groups-rings-fields

[dull ginkgo]

If M in End(M^n) and v in M^n where Mv = 0, then M admits an adjugate Adj(M) so Adj(M)Mv = det(M)v = 0

Does that still imply det(M) = 0 if v isn’t the identity of M^n

[shell pilot]

The answer says that $Aut(Z_1)\approx Aut(Z_2)$ but $Z_1={0}$ and $Z_2={0, 1}$ so it is not one-to-one which means it's not an isomorphism right?

[cloud walrus]

Soap_Opera

[dull marsh]

What's not one-to-one?

[shell pilot]

Any function (mapping) between $Z_1$ and $Z_2$

[cloud walrus]

Soap_Opera

[coral spindle]

Yes, but this isn't the claim

[dull marsh]

Yeah, so there is no isomorphism between Z1 and Z2

[coral spindle]

Nobody said that Z_1 and Z_2 are isomorphic. The question is that Aut(Z_1) and Aut(Z_2) are isomorphic.

These are very different statements.

[dull marsh]

This is the first condition of G and H, now consider their automorphism groups

[coral spindle]

In fact they specifically say in the question that G and H are not isomorphic!

[shell pilot]

Ah, good point. Thank you @coral spindle and @dull marsh

[coral spindle]

Remember the mantra!

Read the friendly question!

The question is your friend, and friendly means nothing else!

[shell pilot]

Haha, so true

Is there a way to calculate the number of automorphisms possible for a group?

For a permutation group, we can just use n! where n=# elements right? But what about other groups like Z_10?

[coral spindle]

There is no easy general method that I'm aware of.

[shell pilot]

e.g. in Z_1 since the only element is 0, is there only 1 possible automorphism which is the identity to itself?

[coral spindle]

As it happens calculating the number of automorphisms of Z_n is extremely easy because of results in elementary number theory

Well. It's easy in the sense that we know a straightforward way to calculate it. Ofc calculating the number of automorphisms of Z_123217387957983721847329714983271463962785963872164987321 is actually quite lengthy

[tardy hedge]

bruh^

u really went for it

[shell pilot]

lol

[coral spindle]

I think it underlines the point

[coral spindle]

For a permutation group, we can just use n! where n=# elements right?
What do you mean by this?

That's rarely the number of automorphisms of a permutation group

[shell pilot]

There are only n! ways of permuting n objects so that is the number of automorphisms available in a permutation group. I'm just asking obviously

[coral spindle]

No this is false.

[tardy hedge]

^that was immediate

You sent that so fast

[coral spindle]

It's something that I know well

[tardy hedge]

Hahaha nice

[coral spindle]

Here's the issue here.

Every group is a permutation group.

[languid trellis]

The number of set automorphisms of {1, ..., n} is n!

[coral spindle]

This is a fact called Cayley's theorem

[shell pilot]

Yes, I actually just learned that. Every group is isomorphic to a permutation group

[tardy hedge]

when i learned cayleys theorem i was like yooo!!! cool!

[vapid vale]

not for n=6

xdd

[coral spindle]

No that's not the statement hk

But yes I'm about to mention this

Even if we take the group of all permutations on n elements -- the symmetric group S_n -- we still don't have Aut(S_n) has n! elements

Let me see if I can remember the complete statement

I believe for n >= 3 and n =/= 6 this is true

But for n=2 this is obviously false, and for n=6 there is a very mysterious (to me) phenomenon

[shell pilot]

Ok, but it's true for many permutation groups?

[coral spindle]

No, it is almost never true.

I can embed a group as a permutation group on as many letters as I like

S_n is a subgroup of S_n+10000

But it has only n! automorphisms, not (n+10000)!

(when n>6)

Remember Cayley: any group is a permutation group.

[shell pilot]

But if the question says what is the order of Aut(S_n) it would be n!. But if it asked what is the order of Aut(S_n+10000) it would be (n+10000)!

[coral spindle]

Provided n>6 yes this would be correct

But this is not what is going on

If I have some permutation group -- i.e., a subgroup of S_n -- it will typically not have n! automorphisms. This is what you were claiming (as written) and it is false.

[shell pilot]

I see what you're saying...I was just saying a general n lol. Not the giant permutation group itself

[coral spindle]

What do you mean a general n

[shell pilot]

Like n could be 1, 2, 3, 4, 5, .......

Well

6+

Or >6

[coral spindle]

And...?

Well this doesn't really matter.

[shell pilot]

|Aut(S_7)| = 7!

[coral spindle]

Okay

[shell pilot]

Same for 8, 9, 10, ....

[dull ginkgo]

Sylow subgroups arriving for silly and fun

S_5 has 6 Sylow-5 subgroups which S_6 can act upon i think

@chilly radish one last question, does the fact that if a matrix K acts on, for R-module M, M^n with nontrivial kernel, then does that mean det(M) = 0, or is that something special for vector spaces

[chilly radish]

No idea, actually

[rotund aurora]

are you asking if K has nontrivial kernel then det K=0?

[coral spindle]

I think we can cook up an example

Let me think

[mighty kiln]

Rings with zero divisors?

[rotund aurora]

the adjugate matrix still makes sense

[mighty kiln]

If rs = 0 then r acts on R with kernel at least (s)

[rotund aurora]

if the ring is commutative

I think you can only say that det(K) is not a unit

well

consider multiplication by 2 in Z/4Z

[coral spindle]

Consider the ring $R = \bZ/4\bZ$ and the module $M = R \oplus 2R$ then note that the map multiplying by $2 \colon M \to M$ has a nonzero kernel, since its image is $2R \oplus 0$

\bigskip
So the diagonal matrix $\operatorname{diag}(2, 1)$ has determinant $2 \neq 0$ but nontrivial kernel on $M^2$.

[cloud walrus]

Boytjie

[coral spindle]

Dammit croq I was writing that up!

And yeah nvm I literally didn't need to look at M^2

My thought process here was that we need something with a nontrivial maximal ideal

Then some module which has structure not viewed by the residue field

Yeah lol

We all got there eventually

[rotund aurora]

btw Arki was first

I was scrolling up so didn't see that you mentioned it @mighty kiln

[coral spindle]

Arki too smart, too op

[mighty kiln]

Fox supremacy

[coral spindle]

This is furry fascism

leave the catboys be

[dull ginkgo]

What if the adjugate is 0

[rotund aurora]

the formula is A * Adj(A)=det(A)*I, it works over any commutative ring iirc, no division is involved in the proof

so if Adj(A)=0 then det(A)=0

[dull ginkgo]

Yes but that makes it so i don’t know if i can gauruntee the existence of a v such that Mv = 0

Or the existence of such implies det(M) = 0

[languid trellis]

Boytije catboy confirmed?

[coral spindle]

I'm not a catboy

I'm a cat MAN

[languid trellis]

Worst superhero ever

[primal beacon]

the two are equivalent

[last locust]

Hi, hope you’re doing alright.
I was solving problems from menini 2004 and got a little stuck on this one

[void cosmos]

nice

[last locust]

I solved it thinking about $\langle x\rangle$, it is isomorphic to $\mathbb{Z}/m\mathbb{Z}$, say the isomorphism is $f$.

[cloud walrus]

𝕋℟𝕀𝕍𝕀𝔸𝕃

[void cosmos]

ok

[last locust]

Because the gcd(m,n)=1, the equation $ny’\equiv x’ (mod m)$ (where x’ is a representative of $f(x)$) has solutions for y’. Therefore $y=f^{-1}(\overline{y’})$ is a solution.

[cloud walrus]

𝕋℟𝕀𝕍𝕀𝔸𝕃

[last locust]

My question is: is there any other more direct way of proving the statement?

[dull ginkgo]

For general modules?

Over a commutative ring?

