groups-rings-fields

[tough raven]

(Let's assume c is non-zero.)

[pastel cliff]

ok im having a bit of a hard time

elements look like cr + aR

[tough raven]

Let K/F be an extension and α1, α2 be algebraic over F in K with minimal polynomials over F f_1, f_2 respectively such that
(I) f_1 splits over K
(ii) for every zero α_1' of f_1 in K, the minimal polynomial of α_2 over F(α_1') splits over K

Does f_2 necessarily split over K?

[pastel cliff]

oop

[tough raven]

Yes

So what are you trying to prove?
That it's cyclic?

[pastel cliff]

yeah i didnt finish bc i didnt wanna interrupt lol

but we need to find some element of that form that generates cR/aR then

in other words some b, such that br = cr + aR right?

ohh wait

we know that since c | a, cb = a

so can we write cr+aR as cr + cbR

idk how useful that might be tho lol

[tough raven]

You want a b in cR/aR st every element of cr + aR of cR/aR, is br' for some r'

or if you'll allow a slightly different notation, $\bar{cr} = b r$, where $x \mapsto \bar{x}$ is the quotient map from $cR$ to $cR/aR$
which is an R-module homomorphism

[cloud walrus]

Raghuram

[pastel cliff]

is the kernel of that map useful

[tough raven]

Later, perhaps

Right now you're trying to prove cR/aR is cyclic, so you need to find the b to use as a generator

[pastel cliff]

i suspect a/c but cant prove it

[tough raven]

It is not

OK, another way of saying a module is cyclic is that it's generated by a single element

Also, if f:A -> B is a module homomorphism, then if S is a generating set for A, then f(S) is a generating set for f(A)

And the quotient map from cR to cR/aR is a surjective R-module homomorphism

[pastel cliff]

so if c generates cR

f(c) will generate cR/aR

so it's really a question of defining that homomorphism

[tough raven]

Almost. f(c) generates f(cR), and if f is surjective from cR to cR/aR then f(cR) = cR/aR

[pastel cliff]

so defining a surjective homomorphism

ok be back in 10

[tough raven]

Cool

[tough raven]

One idea (which is in fact how I came up with the problem) is that K “has enough roots” that there are “as many as possible” F-homomorphisms from F(α_1, α_2) to K.

Specifically, in general there are at most sepdeg(f_1) F-homs from F(α_1) to an extension of F, with equality iff f_1 splits; and then for any such F-hom φ, there are at most sepdeg(min poly of α_2 over F(α_1)) extensions to F(α_1, α_2) with equality iff φ(the min poly) splits.
So there are at most sepdeg(f_1)*sepdeg(min poly of α_2 over F(α_1)) F-homs of F(α_1, α_2) into a field extension of F, with equality iff conditions (i) and (ii) hold.

In particular, we can do the same thing with α_2 first, and get that the number of F-homs is something iff f_2 splits and something else, which would force f_2 to split.

But to do that I need to show that sepdeg(f_1) sepdeg(min poly of α_2 over F(α_1)) remains the same when we interchange α_1, α_2, ie that the separable degree is well-defined for a finite extension.

(I'm trying to use this to prove that separability is transitive, set of separable elements forms an intermediate subfielde etc. so I can't just use those here.)

[tough raven]

It would also be nice if it turned out that for any conjugate γ of α_2 in an appropriate field extension (probably the normal closure of F(α_1, α_2)), there is an automorphism sending α_2 to γ and fixing at least one of the conjugates β of α_1. Then α_2, γ would be algebraically indistinguishable over F(β), hence forcing γ to be in K because minimal polynomial of α_2 over F(β) splits.

[pastel cliff]

sorry i cant really comment on wht you're working on

[pastel cliff]

but i took a quick break, am back and i think this defines the homomorphism!

c maps to c(r + b)?

though im not sure sure how that would work, b is fixed but i dont think r is

[tough raven]

You don't know what b (a generator of cR/aR) is, though
The point of doing it this way and constructing a surjective homomorphism from cR to cR/aR was to avoid going looking for it, and instead get it automatically as f(c)

There's a much more obvious surjective homomorphism – the quotient map.

[pastel cliff]

oh so im trying too hard

so we just send cr \to cr + ar?

[tough raven]

To the coset cr + aR, yes

[pastel cliff]

and then this generates cR/aR right

i was way overcomplicating this

sorry if im being a bit slow too, i appreciate the patience

rather, any element in that coset/equivalence class generats cR/aR

[tough raven]

Yes. That is, f(c) generates f(cR), where f is defined by f(cr) = cr + aR (which we know is a surjective module hom from theory about quotients of modules); and since f is surjective, f(cR) is the full codomain cR/aR, so f(c) generates cR/aR; and f(c) = c + aR.
So c + aR generates cR/aR.

[tough raven]

No, the coset itself is an element of cR/aR, which generates it.

Although IMO, you shouldn't think of the elements of a quotient of an algebraic structure as sets

(even if they technically are in the usual construction of the quotient)

[pastel cliff]

yeah equivalence classes are themselves elements

grappled with that a while ago thankfully

[tough raven]

Right

So now you need to prove cR/aR \iso R/I

for some I

[pastel cliff]

that I is probably the annihilator

[tough raven]

One of the easiest ways to show X/Y \iso Z is to find a surjective homomorphism from X to Z whose kernel is Y

[pastel cliff]

yeah kernel of that surjective homomorphism comes in handy now right

[tough raven]

Here, both sides are quotients, so you could try either way:
(i) find a homomorphism from cR to R/I with kernel aR
(ii) find a homomorphism from R to cR/aR with kernel I

But since I has been left unspecified, the second way is easier, because you can just define I to be the kernel and skip having to prove that

[tough raven]

Yep, in a way the point of quotients is being able to use First Isomorphism Theorem with them

[tough raven]

(and if you do need to know I, you can compute it as the kernel of the homomorphism instead of having to guess)

[pastel cliff]

i have this proof from before too

[tough raven]

Well, then you're done since you already showed cR/aR is cyclic

[pastel cliff]

[tough raven]

In the last paragraph, {r | rm = 0} is Ann_R(m) by definition, right? Not Ann_R(M)

Oh, R is a commutative ring so they're equal (though not by definition)

[pastel cliff]

my only concern then is that the divisibility thing never came into play

[tough raven]

You need the divisibility condition for aR to be a subset of cR

[tough raven]

Got it—we can pick a big field extension (where all the minimal polynomials over F split) so that all the conditions hold whichever order you take, so all the expressions are equal (to the number of homomomsrphisms into this big field). Then we can use the equality.

[echo granite]

Can someone proof this lemma, or make an example, because I didn't get it

[pastel cliff]

why do we say that a free module is "on" a set X

[hot lake]

When you have to count the size of a subset of AxB you can sort the elements horizontally and sum for each row the number of elements in the row or you can sort vertically and sum for each column the number of elements in the column

Are you asking why we use "on" instead of another conjunction

[rotund aurora]

So I wanted to show that the map defined by $(x \colon y) \mapsto (x^3 \colon xy^2-x^3 \colon y^3-x^2y)$ only has one point in the image with two distinct preimages.

