groups-rings-fields

[chilly ocean]

Get all my sneaky little number theory tricks in

[viral bear]

I would probably appreciate it better if I studied number theory

do you know a good elegant way to describe how sylow works btw?

[chilly ocean]

It's really best thought in terms of prime power modulos

Idk, I'd need to think more on that one - I'm still fresh out of Sylow school

[viral bear]

ye thats my main issue with them

its hard for me to remember stuff if I don't understand the deeper "why?"

[chilly ocean]

Yeah, you need to be able to walk it back

https://upload.wikimedia.org/wikipedia/commons/d/dc/Diagram_for_the_First_Isomorphism_Theorem.png

NIce diagram for that little quote

Going to try my hand at category theory from the napkin project

And then do better

[viral bear]

category theory sounds fun

but its also one of those things that probably requires a good amount of context to really appreciate fully

like general mathematical context

[chilly ocean]

Yeah - I think it'd still be in the "abstract nonsense" bin unless I did an appropriately algebraic/topological research degree

[viral bear]

ggs with the transferring of the structures

also my view on category theory

is once I reach the point where I can do the exercises in maclane I am ready to learn the subject

before that probably can wait a bit

[oblique river]

what is the question?

I'm not sure what this is supposed to be proving

so that doesn't look like transport of structure to me

unless i dont know what transport of structure means

transport of structure would be like

suppose that G is a group, and let S be a set

and suppose f: G --> S is a bijection

then we can "transport" the group structure on G to turn S into a group

if g is the inverse of f, then we can define the group operation on S as follows:

s*t = f(g(s)*g(t))

i.e. "go to G, multiply, and come back"

but here, we're defining a completely new group structure on X which is totally unrelated to the one on R

and then we're proving that they're isomorphic

if this were really transport of structure I feel like it should be "there is a bijection R --> X given by a --> (a, a^2), so we can transport the group structure on R to one on X"

I mean there is a wikipedia page

literally called

transport of structure

in any case just forget "transport of structure"

all that problem is doing is proving that R and X (with their defined group structures) are isomorphic

do you know how to show that two things are isomorphic?

great

I mean

what?

they literally just do it in the other order

show bijective then show morphism

ignore that

do you believe/agree that this function is bijective

yes, g

(and hence of course g^(-1) is also bijective cuz it's the inverse of g, which is bijective)

great

now let's show it's a morphism

let a and b be real numbers

what is g(a+b)?

great

now what is g(a) + g(b) (where this + is now happening in X)

i.e. what is (a, a^2) + (b, b^2) using the operation on X they defined

they told you how to add things on X

in fact, they literally did what we are trying to do

yes they did

https://media.discordapp.net/attachments/496784958430380033/534561969286676490/unknown.png?width=932&height=524

dude please

it's right there

HAHAHAHAHAHAHA

ahahahahaha

lol

my b

I thought that was the question hahahahahahahaha

so wait

i'm confused

what's the issue?

you defined a group structure on X

okay

then forget everything i've said so far

well then

I need to know

what question 26 is

cuz the problem said "like in 26"

what?

the problem says "do it like in 26"

ok lemem look

ull have to give me a second

cuz there is a disaster waiting to happen in #math-discussion

so wait, in 26 they give you a set X to work with

but in 27 did you just make up that set X?

really?

so like you could have used any set?

did it have to be isomorphic to R?

but why did you pick that set X

[cloud walrus]

creamy shits:

creamy shits:

[oblique river]

I mean it's not really "in a similar fashion" cuz the group operation is different

yes I understand yoru group structure

I just don't understand what this question is asking

like

why can't i solve the question with the following:

Let X = {(0,0)}. that is clearly a group with identity (0,0).

done

[solar wyvern]

@spark plank

[oblique river]

lol

why didnt you tell me that earlier

you said you made up that X

[spark plank]

nani

[solar wyvern]

u find a nonabelian group with nontrivial normal subgroup?

[spark plank]

didn't we do this yesterday or something?

I did {e, a} and {e, a, b, c}

[oblique river]

lol you should have told me earlier

[solar wyvern]

is that noncommutative?

[oblique river]

okay tyhen yes you defined the correct group structure

[spark plank]

?

[oblique river]

I have no idea what you need the transport of structure for

[solar wyvern]

wait is it @oblique river 🤔

[spark plank]

commutative

with x² = e

[solar wyvern]

so commutative groups have boring normal subgroups

ie

all of them

are normal

[oblique river]

the question literally just asked you to put an abelian group structure on that set. you did that. you are done with that question

[spark plank]

[solar wyvern]

(u should show this)

[spark plank]

nontrivial tho

[solar wyvern]

I asked for a noncommutative one this time tho

😄

[spark plank]

[oblique river]

what @solar wyvern ?

sry I was talking to @chilly ocean

[spark plank]

I haven't even constructed noncommutative group yet smh

[solar wyvern]

hmm

[spark plank]

[solar wyvern]

well do that too while you're at it

[spark plank]

tell me least order of noncommutative group tho

[solar wyvern]

you've shown all the groups of order 1,2,3

[spark plank]

6?

8?

we dun up to 4

[solar wyvern]

well

[spark plank]

and I told for 5 and 7 they unique

[solar wyvern]

do u know why

[spark plank]

proof by brute force

[solar wyvern]

nah, easier way

[spark plank]

[solar wyvern]

you did lagrange right

[spark plank]

ya

oh

oof

[solar wyvern]

(use lagrange)

[spark plank]

kek

[solar wyvern]

then ur done

[spark plank]

dat was pretty ez

[solar wyvern]

yeah

[spark plank]

yay primes

[solar wyvern]

so it can't be 5 or 7

try 6 i guess

[spark plank]

k

bbl8r

[solar wyvern]

otherwise there's definitely noncomm of order 8

(i think there's maybe one in 6 tho)

[bleak abyss]

S_3

[oblique river]

@chilly ocean okay if that's all you want then I told you how to do it already. define a bijection f: R --> X by f(a) = a^2. check that is a bijection. let g be the inverse. then define a group structure on X by x + y = f(g(x) + g(y)). that is, take your elements in X, pull them back to R, add them there, and push them back to X. that's it.

