groups-rings-fields

[spark plank]

when do we get bigger numbres

instead of dis sed polynomial growth

[spark plank]

[sonic current]

it's so weird that there are people who consider rings without multiplicative identity

who are they and what are they up to

say you have a ring and you consider it without its multiplicative identity. to prove the center of the ring is a subring, what would be the most obvious way to show it's not empty?

because it's trivial with the identity

[solar wyvern]

zero is in there tho?

commutes with everything

in both operations

@sonic current

[sonic current]

Oh, sure. Good eye!

[snow dew]

L1 with convolution product is a unitless ring

and it's kinda very usefull

[full blaze]

👌

[snow dew]

having a unit is kinda great tho, and aproximating an unit or building a bigger space (distribuitions) where there is an unit is a thing you see a lot

physicists don't care tho, they just write diracs all over the place like it was in L1

[safe monolith]

Hi, do you guys have any recommendations on material for learning about Abstract Algebra?

Also another question, I'm working on a problem where I'm trying to test whether a binary operator is associative or not and I found this: https://en.wikipedia.org/wiki/Light's_associativity_test

Let a binary operation ' · ' be defined in a finite set A by a Cayley table. (...)

Can this problem be solved without knowing A?

[chilly ocean]

will repeat question in algebra channel

[sonic current]

guys, I have an interesting problem that is supposedly a bit hard

Let A be a non-empty ring such that, for all a in A \ {0}, there is a unique a' such that aa'a = a.

Show that A doesn't have zero divisors, other than 0

the catch: A doesn't necessarily have the multiplicative identity

(I know, but my teacher uses this definition)

so what I did was
aa'a = a
aa'a -a = a -a
a(aa' - 1 ) = 0

which is probably a stretch since we don't know about the existence of 1

aa'a -a = 0 so maybe this is the most we can go

[earnest valley]

Commutative ?

[sonic current]

hmm

good question. I don't think so

[earnest valley]

A proof by absurd ?

[sonic current]

maybe

I think the uniqueness of a' might be important as well

i'm not sure

[sonic current]

oh! a friend solved it

actually it was in a book

it's absurd surely, but a very fancy one

[inner acorn]

Can anyone link me to a proof that the octonions have no zero divisors ??

[inner acorn]

<@&286206848099549185>

I think I can break this problem down into showing any subalgebra generated by two octonions R(a,b) is isomorphic to the quaternions (more importantly, is associative)

but I kinda need help with that qwq

http://etheses.bham.ac.uk/4349/1/Ghatei13MPhil.pdf
page 9 has a theorem on it, but I don't really understand it :<

[inner acorn]

...Okay, understoof most of thm 1.1 (part 2) up till "multiplication by i or j are orthogonal maps" on page 10 (7 lines up)... why is this??

[inner acorn]

figure it out x-x

omg.. that took too much time

[chilly ocean]

the sides of a 13-gon can be colored red or blue. Two colorings of the shape are considered identical if one is a rotation of the other. How many distinct colorings are possible?

[earnest valley]

Hint: group action.

[raw moth]

1/13 (2^13+2*12) = 632

if you know burnside that should make sense

otherwise

¯_(ツ)_/¯

[chilly ocean]

yeah I got it using Burnside

[covert vector]

OK TIME FOR AN EASY LINEAR ALGEBRA PROBLEM

[stone fulcrum]

YAY

[covert vector]

we have W a subspace of R⁴,
W = span{(6,2,-4,-1), (0,3,2,-2)}

and a point x = (-7,-4,7,6)

how to determine closest point in W to x?

Me and company are struggling far too much on this

ok so the strat was

that span is orthogonal

and uhh

[thorny slate]

complete W to a hyperplane by adding a vector orthogonal to x-W

then project

[stone fulcrum]

Pick some vector that goes from x to a point in W. Find the projection onto W. This forms a right triangle with a leg being the shortest distance

[thorny slate]

I guess

yeah I like that better

[covert vector]

Ok what we did now is project x onto W

from Graham schmidt

and so instead of us all being crazy

the web assignment is glitched

and it's wrong

[stone fulcrum]

Why Graham Schmidt? Should be able to find a protection with the dot product

Oh you found an orthogonal vector

[sour plume]

(psssst it's gram schmidt and not graham schmidt (i have nothing more worthwhile than that to contribute))

[errant drum]

TIL ^

[stone fulcrum]

Indeed now I know

[covert vector]

yeah but I will still mispel anyway

[kind dust]

[oak perch]

er ... different type of question than is usually asked ... What is at the forefront of Abstract Algebra research?

[stone fulcrum]

Galois theory, especially when it comes to rings, is incomplete.

Universal algebra is a lot more open ended. What are some other, more exotic algebras? Can they do anything new? If so, can we generalize concepts our regular algebras don't cover?

@oak perch

[spark plank]

math is never complete

[oak perch]

are semigroups still a thing?

@stone fulcrum

[stone fulcrum]

I think so yeah, but they aren't really important

[full blaze]

C-star algebra

[prisma ibex]

I dunno there's a lot of stuff that could be considered abstract algebra that just goes under more specialized names.

Representation theory is a wildly active field. Galois theory is very much active. Certainly a large chunk of abstract algebra research is the commutative algebra side of algebraic geometry.

Lots of homological methods all around.

[viral bear]

speaking of which I really need to study for my galois theory exam

[stone fulcrum]

The identity has order 1 and 1 divides any n. So, the subgroup has identity

The order of the inverse is the same as the order of the element. The subgroup has inverses

@chilly ocean

[raw moth]

there's a nice proof for that last bit in an abelian group

let a have order x

or I guess a simple proof for what is necessary is

let a have order x, then e = (a a^-1)^x = a^x (a^-1)^x =, so order of a^-1 divides x and hence divides n

[stone fulcrum]

You can also think about the cyclic subgroup. If a generates it, then so does a¯¹

[viral bear]

This also can be done pretty easy without assuming it's Abelian, a^-x = e a^-x=a^xa^-x=e, and for any smaller natural number y, a^-y = ea^-y = a^x a^-y = a^x-y != e because a is order x, and 1<=y<x. I think it's valuable to see this because it's a useful property to have generally.

