid olive]

-5x

[south patrol]

Or do you mean it would give you amth else

[solid olive]

you get -5 + 5x

if you sum from 0 to 1

we want -5+x

[south patrol]

It does make sense to tensor over a noncomm ring after all

[simple valley]

Tensoring over R turns an (S,R)-bimodule and an (R,T)-bimodule into an (S,T)-bimodule

[south patrol]

Ye

Sure

Oopsies

Hm

[simple valley]

For commutative R, R-modules are (R,R)-bimodules

[south patrol]

Ye

[simple valley]

G reps are (C[G], C) bimodules at best, no G action on the right

You can tensor over C here but the [G] part being diagonal seems like a tricky bit

Feels almost cartesian comonoidal, something tensoring doesn't do

I did appreciate induced reps as extension of scalars though, but that led to this question

[chilly ocean]

Could anyone give a hint as to how to get started here. Been stuck for some time now

[void cosmos]

what have u treid

tried*

[chilly ocean]

I've tried using results I know, e.g. since x must be in all maximal ideals M, it must be mapped to 0 in all the field quotients R/M, and then trying to find smth there since we're dealing with a field, but failed. Also tried to see if I could maybe show that x must be nilpotent since that would at least give one direction, but also failed.

Idk, I'm having a hard time finding where to begin

[void cosmos]

okay

suppose x is in J(R)

and suppose there exists r in R such that 1+rx is not a unit

start with this ^

[chilly ocean]

Ok. Thanks

Thank you. Solved both ways. Don't know why I didn't think to try the straightforward approach...

[solid olive]

when multiplying polynomials over Z_n you don't need to worry about the degree's correct you only need to reduce the coefficiants

[formal ermine]

wdym?

[solid olive]

for example (1+x+x^2)(3x^2+_x^3) over z_6 is 3x^2+4x^3+2x^4+3x^5 not 3x^2+4x^3+2x^4+3

basically you don't need to take your exponent's mod n

[coral spindle]

The exponents are not affected by the ring, yes.

[solid olive]

okay

[south patrol]

Indeed perhaps I should mention that like

1,x,x^2,x^3,... are always all distinct in R[x] for any ring R

It being Z/nZ or whatever is irrelevant

[torn warren]

Is this a counter-example?

transposition is defined by the cycle with length 2, but $\sigma_2$ is not a cycle, so it is not a transposition?

[cloud walrus]

Witness

[simple valley]

This and hence the evaluation maps need not be jointly injective as is the case over say C

[solid olive]

can someone double check me: wanting to find (1+4x^2+3x^3+2x^4)(5+2x^2+x^3) over Z_6 and I got 5+4x^2+4x^3+4x^4+4x^5+3x^6+2x^7

[summer path]

this is literally just multiplying polynomials out

[solid olive]

yes but after 7.5 hours of math jumping from linear algebra to abstract...my brain is starting to turn to mush so I'm not sure I did my mods right

99% sure I did though

[formal ermine]

in my case I want to know when the radical is not prime. we can use the dimension of the ring here, ye? because if the amount of prime ideals that countian our ideal is larger than the dimension of the ring then the prime ideals can't be ordered in a chain

[simple valley]

So just the cyclic subgroup generated by (1234)? Yea that sounds good

[torn warren]

yes

[torn warren]

is my reasoning correct?

transposition is defined by the cycle with length 2, but $\sigma_2$ is not a cycle, so it is not a transposition?

[cloud walrus]

Witness

[simple valley]

I don't know what the "official" definition of transposition is, but I wouldn't call (13)(24) a transposition

[formal ermine]

transpositions are cycles of order 2 lol

[rotund aurora]

In that case yes

[formal ermine]

epic

thanks

https://media.discordapp.net/attachments/230049395180175361/1061846645538562078/IMG_8619.gif

[torn warren]

Here is the definition from textbook, so according to this def, $\sigma_2=(1,3)(2,4)$ is not a cycle, so not a transposition

[cloud walrus]

Witness

[coral spindle]

No...

Oh right yes I misread

I thought you were saying it is not a cycle therefore a transposition lmao

A transposition is a cycle of length 2, not a product of cycles of length 2.

