groups-rings-fields

[smoky cypress]

well idk Galois theory is at about page 200 or something

I'll get there fast

[chilly ocean]

Can't we do an operation called irrational conjugate like

$(x+y\sqrt{b})' = x-y\sqrt{b}$ then

prove that it has nice properties

[magic owl]

Why do Galois theory when you can move on the homological algebra

To*

[cloud walrus]

Blitzkrieg:

[woven delta]

The nicer way of saying that is that conjugation is an automorphism

[magic owl]

Woke

[chilly ocean]

Yeah, exactly that

No Galois theory needed?

[smoky cypress]

[chilly ocean]

I'm not sure, I'm guessing this works

definition of the conjugate is a bit wonky for sure

[topaz solar]

You don't need galois to show that such a conjugation can preserve the whole "being a zero" thing

but uh
galois is nice because there's more stuff you can do than just take a conjugation in that sense

[chilly ocean]

Suppose that $x_1+y_1\sqrt{b} = x_2+y_2\sqrt{b}$, then $x_1-x_2 = (y_1-y_2)\sqrt{b} \implies x_1 = x_2, y_1 = y_2$. This means we can define conjugation as $(x+y\sqrt{b})' = x-y\sqrt{b}$. ($x, y, x_1, y_1, x_2, y_2\in \mathbb{Q}$)

[cloud walrus]

Blitzkrieg:

[chilly ocean]

It's clear that ' is an endomorphism

[smoky cypress]

i think you got the sign wrong

[chilly ocean]

yes. It doesn't matter though.

[smoky cypress]

yep

i mean that is the conjugate, in hs we've been taught about that

[chilly ocean]

Yeah. And now proceed for a proof which is the same as for complex conjugate being a root of polynomial with real coefficient i. e. $P(z) = 0 \implies P(\hat{z}) = 0$

[smoky cypress]

lol

i know how to do that

so I spotted the similarity

so that's why I wondered is there a generalization of this idea

[cloud walrus]

Blitzkrieg:

[chilly ocean]

it's \overline isn't it

[smoky cypress]

idk maybe

yes

[topaz solar]

$\bar{z}$

[cloud walrus]

Darkrifts:

[topaz solar]

$\overline{z}$

[cloud walrus]

Darkrifts:

[topaz solar]

bar is shorter

[smoky cypress]

ew

it's also thinner

i think one works for multiple characters and one doesn't

[topaz solar]

Compare $\mathrm{Gal}(\bar{Q}/Q)$ and $\mathrm{Gal}(\overline{Q}/Q)$

[cloud walrus]

Darkrifts:

Darkrifts:

[smoky cypress]

yeah

$\overline{xyz}\bar{xyz}$

[cloud walrus]

Whoever:

[chilly ocean]

But looking at the Galois stuff, \bar is better that time

[woven delta]

Oh so the really nice way to talk about conjugation preserving roots is in a much more general sense

So in model theory there's a notion of definability

And it's fairly easy to show that if a set is definable, then a automorphism maps elements of that set within itself

[magic owl]

I was with you until you abstracted all the way to model theory lmao

But that’s pretty cool

Are there no requirements on what category the automorphism is in

[woven delta]

So this is automorphism in a much more limited sense

[tame bear]

Suppose that $R$ is a ring with unity and for all $x$ and $y$ in $R$ we have $(xy)^2 = x^2 y^2$. Prove $R$ is commutative.

[cloud walrus]

Sigma:

[tame bear]

basically spent all day on this dumb problem

just "plug stuff in and get lucky"

[chilly ocean]

btw there are noncommutative rings with $(xy)^3 = x^3 y^3$ for all $x,y \in R$

[oblique river]

I mean that should be a pretty standard argument

[cloud walrus]

hochs:

[chilly ocean]

won't spoil the fun of looking for it

[oblique river]

it's going to be super similar to the similar proof for groups

I agree it's a dumb problem

but the proof is pretty short

ah actually I see why it's tricky, nvm

[tame bear]

i was fucking with xy - yx for like an hour

but unity is vital to the argument

i mean, $(xy)^2 = x^2 y^2$ doesnt mean commutativity if you dont have unity,
so i doubt 3rd powers would prove commutativity

[cloud walrus]

Sigma:

[oblique river]

here's a cool theorem I read recently. It might have gotten brought up in here actually... so maybe this is a repeat but: if in a group G you have (xy)^n = x^n y^n for 3 consecutive integers n, then G is abelian

but 2 consecutive integers is not enough: it's always true for n = 0 and n = 1 in any group

(so this is in some sense a "generalization" of the problems to "prove if it's true for n = 2 it's abelian" and "if it's true for n = -1, it's abelian")

[tame bear]

thats pretty neat

[chilly ocean]

yeah I remember doing these exercises. they're not so revealing ultimately .

[woven delta]

Yeah they felt more like Algebra in the high school sense

[tame bear]

its practice, i cant bang it too much

and i like(d) doing hs algebra 👀

[chilly ocean]

this reminds me of that exercise in topology where using closure and complement and interior operations one can get at most 14 (or was it 13?) distinct sets starting from a subset of a topological space

[tame bear]

that sounds awful

[mild laurel]

It's 14

It's called kuratowski's complement closure problem

It's in Munkres I'm pretty sure

[magic owl]

Is two sufficient if we restrict n away from 0,1 buncho?

[oblique river]

no, I think it still requires 3

I can't give an example

but the proof is basically just manipulate and use the identities and you really need all 3

maybe it happens to be true with 2 but I don't think so

[smoky cypress]

Why does it got to be a ring?

Can it just be a monoid?

For sigma’s question

[magic owl]

Yeah, I don’t think the proof uses the additive part

Maybe you need it

[smoky cypress]

Maybe you do, because sigma wrote xy-yx

Idk still reading topology and abstract

[magic owl]

You do

Just finished the proof haha

That was honestly a fun exercise

[brisk granite]

I don't understand the definition of phi

what is p supposed to be?

Here is the whole thing

is it supposed to be a^p ?

Also, how do I know it has a unique subgroup of order p?

[magic owl]

It's a^p

but in abelian notation

[brisk granite]

what is abelian notation?

[magic owl]

Do you know what an abelian group is?

[brisk granite]

yes

[magic owl]

well we often use + instead of \cdot

[brisk granite]

A group where all elements commute

[magic owl]

for abelian groups

[brisk granite]

ok

[magic owl]

to emphasize the abelian-ness

then if + is the operation

[brisk granite]

I see

[magic owl]

multiplication is repeated addition

so we do pn instead of n^p

I agree that the notation there is bad

[brisk granite]

ok, thankyou

[magic owl]

it should at least have been more clear imo

[brisk granite]

yea, the book could've explained it before using it

[mild laurel]

What book is it

[mild laurel]

@brisk granite

[brisk granite]

idk, I just had someone from another server send it to me

I'll ask him tho

[mild laurel]

Then how do you know the book doesn't explain it before using it wtf

[brisk granite]

he said that

also, he said it's abstract algebra by simon rubenstein

[steep hull]

salzedo

[thorn delta]

This question comes from a section about restricting homomorphisms to subgroups. I came up with a proof for this, but it has nothing to do with restricting homomorphisms. Does anyone have any idea what the question is looking for in a proof?

