groups-rings-fields

[hidden haven]

You should prove it probably

Not sure what you mean by inclusion of partitions

Is it one partition being finer than the other?

[warm holly]

Yep

[warm holly]

Alright thank you

[barren sierra]

Stuck on this

[rustic crown]

find a, b such that ab = 0. if ba = 0 we done
otherwise take x = y = ba?

[barren sierra]

What

Hm

Oh I see

[worldly dome]

@lament yacht off topic but here was the doubt

[wooden ember]

i think you got the wrong channel

[urban acorn]

i think he got the wrong server

[worldly dome]

yeaa sorry :/

[frank fiber]

is every algebraic closed field a noetherian ring?

[upper pivot]

Every field is, as they only have two ideals both finitely generated

[next obsidian]

Hurb

Or you can say

ACC is trivial

Hahaha

[dire bramble]

stupid question just to check if I'm crazy

2Z_8/4Z_8 = 4Z_8 as Z_8 modules correct?

just directly from first iso, using homomorphism x maps to 2x

[next obsidian]

Uhhhh

I think so?

Yeah

Yeah

[dire bramble]

ok good to know that i can in fact calculate a quotient module

[terse crystal]

This is equivalent to every element of G/Z(G) has order at most 2

xy^2x^-1=x(yx^-1)x(yx^-1)=(yx^-1)x(yx^-1)x=y^2 therefore any y, y^2 is contained in Z(G) . Conversely x(xy)^2=(xy)^2x we obtain that xyxy=yxyx

Therefore when G is a finite group, |G/Z(G)| is divided by 2, any group of even order has an element whose order is 2, QED

But I don’t know how to prove it when G is of infinite order

I think it’s false when |G| is infinite, haven’t found counterexample yet

[terse crystal]

Just prove it yourself by definition… given homomorphisms f_i: <X_i ; R_i> to H such that φ_1f_1=φ_2f_2 we have a unique f from <X; R> to H such that the composition of f and the canonical morphism from <X_i ; R_i > to <X;R> equals f_i . Notice that f_i is induced by a map F_i from X_i to H such that F_i (R_i) =e_H so you can construct a unique map F from X to H such that the composition of (the restriction of (the canonical homomorphism from <X_i ; R_i > to <X;R>) on X_i ) and F equals F_i . Then you just check that F(R)= e_H.

Where X is the union of X_1 and X_2, R is the union of R_1,R_2 and {φ_1(g)φ_2(g)^-1: g from G_0}

[terse crystal]

Just saw it I am a dumb 😂

[robust pollen]

It's trivially true that If $K \to L$ is a field extension, then $L \cong L \otimes_K K$ as $L$-algebras, right?

[cloud walrus]

expectTheUnexpected

[robust pollen]

lul, of course it is, I kind of missed a "not" in the exercise

[golden pasture]

i feel like im stoning very hard but if $X$ is a scheme with $X=X_1\cup X_2$, $X_1,X_2$ are nonempty closed sets (i.e. $X$ is not irreducible) and $x\in X_1\cap X_2$, then does $\mathcal O_{X,x}$ have zero divisors

i feel like it is true cuz it is when $X$ is an affine scheme but i cant seem to prove it in general

[cloud walrus]

ari 亲

[golden pasture]

(general=X may not be noetherian)

oh wait

i can just take irred components

[urban acorn]

I know basically no algebraic geometry, but don't you mean X1, X2 are proper rather than nonempty?

[golden pasture]

nonempty proper closed subsets yea*

my proof is prob kinda iffy but it's too early for my brain to function

[gritty sparrow]

this may be stupid but isn't spec(KxKxK) a counter example, where K here is some field. In this case the local rings should be fields hence they won't have zero divisors, but X is reducible and we can take X1 to be the points Kx0xK and KxKx0 and X2 to be the points 0xKxK and KxKx0 with common point KxKx0.

[urban acorn]

So you're saying the counterexample is the KKK spectrum?

[wooden ember]

i suppose they meant bijection preserving inclusion here?

[golden pasture]

ah right thats true

i think what i was hoping to show is more of whenever x is in the intersection of distinct irreducible components then the local ring there has zero divisors

[robust pollen]

Actually, either wording is imprecise in everyday speech, because both can be construed as "It is an inclusion which preserves bijection" or "it is a bijection which preserves inclusion". Of course, only one of those interpretations makes sense, the latter one.

i.e. you should read "inclusion preserving bijection" as "inclusion-preserving bijection".

[wooden ember]

ah yeah inclusion-preserving makes sense

cause i falied to see how the correspondence itself could be an inclusion

[regal carbon]

There is a difference when they write capital letters (groups, subgroups) and small letters (elements inside a group) right?

Just like this

[urban acorn]

if you wanted to you could say g is a group and G is an element of it, that's just not standard and it's more messy and uncomfortable

[regal carbon]

From the above picture sir/ma'am

[regal carbon]

@urban acorn thanks

[robust pollen]

Does there exist a $\mathbb{Q}$ -algebra $A$ such that $\mathbb{Q}(\sqrt[4]{2}) \cong \mathbb{Q}(\sqrt{2}) \otimes A$ as $\mathbb{Q}(\sqrt{2})$-algebras?
I have been banging my head against the wall, but I cannot come up with any solution...

[cloud walrus]

expectTheUnexpected

[rustic crown]

well $A = \mathbb{Q}(\sqrt[4]{2})$ should do the job? Like in general $R \otimes_R M \cong M$.