[torn warren]

sorry for the confusion, what I mean is: to be qualified as a transposition, it should be first be a cycle (next consider the length=2 or not), but here it is not even a cycle, so it is not a transposition (no need to consider the length since it is not a cycle)

[coral spindle]

Yes

[torn warren]

thank you

[torn warren]

what is the name of this fish?

[charred crescent]

if we want to find 2 generators of D_n, can we just find a permutation for a rotation and a permutation for a reflection that can be decomposed into relatively disjoint transpositions?

so for example:
D_4; choose (1,2,3,4) and (2,4) because
(1,2,3,4) = (1,2)(2,3)(3,4), which are all disjoint when compared to (2,4)

[shell agate]

Looks like a grouper

https://commons.wikimedia.org/wiki/File:Georgia_Aquarium_-_Giant_Grouper.jpg

[torn warren]

they are not disjoint

[torn warren]

great

[formal ermine]

how appropriate

does the grouper like group theory?

[torn warren]

Definition: grouper is someone who study group theory

you can reference to this prob @charred crescent

[charred crescent]

they aren't?

well i guess i meant to say irreducible or something

something that describes the fact that (1,2)(2,3)(3,4) sends 2 -> 3 and (2,4) sends 2 -> 4

and you don't get a situation like (a,b)(a,b)

[torn warren]

(a,b)(a,b)=e

disjioint means the elements in different parenthesis are distinct

so (1,2)(2,3) are joint, since they have element 2 in common

actually it is a cycle (1,2,3)

[charred crescent]

i think that's what i meant to say

im not sure the term for the example i gave

[white oxide]

Am I on the right track here? For the $\rightarrow$ direction, G is not cyclic so it is isomorphic to $\mathbb{Z}{(p_1)^{r_1}} \text{ x } \mathbb{Z}{(p_2)^{r_2}} \text{ x } \dots \mathbb{Z}_{(p_n)^{r_n}}$

[cloud walrus]

okeyokay

[white oxide]

Such that $gcd((p_j)^{r_j}, (p_i)^{r_i}) \neq 1$, for $i \neq j$

[cloud walrus]

okeyokay

[white oxide]

I'm having trouble seeing how $G$ can have a subgroup of $\mathbb{Z}_p \text{ x } \mathbb{Z}_p$ though, because if it did its elements would have $2$ components whereas $G$ itself has entries with $n$ components

[cloud walrus]

okeyokay

[white oxide]

Any hint would be appreciated

Oh, maybe I could use the fact that $G$ is of order $(p_1)^{r_1}(p_2)^{r_2}\dots(p_n)^{r_n}$

[cloud walrus]

okeyokay

[rustic crown]

(hi boytjie )

[coral spindle]

Are you aware of these theorems?
(1) Cauchy's theorem: if p | |G| then there is an element of order p in G
(2) The classification of finitely generated Abelian groups

(also hi det )

[white oxide]

no i don't know cauchy's theorem

but i'm aware of 2

i think i got somewhere though

so we know that G has a subgroup of order $p^2$

[cloud walrus]

okeyokay

[coral spindle]

OK, you can use the classification of finite simple Abelian groups

[white oxide]

what is a simple group?

i haven't learned that yet

[tulip agate]

Note that the subgroup is merelyisomorphic to Zp cross Zp.

[coral spindle]

I didn't mean simple

mb

[white oxide]

because if all of the $r_i$ were equal to 1, then $G$ would be cyclic

[coral spindle]

I meant Abelian lmao

[cloud walrus]

okeyokay

[coral spindle]

It would be an absolutely bonkers question if you needed the classification of finite simple groups

[white oxide]

what is a simple group tho

i'm curious

[white oxide]

ohhh okay right

[formal ermine]

no non trivial normal subgroups

[coral spindle]

A simple group is one with no non-trivial normal subgroups

So an Abelian group is simple iff it is isomorphic to Z/pZ where p is a prime

An example of a non-Abelian simple group is the alternating group (provided it's big enough)

[fast stratus]

Can’t think of an example where H != {e} or G… anyone?

[rustic crown]

take any finite abelian group

[formal ermine]

...take G to be any finite group?

aahahahah

det beat me to it

[rustic crown]

hehe :3

Z/2Z x Z

[coral spindle]

An interesting example of an infinite group G for which H = G is the quotient group G = Q/Z.