[final gulch]

what was your proof?

[thorn delta]

I just said that the image of phi must divide the order of G and G'. Therefore the order of Im(phi) = 1. The only such homomorphism is the trivial one

[final gulch]

looks good to me

[thorn delta]

hnnghh but the section is about restricting homomorphisms!1!11

[final gulch]

idk I wouldn't try to overcomplicate it😋

[thorn delta]

alright

[tender mist]

@thorn delta well a restriction is needed to prove that the order of the image divides the order of G, I think

Don't know if you were asked to justify it though

[chilly ocean]

Maybe it's used in another exercise which is more on the topic

[thorn delta]

I noticed later on that its kinda used for something else in the section. It was basically the problem I proved here, but regarding a homomorphism with its domain restricted to a subgroup of G whose order does not share any factors with G'.

The order of the image divides G because the preimage of any element in the image is a coset of the kernel of the hom. The cosets partition G... etc etc

[chilly ocean]

ok. I'm bored so let me give an exercise from algebra

book - Kostrikin

Every subgroup of index 2 is normal (H - subgroup, G - group, then |G|/|H| is the index)

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

[cloud walrus]

Blitzkrieg:

[mild laurel]

@brisk granite There's no book by that title and author

[chilly ocean]

I think something like this might work

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

Blitzkrieg:

[chilly ocean]

and it's pretty clear it's a homomorphism as well

[cloud walrus]

Blitzkrieg:

[chilly ocean]

which is part of reason why it's clear

yeah. I'm done I suppose

because otherwise it would have order 3 which is impossible because y would be of order 6

so simple yet so convoluted

How to prove that subgroup of order 2 is normal

[cloud walrus]

Blitzkrieg:

[chilly ocean]

Not that I know anything about group theory

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

Blitzkrieg:

Blitzkrieg:

[chilly ocean]

that's a bit of exercise

idk what he means by action on left cosets

Wait there is something. Action by left translations

[cloud walrus]

Blitzkrieg:

[chilly ocean]

So let's take any such transitive action and I guess we need to construct H

[cloud walrus]

Blitzkrieg:

[chilly ocean]

We know that H_x are non-empty

[cloud walrus]

Blitzkrieg:

[chilly ocean]

Doesn't seem like it's useful, so maybe

hmm

Take $y$ so that $e\in H_y$. Let $g, h\in H_y$, then there exists $x'$ such that $gx' = y$, hence $ghh^{-1}x' = y$ so $gh\in H_y$

and ability to take inverses

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

Blitzkrieg:

[chilly ocean]

lol

Oh, there is something

[cloud walrus]

Blitzkrieg:

[chilly ocean]

We can probably take arbitrary stationary subgroup as our H and bijection between left cosets and X as our sigma

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

Blitzkrieg:

[chilly ocean]

For $y\in X$, $\sigma(y) = aH$. $\sigma^{-1}\Phi_x'\sigma(y) = \sigma^{-1}(xaH)$

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

[chilly ocean]

I'd have to write sigma more explicitly

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

[chilly ocean]

so actions are in fact equivalent

end

I was chocking on this one a bit, I'm not very familiar to group actions though

Next exercise

[cloud walrus]

Blitzkrieg:

[chilly ocean]

that is

[cloud walrus]

Blitzkrieg:

[chilly ocean]

centrum is set of elements which commute with every other element if I recall correctly

yeah

of course it's normal subgroup

[cloud walrus]

Blitzkrieg:

[chilly ocean]

Take $y$ not in the centrum, then if $y^2\neq e$, $y^2$ generates $Z(G)$. This would mean that $G$ is cyclic, because $G = Z(G)\cup yZ(G)$, contradiction

[cloud walrus]

Blitzkrieg:

[chilly ocean]

If $y^2 = e$, then $p = 2$ and every element of $G$ has order $2$. But then every element of $G$ commutes, contradiction

[cloud walrus]

Blitzkrieg:

[chilly ocean]

Because if every element has order 2, then a group commutes, it can be done as an exercise

https://cdn.discordapp.com/attachments/496784958430380033/614626434153054209/392013939191316491.png

the equality is wrong and that wouldn't mean that G is cyclic, but we can still derive that G has a subgroup of order 2p (namely the one generated by y), from which p = 2

from which G is cyclic anyway

why would y^2 be in centrum 🤔

G/Z(G) is abelian

$G = {x^ky^l: k, l\in{0, ..., p-1}}$ where $x\in Z(G)$

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

[thorn delta]

what is the question

[chilly ocean]

idk I was solving exercise here
prove that group of order p^2 where p prime is abelian using that centrum of p-group is non-trivial

I figured that you need to use that Z(G) and G/Z(G) are cyclic here so G is of the form above (if order of Z(G) is prime)

and so trivially abelian

out of boredom

[tame bear]

If $G/Z(G)$ is cyclic, then $G$ is abelian.
And if $G$ has order $pq$, both prime, and non trivial center, then $G$ is abelian from above.

[cloud walrus]

Sigma:

[thorn delta]

https://media.discordapp.net/attachments/533698807301406721/614574616320540672/unknown.png
Im not really sure what to do on this one. I see that for some z in the intersection of the cosets, you can write an arbitrary h = x^-1 yk, but I am not really seeing how to proceed from there

[mild laurel]

Wait what are these letters

Is h in H and k in K? @thorn delta

[thorn delta]

yes. z = xh = yk for some h in H and k in K.

[chilly ocean]

if $z \in xH \cap yK$ then $xH \cap yK = z (H \cap K)$

[cloud walrus]

hochs:

[chilly ocean]

as a hint

hard to give a hint here without giving it away though

[thorn delta]

grr i don't really see why that is tho

[chilly ocean]

can you show at least one of the subset inclusions?

for example $z(H \cap K) \subseteq zH = xH$ and $z(H \cap K) \subseteq zK = yK$

[cloud walrus]

hochs:

[thorn delta]

im stumped. I know what you mean, just no idea how to justify it

[chilly ocean]

let's start with $z(H \cap K) \subseteq zH$. Can you justify this?

[cloud walrus]

hochs:

[thorn delta]

you can write any element in $z(H \cap K)$ as $zh$ for some $h \in H$.

[cloud walrus]

kxrider:

[chilly ocean]

yes, now what about $zH = xH$? ($z \in xH \cap yK$ as a reminder)

[cloud walrus]

hochs:

[thorn delta]

zh = xh'
z = x(h'h^-1)
zH = x(h'h^-1)H = xH right?

