groups-rings-fields

[toxic zephyr]

but xh2x^-1 not in H means =xh3

[chilly ocean]

Yes

So x = xh2x^-1h3

[toxic zephyr]

h2x^-1=h3
x^-1=h2^-1h3 in H

[chilly ocean]

I think you mean h2x^-1 = h3^-1

[toxic zephyr]

how?

[toxic zephyr]

just multiply by x^-1 on both sides

on the left

[chilly ocean]

From x = xh2x^-1h3 you get e = h2x^-1h3

[toxic zephyr]

I don't want to go there tho

[chilly ocean]

Multiplying by h3^-1 on both sides you get h3^-1 = h2x^-1

not h2x^-1=h3

[toxic zephyr]

$xh_2x^{-1}=xh_3$

$h_2x^{-1}=h_3$

[cloud walrus]

eigentaylor

[toxic zephyr]

why would we do that though?

isn't this much clearer

[chilly ocean]

You said h2x^-1=h3

[chilly ocean]

which seems to contradict this

[toxic zephyr]

how is that not clear from the latex

[toxic zephyr]

how are you getting this in the first place?

shouldn't it be h3 inverse on the RHS

[chilly ocean]

I don't remember

[toxic zephyr]

how is this wrong then?

[chilly ocean]

I might have made a mistake

[chilly ocean]

I think this is correct, sorry

[toxic zephyr]

I have another question. if a group has order r, then how do you prove that the rth power of any element is the identity?

I'm doing the next question and essentially I think the idea is that we want to show that (aN)^r=N for all a in G. given N has index r

[chilly ocean]

Do you know Lagrange's theorem?

[toxic zephyr]

the order of a subgroup divides the order of the group?

[chilly ocean]

Yes

[toxic zephyr]

yeah. but all I remember about the proof is that each coset has the same number of elements

is that used to prove g^|G|=e?

[chilly ocean]

The proof I know of uses it

It is not difficult to prove Lagrange's theorem though

[toxic zephyr]

or oh is the idea that the order of any element must divide the order of the group because the powers form a subgroup?

[knotty badger]

yeah I agree with bequi here

[knotty badger]

mhm

[toxic zephyr]

so |G|=k*o(g) implies g^|G|=e^k=e I see that's simple

[chilly ocean]

Also another way to do 2. is by using that the number of left cosets equals the number of right cosets

[toxic zephyr]

oh. is that only true iff normal subgroup?

[chilly ocean]

It's true always

Otherwise you couldn't use it here because you don't know yet that H is normal

[toxic zephyr]

so how does that do 2?

[chilly ocean]

You have two partitions {H, xH} and {H, Hy}

Can you see how it does it?

[toxic zephyr]

thinking about it
clearly xH=Hy

[chilly ocean]

Is Hy=Hx?

[toxic zephyr]

yeah I think so. as long as Hx is not H then it has to be Hy right

[chilly ocean]

Yeah

[toxic zephyr]

alright then I'm with you

[toxic zephyr]

is there something about how xH=Hx for all x in G that implies H is normal?

[chilly ocean]

Sometimes that's even taken as a definition of normal

[toxic zephyr]

interesting

in your opinion, which proof is better? I think the second one tbh lol

[delicate orchid]

If f is injective (which it is for wreath products) just restrict your reps to H

I’ll answer this in full by characterising the irreducibles of a wreath product once I’ve been awake for longer than 5 minutes

[topaz solar]

Yeah, that’s sorta the part why I was like that’s not generic enough to see what’s going on etc etc

[delicate orchid]

dunno what that means

you're doing the easy direction here

it's much harder (although still not that hard) to take reps of G and H and turn them into reps of the wreath product

wait are your groups finite

[topaz solar]

No

Which is why this is nontrivial at all

(Analytically, that is)

[delicate orchid]

why the fuck you asking me then

I mean what I said is still true

[topaz solar]

Because you are the silly little wreath product man

[delicate orchid]

just restrict to H

[topaz solar]

That’s a good point, but it seems hard to get any nice info like uhhhh irreducibility of the thing on f(H), or having it actually quotient

(When it’s not a wreath)

[delicate orchid]

having what actually quotient, H is a subgroup not a subquotient

[topaz solar]

The wreath is what I care about but because it’s injective into Aut(G) it sorta turbo sucks

For some horrid technical data reasons

[delicate orchid]

if it wasn't injective we would have a serious problem

[topaz solar]

And this has been a thorn in my side for months

[delicate orchid]

you can't deflate reps to subquotients so it's important that H exists in G as a subgroup rather than just in a SES

[topaz solar]

Yeah this is sorta exactly the issue

The thing is a subgroup but like not really

[delicate orchid]

it just really is a subgroup though I'm confused

(1|h) is a subgroup of the wreath product iso to H

and then because f is injective H is iso to its image in Aut(G)

[topaz solar]

Ah yeah that’s not the issue, I meant the f(H) < Aut(G) part

[delicate orchid]

so there's no deflation requirement there either

[topaz solar]

It’s iso to a subgroup but not equal to one, and there’s some goofy analytic troubles

Basically it’s only an issue because this is a question for mentally deranged individuals

Since any way to make this actually work, kinda has to work for any semi direct

And I just hoped some insane rep theorist would have something

[delicate orchid]

why on earth does this matter

[topaz solar]

As in, the issue is some measure pullbacks kinda fall apart, because of factoring nonsense, if the image isn’t particularly nice

[delicate orchid]

isomorphic groups have isomorphic representation rings

[topaz solar]

Which I’ve been bashing my head against the wall on this question for months

[delicate orchid]

but NOT the other way around

[topaz solar]

But basically turns out my argument works iff the image f(H) has extra structural properties I just don’t have on H (a priori)

[delicate orchid]

assume that it does and then prove your result

I won?

[topaz solar]

Unfortunately that’s basically like uhhh

Assume the result, then the result is true

[topaz solar]

I should look at this though

That is, the representations rings, I mean

[frail summit]

Quick question. Suppose $n = p^k m$ for $p$ prime and $m$ relatively prime to $p$. Then over some field $F$ of characteristic p, we have $x^n -1 = (x^m - 1)^{p^k}$, so if I'm not mistaken, a root of unity of $x^n-1$ over $F$ must also be a root of unity of $x^m-1$, so we kind of lose some roots of unity by restricting to the characteristic.
Is there a geometric way to understand this in the complex plane?

[cloud walrus]

Uncle Fyodor

[winged void]

thanks alot for your help

[crystal vale]

I want to prove that if G is a finite group and p^e divides |G| then there is a subgroup of order p^e.

Let G have order p^(n)m where p does not divide m and n≥e.

By Sylow there is a subgroup of order p^n, now use induction so there is a subgroup of that Sylow p-subgroup of order p^e so that subgroup is subgroup of G.

Is it correct?

[rocky cloak]

You're missing a few steps I think. How exactly are you applying induction?

[warm ember]

the sylow group has nontrivial center

[crystal vale]

I am assuming that that result is true for all groups of order < |G|

And if G has order p^n

Then I will use that non-trivial center

[crystal vale]

Yes

[rocky cloak]

Sure, that works

[crystal vale]

So that's true, right?

[long obsidian]

Say P is a polynomial and $\Delta P= P(z+1)-P(z)$ then for the polynomial $zCr=(1/r!)z(z-1)...(z-r+1)$.

How can I show $\Delta( zCr)= zC(r-1)$?

I tried to just write everything out and factor but I got

$\Delta (zCr)= zC(r-1) [\frac{1}{r (r-1)}(z-r+1)(z-r)-\frac{1}{r}(z-r+1)]$

I'm not sure what to do with this. This fact apparently is related to pascals triangle but I'm not sure how

EDIT
Come to think of it I think this is false because it sounds like it's saying a degree (r-1) polynomial arises as the difference of a degree r polynomial and degree (r-1).

But if you add two polynomials with different degrees the degree will clearly be the max of the two degrees.

[cloud walrus]

HausdorffT1

[south patrol]

Both sides are polynomials of degree r-1 with coefficients over Z, so it suffices to check they agree on z = 0,1,...,r-1

and then it is a standard identity on binomial coefficients which yes is equivalent to saying pascal's triangle gives binomials

[arctic trail]

$\rtimes$

[cloud walrus]

Trivial Lemma

[arctic trail]

ah semidirect product

[hollow mica]

I think they’re both degree r but it’s still the same

[south patrol]

Hm zCr is of degree r right so after taking a difference it'll be r-1

[long obsidian]

I don't understand why both sides are polynomials of degree r-1. nC(r+1) is degree r+1 and nCr is degree r. I don't see why the difference doesn't at least have the leading term (1/(r+1)!)z^{n+1}. It's not like nCr has a degree r+1 term to cancel out in the difference.

Am I missing something?

[wooden rain]

Is it true that if a pair of vectors is orthogonal in R^2 with regard to the standard inner product, the pair is also orthogonal with regard to any other inner product in that space? This claim seems pretty obvious because all inner products on R^2 are basically an ellipse

[south patrol]

But its n C (r-1) the

And the difference cancels out the top coefficient

since f(z+1) and f(z) have the same top degree thing

[rocky cloak]

Nope, that's not true.

