groups-rings-fields

[void cosmos]

also question

finding the degree of a specific element over Q

is the usual way just trying to play around it to find a minimal polynomial?

[rustic crown]

you have algos to do it, but for all practical purposes... you just play around

[void cosmos]

yea i meant like on an exam

[rustic crown]

find one poly

hope it's irred

[void cosmos]

yea cool

[rustic crown]

if not, take an irred factor

[void cosmos]

it will be cuz thats how exam problems are made hehe

[glossy crag]

Quick sanity check:

For prime p, a root of unity is a (unique) product of a root of unity of p-power order and a root of unity of order prime to p.

If ord(x)=n, write n as p^km with p\nmid m, then 1=ap^k+bm => x=x^{bm}\cdot x^{ap^k}, ord(x^{bm}) is a power of p and ord(x^{ap^k}) is coprime to p. Uniqueness holds because if xy=x'y' are two such decompositions, then y=(x^{-1}x')y' and the order of y would be the product of orders of x^{-1}x' and y' (the former a power of p), hence divisible by p (if x\neq x').

[void cosmos]

yoooo

if two extension fields are isomorphic as vec spaces

then they are the same XD?

[coral spindle]

No

[kind jacinth]

Can i sinplify like this:

the question is: What is the order of rsr2srs3 in D12?

maybe even further simplified to just s^3 since r^12 doesnt do any rotation basically

[warm crow]

.

[coral spindle]

If that's what you've calculated, then yes.

[warm crow]

Thank you!

[void cosmos]

yo bad question

can someone walk me through the tower theorem argument

for proving Q(sqrt(2),sqrt(3)) = Q(sqrt(2)+sqrt(3))

right is subset of the left

[coral spindle]

Um, not sure why you'd use the tower theorem

[void cosmos]

iddk feels cooler

[coral spindle]

Oh no I do see what you mean

as you were

[void cosmos]

ok

so we know that Q(sqrt(2),sq3) : Q = Q(sqrt(3)+sqrt(2):Q * Q(sqrt(3),sqrt(2)):Q(sqrt(3)+sqrt(2)

right?

we also know that left side is 4

so i want to show that Q(sqrt(3),sqrt(2):Q(sqrt(3)+sqrt(2)) = 1?

that is enough right?

[formal ermine]

you need to also show that [Q(sq2 + sq3) : Q] = 4

this immediately implies Q(sq2 + sq3) = Q(sq2, sq3) though

[void cosmos]

if i show Q(sqrt(3),sqrt(2)):Q(sqrt(3)+sqrt(2) cant be 2 then im done

[formal ermine]

ye

[void cosmos]

ik sqrt(2)+sqrt(3) is in Q(sqrt(3),sqrt(2)) so if i can show this cant be a root of an irreducuible polynomial of deg 2 then im done?

[coral spindle]

Try finding its minimal polynomial

[void cosmos]

doesnt x^2-(5+2sqrt(6)) work tho?

it doesnt

sqrt(6) isnt in Q(sqrt(2)+sqrt(3)) ig

right/

so im done?

[coral spindle]

Are you sure?

sqrt(6) is in Q(sqrt(2) + sqrt(3)), in fact.

[void cosmos]

how tho

lmao

[formal ermine]

Q(sqrt(2) + sqrt(3)) = Q(sqrt(2), sqrt(3))...

[void cosmos]

yea lmao

brain fart nvm

okaay i am going to do it manually

suppose x^2+bx+c is a polynomial who has sqrt(2)+sqrt(3) as a root

[void cosmos]

this has to be reducible tho rihgt?

[coral spindle]

That's not a polynomial over Q.

[void cosmos]

its over Q(sqrt(2)+sqrt(3))

[coral spindle]

Then that's not the minimal polynomial

the minimal polynomial of a over Q(a) is x - a.

[void cosmos]

yes but idk yet that Q(sqrt(2),sqrt(3)) is Q(sqrt(2)+sqrt(3))

like i have to prove that deg 2 wouldnt work

so ig manually should work

yeaa

i got it

another proof check:

K(u) = K(u^2) if K(u)|K is of odd degree

proof: K(u^2) is contained in K(u) --> by tower we have K(u):K(u^2) must be odd

but now consider x^2-u^2 in K(u^2)[x] , this has degree 2 so we know this extension atmost has degree 2 but its odd tho it must be 1 hence equal?

[barren sierra]

dumb question

A[m] is just notation for elements of order d where d | m?

[coral spindle]

WHY DOES EVERYONE SAY THEIR QUESTIONS ARE DUMB/BAD STOP ITTTT

[barren sierra]

yea my bad I yell at people enough for doing that I should stop doing it myself

[coral spindle]

Here, A[m] means the m-torsion elements of A

[barren sierra]

it's just bad habit at this point

[coral spindle]

So that is to say, $A[m] = {a \in A\mid ma = 0}$.

[cloud walrus]

Boytjie

[coral spindle]

The notation is really awful, I know

[barren sierra]

ok so I was right?

[coral spindle]

But it really is literally the definition of ker m

Yeah you are right I was just giving you more context

[barren sierra]

dope

[coral spindle]

tru

[void cosmos]

is this a correc tprof

[summer path]

I simply say that I am dumb to justify my bad questions :)

[barren sierra]

yea could I get a hint for this

[steady apex]

any good online sources for group theory?

[coral spindle]

If you look up the name of any major textbook and "pdf" you'll probably find a copy

I'm partial to Fraleigh, and there seem to be pdf copies on google

[barren sierra]

like ok say C, D are abelian groups and I have some phi: C/mC -> D, I don't really know what the corresponding C -> D[m] should be

[summer path]

Fraleigh is nice; Humphrey is okay I think as well

[barren sierra]

Is there some way I can say G o F = F o G = id_Ab or something?

idk what id_Ab would even look like

or how composing functors really works

[coral spindle]

Nah, adjoints aren't usually inverses like that

it means something a bit different

I'm trying to think about what the correspondence would be

[delicate orchid]

u do one then the other

[barren sierra]

well sure but like actually working with them

[delicate orchid]

yeah so an adjunction basically means there's a bunch of bijections between whatever and whatever, in this case between Ab(A, A[m]) and Ab(A/mA, A), I believe

where these bijections obey some naturality law

[coral spindle]

I think maybe they stated these the wrong way around. I think it should be a correspondence Hom(C[m], D) <-> Hom(C, D/mD), which I think I can see.

[delicate orchid]

I find this commutative diagram particularly enlightening

[barren sierra]

hm yes this definitely clears things up

[formal ermine]

ohhhhhh

[gilded osprey]

I am asked to prove/disprove that if $G/H$ is Abelian, then $G$ is Abelian.

My counterexample is to take $G=D_3$ (known to be non-Abelian) and $H={\epsilon, R, R^2}$, which by having index 2 (in $G$) is normal. $G/H$ then has order 2, so it must be Abelian.

Is my logic correct on this?

[formal ermine]

that makes sense

[cloud walrus]

JacobHofer

[formal ermine]

take any non abelian G and H = G

[gilded osprey]

Oh huh, that works as well too I guess lol

[barren sierra]

but I think that example also works?

[delicate orchid]

S_n/A_n, n > 3 also

[gilded osprey]

Was my logic with D3 still correct tho?

[delicate orchid]

yes

[barren sierra]

yea looks good to me

[coral spindle]

Yes

[gilded osprey]

Awesome, thanks a ton! 😄

Actually glad I used my D3 example, since the next one asks about G/H and H both being Abelian, and so my example works for both problems!