[chilly ocean]

Yes.

so you've shown that $z(H \cap K) \subseteq xH$. Likewise (similar calculation) $z(H \cap K) \subseteq yK$ and hence one of the set inclusions $z(H \cap K) \subseteq xH \cap yK$

[cloud walrus]

hochs:

[chilly ocean]

now let's try the other inclusion. To start, take an element $z' \in xH \cap yK$ and you should show that $z' \in z(H \cap K)$

[cloud walrus]

hochs:

[chilly ocean]

the condition $z' \in z(H \cap K)$ can be rewritten as $z^{-1} z' \in H \cap K$. This in turn being equivalent to $z^{-1}z' \in H$ and $z^{-1} z' \in K$. We know that $z' \in zH \cap yK$, or $z' \in zH$ and $z' \in yK$ in particular.

[cloud walrus]

hochs:

[thorn delta]

do you mean xH?

[chilly ocean]

right, $xH$

[cloud walrus]

hochs:

[chilly ocean]

which we know $xH = zH$ from above

[cloud walrus]

hochs:

[chilly ocean]

likewise $yK = zK$.

[cloud walrus]

hochs:

[thorn delta]

and to say $z^{-1}z' \in H$ is equivalent to saying $z' \in zH$ correct?

[cloud walrus]

kxrider:

[chilly ocean]

Yes.

[thorn delta]

I am confused by what I did to show that xH = zH. It seems like you could use the same argument to show that any two cosets of the same subgroup are equal which is clearly not true...

[chilly ocean]

it comes from the fact (you can check this as a review) that if $z \in xH$ then $zH = xH$

[cloud walrus]

hochs:

[thorn delta]

oh okay i see now i used the fact that z is in the intersection $xH \cap yK$

[cloud walrus]

kxrider:

[thorn delta]

Anyway, I think I will sleep on this. Thanks for helping C:

[chilly ocean]

np. night

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

[chilly ocean]

I have no idea

[cloud walrus]

Blitzkrieg:

[fading wagon]

Did they specify the geometric interpretation?

[final gulch]

by using S^3 ~ SU(2)

you multiply the corresponding matrices in SU(2)

[cloud walrus]

Blitzkrieg:

[final gulch]

do you know how to write down a diffeomorphism from S^3 to SU(2)?

I was trying to write it but I forgot how to make matrices in here 😂

[chilly ocean]

$ a(\varphi, \theta, \psi) := b_{\varphi}c_{\theta}b_{\psi} =
\begin{Vmatrix}
\cos(\theta/2)e^{i(\varphi+\psi)/2} & i\sin(\theta/2)e^{i(\varphi-\psi)/2} \
i\sin(\theta/2)e^{i(\psi-\varphi)/2} & \cos(\theta/2)e^{-i(\varphi+\psi)/2}
\end{Vmatrix}
$

[cloud walrus]

Blitzkrieg:

[chilly ocean]

nah, it's an algebra book

[fading wagon]

but now they are writing it as a vector of 4 numbers

[chilly ocean]

I don't know either, there was no such notation earlier

[fading wagon]

I'm pretty sure they are expressing it in terms of $\mathbb{R}^4$

[cloud walrus]

Element118:

[final gulch]

$(\alpha,\beta)\mapsto \left(\begin{array}{cc} \alpha & -\bar\beta \ \beta & \bar\alpha\end{array}\right)$

[cloud walrus]

Seoin:

[final gulch]

there we go

[chilly ocean]

ok

yes, there is something like that I just realized

[final gulch]

with \alpha = a + bi and \beta = c + di you get a map R^4 -> 2x2 complex matrices

and if (\alpha,\beta) is in S^3 then the image has determinant 1 edit: and is unitary

[chilly ocean]

so the exercise is to use geometric interpretation of SU(2) to calculate something in R^4?

seems a bit odd

that's just direct calculations from now, thanks

[final gulch]

yeah they are just defining a product on the sphere using this diffeomorphism and asking you to compute something using the definition

[chilly ocean]

it's kind of like we're considering an algebraic object and deliver it to a non-algebraic object so it can too have an algebraic structure

its pretty cool

[final gulch]

well SU(2) is already both an algebraic and a geometric(smooth) object
since it is a submanifold of 2x2 compex matrices = C^4

[smoky cypress]

If I want to check the multiplication table of a binary operation is associative, what can I do besides the brute force method ( O(n^3) )

[chilly ocean]

I'm glad you ask because it's

kind of my subject (semigroups)

Look up Light's associativity test

[thorn tree]

I remember there's a way to do it in n^2, though I forget the name

yeah that ^

[smoky cypress]

Oh ok

[chilly ocean]

https://en.wikipedia.org/wiki/Light's_associativity_test

[thorn tree]

what's the operation?

if it's a common one (addition, multiplication, function composition, etc) you can probably get away just by saying "we know this is associative"

[chilly ocean]

nah he probably tries to compute small semigroups

check if they are

[smoky cypress]

Lol

I’m just wondering

This is just merely my curiosity

[thorn tree]

got it

[chilly ocean]

Light's associativity test is the only one I skipped because it was explained badly in my book

[thorn tree]

also note none of this works on infinite groups... you're usually better off writing the expressions out and showing they're the same

[smoky cypress]

It says that this algorithm doesn’t improve on the worst case scenario

[thorn tree]

like expand (a + b) + c and a + (b + c)

[smoky cypress]

Oh

Yeah of course

[chilly ocean]

There's some probability algorithms mentioned on the bottom

[smoky cypress]

If the binary operation is define using some expression then it should be easy

Maybe

[chilly ocean]

by Rajagopalan and Schulman

in general, it's hard to check something is associative

[thorn tree]

it's hard if you do it in a brute force approach

[smoky cypress]

Ok

[chilly ocean]

no

It's hard in general

brute force or not

[smoky cypress]

Lol

[chilly ocean]

i'm the expert here k

[delicate chasm]

🤔

[chilly ocean]

I'm joking

but it's true

[thorn tree]

by "hard" do you mean mathematically hard or just hard for people

[chilly ocean]

I mean that given a magma

it's hard to check if it's associative

usually

[thorn tree]

I don't think an n^3 algorithm counts as mathematically hard

you still haven't defined hard

[smoky cypress]

Lol

n^3 algorithm is really slow in my opinion

But

[delicate chasm]

There's no easy algebraic method for it I'm guessing

Strict associativity is boring anyway :P

[smoky cypress]

I don’t think we’ll ever need a multiplication table that bugger than 100

[thorn tree]

nah, polynomial time is usually considered fast

[delicate chasm]

Coherence data ftw

[smoky cypress]

Well ofc polynomial is good

[thorn tree]

if you think n^3 is slow consider 3^n

[chilly ocean]

it's not boring

[smoky cypress]

Puer is boring

[delicate chasm]

☺

Nah, it's not really boring

Just too hard 😂

[chilly ocean]

Checking for it is hard

Harder than checking for groups anyway

[smoky cypress]

Why is that so?