[rare geyser]

hey i have a question thats probably really easy, but im stuck

its for a review on sets for my analysis class

[stone fulcrum]

[rare geyser]

can you go in voice?

idk how to start with b and d.

i don’t understand how i would begin that proof

give me 1 min

[stone fulcrum]

Hmm. Let me think about how to put it rigourously

If you take the intersection of sets, then map them, that's the same as mapping the sets then intersecting the result

@rare geyser

Hmm. I'm really not sure why lel that's always something I've taken for granted

∀i, y ∈ Ci

Yeah. That's essentially what you said just now

i'th set

That's an easy way to think about it. You can just say the sets are C1, C2, C3, then ∩C is their intersection

But sure, go ahead

[rare geyser]

he was able to help me

thank you! @stone fulcrum

[cyan quarry]

T and S are linear transformations such that T: V -> W and S: W -> V. How do I prove that (S + T)* = S* + T*?

[stone fulcrum]

How do you add two transformations? What does * mean?

[cyan quarry]

I'm boggled myself

[solar wyvern]

@cyan quarry composition

[cyan quarry]

if composition means g(f(x)), then I don't think that is what it means

as I have a different symbol that denotes that

[solar wyvern]

oh is that dual?

[cyan quarry]

yeah

only thing I've found online is http://algebra.math.ust.hk/vector_space/12_dual/lecture3.shtml

nothing regarding the proof

[solar wyvern]

do you know what a dual space is

[cyan quarry]

yes

but the "addition" of linear transformations isn't defined as far as I know?

[solar wyvern]

what have you tried

[cyan quarry]

I proved composition stuff in the previous question

here since V and W are different vector spaces

of unknown dimensions

I can't try to say something with their matrix equivalents

and their transpositions

[solar wyvern]

okay, what's your definition of a dual space

[cyan quarry]

Given a vector space V over a field F, the dual space is the subspace of linear transformations that take V to F

or maybe linear functionals is more correct

[solar wyvern]

how is that a subspace

[cyan quarry]

it's a subspace of all the linear transformations

the linear functionals obey additivity and.. homegenity? (not sure how to spell)

[cyan quarry]

right wait then

so it should be that (T + S)(v) = T(v) + S(v)

(T + S)*(l) = l o (T + S)(v) by definition

[cloud walrus]

creamy shits:

[errant drum]

You should mention that p|n and k|n

[solar wyvern]

i don't see any n's in there @chilly ocean

[gentle pendant]

ord(g)|n <=> g^n=e.

Let g,h in H. Then (gh^-1)^n = g^nh^-n=e^2=e, hence gh^-1 in H and we can conclude H is a subgroup.

[solar wyvern]

gomez got tired of yes O:

[errant drum]

Gomez with his god tier skills 😭

[gentle pendant]

not tired, just showing you can be more concise.

[errant drum]

Tfw you still prove subgroups like a pleb

[gentle pendant]

that's cool, you don't have to do it that way. the important observation is the first line.

[errant drum]

How often do you change your mathecathematiican names

[gentle pendant]

Get rid of inverses in my second line and that is the proof of closure.

Pretty often recently lol.

[errant drum]

Aren't you worried you'll run out xD

[solar wyvern]

[gentle pendant]

All great mathematicians were actually cats, so I think I have a good supply.

[errant drum]

I suppose you'd have to state the identity was in H

[gentle pendant]

Yes, if H is a nonempty subset of G and for every g,h in H we have gh^(-1) in H, then H is a subgroup.

[errant drum]

[gentle pendant]

Nope, you don't need to. The above test contains that information.

Let g=h

[errant drum]

Ahhh that true

[gentle pendant]

That's why it is a tiny optimisation and saves a little time.

But its not something essential.

meow

[spark plank]

interesting problem

[solar wyvern]

oh

@spark plank

u want problem?

[spark plank]

sure

[solar wyvern]

so u know what a normal subgroup is now, I take it?

[spark plank]

uh

[solar wyvern]

um

[spark plank]

aHa' and a'Ha are subgroups for all a?

[solar wyvern]

okay so first show that

[spark plank]

OML ADORABLE SLIMES

[solar wyvern]

either implies the other

you don't need both afaik

and wait

you want aHa' = H for all a in G (assuming a' is inverse)

another way to look at it is that it's a subgroup which is invariant under conjugation by G

[spark plank]

oh okay

[solar wyvern]

so try showing that either one implies both

[spark plank]

and by okay, I mean, okay gimme a few minutes while I slime

[solar wyvern]

also that those are equivalent to aH = Ha

[spark plank]

you will see

[solar wyvern]

i don't tho?

anyways, the next thing to do is show that there's a minimal normal subgroup containing S⊂G any sub_set_ of G

[spark plank]

you will tho

[solar wyvern]

(actually you can do something similar for any subset S to make it into a subgroup (not necessarily normal)

actually u might know that already

[spark plank]

I'm a bit late, forgive me

maybe I'll upload on the Chinese New Year

@solar wyvern wait isn't aHa' pretty trivially a subgroup?

[valid elbow]

I like it

[spark plank]

"so try showing that either one implies both"

wdym

[solar wyvern]

so

how is it trivially a subgroup, and how is that what I asked

@spark plank

[spark plank]

I mean when u multiply elements, the left and right's cancel

And idk wut u asked

@solar wyvern

[solar wyvern]

okay

so in an abelian group that may be true

@spark plank what nonabelian groups do u kno

(preferably on the smaller side)

using something like D_4 or S_3: construct a subgroup, then conjugate by each other element of the group to check for closure.

[spark plank]

I don't remember tbh

[solar wyvern]

there's nothing to remember

[spark plank]

Also holy fork I looked in the mirror

[solar wyvern]

rip SA

[spark plank]

My eyes are hella bloodshot

[solar wyvern]

get some sleep?

