groups-rings-fields

[thick fractal]

🤷‍♀️

Teacher is tryin to teach us upper algebra levels and crap rn so I decided to come back here and look around

[urban acorn]

i didn't say anything technical or refer to any actual algebra knowledge

[faint elm]

I want to show that $\mathbb{C}[S_n]a_{\lambda}b_{\lambda} \simeq \mathbb{C}[S_n]b_{\lambda}a_{\lambda}$ as representation

[cloud walrus]

emme:

[faint elm]

Hint says that the map $\cdot a_{\lambda}$ is the right one

[cloud walrus]

emme:

[faint elm]

But this map carries $\mathbb{C}[S_n]a_{\lambda}b_{\lambda}$ to $\mathbb{C}[S_n]a_{\lambda}b_{\lambda}a_{\lambda}$

[cloud walrus]

emme:

[faint elm]

So the problem now is to show that $\mathbb{C}[S_n]a_{\lambda}b_{\lambda}a_{\lambda} \simeq \mathbb{C}[S_n]b_{\lambda}a_{\lambda}$

[cloud walrus]

emme:

[faint elm]

Do you have any hint?

Nvm got it

[uncut girder]

What are irreducible R[x] modules? Where R is reals

What are non diagonalizable 2x2 matricies

[wind steeple]

[(1 1) (0 1)]

[blissful root]

that's indecomposable but it has the submodule e1

so it's not irreducible

worse one:
[cost, sint]
[-sint, cost]

by the real jordan canonical form, iredducible ones are gonna then be either 1 dim ones, or 2 dim ones where x acts with complex eigenvalues

[fringe nexus]

How do i find splitting field of x^6-4?

over q

is it

Q(primitive 6th root of unity, $3\sqrt{2}$)

wait nvm thats wrong

[cloud walrus]

aaaaaaaaaaaaaaaa:

[fringe nexus]

cube root of 2

[blissful root]

@fringe nexus looks right to me

basically your 6 roots are $\sqrt[3]{2}w^k$ where w is the 6th root of unity

[cloud walrus]

urboz:

[blissful root]

so by dividing two consecutive ones you get w

[fringe nexus]

what degree is this extension exactly?

[blissful root]

uhh

[fringe nexus]

is it 6

im not sure

[blissful root]

I wanna say 9

so

Q(w) is degree 2 over Q

and Q(\sqrt[3]{2}) is degree 3 over Q also

this implies that Q(\sqrt[3]{2}, w) is degree 6

[fringe nexus]

hmm

i have no idea

wait

by w do you mean

primitive root

[blissful root]

ye

[fringe nexus]

its phi(6)

[blissful root]

it's |Z/2^x| |Z/3^x|

and that's uhhh

2?

that's so weird

[fringe nexus]

yes

[blissful root]

oh I see

yeah x^6-1 factors a lot

then it's degree 6

[fringe nexus]

so its degree 6?

[blissful root]

2*3

[fringe nexus]

okie

[north sand]

if H is a (non-trivial) subgroup of a group G, does there always exist a (surjective or at least non-trivial) group homomorphism G->H ?

[fading wagon]

Hmm, the kernel of the homomorphism is isomorphic to a normal subgroup of G, so if G is simple, then H would be G or trivial for it to be surjective

@north sand

[north sand]

thanks

[urban acorn]

@north sand There will be such a (surjective) homomorphism if and only if H is normal. The non-surjective case can be reduced to the same problem with a different H and so this answers your question without a loss of generality.
EDIT: I misread your question at first. The only if direction is correct, but the other direction isn't.

@fading wagon the kernel of such a homomorphism is a normal subgroup of G, it's not just isomorphic to one.

[blissful root]

why does normality matter?

[urban acorn]

Do you know what quotient groups are?

[blissful root]

it's false that having a map G->H means that H is normal

[urban acorn]

nevermind

[blissful root]

when H is a subgroup of G

[urban acorn]

yeah, that's right

i misunderstood the question twice, then

in my defense i drank a lot of wine earlier

and juice

juice is tasty

[blissful root]

nice

[urban acorn]

okay i reread the question

He specified that the homomorphism must be surjective. So he's basically asking if there's always a quotient group isomorphic to each subgroup of a group. Right?

If it doesn't have to be surjective, then it can be isomorphic to any quotient group of the subgroup, but he specified that even in that case it can't be trivial, and so we exclude the isomorphism between the trivial quotient groups.

Someone check if what I'm saying is right.

[blissful root]

what are you saying exactly

[urban acorn]

im just rephrasing his question in terms of quotient groups rather than isomorphisms

[blissful root]

yeah surjective maps are quotients

[urban acorn]

okay good

so i can somewhat understand math right now

[blissful root]

[urban acorn]

sue me

[chilly ocean]

can someone give me an example of haar measure

[blissful root]

lebesgue measure on R^n

e.g. on U(1), lebesgue on [0,2pi)

[woven delta]

[chilly ocean]

i obviously meant aside from that

[blissful root]

how about

embed S3 in a nice way in R^4, induce measure by integration?

[chilly ocean]

so \lambda^3?

is there any example that isnt \lambd^n?

[blissful root]

what's lambda

[chilly ocean]

lebesgue measure

[blissful root]

uhhh

it's not obvious that this is just lebesgue

like

on local charts it's not

it only "kinda looks like lebesgue" globally under a nice embedding on R^4

[chilly ocean]

S^3 is what exactly

[blissful root]

sphere

[chilly ocean]

as a group

i mean

[blissful root]

uhhh

SU(2)

[chilly ocean]

after embedding its a group under addition

in R^4?

and the measure is translation invariant

[blissful root]

no it's not a group under addition

i'm not embedding it as a group

[chilly ocean]

how then

[blissful root]

you have to embed it in a nice way

so that the group action is symmetric

aka it rotates around

with derivative of norm 1

everywhere

and then you can integrate on it

just as a manifold

[chilly ocean]

can you actually give me an example of haar measure on a topological group

[blissful root]

what qualifies as an example?

[chilly ocean]

or just handwavey statement

[blissful root]

I gave you an easy one and you said it was obvious

then I gave you this one and you say its weird I guess

[chilly ocean]

you gave me the lebesgue measure

you didnt give me one

[blissful root]

this isn't the lebesgue measure

[chilly ocean]

you said you can integrate on it

[blissful root]

yes

[chilly ocean]

integrate with respect to what

the measure is what i need

and you didnt give me one

what is the group

what is the measure

[blissful root]

hmm

i mean this is lebesgue measure modulo local charts

but this is true of every haar measure i guess

like, you are integrating with respect to the lebesgue measure in the parametrization

but saying that's just lebesgue is excessive

[chilly ocean]

i found an example

positive reals under multiplication

[blissful root]

take log and this is lebesgue

[cloud walrus]

lévy (sto):

[chilly ocean]

ya

is this your example too?

