groups-rings-fields

[clear obsidian]

I know Aut(A) but what was Sym(A)? Im having trouble looking it up?

[woven delta]

Symmetric group on A letters

[clear obsidian]

ah ok thanks

Does anyone know of some kind of mini encyclopedia and/or pdf of all of the ways of finding the number of albean subgroups or just subgroups of a given group/albean group?

(This is asked very often in the mgre)

[mild laurel]

@uncut girder the prime ideals that belong to an ideal are exactly the ones containing the ideal no?

@clear obsidian all finitely generated abelian groups have a certain form

[uncut girder]

Ok I didnt know that

Because it was in a exercise I skipped

[mild laurel]

I'm pretty sure it's directly in the text?

[clear obsidian]

hard to do every single exercise

[mild laurel]

not home rn but I can check later

[clear obsidian]

mhmm thanks, I can look it up but thanks

[mild laurel]

I wasn't talking to you but

@clear obsidian It's called the classification of finitely generated abelian groups

[clear obsidian]

ah my bad & ty

[uncut girder]

Is it?

Oh shit you're right

❤

@mild laurel

[tame bear]

so the proof i made for the problem i posted has an error
after spending all day on it, i finally fixed it

took 4 pages

[latent anvil]

@placid pond why a p-group? If it's proper take an element outside the subgroup and look at the subgroup it generates

The version I heard was "prove that if a group has a unique maximal subgroup then it's nilpotent". It's put on the 7/10th problem set after a super long problem introducing nilpotent groups to show students that sometimes all the machinery gets in the way

[uncut girder]

@tame bear good job

[hollow stag]

The average of 7 consecutive even integers is 20. What is the median?

[woven delta]

#prealg-and-algebra

[valid elbow]

2(a+a+1+a+2+a+3+a+4+a+5+a+6) = 20

7a + 21 = 10

something something

[hollow stag]

Thanks

[bleak abyss]

@hollow stag please don't ask questions in this channel again

[hollow stag]

No would answer in others, people would write over it and people answer and it takes forever.

[bleak abyss]

Doesn't matter

This channel is reserved for questions about abstract algebra

[woven delta]

I have an abstract Algebra question

Suppose you have a module over an integral domain

[bleak abyss]

@valid elbow along those lines, if someone asks a question in the wrong place please do not answer

[smoky cypress]

Integral domain? That’s a thing??

Oh yeah it is

I remember now

Nvm

[woven delta]

Does 0-> Tor(M) -> M -> M/Tor(M) -> 0 split

I think it does

Tor here just means torsion

Cause you define a section M -> Tor(M) so that a torsion element gets mapped to itself and a nontorsion element gets mapped to 0

But I can't check the details in that map cause I'm at a concert

(The integral domain condition guarantees Tor(M) is a submodule)

@bleak abyss do you know if this is true?

[valid elbow]

owo

[woven delta]

Actually I think this is false now

It's true over PID's, because torsion free = projective

[bleak abyss]

Okay now I'm here sorry @woven delta

0 -> Tor(M) -> M -> M/Tor(M) -> 0 splitting hmm

[woven delta]

If it splits the map has to be the one I described I think

Actually maybe not

Yeah no

Split just means dsd=d

I always forget the order of the maps lmao

[bleak abyss]

I guess if we wanna kill it then we should try k[x,y] or Z[x]

[woven delta]

Yeah

It might be true for Noetherian tbh

[bleak abyss]

Okay so

If this sequence were to split

Then torsion submodules are projective yeah

[woven delta]

Is that true?

I don't think that's true

Why does this imply generally

[bleak abyss]

Err wait torsion submodules of free modules I mean

I'm bad at crucial adjectives

[woven delta]

🤔

[bleak abyss]

Projective = direct summand of free module

[woven delta]

Doesn't free imply torsion free?

Or is my intuition off

[bleak abyss]

Right

Okay I can't think today

[woven delta]

Projective implies torsion free more generally

For the obvious reason

[magic owl]

(In fact the torsion modules don’t even have projective objects in the homological sense iirc)

(By “The torsion modules” I mean the full subcategory or non-torsion-free R modules)

Of*

[woven delta]

@magic owl do you have any nice examples?

Of where this is not split

[magic owl]

Oh I didn’t read the rest of the convo

[woven delta]

[magic owl]

Wait I’m a little confused but also tired

How would your thing split? If m is a generator for the torsion free part of M it gets $m\mapsto 0 \mapsto 0$?

[cloud walrus]

MaxJ:

[woven delta]

Yeah

[magic owl]

So it works always in the trivial case

I think

The trivial case being a torsion module

[woven delta]

I guess I'm assuming sum of torsion free is torsion free

[magic owl]

It is I believe

Bc projectives can’t have torsion

[woven delta]

?

[magic owl]

Wait my arg is bad

But I still think ur right

[woven delta]

Yeah

Idk if its the case

[magic owl]

I always think of splitting as being like, a quality associated w nice epis

[woven delta]

Maybe you could do something with the field of fractions

Idk

[magic owl]

So idk how it would work w inclusion

[woven delta]

So the general form of split

Is dsd=d

[magic owl]

Yeah it’s a left inv

[woven delta]

Notice how that specializes for epi

And for mono

[magic owl]

Ahhh

Ok

Ok then maybe I won’t be of help unless I think about it more lol

[woven delta]

Yeah that's fine

So the reason you need that generalization is you can talk about splittings of chain complexes

Which can be neither epi nor mono

[magic owl]

Yeah lol I actually misread the def of that

Back when I first learned it

Yeah ok the naive maps you were thinking about work

[woven delta]

Why?

[magic owl]

If they are actually homomorphisms

Which I’m not sure they are?

[woven delta]

Yes

That's the problem

[magic owl]

Yeah I’m caught up now haha

[woven delta]

It works if sum of 2 nontorsion is nontorsion

[magic owl]

I can’t break my intuition from the fg modules

[woven delta]

Or 0

Lmao yeah same

Also pids

I had to relearn tensors recently cause of doing tensors in Lee

The vector space case is too simple

[smoky cypress]

first time seeing #groups-rings-fields with 50+ messages

[woven delta]

I guess I'll ask in hopf

I would probably know the answer to this question if I read more about the Tor functor tbh

[magic owl]

I was trying to think of something about the universal property that implies this

But I can’t really ::

😦

If it’s true I feel like it might be true in full generality

[woven delta]

Integral domain is full generality

[magic owl]

Well

[woven delta]

Oh it's not

Oof

[magic owl]

Free objects in categories w biproducts

[woven delta]

Oh ok

[magic owl]

Was what I meant

[woven delta]

Where is there a free object here?

