groups-rings-fields

[fading acorn]

0$ on Lang algebra

grad algebra thing

[barren sierra]

do you mean like a textbook for a proof based linear algebra course?

or have you already taken a proof based linear algebra course?

If the former, then I like Friedberg, Insel, and Spence's Linear Algebra textbook alot

if the latter, most every abstract algebra textbook will cover some multilinear algebra, and that's what I would call "advanced linear algebra" after a proof based course in linear algebra

[karmic moat]

hoffman and kunze is good

[hollow fjord]

I need some hint for the last part of this problem

[tardy hedge]

C(x)R C and C(x)C C are not isomorphic as R-modules because the first is free of rank 4, and the second is free of rank 2?

[rocky cloak]

Notice that as an abelian group O is equal to ZxZ since it's all combinations of 1 and omega.

Then for a finite index subring R, what can the order of omega be in O/R? Which group is O/R then?

[tardy hedge]

Why can't two finite free modules of different rank be isomorphic? In dummit and foote they prove it by reducing to the situation with vector spaces and using the result there, but is there a way of seeing it more directly?

[untold torrent]

Is this statement true?

[rocky cloak]

So this is something that actually isn't for general rings.

But it is true for commutative rings and it is true for Noetherian rings.

For commutative rings you can prove it by reducing to fields as you say.

For Noetherian rings, if R^n = R^m with n>=m, then you can consider the endomorphism where R^n projects down to the first m coordinates and then use the isomorphism R^m -> R^n.

Then this is a surjective endomorphism of Noetherian modules. It is a general fact that a surjective endomorphism of a Noetherian module is an isomorphism (try to prove it yourself, hint:||consider the kernel of f^k as k goes to infinity||)

[rocky cloak]

For commutative rings this can also be proven with the determinant and adjugate matrix if you don't want to reduce to fields

[untold torrent]

How is this true? Please explain in easy way

If I take z6 and at x=3 and d=2

@rocky cloak

Sorry for a ping🙏

[rocky cloak]

So say
{1, g, g^2, g^3, g^4, g^5} is our cyclic group of order 6.

Then the solutions to x^2 = 1 is 1 and g^3, exactly 2 solutions.

In general say g generates the cyclic group G. Then g^k is a solution to x^d = 1 iff (g^k)^d = g^kd = 1 which happens iff kd is a multiple of n which happens iff k is a multiple of n/d. And there are exactly d multiples of n/d between 0 and n

[untold torrent]

Thanks for a beautiful explanation

Your PFP is cool!!

Pika pika😅

[grand yarrow]

is the center of a frobenius algebra a frobenius algebra? if so what's the comultiplication?

[abstract silo]

oh uh what happened

[candid patrol]

C* is algebraically closed

[minor gazelle]

aren't those proof actually routine? after a couple of times they just differ in symbols? like verifying something a group or homomorphism

[rustic crown]

yep it pretty routine, and you'll see such stuff for other algebraic structures as well, e.g. vector spaces, rings, modules, etc. one funny thing would be to generalize all these little propositions as a big proposition :p

[minor gazelle]

Another question is it considered okay ish for rigor if this definition is provided because, with decomposition there is no need to discuss anymore

[rustic crown]

there's one step missing in there

[minor gazelle]

generalized associative property?

[rustic crown]

a priori x^(a-b) = x^a * (x^-1)^(b) only holds true when either one of a or b is 0.

b = 0 is obvious, a = 0 is the definition of x^(-n)

[minor gazelle]

So, but why they have to be 0? I am a bit confused 😵‍💫

[rustic crown]

oh i wrote it badly ><

in the line x^((a+c)-(b+d)) = x^(a+c) * (x^-1)^(b+d) we're using something more than the definition

[rustic crown]

which was this statement where a, b are any non-negative integers.

and we only know why this statement x^(a-b) = x^a * (x^-1)^(b) is true when either a = 0 or b = 0.

we do not know it in general without a small check

[minor gazelle]

I have this, but I dont know if it would justify it

[rustic crown]

not quite

this is like this one annoying that one needs to deal with when definiing integers as well

[minor gazelle]

will case discussion fix it?

[rustic crown]

if you define Z := N ∪ -N ∪ {0}, then it's a lil annoying to verify why the operations on Z are still nice only from the knowledge of operations on N

a stupid way is to just take a million cases and verify for example that + is still associative on Z

[minor gazelle]

So I prove a proposition that x^{mn}=(x^{m})^n will it work?

its like negative part that can't be justified

so if I can justify x^(a-b) = x^a * (x^-1)^(b) then will it work

😵‍💫

[rustic crown]

we don't need to use the other statement, this will follow easily from prop 1.4

consider the two cases a >= b and a <= b

but yea having that other statement doesn't hurt ;3

[minor gazelle]

I am a bit curious is this statement true or provable? proposition x^{mn}=(x^{m})^n it would be quite powerful i feel

[rustic crown]

yep it's true (and provable) :3

[rustic crown]

one can do a simple induction on n to prove it for n >= 0 and then handle n <= 0 from the first case and using the fact that
x^(-n) = (x^(-1))^n = (x^n)^-1

which is proven directly using the definition of being an inverse

[hazy pollen]

hi guys

im Isomorph

but isomorphed

try to decipher my username

display name*

[dim wagon]

I just learnt about this and was wondering if this would still hold if G was finitely generated and I replace all instances of for all g with for all generators g_j?

[dim wagon]

I think doing these would imply the first 3 parts of the original proposition but am not sure about the 4th

[rocky cloak]

You may have some trouble with point 2.

But I think you can fix it by having it be all v instead of just vi.

[minor gazelle]

is it actually always a good idea to break down a problem into several small propositions and then prove the more general statement by using all the small propositions? some propositions are quite trivial, but they always streamline proof. It's so fun and trivialize (also streamline effort for other problems ) . It also make the proof more clean and neat..

[minor gazelle]

is this identity actually true?

[surreal dagger]

yep

as a set

[minor gazelle]

So interesting, I was wondering and it wasnt hard to prove 😭

[hollow tartan]

What book is this. Love the formatting

[surreal dagger]

Fun! Usually you cant expect the union of subgroups to be a subgroup. Maybe you know the situation from e.g. vector spaces already, but if you have subgroups H,K of G then the union of H and K is only a subgroup under certain conditions. You could think about under which conditions the union is a subgroup and on the other hand look for a concrete counter example in e.g. the integers Z when the conditions arent met.

[tribal moss]

(The union of just two subgroups is the easy case here).

[white oxide]

Can somebody explain to me why the paragraph between the two red lines is needed?

[minor gazelle]

Elegant book on overleaf (dead project I think I asked someone to get it)

[minor gazelle]

I got that 🥰🥰 still very interesting even the equality of the sets.
Because G is not necessarily finite or cyclic but we write it as a union of a bunch of cyclic group

[kind temple]

h : G —> H, 𝛟 : F(\coprod_𝛂 G_𝛂) -> H, where F is the free group and \coprod is being taken in Set (so just the disjoint union)

G is a quotient of the domain of 𝛟, and the red parts show that 𝛟 is a homomorphism that descends through the quotient to h.

honestly, i much prefer the diagram chase proof of the universal property of the free product after showing that every group has a presentation

[tardy hedge]

Not sure what u mean by “then use the isomorphism R^m->R^n”

Oh wait

I get it

Nvm i actually dont know why this shows n = m

[rocky cloak]

So you have a projection
R^n -> R^m
just ignoring the last n-m coordinates, composing with the isomorphism R^m -> R^n gives a surjective map R^n -> R^n.

Little exercise: if M is Noetherian and f: M -> M is surjective, then f is also injective.

This means R^n -> R^m is injective so n-m is 0

[minor gazelle]

Algebra is fun, like how to using a bunch propositions to prove everything, with more and more trivial outcomes it accumulates in such a way I can cite different propositions for proving a lot of things.. and helped me with memorizing too.. that’s such a fun game like building sand castle. Trivial outcomes become so non-trivial when combined

[tardy hedge]

Thats why i like algebra too

[minor gazelle]

Like playing legos, I felt this for first time 🥰 and it had been so fun, like more Legos you have the more you can do

[twilit wraith]

Yeah, thats what "lemma" means actually

A more technical statement used to cleanly prove a larger theorem

[misty holly]

I don't understand the notation in the second bullet point

does it mean that H_i is disjoint from the smallest subgroup of G containing the rest of the H_j?

[misty holly]

right

is there some easy way of doing this other than just knowing it/guessing?

I only know three, Z_8, Q_8 and D_4

should I use Cayley's theorem?

oh Z_2^3 is also different

[amber plank]

fundamental theorem of finite abelian groups, plus Q8 and D4 gets you all of them

otherwise - think about direct products and how you can build groups out of Z_2

[misty holly]

We haven't done the fundamental theorem yet

But yeah Z2^3 and Z2 x Z4 are two others I can construct

which makes 5

I can construct the abelian groups using the fundamental theorem of abelian groups, which gives Z8, Z2 x Z4 and Z2^3.

To make the non-abelian groups:
We know there can't be any element of order 8, as then the group would be cyclic and hence abelian.
It needs to have at least one element of order 4, otherwise the group would be boolean and hence abelian (and isomorphic to Z2^3).

Let a be an element of order 4, and let A = {e, a, a^2, a^3}. By Lagrange's theorem, A has two cosets in G: A and G \ A = bA for some b in G \ A.
Then, {a, b} generates G.
If |b| = 2, G is isomorphic to D_4.
If |b| = 4, G is isomorphic to Q_8.