[fast stratus]

I edited

[torn warren]

No, it is wrong, as long as G is abelian and finite, it is always isomorphic to this direct product

[coral spindle]

C\{0} under multiplication is an example for your edited question

[white oxide]

oh sorry i left out what i was going to say after

but yea ik that my fault

[coral spindle]

And in fact in this case, the group H is isomorphic to Q/Z

[torn warren]

it asks you to show there exists a subgroup of G ,not to prove G

[rustic crown]

is there a name for Q/Z

like S^1 but discreter

[white oxide]

Suppose $G$ is not cyclic. Since $G$ is a finite abelian group, $G \simeq \mathbb{Z}{(p_1)^{r_1}} \text{ x } \mathbb{Z}{(p_2)^{r_2}} \text{ x } \dots \mathbb{Z}_{(p_n)^{r_n}}$ Then $G$ has order $(p_1)^{r_1}(p_2)^{r_2}\dots(p_n)^{r_n}$. Observe that if all of the $r_i$s were equal to 1, then $gcd((p_i)^{r_i}, (p_j)^{r_j}) = 1$ for any $i \neq j$ and hence $G$ would be cyclic, a contradiction. Hence, at least one of the $r_i$s is greater than one; for such an $r_i$ corresponding to $(p_i)^{r_i}$, $(p_i)^2\mid(p_1)^{r_1}(p_2)^{r_2}\dots(p_n)^{r_n}$. Then, since $(p_i)^2$ divides the order of $G$ (which is a finite abelian group), there exists a subgroup $H$ of $G$ of order $(p_i)^2$

This is what I have so far

oops wait

edit used hence twice lmfao

i can't type holy shit

[cloud walrus]

okeyokay

[white oxide]

i'm trying to prove that H is isomorphic to Zp x Zp, but the problem is that gcd(p, p) is not equal to 1

i think it has to do something with H not being cyclic, so if i could show that maybe i could show it's decomposable?

well

that's assuming the first half of my proof is correct

[coral spindle]

Pretty sure there is but for the life of me I can't remember what it is...

[tribal moss]

If there were a nice name, I'd expect it to be referenced at https://en.wikipedia.org/wiki/Root_of_unity#Group_of_all_roots_of_unity

[fast stratus]

And isomorphism is?

[rustic crown]

Q/Z --> C* given by q+Z --> exp(2 pi i q)

[cloud walrus]

Witness

[torn warren]

This should be a useful result

[tribal moss]

That doesn't look right. S_3 has order 2·3 but has |S|=4.

[white oxide]

if a subgroup is finite and abelian but noncyclic, is it decomposable?

[rustic crown]

yep. structure theorem for finite abelian group says that every such is direct sum of cyclics

[white oxide]

sweet

i think i got it

[white oxide]

thank you!

[torn warren]

so it satisfies, right?

[tribal moss]

No: n=3 is odd but |S| is not 2.

[torn warren]

wow, thank you for confirming this idea

[torn warren]

I said |G|=2n, not n

[tribal moss]

And |G|=6=2·3=2n for n=3.

[torn warren]

ah, right

I forget

For any group $G$, if $|G|=2n$, define the solution set $S={ x | x^2=e, x\in G }$ , then $|S|$ is even. If G is abelian and $n$ is odd, then $|S|=2$.

[cloud walrus]

Witness

[torn warren]

thank you!

to make it exactly equals 2, it needs Abelian...

but this is only sufficient I think

[past temple]

so given vector spaces U,V, the tensor product U (x) V is the space of all formal linear combinations of u_i (x) v_j, where {u_i}, {v_j} form a basis for U,V respectively

then, given u in U and v in V, u can take the tensor monomial u (x) v by expressing them as a linear combination of basis vectors

but what are some elements in U (x) V that aren't tensor monomials?