[spark plank]

Probably

kek

I swear I only removed like 3-4 hrs of sleep from my daily schedule

I'm sure that's healthy

Anyways ttyl8r

nani

oof

[solar wyvern]

nandemonai

[spark plank]

[solar wyvern]

sekrit group notations

[spark plank]

but gn

[solar wyvern]

left hand side is generators, right hand side is relations (constraints basically)

night

[spark plank]

Also oof tfw not on pc so can't upload Rimuru sleepy pic

[solar wyvern]

[kind dust]

Uhh, what's with the string of Anime pics. xD

[spark plank]

y not

[craggy robin]

yoneda calm down with all this slimy anime

[errant drum]

The identity and inverse follow from the the fact
f(ab) = f(a) * f(b)

You can try prove that $$f(e_G) = e_H$$

[cloud walrus]

Cosmicrays:

[errant drum]

But yeah you don't need to check

Hint:
$e_H = f(e_G)^{-1} * f(e_G)$

[cloud walrus]

Cosmicrays:

[errant drum]

Now combine that with what you already have

H is a group

You know that it must have an inverse

For you to prove that $f$ is a homomorphism you need to have groups:
$(G,\circ)$ and $(H,*)$

[cloud walrus]

Cosmicrays:

[errant drum]

Like you need those groups to exist before you can say that f is a homomorphism

[cloud walrus]

creamy shits:

[errant drum]

👍

Now show $f(a^{-1}) = f(a)^{-1}$

[cloud walrus]

Cosmicrays:

creamy shits:

[errant drum]

$e_G?$

[cloud walrus]

Cosmicrays:

[errant drum]

e_H

[cloud walrus]

creamy shits:

[errant drum]

Yee

[cloud walrus]

creamy shits:

[raw moth]

yes

[stone fulcrum]

Hey Cream. How's it going? Exam soon?

Hey, happy to hear it

Hopefully that's true?

... Hey, uh I'm glad to hear you're all prepared and uhm nothing is wrong

[oblique river]

youre gonna do great

[stone fulcrum]

I'd give you 100

[oblique river]

:P

[errant drum]

Good luck!

[stone fulcrum]

Gib me D4/{e, r²}

[cloud walrus]

creamy shits:

creamy shits:

[hot folio]

whats the best way to check if set is a subgroup of a group ?

i'm thinking: let H be a set, and G a group

H is not trivial

for all a,b in H, ab is in H

the inverse of H is the same as the inverse of G

nah this is not it

welp, the general one is to show that H is a group, and H is a subset of G and the identity element and the inverse of H is the same as G

[raw moth]

H non-empty, a,b in H implies ab is in H, a in H implies a^-1 is in H

[hot folio]

^

thats the one

[spark plank]

H non-empty, a, b in H implies a(b^-1) in H covers both

[viral bear]

I wish someone told me this earlier tbh Parts (1) and (2) are respectively the First and Second Isomorphism Theorems. They are the translation into normal subgroup language of two straightforward facts: restricting a homomorphism to a subgroup yields a homomorphism, and composing two homomorphisms yields a homomorphism

Like understanding these theorems in terms of homomorphisms makes a lot more sense

[errant drum]

Never seen this before

That's much better than the explanation my prof gave which was "one of em is a triangle and the other is a diamond"

[somber bramble]

but G/ker(f) isn't a subgroup of G

im not quite sure how to interpret this statement

[viral bear]

The numbering used here is a bit different than Ive seen in other sources

The first and second are actually the second and third on the normal ordering

I didn't notice this right away when i posted because the definitions of (1) and (2) were written above

[somber bramble]

so they’re saying that $$SN/N \cong S/S\cap N$$ follows from restriction to a subgroup and $$(G/K)/(N/K) \cong G/N$$ follows from composition?

[cloud walrus]

Sascha Baer:

[oblique river]

basically whenever you want to prove a statement of the form A/B = C/D

what you want to do is find a map from A to C

compose it with the quotient map C --> C/D to give a map A --> C/D

prove that map is surjective

and then prove that the kernel is B. then by the isomorphism theorem (A / ker = image) you get A/B = C/D

(or maybe switch the roles of A/B and C/D. i.e. instead of the above, start by finding a map C --> A...)

so for the first one, try finding a map S --> SN to start

and for the second one, start with a map G --> G/K

and follow the steps I outlined

@somber bramble

[hot folio]

$ let \ R \ be \ a \ relation \ in \ \mathbb{R} \ defined \ as: \ \forall \ x,y \ \in \mathbb{R} , \ xRy \ \leftrightarrow \ \exists \alpha \ \in \ \mathbb{R} , \ x = y\ + \ 5 \alpha $

[cloud walrus]

Fibi The Enthusiastic:

[hot folio]

i've never seen a relation like this, like "that uses a third variable alpha"

what i'm supposed to do here?

say i'm checking for reflexivity

$ xRx \ \leftrightarrow \ x = x + 5 \alpha \ \ \alpha = 0 \ \in \mathbb{R} $

[cloud walrus]

Fibi The Enthusiastic:

[hot folio]

so i'm solving for alpha, and alpha must be in R ?

to check if R is symetric:

$ xRy \ \leftrightarrow \ \exists \alpha \ \in \mathbb{R} , \ x = y+ 5 \alpha \ yRx \ \leftrightarrow \ \exists \alpha \ \in \mathbb{R} , \ y = x + 5 \alpha $

[cloud walrus]

Fibi The Enthusiastic:

[hot folio]

what do i do next ?

basically the question is: instruction (or human algorithm) to show that R is relation of equivalence (of this type of stuff where there's a third external variable alpha)

[viral bear]

well to show something is an equivalence relation you need to show reflexivity, symmetry, and transitivity

so you need to show transitivity

[chilly ocean]

hi cant someone help me 😭

can*

[viral bear]

with what

[chilly ocean]

Brian spins a fair spinner numbered 1 - 5 and flips a fair coin.
What is the probability of obtaining a 2 and a head?

