groups-rings-fields

[oblique river]

yep that's right

so now what are algebraic functions on U? well, all our usual polynomials are

but now we have things like 1/x

that isn't defined at 0 but it is defined on U!

[urban acorn]

I see

[oblique river]

in general you can form any rational function of the form f(x)/x^n

[urban acorn]

okay

so they're rings of fractions

[oblique river]

and that ring will be the ring associated to that open set U

that's right

[urban acorn]

over polynomials which are potentially 0 on our closed set

okay

so

the second question I have about schemes

was about the construction of general schemes

so, presumably, a scheme is a topological space such that for each element there is some neighborhood of it homeomorphic to an open subset of an affine scheme? and then also with a sheaf and a requirement for that sheaf to be compatible with the sheaves on the affine schemes

are there requirements like the space being second-countable? like for example is often required of manifolds

so they don't become uninteresting aggregates of egregious amounts of space

[oblique river]

I think of those requirements as more just like

analog of the definition of a manifold

as "locally looks like R^n"

[urban acorn]

so like a manifold is a topological space that locally looks like R^n, a scheme is a sheaf that locally looks like Spec(R)?

[oblique river]

a scheme is a "locally ringed space" which looks locally like Spec(R)

a "locally ringed space" is "a topological space with a sheaf that satisfies some extra property"

[urban acorn]

the problem is for topological spaces I can actually imagine geometrically how for some point the space looks like R^n, and I can't do that for sheaves

I guess that's something that I'll need to learn this more rigorously before I understand

[oblique river]

yeah, so that is a really good question in general and something which takes getting used to in algebraic geometry. for example, what should Spec Z[x] look like?

it's much bigger than just Z. While (x-a) is a prime ideal for each a in Z, there are a lot more prime ideals, like (3) or (7, x^2 + 1)

this is why it's a good idea (in my personal opinion) to learn about the case of curves over C and R first

so that you get a good intuition for how to translate geometry into algebra and vice versa

so that when you start reading things like "let R be a ring and consider the scheme Spec(R)" you have some kind of picture you can go back to

[urban acorn]

I see

[oblique river]

when i think about htis stuff a lot of times i do just squint and try to imagine the case of like, C or R

[urban acorn]

to you, do schemes feel like natural geometric objects the same way that manifolds are?

do you think algebraic geometry is geometry?

[oblique river]

hmm, so in thinking about that question i'm reminded of a quote from von neumann (i think) which is soemthing along the lines of "in math you never understand things, you just get used to them"

chm please

go easy

[urban acorn]

I like that quote, but it's only true in a certain way

[next obsidian]

Just gonna put in my bit then leave, I don't see geometry in this stuff, that Von Neumann quote is definitely how I feel about it

but some ppl def see a lot of geometry, I wish I could see what they see

[oblique river]

ok phew sry chm i assumed yuou were gonna come in and start blasting him with crazy AG stuff

[urban acorn]

lmao

[next obsidian]

That's probably just because I'm not too visually oriented and also how I went about learning it

I think going slow and working up rather than just going Scheme + adjectives Hartshorne exercises brrrrrrrrrrrrrrrrr is better

[oblique river]

so like, I do think there is a spectrum (heh) of geometric visualization

on one end, if you're doing like, varieties over C, then yeah that's very geometric

and there are also theorems which relate like

"properties of V as an analytic space" to "properties of V as a scheme"

so you can get really concrete geometrically there

[next obsidian]

Yeah

[oblique river]

but if your rings are like, not at all like C

[next obsidian]

Complex geometry is like

[oblique river]

then that becomes a little farther from classical "geometry"

[urban acorn]

that reminds me of something someone said to explain to a layman how mathematicians think about higher dimensional spaces

[oblique river]

yeah that's a good analogy I think

[next obsidian]

There's still stuff that's like

reminscient enough that you can kinda make some sense

[urban acorn]

"when you hear 'n dimensional space', you imagine a 3 dimensional space and say 'n dimensional space' in your head very loudly"

[oblique river]

yeah hahaha

like nobody really can visualize 25-dimensional space

but you can visualize geometry in 3-dimensional space

[next obsidian]

like if you're familiar with a bit of manifolds you might know what a bundle is, and they exist in AG too

[oblique river]

and you can prove theorems about how 25-dimensional geometry is similar to/different from 3D geometry

and that kind of helps you "visualize"

[urban acorn]

yeah

[next obsidian]

so you can pretend to know what a like P1-bundle over something is and P1 is something you can... sorta visualize and then bundle is kinda like, something you can visualize

[oblique river]

here is a "visualization" which will make no sense to you but maybe more sense to chmonkey

[urban acorn]

I recently managed to sort of visualize the 3-sphere

[oblique river]

this is "mumford's treasure map" and it's a drawing/representation of Spec Z[x]

https://pbelmans.ncag.info/atlas/mumford-original.png

[next obsidian]

Bruh

I have seen this so many times and know what all the info is

but it does nothing for me

[oblique river]

oh, well I found it helpful

and it kind of informs how I see Z[x]

[next obsidian]

This doesn't really help me think about it very much haha

[urban acorn]

it looks pretty cool visually

[next obsidian]

Yeah for me I see Z[x] as a polynomial ring and start thinking of a lot of polynomials and like the lattice structure of the prime ideals

[oblique river]

I mean, here's how I think it's kind of a helpful visualization

the krull dimension of Z[x] is 2

and here is a visualization which looks 2-dimensional

[urban acorn]

I like the fact that Z[x] represents the forgetful functor

[oblique river]

that's a very algebro-geometric statement!

you already ahve the mind of an algebraic geometer :^)

[urban acorn]

lol

[next obsidian]

"object is just functor"

[urban acorn]

lmao

[next obsidian]

already on your way to defining a redacted because buncho will yell at me

jk, but that's actually useful, just not really much at first IMO

I think a lot of ppl will say AG is __ but it doesn't become really become like __ until a lot of time has passed

this may or may not be true of other fields

[oblique river]

what chm is trying to say is that the perspective "Z[x] represents the forgetful functor" isn't useful until it becomes useful

at which point it becomes very useful

[next obsidian]

yeah

I just didn't get much out of trying to think all categorical and abstracty and functory

until it necessitated it

[oblique river]

yeah

going back to what knida started this convo

I think AG does have a bad reputation as being like

[oblique river]

super fancy and abstract and stuff

[urban acorn]

I only understand like a third of this but lmao

[oblique river]

but really a lot of the ideas are concrete

and rooted in complex geometry

and starting there and building up I think is pretty accessible

[next obsidian]

Buncho UW has a complex geometry class this winter but you need like RG and bundles course and stuff to take it I thinjk

so I am very sad

[oblique river]

aww thats too bad

[next obsidian]

I might still try to take it anyway, but I have wanted to learn complex geometry for a while

It sorta makes sense given who's teaching it and that they taught RG and bundles last year in a topics sequence

but it sucks taht if you didn't take that sequence you kinda are SOL next year

[urban acorn]

I don't know anything yet, but this interplay between the "spaces" and purely algebraic rings fascinates me

[next obsidian]

I feel like GAGA (the thing buncho mentioned which basically translates complex AG to complex geometry / analytic spaces) is like

the purest form of that

it relates things which are like by all means spaces, even without having to stretch your brain at all to consider these weird AG "spaces" as actual spaces

to... well those weird AG "spaces" which are reflecting all this ring info

[oblique river]

me too :^)

[next obsidian]

Like here's a thing that's cool is that for people doing complex AG sometimes it's really beneficial to know a lot of complex analysis because you can just solve the AG problem with complex analysis

And there's tricks you can do to reduce problems to statements over C and that stuff always blows my mind, C is really this magical object IMO

[urban acorn]

it's weird that all of this structure can be obtained just by playing with definitions and rules for manipulating symbols

[oblique river]

formalism moment

[next obsidian]

I feel like if you mean how sometimes things can be easier to solve in different contexts, even though the things are "equivalent" this shows up a lot I think, which is really cool and I like it too

Algebraic groups and Hopf algebras are super related like this as well, and sometimes it's easy to look at something in the algebraic group context, other times the Hopf algebra perspective is better. So you can take your pick on how to solve something, then you just have to translate it back over

[urban acorn]

what are algebraic groups?