[chilly ocean]

Well. I meant more like

if we want to find groups of a given order say

it's pretty easy for a computer

[smoky cypress]

Ok

I see

[chilly ocean]

but with semigroups, it's ultra-hard

[smoky cypress]

Lol

But is checking associativity of a semi group hard?

Well

Nvm

That makes no sense

[chilly ocean]

yeah lol

checking associativity for a semigroup is ultra easy

semigroup is associative

[smoky cypress]

Prove semi groups are associative

[delicate chasm]

Check associativity for a 2-group

It's.. interesting xD

What was the original question? 🤔

[chilly ocean]

good algorithms for checking associativity of a magma

[smoky cypress]

What’s a magma

[tame bear]

set with binary operation

[chilly ocean]

magma = groupoid = set with binary operation

[tame bear]

ha i win

got it first

[maiden ocean]

snipus

[chilly ocean]

I was just typing more info

so I win

[smoky cypress]

Lol

[delicate chasm]

magma=groupoid?

[smoky cypress]

Different name I suppose

[chilly ocean]

yes, it's another name

[smoky cypress]

Sniped

[tame bear]

groupoid is a category with every morphism invertible

[delicate chasm]

^

[smoky cypress]

[chilly ocean]

groupoid is also magma

[delicate chasm]

Different concepts I guess

Maybe

[chilly ocean]

yes

but same name

[smoky cypress]

Now why would you use higher math terms, this is like using peano axioms and recursive definition to calculate 12592*298592

Jkjk

[tame bear]

category theory is nice

[chilly ocean]

well maybe

I didn't get to a point in which it would be interesting for me yet

[tame bear]

higher math does make algos easier

fair

just drops a dozen examples

[chilly ocean]

anyway. Let's do some algebra maybe.

exercise is like this

[cloud walrus]

Blitzkrieg:

[chilly ocean]

So I'm guessing no, because there is quaternions

https://en.wikipedia.org/wiki/Algebra_over_a_field

[cloud walrus]

Blitzkrieg:

[chilly ocean]

has complex numbers as it's subalgebra

so, a plane isomorphic to complex numbers

[cloud walrus]

Blitzkrieg:

[chilly ocean]

because it's just a matter of taking correct basis

and it's supposed to be associative

is it not always associative? I guess not

[cloud walrus]

Blitzkrieg:

[chilly ocean]

it's an algebra with division, but nothing indicates we can take an inverse

by division I guess they mean we can left/right cancel

I guess they mean division algebra

https://en.wikipedia.org/wiki/Division_algebra

[final gulch]

I don't think it's easy to prove? that page says "Later work showed that in fact, any finite-dimensional real division algebra must be of dimension 1, 2, 4, or 8. This was independently proved by Michel Kervaire and John Milnor in 1958, again using techniques of algebraic topology, in particular K-theory."

I mean maybe it's easier to just rule out dimension 3? but idk

[chilly ocean]

It's 4th exercise in Kostrikin's third tome of introduction to algebra

chapter one, classic groups of small 'dimensions'

idk if dimensions is the right word

I heard something like that, but I assume, it must be easy for dimension 3

[final gulch]

I see
honestly I probably have no idea about the subject matter leading up to this exercise

[chilly ocean]

it was just groups of matrices (like earlier) + quaternions

[cloud walrus]

Blitzkrieg:

[final gulch]

how about assume that ij = a + bi + cj and left multiply by i to get -j = (ac - b) + (a + bc)i + c^2j

[chilly ocean]

ohh

[final gulch]

unless I made a mistake then gotta have c = 0 hence a = 0 hence b = 0

[chilly ocean]

now you can use that they are independent

[final gulch]

so ij = 0

oh wait

not c=0, it's already a contradiction at that point

because c^2 can't be -1

[cloud walrus]

Blitzkrieg:

[final gulch]

right

[chilly ocean]

works 👌🏿

[final gulch]

[potent lynx]

yo blitz

where did you learn algebra from

[chilly ocean]

University

you just got to have good teachers

who make the subject really likable

and then you just go from there

I've tried to learn it in highschool, but there was no good resources

I kind of gave up, but on second year of uni, professor of algebra with which I had lessons with really, gave me a second wind

[smoky cypress]

Nice

Imagine learning algebra in high school

[chilly ocean]

It was last class of high school 🤷🏿

[smoky cypress]

Your school teaches abstract algebra?

In high school?

[chilly ocean]

no

[smoky cypress]

Oh

[chilly ocean]

I mean that I was trying to teach myself group theory in the last year of high school

but only resources I found online were bad

[smoky cypress]

Oh

Lol

Well I’m reading jacobson rn

Do you consider that a good resource?

[chilly ocean]

I don't know it

[smoky cypress]

Jacobson’s basic algebra

[chilly ocean]

From group theory I know only the stuff I had in uni

[smoky cypress]

Oh I see

[chilly ocean]

or rings etc

and Kostrikin, though I haven't really read it

it's a good book though, Kostrikin, maybe a bit hard

[smoky cypress]

Ok

[chilly ocean]

but everything is easier when you have an algebraic background

of some kind

probably the most confusing topic in group theory is actions via a group

from my shallow knowledge about it

[smoky cypress]

?

Actions via a group

The symmetry group?

[tame bear]

no

group actions

[smoky cypress]

Lol...

Ok idk what that is

Maybe it’s coming up

I just finished the chapter on groups

Now it’s isomorphism and cayley’s theorem

Pretty useless i feel like

[tame bear]

the symmetry group of a triangle acts on a triangle exactly like you think it does

cayleys is pretty useless

it is interesting to think about how small of a symmetric group is sufficient for any group

using cayleys argument you get that S3 is a subgroup of S6, but its obviously contained in S3

[smoky cypress]

Yeah

It’s pretty interesting

[chilly ocean]

Yes. It will tell you that it is isomorphic to a subgroup of symmetric group, but not the one with least order

[tame bear]

least order is difficult apparently

[chilly ocean]

I like Cayley theorem because it has an analogy in the study of inverse semigroups

everything that's analogous is a bit cool

imo

[tame bear]

cat is entirely analogies

[potent lynx]

i am exactly in high school

you grad school? @smoky cypress

[smoky cypress]

I forgot to change my name

Allow me to introduce myself

[final gulch]

🙄

[potent lynx]

oh fuck really?

