groups-rings-fields

[woven delta]

lol

aight have a gn

[solemn rain]

u2

[outer meadow]

i have a question, how do i get an intuition for groups? i recall in algebra class, there are a bunch of proofs where they seem to pull the argument out of thin air

eg, sylow theorem: oh! just take a normalizer here, oh! just take an index there, oh! then consider multiplying the group by this other group!

[solemn rain]

well

for the big boy theormes

things are just more polished

like they are made like this just to be mor ebeautiful

and like

more short

[stone fulcrum]

That's the beauty of group actions lol. A lot of them really are from thin air. The correct action + the correct thing to look at = HUGE theorems

[solemn rain]

and yea group actions are like

weird like

'just look at this group action'

then u get aph d

thatws it

[patent girder]

Can someone help me with this problem?
For the set {a,b,c,d}, find all possible binary relationships that make this into a partially ordered set. (You must find 19).

[outer meadow]

i think this should basically just be all length>1 strings consisting of abcd

ah no, there are more than that

[patent girder]

@outer meadow doesn't binary relationships mean that it is only between two elements? In that case I can only find 16, can you help me find the other 3?

[outer meadow]

i would ask you to list the 16, but that is probably a pain in the ass. but now actually, i think you can formulaically list them all by considering trees/"forests"

[patent girder]

i got a<a, b<b, c<c, d<d<, a<b, a<c, a<d, b<c, b<d, c<d, b<a, c<a, d<a, c<b, d<b, d<c

[outer meadow]

er, x<x shouldn't count

[patent girder]

isnt it the relfectivity rule?

[mild laurel]

Binary relationship means that its a subset of R x R

where R is your set {a,b,c,d}

[patent girder]

i really dont get the problem, can you help me find the 19 subsets

[mild laurel]

The reflexivity rule requires that x < x for all x in your set

[patent girder]

i was given the conditions
Let S be any nonempty set. A partial order in the set S is a binary relationship ≤ such that it satisfies:
• a ≤ a (reflexivity)
• if a ≤ b and b ≤ a, then a = b (antisymmetry)
• if a ≤ b and b ≤ c, then a ≤ c (transitivity).
For the set {a,b,c,d}, find all possible binary relationships that make this into a partially ordered set. (You must find 19).

[mild laurel]

I mean, idk what to help with, you should go review what relations even are

yeah, and those three conditions must be true for all a,b,c

in particular, you must have that a ≤ a for all a

[patent girder]

yeah my teacher gave them to me and never reviewed them, hasn't responded to me and i feel stuck

[outer meadow]

i think this should be all isomorphism classes of posets of size 4. mm, oh, i'm missing the one where there's a square with nothing inside

in some sense, you could list all relations, and check which ones form posets, but i don't think anyone wants to do that (RxR is size 4x4=16, then there are 2^16 many subsets)

[patent girder]

All i need to do now is find the 3 nonisomorphic elements in (a,b,c,d), the only problem is I can't find them

[outer meadow]

i'm not sure i understand

here's one i can suggestion, you wrote down some inequalities above, what are the corresponding relations, in precise terms?

[patent girder]

in the set the 16 elements that i found were all the isomorphic elements that are shown in your picture, there remain 3 more orders thataren't in those pictures i need to find those 3

[outer meadow]

i'm still not sure i follow. in the picture i drew, i intended eg the top left box as the poset where eg {a<b}, and nothing else holds, similarly, the posets {a<c}, {a<d}, etc.

you wrote down a<a, what is the relation that corresponds to this?

i guess i'm using "<" in the strict sense, whereas you're using it in the non-strict sense

but a≤a trivially, so in some sense there is no point to stating this

[patent girder]

i meant ≤, my keyboard doesn't have thatoption and i wasn't going to copy and paste in 16 times so i wrote >

[chilly ocean]

Does anyone know a simple proof that the signature homomorphism S_n -> {-1, +1} is well-defined?

[solemn rain]

okay so its f(phi) = product(x_phi(i)-x_phi(j)) on i<j<=n

the action of this phi is either the discriminant or negative discriminant

if discriminat = 1

if negative discrminant = -1

f(phi) = 1 if phi is even

f(phi) = -1 if phi is odd

try this

@chilly ocean

[chilly ocean]

@solemn rain I think you forgot to divide by something

[solemn rain]

rly?

are u sure

i dont think so

okayy forget about what i said

f(phi) = 1 if phi is even
f(phi) = -1 if phi is odd

try diff cases

f(phi1)f(phi2) if both are even

[chilly ocean]

how do you show that transposing the x's multiplies it by -1

[solemn rain]

f(phi1)f(tao1) if one is even one is odd

[chilly ocean]

it looks like it's going to be an ugly proof, which I want to avoid

because the proposition itself is very simple and fundamental

[solemn rain]

nah it really sint

at all

[chilly ocean]

ok then can you explain the full proof

[solemn rain]

lmao just show that f(phi1)f(phi2) = f(phi1phi2)

wiht this definition

f(phi) = 1 if phi is even
f(phi) = -1 if phi is odd

isnt this the sgn function?

defined correctly?

i think so

[chilly ocean]

just show that f(phi1)f(phi2) = f(phi1phi2)

how do you show it

also you need to prove that f(phi) = -1 for all transpositions phi

[solemn rain]

thats the definition tho?

for all odd phi

[chilly ocean]

how so

f(phi) = product(x_phi(i)-x_phi(j)) on i<j<=n

[solemn rain]

okay

[chilly ocean]

this is your definition

[solemn rain]

nah forgeta bout that

[chilly ocean]

ok so what's your proof

[solemn rain]

define sgn:S_n ---> {-1,1}

sgn(phi) = 1 if phi is even

sgn(phi) = -1 if phi is odd

now try this

[chilly ocean]

circular logic

i want to prove that "even" is well-defined

[solemn rain]

phi is even if phi(discriminant) = discrimnant

[chilly ocean]

ok but my definition of even is that if it's a product of even number of transpositions

[solemn rain]

yes

thats also true

ofc

[chilly ocean]

so how do you prove they coincide

is what I'm asking

[solemn rain]

ohh

okay sorry my bad

okay so

write out the discriminant polynomial

[chilly ocean]

I want to prove that if it's a product of even number of transpositions, it's not a product of odd number of transpositions

[solemn rain]

the product thingy

well obv?

[chilly ocean]

how so

prove it

[solemn rain]

isnt this trivial?

I want to prove that if it's a product of even number of transpositions, it's not a product of odd number of transpositions
??

[chilly ocean]

then prove it

[outer meadow]

wikipedia gives a proof using polynomials, if you like that

[chilly ocean]

i've seen it and I don't like it

[outer meadow]

what's ur issue with it?

[solemn rain]

well thats how u define it

ig

[mild laurel]

why not? It's not obvious that you can't write a permutation as the product of even transpositions and also as the product of odd transpositions

oops lag but, the point stands

[chilly ocean]

1. Because it's a magic hat solution. Not really a natural way to prove it.
2. The proof that the sign of a transposition is -1 requires some casework. Not natural and ugly.

[solemn rain]

why not? It's not obvious that you can't write a permutation as the product of even transpositions and also as the product of odd transpositions
oh really mb

[chilly ocean]

The proposition itself is very simple and the proof seems like an overkill.

[solemn rain]

i think it works

if u use determinants then.

the sign of a permutation is the determinant of the matrice when its written in array form

[chilly ocean]

I came up with the question when I wanted to find the simplest way to define a determinant

[solemn rain]

maybe thats easier lmao?

[chilly ocean]

to prove it's well-defined

it depends on the fact that sign is well-defined

they're equivalent facts

i've already thought about it

Imagine if you had to explain why sign is well-defined to a fifth grader

if there isn't a nice proof, maybe someone has an explanation why not

[solemn rain]

its technical ig

like sign stuff works for alot of stuff maybe

its ot just made for defining even and odd permutrations

the deriv of abs(x) is sgn(x)

for example

thats a shit explanation but just that

thats my guess

idk tbh

[chilly ocean]

the deriv of abs(x) is sgn(x)

I think that's a different kind of sign lol

That's why I'm interested it: the sign homomorphism is fundamental to linear algebra and pretty much all of differential geometry

as well as every other area that deals with antisymmetric stuff

I can't believe there isn't a simple proof of such a simple fact

[scarlet estuary]

I came up with the question when I wanted to find the simplest way to define a determinant

what do you mean by "simplest"

i'd consider the "pure mathsy" way to be "simple", but not computationally useful by itself

i.e. define it as the unique alternating multilinear map satisfying det(I) = 1

with the motivation for this definition being the multiplicative rule; ie translating the multiplicative structure of matrices into the multiplicative structure of real numbers by using det(AB) = det(A)det(B)

[chilly ocean]

i.e. define it as the unique alternating multilinear map satisfying det(I) = 1

that's what I had in mind too

and to prove it's well-defined you need to show that sign is well-defined

[scarlet estuary]

do you?