For vector spaces they are

[last locust]

I feel like this is not the intended way for proving it

[sly frost]

slightly more direct, but this is fine

since m and n are relatively prime you have a, b such that am+bn = 1, so bn = 1-am, so (x^b)^n = x*x^-am = x

yours is aesthetically better I'd say

although in fact ⟨x⟩ is isomorphic to Z/mZ where x maps to 1

so you're solving ny = 1 mod m

[last locust]

That sound more like it, thank you

[last locust]

Oh yeah, that’s true, I didn’t notice haha

[sly frost]

but this is really the same thing - solving bn = 1 mod m means you have integer a such that bn = 1 + am

and ⟨x⟩ is isomorphic to Z/mZ where x^n maps to n

[last locust]

Yeah, I realize now I wasn’t that far, thank you again

[primal beacon]

Okay if adjM=0. Let k be the rank of M, take a non zero determinant sub square matrix N of rank k(there has to be one), let I be the set of indices associated to it. Since k<=n-2, you can add a column and a row to it

call the new matrix N and the new index set J

then det N = 0 but it has a non zero minor so that adj N != 0, finally take any v thats zero outside J but non zero in J such that adj N. v|_J is non zero (there exists one since adj N is non zero). Finally N.(adjN.v|_J)=0, by extension M.v=0

[primal beacon]

Okay that argument is not complete it's not clear why M.v=0, the reason is that if you take your initial N. You can add any row and column to it, changing the index set J. Suppose you want to know if your i'th entry (in M.v) is 0, then if i is in I you are good, otherwise simply add the i'th column and row to N and do the same argument

[viral halo]

In order to show that the fields Q[x]/(x^2-2) and Q[x]/(x^2-3) are not isomorphic is it enough to just note that the multiplication is different in them? In Q[x]/(x^2-2), if we let theta be a root of x^2-2 the multiplication looks like (a+b(theta))(c+d(theta))=ac+2bd+theta(ad+bc), while in Q[x]/(x^2-3) it evaluates to ac+3bd+theta(ad+bc). Or is some other property needed?

[void cosmos]

no

[primal beacon]

the former has discriminant 8 while the latter has discriminant 12, that's probably a clearer way of saying it

[viral halo]

I haven't learned about that 😦

[primal beacon]

then why is the multiplication by a generator an invariant?

[coral prawn]

(Asked this yesterday but it got buried) I just worked out some proofs of my own for these but I’m not entirely sure if they’re correct - are these statements true?
(1) Given a free group G and a subgroup H, if G is generated by generators g_a and H by h_a, and if g(h_a)g^-1 is in H for all g in G, then ghg^-1 is in H for all g in G and h in H
(2) If (g_a)h(g_a)^-1 is in H for all h in H, then ghg^-1 is in H for all g in G and h in H
(3) (both (1) and (2)) If (g_a)(h_b)(g_a)^-1 is in H (for all g_a generating G and h_b generating H), then ghg^-1 is in H for all g in G and h in H
And lastly, is there any efficient way to compute quotient groups knowing just the generators of G and H (assuming H is normal)? It would seem there must be if (1)-(3) are true.

[primal beacon]

in your argument you are assuming that a square root of 3 has to get sent to a square root of two, but clearly that can't happen since the morphism is a morphism of fields

Instead suppose there is an element a+b.sqrt(2) such that (a+b.sqrt(2))^2=3 and find a contradiction

[viral halo]

oh ok

[primal beacon]

because an isomorphism would mean there is a square root of 3 in Q(sqrt(2))

[coral prawn]

Essentially this just amounts to saying if we know only the generators of G and H, we can find if H is normal in G

[viral halo]

I think I got it,ty

[primal beacon]

np!

[primal beacon]

@dull ginkgo I wrote an argument there

[dull ginkgo]

Thank u

[shell pilot]

What is Aut(Z_1) and Aut(Z_2) and why are they isomorphic?

[coral spindle]

OK so Aut(G) is the group of automorphisms of a group G, I hope this is all clear.

Let's try and work out Aut(Z_1)

What're your thoughts. Talk me through it. What comes to mind.

Just like – what are we looking for, what might we do to find it

🦗

Just gonna shout out to the void here...

!noans

[flat tree]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[tardy hedge]

Boytjie what is your research area?

[coral spindle]

I do representation theory of finite groups, in particular I care about things related to the McKay conjecture and I look at some finite groups of lie type too

[tardy hedge]

Cool

[tropic spade]

In the definition here and the statement of (2) for jordan holder

Should these be proper subgroups?

I'm thinking if not, i can just repeat subgroups in (2) to get a counterexample?

[coral spindle]

I suppose so but this is implied by "... are two composition series ..."

since the trivial group is not a simple group

(I.e., N_i+1/N_i being simple means that in particular N_i+1 is not N_i)

This is why the trivial group is excluded from the definition of a simple group btw.

[tropic spade]

Okay I see it now. So even if I picked a composition series, repeating any subgroup more than once to try and produce a new composition series violates the simplicity part of the definition which forces proper inclusions?

[coral spindle]

Yeah that's right

Like it's the same thing with 1 not being prime lol

[tropic spade]

Ah that makes sense. I was getting really turned around by this worry lol. Thanks!

[coral spindle]

No worries. Now if/when you see simple modules you'll be ahead of others in remembering that 0 is not a simple module :)

[acoustic igloo]

is it hard to write a permutation in a permutation group as a product of generators?

[coral spindle]

Of which generators? Can you be more specific about what you mean?

[acoustic igloo]

maybe in some cases it will be easier than others

generators of the permutation group

[coral spindle]

Your question doesn't seem to have anything to do with permutation groups so it confuses me why you're referring to them

OK let's put it this way

The moves of a rubik's cube generate a group of permutations on the squares of the cube

Now if I have scrambled a cube, solving it is the same as finding a way of expressing that permutation in terms of the generators.

That's pretty hard, isn't it?

I think that answers the question

[sly crescent]

How do we know that there is only one permutation that can solve a given scramble?

[coral spindle]

That's the definition of a scramble. That is precisely a permutation.

[sly crescent]

Update: I figured it out

[barren sierra]

Inverses in a group are unique

[coral spindle]

The rubik's cube group acts faithfully on the squarelets of the cube.

So the element is described by the permutation.

Man I miss Wew. He hated the term 'faithful.' What a king.

[acoustic igloo]

can you tell if a particular permutation is an element of a particular subgroup of a permutation group?

[sly crescent]

I was going to say that there are 4 edge pieces and 4 corner pieces of each color, and they’re interchangeable, but then I remembered that they’re all attached to other colors

[coral spindle]

Given what description of the subgroup?

[acoustic igloo]

hmmm

[barren sierra]

This is in general a hard problem even if you know the generators of the subgroup

[coral spindle]

Otherwise you're basically asking "how do I tell if something is in a set"

[sly crescent]

He returned recently

[acoustic igloo]

i was thinking of generators of the subgroup

[sly crescent]

#alg-top-geo-top message

[coral spindle]

No this is a bit misleading, like it really depends on what the description of the subgroup is. Like if you are given some straightforward predicate for it then this is easy.

But yeah like as you alluded to Spamakin

If you just have generators, it's extremely difficult yes.

[barren sierra]

If you know just the generators I'm pretty sure it's hard

I'm forgetting if it's undecidable hard or not

[coral spindle]

I think there might be some absurd algorithm that can do this in MAGMA with relative speed (for permutation groups specifically)

[coral spindle]

Well we are assuming finite permutation group lol

[barren sierra]

Oh right

[coral spindle]

But yes the infinite analogue has all the hallmarks of an undecidable problem

I'm not sure if this comes directly from the undecidability of the word problem

Maybe

But I can't see immediately how to derive it

(Word problem for finitely presented groups*)

[acoustic igloo]

thanks

[coral spindle]

ChiliLion's been cookin' for a while I'm excited to see what they have to say

[terse idol]

oh oops

[coral spindle]

Haha

[terse idol]

im just trying to work through a problem im having until im satisfied with asking it

[acoustic igloo]

are there lots of big finite groups that have no subgroups?

like prime cyclic groups

[terse idol]

i mean like, aside from the trivial one?

[acoustic igloo]

yes

[sly crescent]

And themselves?

[acoustic igloo]

yeah

[terse idol]

if so then yeah all of the cyclic ones with prime order shouldn't have any subgroups

[coral spindle]

No

[terse idol]

oh really?