[next obsidian]

It's like a Lie group

but for varieties

So a lot of things you know are algebraic groups. A lot of matrix groups for example

[oblique river]

i.e. "it's a group but it's also a variety at the same time, and the structure of G as a group is compatible with its structure as a variety in some technical way"

[next obsidian]

Like SLn, GLn, etc

It's cool because varieties are really geometric because they are embedded in affine space and the group structure is kinda like automorphic ways to shift them around

I mean at the most basic level K^n is an algebraic group because you can literally just add points together

[urban acorn]

aha

I remember hearing something about cubic curves

and finding the point between two points

is that related?

[oblique river]

elliptic curves?

[urban acorn]

maybe

[next obsidian]

Elliptic curves are an algebraic group

[oblique river]

yeah thats an example

it's a variety but it also has a group structure

[next obsidian]

That sounds like maybe what you are thinking of

you can grab two points and kinda like... produce this new point

and there's a point at infinity which is the identity

I forget the process but there's a geometric way to do it by drawing some lines and finding some intersection

[upper pivot]

Yeah draw a line and find the third intersection, and then reflect it along x axis.

[urban acorn]

are the curves symmetric over the x axis?

[oblique river]

well

[upper pivot]

Right so you can reduce cubics to something called Weierstrass normal form, which is

[oblique river]

yeah john is explaining

[upper pivot]

Like every cubic by some rational transformation is of form

y^2= cubic in x

[next obsidian]

you don't even need an x^2 term right?

[urban acorn]

is a rational transformation the equivalent of an affine transformation in projective space?

[upper pivot]

Yeah you can get rid of the x^2

[next obsidian]

I think you only need linear substitutions?

[upper pivot]

I don’t remember but I think that may be true

[next obsidian]

so you replace x and y with polynomials in x and y

or like... introduce two new variables X and Y

set x to a polynomial in X and Y, and y to one in X and Y and then you get something like

Y^2 = aX^3 + bX + c

[upper pivot]

The transformation steps are a bit tedious lol

[next obsidian]

yeah

[urban acorn]

and this is true for C^2? or for a general field? or what?

[upper pivot]

Any field not of char 2 I think?

Something like that lol

[next obsidian]

I think char 3 is an issue too

[upper pivot]

Oh yeah might be

Right yeah it is to get rid of the x^2

[next obsidian]

There's a ton of stuff on elliptic curves

they find use in cryptography

[upper pivot]

But yeah like the theory of this is really cool

[next obsidian]

as in all modern ones are based on elliptic curves I think

[upper pivot]

From a geometric viewpoint

This is just a torus.

[next obsidian]

also

:(

Everything reminds me of her, even elliptic curves

sad moment

[upper pivot]

Sadge

[urban acorn]

sed

[upper pivot]

There’s a really cool function called the weierstrass p function

[urban acorn]

so chm's ex's curves formed an algebraic group?

[next obsidian]

no

she really liked elliptic curves

and I learned everythignn I know about them form her

[upper pivot]

That allows you to go between Torii and elliptic curve, and in fact the elliptic curve inherits the group structure from the torii through this map.

[urban acorn]

oh :(

[upper pivot]

Anyhow I’ll let u guys go back to talking about the general theory lol

[urban acorn]

okay then

can I change ax^2 + bxy + cy^2 + h into Ax^2 + Cy^2 + H through an affine change of coordinate?

[terse crystal]

like I said , yes

[urban acorn]

what are some examples of affine schemes that are really pathological? what is it about the underlying rings? what kind of schemes do - say - integral domains produce?

yeah, but do you have just a simple construction of what A and C would be?

cause I didn't follow the explanation involving orthogonal and symmetric matrices

and it's probably unnecessarily high-powered for what I want

maybe later I'll look into that perspective

[oblique river]

there is a geometric analog of being an integral domain but it's a little complicated

but basically it means that you're irreducible topologically

something like two lines crossing is not irreducible because it has two proper closed pieces whose union is the whole space

for example, if you take the ring C[x]/((x-1)(x-2))

V((x-1)(x-2)) is the union of two points so it's not irreducible

and correspondingly that ring is not an integral domain

[urban acorn]

so, x-1 isn't 0

and x-2 isn't 0

but (x-1)(x-2) is zero

and corresponding to that

the point "1" isn't closed, the point "2" isn't closed", but their union is?

[oblique river]

no, both of those points are closed

that's the whole point here

[urban acorn]

my bad

[oblique river]

the variety is the union of two proper closed subsets

therefore it's reducible

[urban acorn]

okay

[oblique river]

a more complicated example is C[x,y]/(xy)

well, still not too complicated, just 2D instead of 1D

that's not an integral domain

and correspondingly V(xy) is the union of the x- and y-axes

i.e. the union of two proper closed subsets

so you can think of irreducible as like "only having one piece"

[urban acorn]

okay

[oblique river]

and that's the geometric version of being an integral domain

[urban acorn]

are the details of the definition of a sheaf tedious?

[oblique river]

i thought they were a little tedious, but they're all easily-motivated if you think about like, functions on a manifold

or even like, functions on any topological space

basically, let X be a topological space and for any open set U of X, let F(U) be the ring of real-valued functions on U

then F (i.e. the assignment of open set to ring) is a sheaf on X

the sheaf axioms basically say things like

if f: U --> R and g: V --> R are two functions which agree with each other ont he intersection U cap V

then you can glue them together to get a new function (U union V) --> R

(also i'm not saying "continuous" anywhere but i'm thinking of continuous functions)

another sheaf axiom says that

if {U_i} is an open cover of some set U and f is a function on U to R

and if the restriction of f to each U_i is the zero function

then f itself is the zero function

I think both of these are pretty "obvious" statements about continuous functions

and that's basically what the sheaf axioms are

just abstracted a little bit

[urban acorn]

I see

[median pawn]

well phi(f) = phi(g) means f and g are in the same coset, so f(0) = g(0) right
aaaa so every coset has an element of the complex field associated to it
i need to show that T: X/M --> C given by T(f) = f(0) is an isomorphism, right?

[urban acorn]

Would you say that the structure of a scheme describes the notion of a "space that has polynomials"?

[terse crystal]

Being Integral means being irreducible and reduced I think. spec(C[x]/(x^2)) is irreducible while the underlying ring isn’t integral

[urban acorn]

exactly

[oblique river]

yes tha'ts why I said "basically", you'll see the previous thing I said was that "it's a ltitle complicated geometrically"

since being reduced isn't something you can see as a topological space

my omission of detail is intentional

in any case

[oblique river]

I would say that the structure of a scheme is more like "algebraic functions on a space"

need not be polynomials, remember the example from earlier

U = C \ {0}

and f(x)/x^n

[urban acorn]

oh, right

but they're motivated by polynomials

[oblique river]

I mean again I would say they're motivated by rational functions in general

ti's just that the only rational functions which have no poles (i.e. are defined on all of C) are polynomials

but like, you can also look at something like

C[x,y]/(y^2 = x + 1) or something

[urban acorn]

how about on fields that aren't algebraically closed?