[spark plank]

also u need slimier pic

[solar wyvern]

tbh not sure I've really given SA many tools to construct groups tho

[spark plank]

[solar wyvern]

(we will get there)

[spark plank]

honestly brute forcing everything rn

no, we need slime pics now

[solar wyvern]

that's fair enough for sm0l groups

[bleak abyss]

I mean, a lot of groups are just known examples rather than building groups out of others. There are some for sure, in particular semidirect products

[solar wyvern]

dami let SA learn direct products first 🤦

[bleak abyss]

But usually there are "classes" of groups that you just know about

S_n, A_n, D_n, matrix groups, etc

[solar wyvern]

tbh I still feel D_n heretical and weird

before semidirect prods

[bleak abyss]

I mentioned semidirect because that's how you'd get non-abelian groups out of abelian groups

While direct for finite abelian groups is kinda trivial

[solar wyvern]

yeah, the basic tools I can tell SA about wouldn't help SA right now even

so better to just crank out basic examples

[bleak abyss]

And lol I prefer to think about D_n as the automorphism group of the cycle graph rather than rigid motions

[solar wyvern]

oh yeah that

[bleak abyss]

At least it helped me think about it somehow because I could process adjacency better than I could process "rigidity"

[cloud walrus]

creamy shits:

[oblique river]

that part is irrelevant

if all you want to do is transport of structure

then you dont need to write that

[solar wyvern]

isn't Dn a semidirect of Cn ⋉C2?

[bleak abyss]

Should be yeah

[oblique river]

sry let me say it another way: that is the transport of structure

[solar wyvern]

up to proper way of writing it maybe

[oblique river]

that's not a proof of anything

that's just defining what the group law is

[solar wyvern]

@bleak abyss tbh I'm unfamiliar with graph theory so not so comfortable with the Aut(cycle graph)

[oblique river]

we can move to #help-5 if this is too confusing having two conversations here

[solar wyvern]

oh

sorry (I'm done for now)

[bleak abyss]

Rip

So if I remember right, the presentation is $D_n = \langle r,s \mid r^n, s^2, (rs)^2 \rangle$

[solar wyvern]

oh

[cloud walrus]

Daminark:

[solar wyvern]

yeah, but there's another one which is nicer for reasons I think?

afaik presentations quite nonunique

(rs)^2 is the "rigidity" constraint?

[bleak abyss]

As I mentioned earlier, I don't think well about rigidity. But in any event, this gives $srs^{-1} = r^{-1}$. So now, $Aut(C_n) = C_n^{\times}$ since you can send 1 to any unit. Our map is thus $C_2 \to C_n^{\times}$ where the non-trivial element goes to $-1$ (aka acts by inversion).

[cloud walrus]

Daminark:

[spark plank]

G={e,a,b,ab,ba,c}

aa=bb=c;ac=ba;bc=ab;ca=b;cb=a;cc=e

@solar wyvern I think this worls?

As a noncommutative group

nvm

abc = aab = cb = a

=> ab = e

[oblique river]

in general, it becomes basically impossible to try to define a group by just writing down its multiplication table

for groups with more than 5 elements

[spark plank]

try me

[oblique river]

I mean you just tried to do it and failed :P

I think that's evidence enough haha

[spark plank]

just means try more tbh

[oblique river]

you should really learn some examples of groups

[spark plank]

[chilly ocean]

Anyway

Rimuru

[spark plank]

[chilly ocean]

Draw out the table for (Z_31,+)

[spark plank]

[oblique river]

like, that's how it happened historically as well. people didn't just write down the definition of a group and say "ok now time to come up with examples"

[spark plank]

I don't even know what those mean and no spoilerz so bye

[oblique river]

the examples came first

and then people realized "oh wait, these examples all have the same underlying structure... in fact, this structure seems to appear everywhere... let's give it a name!"

[chilly ocean]

Thank you for the history lesson

[oblique river]

uhh

I'm just trying to provide @spark plank with some motivation

[bleak abyss]

@chilly ocean that table is physically impossible to draw

It's uncountable

[chilly ocean]

Then SA should have no problem

[spark plank]

[solar wyvern]

did you try S3

[spark plank]

u say this like I know wut u mean

[solar wyvern]

umm, the bijections on a set with three elements

[spark plank]

I'll try it then...

o rite 6 = 3!

[solar wyvern]

[spark plank]

Hm naisu

So now I need subgroup of this? @solar wyvern

Other than itself

[oblique river]

didn't you want to find a normal subgroup as opposed to just any subgroup?

[spark plank]

O rite

H = {a, b, c}

Where a(x)=x

For all x

[oblique river]

(usually called the identity, but yes go on :) )

[spark plank]

And b(x), c(x) never equal x

[oblique river]

yes that's correct

[spark plank]

Was about to check the other 3 functions

But Lagrange says this must be order 2, 3

And symmetry says this should be it

[oblique river]

what do you mean by "symmetry"

[spark plank]

Like I can check with 2 of the other functions not here

And it covers all those cases

[solar wyvern]

(not sure what symmetry means still)

[spark plank]

Like call the others x, y, z

[solar wyvern]

also there's some handy notation used for this stuff

[spark plank]

If I check {id, x, y} it covers {id, x, z} and {id, y, z}

[solar wyvern]

ah

okay yeah that's fine i guess or is it???

[oblique river]

I mean

{id, x, y} is not going to be a subgroup

[spark plank]

Well ya, I checked

[solar wyvern]

also a bit confused

[oblique river]

let alone normal

[solar wyvern]

those are the points

the group is made of permutations on those points, not the points themselves

[oblique river]

I think x and y are permutations

[solar wyvern]

ah

ok

[oblique river]

I think the permutations are "id, b, c, x, y, z"

which is not good notation hahaha

but we're rolling with it

so can we get back to how you proved that {id, b, c} was normal?