[chilly ocean]

yes

think about it

they're equivalent facts

[scarlet estuary]

cant you just use change of basis matrices

maybe im an idiot

[chilly ocean]

explain what you mean

[scarlet estuary]

hold on ive seen a proof of this before

one sec

[chilly ocean]

it probably implicitly used the fact that sign is well-defined

[scarlet estuary]

oh wait different type of well defined

not well defined wrt a basis, but in general

alright sure

in any case, this proof https://math.stackexchange.com/questions/1988084/uniqueness-of-factorising-permutations-into-cycles doesnt seem that nasty to me

i guess you need to prove that factoring permutations into cycles is possible in the first place but

that should be intuitive

[chilly ocean]

this is not the proof

[scarlet estuary]

ok maybe im misunderstanding then

[chilly ocean]

and the cycle proof I consider ugly

there is a proof using cycles by considering how multiplying by a transposition changes the cycle decomposition

it always adds or subtracts one cycle

hence +1 mod 2

but that also seems like an overkill

although it's the simplest proof I know

[scarlet estuary]

im afraid im not familiar with anything nicer

if there was a really nice proof here, i'd expect determinants in general to be easier to compute tbh

[chilly ocean]

how so?

[scarlet estuary]

in the sense that

a "nicer" proof would most likely arise from making some observation on the properties of the constituent cycles in the cycle transposition

but if we can make some nice observation there

i'd expect that observation to be usable in simplifying det calculations

but there isnt really a way to simplify determinant calculations, like, ever

unless you have zeroes in your matrix entries or already know an inverse

thats just a heuristic thing though

not a formal argument

[chilly ocean]

I would like a proof that doesn't consider the cycle decomposition at all

[scarlet estuary]

yeah i getcha

hmm

[chilly ocean]

Essentially I would like to be able to explain to a 5 year old why if you have n boxes labeled 1 through n, and whenever you switch two boxes you press a button (on/off), why it's true that whenever you return to the same arrangement of the boxes the button will be in the same state

[scarlet estuary]

yeah i figured that much out

when i actually thought about what you were asking

[sorry it's like 3 am]

[delicate bloom]

fun problem I'll think about it, I agree there should be an easy intuitive way to see it past like ordering numbers or whatever

[patent girder]

can someone show me an example of a nonisomorphic partial order?

like can't i always have a function f(s) = s and the relation will not change

[golden pasture]

integers under normal ordering and integers where no 2 integers can be compared

[chilly ocean]

or 1<2<3 and 1<2

[patent girder]

OH OK

[patent girder]

so isomorphism is between two sets, thats wierd since i have a question asking me for all nonisomorphic sets for (a,b,c,d)

[chilly ocean]

nonisomorphic sets for (a,b,c,d)
That means sets that are not isomorphic to (a,b,c,d)

[shrewd halo]

@chilly ocean construct a polynomial in n variables that always negates under a transposition of 2 elements

[chilly ocean]

I assume you mean the polynomial Prod_{i<j} (x_i - x_j)

[shrewd halo]

Is that not proof of the fact you wanted?

[chilly ocean]

not really

how would you prove it negates when you switch 2 elements?

@shrewd halo

[shrewd halo]

@chilly ocean

In this example, we switch a and b and look at all relevant terms which contained an a or a b

[golden pasture]

jus checking if $A=\frac{\mathbb Z}{2\mathbb Z}\times\frac{\mathbb Z}{2\mathbb Z}$ and $\mathfrak p=\frac{\mathbb Z}{2\mathbb Z}\times{0}$, then $A_{\mathfrak p}$ should contain $2$ elements, $0,1$ right cuz like $\frac{(1,0)}x=0$ since $(1,0)\cdot(0,1)=0$ and $\frac1{(0,1)}=1$ since $\left(1-(0,1)\right)(0,1)=0$

[cloud walrus]

Ariana:

[oblique river]

localizing non integral domains is weird

but actually don't you get the zero ring? like consider the element (0,1)/(0,1). On the one hand, that's clearly equal to the identity. But on the other hand, it's equal to 0/1 since (0,1)*(1,0) = 0

So since 1 = 0 you get the trivial ring

@golden pasture

[golden pasture]

yea nonintegral domains areotmvimjcmjoce

for the (0,1)*(1,0) where does the (1,0) come from cuz like A-p={(0,1),(1,1)}

i need sage to check the localizations for me

[oblique river]

so the definition of localization is

equivalence classes of things of the form a/s where a in A and s in A\p

and the equivalence relation is

a/s = b/t if there exists some u in A such that at-bs = 0

so take a = s = (0,1), b = 0 and t = (1,1)

oohhhh wait I see

the u has to be in A\p as well

ahhhhhh

okay yeah in that case I think you might be correct

@golden pasture

sorry!

[golden pasture]

yea idk things that aren't domains are

ill have faith it's correct

time to implement product rings and localization in sage

[oblique river]

hahaha

have fun :P

[toxic zephyr]

Was working on this with a friend and we're not sure about the solution.

We got $p_n(t)=tp_{n-1}(t)-p_{n-2}(t)$

[cloud walrus]

nix:

[toxic zephyr]

But we weren't sure if there was a more specific answer

[worthy kindle]

I think p_n is a Chebyshev's polynomial

[deft torrent]

Guys, I have quite a few questions trying to understand a proof for the Fundamental Theorem for Finite Abelian Groups. Can I ask them here?

[mild laurel]

go ahead

[smoky cypress]

Isn’t it finitely generated abelian groups?

[kindred mist]

For number 12 so far I’ve tried constructing inverse homomorphisms to the monomorphisms, applying commutativity of the diagram & exactness, am I correct that I can’t just add arrows to & from zero on both ends of both sequences? (I can post a picture if it’s not clear what I mean)

[outer meadow]

iirc the way you prove these problems is to just bash definitions

[outer meadow]

i guess i usually just forget about "constructing homomorphisms" and whatnot and just think of these as a puzzle

but my algebra is shit so maybe this is not how you're "supposed" to think of it

[deft torrent]

The lemma is as follows: Let $G$ be an abelian $p$-group of order $p^\alpha$ where $p$ is prime. Then $G$ is a product of cyclic groups.

Notation:
Let $\phi:G \xrightarrow{}G$ such that $\phi(g) = g^p$.
Let $G^p = \operatorname{im}\phi = {x^p : x \in G}$.
Let $G_p = \ker \phi = {g : g^p = 1}$
Let $\overline{G} = G / G^p$

The proof is structured by using induction on $\vert G \vert$. If $\vert G \vert = p$, then it is true that $ G = \langle x\rangle$. It then states that $G^p = \langle x_1 \rangle \times \langle x_2 \rangle \times \cdots \times \langle x_t\rangle$. How is this true? Is it some sort of induction argument?

[cloud walrus]

Stephen James:

[kindred mist]

Thank you Stephen!

[deft torrent]

Ohh i was asking a question

I don't think I answers your question

[kindred mist]

ack I thought you were posting a lemma to my lemma nvm I might be able to answer yours give me a minute

[outer meadow]

presumably the argument goes that G^p is a product of cyclic groups by the inductive hypothesis (G^p has order less than G) and G_p is written as a product of cyclic groups each of order p, then G_p and G^p have trivial intersection, and orders multiply to that of G, so by abelian-ness the product equals G

[solemn rain]

There is only one group of order n iff (n,phi(n)) = 1

is this hard to prove

( phi(n) = |(Z/nZ)^x| , euler totient )

[halcyon siren]

I swear I saw a proof for the five lemma on wikiproof sometime ago

[upper pivot]

oh yeah i encountered this in AT, was just some quick diagram chasing iirc

(well the one i did actually had injectives and surjectives so might not be the same)

[chilly ocean]

hey i am self learning group theory and i have a basic question , is it possible to ask it here?