[delicate orchid]

I mean, they actually are enlightening

[barren sierra]

I don't see how it's the wrong way around and even if it is I don't see the correspondence

[pastel cliff]

https://tenor.com/view/timi-gif-20112884

[pastel cliff]

https://tenor.com/view/agenturleben-agencylife-hannover-kochstrasse-kstr-gif-24117466

[elder wave]

anyway i also think your diagram is quite enlightening sebb

[pastel cliff]

hehe i just think it's goofy that i can go to my friends and be like "look at this math" and show them a bunch of arrows

[topaz solar]

It’s very useful for silly diagrams, like snake lemma or such

[pastel cliff]

globular

the best thing about learning more math is that words start to sound more made up

Frobenius

[south patrol]

Yes, somebody's name

LOL

[pastel cliff]

dummit and foote probably came out alice in wonderland

[topaz solar]

Globular operad is a silly globular set

[topaz solar]

Globular is from globes tho

[south patrol]

Ye

[topaz solar]

I do wonder what’s missing from these contractible globular operad algebras to keep it from being opetope-y

[barren sierra]

Hm I think part of my issue with this is when looking in my texts (Awodey, Rotman)

it's talking about natural transformations

and we have not covered this

[elder wave]

i mean

you kinda need that for the functor iso to make sense

[rustic crown]

yea lol

[barren sierra]

[rustic crown]

wait so was your question answered?

cause the definition of natural transformation isn't really hard lol

[barren sierra]

no my question has not been answered

I am still very stuck

[rustic crown]

okie so what we mean is like

Hom(F(C), D) is a set right

[barren sierra]

right

[rustic crown]

and two sets are bijective if and only if they have teh same rize

[barren sierra]

well we want more than bijection, we want isomorphism right?

[rustic crown]

but this is not enough for some pair of functors to give "point-wise" iso

they have to interact with each other in some way

[coral spindle]

dw about the isomorphism quite yet.

[pastel cliff]

det

[coral spindle]

as it is, an 'isomorphism of sets' is just a bijection

there are some additional conditions happening, but again, don't worry about it yet

[rustic crown]

so yea, just asking isomorhphism of two sets isn't a lot

[coral spindle]

I love stealing det's thunder nom nom nom delicious

[rustic crown]

like another example is like the order preserving bijection from nullstellensatz

knowing just the "bijection" exists isn't good enough

you want to see that actual nice bijection

anyway, until you see and properly understand the definition of natural transformations/isomorphism just think that you wanna define the iso Hom(F(C), D) = Hom(C, G(D)) without definining things which are very specific to C or D. a similar map should be possible to define if you replace C and D with other objects C', D'.

here you wanna give a simple bijection between
Hom(A/mA, B) = Hom(A, B[m])

[barren sierra]

hm ok

[rustic crown]

so do you see a "nice" bijection?

[barren sierra]

not immediately

[coral spindle]

Well I mean let's maybe not focus quite yet on getting a bijection

if I gave you a map f : A/mA -> B, and an element a of A, could you give me an element of B[m]?

I.e. can you think of a nice way of turning maps A/mA -> B into maps A -> B[m]

[rustic crown]

(or an intermediate map A --> B satisfying some properties)

[barren sierra]

honestly no. Not really sure how B[m] relates to A/mA

[rustic crown]

okie forget about B[m] for a minute

how do maps from A/mA --> B relate with maps from A --> B

(iso theorems are your friends )

[barren sierra]

well I know I can turn a map A -> B into a map A/mA -> B

[rustic crown]

oh you can't do that always

[coral spindle]

That's not always true I'm afraid

But the reverse is certainly true (can you see why?)

[rustic crown]

other way is true though, which is prolly what you mean

🔫

[coral spindle]

Are we literally the same person

[barren sierra]

whoops yea

[rustic crown]

.<

yea so recall what's the condition when a map A --> B factors through A --> A/mA

[barren sierra]

wdym condition

[rustic crown]

okie we see that we have a nice function
Hom(A/mA, B) --> Hom(A, B)
given by precomposing with A --> A/mA

i'm asking you what is the image of this function

can you describe it using "m"?

[barren sierra]

oh uh

I'm blanking >_> the image is the morphisms with mA as a kernel right?

[rustic crown]

yep

[next obsidian]

Uh

[rustic crown]

oh wait

[next obsidian]

Is A a local ring

[rustic crown]

lol

[next obsidian]

Otherwise no

[rustic crown]

its an abelian group lol

and m is an integer

[next obsidian]

Lmfao

[barren sierra]

wait no I'm dumb

[rustic crown]

you only want mA to be in the kernel

[barren sierra]

yea

[rustic crown]

not exactly be the whole kernel

[barren sierra]

right right

[rustic crown]

chmonkey does so much ag

[barren sierra]

ag >_<

[rustic crown]

right and mA being in the kernel is same as saying image of A --> B dies when you multiply by m

[barren sierra]

yea

[rustic crown]

which means A lands inside B[m]

[barren sierra]

uh lemme parse that real quick

[rustic crown]

yee take your time :3

[coral spindle]

f : A -> B
f(ma) = 0
so...

[barren sierra]

cause individually that all makes sense but in the context of the problem

I don't necessarily see how mA being contained in kernel of A -> B is the same as saying the image of A -> B is contained in B[m]

[elder wave]

isn't it f=ma

[coral spindle]

it's right here it's right hereeeeee

[barren sierra]

oh right that's how homomorphisms work

_> it's been a long week

[coral spindle]

Tadah you have a map from Hom(A/mA, B) -> Hom(A, B[m])

[delicate orchid]

wtf is B[m]....

[coral spindle]

I defined it above

Now you can try proving it's a bijection!

[barren sierra]

B[m] = {b in B | mb = 0}

[coral spindle]

@delicate orchid

[barren sierra]

should be written [m]B smh

but that looks awful

[rustic crown]

Tor^Z_1(Z/mZ, B)

[delicate orchid]

ah ok, now I get it 😌

[barren sierra]

ok proving this is a bijection shouldn't be too bad

ok so now that we've done this, what are natural transformations and how does this relate (or is that too loaded)

[delicate orchid]

I mean going off your definition you just need to show the hom-sets are iso

[next obsidian]

Bad wew

[rustic crown]

the map we defined is natural, since the same recipe can be used to define a map if you replace A, B with A', B' >.<

[next obsidian]

If you’re gonna do it you should go and prove the useful part

It shouldn’t be hard regardless

[rustic crown]

come back to it later, when you read the actual def of a natural map

[delicate orchid]

I know u need the funny \phi(G(u)\circ whatever) = whatever but u know what... I don't care

[barren sierra]

oh so natural is just a way of saying that it doesn't really matter what the actual set is?

[next obsidian]

It’s a square commuting

[coral spindle]

The wikipedia article is a fine introduction

[next obsidian]

It isn’t hard to verify something is natural ChmonkaS

[coral spindle]

we can talk about the particular naturality condition later

[barren sierra]

what useful part?

[rustic crown]

but i think it takes some time to get used to "why no choices" and "cute square commuting" are related

[coral spindle]

just don't read the nlab page

[next obsidian]

In one way yes but it says more

What are you showing are iso?

m-torsion and something else?

[rustic crown]

yee adjunction

[barren sierra]

original question

[next obsidian]

What two sets specifically

[barren sierra]

For any abelian groups A, B Hom(A/mA, B) iso Hom(A, B[m])

[coral spindle]

Well you've shown that, if you've shown there's a bijection

when you've read up on what a natural transformation is, the folks in #category-theory will be glad to help you understand adjunctions

[delicate orchid]

the triangle laws my beloved

[barren sierra]

right

[next obsidian]

Okay spamakin given an isomorphism f:B -> B’ then this induces an isomorphism Hom(A/mA, B) -> Hom(A/mA,B’) and also Hom(A,B[m]) -> Hom(A,B’[m]) right?