[long obsidian]

Oh my God I was evaluating the difference between nCr and nC(r-1) versus the difference between (n+1)Cr and nCr. For some reason your notation helped me realize that thank you for your help

[rocky cloak]

For any pair of linearly independent vectors, there is an inner product that makes them orthogonal. So conversely, you can always choose an inner product that makes them not orthogonal.

I'm also not sure what you mean about the inner products being "basically ellipses"

[chilly ocean]

If this was the case, automorphisms of vector spaces would have to preserve orthogonality, but they don't

[wooden rain]

Every inner product on $\mathbb{R}^n$ can be represented by a diagonal matrix of positive terms on the diagonal in some basis, and so we obtain a formula $\langle \vb x, \vb y \rangle = \sum_{k = 1}^n \alpha_k x_k y_k$ for $\alpha_k > 0$. I think I confused myself at this step because I wanted to reduce the problem to a smaller set than the whole $\mathbb{R}^n$, and I tried to normalise the value. That's how I obtained the equation $\sum_{k = 1}^n \alpha_k x_k y_k = 1$, which in the case of $n = 2$ is an ellipse

[cloud walrus]

Thingoln

[wooden rain]

I can do it of course, but I'd only obtain an ellipse when $\vb x = \vb y$ 😛

[cloud walrus]

Thingoln

[rocky cloak]

The "in some basis" is the crucial part though.

You're reasoning seemed to be pinned on it being with respect to the standard basis.

[wooden rain]

Right. After changing my basis, I would get an equation for an ellipse, but if the basis was (1,0) and (1,1), then in the standard basis the graph would not be an ellipse

I have a hard time explaining my intuition at times, but I think I got it. Thanks a lot for the help 🙏

[wooden rain]

Come to think of it, I now realise that the class of isometries of euclidean vector spaces would be meaningless if what I said were true

[topaz solar]

ye

[oblique glen]

Please give example of a subset of a ring which is group under addition but not closed under multiplication

[delicate orchid]

Z in R if you mean closed under mutliplication in the ring like an ideal?

otherwise uhhhhh

oh pick any copy of R in C that isn't the real line

[south patrol]

(A line through origin necessarily)

[delicate orchid]

duh

did you think I meant copy setwise?

why would I randomly switch categories like that

[south patrol]

I know you meant that but I was just adding this for clarity for the asker

[delicate orchid]

but here's the asker

[rotund aurora]

evil Wew

[toxic zephyr]

quick question about this. i used that <X> is the set of all finite products of
xi^(ei)
where ei is 0 or 1
is that standard or is there a different definition of <X> i was expected to use?

$\gen{X}=\bdef{\prod_{i=1}^nx_i^{e_i}:x_i\in X, e_i\in\bdef{\pm 1},n\in \bN}$

[cloud walrus]

eigentaylor

[toxic zephyr]

idk if there's something else with the fact that <X> is the intersection of all subgroups containing X or whatever

that i was supposed to use

im pretty sure what i did is right but idk its been a while

[delicate orchid]

nah this proof is very trivial

[toxic zephyr]

okay cool

[delicate orchid]

u basically just put gg^-1s everywhere and that's it

[toxic zephyr]

yeah $g\paren{\prod_{i=1}^nx_i^{e_i}}g^{-1}=\prod_{i=1}^n\paren{gx_ig^{-1}}^{e_i}=\prod_{i=1}^nx_i'^{e_i}\in\gen{X}$

[cloud walrus]

eigentaylor

[delicate orchid]

yur

[toxic zephyr]

cool thanks for the santity check!

[tardy hedge]

yur

[toxic zephyr]

yur!

so i know i asked about this last night (and ofc i could be a little more specific about how Lagrange's theorem implies the rth power is the identity), but i'm wondering what the analogous proof is if i don't use the quotient group?

[south patrol]

Hm tbh Lagrange or smth like it is inevitable - suppose G is finite of order r and N = {e}

[tardy hedge]

Nice

[arctic trail]

Didnt know the latez off the top of my head

[topaz solar]

Fair

[hidden cairn]

can i get a small hint on this q?

the second assertion follows from isomorphism theorems but cant see the way to prove the first one.

oh nvm im just stupid sometimes

the first follows immediately from |PN| | |G| 🤦‍♀️

[toxic zephyr]

could someone give me a hint for this one? i never really formally learned much about HK stuff. like i want to say that |G|=c1[G:H]=c2[G:K] or something but does G necessarily have a finite order? i'm not exactly sure what i want to show and what is sufficient.

im not sure if there's something obvious we can know about the two subgroups if the indexes are coprime. i'm guessing that it means their intersection is trivial. but i'm not sure why exactly that would be.

non-normal subgroups in general just sorta confuse me lol

[arctic trail]

Is this true?

[toxic zephyr]

is what true

[arctic trail]

If one of them is normal, then this would follow easily but

The product of subgroups is generally not a subgroup

[delicate orchid]

I’ve been trying to think of counter examples and haven’t thought of one yet

[toxic zephyr]

oh like the original question?

[arctic trail]

Yeah

[toxic zephyr]

so this was from the first homework from last year. i'm trying to bone up on my group theory before the class starts. so i can't like ask the prof or anything, and it's possible there are mistakes and i'd have no way of knowing

[arctic trail]

Did the professor assume HK is a group?

[toxic zephyr]

i have absolutely no idea

i've been reading my old textbook, and it says that HK is not necessarily a subgroup

[arctic trail]

Yeah exactly

[toxic zephyr]

so maybe the coprime indices imply that it will be somehow?

[arctic trail]

Does HK even partition G when HK isn't a subgroup?

We can define an equivalence relation on the left cosets of G by: hH equiv h'H if there exists g in G s.t. hH = h'Hg

[arctic trail]

Of H*

I meant K instead of G aaaaa

Its almost 2am

And thats the same as hHK = h'HK

[delicate orchid]

Time zone buddies

[arctic trail]

Show the number of left cosets of HK is divisible by the number of left cosets of H and it should follow (?)

[delicate orchid]

The coprime index thing is almost forcing this to be true just by an order argument when G is finite

[arctic trail]

I dont remember if the number of right cosets have the same size

[toxic zephyr]

is G finite? or is it possible it's infinite?

[arctic trail]

As left in general

[delicate orchid]

They wouldn’t specify finite index if it was finite

[arctic trail]

If so, try following this vein of thought while I go to bed

[delicate orchid]

We’re still in the stage of trying to convince ourselves this result is reasonable

Or I am anyway. Mr. lemma seems to have gotten there

[toxic zephyr]

sorry i'm not understanding this lol

so G has possibly infinite order?

[delicate orchid]

It is blindingly obvious that they aren’t imposing a finiteness condition on G

[toxic zephyr]

i was wondering if the fact that H,K are finite index and G=HK might imply G is actually finite regardless

[arctic trail]

Apparently every finite index subgroup contains a finite index normal subgroup

[arctic trail]

No

[toxic zephyr]

like {e}? or does that not count

[toxic zephyr]

rip

i feel super out of my depth here. glad i decided to try to study before the class starts.

can we know that H interset K is trivial?

actually just in general, i dunno what we can really do with cosets if the subgroup isn't normal. like we know that G={H,c1H,...,c(l-1)H}={K,d1K,...,d(m-1)K} where m and l are coprime. but how would we even try to show that H interset K is trivial?

[delicate orchid]

[G : H cap K] <= [G:H][G:K] by second iso I think. And because the intersection is a subgroup of both, it’s index divides both [G:H], [G:K]. But then because [G:H], [G:K] are coprime and they both divide [G:H\cap K] so does the product [G:H][G:K], thus [G:H][G:K] = [G:H \cap K]. Now [G:HK] = |G||H cap K|/|H||K| = |G|^2|H cap K|/|H||K||G|= [G:H][G:K]/[G: H cap K] = 1.

[south patrol]

How do you know HK is a subgroup?

(when you use [G:HK] etc)

But yes once you have [G:H][G:K]=[G:H \cap K] you should be able to conclude [I found a proof which did what you did and then concluded without proof lol]

(v nice wew)

[hidden cairn]

could anyone give me a small hint?

[serene dune]

can i assume G is finite?

gcd. lcm =hk .n

lcm = hk.n

[crystal vale]

I don't know but H is solvable right? And the intersection of the normal subgroup is normal

[serene dune]

then by structure they are h, k sylow subgroup ?

[long obsidian]

If I have a graded module M over a graded ring S then does it follow that for some homogenous elements x in M

then Ann(x) is a proper ideal of S?

I'm reading somewhere that suggests this but I can't tell why.

[rocky cloak]

I mean, as long as x is nonzero Ann(x) is a proper ideal.

[long obsidian]

Why is that for nonzero x in M?

Does it have to do with the grading?

I think 0 is in M0 and for any x in Mk then there should exist some element s in S so that sx is not in M0. But otherwise I can't see why and am not sure if that's why

[rocky cloak]

Well, if M is nonzero it contains a non-zero element. And since M is graded we can pick that element homogenous

And 1 doesn't annihilate x, so Ann(x) is a proper ideal

[long obsidian]

Wow okay got it thank you.