Does this argument work?

Oh I guess we need to see how a and b interact with each other

Since [G: A] = 2, bA = Ab so either ba = ab or ba = a^3 b. It can't be the first otherwise the group would be commutative
Then:
(i) if |b| = 4, G \cong Q_8
(ii) if |b| = 2, G \cong D_4

[misty holly]

I think D_2n is isomorphic to S_n x Z_2

In a regular 2n-gon, we can permute all the longest diagonals, and the Z2 corresponds to a normal/flipped orientation

[coral spindle]

I'm afraid not.

If indeed by S_n you mean the symmetric group

I am rather confused by this question because it's actually quite misleading lol

[misty holly]

yeah on second thought my reasoning does not work at all

we cannot possibly permute all the body diagonals using only rigid motions

[coral spindle]

In fact even worse

these groups don't even have the same size

S_n x Z_2 has n! . 2 elements

D_2n has 2n elements

[misty holly]

lol right

[coral spindle]

so for example for n=7, you are out by a factor of 720

[misty holly]

D_2n has 4n elements but yeah

[coral spindle]

Uh you sure about that?

Oh wait I see

Yeah ok, a bit unusual notation, but no problem

[misty holly]

Here D_2n is defined as the group of rigid motions of a 2n-gon, not an n-gon

yeah

[coral spindle]

I'm not sure if they have any particular statement in mind about D_2n, it's very strange that they put this in the question like this, unless they wanted you to disprove something in general

Oh wait I see the issue.

I think they actually wanted you to prove:
$$D_3 \times \mathbb Z_2 \cong D_{6}$$

[cloud walrus]

Boytjie

[coral spindle]

And they just made a typo.

Try that

[misty holly]

D6 \cong S3 x Z2 is proven in a previous example

[coral spindle]

And you've shown S_3 \cong D_3?

Well then

I think I said too much then!

[misty holly]

Oh

Ohh I see

We can embed two regular n-gons in the regular 2n-gon by connecting every alternate vertex, and applying any rigid motion to a chosen one of those two n-gons gives us an isomorphism from D_2n to D_n x Z_2

Hmm actually

It also depends on if n is odd or even

Like if we flip one of the embedded n-gons it doesn't correspond to a rigid motion of the 2n-gon...

Yeah honestly I cannot figure this out

[coral spindle]

My hint is that this gives you the full picture in general if you notice the right generalisation

[misty holly]

Oh well we can take H = {id, r^n} and K = {id, r^2, …, r^{2n-2}, s, r^2 s, …, r^{2n-2}s}

But I don’t see it physically using the 2n-gon

[rocky cloak]

I can make a reasonable conjecture / true observation for ||n odd||, but I'm not sure if it becomes anything meaningful for ||n even||

[misty holly]

Yeah I think this only works for n odd

[coral spindle]

I have no idea how to see things physically

I just see it abstractly

[misty holly]

Fair enough

[south patrol]

I can make a conjecture about D_2n

Whether it is relevant to the previous discussion is another matter

[misty holly]

Where n is even?

I think we’ve nailed down the case when n is odd

[rocky cloak]

It's a central extension
C2 -> D2n -> Dn

[south patrol]

I was just making a joke dw

[coral spindle]

My conjecture is that it has order 2n

[south patrol]

Controversial

[coral spindle]

I saw that typo jagr!!! I see you!!!

[rocky cloak]

I'm a ninja

[coral spindle]

As ever. By the way, the generalised quaternion groups Q_4n are of order 4n

[south patrol]

Boytjie i had a conjecture and i checked it up to n = 1 million but it failed after that

[coral spindle]

I WILL die on this hill

[misty holly]

What does this mean?

[coral spindle]

Was it that they all have order < 2mil?

[south patrol]

all numbers are <= 1 milliom

[rocky cloak]

Just that C2 is a subgroup of the center with D2n/C2 = Dn

[coral spindle]

I'm not sure kheer has seen quotients yet

[misty holly]

I have sort of?

We’ve done indexes and stuff

[coral spindle]

So you haven't seen quotients, ok

[misty holly]

I’ve seen them outside of the book

[coral spindle]

This could be rephrased as saying that there is a surjective homomorphism D_2n → D_n, and this homomorphism has a kernel isomorphic to C_2

[misty holly]

How do we know this?

[coral spindle]

Uh we can kinda just calculate it

[misty holly]

What would the homomorphism look like

[coral spindle]

OK well

I think you should try that!

[misty holly]

Right lol

[south patrol]

Or geometrically, ||D_2n acts on the regular n-gon by rotations at two times speed ig?||

[coral spindle]

My hint is that the reflection element will be sent to the other reflection element, and the rotations... well, remember that the kernel of the map is going to be {1, r^n}

[south patrol]

Maybe i am silly

[rocky cloak]

Might start with D6 -> D3

[coral spindle]

Nice

This works yh

[south patrol]

Need to be careful w reflections too but hopefullt sorta clear how it works idk

[coral spindle]

There we go, seeing it as a central extension actually explains it better

nice

[south patrol]

Nah like

[misty holly]

So the rotation element goes to the double rotation?

Then r^n is in the kernel

[coral spindle]

Yeah sure, I would say it's really just the rotation generator goes to the... other rotation generator lol

but yes geometrically it's a double rotation

[misty holly]

Yeah right

[coral spindle]

But yeah that's correct

[misty holly]

So that means D_2n / Z2 is isomorphic to D_n?

[coral spindle]

Well you need to be careful about how you phrase that

[misty holly]

Or like a subgroup of D_2n isomorphic to Z2

[coral spindle]

because we cannot talk about quotienting by groups merely by their isomorphism type

yes

that's exactly it, yes

[misty holly]

So does that not give us our result?

[coral spindle]

Which result?

[misty holly]

That D_2n is iso to D_n x Z2

[coral spindle]

No, absolutely not.

Let me compare with something else

There is a subgroup of Z_4 isomorphic to Z_2, namely 2Z_4

Z_4/2Z_4 is a group of order 2, hence is isomorphic to Z_2

Does that mean that Z_4 is isomorphic to Z_2 x Z_2?

[misty holly]

No

[coral spindle]

Exactly

[coral spindle]

c.f. above

[misty holly]

so what information does that relation tell us?

I don't really understand the significance

[coral spindle]

It's not really easy to explain what quotients mean and why they're useful without just doing a bunch of stuff about quotients

I would say just wait until you learn more about quotients, and rest assured they are useful

[misty holly]

I mean it's just an equivalence relation right

[coral spindle]

I don't know what you mean by that

[misty holly]

I'm not sure either

I think it's too late for me to be doing more group theory

[tardy hedge]

Ya ur not wrong there is an equivalence relation put on the elements when ur talking about “quotient” stuff

But the extra part is that the equivalence classes also form a group with the operation being “inherited” from the original group

[thorn jay]

Not "just", it's an equivalence relation that "respects" the group operations

In some way

And even more, the relation is totally determined by any of its equivalence classes

[knotty badger]

it's a congruence in the categorical sense!

[thorn jay]

It's also a congruence in the normal sense

[knotty badger]

what's a congruence in the normal sense

[thorn jay]

The youth today I swear..

[knotty badger]

wait how old are you

[thorn jay]

Probably younger than you

18

[knotty badger]

...

i'm 23

[thorn jay]

It's just a bit that UA is a dead subject only done by old people lmao

[knotty badger]

UA?

i only know UA the fictional high school

[tardy hedge]

Woohoo adult now

Now when i visit netherlands we can party

[thorn jay]

A congruence on an algebra A is an equivalence relation ~ on A such that if f is an n-ary operation on A, and ai ~ bi for i=1,...,n, then f(a1, ..., an) ~ f(b1, ..., bn)

[tardy hedge]

Ong tho if im somehow ever in the netherlands ill hit u up

[knotty badger]

huh ok

the categorical version of congruence seems simpler to me

[thorn jay]

This way it's more obvious that you get an algebraic lattice

Imo

[knotty badger]

a what

[thorn jay]

For example, you can easily create an algebraic structure such that the subalgebras are precisely the congruence of A

[thorn jay]

I am crying

It's a lattice that is isomorphic to the lattice of closed sets of an algebraic closure operator, one where the closure of a set can be decided locally (i.e. Y ⊂ X is closed iff the closure of {y} is a subset of Y for all y in Y)

[knotty badger]

i don't know what these words mean

[thorn jay]

Alternatively, a complete lattice where every element is the join of compact elements

[knotty badger]

@.@

[thorn jay]

Complete meaning that every arbitrary meet and join exist

You working with lattices every day if you're doing algebra; subgroup/ideal/submodule lattices

[knotty badger]

im not an algebraist :P

[thorn jay]

Smh

Logic and computer science uses lattices a lot too

[knotty badger]

oh i like your deltarune bio

[thorn jay]

Oo sounds fun

[knotty badger]

unfortunately i don't do much theoretical computer science

also i'm generally an illogical person

[thorn jay]

Though I don't drink haha

[desert verge]

to me a lattice is just Z^n

[thorn jay]

Not that kinda person

[desert verge]

never too late to start

[thorn jay]

Well you would be WRONG

I'm the kinda person to get addicted very easily so honestly i do not want to tempt fate

[thorn jay]

Life changing game

[knotty badger]

indeed it is

[thorn jay]

This math problem is still
Pissing me off

I am the original

Starwalker

[soft tiger]

I was looking at a post that talked about why we only take quotients of groups with respect to normal subgroups, the idea of a congruence in universal algebra was relevant in the discussion.