[coral spindle]

I want you to try and work out all the groups that have no (nontrivial) subgroups

[acoustic igloo]

😮

[coral spindle]

You've pointed out that all the cyclic groups of prime order have no nontrivial subgroups

Are there any more?

Hint: if you have an element g of G then <g> is a cyclic subgroup of G

[acoustic igloo]

ohhh

so if G is not cyclic then <g> is a proper subgroup of G

[coral spindle]

That's right

So we conclude what from this?

[terse idol]

yknow what im just gonna ask it so someone can give me a direction to start
one of the questions i got wrong on a homework was as follows:
Let p be prime. Find a subgroup H of $G = \mathbb{Z}{p^2} \oplus \mathbb{Z}{p^2}$ s.t. G/H is isomorphic to $\mathbb{Z}{p} \oplus \mathbb{Z}{p}$
My initial answer was H = $Z_p \oplus Z_p$, but that's not even a subgroup of the former (no clue what i was cooking), so im reattempting

[cloud walrus]

ChiliLion

[acoustic igloo]

only cyclic groups can have no subgroups

composite ones have subgroups

[coral spindle]

You're right to say that it's not a subgroup of G, but perhaps it is isomorphic to a subgroup of G. Think about this maybe?

[acoustic igloo]

so it's only prime cyclic groups

[coral spindle]

Yeah exactly. G must be cyclic, and the only cyclic groups with no proper subgroups are those of prime order

In fact these are the only groups of prime order, but we don't need to prove that yet

So it's really like

there are very very few groups with this property, and they're all kinda boring lol

[acoustic igloo]

right, wow, thanks

[coral spindle]

Like just to put this in perspective

How many primes are there of size less than 2000?

Like

not that many right

[acoustic igloo]

right

[coral spindle]

certainly less than 1000 lol

Now

Just groups of order 1024

there are 49,487,365,422

In fact over 99% of the groups of order less than 2000 are of order 1024

(Counted up to isomorphism of course)

[acoustic igloo]

😮

[coral spindle]

So in a very real way there are very very few groups with this property

[dull ginkgo]

Power of 2

makes sense

[coral spindle]

Yeah exactly

Lookin at those 2-groups

It is conjectured that, in plain terms "almost all" finite groups are groups of order 2^n for some n, but I don't believe this has been resolved.

I wouldn't really have my finger on the pulse though.

[dull ginkgo]

99% of the work in that sounds like trying to characterize "almost all"

[coral spindle]

Oh no that's the straightforward part, like they're just saying that the proportion of finite groups less than n that are 2-groups tends to 1

It's just really hard to work with asymptotic bounds on groups

Like afaik most of the arguments for showing bounds just go "here's a bunch of groups we constructed"

[topaz solar]

Well, you’re looking for a word which spits out 0 yeah?

[coral spindle]

I mean yes but the problem is given some set of generators of a subgroup of a group, right

So it's a word but in a different set of generators

Sorry I need to make myself clearer

[terse idol]

if i construct some valid G1, G2 which are both isomorphic to $\mathbb{Z}{p} \oplus \mathbb{Z}{p}$ and which I can direct sum then is $G1 \oplus G2 \cong \mathbb{Z}{p^2} \oplus \mathbb{Z}{p^2}$?

[cloud walrus]

ChiliLion

[coral spindle]

I'm imagining we have a finitely presented group, a secondary finite set of generators of a subgroup, and a word that we want to decide membership of the subgroup for

[coral spindle]

No.

[terse idol]

hmm ok

[coral spindle]

Z_p (+) Z_p is not Z_p^2.

This is very important.

[terse idol]

right, makes sense

[coral spindle]

Look for subgroups of Z_p^2. What are subgroups of Z_n in general?

Maybe even write out some small examples like Z_4

[topaz solar]

Well that’s exactly asking computability of spitting out a word that sends an element to 0, sorta?

[coral spindle]

It may even be simpler to just look at G = Z_p^2 instead of Z_p^2 (+) Z_p^2. The direct sum is a red herring.

[coral spindle]

I don't really see how

Oh no wait

I misinterpreted you

[topaz solar]

Well like, if we can say g is in <x1,…,xn>, then that means there’s certainly a word that gives you g

[coral spindle]

yes that's just computing the inverse, sure

[topaz solar]

So it’s like, not too far from decidability of word problem for <x1,…, xn, g>?

Maybe

That gives a c.e. thing rather than decidable tho ig

[coral spindle]

Yeah I see

Well it's a start

[delicate orchid]

will do boss!

[coral spindle]

WEW!!!!!!!!

WEW HOW ARE YOU

[topaz solar]

For anything in that subgroup, we’d have an algorithm to spit out a word for the inverse from it

[delicate orchid]

sorry I was busy for the past 3 months trying to find subgroups of Z_n

[topaz solar]

I think?

[coral spindle]

WEW YOU HAVE AN ALBUM?????????

[delicate orchid]

I have like 7

[dull ginkgo]

bro makes music

[coral spindle]

Wew be honest with me. Wew. Have you been to any conferences lately.

[delicate orchid]

two conferences

[coral spindle]

My god

[dull ginkgo]

unfortunate, you could be awesome and go to 0

[delicate orchid]

what do u know that I don't

[coral spindle]

Idk

I've only been to one conference recently so I'm losing ig

But I did see a really good talk on fusion systems at it

[topaz solar]

Hmm that might be too much what I’m suggesting, since we just want decidability of being in the group, which implies a partial algorithm for inverses, but if we have a group with undecidable word problem we shouldn’t be able to determine decidably if a word is equal to any word in a subset of the generators necessarily?

[coral spindle]

I was not invited to give a talk :(

Sad times

[delicate orchid]

I ain't gonna give a talk I'm too scared cold and alone in these dark times

[coral spindle]

I love giving talks. It allows me to stroke my massive throbbing ego

[dull ginkgo]

i haven't been to a conference :3

[delicate orchid]

what did they discuss

[terse idol]

i mean if we're looking at subgroups of $Z_{p^2}$ iso. to $Z_p$, every single element of $Z_{p^2}$ except for multiples of $p$ is a generator, and the ones that are multiples of $p$ generate groups of order $p$(?)

[cloud walrus]

ChiliLion

[dull ginkgo]

lol

Okay so S_6 having a nontrivial automorphism isn't as interesting as the other ones not

I wonder if there's a way to narrow it down beyond it having 6 sylow-5 (cyclic) subgroups

[coral spindle]

It really went over my head... but it was essentially a description of a classification result. The guy kinda motivated it by implying fusion systems might help with the classification of FSGs but unfortuantely relied on the classification of FSGs during it

It was a really impressive talk, I wish I understood it better

[delicate orchid]

that's baby shit. PSL_2(4) is isomorphic to PSL_2(5)

[dull ginkgo]

I'm not a group theorist you punk

[delicate orchid]

society if we could find exotics without CFSGs

[coral spindle]

That is pretty fucked up tbh

[dull ginkgo]

okay that's actually kind of neat

[coral spindle]

THE DUALITY OF MAN

beautiful. mods can we pin this

[delicate orchid]

yeah the derivation of the iso is a lot like how u find the exceptional outer automorphism of S_6

it's some bizarre combinatorial coincidence

[coral spindle]

Ah so Sylow bullshit

?

[dull ginkgo]

wait isn't that also iso to A_5

[delicate orchid]

yes

don't worry about that part sweaty

that part is trivial to see if you draw an icosahedron then take so many edibles it'd kill an elephant

[coral spindle]

I'm so glad you're still the same, wew

[dull ginkgo]

or amphetamines if you're studying sheaves

[delicate orchid]

here's a page of cursed bullshit
https://en.wikipedia.org/wiki/Exceptional_isomorphism

[dull ginkgo]

Using the sylow-conjugation action is a silly way to blast through problems using stupid ways

[coral spindle]

See this is why people say that finite groups suck

I just wanna do character theoryyyyy is that so badddddddd

[delicate orchid]

ok then do character theory lil nerd

[terse idol]

ok wait so if i have $H = <(p, p)>$, would tht work?