[spark plank]

hey man, I didn't even know what S3 referred to

[solar wyvern]

fair

[chilly ocean]

Cycle notation tbh

[spark plank]

^ pretty much

[solar wyvern]

yeah commonly you write the set S = {1,2,3} and the permutations as a sum of disjoint cycles

[spark plank]

I checked bHc and xHx (since bc=xx=e)

[solar wyvern]

if it's finitary

[chilly ocean]

(1 2 3){a,b,c} = {c,a,b}

[spark plank]

[solar wyvern]

so

() would be identity

[spark plank]

Well I know about these things but got no notation

[solar wyvern]

(12) sends 1 to 2 and 2 to one, leaves everything else fixed

[chilly ocean]

Position 1 -> Position 2 -> Position 3 -> Position 1

[spark plank]

ic

[chilly ocean]

(1 2 3)

[spark plank]

I thonk

[chilly ocean]

(1 3 2)

You get the pitcher

[solar wyvern]

(123) sends 1 to 2, 2 to 3 and 3 to 1 i'm not 100% sure what jichael doing tbh

[spark plank]

Well anyways just two nontrivial cases to check

[oblique river]

sure

I'll let you get away with it even though you could be more precise haha

[spark plank]

[oblique river]

did you ever prove that all subgroups of abelian groups are normal?

[chilly ocean]

@solar wyvern What ur doin bruh

[solar wyvern]

trying to live in a society

[spark plank]

I ain't doing the same thing twice more when it just letter swapping man

Uh

[oblique river]

also just for the record

you didn't have to actually check that bHc = H

because b and c are in H

[spark plank]

I don't thonk so

[oblique river]

so obviously it's true

[solar wyvern]

[spark plank]

oof kek

[chilly ocean]

Hold up just a second

Since when is b,c in H obvious

[oblique river]

cuz H was defined to be {id, b, c} lel

[chilly ocean]

And since when was closure implied?

[spark plank]

[solar wyvern]

also for normal subgroups, you only care about aHa^{-1}, so bHc doesn't have to be H?

[oblique river]

c = b^(-1)

[spark plank]

^

[solar wyvern]

but just pointing out

[chilly ocean]

b^2 = ?

b^-2 = ?

[spark plank]

Pretty obvious looking at {1,2,3}

[chilly ocean]

[solar wyvern]

not for me

[oblique river]

yo mackson chill, you don't have to write down the entire multiplication table of S_3 to know it's a group

[spark plank]

Like b(x) = x+1 and c(x) = x-1

And 3+1=1, 1-1=3

[oblique river]

that's one way to think about it but cycle notation is definitely more standard

(123) and (132)

[solar wyvern]

oh, have you actually shown S_n has n! elements?

[spark plank]

shhh fork standards

[solar wyvern]

since that's another good ex

[spark plank]

Combinatorics smh

U can't fool me

[solar wyvern]

but why

[spark plank]

[solar wyvern]

well, it's just showing that its elements are permutations like you've always known them

🤷

[spark plank]

can proof by induction too I thonk

But

not rly group problem

[chilly ocean]

@oblique river Thank you for your request to avoid a multiplication table

[spark plank]

I should try to sleep while I feel gud too

[solar wyvern]

ok

so u got a noncomm grp right?

[spark plank]

Reminds me of my series multiplication pic

Ya

[solar wyvern]

did you find any (interesting) normal subgroups?

[spark plank]

inb4 have to find noncommutative normal subgroup via S_6

[solar wyvern]

nah

[spark plank]

[solar wyvern]

there's an order 8 one

[spark plank]

[solar wyvern]

I'm using for a hw ex now I think

[spark plank]

[solar wyvern]

but I'll have that sorted out next time i guess

g'night

[spark plank]

k

Hold up

[solar wyvern]

wat

[spark plank]

S_3 is basically two unique groups of order 3 that connect together with some stuff

Wait no

[solar wyvern]

[spark plank]

Not quite like that

[solar wyvern]

almost

[chilly ocean]

[spark plank]

But rEEE

Kay gn

[solar wyvern]

you'll get there 😃

[spark plank]

[chilly ocean]

Group of order 3 only has one structure

no?

[spark plank]

[solar wyvern]

ya

[chilly ocean]

2,2,2,3,3

[solar wyvern]

darn i gotta review semidirect product stuff myself

but we can probably do some basic stuff with quotients and actions soon, then iso thms i guess

probably a good time to show why one can't get a group if you quotient by a nonnormal subgroup

[chilly ocean]

Probably not as slow as most to pick this up going forward

[solar wyvern]

I think so 😄

I'm trying to keep things pretty concrete too

and finite groups themselves are rather concrete

tbh I didn't really understand actions until like less than a year ago tho

[errant drum]

If you wanted to prove that R is a Euclidean ring you'd need to find a norm function that statisfies those properties so if you wanted to prove something was not a Euclidean ring how would you go about showing no norm function satisfies those properties.

Those properties being Norm(a) <= Norm(ab) and you can write any element in the form a=bq+r

Where norm(r) < Norm(b)

Isn't there an easier way?

[hot folio]

when we're checking if a subset H is a subgroup of G, we check that : for a,b in H, then ab is in H (does ab means a time b ? or a * b, where * is the composition law

[errant drum]

a * b = a <group operation> b

If your group was functions it would composition

Often in group theory we use notation that looks like multiplication probably because we're lazy idk

However multiplication is there to mean group operation

[hot folio]

yeah someone told me that yesterday here and i was confused

thanks

[errant drum]

Np :)

<@&286206848099549185> about 5 comments above

[fringe nexus]

So in lecture yesterday my professor said something about non isomorphic groups of finite size?

Im not really sure what that means

We were ‘filling in’ multiplication tables but I didnt really understand it

Is there something i can read to understand it better?

[whole basalt]

So I'm gonna repost a question I had yesterday

What is the weakest condition such that a set of functions (C→C) forms an integral domain?

I'm pretty sure it's true for analytic functions, and I know it's not true for continuous functions

[oblique river]

@errant drum have you looked at something like this? http://www.maths.qmul.ac.uk/~raw/MTH5100/PIDnotED.pdf

I'm not sure if there's a single method that will work for any ring R

[covert vector]

meromorphic functions work

that's a lot more than an integral domain, those form a field

[chilly ocean]

Hang on - integral domain => ring right?