[golden pasture]

yes

[chilly ocean]

so i have a symetric group S7 and there's two elements alpha and beta in it written in the form of 2x7 matrices , how to i calculate the product between the two elements ? is it simply matrix multiplication?

since it wasn't indicated what the group operation is , just a symetric group S7

[upper pivot]

oh those arent matrices

they tell you how the permutation is working

[chilly ocean]

ohhhhh

[upper pivot]

and group operation is just composition

[chilly ocean]

i never saw that notation before so i got confused , sorry 😅

[upper pivot]

do you see how they are telling you what the permutation is?

[chilly ocean]

so if i have 1 over a 3 it means it's a permutation between 1 and 3?

[upper pivot]

it means the eprmutation sends 1 to 3

[chilly ocean]

ah okay i see i see

i was used to writting them vertically with arrows

xD

and so to compute the product ( if non-abelian ) of alphabeta and betaalpha i have to take the composition for each element?

[upper pivot]

yeah

and also yeah, not abelian

usually these two things will give u distinct things

[chilly ocean]

aha i see , thank you for your help !

[upper pivot]

np

[kindred mist]

I looked at the one on proofwiki, I think its not exactly the same but I think I can adapt it

[tribal pasture]

In this, do we assume that H is a normal subgroup or just a subgroup?

[knotty mason]

it must be normal for the quotient to be well defined

[tribal pasture]

I am guessing over here G/H doesnt mean the quotient, but rather the set of cosets because the author doesn't mention the word normal at all

[oblique river]

you are correct

@tribal pasture

H need nto be normal

and for any subgroup H of G, G/H is a well-defined set

[tribal pasture]

Thanks

[oblique river]

the set of left cosets

H\G would be the set of right cosets

[mint brook]

Lmaooo

That was kinda funny becayse it sounded like you said "you're right, all the functions are continuous"

[smoky cypress]

Why is it funny that banana just merely stated a fact?

[chilly ocean]

Which book did you use for Abstract Algebra?

And has anyone used Serge Lang's? What do you think of it?

I'm thinking of this or Artin

[sharp sonnet]

uh

don't use Lang as your first book

[chilly ocean]

@sharp sonnet Why?

[sharp sonnet]

its hard

not suited for a first course imo, especially for self-study

artin is a nice book

[chilly ocean]

it covers same stuff as artin and others but i see your point

i will try get through artin then

thanks @sharp sonnet

[golden pasture]

lang is like nice if you know the topic roughly and want to poke more for like group stuff imo
it's homological alg part is also pretty nice learnt like half of what i know from there lol

lang as first book is def suicide use like a more normal book, jacobson, dummit&foote, herstein, artin maybe etc.

[sharp sonnet]

exercise: take an arbitrary book on homological algebra and solve all the exercises without looking at the text

[golden pasture]

ye the older editions lol

[stone fulcrum]

exercise: prove open questions in algebra without knowing algebra

[sharp sonnet]

just develop all the theory yourself, its easy

[golden pasture]

pull a galois and develop new theory

+the death part too

[stone fulcrum]

If you don't die in a duel, are you even cool?

[golden pasture]

is dueling myself considered cool

[sharp sonnet]

i would be happy if i could do either

[stone fulcrum]

Only if you lose

[mint brook]

@smoky cypress because it wasn't intentional lmao

[smoky cypress]

[mint brook]

[chilly ocean]

@golden pasture i dont see why he would make a introductory books for undergrads who don't have knowledge in it, only to end up requiring knowledge in the subject prior anyway

[scarlet estuary]

I mean

It's literally published under Graduate Texts in Mathematics

That's not to say it's unapproachable for undergrads, but it's not meant to be an introduction per se

More a reference and deeper dive into the concepts

[chilly ocean]

@scarlet estuary What, no

it's published under undergraduate texts in mathematics

https://www.springer.com/gp/book/9781475768985

[scarlet estuary]

Oh I thought you were talking about Lang's Algebra

Ill admit I forgot Undergraduate Algebra exists

Since no one uses it afaik

[mild laurel]

Ariana was definitely talking about the graduate texts in mathematics one though

[chilly ocean]

Oh seems like people forgot about Lang's book

[scarlet estuary]

Yeah sorry my bad

I don't really have an opinion on his other book, have never seen it used

[chilly ocean]

It basically starts with number theory and then groups, rings, etc

unlike artin which starts with matrices

[sharp sonnet]

i have never seen this book in my life, lol

[chilly ocean]

Hahah

Have a look inside if you want

He got so many books

[latent anvil]

Can someone help me fix some confusion with some noncommutative ring stuff? I'm trying to show that for a left module M over a ring R and for an abelian group G, Hom_Z(M, G) ≈ Hom_R(M, Hom_Z(R, G))

Where the R module structure on Hom_Z(R, G) is (rf)(s) = f(rs)

I tried to define h : Hom_Z(M, G) -> Hom_R(M, Hom_Z(R, G)) by h(f)(x)(r) = f(rx), but it seems like h(f) isn't necessarily linear

Since h(f)(rx)(s) = f(s(rx)) = f((sr)x) while (r h(f)(x))(s) = h(f)(x)(rs) = f((rs) x)

[golden pasture]

oops mb i also thought it was Lang - (GTM) Algebra :p

[latent anvil]

Figured it out, the R module structure on Hom_Z(R, G) is wrong

[cedar pagoda]

Does anyone know if there is a ring homomorphism from C to R?

Or any ring homomorphisms from C to any set that isn't an automorphism?

[mild laurel]

Just take some extension of C and take the embedding map

[cedar pagoda]

can you elaborate?

[mild laurel]

There's a map from C to C[x] for example

[cedar pagoda]

Do you mean like there is a map between C and (C, 0)

I'm trying to find something more interesting i guess

[mild laurel]

Uh what

Do you know what C[x] is

[cedar pagoda]

no

sorry

[mild laurel]

It's a polynomial ring

[cedar pagoda]

Oh okay

that's still sort of trivial though right?

c -> cx?

[mild laurel]

Uh, that's not a ring homomorphism

c -> the constant polynomial c

[cedar pagoda]

yeah, sorry I'm not too good at algebra

I'm specifically looking for a homomorphism that shrinks C if that makes sense?

like mapping to a lower cardinality set

like the group homomorphism of the group (C / 0, *) f(x) = magnitude(x) would be an example

[mild laurel]

You can't

a ring homomorphism from a field must be injective

[cedar pagoda]

Ahh I didn't know this, thanks!

[brisk granite]

Let $f(x)$ be an irreducible polynomial over a field $F$, and let $L$ be its splitting field. Suppose that $K$ is an intermediate field between $F$ and $L$, and that over $K$, $f$ factors as $f(x) = g_1(x)\dots g_n(x)$, where each $g_i$ is irreducible over $K$. Show that all the $g_i(x)$’s have the same degree.

[cloud walrus]

PolyBeanDip:

[brisk granite]

This is wrong when K isn't normal right?

For example, let $L$ be the splitting field of $x^4 - 7$ over $\mathbb{Q}$. Then, an intermediate field would be $\mathbb{Q}(\sqrt[4]{7})$. In this field, $x^4 - 7 = (x+\sqrt[4]{7})(x-\sqrt[4]{7})(x^2 + \sqrt{7})$.

[cloud walrus]

PolyBeanDip:

[mild laurel]

Yeah, that seems right

[brisk granite]

the question or the counterexample?

[mild laurel]

the counterexample

[brisk granite]

ok

also, assuming K/F was normal, then I'm trying to show that I can send any g_i to any g_j using the elements of Gal(L/F)

is this the right approach?

Also, I say that f(g_i) = f(g_j) iff f sends the roots of g_i to those g_j.

[latent anvil]

Let α, β be roots of f in L. I claim there is an automorphism α of L/F sending α to β. Let σ : F(α) -> F(β) be the isomorphism sending α to β and fixing zf. Then we have g in F(α)[x], h in F(β)[x] such that f(x) = (x-α)g(x) = (x-β)h(x). Extend σ to an isomorphism τ : F(α)[x] -> F(β)[x] by applying σ coefficientwise. Then (x-β)h(x) = f(x) = τ(f(x)) = τ((x-α)g(x)) = (x-β) τ(g(x)), so τ(g(x)) = h(x). Since L is a splitting field of g(x) over F(α) and h(x) over F(β), and τ(g(x)) = h(x), we can extend σ to an automorphism σ : L -> L fixing F and sending α to β.