[delicate orchid]

I got the cde triangle tattooed onto my forehead

[cloud walrus]

ChiliLion

[coral spindle]

What do you think? What's the order of this group?

[delicate orchid]

what group are we trying to find subgroups of

[terse idol]

orig. question

[coral spindle]

That was a bad prompt

What I mean to say is

Let's see, let's work it out

What's the order is a good start

Maybe write down the elements in a different way

[terse idol]

H has order p

hmm

[dull ginkgo]

alternating groups really like to be projective special linear groups

[coral spindle]

Ah but we're looking for a group of order p^2 aren't we

[dull ginkgo]

sounds like a sylow action

[terse idol]

we are...

[delicate orchid]

only up until A_9 or somesuch

[coral spindle]

This is a good start so keep at it

You're almost there

[dull ginkgo]

Proving it's the END sounds like the worst

[delicate orchid]

why yes, they do induce isomorphic fusion systems for every prime - what an astute observation!

[dull ginkgo]

what's a fusion system, papa wew

[delicate orchid]

nah it's just an order argument

[coral spindle]

OK I talked about fusion systems once already today that's enough

no more thanks

[delicate orchid]

boytjie what's our supergroup here please respond

[coral spindle]

And also it is 1 fucking am wtf

[coral spindle]

who knows wew, maybe u should spend another 3 months

smell u later

[delicate orchid]

what a nerd

[dull ginkgo]

,rotate you fuck

[dull ginkgo]

oh, neat

[delicate orchid]

this isn't including the saturation axioms but I am not subjecting you to that bullshit

[dull ginkgo]

thank you papa wew for the kuhnowledge

[chilly radish]

What's Inj

[delicate orchid]

set of injective maps

[chilly radish]

And Hom_S

[delicate orchid]

transporter set :trollface:

[chilly radish]

As group homomorphisms?

I bet Bobby defines this stuff earlier in the book

[delicate orchid]

Hom_S(P, Q) := {c_s \in Inn(S) : c_s(P) <= Q}

[chilly radish]

I'll just take a gander

[delicate orchid]

he does, and yes

[chilly radish]

Ah ok makes sense

[delicate orchid]

all fusion systems are subcategories of FinGrp

[chilly radish]

Ye

[dull ginkgo]

omg hi ShiN how is u

[chilly radish]

Hello

We've spoken today lmao

[delicate orchid]

lol

[dull ginkgo]

i don't CARE

[chilly radish]

I'm tired

[dull ginkgo]

I'm going to work on jacobson in a bit

still processing the matrix stuff

[delicate orchid]

yeah that matrix multiplication is hard!

[dull ginkgo]

End(K^n) ~= M_n(End(K)) still making me think

and the K-linear endomorphisms of ring K as a left-module being iso to it's opposite ring

[delicate orchid]

yeah

this makes most sense over F_1 in the burnside category

[dull ginkgo]

F_1

[rotund aurora]

Can there exist a torsion, finitely generated, finitely presented, infinite group?

[delicate orchid]

https://en.wikipedia.org/wiki/Tarski_monster_group

I think they're finitely presented

[rotund aurora]

why do you think so?

[delicate orchid]

yeah good point

hmm

yeah they can't be I take it back

all but one adjective is pretty good though

[dull ginkgo]

lol

[rotund aurora]

I think that what I said is not possible, but I have no idea

[dull ginkgo]

RIGHT

RIGHT MODULE

NOOOOOOOOOOOO

[terse idol]

im having trouble thinking of one that gives me the result i want;
if i have $H =\langle (1, 0) \rangle$ as my subgroup, then im pretty sure the quotient space sort of be represented solely by the second digit, which means its equivalent to $Z_{p^2}$(?), which is a dead end as $G/H$ would be $Z_{p^2}$ by first isom. thrm

i think $H =\langle (1, 1) \rangle$ can be thought of as the difference a-b between (a, b), which im pretty sure still maps to $Z_{p^2}$

[cloud walrus]

ChiliLion

[shell pilot]

Sorry about that....was in an area with no signal

I'm not just trying to get answers. I'm not even in a math course. I am just self studying abstract algebra.

Z_1 is the set {0} with binary operation of addition modulo 1
Z_2 is the set {0, 1} with binary operation of addition modulo 2
So the Aut(Z_1) is simply phi(x) = e right?

[coral spindle]

I have been resummoned

[coral spindle]

OK so

[shell pilot]

Lol

[coral spindle]

the Aut(Z_1) is phi(x) = e

I'm spooked by this language again

Let me be extremely clear

Aut(Z_1) is a set of things. It is not a function, it is not an automorphism. It is a set of automorphisms.

Aut(Z_1) = {phi} where phi(x) = 0, that is correct.

But it is wrong to say what you said there. I am not sure if you have misunderstood something or you are just using language imprecisely, but I need to point this out.

OK, so you are totally correct so far.

I'd like you to think about Aut(Z_2) now

[shell pilot]

I understand that Aut(G) is the set of all automorphisms on the group G

[coral spindle]

Let's simplify. Let G = Z_p^2. I want you to find a subgroup H of G where G/H is isomorphic to Z_p.

[shell pilot]

Verbiage will be better from now on

[coral spindle]

Hint: this is going to be equivalent just to finding a subgroup of G of order p, which (even stronger hint!) you actually already know how to do.

Then we can move on to Z_p^2 (+) Z_p^2.

[terse idol]

i mean the only subgroup of order $p$ should be similar to $\langle p \rangle$ no?

[cloud walrus]

ChiliLion

[coral spindle]

Idk what you mean by similar to

What does that mean

[terse idol]

<2p> <3p> etc

to <(p-1)p>

[coral spindle]

OK so this is true, and more specifically you mean just IS

[terse idol]

sorry my brain is frying 😭

[coral spindle]

The only subgroup of Z_p^2 of order p is <p> = <2p> = ... = <(p-1)p>

[terse idol]

right ,because its equivalent

im tweakin

[coral spindle]

No it's not equivalent, it's literally equal.

[shell pilot]

Aut(Z_2) = set of automorphisms of Z_2.

[coral spindle]

They are equal sets.

They are the same set.

[terse idol]

that's what i meant im just bad at speaking

[coral spindle]

OK

[terse idol]

oopsies

[coral spindle]

So great, you have a subgroup of Z_p^2 that works for this simpler problem, namely <p>

Now can you think about how we might extend this to work for Z_p^2 (+) Z_p^2?

[coral spindle]

If it's not clear, I'd like you to try and determine the elements of this set

I'd like a description of the automorphisms. Hint: think about some nice properties that automorphisms – or more specifically homomorphisms in general – have

(Not the defining property, but some other nice property)

[topaz solar]

Tarski monster my beloved

[coral spindle]

OK but for real I need to sleep now

someone take the wheel

[terse idol]

hmm ok

[dull ginkgo]

Generating sets are a huge help :3

[terse idol]

ok so basically i need all elements that can be represented as (mp, np) for some m,n in this subgroup. No idea how to notate that, but it's a subgroup nonetheless because it passes all the subgroup tests

this is what was happening in my head when i said <(p,p)> teehee

oops

[coral spindle]

There is a nice way to write this subgroup using the same (+) notation

Can you see it?

[terse idol]

pZ_p (+) pZ_p?

[coral spindle]

That’s right

[terse idol]

i see

[coral spindle]

Or wait no

[terse idol]

i am blind

[coral spindle]

pZ_p^2 you mean

Or just <p> (+) <p>

[terse idol]

oh this makes sense to me now

thanks a bunch 💀 i probably walked like half a mile pacing back and forth trying to vizualize thing and figure out what i was doing

[shell pilot]

Well, Gallians book covers homomorphisms after isomorphism so I haven't gotten there yet but what I'm thinking is that for phi to be an automorphism of Z_2 then phi must map 0 to 0 or 1 and phi must map 1 to 0 or 1 right?

So an example of an automorphism could be phi(x) = x + 1?

Another could be phi(x) = x as the trivial one?

[terse idol]

well like

if you map 0 to 1

you neccesarily must map 1 to 0

and vice versa

[shell pilot]

Ok definitely understand that

So then why is Aut(Z_1) isomorphic to Aut(Z_2) if Aut(Z_2) has more than one possible automorphism in Aut(Z_2)?