[oblique river]

on V(y^2 - x - 1) you have this function called y

which you can think of as being equal to sqrt(x + 1)

[urban acorn]

so, for example, consider 1/(x^2 + 1) as a function on R

[oblique river]

so you also kind of get algebraic functions in the mix

yeah so things are funky over non algebraically closed fields

as a function on Spec R[x], x^2 + 1 does have a "zero"

namely, Spec R[x] has a point corresponding to the maximal ideal (x^2 + 1)

and if you evaluate the "function" x^2 + 1 at that point, you get 0

so 1/(x^2 + 1) isn't actually an algebraic function on Spec R[x]

[urban acorn]

oh, okay

that's pretty neat

it does mean though

[oblique river]

while Spec C[x] can really be thought of as just the complex numbers (along with the point corresponding to the 0 ideal which we kinda mentioned earlier)

Spec R[x] is "bigger" than just R

[urban acorn]

that R doesn't cover enough of the spectrum of R[x]

[oblique river]

that's right

R[x] has more maxmial ideals than just (x-a) for a in R

[urban acorn]

lmao

algebraically closed fields are compact spaces

[oblique river]

what do you mean by that? :o

oh wait I see

yeah in analogy with the thing earlier

[urban acorn]

so

yeah

[oblique river]

haha i hadnt thought of it that way but yes in some sense that is true

a field k is in bijection with Spec k[x] (again ignoring this special point (0)) if and only if k is algebraically closed

err, by "in bijection with" i mean via the map a --> (x-a)

[urban acorn]

yeah

[oblique river]

there exists some bijection between R and Spec R[x] becuase both sets have the same cardinality haha

[urban acorn]

lmao

I believe the technical term is

unnatural-abomination-isomorphism

[oblique river]

hahahaha

exactly

[maiden ocean]

i believe we call that "algebraic geometry"

[oblique river]

hahahaha

[urban acorn]

we don't allow hate speech in here

[maiden ocean]

[urban acorn]

anyway

so, given some scheme

we have the ring associated to the entire space

is that functorial?

[oblique river]

just to be clear you can only ask that for affine schemes

but yes there is an equivalence of categories

between commutative rings and affine schemes

[urban acorn]

wait

so for general schemes

[gritty sparrow]

No, you can ask that for any scheme, and the answer is yes it is functorial (we’re talking about global sections right?)

[oblique river]

yes global sections, and yes you are right you can ask for any scheme, that's my bad

[urban acorn]

okay, I was confused

[oblique river]

i was specifically thinking of that equivalence which breaks down if you try to extend to all schemes

[urban acorn]

alright

[oblique river]

sorry about that

[urban acorn]

so, it's still functorial, but not an equivalence?

[oblique river]

yes

[urban acorn]

huh, pretty cool

[oblique river]

as an example, if you look at projective space

the only functions on projective space are constants

like on P^1 let's say

so the "ring of functions" would just be C

or R or whatever your base field is

[gritty sparrow]

Even cooler, it has an adjoint. It is the spectrum functor of a ring

[oblique river]

but like, the functions on a point are also just the constants

[urban acorn]

not a lot of information then

[oblique river]

yes, but the entire scheme knows more than just "global functions"

the scheme knows things like "what are the fucntions on this open set"

[urban acorn]

yeah

[oblique river]

and that tells apart P^1 from just a point

here's another sort of weird example which kind of shows how the scheme structure can remember things that the topology cant

the spectrum of any field K is just a point, topologically

since any field only has one prime ideal

[urban acorn]

right

[maiden ocean]

non reduced rings

[oblique river]

but the schemes Spec C and Spec R (for example) are different

because the functions on Spec C are C

and the functions on Spec R are R

so even though these are both just points, the scheme remembers more

also things like

there is a ring hom R --> C (just inclusion) but no ring hom the other way

which means there is a scheme map Spec C --> Spec R but not the other way

[urban acorn]

what are morphisms between schemes?

[oblique river]

the easy answer is "functions which preserve all of the relevant structures"

[urban acorn]

lmao

[oblique river]

but like once you prove all the things you just see that

for affine schemes at least

morphisms from Spec S --> Spec R are in bijection with ring homs R --> S

like, you can define "morphism of scheme" in some complicated way, and then prove for affine schemes that it reduces to that

[urban acorn]

Is the Zariski topology appropriate to think of as a topology in the same sense that, say, the topology of a ball describes its geometric structure?

[oblique river]

hmm im not sure i know what you mean

[urban acorn]

or is it just an important collection of sets that satisfies properties that happen to be formally analogous?

so, the open set definition for topological spaces

is introduced to formalize some definition

[oblique river]

so the zariski topology by itself is very coarse

[urban acorn]

of spaces that are "squishy" and don't preserve distances, but preserve the way the space is "attached" to itself

[oblique river]

but if your rings aren't topologized in a nice way then there's not really any other option

like let's say you wanted to work with curves over the field Z/pZ

[urban acorn]

and that are the most general setting for continuity

[oblique river]

or over some really weird field

[urban acorn]

so my question was

does the Zariski topology really describe that kind of geometric structure on affine schemes?

or is it just a collection of important subsets for algebraic reasons

[oblique river]

hmm so let me try to give two answers and i hope that some combination of the two is satisfying

[urban acorn]

that happen to formally resemble the closed sets of a topological space as in the initial intuition

[oblique river]

1. it's kind of hard to talk about the geometry of schemes if you only look at the topological space. so i guess like, the topological space by itself is more of a nice template for you to build a sheaf on. that said, there are some theorems about recovering the "usual geometry" on a complex variety from the "zariski geometry" of it, so there is still useful information encoded in the zariski topology

2. sort of in the other direction, I would say that open sets in the zariski topology are actually meaningful in comparison to more general topological spaces or manifolds. for nice topological spaces (hausdorff plus some other kind of separation condition), the closed subsets of X are exactly the subsets of X which are zero sets of some continuous function. Now let X be an affine scheme and replace "continuous function" with "algebraic function" and the exact same statement is true

that's kind of a geometric thing -- what are the zero sets of functions? in the manifold world you talk about continuous functions and in the scheme world you talk about algebraic functions

but in both cases, the open sets (or, well, their complements, the closed sets) capture exactly that kind of geometry of where functions can be 0

[urban acorn]

hmm

it's like a beautiful but sort of unholy relationship between geometry and algebra

but yeah, these are good answers

[oblique river]

i'm glad that they were somewhat useful haha

also

ive been sitting here for a while im gonna get up for a bit

if you have more questions you can DM me or just @ mention me and i'll try to get back to you at some point

[urban acorn]

alright, thanks, I probably will tomorrow cause it's pretty late for me right now

[oblique river]

sounds good, and gn

[bleak perch]

quick question, what are the units in in z mod p, where p is some prime?

[urban acorn]

all non zero elements of Z/pZ are units

i.e. Z/pZ is a field

[bleak perch]

gotcha thanks

[terse crystal]

Figured it out. The former has a surjection to S_3 while the latter doesn’t

[weary terrace]

Hi.
Let G be a group and N a subgroup of G. Can we say that the number of conjugacy classes of G that are contained in N equals the number of conjugacy classes of N?

I mean, obviously each conj class of G contained in N is a conj class of N.

What about the other direction? Does any conj class of N a conj class of G contained in N?

[mild laurel]

No, this isn't true. Consider the subgroup {e, (1 2)} inside of S_3.

[dusty river]

Conjugacy classes of that are still the conjugacy classes of S_3 intersected with it though

It's just that in general they could become finer

Things that were conjugates could no longer be conjugates

I think C_3 inside S_3 is an example. It has 2 conjugacy classes of S_3 contained in it, {e} and {(123),(132)}, but when you look at it as a subgroup then all the conjugacy classes become singletons

[dusty river]

Oh you were giving a counterexample to the converse, nvm

So both are false

[mild laurel]

Yeah sorry I should've specified

[weary terrace]

@mild laurel So,to be certain, a conj class of N will be {(1 2)}, but if we conjugate with the elements of G we have a larger class. Thus the converse claim is false.

What if N=ker \rho for some \rho:G->GL(V) (kernel of a representation)?

then ker \rho is a niormal subgroup

normal

[dusty river]

Since it's a normal subgroup, the conjugacy classes are either disjoint or contained

But after that they could still be finer I imagine?