[chilly ocean]

It's highly doubtful

[smoky cypress]

@final gulch shhhhhh

[chilly ocean]

lol

[smoky cypress]

[potent lynx]

for reals man

u undergrad or what

[smoky cypress]

Of course not

I don’t have a PhD

Lol

[chilly ocean]

I could probably get into ring theory easily

not that it's of use to me

[tame bear]

im learning rings right now

[chilly ocean]

because rings are just semigroups with 0 with additional binary operation of addition

there's just more leverage because of addition

[chilly ocean]

Seems interesting

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

[chilly ocean]

and then it's trivial to see that D(a) = C(a)

and other way is even more trivial, since C(a) is always a subgroup of G

it was easy, but this exercise has 2 parts so maybe the second will be better

[cloud walrus]

Blitzkrieg:

[chilly ocean]

let me see

[cloud walrus]

Blitzkrieg:

Blitzkrieg:

[chilly ocean]

so it's a subgroup

[cloud walrus]

Blitzkrieg:

[chilly ocean]

Sigma

which is slightly stronger than what you said yesterday

[smoky cypress]

Just realized there is an analogous result of cayley’s theorem about monoid

Interesting I guess

[chilly ocean]

What is that result?

https://planetmath.org/cayleystheoremforsemigroups

this one?

[tame bear]

is that stronger? it just looks like the contrapositive

[chilly ocean]

well. You said something like, suppose G has order pq... etc

[tame bear]

oh that, right

[chilly ocean]

ya. I wasn't aware that you have Cayley theorem for semigroups, interesting

with semigroup of transformations in place of symmetric group

[smoky cypress]

Yeah

I just read about it in the intro of this chapter

[chilly ocean]

there's also a similar theorem for inverse semigroups

that they are isomorphic to a subsemigroup of https://en.wikipedia.org/wiki/Symmetric_inverse_semigroup

which is something between a symmetric group and transformation semigroup

[tame bear]

Must every ring with a prime number of elements be commutative.
Prove or find a counter example.
Am i right in saying that, \ if there exist $a,b \in R$ (which dont commute) with $ab - ba \neq 0$ that any element of $R$ is of the form $n(ab -ba)$?

[cloud walrus]

Sigma:

[chilly ocean]

well. Additive group is cyclic (with prime number of elements)

so yea

[tame bear]

kay

just feels like im missing something

[oblique river]

does your ring have unity?

[tame bear]

all rings have unity

[oblique river]

if so: every ring with unity, commutative or not, admits a unique homomorphism from Z

[chilly ocean]

oof

[oblique river]

f: Z --> R defined by f(1) = 1

[chilly ocean]

not all rings, those which have unity are called unitarial rings

[oblique river]

lmao blitz pls no

nonunital rings don't deserve to be called rings

they should be called "nonunital rings"

[tame bear]

technically havent gotten to ring morphisms yet

[bleak abyss]

Manifolds with boundary aren't manifolds

[chilly ocean]

but

[oblique river]

great, then say that the subring generated by 1 is a subring

that subring is clearly commutative

just argue that it must be the whole ring

[bleak abyss]

I think the notion of a non-unital ring isn't useful enough that we should have to specify unit when we have one

[tame bear]

nonunital rings
or rngs as i like to call then
dont exist

[chilly ocean]

[bleak abyss]

If you're doing stuff like Banach or C* algebras then maybe

But otherwise you're usually working with unital rings

Non-unital ones are ideals anyway

[tame bear]

the book doesnt include unity in its definition

shame

im learning Gabriel algebra i guess

[oblique river]

are you using D&F?

[tame bear]

Saracino

[bleak abyss]

How many algebra books are there holy crap

[chilly ocean]

'Non-unital ones are ideals anyway' 🤔

[tame bear]

too many, cuz too few of them are good

[bleak abyss]

I'm pretty sure given a non-unital ring you can stick on a new element 1 and then get a unital ring

And your non-unital ring becomes an ideal in the unital ring

[chilly ocean]

yeah

you can

[bleak abyss]

Tbh the world needs just 3 algebra books

[tame bear]

isnt there a map from R to
Z x R
or something

[bleak abyss]

Artin if you don't know linear algebra

Jacobson if you do

And Lang or D&F for looking things up

That's it

[tame bear]

and aluffi

[chilly ocean]

3 algebra books 🤔 maybe for uni students

[bleak abyss]

Aluffi's exercises aren't that great I'm told, I'm pretty sure anyone who would read that is served well enough by Jacobson or Lang

[chilly ocean]

and it's not like your location isn't a thing

and the language you speak

[bleak abyss]

I read that as "localization" and was confused

[chilly ocean]

let's just end the discussion

[bleak abyss]

Sure, take those and translate them

[tame bear]

https://twitter.com/UnusualVideos/status/1165317554513960960
cant post in cats but this one is good
yeah, i cant say i recommend aluffi much

Unusual Videos 🤔 (@UnusualVideos)

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Retweets

6090

Likes

23230

▶ Play video

Twitter

it has a bit too much faith in the reader

[final gulch]

oh my god what did i just see

[tame bear]

rings are weird
a nonunital ring can contain an ideal with unity

[chilly ocean]

Is that weird?

[tame bear]

yeah

[chilly ocean]

Seems normal. Idempotent elements generete ideals.

Or not ideals but subrings. Those can be ideals as well, seems pretty ok

[oblique river]

what do you mean by "ideal with unity"

cuz surely you don't mean the ideal contains the unit of the ring because you said the ring isn't unital

[tame bear]

it can contain its own unity

that isnt unity for the whole ring

[chilly ocean]

Should be clear

[oblique river]

can't that happen in unital rings?

if R and S are two rings, R is an ideal in R x S

which has a "unity" (1,0)

I was clarifying because I didn't agree with the statement so I wanted to make sure we were using the same definition

[chilly ocean]

Given idempotent e of a ring S, can you have ideal I of that ring, such that e is unity of I.

[oblique river]

yes, idempotents correspond to decompositions of your ring as a product

[chilly ocean]

Huh. That was a guess lol

What do you call a ring with no idempotents then?

[oblique river]

"a ring without idempotents"

maybe you should specify "nontrivial idempotents" because a ring always has 1 and 0

[chilly ocean]

Yes, sorry

[oblique river]

but in any case, a ring has no nontrivial idempotents if and only if its specturm is connected

in the zariski topology

[chilly ocean]

I'm thinking it would take a while to explain to me what that means

[oblique river]

for any commutative ring R, consider the set of prime ideals of R, call this set Spec R

it turns out that you can define a topology on this set in a natural way

it's called the zariski topology

that's about it

[chilly ocean]

So, decomposition is for commutative rings?

[oblique river]

hmm, sure yeah I guess my statement earlier about the spectrum would only be valid for commutative rings

[chilly ocean]

' yes, idempotents correspond to decompositions of your ring as a product' holds in either case?