You literally just say “lol I append f”

[barren sierra]

right

[next obsidian]

And now your maps to B (or B[m]) are maps to B’

You also have these isomorphisms “across”

From like A/mA,B and A,B[m]

But you actually have two of these, one for B and one for B’

Cuz your proof makes a bijection for arbitrary B

[coral spindle]

the diagrams... can you hear them crying to be set free?

[barren sierra]

idk how to draw diagrams using tikz so they shall not be free

[next obsidian]

but you can draw them in your mind's eye right?

[barren sierra]

yea

[next obsidian]

the horizontal ones just chagne B to B'

[rustic crown]

mind's eye

[barren sierra]

I think I see what's going on

[coral spindle]

I'm not sure anyone can, they just use https://tikzcd.yichuanshen.de/ surely

[next obsidian]

and the vertical ones change the like A/mA to A and B to B[m]

so like

you might ask

is the isomorphism across the Hom-sets for B and B' "the same"

because the objects involved on both sides are "the same" (isomorphic)

[barren sierra]

https://q.uiver.app/ this is what I use

[delicate orchid]

is this the set up boss?

[next obsidian]

yeah

so these vertical arrows are basically an isomorphism between the same objects

[barren sierra]

right cause of what we just showed

[rustic crown]

wew likes left to right notation

[delicate orchid]

I do

[next obsidian]

saying that they "are the same" is just saying this diagram commutes

like if you invert one of these isomorphisms you get two ways to go from say, Hom(A,B[m]) -> Hom(A,B'[m])

these being equal is saying this diagram commutes

[delicate orchid]

specifically, doing postcomposition by f first and then mapping through the isomorphism is equal to doing the isomorphism and then doing postcomposition

[barren sierra]

yup

[next obsidian]

so natural is just asking that this diagram always commutes

[barren sierra]

ok this is making a little more sense now

[next obsidian]

it says that you can replace B with B' and things will always work

because when you replace objects by isomorphic ones sometimes maps that you want to line up, don't

[coral spindle]

well there is more to the naturality here...

[next obsidian]

Yes this is true

cuz you can also replace A with an A'

[delicate orchid]

and do PREcomposition!! woah!!

[rustic crown]

[next obsidian]

so it's a little bit more we're asking, but let's only deal with B for now

[rustic crown]

(don't we also need to deal with arbitrary maps B --> B', or do looking at iso suffices? that might not be enough right?)

[next obsidian]

yes it isn't just isos

[barren sierra]

oh wut

funky

[next obsidian]

that's why I said here it says more

but I forgot to remove the iso part yeah

but it says that these isomorphisms you have of these Hom-sets also play nice with maps between the B

[rustic crown]

wew drew the general diagram

[next obsidian]

and this is useful because it says you don't need to keep track of stuff as much

[coral spindle]

The classic example of a natural transformation, by the way, is the natural isomorphism between a group and its opposite

if we have a group G, we can define G^op by simply reversing the operation, but with the same underlying set.

So if a,b are in G^op, then the group operation a * b (in G^op) is defined as ba (the group operation in G)

This is a functor Grp -> Grp (one might call it an endofunctor)

and there is a homomorphism G -> G^op taking g to g^-1. This map is an isomorphism, and a natural map from the identity functor to the opposite group functor.

This is what people mean when they say "a group is naturally isomorphic to its opposite"

[barren sierra]

Right I've seen that before

Although with rings instead of groups but basically the same argument

[coral spindle]

Oh but rings are most certainly not naturally isomorphic to their opposite

I think you may have something else in mind

[barren sierra]

It was something similar

I think matrices over the ring?

[coral spindle]

Ah, that would make more sense

[barren sierra]

Anyways the group example makes sense

So it's basically trying to formalize this notion of natural that comes up alot

[coral spindle]

There is also a classic example of an adjunction that may help you a bit

Let's denote by U : Grp -> Set the so-called 'forgetful functor'

this just forgets that we have a group, and gives us the set of group elements.

Now conversely, there is a nice way to take a set and get a group, namely the 'free functor' F : Set -> Grp

so this takes a set of symbols, as it were, and builds the group with no assumptions, generated by those symbols

(I'm being imprecise here but I'm hoping you've seen the free group before)

Anyway, this is an adjunction, because maps F(X) -> G correspond to a choice of generators, i.e. a map X -> U(G)

this turns out to be natural in the way we discussed above: Hom(F(X), G) <-> Hom(X, U(G)).

Provided the kind of algebraic structure you're looking at is 'nice' enough, there actually always exists such a 'free algebraic structure'

e.g. there is such a thing as a free Abelian group, or a free module

and this is uniquely defined by being adjoint to the forgetful functor

But unfortunately there is no such thing as a free field (huge sad I know)

you've gone quiet

I hope I haven't overwhelmed you

[white oxide]

is it possible to prove that if n is less than 3, then Sn is an abelian group? i.e. can we use the contrapositive here? i'm not going to use it because then it's basically trivial but it seems kinda like cheating if you can

[next obsidian]

Showing that S_n is abelian for n < 3 is not the contrapositive of this statement, this statement is completely logically independent of the problem in the image

The contrapositive would be that if S_n is abelian, n < 3

[white oxide]

ah okay, thanks

[topaz solar]

Free complete lattice on 3 generators

[barren sierra]

Nah nah you're fine I just got busy

[topaz solar]

I always forget the adjoint directions

[white oxide]

can somebody check my proof? i gave an example just inc ase

[topaz solar]

What’s (12)(23) and (23)(12)

[barren sierra]

Why define sigma then tau

[topaz solar]

Well you can consider an element of S3 as an element of S4 yes?

or so on

[barren sierra]

I don't see the relevant of 4 and I don't see why you defined sigma only to say tau = sigma in many more words

[white oxide]

wait you can?

[barren sierra]

Yes

[white oxide]

i didn't know that

[topaz solar]

Permuting only 1, 2, 3 and constant on higher

[white oxide]

okay well that makes the proof a lot easier lol

huh

okay

[topaz solar]

Now take witness of non-abelian and pass it up

[barren sierra]

Also you concluded S4 is nonabelian

Not S_n for n≥3

[topaz solar]

See above the example text

[white oxide]

yea i know i was just doing an example for how you would construct a permutation for Sk+1

[topaz solar]

But yeah see it as bijections from the set of numbers to itself, then clearly you can send that to a bijection on the bigger sets by making it the identity elsewhere

[white oxide]

oh ok but i didn't know that implied it was in S_n+1

regardless good to know, thanks

could we use Cayley's theorem in some possible way here?

like could we construct a group from A

because all that matters is we show that |H| = |A|

or is that a dumb idea

like for example

if A has n elements

could we just map each element in A to an element in Z_n

and so their cardinality would be the same

then we could do work using Z_n

like

Could we define $\tao: A \mapsto \mathbb{Z}_n$ such that $\tao(x_i) = i - 1 \text{ for } i = 1 \dots n$?