[blissful root]

i mean

no

this is the lebesgue measure on R

literally

[chilly ocean]

no

\mu is a different measure

literally

[blissful root]

it's the same up to homeo

the homeo is log

[chilly ocean]

\mu(a,b) \neq \lamda (a,b)

so its not the same measure

as you just said

saying its the same up to homeo is like saying all absolutely continuous measures on R are the same

[blissful root]

what I said is that good examples of haar measures are locally isomorphic to the lebesgue measure

this is globally isomorphic to lebesgue

I thought you'd say this is obvious

as you did for lebesgue on R

[chilly ocean]

youre correct

i guess

i just wanted a concrete example

your example with S^3 had no group and no measure

[blissful root]

S^3 is SU(2)

the measure is given by integration

[chilly ocean]

ok

integration how

give me an example

what is the sigma algebra?

[blissful root]

remember in calculus how you integrate on the sphere?

give local charts, integrate the pullback of the lebesgue measure

[chilly ocean]

jacobi transformation?

integral of images?

[blissful root]

yeah locally it looks like the integral of a jacobian determinant times lebesgue

[chilly ocean]

ya

i know that

cool

continue

[blissful root]

I mean that's the example

but you have to be careful because you can't just pick any homeo SU(2) = S^3 for this

it has to be the standard embedding

or something isometric

[chilly ocean]

i dont know wht that means

how do you equate SU2 to S^3

[blissful root]

ok let me write it

if I dont botch this up

[chilly ocean]

i havent done differential geo

[blissful root]

I think the one on wiki works

https://en.wikipedia.org/wiki/Special_unitary_group#The_group_SU(2)

but yeah you should do diff geo first

I assumed you had

[chilly ocean]

ah

so we have SU2 \to C^2 \to R^4?

[blissful root]

yeah

hopefully this embedding works

but I havent checked

[chilly ocean]

very cool thanks

do you do geo

[blissful root]

AG

[chilly ocean]

makes sense

@blissful root is it possible to construct a haar measure on a group not homeomorphic ro a subset of R^n

or does it also need to be diffeomorphic

to use jacobii that is

[blissful root]

it should be possible

under some conditions, but I'm not familiar with that

I care about lie groups

so that's why I focus on those

locally compact hausdorff is important

but aside from that i don't know if you need more

[chilly ocean]

a lie group is just a topological group that is also differentiable right

like the maps are also differentiable

uh but by definition smooth manifolds are homeomorphic to a vector space

or just locally

ugh idk this shit why am i reading about it

[blissful root]

a lie group has to be a manifold

so it's locally homeo to a vector space

but manifolds can be embedded into R^n

by whitney's theorem

also, yes, a general locally compact hausdorff group has a haar measure

it's enough

but the ones that aren't lie groups are weird

[chilly ocean]

so lie groups are really ncie groups

that allow calculus`?

[blissful root]

lie groups are incredibly nice for many reasons

[chilly ocean]

cus theyre parts of R^n but funky and under other operations

[blissful root]

but yeah being manifolds means you can do calculus on them

[chilly ocean]

so its like sep hilbert spaces being l^2(K)

thank you

[uncut girder]

Given two idempotent commuting linear operators A,B on finite dimension vector space V, prove ker AB = ker A + ker B

[fickle brook]

#linear-algebra

[blissful root]

notice AB: V -> V/(kerA + kerB) is a map

check the kernel

otherwise, more concrete: characterize idempotent operators as projections, therefore diagonalizable.

[uncut girder]

Thanks

[lone niche]

im just concerned with is enough as a proof for this

the dim(P_n) = n + 1

so like f' and so on is of deg(n-1)

[woven delta]

🤔

[waxen iron]

So I was going over my class notes and I found something pretty odd:
"Let G be a group and H a subgroup of G. Then ∀x,y∈G xH∩yH ≠ ∅ ⊻ xH=yH"
I'm reasonably sure it's an "=" not a "≠" , ye?

There's also a big change I'm being dumb and it's correct so I wanted to make sure 😅

[fickle brook]

yes

[chilly ocean]

yes

[waxen iron]

Many thanks!

[chilly ocean]

no problem !

[waxen iron]

Back again 😅

Could ya'll give me the intuition behind this automorphism?
It's a "function" that takes elements in G and takes them to other elements in G?
I'm rather confused

Oh and G is a group

[urban acorn]

think of automorphisms as a formal way to talk about the symmetries of a group

take, for example, the Klein four group.

any distinct non identity elements are not equal

but they are interchangeable in that if you permuted them, if you said a wherever you previously said b and b whereever you previously said a, no one would notice

now think of the integers

under addition

if you switched positive with negative numbers, literally no one would know (note: we are working under addition)

it's another sort of symmetry of the group

the way to formalize this is that the automorphism is the function that tells you how to permute the elements (for example x -> -x for the integers) such that the new permutation of the elements of the group is indistinguishable from the original structure-wise

clearly it needs to be a bijection cause it's a way to permute the elements

and suppose you pick three elements a,b,c

that relate to each other according to the group structure in some way ab=c

then you would want the new elements given by your function to relate to each other in the same way

f(a)f(b)=f(c)

this is essentially the same as saying f(x)f(y)=f(xy) for all x,y in G

that is, f is a homomorphism

so this essentially gives us intuition why isomorphisms from the group to itself are the same thing as symmetries of the group

and the name for that is automorphisms

of course, you can have automorphisms of other things

doesn't even have to be an algebraic object

everything that has a notion of isomorphism has automorphisms - ways in which it's similar to itself

[carmine sentinel]

if i have a ring and i define an equivalence class where a~b iff a =x*b where x is a unit

does this have a name?

[fickle brook]

yeah

a and b are associates

[carmine sentinel]

how would you find the number of classes for a ring?

[wind steeple]

Each class has <= |A^x| elements

[chilly ocean]

@waxen iron This is a general principle in mathematics that could be called "transport and go back". Basically, if x is a transformation of a space X, then gxg^{-1} "does essentially the same thing as x", but after changing your point of view by g.
A concrete example would be if G is a matrix group. If x is a certain transformation of R^n and g a matrix, then gxg^{-1} will be the same "type of transformation as x", but after a change of basis given by g.
For example, if x is a symmetry with respect to a line D, then gxg^{-1} will be a symmetry with respect to the line g(D), if x is a projection onto a plane P, then gxg^{-1} will be a projection onto g(P), etc.