[hidden cairn]

sorry how does this help i cant see

I'm thinking of taking a normal series 1 < A < ... < H, and another one H/H < G_1/ H < ... < G/H. then merging the two, 1 < A < ... < H < G_1 < ... < G, we get A abelian and normal in H. but i feel like this is not enough to conclude A is also normal in G

[crystal vale]

I have no idea, just tried

[rocky cloak]

So I don't know exactly which definition of solvable you're working with, but one is that there exists a sequence of normal subgroups (in G)
1 = G0 < G1 < ... < Gn = G
such that Gi+1/Gi is abelian.

If you're working with another definition you might try to prove this first.

Anyway, given such a sequence consider it's intersection with H

A more common definition is to just require Gi to be normal in Gi+1. To prove the stronger definition you may want to think induction.

[rocky cloak]

A more direct approach could be to try to come up with subgroups of H that are invariant under all automorphisms (hence also conjugation by elements in G). There are some natural ones that come up when thinking about solvable groups.

[hidden cairn]

I'm using this definition with the extra condition that G_i is normal in G_(i+1) as well

[toxic zephyr]

does the fact that their indices are coprime imply that at all maybe?

[hidden cairn]

is the idea behind this to show there exists a minimal subgroup of G contained in H?

[rocky cloak]

Yup

[hidden cairn]

then do we claim H cap G_i is the minimal subgroup contained in H, where i is the largest number st G_i does not contain H?

[rocky cloak]

Yes, go on

Or not sure what your saying in your second half there

But you consider Gi minimal such that HcapGi is nontrivial

[hidden cairn]

so say we have K < Gi cap H. and K normal in G. and nontrivial, then K itself can be inserted into the normal series 1 < G1 < ... < G_(i-1) < K < G_i < ... < G. This contradicts with Gi being the minimal.

this reasoning feels a bit handwavy and prone to error rn but i'll come back to it tmr

[rocky cloak]

I'm not sure where this K comes from or what you're trying to do now

But the point is that Gi-1 cap H = 1, but Gi cap H doesn't

[hidden cairn]

Tried to show Gi cap H is minimal by assuming there exists a normal non-trivial subgroup K < Gi cap H

Then K must either turn out to be 1 or Gi cap H itself

If it is not one of those, K can be inserted into the chain of subgroups (not sure about this ?) and then Gi becomes no longer minimal as H cap K = K is non-trivial

But I guess we don’t know if G_(i-1) < K holds so I can insert K in between G_(i-1) and G_i in the chain of subgroups

[rocky cloak]

It's only minimal in the sense that if you go to Gi-1 and smaller then the intersection is trivial

That's not something to show, that's just how we picked Gi

Then you have this normal subgroup Gi cap H to work with

[hidden cairn]

We want to show Gi cap H is a minimal subgroup of G because we picked i such that the intersection is non-trivial, right?

In a previous exercise, I already showed minimal subgroups of solvable groups are necessarily abelian

[rocky cloak]

Allright, you might be able to do something like that.

I think it's probably easier to just show that Gi cap H is abelian, but you can try that

[crystal vale]

Let G be a finite Abelian group and H be a subgroup of G then prove that G has a subgroup which is isomorphic to G/H.

Any hint?

[rocky cloak]

Is this without the classification of finite abelian groups?

[crystal vale]

This is given in the problem set so maybe we can use that

[crystal vale]

Finite Abelian groups are direct products of cyclic groups, right?

[rocky cloak]

Well, then I guess it really just comes down to looking at what the subgroups and quotient groups of finite abelian groups can be

In particular if you write G/H as a product of cyclic groups, you can try to relate that to G through the map G -> G/H

[crystal vale]

Okay

[delicate orchid]

You don’t need to, |HK| = |H||K|/the intersection irregardless

[south patrol]

But these guys may all be infinite

[delicate orchid]

ok? is mans bothered

[crystal vale]

Let G be a finite Abelian group of order mn, where m and n are relatively prime positive integers. Then show that there exists unique subgroups of order m and n such that G is isomorphic to their direct products.

I used the hint, G_1 = { x in G | x^m = 1 } and G_2 = { y in G | y^n = 1 }. Yes these are subgroups because G is abelian.

Now if I define the mapping from G_1×G_2 to G such that (x,y) -> xy then it is homomorphism and I proved that this is injective and surjective so it is isomorphism.

Is it correct?

[south patrol]

Yes looks good

[crystal vale]

Okay thank you

[long obsidian]

If M is a graded module over a graded ring S and I consider a homogenous element m in M,

Then why is Ann(m) a homogenous ideal of S?

[mighty kiln]

Suppose rm = 0, and consider each homogeneous part of r

[crystal vale]

If G is a finite group of odd order then for any x≠e, x is not a conjugate to x^(-1).

Let x = gx^-1g^-1 for some g≠e in G.

Then, x^-1 = gxg^-1 so x = g^2xg^(-2).
We get x = g^3x^-1g^(-3).

For n = 2k, x = g^nxg^n and for n = 2k +1 x = g^(2k+1)x^(-1)g^(-2k-1).

And G has an odd order so we have x = x^(-1) implies x = e because x^2 ≠ e. So for x≠e x is not conjugate to x^(-1).

Is it correct?

[long obsidian]

So if r was decomposed into homogenous parts as r=r1+r2 in S then r1m and r2m are in different summands, say M1 and M2.

Does it just follow from these terms being in different parts of the direct sum?

[mighty kiln]

Yea

[south patrol]

Yes

Though you seem to have changed m to x lol

[long obsidian]

Whoops lol

Are we using that 0 is homogenous or maybe that 0 is in a particular summand such as M0? I have a worry that 0 of M is in both M1 and M2 .

[rocky cloak]

0 is in everything

[crystal vale]

Any comment?

[long obsidian]

Oh then I can't tell why a non homogenous element of S can't annihilate a homogenous element m of M

Maybe I need extra conditions on M like being f.g. and S being neotherian?

[rocky cloak]

non-homogenous elements of S absolutely can annihilate m.

if r1 and r2 both anihilate m, then r1+r2 does as well

[long obsidian]

So Ann(m) isn't necessarily a homogenous ideal if m is homogenous?

[rocky cloak]

Yes it is

a homogenous ideal is an ideal genereted by homogenous elements

Ideals only consisting of homogeneous elements doesnt exist for most reasonable rings

[long obsidian]

So is the argument more like, say r has homogenous decomposition r=r1+r2 then rm=0 implies r1m=0 and r2m=0 hence the homogenous parts of each r are in the ideal Ann(m).

Which means the ideal can be generated as Ann(m)=<r1,r2,...>

[rocky cloak]

yes exactly

[gaunt bone]

Possible dumb question. Do many people use Category theory in groups-rings-fields research?

[void cosmos]

yeah

algebra in general

for example with rings u can sometimes try to study the ring by looking at the category of modules over that ring

u can for example ask if i have two rings R and S, does R-Mod being equivalent (as categories) to S-Mod give R is isomorphic to S ?( the answer is no )

with groups there is this thing called tannaka duality in which u can (in a way) recover a group by looking at the automorphisms of the forgetful functor that preserve the monoidal product

something like that i am not well versed but this exists and u can just look it up and read about it

but basically u can recover G from Rep(G)

usually tho, the theory itself gets built instead of just being used to solve group theory problems. for example u can read about tensor categories which are categories that are almost like Rep(G) (symmetric tensor categories give or take)

u can study these things for their own sake and ig they are fascinating in a way that i can not appreciate

there aren't many elementary examples at all except for Rep(G). u can read about this more in delignes paper on tensor categories

[delicate orchid]

yes, a lot

for example: cohomological algebra

[gaunt bone]

This may be appropriate for category-theory chat, but are all groups-rings-fields morphisms in category theory invertable? what does one do when they're not, so, can a catagory morphism be uninvertalbe? I thought they must be by definition.

[delicate orchid]

no?

also why are you saying groups-rings-fields lmfao

a category with every morphism inveritble is a groupoid

[gaunt bone]

groups-rings-fields are objects in Cat theory. A little confused, since Categories seem to need invertible morphisms by def, no worries, just thinking a little out loud.

[topaz solar]

They do not require them by definition

Please reread the definition

[gaunt bone]

okay

[delicate orchid]

I can't take you seriously when you keep repeating the channel name LMFAOOO

also they're not objects "in cat theory", they're objects in a category

and if you work with the category of fields I will cry for you

[gaunt bone]

Please speak with some respect

[delicate orchid]

nah

[gaunt bone]

Anyhow, category arrows can be defined to and from each object, I relooked at the def, yes, seems they must not be invertable by def. Please speak with respect

[fossil shuttle]

@gaunt bone The identity morphism c -> c is invertible. Other morphisms need not be invertible

[knotty badger]

a category where every morphism is invertible would be a groupoid

this may have been where your confusion arose…?

e.g. you can think of groups as one-object groupoids

[delicate orchid]

cmon now

not direct at you pseudo but like

you want me to give you respect when you just completely ignore my posts?

[coral shale]

a category with every morphism invertible is a groupoid

[cloud walrus]

HausdorffT1

[arctic trail]

yeah, G is isomorphic to the standard representation, no?