[thorn jay]

Congruences in UA are always relevant to the discussion

If it isn't then I'll make it so

:D

[knotty badger]

mhm, i prefer to understand this through categorical congruences

[thorn jay]

Well it works precisely because a congruence relation is an equivalence relation R that is also a subalgebra of AxA

What is interesting is that considering the congruence as that subalgebra is often fruitful

So maybe you category folk have some good ideas

[knotty badger]

yes that's what i prefer

[thorn jay]

Though there is of course no difference between the definitions

Like, they're not even really equivalent as just restatements of the same thing

[knotty badger]

i see

[junior citrus]

Hey Here is simple task May you solve the problem. That's easy
It takes for two minutes.
a + b = a * b ===>>>> a != b ~~a ,b ??

[thorn jay]

So true

[desert verge]

get out

[thorn jay]

AI ass

[junior citrus]

You are very rude.

[twin merlin]

https://cdn.discordapp.com/attachments/1175386007919661121/1267536003342794904/attachment-7-1.gif

[desert verge]

you are not welcome here

[twin merlin]

hi enpeace 👋

[thorn jay]

She rude on my buster till I swoon

[thorn jay]

Haihai

What's up?

[junior citrus]

Okay I have seen

[thorn jay]

Same

[tardy hedge]

Dafoq

[junior citrus]

what does Dafoq stand for?

[junior citrus]

what does Dafoq stand for?

[desert verge]

who's in ur pfp btw?

is that you

[tardy hedge]

Dafoq stands for dafuq

[junior citrus]

what langualge?

[thorn jay]

Linkedin ahh pfp 💀

[desert verge]

what the fuck

^that's what it stands for

[tardy hedge]

My friend finally convinced me to make a linkedin 😭

[junior citrus]

Okay pls help you

[south patrol]

Please don't spam the same question everywhere

Especially as it seems inappropriate to all the channels it has been posted in (and it is also unclear what it means)

[thorn jay]

Given that you've proven that (-a)b = -(ab)

[desert verge]

?

[south patrol]

ye

perhaps a little cleaner if you just show (-a)b + ab = 0 and then subtract / use uniqueness of inverses (more readable)

No need to use a twice - could just do 0a = (1-1)a = etc

perhaps you want non-unital rings though hm and i wonder if you need to use your proof then

fair

[rocky cloak]

0*a = (0+0)*a = 0*a + 0*a

[south patrol]

yeah lol

that would be my approach too heh

[rocky cloak]

Subtract 0*a

Isn't this circular.

Like aren't you using (a-a)*b = 0 at the end?

Or am I misunderstanding the steps?

[thorn jay]

It is lol

[rocky cloak]

Yeah, if x=y then
f(x) = f(y)

[thorn jay]

This seems so convoluted it's just the cancellation law for groups 😭

0 + 0a = 0a = 0a + 0a => 0 = 0a

It works because of associativity

[tribal moss]

Almost everything about addition is to some extent "because of associativity", so it might not be terribly enlightening to single it out here.

Once you have 0a = 0a + 0a:
First add -0a on both sides, giving 0a + -0a = (0a + 0a) + -0a.
Then associativity lets you rewrite to 0a + -0a = 0a + (0a + -0a).
Then, by the defining property of minus, 0a + -0a = 0, so we have 0 = 0a + 0.
Finally, by the defining property of 0, the right-hand side of that is just 0a.

... where -0a means -(0a).

[thorn jay]

In general this is called the cancellation law:
a + b = c + b => a = c

Proof is much the same

[tribal moss]

Well yes -- though it's arguably even more fundamental than functions, really. If you want to be fancy you can call it the "substitution property of equality".

For example, if we know a=b and a>c, then we can conclude b>c.
There's no function around there, but it's still the same phenomenon: when we know a=b we can replace one or more instances of a with b in some fact we know to be true, and get another true fact.
In the case of functions, we may know a=b and f(a)=f(a) -- because everything equals itself -- then we can replace the second a in f(a)=f(a) with a b and get f(a)=f(b).

It's not particularly important at this level that you keep track of the formal names for all these manipulations, just that you get enough experience with using them, whatever they're called.

[rocky cloak]

If a thing equals another thing, then they're the same thing

I would think the problem is with understanding real numbers, not really with equality

Real numbers are kinda wack

[tardy hedge]

is that why you study algebra

[rocky cloak]

Well, this is a property of real numbers.

If a nonegative real number is less than 1/n for all n, then it must be 0. This property is called the Archemedian property.

Exactly how you demonstrate this property depends on exactly how you define the real numbers.

[south patrol]

well a funny thing is like

What do you mean by 1.99... in the first place

[merry harness]

because real numbers are complete

actually maybe thats not the correct way to say it

[south patrol]

i would say it is the archimedean property. there are "no infinitesimals"

well this is what jagr is saying

[merry harness]

because the real numbers are dense. and note two numbers can be arbitrarily close/equal but have different decimal expansions and are equal, because complete

[south patrol]

dense in themselves do you mean

[merry harness]

Yeah

So there's the notion of arbitrarily close

But that's equivalent to equality by archimedean principle, i think

Wait that was already mentioned

[merry harness]

yee this

Given a < a'' in R, there always exists a' s.t. a < a' < a''

R is dense in R

👍

[merry harness]

So if they are arbitrarily close, i.e. a' doesn't exist, then they must be equal by contradiction

Also note that 0.9... is literally 1

It's just another way of writing it

So equality it kind of a given

[tribal moss]

Perhaps another way of saying it would be that we invent the real numbers for the particular purpose of filling out the gaps between the rationals. But we don't want more real numbers than we need for that. So if we have two reals that don't have a rational strictly between them, they should have been the same real to begin with.

[misty holly]

Hello it's me again

I've proved D_2n is isomorphic to D_n x Z_2 for odd n, as D_2n is the internal direct product of the subgroups H = {id, r^n} and K = {id, r^2, ..., r^{2n-2}, s, sr^2, ..., sr^{2n-2}}; H is isomorphic to Z_2 and K is isomorphic to D_n

How do I find out anything about D_2n when n is even?

[rocky cloak]

I mean, how you find out anything is look at examples and follow your nose.

The question here I guess is what you would consider a finding.

Is it for example true that D2n is not isomorphic to C2 x Dn for n even? If so can you prove that?

[misty holly]

Yes they are not isomorphic, D2n has an element of order 2n while C2 x Dn does not

because all elements of Dn have order at most n, and since n is even elements C2 x Dn have order at most lcm(2, n) = n

[rocky cloak]

So that would work as a conjecture you've proven right

"D2n is isomorphic to C2xDn if and only if n is odd"

[misty holly]

Yes

But what happens for n even?

[rocky cloak]

Then it is not isomorphic up C2xDn

[cedar vault]

We get a semidirect product

[misty holly]

Oh I haven't learned about those yet

[rocky cloak]

Dn is in general a semidirect product of Cn by C2, but it's sort of unrelated to what you proved so far (and to whether n is even)

[misty holly]

So is there no "simple" decomposition of D_2n when n is even?

[rocky cloak]

Not in terms of direct products

[rocky cloak]

One way to define direct product is that G is the direct product of subgroups H and K if H and K have trivial intersection, G = HK and H and K are both normal subgroups.

A semidirect product is then the same except you only require one of the groups to be normal.

So it's like "half" the definition of direct product.

[misty holly]

Do we really need H and K to be normal subgroups? We just need their elements to commute with each other
does that equate to being normal?

[vocal pebble]

Yes, fix h in H, then hK = Kh, same for fixed k in K then any element being a product of elements in H and K gives you the equivalence

[rocky cloak]

Yes that is in some sense a stronger statement really, but it is equivalent in this case

[misty holly]

if g = hk then h'g = gh' if and only if h'hk = hkh' = hh'k if and only if h'h = hh'

[rocky cloak]

K being normal just comes down to hk = k'h for some other k' in K, so having hk = kh is stronger

[vocal pebble]

Are you trying to prove H is normal here?

[misty holly]

I'm trying to find the conditions when H is normal if G is an internal direct product of H and K

[rocky cloak]

One way to think is if H is normal, then for every h in H and k in K there exists an h' in H with
kh = h'k
if h' is not equal h you get
h^-1 k h = (h^-1 h') k

Since h^-1 h' is not in K (since it's in H and isn't the identity) the right hand side is not in K.

That means K isn't normal since then you should have h^-1 k h in K

[analog tusk]

ive been looking for some conditions about when polynomials in multiple variables will factor completely into linear factors (over, say, C). i wasnt able to find anything about this online, but i was able to produce a counterexample xy+z that cant be factored. does anyone know of any conditions that guarantee complete factorization?

i was able to find some conditions on mathoverflow actually 😅 not sure how i missed this until now

if p is homogeneous of two variables, then it can be factored

a bit stronger of a condition than i was hoping but 🤷‍♀️

[rocky cloak]

Being homogeneous in two variables is essentially just being a polynomial in one variable, so that shouldn't be surprising.

But I also found some condition while searching around.

For polynomials in two variables, there's a matrix called the Ruppert matrix, and the nullity of the matrix is equal to the number of irreducible factors of the polynomial.

So if the number of factors equals the degree, then those factors would need to have degree 1

[analog tusk]

i see, ill look into that

tyty

[tribal moss]

This seems to lead straight to algebraic geometry.