[rustic crown]

sum of two monomials :p

u1⊗v1 + u2⊗v2 for example

if either of U or V is one dimensional, then clearly the tensor product contains only monomials (of not necessarily basis vectors)

[past temple]

ah okay, its bc u can only combine two tensor monomials if they agree in one of their entries

also a basic question, but a representation U of a group G is defined as a homomorphism from G -> GL(U)

but to check that smth is indeed a representation, does it suffice to check that u have a map G -> End(V) such that identities get mapped to identities, and

that the group homomorphism axiom is satisfied?

because the image of such a map should also be a group

meaning that it would be contained in GL(U)

[south patrol]

Sure, you're basically checking it's a monoid homomorphism and then those send units to units in general

[past temple]

gotcha

[past temple]

also incredibly basic

and im stunned at myself for not immediately knowing why this is true

but given vectors v, w in V

how can one prove that there exists a linear endomorphism T:V -> V such that Tv = w?

[south patrol]

Tbf I think this is kinda non trivial in general since you have to assume you can extend maps or take direct complements, I believe

But basically it'd be: take v in V, extend to a basis of V and then consider the map V -> V sending v to w and every other basis vector to 0 (or whatever)

Or, basically equivalently, pick a direct complement to <v> and then consider the map V -> <v> -> W, the first being the projection and the second sending v to w

Ofc with finite dim spaces you don't need anything choicey

Maybe there's an easier method which I don't know though

[past temple]

hm ok, i guess its harder than i thought

[tribal moss]

You don't need to extend bases -- having just one basis for V is enough. Pick a basis vector that appears in v with a nonzero coefficient, map that basis vector to an appropriate multiple of w, and every other basis vector to 0.

I wonder what the choice strength of "for every k-vector space V and every v in V, there is a linear functional f: V -> k such that f(v) != 0" is.

[tribal moss]

http://karagila.org/wp-content/uploads/2016/10/axiom-of-choice-in-analysis.pdf states (corollary 39) that it is consistent with ZF+DC that there can be a nontrivial real vector space that has no nonzero linear maps to R.

[tribal moss]

Oh, and this implies that Pdk's problem does need choice: Karagila shows that there are models of ZF+DC where $(\ell^\infty / c_0)^*$ is trivial. But $\ell^\infty / c_0$ certainly is not itself trivial. If we let $a$ be a nonzero element of $\ell^\infty / c_0$, then there cannot be any linear $T: \bR \oplus (\ell^\infty / c_0) \to \bR \oplus (\ell^\infty / c_0)$ such that $T(0,a) = (1,0)$.

[cloud walrus]

Troposphere

[past temple]

i see

here's kind of another question

so it's well known that the 1-dimensional representations of a finite group G are the same as the representations of G/[G,G]

and since G/[G,G] is a finite abelian group, by FTGAG it decomposes as a direct product of cyclic groups

it's also known that any 1-dimensional irreducible representation over a field F of a cyclic group is identified with the kth roots of unity in F, where k is the order of the cyclic group

so once we decompose G/[G,G] into a direct product of cyclic groups, and associate to each of those factors a root of unity

what is our resulting representation of G/[G,G]? (and therefore of G)

[south patrol]

True, though if you assume existence of a basis then you can easily just swap smth out for v right? (assuming v non-zero)

[tribal moss]

Um. Fix a decomposition of G/[G,G] and furthermore fix a generator for each of the cyclic factors.
Then the representations correspond exactly to a choice of k-th root of unity for each of the generators (where k is the order of that generator). That gives exactly |G/[G,G]| different one-dimensional representations, like there should be.
Is there more to your question than that?

[tribal moss]

Oh yes, good point.

[past temple]

ok yeah thats exactly what i thought, thanks

[past temple]

ok, so now im considering the permutation representation of S_n in F^n, where char F = 0

lets define F_0^n = {vectors whose entries add up to 0} and F_const^n = {vectors whose entries are all equal}

I'm currently trying to prove that F^n = F_0^n (+) F_const^n as representations

but im struggling to come up with the explicit isomorphism

[south patrol]

Isn't it just the identity?

Or rather, it's a direct sum decomp into G-stable subspaces, so it's a direct sum decomp of reps

[tribal moss]

Each of your direct summands is an invariant subspace under every permutation matrix.

[tacit hemlock]

How would you even start this Q?