[chilly radish]

Rep(G) is the category of representations

Moamen is talking about tannaka-krein duality

[dawn quest]

How come when I ctrl+f groups-rings-fields in any algebra textbook I get no results

I thought they would be a pretty big deal

Please be respectufl

[hollow mica]

i'm taking a class on groups-rings-fields next semester and i'm kind of scared because I've only studied them individually and not as a single object. is there anything i can do to prepare

[arctic trail]

what do you mean 😭

[rocky cloak]

Are you trolling?

[dawn quest]

I will not be repeating myself

[rocky cloak]

Well, in case you're not. Groups, rings and fields are 3 different things

And I guess ctrl+f random text books is probably not a good methode to look stuff up in general

[toxic zephyr]

that's also just not how ctrl+f is supposed to be used
if you ctrl+f "angles-circles" or "angles circles" in a geometry textbook, you're not going to get anything because how do you write a sentence with "angles circles" in it? even though it's 100% guaranteed a geometry text will talk about angles and circles. it won't pick up "angles, circles" either.

try "group", "ring", and "field" separately.

[tardy hedge]

does anyone here like combinatorial structures

it seems like such an unpopular area of math lol

[delicate bloom]

I imagine those people hang out in #combinatorial-structures

[tardy hedge]

Fax dude fax such fax

at this point its weird going to those uncharted servers

im so used to the nice familiar faces here

in algebra land especially

[delicate bloom]

[tardy hedge]

lul

jagr potato wew are legendary

hall of fame status

[white oxide]

for 2a could we just pick (s, t) for each orbit O_{s, t} since the orbits partition M?

[toxic zephyr]

two questions

• any hints for proving that f is surjective?
• for the proof that ker(f)=MnN, the only iffy step is that n in M iff mn in M is a valid step, right? the => direction is clear, but if mn in M, then mn=m' implies n=m^-1 m' in M so the <= is also true right?

also, isn't this kind of like the chinese remainder theorem?

[crystal vale]

The injective part is correct

[toxic zephyr]

does this work for the surjective proof?

[crystal vale]

Yes

[toxic zephyr]

awesome

okay cool thank you so much

[crystal vale]

In general take g_1^(-1)g_2^ = mn then g_2n^(-1) = g = g_1m now f(g) = (g_1, g_2)

[crystal vale]

Don't forget to prove that it is homomorphism

[toxic zephyr]

ooh right

[south patrol]

Seems oddly underrepresented here

Not sure why

[toxic zephyr]

so i haven't taken a class that covers this level of group theory yet (doing this to prep for the class), can anyone point me to what results/theorems i would want to look into to prove these?

[topaz solar]

There’s possibly a bit you can get out of some in foundations channel, but yeah, it’s certainly not too popular here

[south patrol]

It is fun tbh lol

I am encountering more combinatorics lol

[vapid vale]

do you know that A_n is simple for n >=5

[toxic zephyr]

honestly, i dunno what a simple group is

[vapid vale]

ah

a simple group is one with no proper nontrivial normal subgroups

[crystal vale]

I have the same question problem 2

[vapid vale]

i think you should take this fact for granted (it appears in most textbooks) – with that fact its not so hard of a proof

[toxic zephyr]

so... to use that i assume we would need to know that A_n is sort of... maximal i guess in Sn?

[crystal vale]

So if any H is a non-trivial proper normal subgroup then it is || normal subgroup of A_n ||, right?

[south patrol]

That is clear though

[vapid vale]

it is an index 2 subgroup

[crystal vale]

And there is no proper subgroup > | G |/2

Thank you @vapid vale

[toxic zephyr]

is that obvious because |An|=n!/2?

[south patrol]

Yes

[toxic zephyr]

i see

okay

then that makes a lot of sense

[south patrol]

A_n is the kernel of a surjection S_n -> Z/2

Etc

[toxic zephyr]

and that's the sign homomorphism right?

[vapid vale]

yes

[toxic zephyr]

okay okay i feel like i'm getting it

so we know that A_n can't be contained in any other normal subgroup than S_n itself. but how do we know there is no other proper normal subgroup that, though perhaps intersects nontrivially with A_n, is not properly contained in A_n? is it that k | n! and k != n!/2 implies that k | n!/2?

[vapid vale]

consider their intersection

[topaz solar]

This is the part where this is relevant

[crystal vale]

I forgot this part in my proof

I want to prove that in a finite group of order pqr, p<q<r are distinct prime numbers.

Then if q doesn't divide r-1 then q Sylow - subgroup is normal in G.

We have the result that there exists a normal subgroup N of order qr. Now take Q as Subgroup of order q such that Q \subset in N.

Since q does not divide r-1 so Q is normal in N. Then for any g in G, gQg^-1 \subset N, and which is q Sylow - Subgroup of N.

So Q = gQg^-1 for all g in G.

Is it correct?

[crystal vale]

For any Q q Sylow - subgroup, we have N such that N = QH, where H is a r Sylow-subgroup. And H is normal and N has index p thus N is a normal subgroup of order qr

[dim widget]

Yes that’s fine

[crystal vale]

Okay thank you ❤️

[delicate ledge]

can someone help me prove that given a cyclic normal subgroup H of G, every subgroup of H is also normal?

i think we need to use that every subgroup of a cylic group is normal, correct?

[toxic girder]

are all functions morphisms?

[warm ember]

no

[warm ember]

every subgroup of a normal subgroup is normal

[lapis latch]

Mhh no

[crystal vale]

Every subgroup of a cyclic group is cyclic

Now let that cyclic normal subgroup is < a > now for g in G, gag^-1 = a^s, for some s

Now take any subgroup of < a > it is of form < a^k >

Now find ga^kg^-1

|| ga^kg^-1 = a^(sk) which is in < a^k> ||

Now can you show that for any element in < a^k >, ga^kmg^-1 in < a^k >

[tardy hedge]

Wat

[umbral plume]

yeah i can only think of a few people

i will probably start posting there soon once i get off my ass

combinatorics 🔛 🔝

[toxic girder]

i apologise i see my mistakes now

[cobalt heath]

My brain is broken. So for module homomorphisms f, g : M -> N,
ker(f) > ker(g) implies f factors through g?

What is the intuitive way to see this?

[white oxide]

any ideas?

[topaz solar]

M -> M/ker g -> N, since ker g < ker f?

[cobalt heath]

I have issues seeing how this gives the factoring

It should be obvious, yea but..

[topaz solar]

I dunno what exact sort of factoring you’re looking for

Which I probably should but uhhh oops

[cobalt heath]

If I recall correctly, f = h \circ g for some map h : N -> N.

[mighty kiln]

If M → N sends A ⊂ M into B ⊂ N you induce M/A → N/B

[serene dune]

its an induced property by group elements, which just fix the whole subgroup. a portion of the subgroup may not be fixed

[mighty kiln]

So you get M → M/ker(g) → M/ker(f)

I don't think M/ker(g) → N and M/ker(g) → M/ker(f) → N are the same tho?

[topaz solar]

Well, presumably one is g and the other is f

I was thinking of this, but I didn’t see a good way to extend q: M/g -> M/f to a total N -> N, after identifying w/ submodules of N

At least without some free generation of PID-ish shenanigans

[mighty kiln]

I don't think free over PID suffices

You probably need like vector space

Cuz like 2,3: Z → Z

[topaz solar]

Yeah basically

Things do factor over mod g, mod f but I couldn’t see any way to get the desired thing without some very strong homogeneity

Which I’m not sure would really work nicely anyhow there, but

If you could get something like M/g as a direct summand you could do a bit more

(Which this is what I meant by free generation, like in the image then extending to the whole thing)

[cobalt heath]

Ah, so this only works for special case where e.g. g is surjective.
I thought it worked in general, since a material mentioned like it does.

Resolves some of my confusion, thanks!

[crystal vale]

Let G be a finite group which has only one automorphism. I proved that it is Abelian and every non-identity element has order 2.

How can I show that G has at most order 2

We can say that G is 2-group

[serene dune]

aut sends generator to generator

when the order is 2 u can explicitly show there is only identity mapping possible

[crystal vale]

Yes

[serene dune]

where as more than 2 elements will have more than 1 non trivial generator

tho i have no proof for it

but conjecture at the moment, lol

[crystal vale]

I seen its proof

[serene dune]

alr

[crystal vale]

Then we contrapositive it

[serene dune]

then a sawpping possible leads to nontrivial aut

oh yeah

[crystal vale]

But I don't want that

But then G is isomorphic to (Z/2Z)^n

If n > 1 then we have non-trivial automorphism

[serene dune]

i mean there are non cyclic groups

aut(V4) \cong S3

[crystal vale]

G is Abelian and every non-identity element has order 2 so by the fundamental theorem of Abelian group

[serene dune]

oh i was talking about most general case

didnt notice the abelian part

but yeah it looks fair

[crystal vale]

Thank you

[serene dune]

need this

lol

[crystal vale]

But if I have that then contrapositive

Without it we can prove it

[serene dune]

i just need that in general

alr i will try to prove

@crystal vale have u taken topology ?

yet

[crystal vale]

Yes

[serene dune]

is there any spam-esque channel wehere we can talk without having to stick to channel protocol

[crystal vale]

DM

[crystal vale]

if A is a mapping from group G to G such that G is one-one and Ah_g = h_gA for all g in G, where h_g(x) = gx, then prove that A = h_k, for some k in G. I proved that A(x) = xA(1).
any hint?