[oblique river]

oh, yes I think that is true for noncommuative rings

as well

for any ring R, if e is an idempotent then so is (1-e)

and you look at the ideals generated by e and (1-e) and those are the rings that make up the product

although maybe that's hard if you're not commutative

cuz of right vs left vs 2-sided ideals

I think the conclusion is: don't ever work with noncommutative rings

[chilly ocean]

So, maximal connected subsets of ring in zariski topology make up decomposition basical

[oblique river]

idk what simple means

but something like that is true

[chilly ocean]

I wrote wrong

[oblique river]

ah okay, yes that is correct

[chilly ocean]

This zariski topology seems really interesting

[oblique river]

it is :)

I mentored an undergrad who was learning about it this summer

so I've been thinknig about it recently

[chilly ocean]

Thank you.

[woven glade]

So I have this question for my abstract algebra class that is:
suppose we have two functions f: X --> Y and p: X ---> Q.
Find a necessary and sufficient condition under which "f factors through p",
ie such that there exists a function F: Q --> Y such that
f = F * p, where * means composition of functions.

I'm new to algebra as this is my first time taking this course, and I'm confused on where to start. I'm trying to think about what condition will this function F exists. I

[cloud walrus]

Blitzkrieg:

[chilly ocean]

i. e. p(x) = p(y) then f(x) = f(y)

one way is clear

suppose that p(x) = p(y) implies f(x) = f(y) and try to prove the other one

ok. Enough hints

I have that listed as a lemma in one of my books, I just copied it tbf

functions on sets have a lot of similarities to linear operators on vector spaces, and by that I mean that number of elements in a set plays similar role to dimension of a vector space etc. At least for functions from the same set on itself and operators from the same set to itself.

I wonder if there's a theory that emphasizes this relation. Like for example maybe category theory

[woven delta]

Yes

Vector spaces are cardinals from the categorical perspective

Assuming AOC

@chilly ocean

[bleak abyss]

You'll want adjoint functors for this

Fix a field k

Given a set S, you can form the vector space of formal linear combinations of elements of S, call that k\langle S\rangle

So this is a vector space with S as a basis

And then given a vector space, you can just think of it as a set, and forget that there are other operations involved

The first guy is called the free functor, the second is the forgetful functor

The thing is, these are adjoints

Meaning

Let's call F the free functor and U the forgetful functor

Then Hom_{Set}(S,U(V)) = Hom_{kVect}(F(S),V)

This says that if V is a vector space, then set functions from S to V are the same as linear functions k<S> -> V

In day to day language, this means that a function on the basis of a vector space extends linearly to the whole space

[woven delta]

@bleak abyss did I trigger you with that assuming AOC remark?

[bleak abyss]

Yeah I'm about to do this to you

[woven delta]

😻

[uncut girder]

Nice

I used to make stick figure fights in flipnote

Not that good)

[clear obsidian]

Let X be totally ordered. so is connexity equivalent to trichotomy?

(I would think it is clearly since were comparing a <=b or a>= b to: a < b, a > b, or a = b)

[chilly ocean]

Did you mean for X to be partially ordered?

[woven delta]

Connexity?

[cloud walrus]

Blitzkrieg:

[chilly ocean]

but saying that connexity is equivalent to trichotomy is wrong, because it implies we mean the same relation, the one which partially orders the set X

I dont think this has anything to do with abstract algebra though

I didn't specify but everyone knows what < means

[chilly ocean]

So far, Abstract Algebra is going great. Though I find myself overthinking these 1-4 line proofs so much.

[stone fulcrum]

The early proofs are the most fun ones

[fading wagon]

Don't worry about overthinking them. Just make sure you don't underthink them

[smoky cypress]

What are the uses for cycles and transpositions?

Like these things about permutation is pretty intuitive, but I don’t really see a lot of applications

[bleak abyss]

Are you asking about why S_n in general is important, or why cycles and transpositions are good for understanding S_n?

[smoky cypress]

Why cycles and transpositions are important

I know permutations are important for sure

[bleak abyss]

So, the idea is that these guys both generate S_n

Every permutation can be written pretty much uniquely as a product of disjoint cycles

[smoky cypress]

Yeah but that’s pretty obvious to me

Like

I’ve figured this out when I played around with permutations

[bleak abyss]

So then the idea is that oftentimes you can find things out about general permutations by just trying them for cycles, and cycles are sorta clear in your mind

For example, what's the order of a permutation?

[smoky cypress]

Oh i c

[bleak abyss]

Well, write it as a product of disjoint cycles

Disjoint cycles commute

[smoky cypress]

The lcm of all the orders of the cycles

[bleak abyss]

Exactly

Are you familiar with conjugation in a group?

[fading wagon]

and when we have a symmetric group $S_n$, it can act on a set of n elements

[cloud walrus]

Element118:

[smoky cypress]

Nope

[fading wagon]

Conjugation for symmetric group is basically doing a renumbering

[smoky cypress]

[fading wagon]

So the cycle structure stays the same

[smoky cypress]

Ok

[fading wagon]

$g^{-1}xg$

[smoky cypress]

[cloud walrus]

Element118:

[smoky cypress]

Renumbering seems like applying a permutation to every element in a permutation group

[bleak abyss]

So, you say two elements x and y are conjugate if there's some g such that g^{-1}xg = y

For example, take the group of invertible matrices. Two matrices are conjugate iff they're similar

[smoky cypress]

Uhhh

Similar matrices?

[fading wagon]

yeah, that's the definition

[smoky cypress]

Ok

[fading wagon]

If A, B are similar matrices, then $A=P^{-1}BP$ for some P

[cloud walrus]

Element118:

[smoky cypress]

i havent learned that but ok

It’s just true by definition

[bleak abyss]

The notion of similarity makes sense even for non-invertible matrices, and it turns out two matrices are similar if and only if they're representations of the same linear transformation in a different basis

That idea is pretty simple

If P is invertible, then it sends one basis of R^n to another

[smoky cypress]

Ok

[bleak abyss]

(where R is any field 🙃

[smoky cypress]

I kind of see it

[bleak abyss]

So the idea is push your basis across, apply the linear map, revert to original coordinates

[smoky cypress]

But it’s pretty high level and I need to learn linear algebruh

autocorrect automatically corrected it to bruh not my fault

[bleak abyss]

Yeah that'd be important. Artin does both algebra and linear algebra for what it's worth

Nah I'm holding you accountable for that

But yeah so

One nice thing is that a lot of linear algebraic things are invariant under similarity (characteristic and minimal polynomials in particular)

[fading wagon]

and if you do linear algebra and abstract algebra, we have representation theory

[bleak abyss]

So you can say "Define the blah invariant of a linear transformation by picking a basis and doing this to a matrix", and the point is that this is well-defined

[smoky cypress]

Wait so artin’s book has both linear algebra and algebra?

In the same book?