[tulip hearth]

Isn't that called "conjugation of x by g"?

https://en.m.wikipedia.org/wiki/Conjugacy_class

[woven delta]

It's called an automorphism

[tulip hearth]

i think it's more specific than that, automorphism is just bijective homomorphism of thing with itself

[woven delta]

No, this is the big idea of automorphisms

Well one of the big ideas of automorphisms

[urban acorn]

yes, in a group, this is called conjugation

furthermore, the map sending any element to its conjugate by a fixed element is always an automorphism

automorphisms which can be expressed in that form are called inner automorphisms

the inner automorphisms are a subgroup of the group of all automorphisms of the group

the quotient of the automorphism group by the inner automorphisms (which I think are always normal) is called the outer automorphism group

the inner automorphism group is isomorphic to the quotient of the group by its center.

[woven delta]

Who are you explaining this to lol

Everyone knows this

Except for maybe @upper pivot

[upper pivot]

smh bulli

[waxen iron]

@urban acorn and @chilly ocean Many thanks for the explanations!
I've been digesting them and I think I got the feeling for things!

[urban acorn]

who are you explaining this to lol everyone knows this
many thanks for the explanation!
@woven delta get destroyed

oh wait that was in response to something else, im a clown

i said it in reply to what nox and you said "isn't that called conjugation?" "it's called an automorphism"

although i said much more than i needed to

[carmine sentinel]

@wind steeple how does this help?

for Z/nZ the number of classes is the number of factors of n right?

[wind steeple]

idk

I see that in an integral domain, the cardinal of a class is exactly |A^x|

why would the number of classes in Z/nZ be the number of factors of n ?

[carmine sentinel]

because for each factor, the set of associates is disjoint

like if n=6 there are 4 sets {1,5} {2,4} {3} {6}

[wind steeple]

yeah but lucky you to have exactly the number of divisors of 6

I don't see where the number of divisors occurs here

it's obvious that each class it's disjoint since it's an equivalence relation

[carmine sentinel]

for each factor k|n, the set of all x such that gcd(x,k) = k forms a class

[wind steeple]

that's false

for n = 8

k = 4

x = 4 is in the class of k

and also in the class of 2 by your proposition

that's false since classes are disjoint

[carmine sentinel]

the classes are {1,3,5,7} {2,6} {4} {8}

[wind steeple]

yeah and that's refuting your assumption

[carmine sentinel]

how so

[wind steeple]

k = 2 :
the set of all x st gcd(x,2) = 2 is {2,4,8}

k = 4 :
the set of all x st gcd(x,4) = 4 is {4,8}

by "for each factor k|n, the set of all x such that gcd(x,k) = k forms a class"
each are classes. Since they're not equal they're disjoint, but it is not the case then your sentence is false

[carmine sentinel]

sorry i mean gcd(n,x) = k

[wind steeple]

ok

that's right then

prove it now

[meager flint]

Let V,W be finite dimensional complex representations of a finite group G. I'm supposed to explain how H:=Hom(V,W) is a natural representation of G, but I don't even get what Hom(V,W) means in this case because representations are homomorphisms???

[wind steeple]

what's your definition of representation ?

[meager flint]

a homomorphism from the group G to the group of invertible linear operators on a vector space V

[wind steeple]

ok

it this case, V is a vector space

in your sentence , they call V a "representation" but it is a verbal abuse

actually, V is the vector space of your representation

[blissful root]

the question is: you have maps G -> End(V) and G->End(W), find a natural map G->End(Hom(V,W))

[meager flint]

End meaning what?

[blissful root]

endomorphisms

[meager flint]

Oh

Okay

Thanks y'all

fr

[carmine sentinel]

yeah im not sure how to do this rigorously

[blissful root]

what are you trying to do?

[carmine sentinel]

because i want to apply this to polynomial rings Z[x]/(p(x))

[wind steeple]

you don't need polynomials to prove it

[carmine sentinel]

like i want to show the number of classes is the number of factors of p(x)

[blissful root]

classes of what

[carmine sentinel]

associate classes

[blissful root]

hmm

[wind steeple]

that's true here too

it's true for every PID

[blissful root]

in general this is a tricky issue but for PIDs it works out nicely

cuz it's gonna be precisely the ideals

[bleak abyss]

Wait what's the problem?

[blissful root]

and the ideals are given by factors of p(x)

find the associate classes of a ring

[carmine sentinel]

what if its not an integral domain?

[blissful root]

it's hard

why are you trying to find those classes?

[bleak abyss]

As in like, elements which differ by a unit?

[blissful root]

yeah

i guess you're precisely asking to classify all principal ideals

[winged belfry]

hi

remember me

@bleak abyss

[blissful root]

which is somewhat interesting

[carmine sentinel]

but this is not the same as principal ideals right?

[blissful root]

it is

because two elements are in the same associate class iff they generate the same principal ideal

[bleak abyss]

Star2825 I remember you and unfortunately each time you ping me this way my memory of you becomes slightly less positive

[winged belfry]

oh..

[carmine sentinel]

thats only true for integral domains

[winged belfry]

rip

[blissful root]

how so

I'm not using being a domain anywhere

x = uy <=> y = vx <=> (x) = (y)

[winged belfry]

sorry about that

[blissful root]

hi @bleak abyss remember me?

[carmine sentinel]

https://math.stackexchange.com/questions/355994/two-principal-ideals-coincide-if-and-only-if-their-generators-are-associated

[winged belfry]

lol

@blissful root I dont think that was a good idea to do

but good luck

[blissful root]

whaaaaaaaaaaat

[bleak abyss]

Problem is this

[blissful root]

ok what am i doing wrong

[winged belfry]

lol

[bleak abyss]

Let's say a = bc and then b = ad

[winged belfry]

he literally just told me that

[carmine sentinel]

then a = adc

[blissful root]

go away star

[bleak abyss]

a = acd so a(1-cd) = 0

[blissful root]

we're trying to do math here

[winged belfry]

oof..

ok

[bleak abyss]

But if you're not an integral domain

You can't use this to conclude cd = 1

[carmine sentinel]

yeah

[blissful root]

i don't see where I'm using this, im so bad

[bleak abyss]

So it is true that if x divides y and y divides x

[blissful root]

if x and y are associates then they generate the same principal ideals

that part's fine

[bleak abyss]

Then (x) = (y)

[blissful root]

if x and y generate the same principal ideals then

ahh

this part isn't fine

[bleak abyss]

The point is being a multiple by a unit is stronger

[blissful root]

yes

I see now

[carmine sentinel]

but im trying to see why this is fine for Z/nZ or Z[x]/(p(x))

[blissful root]

so it's a finer decomposition than principal ideals

because they're PIDs

well no

not for p(x) not irreducible

do you mean p(x) irreducible?