[hot folio]

i know that ixseft

i just don't know how to work with this type of relations

usually its x = y + number

[viral bear]

probably should ask in a different channel @chilly ocean this is for abstract algebra, that is a question about probability

[hot folio]

or sth

idk how to deal with alpha

[viral bear]

anyway I just said that to start since you didnt mention that explicitly

so to be transitive means that: $xRy \land yRz \Rightarrow xRz$

[cloud walrus]

ixsetf:

[viral bear]

I would start by writing down what that means for this particular relation

[hot folio]

ok, lets work for R to be symetric

if i understand that then its not hard to figure out the rest

i already wrote it

https://cdn.discordapp.com/attachments/496784958430380033/534432952961073152/433735907225501696.png

[viral bear]

oh wait I thought you already solved it

hint

use first equation to solve for y

[hot folio]

$ y = y + 5\alpha + 5 \alpha = y + 10 \alpha $

[cloud walrus]

Fibi The Enthusiastic:

[hot folio]

so 10alpha = 0 then alpha = 0

[viral bear]

uh not quite

so in symmetry

you have the top part

$ xRy \leftrightarrow \exists \alpha \in \mathbb{R} , \ x = y+ 5 \alpha$

[cloud walrus]

ixsetf:

[hot folio]

xRy must be the same as yRx

[viral bear]

that is the condition for symmetry yes

your starting point is xRy

you need to show this implies yRx

the implication becomes equivalence because x and y are arbitrary

the hint is to solve for y starting with this equation

[hot folio]

so xRy is $ x = y + 5 \alpha \ and \ we \ need \ to \ show \ that \ it \ implies \ yRx \ which \ is: \ y = x + 5 \alpha $

[whole basalt]

I've got a question about functions

[hot folio]

like this ?

[cloud walrus]

Fibi The Enthusiastic:

[whole basalt]

What are the weakest conditions under which a set of complex valued functions form an integral domain with respect to poinwise addition and multiplication?

[viral bear]

right

keeping in mind that the two alphas need not be the same

[whole basalt]

(Whenever someone gets a chance though, no rush)

[hot folio]

$ x = y + 5 \alpha \ y = x - 5 \alpha \ put \ \alpha_2 = - \alpha \ y = x + 5 \alpha_2 $

[cloud walrus]

Fibi The Enthusiastic:

[viral bear]

correct

[hot folio]

oh i see, the thing i needed was the two alphas need not be the same

thanks ixsetf 👌

[viral bear]

np

one trick for something like this is to think through concrete examples

for example 0R1

because 0 = 1 + 5(-0.2)

so for symmetry you would need that 1R0

so 1 = 0 + 5x

thus x (alpha 2) must be 0.2

[chilly ocean]

Yes

Wat is dis

[viral bear]

this question needs words

transpoptart?

(joke)

but yeah idk it looks like you have a handle on what you are doing

[chilly ocean]

(x,x^2) -> (x,x^4)

[cloud walrus]

creamy shits:

[chilly ocean]

Oh, then you got it fam

Just gotta show that x -> (x,x^2) is isomorph

[stone fulcrum]

I really don't think "transport of structure" is a real thing. I kind of think your professor is an applied mathematician who thought he could do group theory, and this problem is a weird amalgamation of the two fields. You generally need to show that something is a group before you can do any group theory onto it

[chilly ocean]

Tbh his prof sounds like a bit of a mathmagician

[viral bear]

transport of structure I think is real but this seems like a weird context to apply it

[cloud walrus]

creamy shits:

creamy shits:

[viral bear]

the wikipedia page for transport of structure talks about isomorphisms between vector spaces

and defining an inner product on the later by the isomorphism

[chilly ocean]

Heh, instead of getting caught up in the minutae

Are you just looking for

,$ f:(\bR,+)\to(X,\oplus)?

[cloud walrus]

Jichael Mackson:

[chilly ocean]

s.t. f is bijective

Oh right because Ker(f) = e iff f is bijective - das smort

[viral bear]

have you learned the definition of a homomorphism

ok thats good at least

basically the idea of this concept he's trying to give you

is that most properties of groups are preserved under isomorphism

yep

[chilly ocean]

Basically isomorphism just relabels your elements

[stone fulcrum]

Isomorphisms preserve everything. Homomorphisms are lazier

[viral bear]

everything*

meaning everything that isn't dependent on your choice of labels

[chilly ocean]

This weirdly sorta becomes the foundation for algebraic topology . Trying to find "related bendy functions" that work up to homotopy.

Nice sneak prof.

You actually might want to try to get through the first few pages of Hatcher

This problem gets a lot more interesting.

Probably what says "define:" no?

[viral bear]

nah

hes describing a procedure

so he's just listing steps

so everything is the definition

he's defining a procedure

and those steps together make up the procedure

the place where he proves this procedure works doesnt seem to be there

but its very plausible it is proved elsewhere

this is probably one of the intuition traps that is easy to fall into when youve done a lot of math

its one of those things that is going to seem "obvious" to him

so he probably though he could skip proving it

its not too hard to prove for any particular property

the idea is like

if I take a group

and that group has some property defined independent of the labels

like for example abelian is defined as $\forall (g,h \in G)(gh =hg)$

[cloud walrus]

creamy shits:

ixsetf:

[viral bear]

forgot the name of the forall

i dont have to write it that often in proofs

anyway

because it is just referring to a generic g and h

and doesn't consider the specific labels that the group uses

when you "relabel" the elements of a group by isomorphism

the property of Abelian-ness doesn't change

if you take any property like that

like order of the group

or the order of the center

or whether the group is simple (if you dont know this yet you can ignore for now)

then a relabeling wouldn't impact that property

?

all groups have a trivial subgroup

a simple group has no normal subgroups that arent trivial

if you dont know what that is you can ignore it

[chilly ocean]

In a simple group

ye

[viral bear]

but yeah like

that is 100% not a rigorous argument, but you don't have the tools available in that class to do a rigorous argument for more than one property at a time

ok looking at this more carefully that last line is pretty garbo

the part about $\forall A, B \in X, A + B = f^{-1}(f(A)*f(B))$

[cloud walrus]

ixsetf:

[viral bear]

like its highly unclear what the take away from that should be

[chilly ocean]

Hang on, isn't this just screwing with homomorph defs in the isomorphic case?

[viral bear]

yeah it is but he didn't make that clear at all

[chilly ocean]

I'll give you that

I had to thonk for a minute

[viral bear]

its also called Define

which its not

[chilly ocean]

I mean it sounds like he more poorly explains a single concept badly in a lot of ways

Which is just symptomatic of intuition overload. He wants to go jedi with people who've never seen this stuff before.