[past temple]

yes they're both invariant subspaces, but that doesnt necessarily mean we have a decomposition

[tacit hemlock]

I have: 18 | G and 24 | G, by Lagrange's Theorem, 72 | |G|, but I don't think that gets me anywhere

[past temple]

for example, given any vector space V and a G-invariant subspace U, V/U is also G-invariant

and V = U (+) V/U as vector spaces

[tacit hemlock]

a hint anyone?

[past temple]

but not necessaarily as representations

[past temple]

use lagrange's theorem

[south patrol]

Uh yeah it does right?

[tacit hemlock]

I did

[south patrol]

V/U isn't a subspace though...

[past temple]

but it can be identified with a subspace

[south patrol]

I don't really understand how that links to this though

[past temple]

namely, we choose a basis for V, say v_1,...,v_k such that U = span(v_1,... v_n), n < k

[south patrol]

I mean yes, agreed

But having a decomp into G-stable subspaces does indeed give a decomp as G-reps

It's stronger than having a G-stable subspace and taking the quotient cause not every G-stable subspace admits a G-stable complement

as your example kinda shows i suppose

[past temple]

ok see yeah thats what i was getting rly confused about

because i thought that if U is a G-stable subspace, then V/U is a valid representation of G, that can be identified with the complement of U

[south patrol]

Yeah I mean identifying V/U with a subspace is kinda "messy" anyway

[tribal moss]

There's no "the" complement if all you have is V and U.

[south patrol]

In particular "the" complement isn't a thing

lol beaten to it

[past temple]

wait so

even though V/U is a representation

and V = U (+) (some identification of V/U with a subspace of V), as vector spaces

that identfied subspace isn't necessarily G-stable?

[tribal moss]

Correct.

[past temple]

even though V/U is.

thats very interesting

ok well ig that answers my question, since F_const and F_0 have dimension 1 and n-1 respectively (as long as char F = 0)

and have trivial intersection

[south patrol]

I think yeah I mean to me it's that identifying V/U w a subspace is messy and relies on you just picking a direct complement

Usually when such arbitrarily choices are made, you can't really expect it to behave well with the additional algebraic structure

And yeah also there are examples where a G-stable subspace U has no G-stable complement (though for char 0 and other cases this is fine; Maschke's theorem is the result you want) and so in particular that implies any subspace you identify V/U with cannot be G-stable

[past temple]

right

[tribal moss]

For example suppose we have the two-dimensional representation C_2 where the nontrivial element of C_2 negates the x-coordinate of a vector.
Then the span of (0,1) is a G-stable supspace, and the span of (1,1) is one direct complement of it, but the latter span is not G-stable.

[past temple]

but the complement produced by maschkes theorem is much more involved

namely i think its produced by taking the kernel of a projector?

ill have to review in more depth

[past temple]

i see.

[south patrol]

Yes, you take an arbitrary projection operator V -> U and make it into a G-equivariant projection operator (that's how i think about it)

By averaging

Have you looked at hom representations at all? It turns out this is special case of something lol

[past temple]

like, if u have representations U, V of G

then the reprsentation Hom(U,V) produced by g * phi = g_V^-1 phi g_U

[south patrol]

well i think it should be like (g.φ)(v) = g.φ(g^-1.v) but yes

Anyway, so it turns out - provided #G is invertible in your field - that if you define e = 1/#G Σ g, then the map v -> e.v defines a projection V -> V^G

(where V^G is the set of elements v with g.v = v for all g)

[south patrol]

Idk if you've seen this?

But the point is you just average stuff lol

In the case of Maschke's theorem, we take a projection π, which is an element of Hom(U,V), and then project it onto Hom(U,V)^G

and Hom(U,V)^G is exactly the G-equivariant maps U -> V

Eh I guess that is overcomplicated for proving Maschke but shows this is part of more general theory

[gleaming condor]

\ \

[tacit hemlock]

Is the intersection of 2 strict subgroups also a strict subgroup?

I know that in the case of H ≤ G and K ≤ G, H⋂ K ≤ G

so is H<G and K<G => H ⋂ K < G also true?