What's the operators on the functions?

[covert vector]

yes

pointwise multiplication

well actually just multiplication I guess

for my meromorphic example, I allow this type of thing to occur:

f(x) = 1/x

g(x)=x

fg(x)=gf(x)=1

@chilly ocean

[chilly ocean]

Gotcha gotcha

And then just normal addition from real intuition?

[covert vector]

yes

[chilly ocean]

Alright then, I was just wondering if it was addition and function composition

Which would be weird but interesting

[oblique river]

that wouldn't form a ring

[covert vector]

nah, function composition rarely works

[oblique river]

because composition doesn't distribute over addition

[chilly ocean]

He's got a point

[oblique river]

unless you're working in a very limited setting, like linear functions

[covert vector]

ya

[chilly ocean]

Alright then

Hm

[covert vector]

my idea for @whole basalt is to find a set of functions that are Nonzero for all but finitely many points in C

that is closed under addition

[oblique river]

you can just also force it to happen, like

[covert vector]

oh like localize something

[chilly ocean]

does integration count as abstract algebra?

[oblique river]

the set of continuous functions f: C --> C s.t. the zero set of f is bounded

[covert vector]

no

[chilly ocean]

#calculus

ty

[oblique river]

or the set of cts functions f: C --> C s.t. the zero set of f contains no open subset

[covert vector]

ya but we still need closed under addition

[oblique river]

(well maybe that last one would fail under addition)

oh wait

they both do

hahaha

rip me

[covert vector]

:P

I'm saying to forget continuity !! scroo that

[oblique river]

yeah scroo that

[covert vector]

here is an incredibly forced example:
functions C→C where every Nonzero function has all finite fibres

[oblique river]

are you sure that will work? i'm worried that there could still be some "cancellation" in the finite fibres part

[covert vector]

hmm

[oblique river]

like multiple finite fibers could somehow come together to form some alrge open set

[covert vector]

I think it is fine 🤔

hmmm

no it should be fine, if they're all finite

[oblique river]

like consider the function f(z) = z if Re(z) <= 0 and = z+1 if Re(z) > 0, and g(z) = -z if Re(z) <= 0 and = -z + 1 otherwise

then f+g is identically 0 on the left half plane

[covert vector]

:o

[oblique river]

but both functinos have finite fibers (in fact, f is injective and g has no fibers of size larger than 2!)

so even "uniformly bounded fibers" won't work

[covert vector]

dam!

rip xD

[bleak abyss]

You call?

[covert vector]

hi dami !!

[oblique river]

I feel like unless you wanna try to use the axiom of choice in some clever way

you're not gonna be able to get any examples that don't use some kind of regularity condition

like "meromorphic"

[covert vector]

😐

[oblique river]

or, you're not gonna get any "big" examples.

[covert vector]

ya I said mero earlier

[oblique river]

I know

I'm just saying i'm not sure how much better one can do if you don't use any regularity

[covert vector]

Ok

[chilly ocean]

Hang on, C is a field. So doesn't that mean that any two functions that do make a zero divisor would be 0 on the support of the other function? ie. all that is required is disjoint supports for any two functions for it not to be an integral domain?

[covert vector]

yes

[chilly ocean]

So super-weak condition, assuming the ring is generated, bare minimum is non-disjoint supports of generators?

[covert vector]

the problem is addition

wait 🤔

generated means what again

[chilly ocean]

Ah, I start with a few functions

And I just say "hey lets apply operators arbitrarily"

If a+b isn't in my ring - it is now

[covert vector]

oh ya that'll be fine probably, just that it'll be a smol ring

[chilly ocean]

Not weak enough?

[covert vector]

I mean if you have finite generating functions your ring will be at most countable

[oblique river]

but the issue is that the support of the generators could be nondisjoint, but maybe their sum is 0 everywhere

[chilly ocean]

^ okay

[oblique river]

err, I guess tha'ts not an issue

but like

the example I wrote above

that f and g. their supports are open in C

(open and dense)

but their sum is 0 on the left half plane

so (f+g)(z) * (f+g)(-z) = 0

oh wait. I guess you don't have to contain (f+g)(-z)

hmm maybe you're on to something but it still feels like a counterexample will arrive pretty quickly

wait, no I just gave you a counterexample

[chilly ocean]

It usually does but I'm just thinking in terms of what makes products 0

[oblique river]

take the ring generated by my f(z), g(z), f(-z), g(-z)

those all have nondisjoint support but there are zero-divisors

[chilly ocean]

like consider the function f(z) = z if Re(z) <= 0 and = z+1 if Re(z) > 0, and g(z) = -z if Re(z) <= 0 and = -z + 1 otherwise

Just for my own sanity

[oblique river]

yeah sry

[chilly ocean]

okay, okay, hm

[oblique river]

the whole issue is that adding functions can fux with the support real bad

[chilly ocean]

For sure

Hang on, there's a disjoint support in the generators, no?

Support of f(z) and f(-z) are disjoint

[oblique river]

what? f(z) and f(-z) are only zero on the imaginary axis

their support is everything else

[chilly ocean]

Oh darn, got my wires crossed

Sorry

Yep, the counterexample holds

[chilly ocean]

Okay I'm going to think of a "no fuck with my supports" clause.

[bleak abyss]

Lmao

[chilly ocean]

Alright, any two generators f,g of this ring may not send a given X of C to both and A and -A. Where -A is just A except all the elements are now negative.

Thus - no support fucking - ever.

I think we can weaken it a little by saying that X would have to be on the support of f and g

Hang on I'm dumb f(z) = -1, g(z) = z (if |z-1| < 1, 0 otherwise)

g*f*g = 0

Maybe I should confine my mathematical diahorrea until I have an actual answer

Wait wut

Guys, be real with me, am I brain damaged?