[hazy pollen]

zangish

[sage marsh]

The basic concept that D_{2n} = D_n \times \mathbb{Z}_2 for odd n is based on how the rotating portion functions with respect to the reflecting portion equal both portions complement one another nicely. However, for even n, r^n transforms into a reflecting portion from a central rotating one and therefore cannot be so conveniently decomposed into a direct product.

[white oxide]

Why can we use Theorem 68.4? Don't we need to know that the groups i_{\alpha}(G_{\alpha}) generate G?

[rocky cloak]

What is the extension property of 68.3? Does that perhaps entail them generating the groups?

[rocky cloak]

So the uniqueness of this h would guarantee that it generates.

But it seems uniqueness is not part of the property...

It's a little unclear what is meant to me

But for example if G is the free product times something else, then you still have this property (*) without having uniqueness

[misty holly]

20. should be true for any prime p, since there must be some element of order a multiple of p. Let |a| = pk, then |a^k| = p so that <a^k> is a subgroup of order p

I don't think 21 is true but I can't really think of a counterexample

oh A_4 might work

[candid patrol]

Yes but how do you know that such a element exists ?

[candid patrol]

It is a crontrexemple yes

Let H be a subgroup of order 6 in A4, then H is normal, so you can check pi : A4 —> A4/H

[misty holly]

the order of G is the lcm of the orders of all the elements in G

[misty holly]

Why must H be normal?

We haven't really covered normal subgroups and quotient groups yet

I think I can prove it by construction though

[candid patrol]

[A4:H] = 2

[misty holly]

oh right

[candid patrol]

Ok why not, if there is such a think in your lesson then your proof is correct

[misty holly]

Well we have |a| is a multiple of |G| so that is a corollary of this

[rocky cloak]

The order of G is a multiple of that lcm at least

The lcm of the order of all elements is called the exponent of the group

[misty holly]

oh?

then my logic doesn't work

[rocky cloak]

Like C2xC2 has order 4, not 2 for example

[misty holly]

so then a group of order 3k doesn't necessarily need to have an element of order 3, at least not by my logic

since the lcm of the orders can be k, or even a factor of k

[rocky cloak]

Yeah, the fact that a group of order pk has an element of order p for a prime p is Cauchy's theorem

It's a nontrivial thing

[misty holly]

is this true for nonabelian groups too?

[rocky cloak]

Yes

But it is important that p is prime

[misty holly]

right

well

if the order of a is not divisible by 3 then the quotient group G / <a> has order divisible by 3

[rocky cloak]

Careful that the quotient group is only defined if <a> is normal

[misty holly]

yes I'm only considering abelian groups here

[rocky cloak]

I see

[misty holly]

so by induction I can assume G / <a> has an element b of order 3

so any element in the equivalence class of b in G has order 3?

[rocky cloak]

Not quite. But if b has order 3 in G/<a>, what can the order of b be in G?

[misty holly]

uhm

well b can't be in both G/<a> and G right

Do you mean the equivalence class of b vs b?

[rocky cloak]

Yes, that's what I mean

Fix some b such that [b] in G/<a> has order 3

[misty holly]

So that means b^3 is in <a>?

[rocky cloak]

Yes

[misty holly]

oh then if |<a>| = k then b^3k = e so b^k has order 3, and b^k != e because b^k is not in the same equivalence class as e since k isn't a multiple of 3

okay nice

[rocky cloak]

That's it. Now there is some care to be taken if k isn't relatively prime to 3, but you can remove that case when choosing a

[misty holly]

quotient groups are pretty cool

[tardy hedge]

ya i remember finding them cool too when i first saw it

[candid patrol]

Fact, mb I didn’t read carfully

[misty holly]

This seems like a really convoluted way to prove the lemma and I don't understand the motivation behind some of the steps taken

Is there an easier way to prove this?

Or a more insightful method I should say

[karmic moat]

https://kconrad.math.uconn.edu/blurbs/grouptheory/Ansimple.pdf

this might be helpful but idk

that being said the proof of your original statement is almost always ugly afaik

when my prof did this he said "this is my least favorite proof that i have to do once a year. if you had beer with me i could tell you the proof for any result from this course except this proof"

[misty holly]

lmfao

how did anyone even discover this

[karmic moat]

just rote computation really

i mean galois had this idea

[thorn jay]

People reaallyy care about permutation groups

[karmic moat]

but yk galois was a freak

[misty holly]

freak in the sheets

[thorn jay]

Galois is always to blame

[karmic moat]

and/or adore

[thorn jay]

Not mutually exclusive

[misty holly]

what is this fascination with 3-cycles though?

[thorn jay]

:3

Generators of An I presume?

[karmic moat]

yeah

[karmic moat]

this blurb is really nice

[misty holly]

this proof makes no sense to me

[misty holly]

I'll read through it

hopefully it gives better insight

[karmic moat]

lol if u really wanna see something that makes u question group theorists' minds/sanity

look at the classification of finite simple groups project

[misty holly]

yeah I've seen videos about that lmao

this proof seems the nicest

The core idea is still showing N has a 3-cycle though

[karmic moat]

yeah

[misty holly]

where did this claim come from?

Are all cycles of the same cycle type conjugate in Sn?

oh wait we did this in a previous claim

this should work

[coral spindle]

You should indeed know what the conjugacy classes of S_n look like, due to that proposition

[vivid kestrel]

In representation theory we essentially study how the group algebra kg acts on its simple regular left modules, what if we replaced it by right modules, would we get the same characters

[mighty kiln]

Is that just the representation theory of the opposite algebra

[rocky cloak]

You have a canonical equivalence between right and left G-modules.

If M is a left G-module then
mg = g^-1 m
makes M into a right G-module

In general like Arki says right modules are left modules of the opposite algebra.

But kG is isomorphic to its opposite by g |-> g^-1

[vivid kestrel]

Ah, wonderful, thanks to you both

[celest cosmos]

D'où tu parles anglais toi

Double jeu

[candid patrol]

Let me cook : )

[thorn jay]

Don't burn the kitchen down

[white oxide]

Let $A$ be a local ring with maximal ideal $\mathfrak{m}$, $M$ and $N$ $A$-modules. To show that the map $f: k \times (M \otimes_A N) \to (k \otimes_A M) \otimes_k (k \otimes_A N)$ is well-defined, does it suffice to show that $f$ is bilinear in the second entry? For suppose that $(x, s) = (y, t) \in k \times (M \otimes_A N)$. Then $x = y$, and $s = t$ if and only if $s - t$ is in the submodule generated by all "bilinear" expressions. This happens if $f$ is bilinear in the second entry. (Correct me if I'm wrong)

[cloud walrus]

okeyokay

[tardy hedge]

yea maybe you can fix k and then show the map fk : M x N -> (k(x)A M) (x)k (k(x)A N) is bilinear

[sage sand]

trying to understand how group presentatino works, if i define $G=\langle a,b| a^3=b^6=1 \rangle$. How do I know that $a\not=b^2$

[cloud walrus]

Former Rank 7 LLORT AJNIN

[tidal schooner]

The equation $a=b^2$ doesn't follow from the equation $a^3=b^6=1$ (morally, you can't ``cancel exponents''). Look at the group $\bZ/3\bZ\times\bZ/6\bZ$ for a counterexample

[cloud walrus]

harmacist

[sage sand]

i think im not really sure how group presentations are defined, how would we check that $a=b^2$ doesn't follow from the equation $a^3=b^6=1$? Is it right that to check this, we find an example of a group which satisfies the latter but not the former?

[cloud walrus]

Former Rank 7 LLORT AJNIN

[tidal schooner]

Yes; check my previous message for an example

[sage sand]

ah okay, this will mean that if i want to check that some given equation is true, I have to show there is no example of a group which satisfies the relation but not the equation?

[sage sand]

would this be the same as checking whether an equation follows from the relations?

[tidal schooner]

Yes, it suffices to show that the equation you want follows from the relations. That shows that no counterexample exists

Edited for clarity

[sage sand]

this feels kind of obvious but is there a better way to claim that a=b^2 doesn't follow from a^3=b^6?

i know its clear enough jsut by looking at it but Im wondering if there is something nicer beyond just eyeball testing the two

[tidal schooner]

If you want to do it rigorously then finding a counterexample is usually the best approach

[sage sand]

ah okay, its starting to make sense now

[tidal schooner]

(Side note: If you know about free groups and quotient groups, technically a group presentation is defined as a certain quotient of a free group. In your example, we start with the free group $F_2$ with two generators $a,b$. Then, we let $N$ be the smallest normal subgroup containing the element $a^{-3}b^6$. The group in your presentation is then the quotient $F_2/N$. The claim here is that $N$ doesn't contain the element $a^{-1}b^2$.)

[cloud walrus]

harmacist

[south patrol]

A way to make it precise is that if you find any group H with elements a,b where a =/= b^2 but a^3 = b^6 = 1, then theres a map G -> H sending x to a and y to b which sends xy^-2 to something nontrivial

[sage sand]

ah okay, this is nicer but i feel that i run into a issue when i try thinking about the free group, for example, if i have the free group generated by just 1 element g and i want to show that g^2\not=g, then in my search for a counterexample, the counterexample would just be the free group itself but this is circular right? or am i misunderstanding how the free group is defined

[south patrol]

You can make easier counterexamples

[south patrol]

Like Z/3 in this case

[sage sand]

oh okay i see it, we are allowed to add more relations, thank you

ah okay, i think I have got it, i think my original assumption was that the group presentations are how all groups are "actually" defined but this doesn't seem to be the case

thanks @south patrol @tidal schooner

[misty holly]

What are some ways to check the normalcy of a subgroup? Checking for gN = Ng gets quite tedious when there's 8 subgroups

[south patrol]

Usually/often it is easiest to first compute the conjugacy classes. Then use the fact that every normal subgroup is a union of conjugacy classes

[misty holly]

the subgroups of order 4 are trivially* normal because their index is 2

[south patrol]

And use Lagrange's theorem and things

[misty holly]

Oh?