[viral bear]

yeah intuition overload is one of the worst mistakes you can make

short of like doing drugs in class or something

[chilly ocean]

Tbh - I'd actually tell your prof this - he needs to condense his notes for noobs. Just follow Artin's proofs or smth ffs

[viral bear]

is he roughly following a textbook?

[chilly ocean]

Did he mention a text to follow at the beginning of the class?

[viral bear]

that was opposite day

[chilly ocean]

You'll probably survive better if you treat them as textbooks and then just follow his weird little procedures

[viral bear]

you should read that as do use them as textbooks

[chilly ocean]

Tbh - I'd get mad for college tuition

This is bad teaching

We're slowly ironing out definitions and proving

[viral bear]

my main question is how people like this both get PhDs and jobs

[chilly ocean]

But we're making up for a full undergrad class lmfao

(((escape into academia on research merit)))

Anyway - you already know what I suggested - Fraleigh for super friendly approach, Artin for a bit more aggressive, D&F if you're feeling like the math queen

[viral bear]

hes probably not very smart

otherwise he would teach better

[chilly ocean]

Tbh - I have met guys who don't give a fuck about the teaching

[viral bear]

this doesnt give me that sort of vibe though

if they were super smart and didnt care

[chilly ocean]

Bruh

[viral bear]

they would just say to define an isomorphism between the groups

and that preserves structure

[chilly ocean]

Legit

"If you wanna prove it - look it up on Math.SE you lazy fucks"

But that reading list is ecclectic as heck

coding theory

combinatorics

[viral bear]

what the fuck is this course called

[chilly ocean]

Wat

Discrete math with graph theory

This is like an algebra 1 course right?

[viral bear]

my impression is that this might be one of those half and half courses

did you learn what a graph was

[chilly ocean]

What the actual course title/description?

[viral bear]

then half of those books dont apply to literally anything

[chilly ocean]

Right

[viral bear]

ok so it is a half and half

[chilly ocean]

So group theory was pretty much sem 1

You're expecting discrete math in sem 2?

[viral bear]

oh is this a yearlong course?

[chilly ocean]

Alright then

Peano's axioms mayb

[viral bear]

probably counting as in combinatorics thing

[chilly ocean]

That would make more sense

[viral bear]

enumerate number of ways to do something

[chilly ocean]

Computability theory intro

[viral bear]

why?

[chilly ocean]

It's actually taught really badly in the UK

[viral bear]

oh nice

[chilly ocean]

What is it that math olympians use?

The nine methods?

[viral bear]

idk

[chilly ocean]

Any who

You literally learn combinations and permutations and they're important for something

And then: the end

That's the best you're getting from highschool

[viral bear]

ye thats too bad

[chilly ocean]

No stars and bars or anything

[viral bear]

but like imo, combinatorics is a really easy subject to get intuition for

if its taught remotely close to right

[chilly ocean]

Combinatorics is super cool - but I never invested the energy

And I haven't used it except in probability theory in my entire degree

So Idk what's going on there

[viral bear]

I did a lot of graph theory but only a bit of general combinatorics

that said a lot of ideas that come out of combinatorics are close to physical concepts

[chilly ocean]

Graph theory was available but seeing as how I would've graduated without doing ring theory - I figured I'd pass.

[viral bear]

I actually never covered rings in my undergrad

[chilly ocean]

Arguably enumeration is the most natural part of math out there, so it makes sense.

[viral bear]

algebra was 100% focused on groups

ring and field theory I just figured out as needed for the classes when I got here

(UK msc)

[chilly ocean]

Hm. It irks me that I'll be graduating and we will have only touched upon algebra

Are you a PhD now then?

[viral bear]

no doing msc

[chilly ocean]

Oh right

What's your thesis?

[viral bear]

the way it works here is you write your thesis over the summer

you dont pick topics until feb-march

[chilly ocean]

Ah

And UK MRes - is there any point if you didn't go to Oxbridge?

And what are you considering for a topic, etc. etc.?

[viral bear]

well I'd like to do something that extends my undergrad project in some way

to get something thats not on "the list"

I have to pitch the idea to someone

so once im done with exams I'll look for someone who has research interest closest to what I want to do

[chilly ocean]

Fair enough - are you considering taking it to a research degree afterwards?

[viral bear]

probably not, I think next year I will go try to get a job

my interest in writing software is now fairly even with my interest in math

and the quality of job opportunities for something like that seem better than they would be if I was trying to do pure mathematics in academia

that said if I end up not liking how things end up with that path

I will try to apply for something next year

[chilly ocean]

Fair does. Sounds like a good plan of action.

I'm just looking at various MSc's flip-flopping between geometry and biomath

[viral bear]

theres also the benefit with that that if I do succeed with job stuffs then I can also pay down student debts

best option I'd say depends on what you want to do after

[chilly ocean]

Math, not starve, my needs are fairly simple. But I worry my work ethic is crap and I'll have to let reality take the wheel

[viral bear]

well geometry and biomath are both math

and if you purchase food you will not starve

[chilly ocean]

Ah-hah

[viral bear]

but ye like

im pretty sure like

the quality of your research is going to matter a lot more than which of those topics you choose

in regards to the % chance of not starving

[chilly ocean]

Here's hoping I'll know what I'm doing 😬

[viral bear]

quality of research is a function of like

effort, aptitude, luck

so like

try to choose whichever of those maximizes your aptitude

and your motivation

since that will influence your effort

if one of those makes you more lucky somehow you can also pick that one

[chilly ocean]

Biomath is due some pretty big breaks in luck

But that's only because the biology has come far enough to make it formalizable.

[cloud walrus]

creamy shits:

[chilly ocean]

g(a,b) -> (a+b,(a+b)^2)

g^{-1} (a+b,(a+b)^2) -> (a,b)

It's not 1-1 for starters

2-1, 3-2, 4-3

Yours ain't

let a=2,b=-1, and a=3, b=-2

They shouldn't be mapping to the same point

[viral bear]

when he asked you to prove uh

[cloud walrus]

creamy shits:

[viral bear]

the abelianness of that set

did he give you an operation

[chilly ocean]

Actually

hol' up nigga

[viral bear]

heres something you can do if you want to piss him off

[chilly ocean]

Why's your function mapping pairs from IR?