[whole basalt]

Well I got a chance to read what's been said

I never really understood what meromorphic means

[covert vector]

its basically fractions of holomorphic functions @whole basalt

or fraction of analytic functions

where denominator is not zero too much

[chilly ocean]

Alright, thought about it some more, let S(f_i) be the support of the function f_i. We generate a ring with a set of C->C functions, we add the stipulation that S(f_i) n S(f_j) be non-empty and that for any S(f_i)n S(f_j) that there is no f_a and f_b such that f_a:S(f_i)n S(f_j) -> A and f_b:S(f_i)n S(f_j) -> -A (where -A is just all elements in A multiplied by -1). This gives our ring two properties:

1. S(f_i . f_j) = S(f_i) n S(f_j)

As where f_i(z) = 0 (ie. z in C-S(f_i)) then f_i . f_j = 0 similarly for f_j.

2. S(f_i + f_j) = S(f_i) u S(f_j) - B(f_i,f_j)

Where B(f_i,f_j) where f_i: X -> A and f_j: X -> -A for some proper subset X of S(f_i.f_j). (As by our second stipulation, there can be no f_a:S(f_i.f_j)-> A and f_b:S(f_i,f_j) -> A)

Since B(f_i,f_j) is always a subset of S(f_i,f_j) there is always a non-disjoint support between sums of functions. And so - the support of any sum or product of two functions is always non-empty.

I think this covers it.

So what is it? It's basically just FIP for the support of functions and a stipulation that no pair of functions may map any of the support intersections to a set and its "negative". I think the set notation shows that this works out because of the completeness of complex numbers

I'll try to break it now.

[errant drum]

@oblique river Thanks it turned out I was overcomplicating the question but that was a good read 👍

[chilly ocean]

Bollocks

I have got to screw down my verbal diahorrea

Right,

1. S(fg) = S(f) n S(g)
2. S(f+g) = S(f) u S(g) - B(f,g)

Where f:S(fg)->X, g:S(fg)->Y, h:AxB -> AxB h:(a,b)->(a,b).

B(f,g) is s in S(fg) such that h(S(fg),f(S(fg)))=h(S(fg),-g(S(fg))) FIP on S(f), S(g), we require that B(f,g) < S(f) n S(g)

And das it

I don't think you can get any weaker than that

[chilly ocean]

@covert vector I'd appreciate a sanity check

[covert vector]

later

[chilly ocean]

Thanks!

so how do I find the other way to unscramble the board in 3 moves?

Ive been at this for like 2 hours

What moves are legal?

Because I have a way but it requires you to flip the middle column

@chilly ocean

Well assuming your just spinning rectangles

You limit yourself seriously in terms of possible paths by just getting rid of "upside-down" squares

So there's the softer problem of

U U U     R R R
R U R --> R R R
R U R     R R R

[covert vector]

ooh cool problem

I don't understand exactly what the set up is tho

[chilly ocean]

so you can only use 3 spins. Each spins it 180 degrees

you want to get 123,456, 789 right side up

[covert vector]

ohh

[chilly ocean]

the problem is if you get rid of the upside downs, the numbers still need to be in correct order

so you are saying just try to get them in the right orientation first cause that will limit the number of options

[covert vector]

what about
{123} → {23,56,89} → {2,5,8}

hmm

ok that doesn't work

cuz it flips a multiple of 3 tiles

so how do we make this abstract algebra

[chilly ocean]

tried labeling them from the initial part using a two way system where orientation= up or down

so I think starting like this

which should turn into that

Tbh - I might have just written a computer program which spams the 1+4+4+9+9+12 possible moves

== (1+4+4+9+9+12)^3

[mellow vapor]

59‚319

[chilly ocean]

Eh - not that bad

[viral bear]

1+4+4+9+9+12=6^2 btw

you select some continuous section of rows (4C2) and same for columns

[chilly ocean]

6^2 = 39

for all possible spins you mean?

[viral bear]

oh wait how did you get 39

[chilly ocean]

== 1+4+4+9+9+12

[mellow vapor]

39

[viral bear]

yeah but how did you get that there were 39 options

[chilly ocean]

I got 36

1 3x3, 4, 3x2, 4 2x2 , 9 3x1, 9 1x1, 12 2x1

1+4+4+6+9+12

[viral bear]

6 3x1

[chilly ocean]

O yah

😄

[viral bear]

but yeah like, its easier to break it up as the product of row and column choices

since you can also generalize to larger squares

nxn square should have ((n+1)C2)^2m options after m moves

[solar wyvern]

where's the algebra

[viral bear]

im trying to think if you can apply burnsides lemma to this

[chilly ocean]

Probably

[viral bear]

thats the best shot at making it algebra

these permutations would also form a group, but you can probably generate any permutation from it so that part isnt super interesting

[chilly ocean]

Im not sure how to use burnsides lemma. This is second day taking abstract

[viral bear]

how did they motivate this exercise btw?

this seems outside what I normally consider abstract algebra

[chilly ocean]

Yeah its IBL

[viral bear]

idk what ibl stands for

Inquiry-based learning?

[chilly ocean]

yea

[viral bear]

so this is based on something someone asked about in class?

[chilly ocean]

nope

[viral bear]

idk what inquiry based learning actually is btw

[chilly ocean]

its a version of learning where you are giving problems sets and pretty much expected to not be a able to do most of them. Then you go to class and see what you did wrong. That's basically how you learn. But it would be nice to solve it if I can

Then you use what you learned and corrected to study for the test

[viral bear]

yeah sounds fun

I probably would have learned more from something like that

[chilly ocean]

yeah, but it's like communism it works really good in theory, but not application. Since the class is mostly scattered age groups we don't work together as much as we are supposed to.

Like Im hoping it works out. But there are just so many mediating variables that could make this a very hard stressful year

[viral bear]

hmm

anyway I am going to write some groupish things about that puzzle

[chilly ocean]

ight thanks man

[viral bear]

the set of permutations of numbers and orientations is isomorphic to $S_9 \times \bZ_2^9$

[cloud walrus]

ixsetf:

[viral bear]

the original square written in cycle notation (for the S_9) part is (12983)(56)

when looking only at the symmetry part, 1x1 acts as identity, 1x2 and 1x3 are a transposition, 2x2 is a pair of transpositions, 2x3 are 3 transpositions, and 3x3 is 4 transpositions

[chilly ocean]

O

yeah I remember those are linear

[viral bear]

what is linear?