I haven't really studied conjugacy classes in depth yet but I think I understand

[misty holly]

Am I right in saying that any normal subgroup of G of order 2 must be generated by an element in the center of G?

Since if N = {1, h}, then g1g^-1 = 1 is in N for all g in G, so we just need ghg^-1 = h or gh = hg for all g in G

[misty holly]

Then with the group representation D4 = <r,s | r^4 = s^2 = 1, srs^-1 = r^-1>, {1, r, r^2, r^3}, {1, r^2, s, sr^2}, {1, r^2, sr, sr^3} and {1, r^2} are the only normal subgroups

[white oxide]

Is the map $f: k \times (M \otimes_A N) \to (k \otimes_A M) \otimes_k (k \otimes_A N)$ given by $x \times (m \otimes n) \mapsto (x \otimes m) \otimes (1 \otimes n)$ bilinear?

[cloud walrus]

okeyokay

[white oxide]

Is this an error? Should it be a \subseteq a^{ec}?

Nvm I guess we always have that

Why is that second iff true then?

[tardy hedge]

because if x is in a^ec then by (2) x is in some (a:s) so sx is in a. But also a^ec is in a so x is in a

so sx in a for some s in S -> x in a

[hazy pollen]

isomorph

[tardy hedge]

Yes isomorph 💕

[thorn jay]

ism

[crisp roost]

are there any examples of sets with the cardinality of the power set of the real numbers, that form a group? (or generally, any infinite set with a cardinality above the real numbers)

[velvet hull]

You can take any “sufficiently” big subset of functions from R to R and that would be something you’re looking for

Well, R to R or C to C or R^n to C^m etc

[rocky cloak]

For any infinite set X, the F2-vector space of dimension |X| also has cardinality |X|

But maybe that's boring

[velvet hull]

So real-valued functions under addition, complex valued functions (that are nonzero everywhere) under multiplication

[rocky cloak]

Bijections of R under composition

[crisp roost]

is there such a method of construction for sets of arbitrary cardinality for rings and fields as well?

(i was trying to find a set the cardinality of P(R) that has a well defined addition/multiplication)

[rocky cloak]

Sure, the polynomial ring or field of rational polynomials will work

[velvet hull]

Fields are characterised by cardinality and character up to isomorphism, and a field exists for every (infinite) cardinality and character

[rocky cloak]

Algebraically closed fields are characterized by characteristic and transcendence degree.

For uncountable fields it's cardinality is equal to its transcendence degree

[crisp roost]

whats a transcendence degree

[velvet hull]

Does uniqueness fail if you drop the alg closed assumption

Oh yeah it does

LMAO

Okay

[rocky cloak]

Size of transcendence basis (maximal algebraic independent set)

[crisp roost]

i think i'm gonna need to do a bit more reading to understand but thx

[rocky cloak]

A set is algebraically dependent if it satisfies some polynomial equation.

And independent if not.

[marsh scaffold]

Wait most probably #advanced-algebra is the right place for this

[leaden geyser]

What resources or links do you recommend for for Group theory

[ashen heron]

theres a quick review here- #book-recommendations message of a whole bunch of popular algebooks

[covert cliff]

is anyone familar with this notation? I am confused as to what O is supposed to be? Is it a subgroup?

[thorn jay]

Open subset?

[covert cliff]

oh maybe

[charred iris]

I've seen $\overset{\circ}\subseteq$ used to mean 'open subset of' so yeah I would assume this is just a variation in it

[cloud walrus]

Edward II

[covert cliff]

why not just say open at that pint?

[charred iris]

why have a symbol at all and not just write `subset of' each time

[covert cliff]

because 'subset of' is considerably longer than a 4 letter word

and the notation for subsets is pretty universal

[charred iris]

yes, but at some point even 'open' gets pretty annoying to keep rewriting (and while this is not the case here, it does mean you might need to have longer sentence structure to get the word 'open' into a reasonable place while trying to introduce other conditions)

[covert cliff]

maybe, but if thats the case I think for unconventional notation people should mention it in a preface or something

[south patrol]

Ugly symbol

[covert cliff]

this too

I also have a feeling that if were to hand write this then writing that symbol would get annoying faster than writing 'open' after intializing the subset

[thorn jay]

I swear

If you use U ⊂ G then everyone will know what you're talking Abt anyways

O also looks ugly

Way too round, looks like it'd roll away

Never use o or O on its own in math

Always make it have like parenthesis or square brackets

[misty holly]

If $G$ is finite then $G/H\cong \bZ_n$ where $n=\frac{\abs{G}}{\abs{H}}$

[cloud walrus]

kheer257

[misty holly]

what happens if G is infinite?

[paper flint]

Do you know of an infinite cyclic group?

[misty holly]

Z?

[paper flint]

Yep; what are its subgroups? What do the quotients look like?

[misty holly]

I guess the same

Z/nZ for some n

[paper flint]

The notation already suggests the answer here 😛 try proving this

Also, this is good enough - Z is the only infinite cyclic group (try proving this as well if you haven't already)

"only" in the "only one up to isomorphism", as is the norm in algebra

[misty holly]

Yeah I forgot I proved that earlier

So if G is cyclic and H is a subgroup of G then G/H is isomorphic to Z/nZ where n = [G: H]

[paper flint]

In the trivial case where G=Z, H={0}, you'd not want to describe n as the index of H in G maybe (the Z/nZ description still makes sense though)

[misty holly]

hmm

then [G: H] = inf

so Z/infZ is just {0}

[paper flint]

No, the quotient should be isomorphic to Z; it's trivial when you quotient by Z itself

[misty holly]

yeah

that's why it doesn't work

[paper flint]

Well, "infZ" doesn't appear to be well-defined to begin with

[misty holly]

yeah fair enough

but an infinite cyclic group can't have finite nontrivial subgroups

so H = {0} is the only case which fucks up

[coral spindle]

No

N is not a subring of Z

you need additive inverses, and then you're done.

[tidal schooner]

No, because that doesn't show that it's an additive subgroup. We have to also show that it's closed under additive inverses

But if you replace "closed under addition" with "for all x and y in R, we have x - y in R," then that is enough. That's called the subring test theorem

[tribal moss]

In most contexts you'll also want it to contain 1 -- (though that depends on whether a ring has to contain 1 for you).

[karmic moat]

Are there contexts where ppl dont use rings with 1

[thorn jay]

Wanting to define the UA commutator without having the boring result that abelian rings are simply the zero ring

[tribal moss]

Some function spaces don't contain the constant 1 function (for example due to requiring the functions to vanish at infinity), but otherwise behave well enough under pointwise multiplication that it makes sense to use ring-derived tools for them.

[white oxide]

Are we allowed to interchange the rings that $(M \otimes_A N) \otimes_B P$ and $M \otimes_A (N \otimes_B P)$ are modules over? For instance, I wish to construct a bilinear map $(M \otimes_A N) \times P \to M \otimes_A (N \otimes_B P)$, and use the universal property to factor through $(M \otimes_A N) \otimes_B P$. This forces me to consider $M \otimes_A (N \otimes_B P)$ as a $B$-module, via $b \cdot (m \otimes (n \otimes p)) = m \otimes (nb \otimes bp)$. Similarly, I must consider $(M \otimes_A N) \otimes_B P$ as an $A$-module when constructing an inverse. This feels morally incorrect, since after all isomorphic modules must be over the same ring.

[cloud walrus]

okeyokay

[south patrol]

non-unital rings also often appear if you are interested in augmented algebras (an augmented k-algebra is a k-algebra A equipped with a map A -> k such that the composite map k -> A -> k is the identity). Augmented k-algebras are in one-to-one correspondence with non-unital k-algebras, but the latter feels like it has "less data"

[south patrol]

You can construct the maps as a map of B-modules or A-modules and then check that the structures on each side match up under those maps (for example)

I think it is cleanest though to formulate a universe property for each side

[white oxide]

I see

Yeah what I'm doing is that I'm interchanging the rings they're over in order to obtain the maps

Then I just choose a ring to view them both over and compose those maps and show that they're the identity

[white oxide]

How does M_k and N_k being vector spaces over a field imply that M_k or N_k is zero?

[velvet hull]

The tesnor product of vector spaces is straightforward “multiplication” of vectors
In particular in finite dimensions dim(V tensor W) = dim(V) * dim(W)

[tribal moss]

Am I right that the correspondence from non-unital to augmented algebras is left adjoint to the forgetful functor from unital to non-unital algebras?

[white oxide]

What do you mean by multiplication of vectors?
Do you mean that, if we replace $\otimes$ with $*$, we have $x_1 + x_2 * y = x_1 * y + x_2 * y$, so that it mimics the distributivity property for multiplication?

[cloud walrus]

okeyokay

[white oxide]

Also, is the result you cited about the dimension easy to prove? Or is it better to read a proof of it (I've spent 3 hours on this exercise already lol and I want to move on)

[south patrol]

Hm this is interesting - I wonder what the best way to phrase this is. But yes, the functor {Unital} -> {Non-unital} has a left adjoint which informally sends A to A (+) k, and then this object can be endowed with an augmentation (in the "obvious" way) such that it is the corresponding augmented algebra to A

[tribal moss]

Thanks, then I think I reconstructed the correspondence correctly.