So suppose that $(x^3, xy^2-x^3, y^3-x^2y)=\lambda(z^3, zw^2-z^3, w^3-z^2w)$. From the first equation, it follows that $x=\varepsilon \lambda^{1/3}z$ where $\varepsilon^3=1$. With this and the second equation, it follows that $y=\pm \lambda^{1/3}\varepsilon w$, and substituting both expressions into the last equation, we obtain
[
\pm(w^3-z^2w)=w^3-z^2w
]

[pastel cliff]

kinda?

like what's the point of the set X i guess

[rotund aurora]

If the sign in the last expression is positive, then that relation is trivial

[hot lake]

Is like a basis for a vector space

[rotund aurora]

idk how to show that, unless I missed something

[hot lake]

Basis are occasionally useful

So is the same for free modules

[pastel cliff]

i understand that a free module is one that admits a basis

[cloud walrus]

Croqueta

[hot lake]

And more generally, free objects on a set

[pastel cliff]

this is the defn im given

[hot lake]

Is saying you build the object as freely as possible from that set

[pastel cliff]

is X just an index set?

[hot lake]

Yes

[pastel cliff]

oh okay

and in the infinite case we treat it kinda like a direct sum im assuming

[hot lake]

Yes

[pastel cliff]

otherwise it wouldnt have a basis(?)

[hot lake]

It maybe would

But that would be too big

You want your free object on X to be generated by X

Ah infinite product instead of an infinite direct sum would have some elements not generated by X

[pastel cliff]

ok got it

[hot lake]

And in your definition

You get a canonical inclusion

Of X into R^X

So you can abuse language and talk about X as a subset of R^X

[pastel cliff]

map each element to the basis element at each index?

[hot lake]

Map each element to the sequence that is 0 everywhere except on that element where it is 1

[pastel cliff]

det

but also an adjacent quesiton

[rustic crown]

(okie one tiny thing... this is non-standard notation >.< most people would use R^{⊕X} or R^{(X)} for this. And R^X would be the product without requiring finite support)

[hot lake]

Ah yeah I didn't notice they use the same notation as the infinite product

That's awkward lol

[pastel cliff]

i have this thm

why is that second part necessary

doesnt the basis already necessarily span?

so "adding" basis elements shouldnt matter

we already have a spanning set

[tough raven]

It's only given to contain the first n-m elements of the existing basis.

w_1, …, w_m has been added after removing v_{n-m+1}, …, v_n.

[pastel cliff]

oohhhhhh

i see

didnt notice that detail

[rotund aurora]

How advanced is the Riemann Roch theorem? What is the background?

[pastel cliff]

this is a bad start isnt it

can i literally just say "if any r_i is not = 0, it is clear that the triple will have a nonzero index and therefore not be equal to (0,0,0)"

also idk why i wrote x_i

[blissful moon]

What does it mean for a projection to be minimal? In particular, what does it mean if i say x is a minimal projection on a finite dimensional matrix algebra?

[formal ermine]

if x - p is a positive operator for any non zero projection p then x = p

[chilly ocean]

are these 2 examples the same? I am thinking that invertible linear maps on itself is the automorhpisms on the category of Vect (P.S. i dont know category theory that much)

[pastel cliff]

det

[rustic crown]

the category would be Vect(k) and yep, the first is a special case of the second :3

[chilly ocean]

aight lets go

[rotund aurora]

Express the projective plane curve V(x^2z-xy^2+xz^2-2xyz-x^2z-2yz^2) with respect to the reference system {(1 : 1 : -1), (1 : 0 : -2), (1 : 0 : 0), (0 : 1 : 0)} What does this mean?

[pastel cliff]

projective plane smells of alg topology

[rotund aurora]

projective plane smells projective stuff

[rustic crown]

i just learned what Riemann Roch is, so don't have a good intuition with it... but the proof wasn't so hard given enough time. once you have developed sheaf cohomology, defined line bundles, kahler differentials and computed cohomology of projective space, it should be pretty accessible.

[formal ermine]

what's S

[rustic crown]

a letter

[formal ermine]

[rustic crown]

[pastel cliff]

i know this is a bit long but can u check rq pls det

it's nothing hard

[rustic crown]

[pastel cliff]

it's just checking two sets are the basis

of Q^3

and that that matrix at the bottom is the change of basis matrix

the last part is the only one im uncertain about

[rustic crown]

there is still something to be done

the argument you give doesn't work over Z

so what you have used is that if the dimension is 3, a linearly independent set of size 3 is automatically a spanning set

the linear independence check looks correct

[pastel cliff]

oh quick tangent

cuz im forgetting my lin alg

when do dimension and rank differ

[rustic crown]

dimension is used only over fields

rank for more general rings

if the ring is commutative that is

[pastel cliff]

all rings are commutative

[rustic crown]

because R^{⊕X} = R^{⊕Y} would imply |X| = |Y|

over Z, you're not sure if any element of Z^3 can be written as a Z-linear combination of g1, g2, g3. you know it can be always written as a Q-linear combi from vector space theory

like simplest example would be consider the Z-module Z

and g = 2 gives you a linearly independent set of size 1

but it doesn't span

over Q everything works like you expect

[pastel cliff]

but if it's a basis in Q3

consisting of integers

maybe im thinking too simply about it idk

is the matrix at the bottom right tho

[rustic crown]

idk the matrix looks "too positive"

what's gamma_{A, B} supposed to do?

[pastel cliff]

yeah i kinda just multiplied shit together lol

[formal ermine]

why is the zero ring not an integral domain

[rustic crown]

because it is not

[formal ermine]

why

[pastel cliff]

gamma is homomorphism $Hom_R(R^m, R^n) \to Mat_{m \times n}(R)$

[rustic crown]

why do you want it to be lmao

[cloud walrus]

sebbb

[formal ermine]

so that the ring itself is a prime ideal

[rustic crown]

but you want to be able to take field of fractions of integral domains

[formal ermine]

it feels stupid to me that the extension of an ideal in Z to Q is not prime

[rustic crown]

well because Q hardly has any ideals

[pastel cliff]

det what am i missing

is it not a basis of Z?

[chilly ocean]

i do not understand this diagram

[thorn delta]

lang

[chilly ocean]

like okay? what does it mean

[rustic crown]

how do you define the change of basis matrix?

[chilly ocean]

yes

[rustic crown]

sage: A, det(A)                                                                 
(
[3 1 0]   
[1 3 1]   
[1 1 1], 6
)
sage: B, det(B)                                                                 
(
[1 0 1]   
[2 1 0]   
[0 2 1], 5
)

[pastel cliff]

he has a book on every topic under the sun it seems

[rustic crown]

doesn't look like bases to me.

[pastel cliff]

hahaha det typing "det"

[rustic crown]

[chilly ocean]

this is the previous part

[rustic crown]

if they were basis, it would have had det +-1

[chilly ocean]

but idk how this diagram depicts what is written here where did the isomorhpisms come from

[pastel cliff]

along with this

we havent defined det yet technically

poor det

[rustic crown]

oh okie

so gamma sends g_i to something and you wanna write g_i in terms of f_j

[thorn delta]

I had this exact same question a few years ago trying to read lang. I don’t think it’s anything deep. Just an occasionally useful shift in perspective

[chilly ocean]

but i do not understand the shift at all

[thorn delta]

Maybe someone else has more wisdom on the matter tho

[hot lake]

it's saying that if 0 -> G' -> G -> G'' -> 0 is exact then G' is isomorphic to ker g and G'' to G/ker g = G/im f (= coker f)

the map f : G' -> G in the sequence gives an isomrophism G' -> ker g by restricting the image

[chilly ocean]

coker?

