groups-rings-fields

[coral spindle]

?

[dull ginkgo]

Show the tensor product is trivial?

[coral spindle]

Yes

And we win here

[dull ginkgo]

That works lol

[south patrol]

Minor typo

Gotem

[coral spindle]

OH NOES

[south patrol]

Wait no

Sorry

[coral spindle]

OH YESS

[south patrol]

Division by 2 is a crime

[coral spindle]

it is

[south patrol]

You can never divide by primes

[coral spindle]

Wise

[languid trellis]

so a (x) b = a (x) b(x^2+1) = a(x^2+1) (x) b = 0 (x) b = 0

weow

and we need k[x] module structure to be able to move stuff around on the inside of the product

[dull ginkgo]

Can we uh

Actually do that’s

Yeah the x isn’t in the ring

We’re censoring over

*tensoring

[languid trellis]

actualyl ur right

hm

[dull ginkgo]

(x-1)(x+1) = x^2 - 1

Oh i see

you add 1

[languid trellis]

i suppose we can have (x^2+1) = 1 in Q[x]/(x-1) in the same way we can reasonaly say 1 = 3 in Z/2Z because they're representing eq. classes on Q[x], resp. Z

so I stand by that

it works

if in doubt, the tensor product is 0

[coral spindle]

Yeah that's right

[dull ginkgo]

Isn’t like

R/I (otimes)_k R/J ~= R/(I + J) if R is a ring containing the ring we’re tensoring over or some shit

Can we still prove the tensor product is 0 in char 2 or nada

[coral spindle]

Hm

Well (x-1) contains (x^2+1) in that case

So I think not

[dull ginkgo]

Yeah

Welp back to being brain melted by noncommutative matrix shit and opposite rings

[languid trellis]

sounds cringe

[dull ginkgo]

Endomorphism ring of R as a left R-module is isomorphic to it’s opposite ring :3

[main flower]

What did you do?

[main flower]

Thank you! And yeah this makes sense I think the only one I’m having trouble understanding is the one generated by the four cycle that fixes all the vertices. I’m not sure why all the edges aren’t the same.

But I’ll take some time later to reason through it and I’ll probably be able to figure it out.

[acoustic igloo]

the rotoreflection?

i got messed up by that one too

[main flower]

If you mean the one in the lower right hand corner then yes. This problem made a lot more sense once I realized my mistake and it was talking about edges instead of faces.

I bet if I draw it out, it will make sense.

[acoustic igloo]

the white edge never intersects the plane of reflection

the red edges do

[main flower]

Oh yeah of course! Thank you.

[dull ginkgo]

A bit confused by 1. Here

How are we saying f_i = some sum over e_j

If they lie in different modules

[crystal vale]

How many homomorphisms are there from Z/2Z × Z/2Z to the S_3?

Since G/ker isomorphic to subgroup of S_3. But if ker is trivial then there is no subgroup of order 4 in S_3.

So if the kernel has order 2 then there are only 3 subgroups of order 2 in S_3.

And if the kernel is the same as G then mapping is trivial.

So total 4

[dull ginkgo]

e_1 and e_2 are involutions that generate Z/2Z^2

[crystal vale]

Yes

[dull ginkgo]

The image of them must have order dividing 2, thus either 1 or 2.

So they must be sent to involutions in S_3

Or the identity

So i think there’s more than 4

Restating

R^n and R^m aren’t the same module here

So how are we saying f is equal to some of e in a different module?

Unless it’s like, the image of f_i or smth

[south patrol]

Where does it say this is the case

Wait

[dull ginkgo]

Like it just seems like an abuse of notation or smth

[south patrol]

Idk what bit you mean

[dull ginkgo]

I can’t parse what it’s intent is

[south patrol]

Oh I'm blind sorry lol

[south patrol]

To me it kinda feels like a mistake lol

[dull ginkgo]

I think it’s meant to be an image or smth?

like mu(f_j)

Of mu from R^m to R^n

With the associated m x n matrix A

I’ll do the problem under that assumption lmao

[dull ginkgo]

Should i go about the problem using a base we know a priori for R^m

And then show f is another base?

[delicate orchid]

https://tenor.com/view/erm-what-the-freak-puppy-stare-cute-reaction-gif-15215675160072457679

[dull ginkgo]

This definitional bullshit is really fucking with me

[crystal vale]

How?

[dull ginkgo]

I know for a fact there is more than 4

[delicate orchid]

there should be 10 I think

[dull ginkgo]

Yeah

My combinatorics answer gave 10

[delicate orchid]

3 possible choices of where to send each generator of C_2^2

+1 for the identity

[dull ginkgo]

Yeah

[south patrol]

under what map

[dull ginkgo]

Some morphism

This problem is confusing me heavily

[south patrol]

Well it won't be ome random map

[dull ginkgo]

Because matrices are relative to bases

[south patrol]

I think it is just very badly written

[dull ginkgo]

And we don’t know f is a base a priori

So we can’t say shit about the matrices, no?

Unless we assert it’s like an image

[crystal vale]

Got it, thank you

[delicate orchid]

nah I think this is bull. See e_1 and e_2 commute but the centralisers of each involution in S_3 are trivial

so their images can't commute

[dull ginkgo]

Because that sum doesn’t make sense because they are in different modules

Ergo what the fuck is this problem asking me to do

[crystal vale]

Yes

[delicate orchid]

I agree with your answer now, Mr. Notknow

[dull ginkgo]

Should i skip this problem?

[crystal vale]

I just agreed with @dull ginkgo answer 😂

[crystal vale]

Now I got it your point

So both e_1 and e_2 must be mapped to the same transposition

[dull ginkgo]

OH YEAH

if they get mapped to different ones

Then their image’s product would be a three cycle

But the product in Z_2^2 is an involution

So that’s a contradiction

like (1 2) (2 3) = (2 3 1)

[delicate orchid]

(231)
I'm going to cry

it's correct but just write (123) ffs

[dull ginkgo]

Composition order

[delicate orchid]

no, it's not composition order

you just put the 1 first

[crystal vale]

For a p-group of order p^4, assume the centre of G has order p^2. How many conjugacy classes of G?

Any hint?

[delicate orchid]

G/Z(G) = Inn(G), orbit stabiliser?

[dull ginkgo]

Conjugacy classes are orbits under conjugation

[delicate orchid]

I have another hint if you need it

[dull ginkgo]

So your non central elements, p^4 - p^2 = p^2(p^2 - 1) = (p - 1)(p + 1)p^2 many, are divided amongst your remaining conjugacy classes

[delicate orchid]

or this guy could just say it anyway

[tardy hedge]

I feel like this should have been its own prop/theorem with a proof. Seems important and it’s taking me a while to think about

[crystal vale]

Can we consider 0 as the prime element in ring Z ?

[dull ginkgo]

@crystal vale what can you say about the size of a conjugacy class in a p-group?

[delicate orchid]

long winded way of saying that D has to be injective

*projective

[crystal vale]

Divisor of p^4

[dull ginkgo]

And what are those divisors?

[tardy hedge]

Oh ok, yea they introduce projective modules on the next page

[crystal vale]

p, p^2, p^3, p^4

1

[dull ginkgo]

Yeah

[dull ginkgo]

I think you can use this

I think?

[delicate orchid]

you can

but you need something else to narrow it down to one case

why can't there be one conjugacy class of size p^2? that is the question you must now answer

[dull ginkgo]

||p-nary representation?||

[delicate orchid]

no idea what that means

[delicate orchid]

I take it back this might still be possible

[crystal vale]

No because we define prime as non-zero element

[delicate orchid]

you know that Inn(G) is C_p x C_p is what I was getting at

because if it was cyclic then G would be abelian

[dull ginkgo]

I think?

You can consider conjugacy classes with size p^k

[delicate orchid]

don't be a slave to definitions, the idea of the zero ideal being prime is important

[dull ginkgo]

And i don’t think there can be more than p-many

For a given p^k

[delicate orchid]

k = 0 then pick C_{p^n} for a gigantic n

[dull ginkgo]

Oh shit

Then my method doesn’t work

Big sad

[crystal vale]

Yes I thought but then I am confused

[tardy hedge]

What is F| psi(L)

F restricted to psi(L)?

but psi(L) is in M, and F is D -> M

[dull ginkgo]

Both of these make NO SENSE

Relative to “a basis”

But we need two!!!

Oh wait misread :3

[crystal vale]

I used

[crystal vale]

I don't get it

[tardy hedge]

We all be confused rn

Lol

[delicate orchid]

Miz wrote out the exact same stigma that very explicitly I’m not sure what you’re not getting

*argument

iPhone 6 on that good zaza

[crystal vale]

If I define action on X such that a(g,h) = (aga^-1, aha^-1).

But if G is abelian group then | X | = | G| × | G|, right?