[bleak abyss]

Since if you choose two different bases, you'll get similar matrices, so they share it

Yup

[smoky cypress]

Ok I’ve actually always thought that’d be a good idea, because linear algebra seems like just the study of a homomorphism

[bleak abyss]

Anyway so, conjugacy ends up being a super important idea in group theory

[smoky cypress]

Between vector space

Ok

[bleak abyss]

For example, conjugate elements have the same order

Actually so one good exercise is to show that conjugation by any element is an automorphism of a group

[smoky cypress]

Not sure what automorphism is yet

[bleak abyss]

An automorphism of a group G is an isomorphism G->G

[smoky cypress]

But what is the definition of conjugation?

Ok

[bleak abyss]

So for example, identity is an automorphism

[smoky cypress]

Yep

Trivial one

[bleak abyss]

The map sending an element to its inverse is an automorphism iff the group is abelian

[smoky cypress]

But group is invertible tho

[bleak abyss]

I said abelian not invertible

Smh learn to read

😛

[smoky cypress]

[fading wagon]

groups are not invertible

[bleak abyss]

Anyway so, the idea there is that (gh)^{-1} isn't necessarily g^{-1}h^{-1}, it's h^{-1}g^{-1}

[smoky cypress]

Your mom isn’t invertible

Ok

[bleak abyss]

When are those guys equal for all g and h? When G is abelian

Now the exercise is this

[cloud walrus]

Dami:

[smoky cypress]

🤔

I need to go rn but 30 min later probably will try this

Seems reasonable though

Ah it’s pretty trivial

[bleak abyss]

Yup

And this means conjugate elements share a lot of properties. For example, they must have the same order

[smoky cypress]

It’s a homomorphism is pretty clear

Oh

That’s right

Isomorphism preserves a lot of stuff

[bleak abyss]

Yup

And there's a lot more going on, conjugation is one of the most important things out there in group theory

Now: given an permutation, it has a cycle type

So when you write a group as a product of disjoint cycles, I'm gonna put a requirement here

First off, write all the fixed elements as 1-cycles

So (134) will be written as (134)(2)

[fading wagon]

Well, the missing one-cycles will remain missing one-cycles anyway

[bleak abyss]

Sure but I'm gonna talk about cycle type so I want to keep track of everything

Second, I'm gonna order the cycles in a certain way (they commute so yeah I can do that): longest cycles first

So now the cycle type of a permutation is just the list of the lengths of the cycles in order (this is now unique)

So (13489)(25)(7) has cycle type (6,2,1)

Prove: two permutations are conjugate in S_n if and only if they have the same cycle type

[smoky cypress]

Wait what?

Two permutation σ and γ are conjugate under that certain kind of automorphism f if f(σ)=γ right?

Since conjugate is a map

[chilly ocean]

those are called inner automorphisms

[smoky cypress]

[chilly ocean]

two elements of a group a, b are called conjugate (or that they belong to the same conjugency class (?)) if there exists an inner automorphism f such that f(a) = b

conjugency, I'm not sure if you spell it like that

[smoky cypress]

No more definitions I’m already in a sea of words that I don’t understand

[fading wagon]

An inner automorphism $f$ of a group $G$ is:\

1. Pick an element $g \in G$ \
2. $f(x)=g^{-1}xg$.

[smoky cypress]

But ok

[chilly ocean]

Additionally, relation of being a conjugate is an equivalence relation. Abstraction classes of this relation for the group S_n are very simple as dami already mentioned. You should to have your way with the exercise he gave you.

[smoky cypress]

Ok

I see

Since all groups are a perm group

I guess this is pretty important

[fading wagon]

all finite groups are isomorphic to a subgroup of a symmetric group

[smoky cypress]

Also @fading wagon put {-1}

Yeah

[fading wagon]

welp, forgot to proofread

[smoky cypress]

Also maybe put $ around the g, f, G

Lol

[cloud walrus]

Element118:

[woven delta]

@fading wagon not just all finite groups

All groups in general

[fading wagon]

Okay, then you need to introduce infinite symmetric groups

[oblique pecan]

I just need answer cause my Brother asked me as a test

[fading wagon]

@oblique pecan prealg

[oblique pecan]

K

What do u mean

[fading wagon]

this isn't abstract algebra

[oblique pecan]

K

[fading wagon]

probably should move to some #❓how-to-get-help channel

(try not to target alpha)

[oblique pecan]

K

[woven delta]

Sym(S) is the set of all bijections of S @fading wagon

There shouldn't be a distinction between finite and infinite symmetric groups

[fading wagon]

there is already a distinction, one is finite and the other is infinite

[woven delta]

I mean in the way it's taught

[fading wagon]

but sure, let's call all of them symmetric groups

[woven delta]

[fading wagon]

I suppose infinite symmetric groups don't have alternating groups?

(as a subgroup)

[woven delta]

No I don't think so

The point of cayleys theorem is that you can view every group as acting on its underlying set

Which gives an injection into Sym(G)

So every group is a subgroup of a symmetric group

[fading wagon]

if you put it that way

sure

[woven delta]

For example you can view the element 1 of Z as the bijection f(x) = x+1

And so on

[clear obsidian]

I was just asking about order relations here because I was constructing the proof that theres always a maximal ideal using Zorn's lemma

I guess maybe elementary number theory would have been better to ask that in idk

[mild laurel]

Is this a question?

[clear obsidian]

the second thing yeah is

[mild laurel]

Here's fine

[winter vigil]

How do I prove that f is irreducible, if and only if f(x+n) is irreducible.
I tried induction for the "only if" direction, as I am not sure if it makes sense or is the method.
Basically, the base case of n=0 is proved.
Then for n, I said either f(x+n) is irreducible, then we are done. If it is not, then f(x+n) can be written as a product of smaller degree polynomials. But as f is irreducible, deg f is the minimum. Therefore there does not exist polynomials that have a smaller degree than f. Hence f(x+n) must be irreducible.
Does this work ? If so I'm going to attempt the other direction.

[chilly ocean]

f(x) is reducible if and only if f(x+y) is reducible

[winter vigil]

U mean, prove this instead ?

[chilly ocean]

This is equivalent

[winter vigil]

May I ask, how do I know when to change the question ? I know its asking the same thing though

[chilly ocean]

Intuition perhaps 🤔

I don't know myself

[winter vigil]

Oh ok thanks, I will try that

But may I ask is there any flaws in that proof ?

Or it just doesn't work in general ?

[oblique river]

If f(x+n) = g(x)h(x) then f(x) = g(x-n)h(x-n)

[chilly ocean]

yeah. Now comment why g(x-n) and h(x-n) should be non-constant

[oblique river]

(are you asking me to do that or telling otoro to do that? :P)

[chilly ocean]

n kind of indicates we have Z in our ring, so I would change that to arbitrary y from our ring

[winter vigil]

Why is it both though ? As f is irreducible, isn't only either g or h is non-constant ?