[rustic crown]

if A and B are these matrices, then we want B^(-1)A right?

[hot lake]

sorry, cokernel is another thing

[chilly ocean]

how in the hell is G' iso ker g

[hot lake]

the exactness says that ker f = im (the map from 0 to G') = 0 and im f = ker g

so f is injective

and its image is ker g

so if you define f' : G' -> ker g by f'(x) = f(x)

then f' is an isomorphism

[thorn delta]

also, by "shift in perspective" i mean for example that most reasonable statements about one converts to equivalent statements about the other exact sequence

[chilly ocean]

by 0 you mean {e} in grp?

[hot lake]

yeah, in an exact sequence, a 0 means the trivial group

there is only one possible group morphism from 0 into G, and also from G into 0

so those maps are also called 0

and when we say im f = 0 or ker g = 0

we mean they are {e}

[chilly ocean]

aahhh its starting to make some sense

[thorn delta]

oops i think i misunderstood when you asked

like okay? what does it mean

[hot lake]

to get an isomorphism from G'' to G/ker g, for a given x in G'', g is surjective so you pick y in G with g(y) =x, then you map x to the projection of y modulo ker g

[chilly ocean]

i was responding to the book

[thorn delta]

yea, but by "what does it mean" i thought you were asking why we care more philosophically speaking

[hot lake]

or rather you just map x to g-1({x}) since that's an element of G/ker g

technically

so any short exact sequence 0 -> G' -> G -> G'' -> 0 is "isomorphic" to a short exact sequence 0 -> H -> G -> G/H -> 0

[chilly ocean]

ah so for any exact sequence 0 -> G' -f> G -g> G'' -> 0 it is the same (isomorphically) as 0 -> ker f -> G -> G/ker f -> 0

[hot lake]

yeah

[chilly ocean]

yeah i was writing that

[hot lake]

we wrote the same thing at the same time lol

[chilly ocean]

ye

aight thanks so much

[hot lake]

also sometimes maybe people write 1 instead of 0

it's still the trivial group

[coral spindle]

det gave you background for the big huge complicated Riemman–Roch theorem. The basic version (that Riemann and Roch actually proved) doesn't have much background. If you know what a plane projective curve and a riemann surface is, you're halfway there. Kirwan's book covers everything.

[rotund aurora]

det overkills everything

xDD

thanks, Ill look into that book

[coral spindle]

And just for the future

that's more an algebraic geometry topic

[rotund aurora]

you mean that I should ask in #algebraic-geometry next time or that I should be aware that its alg geo (because I know its alg geo, but I usually ask here so I just asked here lul)

[coral spindle]

The people who hang around there are just more likely to be able to answer

[rotund aurora]

👍

[rustic crown]

oh there are multiple riemann rochs?

[coral spindle]

and also something something this channel is overburdened something something

[coral spindle]

well one is a generalisation of the other

But the original formulation didn't use sheaves, for example

[rustic crown]

[spark crow]

quick question, i'm trying to show that $f$ is a homomorphism
Assume $G = (\mathbb{Z}, +)$ and $H = (\mathbb{Z}/3\mathbb{Z}, +)$.
I define my Homomorphism $f: G \to H; 3n+x \mapsto x$ where $x \in {0,1,2}$
Now i can easily show that $0_G = 3 \cdot 0 + 0 \Rightarrow f(0_G) = 0_H$
But how would I go to show that $f(x +_G y) = f(x) +_H f(y)$

[coral spindle]

Or perhaps I should say they didn't realise it used sheaves

We cannot answer this question without more information, achilles

[tribal moss]

Like Gauss and Euler didn't realize their work was really an aplication of ZFC.

[coral spindle]

What you've written didn't make sense

defining f : Z -> Z by 3n + x goes to x? This is not well-defined.

[rustic crown]

i thinkt aht means 3n+0 goes to 0, etc

all 3 said at once

[coral spindle]

Right, so this is in fact a quotient Z/3Z not the set exclusion Z\3Z

[spark crow]

yes i wanted to shorten 3n+0 -> 0, 3n+1->1, 3n+2->2

[rustic crown]

oh oopsie

didn't even see a backslash lol

[coral spindle]

yes it was updated to include the quotient

[spark crow]

sorry I meant Z/3z

[cloud walrus]

achilles199703

[coral spindle]

So you are defining Z/3Z to be simply the set {0, 1, 2}

instead of the usual definition of a quotient group?

[spark crow]

i am actually not sure on the notation but i thought using 0,1,2 is acceptable as they are representatives

[coral spindle]

Do you know what a quotient group is or is this just notation you've been told?

I just need a yes or no answer here

[spark crow]

hmm, i'd say yes

[coral spindle]

OK

So your map is actually sending x to x + 3Z

after which the proof is trivial: f(x + y) = (x + y) + 3Z = (x + 3Z) + (y + 3Z), both equalities being by definition

[spark crow]

the qutient group Z/3Z = {3n+0, 3n+1, 3n+2}

[spark crow]

oh, yeah that makes sense

i knew i was making a stupid mistake thank you

[coral spindle]

Notice how choosing representatives made this more complicated!

@hidden haven any particular meaning to that emoji I should know?

[hidden haven]

Expecting a yes or no answer to an X or Y question

[coral spindle]

Lol, good point

[hidden haven]

Born a logician 😌

[formal ermine]

what's an example of an ideal whose image under some ring homomorphism (from one ring to a different one) is also an ideal

[coral spindle]

0

[formal ermine]

I was just about to say

not the zero ideal

[coral spindle]

The whole ring, and the identity

[hidden haven]

If ring homomorphism is surjective this is true for any ideal

[formal ermine]

if it's not surjective?

what's the identity

[coral spindle]

the identity homomorphism

[formal ermine]

from one ring to a different one

[coral spindle]

if it's not surjective it may not be closed under multiplication by scalars

[coral spindle]

Give it a different name, must be different

[formal ermine]

[coral spindle]

But fr

[blissful moon]

If I have a *-isomorphism f : M_n -> M_n that maps x to yxy^{-1} (i.e., the conjugate given by y in GL_n). Is it true that y* = y^{-1}? (so is it true that y in U_n as well?)

[coral spindle]

apply any automorphism

[spark crow]

$f: G \to H \quad x \mapsto \begin{cases}[0] &\text{if}x=3n\ [1] &\text{if}x=3n+1\ [2] &\text{else}\\end{cases}$

[formal ermine]

cases

[coral spindle]

Can you tell me what a *-isomorphism is?

[blissful moon]

bijective *-homomorphism

[cloud walrus]

achilles199703

[spark crow]

So if I would want to use representatives it would be better to do it like this, or am i wrong

[coral spindle]

and what is a *-homomorphism?