But the number of Conjugation classes in G is |G| × | G|, right?

[crystal vale]

How G/Z(G) = Inn(G) ?

[delicate orchid]

That’s a very standard result

[crystal vale]

Inn(G) is a subgroup of group of automorphism, right?

[delicate orchid]

Consider the homomorphism sending g -> (x -> gxg^-1) and then apply first iso

[crystal vale]

Oh wait

Yes got it

I thought how they are equal

It isomorphic

[delicate orchid]

nobody likes a pendant

you know what I meant

[tardy hedge]

Maybe he didnt

[delicate orchid]

then that's a skill issue

[tardy hedge]

Lol

[crystal vale]

🥲

[still dew]

Number of conjugacy classes in what? The group?

Oh wait yove written G

Mybad I'm blind sometimes

Why is the number of conjugacy classes in G |G|x|G|?

It's just |G| for abelian G

[delicate orchid]

X = {(g,h) \in GxG : gh = hg} ||= {(g,h) \in GxG : h = ghg^-1}|| ||=> (g, h) \in X <=> g \in C_G(h)||. ||Apply burnside's fixed point theorem|| 3 different levels of hints

[crystal vale]

But I defined operation on X so number of conjugacy classes in G means ?

[still dew]

Then you're talking about orbits here

And in the question they mean conjugacy classes in general

[crystal vale]

Oh

Got it your point

Then let me think

[sly rain]

Theyre called conjugacy classes under the conjugation Action i think

[still dew]

Yeah but he's calling his orbits conjugacy classes

And getting confused with the conjugacy classes given in the problem

[crystal vale]

What does it mean by conjugacy classes in G ?

[still dew]

This

[delicate orchid]

the conjugacy classes of G

you can't let a slightly different preposition throw you off

[crystal vale]

You mean when we define action on G by Conjugation

[delicate orchid]

what

are you under the presumption that "orbits" and "conjugacy classes" mean the same thing

[crystal vale]

When operation is Conjugation

[tardy hedge]

Still stuck on this blurb if anyone can help explain it

Im confused on F | psi(L) = psi’(g) and how it all relates to the exactness and the bijection of F and f,g etc

I was told that this blurb does not make sense from some other guy

[crystal vale]

Got it, thank you and || you mean C_G(h) as stabilizer of h ||

[delicate orchid]

I do

[crystal vale]

Yes that notation is the center of h , which is the same as the stabilizer of h under Conjugation

[delicate orchid]

centraliser, not centre. Otherwise yeah that's right

[crystal vale]

Yes

[tardy hedge]

Poo

[still dew]

X^N denotes fixed points of X under N action

N/X denotes set of orbits

Now I have two issues with (a)

If they're saying that there is a g action on X^N making the inclusion map g map

Then using that logic i(g*x)= g.i(x) -> g*x = g.x

But X^N is not stable under G action is it?

[rocky cloak]

It seems maybe wohever wrote the excerpt mixed whether F was D -> M or M -> D midway.

I don't think there is a super natural way to express F in terms of f and g anyway. Like I think it's better to just think about it like every f: D -> N factoring through M -> N

[noble lynx]

can grobner basis help for computing free resolutions over quotients of polynomial rings

[delicate orchid]

the entire point is that you need to construct a new G-action on X^N

but yes you're correct, X^N is not always invariant

[still dew]

Yeah but that new g action must obey the induced one

* is the new action

[still dew]

Is t this true

[delicate orchid]

I feel uncomfortable claiming that equality holds

well no it definitely does hold for x in X^N you're right

[still dew]

I mean * is a g action

Yeah

[delicate orchid]

the idea here is to split G into cosets of N with a transversal {t_1, ..., t_k} and then, given a g \in t_iN, g*x := t_in.x = t_i.x = g.x

at least I think so

let x in X^N, n.(g.x) = ng.x = gg^-1ng.x = gn'.x = g.(n'.x) = g.x, where n' = g^-1ng \in N

so X^N is indeed invariant under the original G-action

cool, everything works

[still dew]

Wait sorry where did we get that

[delicate orchid]

here

[still dew]

From what I understand any element of G can be if the form some ng g is from your choice of transversals

[delicate orchid]

yus

[still dew]

So you've used normality to get that this is basically just an action of the transversal

[delicate orchid]

yus

[still dew]

And why is this action in X^N

[delicate orchid]

you only have g*x = g.x on elements fixed by N, which as you pointed out is required for the inclusion to be a map of G-sets

also I don't think it would be well defined on a larger set

like, determining which coset a product of two elements of G lie in is highly non-trivial (google "transfer") and almost always won't give you a group action on X

[delicate orchid]

like I am very much using the fact that x is fixed by N in the "t_in.x = t_i.x = g.x" step

[still dew]

Okay but where exactly are we mapping a g*x , g is a representative

[tardy hedge]

Ok yeah, thanks. That surjectivity is the only new thing here I suppose. The overarching question here from what i understand was, to what extent do properties of L and N imply related properties of M. All homomorphisms from D to L are related to homs from D to M, since L is a submodule of M, so that one is kind of trivial. So the main problem here is what D->N can be lifted to D->M, and that Hom sequence is exact iff every D->N can be lifted

@rocky cloak

Was in reply to jagr

[delicate orchid]

g*x is defined to be t_i.x, so we send it to t_i.x

maybe try working through it yourself, I kind of gave the entire thing up really quickly to assure myself that it worked

[still dew]

Okay

Well thanks for the help @delicate orchid

[delicate orchid]

no worries

[tardy hedge]

I will now learn projective modules

[still dew]

Also there are problems like what if X^N is empty or stuff

Isn't that a possibility

Group actions on empty set

[delicate orchid]

oh god

uhh

good point!

I think the condition vacuously true?

[still dew]

I guess

This excercise is actually very unintuitive to me

[delicate orchid]

but there isn't like... a map into the empty set

so G x empty = empty -> empty can't even exist

[still dew]

Ah true

Then waddfck

[delicate orchid]

wait we're being REALLY stupid

1 \in N

right?

[still dew]

Yeah

[delicate orchid]

no that doesn't work

yeah u raise a really good point. I legit think this exercise didn't even think about that

[still dew]

Isn't this a valid point

Like solid af

Lenstra done fcked 💀

[delicate orchid]

assume X^N is non-empty 😹

[still dew]

Naww

[rocky cloak]

Yeah, so basically, lifts of D->N to D->M don't always exist, but if they exist then they're unique up to maps D->L.

So assuming you make some choice of lifts for each D->N, then you can describe every D->M as a combination of D->N and D->L

[dark wave]

For those of you who are interested in group theory, I highly recommend The Finite Simple Groups by Robert Wilson if your university gives you access. It is very understandable for the topic it is on (at least the first 2 chapters are, I just started chapter 3) and is a very interesting read.

[tardy hedge]

Oh so projective modules seem pretty powerful

[tardy hedge]

If you have A,B,C in a short exact sequence, with maps psi and phi, and its a split sequence, are psi and phi necessarily the inclusion and projection maps?

[south patrol]

You seem to be slightly misunderstanding what a split exact sequence is

There is no sensible notion of "the inclusion" etc in general

The point is that by definition a split exact sequence is equivalent to the inclusion then the projection

So the answer is "yes by definition" or "that doesn't make sense" depending on what you mean by your question

[tardy hedge]

Oh ok yea i see

In dummit and foote they never showed how they are equivalent sequences

[soft wyvern]

and how can I relate the polynomial t^3+t^2+1 with 1???

[rain grove]

Let $H_i$ subgroup of group $G_i, i = 1, \dots ,s$. We can see that $H_1 \times \dots \times H_s$ pis a subgroup of $G_1 \times \dots \times G_s$. Does every subgroup of $G_1 \times \dots \times G_s$ have that structure?

[cloud walrus]

OHHELLNAH

[rain grove]

So i checked the solution on it says "No. ${(g,g) | g \in G}$ is a subgroup of $G \times G$". Ok but doesnt that have the same structure because G is still a subgroup of G?

[cloud walrus]

OHHELLNAH

[rain grove]

Can someone help me understand this

So we first choose H_i for i = 1..s and then claim every subgroup of G_1 x...x G_s is in the form on those choses H_i?

Thats what the question is asking?