C_3 in S_3 is normal, not sure if it's a kernel of a representation

[mild laurel]

Any normal subgroup is the kernel of some representation

[dusty river]

Oh then that's a counterexample

[mild laurel]

Just look at the regular representation of G acting on G/H

[dusty river]

Idk any representation stuff lol

[weary terrace]

@mild laurel its kernel is H

?

[mild laurel]

Yeah

[weary terrace]

Oh sorry for a minute I thought you were answering me

@hidden haven thanks!

[novel parrot]

i have aquestion

if j is number of initial zeros

why is j1 =1

j2 =2

is this maybe .. a typo

[mild laurel]

j1 is the number of initial zeroes plus one I guess

Look at part (a) again, they define j1 to be the position of the first nonzero entry in row 1

[novel parrot]

i thought j1 was a column

[mild laurel]

I mean sure, j1 specifies which column the first nonzero entry in row 1 is

Which is why j1 = 1

[novel parrot]

okey

yup

[weary terrace]

Say G (a group) has n\cdot l conjugacy classes and G/N \cong Z/nZ (N is a normal subgroup of G). I want to prove N has l conjugacy classes.
So I know that G is a disjoint union of its n \cdot l conjugacy classes and that N has n cosets (equivalently conjugacy classes, since it's Z/nZ). G also equals a disjoint union of the cosets of N. How do I connect the dots?

[crimson falcon]

Oh nice, gonna do that, thanks a lot for the exercices and the effort really appreciate it!

[novel parrot]

i have got to question 19 now

im looking at the second matrix

but i am getting something slightly different

i have 5 in row 2 column 5 instead of 2

[weary terrace]

Up

[terse crystal]

There are 3 conjugacy classes of S_4 that are contained in A4 but A4 has 4 conjugacy classes. And all 3 cycles form one conjugacy class of S_4 but also form 2 conjugacy classes of A_4

[terse crystal]

Difficult to read...

[terse crystal]

I also got 5

[weary terrace]

I'll try latex:

Say G (a group) has $n\cdot l$ conjugacy classes and $G/N \cong \mathbb Z/n\mathbb Z$ (N is a normal subgroup of G). I want to prove N has $l$ conjugacy classes.
So I know that G is a disjoint union of its $n \cdot l$ conjugacy classes and that N has n cosets (equivalently conjugacy classes, since it's $\mathbb Z/n\mathbb Z$). G also equals a disjoint union of the cosets of N. How do I connect the dots?

[cloud walrus]

RiesZ

[weary terrace]

Is it more readable now? Or did you refer to the wording?

[terse crystal]

If N is not trivial then the number of conjugacy classes is greater than 1

And btw I don’t know what do you mean by saying N has n cosets it makes sense to me if you say G/N has n cosets

[terse crystal]

Sorry I mistook L as 1

[weary terrace]

Best said |G/N|=n

Is it correct to say that if $g_i g_j^{-1}\in N$ then $C_{g_i}=C_{g_j}$ , where $g_i,g_j\in G$ and $C_{g_i},C_{g_j}$ are the conjugacy classes of $g_i,g_j$ respectively?

[cloud walrus]

RiesZ

[weary terrace]

If so, we have a bijection between the set of conj classes and cosets

[terse crystal]

I am constructing a counterexample. If a group G has n conjugacy classes, then the direct product of G and itself has n^2 conjugacy classes but not 2n conjugacy classes

Making sense?

GΠG/G is isomorphic to Z/2Z

I just need to find a G here who has even number of conjugacy classes then the counterexample is found

G=A_4

[weary terrace]

Is it isomorphic to Z/2?

Did you mean a direct product here?

[terse crystal]

Yes

[terse crystal]

Yes , there exists only one group that has order 2 which is Z/2Z

[terse crystal]

G can be viewed as a normal subgroup of GΠG,G={(g,1):g from G}

[weary terrace]

Sorry I must be missing something basic.. |A_4||A_4|/|A_4|=|A_4| \neq 2

Did you mean N instead of G in the denominator?

[terse crystal]

I made a mistake,rethinking

[terse crystal]

It might be correct I am proving that G is isomorphic to the direct product of N and Z/nZ since if N has j conjugacy classes N Π Z/nZ has nj conjugacy classes.

[weary terrace]

That's a cool approach, didn't think about it

[terse crystal]

It is isomorphic right? G/N is generated by a element xN of order n then I map (h,k) to x^k multiplied by h where h is from N

[weary terrace]

Seems easy though:
each g\in G equals g_i\cdot n for some n\in N and g_i is a transversal of G/N

I hope lol

sounds right

I think its a morphism too

thanks very much! I'll dig into it

[terse crystal]

Not done yet...

[weary terrace]

how come?

[terse crystal]

It is correct if I can find such x which has order n itself and x is contained in Z(G)

Other wise that map won’t be a homomorphism

[weary terrace]

Is it not enough to say that x\in Z/nZ? as a transversal of course

It follows from the isomorphism G/N=Z/nZ

I would, however, check if the map is well defined

[terse crystal]

Not enough you see the map mapping (k,h) to x^k h is a homomorphism means that x itself has order n and x^khx^st=x^(k+s)*(ht)

So you need to find a x having order n such that G/N is generated by xN also x is commutative with any element of N

So your statement still could be false I need to try to find counterexample again

[weary terrace]

Oh I see.. Like that:
https://math.stackexchange.com/questions/2768685/is-g-isomorphic-to-fracgh-times-h

[terse crystal]

Yes so generally those representatives from cosets of G/N themselves should form a group isomorphic to G/N also those representatives should be commutative with any element of N.
only in that case G is isomorphic to the direct product of N and G/N

[weary terrace]

So if I get it correctly, I have to add more info in order to make my claim correct?

[terse crystal]

Yes the condition I gave is correct I think and I think I have found a counterexample of the original statement

S_8 has 22 conjugacy classes

Wiki said A_8 has 14 conjugacy classes

If we believe Wikipedia then that is a counterexample

[terse crystal]

If this holds then the number of conjugacy classes of G is equal to the multiplication of the numbers of conjugacy classes of G/N and N respectively.

[weary terrace]

Thank you for the effort @terse crystal, I appreciate it.
I do have extra info but I'm not sure if it helps here.
N=ker \rho where \rho:G->GL(V) is a representation of G.

So the original question I'm trying to answer is:
Let $G$ be a group, $\rho:G \rightarrow \textrm{GL}(V)$ a one-dimensional representation of $G$ and $N=\ker \rho$. $G/N \cong \mathbb Z/n\mathbb Z$ and $G$ has $n\cdot l$ irreducible representations, $l\in \mathbb N$. Prove that $N$ has $l$ irreducible representations.

[cloud walrus]

RiesZ

[terse crystal]

clearly false...

Since I can use the same counterexample

Notice that sign is an one dimensional representation of S8 whose kernel is A8

Where sign is a map from G to nonzero elements of C (=GL(C)) mapping even permutations to 1 odd permutations to -1

[weary terrace]

Yeah, very nice.

I have extra info regarding tensor powers of \rho but I feel like I've been spamming here all day 😛

[frank fiber]

why this series is normal?

[chilly radish]

Group generated by the commutator of 2 normal subgroups is normal, then by induction all the gammas are normal I think

[wispy pivot]

Is there a list of curated exercises from D&F that a college course assigned for their group theory course? Looking to self-study D&F but given that it has an overabundance of problems I'm looking for some guidance on which ones to prioritize.

I found this example for Harvard's Rings courses (Math 123) that uses D&F. But from my understanding their groups course doesn't assign problems from the text.
https://svasey.github.io/academic-homepage-may-2020/123-spring-2020/

[unreal portal]

what parts of dummit and foote in particular? The book is massive and has enough material for like 3 courses

https://www.math.ucdavis.edu/~anne/FQ2001/250A.html

this uses dummit and foote, but they have custom exercises. But they also have the exercises and solutions on the page

https://math.dartmouth.edu/~auel/courses/350f17/syllabus.html

this also looks good, they give specific problems from D&F to solve

If you google "syllabus dummit and foote" I'm sure you can find a bunch of others

[wispy pivot]

hey thanks this looks really good

[wispy pivot]

hmm kk i did try once but didn't find anything helpful in the couple page...