[carmine sentinel]

nope

[blissful root]

then use CRT

and use that to reduce to the PID case

[carmine sentinel]

should i split Z[x]/(p(x)) into a direct product?

[blissful root]

yeah

so this reduces to the case where p(x) = q(x)^k for q irreducible

and then uhh

then it looks like it's still true

[carmine sentinel]

hmm

[wind steeple]

I think you can inject Z in its fraction field and study bezout relation here

[blissful root]

in the end it does look like it's all the factors of p

regardless of not being a domain

[wind steeple]

wait

no that's false

[blissful root]

it is?

[bleak abyss]

The existence of x and y in Z is harsher than the existence of x and y in Q lmao

[wind steeple]

we don't have "for every Q|P, the class of Q is the polynomials X that satisfy gcd(P,X) = Q"

by degree

[blissful root]

no?

you got an example?

[bleak abyss]

Wait maybe I was thinking of something else, Z into Q or Z[x] or what?

[blissful root]

see, associate classes are finer than principal ideal classes, so we can look at principal ideals first

and that's precisely the decomposition into gcd(P,X) = Q

for Q|P

now the question is if any of those can split

[wind steeple]

idk, if gcd(X,P) = Q and X has an higher degree than Q, then they can't be associated

[blissful root]

why

of course they can

you are doing things mod p

[wind steeple]

mh yes, but is the degree of X is smaller than P

[blissful root]

what

[wind steeple]

oh

ok no nevermind

[blissful root]

yeah degree doesn't work right anymore

I don't think the classes split. let's see
x = qa
y = qb
a,b coprime to p
does it follow that x = uy?

elements coprime to p are precisely the units

so yes

this works

this proof more generally works for any quotient of a domain

[wind steeple]

I what only thinking that invertible elements of Z[X]/p(x) were -1 and 1 lel

[blissful root]

so here's the proposition

[wind steeple]

mb

[blissful root]

Lemma (urboz): The associate classes in any quotient of a PID are given by prime ideals

@carmine sentinel

is it actually though?

I think so

[wind steeple]

Z[X] isn't a pid

[blissful root]

lmfao

ok hang on

ok scratch what I said

I was working over k[x] for some reason

im getting confused now

[carmine sentinel]

same

[blissful root]

try using gauss lemma to go from this argument for Q[x] to Z[x]

it should work

[carmine sentinel]

wait why are associate classes ideals

[blissful root]

the claim is that x and y are associate iff they generate the same principal ideal

which in the Q[x] case is iff their gcd with p is the same

[carmine sentinel]

oh ok

[blissful root]

but in the Z[x] case should be iff their gcd with p is the same and their ideal in Z is also the same

which should follow by gauss lemma

I'll leave it here cuz I have to grade lol but let me know if I'm wrong eventually

[carmine sentinel]

nah this works thanks

[wise grove]

@carmine sentinel are you the same iceman from r/math who was telling a first year undergrad to attend the dinner after seminar talks?

[bleak abyss]

Fuckin 'exposed

[woven delta]

Someone at Stony Brook got rekt for doing that

Got kicked the fuck out

[blissful root]

tell us more

what happened @woven delta

[woven delta]

She tried to eat the food

From the Algebraic Geometry group tea or something

[blissful root]

oh she sneaked into the seminar to eat?

[woven delta]

And she's super conspicuous

Nah

She was trying to attend the seminar

But she also didn't know any math

[blissful root]

lol

[woven delta]

It was like a week into her first semester

[blissful root]

wow

[woven delta]

She's very interesting honestly

[blissful root]

but how did it all play out?

[woven delta]

She corresponds with Serre

[blissful root]

woah

[woven delta]

Yeah

She sent him a bunch of letters

[blissful root]

why

[woven delta]

And he replied

[blissful root]

he's like 93

[woven delta]

Idk

He's very encouraging apparently

[blissful root]

i'd imagine

but ok what happened with the food?

she approached it and then what

[woven delta]

She just got kicked out

Idk the whole story

I wasn't there

[blissful root]

oh

[woven delta]

She told the story to us

[blissful root]

sad

[woven delta]

Yeah

I would have loved to see it

[blissful root]

same

I have gone to a total of one (1) post seminar dinner

oh well technically I guess two

but one doesn't count cuz it's a group that meets all the time

and we usually have pizza

[woven delta]

Ah ok

I haven't gone to any real dinners

The seminars I go to are usually pizza ones

Or during the day

[blissful root]

I don't usually go out of my way to talk to mathematicians

that's probably bad

i just talk to people who I have business with

[woven delta]

Networking is a thing

That's probably important

I'm really bad at asking questions during seminars

[blissful root]

yeah it's terrifying

i just keep my mouth shut

I had a really good question one time that I didnt ask

i dont know why

[woven delta]

That sucks

[uncut girder]

@woven delta bro

I know who you're talking about

The other day I attended a colloquium

And Dennis announced the dinner that would be subsidized

Uhh I didnt go cause I'm an undergrad XD

[woven delta]

You probably could have gone though

Cause your a 3rd year

And you know people in the department

Why is Dennis in like 3 of the seminars I go to

Fuck

[uncut girder]

Oof

[waxen iron]

I'm trying to pull together all that I've learned so far in Group theory. My teacher is great but he doesn't follow a textbook and doesn't talk about some alternative definitions or relations between topics.

So with that being said

Is this true?

Where N is a normal subgroup of G

[fickle brook]

,rotate 90

[fickle brook]

well as long as G is finite

i guess it is true

[iron cedar]

@waxen iron there are probs many online group theory books

/ notes

[waxen iron]

Thanks Ann!
Just wanted to make sure, it needed to be true for me to understand the definition correctly

I've looked around for some, but I think I'll do with just class notes. The only problem with class notes is notation and definitions, the teacher sometimes defines things in unconventional ways for the sake of intuition.