[cloud walrus]

okeyokay

Could we define $\tao: A \mapsto \mathbb{Z}_n$ such that $\tao(x_i) = i - 1 \text{ for } i = 1 \dots n$?
```Compilation error:```! Undefined control sequence.
<recently read> \tao 
                     
l.57 Could we define $\tao
                          : A \mapsto \mathbb{Z}_n$ such that $\tao(x_i) = i...
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.```

[white oxide]

For $x_i \in A$

[cloud walrus]

okeyokay

[white oxide]

basically can we rename the elements

[topaz solar]

Why not just number the finite elements of A

And then do the obvious thing of order n

It’s a cyclic subgroup too

[white oxide]

wdym

also if G is a cyclic group isomorphic to G', G' then cyclic?

[barren sierra]

Ye

Try to prove it

[white oxide]

oh right i'm stupid i did

it has to do with the homomorphism property right

[topaz solar]

Or

I mean

No need to do any weird isomorphisms

You just need one permutation on A of order |A|

[white oxide]

oops you're right

well ig at least with my method i don't have to show that it's cyclic

or supply a generator

[sonic coral]

these seem pretty simple. i just wanna make sure i’m not overlooking anything small in these proofs if that’s fine

[topaz solar]

Easiest way is to just give an explicit one tbh

[white oxide]

well i wrote a proof but it's a bit hand wavy and i don't know if it's correct

but yeah i'll probably do the problem a third time that way

[topaz solar]

If you want my proof, I’d just enumerate the elements of A, then (a1 a2... an) an n-cycle in A

[torn warren]

what does this sentence mean, why turn over and can rotate another 4 position?

[rotund dragon]

if you take an n-gon, you can flip it over first and then rotate it

e.g. take a square paper napkin or something similar and flip it over then rotate it

[torn warren]

do you mean to do a mirror first?

do you mean this? @rotund dragon

[rotund dragon]

yeah

[torn warren]

even I flip respect with the diagonal line, then rotate, it gives the same 4 positions as flip with the middle line

[sonic coral]

Im working too hard on these right? The center is abelian, and every subgroup of an abelian group is abelian, and every subgroup of an abelian group is normal. Hence every subgroup of the center is normal?

Similar argument for the third one

[tender wharf]

yep

really you don't need to do too much because it follows by definition of center

[sonic coral]

for which?

[tender wharf]

for all of them

actually, why didn't you first prove that every subgroup of the center is normal

then if G is abelian, Z(G) = G

so every subgroup of an abelian group being normal follows as a corollary

[sonic coral]

I have done this

[tender wharf]

ah okay then yeah

I mean the proof looks fine to me

(unfortunately the lighting made it difficult to read so I only skimmed it)

[sonic coral]

The first proof is fine, and the argument i typed for the other two is probably better right?

[tender wharf]

congrats, you discovered normal subgroups

[tender wharf]

"typed"
yeah, the arguments are all fine, although slightly long winded

[sonic coral]

ill get there I hope, albeit very slowly

[tender wharf]

for the third proof, notice that Z(G) is a subgroup of Z(G) so you don't have to do anything.

really that's a corollary needing no proof

[sonic coral]

true yeah

[solar glacier]

Is 5 first part 2? Since the minimal polynomial has degree 2

[cloud walrus]

MyMathYourMath

[torn warren]

ah, I haven't learned that part yet, but I will remember this term first @tender wharf

[solar glacier]

Are you taking algebra rn

[torn warren]

I am learning it now, self learning 🙂

[solar glacier]

Ahh

[torn warren]

this is my solution 🙂

[shell ravine]

Let p and q be primes with p > q. Prove that a group of order pq has at most one subgroup of order p.

[shell ravine]

need help w this

[torn warren]

Lagrange thm, if there is a subgroup H of G ,then, |H| divides pq

if |H|=p, then since p is prime, and p>2, it means H is cyclic

[chilly ocean]

Why RA is left ideal? (A is a subset of R)

[formal ermine]

sylow

@chilly ocean @torn warren lagrange doesn't help here

[torn warren]

right, it can't exclude another subgroup with the same order p

but that's where I learned so far 🙂

still far away from Sylow

[formal ermine]

ye but it doesn't really help

[torn warren]

wow, Sylow is at chapter 36, I am at chapter 10

[chilly ocean]

What's the textbook

[torn warren]

Fraleigh

hopefully I can see Sylow before April

[chilly ocean]

Help this plz

[barren sierra]

what is the definition of RA?

@chilly ocean

[chilly ocean]

{ra|r and a respectively elements of R and A }

[barren sierra]

not quite

[chilly ocean]

R has 1

[barren sierra]

$RA = \left{ \sum_{i \in I} r_i a_i | r_i \in R,~ a_i \in A \right}$

[cloud walrus]

Spamakin🎷

[barren sierra]

it's sums

not just single ra's

this should make it easier to see why it's a left ideal

finite sums btw

[prisma shuttle]

if we let V_i denote the vector space associated to a vertex V in a path algebra for an indexing set I

and we let V denote the vector space of a representation of the path algebra itself

then is V = direct sum of all the V_i?

[white oxide]

Can somebody give me a hint for the forward direction of this proof?

If (1, 1) is a generator of $\mathbb{Z}_n \text{ x } \mathbb{Z}_m$, then the number of summands that gives the identity is $mn$ if and only if $gcd(m, n) = 1$

[cloud walrus]

okeyokay

[white oxide]

so i initially assumed for contradiction that gcd(m, n) = d > 1

so there exist integers r and s such that mr + ns = d

so m/d and n/d are coprime

and i kinda need a hint from there

oh wait i'm stupid

then we can just take mn/d

ok ignore everything

[topaz solar]

Didn’t you recently do a problem about |ab|=|a||b| kinda thing?

[white oxide]

nah

[topaz solar]

Ok it was someone else then oop

[faint fractal]

can someone help me understand the proof to this?

I don't exactly understand how it works

[timber oxide]

Trivially, $N=\bigcup_{n\in N}c(n)$ as $c(n)\subseteq N$ for all $n\in N$.

[cloud walrus]

Timwestlund

[faint fractal]

this means nothing

[timber oxide]

A normal subgroup is defined as a subgroup that is closed under conjugation, hence the conjugacy class of any element in N is contained in N.

[faint fractal]

why does it need to be contained in N?

can you show me an example? Like if we're working in S4, why is it that if we only have a few elements in a conjugacy class, it can't be normal

[timber oxide]

The conjugacy class of n is ${gng^{-1},g\in G}$, for any $n\in N$ we have that $gng^{-1}\in N$.

[cloud walrus]

Timwestlund

[kind jacinth]

List all cyclic subgroups of D5 and D6. Argue that you have found all cyclic subgroups

how do i solve this?

so I know for d5 = {e, r, ..., r^4, s, sr, .. sr^4} where r is rotation and s is reflection

so <e> = {e}
<r> = {e, r, ..., r^4}
<r^2> = {e, r^2, r^4}
<r^3> = {e, r^3}
<r^4> = {e, r^4}
<sr> = {e, sr} (and similar thing for sr^2, sr^3, sr^4)

<s> = {e, s}

would the answer be there are 9 cyclic subgroups?

but idk what to conclude from this

is it that r^3, r^4 and all reflections are cyclic subgroups?

[timber oxide]

If we work in $S_3$ we have $c((1,2,3))={p(1,2,3)p^{-1},p\in S_3}={(1,2,3),(1,3,2)}$. This is a subset of $A_3$, which is the only normal subgroup of $S_3$, which is expected as $(1,2,3)\in A_3$. I don't understand the second part of your question.

[cloud walrus]

Timwestlund

[frank cosmos]

Does anyone know if the group correspondence theorem is in Lang algebra

[shell agate]

<r^2>=<r> actually; since r^5=e, r^6=r, and so <r^2> contains r.

In fact this is true for all r^n (in D5)

So:
— if you got one rotation, you got them all
— if you got one symmetry*rotation (sr or sr^n), you got them all (similar argument)
That makes 4 cyclic subgroups: {e}, <r>, <sr>, <s>.

[kind jacinth]

right, i have to fix the sets as i have multiple mistakes now that i think about it

[shell agate]