If thats the case then yeah I can prove that

[willow cipher]

im doing this question. i proved A has a Q-vector space structure. End(A) must contain a subring isomorphic to Q and then u use this subgroup to define a left multiplication on A. i cant prove it must have dimension 1. there this answer on stackexchange aswell but i dont really understand it https://math.stackexchange.com/questions/2673535/let-a-be-an-abelian-group-such-that-end-aba-is-a-field-of-characteristc

[sly frost]

if |h| does not divide |g| here, that means |h| has to have a prime factor with an exponent larger than its exponent in |g|

[tardy hedge]

Woah. Who are you people

[sly frost]

so if that prime factor is p and |h| = p^n ∙ r without that exponent and |g| = p^i ∙ s then g^(p^i) has order s and h^r has order p^n and gcd(s, p^n) = 1, so g^(p^i) ∙ h^r has order s ∙ p^n according to the exercise above, which is larger than |g|, so we have a contradiction

from there it's very easy to prove that (Z/pZ)* is cyclic with that hint in the exercise

[sly frost]

no because both the coordinates have to be the same in (g, g), so that's not the same as G x G

[kind temple]

yea i ended up doing this following the exercise

[sly frost]

right, although the hint basically solves the problem here

the toughest for me was proving that Z/pqZ* is not cyclic for odd primes p, q

[crystal vale]

I need to prove that if G is a finite group of order > 2 then it has non-trivial automorphism.

If G is a non-abelian group then there x and y such that xy≠yx. So let's map x->yxy^-1 then it is automorphism mapping but not identity mapping.

And if G is a finite Abelian group, then if there is an odd prime divisor of |G| then take mapping x-> x^-1, then there exists atleast one x such that x^2 ≠ e.

And if there is no odd prime divisor of |G| then |G| is isomorphic to (Z/2Z)^n for some n in N. Then map e_1,...,e_n are generators just non-trivial permute e_1,...,e_n, then it is automorphism which is non-trivial.

Is it correct?

[quartz wind]

why not just a permutation on its elements?

[crystal vale]

Need automorphism

[quartz wind]

oh i see nvm

[lone niche]

Why do you want to relate them?

[willow cipher]

for the finite abelian case, note that Aut(Z_n) is ismorphic to the unit group of Z_n. the euler phi of n is never 1 for n > 2 so for each Z_n, there exists non-trivial automorphisms.
if there exists non-trivial automorphisms f,g on G,H resp, then (f,g) is a non-trivial automorphism on G x H
and then u get what u want by the classification theorem for finite abelian groups

[soft wyvern]

can we talk in private?

[shadow lance]

:0

I didn't know know this part of the server existed

[crystal vale]

Oh

But can you check if my proof is correct or not?

[willow cipher]

yeah its fine

[lone niche]

If it's related to your question, it's better to ask it here so that someone else might chime in too in case they have other perspectives, or I say something that's wrong

[soft wyvern]

ok

i dont understand

[tribal moss]

It (Notknow's property) is true for infinite groups too.

[soft wyvern]

why this polynomial is generated by 1

[crystal vale]

Which result?

Oh

[crystal vale]

Okay, thank you

[lone niche]

That polynomial is actually zero in the quotient module.

[tribal moss]

Was this your original question?

[soft wyvern]

yes

[soft wyvern]

so, every module of the form k[t]/p(t) is cyclic?

[tribal moss]

"Generated by" for a module means that you can get everything in the module by multiplying the generator by some element of the ring.

[soft wyvern]

for every polynomial p(t

[tribal moss]

Every R-module of the form R/I where I any some ideal, is cyclic.

[lone niche]

As a k[t]-module, yes.

You need to specify the ring to prevent confusion I think.

[tribal moss]

The professor's argument is that if [f] is any residue class in R/I then f·[1] = [f·1] = [f], and therefore [f] is in the span of [1].

[crystal vale]

Is there any result on automorphism such that aut(H^n) is isomorphic to aut(H)^n ?

[soft wyvern]

the statament only say let k a field and A=k[t]

[soft wyvern]

ooh, i see

[tribal moss]

They're not even the same size in the finite case ...

[willow cipher]

its discussed here what u can realise Aut(G \times H) as
https://math.stackexchange.com/questions/1236571/automorphism-group-of-direct-product-of-groups

[crystal vale]

I need to find Aut((Z/pZ)^n) where n is between 1 and p

[soft wyvern]

and what about the module $k[t]/(t)\oplus k[t]/(t)$

[cloud walrus]

jorge.534s

[tribal moss]

If p is prime, then that's linear algebra: invertible n×n matrices over Fp.

[crystal vale]

Okay, I am interested in the size of Aut(H) in a finite case ?

[tribal moss]

What do you know about H? I may have lost some context.

[crystal vale]

In general if H is a finite group then what will be Aut(H)?

[tribal moss]

I don't think you can say anything other than "it's the automorphism group of H" unless you know more about H than just "it's a finite group".

[lone niche]

What do you think about this module?

[crystal vale]

Oh their size must be finite, yes

[soft wyvern]

I know this is not cyclic, because if it is cyclic, it should be ismorphic to k[t]/f(t) for some polynomial f(t)

but as t is 0 in k[t]/(t), then t is 0 in k[t]/f(t)

and this happends iff f(t)|t so f(t)=1 or f(t)=t

this is correct?

[tardy hedge]

Why should it be isomorphic to k[t]/f(t)

Im not saying ur wrong im just wondering

[crystal vale]

Thank you, but where can I learn more about Automorphism on Direct products of the group?

[lone niche]

I don't follow this part. Is f(t) the same polynomial considered in the previous isomorphism?

[lone niche]

https://math.stackexchange.com/questions/1779875/each-cyclic-r-module-is-isomorphic-to-an-r-module-of-the-form-r-j

[soft wyvern]

is my professor´s argument

[soft wyvern]

yes

[tribal moss]

Yes, that sounds correct.

[soft wyvern]

t acts as 0 in k[t]/(t)\oplus k[t]/(t), so t acts as 0 in k[t]/f(t)

[tribal moss]

Well, more or less correct -- you have shown that the ideal (f(t)) is either (t) or (1), not that f itself is one of those two polynomials.

[soft wyvern]

so here is the contradiction but i dont get it

[soft wyvern]

yess

the guideline says if f(t)=(1) the k[t]/(t)\oplus k[t]/(t) is isomorphic to (0), but this is absurd

[tribal moss]

Right.

Though that should be "... if (f(t)) = (1)".

[soft wyvern]

exactly

[tribal moss]

It's not clear to me if it's you or your source material that keeps leaving out the brackets around principal ideals ...

[soft wyvern]

its my source material

but

if (f(t))=(t)

then k[t]/(t)\oplusk[t]/(t) is isomorphic to k[t]/(t)

[tribal moss]

Which is also absurd if you view them as vector spaces over k -- one has dimension 2, the other has dimension 1.

[soft wyvern]

but by first isomorphism theorem k[t]/(t) is isomorphic to k, then k is isomorphic to k\oplus k

[willow cipher]

on that post they recommend reading these
https://link.springer.com/article/10.1007/s00013-005-1547-z
https://link.springer.com/article/10.1007/s00013-008-2653-5

[soft wyvern]

ahh, i understand

so this is the method to check if a direct sum of modules is a cyclic module?

suppose that is isomorphic to R/I and get a contradiction?

[tribal moss]

Well, yes, since "isomorphic to R/I" is more or less what "cyclic module" means. (Or close enough that it could be the definition).

[crystal vale]

Okay thank you ❤️

[tribal moss]

You could also have gone the other way around and said: k[t]/(t) is isomorphic to k, and gets a k[t]-module structure by saying that scalar multiplication by t always produces 0.
Then whichever element of k² we try to make a generator cannot be, because no matter what from k[t] we scale it by, it will still be parallel to the original purported generator.

[soft wyvern]

the last question : if p(t) and q(t) are coprimes, then k[t]/p(t)\oplus k[t]/q(t) is isomorphic to k[t]/p(t)q(t) by chinese remainder theorem and then is cyclic as the first example

[cobalt heath]

Is there ever a formula for determinant of n×n block matrix?
So that I can analyze it combinatorially.

Like, I would love it as linear combination of determinant & trace of product of a few matrices.

In my case, I only need to consider the case where each component is in G <= SL(2, C),
where G = <A, B1, .., Bk>,
B1, .., Bk commutes with each other.

I wish there was a determinant-preserving map
$$GL(k^n \otimes W) \to GL_n(GL(W))$$

[cloud walrus]

Absta

[crystal vale]

If I take G = D_4 × Z then we can treat {e}× Z, where e is an identity element in D_4 , as a subgroup of G, right? Then its index is 3?

And then there exists an element in G such that a^n not in that subgroup, right?

We can take a = (r,1) then a^3 not in {e} × Z, r^4 = 1

[mighty kiln]

Its index is 4

[crystal vale]

Oh wait

Let H be subgroup {e} × 3Z

I see

But what is the index of H ?

[topaz solar]

what do you think it is

[crystal vale]

24?

[topaz solar]

why

how do you determine index

[crystal vale]

Number of distinct cosets of H, so one is e× 3Z , next we can work on e×{3Z +1} and e× {3Z +2}, similarly we can work on e

[topaz solar]

those cosets you list dont seem distinct

[crystal vale]

e × 3Z and r × 3Z are different?

