groups-rings-fields

[latent anvil]

That I don't actually want help with

If G is a group, then G is not the union of conjugates of any proper subgroup of G of finite index

If H is a subgroup, then G is the union of all H^g iff H contains an element in each conjugacy class of G

Okay so new argument

Let H be a finite index subgroup, with coset representatives g1, ..., gn

we have a surjection p : union_i {i} × H^gi -> union_{g in G} H^g

This has a split map j going the other way

We also have bijections hi : H -> H^gi

And so a bijection h : union_i {i} × H -> union_i {i} × H^gi

And finally a function k : union_i {i} × H -> G given by k(i, h) = gi h, which is a bijection by lagrange

So k ° h^(-1) ° j is a map union_{g in G} H^g

Is it the inclusion?

If it is, I'm completely solid

This would mean k(h^(-1)(j(x))) = x, or j(x) = h(k^(-1)(x))

Maybe that's the trick? Choose the map j spiritually

*specifically

Show its an inverse for p

Conclude its not surjective?

[cobalt quarry]

I just gotta say,
the first time my mind was truly exploded
in maths
was when I realized
that empty functions exist

And now I am using empty functions daily to construct counterexamples

Madness!

[paper birch]

Why exactly can we map G to S_x by conjugation? Why is this valid

[chilly ocean]

reminder that sylow isn't pronounced in any way similar to "silo"

conjugation gives you an eq rel

[paper birch]

I actually have a Norwegian friend whom I asked how to pronounce it, so that helped

[chilly ocean]

you are essentially projecting

[paper birch]

Ahhh, of course

[chilly ocean]

if H and K are conjugate subgroups, then [H]=[K]

[paper birch]

I see, thanks a lot

[chilly ocean]

np

[somber bramble]

til sylow is a norwegian name, I kinda assumed it was polish or sth

well, doesn’t change how I pronounce it, Norwegian y is closer to German i than to ü (which is essentially Norwegian u)

[paper birch]

Yeh according to my Norwegian friend it is a very un-norwegian name

[simple agate]

I'm finding the GCD of two polynomials over some ring K[X] via the euclidean algorithm. however I get to a step where I have to divide via a non-monic polynomial and was wondering whether it suffices to do it as "normal" by noting that the leading coefficient is a unit in K?

[wind steeple]

Yes

[simple agate]

thanks

[simple agate]

if I have a ring A with ideal I and also an ideal N of A/I how can I interpret the set ${a \in A : a + I \in N}$

[cloud walrus]

alex:

[simple agate]

in my head I feel like it should be a + I is a subset of N

[mild laurel]

I'm not sure what your confusion is

N is just an ideal of a ring

Sure the elements of A/I look like a + I, but N is still just an ideal

[simple agate]

oh I think I got confused by the definition of A/I. I need to refer back to the structure of that

the idea of an ideal of a quotient is seeming a little too abstract for me right now

[mild laurel]

It's really not anything too crazy

Sure the elements of A/I might look a little weird

But they're just elements of a ring like normal

[simple agate]

yeah I just thought that a + I is a one dimensional set and so is N

so it was weird to say a + I is in N

[mild laurel]

Not sure what one dimensional means but

[simple agate]

as in I thought N contained elements of I and not copies of I

[mild laurel]

Sure

[steep hull]

@latent anvil I found a sol to the finite case. Notice by Orbit-Stabilizer that there are |G|/N_G(H) different conjugation actions on H. In all these different actions, there are at most |G|/N_G(H)*|H|<=|G| elements as every element of |H| is an element of the normalizer, where equality can hold iff |H|=N_G(H). Making this assumption, we see that in these different actions (associated with left cosets of N_G(H) in G), the identity element appears every time, so we must subtract it |G|/N_G(H)-1=|G|/|H|-1 times from the count. As |H|<|G|, we can never obtain precise equality, so we’re done.

[brazen frost]

If I have two finitely generated abelian groups A and B, how would I go about computing Hom(A,B)?

[latent anvil]

@steep hull is this for the thing about union of cosets of a proper subgroup?

That's the proof in the finite case I mentioned

The problem on my homework is specifically for infinite groups as well

[steep hull]

@latent anvil Yeah. Wait, could you modify my proof to fit the infinite case (since H having finite index implies N_G(H) has finite index)?

[latent anvil]

I haven't been able to easily

I don't have a solution for this problem yet

[cloud walrus]

iamtim:

[upbeat burrow]

I can clearly see the ideals are prime.

[cloud walrus]

iamtim:

[upbeat burrow]

I can't see it for any of the other ideals though.

[chilly ocean]

@brazen frost fix a minimal set of generators for A and use them to construct the homomorphisms

[brazen frost]

That makes sense thanks!

[magic owl]

ok team

time for some problem solving

Let G be a free group

Then what do we know about Z[G]?

Well, it's elements look like $\sum n_ia_i$

[cloud walrus]

MaxJ:

[magic owl]

So additively

It has a basis in all of the elements of G

Wait this is weird

Ok ok so what is ZG in general

It’s the free Z-module on the elements of G

As a ZG-module it is just the free ZG module

On one generator

ok then lets think about the homology perspective

we've got Z^n for our homology

How does G act on this? Well, up to g-orbits, we are looking at r different loops

Ok wait so take the generators of the homology

And identify them if they share a g-orbit

Then picking one representative is a basis

How many orbits are there?

so I have n H_1 generators (in the obvious way on n-circles)

these lift to r H_1(Y) generators

ok so lifts have all the same cardinality

put that its like

k

then we got k*r generators for the cover

The G-action permutes fibers

so everything within the fiber

is identified

so why is that not Z[G]^r?

well, lets think about what the trivial Z module is

what might generate

this trivial Z module

its something that has a trivial G action and a generator

but what does G act trivially on?