Now I claim any automorphism σ : L -> L fixing F satisfies σ(K) = K. Let θ in K be arbitrary, and let m the minimal polynomial of θ over F. Then 0 = σ(0) = σ(m(θ)) = m(σ(θ)). Since m has a root in K, it splits over K (this is where we use normality), and thus σ(θ) in K. Thus σ(K) <= K, and by the same logic applied to σ^(-1), σ(K) = K.

Now finally suppose g is the minimal polynomial of α. Let h be the polynomial obtained by applying σ to the coefficients of g. Then h(β) = h(σ(a)) = σ(g(α)) = σ(0) = 0. Note that g' is in K[x], and h is still irreducible (if it factored we could apply σ^(-1) to the factorization to get one for g), so h is the minimal polynomial of β. By how h was defined, deg h = deg g.

If you start with g_i and g_j you can choose α to be a root of g_i and β a root of g_j, and then g_i = cg for a nonzero constant c and g_j = c'h, which means deg g_i = deg g = deg g = deg g_j

@brisk granite way too in depth solution here

Tl;dr is to choose roots of the polynomials in L, get an automorphism of L/F swapping them since L/F is a splitting field, deduce that that automorphism sends K into itself by normality, and then conclude that applying the automorphism to the coefficients of g_i gives g_j, so they have the same degree

[brisk granite]

"Since L is a splitting field of g(x) over F(α) and h(x) over F(β), and τ(g(x)) = h(x), we can extend σ to an automorphism σ : L -> L fixing F and sending α to β. "

what theorem are you using here

[brisk granite]

Since I'm not rly sure what you're invoking, I used a different proof. Consider the polynomial g(x) = \prod_{sigma in Gal(L/F)}(x - sigma(α)). This polynomial is invariant under the elements of Gal(L/F) and hence belongs in F[X]. Then, f | g, and, hence, beta must be a root of g. Which means there exists sigma such that sigma(alpha) = beta

[latent anvil]

If you have a field isomorphism σ : F -> F' and polynomials f(x) in F[x], f'(x) in F'[x] such that when you apply σ to the coefficients of f you get f', then for any splitting fields K/F for f and K'/F' and f' you can extend σ to an isomorphism K/F -> K'/F'

This is an important result

The proof is to inductively build up K and K' by adjoining one root at a time and use the fact that if f, g are irreducible and α, β are roots of them in some extension, you can extend σ to an isomorphism F(α) -> F'(β) sending α to β

@brisk granite I don't believe your proof since it assumes L/F is Galois, but L/F might not be separable. Take F = F2(t) and L to be the splitting field of x^2 - t. Then Gal(L/F) is trivial since x^2 - t has at most one root in any extension.

At least, it assumes Galoisness if I read it right

[latent anvil]

Here's the proof of the lemma I used.
Suppose you have a field isomorphism σ : F -> F' and polynomials f(x) in F[x], f'(x) in F'[x] such that when you apply σ to the coefficients of f you get f', then for any splitting fields K/F for f and K'/F' and f' you can extend σ to an isomorphism K/F -> K'/F'.

We proceed by induction on [K : F]. If it's 1, then K = F, so f is a product of linear polynomials in F[x]. Applying σ to that gives the same sort of thing, so K' = F as well. Then σ works as its own extension.

Suppose instead that [K : F] > 1. Then then f has a root α in K that's not in F. Factor f(x) = (x-α) g(x) for g(x) in F(α)[x]. The minimal polynomial m of α over F divides f, so σ(m) divides f'. Choose a root β of σ(m) (so β is also a root of f'). Then we can factor f'(x) = (x-β) g'(x) for g' in F'(β)[x] . It's easy to see that K is a splitting field of g over F(α) and K' is a splitting field for g' over F'(β). Further we can extend σ to an isomorphism τ : F(α) -> F'(β) sending α to β. Then (x-β)g'(x) = f'(x) = τ(f(x)) = τ((x-α) g(x)) = (x-β) τ(g(x)). Thus τ(g(x)) = g'(x). Since α isn't in F, [K : F(α)] < [K : F]. Thus by induction we can extend τ to an isomorphism K -> K'

[chilly ocean]

could someone please recommend a good introductory group theory textbook, which is available online

[toxic scaffold]

you mean for free?

(legally)

I only have looked at $$ books like Lang, but you could try this http://abstract.ups.edu/aata/aata.html @chilly ocean

can't vouch for it in any way

[sharp sonnet]

(just download from libgen)

[golden pasture]

yea just libgen them

(irl jacobson is like really cheap for it's scope)

[woven delta]

@mild laurel what you said about homorphisms from fields being injective isn't strictly true

there is a trivial counterexample actually

[mild laurel]

It's true if you require ring homomorphisms to send 1 to 1 right?

[latent anvil]

no

you can map into the zero ring

[mild laurel]

Oh yeah right

[woven delta]

the joke was the word trivial

[bleak abyss]

Math patch notes: 0 is no longer even a ring

Lol jk

[latent anvil]

someone tried to tell me that once

They said Aff doesn't have an initial object

And were adamant that 0 isn't a ring

[bleak abyss]

You're adamant too?

[latent anvil]

You saw nothing

Anyways this person is like in year >= 5 of an AG PhD and knows more math than i will for many years

Which makes it even more bizarre

[bleak abyss]

I mean these things are philosophical

Like in principle you can prove your theorems and be like, nah these considerations pale against the a e s t h e t i c of having/not having 0

[latent anvil]

Yeah conventions always seem way more important than they actually are

People don't actually reason at the level of definitions

[bleak abyss]

The correct answer is what I do

Namely have conflicting conventions based on what's more convenient at that very second

[latent anvil]

Mood

[bleak abyss]

Does \mathbb{N} contain 0 or not? Depends on what I'm doing with it

Ambiguity is power

[scarlet estuary]

yeah i let it stay ambiguous and even change up my preference mid-argument

this turns out to be convenient in many proofs

example:

We have that 0 is a natural number. However, 0 is not a natural number. So we have a contradiction, and thus by explosion, RH.

[woven delta]

just say $\omega$ like a chad logician

[cloud walrus]

Liquid:

[outer meadow]

obligatory $\cup \omega \cup$

[cloud walrus]

weast:

[latent anvil]

just identify them lol

There's an obvious bijection between the positive and the nonnegative integers

so you can say they're equal without any trouble

[chilly ocean]

Yeah, I probs just go with Lang’s Undergraduate Algebra, seems accessible

[latent anvil]

lang has an undergrad algebra book??

Woah

[chilly ocean]

@latent anvil https://docs.google.com/file/d/0B1pXoUPGwCetajA1Sk50cmNOaDA/edit well at least its called that and terms of content it seems accessible for the undergrad

[eager willow]

I'm trying to decipher a proof but I'm missing something:\

If $K$ is Galois over $F$ and $G = Gal(K/F) = G_1 \times G_2$ iss the direct product of two subgroups $G_1$ and $G_2$, then $K$ is the composite of two Galois extensions $K_1$ and $K_2$ of $F$ with $K_1 \cap K_2 = F$.
Let $K_1$ be the fixed field of $G_1\subset G$ and $K_2$ for $G_2 \subset G$. Then $K_1 \cap K_2$ is the field corresponding to the subgroup $G_1G_2$.

[cloud walrus]

datorangeguy:

[eager willow]

I get that much... but the next claim is that $G_1G_2 =G$ which I don't understand

[cloud walrus]

datorangeguy:

[mild laurel]

Do you understand what the notation G_1G_2 means?

[eager willow]

Yes it's ${g_1g_2 : g_1 \in G_1, g_2 \in G_2}$ but how is that all of $G = G_1 \times G_2$?

[cloud walrus]

datorangeguy:

[mild laurel]

And what do elements of G_1 x G_2 look like

[eager willow]

Well they should be ordered pairs, but that doesn't really make sense if each entry in the pair is itself an element of the group of ordered pairs?