[viral bear]

prove they have the same cardinality

use axiom of choice to define a bijection

[cloud walrus]

creamy shits:

[chilly ocean]

|R| = |R^2| ?

[viral bear]

and define the rule by that thingy of $f^{-1}(f(A)*f(B))$

[chilly ocean]

Oh wait

[cloud walrus]

ixsetf:

[chilly ocean]

It's a curve in R^2

Alright

Now that son

Is bijective

Both onto and one-one

Yep

Yep yep

You'll be using info like (-2,4) =/= (2,4)

[viral bear]

im going to point out that this is a stupid exercise

because without defining a rule you can do like basically anything you want

[chilly ocean]

Ye

lololol

The thing is bijective - but y'all gotta show it's a homomorph

g(x)

Then ur don

[viral bear]

I think thats the point of "transfer of structure" or whatever

you do the procedure thing he said

[chilly ocean]

You wanted isomorphin

You got isomorphin

Don

[viral bear]

then you get isomorphing by default

[chilly ocean]

Cuz the info needed to make a bijective map is contained in x

[viral bear]

have you heard of the axiom of choice?

[chilly ocean]

All you've done is added some flavor to the real line and stuck to it IR^2

[viral bear]

well rough version of axiom of choice

actually nvm you dont need it

[chilly ocean]

Cartesian product of non-empty sets is non-empty is a good one imo

[viral bear]

Consider a group G and a set X

with |G| =|X|

then by the definition of cardinality you have a bijection from G to X

because you are defining your own rule

[chilly ocean]

@chilly ocean Ye

[viral bear]

you can use that bijection to define it

[chilly ocean]

f(x) = f(y) implies x = y is def. of injective

[viral bear]

ye

[chilly ocean]

Yep

[viral bear]

I think i get what he was trying to say in those lectures now btw

[chilly ocean]

Surjective just show the domain and codomain are same cardinality kek

[viral bear]

so like

if you have a bijection between G and X

say f

for any a, b in X

you have some g, h in G with f(g)=a and f(h)=b

then you can define your rule for a and b as a+b = f(g+h)

yeah exact

[chilly ocean]

👌

Nop

(-2)^2 = 2^2 = 4

For every x there is a unique (x,x^2) tho

[viral bear]

ye

but yeah like

once you have the bijection

you can find your g and h by the inverse map

so f(g+h) = f(f^-1(a)+f^-1(b))

which is what he was trying to say on the slide

so you can then use a+b = f(f^-1(a)+f^-1(b)) as your rule

and you get a homomorphism out of your bijection "for free"

you get what i mean?

[chilly ocean]

Less operations

More homomorphin

Usual way of thinking of homomorph is f(a+b) = f(a)*f(b). So he flips it on its head

Eyup

Without having to use the * operator

It's a nice lazy approach tbh

[viral bear]

(it would have been nice if like, he explained any of that though lol)

[chilly ocean]

To show there's a homomorph

Ye - it comes from an abused version of the homomorph definition

[viral bear]

so above I defined a and b as f(g) and f(h)

so if you define a+b = f-1(a) * f-1(b)

[chilly ocean]

Ye

[viral bear]

you basically are writing the definition of homomorphism as your rule

so as like a simple example

Z2 consists of the set {0, 1}

with an operation sending 0*0=0, 0*1=1, 1*0=1, 1*1=0

[chilly ocean]

\ for esc

[viral bear]

ye forgot lol

[chilly ocean]

\*\ * eg

[viral bear]

\\\

[chilly ocean]

You know what

fuk it

[viral bear]

so lets say I want to transfer of structure to {a,b}

then i define f mapping 0 to a and 1 to b

I can simply swap the symbols in my definition of the rule

so instead of 0*0=0

I have a*a=a

and 0*1=1 is a*b=b

you see whats going on there?

[solemn hollow]

[viral bear]

right

so how do you know how to relabel?

given b*a, how do you know what that is?

well you know the inverse of f applied to b is 1, and the inverse image of a is 0

so thats where the f^-1s come in

you then apply the original group operation to 1 and 0

and get 1

but you have to translate that back into {a,b}

so you apply f to everything

[chilly ocean]

b*a = f(f^-1(b)+f^-1(a))

Take it slow

f(f^-1(b)+f^-1(a))

=f(1+0)

=f(1)

=b

[viral bear]

by doing that you are getting a relabeling of the result

[chilly ocean]

"transferance of structure"

Only can do this because f is bijective tho

[viral bear]

since otherwise no f^-1 exists

[chilly ocean]

injective might suffice for inverses - but you can't guarantee closure of the group on the codomain

[viral bear]

uh it has to be bijective

because if its not surjective the inverse isnt defined over the whole domain

[chilly ocean]

Inverse can just be defined over image no?

[viral bear]

it can sort of

I dont think thats the standard definition of inverse though

[chilly ocean]

I'll have to check

I just take inverse to mean - I know how to get back to my preimage

[viral bear]

if you dont require surjectivity

you lose the ability to do things like f^-1(f(x))

because the codomain of f is larger than the domain of f^-1

[chilly ocean]

Riiight

Okay, gotcha

I was thinking that no surjective => codomain of f < domain of f^-1

[viral bear]

fun pattern btw

since you can define a function as a relation with some properties

a function being surjective means that the relation xRy that defines it is a injective function when transformed into yRx

[chilly ocean]

By the same R?

[viral bear]

I meant the relation defined by swapping the order

so like aR'b = bRa

[chilly ocean]

So creates an equivalence relation on equivalence relations

[viral bear]

then R' is an injective function exactly when R is a surjective function

what do you mean?