[chilly ocean]

hey folks i am thinking on writing a introductory blog post on group theory. I will take a different approach and first introduce some category theory with concepts of homomorphism&isomorphism. Then since groups are nothing more than a collection of endomorphisms (automorphisms or symmetries) of some mathematical object, i will easily and intuitively define them.

What is your thoughts?

Would it be too confusing for an introductory text?

(of course i am talking about concrete groups that acts on something here. After that i will define abstract groups as the algebraic structures conforms to the definition of a group but not acts on something. )

[magic owl]

Learning groups from a purely categorical viewpoint as a first pass

would be very weird

Groups are generally people's first intro to abstraction and axiomatization

maybe Linear Algebra

making this too abstract too fast is unadvisable, in my opinion

I would think that writing this post for people who already know group theory to learn it from a different perspective would be more interesting

[chilly ocean]

this is a better idea

traditional way textbooks introduces the topic is often very non intuitive or completely focuses on the geometric side of the topic

[magic owl]

Yea I agree that understanding it categorically is way more helpful

[chilly ocean]

but there are also symmetries of abstract objects

[sharp sonnet]

aluffis books does exactly this?

[magic owl]

But its a hard intro

Aluffis book is pretty decent about this

[sharp sonnet]

he defines categories and then defines a group as a groupoid with a single object

but tbf it's labeled as a joke

[magic owl]

tbf that's not the perspective anyone is talking about

focusing on groups via their homomorphisms is different that categofication of groups

[chilly ocean]

for example: i will maybe introduce graphs and isomorphisms between them. Then i will introduce automorphisms (symmetries). And i will show those automorphisms forms an algebraic structure

namely a group. Then i will generalize it

[latent anvil]

@magic owl I strongly want to never think about Z[a free group]

That shit sucks

[magic owl]

Lmao

[inner acorn]

Is there a standard letter we use to denote morphisms between magmas?

[gentle pendant]

🐱 : A -> B

[inner acorn]

xD

I'll just use f: M -> N lmao

[gentle pendant]

yeah just use f lol

[gentle pendant]

poor armless kitty

and legless

😿

[bleak abyss]

Okay so now

There are two definitions floating around for an abstract variety

[chilly ocean]

watching

[bleak abyss]

One is charts and one is ringed spaces

Gonna show they're equivalent

[chilly ocean]

Ok

[cloud walrus]

Daminark:

Daminark:

Daminark:

[bleak abyss]

Like okay to be sure I guess we can just define what it means to be a regular function when you're an open subset of a chart in this way

And then given a general open set yeah you're regular if you're locally regular

So okay this is obv a sheaf of functions

[stone fulcrum]

obv

[cloud walrus]

Daminark:

[bleak abyss]

Takes a sec to write out but really it's easy

So now in reverse

Reverse should be giga trivial

Since like

You already have the candidate charts

Just gotta say transition maps are regular

But if we're talking algebraic subsets then coordinate functions are regular

Okay I guess I should be explicit here, the point is that like, if you restrict an iso of ringed spaces to appropriate subspaces you get an iso

So it suffices to say, let's say you have a map between affine sets that's an iso as ringed spaces, is it regular?

[cloud walrus]

Daminark:

[bleak abyss]

So okay ringed spaces and charts are equivalent

[chilly ocean]

(If you write gg, just add QED with a black box)

[bleak abyss]

Lmao

Alright so now

[twilit pawn]

hey guys, I need to check I haven't gone insane over something simple

[bleak abyss]

You're probably insane anyway but go for it

[twilit pawn]

any non-zero v is an eigenvector of H here right?

with eigenvalue equal to, uh, H?

[bleak abyss]

Wait wut

Okay lemme think here

V is a U(E,H)-module

So for each element of U(E,H) we have a linear endomorphism of V

We don't really have any specific data about H

Also eigenvalue here would be a complex number

Not H, H is just a symbol

[twilit pawn]

wouldn't an eigenvalue here just be a member of the ring it's over?

[bleak abyss]

Nope

[twilit pawn]

which in this case is U(E,H)

[bleak abyss]

So like

A U(E,H)-module is a C-module, aka a complex vector space

Right?

[twilit pawn]

ok yes this seems true

or wait

[bleak abyss]

You just restrict the action

Like, C is considered a subset of U(E,H)

[twilit pawn]

I agree with that for sure

[bleak abyss]

So the point is that a U(E,H)-module V is a complex vector space. So now you ask how E and H act on it

Turns out the actions of E and H are linear

[twilit pawn]

I'm not sold on this. U(E,H) is an algebra and thus a ring. So V is some module over this ring.

[bleak abyss]

Yeah. So V happens to be a complex vector space, right? That's part of the module structure

Now I'm gonna define a function T_E:V->V

[twilit pawn]

U(E,H) happens to be a complex vector space

[bleak abyss]

Given by T_E(v) = Ev

[twilit pawn]

V is a module over this complex vector space

I guess if you restrict the ring, you turn V into a C vector space

[bleak abyss]

Yes

That's what I'm saying

[twilit pawn]

ok, so the eigenvalues are specifically the ones over C?

why not the ones over the whole ring?

So this is a like a general point. If you have a module M over a ring R, and a field F is inside R

you can restrict the scalars you're considering and think of M as an F vector space

but I would still think the eigenvalues on operators M -> M ought to be in R

[bleak abyss]

So like

Eigenvalues as modules in this case ends up being trivial

But do you agree that once V is a complex vector space, multiplication by E is linear?