We have {r^is^j } × {3Z + k} cosets for i = 1,2,3,4, j = 1,2 and k = 0,1,2

[topaz solar]

yes

[crystal vale]

I think all are distinct cosets

[topaz solar]

every element of D4 gives a coset

[crystal vale]

Yes

Then how many cosets do we get?

[topaz solar]

24

you are correct

[topaz solar]

how did you get 3 here though

[crystal vale]

Mistake

[topaz solar]

indeed

that has index 8, if D4 has 8 elements

[crystal vale]

Yes

Let G be a group and H be a subgroup of finite index n, then for any g in G g^n in H, I don't know this statement is true or not therefore I want counter example but I showed that g^k in H for some 1≤ k ≤ n

This statement is true when H is normal , I mean normal is sufficient condition

[topaz solar]

this is true

Ah wait, its a subgroup of finite index, not that the group is finite

Nvm I'm just silly and blind

[topaz solar]

Yeah, being normal would do it

[topaz solar]

So gH, g^2H, g^3H, ..., g^n H is a list of n cosets, that's all of them, hence at least one pair is equal and we can move g around, so one is H, so at least one of these powers is in H

[surreal sluice]

Assuming $K$ and $L$ are subfields of some large field. I was trying to see if $!$ will exist?

It is clear that it will be unique if it does. But I am finding it difficult to show it is well-defined once we define it using representation of elements of KL as rational K combinations of elements of L. (Although, I am not sure if the main statement is true tho) (This is a stronger version of Corollary 7.16 in Milne's Field theory notes.)

[cloud walrus]

J

[topaz solar]

I really should sleep more

[topaz solar]

If we do like, a square with Q extended by both Q(x) and Q(y), the KL = Q(x, y) yeah? but what if Omega is Q(x) with sigma(y) = x, rho = id?

[surreal sluice]

Thank you. That explains the closure requirement on omega.

[crystal vale]

Yes but can we say that x^n in H for all x in G

[long obsidian]

What is the integral closure of Z/4Z? I'm confused cause it sounds like every ring has an integral closure. But what about rings that have zero divisors

[topaz solar]

I'm trying to think if theres any counterexample, but clearly you have to have something noncommutative

(Since it needs to not be normal)

I really need to get better at these algebraic things tbh

[rocky cloak]

Typically you talk about the integral closure of a subring.

For an integral domain R one usually just says the integral closure of R for the integral closure of R in it's field of fractions.

So to ask about the integral closure of Z/4Z you first need it to be a subring of some larger ring

[topaz solar]

What was the definition of integrally closed again, solutions to monics?

[rocky cloak]

The integral closure is the set of roots of monic polynomials. And an integral domain is integrally closed if it's equal to its integral closure

[crystal vale]

Yes

That's why I take D_4 × Z but no it doesn't work

[rocky cloak]

Look at C2 inside S3

[crystal vale]

Means if I take H = (12) then (23)^3 = (23) does not lie in H, which is right

Thank you ❤️

[topaz solar]

I truly am the silliest of billies to not see it

[crystal vale]

Let G be a finite group of order 2k, where k is odd. Let a in G which has order 2 then T_a : G->G such that g->ag. Any hint to prove that T_a is an odd permutation?

Yes (T_a)^2 = T_e

So T_a ≠ T_e so T_a has order 2

[crystal vale]

I got it but if someone has any good observation then I am interested

[rocky cloak]

T_a is the product of k transpositions and k is odd

[crystal vale]

Yes that's what I want to prove

First it shows that it has products of disjoint transposition

Then we observe that (x g(x) ) is forms of that transpositions

[rain grove]

I don't understand

[crystal vale]

See they contain (a,a) not (a,b)

[rocky cloak]

Pick representatives of G/(a), then Ta transposes them

[crystal vale]

So they are not the same as G×G

In G×G you have for a≠b (a,b) but in case of H = { (g,g) | g in G } , (a,b) not in H

[rain grove]

And why is that NOT the same "structure" ?

[crystal vale]

Since you take G_1 = G_2 = G

And H is a subgroup of G×G such that H = {(g,g) | g in G }

Now have you H_1 of G_1 and H_2 of G_2 such that H = H_1 × H_2? and here G≠{e}

[rain grove]

Ohh I see

ty

[crystal vale]

Sorry I don't get your point but let g(a) is a representative of G/(a) then T_a maps to g(a)

[rocky cloak]

Like you pick representatives
x1, x2, ..., xk
Then T_a is
(x1, ax1)(x2, ax2)...

[crystal vale]

Got it, thank you

Let G be a finite group. Let N be a normal subgroup such N and G/N have orders relatively prime. Let f be automorphism of G. Prove that f(N) = N.

Since |f(N)| = | N| so if we prove that f(N) contained in N then f(N) = N.

Let there exist n in N such that f(n) not in N then f(n) + N in G/N.

Let |G/N| = s and | G | =t,

Then f(n)^s in N, by since they are relatively prime so we can write n = n^(as+bt) then f(n) = f(n)^(st) implies f(n) in N. Contradiction so for all n in N f(n) contained in N.

I think there is some mistake

[rain grove]

Let G a group and H and K subgroups. Is it possible that H and K have some common elements as subgroups and H is not subset of K and K is not subset of H?

[topaz solar]

yes

consider the even integers, and the multiples of 3

they both contain multiples of 6

[still dew]

I haven't read most of it but I guess you have the idea , basically no element outside of N can have order that shares divisors with order of N , to see this just consider the projection map
G -> G/N there is a relation between ord(a) and ord(pi(a))

Use that with the fact that N and G/N have coprime orders

[rain grove]

So for group additive group $(\mathbb{Z},+)$, that can be subgroup $H=(2\mathbb{Z},+)$ and $K=(3\mathbb{Z},+)$?.

But then their union $H \cup K$ is also a subgroup right?

[crystal vale]

No

[cloud walrus]

OHHELLNAH

[crystal vale]

HUK contains the element which is divisible by either 2 or 3

But if you take 2 and 3 in H union K then 2+3 not in H union K

[rain grove]

Okk I see

[rigid prairie]

if I have element a in the group G that is conjugate only to itself. I don't understand way for any g in G we get that gag^-1=a.

the definition of conjugate talk only about the existens of some g in G and not for every G. is it because of the coset of a?

conjugate is an Equivalence relation so if a~b so b is in R(a)=aH

[south patrol]

This is exactly what it means to say a is conjugate only to itself

The elements conjugate to an element b are (by definition) those of the form gbg^-1 for g in G

[rigid prairie]

aH={a}={gag^-1|g in G} so for all g in G we have that a=gag^-1
something like this?

[south patrol]

I'm not sure what you mean by aH and stuff here

[shell pilot]

Let G=S_3. Let H = <(12)> and K=<(13)> be subgroups of G. Then I wrote out that H={e, (1 2)} and K={e, (1 3)} but how do you compose HK? I am trying to show an example of two subgroups of a group that together (composed) are not a group.

I thought about HK = {e, (1 2), (1 3), (1 3)(1 2)} but not sure about the last element in that set

[crystal vale]

Hk= {hk | h in H and k in K}

Now (13)(12) = (123) but {e, (12),(13), (123)} is not subgroup

And in this case HK ≠ KH

[shell pilot]

Thank you! So it's not like function composition. It's just a simple "multiplication" of the elements

[crystal vale]

Yes

[shell pilot]

Is this HK incorrect then on StackExchange?

Should that last element be (1 2 3)?

[crystal vale]

This is convention

Someone operates right to left and someone operates left to right

So here (12)(23) work as first 1 goes to 2 then 2 goes to 3 so 1 goes to 3

Some authors use composition as fg(x) = g(f(x)), (x)fg= ((x)f)g maybe you can find this notation in some books

[shell pilot]

Ok, thank you!

[tardy hedge]

Ohh projective modules ultimately coming from free modules makes sense

Those free objects are a cool powerful tool

[willow cipher]

can i get some help on this?

[rocky cloak]

Well, you know A is a Q-vector space, so all you have to do is show that if A has dimension more than 1, then the endomorphism ring is not a field.

Several ways to do this, you could find to endomorphisms that don't commute, or two nonzero endomorphisms that compose to zero, or an endomorphism without an inverse. The world is your oyster here.

[hidden cairn]

how does (5) imply (2)? Shouldn't it say gNg^-1 = N instead of \subseteq?