[south patrol]

The point is that given a basis {v_i} of V and {w_i} of W, the set { v_i (x) w_j} is a basis of V (x)_k W

[velvet hull]

Let B be a basis for V and B’ be a basis for W, then using bilinearity you can prove that B tensor B’ is a basis for the tensor product

Sniped

[south patrol]

(This works in arbitrary dimensioms)

But also if you know a little more, then all vector spaces are flat modules. So any non-zero (hence injective) map k -> M induces an injective map N -> M (x) N

[tardy hedge]

when okey asked that question I thought to replace Mk and Nk with k^n and k^m and then distribute tensor product over direct sum to maybe show something

[south patrol]

That also works

Even in arbitrary dims

(Though ig here things are fg anyway)

[tardy hedge]

k^n (x)k k^m = direct sum k^n m times

[south patrol]

I love Nakayama.

[tardy hedge]

if direct sum k^n m times = 0 could you say from there then either n or m is 0

so dim of one of them is 0 originally

[south patrol]

I guess this statement is also true over arbitrary rings right? Since we can localise

Though i gusss at this point in the book that may not have been introduced idk

Ye, Indeed it is k^nm

[tardy hedge]

yea so n or m is 0

[velvet hull]

Like in finite dimensions, you can think of v tensor w to be the vector (w1v w2v …. wnv), where it consists of n copies of the vector v stacked on top of each other

The structure of the tensor product of vector spaces is very concrete because vector spaces are just that nice

There is no degeneracy anywhere

[white oxide]

Sorry I'm probably tripping but why are you assuming that x is in a^ec?

[tardy hedge]

showing a^ec subset a implies (sx in a for some s in S -> x in a)

[white oxide]

don't we just assume that sx is in a and nothing else? if x is in a^ec and we're assuming that a^ec \subseteq a then x in a is immediate

[tardy hedge]

ya i guess i phrased it badly

[tardy hedge]

ya this one

[white oxide]

Anyways yeah makes sense

cuz we use ii)

ty goat

Atiyah Macdonald is so goated

[tardy hedge]

I agree

you can learn a lot from so few pages

[white oxide]

yeah each page takes me at least 2 hours tho lol

[tardy hedge]

Yes i have a strong connection to each page because of that lol

[white oxide]

what did you need comm alg for? algebraic geometry? number theory?

[tardy hedge]

for combinatorial algebra for masters thesis

I would like to learn algebraic geometry someday, not sure when that would be though

[dull ginkgo]

So wait does like the automorphism-stabilizer Galois correspondence like boil down to like Artin’s Lemma and the definition of a Galois Extension

[pallid jay]

after banging my head against artins book for so long, going through dummit&foote feels like running without sand in my shoes

[dull ginkgo]

Jacobson:

[pallid jay]

whats that

[dull ginkgo]

Jacobson’s Basic Algebra I

The book I used during HS and completely FUBAR’d me in terms of viewing LA in scientific contexts

[misty holly]

If H is a subgroup of G of order k, then gHg^-1 is also a subgroup of G with the same order for all g in G. But if H is the only such subgroup gHg^-1 = H i.e. H is normal

If g generates a normal subgroup of G, then for all x in G, x<g>x^-1 = <g> = {g^k: k in Z}

Then, xC(g)x^-1 = {xyx^-1 in G: yg = gy}

Not too sure what to do after this

I need to find a bijection between xC(g)x^-1 and C(g)

[coral spindle]

I'm not sure you do need to find a bijection?

In any case a bijection is easy to write down

No, rather you need to show equality

[misty holly]

I mean I need to prove they're the same sets

[coral spindle]

Yes

OK, so a good way to do this might be to show that if c is in C(g)

then so is xcx^-1

[misty holly]

ahh right

[coral spindle]

And what property would xcx^-1 need for that to be true?

[misty holly]

xcx^-1g = gxcx^-1

[coral spindle]

(This is not quite sufficient yet, because that only shows that xC(g)x^-1 is a subset of C(g), but the other way around is similar)

So do that now

[misty holly]

so xc (x^-1gx) = g(xc), and we know (x^-1gx) = g^k

[ (xcx^{-1})g = xc(x^{-1}gx)x^{-1} = x(x^{-1}gx)cx^{-1}=g(xcx^{-1}) ] as $x^{-1}gx \in \left<g\right>$

[cloud walrus]

kheer257

[misty holly]

Wait no

We still need to show that if c commutes with g then c commutes with g^k for any k

but that should be easy

I feel like this proof would be easier to write out if we considered the centraliser of a general subgroup instead of just an element

[coral spindle]

It's not hard if you remember our assumptions. What does it mean for <g> to be normal?

[misty holly]

xgx^-1 = g^k for some k

[coral spindle]

So now you can do it by induction

[misty holly]

I mean we can just take the equation xgx^-1 = g and raise it to the k'th power

to get xg^k = g^k x

[coral spindle]

Yeah!

That's right good job

[misty holly]

Thank you!

Should I make a thread here? It feels like I'm inflating this channel with my own doubts

[coral spindle]

There's no need to but you can if you want. I always think it makes people less likely to see your questions but idk

[misty holly]

Yeah that's true

If I do make a thread can I advertise it in this channel if I ask anything there?

I've seen some people do that

[coral spindle]

Sure just don't get spammy about it lol

I trust you tho

[misty holly]

Lol of course

I think I'll do that then

kheer's abstract algebra arc

[arctic trail]

It seems to me that it's true that for a finite solvable group

Every Sylow p-subgroup is complemented, by Hall's theorem

is this true?

[crystal vale]

Any hint for counterexample?

[fading acorn]

$\Z/n\Z$ when $n$ is not prime ig

[candid patrol]

Yes

take n = 6

[tribal moss]

That gives a ring that's not a domain, at least, but not immediately the a and b of a counterexample.

In fact I don't think there is any counterexample in Z/6Z.

[tardy hedge]

isnt there like two different elements that generate same cyclic group but they wouldnt be associates

Idk tbh i reallyyy need to review these basic rings and properties lol

[tribal moss]

I suspect a counterexample needs to be non-commutative.

[fading acorn]

oh yeah right

mb

[arctic trail]

ugh constructing counterexamples is harder than I thought

Maybe R is Z[e] with e = e^3

and taking a = e and b = e^2

[fading acorn]

how about this? ring A of all polynomials vanishing at 0
then obviously x or -x generates A but there doesnt exist any element ux=-x for every u in A because -1 \notin A

idk

[arctic trail]

I assume rings need to be unital

[arctic trail]

Or more well behaved R = F[x]/(x-x^3)

F[x]x = F[x]x^2

I suspect any example must have non-trivial Jacobson radical

Or maybe a projectivity / injectivity condition somewhere

[tribal moss]

Perhaps the ring of vector space endomorphisms of R^infty, with a(x,y,z,...) = (y,z,...) and b(x,y,z,...)=(0,y,z,...).

[arctic trail]

isn't this needlessly complicated?

[tribal moss]

Perhaps. Do you have something simpler?

[arctic trail]

isn't this an example?

a = x, b = x^2

[tribal moss]

Hmmm.

[crystal vale]

Yes there is no counterexample in Z/6Z

[crystal vale]

Yes otherwise take R = 2Z and a = 4 and b = -4

[tribal moss]

Are you sure there's no unit u with x²=ux?
(I can't show one, nor prove immediately there isn't one).

[tribal moss]

Yeah, otherwise the "there is a unit u ..." condition doesn’t make sense.

[crystal vale]

So this works?

[tribal moss]

I'm not quite convinced there's no u there either yet. Just thinking aloud.

[tribal moss]

Then a=(1+x-x²)b, and u=1+x-x² is a unit because ux²=x, so u(1-x²+x^4)=1.

[crystal vale]

Okay Ra = Rb, but why do you think there is no such u?

[tribal moss]

The other way around seems to be easier:
a(x,y,z,...)=(0,y,z,...) and b(x,y,z,....)=(y,z,...).
If a=ub then u has to be the shift-right operator to give the right output for all inputs, and that is not invertible.

[vast verge]

I don't understand how this theorem is applied

[crystal vale]

No, you have to use the fact group of order p^2 is abelian

[vast verge]

I'm a little rusty, sorry. I don't remember coming across this

Oh it's Theorem 9.8 instead

[crystal vale]

Yes got it, thank you

How do you get this?

[tribal moss]

I think all I can say is inspired guesswork.
The problem said Ra and Rb instead of the more common aR and bR, which made me guess they were probably one-sided ideals and R would need to be noncommutative.
Further, if Ra=Rb then there must be t and u such that a=ub and b=ta, and the only task is whether t and u can be prevented from being inverses. Still, though, since tub=b snd uta=a, they are somewhat like inverses.
And then, End(R^infty) is my favorite noncommutative ring with something like almost-inverses in it...

[tardy hedge]

Oo i should study End(R^infty) . I should take a look at more weird examples like that

[glad osprey]

a(x, y, z) = (0, x, y, ...) and b(x, y, z) = (y, z, ...) are also weird because they are linear functions that are injective/surjective respectively, but not bijective

[dull ginkgo]

So uh, assume R is commutative, and Ra = Rb, but a =/= b
Then there exists an x, y in R such that: xa = b, yb = a, thus (x - 1)a =/= 0, and (yx - 1)a = 0

So uh, assume R = Z/4Z[X,Y]/(XY - 1)
then if x = X, y = (Y + 2), a = 2
Then 2(xy - 1) = 2(X(Y+2) - 1) = 2(XY + 2X - 1) = 4X = 0
2(x - 1) = 2(X - 1) = 2X - 1 =/= 0

Then a = 2, b = 2X would work I think?