[quiet pelican]

I don’t think what you’re given is true unless you know G is abelian
I think you should have A = x -> xg for some g (which you’ve already proven with g = A(1))
To prove it, if A = h_k and Ah_g = h_gA then kg = gk (evaluating at 1)
In particular, if G = S_3 there is no such non-identity map
But right multiplication by (12) is such a map, contradiction

[knotty badger]

Yeah this is yoneda stuff

A map that commutes with all left-multiplication must be right-multiplication

Via following where the identity goes

[crystal vale]

This is Herstein' s exercise maybe I stated it wrong but I think it is a typo

[knotty badger]

I’m very confident it’s a typo, yeah

The formula A(x) = x A(1) is correct

This is essentially applying the yoneda lemma

[crystal vale]

But I didn't apply A is an injective

[knotty badger]

I don’t see why you’d need to

[quiet pelican]

You don’t need it

[knotty badger]

Indeed you can deduce it from this

[quiet pelican]

G-equivariant endo morphism implies bijective

[knotty badger]

Because right-multiplication is bijective

[knotty badger]

Like to be clear it is a good instinct to have of “hm, I haven’t used up all the assumptions”

It’s just that in this case, one of the assumptions was genuinely unnecessary

[hidden wind]

we doing a category theory crash course in class

[serene dune]

category for working mathematician 🤦‍♂️

Splitting a ring by introducing fractions may result in a non-ring due to loss of closure under multiplication, but splitting a field preserves its field structure since fields are closed under division by non-zero elements.

am i right ?

[mighty kiln]

What is splitting

[serene dune]

$R[\alpha]$

[cloud walrus]

yeshua

[serene dune]

i forgot to say after applying evaluation homomorphism

[chilly radish]

I think you're confusing the term "splitting field". "Splitting a ring" is not really a thing, and the sentence you wrote doesn't make a lot of sense anyways

[white oxide]

anyone know if this is okay?

also what do they mean by "determine the restriction of the equivalence relation to its closure"

do they mean find the closure of the set of representatives

[rocky cloak]

No, like take the closure of the fundamental domain, then describe the equivalence relation restricted to that. I.e. which points in the closure are in the same orbit

As for the first part, saying pick (s, t) from the orbit of (s, t) isn't really specifying a choice.

You're supposed to pick one thing from each orbit

[white oxide]

oh okay got it

would picking the s value that'sin [0, 1) and the corresponding t value work

[rocky cloak]

Yeah, that would do it

[white oxide]

like $(s - \floor{s}, t^{-\floor{s}})$

[cloud walrus]

okeyokay

[white oxide]

oh okay cool

[trail harness]

Can someone help me find a set of 3 generators for the ideal ${f(X, Y) \in \Z [X, Y] : f(a, b) \equiv 0 \pmod{p}}$ for $p$ prime and $(a, b) \in \Z^2$? \
\
I was trying by cases: if $p | a$ and $p | b$ then this ideal is generated by the constant polynomial $p$, but if $p | a$ and $p \nmid b$, we know that $f(X, Y) = a_n(Y)X^n +...+ a_1(Y)X + a_0(Y)$ and since $p | a$, we get that $f(a, b) \equiv a_0(b) \pmod{p}$ and because $a_0(Y) \in \Z [Y]$, then $a_0(Y) = b_mY^m +...+ b_1Y + b_0$, and then idk how to continue

[cloud walrus]

アポーロ

[coral spindle]

You're trying to work "top down" where you try to find a set of generators by sifting through the definition of this ideal, but this isn't a good way to do this because there are many ways to generate this ideal which you must choose from.

Try this "bottom up" approach: name a few small elements of this ideal (perhaps try to name a couple that are degree 0 and 1) and then try showing they generate the ideal.

I'll even give you a headstart: the constant polynomial p is in this ideal. Can you find a couple of degree-1 elements that, together with p, might possibly generate the ideal?

Come up with good guesses and then give your best shot at showing they generate the ideal.

[arctic trail]

I think it's easier to reduce modulo p

${f(x,y) \in \mathbb F_p[x,y] : f(a,b) \equiv 0 \pmod p}$

[cloud walrus]

Trivial Lemma

[arctic trail]

Now this looks like a problem in algebraic geometry, probably

[arctic trail]

then just include p

[trail harness]

Maybe p (p must be on this set of generators), pY and other polynomial of deg 1, because in the case where p | a and p not divides b, we can restrict the study to that Y polynomial b_mY^m +...+ b_1Y + b_0. We can divide each b_k by p (so b_k = pq_k + r_k and the polynomials p and pY take care of the pq_k part, r_mY^m +...+ r_1Y + r_0) I suppose that the polynomial 1 - cY may be the third generator, where c is the inverse of b modulo p, but idk how to prove it

[coral spindle]

OK so pY is not a great guess because that's already in the ideal generated by p

You're again trying to do things top down

I really want you to just list a few elements of this ideal

Don't try to be super smart about it, literally just name a few elements

Just throw them out, go ahead, tell me some

[trail harness]

lmao you're right

[coral spindle]

I hope so!

[arctic trail]

you can wlog assume a = b = 0

since it's just a translation

[coral spindle]

Let's keep it simple.

[arctic trail]

tf you mean keep it simple

this simplifies everything

if you have p(x,y) and p(a,b) = 0 then q(x,y) = p(x+a,x+b) satisfies q(0,0) = p(a,b) and p(x,y) = q(x-a,y-b)

and the modulo p reduces everything further cause now you're looking for 2 generators not 3 (one of the generators goes to 0 when you quotient by (p)).

[coral spindle]

Let's let apollo do the question maybe

[arctic trail]

I just think your approach is horrible 🤷‍♂️

[coral spindle]

OK

[arctic trail]

cause here it's like "which polynomials are congruent to 0 (mod p) at 0,0"

[trail harness]

Ok, p, 1 - cY (where c is the inverse of b modulo p), 1 - dX (where d is the inverse of a modulo p, assuming that p does not divide p) and I think that every 1 degree polynomial in this ideal is generated by these 3 polynomials...

Let Ax + By + C be a polynomial s.t. Aa + Bb + C = 0 (mod p), then C = -(Aa + Bb) (mod p), then:

Ax + By + C = -Aa(1 - dx) - Bb(1 - cy) (mod p)

Then if you have Ax + By + C, you can divide A, B and C by p, the p generator take care of the p part and then the 1 - cY and 1 - dX take care of the remainder, it's right?

[coral spindle]

Yeah this is looking much better to me

Here's an even simpler thing

Instead of 1 - cY, why not Y - b

And similarly X - a

[arctic trail]

(which in the case a=b= 0 those polynomials are X,Y and p(0,0) = 0 iff the constant term is 0.)

[coral spindle]

Your approach works I think (I would have to inspect the details a bit) but yeah just having Y - b and X - a simplifies it even more

[trail harness]

Sure, idk why did I forgot X - a and Y - b lmao

[arctic trail]

it does not work in the case b = 0 or a = 0

[coral spindle]

Haha you were trying to be too smart about it!

But yeah you also don't have inverses mod p when a or b are divisible by p

So this is just the right way

See it's not too hard if you start with an idea of what the generators might be :)

[white oxide]

would it just be something like (0, t) ~ (1, -t)

[trail harness]

But I just have proved that it generates the set of one degree polynomials in this ideal, how can I prove that it generates the whole ideal? Maybe using the fact that f(a, b) = 0 implies that the polynomial f(X, b) = 0 has a root in a and then it is divided by (X - a) or something like this?

[coral spindle]

Well as it happens this is a question that has come up a few times recently lol

Let me give you the lowdown

[arctic trail]

every polynomial p(x,y) has a decomposition as a sum of monomials of the form (x-a)^k(y-b)^r (k,r >= 0)

[coral spindle]

We know that p, X-a, and Y-b are all in the ideal. So we'll use them from now on.

Let's say that some polynomial f(X, Y) is in the ideal, so f(a,b) = 0 mod p

Now f(X, Y) is in the ideal iff f(X, Y) + g(X, Y) is in the ideal, where g(X, Y) is some element of the ideal I've chosen, right?

[trail harness]

Hmmmm, that's true, and k, r > 0 if p(a, b) = 0?

[coral spindle]

So in particular, since I know that X - a and Y - b are in the ideal, I can always manipulate it so that f(X, Y) is in the ideal iff a certain degree-0 thing is in the ideal

Can you see how I might have that happen?

[arctic trail]

if p(a,b) = 0 then the "lone-some monomial" the one with k = r = 0, has to be divisible by p

Hence p(x,y) is in the ideal generated by (p,x-a,y-b) if p(a,b) = 0 (mod p)

[arctic trail]

The proof for this is simple.
the polynomial q(x,y) = p(x+a,y+a) admits a representation as a sum of monomials of the form x^k y^r (k,r >= 0) (because it's a polynomial)
so p(x,y) = q(x-a,y-a) can be written as the sum of monomials of the form (x-a)^k(y-b)^r (k,r >= 0)

by simply replacing x-> x-a and y -> y-b in the decomposition of q

[trail harness]

Yeah, this is true because X - a and Y - b are monic, then it works

[coral spindle]

Yeah exactly

[trail harness]

But how exactly this proves what we want? This says that f(x, y) is in the ideal iff a certain constant is a multiple of p, but what is this constant exactly?