For D6, i'll let you work it on your own, but note that <r^3> is not equal to <r> in D6.

[kind jacinth]

so with either of r^2, r^3 ,...i can generate the whole set?

(for D5)

[kind jacinth]

why dont we consider <r^2> a cyclic group as well for example?

oh nvm we just choose the smallest which is r.

[shell agate]

<r^2> is a cyclic subgroup

but it is exactly the same as <r>

[kind jacinth]

i see

thanks! ill attempt D6

[kind jacinth]

for D6 is it {e}, <r>, <r^2>, <r^3>, <s>, <sr>?

<r^4> = <r^2> and <r^5>=<r>

[shell agate]

almost

what is <sr^2>?

[kind jacinth]

oh its {e, sr^2, r^4, s, r^2, sr^4}?

and yeah that would be different from sr

[shell agate]

indeed

<sr^3> as well is different

[kind jacinth]

and after that by argue they want me to prove?

or is the expansion of all enough? (like stating the sets generated by r, r^2, ...., sr, sr^2, ..

[shell agate]

Yes, if you list all the possible generators and show that they are accounted for

[kind jacinth]

alright thanks!

[cloud walrus]

ActiveChapter

[novel parrot]

So M = Ry1 + Ker(v)

i dont understand why the sum is direct (uniquely sum-able)

the last line

I understand that they show that Ry1 cap ker(v) = {0}

[quiet pelican]

Is there a nice way to prove that X^3 - 9 is irreducible in Q[X] that isn’t just unique factorisation of integers?

[sage lodge]

suppose that it was, then one factor must be linear, which implies that it has a rational root, which is false

[quiet pelican]

The way to prove the last bit is unique factorisation

[chilly ocean]

I’d shift x->x+1 and check if it’s Eisenstein at a prime

Sometimes works

[quiet pelican]

I suppose Eisenstein ish method does boil down to unique factorisation in the end, even though it’s closer to what I want

[sage lodge]

what do you mean?

you have the rational roots theorem

can just check that \pm 1, 3, 9 are not roots

[quiet pelican]

Why didn’t I check the x + 1 shift?
I’m sure I did
But it does work

[quiet pelican]

Wait

No

[chilly ocean]

I don’t think it works

But theses similar ideas that might

In any case it’s degree 3 so maybe move to Z then to Z/p and check all cases for some random primes

Like x^3-9 is irreducible in Q iff irreducible in Z, then it’s not irreducible in Z if it’s not irreducible in Z/p for some p

So idk try p =2,3,5,7,.. your computer can find a p that verifies it’s not irreducible mod that prime and that proves it

Since irreducible iff has a root since degree is 3

I think that works

[quiet pelican]

It is irreducible
But it’s not irreducible mod 2

[chilly ocean]

It just has to be irreducible mod some p

2 doesn’t work, maybe 3 or 5 or 7

But once u reduce there’s only finitely many things to check

Because u just check for roots

[quiet pelican]

None of those work

[pliant forge]

given f:X to Y and g: Y to X such that gf = id_X, then we know Y = Im f + Ker g (direct sum). I feel very blind atm but is it the case that Y \cong Ker f + Im g?

[spark crow]

so i am trying to understand how symetric groups work and try to calculate the cayley table however, my symetric group ends up abelian because i am probably using cycle notation wrong. So for me (12)(23) = (23)(12). As the symetric group S3 is isomorphic to the dihedreal group D3 of order 3 i tried to do the same there and i understand that the reflections remain fixed in space. However i dont see how this works with cycle notation. What am I missing?

[summer path]

These are some stacked questions...

[sage lodge]

why does this require unique factorization? @quiet pelican

[summer path]

I don't think 2 is a cubic residue mod 7?

[quiet pelican]

Oh yeah
Brain not working

[summer path]

(12)(23) sends 1 to 2; (23)(12) sends 1 to 3, so no they are not equal

[formal ermine]

are two rings isomorphic iff their prime spectrums are homeomorphic?

I know the => direction is true

but what about the other one

[spark crow]

weird i think thats wrong, double transpositions should be the same as 3 cycles, so now i get (12)(23) = (123) and (23)(12) = (132)

and yes i know they are not equal but my question was probably poorly phrased

[elder wave]

i don't really know but i feel like this isn't true

[summer path]

If you want to show s3 and d3 are isomorphic, you can do it via cayley table as you mentioned or you can kind of just look at the definition of Dn and try to write down a group homomorphism to Sn

[elder wave]

maybe look at local rings

[spark crow]

probably better phrasing would be, i try to look at the group D3 and if i draw a triangle with numbers denoting each point starting at the top going clockwise. if i do (123)(12) i get the same result as if i would do (12)(123), however as the reflections axis are fixed in space i can't just label the reflection between 1 and 2 just (12) cause after a clockwise rotation it would just transpose (13)-

[elder wave]

ah nvm my argument wouldn't work

[spark crow]

no i dont want to show that these groups are isomorphic, I actually want to show that the word problem for S3 can be calculated by a circuit, however i don't know how to encode it properly because i heavily rely on my drawn picture right now, and don't see a way to currently encode this, without making it accidentaly an abelian group

[formal ermine]

this would only "make sense"

because otherwise we wouldn't be able to like

go back from geometry to algebra

[elder wave]

since you'd get two point spectra and there could be different topologies

[summer path]

Im not sure what a circuit is

[spark crow]

not important but its kind of like a program using only boolean gates

[summer path]

I see

Well you can generate S3 (or D3) by elements (12) and (123) noting that (12)^2 = e and (123)^3 = e, and (12)(123)(12) = (132) = (123)^{-1}, I'm not sure if this helps you though

[formal ermine]

answer is no. timo showed me a counterexample in dms

[elder wave]

I'm gonna put this here in case i'm wrong for someone to correct me: Fields all have spectra that consist of one point, those should be homeo but not all fields are iso as rings

[spark crow]

ah yeah this definition is the one that started this whole mess :), i dont know why i just think i maybe misunderstood one singular thing which has spiraled down the line, and i just can't figure out my real problem

[summer path]

I also don't think Eisenstein will work here at least not of shift ax+b

[spark crow]

ok maybe i can formulate my problem clearly this time. if you have (12)(123) = (123)^{-1}(12)^{-1} you can then multiply both sides with (12), (12)(123)(12) = (123)^{-1}, till here i understand it but now i should be able to multiply again with (123) resulting in (12)(123)(12)(123) = e, so i dont know where the error is now plugging in 1 i get (1->2->1->2->1) so 1 -> 1
plugging in 2 (2->3->1) so 2 -> 1 which is obviously wrong, and 3 (3->1->2->3) so 3->3 is alright. so where is the error. (123)(132) = e should be correct. so multiplying by (132)^{-1} definitly is (123) = (132)^{-1} so i am probably doing a simple mistake.

Probably important: if presented (12)(123) i first calculate (123) then (12) as those are to my understanding just permutations so if f=(12) and g=(123) i would calculate f dot g.

[empty rose]

you forgot to apply the last (12) to 2

[spark crow]

thanks

[empty rose]

start at 2, apply (123), you get 3, apply (12), you get 3, apply (123), you get 1, apply (12), you get 2

[spark crow]

and it is such an obvious mistake, crying

[empty rose]

well... there are two types of "obvious"

there are some problems that you can just look at and you immediately know the solution, it's obvious how to solve it

there are some problems where if you're told the solution, it's easy to check that it's correct

i told you exactly what the mistake was and you can easily see that i'm right, but that doesn't necessarily make it easy to find it in the first place

[spark crow]

i know, but its literally like calculating 1+2+3 and your wondering why it is 5 because you calculated 2+3 and forget to add 1 in the end

[empty rose]