[fickle tapir]

for closed under addition, if x,y in S then x in F_i and y in F_J. if x and y are in the same field, then we know its closed under addition and multiplication

but what if F_I != F_J?

[maiden ocean]

Is this proof even correct...?

How can we say that a_i is in A_n-ki

we dont know anything about what xi or ai xi is

because just because the sum of the a_i x_i is dim n doesnt mean that each individual a_i x_i is necessarily dim n

[oblique river]

a and x_i are both homogeneous

and wlog we can take the a_i to also be homogeneous

just split each a_i into its homogeneous parts

so a_ix_i is also homogeneous. if it's not of degree n it must be cancelled out by some other term which is also not of degree n

(or some other collection of terms)

basically just, you can ignore all of the terms which aren't degree n because you know they must all cancel out with each other

[cloud walrus]

add
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[novel parrot]

is the first ordering x1 > x2 > ... > xn?

or what does first choose an ordering actually mean?

<@&286206848099549185>

[unreal portal]

The first order can be any permutation of x1 through xn. You first sort by total degree, then by this lex order on the variables

[novel parrot]

ok

and if degree is same

[unreal portal]

that's when you use this first order you chose

it's basically the tiebreaker

[novel parrot]

ok

i dont understand ordering

we look at the variables backwards

and the first smaller exponent is our larger monomial yes?

[unreal portal]

so you go backwards through the variables until you find a variable where the corresponding exponents aren't the same

[novel parrot]

ok and which is our larger monomial

after checking backwards

m1 = x_1 > x_n = m2 because m1 has 0 for exponent on x_n yea?

[unreal portal]

doing it in words like this confuses me

this is how my notes has grevlex defined

these vectors just correspond to the exponents, so (4,7,1) corresponds to $x_1^4x_2^7x_3$

[cloud walrus]

cgodfrey

[novel parrot]

ok yea in the second example the total degree is same

and the last position is smaller

so we let that monomial be larger

[unreal portal]

yeah

[novel parrot]

okie

thank you

[unreal portal]

yeah the wording in your screenshot is a little confusing

[novel parrot]

yup lol

nice notes btw

$x_1x_2^3 >_{grevlex} x_1x_2^4$

[cloud walrus]

ActiveChapter

[novel parrot]

$x_1x_2^4 >_{graded lex} x_1x_2^3$

[cloud walrus]

ActiveChapter

[novel parrot]

these are correct right?

@unreal portal

[unreal portal]

I think this one is wrong. Both grevlex and grlex first sort by total degree, so both would have $x_1x_2^4$ be greater

[cloud walrus]

cgodfrey

[novel parrot]

oh yeha

[unreal portal]

to give a fuller example, this is what you would get ordering a polynomial under each ordering

(ignore the typo in the second term of grlex)

[novel parrot]

[unreal portal]

anyone recognize valuation or how tf they get -1 and 0 for these examples

googling valuation of laurent polynomial doesn't come up with anything useful

looks like it's the smallest exponent

[mild laurel]

yes thats right

[cursive temple]

im not sure how to prove that given y'^2 = 0 then yx'^2 = 0 for some nilpotent x'

i guess i could just write y' = f(x) + g(x)y and x' = f(x_1)+g(x_1)y_1 and then just try to smash everything together

i just hope there is a nicer solution

[cursive temple]

All nilpotent x' i mean

[novel parrot]

How can i show that two minimal basis have unique amount of generators?

$(LT(h_1), \cdots , LT(h_m)) = LT(I) = (LT(g_1), \cdots, LT(g_n))$

[cloud walrus]

ActiveChapter

[novel parrot]

how to show thaat n = m?

<@&286206848099549185>

$LT(h_1) = q_1LT(g_1) + q_2LT(g_2)$ and $LT(h_2) = q_3LT(g_2)$

[cloud walrus]

ActiveChapter

[novel parrot]

LT(h_2) doesnt divide LT(h_1) here right?

[thorny flame]

How do I tell if a permutation is odd or even?

[dusty river]

Write its cycle structure. An n cycle is odd iff n is even

[thorny flame]

what is n representing?

[scarlet estuary]

the length of the permutation in cycle notation

[thorny flame]

(3 2 1) would be 3?

[dusty river]

Yes

[thorny flame]

so (3 2 1) is even?

[dusty river]

Yes

[thorny flame]

Is there a proof or is it true by definition?

[dusty river]

If you decompose it into a composition of 2 cycles l

You'll get an even number of them

[scarlet estuary]

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

for example

[thorny flame]

(3 2) (2 1) (1 3)

ohh.. I see

[scarlet estuary]

this can be proved via a very quick inductive argument

using whatever definition of parity you prefer

[dusty river]

Counting inversions

[thorny flame]

What about permutation of 1 to 3

[dusty river]

(1 3)?

[thorny flame]

it's a cyclic permutation

[dusty river]

(1 2 3)?

[thorny flame]

3 -> 2
2 -> 1
1 -> 3

[dusty river]

Are you asking why what namington wrote works

[thorny flame]

yes

[dusty river]

Try applying those 2 transpositions on a 3 element set

Or like symbolically composing

[scarlet estuary]

it seems you forget how products/compositions of cycles work

[dusty river]

Whichever one you prefer

[scarlet estuary]

when i write (3 2)(2 1)

these are evaluated right-to-left

so lets follow each element

1 is sent to 2 by (2 1) and then 2 is sent to 3 by (3 2), so 1 is sent to 3

2 is sent to 1 by (2 1) and then 1 is unaffected by (3 2), so 2 is sent to 1

3 is unaffected by (2 1) and then 3 is sent to 2 by (3 2), so 3 is sent to 2

hence we have 1 -> 3, 2 -> 1, 3 -> 2

this is equivalent to (3 2 1)

[thorny flame]

oh right
1 - > 1 -> 2
2 -> 3 -> 3
3 -> 2- > 1

[scarlet estuary]

in your case, (3 2) (2 1) (1 3), this is actually a different permutation

for example, 1 is sent to 3, unaffected, and then sent to 2

so 1 -> 2

2 is unaffected, then sent to 1, then unaffected

so 2 -> 1

3 is sent to 1, then sent to 2, then sent to 3

so 3 -> 3 (ie 3 is unchanged)

so (3 2)(2 1)(1 3) actually is just (2 1)

[thorny flame]

right

1 -> 1 -> 2 -> 2
2 -> 3 -> 3-> 1
3 -> 2 -> 1 -> 3

Thanks

[scarlet estuary]

careful

products of cycles are read right-to-left

not left-to-right

(by 99% of textbooks at least)

it doesnt matter here but it does in some cases

[thorny flame]

I'm israeli so we read everything left to right.
So it may explain it.

[scarlet estuary]

well, it may be different in different sources

but i dont think ive seen a textbook that uses left-to-right

¯_(ツ)_/¯

[chilly radish]

Just fyi we still use the same convention (right to left) in israel

But uhh, hebrew is written right to left aswell

Cycle multiplication is just funny looking function composition

[scarlet estuary]

the confusing thing is that cycles are internally read left-to-right but externally composed right-to-left

it makes some sense when you realize its just a way of representing function composition

but its still kinda weird

[thorny flame]

כן, טעות שלי

Right, so is it only the perumtation right to left or also the action sequence

?

[chilly radish]

חחחחח

[thorny flame]

for example ( 1 2 3 ) would be (1 2) (1 3)
And so the actions would be
1 -> 3
2 -> 1
3 -> 2

And the latter
1 -> 2 -> 2
2 - > 1 -> 3
3 -> 3 -> 1

Or am I making a mistake?