[magic owl]

Wait, adjoints preserve the (co)limits, so maybe biproducts work out

Oh I guess you’re right fuck

I keep associating torsion free w free

It’s true free objects I think but yeah

[woven delta]

Yeah sum of two nontorsion is not necessarily nontorsion

[magic owl]

Ah ale

Alright*

[woven delta]

Did you see the example 12 gave?

[magic owl]

Hopf? Gimme a sec

Ahh yeah ok

[woven delta]

@magic owl isn't every element killed by y^2 in that example lmao

[bleak finch]

Let $T$ be a linear transformation on the vector space $V$ over the field $F$. Show how to make $V$ into an $R$-module in a natural way, where $R$ is the polynomial ring $F[X]$.

[cloud walrus]

PPrincipleTwo:

[bleak finch]

Why did the problem writer give us $T$?

[cloud walrus]

PPrincipleTwo:

[magic owl]

Uhhhh

No idea

I don’t think you need it

And if the construction is natural you def shouldn’t need it

[bleak finch]

I don't need it

I don't need it

I DEFINITELY don't need it

[magic owl]

I NEED IT

[bleak finch]

PIIIINKY PIIIINKY

[mild laurel]

Wait you do need it?

if v is a vector in V, then what's xv?

@bleak finch

[bleak finch]

its v with each component multiplied by x

[mild laurel]

is that another vector in V

[bleak finch]

no

[mild laurel]

Then?

@bleak finch so are you sure this works

[bleak finch]

It might be an element of V' where V' is an F[x]-Module

[mild laurel]

the fuck is V'

[bleak finch]

An F[x]-Module

[mild laurel]

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

Read the problem again?

You're trying to make V into an F[x] module

[bleak finch]

In order to do that

Do I need to find a way to make scalar multiplication work for F[x]?

Is that sufficient or do I need to do more?

[mild laurel]

Idk you tell me what you need to make something into a module

[bleak finch]

An abelian group and scalar multiplication over that abelian group

[mild laurel]

So then the answer to your question is

[bleak finch]

yes

[mild laurel]

Then how do you do it

[bleak finch]

well the problem is that the abelian group underlying V (called A) as an F-module is a subset of the possibly abelian group (called A?) underlying the F[x]-Module. Just because I know that A is abelian doesn't tell me that A? is abelian.

[mild laurel]

what

You're making V into an R-module, which means that you're making A into an R-module

[bleak finch]

V is already an F-module, because it is a vector space.

A is not an F-module because it is an abelian group.

The new R-module built from V is going to have an abelian group A' that has A as a subset.

[mild laurel]

No

You can make A into an F-module in a very obvious way, by essentially making into V

Okay I think I understand why you're confused

People use very loose language here

It makes no sense to make V into a module, because V is not an abelian group as you noted

[bleak finch]

V is already an F-module

[mild laurel]

What they mean is to make A into a F[x] module

Right, so it makes no sense to make V into an F[x] module because V is already a module, and thus not an abelian group

[bleak finch]

Now let's take the underlying abelian group A of V.

I know that if I have scalar multiplication defined by the subset of F[x] with degree zero, that is well-defined, because it results in another element of A.

But if I take (x + 1) * a? That is an element not in A.

[mild laurel]

Sure

But there's no reason x * a couldn't be a + a

In other words, you could have the "action" of x on a be multiplication by 2

[bleak finch]

If A is a finite group then yes there is a reason why it can't be

The cardinalities may not be compatible

[mild laurel]

what does that even mean

[bleak finch]

Hold on.

I think I understand why I was given T.

Take a map that for instance has (x^2 + x + 1) * a := T^2a + Ta + a

[mild laurel]

check that this indeed satisfies all the properties of a module

[bleak finch]

Now the problem is straightforward.

[mild laurel]

yep

[bleak finch]

I just have one question

If M' is an R'-module that is an extension of the R-module M, is it permissible for M' and M to share the same abelian group?

$1$ exists. Associativity is satisfied because polynomial multiplication maps to the composition of linear combinations of linear transformations...which is associative. Because $T$ is linear, the right distributive law is satisfied due to $T$ being linear and the left distributive law is true by definition. Therefore this is an $F[x]$-module.

[cloud walrus]

PPrincipleTwo:

[bleak abyss]

Okay time for the coker version of the snake lemma

I verified exactness everywhere except where d is involved

First gotta show d is well-defined. The first choice made was x'' = v(x)

[cloud walrus]

Dami:

[bleak abyss]

And I think that's it, u' is monic so the choice of y' is unique.

Now exactness

[cloud walrus]

Dami:

Dami:

[bleak abyss]

Oh

[cloud walrus]

Dami:

[bleak abyss]

So that's exactness at Ker(f'')

Okay gonna repost

Because I don't believe in scrolling

Okay time 4 exactness @ coke r f'

[woven delta]

Which book is this?

[cloud walrus]

Dami:

[bleak abyss]

It's Atiyah-Macdonald

[woven delta]

Oh ok

Did you see the algebra stuff from last night btw?

In hopf

[cloud walrus]

Dami:

[bleak abyss]

And I'm done with snake lemma since I did everything not involving d at home

I noticed it happened but never went and carefully read it/thought about it

[woven delta]

I think it's fairly interesting/useful for intuition

[bleak abyss]

Might check it out at some point soon

In the meantime

Dimension is an additive function on finite-dim k-vector spaces

That's just rank-nullity

Okay to elaborate slightly on this prop

The split is

0 -> N_0 -> M_0 -> N_1 -> 0 -> N_1 -> M_1 -> N_2 -> 0 -> ...

This alternating sum cancels -\lambda(N_1) from the second SES with \lambda(N_1) from the one before, and so on

So yeah gg

Ah now time for tensor products

The thing I don't know

[smoky cypress]

Any textbook suggestion for abstract algebra

[mild laurel]

Depends on how comfortable with proofs you are

[smoky cypress]

I’m really comfortable

[mild laurel]

The standard recommendation is Dummit and Foote

[smoky cypress]

Ah ok thanks

I would go with the standard recommendation

[mild laurel]

Fraleigh has a pretty popular one too, I think the one by Rotman is pretty talked about

[smoky cypress]

i don’t really like Fraleigh’s

Lol

Wait Dummit and Foote wrote a book together?

[mild laurel]

Hm? Why?

[smoky cypress]

I thought those are two books

But apparently the book’s author was Dummit and Foote

[mild laurel]

Oh yeah, its two authors

[smoky cypress]

Omg the notation used for gcd in dummit and foote is distusting

(a, b)

Who does that

[bleak abyss]

Everyone

When you learn about ideals you'll see why this is good notation

[smoky cypress]

Ok

[magic owl]

@woven delta getting into tor myself

Can you give me a hint about how we think about the fixed module for the tensor?

I’m confused as to why Tor(A,B) is defined with two modules I guess

[woven delta]

How would you define it with one module?