[formal ermine]

the extension of nZ from Z to Z[i] is nZ[i]?

[blissful moon]

a ring homomorphism, is linear, and has the property that f(y*)=f(y)* for all y

[spark crow]

i am not familiar with this notation, what does Z[i] represent

[coral spindle]

And what does that notation * mean

[formal ermine]

gaussian integers

[coral spindle]

is that conjugate transpose?

[formal ermine]

equivalently Z[x]/(x^2 + 1)

[blissful moon]

it's an involution. since it's in a matrix algebra M_n, i guess you could take it to mean conjugate transpose

[spark crow]

ok but i use [0] as represantative for the class 3Z+0

[formal ermine]

this is not related to your question

it's a question of my own

[spark crow]

oh sorry my bad

[tribal moss]

In the sense of "the smallest ideal of Z[i] that contains nZ"? If so, yes.

[formal ermine]

it's the ideal generated by the image of nZ (which is just nZ)

so yeah

ok thanks

[sharp imp]

Does anyone know what's going on with that Z_2 part of the automorphism group of A_6? I'm trying to get an intuition for how it got there

[rotund aurora]

How do you compute discriminants of polynomials quickly? If A is your polynomial, would you use the characterization of the discriminant of A in terms of the resultant of A and A' ?

[formal ermine]

can't you also express it in terms of its roots

[rotund aurora]

yeah

mmh actually

Res_x(A,A') is a (2n)x(2n) determinant, where the entries are like the coefficients of A, so I guess its easier than just calculating roots and multiplying their differences lol

but idk if there are other methods for calculating the discriminant

[spark crow]

last question what law or definition is used for the second equality, it feels like a distributive property, but i am uncertain why exactly this works, even though i see why it works

[coral spindle]

It is simply defined

That's how we define the operation in the quotient group

Are you sure you're familiar with quotient groups? Maybe you should look back at your book/notes.

[spark crow]

thats what im doing right now, thanks

its literally in my notes, but as (aN)(bN)=(ab)N but as we use + in Z/nZ the implicit * can be replaced with +, thanks again

[charred moat]

This may be a silly question but where does the fact that $dim(R[\lambda]/(f(\lambda)) = degf(\lambda)$ come from?

[cloud walrus]

Modeka

[spark crow]

So if I have G=(Z,+) and H=(Z/3Z,+) and my Homomorphism f:G->H, then ker(f)=3Z, if I understand this right. As ker(f) is a normal subgroup of G, i can now build the qutient group G/ker(f) which in this example would be Z/3Z if i made no mistake which in turn is again H. As far as I understand G/ker(f) is isomorphic to f(G), but must not necessarily be identical. So my question is, has anyone a better example where G/ker(f) and f(G) are only isomorphic and not exactly the same?

[lethal dune]

what do you mean same if not isomorphic?

[lethal dune]

as sets? you can get plenty of examples like say Z_4 and <i>

[coral shale]

surely most examples will have them being not the same

[spark crow]

i meant literally identical or the exact same and not only isomorphicly identical

[coral shale]

The point being you start with some surjective homomorphism phi: G -> H for some G and H

The only case where G/ker(phi) coincides with H

is if you chose H = G/ker(phi) and phi = pi

in the first place

[charred moat]

maybe equal as sets or equal more strongly in the sense that both their underlying set and underlying structure is identical

[primal tusk]

does anyone know how to prove part d?

Basically I have found f1,f2,f3 which correspond exact to the minors of the matrix in part e

[coral spindle]

You haven't shown us what R_4 is

[primal tusk]

sorry

so i know that (f1,f2,f3) is a subset of ker(phi)

but i have no idea how to prove that ker(phi) is a subset of (f1,f2,f3)

$f_{1}=x_{2}x_{4}-x_{3}^{2}\ \ \ \ \ f_{2}=x_{1}x_{4}-x_{2}x_{3}\ \ \ \ \ \ f_{3}=x_{1}x_{3}-x_{2}^{2}$

[cloud walrus]

Bradley

[tawny gust]

Maybe you should try to reformulate your question because it is all over the place

[chilly ocean]

you can type normally and just put the dollar signs around the math. you don't need to put dollar signs around the entire message and then do some weird text stuff

[primal tusk]

it was like not sending when I tried to do it normally

its suppsoed to say "such that ker(phi) = (f1,f2,f3)"

in the last picture but its cut off for some reason

man i dont even know latex i just want to ask a question 🥲

[void cosmos]

@rustic crown you were right

[rustic crown]

uwu?

[void cosmos]

I emailed the prof for what text she used and it was hungerfords undergrad algebra

and guess what an euclidean norm is there

uwuwuwuw :3

[rustic crown]

d : R\ {0} --> N

[void cosmos]

not that

I meant that it had n(a)<n(ab)

as an axiom

[rustic crown]

oh right

[void cosmos]

uwuuwu

[chilly radish]

ah yes, d for "euclidean"

or is it d for "norm"

[rustic crown]

degree

or v

for valuation

[chilly radish]

stop ruining my joke

[void cosmos]

^_^

[rustic crown]

gomen

[void cosmos]

<_>

[rustic crown]

det bad at jokes

.<

[chilly radish]

you mean \nu for naluation

[rustic crown]

lol

[void cosmos]

.<

^.^

god i love det lmfaaaoo

^,^

[rustic crown]

^w^

[void cosmos]

Nyuu

nyu

[rustic crown]

iibui

[formal ermine]

what have I walked into

anyway

what does this mean

the red part

[rich patrol]

Your homomorphism is being decomposed into two maps p and j

[formal ermine]

so like

[rich patrol]

p is subjective (Why?)

[formal ermine]

f is uniquely determined by p and f?

[next obsidian]

p is just f but with a different codomain

j is the inclusion of a subset into B

[formal ermine]

why are we doing this

[abstract orbit]

is there a bijection between multiplicatively closed subsets S in a ring R and prime ideals P given by P to R\P=S?

[next obsidian]

No

Take {1,f,f^2,…}

There’s a bijection between sets of primes and saturated multiplicative sets though

Google around about what a saturated multiplicative set and you should be able to find a statement

[tawny gust]

How does one go about proving that an element is not in an ideal?

If it is generated by multiple elements

[rustic crown]

do you have a specific example in head?

you could try to show it's non-zero in the quotient

but dunno what else you could do in general

[rotund aurora]

An ideal of R is a submodule of the R-module R, so one way is via linear algebra. For example, to show that 2 is in the ideal (3,5) (in Z) you would have to show that 2 is a Z-linear combination of 3 and 5, that is, to find integers x,y such that 3x+5y=2

[white oxide]

to show that G is finite, could you use the fact that the only subgroups of G are the trivial subgroup and the improper subgroup or G itself, and then the order of G must divide the order of G so G is finite?

[rustic crown]

nu

[chilly radish]

No, since that's a priori assuming the order is finite

To clarify, to use Lagrange's theorem in the formulation of "order of subgroup divides order of group" you need to know that your group is finite to begin with

Since otherwise "divides" is meaningless (well it has meaning with cardinal arithmetic but that's irrelevant to this formulation really)

[white oxide]

oh ok

i have no clue how to show that it's finite then

[rustic crown]

given an element can you construct a subgroup using that?