[chilly radish]

The cycles themselves are left to right, but their composition is right to left

So the first one would simply be 1->2->3

Second would be
1->3->3
2->2->1
3->1->2

[thorny flame]

huh

[chilly radish]

And the last one is equivalent to (321)

If you wanna decompose (123) into transpositions you could do (12)(23)

[thorny flame]

I see thanks

[chilly radish]

Ok so i'm trying to prove the five lemma, I have a proof but it's a bit messy and idk if there's a more succinct proof. Anyways I also wanna make sure i'm not being dumb. Suppose there is a commutative diagram of groups with exact rows
$$\begin{tikzcd}
{A_1} & {A_2} & {A_3} & {A_4} & {A_5} \
{B_1} & {B_2} & {B_3} & {B_4} & {B_5}
\arrow["{i_1}", from=1-1, to=1-2]
\arrow["{i_3}", from=1-3, to=1-4]
\arrow["{i_4}", from=1-4, to=1-5]
\arrow["{i_2}", from=1-2, to=1-3]
\arrow[from=2-1, to=2-2]
\arrow[from=2-2, to=2-3]
\arrow[from=2-3, to=2-4]
\arrow["{j_4}"', from=2-4, to=2-5]
\arrow["{f_5}", from=1-5, to=2-5]
\arrow["{f_4}", from=1-4, to=2-4]
\arrow["{f_3}", from=1-3, to=2-3]
\arrow["{f_2}", from=1-2, to=2-2]
\arrow["{f_1}", from=1-1, to=2-1]
\arrow["{j_2}"', from=2-2, to=2-3]
\arrow["{j_3}"', from=2-3, to=2-4]
\arrow["{j_1}"', from=2-1, to=2-2]
\end{tikzcd}$$
Suppose $f_2,f_4$ are surjective and $f_5$ is injective, then $f_3$ is surjective.

[cloud walrus]

ShiN

[chilly radish]

Let $y \in B_3$, consider $j_3(y)$. $f_4$ is surjective, so there is some $z \in A_4$ such that $f_4(z)=j_3(y)$. From exactness $j_3(y)\in \text{ker}j_4$, so from commutativity we have
$$j_4f_4(z)=f_5i_4(z) \Rightarrow 1=f_5(i_4(z))$$
Since $f_5$ is surjective, it follows that $i_4(z)=1$, then from exactness $z \in \text{ker}i_4 = \text{im}i_3$, so there exists some $w \in A_3$ such that $i_3(w)=z$, then we have that $f_4(i_3(w))=f_4(z)=j_3(y)$, then from commutativity
$$f_4i_3(w)=j_3f_3(w) \Rightarrow j_3(y)=j_3(f_3(w)) \Rightarrow j_3(y\cdot f_3(w)^{-1})=1$$
Again from exactness, this implies that $y\cdot f_3(w)^{-1} \in \text{im}j_2$, so there is some $\ell \in B_2$ such that $j_2(\ell)=y\cdot f_3(w)^{-1}$.

From the surjectivity of $f_2$, it then follows that there is some $\ell' \in A_2$ such that $f_2(\ell') = \ell$. Finally, from the commutativity of the diagram we have:
$$j_2f_2(\ell') = f_3i_2(\ell') \Rightarrow y\cdot f_3(w)^{-1}=f_3(i_2(\ell')) \Rightarrow f_3(w\cdot i_2(\ell')) = y$$
So finally $f_3$ is surjective as required.

[cloud walrus]

ShiN

[burnt trout]

Wow-

[chilly radish]

Oop I just noticed that the final multiplication is reversed but it doesn't really matter

[lunar spruce]

How do I conclude from right side that f(b^-1a)=e? This is a proof of homomorphie theorem

[south patrol]

Is f here a group homomorphism?

[lunar spruce]

Yes

I mean I can say that f(a^-1b)=f(b^-1a) but not that its neutral

[south patrol]

in general we have that f(g^-1) = f(g)^-1 if f is a homomorphism

and if f(a) = f(b) then e = f(a)f(b)^-1

hopefully that gets you started

[lunar spruce]

Ahh ok got it.

thanks

🦢But I got it now.

[south patrol]

nice

[dusty river]

That's pretty standard (I only looked at the outline not the details but they look kinda correct)

But what's not standard is

5 lemma for non abelian groups

[chilly radish]

Yea?

I mean it was in the context of the category of chain complexes of abelian groups

But I read it holds in Grp aswell

So why not

[dusty river]

Ye but

[chilly radish]

I mean yea I prefer additive notation too

[dusty river]

Talking about exactness in non abelian categories

I saw the 1 and got confused lol

[chilly radish]

I mean exactness can be a thing in not-abelian categories right?

Why is that weird to consider

[dusty river]

Idk I just never saw it before but like not all monics are kernels, since not all subgroups won't be normal, and you can't always take cokernels

Also in abelian categories, an equivalent definition of exact is cokernels being coimages

So you would have to make a choice while defining exactness for non abelian categories, as to which one of the 2 definitions you want to keep

Idk I'm just making stuff up

[chilly radish]

Uh ok

Fair enough

[chilly radish]

Ok just checking i'm not dumb again. Suppose I have abelian groups $A,B,C$ such that $\sfrac{B}{A}\cong C$, then I have $B \cong C \oplus A$ since $B \cong \sfrac{B}{A}\oplus A \cong C \oplus A$ right?

[cloud walrus]

ShiN

[chilly radish]

or wait

i'm not sure that $B \cong \sfrac{B}{A}\oplus A$

[cloud walrus]

ShiN

[chilly radish]

feels like it should be tho

[dusty river]

It's not, take B = Z and A = 2Z

[chilly radish]

shit

right

[dusty river]

It does happen in certain nice situations for example when B/A is a free abelian group or more generally any projective abelian group

[chilly radish]

I'm trying to show smth in homology

i'll try thinking about it a bit more than

then*

[ionic ingot]

simple question, just want to clarify but - when asked to give examples a subgroup, and it isn't clarified whether it is cyclic or abelian or non-abelian, is there something we should assume?

for example if asked to give an example of "a subgroup of order 7 in the symmetric group S7 of degree 7" i would say H = < (1 2 3 4 5 6 7) >

would that fit the condition?

[dusty river]

Yeah any subgroup good

[ionic ingot]

great

how could i come up with an example of a subgroup of order 14 in the symmetric group S7 of degree 7?

[dusty river]

What does degree 7 mean here?

[urban acorn]

that it constitutes the full set of symmetries of a set of 7 elements

[south patrol]

i assume you could just remove the 'of degree 7' there lol

[dusty river]

[viscid pewter]

consider the factors of 14

[urban acorn]

I wonder what <(1 2), (1 2 3 4 5 6 7)> is

[south patrol]

that is S7 i believe

[urban acorn]

idk if it's of order 14

oh

[viscid pewter]

actually yeah it doesn't work

[dusty river]

Ye S7

[urban acorn]

welp

[viscid pewter]

in that case i don't think it exists

[south patrol]

because (1...7)^-1(12)(1...7) is either (23) or (71) depending on your convention for permutations i think lol

just keeps cycling round to get all transpositions

[dusty river]

What is the number of conjugates of the sylow p subgroups in an order 14 group

[urban acorn]

is <(1 2 3 4 5 6 7)> maximal in S7?

[dusty river]

I don't wanna recall the theorem statement

For sylow

[ionic ingot]

essentially, yeah. it's just there to reaffirm the S7.

[dusty river]

If we can prove that one of the 2 things is normal then we'll have a semi direct product

[south patrol]

the number of 7 subgroups is isomorphic to 1 mod 2 i think and the number of 2 subgroups is 1 mod 7 iirc lol

[dusty river]

And I think a semi direct product shouldn't be possible

[ionic ingot]

i am so confused rn lol

[south patrol]

using sylow is probs a bit ott here aha

[urban acorn]

1 I think

for n_7

[viscid pewter]

well <(12)> is surely normal, i think?

i haven't even touched group theory in ages

[urban acorn]

and 1 or 7 for n_2

[dusty river]

Wait so if group is m*p^n, then sylow p subgroup has number of conjugates 1 mod p and a factor of m right?