[twilit pawn]

oh yeah. and if it's a complex vector space there is an eigenvalue

so that's good news

[bleak abyss]

So yeah the point is that really, a U(E,H) module is just a vector space with two linear maps T_E and T_H which satisfy the relation E and H satisfy

The data is the same in each case, and the more useful thing is to consider (complex) eigenvectors of the linear maps T_E and T_H

[twilit pawn]

so none of this has been in lecture btw, I'm mostly trying to google to get definitions. Why are complex eigenvalues better?

[bleak abyss]

Lol classic Ginzo

But yeah doing eigenbusiness over modules is hard and in this case it's not even interesting

Like, it's analogous to just saying that \lambda is an eigenvalue of the linear map C^n -> C^n given by x->\lambda x

And everything is an eigenvector

Like yeah sure whatever

But knowing that v is a complex eigenvector of the map x->Ex is information, knowing that it's a "U(E,H)-eigenvector" isn't

[twilit pawn]

well I can take your word for that I guess. in this case you just need to show there's a complex eigenvalue which is easy

and I got the hint part so I guess I'm good to go

[bleak abyss]

👍

[worn ginkgo]

I'm having a little bit of trouble with this, because I don't know what the matrix form of these hermitian forms look like

If this should go in #linear-algebra let me know

but

[bleak abyss]

I mean do you need a matrix form? Just use the definitions for this

[worn ginkgo]

What is the equivalent condition for something to be hermitian/nondegenerate using just the bilinear form?

[bleak abyss]

Really this just boils down to saying that if A is any matrix, there exists some matrix B such that tr(\overline{A}B) ≠ 0

[worn ginkgo]

positive devinite is obvious

[bleak abyss]

Okay so in general

A bilinear form <x,y> is non degenerate if <x,y> = 0 for all y => x = 0

[worn ginkgo]

right, naturally

I get that

is that all I'm doing here lmao

I'm really sorry, I'm very slow today.

[bleak abyss]

Or at least that's how I'd define it, is your definition as a matrix form saying that x^TAy is non-degenerate if A is non-singular?

Yeah that's all

[worn ginkgo]

ok, how do I show it's hermitian

[bleak abyss]

That'll boil down to properties of trace

[worn ginkgo]

I see

OH

[bleak abyss]

Tr(\overline{A+A'}B) = Tr(\overline{A}B + \overline{A'}B)

And then trace is additive

Etc

[worn ginkgo]

yes

wow I'm stupid

thank you

[bleak abyss]

Lol it happens to everyone, no problem!

[bleak abyss]

Okay so

[cloud walrus]

Daminark:

Daminark:

[latent anvil]

Dami mom says it's my turn on the algebra channel

Let Q be the quaternion group

[bleak abyss]

👀

[latent anvil]

Then Aut Q acts on the nontrivial conjugacy classes

Giving a homomorphism φ : Aut Q -> S3

[bleak abyss]

What's she gonna do, ground me?

Oh wait

[latent anvil]

The kernel of this contains Inn Q, and in fact is exactly Inn Q

And it's surjective because of α(i) = j, α(j) = i, a(k) = -k and β(i) = -i, β(j) = k, β(k) = j

Two distinct transpositions generate S3

Now consider the embedding of Aut Q into S6

More formally, let it act on the elements {i, -i, j, -j, k, -k}

This gives an injection ι : Aut Q -> S6

This restricts to an isomorphism <α, β> ≈ <ι(α), ι(β)>

And ι(α) = (i j)(-i -j)(k -k) and ι(β) = (i -i)(j k)(-j -k)

Note that α(β(i)) = -j while β(α(i)) = k

So α and β don't commute

Thus <ι(α), ι(β)> is the subgroup generated by two cycles of type [2,2,2] which don't commute

Choose any outer automorphism δ of S6

By a result I proved last week, δ is inner iff it preserves the transpositions

So it swaps [2,2,2] with [1,1,1,1,2]

Then δ(ι(α)) and δ(ι(β)) are two transpositions which don't commute

And thus share an element

So they generate an S3

So δ ° ι gives an isomorphism between <α, β> and an S3

Let H be a subgroup of Aut Q isomorphic to S3

Then since [Aut Q : Inn Q] = |S3| = 3!, we have [Aut Q : H] = |Inn Q| = 4

So the action of Aut Q on H by left mul gives a homomorphism Aut Q -> S4

[cloud walrus]

Daminark:

[latent anvil]

Call this ψ : Aut Q -> S4

We want to show ψ is an iso

It suffices to show it's injective, so it has trivial kernel

The kernel is contained in H

And normal in H

So it's trivial or A3 = C3 or H = S3

[bleak abyss]

Okay you paused for a sec I'm gonna check one last thing then

[latent anvil]

What is this?

It looks like AG

[bleak abyss]

Yeah

[latent anvil]

LRS => variety?

Sorry

=> charts

[bleak abyss]

Defining a sub-pre-variety

[latent anvil]

Oh no

[bleak abyss]

I already finished the ringed space locally blah = charts

[latent anvil]

Ah okay

[cloud walrus]

Daminark:

[latent anvil]

Okay so

The kernel can't be H

Because H can't be normal

Since if it was, Aut Q would be the internal direct product of H and the Inner Auts

But those don't commute

This also says there's some γ in Aut Q such that H^γ ≠ H

Now in the case K = A3 in H

We have K^γ = K

But the conjugate of that A3 is the A3 in H^γ

So H intersect H^γ contains K

And by maximality of A3 in S3, it is equal to that intersection

So 2 = [H : K] = [H H^γ : H^γ]

What is H H^γ?