[white oxide]

but the only subgroups of G are the trivial subgroup and the improper subgroup right

since there are no nontrivial proper subgroups

are you saying that i can use some element to construct G itself?

[rustic crown]

yep, so if you can build a subgroup out of some element, then that hypothesis would ell you a lot about your group!

[white oxide]

or use the fact that if G is cyclic so then it's of prime order

[rustic crown]

that's not true >.<... but its converse is

[white oxide]

right my b

[rustic crown]

so how do you show a group of prime order is cyclic?

[white oxide]

so nvm

first you have to identify a generator right

and then for any g in G you have to show that if a is a generator then g = a^n for some integer n

[rustic crown]

yep, and how do you pick this generator?

[white oxide]

well it can't be the identity

[rustic crown]

and how does order being prime help

[white oxide]

so let it be some element a

oops

i don't know why my messages didn't send

anyways

maybe you could use the fact that the order of an element of a finite group divides the order of the group

and set up a contradiction of some sort

[rustic crown]

yep, that's the idea.

but do you recall how you proved that?

lagrange's theorem says that order of a subgroup divides order of the group

[white oxide]

order of an element is the same as the order of the cyclic subgroup generated by that element

so it follows immediately from lagrange's theorem

[rustic crown]

yep

so given an element you can consider the cyclic subgroup generated by it lol

[rustic crown]

now 🙈

[white oxide]

oh so a generates G

so G is finite?

[rustic crown]

not quite, it can still be an infinite cyclic group

but do you see why that would be a problem?

[white oxide]

tbh i'm not sure where the contradiction is in that

besides the fact that the order of it must be divided by the order of itself right since it is it's own subgroup

but that was what i suggested before and apparently it doesnt work

OH WAIT

is it because it is of finite order

bercause a^m = e for some positive integer m

or something

[rustic crown]

no i mean

Z does have non-trivial subgroups

so it can't be infinite cyclic

the argument you gave was that if a is any non-identity element then <a> must be the whole group

so every non-identity element generates your group

that's not the case for Z

since <2> = 2Z

[white oxide]

ah

true

no fucking clue then

oh

Z has nontrivial proper subgroups

so that's the contradiction i suppose

because if G was of infinite order then it would be isomorphic to Z which has nontrivial proper subgroups

my bad didn't see that

[prisma shuttle]

wut is a nilpotent homomorphism

i have an intutiive guess as to wut it is but i couldn't find it explicitly defined anywhere so i just wanted to confirm

[next obsidian]

What’s the context

[chilly ocean]

what do you think it means

[prisma shuttle]

phi_s composed n times, for some postiive integer n, is the 0 map?

[uneven sundial]

i have question sry if im interrupting. i understand b) but not a)

[chilly radish]

Yep

[charred crescent]

can someone help me understand the abelianization of a group?

[prisma shuttle]

this is just the quoteint of a group by a commutator subgroup

so essentially wut u can think of it is like

if we denote the commutator subgroup by [G,G]

then the abelinization of G is the group G/[G,G]

and we define the quotient by the relaiotn xy=yx

which is where it gets it name "abelinization" from

[charred crescent]

im just having a little trouble visualizing what happens when you mod out by the commutator

[charred crescent]

so you're saying a is related to b if and only if aba^-1b^-1 = e ?

[prisma shuttle]

do u know the def of a commutator subgroup

u might also know this or see this in texts as a derived subgroup

thats another name for it

[charred crescent]

i probably couldn't recite it

[prisma shuttle]

basically yes

are u familiar with rings?

[charred crescent]

somewhat

[prisma shuttle]

if u think about modding a ring out by an ideal

like R/I right

[charred crescent]

okay no i don't know that much

[prisma shuttle]

u can essentailly think of this as setting all the elements in the ideal equal to 0

oh ok

well u don't need to understand rings

essentially the commutator subgroup is just generated by all eelmetns of the xyx^-1y^-1 (which matches the form u wrote)

and wut "modding out" means is just setting all the thing sin the set ur modding out by to the identity

whch here is e

so basically ur just setting everything of the form xyx^-1y^-1=e

aka xy=yx (which is where it gets it name from)

so yes wut ur thinking of is correct

[charred crescent]

so a is related to b in that sense if aba^-1b^-1 = e

[prisma shuttle]

oh ur asking like

when is a equivaelnt to b?

[charred crescent]

thus G/[G,G] forms a partition of G

[prisma shuttle]

in this new abelinzation of the group?

[prisma shuttle]

is this wut ur asking

[charred crescent]

no im just trying to understand the concept of abelianization in general

[prisma shuttle]

i think u have the right idea

at least from wut ur saying

[charred crescent]

okay, but then using the tool of abelianization to solve a problem is a different story

like i think i understand the definition but i need more time to really let it sink in

thank you

[prisma shuttle]

yea just leanring math in general u need time to get used to the defs and use them in problmes

ima sure it'll come natural to u after doing a couple problems

gl

[charred crescent]

okay one last question lol

how do i know if a subgroup containing all commutators is the smallest one that contains all commutators

[prisma shuttle]

remember that the commutator subgroup is defined as the group generated by the elements

"generated" by def means its the smallest one

[charred crescent]

hmm, can you speak on why that is? or is that just truly implicit in the definition of generated ?

[prisma shuttle]

this is just the definitino of the set being generated

its the smallest one that contains it and satisfies all the conditions

[charred crescent]

okay, i think i understand

thank you

[prime sundial]

for a group to be solvable, do we need the derived series to reach the trivial subgroup in a finite number of steps? is it even possible for the derived series to reach the trivial subgroup in an infinite number of steps?

[next obsidian]

Doesn’t make sense really

It has to be finite

(You could do it “infinitely” by intersecting all of them and this could theoretically be trivial even if it isn’t at any finite step, but this is not the definition of solvable)

[sonic coral]

so i’m trying to go through all the groups of order less than 102 and show that they are either cyclic or not simple (excluding 60)

i’ve knocked out most of them. i still need to use some of sylows theorems to see if i can knock out a few more

but my question is, which orders will i have to look at individually, if any?

i feel like the highly composite numbers will be the difficult ones, like 24 and 36, unless i can get those using a sylow theorem

[next obsidian]

I mean this depends entirely on what classes of groups you manage to systematically show aren’t simple

If you manage to prove burnside’s theorem you automatically handle everything with at most 2 prime factors

But you aren’t going to be proving burnside’s theorem

I think it’s reasonable to get pqr, and p^aq^b where a,b are <= 2 (and p-groups lol)

Maybe p^2q^2 might be hard

[sonic coral]

he gave me a few that would be helpful, i will be proving them in the coming days but i’ve went ahead and knocked out the orders that they would get rid of

and i got sylows theorems today so they are included as well but not listed

i’ve proven lemma 1, and he did lemma 3 at some point during lecture for me

how does lemma 6 help me?

[livid willow]

I know I did a monke calculation, but I think this should be correct enough to check that given statement is false?

For nZ to have that many (23) maximal ideals, it should be this large, correct?