i mean aren't all mistakes like that? anything you can do that's incorrect is eventually just "you lost track of something and didn't do a step you should have"

[spark crow]

true, in a sense

[empty rose]

it's not really the same as 1+2+3 because, these aren't numbers, they're permutations of 3 things
if you have more experience with numbers, like most people do, then permutations are harder to deal with

...and there is also just more steps here, which makes it easier to do one of them incorrectly

[spark crow]

its just a lazy mistake, i now realize i just shouold have checked with either one-line or two-line notation, which would have been clearer i guess

but thanks for cheering me up 🙂

[empty rose]

:)

[spark crow]

and now my other notation also makes sense if i define D3 = <r,s | r^3 = e, s^2 = e, srsr = e>, then rs = sr^2 should be correct. I don't know how to confidently calculate it yet but if thats right i have the right direction

[opal atlas]

I got stuck with this problem in my book. I saw this and straight away my brain decided to not understand anything. Is there someone here who could explain to me how to approach a question like this?

[spark crow]

seems to like the question is incomplete, assuming m,n are intergers (her is missing the set m,n belong to) what would d = (m, n) represent? if d were just a normal tuple the set at the bottom seems strange as why would n dash be included? Further in the set definition what does a in m mean. if the assumption m is integer was correct this would be a semantic error. I don't know what /n should refer to so no clue.

Typically if you want to show that a set A is a true subset of B, you would first show A is subset B or rather forall x in A follows x in B, and then you would pick an Element in B, which is not in A showing inequality of the given sets

[opal atlas]

Thank you so much. So it is not my fault that I am very confused. I am gonna see if I missed something somewhere in this chapter that could give some more information

[spark crow]

Well I could be wrong, cause I don't know either but out of context this just looks wrong so, maybe in some prior chapters there were made some definitions like, "from here on we will refer to gcd(m,n) as (m,n) for simplicities sake", however i find that unlikely, but in the given Context as it is taking about groups and subgroups i would think that they may be talking about subgroups of (Z/nZ^{x}, *) i'm not sure on the notation here but meaning all invertible classes under multiplication maybe. So all in all this question strikes me as very odd. But as i implied before I am just learning this stuff myself. So good luck to you.

[opal atlas]

Thank you!

[spark crow]

Given D3 = <r,s | r^3 = e, s^2 = e, srsr = e> as the dihedral group of order 3, rs = sr^2, i now have a calculation which i think works, but would there be a better way then

rs = rs(e) = rs(srsr) = r(ss)rsr = rrsr = rrs(e)r = rrs(srsr)r = rr(ss)rsrr = r^3sr^2 = (e)sr^2 = sr^2

[formal ermine]

what book is this from? it should be prove not proof

[opal atlas]

That's my fault, i translated it to English. 😉

[spark crow]

it should also be "Prove that .... is a subgroup" i think

[formal ermine]

that notation is weird

and just the question in general

"m, n in *nothing*"

[opal atlas]

I searched a bit through the chapter, and I am guessing that this is what they meant.

[formal ermine]

is that a proper subset

[frank cosmos]

Why does A/H=B/H -> A=B

[novel parrot]

I think they are isomorphic, not equal

[frank cosmos]

A/H is equal to B/H as a set

A and B are subgroups of the same group G

[novel parrot]

oh

idk

Rank is defined as the maximum length of a linearly independent list of elements

if $rx + b0 = 0 = rx =0$ since x is torsion r non zero exists such that rx = 0, so not linearly independent

simple as that right?

[spark crow]

hmm in this context i would have assumed "your set" <= Z/nZ my books use U <= G to denote subgroups, it is strange as your set only contains multiplicative inverse integers under mod multiplication, so all elements in your set would be regular integers, however the elements in Z/nZ are not integers but themself sets of integegers better known as congruence classes

[cloud walrus]

ActiveChapter

[torn warren]

@white oxide

[formal ermine]

isomorphism theorem

[novel parrot]

can you check my proof?

[frank cosmos]

how to prove without

im using this to prove the isomorphism theorem

[formal ermine]

where is it?

[novel parrot]

.

[novel parrot]

.

and any x from TorM, there is non zero r such that rx = 0

so not 1 element can be linearly independant

so rank 0 for TorM

then for next part

rank(M) = rank(M/TorM)

if M has n length linearly independant list, then its also L.I in M / TorM (no elements of that list can be in TorM)

and M/TorM cant have any larger rank, if M/TorM has rank m > n, then that list is also L.I in M

so basically sps that {s_i} is L.I in M/TorM

then sum{a_is_i} = 0 implies that a_i is 0

in M/TorM that means that a_i is in TorM

so translated to M that is, sum{a_is_i} = x where a_i are torsion and x is torsion

so multiply by ring elements to make it = 0

and see its L.I in M

@formal ermine makes sense ?

[opal atlas]

This is from is syllabus written by my lecturer. I think that is why some things just don't make sense

[pliant forge]

below im computing the radical of the representation of the given quiver.
its not k^2 i suppose because the of the relations its quotiented by?

[barren sierra]

I want to show that for some m the functor F: Ab -> Ab mapping A to A/mA preserves coequalizers. That's is what I have but I'm stuck tbh

Can I say that square involving B and C commutes, then get a map Q -> Z

Then project that down to a map Q/mQ to Z?

I don't think so right?

[delicate orchid]

do we not have a map from Q to Z by universal prop of the coequaliser?

[barren sierra]

We do yes

Didn't add that to picture for readability

Well uh

Do I?

[delicate orchid]

pi_c \circ h induces the map no?

[barren sierra]

Can I get that h o pi_c o f = h o pi_c o g?

[delicate orchid]

main thing is if you have a commutative diagram and shove the entire thing through a functor the result is commutative by functoriality - that would be the route I'd go

[barren sierra]

Can I do that? I thought I'd need to start with C/mC to Z, not like C/mC to Z/mZ which is what I'd get from throwing the diagram into the functor

[delicate orchid]

ah good point the functor might not be essentially surjective

not sure then

[barren sierra]

Yea

Hence me being stuck lol

[delicate orchid]

I think so, this diagram commutes for all Z by assumption

I write my compositions backwards to you btw

apologies

[barren sierra]

Ya I got that dw

Uh

Wdym by assumption

I don't see how we can assume this

[delicate orchid]

we're assume (Q, q) is the coequaliser of B, C yeah?

[barren sierra]

Yea

[delicate orchid]

ok then this diagram commutes by the universal property

[barren sierra]

But we start with h o F(f) = h o F(g)

That's our starting assumption

[barren sierra]

How do we get this from that

[delicate orchid]

right so we can't assume it's the coequaliser? I'm confused

[barren sierra]

That dotted arrow you drew exists if h o pi_c o f = h o pi_c o g

But we don't necessarily have that, we start with h o F(f) = h o F(g)

[barren sierra]

I guess what I'm asking is does the square here commute

[delicate orchid]

I see I see ok

[barren sierra]

If it does then I believe you

[frank cosmos]

Why is the bottom thing true ??

[barren sierra]

But I don't think it necessarily does?

[delicate orchid]

well C/mC is a quotient of C isn't it? can we do something with that

[frank cosmos]

=H

because I can pick a coset representative in [h] that is not in h?

[delicate orchid]

no, you can't - it's H/N not G/N

[frank cosmos]

talking to me?

[delicate orchid]

yeppers

if it was gN \in G/N you would be right

[frank cosmos]

im finding elements in G such that [g]=[h] for some h in H right?