[terse idol]

where did you find that

lol

im pretty sure its not true(?)

[shell pilot]

It's in the solutions for Contemporary Abstract Algebra by Gallian

[coral spindle]

There’s a pretty big difference between having two possible automorphisms and actually having two

You have just listed bijections, not checked that they are automorphisms. Do so!

[terse idol]

is U(1) = U(2)?

i forgot how it was defined

[coral spindle]

They are isomorphic. I wouldn’t call them equal

[terse idol]

hmm i see

i will go eat before i spread any more misinfo 🙏 also boy didnt you say you were sleeping...

go sleep!

[topaz solar]

Have I detected unitaries

[sly crescent]

What is U(n) here? It clearly isn’t a unitary group.

[shell pilot]

If they are bijections then I just need to show that there is operation preservation, right? So for phi(x)=x+1 we need to show that phi(ab)=phi(a)phi(b)

phi(ab)=ab+1 so it is not an automorphism since ab+1 does not equal a+1+b+1

[coral spindle]

Group of units of Z/nZ

[dull ginkgo]

Do you know what “finitely generated” means is?

[coral spindle]

Gl miz

[dull ginkgo]

They are isomorphic because they only have one possible isomorphism.

There is only one group up to isomorphism of order 1

[shell pilot]

If the "identity" mapping counts as an automorphism of Z_1 then why doesn't it count as one in Z_2?

[dull ginkgo]

Well there is a single identity morphism per group

That is an automorphism, the identity amongst them

They are distinct “things•

[shell pilot]

So Z_2 has one automorphism as well as the 0->1, 1->0

[dull ginkgo]

That’s not an automorphism

A homomorphism, a well-structured map that respects the group structure, must fix the identity

Namely 0

[topaz solar]

That is not an automorphism because f(1+1) = f(0) = 1 but not f(1)+f(1)=0

[dull ginkgo]

That is a bijection of the underlying set, yes (there are two set bijections) but ONE automorphism

[shell pilot]

The ONE automorphism being the identity which is the same as the one in Z_1

[dull ginkgo]

Well not the same

[shell pilot]

Well

[dull ginkgo]

Yes they are both identities

[shell pilot]

Sorry...verbiage again

[dull ginkgo]

Well i see what you’re saying

Same type of morphism, no

Yeah i see what you’re trying to say

[tardy hedge]

Do yall ever get depressed

All ive been doing for 2 weeks is studyibg math

[dull ginkgo]

I am highly medicated and still so

[tardy hedge]

Sorry im saying it here because you guys are familiar to me

Im sorry if its off topic but pls bear with me for a bit

[topaz solar]

I have not been studying much the past two weeks

[tardy hedge]

Yeah idk ive just been having a tough year

Lots of like, needs on the hierarchy not being met due to circumstances

Anyway sorry this is quite random but just wanted to say something here

[dull ginkgo]

I feel ya

[tardy hedge]

Cause ive been here a LOT the past 2-3 weeks

Thanjs

[dull ginkgo]

A lot of us do

[tardy hedge]

Thanks

[dull ginkgo]

@shell pilot anything else you are stuck on?

[shell pilot]

That helped a lot, I really appreciate the help!

Now I see why 0->1, 1->0 is not an automorphism

[dull ginkgo]

It is a bijection, not an automorphism of groups

[rotund aurora]

what is a direct proof of this fact

[tardy hedge]

What. The. Hell

Using algebraic structures to visualize geometry in like an isomorphism?

Is that right? What field of math is this

I am a little high right now, but even still, thats genuinely so cool to me

Sorry, random i know

[dull ginkgo]

Analytic structures often have an algebraic side to them

[tardy hedge]

Wow i never really learned that before

I need to learn that but idk if i even have the chance to

[dull ginkgo]

Like the entirety of functional analysis basically

[tardy hedge]

Wow

[dull ginkgo]

There’s sometimes cool problems that can be done with it, there was a problem i did in Jacobson that combined that idea

[tardy hedge]

Math is so weird

[dull ginkgo]

This

[tardy hedge]

YOOO

Ok i can understand a lot of it so i gotta read that later

[dull ginkgo]

Okie :3

[tardy hedge]

Why we like math

*do

[dull ginkgo]

Lie groups are also an example

The entirety of harmonic analysis

[tardy hedge]

Yeah i need to learn that stuff and discover what my true math interests are

Unfortunately i got into the good stuff too later

Like, anything is possible, but i shied away from this stuff in undergrad

I am too much of a perfectionist

It would overwhelm me too easily

[tardy hedge]

In terms of i can always keep learning and see where i want to go with it, but im still quite far off

[topaz solar]

And dually

[dull ginkgo]

Same

But i have the sorta luxury of doing it as a hobby and being able to follow my own constraints

At the cost of less motivational resources and more paranoia about my performance

[agile burrow]

I don't know of a proof that doesn't use some Lie theory. The sort of key thing to understand here is the exponential map, which goes from the Lie algebra to the group. Under the given hypotheses, it ends up being a surjective group homomorphism with discrete kernel, so by first iso theorem for Lie groups, G is a compact quotient of R^n by a discrete subgroup, hence a torus.

[rotund aurora]

Unsurprisingly, you could say this is Lie group theory. But I guess this is part of topological/analytic algebra (analytic for the smooth part). I am a fan of topological algebra, but I admit analytic algebra sounds weird.

[tardy hedge]

Oh wow I just kind of understood what u meant a bit with analytic algebra

i mean tbh my analysis is so lacking but like

what is that, using metric space theory to get properties of algebraic structures?

does it use measure theory?

[acoustic igloo]

The dihedral group D_2n with 2n elements is isomorphic to a semidirect product of the cyclic groups C_n and C_2.
does this mean that either C_n or C_2 isn't normal in D_2n?

[acoustic igloo]

is a group the direct product of its composition factors?

[mighty kiln]

Cn is normal

Since the action of D2n on the orientation of polygon has kernel Cn

[acoustic igloo]

i'm not sure i follow but

[mighty kiln]

A polygon has two orientations {0, 1}, and each element of D2n will either flip the orientation or not, therefore permuting {0, 1}

[acoustic igloo]

i think there's some confusion apparent in how i posed my question

oh that's what you mean by orientation

i think i see

[mighty kiln]

No

Not even semi-direct product I think

[acoustic igloo]

i'm confused because like, on one wikipedia page i see Z_12 = Z_4 x Z_3
but i also see elsewhere that C_12 has composition factors C_3, C_2, and C_2
and i see that Z_3 x Z_2^2 is different from Z_12

[mighty kiln]

"Composition factors" refer to factoring via short exact sequence

0 → H → G → K → 0 can be considered "factoring" G into H and K

If G → K has right inverse then it's a semi-direct product

If H → G has left inverse then it's a direct product

[chilly radish]

Extensions are very complicated, (un)fortunately

(And building a group from its composition factors is just performing repeated extensions)

[acoustic igloo]

does Z_3 x Z_2^2 have the same composition factors as Z_4 x Z_3?
so like, these are more detailed descriptions than the composition factors?

Z_2^2 is neither a direct product nor a semi-direct product

i guess i need to look at extensions

[mighty kiln]

Yes there is an SES 0 → Z/2 → Z/4 → Z/2 → 0

Which is just quotienting out the subgroup {0+4Z, 2+4Z} ⊂ Z/4

[acoustic igloo]

thanks

yeah extensions is what i needed to look at

[void cosmos]

is using the Tensor-Adjoint isomorphism enough to prove that the tensor product of projective modules is projective over a comm unital ring

i just said Hom(M tensor N,-) is exact cuz its just Hom(M,Hom(N,-))

literally

thats all i did

does it work

[rocky cloak]

That works. You can also just use that projectives are summands of free modules, and that tensoring with free modules is just direct sum

[void cosmos]

yeah i didnt think of that

when there are many like characterizations of shit u sometimes get tunneled vision into one

hahah i got lucky ig that it works i wouldnt have thought of the sumand of free

mod

ty

[crystal vale]

They define contraction such that f: A->B be a ring homomorphism, if b is an ideal of B, then f^(-1)b is always an ideal of A, called the contraction of b.