[rustic crown]

yep, but you're going for a much sharper lower bound.

you don't need to do a monke calc to see that this statement is false

if nZ is contained in exactly 23 maximal ideals, then so is n^100Z

[novel parrot]

does someone know a good way to decide if a matrix is non singular or not?

what ive been doing is computing its homomorphism and then deciding if its bijective

[rustic crown]

compute det

[novel parrot]

det 0 ?

if non singular

idk determinant

[rustic crown]

so you over a field or comm-ring or int-domain?

[novel parrot]

field

for now

[rustic crown]

det = 0 if and only if singular then

[novel parrot]

oh

[rustic crown]

over a comm-ring, the map is an isomorphism if and only if det is a unit.

(idk how singular is defined really for general comm-rings)

[novel parrot]

in a comm ring does a matrix corresspond to homomorphism still ?

only in certain comm rings right

like noetherian

[chilly ocean]

Compute @rustic crown

[rustic crown]

a matrix is just a data of a bunch of maps

so in any category with finite products and coproducts
a map coprod(A_i) --> prod(B_j) is given by a matrix of maps A_i --> B_j

so even if R is non-commutative the category R-mod is additive which means products and coproducts agree. and a map R --> R is determined by where it sends 1, so you write a map R^n --> R^m as a matrix in M_{m x n}(R).

but det only works nicely with commutative rings >.<

[rustic crown]

so yea this can be generalized quite a bit :p

[novel parrot]

lol

[next vine]

Apologies for this simple question and I'm not sure if it's in the right place. I did a search for quaternions and this channel came up as where it was discussed previously.

I have a body which has a rotation q_H with a position P_H, with another point P_L. I want to translate the vector P_H -> P_L such that the body has no rotation. To do this I got the inverse of the rotation q_H^-1 and transformed the vector P_H -> P_L by this inverse.

To transform, I multiply the inversed rotation by a new quaternion created from P_H -> P_L with no real component then multiplied the result by the conjugate of the inverse. I then take the vector part of the resulting quaternion as the transformed vector where there is no rotation (q_0).

When I have another rotation, I can transform the vector by performing the same operation as before but without inversing the rotation.

The idea is that some time in the future where I have some rotation, I can then transform this vector by another rotation to get the position of P_L.

I there something I am doing wrong or are there any shortcuts I could take?

[novel parrot]

any idea how to solve this?

if B had finite elements then i know how to do it

so essentially B is a finite dimensional vector space over k and int domain => B is a field itself

if we fix an $r \neq 0$ and make the map $x \mapsto rx$

[cloud walrus]

ActiveChapter

[novel parrot]

why is it surjective ?

[quiet pelican]

It’s an injective linear map on a vector space

[novel parrot]

but i need surjectivity too right

[quiet pelican]

By linear algebra, an injective linear map of a vector space into itself is surjective

[novel parrot]

oh

[rustic crown]

finite dimensionality is the key :3

[novel parrot]

thanks guys

[quiet pelican]

Yeah
Finite dim vector space

[livid willow]

Oh I see, thanks

[novel parrot]

whats the difference between vector space and linear vector space?

for example Q vector space and Q-linear vector space

[thorn delta]

nothing

[pliant forge]

why is it that radM + kerf = M where f: M \to N when f(m) != 0 for some m \in radM

[novel parrot]

so Q vector space = Q-linear vector space?

[thorn delta]

yeah

[novel parrot]

okay

[pliant forge]

$m \in \text{rad}(M) \iff f(m) = 0, \forall f \in \text{Hom}_R(M,S)$ for any simple $S$. Then apparently, if $g:M \to N$: $g(\text{rad}(M)) \subseteq \text{rad}(N)$ follows immediately. Why is that so?

[cloud walrus]

Eso

[rustic crown]

say f : N --> S for S simple. and m in rad(M). need to show that g(m) dies under f. (do you see what that's true?)

[pliant forge]

i have showed the equivalence

wait

[rustic crown]

what's the context here? what if i take f = id_M?

[pliant forge]

idk i didnt understand the hint so i proved it with some other way

[pliant forge]

cant see it

[rustic crown]

Well, so we need to show (fg)(m)=0, and fg is a map to a simple, so would kill the radical

[rustic crown]

me no understand it either. How do you define small?

[pliant forge]

A \subset B is small in B if A+X = B implies X = B for every submodule X of B

[rustic crown]

And sowwy, one last thing... Whats the definition of rad that you're using?

For me that statement was a definition

[pliant forge]

intersection of maximal modules

[rustic crown]

Hmm I see

So I think the hint is specifically for f:M-->S

And not any arbitrary M --> N

[pliant forge]

yes i know

[rustic crown]

Oh I got confused a little by that

[pliant forge]

oh shoot my bad

[rustic crown]

So if radM does die under f:M-->S

Then the image of it would be a submodule

Which means it's all of S

Image is (rad M+ker f)/ker f

And it's preimage is rad M + ker f

Which must be all of M

Since the image was all of S

Now if you know rad M is small then, it says ker f = M right?

Which is a contradiction

Since we assumed rad M doesn't die under f = 0

[pliant forge]

ye ic thx

[winged harness]

could someone explain what is meant here by the last line?

this is algebraic topology

" \beta_h is simply conjugation by [h]"

what does that mean?

oh so conjugation by [h] of some [x] is basically just [h][x][h]^-1

but im still confused

[pliant forge]

lol im slow but just figured out that since the composition would also be a hom fg from M to S this would imply fg(m) = 0 as m in radical of the domain of fg. Thus if g(m) != 0 then f(g(m)) = 0 <=> g(m) \in rad(N)

[pliant forge]

about what?

[winged harness]

the last sentence

my group theory is a little bit behind

[pliant forge]

would that imply maybe [h][x][h]^-1 = [x]?

[winged harness]

do u know what's being talked about or should i define stuff?

this is prob a dumb q but how do we know that

and how does that imply [h] is central

like we wanna show [h][x] = [x][h] right

[pliant forge]

B_h is the identity right

so B_h(x) = x

[winged harness]

yes

[winged harness]

oh so just right-multiply by h inverse

[pliant forge]

ig its the same thing

ye

[winged harness]

oh i get it yeah lol

brain fart

[sonic coral]

my professor also told me this yesterday, if a map from a finite set to a finite set is injective then surjective comes for free, why?

wouldn’t the sets have to be the same size?

[rustic crown]

yea they need to have the same size

that's referred to as the pigeon hole principle

but here we're using some other finiteness result

which is rank-nullity

[sonic coral]

oh, yeah i’ve learned about it but never thought about it that way

i can believe it though now that i realize they have to be the same size

[rustic crown]

yea, that's usually stated a little differently, if X and Y are finite sets with |X| > |Y| then a function X --> Y can never be injective.

ig that's the most usual statement but ofc this can be converted to the above easily

[sonic coral]

yeah i can see that, thank you

[rustic crown]

if f : V --> W is a linear map between vector spaces of same finite dimension, then injectivity and surjectivity are equivalent

because rank f + null f = dim V

surjective iff rank f = dim V and injective iff null f = 0

[sonic coral]

is that one of the matrix equivalences

[rustic crown]

what's that

[sonic coral]

idk what it’s most known by

but like the list of 20 or so equivalence’s for a matrix to be invertible

[rustic crown]

rank nullity is just another version of first-isomorphism theorem

because you get an isomorphism V/ker f --> im f

if you take dimensions, it gives dim V - null f = rank f

[rustic crown]

yea ig you can say that

[sonic coral]

just trying to make connections

[chilly ocean]

Is there something like an inverse for quotient groups? Like if G/H is a simple group and H is a known normal subgroup of G, can I "multiply" G/H by H to get G back?.