I am currently thinking of it in terms of sets from an euler diagram

[agile burrow]

Sure, H \cap K is a subset of H, which is a proper or strict subset of G

[tender wharf]

please don't double post this to the help channels next time

[void cosmos]

yo quick fact check

splitting field of x^4+x^2+1 over Q is just Q[sqrt(3)i] right

[coral shale]

y^2 + y + 1
y = 0.5(-1 +- rt(3)i)

dont u need to square root?

the field u wrote has degree 2

which makes no sense surely

[coral spindle]

That would make no sense if the polynomial were irreducible

however it is equal to (1 - x + x^2) (1 + x + x^2)

and indeed it's sufficient to add sqrt(-3)

[void cosmos]

ty

[tawdry apex]

Hello I'm really struggling with Galois theory.
The Problem states that I should use Kummer theory to prove that
(x^7+\theta) is irreducible in (\mathbb{F}_2(\theta)) where (\theta) is a root of (x^3+x+1\in\mathbb{F}_2) )
My current attempt is I've shown that powers of (\theta) are distinct and that (\theta^7=1) which is helpful, since now I have a primitive 7th root of unity in the base field.
Now I believe I need to show that (\theta) is not a 7th power in (\mathbb{F}_2(\theta)) and that would imply that say if (\phi) is a root of (x^7+\theta), then ([\mathbb{F}_2(\theta,\phi) : \mathbb{F}_2(\theta)]=7)

My questions are:
Would showing all of this be sufficient?
How do I approach showing that (\theta) is not a 7th power of an element?

[cloud walrus]

Marci

[rustic crown]

idk how you proceeded lol, but yea showing theta is not a 7th power is enough

and to do that, the simplest way i see is to use some finite fields magic

[simple valley]

just check the other 6 potential elements of F_2[theta]

5 even

[rustic crown]

theta lives in the group F2(theta)*... which has order 7. so if its a 7th power it would be 1 :p

[formal ermine]

it's a finite field, just check every possibility

oh

mniip beat me to it

[simple valley]

it would appear that $a^7 = 1$ unless $a=0$

[cloud walrus]

mniip

[simple valley]

oh duh of course

the powers of theta are all distinct, and there's 7 of them, so the multiplicative group of the field is the powers of theta, a cyclic group of order 7

[rustic crown]

how do you use kummer theory tho

what even was kummer theory lol

[simple valley]

I have no idea what that is

[rustic crown]

iirc something to do with cyclic galois extensions

but i think you needed to know the degree before hand there

anyway, i would have done it like this... say k = F2(theta) and F = splitting field of x^7-theta over this. since char is 2, this is separable, so F/k is galois. if phi is a root of x^7-theta, then other roots are clearly phi * theta^i and so F = k(phi).
Now you can use this to define the homomorphism
Gal(F/k) --> Z/7Z given by sigma --> i where sigma(phi) = phi * theta^i.
since phi generates F, this is injective. and so Gal(F/k) is either 1 or Z/7Z. it can't be 1 since theta is not a 7th power so [F:k]=7. which gives the irreducibility of x^7-theta

[lapis plume]

Let $n$ be a fixed integer, we define

$$\mathbb{Z}_n={0,1,2,\cdots,n-1}.$$

and

$$\mathbb{Z}^*_n={a\in\mathbb{Z}_n:\gcd(a,n)=1}.$$

Define a binary operation $+_n$ on $\mathbb{Z}_n$ by

$$a+_nb=\text{remainder of }a+b\text{ divided by }n$$

for $a,b\in\mathbb{Z}_n$.

Define another binary operation $\times_n$ on $\mathbb{Z}_n$ by

$$a\times_nb=\text{remainder of }ab\text{ divided by }n$$

for $a,b\in\mathbb{Z}_n$.

Given that $(\mathbb{Z}_n,+_n)$ and $(\mathbb{Z}^*_n,\times_n)$ are groups. But are they isomorphic?

I know $(\mathbb{Z}^*_n,\times_n)$ is cyclic since $3$ is a generator, so I believe they are isomorphic but I cannot find a suitable group isomorphism.

[cloud walrus]

Trenton

[tawdry apex]

I have no clue that was just stated in the question and confused me even further lmao. Thank you so much though I didn't think to check possibilities cause I thought it'd be too much effort but the multiplicative group thing works and is probably what the question is looking for. I also very much like your argument I didn't realise phi generated F but that makes a lot of sense

[formal ermine]

no

take eg 6

Z6 will obviously have 6 elements

how many numbers are there coprime to 6?