And what makes you think it would make sense with one module only

By fixed module for the tensor you mean B?

@magic owl

So you have a free resolution of a module A and a module B

Tor(A, B)^i is the homology of the chain complex you obtain by tensoring the free resolution with B and cutting off the last nonzero term of the complex

[magic owl]

I’ve always seen it as Tor(M) and I guessed I just kinda

[woven delta]

So let's show that this is dependent on both A and B

[magic owl]

Intuited and blackboard it

Blackboxed*

[woven delta]

So if A or B are free

[magic owl]

How would we pick A and B if we wanted the result to be the torsion part of an fg group

For example?

[woven delta]

What happens?

So we would let B be Q/Z if we are working in Z-module

Oh you're thinking of the torsion subgroup

Do you have a copy of Hatcher?

[magic owl]

Yeah

[woven delta]

You should do section 3.B I think

It's pretty straightforward

For the Z-module case

And it motivates Tor via the universal coefficient theorem

[magic owl]

Honestly I need to review tensor too

[woven delta]

I hear

I had to review tensors like last week

My intuition was too clouded from tensors over a field

[oblique river]

Is there some confusion about what Tor can mean?

If you have an R-module M, then Tor(M) is the torsion submodule

if you have two R-modules A, B, then Tor_i^R(A, B) is the i-th left derived functor of tensor product

they are related: If R is nice (or many for any ring R?), then Tor_1(R, A) should be the torsion part of A, or something like that.

@magic owl

[magic owl]

Ohhhhh

Ok I guess I made assumptions when reading the notation

[oblique river]

actually scratch the lsat thing I said, Tor^R_i(R, A) = 0 for all i > 0

what is true is Tor_1(R/(u), A) = A[u] = the u-torsion in A

at least if R is an integral domain

[magic owl]

Is there an intuitive reason why they are both called tor if they don’t seem to match up that well?

Are derived functors always defined over projectives rather than objectives?

Injectives*

Is there a notion of coderived? functors

[oblique river]

It's called tor because of the last message I sent

it's related to torsion in modules

Tor is measuring the failure of tensor product to be left exact

and torsion is kinda what makes that happen

which is how they're related

and no, some derived functors are defined using injectives

Ext for example

(well, ext can also be defined using projectives)

I don't know what "coderived" would be.

[smoky cypress]

i feel like the exercises in Foote and dummit are much better

I like this book better than Freleigh’s

Thx @mild laurel have a nice day

[clear obsidian]

So I know that isomorphic groups have to have same order & cyclic groups same order are isomorphic, but if they're non cyclic then they don't have to be isomorphic right?

which is why like Z_30 x Z_30 and Z_15 x Z_60 arent isomorphic right?

[woven delta]

A easier example is Z_4 and Z_2xZ_2

[smoky cypress]

Quick question, if two groups are cyclic and are infinite, and they have the same cardinality, are they isomorphic?

[woven delta]

Yes

There is exactly one infinite cyclic group

[smoky cypress]

Oh

[tame bear]

^ it has to be countable

[clear obsidian]

Z?

[woven delta]

Oh I misread what you said sigma

Lmao

[tame bear]

i do that all the time

[clear obsidian]

nvm Z isnt cyclic right?

[tame bear]

it is

[woven delta]

It is

[clear obsidian]

so it is Z

[tame bear]

iso to Z

[clear obsidian]

so is there exactly 1 up to isomorphism or litearly exactly 1?

[tame bear]

up to iso

[woven delta]

Up to isomorphism

[clear obsidian]

thanks

[woven delta]

But that's all we ever talk about

Equality is a myth

[tame bear]

you can just change the names of the elements in the set to make a different group

which is lame

[smoky cypress]

oof

[woven delta]

I only work in categories mod isomorphism, so they are in fact the same object

To me

[tame bear]

ok skeleton boy

[clear obsidian]

ok yeah we dont really care about equality in abs. alg. now I remember

[tame bear]

cant 🅱ost in chill, but i made my collatz checker more than 100x faster than before

[woven delta]

Lmao

[clear obsidian]

what software is the collatz checker?

[woven delta]

Stop trying to solve collatz and do some local cohomology

[tame bear]

i just wrote it in python

[clear obsidian]

hmm cool

[tame bear]

i need to learn alg before i do coho

only 100 lines

[clear obsidian]

reminds me that I should prob stop trying to solve zeta(3)

although I mostly did that by a few weeks ago

[tame bear]

yeah, cuz im gonna solve closed form of zeta(3)

[clear obsidian]

well I already found a meromorphic function that is holomorphic except for the whole numbers so I might not be that far from it (although still quite far probably)

is super messy though, but thats kinda hard to avoid probably

[stone fulcrum]

@clear obsidian
Zn × Zn is isomorphic to Z(n×m) iff n and m are coprime

So Z2×Z2 is not Z4
But Z2×Z3 is Z6

[clear obsidian]

Thanks! that makes sense

[clear obsidian]

what does this mean? so they're proper subgroups G?

[tame bear]

just subgroup

not proper

[clear obsidian]

ok thanks

[smoky cypress]

gotta love this notation when you first see it

[clear obsidian]

ya seriously

umm also, I guess H < G means h is a proper subgroup of G?

it makes sense because it has the analogous result for x in R (a <= b and b <= a implies b = a)

[smoky cypress]

yes

< denotes a proper subgroup

[clear obsidian]

thanks!

[thorn delta]

its not quite clear to me how to use the counting principle to prove lagranges theorem. It seems like you would have to prove that every subgroup can be written as a coset, but its not clear to me how to do this either.... can anyone shed some light on this?

[shrewd halo]

Do you just want to understand lagrange’s theorem? I can’t really help with the counting principle stuff

[mild laurel]

What do you mean by counting principle

You show that you can split your group up into cosets

[thorn delta]

im pretty sure i understand what lagranges theorem is, i just don't quite see why its true. I must be missing something simple because this made sense to me a week ago.

Counting formula
|G| = |H|[G : H] for H a coset of G.

right but why does a theorem of cosets extend to subgroups?

[mild laurel]

That should say for H a subgroup of G right

[thorn delta]

oh wait hold on a second

yea i think thats what I mixed up. I think i got it now thanks

[tame bear]

in case anyone was wondering, Cayleys better theorem states that a group G is isomorphic to the Automorphim group of its Cayley graph
however a Cayley graph is both edge colored and directed
Cayleys best theorem states that a (finite) group G is isomorphic to the Automorphim group of some (finite) simple graph, ie no colors, directed edges or multiple edges
even further there's Frucht's theorem https://en.wikipedia.org/wiki/Frucht's_theorem

for any finite group G there exist infinitely many non-isomorphic simple connected graphs such that the automorphism group of each of them is isomorphic to G.

there are other generalizations for infinite, finitely generated groups, but who cares

[smoky cypress]

[stone fulcrum]

Oh wow, I've got some reading to do

[tame bear]

Groups, Graphs and Trees is the book that uses the term "Cayley's better theorem"

[brisk granite]

Could I have some help with this problem?