[rustic crown]

that's called the extension problem

which is not easy

[chilly ocean]

The name does not inspire confidence

More specifically, I'm asking this in the context of constructing a group from a composition series

I'm given simple groups, and the question asks me to construct a group from these simple groups with composition factors isomorphic to the simple groups

[rustic crown]

you can always take direct products to get one example

but classifying all such is not an easy task

As mentioned earlier, the first half has been settled, although the complexity of the work
leading to its solution justifies some skepticism concerning the absolute correctness of the proof.
The status of the second half is, as far as I know, (even) murkier.
this was one of the footnotes when i was learning this material...

here by first half of the classification problem, he means classifying all simple groups. and second half refers to understanding how to put together these simples to get all finite groups

[chilly ocean]

interesting

[rustic crown]

given groups N and H, you wanna classify all exact sequences
1 --> N --> G --> H --> 1

there is a special case when you can easily classfy these

which is when G --> H admits a section

that is, if G is a group with a normal subgroup N and another subgroup H such that HN = G and H n N = 1 then G is a semi-direct product of N and H

ig in the abelian situation this has a very satisfactory answer as well... by stuff called Ext-functors

.<

did you want to hear something else?

[chilly ocean]

Thanks for explaining :) I was mostly asking to solve this problem (and was trying to do that right now)

[chilly ocean]

I have to admit, that I'm still not getting any further though. More concretely, the question is this: Let $G_1, G_2, G_3$ be simple groups. Construct $G$ with composition factors isomorphic to $G_1, G_2, G_3$, i.e. construct $G$ such that ${e} = B_0 \trianglelefteq B_1 \trianglelefteq B_2 \trianglelefteq B_3 = G$, where $B_i / B_{i - 1} \cong G_i$ for all i.

[cloud walrus]

MatrixMaker

[chilly ocean]

I think I understand the question, but I don't see how I get get B_2

[lethal dune]

try direct product

[chilly ocean]

Hmm, that makes sense, but why does it work?

[void cosmos]

yo

in a PID every element that is not a unit can be written as a product of irreducibles proof:

suppose a is not a unit

and a =xy , if both are irreudicbles then we are done now say x is not irreduicible --> x=fg for f,g. if both are irreducibles then we are done but then if not then say f is reduibcle ... and so on

but know for every like step we have an ideal containment

so for examplee a =xy --> a in (x)

(x) in (f)

(f) in (...)

by inclusion this is an ascending chain that has a maximal element (the union of such ideals) therefore this process terminates by zorn's lemma

and thats it? we are done

??#@

[elder wave]

this part here is a bit off

[void cosmos]

yea i figured

[elder wave]

notice that you didn't use the fact that your ring is a PID anywhere

[void cosmos]

hmmmmmmmmmmmmmmmmmmmm

yeaa

[elder wave]

you do take the union of your ideals

this union is generated by one element

[void cosmos]

yea so thats why it terminates

[elder wave]

this element has to be contained somewhere

[void cosmos]

yea yea

[elder wave]

and you get a contradiction

[void cosmos]

super smrat

so what would be better to write

as in like for grading

[elder wave]

dunno

[void cosmos]

(the union of such ideals is an ideal , since this is a pid this is generated by one element hence we can keep reducing untill we get to this element? lmfao)

[elder wave]

idk what you mean

you assume your element cannot be written as a product of irreducibles, you get a strictly ascending ideal chain, take the union, this union is generated by one element, this element has to be contained in one of the ideals and that's a contradiction to it being strictly contained in the next one

if you want to show that every PID is factorial you can then use that prime and irreducible coincide in a PID and get the uniqueness of your factorization

[void cosmos]

yea

yea so the union of these ideals is an ideal generated by one elment

but by containments this one element must be contained in atleast one of the ideals

which is a contradiction

is this correct

cuz (a) is just the union

[white oxide]

can somebody help me on how to start this problem?

i wrote down my givens:

there are two left cosets of H, and letting H have order m and G having order n, 2 - n/m

other than that i'm stuck

it seems equivalent to proving G is abelian

[void cosmos]

just write it down

like try an example

u know that this has index 2

which means that the set of cosets of H in G just has 2 elements

write that down

also remember that u have 1 coset which is the identity

which is just H

so ur left with one

[white oxide]

wouldn't this just be the set of left cosets

[void cosmos]

we only have 2 cosets

1 of them is H

so not coset(s)

just 1 coset

but now is it left or right

[white oxide]

how do you know this, isn't the index of a subgroup H defined to be the number of left cosets of H in G

[delicate orchid]

the identity times H = H is always a coset

[void cosmos]

^

[white oxide]

oh ok

[void cosmos]

H is a left coset

[white oxide]

and that is a left coset

[void cosmos]

okay so we have 1 more

left

so now

you know how cosets look like

we have two choices

for example lets consider aH

if a is in H then we know this is just H

right?

[white oxide]

how

[void cosmos]

what does aH mean

[white oxide]

all elements ah for a fixed a in G right and h in H

oh wait

wait no

yeah i still dont see how it's H

because a is not the identity

[void cosmos]

aH = {a*h | h in H}

right

?

[white oxide]

yeah

[void cosmos]

now say a is in H

now H is a subgroup

and we know its closed

so literally a wont act on elements on H other than likee just shifting

like it wont go out of H

cuz H is closed

[white oxide]

yeah but for a counterexample you can take H = {0, 3} which is a subgroup of Z_6 and then say the left coset 2 + {0, 3} = {2, 5} which is not equal to H

[void cosmos]

yes

cuz 2 is not in H

i said assume a is in H

consider 3+{0,3}

[white oxide]

oh

[void cosmos]

what do u get

[white oxide]

ok i see

[void cosmos]

ok

[white oxide]

but what's stopping you from choosing some a not in H

[void cosmos]

nothing

we are going to do that rn

suppose a is not in H

then we know that aH is not H

right

[white oxide]

yeah

[void cosmos]

so whats aH then

here is a hint

what even is the point of cosets

like say we have a group G

and we say A is the set of cosets of G

what does that mean like

what does that do to G lmfao

[white oxide]

just a partition of G right

[void cosmos]

yes

exactly

but bro we only have two

partitions

no

sorry

two cosets

and one of them is H

so the second one must be what

so that the union of both would be G?