It properly contains H

And [Aut Q : H] = 4!/3! = 4

So 2 = [H : K] = [H H^γ : H^γ] = [Aut Q : H^γ]/[Aut Q : H H^γ] = 4/[Aut Q : H H^γ]

So [Aut Q : H H^γ] = 2, and |HH^γ| = 12

[cloud walrus]

Daminark:

[latent anvil]

It fits strictly between H, H^γ and Aut Q

[bleak abyss]

Now good night

[latent anvil]

Night!

I am doing all this to avoid like

5 lines

Of computation

Well the bit about the kernel at least

[chilly ocean]

ChatJax test: $\mathbb Z_n$

[cloud walrus]

Anyone_Someone2018:

[latent anvil]

By normality of Inn Q, HH^γ (Inn Q) is a subgroup of Aut Q, and clearly it contains HH^γ

By maximality, this means that it is either the whole group or just H H^γ

In the second case, Inn Q <= HH^γ, so Note that 2 = [Aut Q : H H^γ] = [Out Q : H H^γ/Inn Q], but also the cosets of H generate Out Q, so this is impossible

In the first case, Aut Q = (HH^γ)(Inn Q), so [Aut Q : Inn Q cap H H^γ] = 2[H H^γ : Inn Q cap H H^γ] = 2[Aut Q : Inn Q] = 2 * 3!

So |Inn Q cap H H^γ| = 2

Let x be the nontrivial element of this intersection

Then x can't be in H

And <x> is normal in HH^γ, and H is normal in H H^γ (index 2)

So HH^γ is an internal direct product of <x> and H H^γ

Oh ugh

I can't eliminate this case without thinking about it

Because at this point I'm classifying semidirect products

And there's a map S3 -> Aut(C2×C2) with image of order 2

Okay, new goal

What's the easiest way to show that Aut Q doesn't have a sylow 3

[fathom whale]

Hey

I need assistance on these 3 questions can anyone help

[mild laurel]

This is not the right channel for this material

[fathom whale]

Which channel should I go to?

Sorry I just joined

[bleak abyss]

#prealg-and-algebra

This is very much not abstract algebra

[tribal pasture]

Why cant 1+1+2+2+2+2 be the class equation of a group of order 10?

Can I argue this? Take an x whose conjugacy class is of order 2. Then we have Z(x) = 5. Since Z ≤ Z(x) and since Z(x) must be cyclic thus either Z = Z(x) or Z = {1}. If Z=Z(x) then |Z| = 5 but we are given that |Z| =2. Contradiction. Thus the given class equation cannot describe a group of order 10.

[latent anvil]

Here's another way to see it maybe?

Z(G) = C2

Take a 5-sylow P

Then [G : P] = 10/5 = 2

And they intersect trivially by order concerns

So G ≈ C2 × C5

But this is not the class equation of G

[tribal pasture]

Yeah I am not allowed to use Sylow Theory

Is my solution correct? @latent anvil

[latent anvil]

You don't need sylow theory

Do you know any group of order divisible by 5 has an element of order 5?

Sorry I'll read your actual solution lol

[tribal pasture]

Thanks 😅

[latent anvil]

Yeah, looks right

[tribal pasture]

Thanks!

[mild laurel]

Uh wait

Why can't Z = 1?

[latent anvil]

Because there's another element whose conjugacy class is trivial

Right?

[tribal pasture]

@mild laurel Because our class equations has 1+1 which means |Z| = 2.

[mild laurel]

Ah I see

Since elements with trivial conjugacy classes commute with everything else

[latent anvil]

Yup

[tribal pasture]

Let N be a normal subgroup of a group G. Suppose that INI = 5 and that IGI is an odd integer. Prove that N is contained in the center of G.

How would I prove this?

[latent anvil]

Well

It's not true

You need to make a lot more assumptions

@tribal pasture

[tribal pasture]

That is a problem pasted verbatim from Artin Algebra

[latent anvil]

Oh sorry

I misread

I read "Prove that |N|=5" lmao

[tribal pasture]

Lol

[latent anvil]

So either N is contained in the center or it intersects it trivially

[tribal pasture]

I think the fact that G is odd rules the latter case out

[latent anvil]

Why?

[tribal pasture]

Nooo idea. But otherwise it would be stated in the problem

[latent anvil]

Oh sure lol

I don't see how to prove this

Immediately

[tribal pasture]

What about with Sylow theory?

[latent anvil]

Still no

No idea about the prime factors of |G|

[tribal pasture]

cries

[latent anvil]

What happens when G = D5 and N = <r>?

That's normal

But not central

Oh odd

Lol

Sorry

[tribal pasture]

Yeah

[latent anvil]

So maybe we can like force an element of order 2?

Let x be a generator of N

[tribal pasture]

I was tryna go for a proof by contradiction by taking an x in N notin Z and use the fact that N =<x> and then somehow use the fact N is in Z(x) and Z is also a subgroup of Z(x)

[latent anvil]

So let's look at Z(x) for sure

That's what I was thinking

[tribal pasture]

Z(x) must divide G so Z(x) is also odd

[latent anvil]

It's order is the size of G over the conjugacy class size

The conjugacy classes of x must be 1 or 3

So why can't it be 3

[tribal pasture]

Why not 5?

[latent anvil]

Well it's contained in N

But it can't be conjugate to 1

x1x^(-1) = 1

[tribal pasture]

Oh so because N is normal we have that for all g gxg^-1 = x^k for k in {0,1,2,3,4}. We know that |C(x)| must divide |G| this |C(x)| cannot be 2,4. If |C(x)|=5 then there is a g such that gxg^-1 = 1. That is a contradiction. Thus either 1 or 3. Correct?