[delicate orchid]

if g wasn't in H then gN would contain e*g = g, which cannot be in a coset of H (as g isn't in H)

so the set can be at most H

[frank cosmos]

why not

[delicate orchid]

the cosets in H/N form a partition of H, i.e. they are all properly contained within H

[frank cosmos]

oh right they are disjoint

[barren sierra]

Well the more important part here is containment

Not necessarily disjointness

[frank cosmos]

ye

[barren sierra]

We have the canonical projection pi_c but idk what else we can do.

Do I have F(f) o pi_B = pi_C o f?

[delicate orchid]

ughghugguhg there's an adjoint as well if I remember your previous question

main question is wtf is F(f)

is it just f -> f+mC lol

[frank cosmos]

so lets say if X is a normal subgroup and A and B are two not-nessecarily normal subgroups containing X, then A/X =B/X implies A=B since if y is in A but not B, then y is in [y] but is not in any coset of B/X which is a contradiction.

is this proof correct

[barren sierra]

I think

[delicate orchid]

ok then just check this lol

like, see if it sends the same element in B to the same thing in C/mC

[barren sierra]

I think it's true but idk if I'm doing something dumb. seems to be true

[delicate orchid]

I'm pretty sure it's true as well

[barren sierra]

Ok ok then I think I can try to go from here

Cause then universal property kicks in

And I get more stuff to play with

[delicate orchid]

omg we're idiots

[barren sierra]

I may be but you aren't

[delicate orchid]

there's a diagonal map induced by the universal property of the quotient

which is unique

and thus equal to every possible composition

[barren sierra]

Omg

[delicate orchid]

B -> B/mB -> C specifically

[barren sierra]

Yea

[delicate orchid]

maybe we could, Wew Lads... maybe we could

[barren sierra]

Accurate title for sticker

[frank cosmos]

wew?

[barren sierra]

I think that looks ok??

[frank cosmos]

ty

[barren sierra]

Wait but I can't necessarily turn a map Q -> Z into a map Q/mQ to Z

That was my other question cause of that square commuted then I knew we could get a map Q -> Z but that's not what I want at the end

[delicate orchid]

So you have another funny commutative square on the right

Not sure why I replied to that diagram oh well

Which then commutes in a way completely analogous to the square on the left

[barren sierra]

Oh

[delicate orchid]

And then, induced map to Z from Q being the co equaliser (I will not let this one go) and induced map from Q/mQ to Z make a little commutative triangle I think

[barren sierra]

Not let what go lol

[delicate orchid]

U keep denying the existence of the funny map Q -> Z

[barren sierra]

It was more I couldn't figure out how to justify it's existence

At the time

[delicate orchid]

You wrote at the top “q is the coequaliser of B, C”

[barren sierra]

Funny map exists if these compositions exists are equal

So showing those compositions are equal reduced to remembering that square at the top left commutes

[delicate orchid]

It exists regardless ^ it might just not play nice

[prisma shuttle]

is a path algebra the direct sum of the vector spaces of each of its vertices

[coral spindle]

No

As a vector space, it has a basis given by the paths (including the trivial paths), not the vertices

But this ignores the algebra structure

[fast stratus]

Can someone give me a counterexample when G is infinite

[narrow bronze]

G = ℤ and S = ℕ

[fast stratus]

Looks good?

I still didn’t understand the hint… what author is suggesting

[molten viper]

So, a vector space is a field equipped with a scalar multiplication, which seems similar to a module as I understand them, is there a relationship here?

[formal ermine]

a module is a vector space but over a ring (you can't divide by scalars)

so every vector space is a module

but not every module is a vector space

[molten viper]

Ahhhhh

Modules are the weak link in my understanding

[coral spindle]

A vector space isn't a field equipped with a scalar multiplication, no. A vector space is an Abelian group with scalar multiplication from some field.

When we talk about vector spaces, typically we fix a field. After that they are Abelian groups with scalar multiplication from the field that we've chosen.

[molten viper]

Hm

I’m a bit confused now

So it’s a module over an Abelian group and a field?

[coral spindle]

No!

Fix a ring R. An R-module is an Abelian group with multiplication from R.

A vector space is a k-module where k is a field.

Is this clearer?

[molten viper]

Could you give an example?

Like a Q module ig

[coral spindle]

What's your favourite Abelian group?

[formal ermine]

the trivial one

[molten viper]

Let’s go with Z_5

[coral spindle]

OK

Z_5 is a Z-module

The multiplication from Z is just repeated addition.

I will denote the elements of Z_5 as [0], [1] etc

Then 2[1] = [1]+[1] = [2]

This is the Z-module structure of Z_5.

[molten viper]

I see

So we take the elements of the group, and then “apply” multiplication from our favorite ring

[coral spindle]

Yes

There are certain rules for the scalar multiplication; look up the definition of a module to see them.

Not sure why I'm getting comments from the peanut gallery; Z-modules are the simplest possible example of a module.

[sonic coral]

since Cauchys Theorem guaranteed an element of order p that divides G, does that also guarantee a subgroup of order p, namely the cyclic one generated by said element?

[coral spindle]

Yes

[formal ermine]

lol

I mean I was just amused by

"give me your favorite abelian group"
responds with Z module

[coral spindle]

Hey, they don't know that Abelian groups are just Z-modules yet 🙊

[prisma shuttle]

why are the irreducibel representations of a path algebra just the one dimensioanl representations

i've been trying to prove this claim for quite a bit of time

ima like 90% sure its true

clearly 1 dim --> irreducible

but the otehr direction i can't seem to get

[torn warren]

the first line should be $\forall a\in S$, the last line should be $a^{p-1}\in S$

[cloud walrus]

Witness

[narrow bronze]

Consider the map G → G; x ↦ aᵢx

[fast stratus]

I meant to write… Suppose a in S

[rotund aurora]

What's the method you guys use for calculating the characteristic polynomial of a matrix?

By hand I mean

[sonic coral]

depends on the size

[rotund aurora]

3x3

or 4x4

[sonic coral]

there’s a formula. although it’s probably just as much work as the standard way to get it

[rotund aurora]

actually

[south patrol]

3x3 is okay by just Leibniz formula, 4x4 I never do lol

Idk why you'd ever need to do a 4x4 by hand unless it's a special case or you have sadistic teachers

[rotund aurora]

nah you are right

I have never done a 4x4

xDD

[rotund aurora]

Leibniz is sadistic tho?

unless you mean rule of sarrus

[summer path]

I've had to take of 4x4 matrix for an ode exam before, the numbers weren't very nice either but at least they were integers I guess

[rotund aurora]

oh like calculating 4x4 determinants I have done in exams, but just integer type determinants, no transcendental x's involved (which is just doing Gauss-Jordan)

[rotund aurora]

for 3x3 characteristic polynomial, you have two coefficients, already, which are the trace and the determinant

the other coefficient Im not sure how to obtain it from the original matrix in a straightforward procedure (it is related with minors and so on)

i think a (-1)^k might be missing there

https://en.wikipedia.org/wiki/Faddeev–LeVerrier_algorithm

Ill go to sleep

will work on this tomorrow, calculating characteristic polynomials is no joke

[torn warren]

but this is equivalent to the author's hint, right?

I didn't get the hint either

[south patrol]

Isn't rule of sarrus longer lol

Idk the poitn of rule of sarrus

Oh sorry

I meant Laplace expansion

[rotund aurora]

Ah ok yeh

[narrow bronze]

The map is bijective

[rotund aurora]

Leibniz is

But its sometimes useful tho

[south patrol]

hm