I would find your own example but idk if this works

[dull ginkgo]

Hom fuckery my beloathed

[tribal moss]

For clarity, my a was a(x,y,z,....)=(0,y,z,...) so it's not injective -- it replaces the x with 0 and leaves everything else in place.

[glad osprey]

Oh, I misread

[tribal moss]

However u(x,y,z,...) = (0,x,y,z,...) was the function that I argue must be the only solution to a=ub.

[tardy hedge]

What are you studying nowadays?

[dull ginkgo]

Functional analysis and Finite Element Methods

[tardy hedge]

Oh nice, i thought u were an algebra person

[dull ginkgo]

i thought too

[tardy hedge]

Finite element methods sounds like some numerical thing lol

[dull ginkgo]

it is yes

[glad osprey]

Yep I guess you could also argue that since b(x, y, z) = (y, z ...) is surjective, there cannot be a unit u such that a = ub, since a isn't surjective, right?

[tardy hedge]

Idk anything about functional analysis

[dull ginkgo]

not far off from me

[glad osprey]

Oops, meant to reply to this

[tribal moss]

Hmm, yeah, I think that argument would work too.

[glad osprey]

Btw, what's the terminology for a and b in case a = ub for a unit u? Associates or something? It's been like half a year since I did any ring theory

[dull ginkgo]

Associates yes

[tardy hedge]

Yes, associates, theyre great pals

Personifying math is funny

[crystal vale]

I see

[dull ginkgo]

This is literally the simplest comm example I could think of

[crystal vale]

How do you conclude that (x-1)a ≠0 ?

[dull ginkgo]

xa = b, so if (x-1)a = 0 then xa = b = a

[crystal vale]

Yes

[dull ginkgo]

Pondering about the idea behind a Lie Algebra

Is the gist of a Lie Algebra the fact that left/right multiplication is a derivation?

Like let’s say we have a ring R, then we can consider the submodule of End(R) {Abelian group endomorphism module} Der(R) = {f in End(R) : f(xy) = f(x)y + xf(y)}, then [x,y] = fg - gf on Der(R) is a bilinear operator on Der(R) right

[oblique iron]

I think defining Lie algebras without their original geometric interpretation as motivation is somewhat abstract and hard to come up with, but this indeed is basically the motivating example for their definition (the two main ones that made people care about them historically were differential operators on the ring of smooth functions over R^n, where composition is sequentially differentiating, and matrices where [,] is the usual commutator)

[rocky cloak]

b = Xa, and X is a unit.

[dull ginkgo]

Shit yeah sorry

[dull ginkgo]

This is curious to me because it seems the whole derivation focus of Lie algebras stems from the classic derivative

Does the general algebraic form of them (over other fields / rings) stem from some similar notion of differentiation over groups or smth

[oblique iron]

you can show that every (finite dimensional) Lie algebra over R arises as a certain set of derivative operators on a special class of group called Lie group (essentially, a manifold is a type of topological space where differentiation makes sense. A Lie group is a group that's a manifold at the same time, so differentiation makes sense), where the derivative operators satisfy certain conditions making them compatible with the group operation. this is the content of Lie's third theorem, and is half of the Lie algebra-Lie group correspondence

If you algebraify all of this to remove the calculus, if I recall correctly you get a similar something over arbitrary rings

[dull ginkgo]

Yeah I am aware of the Lie group defn by considering the derivation on the tangent space of the identity

I assume you would need to instead employ something like varieties or smth (Zariski cotangent) to cast it in an algebraic language

[oblique iron]

In such a context, the keyword is "algebraic group", and I believe they do have a similar Lie algebra available to them

but I only work over the real or complex numbers, so google would answer better than I would

[dull ginkgo]

That’s fair

I have very little Lie group knowledge unfortunately

I wonder if you can derive the lie bracket for something like a semidirect product

[oblique iron]

Also, I was referring to the space of left-invariant vector fields on a Lie group, and not the tangent space to the identity actually (though these are naturally isomorphic)

[oblique iron]

Indeed you can, it essentially involves just taking derivatives of all the operations you have explicitly.

[dull ginkgo]

I was going to try to derive it myself

[oblique iron]

Oh totally fair, though if your Lie group knowledge isn't so good, this is probably not so easy

[dull ginkgo]

Originally I was going to see what happens when you “take the derivative” of a conjugated variable but that seems to be invariant though I might have done it wrong

[oblique iron]

(in that case, you can google the MSE thing I linked afterwards)

In particular, you should look up the adjoint action of a Lie group, it will help with the conjugation

[rocky cloak]

You can reframe the whole story in terms of algebraic groups.

There you can define the tangent space as homomorphisms from the coordinate ring to the dual numbers. So an algebraic group has a corresponding lie algebra

[dull ginkgo]

Is like an abelian variety an algebraic group?

[rocky cloak]

Yeah

[dull ginkgo]

Ah I see

[dull ginkgo]

does the Zariski cotangent <-> derivation duality hold there

[oblique iron]

Thanks jagr, does the whole story work for Lie groups over R? They are then analytic so I guess you have something of real-analytic geometry, but I don't know how much of that story will carry over (without passing to like, smooth group schemes)

[dull ginkgo]

The context was seeing if there is a reference frame “free” form of Newton Euler that can be derived via the Lie group structure of the Special Euclidean group and considering Energy on that

[rocky cloak]

I think you have to sort of distinguish if you're working analytically or algebraically, but I think it should work either way.

I'm not expert on lie theory

[dull ginkgo]

Because it seems rigid body motion is like, an immediate application of Lie theory

Every derivation I’ve seen was choosing a frame where it reduces to SO(3) instead of SE(3) which is the semidirect product of the normal translation subgroup and SO(3)

[oblique iron]

ok, a quick google search shows this is not so easy actually, the usual suspect counterexample of the universal cover of SL(2, R) turns up. thanks for the helpful comments!

[rocky cloak]

Yeah not every real lie group can be nicely turned into an algebraic group

[tribal moss]

I don't think that works as a counterexample.
With a = 2, b = 2X, we have a = Yb and b = Xa, and X, Y are both units since XY=1.

(Oh, that was noticed already, sorry).

[tribal moss]

As an abstraction of my earlier example, let R be any ring that has a pair of one-sided inverses, that is, pq=1 but qp != 1.
Then pqp=p, so Rp=Rqp. However, for any u with up=qp we have u = upq = qpq = q, and q cannot be a unit since it has a strictly one-sided inverse.
(If there is r with qr=1, then qp = qp1 = qpqr = q1r = qr = 1, contradicting the assumption that qp != 1).

[crystal vale]

Yes and qr = 1 and pq = 1 implies r = p

[rocky cloak]

This should work as a commutative example I think

[tribal moss]

Hmm, is it clear that y and xy are not related by any unit?

[rocky cloak]

It's not immediately clear, but I believe it's true

Like anything that's a unit would need to remain a unit when you quotient by something. So for example quotienting by y shows that the x-terms of a unit must be 0.

So the only elements that can be units are of the form
a + f(y)y + g(y)yx
with a nonzero.

Then multiply this by y you get
ay + f(y)y^2 + g(y)y^2 x
And this cannot equal yx

Further looking modulo x+1 and x-1 we see that the only units are the units of k (assuming characteristic not 2).

[tribal moss]

Okay, that sounds convincing. And here I had already begun to look for an argument that a counterexample cannot be commutative. :-)

[tribal moss]

Interesting that this proposed (but non-working) example is almost the same as yours, just with y and x identified.

[thorn jay]

Things get fucky with two variables

[rocky cloak]

I guess morally, what's going on is that in k[x, y]/(x^2 - 1), x is a unit. Then you just have to tweak the example enough so that x is barely not a unit anymore.

As far as (y) is concerned k[x, y]/(x^2 - 1) and k[x, y]/(y(x^2 - 1)) look the same, but in the latter x is no longer a unit

[tribal moss]

Makes sense.

I suppose we could just have winged it from the beginning and arrived at Z[x,y,z]/(zyx-x) with (yx)=(x) as a "universal" counterexample.

(Since this maps into every commutative example of two elements generating the same ideal, so if there's a unit relating yx and x in Z[x,y,z]/(zyx-x), that would kill every proposed commutative counterexample).

[rocky cloak]

Yeah, that's a good observation

[chilly ocean]

Hi guys, I need help with the name of a concept

As I'm from Spain, I learn math in spanish, and I would like to know what's the name of a concept related to rings in english

I know it as "dominio de integridad" and it's basically a commutative ring in which 1 is not equal to 0 and that does not have divisors of 0

So I would like to know the name of this in english

Nevermind

It's integral domain

[rocky cloak]

Protip for translating the more common math terminology.

Go to the Wikipedia page, in this case
https://es.m.wikipedia.org/wiki/Dominio_de_integridad

Then change the language to English

[chilly ocean]

Oh, yes, I forgot about that. Thank you

[true bolt]

huh til

[rocky cloak]

Not everything has a Wikipedia page of course, but it works alot of the time.

I guess the better option is to find a math terminology dictionary.

I know one for Norwegian--English, but I'm sure many others exist

[chilly ocean]

After learning Galois theory, what can I learn next about algebra?