[mighty kiln]

Well if you move g^-1 and g to the other side you automatically get N ⊂ g^-1Ng

[hidden cairn]

🤦‍♀️ thank you

[languid trellis]

For whatever reason, I'm struggling with this exercise. I've concluded so far that simple tensors have unique representation as such a sum because we have (m⊗n) = (m ⊗ r1e1 + .... rnen) = (mr1 ⊗ e1) + ... + (mrn ⊗en). The uniqueness is clear because N is the free R-module of rank n. I'm not really sure why I'm stuck here, but I don't know what the next step is. Any help would be much appreciated

Oh here's an idea. Suppose that representation isn't unique. Then we have (m1 ⊗ e1) + .. + (mn ⊗ en) = (m'1 ⊗ e1) + ... + (m'n ⊗ e'n). That implies that (m1 - m1' ⊗ e1) + ... + (mn - mn' ⊗ en). Claim: mj - mj' = 0 for all j.

If not, then we have at least one index j such that mj - mj' is nonzero. Then

(m1 - m1' ⊗ e1) + ... + (mn - mn' ⊗ en) = -(mj - mj' ⊗ ej), which contradicts the uniqueness of representation of simple tensors given above. So mj-mj' = 0 for all j and 10(a) follows.

[lone niche]

I think you proved that if the representation isn't unique, the representation isn't unique

[rocky cloak]

One thing that is often useful here is to use the universal property of tensor products. I.e. construct a bilinear map from MxN that maps (x, ei) to 0 for i different from j. And to something nonzero for i=j. Then you can reason about when such sums are non-zero at least

[languid trellis]

Is such map even possible? Consider fj : M x N -> L a bilinear map s.t. fj(x,ej) = l, fj(x,ei) = 0 (i neq j). then l = fj(x1 + x2, ej) = fj(x1, ej) + fj(x2, ej) = 2l. So, l = 0.

oh perhaps this map can't be a simple indicator. it should probably depend on the element of M as well

in some way

I don't quite understand what you're asking for jagr. In particular if you're asking for one map or a family of them

[tardy hedge]

For the universal property of free modules, do we know what the kernel of that universal map is, or would that depend on the module we ‘re mapping into

Referring to this

Whats the kernel of F(A) -> M?

It will include all the phi(a) = 0’s plus maybe more?

[rocky cloak]

Yes, it would have to depend on the element of M as well. I suggest you ponder a little more, but the map I'm thinking about is
||MxN -> M, f(x, ei) = x and f(x, ej)=0||

[lone niche]

I think free modules are badly explained in D&F, and I would recommend taking a look at Aluffi's chapter 0 for them. The main idea is that one can think of a free module on a set A, as taking a direct sum of copies of R indexed by A. And thus you determine the map phi by where the basis elements are sent (varphi). It is explained in much more detail in Aluffi.

[tardy hedge]

Determine the map psi based on where the basis’ elements are sent? Is that what u meant?

Basis elements are mapped by phi

[rocky cloak]

It depends a lot on the module and the map phi. I don't think there's a particularly simple way to describe it

[tardy hedge]

Thanks

[rocky cloak]

If you label the elements mapped to be phi: A -> M, then the elements of the kernel are usually called the relations between those elements.

[lone niche]

I think the morphism on the right is a $\Phi$ and diagonal a $\varphi$.

[cloud walrus]

kibocchi

[tardy hedge]

Ohh ok, i guess im not totally familiar with the letters still lol

[tardy hedge]

I feel like im comfortable with d&f explanation, i think they just take a really abstract viewpoint with the elements of the free module being those set functions. I liked that point of view though

But i will check out aluffi’s explanation too, thx

I guess ur right maybe its kind of confusing the crucial point that free modules are like R^n

What about the kernel of phi here? We do know exactly what the kernel is here right? Is it all representations of 0 in M as a linear combo r1m1+r2m2…+… ?

[rocky cloak]

That is exactly what it is.

But of course determining what those are is pretty complicated

[willow cipher]

how do u do that

[rocky cloak]

Well, if you pick a basis, then you can map the basis vectors wherever you want. From there you just play around, it's easy to construct examples

[soft wyvern]

Why in a morphism $f:V\otimes V\longrightarrow K$ , if $f(a\otimes b)=0$ then $a\otimes b= 0$?

[cloud walrus]

jorge.534s

[soft wyvern]

no, the full question is: Let k a field and let V:= k[t]/(t) a k[t]-module. Prove $V\otimes V\neq (0)$

[cloud walrus]

jorge.534s

[soft wyvern]

in the solution, my professor proved $\overline{1}\otimes\overline{1}\neq 0$

[cloud walrus]

jorge.534s

[rocky cloak]

I think maybe the thing you're thinking of is that

f(a(x)b) =/= 0 implies a(x)b =/= 0.

Simply because 0 is always mapped to 0 by linear maps

[soft wyvern]

by the universal property, there is a morphism $f:V\otimes V\longrightarrow k[t]$, and then get $f(\overline{1}\otimes\overline{1})\neq 0$ so $\overline{1}\otimes\overline{1}\neq 0$. I dont understand the last step.

[cloud walrus]

jorge.534s

[soft wyvern]

so this is an equivalence?

[rocky cloak]

I'm not sure what "this" you're referring to, but there are no equivalences here.

Linear maps satisfy f(0) = 0, so if f(x) is nonzero, then x cannot possibly be 0.

[soft wyvern]

i mean, if f(a(x)b)=0 then a(x)b=0 too

[coral spindle]

Well this is simply false in general so this is not going to help.

[rotund aurora]

Is there any relation between the derived length of a group and its minimal number of generators?

[coral spindle]

Oof well I mean no

because an Abelian group can have arbitrarily many generators but derived length 1

But if you look at the length of a group – that is, a length of the derived a composition series – we do get something

since every finite simple group is generated by 2 elements

G is generated by 2l elements

[rotund aurora]

you meant the composition series, right?

[coral spindle]

Woof! What a typo

Indeed I mean a composition series

and certainly not 'the'

The fact that every fsg is generated by two elements is mysterious. I think the current understanding of this comes soley from the classification of fsgs and ongoing research aims to show it without the classification

[rotund aurora]

Right, but I am more concerned with the other direction. That is, if G is generated by "few" elements, can its derived length be "large". By the way, I am assuming that G is solvable. Also, I am mainly interested in the case that G is a p-group, maybe the results are nicer for p-groups

[coral spindle]

Right you want to bound the derived length in terms of the number of generators

Well I don't know maybe someone else can comment

[rotund aurora]

There is this result. It doesn't say anything on the number of generators tho, but it's just to show that maybe there are results

[rotund aurora]

Note that the minimal number of generators is the dimension over Fp of G/Phi(G) where Phi is the Frattini subgroup

[rotund aurora]

Honestly, I just want to know the following: Does there exist a p-group with derived length >=5000 that is generated by 4 elements?

the number 5000 is just an example ofc

[coral spindle]

I feel like magma could tell you this

[rotund aurora]

I assume this is very true, because of what you said about simple groups. But I'm not sure how one could show this

[coral spindle]

I honestly don't think it would follow, just bc p-groups are nasty little buggers

[rotund aurora]

check the above theorem. The order of such a group would be >=p^(2^5000+10000-2), I don't think magma can handle this

[coral spindle]

Idk I think there are nice things that the computational algebraists (read: saints) have worked out for us

But yeah Idk the answer to this and I don't really know how one might find one

[rotund aurora]

yeah, I just mentioned it as an heuristic

what even is an easy way to construct p-groups of very large derived series length (the order or anything else does not matter, as long as it is a p-group)?

[acoustic igloo]

The set of permutations that fix the first element should form a subgroup, right? For example
(234) and (243) would be elements but (1234) wouldn’t. But is this a stabilizer subgroup?

I feel like it should be, but when (234) acts on a set, it moves every element of the set - for example, it moves 1234 to 1342

so 1234 is not fixed by (234)

but hold on i feel like there might be a way to define the group action such that this would be a stabilizer subgroup

[rotund aurora]

btw, isn't the composition series length of a group of order p^n just n (or n+-1, whatever)? Because the only simple p-group is Z/pZ

[acoustic igloo]

like, can you say the set is {1, 2, 3, 4} and if g=(234), then g(1)=1, g(2)=3, g(3)=4, g(4)=2 ?

yeah that seems to make sense...

[rotund aurora]

that's the definition, yes

[acoustic igloo]

oh, um

but it's up to you to define, isnt' it?

you could say the set is {(1, 2, 3, 4), (2, 1, 3, 4)...} and g((1, 2, 3, 4)) = (1, 3, 4, 2)

idk maybe the other way is more standard?

i got confused because i wasn't sure if there's a standard way

[shell pilot]

When looking at A_4 as the alternating group of even permutations of {1, 2, 3, 4}, why is (1) an element? Technically the identity is (1)(2)(3)(4) right? Is it just written as (1) to show that it is the identity? So really it has 0 two cycles which makes it an even permutation?

[tender wharf]

identity can be written as (12)(12)

[shell pilot]

Any specific reason why they use (1)?