I've done everything up till part e, but Idk how to proceed

I think that if I have an element $\tau = \alpha \beta$ where $\alpha$ and $\beta$ are disjoint cycles of different sizes, then, if I pick an $n$ such that $|\alpha| = n$, $\tau^{n}$ fixes at least one element, and isn't the identity

[cloud walrus]

∀ScoopityPoop, Scoop ≠ Poop:

[brisk granite]

Now I don't know what to do when $\tau$ consists of multiple cycles of the same size

[brisk granite]

that are disjoint

[thorn delta]

so I think you can prove part a by showing that there is a bijection between between the right and left cosets of a subgroup. What is the best way to make it "rigorous?" I just kinda wrote

Let y: {xH : x in G} --> {Hx : x in G} such that x_n H mapsto H x_n where x_n in G.

Is this sufficient?

[mild laurel]

You need to show that that function is well defined

And of course show that it is actually a bijection

[thorn delta]

oh yeah... will do.

[static robin]

need some help with this thanks!

i can prove that f is a regular element of (F(E,E),o) with f a bijection.

i can prove that f is injective by f being a regular element

what i can't prove is that f is surjective.

i can rule out one case and that is when E is a set with one element, so necessarly f is surjective.

[woven delta]

What is a regular element?

[static robin]

x is said to be a regular element when it verifies the following : a * x = b * x => a = b and x * a = x * b => a = b

[woven delta]

Oh ok

[static robin]

i think there is another terminology for it but idk

[woven delta]

Left and right invertible

[static robin]

oh inverse !

yes !!

[woven delta]

So you just have to pick a particular function

To show it's surjective

I think

One sec

[static robin]

i'll check it out

[woven delta]

Suppose it's not surjective, and some point p is not in it's image

Consider a and b functions which agree on every point but p

Then a * x=b * x

@static robin

Does that make sense?

[static robin]

Yeah.. kinda ?

[woven delta]

Cause a and b agree on the image of x

For injective you should have done something similar

[static robin]

for injective i took two constant functions g & h and composed them with f : f o g = f o h and since f is invertible i just eliminated it and then i'm left with g = h so x = x'.

[woven delta]

🤔

I'm not sure that works

[static robin]

is that right ?

[woven delta]

Actually yes

It's good

[static robin]

phew.

i got this : we can construct two functions : g & h so that for all x in E g(x) = h(x) <=> x belongs to Im f

and by that g o f = h o f since f is inveritble then, g = h => Im f = E

and if that's true then f is surjective.

but i still can't fully grasp this idea of putting that condition on g and h

[woven delta]

You can always construct such functions

Just take f to be a constant function and change a single value of f

And the second one is g

[static robin]

Got it ! Thanks !

[tame bear]

is xH to Hx even well defined?

from earlier

[woven delta]

What do you mean?

[tame bear]

is the function that takes the coset xH to Hx even well defined
ive gotten
xH = yH => Hx^-1 = Hy^-1
but not Hx = Hy

[static robin]

i think i interupted the other problem, the one before mine :(( sorry

[tame bear]

you didnt, it was an hour before

[static robin]

oh ok :))

[woven delta]

There's a correspondence

Between cosets

It is well defined

xH=yH means there are elements h, h' so that xh =yh'

So (xh)^-1 = (yh')^-1

Oh I see your issue

[tame bear]

its not big deal with regard to the question
just change the map to xH to Hx^-1

[woven delta]

Yeah

[tame bear]

boom, youre done

[woven delta]

The previous map was definitely well defined though

You just gotta do a bit more work

xh =yh' means xy^-1 = h'h^-1, which means xy^-1 is in H

This tells you Hxy^-1 = H

And you're done

[tame bear]

right, my obsession with only using cosets and not the elements got in my way

wait, shouldn't that be y^-1x = h' h^-1

[woven delta]

Doesn't matter

But no

Wait you're right

But it still shouldn't matter

🤔

Yeah this isn't good

Ah well

[tame bear]

:panic:
oh well

[woven delta]

Yeah I don't think the map works actually

At all

🤔

Ah well, I shouldn't spend time on this sort of thing

[bleak abyss]

Okay sanity check

If a matrix over Z/p^2 is invertible, then when we reduce it mod p it's still invertible, and reducing mod p should be multiplicative because yeah in coordinates and shit

So we have a group hom GL_n(Z/p^2) -> GL_n(F_p) that's surjective, just gotta find the kernel

If A reduces to the identity mod p, then det(A) mod p is 1, so det(A) is 1 or p+1, both are units in Z/p^2

So anything reducing to the identity mod p should be invertible itself. Then there are 2 options for each of the n^2 entries

So that should be 2^{n^2}|GL_n(F_p)|

Does this sound right?

[oblique river]

I would just note that a matrix with coordinates in Z/p^2 is invertible iff its reduction mod p is (which you've essentially done)

but it seems like you're forgetting 1+2p, 1+3p, etc

[bleak abyss]

Oh

Right

p^2 \ne 2p

[oblique river]

unless.....

[bleak abyss]

p=2

[oblique river]

🤢

(that's how I feel about p=2)

[bleak abyss]

Yeah it keeps getting in the way

So yeah it's actually p^{n^2}|GL_n(F_p)|

That looks nicer

[thorn delta]

on part a, I know that whenever every left coset is equal to its respective right coset, (i.e. aH = Ha) then H is a normal subgroup, but I'm not totally sure how to approach it here.

With index two, you can name two specific cosets, say x_1 H and x_2 H, but it is not clear to me that the two right cosets of H can't partition G differently. Any hints?

[brisk granite]

One coset that you will always have is the subgroup it self. So, H is a coset

This means that x_1 H and x_2 H have to contain the elements of the group that aren't in H

[fading wagon]

#help-4 for a group theory question

[golden pasture]

fwd qn here?

[fading wagon]

For $n\geq 4$ and any subgroup $G \leq S_n$, we can find a set of generators of G with $\frac{n}{2}$ elements at most.

[cloud walrus]

Element118:

[placid pond]

i can only get n log log n

too bad

up to some constant

[thorn delta]

@brisk granite so 1H=H1=H is one of the distinct cosets and the other would contain elements of of G not in H hmmm

Oh I see. Since we know 1H=H1, and the other coset would be [everything else], xH=Hx

[fading wagon]

@placid pond n log log n sounds interesting. How did you get it?