[latent anvil]

Yup

If it's 1 we're done

If it's 3 there's a nontrivial element of N outside its conjugacy class

Which must have trivial conjugacy class

So it's central

And generates N

@tribal pasture yeah?

[tribal pasture]

Why would the nontrivial element must have a trivial conjugacy class?

[latent anvil]

Its conjugacy class is contained in N

[tribal pasture]

Ah.... And it cant be 5 as per the above argument. And if it was 3 then it would share a point with our x and thus be conjugate. And consequently, lie in its conguacy class contradicting our assumption.

[latent anvil]

Sure

But like

It's doesn't contain 1

And it's disjoint from the conjugacy class of x

So there's 5 - 3 - 1 possible elements of N in there

[tribal pasture]

Okay so I follow you up till generates N

[latent anvil]

N is cyclic of prime order

[tribal pasture]

Agreed. So it can be generated by any of its elements

[latent anvil]

Exactly

[tribal pasture]

But why do we have a contradiction from that

[latent anvil]

We have N = <y>

Where y is central

So N is central

[tribal pasture]

central means?

[latent anvil]

Contained in the center

[tribal pasture]

Ohhh perfect

But why do I not end up having a contradiction with the assumption that suppose C(x) =3?

[latent anvil]

We do lol

but it doesn't matter

[tribal pasture]

So like I get that C(x) =3 implies there exists a y such that C(y)=1. But where is the contradiction for this assumption?

Yeah I get that it doesnt matter. Just wanted to know for my own knowledge

[latent anvil]

We assumed |C(x)| = 3

We did cases on whether it was 3 or 1

[tribal pasture]

Indeed. So we should have a contradiction with the statement that C(x) =3 to ensure C(x) is always 1

[latent anvil]

Right, exactly

The contradiction is that if C(y) = 1

Then our subgroup N is central

So x is in the center

So C(x) = 1

[tribal pasture]

😭 me was hoping for a more direct contradiction. But Thanks! That helped a lot.

[latent anvil]

Sorry fam

[tribal pasture]

Can I say something about C(x^k) if I know C(x)?

[latent anvil]

I don't think so

Like, if x and y are conjugate so are x^k and y^k

But it can be bigger

Or smaller

In size

[chilly ocean]

@tribal pasture A quick argument goes by saying that G acts on N by automorphisms via conjugacy (this is exactly saying that N is normal). This action is encoded in a morphism f: G -> Aut(N).
Now, it is easy to see that Aut(N) has order 4 (N is cyclic, isomorphic to Z/5Z and you can check that automorphisms of Z/5Z are exactly given by multiplication by an element of (Z/5Z)* ~ Z/4Z).
Moreover, since G has odd order, f(G) must have odd order (first iso theorem).
So f(G) has odd order, but its order also divides the order of Aut(N) (Lagrange), which is 4. Hence f(G) = {1}, hence f is trivial.
In other words, the conjugacy action of G on N is trivial, hence N is central

[tribal pasture]

What does it mean to act on something

[chilly ocean]

Ah, well saying that a group G acts on a set X means that you can associate to each element of G a "transformation" of X (in the most general case, a permutation, but sometimes something more restrictive), in a "compatible" way (ie. applying h, then g is the same as applying gh)

For example the group (IR, +) acts on IR via translation

Another example is, the symmetric group S_n acts on {1, 2, ..., n} by permuting the elements

And, in our case, a group G acts on itself by conjugacy

(Tfw when someone in my class says that central and centralizer are the exact same thing and I'm like No)

lol

well one is a noun and the other an adjective

that should be an indication

[tribal pasture]

I see thank you @chilly ocean

[chilly ocean]

I hope that was somewhat clear, but if you have not seen this, there is probably another way to do it

actually yes

[tribal pasture]

We actually used a different method above

[chilly ocean]

I see, very nice

[tribal pasture]

Suppose that the centralizer Z(x) of a group element x has order 4. What can be said about the center of the group?

All I can deduce is Z is a subgroup of Z(x) and thus |Z| =1,2,4

[latent anvil]

@chilly ocean that's very slick

It works for any odd prime, right?

@tribal pasture If |Z| = 4, then x in Z(x) = Z, so in fact Z = Z(x) = G

Z(x) = G iff x in Z

I think the case |Z| = 2 is pretty special too

[chilly ocean]

Thanks ! unfortunately it does not generalize to any odd prime. What makes everything work is that the order of G and that of Aut(N) are coprime

This argument, with the sole assumption that the order of G is odd, generalizes to the case where p is a prime such that p-1 is a power of 2 (ie. a Fermat prime)

[latent anvil]

Ah right, I see

Odd prime isn't the same as fermat prime

[chilly ocean]

yup

all primes except 2 are odd, but we know very few Fermat primes

that being said, you can probably tweak the assumptions on the order of G and the order of N to get other centrality criteria

o(G) odd is a very general assumption

[latent anvil]

Yeah you could

But it'd get hairier

[latent anvil]

does anybody know where I could find a table of similarity classes of GL(4, F2)?

I'm doing it with rational canonical forms

calculating all prime powers in F2[x] of degree <= 4

but it's error prone

[latent anvil]

does anybody know how I might look at degree 4 representations of A7 over F2?

maschke's theorem doesn't apply 😦

[ripe sable]

if i define a ring that spans all comlex numbers and addition is defined as a + b = 4 multiplication as a*b = 4 where a and b are in the ring is that valid

[mild laurel]

What do you think?