[tender wharf]

well you basically have to write something down

but i could choose (2)

it's just simpler to write (1)

[shell pilot]

Ok, and then the three cycle permutations can be written as non-disjoint two cycles, correct?

[tender wharf]

in cycle notation anything thats not written is assumed to be fixed but then if we follow that for the identity we can't write anything which... is bad

so we write something

[tender wharf]

you can always decompose a permutation into a product of 2 cycles yes

not necessarily in a unique way

[shell pilot]

Ok, thank you very much!

[rain grove]

Why is C(S) = C(C(C(S)))

[coral spindle]

What does C(S) mean

[rain grove]

Centralizer, so if $S \subseteq T$ then $C(S)$ are all $t \in T$ that commute with $S$

[cloud walrus]

OHHELLNAH

[coral spindle]

So you in fact mean C_T(S)

But sure I mean

[rain grove]

I managed to prove the $C(S) \subseteq C(C(C(S)))$

[coral spindle]

Yes

[cloud walrus]

OHHELLNAH

[rain grove]

but idk how to do the other side

For this side I first had to prove $S \subseteq C(C(S))$ and then also $S \subseteq T \Rightarrow C(T) \subseteq C(S)$.

[cloud walrus]

OHHELLNAH

[coral spindle]

Do you have any more assumptions on the group T or is this all?

Is it finite, for example?

[rain grove]

It's not specified

[coral spindle]

So there are no additional assumptions.

[rain grove]

Show that .... and that from ... it follows ... From here show that ....

[cloud walrus]

Boytjie

[coral spindle]

Lol this is a bad proof

[tribal moss]

Isn't it true in general that A subseteq C(C(A))?

[coral spindle]

But more straightforwardly

If x centralises CC(S) then it must in particular centralise S so CCC(S) is a subset of C(S).

Man I had such a mental block with this that I just gave it away. Sorry, I'm not being a good pedagogue.

[rain grove]

I mean its ok I still don't understand lol I need to think a little

but ty

[coral spindle]

Note that S \subseteq T implies C(T) \subseteq C(S), NOT C(S) subseteq C(T)

[tribal moss]

You can argue directly: Suppose a in A. Then by definition of C(A), a commutes with everything in C(A). But that means that a in C(C(A)).

[rain grove]

ohh wow yeah

[coral spindle]

So actually the harder part is showing C(S) subset CCC(S)

Lol

[tribal moss]

That's why I used a different letter and said A subset C(C(A)). Just set A = C(S).

[coral spindle]

Mhm

[shell pilot]

If there are 3 left cosets of H in A_4, then does how do I know there are 6 left cosets of H in S_4? I know that |S_4|/2 = |A_4| but not sure how that applies to the order of the left cosets

I also know that by Lagrange's theorem, |H| divides |G|

[coral spindle]

There is another part to Lagrange's theorem

[shell pilot]

Oh, the theorem has a second statement that says left (right) cosets of H in G is |G| / |H|

[coral spindle]

Yes.

[shell pilot]

Thanks!

[shell pilot]

$a, b \in G$ and $H, K \le G$. If $aH=bK$ prove that $H=K$. So far, I have $aH=bK \Rightarrow H=a^{-1}bK$ and by properties of cosets, $a^{-1}b \in K$ How can I conclude that $a^{-1}bK = K$?

[cloud walrus]

Soap_Opera

[coral spindle]

Hint: It's a coset of K and it must contain a special element of K

[shell pilot]

If an element of one coset is in another coset, then they are equal?

[crystal vale]

Yes

[shell pilot]

$aH=H$ if and only if $a\in H$

[crystal vale]

Yes

[cloud walrus]

Soap_Opera

[shell pilot]

Thank you both

[crystal vale]

But why a^-1b in K?

[shell pilot]

Oh

[crystal vale]

And this a and b are arbitrary?

[shell pilot]

Because of the property $aH\le G$ if and only if $a\in H$

[cloud walrus]

Soap_Opera

[shell pilot]

Yes, arbitrary a, b in G

[crystal vale]

Then why you just don't take a = b = e ?

[shell pilot]

I just figured that wouldn't have been an acceptable answer, but I guess it doesn't say you can't use that haha

Oh, and also, the question asks to prove so I guess an example doesn't count as a proof

[crystal vale]

example?

[shell pilot]

By letting a=b=e that is just an example of aH=bK.

It says to prove, so I guess it would need more than just that example

[crystal vale]

No it is not an example

If it is given that aH = bK for every a and b in G then you just take a = b = e

[crystal vale]

Actually they want to tell us that a and b doesn't matter here

[tardy hedge]

H and K are subgroups

[crystal vale]

So a and b are fixed by question

[shell pilot]

I just figured it was asking for a formal proof, so I did a proof, but if it really is as easy as using the identity element, then that is super easy

[crystal vale]

No

[shell pilot]

Ok, so not "arbitrary" a and b

[crystal vale]

It is not same as aH = bK for all a b in G

[shell pilot]

Ok, understood

Thanks!

[shell pilot]

If $R^+\subseteq H\subseteq R^*$ then we can say that $R^+ = H$ or $R^+\subset H$ which could mean that H can have an element less than 0 right?

[cloud walrus]

Soap_Opera

[south patrol]

Yes

Well it has an element less than 0 iff it is not R^+

[crystal vale]

Why?

[south patrol]

By definition of R^+

(Here I am assuming H contains R^+)

[crystal vale]

Got it, thank you

[rotund aurora]

what is the minimal number of generators for the group of upper triangular nxn matrices with unit diagonal over F_p?

[crystal vale]

Order of matrix?

[rotund aurora]

nxn

[icy totem]

I cant understand the definition of "monomial ideal", to me it just looks identical to the concept of Ideal

Because the definition of monomial ideal is basically an infinitely or finitely generated ideal, but all Ideal are finitely/infinitely generated

As far as I know

Also im studying monomial ideals in K[x1, ... xn], and every ideal in this context is finitely generated

So I really cant understand the difference

Someone willing to explain?

[dull ginkgo]

It's generated by monomials

not generated by polynomials that might be sums of them

Let me think of a simple example

Lets say we have the bivariate polynomial ring over the integers

and consider the principal ideal (X^2 + 2Y)

every element in that ideal is of the form p(X,Y)(X^2 + 2Y) right? but we can't necessarily say it's generated by monomials (yes we can write every element as a sum of monomials ofc, but not the same monomials)

[icy totem]

Ah ok thanks

Then an ideal isnt necessarioy generated by monomials right?

[icy totem]

Because this is a counterexamplr

[dull ginkgo]

Yes

In fact this is true of R[X]

[icy totem]

Ok I got it thanks

[dull ginkgo]

Consider (2 + x) in Z[x]

[dull ginkgo]

A more intuitive way to write it

Let’s say we do have a monomial ideal

Generated by monomials (m_1 … m_n)

Then if we have f in the ideal

If we break f up into monomials

each of the monomial summands

Must be a multiple of one of the generators :3

[icy totem]

Ah yes the book im studying from talks about that

Now its clear

[dull ginkgo]

E.g the ideal is a sum over the principal ideals

Every element is a sum of polynomial-scaled generating monomials :3

N AND NOW J

[coral spindle]

big day for letters

[surreal sluice]

Can someone explain why when completing a topological group it is important that the group be first countable?

[dull ginkgo]

I am unsure tbh but at first glance we probably want countable neighborhood basises so we can lift the group operation to the Cauchy sequences

Well Cauchy defined in a weird way here

I wouldn’t say weird actually, just generalizing how you’d say something is Cauchy in R :3

like for N > n,m |x_n - x_m| < epsilon

Instead we can say

N > n,m implies x_n x_m^-1 is in U

Where U is a neighborhood of the identity

I have done basically nothing with topological groups but this is what immediately jumps out to me

So countable neighborhood basises

Kinda is like epsilon = 1/n

:3

[surreal sluice]

This works without countability assumption, as here:

[dull ginkgo]

oh okie :3

[dull ginkgo]

But you need to describe a topology on the Cauchy sequence group

[rotund aurora]

shouldn't it just be the subspace topology of the product topology

[dull ginkgo]

Hm yeah i guess so

[crystal vale]

Can someone give me an example such that f is irreducible in Z[x] but not in Q[x] and f has degree 2

[south patrol]

Well basically take smth irreducible in Q[x] and multiply it by an integer so it lies in nZ[x] or smth

[dull ginkgo]

Yeah

[south patrol]

5(x^2+1), say

[crystal vale]

Yeah

Thank you

But is 5(x^2+1) irreducible in Z[x]

[rotund aurora]

the completion should be complete no? maybe if you have too many open sets it's not complete? But idk

there is nothing wrong with the definition tho

[crystal vale]

I think it is reducible in Q[x]

[coral spindle]

Really? If a quadratic polynomial is reducible over a field then it must have a linear factor, i.e., a root. So what's the root that it has?