[placid pond]

@fading wagon the number of primes dividing n!, counted with multiplicity, is n log log n + O(n)

so it's just a trivial argument

[fading wagon]

ah okay

[smoky cypress]

For 24

Are commuting elements a and b just two elements in a group such that ab=ba

[shrewd halo]

yes

[smoky cypress]

Ok

For two elements a, b in a group, is |a||b|=|ab|?

I can see that if the group is cyclic then this is true

Or maybe not

Nvm

[shrewd halo]

what do you mean by the absolute value?

[smoky cypress]

The order of an element

Well I mean, a and b are elements of a group

So | | denotes the order I believe

[placid pond]

@smoky cypress yeah, that's totally false

sorry

[smoky cypress]

Yeah I realized

Even if G is cyclic

[placid pond]

if the group is abelian and gcd(|a|, |b|) = 1, then it's true

well, you can weaken the first condition to [a, b] = e

so a and b commute

[smoky cypress]

Um

What is []

[placid pond]

commutator

aba^(-1)b^(-1)

[smoky cypress]

[placid pond]

so it's the same thing as ab = ba

[smoky cypress]

Oh

But that is equivalent of saying a and b commute

Or that’s what you said

Lol

Ok

[chilly ocean]

@placid pond that's a bit redundant

[placid pond]

what is

[chilly ocean]

never mind, I realized that you wanted to point out that a and b commute, and not necessarily the whole group, then it still holds

[placid pond]

yes

[smoky cypress]

I hate the notation for (a, b) for gcd

Ugh

[bleak abyss]

I mean in algebra it makes sense

You know what an ideal is?

[smoky cypress]

nope

reading abstract algebra recently

[bleak abyss]

So let's say R is a ring

A subset I is an ideal if it's a subgroup and if x\in I and r\in R => xr \in I

[smoky cypress]

for all r?

what's the intuition

[bleak abyss]

Think of multiples of a given element

[chilly ocean]

@bleak abyss and what about gcd

[bleak abyss]

Say multiples of 2 in Z

[smoky cypress]

oh ok

i can see that

[bleak abyss]

Sum of even numbers is even, anything times an even number is even

Turns out to be very important, and there are many examples that don't look like this

[smoky cypress]

wait you said the ideal is a subgroup, but under which operation?

[chilly ocean]

also very important in semigroup theory

[bleak abyss]

Addition

[smoky cypress]

ok

[bleak abyss]

I mean a ring isn't a group under multiplication

And an ideal isn't a submonoid since it doesn't contain 1

[chilly ocean]

yes

[smoky cypress]

i see that the ideal is closed under multiplication

[bleak abyss]

It is closed under multiplication, but if an ideal contains 1 then it's just the whole ring

Since x = x*1 \in I since 1\in I

[chilly ocean]

Yeah. Since SI = S = IS then (S - ring, I - ideal)

[bleak abyss]

So for the most part ideals aren't submonoids

Anyway the point here is

Ideals play the same role in ring theory that normal subgroups do in group theory

If x\in R, the even numbers example does have an analogue

Namely, you just take the set of "multiples" (in R) of x

That's denoted sometimes as xR but also common as (x)

[chilly ocean]

Ah. The ring generated by x

[bleak abyss]

The ideal generated by x

(Similar notation to the brackets used for span and generating sets in group theory)

Now if I'm using (x) notation, which is p standard

[chilly ocean]

it doesn't have to be an ideal, does it

[bleak abyss]

(S) is usually defined as the ideal generated by S, though it's all context dependent

You could look at the subring but that's way less common, and if you're doing it you wanna use a different notation

So anyway yeah (S) is the ideal generated by S (intersection of ideals is an ideal, so this notion makes sense, prove it)

[chilly ocean]

Are you assuming the ring is abelian?

[bleak abyss]

For now yeah I don't feel like thinking too hard

[chilly ocean]

well, could of just took $RxR\cup x$ instead

[bleak abyss]

I don't see what that would've done for us here

[cloud walrus]

Blitzkrieg:

[bleak abyss]

Wut

[chilly ocean]

ideal generated by x

[bleak abyss]

Okay wait lemme finish my story so you can see where I'm going

[chilly ocean]

you forgot to add x

ok

[bleak abyss]

x = 1x

Anyway so

Let's assume we're chill with (S) = ideal generated by S, which we are

[chilly ocean]

does it has to be a ring with identity though

[bleak abyss]

Rings without identity are the antichrist

They're also ideals in rings with identity

[chilly ocean]

nah. Just for category theorists

[bleak abyss]

Also for number theorists

I guess maybe people who care about Banach algebras do it since C_0(X) doesn't have unit when X isn't compact

And maybe from the cat theory side you could make the case that if you don't require homs to preserve 1, then 0 is a 0 object

[chilly ocean]

you can always attach identity anyway

[bleak abyss]

But also who cares about that?

Yeah exactly so I find it better in any context I see to just roll with identity

I've yet to lose anything for it

[chilly ocean]

ok, continue

Pleas?

[bleak abyss]

Yeah sorry I was out for a sec

So to have everything in one place

Let R be a (commutative, unital) ring and let S be any subset

Since any intersection of ideals is an ideal (exercise), we can speak of the smallest ideal containing S, namely the intersection of all ideals containing S

Call that (S)

In Z, the ideal (m,n) happens to be the ideal (gcd(m,n))

So that kinda makes the notation more forgiveable

[chilly ocean]

oh, I see

[bleak abyss]

@smoky cypress just focus on the last few messages lmao

[chilly ocean]

addition was a bit pointless imo

[bleak abyss]

Wut

[chilly ocean]

we didn't use it

[bleak abyss]

I mean sure but (x) is only used in ring theory really

In group theory this is the subgroup xZ

[chilly ocean]

semigroup theory, principal ideal

its the exact same

[bleak abyss]

I mean I feel number theory likely doesn't care about semigroups too much, tbh I'm not familiar with the pure algebra of semigroups

Chances are semigroup theory isn't established enough to even have standardized notation

[magic owl]

What is this obsession w semigroups lol

[bleak abyss]

I feel like semigroups are pretty much only a thing in like, maybe dynamics

Since you can think of a dynamical system as a semigroup action

[chilly ocean]

J(x) is notation for principal ideal in semigroup theory, but it can be (x) if you want

idk, but so far I've done every exercise from semigroup theory I could, so I feel a bit in love

[bleak abyss]

I mean sure you can make any notation anything if you want, my point is that because ring theory is important in number theory and ring theory commonly uses (x) notation, it makes sense for number theory to write (m,n) for the gcd since in ring theory that actually works

[chilly ocean]

yes. Why was I arguing anyway

[bleak abyss]

The notation from semigroup theory doesn't really have any reason to influence number theory, and if J(x) is what's common then empirically we know it doesn't. If not then chances are it borrowed it from ideal theory

That's what I'm wondering

But yeah so idk if the algebra of semigroups is at all useful, as I said semigroup actions are basically dynamical systems but I have no idea if there's any algebraic fact about semigroups that really tells you something with content about dynamics

[magic owl]

Is gcd some kind of categorical limit?