[mild laurel]

Yeah, it should really say isomorphic to

Rather than equals, but

[eager willow]

the text is dummit and foote and this is the converse to the proposition that I understand the proof for, that $Gal(K_1K_2 / F) \simeq Gal(K_1 / F) \times Gal(K_2 / F)$

[cloud walrus]

datorangeguy:

[mild laurel]

Uh sure

The point is that the elements of G_1 x G_2 are ordered pairs (g_1, g_2)

so every element g of G can be written as some ordered pair (g_1,g_2), but we have that (g_1, g_2) = (g_1, e)(e, g_2) where e is the identity element of G

[eager willow]

aaah I was way overthinking it. Thank you!

[chilly ocean]

could you deduce important and unique properties of finite groups by looking at their graph representation?

[chilly ocean]

the lecture series Visual Group Theory had an alternative definition of a normal subgroup based on graphs
commutativity is another example

[soft elm]

@latent anvil yeah and its free on springer

[tribal pasture]

Should the following corrections be present
"Case 2: Let us consider x = 0"
"Case 3: Let us consider v = 0"

[mild laurel]

@chilly ocean are you talking about the Cayley graph

[chilly ocean]

@mild laurel I guess so, I’ve just started doing group theory, so I am unaware of most of the things, but for the exercise I was basically writing out Cayley tables and generating sets for 8 first orders and I was trying to draw some sort of graph for each group to see in more visual way how elements connect with each other, and I was thinking maybe you can assign to the group an unique graph, which will in visual way give you all the important structure of the group

Is that basically a Cayley graph?

[mild laurel]

I mean, you haven't given any details on how you're actually making this graph so idk

[nocturne quiver]

Hey guys! If a position vector (considering x and y direction) is given in terms of time r(t), and another vector is given in terms of x and y direction such as w(x, y), can I substitute the position vector components to the w vector?

[cobalt pilot]

Another basic question here. I am to show that if $I=(x,2)$ and $J=(x,3)$ then $IJ=(x,6)$. I've tried to go strait to the definitions of generated ideals and product between ideals. So elements in $IJ$ should be on the form $(ax+2b)(cx+3d)$ where $a,b,c,d \in Z[x]$, but this gives me terms with 2 and 3 in them and I can't see how I could get the same in $(x,6)$.

[cloud walrus]

Xiphias:

[mild laurel]

you should show inclusion both ways

show that x \in IJ and show that 6 \in IJ

that will show that (x,6) \subseteq IJ

and do the reverse

[cobalt pilot]

Well, what I did was starting on the reverse first it seems and that's where I got stuck

I can't find an element in (x,6) that matches the one I made in IJ

[mild laurel]

multiply it out, what do you get

[cobalt pilot]

Well, I can merge some elements from Z[x] since they are general, but I get something in the ballpark of ax^2+x(3a+2b)+6d

[mild laurel]

Yeah

So is ax^2 in (x,6)

and is x(3a + 2b) in (x,6)

[chilly ocean]

@mild laurel it’s like that, they are similar to Cayley graphs, mine are just more messy

[mild laurel]

these are exactly cayley graphs

Or maybe, special cases of cayley graphs

[cobalt pilot]

ax^2 is in (x,6) just fine, but I'm struggeling a bit more with x(3a+2b)

[mild laurel]

well, 3a + 2b is just some element of Z[x]

so set f = 3a + 2b or something, then your term is just fx

[chilly ocean]

so could you study which groups are possible by looking at the isomorphisms of graphs that are representing certain groups? For example, can you justify that there is only one group of order 5 ( ie cyclic group z5) by looking at the graphs of order 5 and finding isomorphisms between them?

or is it at some point much harder analysing graphs of high order, rather than using algebraic tools of group theory alone?

[mild laurel]

It's just not a very useful way to look at things I think

One reason for this is that the graph isomorphism problem is NP complete and so the question of whether or not graphs are isomorphism isn't an easy one

Cayley graphs in general though are important, and you can use talk about group actions on graphs to say a lot of things about groups

maybe one example is proving that subgroups of free groups are free

The usual proof requires some algebraic topology, but you can also do it by looking at group actions on graphs

[chilly ocean]

cool, thnks for answering

[sharp sonnet]

One reason for this is that the graph isomorphism problem is NP complete and so the question of whether or not graphs are isomorphism isn't an easy one
@mild laurel that is not true

it is not known whether GI is NP-complete

and as a side note: there is currently research done into solving graph isomorphism for cayley graphs by studying the underlying groups

which is kinda the opposite idea, you study the graph by studying the group for which it is a cayleygraph

but its fruitful sometimes and somewhat recently lead to the solving of GI for cayleygraphs of cyclic groups

by solving i mean there is an algorithm that can decide whether 2 cayleygraphs generated by cyclic groups are isomorphic in polynomial time

[cobalt pilot]

I'm a bit off and on today, busy day. 3a+2b is indeed just an element of Z[x] but if a=b=1 then it's 5x and that's not in (x,6) ?

[cobalt pilot]

oh, I'm learning about products of generators making the generators for the product ideal now. I'll probably just use that instead

[silver prism]

Do no periodic functions have range R?

[scarlet estuary]

this is not abstract algebra, but in any case, the example is "no"

in fact, a continuous periodic function cannot have range R

since periodic functions are bounded, i.e. can't be surjective onto R

[silver prism]

And what about domain?

@scarlet estuary Do they all have domain R

[scarlet estuary]

as mentioned, this is not abstract algebra

ask in an appropriate channel

[silver prism]

Which one?

[scarlet estuary]

#prealg-and-algebra

or any questions channel

[silver prism]

Ok thanks

[cloud walrus]

All functions are continuous:

[mild laurel]

@cobalt pilot 5x is definitely in (x,6)

It's in (x) as well

[chilly ocean]

Hey i have a question, how can i find the order (phi(n)) of the ring Z/nZ

[cobalt pilot]

@mild laurel yea, it made more sense after I saw the proof for product of ideals of generators becoming the ideal generated by the product of the generators (that was hard to type...)

Thanks for the help today!

[mild laurel]

Yeah, that's an important fact too

@chilly ocean what do you meam

The order of the multiplicative group of units?

[chilly ocean]

Yep

[mild laurel]

Sorry, you mean the multiplicative group of units right

[chilly ocean]

Yep that's it

I was thinking of findong the number of coprimesto n

Coprimes

[mild laurel]

That's right

[chilly ocean]

Since that's the intuition i had

But i can't prove it

Is it because the coprimes are the generator of the whole group?

Necause it's cyclic

[mild laurel]

It's because if m is coprime to n, then m is invertible mod n

[chilly ocean]

Ohhhhh

Right i forgot about invertibility, my bad haha

Thank you!

Also are the only elements 1 mod n the coprimes to the power of the order

Or is the converse of euler's theorem not true

[mint brook]

that's not exactly the converse

the converse would be if you have some a to the power phi(n)=1 (mod n) would that mean that a and n are coprimes , and the answer is yes they would be , thus the converse is true

[zenith vector]

Hey, does anyone know how to show that the ring of algebraic integers in Q[sqrt(-11)] forms a UFD?

[mild laurel]

First, it'd be helpful to figure out what the ring of integers of that field is

[cobalt pilot]

@mild laurel Hey, I did another similar problem just to get the hang of it. Let me know if you have time to proof check what I did. So I made $I=(x,3)$ and $J=(x,5)$. Then $IJ=(x^2,3x,5x,15)$ before potentially removing redundant terms. I see that $5x-3x=2x$ so I can replace $5x$ with $2x$ but it seems that $3x$ needs to stay. I did a test with the inclusion proof. So an element in $IJ$ is on the form $sum (ax+3b)(cx+5d)$ which terms multiplies to $acx^2+3cbx+5bdx+15bd$. This is quite obviously in $(x^2,2x,3x,15)$ as $2x+3x=5x$ and the converse is just as obvious. Sounds about right?

[cloud walrus]

Xiphias:

[latent anvil]

looks right to me

can you simplify that answer further?

[cobalt pilot]

Yes of course. Man I'm blind. Now that I have 3x and 2x it's just (x,15)

it seems that if (x,a) (x,b) where a,b don't have any common factors I can always go (x,ab) for the product

should not be too hard to prove I suppose

Trying to make 1 with numbers without common factors can be fun. Here's 25 and 9: (25-9)-9-(9-((25-9)-9))-(9-((25-9)-9))-(9-((25-9)-9))

[latent anvil]

hah yeah

Do you know bezout's lemma?