[ripe sable]

perhaps

[mild laurel]

Have you tried proving it?

[latent anvil]

what are the ring axioms?

[ripe sable]

i think because every element is an additive identity there is no additive inverse if that makes sense guys

[latent anvil]

Your last bit is correct

but they're not additive identties

that would means a = a + b

[ripe sable]

oh yea

ok so this is not a ring

[latent anvil]

Yup

Because there's no 0

[ripe sable]

yea

[latent anvil]

also no additive inverses

well I guess you need 0 first

so nvm

[ripe sable]

so what else could this be

[latent anvil]

Uninteresting?

It's a semigroup

But not really anything beyond that

And not an interesting one

[bleak abyss]

Hey there motherfuckers

It's Daminark time

Okay so I'm still not convinced by one thing I said last night but now that I'm more awake I'll do it

[cloud walrus]

Daminark:

Daminark:

Daminark:

[bleak abyss]

Okay I'm back

[cloud walrus]

Daminark:

Daminark:

[bleak abyss]

Okay now I'm at least clear on the thing I wasn't sure about

Well I know what it is that's subtle, can we actually glue shit together? The answer is I don't know but prob shouldn't burn so much time on this point

[cloud walrus]

Daminark:

Daminark:

[bleak abyss]

Prob should start playing a bit faster/looser from here on atm

Like now I'm verifying (or loosing patience with) every detail but

I just need to get caught up to class

I can read through Milne or something to do everything super super properly later

Anyway I think I'm gonna get some pizza real quick

[potent lynx]

.

give me like 10 mins

what avg time would you give

for the problem

[latent anvil]

I would say 10 mins

[potent lynx]

yea

[latent anvil]

this is not supposed to be a tough problem

[potent lynx]

ye ayea

[latent anvil]

but you haven't seen rings before

[potent lynx]

yea

[latent anvil]

so don't worry if it takes a bit

[potent lynx]

ok 1 more thing im sorry

i am p stupid can you l;ike

giv me the EXACT definitions

of a unit

so like

x is a left unit of y iff x*y=1?

thats an exact definition?

[latent anvil]

not quite

so x is a left unit means there's some y such that x*y = 1

[potent lynx]

oh yea

ok

[latent anvil]

in this case y is a right inverse for x

[potent lynx]

yeaa

[latent anvil]

R, S rings
f : R -> S
for all x, y
f(x + y) = f(x) + f(y)
f(xy) = f(x) f(y)
f(1) = 1

suppose a in R is a left-unit, with right-inverse b. So a*b=1
show f(a) is a left-unit in S with right-inverse f(b)

there's the problem

[potent lynx]

tysm

bad question

what notation should i use

for multiplication

*?

or just ab

oh you used ab

ok nvm

[latent anvil]

yeah, that's fine

as long as you know what it means

[potent lynx]

ok

so i think i did it but idk

you here?

[latent anvil]

yee

[potent lynx]

it took me like a line lmao so i think its wrong

but ok

[latent anvil]

it's probably correc

[potent lynx]

suppose everything u just said

like r and s are rings and so on

[latent anvil]

I said 10 minutes so yo wouldn't worry

yeah

[potent lynx]

let a in R be a left unit with right inverse b

---> a*b = 1

[latent anvil]

yup

[potent lynx]

take f of both sides

[latent anvil]

yee

[potent lynx]

f(a*b) = f(1)

f(a)f(b) =1

f(a)*f(b) =1

[latent anvil]

that's exactly right

[potent lynx]

and f(a) is left unit

nice

cool

i just unfolded defintions ig but thats ncie

nice

[latent anvil]

this is useful though!

that's like

60% of the proofs in intro algebra

[potent lynx]

oh

[latent anvil]

(before things get interesting lol)

[potent lynx]

XD

why was i struggling

with

permutation groups

that was like the hardest shti ever

[latent anvil]

those are trickier

[potent lynx]

proving cycle stuff

[latent anvil]

so do you know the definition of what it means to be an epimorphism?

once again, it's context dependent

[potent lynx]

i think like

a morphiism that you can do

left cancellation i think

[latent anvil]

right cancellation

[potent lynx]

oh fuck

yea

[latent anvil]

here's how I remember it

in Set, epis are just surjections

so if f : X -> Y is surjective and a compose f = b compose f, then a and b "see" all of Y

since all of Y is in the output of f

does that make sense?

so a = b

[potent lynx]

yeaa

[latent anvil]

Can you prove the epimorphisms must be surjections?

In set!

so to make something clear

we can talk about epimorphisms in a fixed category

in this case, we're looking at Set, the category of sets and functions

so an epimorphism is a function f : X -> Y between two sets

which has the right cancellation property with respect to functions

does that make sense?

[potent lynx]

yea

[latent anvil]

So try and prove that an epimorphism in the category of sets is surjective

[potent lynx]

ok

should i do it here or on papepr?

paper*

[latent anvil]

on paper

this requires some more thinking

[potent lynx]

ok

[latent anvil]

i have a hint in mind if needed

[potent lynx]

so i assume that phi is an epimorph

[latent anvil]

but give it some effort

[potent lynx]

and show that phi is surjective

[latent anvil]

yup

in here is also fine btw

also if I don't respond just @ me

[potent lynx]

where should be the other maps

go from to?

like g_1 or g_2

?