[coral spindle]

Well can you figure out what it is?

I'll tell you this

Using the fact that X-a is in the ideal is almost the same as setting X to be a.

[trail harness]

Looks like division, maybe the remainder of something?

[coral spindle]

Hmm not quite

Or well now that I say that, I suppose it is technically the remainder lol

f(X) = (X-a)q(X) + r, so what's r?

[trail harness]

It's the remainder of f(x) in the division by x - a. Then r is a constant

[coral spindle]

OK I'd like to see more of your thought process btw, please talk instead of waiting so long

[coral spindle]

Yes but in particular, you can find the value of r!

Let me write this again

f(X) = (X-a)q(X) + r, so what's r?
Is there a nice value of X that you could substitute in to get the remainder r?

[trail harness]

I was trying to write the constant term of f(x) in terms of the constants in the rhs

[trail harness]

Oh, r = f(a)

[coral spindle]

OK I would really have liked to have known you were doing that!

Even if you're wrong, I wanna see what you're thinking

[coral spindle]

Exactly, yes

So ofc we have two variables

So what do we get once we reduce f(X, Y) like this?

This takes a little bit of thought

Remember what I was saying, I wanna hear your thoughts even if you're totally wrong

[trail harness]

The remainder should be f(a, b) because f(x, b) = (x - a)q(x) + r(b) dividing f(x, b) by x, then if you treat b as a variable, we get f(x, y) = (x - a)q(x) + r(y), if you divide r(y) by (y - b) we get that f(x, y) = (x - a)q(x) + (y - b)q'(x) + r', and then f(a, b) = r'

[coral spindle]

Perfect!

So we know that f(X, Y) is in the ideal iff f(a, b) is

Can you finish things now?

[trail harness]

Sure, if f(a, b) is in the ideal, bc it is a constant, p | f(a, b) and then f(x, y) is generated by (x - a), (y - b) and p

[coral spindle]

Yeah exactly

Since we know p | f(a,b) by definition, we know that f(a, b) is generated by p

So we're done!

[trail harness]

Thank you very much! My brain runs at 5 fps with algebra xD

Specifically when it involves polynomials of several variables

[coral spindle]

No worries

[tardy hedge]

I am doing a reading course in commutative algebra using atiyah macdonald

I will definitely be needing yalls help most likely lol

Actually, has anyone gone through that book? Any thoughts?

I think I am going to be covering most of the book and doing the exercises

[hidden wind]

And you can think of rings as being in some sense a sort of generalization of groups.
— Borcherds

wait... what?

[arctic trail]

what?

[static glen]

I also would not really say that.

[hidden wind]

https://youtu.be/TBtG2IldX4M?si=LvIbQ6EBM4u89TJg

skip to 30:08

[coral spindle]

borcherds makes throwaway comment referring to how rings have an Abelian group structure
people on math server lose their minds

[hidden wind]

also how he so cute

[arctic trail]

you should've added context with the message 😭

[hidden wind]

but it was funnier this way mwahaha

[arctic trail]

the quote doesn't make sense without context

[hidden wind]

[tardy hedge]

Lmao

True tho

He a cutie patootie

He’s like winnie the pooh

[arctic trail]

true

[brisk oak]

Thing I edited some time ago 😅

[hidden wind]

beautiful

[void cosmos]

quick problem verification : if K is some extension of a field F and we have F(a) = F(a^2) ( a is an element in K ) then is a algebraic over F?

why do we need F(a) = F(a^2) ?

can't i say F(a) over F(a^2) has degree 2

but a^2 is in F(a) so a^2 =a_1+b_1*a for a_1 and b_1 in F, rearrange that and there u have ur polynomial?

[coral spindle]

a^2 is in F(a) so a^2 =a_1+b_1*a ...
This doesn't follow.

Consider the field of rational functions k(X)

Clearly X^2 is in k(X) but it is transcendental over k.

[void cosmos]

i see

yeah im stupid mb

tysm

[coral spindle]

Np

[void cosmos]

yeah fuck if a is transcendtal then a in F(a^2) means that a =f(a^2) for some rational function f

and we are done cuz i can just clear denominators and that's it

is that correct?

[south patrol]

That looks good yes

Or it is just clear that F(x) != F(x^2) for transcendental x

[void cosmos]

yeah

thank you

[cloud walrus]

yeshua

[serene dune]

i mean i can show this is why the additive group underlying a ring is abelian.
but is there more satisfactory aha type explanation ?

[crystal vale]

I think this is not the correct solution because how you show under additive operation it is Abelian?

I think you need unity

[serene dune]

arrangement is different there

leading to a contradiction

[crystal vale]

What contradiction?

[serene dune]

lhs is same

u meant i should do (1+1)(a+b)

[crystal vale]

I don't see how it gives you a + b = b + a

[serene dune]

same thing isnt it ?

ab is something

ba is also something

arbitrary

[delicate bloom]

you can't use that to show 0+1=1+0 for instance

generally you're asserting x=ab and y=ba has a solution (a,b) for every (x,y) which isn't true

[serene dune]

that's vacously true coz 0 is identity

i don't need to show that

[delicate bloom]

the point is you said "arbitrary" so I just had to show one single example

[serene dune]

but i think this takes care of your concern

[serene dune]

for the thing i need to prove its still arbitrary yeah

in anyways my question was rather different
im looking for more grandiose explantion

[warm ember]

addition being commutative is an axiom

[delicate bloom]

if you let the addition of a ring be a nonabelian group then it sounds like distributivity gets messed up and you are stuck with a near-ring. I don't know if that train of thinking would be satisfying or not for ya

[crystal vale]

But if you have unity in the ring then you can avoid it

[warm ember]

o

[warm ember]

that works

[serene dune]

yeah

[serene dune]

i mean somehow u cant still avoid for rings that arent CRU apparently

[warm ember]

commutativity doesnt do anything here

if you have an element x that permutes the ring then that works too i think

[gentle crow]

is there a more widely used term for the concept i highlighted in red?

i tried to google "algebraic property isomorphisms" but not very many relevant results show up

[fresh pawn]

I need help, can any of you help with this?

[fresh pawn]

I was linked here

[rocky cloak]

I'm not entirely sure how the proof goes, but the fact that injective modules decompose into indecomposables ones, and that the indecomposable ones are exactly the injective hulls of R/p for p a prime seems relevant.

[arctic trail]

lmao

[south patrol]

Why lmao lol

[cloud lynx]

is (X,X+1) in Z[X] just Z[X]=(X,X+1)? or is it Z?

in my solution is (X,X+1)=Z but I think it must be Z[X] and not Z

[coral spindle]

Well that ideal of Z[X] contains X and X+1, so it contains their difference, namely 1.

So what ideals of any ring R contain the unit?

[cloud lynx]

Z

[coral spindle]

OK let me step back for a bit

Let R be any ring

Let J be an ideal of R

Suppose J contains 1

[cloud lynx]

yes

then J=R

[coral spindle]

Great

[cloud lynx]

so its Z[X] right?

[coral spindle]

So going back, we have that (X, X+1) is an ideal of Z[X] that contains 1

So yes, it's Z[X].

[cloud lynx]

my thoughts were similiar because because 1* g for every arbitrary g in Z[X] means that (1)=Z[X]

in the solution they also said that this is no proper Ideal. What do they mean? because its obv a Ideal

thank you Boytjie

[rocky cloak]

"Proper ideal" means ideal not equal to (1)

[cloud lynx]

ahh ok thank you

I have the Ideal (X^2+1,X+2) in Z[X] and i have to look wether its prime or normal. So (X^2+1,X+2)=(5,X+2) then Z[X]/(5,X+2) isomorph to (Z/5Z)[X]/(X+2)

in the solution they just showed that X+2 is irreducible but I think its easier to use the fact that X=-2 so the factorring is just Z/5Z

my question is why the solution dont simplify further instead of looking wether the polynomial is irreducible

doesnT*

is it because sometimes we dont know what the simpliest Ring is? so its easier when we have a polynomial to look at?

[crystal vale]

<x> is a maximal ideal in R[x], right? Because R[x]/<x> is a field

[cloud lynx]

if R is a field yes

if not then no

R[X]/(X) is ismorph to R

[crystal vale]

Yes R is for real numbers

[cloud lynx]

yes then its a field

[crystal vale]

So <x> contained in <x,2>, right?

And it is proper subset

[unkempt stream]

No, vice versa

In general, if A is a subset of B, then <B> is a subset of <A>

[crystal vale]

But 4 in <x,2> it is not in < x>

[cloud lynx]

no (X) is in (X,2)

pretty sure its a subset

(X,2)=(X)u(2)

[crystal vale]

So (x,2) = R ?

[cloud lynx]

no

[crystal vale]

<x,2> is same as { xp(x) + 2q(x) } ?

[cloud lynx]

lol wait isnt every non zero unit a generator of its ring?

so (2) over R is just (2)=R?