[woeful flint]

Guys is $\alpha = \sqrt{2 + \sqrt{3}}$ constructible. I say it isn't. My reasoning is its minimal polynomial is $x^4 - 4x^2 + 1$ which is irreducible over $\mathbb{Q}$. Therefore $\alpha$ is not the result of finitely many quadratic extensions of $\mathbb{Q}$ and so it isn' constructible. Is this the correct reasoning?

[cloud walrus]

Kraft Macaroni:

[woeful flint]

Update: I realized $\alpha$ is actually construcible and quite easily so. But how come what I've written above is wrong?

[cloud walrus]

Kraft Macaroni:

[cobalt pilot]

@latent anvil I have not, but I expected something like that existed based on what I've worked on today. Thanks!

[wind steeple]

it is not because your polynomial is irreductible that your extension hasn't subextensions @woeful flint

[woeful flint]

Oh whats the criterion for it then

Like visually it makes sense how you can construct alpha

[wind steeple]

yeah it's obviously constructible because there is only square roots of integers

you're searching a criterion on a polynomial to have its extension without subextension ?

it is not easy I guess

[woeful flint]

like ill give an example

[wind steeple]

you have sufficient conditions btw, like if your polynomial is irreductible and has a prime degree

[woeful flint]

$\sqrt[3]{2}$ is not constructible as its minimal polynomial is $x^3 - 2$ and so $[\mathbb{Q}{\sqrt[3]{2}}: \mathbb{Q}] = 3$

[cloud walrus]

Kraft Macaroni:

[woeful flint]

i can see some sort of difference in argument by a corollary now that it think of it

[wind steeple]

yeah but this isn't really a matter of subextension, you have another necessary condition on constructible numbers which states that every constructible number has a power of 2 degree

[woeful flint]

Ah i think i get you

Yeah thats a good point

[wind steeple]

btw you can find number of degree 4 not constructible

[woeful flint]

So im just guessing it might be hard to explicitly find field extensions which work

[wind steeple]

like something with cube roots

[woeful flint]

i get you

just trying to wrap my head around

one final question

is there a method to say compute $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{3},\sqrt[3]{5}) : \mathbb{Q}]$?

[cloud walrus]

Kraft Macaroni:

[woeful flint]

the square brackets means degree

[mild laurel]

do it in steps

calculate $[\bQ(\sqrt[3]{2}): \bQ]$ and $[\bQ(\sqrt[3]{2}, \sqrt[3]{3}) : \bQ(\sqrt[3]{2})]$

[cloud walrus]

Zopherus:

[mild laurel]

and the last one, and then multiply them

[gritty fractal]

is the converse of the first isomorphism theorem for rings true? that is if u have a homomorphism between the additive group of two rings f: R -> S, and ker(f) is an ideal in R, is f always also a ring homomorphism? and ofc why/why not?

[latent anvil]

No. Pick an arbitrary function R\{0} -> S\{0} and extend it by sending 0 to 0

[gritty fractal]

huh?

that doesnt neccesarily mean its an hom. between the additive groups though

[latent anvil]

yeah exactly

but it's kernel is an ideal

more specifically it's the zero ideal

[gritty fractal]

oh sorry i mightve forgot to say but one of the requirements is that f is a hom. between the additive groups

[latent anvil]

Ah sorry I misread what you said

You're thing would imply that an injective group hom between any two groups is a ring hom

Do you require ring homomorphisms to preserve 1?

[gritty fractal]

yes

[latent anvil]

Then look at cyclic groups

There's a lot of automorphisms Z/nZ -> Z/nZ but only one will preserve 1

The rest will probably screw up multiplication as well

[gritty fractal]

yeah i thought it might have something to do with automorphisms but i couldnt quite put my finger on it

thx

[latent anvil]

Well this isn't because they're automorphisms

The thing you're saying is false in most cases

This was just the easiest example for me to write down

[gritty fractal]

ohhh

well imma go check the cyclic group thing, hopefully it works

[ornate arch]

can someone help me with this question about hyperplanes.
How can I prove that a subset H = {(x1,....xn) in F^n | a1x1+a2x2...+anxn=0} of F^n is a hyperplane IFF ai = 0?

[latent anvil]

what's your definition of a hyperplane? I would call { (x, y, z) in R^3 : x = 0 } a hyperplane but it doesn't seem to satisfy your condition

[ornate arch]

@latent anvil it is definied in my textbook as: a subspace of V of dimension n-1 (when V is a vector space of dimension n)

the hint i got is that it has osmething to do with linear dependence to show that dim(H) = n-1 >?

[oblique river]

the problem is that what you've written is false

shamrock gave you a counterexample

so try to fix your statement first

[oblique river]

lol that's not what you wrote

[ornate arch]

i just reread it

=/ 0

[oblique river]

try again now that you've reread it

[ornate arch]

yeah

@oblique river Since H = ker(a1,...,an). We know that through rank nullity, n=rank(a1,..,an) + dim(ker(a1,...,an)

[oblique river]

you tell me lol

(also it's a bit odd to write both => and IFF)

[ornate arch]

oh ok

[oblique river]

also you're missing an =

you say "IFF rank(a1, ..., an) IFF"

but that's not a statement

[ornate arch]

IFF dim(H) = n - 1 =>rank(a_1, ..., a_n) = 1 => (a_1, ..., a_n) is not the 0 vector.

there

[oblique river]

I thought the first way was better lol

but yeah that's the logic

see look you didnt need help after all

[ornate arch]

i was just trying to prove =0 and not =/ 0

for like an hour

[oblique river]

maybe you should have reread the statement earlier than an hour

:P

[ornate arch]

probably

How could I show that all hyperplanes in F^n have that form?

@oblique river

[oblique river]

I mean it depends on what facts you already know

the vector (a1, ..., an) is the orthogonal complement to the hyperplane

[ornate arch]

did not know that, but also dont see what to do with that

[oblique river]

"take the orthogonal complement and look at that vector"

[ornate arch]

of what

[oblique river]

of the hyperplane

[ornate arch]

but how can we just 'know' that the vector (a1, ..., an) is the orthogonal complement to the hyperplane

[oblique river]

do you know the definition of orthogonal complement

[ornate arch]

subspace where the dot product = 0 no

dont see what it has to do with this

[oblique river]

look longer lol

what is the dot product

[ornate arch]

wdu mean

[oblique river]

alright let's practice some problem solving techniques. i've told you that the hyperplane and the vector (a1,...,an) are orthogonal complements. and you are asking how to show that. so look at the definition fo the hyperplane in part a and look at the definition of orthogonal complement

and match up the terms

[tribal pasture]

Can somebody tell me how is the action xW defined here? I know that it is an application of some representation p, but I cant figure out if its supposed to be p^s or some other representation

[ornate arch]

Ok i see, a1*x1 .... = 0

[oblique river]

tbh "problem solving" is more important than just knowing definitions

[tribal pasture]

These are the implicit definitions

[ornate arch]

i understand how theyre related, but i still dont see what this has to do with showing that all hyperplanes have the same form in Fn

[oblique river]

wdym

start with a hyperplane H

take its orthogonal complement

that's 1-dimensional so take a nonzero vector

that's your (a1,...,an)

that was the whole point of this

[tribal pasture]

@oblique river If you can take a look once you're finished with dame, it would be very helpful

[oblique river]

sorry not interested in sorting through representation theory atm

i'm gonna head off in a minute anyway

maybe later

[ornate arch]

Hint: One can find a1, . . . , an as a solution of a homogeneous system of n − 1 equations in n variables

i think im supposed to do it that way

[oblique river]

you should have said that a while ago lol

i'm gonna eat dinner though so gl

[ornate arch]

<@&286206848099549185> ?

can anyone herlp\

[chilly ocean]

did u solve the other question

[ornate arch]

i did

@chilly ocean

[chilly ocean]

what did you change to get the correct answer

[ornate arch]

nothing was changed, what we had at the enx was the right answer

[chilly ocean]

ok

good to know

[ornate arch]

i figured it out, no worries

[chilly ocean]

can anyone give me a hint on this?

[fringe nexus]

a homomorphism maps the identity to the identity

[chilly ocean]

is it |H|<|G|?

idk

[fringe nexus]

well

you map x^i to y^i

that implies that (x^i)^2 is mapped to (y^i)^2

right?