So when I look online for help I get these weird definitions and have trouble putting them together 😅 @iron cedar

[iron cedar]

@waxen iron cambridge group theory lecture notes

if it helps

[waxen iron]

Going to check it out!
Many thanks!

[wise grove]

@woven delta oof

I only went to the dinners if they were some pizza dinner in the math lounge

Perks of taking grad classes is grad students offering you alcohol at Friday night wine and cheese

[uncut girder]

I just take all the cheese

Yum

[fringe nexus]

If I have a real symmetric Bilinear form where B(w,u) = 0 and B(u,w) = 0, while B(u,u) >0, is it true B(w,w) <0 ?

[fickle brook]

no

take B to be the inner product of R^n and u and w to be any two orthogonal nonzero vectors

with n > 1

[waxen iron]

If a non-empty set S satisfies (a) and (b) show that (S, . ) is a group
I think I've showed associativity but I can't figure out how to show inverses and the neutral element.
I can never do this sort of thing

(Oh and S is finite)´

[wind steeple]

Consider the function x->ax for a in G

[chilly ocean]

Don't you have to swap the values?

I forgot how to do this but I know with inverse functions you swap the x or something

[rocky thistle]

Yeah, you get x=(y+2)/3, so y=3x-2.

@chilly ocean

If f(x)=y.

[chilly ocean]

thank you so much

how did you transition the fraction to y=3x+2

@rocky thistle

[rocky thistle]

3x=y+2, so y=3x-2.

This is abstract algebra?

[chilly ocean]

where did you get 3x=y+2

[rocky thistle]

Multiply by 3.

[chilly ocean]

thanks man

[rocky thistle]

No problem!~

[chilly ocean]

@rocky thistle is this right

[rocky thistle]

It should be xy-6x.

[chilly ocean]

where did you get the extra x from

@rocky thistle

[rocky thistle]

You distribute the x when you multiply both sides by y-6.

@chilly ocean

[golden pasture]

defo not (abstract) algebra btw.-.

[rocky thistle]

@golden pasture Yep. @chilly ocean This should probably be moved to #prealg-and-algebra

[rose oxide]

can anyone help?

[woven delta]

@rose oxide do you know what a field automorphism is?

[rose oxide]

where a group is isomorphic on itself?

[woven delta]

Oh, you probably don't know what a field is, nevermind

[rose oxide]

I am certain A is true

well no, B must be true

[woven delta]

Anyway, these are all easy things to show

Do you know what complex conjugation is?

[rose oxide]

maybe if I see it

it rings a bell

I know what z bar is

[woven delta]

a+ bi gets mapped to a-bi

[rose oxide]

yeah

[woven delta]

Oh lol

[rose oxide]

oh z bar is the complex conjugation

lol

🤦‍♂️

[woven delta]

Anyway, if you add two complex numbers

[rose oxide]

right

[woven delta]

And then conjugate them

Is it the same as if you would have conjugated then added?

[rose oxide]

yes?

i feel like it is

a+bi + c + di = (a+b)+(c+d)i the conjugate is (a+b)-(c+d)i
a-bi + c - di = (a+b)-(c+d)i

yeah

they are

well

if they are the same

they are homomorphic

so A is true

[woven delta]

Homomorphic isn't a thing lol

[mild laurel]

I mean, homomorphic is an equivalence relation

[woven delta]

On what?

[mild laurel]

groups

everything

[woven delta]

In what way is it an equivalence relation on groups?

Obviously isomorphism is, but I don't see how homomorphisms would give you one

Unless you want everything to be related

In which case sure

[mild laurel]

yes

[woven delta]

Is this a shit post?

[mild laurel]

Was what I said wrong

[woven delta]

Saying A and B are related if there is a homomorphism from A to B?

Is that what you wanted?

[mild laurel]

yeah lmao

[woven delta]

So everything is related

And this fails for rings

[mild laurel]

Yeah I realized that

[woven delta]

So

[uncut girder]

0 homomorphism exists

[chilly ocean]

which ring does B_cris refer to

[urban acorn]

wow, what an interesting partition

let's publish 2,000 pages studying it

I think I can show an amazing link to composition series

the first and last groups in a composition series of a group with no sporadic groups in its composition series must always be homomorphic

[chilly ocean]

@urban acorn literally what

[urban acorn]

be more specific in your question

[mild laurel]

Maybe you should be more specific, what the hell are you saying

[magic owl]

Wtf does homomorphic mean

Grp is a connected category

[blissful root]

congratulations you proved it @magic owl

[magic owl]

Lmao

[bleak abyss]

3 years ish

Assuming you go about 10 pages/minute

[woven delta]

Lmao

How long would it take to finish Lee @bleak abyss

[bleak abyss]

Pretty similar

[uncut girder]

Wait really

DF takes 3 years?

[blissful root]

you shouldn't look to "finish it"

it's a big book

[bleak abyss]

I mean as I said depends on the pace

If you're doing 10 pages per hour, 3 hours a day

Which is reasonable

Then yeah it takes about 3 years

I mean D&F is like

Long

But really really easy

Interesting doesn't mean difficult

I mean easy in the sense of like

Like compare to Rudin where you have to think really hard about the stuff while you're reading

I mean even compared to Baby

I haven't read Papa myself actually, I used a different book

D&F is practically storytime followed by exercises

Wait what? I tried to read Rudin in my first year and it was tough, kinda got stonewalled a few times even, when I was trying to learn group actions I started reading D&F chapter 1 (also first year) and it was the smoothest thing ever

Took too long to get to the point which is why I read other stuff but like

D&F is the opposite of terse

I mean idk say for Galois theory I also felt like it spelled all the details out, at least among the stuff I glanced through in chapters 13 and 14

I never tried systematically reading it because the number of words is comparable to the number of atoms in the observable universe

I mean nobody reads the later chapters

At that point you read specialized books on the matter

So like, the part of D&F which matters is chapters 1-14

And what I've seen of those chapters makes it seem locally easy to read, the hard part is actually getting through it all

[waxen iron]

"Show that if H is a subgroup of G and [G:H]=2 then H is a normal subgroup"

All I'm showing is tears because the test is next week and I can't solve half of the exercises

Proofs fuck me up so bad it's not even funny

[steep hull]

Think about cosets

[waxen iron]

Reforming: The number of left and and right cosets is the same

[steep hull]

Yes, and we’re trying to show that gH=Hg for each g in G. If g is in H, then it’s trivial. What happens if g isn’t in H?

Do you understand why this is a valid proof technique?

[waxen iron]

Then gH and Hg must divide the group in the same number of cosets
But how does that show they're the same ?