[latent anvil]

that's a good question

check the definition of epimorphism in Aluffi

or w/e

f : X -> Y has to satisfy this property for any set Z and maps g, h : Y -> Z

so we don't know ahead of time which Z to look at

[potent lynx]

yea

so lets just unfold the definition ig

g o f = h o f ---> g = h

now what i want to show is

for all y in Y there exists x in X such that f(x) = y right?

im rusty with basic stuff sorry

[latent anvil]

yeah, that's exactly right

[potent lynx]

so ig take y in Y

ok i think i got it

i am not sure

u here?

@latent anvil

[latent anvil]

Yup

give it to me

@potent lynx

[potent lynx]

ok let me just check something xd

oh fuck my argument is wrong

[latent anvil]

🎊 🎊 🎊

congratulations!

[potent lynx]

what

[latent anvil]

that's how you learn math

[potent lynx]

lmao

ok

[latent anvil]

What was the wrong argument?

[potent lynx]

thats a first step

[latent anvil]

just out of curiosity

you don't need to tell me

[potent lynx]

i like wrote something wrong of the definition

i really couldnt how to do it so i just tried contradiction

konw*

know*

so i assumed epimorphisms are not sujrective

now i wrote what follows from ^ irong

wrong*

and tbh i really dont knwo where to go at all lmao

i will try again

from the start

[latent anvil]

what did you do with the assumption?

[potent lynx]

assume epimorphisms are not surjective

[latent anvil]

if it's untrue that all epis are surjective, then...?

[potent lynx]

then there exists atleast one y in Y such that there does not exist an x such that f(x) = y

idk it just logically looked trash and im p bad at proofs so lmaoo

i didnt know what to do with this

[latent anvil]

Yeah, there is an issue

but it's almost right

actually nvm, I'll say that's 100% right

[potent lynx]

what how

[latent anvil]

there's just like a phrasing issue

so you want to say this:

let f be an epimorphism

assume f is not surjective

yeah?

so you're not assuming that "epimorphisms are surjective" is false

it's all logically equivalent but you need to phrase it right

anyways

so you know there's some y in Y such that y is not f(x) for any x

right?

[potent lynx]

yes\

[latent anvil]

where can we go from here?

and we know f is epi

also, I have good news

[potent lynx]

what

[latent anvil]

the hint I was going to give was "try to prove it by contradiction"

[potent lynx]

lol

im so lucky lmao

[latent anvil]

?

[potent lynx]

so

so you know there's some y in Y such that y is not f(x) for any x
right?

[latent anvil]

yu

*yup

[potent lynx]

y is in Y right?

[latent anvil]

by failure of surjectivity of f

[potent lynx]

and like f(x) is in Y right?

[latent anvil]

Yes

well hang on

what is f(x)?

[potent lynx]

f: X--> Y

[latent anvil]

for any given x (in X), f(x) is in Y

but you need a particular x to say it

[potent lynx]

oh fuck

so i cant just say

f(x) is in Y

[latent anvil]

nope

you need a quantifier on x

[potent lynx]

that would be assuming its surjective

XD

right?

[latent anvil]

it's a "free variabl"

no, there's a subtler thing here

so you know that f(x) is in Y for any x in X

that's just what it means for f to be a function from X to Y

but f(x) doesn't mean anything unless you've defined x somehow

[potent lynx]

yea

[latent anvil]

like saying "for any x in X"

[potent lynx]

so this ragument is trash

ok

so back

so you know there's some y in Y such that y is not f(x) for any x
right?

[latent anvil]

sorry

We do

[potent lynx]

ok

y != f(x) for any x in X

i am tempted to just say g(y) = h(f(x))

but that would be super wrong

ig

[latent anvil]

What's h?

[potent lynx]

function from Y to Z

byea

yea

f(x) is not in Y

[latent anvil]

We need a particular function h

[potent lynx]

yea so its wrong got it

uh oh

[latent anvil]

f(x) is in Y for any x in X

[potent lynx]

yea

[latent anvil]

So to use the fact that f is epi, we need a particular set Z and particular functions g, h : Y -> Z

does that make sense?

[potent lynx]

yea ig

but how would i find these

functions

or setrs

sets*

[latent anvil]

good question

thonk

maybe try examples, maybe try and convince yourself that this should be true without maybe a 100% formal proof

[potent lynx]

ok i am going to try it on paper and try my best

so whats like the outline of the thing we are trying to prove?

if f: X-->Y is an epimorphism

g:Y-->Z and H y--> Z

or no

we construct g and h later

ig

[latent anvil]

Yup

Outline is:

f : X -> Y epi
assume f not surjective
get y in Y such that y != f(x) for all x in X
find a good g,h to contradict the assumption that f is epi

okay so maybe this is a strategy:

suppose we had the right g, h

how could we use that to find a contradiction?

[potent lynx]

they wouldnt equal each other

[latent anvil]

what information would that give us?

yeah, exactly

you need g != h but g compose f = h compose f

does that make sense?

that would contradict the assumption that f is epi

[potent lynx]

g: Y --> Y

g(y) = y

?

[latent anvil]

y is a particular element, right?

Like we fixed an element and called it y

[potent lynx]

yea

[latent anvil]

But there it looks like you're using y as a
variable

[potent lynx]

the y that contradicts urjectivity

[latent anvil]

Yep

So do you mean the function g(t) = t?

[potent lynx]

yea

[latent anvil]

Okay, so see need h to be a different function than g

What does that tell us?

[potent lynx]

h(t) != t lmao

[latent anvil]

Well hold up

Is that for all t?

Or for some t?

[potent lynx]

for { t | g(t) = t }

?

liek

like

idk tbh

im lost

[latent anvil]

That's okay

So two functions are the same iff they're equal on all inputs

[potent lynx]

yea

[latent anvil]

So they're different if they disagree on some input

Right?