[chilly ocean]

linear algebra

[viral bear]

nah this is from abstract

[chilly ocean]

o ok

[viral bear]

anyway this means that your original square is an odd permuation

and therefore its identity is also odd

so you need to use 1 or 3 odd moves

[chilly ocean]

1 or 3 odd number cubed moves?

[viral bear]

in the case of the presented solution they are using 3 odd permuations

[chilly ocean]

so like 1x3s

or 1x1s

[viral bear]

no uh

so I listed the number of transpositions a move does

its odd when its an odd number of transpositions

and even when its an even number

[chilly ocean]

o ok

[viral bear]

that means that the odd ones are 1x2, 1x3 and 2x3

[chilly ocean]

ight

[viral bear]

here is another fact that can help here https://math.stackexchange.com/questions/2235010/prove-or-disprove-the-minimum-number-of-transpositions-needed-to-decompose-si

you technically have (4)(7) as cycles in your square as well

[chilly ocean]

ight so I tested all 2x3 for starting point and they don't work at first

[viral bear]

hmm

anyway

the fact this gives you

is because you have 4 cycles (including the two trivial ones)

you need at least 5 transpositions to get a solution

so if you are using 3 odd ones, you need to use at least one 2x3

[chilly ocean]

hm

o

[viral bear]

this has implications for the 1 odd 2 even too

[chilly ocean]

but that could be later

[viral bear]

but it would take longer to list them

[chilly ocean]

cause at the beginning it isn't possible

so how do I know I have 4 cycles?

like I know the 4, 7 are cycles cause they are in the right place

but for the others ones im unsure

[viral bear]

so for example I wrote (56)

if you look at where the 5 should be there is a 6

so then you look at where the 6 should be and theres a 5

you looped back so thats your cycle

with the longer one you start with where 1 should be, you see a 2, you look at where 2 should be, you see a 9 (upside down), you look at the 9 you see an 8 etc

[chilly ocean]

o I see

so like (4)(7)(56)(893)(12)

idk

[viral bear]

Well you almost got it

The 2 doesn't go back to the 1

[chilly ocean]

hmm so (4)(7)(21389)(56)

[viral bear]

hmm not quite

[chilly ocean]

o

fixed it I think

[viral bear]

is this the square you are looking at?

[chilly ocean]

yeah

[viral bear]

ok so to start off the long one

you look at where the 1 should be

you see there is a 2 there right?

so 1 -> 2

2 is now at the end of the chain

so you look where 2 should go

[chilly ocean]

o ok

[viral bear]

you know what the next one should be?

[chilly ocean]

(4)(7)(13892)(56)

[viral bear]

yeah that works

[chilly ocean]

ight sweet

so how do I use this?

[viral bear]

there are some deep things related to this kind of thing

the one I applied initially was parity

a permutation can be either even or odd

its even if you can get from the identity (the solution square here)

in an even number of swaps

and odd if it takes an odd number of swaps

so like if I start with

123
456
789

I could swap the 1 and the 2

and I would get

213
456
789

that square would be odd because it took you 1 swap to get to it

if I flipped some other two numbers

say 1 and 9

then it would become even again

293
456
781

note that the two swaps led to a 3 cycle

(129)

if I did a third swap I could create

293
456
718

thus getting (1298)

of course you dont need to add to a cycle

I could swap 56 now and get

293
465
718```

which would result in (1298)(56)

[chilly ocean]

hmm

so I want to get one that has 6 cycles

by using that method

[viral bear]

Tbh this is probably too much new stuff to take in in one sitting

[chilly ocean]

yep

[viral bear]

Eventually you'll be able to figure it out though

[chilly ocean]

welp finally got it

only took 5 hours

10 more problems to go now lol, End my life

[viral bear]

You didn't want me to like, give you the other simplifications did you?

Since there were other ways you could have narrowed that down

[chilly ocean]

its all good I got it all done, but today was rough

[chilly ocean]

A morphism on an abelian group?

Well just consult the definition of a morphism

f(a+b) = f(a)°f(b)

And f(b+a) = f(b)°f(a)

Under which circumstances would this be true?

Oh and f(a+b)=f(b+a)

Since you know, abelian

@chilly ocean

Eh remember that they're operators

We use addition as a short hand for a commutative operator on a group and multiplication for a noncommutative operator

So given that we have morphism from an abelian group

It follows that the morphism must map to another abelian group - does that follow?

Yep

Yep

You just want to know if a morphism from an abelian group is always bijective

It follows from first isomorphism theorem

[errant drum]

I suppose that's a more general answer

[chilly ocean]

Hey cosmic can you answer my question on free products?

[errant drum]

No because I'm nowhere near as smart you

[chilly ocean]

Bruh

Alrighty

[earnest valley]

$f:\mathbb{Z}/15\mathbb{Z}\to\mathbb{Z}/15\mathbb{Z}, n\mapsto5n$ abelian and not bijective.

[chilly ocean]

Actually goofed

[oblique river]

@chilly ocean

in your answer

to that last question about cyclic groups

why are you writing f^(-1)? you don't know that f is invertible

what? but what if it's not?

what if G1 = Z/2Z and G2 = Z/4Z

and the map is defined by f(0) = 0, f(1) = 2

or potentially even worse, what if G2 isn't even cyclic. What if G2 = Z x Z/2Z and f is the function f(0) = (0,0), f(1) = (0,1)?

so?

that doesn't tell you that f is an isomorphism

in my first example above, you know that f(1) = 2

great

but that doesn't mean f is surjective

but you're wrong lol

f need not be bijective

i literally just gave you two examples of the form "homomorphism from a cyclic group into some other group"

neither of which are bijective

w-what?

you can be a homomorphism without being bijective

for any groups G1 and G2, the map f: G1 --> G2 defined by f(g) = e for all g in G1 (where e is the identity element of G2)

is always a morphism

but it's never bijective (unless G1 and G2 are both the trivial group)

are you confusing "homomorphism" with "isomorphism"?

the reason I ask is because isomorphisms are necessarily bijective, but homomorphisms don't need to be bijective