[chilly ocean]

I mean, ring theory would have similar notation if they didn't consider just commutative rings

[magic owl]

Like if we make morphisms divides

[bleak abyss]

I mean, they don't

Non-commutative algebra is 100% a thing

[magic owl]

I think it would be a limit

[chilly ocean]

it would be a directed limit, just be maxj, if you set it up correctly with "divides" as arrows

[magic owl]

Yeah that’s what I came to

[chilly ocean]

actually inverse limit. directed limit for lcm

Ik, but (x) is clearly a notation from people who only consider Abelian rings

[bleak abyss]

I mean, it's the notation of commutative algebra

[magic owl]

It should be the limit is the strict categorical sense

As opposed to Colin it

[chilly ocean]

sure

[magic owl]

Colimit

[bleak abyss]

Turns out, low level number theory makes WAY more use of commutative algebra than non-commutative

[magic owl]

But I mean that’s up to taking the opposite category anyway

[chilly ocean]

yes

[magic owl]

And there’s no real rhyme or reason to which direction we make our arrows

[bleak abyss]

So it's more likely that it'd borrow their notation

[magic owl]

Wait no is there?

[chilly ocean]

Well, we distinguish between left right and double-sided ideals

[magic owl]

Composition might get wonky

[bleak abyss]

Idk connections of NT to non-commutative algebra. I'm sure something's there since you care about representation theory

Yes we do, you're completely missing my point

[magic owl]

No it works

[chilly ocean]

some parts of more advanced NT would use substantial amount of noncommutative algebra as well. up to class field theory, some basic arithmetic geometry etc., much more commutative algebra

[bleak abyss]

My point is that there are two almost different subjects

Detailed study of commutative rings

Detailed study of noncommutative rings

[chilly ocean]

sounds fun

You do, but mostly you don't

[magic owl]

Commutative rings ❤️

Give me those s h e a v e s over a base field

[bleak abyss]

Since commutative rings don't have to worry about left vs right vs double sided ideals, it makes sense for them to just talk about the ideal generated by a set

So they don't have a reason to use the more nuanced notation of noncommutative algebra

Hence it makes sense that the two subjects have different notation

[magic owl]

Although sheaves in AG probably care more about rings in general

[chilly ocean]

Well, if you count bi-ideals and quasi-ideals, we have 5 types

sheaves in AG care more about commutative rings and "glueing them" with the organizational tool called sheaves

and sheaf cohomology

[bleak abyss]

Now turns out, it makes more sense for elementary NT to use the notation of commutative algebra than noncommutative algebra, since it's influenced a bit more directly by non-commutative algebra

Hence it makes sense for NT to use (m,n) = gcd

[magic owl]

I wasn’t sure about the commutative part

[bleak abyss]

I don't see what's the problem here

[chilly ocean]

there's quite a lot.

[bleak abyss]

quasi-ideals? What's that?

Actually will I care?

[magic owl]

Lol

[chilly ocean]

(of comm. ring theory at least initially. most of EGA and/or hartshorne level AG assumes that rings are commutative with 1)

Probably not. I was just saying we do

[bleak abyss]

Sure and people who study aleph_{17} in detail care about aleph_{17}, I just don't see what it's adding to this discussion

I just answered an easy question and this discussion is spiraling into madness

[chilly ocean]

Yes, let's stop

[bleak abyss]

Good

Okay hochs I haven't noticed you before but eyyy you like NT?

[chilly ocean]

I used to do so much NT and AG / arithmetic geometry some years ago. I don't do much of that nowadays since I work in the industry, but I still like hearing about them time to time 🙂

[bleak abyss]

(Maybe I did notice you before and my memory of internet people is just bad in which case I'm sorry 😭)

Ah nice

[chilly ocean]

doubtful, since I basically just joined very recently from a recommendation at IRC

[bleak abyss]

Yeah idk much NT or AG at the moment, beyond kindergarten stuff plus vague impressions

But there's a decent chance I'll go into it in the future

My top adviser choice atm is principally an arithmetic geometer

[chilly ocean]

oh cool

which ones do you have in mind?

[bleak abyss]

(Assuming I continue with NT/AG and like it, obv if I start working through Neukirch or smth and end up hating the subject I'll switch but...)

[chilly ocean]

Idk how true it is but I heard from one doctor from America, that number theory will probably be the biggest object of studies in recent years

[bleak abyss]

Jordan Ellenberg is my top choice

Followed by Laurentiu Maxim

Dima Arinkin

[chilly ocean]

oh yes I have a friend who works with him currently at wisc

[bleak abyss]

Those are my top 3, I have a few other possibilities in mind for sure but prob gonna end up working with one of the three

[chilly ocean]

(ellenberg that is)

seems pretty happy with him

[bleak abyss]

(Maxim does like, topology of singular varieties pretty much? Arinkin does geometric rep theory)

Yeah Ellenberg seems super hype. I haven't started working with him, I'm just going into first year now

But my impressions of NT and AG make them seem fun, and also Ellenberg himself seems very broad, thinking also about rep theory and topology which I love

So yeah gunning for him unless he has too many students (which may happen, NT at Wisc is suffering a bit since Melanie Wood went to Berkeley and Nigel Boston retired just now, and Tonghai Yang is chair and will soon go on sabbatical, so yeah Ellenberg is kinda the main guy, maybe Simon Marshall if you count him)

How are you liking your current industry work?

[smoky cypress]

uhhhh

this is too much

i was gone for a second

then

there is like 100+ messages

[chilly ocean]

not as much as I like NT and AG, but I do need some income 🙂

[bleak abyss]

Lol that discussion got derailed hard, the messages from "So to have everything in one place" to when I pinged you are what matter

[smoky cypress]

lol ok

and also

I'm reading Dummit and Foote's Abstract alg

I'm probably will get to those stuff fast

[bleak abyss]

And yeah fair, it's prob a bit early to be thinking about this but the prospect of getting btfo'd by academia are p real so industry is on the back of my mind

[smoky cypress]

Since I already know about groups and rings

[bleak abyss]

Also ughhhhhh D&f is such trash

[chilly ocean]

i liked D&M. maybe had a bit too much on sylow stuff though

[smoky cypress]

?

what

[bleak abyss]

There are two types of people

[smoky cypress]

i like this book

[bleak abyss]

D&F has everything which is good but like

It's made out of the active ingredient of sleeping pills rather than out of paper

It's so boring

[smoky cypress]

oh

[chilly ocean]

I used to have Matsumura's Commutative Ring Theory book next to my bed too to help with sleep