To show every prime ideal of A is the contraction of a prime ideal of A[[x]].

I need to show if f: A-> A[[x]] is a ring homomorphism then for every prime ideal p in A there exists prime ideal q in A[[x]] such that f(p) = q, right?

If it is right, then if I take identity mapping then for prime ideal p in A we can choose q = p such that p is an ideal in A[[x]], right?

[rocky cloak]

Not quite, you need to show that for every p there is a q such that f^(-1)(q) = p.

Note that if p is an ideal in A, then it's not an ideal in A[[x]] (unless p=0)

[crystal vale]

Yes 🥲

But can we generate prime ideal in A[[x]] by the set p ?

[rocky cloak]

You can yes, but there is something to check, whether the ideal you get is still prime or not

[crystal vale]

Yes we can generate ideals but can we generate prime ideals ?

What if q is generated by p and x ?

[rocky cloak]

That would do the trick yeah

You can't generate prime ideals in general, because the intersection of primes need not be prime.

But you can generate an ideal and hope/check that it's prime

[crystal vale]

<x> is the prime ideal in A[[x]], right?

[rocky cloak]

It is prime precisely iff A is a domain

[crystal vale]

Ah, I thought so

[crystal vale]

Yes because if it is not domain then there exists non-zero element a and b in A such that ab=0 but a and b not in <x>

I am not sure about how the ideal looks generated by p.

<p> = { finite sum of a_nb_n | a_n in A[[x]] and b_n in p }, is it correct?

[crystal vale]

Let A be a ring and R its nilradical. Then if A/R is a field then A has exactly one prime ideal.

If A/R is field then R is maximal ideal in A.

Let p_i are prime ideals in A then R is the intersection of all p_i.

Now since p_i ≠ (1) there exists maximal ideal M_i such that p_i contained in M_i.

But it shows that R contained in M_i implies that R=M_i. Thus p_i = R for all i.

Is it correct?

And here ring is commutative with unity

[south patrol]

Yes, though you don't need to invoke maximal ideals. Just note R is maximal (as you said) and for any prime p, R is contained in p so R=p

[crystal vale]

Got it, thank you

[tardy hedge]

Dont you love when youre tryna sleep and then suddenly think of something that makes a concept you were trying to learn make a lot more sense

Woo!

I was thinking more about split sequences , splitting homomorphism and i understand that concept better now

I was confused the first time i saw it

[crystal vale]

In a ring A, let S be the set of all ideals in which every element is a zero divisor. Show that the set S has maximal elements.

But I don't understand how S can be non-empty? We don't count 0 as zero divisor

[delicate orchid]

no other assumptions on the ring A?

and maybe they just mean every non-zero element is a zero divisor

[crystal vale]

A is a commutative ring with unity

[delicate orchid]

meaning the zero ideal is included in A vacuously

[crystal vale]

Got it

[crystal vale]

Thank you ❤️

[delicate orchid]

I mean I don't know for a fact that's what they mean but it's the only way that their claim makes any sense

[crystal vale]

Yes

[rocky cloak]

You can also just allow 0 to be a zero-divisor

[crystal vale]

But the book does not allow

[rocky cloak]

Simplifies things a bit

[dull ginkgo]

A zero divisor kind is like the failure of (0) being a prime ideal

[rocky cloak]

Don't be a slave to the book

[crystal vale]

Okay

[tardy hedge]

😵‍💫

[dull ginkgo]

What do you mean “maximal elements”?

[tardy hedge]

I wonder, are there any other big math servers like this?

That are active like this one

[crystal vale]

Not big but yeah there are

[rocky cloak]

SoME is pretty big I guess

[tardy hedge]

This is like my damn home 😂

Are there invite link u can send me?

I really do appreciate resources like this. You guys are so helpful

[crystal vale]

Maximal element with respect to inclusion order

[crystal vale]

There is one more server, if you want to join I can invite you

[tardy hedge]

Thx

[dull ginkgo]

Inclusion order isn’t over elements, it’s over subsets?

[crystal vale]

Yes so S contains ideals

[dull ginkgo]

Very confused what you mean here

[delicate orchid]

S is a set of ideals

[dull ginkgo]

Okay and so S cannot contain another ideal?

[delicate orchid]

what?

[crystal vale]

Actually the book defined 0 as a zero divisor so my mistake

[dull ginkgo]

You’re not making sense here

[delicate orchid]

dude S is the poset in question here

the elements of S are ordered under inclusion

[crystal vale]

But I don't think it makes any difference

[dull ginkgo]

Oh S has maximal elements

Assuming Zorn’s lemma or whatever choice equivalent of choice

That seems like the easiest way

||Because it’s chain complete. If we have an increasing chain of ideals where each element is a zero divisor, then the union over all of them has the same property, because every given element in that union must lie in one of those ideals and must be a zero divisor, no?||

[delicate orchid]

wow that's boring but yes I think that works

[south patrol]

It is always Zorn and taking unions

[crystal vale]

Taking union and union is an ideal and if a in union then a belong to some ideal I then there exists b such that ab = 0, so every element of that union is a zero divisor so union of chain element is in S. By Zorn's lemma, there is a maximal element, right?

[dull ginkgo]

This

[delicate orchid]

"tHiS" goofy

[dull ginkgo]

i am goofy yes

[south patrol]

Goofy

[marsh scaffold]

Isn't left, right action just a convention or is there more to them than that

Like what do they mean that a left action is a right action in (b) here

From what I understand an action is just a map from G to sym(X) that is a homomorphism

So how are left and right actions different

[delicate orchid]

right action is an antihomomorphism from G to Sym(X), but really there isn't that much difference at all

like as you prove in part a, any left G-set is isomorphic to a right G-set

[marsh scaffold]

Yeah but I dont understand what I have to prove in (b)

[delicate orchid]

you need to prove that the action x.g := g.x is a well defined group action iff [G,G] acts trivially

where ":=" means "defined as being equal to"

not sure how standard that notation is

[marsh scaffold]

So I already have a right action given to me?

[delicate orchid]

well your notation is strange to me because everything seems to be acting on the right anyway

[marsh scaffold]

My notation, you mean the notation in the problem?

[delicate orchid]

yeah

[marsh scaffold]

Wait so what is the problem with just saying that define a right action by mapping x Sigma to sigma x

[delicate orchid]

so I'm guessing you have weird definitions of left/right actions - something along the lines of "a left group action is a map GxX -> X such that x(ab) = (xa)b and a right group action is a map GxX -> X such that x(ab) = (xb)a"

?

which is idiotic and I will not be adopting

[marsh scaffold]

Yea

[delicate orchid]

to solve this problem, you basically need to see what you can say about x(ab) and x(ba) using the fact that [G,G] acts trivially

[tribal moss]

The distinction is between things that satisfy f(g,f(h,x)) = f(gh,x) and things that satisfy f(g,f(h,x)) = f(hg,x).

[delicate orchid]

yeah but like... write left group actions with the group elements on the left? like everyone else in the literaure?

you would never do this for left/right modules

[tribal moss]

(I may have missed some context; my goal was to explain to Redacted that it's not just a notational difference).

[marsh scaffold]

Wow , it's just a lot more times clearer now ig

[delicate orchid]

ah sorry, my mistake - I thought you were replying to me

[marsh scaffold]

Yeah now with this brilliant notation it's clear to me why commutator acts trivially

[delicate orchid]

it's even clearer if you do it the standard way:
left group actions are maps GxX -> X with (ab)x = a(bx)
right group actions are maps XxG -> X with x(ab) = (xa)b

[marsh scaffold]

So how would left group actions become right

[delicate orchid]

like this

[marsh scaffold]

Like idk why this is somehow uncomfortable to me

[delicate orchid]

but if you can see it clearer with tropo's explanation then just ignore me

[marsh scaffold]

Cause x is an element of the set

[tribal moss]

I think what can make that notation confusing is that you're using the "." to denote two different operations, and defining one in terms of the other.