[potent lynx]

oh

i want them to disagree

[latent anvil]

Yup

[potent lynx]

on the f(x)

[latent anvil]

g and h

No, you don't

So you want to find a counterexample to g°f = h°f implies g = h

If the assumption there is false, then the implication overall is automatically true

To find a counterexample, you need something which satisfies the assumption but not the conclusion

[potent lynx]

yea

[latent anvil]

i.e. g°f = h°f but g ≠ h

What's g°f?

[potent lynx]

f

[latent anvil]

Yup

and we figured out g ≠ h means that t ≠ h(t) for some t

Right?

[potent lynx]

yea hopefully

now if t != h(t) for some t

then h(f(x)) != f(x) for some x

right?

[latent anvil]

h(f(x)) ≠ f doesn't really make sense

[potent lynx]

yea

now does it make sense?

[latent anvil]

That's not necessarily true

So where does t live?

[potent lynx]

in Y

[latent anvil]

Think about that

Why does h(t) ≠ t (for some t in Y) not imply h(f(x)) ≠ x (for some x in X)?

[potent lynx]

cuz f(x) !=y for some x in X?

[latent anvil]

Almost

Cuz there's some t (namely that element y we chose earlier) which is not of the form f(x) for any x in X

[potent lynx]

yeaa

[latent anvil]

So how can we use this?

Let's recap

We defined a function g

And we have some thoughts about what an h would need to do

What can we say about h if we want g and h to be a counterexample?

[potent lynx]

h cannot equal g for some t

[latent anvil]

(a counterexample to "f is an epimorphism")

I know what you mean, but it sounds slightly better to say "h(t) must not equal g(t) for some t"

So that's correct

And we thought that this might be impossible

Because we could choose x so that t = f(x)

But we realized that if t = y, then that's not true

Right?

[potent lynx]

yes

so ig

ok

im stuck again xd

[fickle brook]

what problem are y'all doing again

[potent lynx]

proving epimorphisms are surjective

[fickle brook]

epimorphisms in what category? Set?

[potent lynx]

yes

[latent anvil]

Set

[fickle brook]

ugh. ok. cat theory isn't my thing.

pass

[potent lynx]

i think if we let

h(t) = y for some t values

no no if we let

idk fuck

[latent anvil]

Let's think about it

Simple case

What if Y = {0,1}

And y = 0

[potent lynx]

recap

f : X ---> Y

[latent anvil]

Yup

[potent lynx]

g and h : Y---> Z

whats g again?

g(t) = t right?

for t in Y

[latent anvil]

Yup

[potent lynx]

ok now

[latent anvil]

We don't know what h is

[potent lynx]

yea

[latent anvil]

But we know constraints that it must satisfy

[potent lynx]

Y = { 0 , 1 }

t can be 0 or 11

now u want t =0

[latent anvil]

No I'm saying what if y = 0

[potent lynx]

what if y = 0

[latent anvil]

Where y is the element we chose earlier

[potent lynx]

oh

[latent anvil]

We could also do y = 1, same thing

[potent lynx]

if y = 0

[latent anvil]

By symmetry

[potent lynx]

and f(x) != y for all x

f(x) != 0 for all x

f(x) != g(0) for all x?

[latent anvil]

This is true, but I'm not sure where you're going with it

Our goal right now is to construct h

[potent lynx]

ok

well

Z also has to be { 0 , 1 } ?

[latent anvil]

Yeah, by how we defined g

[potent lynx]

we want h not equal 0

for some t then

so that h != g for some t right?

[latent anvil]

We want h not equal to 0 at y=0 specifically

[potent lynx]

ok

we are ging to have to define h

piecewisely

right?

[latent anvil]

Yup

[potent lynx]

ok so

let me try it

h: Y ---> Z

h != 0 for t= 0

h =0 for t!=0 ?

t*

[latent anvil]

I think you want t!=0 in one of those cases

[potent lynx]

yea ayea

sorry edited

didnt catch it

[latent anvil]

So we need a specific value for h(0)

[potent lynx]

1

what else could it be

?

[latent anvil]

Yup

Does this work?

We need to check two things

[potent lynx]

that g!= h for some values of t

and g(f) = h(f)

[latent anvil]

Compose

But yeah

[potent lynx]

g o f* yea

g != h for t = 0

[latent anvil]

Yup

[potent lynx]

whats f(0) tho

oh

wtf

g(f(t)) = h(f(t)) by def

[latent anvil]

Will f(0) might not make sense

[potent lynx]

yea

cuz

its not there

[latent anvil]

Because we don't know 0 is in the domain of f

[potent lynx]

yea yea

[latent anvil]

Right

[potent lynx]

i gotta a bad qeustion (assuming all the other questiosn wrent bad)

for the defintion of the epic

[latent anvil]

I have a quibble first

What do you mean by g(f(t)) = h(f(t)) by definition?

[potent lynx]

what

[latent anvil]

There's a little bit more subtly

[potent lynx]

did i saay that

yea yea

assume

[latent anvil]

Since you defined h piecewise

[potent lynx]

g(f(t)) = h(f(t)) for t != 0

[latent anvil]

Not quite

t doesn't live in the domain of f

g(t) = h(t) for t ≠ 0

Right?

[potent lynx]

yes

t is in y

[latent anvil]

y isn't a set

Oh sorry

Y

[potent lynx]

yea yea

Y*

mb

so its done

[latent anvil]

Is it?

[potent lynx]

h =! g

but g(f) = h(f)

[latent anvil]

Wait hang on I think I got confused