[coral spindle]

Z_n* and Z_n are never isomorphic as they have different orders, but it is an interesting question to ask whether or not Z_n* is a cyclic group. See: https://math.stackexchange.com/a/373711/825884

What you say about 3 being a generator is false.

Typically, 3 is not even contained in Z_n*

[lapis plume]

oh yes I think my question is problematic

But how about $(\mathbb{Z}_6,+_6)$ and $(\mathbb{Z}^*_7,\times_7)$?

[cloud walrus]

Trenton

[lapis plume]

In this case, $3$ is really a generator then

[cloud walrus]

Trenton

[lapis plume]

and the order is the same, which is $6$

[cloud walrus]

Trenton

[coral spindle]

Did you read the link that I sent?

It tells you exactly the answer you're looking for.

[lapis plume]

Ok let me have a read

[coral shale]

Let $G$ be a finitely generated group and $S_G = \{s_1, \ldots, s_n\}$ a generating set.
\bigbreak
Let $F_n = \langle x_1, \ldots, x_n\rangle$ and $S_{F_n} = \{x_1, \ldots, x_n\}$.
\bigbreak
Let $f: S_{F_n}\to S_G$ be a bijection.
\bigbreak
Let $g: F_n \to G$ map words $x_{i_1}\cdots x_{i_k}$ to $f(x_{i_1})\cdots f(x_{i_k})$.
\bigbreak
Claim: $g$ is a surjective homomorphism, hence $F_n / \ker g \cong G$

[coral shale]

@rustic crown also, can i check this is sound, as an explanation

explaining what relations are in a finitely generated free group basically

[rustic crown]

yee

[coral shale]

uhhh

finite relations too specifically or

[rustic crown]

ker g are all the relations

[coral shale]

so ker g can be infinite

and my explanation still ok

[rustic crown]

more often then not it will be infinite

[coral shale]

oh, I think I meant a little differently

'finitely presented'

theres this idea

[rustic crown]

yep

[coral shale]

Im tryna rephrase some ideas in https://kconrad.math.uconn.edu/blurbs/grouptheory/wordproblem.pdf

basically.

[rustic crown]

if G is finitely presented, then you can find finitely many relations such that the normal subgroup generated by them is ker g

[coral shale]

yh

cus he goes into this explanation

and I barely see the point

[rustic crown]

the point is that finitely presented groups can be described (presented) using finitely many generators and relations, so even though these groups maybe huge you only need finite data to work with it

[coral shale]

so ker g are the set of all relations, but then we are interested when this can be finitely generated

but that is normal generated, not just group

so like <<R>>

[rustic crown]

right

[coral shale]

And when there exists finite R, we have a finitely presented group

[rustic crown]

yeep

[coral shale]

oki

[rustic crown]

usually these notions behave very nicely

det doesn't have good intuition with those proofs though, to me they feel like some work which shows the nice properties

[delicate orchid]

"u can write out the presentation before the universe explodes"

[coral shale]

Yh I kindof skimmed this stuff 1st time round

and felt its like 'notation'

[rustic crown]

for example, if a group is finitely presented, and you have a list of finitely many generators, then the relations with respect to these generators are also finitely normally generated

idk if this is true or not, but i want it to be true xD

i remember this being true at least for commutative rings

[coral shale]

finitely presented should be a subclass of finitely generated

[rustic crown]

so like finitely presented says there exists one set of generators such that relations are finitely normally generated

[coral shale]

isnt this the definition?

[coral shale]

yeah

oh ok

ur claiming all choice of generators

[rustic crown]

i'm saying once you somehow know that a group is finitely generated, then any set of finitely many generators is also nice

yea something like "finitely presented is always finitely presented"

[coral shale]

ok.

[rustic crown]

is this true for groups?

[coral shale]

is this solved 😶

i think ive heard some of these word problems are undecidable? not sure

[agile burrow]

This feels not true but idk a counter example off the top of my head

I mean it's kind of equivalent to word problem on the relators of your group, right?

[coral shale]

btw --- going back,

Checking --- formally:
<<R>> = smallest normal subgroup containing R
= intersection of all normal subgroups
= closure under multiplication ; inverse; conjugation by containing group

that right