To be honest, not really

[steep hull]

A subgroup H is normal in G iff gHg^(-1)=H for all g in G. Also, if g is not in H, then gH cannot contain an element of H, so it must be the rest of G (as there is a bijection between H and gH and H has index 2). Similarly, Hg is the rest of G, so they’re the same.

(This covers infinite groups as well)

[waxen iron]

Ooh, right, I was thinking about g=e but obviously if g is the identity it's in H

Many many thanks!
One down, only 30 more exercises I can't solve to go!

[steep hull]

Good luck and don’t be afraid to ask more questions

[waxen iron]

I generally spend 15 minutes looking at a given exercise
Then move on, come back after a while to try to solve it, if after 15 minutes I don't get anywhere I ask

[tribal pasture]

Same dude, test next week, have 0 motivation to solve exercises.

[languid osprey]

If G is a group and, $g \in G$ and $gh \in G$, then does that mean that $h \in G$ ?

[cloud walrus]

Abrar:

[languid osprey]

Oh right it must be true

Because $g^{-1}gh \in G$

[cloud walrus]

Abrar:

[clear obsidian]

what is this # operation here?

[bleak abyss]

Connected sum

You chop a disk out of each space and glue along the boundary (so (S^1\times S^1) # (S^1\times S^1) is the two-holed torus)

[clear obsidian]

ok that makes sense! Thank you!

[chilly ocean]

Anybody home

[chilly ocean]

Show that if group G has 20 elements and it doesnt have an element of order 4 then there exists element of order 10. Any ideas how to do it?

[mild laurel]

What have you tried

[lone moon]

Do you know Lagranges Theorem?

[clear obsidian]

G doesnt have a subgroup of order 4, start there I think...

correct?

[mild laurel]

no that's false

[tribal pasture]

How can I find problems specifically of this form? Any keywords?

[chilly ocean]

To my question: I have lagrange, cauchy, sylov theoerm.

[mild laurel]

Try using them

[chilly ocean]

I tried

There is a hint show that there exists a normal subgroup of order 5 but couldnt do that either

[mild laurel]

Try that

Can you show that there exists a subgroup of order 5?

[chilly ocean]

its from cauchy

theorem

yes

But normal one idk how

[mild laurel]

Think about Sylow's theorem

And think about which one might be relevant to the normality condition

[chilly ocean]

ok will try

[chilly ocean]

What is a rigid motion on a polygon?

[stone fulcrum]

Take some polygon (equal sides, equal angles). Imagine it's sitting in a hole that it fits into perfectly. Pick it up, do anything you want with it, put it back down into the hole. That's an "algebraic action" on the polygon

Important bit is that it should rest in the same spot after the motion.

[chilly ocean]

Okay then

Consider a tetrahedron with vertices 1234 sitting in a 4d hole

Reflecting across the plane cutting through the edge 3-4 hitting the opposite edge perpendicularly leaves a tetrahedron in the same place, right?

[stone fulcrum]

Why a 4D hole?

The hole analogy doesn't work with a 3D shape lol

Just, should look the same before and after motion

[chilly ocean]

Reflecting across the plane cutting through the edge 3-4 hitting the opposite edge perpendicularly leaves a tetrahedron in the same place, right?

[stone fulcrum]

There's infinitely many such planes

"hitting the other edge perpendicularly" oh I see

[chilly ocean]

For any skew lines, there is only one plane that passes through the firt line and hits the second line perpendicularly lol

Now you made me feel like my high school geometry class I hope you're happy young man

[stone fulcrum]

Fair, yes the tetrahedron would look the same before and after

[chilly ocean]

Then (12) is by definition a rigid motion.

[stone fulcrum]

Well, one could say it's an algebraic action but we don't normally talk about reflections

Like I said, these are things that should happen if you pick up and place down the object

I feel like maybe I need some context lol. Is this from a group theory perspective?

[chilly ocean]

Yes

[stone fulcrum]

There's nothing "un-group theory" about reflections but we don't normally bring them up for simplicity

[chilly ocean]

It's easy for me to look at a pentagon and find the actions (permutations) that keep its chakras well-aligned.

They're rots and flips through an angle bisector

But for a tetrahedron, there should be rots and something else

[blissful root]

you have 3 axes of rotation

for a total of 12

and 1 reflection if you want

[chilly ocean]

Why isn't every vertex an axis of rotation

[blissful root]

you could think about it that way

but then you get redundancy

aka you can generate the other rotations just by these three

it's enough to pick 2 vertices as axes of rotation

and one other rotation

which is like diagonal

[stone fulcrum]

It's pretty hard to talk about the groups of 3D shapes without certain theorems, as you do get redundant elements

[blissful root]

what theorems

[chilly ocean]

Okay so I'm missing something.

[stone fulcrum]

If you know the orbit-stabilizer theorem you can at least get the group size easily

[tribal pasture]

Prove that if |G| = p^n then G contains an element of order p.
Assume this as inductive hypothesis. Consider |G'| = p^{n+1}. Either G has subgroups or it doesnt. If G has a subgroup then it has an order of p^k for k < n+1 then by inductive hypothesis it contains an element of order p.
If G doesnt have a subgroup then it is a cyclic group and thus the element p^n has the order p.
Is this correct?

[blissful root]

If G doesnt have a subgroup then it is a cyclic group
??

[tribal pasture]

If G has a subgroup the subgroup must divide the order of G. Thus the subgroup has the order p^k for k < n+1. By induction then we have that the subgroup contains an element of order p and hence the group.

[blissful root]

that part's fine

[wind steeple]

Nope

<n+1 is false

It's <=n+1

[blissful root]

it's <

subgroup means proper

[wind steeple]

G is a subgroup of G

No

[blissful root]

yes

[wind steeple]

Proper subgroup means proper subgroup

[blissful root]

by subgroup he clearly means proper

[wind steeple]

But ok

[uncut girder]

How do you prove v(x)v' = v'(x)v implies v=av'

[mild laurel]

what's v(x)

[uncut girder]

v(x)v' is an element of the tensor product

[mild laurel]

smh $\otimes$ exists

[cloud walrus]

Zopherus:

[uncut girder]

im at a loss

omg this is so easy

[simple agate]

if I want to show $\mathbb{R}[X]$ is not finitely generated as an $\mathbb{R}$-module, can I assume by contradiction that $\exists S[X] \subset \mathbb{R}[X] : S[X] = {f_1, \dots, f_k}$ for some polynomials $f_i$. Then let $n := \max_{1 \leq i \leq k}\deg{(f_i)}$ and $g(x) = x^{n+1}$ then $g \in \mathbb{R}[X]$ but $g$ cannot be constructed additively from the elements of $S[X]$?

last line is really badly worded and I would improve it, just trying to get the logic right

[cloud walrus]

alex:

[lone moon]

That sounds like a valid reason

You don't need S

just assume that it is finitely generated and you get this contradiction

[simple agate]

thanks!