[chilly ocean]

yes

[fringe nexus]

now

(x^i)^2 = x^(2i)

what happens when m|2i

[chilly ocean]

idk

[fringe nexus]

huh

what is x^m

where m is the order of the group

[chilly ocean]

x composed with itself m times?

[fringe nexus]

yes

well we have that x^m = 1

[chilly ocean]

yes

[fringe nexus]

and homomorphisms map identity to identity

that should be all you need

[chilly ocean]

Group G of order 8 with 2 generators x and y such that x^4=y^2=e and xy=yx - is that dihedral group?

also, you can have abelian subgroups in non-abelian groups, right?

[stone fulcrum]

I'll use ' for inverse
Every dihedral group satisfies xy = yx'. But that's not what is here.

Also yes you can have an abelian subgroup of a non-abelian group.

[chilly ocean]

direct product of Z2 and Z4 then? 🤔

[stone fulcrum]

That's exactly it!

Actually maybe there's even more, I'm not sure

[chilly ocean]

more products?)

[stone fulcrum]

The dihedral group of 8 elements isn't abelian

More groups

[chilly ocean]

Z4xZ2 is (a,b st 4a=2b=0) so it seems like it is exactly that

i mean it doestn look like cyclic

what else is left?

it says there is some group E8

but i dont know what it is exactly

[stone fulcrum]

Nah it's not cyclic, since the only cyclic group of 8 elements is Z8, and Z2×Z4 ≠ Z8

I have no idea what E8 is either haha

[chilly ocean]

I found it on groupprops.wiki that it is a group of order 8 and google says that it is a Lie group and it looks scary😯

[upper pivot]

E8 is apperantly some hard core lie group of order 248

[chilly ocean]

thats definitely not group i am looking for lol

its direct product of z4xz2 then

[upper pivot]

yeah, do you see why

[chilly ocean]

because there is a cyclic subgroup of order 2 and what seems like subgroup of order 4 like klein group🤔

i mean i guess i can just write out direct product and simply compare with what i have to verify

[upper pivot]

i mean i'd try to use intuition too

so you have $\langle x,y: x^4=y^2=1 , xy=yx\rangle$

[cloud walrus]

JohnDoeSmith:

[upper pivot]

look at the generators for Z/4 x Z/2

for example, and see if this makes sense there

[stone fulcrum]

Is Z2×Z2×Z2 generated by two elements? I'm thinking no

[golden pasture]

show that the subgroup generated by 2 elements have at most 4 elements

[chilly ocean]

@golden pasture 4 elements - e,x,y and xy, right?

[golden pasture]

yea

[chilly ocean]

i am wondering - could you define group using ternary operation? would it be much different from binary operation on set, or could you always breakdown that ternary operation into binary relations?

i was just thinking how would like 3d cayley table look like and whether it would make sense at all😐

[solemn rain]

f:G^3--->G

and just put on this associativity and such and such ..

ig

[sharp sonnet]

but how do you reproduce identity and inverses

you could create a 3d cayley table for a group, just you wont get any new information i guess

[chilly ocean]

so would groups where operation is ternary be isomorphic to just normal groups with binary operation?

[sharp sonnet]

i am not sure how you would define sth like an identity or inverses for a ternary operation

my point was, if you already have a group you can define a ternary operation that is just group operation twice

ofc this does not yield new information, so why would you

[chilly ocean]

yeah i see what you mean

[chilly ocean]

This channel ok for alg geo?

[chilly ocean]

Prove that a morphism $\phi: X \mapsto Y$ between two affine varieties is an isomorphism of $X$ with
a subvariety $\phi(X) \subseteq Y$ if and only if the induced map $\Phi: k[Y ] \mapsto k[X]$ is surjective.

[cloud walrus]

Azu:

[chilly ocean]

My attempt in the forward direction so far:

Since $\phi$ is a morphism, it is regular in every point $X$

[cloud walrus]

Azu:

[chilly ocean]

$X$ is open and dense in its own Zariski (subspace) topology

[cloud walrus]

Azu:

[chilly ocean]

Therefore, $\phi$ is a polynomial map by a theorem

[cloud walrus]

Azu:

[chilly ocean]

Similarly, $\phi^{-1}$ (which exists because $\phi$ is an isomorphism) must be a polynomial map

[cloud walrus]

Azu:

[chilly ocean]

I don't know how to proceed from here

@mild laurel hey man... can I please ask for your assistance? I know you're really good at this

sorry for the ping

[latent anvil]

Did you get this resolved Azu?

[chilly ocean]

@latent anvil yes I did

If you want, I can share the solution

[latent anvil]

Nah, I'm okay

Thanks for the offer

[sterile eagle]

Hi I need help with an exercise on minimal polynomials:

θ ∈ C root of x^3 + 3x^2 −3 and α = θ^2 + θ. Determine:
-the minimal polynomial of α in Q.

I have no idea what how to start this

[sterile eagle]

<@&286206848099549185>

[knotty mason]

try to find a relationship between 1, alpha and alpha^2

reduce theta^3 using the x polynomial

[sterile eagle]

hm I got α^2 = 3(α-1)

do I use this for anything?

[knotty mason]

thats the min poly

[sterile eagle]

I'm very confused

so α^2 -3α -3 is the min polynomial?

[knotty mason]

yes

[sterile eagle]

oh very nice thx

[chilly ocean]

let H=<x> and suppose |H|=6. Then how does x^5 generate H? from my understanding H={x^{-5},...,x^{5}} or something like that

I think I'm conceptually missing something here

[stone fulcrum]

H is a cyclic group of 6 elements, H is Z6

In a cyclic group, any element that is coprime to |H| generates H

I mean, given that 5 generates Z6

[chilly ocean]

we haven't covered what Z6 means yet

[stone fulcrum]

Basically the integers mod 6

It's the group
0,1,2,3,4,5
with modulo 6 addition

Any element coprime to 6 generates it.

[chilly ocean]

we haven't covered modular arithmetic either yet

[stone fulcrum]

Oof haha, then let me think of how to convey this

[chilly ocean]

ty

[stone fulcrum]

Basically xⁿ generates the cyclic group of m elements, given that n and m are coprime

That's a proof I found online haha

[chilly ocean]

ok

sorry but what does that mod n thing mean

[stone fulcrum]

I wish I had a simpler explanation sry. It's not an obvious fact I'm just very used to it

[chilly ocean]

do you think it’d be helpful to do a bit of number theory before this?

[stone fulcrum]

ra = 1 (mod n)
Means
ra - 1 is a multiple of n

And yes lol. I'm suprised your course didn't start with the basics

[chilly ocean]

ya interesting all right ty

[stone fulcrum]

Oof Yeah that proof is assuming you're working in Zn as well

Hrm

[smoky cypress]

\begin{align*}
(x^5)^0&=1\
(x^5)^1&=x^5\
(x^5)^2&=x^{10}=x^6x^4=x^4\
(x^5)^3&=x^{15}=(x^6)^2x^3=x^3\
(x^5)^4&=x^{20}=(x^6)^3x^2=x^2\
(x^5)^5&=x^{25}=(x^6)^4x=x\
(x^5)^6&=x^{30}=(x^6)^5=1\
\end{align*}

[cloud walrus]

Whoever:

[smoky cypress]

All elements of H can be written as power of x^5, therefore x^5 generates H

@chilly ocean

[chilly ocean]

damn

thanks that's what I was looking for

[smoky cypress]

But

Just a note

This process can be generalized to what Kaynex said above

If you have a cyclic group H of order n with generator x, then for any integer m with gcd(m,n)=1, we have that x^m is also a generator of n

[stone fulcrum]

Oh yeah fuq should have just showed the computation haha

[chilly ocean]

ok thanks both of you

[chilly ocean]

Hello

[vital quail]

[chilly ocean]

Suppose e is in G

and H is a subgroup of G

[vital quail]

is G a group

[chilly ocean]

yea

[vital quail]

you shouldn't use e for an arbitrary element kek

[chilly ocean]

how do I prove that the same e is in H

[vital quail]

standard notation for identity

[chilly ocean]

e is identity

[vital quail]

lol oh

i mean

[woven delta]

Lol

What is the definition of a subgroup?