[urban acorn]

yeah

[dusty river]

Not in S7

[south patrol]

i'm pretty sure <(12)> is not normal as a normal subgroup must be a union of conjugacy classes

[ionic ingot]

can it be H = < (1 2), (1 2 3 4 5 6 7) >

[urban acorn]

yeah it's not normal

[viscid pewter]

oh

[dusty river]

Nope

[south patrol]

and the conjugacy classes of S7 are just permutations of a given length

[ionic ingot]

why not?

[dusty river]

That is all of S7

[south patrol]

that is S7

[viscid pewter]

that generates the whole thing, yeah

[dusty river]

Because you can do a bubble sort kinda thing

[viscid pewter]

i don't think it's possible

[dusty river]

To rearrange everything as needed

[viscid pewter]

because it's just like

the group can't be cyclic

but if it's not cyclic then it's surely gotta have an element of order 2 and an element of order 7?

[ionic ingot]

i don't think it matters whether its cyclic or not

[urban acorn]

You should be looking at the way <(1 2 3 4 5 6 7)> sits in S7

[ionic ingot]

it just says subgroup

[viscid pewter]

yeah i know

but i'm considering both cases, cyclic and non-cyclic

[urban acorn]

okay, I figured it out

[ionic ingot]

hmm okay waiy

[urban acorn]

I found a subgroup of order 14

[viscid pewter]

fuck

[urban acorn]

It's the dihedral group of order 14.

D7

[viscid pewter]

wait

yeah i see it

[urban acorn]

and it's the permutations of the 7 vertices

[south patrol]

yeah i was going to say, usually there's a sexy interpretation for these sort of things

[viscid pewter]

ok i'm bad ignore me

[urban acorn]

[viscid pewter]

i should have seen that

[south patrol]

permutations of 7 vertices having a subgroup corresponding to the symmetries of a heptagon ig?

and then you can find generators by drawing a diagram or something, lol, if you're lazy like me

[dusty river]

Ye it's isomorphic to <(1234567), (12)(34)(56)>

Nice

[urban acorn]

yeah

[dusty river]

Or like something along those lines

Actually not isomorphic to that

The cycle should be (7135642)

[viscid pewter]

def not, that's not disjoint cycles

[dusty river]

Ye

[viscid pewter]

no?

[urban acorn]

<(1234567), (27)(36)(45)>

[dusty river]

Oh ye probably a typo

Ye

[south patrol]

yeah typo sorry

[ionic ingot]

so this is a subgroup of order 14 of S7?

[urban acorn]

yeah

[south patrol]

visualising a heptagon properly did not go well

[ionic ingot]

i see

(1234567) can also be written as (), yeah?

[viscid pewter]

nononono

[urban acorn]

I have not seen that notation before.

[south patrol]

i've not seen that notation myself, i thought () would refer to identity if anything

[viscid pewter]

yeah

it's not identity

[urban acorn]

same

[viscid pewter]

(1234567) is 1 -> 2, 2 -> 3, etc.

[urban acorn]

but i'd just write the identity as 1

[south patrol]

i write identity as e ¯_(ツ)_/¯

[dusty river]

(123) means 1 → 2 → 3 → 1

[dusty river]

But the group is abelian

[viscid pewter]

it's not?

in what world

[south patrol]

the username

[viscid pewter]

am i missing a joke

[south patrol]

look at the username of the person coldilocks responded to lol

[dusty river]

"all groups are abelian"

[ionic ingot]

ahhh OOP yeah identity y'all are right, sorry

[viscid pewter]

oh fuck

i'm blind

i hate this

[ionic ingot]

thank you!!!

[south patrol]

i wonder when learning basic cat theory is useful for abstract algebra lol

it seems many theorems have cute interpretations but then idk if that's actually helping me or if i'm just going 'lol diagram cool'

[dusty river]

It's really useful in Atiyah Macdonald

[south patrol]

Oh sure, commutative algebra?

[dusty river]

Ye

[south patrol]

Yeah I'm not quite there yet but that's cool

cheers

[ionic ingot]

i know this is a group, but i'm pretty sure it's not an abelian group because matrix multiplication is not commutative, but i think it's best to find examples for these kind of things. can anyone help me with an example of matrices A and B?

[urban acorn]

I'm just starting to learn category theory, but I find it's a very nice tool in a more organizational way, and for thinking of things in a slightly different perspective

like, you don't have to worry about the internal construction of what R[x] looks like with formally introducing indeterminates and such

[south patrol]

Sure, that's cool

[urban acorn]

and you don't need to worry about constructing F_n as much

because it's simpler to see them as having certain universal properties

[south patrol]

in what way? sorry i'm too noobie to see how that would help with stuff like F_n

ah

[urban acorn]

a morphism F_n -> G is the same thing as an n-tuple of elements of G

That characterizes F_n up to isomorphism

[viscid pewter]

(1, 0; 0, 1)

(14, 0; 0, 14)

[south patrol]

oh sure, that's cool ^

[ionic ingot]

but then 0 isn't part of the {1, 4, 7, 13} ? or is it considered to be because modulo stuff

[viscid pewter]

?

[ionic ingot]

and 14 isn't part of modulo 15 i'm p surw

[viscid pewter]

what are you on

[urban acorn]

everything is part of modulo 15 lol

[cloud walrus]

Ravage

[urban acorn]

it just collapses some time, so it's not so much that 17 doesn't exist mod 15, but rather it does exist, but it's redundant to write it because it's equal to 2 which is simpler to write

[south patrol]

that is the identity which has determinant 1 which is in {1,4,7,13}

[urban acorn]

14 does exist, and in a very nice way: it's equal to -1. In particular, it is a unit.

[viscid pewter]

where tf are you ppl getting {1,2,3,4} from

what does that even relate to

[viscid pewter]

yeah, this is how i found examples so fast, because i was thinking (-1, 0; 0, -1) squared = i

[ionic ingot]

fuck wait

fixed it

[viscid pewter]

well

identity has determinant 1

1 * 1 - 0 * 0

[ionic ingot]

yes

[dusty river]

One use that you might be able to appreciate is that an abelian group A isn't just isomorphic to 0 ⊕ A, it's also naturally isomorphic. This means that any exact sequence of abelian groups remains exact when you take ⊕ of every group in it with the trivial group with maps replaced with appropriate thing on the new component

This is kind of a trivial example

[south patrol]

lol nice

[dusty river]

But naturality of isomorphisms preserves exactness

Which is really good

[south patrol]

i am realising there's plenty more stuff to learn first though, I've only going into second year aha

[ionic ingot]

okay wait but in this case AB = BA but i'm trying to prove that's false

[viscid pewter]

oh, right

[ionic ingot]

that it's not always equal

[south patrol]

Try matrices with just elements 0 or 1

Pretty sure even some of those don't commute

But don't want to give it all away so I'll leave it at that

[urban acorn]

maybe try (1 0; 0 2) and (0 1; 1 0)

I haven't even checked if they're in the group

but my intuition is that they are and they don't commute

[viscid pewter]

determinants are wrong

[ionic ingot]

maybe ( 1 0, 0 1) and (1 1, 0 1)

[viscid pewter]

first one isn't in the group

[viscid pewter]

identity always commutes

watch this tho

[south patrol]

the first matrix is the identity and so it'll commute with each matrix

[ionic ingot]

ah fuck

[viscid pewter]

A = (1, 1; 1 0) and B = (1, 0; 1, 1) don't commute

[gritty sparrow]

Wouldn’t have expected this from you given your username

[viscid pewter]

and both have determinant 1

[urban acorn]

okay then, change the second one to (1 0; 0 4)

@gritty sparrow lmao

[viscid pewter]

that's a diagonal matrix, surely it commutes?

thought they were the centre

[south patrol]

Well that just proves these matrices don't form a group /s

[gritty sparrow]

Diagonal matrices may not commute

[urban acorn]

the center are diagonal matrix where all the entires across the diagonal are the same

[viscid pewter]

really?