[chilly ocean]

R ~ R' when aR'b = bRa

[viral bear]

uh sort of

these arent equivalence relations btw

they are just normal relations

[chilly ocean]

Yeah I thought it was a little weird

[viral bear]

A relation between two sets is a collection of ordered pairs containing one object from each set. If the object x is from the first set and the object y is from the second set, then the objects are said to be related if the ordered pair (x,y) is in the relation.

from the first website on google

[chilly ocean]

Ahhh

And equivalence just stipulates stuff like

for each x, (x,x) has to be in there?

[viral bear]

yeah an equivalence relation is a relation that is reflexive, symmetric, and transitive

[chilly ocean]

Alright, this illuminates stuff a bit more

[viral bear]

a function f:A->B is a relation with the property that for all x in A, there exists a unique y in B such that (x,y) is in f

[chilly ocean]

So aRb defines an injective function being a relation with the added condition for each x there is a unique y such that (x,y) is in the relation?

Right.

[viral bear]

I think the wording got scrambled there a bit

there are too many uniques to make sense

updated version is the definition of a function

[chilly ocean]

Right

[viral bear]

that definition is just a function in general

[chilly ocean]

Okay, because we can still have (2,1) and (3,1) in there just fine

[viral bear]

to be injective you need that for each y in there is a unique x such that (x,y) is in the relation

[chilly ocean]

Okay

[viral bear]

and to be surjective you have that for all y there exists some x such that (x,y) is in the relation

[chilly ocean]

Right

xRy = yR'x means

If xRy is injective

Then its inverse is surjective?

[viral bear]

I think you actually caught me making a mistake here tbh

[chilly ocean]

I mean I'm having a lot of fun learning about relations

We just get taught about equivalence relations for proofs

[viral bear]

I'll drop the actual fact that is related to these though

to make up for the first one being a dud

[chilly ocean]

Dawh

and yay

They be related in the relation of my heart

[viral bear]

a relation that is both a function and an equivalence relation is a bijective function

[chilly ocean]

Okay

[viral bear]

thats an if and only if

[chilly ocean]

Hang on, but if xRy then I need (x,x) to be in the relation for all x?

[viral bear]

god damn it two duds

i blame the fact that its been like 3 years since I did this

[chilly ocean]

I'd get a straight line on every bijective function

if f:IR -> IR

Anyway this is neat

@chilly ocean You are ready for AT my chile, come

[viral bear]

alright i should study more and stop getting things wrong about functions

[chilly ocean]

Some how I learnt pointset without relations

Follow his procedure

It's right

Since g is bijective

[viral bear]

relations are important foundationally but they arent that important in like actual research fields

other than equivalence relations

[chilly ocean]

Which we use for proofs

[viral bear]

ye

[chilly ocean]

Set theory abstractions are always nice though

Gives you more language

[viral bear]

but ye im off

[chilly ocean]

Right

So a+b = f(f^-1(a)*f^-1(b))

Or

Btw

,$ a\oplus b

Do what

[cloud walrus]

Jichael Mackson:

[chilly ocean]

On your group (X,o+)

Right

Would it not be easier to give it

(a+b,a^2+b^2)?

(If you have any choice)

Right we'll just use this then

,align (a,a^2)\oplus (b,b^2) &= f(f^{-1}(a,a^2)+f^{-1}(b,b^2))\ &=f(a+b)\&= (a+b,(a+b)^2)

[cloud walrus]

Jichael Mackson:

[chilly ocean]

Yep this works

Because those are the elements of $(X,\oplus)$

[cloud walrus]

Jichael Mackson:

[chilly ocean]

,$ a\in (\bR,+), (a,a^2)\in (X,\oplus)

[cloud walrus]

Jichael Mackson:

[chilly ocean]

So

We show that the homomorphism is there

Since 1) f is bijective

2. Is satisfies the definition of a homomorphism

Although a butchered one made especially under the presumption that you've found a bijective morphism

So yeah - you did the thing

Isomorphism = Bijective Homomorphism

And we're done

What's got you confused?

Well homomorphism is $f(a+b)=f(a)*f(b)$ right?

[cloud walrus]

Jichael Mackson:

[chilly ocean]

If isomorphism

Then bijective

So

$f^{-1}(c*d) = f^{-1}(c)+f^{-1}(d)$

[cloud walrus]

Jichael Mackson:

[chilly ocean]

So

$c*d=f(f^{-1}(c)+f^{-1}(d))$

[cloud walrus]

Jichael Mackson:

[chilly ocean]

Normal way is just

1. Prove is homomorphism

2. Prove is bijective

This way is just

1. Prove is bijective
2. Prove inverse is homomorphism
3. Therefore is isomorphism

The latter way has an advantage since you only have to think in the original group operation

That is + from (IR,+)

Rather than fuck around with o+ and +

No explicit computation is required for o+ and thats its advantage

Just show that each (x,x^2) has been mapped to by x

[cloud walrus]

creamy shits:

[chilly ocean]

Yep

P much.

If you wanted to get rigorous you could a double inclusion between the image of g and intended codomain.

Ye

,$ A\subset B, B\subset A \implies A=B

[cloud walrus]

Jichael Mackson:

[chilly ocean]

normal subgroup

And then you're there kiddo

cosets of*

But ye, das fine

Anyway - we should do fundamental groups some time

Is easy, trust

I mean I'm hardly surprised - it sounds like the teaching is trash

At least you were able to get out there and find other ways to learn

It's kinda ironic that you have to get a level of intuition from doing a good course in group theory - to follow his course in group theory kek

Anyway yis - don't let this put you off algebra - it's very pretty when it works.

Seriously recommend Fraleigh

[cloud walrus]

creamy shits:

[chilly ocean]

Alright

Line 1

Bijection is the cause of your implication - state this explicitly

Line 2 you show that applying the inverse function and original groups operator that you yield the correct product using oplus

So all definitions are satisfied and thus the morphism is valid.

To get it more rigorous in your head

If the second line wasn't true - then there'd be no inverse homomorphism

Therefore no isomorphism.