[prisma thunder]

Is there a way to prove the algebraicity of an element when it's not over the rationals? For example, I am attempting to show e is algebraic (or not) over the field of periods P (i.e. [P(e) : P] is finite), and I have no idea where to start.

[blissful root]

what's the field of periods?

[waxen iron]

I need to show this and I'm kinda lost

[blissful root]

check the order of both groups

[waxen iron]

It's not the same, but here's the thing

Wait

[blissful root]

what is this map?

oh you are taking x mod 4

for each of them

i mean this isn't even surjective

[waxen iron]

But it's an homomorphism correct?

[blissful root]

sure

[waxen iron]

Okay then it makes sense

I was mixing the homomorphism and isomorphism symbols

Many thanks!

[blissful root]

yw

[waxen iron]

Okay I was going over my notes and my teacher said that "G is cyclic=>G is abelian"
He didn't show a proof, he said for us to take it as fact (He did this in the beginning when we didn't have the proper tools to understand the demonstration)

But now trying to show this myself I'm having a hard time doing it

[wind steeple]

write down definitions

[pearl mural]

alternatively, we know something about all finite cyclic groups, and we know something about every infinite cyclic group

and if you know properties about those, then you can determine why cyclic groups must be abelian

[wind steeple]

you don't need to take about isomorphism, it's pretty trivial with only the definitions...

[waxen iron]

Okay so: If G is cyclic then there's some g in G such that <g>=G

Or g^n=1

for some n in N

And obviously if G is abelian then ab=ba for a,b in G

[pearl mural]

@waxen iron if G=<g>, then each element in G is equal to g^k for some k

so say you have two elements, a=g^k, b=g^j

what's a*b

what's b*a

[waxen iron]

Ooooooooooooooooooooooooh

[pearl mural]

@wind steeple yeah you're right -- I just like reducing things to Z_n and Z as soon as possible

makes it easier to think about

All cyclic groups of order n are isomorphic to Z_n, all infinite cyclic groups are isomorphic to Z

Z_n here \cong Z/nZ

[waxen iron]

"All cyclic groups of order n are isomorphic to Z_n"~It's funny, my teacher showed the opposite, that all Z_n groups are isomorphic with the cyclic group of order n @pearl mural

[pearl mural]

lol

I mean if a\cong b then b\cong a so it's the same thing

[waxen iron]

Yeah, I just though it was mildly amusing that you said the exact inverse

Anyhow, many thanks!

[pearl mural]

np

[fading wagon]

All cyclic groups of order n are cyclic groups of order n.

[urban acorn]

All cyclic groups are finite.

[magic owl]

Woke

[urban acorn]

exactly

[magic owl]

$Z/\infty Z$

[cloud walrus]

MaxJ:

[urban acorn]

that's a conspiracy maintained by the government

also

cough $\frac{\bZ}{1\bZ}$ cough

[cloud walrus]

Intel:

[woven delta]

👀

That's the trivial group

You want $\bZ / 0\bZ$

[cloud walrus]

Liquid:

[tribal pasture]

If I have xyz as a element in a group, I cannot claim right that xy is also an element of the group?

As in does xyz in G imply z^{-1} is in G?

[wind steeple]

xy in G iff z in G

[tribal pasture]

I dont think it answers my question because it neither negates nor confirms whether xyz in G => z^{-1} in G <=> z in G holds true.

[mild laurel]

Uh what

z^{-1} in G <=> z in G is always true

[tribal pasture]

Yeah I am concerned about the first implication though as requested in my first quesiton

[blissful root]

ab in G does not imply b in G

[mild laurel]

It literally doesn't matter

If the second thing is always true

Then the overall thing implication will always be true

[tribal pasture]

@blissful root Okay perfect. thanks!

[rapid tinsel]

Hello!
If I have an abelian group G and a subgroup H, is there a quick way to show that G/H has order 3? E.g. to show that G/H has order 2, I just have to show G and H are different and any two elements in G \ H sum to give an element in H

[tribal pasture]

Are G and H finite?

[rapid tinsel]

No

[mild laurel]

Vaguely the same thing will work

Take two elements a,b in G\H such that a - b is not in H, and then show that a + b is in H

[rapid tinsel]

Ooh nice

Thanks!

I suppose the same trick wouldn’t work as well for order 4

And it doesn’t generalise well for any n

[mild laurel]

Yeah, I mean

There begin to be two groups of order 4 so

[rapid tinsel]

Yeah but the order 3 trick is nice

[magic owl]

Ok so

I have two copies of the same graded ring Z[x]/<x^2> and Z[y]/<y^2>

I want to take their tensor product

It makes sense that this is going to end up being Z[x,y] with some relations

It makes sense that we keep the relations x^2,y^2=0

But we also (apparently) get a anti-commutativity relation xy=-yx

[magic owl]

Nevermind

I am a dumb bitch(tm)

[fringe nexus]

Let N be even integer, and find the kernel of the action of $D_{2n}$ on the sets consisting of pairs of vertices of a regular n-gon.

[cloud walrus]

aaaaaaaaaaaaaaaa:

[fringe nexus]

Why isn't this kernel simply just the trivial kernel?

[mild laurel]

What does it mean by pairs of vertices of a regular n-gon?

[fringe nexus]

like you have the pair (i,n/2+i)

if you label the vertices clockwise starting with 1

[mild laurel]

So opposite vertices?

[fringe nexus]

oops it should be opposite vertices

yea

[mild laurel]

Ah n is even, that's why you can do that

if I'm not mistaken

[fringe nexus]

hmm

it doesn't

[mild laurel]

nope im dumb

yeah

Rotation does though right?

180 degree rotation?