[urban acorn]

for the record, they commute with each other, but not necessarily with other elements of the group

[south patrol]

yeah

[viscid pewter]

oh ok

welp, got all of that completely wrong

[dusty river]

Saketh you missed lean today

[ionic ingot]

i'm confused?

[ionic ingot]

so this doesnt work?

[viscid pewter]

no, it does work

[gritty sparrow]

I was asleep

[viscid pewter]

i was saying someone else's example didn't work but it does

[ionic ingot]

aight bet

[viscid pewter]

AB = (2, 1; 1, 1); BA = (1, 1; 1, 2) if i can multiply things mentally

ok i can

[ionic ingot]

i got AB = (2 1, 1 0) and BA= (1 1, 2 1)

[rancid narwhal]

Is the map $g \mapsto (x \mapsto gxg^{-1}) where $g \in GL$ surjective? Was unsure. Trying to use the first isomorphism theorem to show the isomorphism between PGL and GL(non-zero multiplies of identity).

[cloud walrus]

Ravage
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[urban acorn]

to ask whether a map is surjective you have to specify the codomain

If the codomain is the automorphism group of GL, then the answer is generally no

when I googled PGL it was just defined as GL/Z(GL)

so, what definition of PGL are you working with?

[south patrol]

Are there any nice things you can do once you know about "the" (up to isomorphism/permutation) composition series of a group? The notes I've been reading just prove Jordan-Hölder and leave it at that, lol

Ig maybe it's just something you can use to show groups are not isomorphic and maybe useful in galois theory or smth

[oblique river]

It is useful in galois theory, and more generally whenever groups pop up

it's useful to know if a group is solvable or not

which means that all of the terms in its composition series are cyclic

you're right that there's not too much you can say just about solvable groups, but they are important when getting applied

[south patrol]

Cheers, yeah I'm aware of the link to solvable groups and ig it's natural this would lead to them, but thanks - idk any Galois theory and I suppose I'd see the series more when I do some of that :)

[oblique river]

yep! in fact, galois theory is where the name "solvable group" comes from. it has to do with solving polynomials using radicals.

i.e. "if p(x) is a polynomial then when can we express the roots of p(x) just using basic operations and radicals"

the quadratic formula is how you do it for quadratic polynomials, but not all polynomials can be solved using radicals like that, and that's got something to do with solvable groups

[south patrol]

yup sweet :)

ironically i learnt some stuff through Arnold's famous 'abel's theorem in problems' and was introduced to those ideas and solvable groups too early, very cool to see it again now i actually know some theory lmao

i still don't quite know the link to Abels' theorem with Galois theory but it's cool hehe, cheers

(nice profile picture)

[oblique river]

thank you :^)

you'll get there -- for better or for worse there is a lot of background in galois theory that you need to build up before you get to solvable groups

but i think the journey is very nice

[south patrol]

indeed, at my uni it's third year i believe and i'm only going into second lol

[south patrol]

sweet :)

[oblique river]

I also took it as a 3rd year student

[south patrol]

oh sure, cool

[maiden ocean]

@oblique river uhh so i just read ur explanation of AMs proof of some stuff about graded rings yesterday and im a little confused

[oblique river]

[maiden ocean]

is there a reason that, say, a + b = x in A_n implies a and b are themselves homogeneous

[oblique river]

no

for example (x + 1) + (x - 1) = 2x is homogeneous but neither summand is homogeneous

but you can always break up a and b into their homogeneous parts

like a = sum a_k and b = sum b_k

[maiden ocean]

i guess i dont see why in generality we can say that the non homogeneous terms have to cancel

Like in polynomials it makes sense

But for a general graded ring

[oblique river]

the degree-k part of a + b is just a_k + b_k

a general graded ring A is still the direct sum of each of its pieces

meaning every a in A can be written as a sum of a_k where a_k in A_k

and if you add two things in the same homogeneous piece, you stay in that homogeneous piece

[maiden ocean]

Do we necessarily assume the A_i are disjoint

except at 0

[oblique river]

yeah like, A is a direct sum of the A_i

[maiden ocean]

because otherwise it seems like you could have like A_1 subset A_2 so something could be of degree 1 and 2

[oblique river]

like writing somthing like A = B \oplus C necessarily means that B and C (as subsets of A) intersect just at 0

I mean, you could take the direct sum of something with itself

A \oplus A

[maiden ocean]

Oh ok this makes more sense lol

[oblique river]

but then you distinguish between the first copy of A and the second

like $\bR^2 = \bR \oplus \bR$

[cloud walrus]

Buncho Bananas

[maiden ocean]

I guess i was thinking like sum of ideals

But its not that

[oblique river]

yeah, except in some really degenerate cases, the homogeneous pieces aren't going to be ideals by themselves

like, $A_1$ isn't going to be an ideal of $A = \oplus_{i \geq 0} A_i$, but $\oplus_{i > 0} A_i$ will be

[cloud walrus]

Buncho Bananas

[oblique river]

(again unless it's a very degenerate scenario)

[maiden ocean]

Ok yea

this makes sense

thanks lmfao

[oblique river]

yeah np

[maiden ocean]

[oblique river]

I know that like, it's scary to tell yourself "whenever i see a Z-graded ring I'm going to imagine polynomials" but honestly that's not a bad strategy

i mean, you do have to be careful, not everything is exactly like polynomials. for example you could just not have any terms of degree 1

the ring $\bC[t^2, t^3]$ is a graded ring with no elements of degree 1

[cloud walrus]

Buncho Bananas

[oblique river]

(it's the ring of polynomials without a linear term)

[maiden ocean]

I see

[oblique river]

that might feel like a weird ring but it's actually isomorphic to $\bC[x,y]/(x^2 - y^3)$ which maybe looks more reasonable

[cloud walrus]

Buncho Bananas

[oblique river]

the isomorphism is x = t^3 and y = t^2

[maiden ocean]

Yeah i mean that makes sense, affixing t^2 gets you polynomials of degree 2 and affixing t^3 gets you polynomials of degree 3 so the graded structure is given by constant, homogeneous polys of degree 2, and homogeneous polys of deg 3

[oblique river]

that's right

(and then products of them can have higher degrees)

but yeah one of the nice things about graded rings is that you can in many cases just pretend that everything is homogeneous

like in this case, we have "x = a sum of terms" and if x is homogeneous we can WLOG assume that all of the terms on the right are also homogeneous, cuz each term on the right can be written uniquely as a sum of homogeneous things

and then since we know that each homogeneous piece is "disjoint" from the rest, you automatically know that all of the terms on the right which aren't of the correct degree must cancel

[maiden ocean]

Yea

[oblique river]

[maiden ocean]

ok now i can finally stop being lazy and finish this chapter

[oblique river]

hahaha

good luck!

[ivory dust]

Any thoughts?

https://cdn.discordapp.com/attachments/845757271313219584/875177059985535026/unknown.png

First of all suppose H is not trivial {e} group w order 1 otherwise we are done,
gcd(ord(G)/ord(N), ord(H)) = 1 and ordN and ordH divide ordG by lagrange
means ordN and ordH share common factor

is what i have so far

oh ord(G/N) = ord(G)/ord(N) aswell (not sure if this is thm or just property of coset order)

[dusty river]

Yeah this is pretty much Lagrange, this is how it's proved

Suppose there were some element x which were in H but not in N

You can do what you said with this x and find a common factor

Well not exactly what you said

But look at the order of x in all the groups involved and you'll get something

[ivory dust]

okie thanks

[chilly ocean]

What is this question asking, and can someone help me explain the process of the proof thanks

[stone fulcrum]

Hard to answer this as you do have a proof in front of you. Do you know what it means for g to be a generator? Do you know what "f(gh) = f(h)" means?

[chilly ocean]

@stone fulcrumPlease don't post anything if you don't have anything meaningful to say

g is a generator if every other element of G is a power of g