I'm still reviewing what I learn in my math major, but when I reach that point, I would like to know how to continue

I know there's algebraic number theory

But I don't know anything else, so which path would you take?

[thorn jay]

You can go a little broader, take commutative algebra or representation theory if you haven't already and want to continue with the algebra

[chilly ocean]

Okay

[chilly ocean]

What would be a good resource to learn about it?

[thorn jay]

I'm not sure about comm alg (I'm learning it fairly nonstandardly using a book that also immediately covers algebraic geometry), but for group reps, Serre's "Linear Representations of Finite Groups" is really good

[south patrol]

A standard thing for commutative algebra which I would recommend is the book by Atiyah and Macdonald

Lots of good exercises which go further than the main text

I second this recommendation of Serre's book for reps

[chilly ocean]

Thank you for your answers

[thorn jay]

I've been McLovin ™ it

[coral spindle]

In addition to Serre's book that enpeace and potato recommended, another good book for learning rep theory is James and Liebeck

[thorn jay]

If it doesn't help then sure

[delicate orchid]

is someone forcing you to read every example and every word in the book. Do I need to call in a SWAT team?

[thorn jay]

That just means you grasp the concepts well lol

Dummit and Foote is just a very detailed book covering many topics of algebra

Serre's rep theory book which is great is only like 174 pages, and even less of actual math, with the compromise that it is very dense

Not all AA literature is long, dummit and Foote is uniquely long in that regard

[crisp knoll]

Which book is a good suggestion for commutative algebra

[karmic moat]

Group theory or ring/module theory?

For the latter, Atiyah-Macdonald

For the former idk

[neat violet]

Clark is v concise, might be more to your liking (if he covers the topics you want) https://books.google.com/books/about/Elements_of_Abstract_Algebra.html?id=B6LDAgAAQBAJ

Hth

[thorn jay]

The famous commutative algebra studying nonabelian groups

[karmic moat]

Typo

[thorn jay]

Else it would again just be modules over a PID

(abelian groups I mean)

[fading acorn]

atiyah

[chilly ocean]

I reviewed adjunctions of fields

This is making me feel pretty happy, to finally understand algebra

I felt so useless in my major, but finally I've found peace

And all thanks to Kostrikin

[thorn jay]

One can only hope to understand algebra...

[south patrol]

Wdym by this

[thorn jay]

Adjoining an element maybe?

[south patrol]

Sure ye, just not rly common notatiom

[chilly ocean]

I'm not familiar with math in english

And I sometimes do direct translation from spanish, so it may not be the name that should be said

[coral spindle]

We might say field extensions, if I'm understanding what you mean

Such things happen :)

[thorn jay]

Dutch has very weird naming conventions regarding homorphisms and isomorphisms

I refuse to learn them, as such I only do math in english

[kind temple]

what’s weird about them

[next obsidian]

Dutch

De hoomomorefism

[chilly ocean]

Yes

But I'm meaning specific field extensions

Like if you extended a field F with just an element a, and you write F(a)

[coral spindle]

adjoining an element to a field, sure

extending by a single element

[thorn jay]

Isomorfie and isomorfisme mean different things??

My teacher told it to me but I forgor

[kind temple]

do the other versions of morphism also use the fie and fisme endings?

[thorn jay]

I think so

[kind temple]

oh man

[thorn jay]

Except morphism

That's just morfisme

[kind temple]

bruh

[thorn jay]

Never heard it been called a morfie

[kind temple]

i would just start calling them ifies and ifismes

[thorn jay]

I just stopped trying to do it in dutch

[kind temple]

lol

[dull ginkgo]

morphene

[kind temple]

morfiend

[thorn jay]

A simple monoid is not necessarily a group, right?

Simple being that it has no nontrivial congruences, or every homomorphism is constant or an embedding

[tribal moss]

{0,1} under multiplication would be such an example, I think.

[thorn jay]

Right

[velvet hull]

You got some of that morfie? I could use a hit

[balmy vector]

is it true that in a solvable group G (not necessarily finite), any nontrivial normal subgroup must contain a nontrivial abelian normal subgroup?

i.e. if H is normal in G and not 1, theres a subgroup K of H thats normal in G and abelian (and not 1)

[rocky cloak]

Sure, just consider the derived series intersected with H

[balmy vector]

sorry - how would that result in an abelian subgroup though?

[rocky cloak]

A group N is abelian iff [N, N] = 1

[balmy vector]

ohhh i get it now

thanks!

[kind temple]

is there a way to construct the free group of a set using the free monoid of a set?

i feel like this would lead to a cleaner description of the construction of the free group

if you are given a set S, you can look at the free monoid F_m(S U S^-1).
i want to quotient by the smallest equivalence relation on F_m(S U S^-1) containing (1,1), (ss^-1,1), and (s^-1s,1) for all s in S or something like that

but um. idk. this is just a brainstorm. i need to think about it some more

[tribal moss]

That ought to work -- it's basically a rephrasing of the usual "concrete" construction of free groups anyway, just saying "free monoid" instead of "strings".

[kind temple]

i guess its a nice repackaging for me?

[tribal moss]

Your equivalence relation must also be closed under "if a~b then ac~bc and ca~cb".

[kind temple]

do you think there are any ways to use the universal property of the free monoid and figure out the right maps to look at instead of constructing the equivalence relation?

[tribal moss]

(Which is implicitly necessary for using it to make a monoid quotient, of course)

[kind temple]

oh, you just mean that it is compatible with concatenation

yea, okay, i wasn’t thinking of that

[kind temple]

i have had a couple ideas with this, for example, you want to be able to get point-wise inversion of a list, which is the image under the free functor of the map taking s to s^-1 and s^-1 to s on S U S^-1, and you also want reversal, which can be defined directly on the free monoid

[tribal moss]

Not sure about that -- the condition is more about how groups differ from monoids than about the monoid world itself. But you'll probably need to use the universal property of the free monoid to prove that the group you end up with is actually free.

[kind temple]

i had one more question actually related to this

so there are adjunctions Set <=> Grp, Set <=> Mon and Mon <=> Grp.
the composition of the free functors Set -> Mon and Mon -> Grp should be the free group functor Set -> Grp, right?

or at least naturally isomorphic to it

idk, i haven’t looked at the adjunction Mon <=> Grp yet

[knotty badger]

yes this is true

this follows because group homomorphisms are the same as homomorphisms of the underlying monoids

[kind temple]

which part, equality or natural isomorphism?

[knotty badger]

natural iso probably

equality of functors is a little difficult to achieve in practice

[tribal moss]

Also, adjunctions compose, and the forgetful functors Grp to Mon to Set clearly fit together.

[cursive spindle]

[kind temple]

i haven’t looked at compositions of adjunctions yet either, i probably should. so much stuff backlogged for me

[south patrol]

There is not too much to it fortunately

Hom(LL'x, y) = Hom(L' x, Ry) = Hom(x, R' Ry)

[knotty badger]

yeah the homset def allows you to compose adjunctions really easily

[kind temple]

yea, i figured as much, just some getting used to after i think about it for a bit

[knotty badger]

really just $(AB)^T = B^T A^T$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[cursive spindle]

transpose moment

[knotty badger]

fr

[thorn jay]

I shouldn't be this surprised that localisation generalizes nicely to a general context of algebraic structures with a distributive monoid operation, after all it is nothing more than localisation of a (commutative) one-object V-enriched category where V is a variety of algebras

[kind temple]

i wish i understood this last point 😭

anyways, thanks all!

[muted heron]

How should one think of the quotient \tilde{W} / W_aff? Where \tilde{W} is the extended affine Weyl group

[muted heron]

https://tenor.com/view/jfk-clone-high-i-like-your-funny-words-magic-man-jack-black-gif-18659433

[thorn jay]

Although it seems it has turned my universal algebra into monoid theory

[maiden crater]

Suppose a finite set G is closed under an associative product and that both cancellation laws hold in G. Prove that G must be a group.
Is there anyway to do this by just chosing the right elements?

[velvet hull]

Show that G has a left identity, show that G has a right identity, show that G has a total identity and every element has a two sided inverse. Then G is a group

[cursive spindle]

I like to think of elements in this quotient as the deck transformations

[rocky cloak]

It's worth noting that the empty set satisfies the above without being a group

But as an actual hint: a function on a finite set being injective implies it's bijective

[muted heron]

Can you elaborate more on this?

[cursive spindle]

The quotient is the set of elements in \tilde{W} of length 0. Any element inside the quotient acts on the fundamental alcove by some automorphism and its action on the fundamental alcove will induce a permutation of the walls of the fundamental alcove.

[muted heron]

hm

Length here meaning the usual length defined on the words?

[cursive spindle]

yes

[muted heron]

hm okay

[minor gazelle]

Hii~ good afternoon~ I wonder if this way of showing the multiplication of normal subgroup is valid

instead of using representative, I showed the sets are equal ...

[cyan skiff]

Your proof is valid since you are only using the def of normality and some of the properties of cosets. But in the first sequences of statements why did you end with an inclusion rather than an equality $(aN) (bN) = a(Nb) N = abNN = abN$ the last step follows from the closure of N. Also, I don't think this is what the proof is actually asking for. You need to show that the function is well-defined which means if a = b (any two arguments are the same) then f(a) = f(b) ( there images should be the same)

[cloud walrus]

Taaha_Tariq

[minor gazelle]

That’s a part I had doubt though I tried to justify it is well-defined by manipulating the image of the multiplication.. I felt it’s a bit off too I tried to use a double inclusion to see if I can escape the circularity…