Which would be pretty weird

It just means that it satisfies the definition of homomorphism

,$ f^{-1}(a\oplus b) = f^{-1}(a)+f^{-1}(b)

[cloud walrus]

Jichael Mackson:

[chilly ocean]

^ is true if there is an inverse homomorphism

If we assume this is true and knowing f is bijective:

,$ a\oplus b = f(f^{-1}(a)+f^{-1}(b))

[cloud walrus]

Jichael Mackson:

[chilly ocean]

We confirm it is true by computing $a\oplus b$

[cloud walrus]

Jichael Mackson:

[chilly ocean]

Using the RHS

We know what $a\oplus b$ should be

[cloud walrus]

Jichael Mackson:

[chilly ocean]

If we recover it from the RHS - we confirm that f^-1 is indeed a homomorphism

Yep

You defined it

,$ a=(c,c^2), b=(d,d^2) \implies a\oplus b = (c+d,(c+d)^2)

[cloud walrus]

Jichael Mackson:

[chilly ocean]

That the definition of $\oplus$

[cloud walrus]

Jichael Mackson:

[chilly ocean]

What $a\oplus b$ is - is known

[cloud walrus]

Jichael Mackson:

[chilly ocean]

What $f(f^{-1}(a)+f^{-1}(b))$ is - is not

[cloud walrus]

Jichael Mackson:

[chilly ocean]

If the inverse is homomorphic and f bijective - they are supposedly equal

We find by computing the latter - that they are in fact are

And since we already know f is bijective - this means that the inverse is homomorphic

If the inverse is homomorphic - then f must be isomorphic.

What doesn't follow from my explanation?

You get to pick it, right?

Actually

You have.

[cloud walrus]

creamy shits:

[chilly ocean]

Alright

Is oplus defined in 26?

Does it follow from that definition that, $A\oplus B =(a+b,(a+b)^2)$?

[cloud walrus]

Jichael Mackson:

[chilly ocean]

This diagram was on the question paper?

[viral bear]

that diagram seems very wrong btw

[chilly ocean]

Kk

Well if you're using that to define ao+b

[viral bear]

imagine if you had A and B in a horizontal line

[chilly ocean]

Yeah that won't be (a+b,(a+b)^2)

[viral bear]

then A+B would be O

actually nvm thats fine

[chilly ocean]

That'll be the solution to x^2 = (b^2-a^2)/(b-a)x

[viral bear]

imo just skip the diagram if you arent required to provide it

[chilly ocean]

O wait

(b-a)(b+a)/(b-a) x = x^2 => a+b = x

=> A o+ B = (a+b,(a+b)^2)

Kk

So is defined

[viral bear]

good job tbh

good geometric reasoning

[chilly ocean]

Right?

Damn, I just would've skipped straight to the symbol pushing

So ye

o+ is defined by solving that thing

:l

Nigga

You defined it

What was the point of this run around?

I'm angry

And hungry

Brah this is the transport of structure

You show that the groups are isomorphic and that you can build the latter groups structure from the first group and a bijective mapping between the two who's inverse is a homomorphism

That's the entire strategy

This is the point of skipping actually using the latter's operation as you normally would in a: 1) show homo 2) show homo bi

This is

1. find bi map, 2) show backward bi map is homo 3) imply isomorph

Yeah

In way - it's pretty elegant

But it's just a different/less direct strategy to achieve the same thing

find an isomorphism

[viral bear]

I think you can actually simplify the proof a bit if you go backwards with step 2 btw

you start with the definition of homomorphism

and then do iff steps to construct your operation

then you have a bijective homomorphism

thus isomorphism

[chilly ocean]

It's interesting to note that it doesn't matter which direction you decide to prove is homomorphic - you'll arrive at the same conclusion.

[viral bear]

ye thats worth noting

yeah

its slightly deeper than that

basically a bijective homomorphism implies that the inverse map also has homomorphism properties

f bijective => (f(a+b) = f(a)*f(b) <=> f(c*d) = f(c)+f(d))

where statements hold for all a b, and all c d

[chilly ocean]

H = G/Ker(f)?

[viral bear]

1st iso theorem was always the easiest for me to remember

[chilly ocean]

Some texts switch 2nd and 3rd 👀

Nice fact tho if Ker(f) = {e} => G = H

[viral bear]

f(gxg^-1)=f(g)f(x)f(g^-1)=(because x in kernal)=f(g)f(g^-1)= e, => gxg^-1 in kernel

[chilly ocean]

Ye

Quotient group is group itself

[viral bear]

yeah the whole group is normal

[chilly ocean]

G/{e} = G

H = G/Ker(f) etc.

[viral bear]

reposting because I still think this is cool
Parts (1) and (2) are respectively the First and Second Isomorphism Theorems. They are the translation into normal subgroup language of two straightforward facts: restricting a homomorphism to a subgroup yields a homomorphism, and composing two homomorphisms yields a homomorphism
the first and second are actually second and third

[chilly ocean]

Proof of First Isomorphism theorem tho?

t!wiki restriction map

[fossil mango]

📖 | ** https://en.wikipedia.org/wiki/Restriction_map **

[chilly ocean]

https://en.wikipedia.org/wiki/Restriction_(mathematics)

Le oof

[viral bear]

they didnt consider the first one to be good enough to exist i guess

[chilly ocean]

Got a function f:A->B, got a subset of A, C: f|_C is then the restricted function from C -> B

Where f|_C = f(x) , x in A as it would be in f:A->B

[viral bear]

yepo

[chilly ocean]

Homomorphism is just a function - so the magic is, restrict the homomorphism on your group to the subgroup - it's still a homomorphism.

[viral bear]

this is from Stewart btw

[chilly ocean]

Good old Stew

[viral bear]

you said you were going to pick that up right?

[chilly ocean]

Galois theory specifically

[viral bear]

ye thats the one

[chilly ocean]

Oh noice

I've heard good things

[viral bear]

ive read good things

although tbh I wish I gave up on the lecturer earlier

the class this was for was spilt between like group theory and galois theory

and the TA taught the galois theory bit

[chilly ocean]

@chilly ocean See if a restricted homomorphism would still satisfy homomorphism properties.

Group theory and Galois theory :0

I would've imagined that the group theory would be a prereq

[viral bear]

ye the group theory was fine

its like sylow theorems and stuff

[chilly ocean]

Ah

Sylow theorem is da bes tbh