[smoky cypress]

why is that so

im not sure what you mean by that

[bleak abyss]

Idk I feel like it's written as a sequence of facts

If you can read it then definitely read it lmao

[smoky cypress]

oh oh oh

[bleak abyss]

Just that I've literally fallen asleep IRL with that book

[smoky cypress]

i would agree

yes

[bleak abyss]

Since he just takes so long to say anything

And his chit chat just makes it seem like he doesn't like the subject either but was told at gunpoint to talk about it

[smoky cypress]

I really like the kind of book just list out facts

Well

That's a bit extreme for me to say that

[bleak abyss]

Jacobson Basic Algebra is one I used a bit of for qual review

It's real good

[chilly ocean]

yeah I used Jacobson as well (Vol 1 and some of 2)

(for the quals long ago)

[smoky cypress]

I like to have two books, where one is super long and talkative about intuitions, and the other is just a list of facts

[chilly ocean]

you might like Eisenbud's Commutative Algebra with View Towards Algebraic Geometry

paired with Matsumura's more dry, bourbaki approach

[smoky cypress]

haha

[chilly ocean]

for commutative ring theory anyway

[smoky cypress]

but i don't know algebraic geometry

[bleak abyss]

Contains a lot of non-standard content (Vol 1 starts with the standard group/ring/module/Galois theory, but then goes into a lot of detail on a bunch of stuff like classical groups, quadratic forms, etc, and Vol 2 is more on advanced topics), and also I feel his treatment is the most efficient for minimizing repeated work on axiomatics

Like when he does group theory he throws in a lot of stuff on monoids

[smoky cypress]

I have Jacobian too

[bleak abyss]

e.g. he defines congruence on a general monoid as just an equivalence relation such that multiplication descends to equivalence classes

He then proves that if you're a congruence on a group, then you're quotienting out by a normal subgroup

But does some stuff in the generality of congruences on monoids

You're thinking, okay whatever what's the point of that? Thing is, when he goes to ring theory, a ring is a group that's also a monoid with distributivity holding

And a lot of the ring theory stuff is basically just double citing the monoid result

Which I at least appreciated, sped things up a bit

[smoky cypress]

This book?

[bleak abyss]

Artin is supposed to be good for getting someone who doesn't care about algebra to care about algebra but idk it myself

Yeah that

But books don't matter that that much

[chilly ocean]

categorical unification when done right is a beautiful thing

[magic owl]

Who learn GT without knowing what a Cartesian product is

[chilly ocean]

grothendieck was a master at this too

[bleak abyss]

D&F has all the content you need

So work through it

[smoky cypress]

lol max I first learned abstract alg without knowing cartesian prod

@magic owl

[magic owl]

Lmao

[bleak abyss]

I just personally get too bored, I prefer talking to symbols (e.g. module is ring hom R->End(M) instead of listing out symbolic axioms) and don't like dragging

Lol it's funny how I hate so many standard books with a strong passion

[woven delta]

I need something to motivate my Algebra

[bleak abyss]

Hatcher and D&F

[woven delta]

That's why I like AT

[bleak abyss]

So I got into algebra basically through Sylow theory

[woven delta]

Oof

[bleak abyss]

My intro was in this REU paper I wrote summer after my first year

[magic owl]

Oof

Sylow is horribly dry imo

DF treatment of sylow is top tier boredom

[woven delta]

I don't like the problems in algebra

[bleak abyss]

I was like, okay algebra sounds hype, and at the time I was in this apprentice program so we had a 5 week class on linear algebra and graph theory

[smoky cypress]

Ok @bleak abyss this might be too much to ask, but can you give me like some books on abstract algebra and compare them a little bit?

[woven delta]

I need some reason to care about Algebra

[smoky cypress]

just like 3 or 4

xD

[bleak abyss]

And they say in the description that apprentice topics ideally should relate to lectures

[magic owl]

I find categorical unification makes me interested in algebra

[bleak abyss]

So I was like okay, algebra sounds fun, linear algebra is associated, and there's some graph theory

[magic owl]

Like I like learning how the results come together in the more general theory

[bleak abyss]

So what should I do my thing on? Group actions on graphs

[magic owl]

Like first isomorphism in abelian categories

[woven delta]

[bleak abyss]

But I didn't know groups so the first step is to learn groups

[chilly ocean]

a lot of the modern post-grothendieck-revolution algebraic geometry requires that you know tons of algebra, if you care about algebraic geometry

[woven delta]

Yeah but that's a good reason to study it

Can you imagine being a group theorist?

I would rather work in industry

[smoky cypress]

i would want to be a group theorist

tbh

[bleak abyss]

But eventually my adviser was like wait up you don't actually have to do graph theory

And said he felt Sylow would be a good holy grail for my paper

[magic owl]

Whoever learns only group theory “I want to study group theory”

[bleak abyss]

Turns out I like combinatorics

So yeah Sylow was super fun for me

I mean tbf representation theory

[smoky cypress]

why you gotta say the truth

[bleak abyss]

It's so hype

[woven delta]

Rep theory of groups seems pretty cool

[chilly ocean]

liquid how 'bout a study of some specific lie groups, their representations, or heck even the mysterious absolute galois group Gal(\overline{Q}/Q)

[bleak abyss]

Git fuckin sniped you nerd @woven delta

[woven delta]

Goddamn

I don't know if I care about Geometry either tbh

[magic owl]

I like my algebra applied but I hate specific algebraic structures lol

[bleak abyss]

Lol so, when we were naming the channels in advanced math we kinda got stuck on naming the NT channel

[chilly ocean]

hell breaks loose in trying to attach meaningful representations to the absolute galois group (e.g. fontaine mazur)

[bleak abyss]

The current structure was in response to how we had "advanced geometry" as a channel, meant for like DG and all, and 9th graders felt they were advanced rel 8th graders so they'd ask in there

[woven delta]

I remember that

It was pretty bad

[chilly ocean]

that's kind of cute

[woven delta]

I don't think undergrads usually do much geometry

[bleak abyss]

So we were like okay "Advanced X" doesn't prevent high school questions well enough, so "Advanced NT" won't cut it. Eventually decided that Galois theory was close enough to put it there. It's not optimal since Galois theory is also algebra

[woven delta]

Outside of certain places

[bleak abyss]

But at one point I was thinking of titling the channel "Representation Theory of Gal(\overline{Q}/Q)"

[chilly ocean]

that sounds awesome

[woven delta]

@chilly ocean are you a grad student?