(in fact most of them aren't!)

also, this problem doesn't ask you to show that f is bijective, it only asks you to show that it's a homomorphism

you are off to a good start. the point is that you only know that f(a) = b. so what is f on all of G1? Well, if it's gonna be a homomorphism, we better have f(e) = e. Ok good start. But what about other elements of G1? Well, since a generates G1, we know that any element of G1 is of the form a^k. So if f is gonna be a homomorphism, we better have f(a^k) = f(a)^k = b^k

so now we've completely determined f: f(a^k) = b^k. all that's left is to check that it is in fact a morphism

w-what?

the question asks you to prove that f is a morphism!

oh, I see

it's phrased slightly differently

sure, you're given a morphism

and asked to show that it's completely determined by f(a)

[cloud walrus]

creamy shits:

[oblique river]

another way of phrasing that is: If I know f(a), does that tell me what all of f is?

yes

[cloud walrus]

creamy shits:

[oblique river]

the question never asks you to "recover a"

idk what that face is, the question literally doesn't say that lol

but that's a different problem lol

[cloud walrus]

creamy shits:

[oblique river]

yes

you should also probably say in words that "every element of G1 is of the form a^k"

where by a^k I mean all those triangles

uhhhhhhhhh

whether or not G2 is finite is completely irrelevant

and G1 is either Z/nZ or Z

so it might be finite or not

but that doesn't change anything

it's cyclic

so by definition

every element in G1 is of the form a^k for some k

if your worry is about choosing k negative

then, well, dont worry cuz the argument is literally the same lol

f(a^k) = b^k

cuz f(a^k)*f^(a^-k) = f(e) = e

so if f(a^k) = b^k

then f^(a^-k) is the inverse of b^k

which is b^-k

i have to go now so gl

[bleak abyss]

@covert vector random algebra problem of the day

Is (x-1) a maximal ideal?

[covert vector]

where

what ring

[bleak abyss]

Polynomial ring over complex numbers

[covert vector]

yes

[bleak abyss]

....in two variables HAH

[covert vector]

-_-

[spark plank]

@solar wyvern say are prime generators elements of a group such that p and e are the only elements that multiply to p?

[solar wyvern]

@spark plank context?

[spark plank]

nope just random question

feel like u or woog mentioned something about primes in groups

[solar wyvern]

honestly I haven't heard "prime generators" used wrt groups per se, but I probably would use "prime generator" to mean an element of a ring which generates a prime ideal

[spark plank]

nani

[solar wyvern]

for example the integers form a ring under addition and multiplication, and the (signed) prime numbers and 0 individually generate prime ideals

[covert vector]

I am woog

ya prime generator havent heard of

[solar wyvern]

prime ideal means that it's a additive subgroup, but closed under multiplication--so for the ideal in Z generated by p a prime number, we'd often write this \

$$(p) := p\mathbf Z = {ap \in \mathbf Z: a \in \mathbf Z}$$

[cloud walrus]

flimflam:

[solar wyvern]

so like

[covert vector]

I thought this was originally a group question tho

[solar wyvern]

for $a,b \in (p) = p\mathbf Z$, and for $r \in \mathbf Z, ra \in (p) = p\mathbf Z$

that's the ideal part, the prime part means

[chilly ocean]

Sounds ideal

🥁

[solar wyvern]

We say $I \subset R$ is an \emph{ideal} of a ring $R$ if if it's a additive subgroup, but closed under multiplication--so for the ideal in Z generated by p a prime number, we'd often write this \

$$(p) := p\mathbf Z = {ap \in \mathbf Z: a \in \mathbf Z}$$

for $a,b \in (p) = p\mathbf Z$, and for $r \in \mathbf Z, ra \in (p) = p\mathbf Z$

We say an ideal $\mathfrak p \subset R$ is a \emph{prime ideal} of $R$ if for any ideals $I,J \subset R$, we have \

$IJ \subset \mathfrak p \implies I \subset \mathfrak p \lor J \subset \mathfrak p$

[spark plank]

[cloud walrus]

flimflam:

[solar wyvern]

It's a lot at first, but there's a nice tie-in to groups which you'll get soon enough

[stone fulcrum]

I know the definition as "I is prime if for ij ∈ I, then i ∈ I or j ∈ J"

[solar wyvern]

that's uggo tho

[stone fulcrum]

K fine JEEZ

[solar wyvern]

(they're equivalent for commutative rings afaik, and if you understood any of this that's a good exercise @stone fulcrum @spark plank )

ncomm rings are spooky and I feel as thought there are issues where I'd need to be more careful tho

[spark plank]

I feel like the definition of prime is weird

[stone fulcrum]

You need to use the idea of multiplication of ideals for that one, which is more tech than required

[solar wyvern]

@stone fulcrum ultimately I think it's nicer in the long run

[stone fulcrum]

Maybe I'm just scared of that

[solar wyvern]

it might be easier to manipulate things for the element def

but it's good to get used to both

[stone fulcrum]

Still fighting with commutative algebra

[spark plank]

I'd feel like you'd call a subgroup prime if it doesn't have any subgroups other than {e} and itself

[solar wyvern]

because ideals are just R-submodules, and abelian groups are Z-modules

[stone fulcrum]

I feel like I should have learned what modules were first

[spark plank]

kek

[solar wyvern]

😄 yeah, it's sorta weird like that

but after you figure out rings/modules

that's basically it

@spark plank I'm thinking if you'd actually call a group prime ever, but the condition you noted is actually quite stronk

[spark plank]

[solar wyvern]

if it doesn't have any nontrivial normal proper subgroups, those are called simple groups.

these are much more interesting

well, think about it, what sort of groups do you know which don't have any nontrivial subgroups

[spark plank]

{e, a, a²}

[solar wyvern]

think about divisibility

[spark plank]

nani

3 elements?

3 is prime?