[crystal vale]

No sorry I mean it is reducible in Z[x]

I need an example such that f is irreducible in Z[x] but not in Q[x]

[south patrol]

That is impossible

[crystal vale]

Unless f has degree 0

[south patrol]

Well okay sure by the convention that units aren't irreducible

[crystal vale]

But they say that if f is primitive and f is irreducible in Z[x] then it will be irreducible in Q[x]

So I think they want to use primitive somewhere

[rotund aurora]

I don't understand this comment

[dull ginkgo]

It’s very uninformed so i would ignore it

I’ll delete it

[south patrol]

Hm irreducible in Z[x] should imply primitive anyway

Since if some prime divides all the coefficients then that gives you a proper factor

[crystal vale]

No maybe because 5 is irreducible in Z[x] not primitive

[south patrol]

Okay I guess that is the only exception

[crystal vale]

Yes non-zero constant are not irreducible in Q[x]

[rotund aurora]

I never dealt with this concept of completion of a group. Maybe try to see what your book does with it, and try identifying whether some proof uses first countability

[inner harbor]

Heyy

I'm stuck

Trying to prove that Gal(Q(i,fqrt(p))/Q) is isomorphic to D4

p is prime

And fqrt is ^1/4

[lone niche]

I think there's a few ways to do this. Have you computed the order of the extension?

[inner harbor]

It's 8

Well it matches that noice

I thought we could take two morphisms s and r and verify that srsr=id

Where s has order 2 ans r order 4

[topaz solar]

Well, is the Galois group commutative?

Because like, how many order 8 groups are lying around, after all

[inner harbor]

Well we have possibly D4 and H8 here

But how do we choose

[topaz solar]

Well like, is this Galois group commutative

[inner harbor]

Well I guess not cause D4 is not

[lone niche]

I think at this point we can check the group structure directly, since we have an automorphism that moves i->-i, and another automorphism that moves the fourth roots of p. We notice that the automorphism that moves i->-i also moves some of the fourth roots of p.

Since the degree of the extension is 8, and by building a tower of fields by first adjoining the fourth roots of p, and then i, first adjoining square roots of p and i, and then adjoining the remianing 4th roots, the galois group should be determined entirely by these two generators

alternatively i think we can also count number of subgroups of different order

[rotund aurora]

I think a sequence of the type 1-->G(L/K)-->G(L/Q)-->G(K/Q)-->1 is always split?

[rocky cloak]

Take L to be Q adjoin a fifth root of unity. Then G(L/Q) = Z/4. And let K be the real part of L. Then the sequence is
2Z/4 -> Z/4 -> Z/2

[dull ginkgo]

:3

[grizzled spindle]

Is this doable by induction easily with these recursion formulas? This is in a field over the reals and < , > is a bilinear form. The other notation isn’t relevant to solve the problem, just the inequalities

I haven’t really done 2 variable recursion so I’m somewhat loss for this

[rotund aurora]

ah, okay I see my mistake. But yeah, I did not believe it, but I didn't see an immediate counter-example

like the action on G(K/Q) is given by zeta-->zeta^3 or zeta-->zeta^2, which is not complex conjugation, but when squaring you get complex conjugation

I think in the example of Moh you could see it splited, K being Q(i), you can lift the nontrivial element from G(K/Q) to G(L/Q) by complex conjugation, which is of order 2 in both groups

[lone niche]

Another method is that for a polynomial of degree n, the galois group must be isomorphic to a transitive subgroup of Sn. And for S4, the only non abelian transitive subgroups are D4, A4, S4. And again by degree considerations it must be D4.

[acoustic igloo]

i'm trying to understand extensions and stuff - i'm looking at the quaternion group Q_8 - am i to understand that Q_8 is an extension of some smaller groups? But what kind of extension?

[coral spindle]

Idk what you mean by that exactly

Yes, Q_8 has a normal subgroup which is isomorphic to C_4

So it is an extension of C_2 by C_4

That's really it, it's hard to say very much else unless we talk about group cohomology, and even then that is not always helpful

[acoustic igloo]

is it a semidirect product?

[coral spindle]

No

It is not a split extension

[acoustic igloo]

is it a central extension?

[coral spindle]

I'm not super familiar with central extensions, but if I'm not mistaken, no it isn't – the normal subgroup is not central

[acoustic igloo]

okay

thanks

[coral spindle]

It is a central extension of C_2 x C_2 by C_2 though

But the extension C_4 → Q_8 → C_2 is not central

[acoustic igloo]

oh

a group can have different short exact sequences

[coral spindle]

Of course

A group may have many normal subgroups which induce many quotients. I hope this isn't a surprise!

[acoustic igloo]

yes... i'm not sure why i was surprised

i guess because C_2 was in both of them

is the semidirect product of two groups unique?

[coral spindle]

No

[acoustic igloo]

Ah!

[coral spindle]

a direct product is a semidirect product

Are you aware of the data required to build a semidirect product?

Like have you seen semidirect products constructed

[acoustic igloo]

no

[coral spindle]

OK well the data needed is an action of one group on another by automorphisms

That action determines the product

It could be trivial, which produces the direct product.

There are typically many ways that one group acts on another and therefore many different semidirect products of those groups.

The wikipedia page explains how the 'outer' semidirect product is constructed fairly well

[acoustic igloo]

oh i see... okay on the list of small groups on wikipedia they sometimes show like Q_12 = Z_3 ⋊ Z_4
this means Q_12 is a semidirect product of Z_3 and Z_4, but doesn't specify the action?

[coral spindle]

Well let's work through this

What's Aut(Z_3)?

[rotund aurora]

Sorry to bother, but do you happen to know about this @coral spindle ?

[coral spindle]

(In this example, Z_4 is going to act on Z_3. That's indicated by the direction that ⋊ is pointing)

[coral spindle]

I don't know off the top of my head but I'll give it a look in a bit

It really ought to be n-1, but I will think about it properly soon

[acoustic igloo]

i don't really know, just that Aut(Z_3) is a morphism from Z_3 to Z_3?

i'm not sure if it's unique

[coral spindle]

Aut(Z_3) is the set of all automorphisms on Z_3.

[acoustic igloo]

okay

[coral spindle]

I'm surprised you don't know this, this is extremely important to know about

OK

So I'll just say this instead

Aut(Z_3) is isomorphic to Z_2

[rotund aurora]

I didn't really verify it thoroughly, but I think the subgroup where the diagonal above the main diagonal is zero is normal. When you quotient, you get something isomorphic to F_p^(n-1), so the number of generators ought to be >=n-1

[coral spindle]

So a semidirect product is going to consist of some map f : Z_4 → Z_2

Now furthermore if we're writing Q_12 = Z_3 ⋊ Z_4, we're certainly not talking about an Abelian group here

I.e., this is not a direct product

So we have some nontrivial map f : Z_4 → Z_2 which determines this semidirect product

And in fact there is exactly one nontrivial map of this kind.

So in fact the semidirect product is uniquely specified by writing Z_3 ⋊ Z_4.

This is quite common to do; often when people write this there will be more than one such map, but they will all produce the same group up to isomorphism.

[acoustic igloo]

ok this is a lot to think about

thank you

[coral spindle]

Well you are missing context so I'm not surprised.

Are you working through a textbook?

[acoustic igloo]

no

[coral spindle]

You should work through a textbook

I like Fraleigh personally. People here seem to like Artin but I haven't tried it. People also often talk about Dummit and Foote.

[rotund aurora]

note that Artin doesn't touch extensions nor semi-direct products (iirc)

[coral spindle]

But I'm sure it covers the notation Aut(G)!

[acoustic igloo]

oh i have fraleigh

[coral spindle]

So right let me think about this

[rotund aurora]

note this

[coral spindle]

Croq do you happen to know how this group corresponds with a root system?

[rotund aurora]

no

[coral spindle]

But yes what you wrote here is totally correct

Well ok so

When we work with an ACF of characteristic p instead of F_p, you're looking at something called the unipotent radical of a Borel subgroup

And it turns out that these have nice properties related to the root system of the reductive group

But the long and short of it is that yes, indeed, the composition factors look like F_p^{n-1}, F-p^{n-2} etc in this case (unless some weird shit happens in A_n, but I don't think it does)

Chevalley's commutator formula allows you to calculate this kind of thing if you need something to google

But yes I do believe it should be exactly n-1 generators, since as you say the quotient needs at least n-1 and if I'm not mistaken (I will have to look up Chevalley's commutator formula for type A_n) you can just choose the matrices with entry 1 in the obvious squares and 0 elsewhere

So what I mean is

if $e(i,j)$ is the matrix with $1$ in entry $i,j$ and zeros elsewhere, I think that the group should be generated by matrices of the form $I + e(i,j)$ where $i-j=1$

[cloud walrus]

Boytjie

[coral spindle]

This actually shouldn't be too tricky to show inductively lmao

I think there miiiiiiiiiiiiiiiiiiiight be issues if p=2 but I would need to work this out on paper

But yeah the subgroup you were looking at was the F_p-span of everything of the form I + e(i,j) where i-j > 1

That was badly phrased but I hope the intention is clear

[rotund aurora]

interesting stuff

indeed the subgroup of upper triangular matrices is the first example they give in Wikipedia

[coral spindle]

Oh for Chevalley?