[delicate orchid]

true

I'll just latex it

[marsh scaffold]

Okay thank you @tribal moss @delicate orchid this was quite insightful to me

[marsh scaffold]

Also this antihomomorphism thing

[marsh scaffold]

From here also I believe we can get that commutator acts trivially

[delicate orchid]

the commutator doesn't always need to act trivially at all

consider G acting on itself via right multiplication

[tribal moss]

(If [G,G] acts trivially, then the action G->Sym(X) factors through the abelianization of G, which mean that f(gh,..) and f(hg,...) must be the same permutation).

[delicate orchid]

unless you're saying that assuming the action is both left and right then the commutator must act trivially? which is true

[delicate orchid]

there's a much easier way of seeing it, x(ab) = (xb)a = x(ba) = x([a,b]ba)

I suppose I should write it clearer:
x(ab) = x([a,b]ba)
x(ab) = (xb)a = x(ba)
<=> x([a,b]ba) = x(ba) <=> x[a,b] = x

this is basically the decategorified version of what you said I'm aware

[tribal moss]

Ngl, this notation looks confusing to me too. If we're talking about two distinct actions in the same argument, then I think they ought to have more notation than just juxtaposition.

[delicate orchid]

issue with that is I'd have to latex it

but the entire point is it's the same action

there's only one action

[tribal moss]

Fair, I should have read the original problem more closely.

[marsh scaffold]

This is more of an off beat question but for this excercise 2.4 is there a way to think about what exactly is the action that is defined the way it is here doing to the function

Ps: I was able to get through all the verifications asked in 2.4 I was just looking if there is some nice interpretation to think about this

[main flower]

Hello I just wanted to ask about the coloring of a tetrahedron. I know the symmetries of a tetrahedron are isomorphic to S_4 and that I need to use Burnside's Lemma, but I am unsure of how to apply it. Can I calculate the colorings of the faces as permutations of S_4 and the colorings of the vertexes as permutations of S_4 and multiply them together? I was thinking something like this. (Work in pdf)

[soft wyvern]

Hi, i have a doubt about cyclic modules. I was trying to prove that $K[t]/(t^3+t^2+1)$ is a cyclic $K[t]$-module. My professor told me this module is generated by the class of the constant polynomial 1, so is cyclic, but I dont understand why

[cloud walrus]

jorge.534s

[chilly radish]

This is why modules are better. Right vs left matters

[delicate orchid]

group actions are modules over F_1 they're the same structure. Keep coping

[chilly radish]

F_1 isn't a ring

[delicate orchid]

and you are not swag

[chilly radish]

I'm a commutative bioperad

[coral spindle]

Ah yes this thing that doesn't exist. It's not a ring

[delicate orchid]

I'm a symmetric biset

[coral spindle]

I support your journies

[rotund aurora]

What is the absolute Galois group of F1

[chilly radish]

Doesn't exist yet

The category of Motives not existing never stopped anyone

[delicate orchid]

you see chat it's funny because

[long obsidian]

How do you define the exterior algebra of a module?

[coral spindle]

Tensor algebra modulo anticommutativity

[chilly ocean]

how do you define it for vector spaces?

[tropic spade]

For 9 here I am stuck. I've worked it down to r=2 so that I have two composition series 1 < N < G and 1 < M < G which I have isolated to the case (N cap M)=1 and N, M are distinct. Then I showed G=MN. I see that by second iso this should give me G/N iso to M and G/M iso to N I think. Now I need to show the bit about isomorphisms between composition factors from Jordan Holder but I'm not sure how that follows?

Feels like I'm missing something easy here.

Oh yah here is how df states jordan holder fwiw

[dull ginkgo]

noncommutative matricies fry my brain

[rotund aurora]

You are done

Note that (2) says that you can permute the composition factors

You could take N, M simple and consider the product NxM. You have the series 1-N-NxM and 1-M-NxM

[tropic spade]

Oh wait is the permutation just NM/N -> M/1 and N/1 -> MN/N?

Okay yah I see it

Man I am dumb as hell sometimes lmao.

[dull ginkgo]

Hey chat once again returning to this

If we have a ring R

We can view R as both a left and a right module over itself

but is End_R(R) for RIGHT module morphisms isomorphic to R, but isomorphic to R^op for left module morphisms?

a_r(b_r(x)) = a_r(xb) = a_r(x)b = xab

but a_l(b\l(x)) = a_l(bx) = b a_l(x) = bax

Actually i need to think a bit more about left vs right modules

Let G be an abelian group, and R a ring

Then can G be equipped with a left module structure for a morphism from R to End(G), and a right module structure for a morphism from R^op to End(G)

a_r(b_r(x)) = a_r(xb) = x(ba)

a_l(b_l(x)) = a_l(bx) = (ab)x

[chilly radish]

I told you this

[dull ginkgo]

I think i misremembered that, thanks for the clarification

So it’s because (R^op)^op = R

[chilly radish]

Morally yea

[acoustic igloo]

i don't think you can just multiply them, because there could be a vertex coloring and an edge coloring that could be combined in more than one distinct way

perhaps you could use a group that represents the symmetries of both the edges and the vertices?

hm no, i don't know

maybe the group should act on the set of both vertices and edges

you only counted how many vertices were fixed, not edges

oh but then the bases of the exponents wouldn't match

[lone niche]

In general for a ring R with 1, R will be will be a cyclic R module over itself (r*1=r for any r). But then it's also easy to check that the image of that generator generates the quotient.

[main flower]

Ah I’m so dumb I thought it was faces and vertexes not edges and vertexes. I’m going to have to go back to the drawing board.

But S_4 should act on edges and vertexes because it’s just a representation of a symmetric change in the tetrahedron.

[acoustic igloo]

i think i got it

[crystal vale]

If I want an example such that A group G with a normal subgroup H such that the factor group G/H is not isomorphic to any subgroup of G.

So if we take G = Z and H = 5Z, then there is no Subgroup of Z which is isomorphic to Z/5Z, right?

[south patrol]

Yes, good

[sly frost]

any nontrivial subgroup of Z will have infinite order and any quotient of Z by a nontrivial subgroup will have a finite order

[crystal vale]

Yes, thank you

I don't know if my proof is correct or not

If p_1,..,p_n are primes which divide |G| no other prime divide |G|.

Since there exists an element of order p_1 then take K_i is a cyclic group of order p_i.

Then what if I take H_i as a direct product of K_1, K_2,...,K_(n-i+1)?

[south patrol]

Hm is that necessarily a subgroup?

It's easiest imo to use induction and the "correspondence theorem"

which i think is smetimes called 4th iso or smth

[crystal vale]

Oh

[crystal vale]

Induction on what?

[south patrol]

Size

[crystal vale]

Okay, thank you

[languid trellis]

Does "quotient field" mean field of fractions?

[delicate orchid]

tf does Q/R mean here

[languid trellis]

quotient module?

[delicate orchid]

hmm. I'll allow it just this once

[languid trellis]

It's generalising (Q/Z) \otimes_Z (Q/Z) = 0

i imagine

[delicate orchid]

yeah true

[south patrol]

Yes in this context

[dull ginkgo]

What is Q/Z as a quotient module actually

Rational subgroup of the circular group?

[languid trellis]

Q/Z is S1

wait

[coral spindle]

No it's not!

[languid trellis]

R/Z

R/2piZ or something

[coral spindle]

Yes it's the torsion subgroup of the circle group

[languid trellis]

its close enough

[coral spindle]

One is countable, the other is uncountable

[languid trellis]

yeah they're both infinite. same thing

[dull ginkgo]

So this problem would be equivalent to showing every pair is a bilinear combination

?

[uneven jackal]

with a difference being also that Q/Z has much more automorphisms than S^1 as a topological group

[coral spindle]

Q/Z is not topologically cyclic but R/Z is

[delicate orchid]

this. So much this

[dull ginkgo]

How would you show a tensor product is trivial

[delicate orchid]

look at elementary tensors and show they're all zero

[coral spindle]

Um 🤓 because the splooge-zucker spectral sequence vanishes on the 69th page we can clearly infer

[dull ginkgo]

Elementary tensors?