H union (what) = G

given H is a subgroup ( H is subset )

[white oxide]

so two subgroups H make up G right

and we have one being H itself

i have no clue dude maybe the coset which is disjoint from H itself

[void cosmos]

try it

with sets

{1,2,3,4} = G

H={1,2,3}

what do ud o

[tribal niche]

H is a subgroup of G and a coset is an element of G/H, which are represented by [a subset of] elements of G, but are usually denoted aH, bH, etc

[void cosmos]

H union what is G

[white oxide]

{4}

oh wait

it's just exhaustive

so yeah it would have to be what's disjoint from H

still tryna see how that's a right coset though

oh wait

yeah nvm

no clue

bc the index of a subgroup is defined to be the number of left cosets of H

so idfk how i can get from there to a left coset being a right coset

[void cosmos]

its okay

so now

we know that aH must be G-H

right

[white oxide]

what is G-H

[void cosmos]

elements in G that are not in H

[void cosmos]

this

[white oxide]

oh you mean G/H

[void cosmos]

no

its an algebra course bro

u cant do that

like no way relative complemeent is this way

in your algebra course

[white oxide]

idk what that means

but ok i'll trust you

okay anyways

[white oxide]

are you trying to get that it just leaves one subset with one element

[void cosmos]

forget about algebra

if ur doing set theory

the relative complement of B wrt to A is just elements in A that are not in B

this is denoted sometimes either by A-B or A/B in some textbooks

but in algebra ur going to deal with quotients soon

which are alwaays written with / notation

quotients are not relative complements

(lmfao ig)

so thats what i meant

so now

can u decompose G into right cosets @white oxide

just like as we did with left cosets

[tribal niche]

i have never seen it denoted A/B it's usually the other slash A \ B

[void cosmos]

yes thats what i meant mb

[white oxide]

how would you know that there are two right cosets?? i thought the index of a subgroup refers to how many left cosets there are, so for all we know there could be much more than 2 right cosets or just one

i'm so fucking lost jesus

is it just like a question you assume that a not equal b and you try to prove a equal b or some shit

[void cosmos]

no

you just know that these cosets partition your group

thats all u need to know

now

u are right to be lost

and what ur saying makes 100% sense

but

let me like

finish it up for you

forget all this shit

let G be a group

define f(g) = g^-1

now with this we can natural project this hsit and induce the map aH--> Ha^-1

i want you to prove that this is a bijection

hence the number of left cosets is the same as the right

first tho

take this for granted

and then do ur problem

[void cosmos]

then comeback and do this

[white oxide]

oh you're right

ig it is true that the number of left cosets is the same as the numbe rof right cosets

so i had to prove that before i did this problem

ok thanks

great

[void cosmos]

yea makes sense

[coarse stag]

im kind of lost on this, can someone explain to me what the difference between a composition and principal series are? For a principal series, if the factor groups must be simple, then H_i must be normal in H_{i+1} and thus be a subnormal series with simple factor groups, which means its a composition series. Am I missing something?

[void cosmos]

key word is normal series

and subnormal series

[coarse stag]

but doesnt the criteria of factor groups being simple mean that the normal series has to also be a subnormal series?

[tribal niche]

the opposite may not be true

[delicate orchid]

does a subnormal series have to be a normal series?

[coarse stag]

Yes... unless im tripping

[cloud walrus]

monkeman

[coarse stag]

shouldnt it be the other way arround?

[delicate orchid]

A subnormal series only requires each group in the series to be normal in the next one, a normal series requires that and each group be normal in G

if I remember rightly

[tribal niche]

no, because as you said here, all normal series are subnormal series, but you never showed/stated if all subnormal series are normal series (because they aren't)

[white oxide]

could you define $\phi(ag) = (ag)^{-1}$

[cloud walrus]

okeyokay

[white oxide]

which would be equal to g^-1a^-1

[coarse stag]

A subnormal series is a series of subgroups of G such that H_i is normal in H_{i+1} right?

[white oxide]

wait no i'm stupid asf

[tribal niche]

nope this is the right idea

but you want to work with elements of H, if that is unclear

[delicate orchid]

yes, but a normal series also requires each H_i to be normal in G

it's got an extra condition on it

[void cosmos]

it's not eeasy

take ur time with it

[white oxide]

right but then wouldn't aH not be equal to Ha^-1

sorry

wrong phrasing

like

it would be insufficient to prove that the number of them is the same

because don't we have to prove aH and Ha

have the same number or smt

[tribal niche]

every element of a group (i.e. H) has an inverse

[white oxide]

not aH and Ha^-1

[delicate orchid]

doesn't matter, find an inverse map, it's a bijection

then the sets are the same size

[white oxide]

oh okay

right every element has an inverse

[coarse stag]

ah I assumed that H_i being a subgroup of H_{i+1} which is normal in G implied H_{i} is normal in G

[tribal niche]

no, and this is unfortunate tbh but this is not true

[white oxide]

to show injectivity would you take $\phi(ax)$ and $\phi(ay)$ for some $x, y \in H$

[cloud walrus]

okeyokay

[white oxide]

and then try to show x and y are the same

which is almost trivial

[tribal niche]

the simplest counterexample is that G is a normal subgroup of itself, but any subgroup H which is not normal is a subgroup of a normal subgroup of G (which is G itself lmao), but H is not normal

[coarse stag]

ok

[delicate orchid]

I wouldn't bother with injectivity/subjectivity, just find a well-defined inverse to phi

[coarse stag]

Also I thought the criteria for normal series is just that H_i < H_{i+1} and not necessarily normal to H_{i+1}

[tribal niche]

well a normal series has to be subnormal

[coarse stag]

yeah ok

nvm I think the book fucked up lol

[tribal niche]

it might be true that if a subgroup H' of a normal subgroup H is normal to G, then it is also normal in H, but i have to verify this

this seems true, so the book's definition is correct, but it is slightly confusing. it might have been clearer if they stated that a normal series is subnormal but maybe that's an exercise, who knows

[coarse stag]

yeah i think it is actually

by conjugation

[tribal niche]

yeah

[white oxide]

also wouldn't this just be showing that the number of elements in a left coset and right coset are teh same, not that the number of left cosets and right cosets are equal

[round sandal]

What does Aluffi mean by the negation symbol and this arrowhead that are present in some exercises? I couldn't find him mentioning them in the first pages of the book.

It's algebra chatper 0

[delicate orchid]

I thought that's what you were doing

[white oxide]

wouldn't this prove that the number of elements in a given left coset are the same as the number of elements in a right coset and not that the number of right and left cosets are the same

[white oxide]

oh nah

my b

[delicate orchid]

oh right the index two thing

[tribal niche]

probably some weird indication of problem importance/difficulty... check the beginning of the textbook

[white oxide]

ig you could define it from the set of left coset groups to the set of right coset groups but we haven't gone over coset groups yet

so i have no fucking clue

[void cosmos]

either difficulty or its going to be referenceed soon

referenced*

[round sandal]

Thanks!

[delicate orchid]

let g not be in H, then the set of left cosets is {H, gH} since it's index two, consider the right coset Hg - this is either equal to H or gH, can you see why that's true?

[white oxide]

yeah

[delicate orchid]

ok, see if you reason why Hg cannot be equal to H, and therefore must be equal to gH

[white oxide]

No yeah I got that part

It makes sense now

I’m just trying to prove why the number of left courts and right closets are the same

Oops autocorrect

Lmfao

[delicate orchid]

ah ok ok

G \cong G^{op}

[void cosmos]

show that f(gH) = Hg^-1 is a bijection

let me do 1 part for u