Nice

[rotund aurora]

no, for Borel subgroup

[coral spindle]

Right right

Yeah in general we even fiddle with the definitions of classical groups to make this nice

[rotund aurora]

I could not find the chevalley thing tho

[coral spindle]

Like the symplectic group is defined independently of the choice of symplectic form, but algebraic group people will choose a different one to what differential geometers do, just so that the maximal torus and Borel subgroup can be chosen nicely

[rotund aurora]

is a standard reference for all of this Milne's book on algebraic groups? I am not familiar with any of this, but it sounds interesting

[coral spindle]

I don't know if there's a standard agreed-upon reference. I've been recommended Springer and Humphreys – I think Springer's book is good – but if you just kinda want to get started with algebraic groups quickly, then I recommend Malle & Testermann's book which is a good reference.

They also include an amazing collection of results on finite groups of lie type, which is really what we care about if I'm feeling cheeky

[rotund aurora]

btw if G and H are (finite) solvable groups then the (regular) wreath product G wr H is also solvable. Can you say anything on the derived length of G wr H in terms of that of G and H?

I mean more specifically a lower bound

If Im not mistaken, an upper bound is just dl(G)+dl(H)

[coral spindle]

Oh I worked this out a while ago

Oh no wait I assumed that H was Abelian. Nevermind.

[rotund aurora]

That's still interesting

[coral spindle]

Let me get my notes

So if $H$ is Abelian -- so here I mean the group $H$ acts on $G^n$ by permutations -- then you can argue that there's a subgroup of $(G \wr H)' \leq G^n$ which is isomorphic to $G$. I think you have to look at some diagonal embedding stuff. So you do get that $dl(G \wr H) \geq 1 + dl(G)$

[cloud walrus]

Boytjie

[coral spindle]

I was interested in this because I wanted to construct groups of arbitrarily large derived length

And I only checked this really for H = Z_2, but I am confident that the argument generalises.

[rotund aurora]

yeah, that's what I'm trying to do

[coral spindle]

Aha well I have kinda spoiled things

If H is non-Abelian this would get very messy and I can't see a straightforward way

But yeah one way of getting groups of high derived length is to keep on wreathing by Z_2

[rotund aurora]

no, because I also want to keep the number of generators bounded

[coral spindle]

Right ofc

[rotund aurora]

and when you take wreath products, the number of generators goes up by at least 1

[coral spindle]

I will say that I think you want nonsplit extensions if you want to keep the number of generators low. Like you want some generator of a group not to lift perfectly, so that you get some extra elements for free by choosing a good lift

[rotund aurora]

well I saw this result for nilpotent groups, but I'm mainly interested in p-groups

well it is possible that when taking just one wreath product the number of generators stays the same, but if you keep taking wreath products, it will be unbounded

[coral spindle]

Wow that's a strong result

Though I guess some of the strength is presumably in the fact they're using a very particular action for the wreath

[rotund aurora]

here is the paper if you want to check it out, it's short but I haven't read it https://archives.maths.anu.edu.au/people/Kovacs/K071.pdf

also, are there nice bounds for the size of the commutator subgroup of a p-group G?

[coral spindle]

if |G| = p^n then 1 <= |G'| <= p^{n-2}

[rotund aurora]

but for example, I don't think it can equal p^(n-2) when n is large?

[coral spindle]

Oh hm

You know what I don't know

p groups more like piss groups

[rotund aurora]

you can embed any (finite) group into a (finite) group generated by 2 elements right

oh yeah

Sn

[acoustic igloo]

yeah the symmetric group can be generated by 2 elements

[rotund aurora]

but the Sylow of Sn needs lots of generators I think

[coral spindle]

Yes S_—nvm lmao

[coral spindle]

Right it's a product of things of the form C_p wr C_p wr ... wr C_p

So it should need a lot in general

I can give a more precise description if you'd find it helpful

[rotund aurora]

what else could you say?

[coral spindle]

The precise description in terms of n

But yeah the long and short of it is that it's probably not going to be a good example for what you want

[rotund aurora]

I'm just wondering if any p-group can be embedded inside a p-group generated by 2 elements

every p-group embeds in a Sylow of S_n, but C_p wr ... wr C_p doesn't seem easy to handle

[acoustic igloo]

does an automorphism need to send elements to elements of the same conjugacy class?

or is that something different

[rotund aurora]

also, maybe you can write a stupid presentation, but then idk if the resulting thing will be finite. Consider the free group generated by x,y, say the p-group G has p^n elements, then you can consider p^n "independent" words in x,y and basically copy the multiplication table of G. Then try adding more relations and hope that you get a finite p-group?

[coral spindle]

No

See: automorphisms of Abelian groups

[rotund aurora]

crazy counter-example

[coral spindle]

[coral spindle]

c.f.

[acoustic igloo]

haha

why is that crazy lol

[coral spindle]

I am self-conscious that I am doing the mathematician thing of "For example, take a nonspecific thing"

[acoustic igloo]

oh, i don't mind lol

[rotund aurora]

If you have time, yeah, I am interested

[coral spindle]

Oh like Sylows of S_n?

Sure

Let $n = a_0 + a_1p + a_2p^2 + \cdots + a_np^n$ be the $p$-adic expansion of $n$. So in particular $0 \leq a_i < p$ and this is the only such expression for $n$ of this form.

[cloud walrus]

Boytjie

[rotund aurora]

I know how to calculate v_p(n!) if you are going there

[coral spindle]

Oh yeah well this is helpful for both

Then writing $\wr^k C_p$ for the $k$-fold wreath product, the Sylow $p$-subgroup of $S_n$ is isomorphic to $(\wr^1 C_p)^{a_1} \times (\wr^2 C_p)^{a_2} \times \cdots \times (\wr^n C_p)^{a_n}$

[cloud walrus]

Boytjie

[acoustic igloo]

but i thought abelian groups only have 1 conjugacy class - so an automorphism would take every element to the same conjugacy class

[coral spindle]

Hm is there a big wreath product symbol? This works fine ig

[coral spindle]

There is only one group with a single conjugacy class

I think you need to think about what the conjugacy classes of Abelian groups look like

[acoustic igloo]

oh i misread

there is one for each element

[coral spindle]

That's right

[acoustic igloo]

sorry

[coral spindle]

(To clarify: \wr^1 C_p is just C_p)

[rotund aurora]

ah, that's nice. I only knew it for S_p^n

[coral spindle]

Yh it's convenient

It also tells you the normaliser quite easily too

[rotund aurora]

no it doesn't

[coral spindle]

:kek: it's not too bad after this

[rotund aurora]

I mean you can't just look at an abstract isomorphic copy of the Sylow to determine its normalizer

[coral spindle]

Oh yes

Yes I see what you mean

In my head this is the same as the actual Sylow lol

[rotund aurora]

oh yeah

cuz like G wr H acts on G x H right

[coral spindle]

Yeah it's just a bunch of p-cycles

[rotund aurora]

so wr^n Z/pZ acts on {1,...,p}^n=[p^n], for the case S_p^n

[coral spindle]

You just need to add the extra a_0 '1-cycles' haha

But yeah like the normaliser mod the Sylow looks like S_{a_0} x S_{a_1} x ...

[sonic coral]

I know that every group of order pq for primes p < q where p does not divide q-1 is cyclic. i also know that a group of order n is cyclic is (n,phi(n)) = 1. what is the first instance (if any) where the (n,phi(n)) = 1 result realizes a cyclic group that isn’t found with the sylow theorem result

i feel like one should exists, and there would need to be 3 prime factors

[coral spindle]

So you're asking for n with phi(n) coprime to n with at least 3 prime factors

[sonic coral]

yes

[coral spindle]

Like yeah just look for it man

In fact literally the smallest result works doesn't it?

2 * 3 * 5

Nah lol my bad

[sonic coral]

it can’t have a 2 i don’t think

[coral spindle]

3 * 5 * 17, there ya go

[sonic coral]

3 5 17 works actually

[rotund aurora]

I think I see better now why this is true from this drawing. Like the normalizer modulo the Sylow should permute this trees (there is only one in the case S_p^n)