[coral spindle]

Yes

[crystal vale]

So (x,2) is R?

[cloud lynx]

then you re wrong (x,2)=R[X]

[coral spindle]

I'm not sure why you have to specify nonzero. Every unit generates the ring, as an ideal yes

[cloud lynx]

yes

R[x] and not R right?

Because its an Ideal in R[X] not R

[coral spindle]

It's the whole ring, for whatever ring it is an ideal of.

[cloud lynx]

ok

my brain is fried

if I have a galois group of order 4 there exists 2 groups with order 4 Z4 and Z/2Z x Z/2Z how do I know to which group the galois group is isomorphic to?

[coral spindle]

You cannot distinguish them without more information

One way would be to look at the relevant roots.

[cloud lynx]

mmh for example the galosi group of Q(sqrt(2),i)/Q

it has order 4

[coral spindle]

Well the important values are $\pm \sqrt 2$ and $\pm i$. Clearly these two pairs are permuted amongst each other, and this action is faithful.

[cloud walrus]

Boytjie

[coral spindle]

So think about what the group looks like as a permutation group.

[cloud lynx]

öhmm

what does faithful mean?

[coral spindle]

I mean it's a permutation group.

[cloud lynx]

yes so?

[coral spindle]

The elements of the Galois group are entirely determined by what they do to these values

Can you write down all the possible permutations (hint: you can)

And therefore you know what the group is

[cloud lynx]

so by looking what kind of subsets the permutations generate?

[coral spindle]

Try it and see

[cloud lynx]

sooo

y,x,c,v are my permutations

y= id

x(sqrt(2))=-sqrt(2) , x(i)=i

c(sqrt(2))=sqrt(2) , x(i)=-i

[coral spindle]

OK you don't have to list them for me

[cloud lynx]

v(sqrt(2))=-sqrt(2) , x(i)=-i

[coral spindle]

I can see you've got the right idea

[cloud lynx]

so if we look at the order

[coral spindle]

So now you should be able to see what group that is

[cloud lynx]

we have 3 subgroups with order 2

and one is just the trivial group

so Z/2Z x Z/2Z

[coral spindle]

Yup

[cloud lynx]

mmh then its actually kinda unnecessary to look to what group it is isomorph to

i thought it would be helpful to know so i dont have to determine the subgroups

[coral spindle]

But that's literally what you did, you just looked at it and saw that it's not option 1 so it's option 2

[cloud lynx]

yes but I thought we can see it easier by just looking at the Order so we dont have to determine all Subgroups by hand you know what i mean? because if G is isomorph to a well known group we also know its subgroups. so we dont have to think about the subgroups. that were my thought

my tutor said it is very helpful to know to which group the galois group is isomorphic to

[serene dune]

i think it's A|B

[gentle crow]

would "properties preserved under isomorphism" encompass "algebraic property" as defined in the screenshot?

is there a handy reference listing some common properties preserved under isomorphism?

[serene dune]

for any commutative ring R, the units of R[x] are same as the units of R

and if R is a field (here), its more convenient

[serene dune]

name story different name XD

[serene dune]

whatever u can find,
if we are talking about groups here
from cayley table to orders of each element

i mean the algebra is taking place over a set

[coral spindle]

Miz I'm not sure what you meant here, this is just incorrect if you're talking about ideals.

I can't think of what meaning of angle brackets you could be referring to that would make this true

[coral spindle]

2x+1 in (Z/4Z)[x] would like a word with you

If you mean commutative domain you'd be right.

[serene dune]

oh yeah mb

its ID

[unkempt stream]

I wrote it backwards by accident,

But my intent was that larger generating sets generate larger ideals

[long obsidian]

For a ring S and a prime p does it follow that $(S/p)_p=0$? Where the subscript denotes localization at the prime

[cloud walrus]

HausdorffT1

[coral spindle]

I can only assume you mean localising at the ideal of S/p corresponding to the prime p of S, namely just the zero ideal of S/p. So you're really asking if the localisation of an integral domain at zero is zero. So no, that's frac(S/p).

I assume you mean commutative rings. Noncommutative localisation gets narsty

[long obsidian]

Yeah I meant commutative rings. Okay thank you

[unkempt stream]

ORE LOCALIZATION JUMPSCARE

[arctic trail]

you can do it through subgroups

it's always helpful to write the diagram (I don't know the name?) of the subextensions and subgroups of the galois group

[arctic trail]

the galois correspondence thingy

I forgot the name

this

Z/4Z only has 3 subgroups (itself, 2Z/4Z, and 0) so it has to be Z/2Z x Z/2Z

[ember field]

If someone says "Consider K[x] where deg(x) = 3" then it has to mean that x is some element in a polynomial ring k[t] where t is indeterminate, no?

[coral spindle]

You can in particular see this as the subring K[t^3] of K[t] where we simply call t^3 the name x, yes

But really what's happening is we're talking about graded rings, and this person is simply telling you what the grading on K[x] is

Graded rings can be seen as a way to generalise the idea of degree.

[ember field]

And the grading is that k[x]_i = elements of degree 3*i, or no?

[coral spindle]

The grading ought to be the one such that k[x]_3 is the monomials :)

because those are the things of degree 3

[hidden wind]

oh is this Ore as in Øystein Ore?

[rocky cloak]

That's right

[hidden wind]

i’ve had a very brief look at some article of his on sone algebra stuffs or other, but i don’t understand the basic objects, that of a lattice

[toxic zephyr]

every element of Sn can be written as a product of transpositions right?

[vapid vale]

yes

[toxic zephyr]

there's probably an easy way to find the transpositions necessary to write a generic element isn't there?

[coral spindle]

There is

You know about the cycle decomposition, yes? Try using that

[toxic zephyr]

cycle decomp? if so, not by name

oh actually yes. that's what I've been doing I think

[ember field]

When you say monomials here, you just mean the elements of k[x]_1, no?

[coral spindle]

I mean that x is in k[x]_3

Because we defined deg(x) = 3

[ember field]

so maybe k[x]_i is monomials of degree i if i is a multiple of 3 and 0 if not?

[coral spindle]

That's correct

Or really it's the span of those monomials, potato potato.

[toxic zephyr]

so if we have (a1...ak) is that equal to
(a1ak)...(a1a2)?

[ember field]

but these are monomials in x, yes?

[coral spindle]

Yes

[vapid vale]

yes

nested yes

[coral spindle]

Well where does it send a_1?

[ember field]

tysm

[coral spindle]

Oh you're using right-to-left composition mb

I default to left-to-right when I see cycles...

[vapid vale]

in fact you can also show that you only need adjacent transpositions

[toxic zephyr]

interesting

I read something about decomposition into commuting cycles (I think)? is that different from decomp into transpositions? I would think so, I imagine commuting transpositions is asking too much

[toxic zephyr]

right to left is the only way that makes sense to me lol
I need my function intuition haha

[vapid vale]

well disjoint cycles commute

so yes

[toxic zephyr]

so can you write like (234) as a product of disjoint cycles?

[vapid vale]

that is a disjoint cycle

[toxic zephyr]

or disjoint transpositions I mean

[coral spindle]

No

[toxic zephyr]

okay that's what I thought

[coral spindle]

Because that would then be a disjoint cycle decomposition

And that's unique

[toxic zephyr]

oh!

[coral spindle]

And you've already written it as (234)

[toxic zephyr]

okay that's really good to know, and makes a lot of sense

[ember field]

why is this not a grading tho?

[coral spindle]

Maybe it's worth proving that the disjoint cycle composition is unique. I'm surprised you haven't seen this bc it's very important for the symmetric group

[toxic zephyr]

I probably did but it was over 2 years ago. I'm studying up algebra before I start a grad course in it in a few weeks.
but intuitively it makes sense.

[coral spindle]

The i is the degree by definition.

And we required deg(x) = 3.

So we need k[x]_3 to contain x.

That's precisely what is being required when we say deg(x) = 3

[toxic zephyr]

rn I'm trying to prove that Sn=<(12),(23...n)>
my thought process is we just need to show every transposition is in that subgroup. I thought maybe writing the second one as a product of transpositions might help, and I'm trying it concretely in S4

[vapid vale]

is it (23...n) or (123...n)?

[coral spindle]

Hint: what does conjugation do in S_n

[ember field]

I thought this meant x was like some degree 3 polynomial like x=t^3+t and we were taking the ring generated by x which is K[x]

[coral spindle]

This is what we've been discussing the whole time, I said this up here.

[toxic zephyr]

it says (23...n) but idk could be a mistake

[coral spindle]

I think the 'classical' result is (1 2) and (1 ... n) but if my quick in-the-head check is right (might be wrong tbh) this also works

[vapid vale]

yeah i think you can get 123..n from 23..n

i think 12...n is much more illustrative

[toxic zephyr]

oh yeah I was able to show (rather, by messing around I found) that (12...n)=(12)(2...n)

[void cosmos]

1. is true

[toxic zephyr]

is it something obvious? I mean we only need to think about conjugation by transpositions right?

[coral spindle]

well why not start by conjugating the transposition by the other thing

You know what kind of thing that produces...