[scarlet estuary]

so you want to show that the identity in subgroups is the same

[chilly ocean]

I know there exists an identity in H

but how do i show the identity is same

yea

[vital quail]

so take your e_G and e_H

[woven delta]

If your subgroup is closed under multiplication and inverses

Then it's immediate that the identity for the whole group is inside H

And assuming you've already proved that there can only be 1 identity in a group

Then you are done

[chilly ocean]

why is it immediate? Like how to show that

"immediate that the identity for the whole group is inside H"

[woven delta]

Suppose g is in H. Then g^{-1} is in H and so gg^{-1}= e is in H

A nicer fact that you might want to know is that the only element which fixes any other element is the identity

[chilly ocean]

ic

Can i prove it like this

so e_h is identity in H

so returning to G I have e_h*e_h = e_h, which I presume is true because G inherits operation of H

then there exist inv(e_h) in G also

so I can get e_h = e_g

[woven delta]

Yeah sure

This is not a very hard fact to show

In fact I don't even consider it to be something you have to show tbh

[chilly ocean]

the part I want to explain better is why e_h*e_h = e_h is true in G, like a concise explanation

[woven delta]

Well what is your definition of a subgroup?

[chilly ocean]

I mean its a subset that satisfies the same axioms 1)associative 2)identity exist 3)inverse for all element and 4)the binary operation is closed

[woven delta]

Okay so the group multiplication HxH \to H is just the restriction of the multiplication GxG \to G

That's why

[chilly ocean]

ic, thanks

[woven delta]

So if ab=c is true in H, it is true in G also

[chilly ocean]

Also if an operation of a group is inherited, no matter if the new subset is subgroup or not, its associative right?

[woven delta]

Yes

So being a subgroup is analagous to

1. the subset being nonempty
2. the image of the restriction to HxH of the multiplication function is contained inside H
3. same with the inverse function

[chilly ocean]

what do you mean by 3)?

[woven delta]

So the inverse is a function G\to G

[chilly ocean]

oh isee

thanks

[chilly ocean]

let G,H be cyclic groups, generated by elements x,y. Determine the condition on orders m,n of x and y so that the map x^i \mapsto y^i is a group homomorphism.

can anyone give me a hint? not sure how to go about this problem

[teal perch]

At first glance (I just woke up) maybe you should have a think abt the fact that a homomorphism has to send the identity to the identity

and if $\phi$ is such a homomorphism then $\phi(x^m)=\phi(x)^m=y^m$

[cloud walrus]

euclid:

[chilly ocean]

so maybe if ord(G)>ord(H) or something like that?

wait no the other way around

@teal perch

[teal perch]

oh actually first you should make sure it’s well-defined

ie if x^i=x^j then y^i = y^j

which will give u some restrictions on what divides what

[chilly ocean]

idk how to check that though

[teal perch]

it’s ok! let’s say x^10=1

[chilly ocean]

ok

[teal perch]

then x^11=x, so under our map we want y^11=y

this is only possible if y^10=1, i.e. ord(y) divides 10

[chilly ocean]

wait let me think

if x^10=x how does x^11=x?

wait sorry

if x^10=1 how does x^11=x?

[teal perch]

multiply by x on both sides

[chilly ocean]

lmao bruh moment

sorry

[teal perch]

no worries!

[raw moth]

Iirc the condition is ||m divides n|| or something like that

[teal perch]

thats where we r going yeah

[chilly ocean]

ya I saw the answer but didn't really understand how they came up with it

[teal perch]

so that concrete example for ord(x)=10 up there

shows you that ord(y) has to divide ord(x)

[chilly ocean]

wait one sec how is y^10=1 the same as thing ord(y) divides 10

[teal perch]

no wait

one second

i confused myself

[raw moth]

Note that y^(k ord(y)) = 1 for all integer k

And also that the order is defined to be the smallest positive power

[chilly ocean]

I can see how y^(k ord(y)) = 1 is true

[raw moth]

If ord(y) didn't divide 10 we could find a smaller positive power

[chilly ocean]

wait sorry why is that?

[raw moth]

By taking k ord(y) < 10 < (k+1) ord(y)

[teal perch]

yeah im right i checked

[raw moth]

we have y^10 = y^(k ord(y)) = 1 so taking inverses, y^(10 - k ord(y)) = 1 also

But 0 < 10 - k ord(y) < ord(y)

[chilly ocean]

ok I have to digest this ty everyone

[teal perch]

equivalently, if ord(y)=m then y^10=1 means 10 is 0 mod m

which means either ord(y) = 10 or 10 is a multiple of ord(y)

[chilly ocean]

Let G be a finite group, a theorem says if H is a subgroup of G then |H| divides |G|

my question is it necessary that for all divisors k of |G|, there always exist a subgroup H with k elements? Furthermore, is H unique?

(i.e. whether the converse of the theorem is true)

[sharp sonnet]

no and no

for the first, the smallest counterexample is A_4, which is a group of order 12 but has no subgroup of order 6

for the second consider Z_4 x Z_2 or something

where Z_n is the cyclic group of order n

it has subgroups Z_4 and Z_2 x Z_2 which are both of order 4 but non-isomorphic

@chilly ocean

[chilly ocean]

Oh ok, thanks

Also if a has order(a) = k, I went with a proof by contradiction to show that all elements in <a> is distinct, is there a direct way?

or something im missing

i'm trying to prove some easy theorems in the best ways possible

[stone fulcrum]

They're never distinct. For example, a^(k+1) = a

I'm thinking you mean something else?

[chilly ocean]

Oh yea

what I mean is {a, a^2, ..., a^k}

I mean technically is <a> = {a, a^2, ..., a^k} if its a set rather than a multiset

So what im trying to prove is actually

if ord(a) = k, then |<a>|=k

[stone fulcrum]

What's your definition of order?

[chilly ocean]

for a in G, my books says ord(a) is smallest n where a^n = e

and ord(<a>) = |<a>|

[stone fulcrum]

Suppose there was a repeat such that
a^n = a^m
For n,m < ord(a)

[chilly ocean]

yea thats how i did it

i guess this is a good way?

Also I'm confused about what happens if n is negative, like lets say ord(a) is n, but then there is m<0 where |m|<|n|

such that a^m = e

why is it that we don't consider m as the order, or is that case even possible?

[smoky cypress]

well if a^m=e then a^{-m}=e too

But then -m<|n|

[chilly ocean]

oh yea

[smoky cypress]

which contradicts the definition of order

[chilly ocean]

Is it true that only when the order doesn't exist that negative powers might become relevant

[smoky cypress]

Oh wait n is not the order of a

Sorry

[chilly ocean]

No your idea is right, I got the proof

[sharp sonnet]

i mean if a^k = 1, then a^(k-1) = a^(-1)

so it suffices to look at positive powers

[smoky cypress]

are you saying an element has infinite order

[chilly ocean]

Yea, but in the case when the order doesn't exist

[sharp sonnet]

this only works for finite order, ye

[chilly ocean]

e.g. (Z,+) i noticed in this case negative is necessary

[smoky cypress]

Yeah

Also (Z,+) is the only infinite cyclic group, up to isomorphism

[chilly ocean]

we just started isomorphism so I'm not good at it, but what are structures that are isomorphic to (Z,+)?

[smoky cypress]

any infinite cyclic group

[chilly ocean]

Oh i see

thanks

[smoky cypress]

You can try proving it

[chilly ocean]

yea i probably will encounter it soon

[smoky cypress]

Ok

[chilly ocean]

How about iii), you know what it's trying to ask?

up to isomorphism, find all groups of order p

[smoky cypress]

It's asking you how many groups are there that are not isomorphic to each other

Of order p

[chilly ocean]

how can i do this

don't know how to approach

[smoky cypress]

Do you know how to do (i)?

[chilly ocean]

yea i did i) ii)

well, let G and G' be two groups of order p and let a and a' be their generators

what's the first non-trivial homomorphism G -> G' that you can think of

does a maps to a', a^2 maps to (a')^2 work?

its my first q ive done about homomorphism so dik

yeah that's a homomorphism

idk

well, do you think it's an isomorphism

yea

its bijective

great

so how many groups of order p are there

1 but i don't know why there can't be 2

little confused

like you could map a to (a')^2

etc.

that doesn't matter though
there only needs to be one isomorphism for them to be isomorphic

Oh