[bleak abyss]

Since my grad student mentor the second time in the REU was a number theorist and told me once that that's modern NT

[chilly ocean]

not anymore. working in the industry nowadays

[woven delta]

Oh cool

[chilly ocean]

but I love hearing about these things still 🙂

[bleak abyss]

Tbh I don't know anything about Galois reps aside from Tate module is a thing that exists

But yeah hope to learn soon

[magic owl]

Me to me: brush up on the details of the many fields you’ve learned in a cursory way

[bleak abyss]

Nerd shit

[magic owl]

Also me to me: i n f t y c a t e g o r i e s

[bleak abyss]

Be like me

[woven delta]

I'm down to specialize already

[bleak abyss]

Learn everything poorly but j u s t enough to survive for another day

[woven delta]

I'm more down to work with people than to work with textbooks these days

[magic owl]

Should I bother with complex dami

I’m slotted tontake it rn

But I feel like I won’t like it

[bleak abyss]

Who's teaching?

[magic owl]

Let me see

[bleak abyss]

But yeah I'm not gonna specialize yet

[woven delta]

I want to find an advisor by the end of the year and finish quals

[bleak abyss]

NT/AG is my tentative eventual field of study, but topology and rep theory are very close seconds

So I'll want to get a clearer picture to decide

[woven delta]

How do you guys deal with burnout btw?

I've had it before, but I've always gotten over it

[magic owl]

Looks like instructor isn’t listed

[woven delta]

But I have no idea how

[magic owl]

I just start learning something new

[bleak abyss]

Also even if I choose one now, in many fields there's still a lot I'd like to learn more of. In topology, for instance, I wanna learn... honestly basic AT properly, definitely stuff surrounding K-Theory, characteristic classes, vector bundles, some difftop

[woven delta]

That's fair

[magic owl]

But I have the freedom to do that

[bleak abyss]

Spectral sequences, maybe simplicial stuff

[woven delta]

I should play video games

[bleak abyss]

Even if I don't end up going into that area

[woven delta]

Unfortunately the semester starts next week

So I don't have much time to relax

[bleak abyss]

Yeah burnout I feel is best handled by taking some time to be very unstructured/understructured

Maybe abandon math for a short period of time, but even if you don't, just do math very casually/aimlessly, just for fun

Once math no longer has too much of an association with deadlines over your head and brute force work then try to organize yourself a bit

Nice

[woven delta]

Aside from my disgusting teaching schedule

Lmao

[bleak abyss]

F

Yeah hopefully I pass this topology qual (and that I passed algebra today)

If so then things will be smooth re technical matters I hope

[woven delta]

Something really weird happened like 30 minutes ago as I was walking home

[bleak abyss]

Side note: we should migrate to general

[woven delta]

Yeah

[clear obsidian]

so groups of prime order are cyclic

does this mean that theres only 1 group order p up to iso?

wait a sec I guess it does nvm

bc cyclic --> albelian so order p^1 and the partitions of 1 is just 1 so theres 1 structure

[magic owl]

If you have two cyclic groups of the same order

Map the generator to the generator for an iso

[clear obsidian]

The generation by 1 element makes it easy yeah

so if I want to calculate number of groups of a large composite order, say like 460 then theres no easy way right?

[chilly ocean]

run a program

[clear obsidian]

im prepping for the math gre but yeah true

[chilly ocean]

there are whole libraries of such things

[clear obsidian]

so it just makes structures and finds iso's right?

[chilly ocean]

I'm not sure if they go as far as order 460 though?

[clear obsidian]

I guess you could just get it to start making arbitrary structures represented by matrices

[chilly ocean]

lists every group (up to isomorphism)

not really

actually, maybe

[clear obsidian]

I mean obviously they need to be groups so that narrows it down a lot

but how many computations does it take to check for group along the way instead of just making full structures arbitrarily

[chilly ocean]

Yeah, I guess this is how they find them, more or less

[clear obsidian]

e.g. generate an arbitrary nxn matrix

or whatever2 dimensional commands that would serve this purpose

of course with small numbers its easy but gets progressively more complicated I assume

(acknowledging that complexity here may be vague)

[chilly ocean]

https://www.gap-system.org/Manuals/doc/ref/chap50.html

https://www.gap-system.org/Manuals/pkg/SmallGrp-1.3/doc/chap1.html#X7C16EA1580AC7586

of order at most 2000 expect 1024, so all groups of order 460 should be there

idk why 1024 isn't there 🤷🏿

[stone fulcrum]

There are literally millions of groups of order 1024

They probably were just like "nah"

[chilly ocean]

why 1024 though

[magic owl]

Honestly

Fg ab groups

[stone fulcrum]

Highly divisible? That's my guess

[magic owl]

Are the only ones that I like

[chilly ocean]

ah

[magic owl]

1024 is super-composite

[tame bear]

most finite groups are of order 2^n

[magic owl]

Which doesn’t by itself guarantee lots of subgroups

But yeah that’s the intuition

[tame bear]

https://golem.ph.utexas.edu/category/2012/11/almost_all_of_the_first_50_bil.html

[woven delta]

Lmao

https://www.amazon.com/Groups-Order-Senior-J-K-Hall/dp/B002GQQ11S

[chilly ocean]

this

[smoky cypress]

bad latex

[magic owl]

Groups of order n (n=0)

[smoky cypress]

lol then you don't even have identity

[woven delta]

[tame bear]

inverse implies identity

and all elements have inverses

[woven delta]

Groups are pointed

So you need one element

At least

[magic owl]

That’s why my book is so short liquid

[chilly ocean]

🤔 If we look at groups as universal algebras, then they must contain a 0-nary which is the identity

[magic owl]

Huh

I mean sure

But that’s pretty redundant

“Groups are defined to have identity and thus must have identity”

[smoky cypress]

If $P(x)$ is a polynomial with rational coefficients, $a,b\in\bbQ$, and $\sqrt{b}\in\bbR\setminus\bbQ$, if $P(a+\sqrt{b})=0$ then $P(a-\sqrt{b})=0$

[topaz solar]

That sentence parses but it shouldn't need to be said

[cloud walrus]

Whoever:

[smoky cypress]

is there a generalization of this?

[topaz solar]

ever heard of galois shit?

[smoky cypress]

Ik galois theory exists but haven't actually read anything about it

[topaz solar]

from what I remember, it's relevant

[tame bear]

the zeros of a real coefficient polynomial are symmetric about im = 0

[chilly ocean]

Ik, I learned that few hours ago though lol.

[smoky cypress]

No idea what sigma said but sure

[tame bear]

😠

[topaz solar]

construct $k \in \mathrm{Gal}(\bar{\bbQ}/\bbQ)$

[cloud walrus]

Darkrifts:

[tame bear]

if i is a zero of a polynomial with real coefficients, then so is -i

[topaz solar]

*k not being the identity or the loser conj boy

[chilly ocean]

he said that conjugate is a root as well

[smoky cypress]

yes

i can see that

[tame bear]

its related

[smoky cypress]

yeah that what I felt

I've heard about splitting fields, kind of

um