[vapid vale]

i would think about adjacent transpositions

[toxic zephyr]

yeah I did find that (12)(234)(12) (which is conjugation) gives (134)=(14)(13)
but it didn't ring any immediate bells for what that does

[void cosmos]

I think he is saying that two elements are conjugate iff they have the same cycle structure. so if you wanna show somehting is of a specific form (say transpositions) then u should conjugate it with them

[coral spindle]

Not the long cycle by the transposition

Try the other way around

Conjugate the transposition by the long cycle

Because (and this is important) conjugates of something with a certain cycle decomposition will have the same kind of cycle decomposition, i.e. the lengths of the cycles will be the same!

You've been saying you want to get all the transpositions, so...

Is there a chance you don't know the trick for calculating conjugates in S_n

[toxic zephyr]

OH is this like the alternating group is normal?

[ember field]

degree appears to me to mean two different things, the degree of the element x inside some polynomial ring or it means its in the 3rd part in the grading

[coral spindle]

Uh sure this is a corollary

[toxic zephyr]

or did I miss the mark lol

[ember field]

when someone says x is of degree 3 Im not sure which they mean

[coral spindle]

Is this a no, taylor?

[toxic zephyr]

okay so conjugating the long cycle by the short cycle gives another transposition.
and repeated conjugation gave all the transpositions (1a)

[coral spindle]

Yes indeed

If you know the trick this is immediate

[toxic zephyr]

oh yeah no not really

[cloud walrus]

Boytjie

[coral spindle]

So in particular

[cloud walrus]

Boytjie

Boytjie

[coral spindle]

So you've got all transpositions (1 a)

See if you can do something similar to get all (a b)

[void cosmos]

(1a) generate S_n as a varies

[coral spindle]

Extension task: use this to describe the conjugacy classes in S_n

[void cosmos]

eigentaylor is done

cuz the <{(1a) | a in {2,3,..,n}}> contains the transpositions

(ij) = (1i)(1j)(1i)

[coral spindle]

That's what Taylor said she wanted to try showing, yes.

[void cosmos]

no i was just asking for this cuz i didnt knwo why she had to do that

but ig it was an extension exercise ig

now i have a stupid question after eigentaylor is done

about fields 😄

[toxic zephyr]

oh that makes perfect sense. so if I conjugate (1b) by (12) then that should give (2b).
and then conjugation of (12) by (1a)(2b) gives (ab)?

not sure if that's more convoluted than it needs to be

[toxic zephyr]

but this makes so much sense. reminds me of change of basis a little bit. unsurprising since both involve conjugation

[vapid vale]

i think its illustrative to go between these various generating sets of Sn

imagine one of these toys but only one circle, and the rotating part only allows you to swap 2

that is essentially (12) (swapping) and (12..n) (spinning the circle)

so its clear that thats jsut the same thing as all adjacent transpositions

and its not hard to visualize any transposition

[coral spindle]

This is the most based explanation

[old sphinx]

What is cosets and isomorphism

Need help to determine its defintions

[topaz solar]

Don’t forget about permutation matrices

[toxic zephyr]

oh yeah of course lol

[topaz solar]

Sometimes analogy is more than analogy

[fading summit]

someone help please... am i stupid or is it too late in the evening or is this actually kinda difficult...

Let G be a group and let a, b ∈ G be elements with o(a), o(b) < ∞.

1. Prove that for all k ∈ Z, o(a^k) divides o(a).
2. If ab = ba, prove that o(ab) divides lcm(o(a), o(b)).
3. Give an example of a concrete group G and a concrete element a ∈ G with o(a) = o(a^2).
4. Give an example of a concrete group G and two elements a, b ∈ G such that o(a), o(b) < ∞, and o(ab) ≠ lcm(o(a), o(b)).

[willow cipher]

1. think about the subgroup generated by a
2. think about how (ab)^n = a^n b^n given the condition
3. think of the simplest kind of groups
4. given youve done 2, think of a group that has two elements that dont commute with eachother. matrix groups might be useful to think about

[fading summit]

oh wait yeah, #1 is easymode. but for #2 and #4, my brain is cramping on what lcm(o(a), o(b)) means? i mean, i cant intuitively think it through...

and for #3, i suppose G = Z_2 and a = [2]_2 would work? ...is it really that simple?

[willow cipher]

intuitively u want a value n such that both a^n and b^n vanish since (ab)^n = a^n b^n. so we want both a^n = 1 and b^n = 1 at the same time. one thinks instantly about the lcm of their orders

[willow cipher]

i would suppose the intent of the question was to find a non-trivial element a in G

since every group G has that property where the order of its identity is equal to the order of the identity squared

note [2]_2 = [0]_2 which is the identity in Z/2Z

[fading summit]

lol yeah...

okay thanks, i'll think it through some more

[chilly radish]

nontrivial examples would be an element of order coprime to 2 (e.g. 3) or infinite order

[true summit]

Imma just throw this here too in hopes someone has a trivial answer. Is there a neat way I can figure out whether a 20 cycle and B=(1 4)(2 3) generates S_20 or a subgroup of that without brute force looking for a single transposition?

[true summit]

It’s actually this but instead of 2 it swaps 4 at once

Trying to figure out the logic that that also generates all of S_20

I wouldn’t be surprised if there was a way to find A_20 as a subgroup then use Lagrange to double that to S_20 or something but I don’t know it

3 weeks into algebra has me lostttt

[wild jasper]

I'm tyring to solve 17(a) but I have a few questions. According the Fundamental Theorem of Galois Theory, there is only a one to one correspondence between isomorphsms K -> X where X is an algebraic closure of F contaning L, not just embeddings in to an algebraic closure of F.
Also, what does one to one correspondence mean in this case? Injective or bijective?

[south patrol]

One to one correspondences are bijective

[wild jasper]

thanks

[hidden wind]

i wish everyone would just adopt bourbaki’s terminology already

[wild jasper]

I got the other question cleared up by rereading the proof of the Fundamental Theorem.

But I still have no idea where to start, so would appreciate a hint

[hidden wind]

and retroactively rewrite history/old textbooks while we’re at it

[brisk oak]

Take an arbitrary element from the galois group. What happens if you apply it to this "N_(K/F)(a)" ?

[wild jasper]

ahha

thanks!

[wild jasper]

Can L/F be taken to be a finite extension?

[true summit]

Gonna bump this :/

[wild jasper]

I want to define N_{K/F}(alpha) = \prod_{\sigma \in Gal(K/F)} sigma(alpha) and since these are coset representatives we are told to show that it is well defined, but don't really know what that means in this case...

[brisk oak]

Oh.. this will only work if K/F is normal I guess

[wild jasper]

I'm tyring to show that if K/F is a Galois extension with Galois group isomorphic to the Klein 4-Group, then K=F(sqrt(D_1),sqrt(D_2)) where D_1, D_2 and D_1D_2 in F but are square free in F.

I have tried looking at which subfields K must have, but can't draw any conslusions from that

[brisk oak]

Hmm well subfields of degree 2

[wild jasper]

indeed

[brisk oak]

So of the form F(sqrt(x))

[south patrol]

I believe you need characteristic not 2 but yeah

Actually maybe that is just for the converse

[wild jasper]

ah! So if K was a splitting field of an irreducible polynomial it would only have subfields of degree 4

[south patrol]

Lol

[wild jasper]

?

[dim widget]

yah

[south patrol]

Deleted message dw

[wild jasper]

F has char != 2

[south patrol]

Okay cool

[wild jasper]

so is my reasoning correct?

[dim widget]

Every finite galois extension is the splitting field of an irreducible polynomial

that's one of the definitions

[wild jasper]

In part a I shoed that the klein 4-group was the Galois group of (x^2-D_1)(x^2-D_2)

[south patrol]

Well you'll want to just unwind the Galois correspondence and consider subfield corresponding to nonzero elements

And play around with that

In fact that is essentially the whole solution oops

[wild jasper]

now I am supposed to show that if an extension K/F has the klein 4-group as its Galois groupo, then K=F(\sqrt{D_1},\sqrt{D_2})

[south patrol]

Yes

[dim widget]

a field can be the splitting field of many different polynomials

[wild jasper]

so I have that the subfields are extensions of order 2, then, K must be the splitting field of a polynomial with two irreducble factors of degree 2?

[gusty swallow]

hey guys my teacher in my linear algebra class mentioned this formula for permutation matrices and how it comes down to something with the symmetric group. I know some abstract algebra and tried to prove it but got nowhere with strong induction. The formula is that given some $\sigma\in S_n$,
$$
\mathrm{sgn}(\sigma) = \prod_{1\leq i<j\leq n}\frac{\sigma(j)-\sigma(i)}{j-i}
$$
and sgn is 1 if $\sigma$ is even and $-1$ otherwise

[cloud walrus]

BigBoyConst

[south patrol]

Well ig, though that is not needed

The heart of this is seeing what degree 2 extensions of F look like

[wild jasper]

yes

I just don't see it

can you point me to a theorem I should look at? (Aside from the Fundamental theorem)

[dim widget]

can you describe what degree 2 extensions of F are when F has characteristic not 2

[wild jasper]

they are F(sqrt(D_1)), F(sqrt(D_2)) and F(sqrt(D_1D_2))