[chilly ocean]

We want it reducible

suppose that f(x+n) is reducible... then for some non-constant polynomials g(x), h(x), f(x+n) = g(x)h(x)

[oblique river]

for any ring R there is a unique ring homomorphism from Z to R

[winter vigil]

Well if it is reducible, then f(x) would have factors of smaller degree

[oblique river]

so you can always talk about integers living in R

(for any unital ring at least -- please never work with nonunital rings)

[chilly ocean]

ah. I always think of non-unital rings

[oblique river]

also the problem was only about n, so there's no need to make it more complicated

[winter vigil]

I'm sorry, I haven't learnt rings yet, its the next chapter 😅

[oblique river]

what is a polynomial to you?

[winter vigil]

What I think of is a_0 + ... + a_nx^n

[chilly ocean]

ok. With what coefficients

normally coefficients would be from a fixed ring (rng? 🤔 )

[winter vigil]

For now, is integer coefficients

[oblique river]

okay great

you want to prove that if f(x) is irreducible then so is f(x+n). so we'll do the contrapositive. suppose f(x+n) is reducible, then f(x+n) = g(x)h(x) for some g(x) and h(x) BOTH nonconstant

then f(x) = g(x-n)h(x-n). what can you say about g(x-n) and h(x-n)?

[winter vigil]

Well firstly, deg g + deg h = deg f, as g and h are non constant, substituting (x-n), would result in the same degree for g and h.
Secondly, they both translated n units rightwards ?

[chilly ocean]

This has more to do with the fact that degrees of g(x-n) and g(x) stay the same

[winter vigil]

Well with the substitution, the degrees of g and h remains unchanged. Which implies f is still the same degree as f(x+n). And as g, h is nonconstant, the coefficients remains unchanged as well

Same goes for f

[tender mist]

Well it has sense, deg(g(x-n))=deg(g(x))>0 so g(x-n) is non-constant

[chilly ocean]

yes

[cloud walrus]

Blitzkrieg:

[chilly ocean]

etc

[winter vigil]

Yes, so the degree still remains the same

Does that implies that it is reducible or something more is needed ?

[chilly ocean]

em

f(x+n) = g(x)h(x) for non-constant g(x), h(x) then f(x) = g(x-n)h(x-n) and g(x-n) and h(x-n) are non-constant

i. e. f(x) is reducible

[winter vigil]

Is it possible to say that since, deg f, deg g and deg h remains unchanged and that f(x+n) is reducible, f is reducible, since g and h are non-constant

[chilly ocean]

stop overcomplicating stuff

it's simple

[winter vigil]

Wait let me simplify.
Basically, f is reducible, because g and h remains nonconstant ?

Sorry 😅

[chilly ocean]

after a shift, yes

[winter vigil]

The reason being same degree ?

[chilly ocean]

that's one of the possible explanations, yes

Or you could say that

if g(x) is constant then g(x+n) also has to be constant

which is simpler to explain

[winter vigil]

As for irreducible right ?

[chilly ocean]

Just stick with what you got

[winter vigil]

Ok, thanks

[chilly ocean]

Why are non-unital rigns considered bad?

[mild laurel]

https://math.mit.edu/~poonen/papers/ring.pdf

[chilly ocean]

why is there no natural way of multiplying elements in R/Z ?

[smoky cypress]

The formula $(ab)(ac_1\cdots c_hbd_1\cdots d_k)=(bd_1\cdots d_k)(ac_1\cdots c_h)$ for cycles, what is happening in this formula?

[cloud walrus]

Whoever:

[chilly ocean]

a and b are being switched

a goes to c_1
c_1 to c_2
etc.
c_h goes to b, which in turn goes to a
then b goes to d_1
...
d_k goes to a which goes to b

assuming the convention for multiplication of permutations for you is like we usually take composites of a function

different question, why would we want to multiply elements in R/Z when Z isn't even an ideal of R

and how does that benefit the argument that it's strange to take direct sums of rings

I don't get it. There's almost no argument for why rings should have an unity

But I guess it's helpful to denote them just as rings, when you work in an environment when it is often the case that unity is needed or necessary

[final gulch]

we need to retire the term ring since it is so ambiguous due to this controversy
everyone should write r1ng or rng from now on

[smoky cypress]

rlng

[chilly ocean]

seems good to me

[smoky cypress]

Is it true that two cycles commute iff they're disjoint?

[thorn delta]

well I know disjoint cycles commute, but I don't know whether there are examples of non-disjoint cycles that commute off the top of my head

[potent lynx]

yes @smoky cypress

[smoky cypress]

Ok

[mild laurel]

@potent lynx No

This is not a true statement @smoky cypress

[potent lynx]

oh fuck really i thought i proved that b4

[smoky cypress]

Rip, what is a counter example?

[mild laurel]

Think, it's not too hard to come with one

[thorn delta]

(12)(21)=(21)(12) i suppose

[mild laurel]

Or even (1)(1) = (1)(1)

okay maybe you don't count this as a cycle

[final gulch]

any 2-cycle squared?

[smoky cypress]

hm ok

so the inverse of an cycle

[mild laurel]

you can even get counter examples where one cycle is not just a multiple of the other

[cloud walrus]

Whoever:

[smoky cypress]

there is a homomorphism between any two groups

[chilly ocean]

A poem in Polish about searching for an ideal in a field:
Oj próbuję ja sobie nocami pomału
W niezerowym ciele szukać ideału
Lecz mam taki pech fatalny
Że co znajdę to trywialny

could you translate it

I try in night little by little
To search for an ideal in a non-zero field (non-zero, I guess they mean non-trivial or something)
But I have such bad luck
That every I find is trivial

lol

very nice poem

[smoky cypress]

Ideals

[somber bramble]

algebra exam tomorrow

I’ve never been so prepared for an exam

as far as I can tell, I can answer literally every question he can ask

(we got a list of questions)

[smoky cypress]

Prove that any group with order 20449 is abelian

That's a question I know about

It's pretty trivial I think

@somber bramble

[somber bramble]

is 20449 prime?

it is not

but it’s of the form p²q² where p and q are prime

[smoky cypress]

Yes

[somber bramble]

and I suspect you can somehow show that all groups with such an order are abelian

I think we showed that groups of order p² are at some point?

[smoky cypress]

Yeah I think

Um

[somber bramble]

or sth like that

[smoky cypress]

I don't know about that 😅

[somber bramble]

I vaguely recall sth like that

well, it’s true

just googled it

[smoky cypress]

Lol

I think you show that any group of order 20449 is isomorphic to the direct product of two abelian groups

[placid pond]

@smoky cypress i assume both sylow subgroups end up being normal just by the sylow theorems

i didn't do the computation but if that's not true the statement would be false anyway

[smoky cypress]

Yeah something about sylow theorem

[placid pond]

since you could make up some semidirect product

and ofc you know that every group of order p^2 is abelian