[solar wyvern]

hint: which finite groups can't have nontrivial subgroups, by lagrange

[spark plank]

idk

prime orders

[solar wyvern]

yes

thinking if you can prove those are the only ones which satisfy your condition

[spark plank]

fock I was just gonna ask that

[solar wyvern]

if you restrict to abelian groups, you should definitely be able to show that the ones with only trivial subgroups are of prime order

otherwise....I think you should still be able to prove the generaler case for finite maybe nonabelian

[spark plank]

[solar wyvern]

using cyclic subgroups of your group

[spark plank]

hm

I feel like

nvm

[solar wyvern]

so if any element has order >1, you'll have a nontrivial subgroup right?

[spark plank]

but how do I know it doesn't have order ∞?

wait if it has order ∞ that's still a subgroup

[solar wyvern]

you don't I feel as though you can probably allow infinite groups as long as your generators are finite but I haven't really told you about generators/relations/presentations so I'm not sure it's fair to bring up

[spark plank]

🤦

wait no

if my group is finite

then {aⁿ : n in Z} is a subgroup iff |<a>| is finite

wait no not that

but obviously I can't make more elements than the original group

so this is finite

Hm

[solar wyvern]

if you have inifnite order elements can your group be finite?

[spark plank]

nah

[solar wyvern]

[spark plank]

takes infinitely elements from finite set

[solar wyvern]

seems legit

[spark plank]

invokes axiom of magically appearing elements

hm so I need to proof that either this is a trivial subgroup or this is a subset when |G| is not prime

[solar wyvern]

so in the case of C_p--the cyclic groups of prime order--all your elements aren't order 1 obviously

so you have to show a subset of your group generates (or doesn't i guess) a proper subgroup

[spark plank]

ya it's pretty obvious from Lagrange that prime implies trivial is ez

I'll thonk over foods

[solar wyvern]

the other way is a bit more nuanced but not bad

[spark plank]

nvm food not done

so if there exists a in G such that |<a>| = |G| = pq then |<a^p>| = q

so really all I need to do is prove the existence of a such that <a> is a subgroup and a doesn't equal e

I assume I'm proving this for all groups

actually wait

<a> is always a subgroup

🤦

I'm smort

So either <a> forms a proper subgroup or |<a>| = |G| = pq when G has non-prime order and <a^p> forms proper subgroup

@solar wyvern

[solar wyvern]

is p,q prime here

@spark plank

[spark plank]

p, q are not 1

don't think it matters what they be

[solar wyvern]

if |<a>|=|G| what does that tell you about G?

ie, when is <a> not a proper subgroup

you mostly have the details together but the wording a little confusing 4me

[spark plank]

uh

<a> = G

I don't think it matters tho?

@solar wyvern

cuz <a^p> is a proper subgroup of G

[solar wyvern]

sure, iff q only divisible by pm 1

and itself

[spark plank]

I mean

|G| = pq

where p and q are not 1

and wdym? @solar wyvern

that's true for all q

[solar wyvern]

dwai too much, but I was just nudging at the fact that in the ring of integers, the group of units--that is, the subgroup closed under multiplication-- is {-1,1}

in the Gaussian integers, the group of units is {±i,±1}, in the complex numbers it's the unit circle, and in the reals it's again {±1}

i guess it's a group too so yeah

[spark plank]

[solar wyvern]

in the ring of n by n square matrices, you'll have the group of units GL(n,k) called the general linear group (where k is typically a field, but you can consider GL(\mathbf Z, 2)just as easily)

i think the ring of smooth functions on C (with convolution as multiplication) may have an interesting group of units but idk

@gentle pendant

[spark plank]

:

[solar wyvern]

umm, back to the point, since you can divide anything by invertible elements, having the property you can only factor out invertible elements is very useful

got an extension on hw so thinking whether to go to sauna

feeling mucky

[spark plank]

nani

wdym factor out invertible elements

[solar wyvern]

so like

[spark plank]

nani

wait wut

[solar wyvern]

for 5, you can factor out -1

[spark plank]

groups tho

[solar wyvern]

since 5 = (-1)(-1)(5) - 5

[spark plank]

uh okay

[solar wyvern]

yeah, sorta digression, but factorisation will be important soon enough

a lot of nice things about abelian groups ultimately can be rephrased as abelian groups being Z-modules, and Z being a ring you can do unique factorisation over

it's ultimately nonessential to understand the module side completely atm, but it's good to mull over

[spark plank]

[solar wyvern]

if you're interested in the countably infinite case, there's https://en.wikipedia.org/wiki/Nielsen–Schreier_theorem

(it's a bit involved, but more combinatorics/graph theory)

ie milne states with a reference to proof elsewhere

if you wanna continue to quotients/direct products that'd be cool too

[spark plank]

😩

[solar wyvern]

there's also more graph stuff like https://en.wikipedia.org/wiki/Frucht's_theorem

i don't think this in milne tho

You'll def run into https://en.wikipedia.org/wiki/Cayley's_theorem soon tho

[gentle pendant]

what do you mean by units if convolution is multiplication? @solar wyvern

[oblique river]

@solar wyvern going back to something you said a while ago

considering the real numbers as a ring

the units are all nonzero reals

similarly if you consider C as a ring, the units are all nonzero complex numbers

for the ring of smooth functions on R, the units are all the functions that are nonzero everywhere.

if you're working with analytic functions on C, then the units are just the nonzero constant functions

another note: it's not really the fact that Z is a UFD that makes the module theory nice, it's the fact that it's a PID. Being a PID is stronger than being a UFD. For Dedekind domains (including Z) they're equivalent, so what you said isn't incorrect, maybe just a little misleading. There's also a module theory over general Dedekind domains that is very well understood

[solar wyvern]

oh yeah, units of reals are nonzero

since real numbers are a field

also I goof on Z thing 😦

[oblique river]

its ok it happens :)

[solar wyvern]

Actually wrote PID first but then thought that might scare SA

euclidean domain is stronger than PID tho right? (although you're right it's PID which basically implies all the good stuff)

[oblique river]

yes

[solar wyvern]

[oblique river]

euclidean --> pid --> ufd

tbh I dont know of many things that use being euclidean

i'm sure they're out there

[solar wyvern]

also PID equiv to all the ideals are cyclic submodules? (this should just be nomenclature)

[oblique river]

but it's not relevant to me very much