[fringe nexus]

oh

i guess i wasn't sure if (n/2+i,i) = (i,n/2+i)

aaaa

ok it makes sense

but apparently n=4 is special case

[mild laurel]

Right

Reflections do preserve pairs there

That's the only case I checked for reflection whoops

[fringe nexus]

like if i get a question like this

i just have to see that 4 is a special case?

oof

[mild laurel]

I mean, the dihedral group only has reflections and rotations

So you try reflections for small cases, see it works for 4 and not higher

[fringe nexus]

Let $G_a$ be the stabilizer of a in G and consider $C_a = {ga |g \in G}$ I'm trying to show that $|C_a| = [G:G_a]$ by constructing the map from $C_a \xrightarrow{} G/G_a$ defined by $g*a \xrightarrow{} gG_a$

[cloud walrus]

aaaaaaaaaaaaaaaa:

[fringe nexus]

so the kernel is obviously just $1*a$, and so this is injective, but I'm not sure about surjective

is a better way of proving injectivity something like $gG_a = hG_a \xrightarrow{} h^{-1}g \in G_a \xrightarrow{} h^{-1}ga = a$ and so $ ga = ha$

[cloud walrus]

aaaaaaaaaaaaaaaa:

aaaaaaaaaaaaaaaa:

[tribal pasture]

Why cant a =0 in the centre?

[mild laurel]

a = 0 isn't in GL_n

[tribal pasture]

Oh yes smh

[lone moon]

remember that the GL_n is only a group with respect to multiplication

[tribal kiln]

Anyone on for a quick q?

[mild laurel]

I'm listening

[tribal kiln]

Say I have Z union infinity

and everything is closed p.t. it is finite or contains {infinity}

and open is that ur compliment is closed

Every cover is open

[mild laurel]

Uh, this isn't abstract algebra

[tribal kiln]

Where should I put this then lol?

[mild laurel]

#point-set-topology

You don't know that this is topology?

[tribal kiln]

Yes lol

I didn't see that you could expand advanced math

thnx

[tribal pasture]

Let N be a normal group of order 5 inside a group G with an odd order. Show that N is contained inside Z where Z is the centre of G.

Since N is finite prime, we have that N = <x> for x ≠ 1. Since C(x) | G, we have that C(x) = 1 or 3 or 5.
If C(x) =5, then since gxg⁻¹ ∈ N, we have that there exists a g such that gxg⁻¹ = 1 which implies x = 1 and thus a contradiction.
If C(x) =3 then ∃x' ∈ N\C(x) such that C(x') ≤ 2. Since C(x') | G thus we have that C(x') = 1 which implies x' ∈ Z. Since Z is a group, thus we have x = (x')ⁱ ∈ Z.
If C(x) = 1 then x ∈ Z trivially. Thus shown.
Is this correct?

[tribal pasture]

<@&286206848099549185>

[teal perch]

By Z do you mean the center Z(G)?

[tribal pasture]

Yes

[random quartz]

Does anyone know Binary Algebra ?

[devout spire]

How can i find at which order a permutation p is going to be equal to a permutation q
so like for which k in p^k = q

[wind steeple]

Don't you mean, find n s.t. p^n = k ?

[devout spire]

sorry i fixed it ^ @wind steeple check it again now

normally this can also be expressed like
p^k = q^1
because that's to what i am interested

but i dont know where to start and how to proceed

[wind steeple]

You can decompose p and q into disjoint cycles décomposition

[topaz solar]

Alright, i've managed to get myself stuck a slight bit and could use some help

Say a sequence $a_n$ converges in order to $a$, denoted $a_n \to a$ (sometimes with a $\mathfrak o$ superscript on the $\to$, not sure how common since i'm not exactly an expert) iff there exist monotonic sequences $s_n, t_n$ such that $s_n \leq a_n \leq t_n$ and $\sup s_n = \inf t_n = a$
$\S$ay two sequences are equivalent $(a_n \sim b_n)$ iff $b_n - a_n \to 0$

If I have 4 sequences $(a_n), (a_n'),(b_n), (b_n') \in R^\mathbb{N}$, where $R$ is a partially ordered ring, and both $a_n \sim a_n'$ and $b_n \sim b_n'$ hold, can i say that $a_n b_n \sim a_n'b_n'$?

[cloud walrus]

Darkrifts:

[cloud walrus]

tree3:
Compile Error! Click the reaction for details. (You may edit your message)

[topaz solar]

@steep hull only problem is a_n etc need not converge, just a_n - a_n' converges (equivalent to the two series have the same point to which they converge, provided they do so). I'm slightly lost on how you bounded it but i am smoothbrain so that doesn't surprise me. Might've given me an idea to work it, if this doesn't fully work once i figure out exactly what you meant

i can definitely bound it more than i've been thinking

[steep hull]

Whoops, I’m bad at reading English

[topaz solar]

nah, i'm bad at writing it

the thing i'm working in is really not all that nice

needn't be complete, etc

but uh, I had an idea to try and exploit that $a\leq b, c\leq d \vdash ad + bc \leq bd+ac$ works, but didn't quite have the right idea

[cloud walrus]

Darkrifts:

[topaz solar]

i only looked at the lower bound, which (when you do that) should end up going to 0, and because a duality argument, the upper bound does as well

so you bound your difference in to 0

[steep hull]

Hmm, I think I have an idea

[topaz solar]

very messily, of course

but uh yeah should work

[steep hull]

I was going to write a_n=a’_n+r_n and the same for b_n and do some computations

[topaz solar]

though this bounds a specific thing, not sure yet if i can use it to prove the normal thing

yeah, but uh
consider a sequence 1, 2, 3, 4 etc

that obviously goes off to infinite

but uh
you can have a sequence which also approaches infinity in a very similar way without being an arithmetic sequence

or any other such thing

[steep hull]

Yeah

[topaz solar]

like 1, 3/2, .... approaching 2, and the constant sequence 2, 2, 2, 2

or, alternatively, one approaching sqrt(2) over the rationals

[steep hull]

Writing a_n=a’_n+r_n and doing the same for b_n should work. Let me try writing it up

[topaz solar]

i'll look at it, beats me if it'll work

oh and uh, i have that $\alpha \beta + \alpha'\beta' - (\alpha\beta'+\alpha'\beta)$ should converge to 0, not a properly formally done proof though and assumes that your sequences $\alpha, \beta$ don't run off to some infinitely large elements

[cloud walrus]

Darkrifts:

[steep hull]

There may be some trouble with infinity

[topaz solar]

well ofc there is, that always busts things up, but luckily your bounding sequences s_n, t_n can be chosen to be infinitely large in only finitely many terms

i think

then again, I'm smoothbrain, take anything I say with a grain of salt should it seem sketch at all

[steep hull]

Take (n+1/n)^2-n^2

[topaz solar]

division is nasty

this is a partially ordered ring, which makes it very annoying