Do you mean tensors of the form a (x) b

[delicate orchid]

yeah, the elementary tensors

[languid trellis]

elementary tensors? you mean an arbitrary element of the tensor product

[delicate orchid]

no I don't

[coral spindle]

No

[languid trellis]

z

zzzz

[delicate orchid]

WAKE UP

[south patrol]

Why are people hating on wew rn

He said smth correct/good

[coral spindle]

I'm not hating on wew he's a king

how dare u

[south patrol]

😭

[languid trellis]

yeah that's not his job is it

[south patrol]

Also like 1) ||in fact Q/R (x)_R Q = 0, which implies your result (why?)|| 2) you can view|| Q/R (x)_R Q = (Q/R)[S^-1] where S = nonzero elements of R, and it is clear this guy vanishes||

[delicate orchid]

uh oh!

[dull ginkgo]

I need to learn about tensors

[coral spindle]

oy gevalt this message is fucked up

[dull ginkgo]

I forget how you show a tensor is 0 outside of using the “bilinear kernel” construction

[coral spindle]

😭 it's being carved up

[delicate orchid]

u literally just move shit around

[dull ginkgo]

Very helpful (not at all)

[south patrol]

The hard part is showing smth is nonzero

[south patrol]

It is lol

[delicate orchid]

start with a/b (x) c/d

[languid trellis]

r balanced map moment

[delicate orchid]

show it's zero

[dull ginkgo]

Move around how wtf you mean

[south patrol]

Z/2 (x)_Z Z/3 = 0

[delicate orchid]

then move shit around

[south patrol]

Because 1 (x) 1 = 3 (x) 1 = 1 (x) 3 = 1 (x) 0 = 0

[dull ginkgo]

Yes but when is a (x) c 0

[languid trellis]

da /dc (x) c/d = a/dc (x) 0 = 0

[south patrol]

Real

[dull ginkgo]

????????????

[languid trellis]

its just multiplication. when a or c are 0

[south patrol]

Think about it lol

[delicate orchid]

miz have you ever actually worked with tensors

[dull ginkgo]

Why did 3 just become 0 wtf

[delicate orchid]

BECAUSE IT'S Z/3Z

[dull ginkgo]

Not for a while

[delicate orchid]

LOCK IN

[dull ginkgo]

OH

[delicate orchid]

LOCK INNNNN

[dull ginkgo]

Thought we were working in Z

[south patrol]

But you were fine with 1 = 3... curious

[delicate orchid]

well yeah that's just obvious

[coral spindle]

curious

[languid trellis]

we're working with the Z-modules Z/3Z and Z/2Z

....

[dull ginkgo]

oh okay i misread

[south patrol]

Lmao

[languid trellis]

so its kinda Z

[coral spindle]

I'm seeing all these Zs and I'm thinking ZZZZZZZZZZZZZZZ

[south patrol]

Every abelian group is kinda Z

[languid trellis]

real x2

[dull ginkgo]

Annihilater intersection bullshit

Who cares

[south patrol]

What

[dull ginkgo]

Can’t spell

[coral spindle]

Now do the same for R = k[x] and R/(x^2+1) (x) R/(x-1) [tensoring over k]

[south patrol]

Oh I guess you are saying like

2 and 3 both kill it

[dull ginkgo]

Yes

[south patrol]

So 1 kills it

[delicate orchid]

https://tenor.com/view/no-i-dont-think-i-will-captain-america-old-capt-gif-17162888

[languid trellis]

surely thats 0

[south patrol]

Rahh

[coral spindle]

ok prove it bih

[south patrol]

Almost every time someone writes = k[x] I just work with k[x]

[coral spindle]

What was the method for Z/2Z (x) Z/3Z that generalises

[south patrol]

My brain gets confused when I see R/(x^2 + 1)

I mean didn't swiftee solve the problem lol

[coral spindle]

Idk why I did it like that, it came to me in a dream a bit like your mom

[languid trellis]

if k= R then this is the free module of rank 4?

[south patrol]

Bit inappropriate, especially depending on how you read this

[coral spindle]

But I just said R = k[x]

[dull ginkgo]

Define the annihilator of an element of a module. Let’s say we have two R-modules A and B, then a (x) b = 0 if the annihilators of an and b have nontrivial intersection no?

[south patrol]

Well it's more that the sum is R

That matters

[coral spindle]

This may have been deliberate... I can delete if the mod hat says no

[south patrol]

Oh okay sorry yeah

I get it

[dull ginkgo]

Might actually be an iff lol

[languid trellis]

ok i have no idea then

its probably 0

[coral spindle]

So what if I pointed out that 3 - 2 = 1 is what makes the Z/2Z (x) Z/3Z thing work

[languid trellis]

uh

[dull ginkgo]

oh yeah, coprime ideals

[coral spindle]

We add 2 – which is doing nothing in Z/2Z – then we transfer over the tensor, whence it becomes 0 because 3 is nothing in Z/3Z

So this really works because 3 - 2 = 1

[languid trellis]

yes bezouts means we can move stuff around. Z/mZ (x) Z/nZ = 0 for m,n coprime

[coral spindle]

Bézout yes exactly

Big hint

[languid trellis]

oh i should do polynomial division

probably

[coral spindle]

Well you could but you can also just spot a Bézout identity

[south patrol]

So what I said

Boom

[coral spindle]

Lol

[south patrol]

Chungus

[coral spindle]

Hint: difference of squares

[dull ginkgo]

Assume Ann(a) and Ann(b) be coprime ideals in R

then there exists x in Ann(a), y in Ann(b) such that x + y = 1

a (x) b = (x + y)(a (x) b) = (ax) (x) b + a (x) (by) = 0

actually probably can use ideas like this to find annihilators of tensor products

[languid trellis]

hold the phone

hold the phone

[coral spindle]

No I hate phones

grrr

[languid trellis]

i've forgotten how to work with k[x]

good lord

but we should be able to move stuff from the left hand side to the right and by considering degrees it should vanish

[dull ginkgo]

@south patrol Lets say we are working with R modules A and B

If Ann(A) = I, and Ann(B) = J, then is I + J a subideal of Ann(A (x)_R B)?

[south patrol]

Cursed notation

[dull ginkgo]

I’m on mobile

Sad face

[south patrol]

Sure

No I meant like

"The R module A"

[dull ginkgo]

Is that actually an equality

I don’t think so?

r(a \otimes b) = ra \otimes b = a \otimes rb = 0

For each a,b

[south patrol]

Probably not an equality no

Actually we already have an example

Q/Z (x)_Z Q = 0. Annihilator of Q is 0

But Q/Z is not 0

[dull ginkgo]

Thanku

So let a/b \otimes c/d in the above product

Can’t we use denominator gcd lo

[languid trellis]

x^2 + 1 = (x-1)(x+1) + 2 is the identity I assume you're referring to? I don't see where this could go. My idea is to consider ax+ b mod (x^2+1), divide ax+b by (x-1) to get ax+b = p(x)(x-1) + r(x), then I wanted to move the (x-1) over but that only works if r(x) is 0

[coral spindle]

We're taking the tensor over k

I guess I do need to assume it's of char 0

why not just say k = Q

[dull ginkgo]

u = a/b \otimes c/d
(bd)u = da \otimes bc = 0

Actually I’m stupid

Factor out the 1/bd

as a Frac(R) vector space

[coral spindle]

You are right and this is bugging me, I'm suddenly having a mental block

Oh lmao yes I am taking the tensor over k like an eejit

when in fact we need it over k[x]

Waoh

[languid trellis]

ok. so we are considering k[x]/ (x-1) as a left k[x]-module, and the idea of the tensor product is to multiply by elemts of k[x]/(x^2+1)

how does it being a k[x] module change the situation. (i am writing to myself here)

[coral spindle]

In the module $\bQ[x]/(x-1)$ we have $1 = 1 + (x-1)(x+1)/2 = (x^2+1)/2$

[cloud walrus]

Boytjie

[coral spindle]

Like this is almost too much to say lol

[dull ginkgo]

Do what here btw?