I think it's useful occasionally for conceptual things i guess?

Like you can prove that det is multiplicative using it

And lol, Stickelberger theorem

[rotund aurora]

Problem: given 1000 cows of positive weight, prove that you can always remove one cow so that the remaining 999 cows cannot be split into two groups of total equal weight

[south patrol]

Tf

[rotund aurora]

xD

[south patrol]

Oh i assumed that was a meme initially

Is this actually an interesting question

[rotund aurora]

No idk

[south patrol]

hm

[rotund aurora]

But I had fun when I solved it

[south patrol]

so yes xd

I'll have a think cause I feel like i've seen this before but never actually tried

[rotund aurora]

And you use leibniz

[south patrol]

Oh what lol

Oh like to show that a certain determinant is non-zero or smth? idk

[rotund aurora]

Yeah

[south patrol]

hm interesting

[rotund aurora]

You have some weird system of linear equations

[south patrol]

and i assume this is true for any n rather than 1000 lol

ye

[rotund aurora]

And you always have the trivial solution

But you want no more

[south patrol]

or do you need any hypotheses on stuff

like mod 4 etc

[rotund aurora]

So det should be nonzero

[rotund aurora]

For n even yes

For n odd not necessarily true

[south patrol]

Ah okay thanks

Okay tbf you have nerd sniped me cause i have hw iminently due

but thanks for the problem

[prisma shuttle]

does anyone get this

or know a better place i can ask this question

[torn warren]

yes, it is bijective, then how is related to the S and the inversed element ?

[narrow bronze]

If it is bijective, in particular it is surjective and e is an element of the image

[runic turtle]

anyone here can help with logarithims?

[torn warren]

yes, but how to conclude the inverse is in S ?

if $a\in S$, and $a\neq e$, we can find $x\in G, ax=e$, but how to show $x\in S$ ?

[cloud walrus]

Witness

[torn warren]

it needs to show this map is bijective from $S\rightarrow S$, then it is done, but right now we only have $G\rightarrow G$, which is not enough

[cloud walrus]

Witness

[torn warren]

since S is finite, to show it is bijective, we only need to show it is one-to-one

so it is done

[narrow bronze]

Yes

Wait no

[torn warren]

$\forall x,y\in S, ax=ay\rightarrow x=y$

[cloud walrus]

Witness

[narrow bronze]

You also need that the image of S is S

[torn warren]

S is finite, so one-to-one implies onto, right?

[narrow bronze]

Which you have because S is closed by the product operation

[narrow bronze]

But you don't know that S -> S
You only know S -> G

Perhaps S could be mapped to a different set of the same cardinality

[narrow bronze]

But this

[torn warren]

but to prove it is onto, it needs the condition that the inverse is in S

so we get a logic loop

[narrow bronze]

No

First, we have $f(S) \subset S$ because $S$ is closed. Then, we use that $f$ is injective to show that $|f(S)| = |S|$. It follows that $f(S) = S$

[cloud walrus]

d឵

[torn warren]

is there a way to directly prove it is onto?

[tender wharf]

yes, such people do exist. whether they want to help you is a different question

now tske yourself to #prealg-and-algebra

[tender wharf]

I guess if you have the definition of f?

[torn warren]

this is the op, the f is defined as for a fixed $a\in S, f(x)=ax$

[cloud walrus]

Witness

[torn warren]

wait, what if we remove the given condition that $e\in S$, is this hint method still working?

[cloud walrus]

Witness

[torn warren]

without $e\in S$, the cyclic group method still works

[tender wharf]

you can observe it behaves a little like a left coset

[cloud walrus]

Witness

[torn warren]

yes, similar

[tender wharf]

I don't see any f defined in your question

that you posted

[torn warren]

it is an equivalent way to the author's hint

just rewrite author's hint

[white oxide]

say we let $H = {h \in \mathbb{Z} \mid r_i | h \text{ for } i = 1 \dots n}$where the $r_i \in \mathbb{Z}^+$.

[cloud walrus]

okeyokay

[white oxide]

would the members of H be multiples of r_1r_2....r_n?

why is the generator of that group the least common multiple of all of the r_i

because my idea was that

since the subgroups of Z are precisely nZ for n in Z

this subgroup has to be of the form nZ

so my thought was that each member of this subgroup are of the form r_1r_2....r_n(k)

so it's equal to r_1r_2....r_nZ

but the thing is i don't know if that's true

i don't see why this is true

[tribal moss]

Do you agree that by definition your H contains exactly all the common multiples of the ri's?

[white oxide]

yes

[tribal moss]

And have you proved that a nontrivial subgroup of Z is always generated by its smallest positive element?

[white oxide]

oh no i havent

that would make sense then

i should prov ethat

[tribal moss]

I would assume that is what "our work on cyclic groups" refers to.

[white oxide]

yeah that would make sense

ok i'll prove it, ty!

[vagrant zinc]

Guys, could someone explain this theorem to me with an example? , pls

something more basic please

[south patrol]

Indeed there is an embedding F_{p^m} into F_{p^n} iff m|n

So yes

Oh sorry was dumb

Yes in your case

Lol

Basically you can take the set of roots of x^9 = x in any field of order 81

And that'll be isomorphic to F

If you look at the construction of finite fields of order p^n then you'll see whether this comes from

Np!

[tribal moss]

You can also go the other way: Find an element c of F_n that is not a square (they can't all be squares except in characteristic 2), and adjoin a square root of that. Then you have a field with n^2 elements.

[south patrol]

Oh yeah true here lol, good point

[vagrant zinc]

[south patrol]

Yes sorry sure so this is actually almost a definition basically

Let me give an example for Z/nZ

Well, it's basically the general case

[tribal moss]

The "(n,k)" must mean gcd(n,k).

[torn warren]

it should mention cyclic group @vagrant zinc

[south patrol]

But note that the order of k mod n is simply the smaller positive a such that ak is divisible by n. For such a, ak is the lcm of n and m. So a is lcm(n,m)/k, which can be seen to be equal to n/gcd(n,m) if you look at prime factorisation, for example

Though there will be more

[south patrol]

Why?

This only depends on the subgroup generated by a

Which is always cyclic and order n by hypothesis

Maybe you read it as saying G has order n rather than a?

[torn warren]

it says G has order n, didn't say G=<a>

[south patrol]

No, it says a in G has order n

So a has order n

Arguably not the best writing, but yeah

[vagrant zinc]

Ok I was taking the example of Z_4 under the operation sum and then I didn't understand, I know you are talking about the order of the generator of the group or subgroup but in this case I am taking it as the group, but then one I don't understand
as it would be the comparative with respect to the gcd(n,m)

I don't know, if what I said is wrong, sorry for being wrong.

[tribal moss]

Concrete example, perhaps?
2 has order 39 in Z/78Z.
Its 6th power is 12, which has order 13 in Z/78Z. And 13 happens to be 39/gcd(39,6) = 39/3.

[torn warren]

you can first prove <a^k>=<a^d>, d=gcd(n,k) @vagrant zinc

for <a^d>, its order is n/d

so as the order for <a^k>

[barren sierra]

Consider a cyclic group $G$ of order $10$ . Suppose it's generated by an element $a$, so we have that the order of $a$ is $10$. Let's consider $a^2$. We have that $\frac{10}{\gcd(2, 10)} = \frac{10}{2} = 5$ and indeed we have $\left(a^2\right)^5 = a^10 = e$. Also note that for all $m < 5$ we have $\left(a^2\right)^m \neq e$ so indeed the order of $a^2$ in $G$ is 5.

[cloud walrus]

Spamakin🎷

[barren sierra]

oh shit someone already was helping >_>

[barren sierra]

yea it's awful notation that's quite popular

(x, y) for gcd and [x, y] for lcm

[vagrant zinc]

Sorry, I was organizing my desk

Thanks guys, now I understand

❤️

[barren sierra]

People helped me with this yesterday 😭

How do I show this is surjective

this is an injective map, I showed that