[cloud walrus]

det

[robust pollen]

A is just a Q-algebra, the tensor product is over Q

so in particular A would have to be a two-dimensional vector space over Q

[wooden ember]

am i correct in saying that none of these are ring homomorphisms?

seems weird for them to give a question with only negative answers but i dont see my mistake if there is one

[rustic crown]

yea that's i was wondering.. you wrote 'as Q(sqrt(2)) algebras' and i was confused

[hidden haven]

isomorphic as Q(sqrt2) algebras

[rustic crown]

yep

[robust pollen]

The point is probably that for the first two you need non-multiplicativity, while for the last one non-additivity

[robust pollen]

Indeed 👍

[rustic crown]

can we then just try a generic Q(sqrt(d)) and check if it after tensoring gives Q(2^1/4)

[wooden ember]

aight thanks just wanted to make sure

[hidden haven]

does it have to be a field extension?

[robust pollen]

No, A can be just an algebra

[hidden haven]

ye I am saying that we don't know if it is Q(sqrt(d)) a priori

just from 2 dimensional

[rustic crown]

Ah right getting the inverse of that might be a problem

[hidden haven]

well there might be an argument along the lines of the tensor not being a field if A isn't

that would make sense given that the multiplication is componentwise

[rustic crown]

i'm wondering if we just use the extension of scalars thing to Q[x]/(quadratic poly)

so Q(sqrt(2))[x]/(same thing but in Q(sqrt(2)))

for this to be like Q(2^1/4), i think the poly should be something like x^2 - sqrt(2) but that doesn't live over Q, so won't work

i feel we can formalize this

[hidden haven]

Not sure what you're doing lol

[rustic crown]

[robust pollen]

OT: isn't my reaction Urs Schreiber? Lel

[rustic crown]

need to review more tensor product

[hidden haven]

It is called urs yes

[robust pollen]

But anyway, i believe that there does not actually exist such an A.

[robust pollen]

I once saw him in person and was too starstruck to talk to him

[hidden haven]

I don't even know who he is

[rustic crown]

same lol

saketh is typing for too long... i'm scared

[robust pollen]

The guy who founded nLab boiis

[hidden haven]

Poor question is gonna get destroyed

[robust pollen]

Kcuf

Oh.

[gritty sparrow]

How do I edit a post to reply to the right q

[hidden haven]

Bruh which question are you answering

Lol

You can't change replies

Do it separately

[gritty sparrow]

Wdym?

[hidden haven]

You can't change your message now to be replying to someone

Just delete and repost maybe

[gritty sparrow]

Cool

[gritty sparrow]

This proof is a little sketchy but I think it what you are saying is true in arbitrary schemes. If X1 and X2 are irreducible components and x is in the intersection, consider any open affine neighborhood U around x. U \cap X1 will be dense in X1 and X1 \cap X2 is a closed subset of X1 hence U cannot be contained inside X1 \cap X2 so there is some point of U that is in X1 but not X2. Similarly there is a point of U in X2 but not X1. Essentially this shows that when we restrict to U, these will prove that U is not irreducible, and will infact be irreducible components in U. Then we have reduced to the affine case, and there the proof should be straight forward

[golden pasture]

oh right i completely didnt consider a topological argument zzz

[rustic crown]

what i was saying was a 2-dim algebra over Q will definitely look like Q[x]/(quadratic poly)... now we need to tensor this with Q(sqrt(2))

[golden pasture]

was just wacking on local ring oof

[robust pollen]

Sure, i actually tried to solve this with Mathematica, but it's still going 💩

Like, look for conditions on the quadratic polynomial so that such an isomorphism is true, i.e. there has to be an element of order 4

[rustic crown]

so right exact thing
Q[x] --> Q[x] --> A --> 0
where the first map is multiplication by the quadratic poly with coeff in Q
tensoring by Q(sqrt(2)) should give an exact seq again
Q(sqrt(2))[x] --> Q(sqrt(2))[x] --> Q(2^1/4) --> 0

[hidden haven]

why is Q(sqrt2) tensor Q[x] = Q(sqrt2)[x]?

[rustic crown]

yee afair

[hidden haven]

seems believable

but what do with the sequence

[rustic crown]

so now that quadratic poly can't be reducible over Q(sqrt(2)) else we don't even get a field

after translation it should look like x^2 - d

[hidden haven]

right

[rustic crown]

but this would give Q(sqrt(2), sqrt(d))

as d is not like 2, else x^2-d is reducible over Q(sqrt(2))

[hidden haven]

seems legit like

[rustic crown]

but this is galois... and Q(sqrt(2^1/4)) isn't 😛

seems correct to me

[hidden haven]

lol

[wooden ember]

Z[x]/x^3 gives the polynomials with degree at most 2 with integer coeffs right?

and in general Z[x]/x^k is the polynomials of degree at most k-1?

[robust pollen]

If it's reducible the ideal would not be maximal, and then A would have zero divisors, which would (if such an iso existed) imply that Q(2^1/4) has zero divisors, absurd.
So then the rest of the proof would establish that Q(2^1/4) is biquadratic, hence Galois. But the splitting field of x^4-2 has degree 8, so this can't be true. Did I get it right?

[hidden haven]

ye

[robust pollen]

Tight!

[hidden haven]

yes but with multiplication that loops the degrees around

[wooden ember]

yeah right

thanks

[urban acorn]

in general when f is a monic polynomial of degree n, the polynomials of degree strictly smaller than n in R[x] constitute a complete system of unique representatives for the elements of R[x]/(f)

[wooden ember]

thought so yeah

not sure i see what happens when f isnt monic though

[urban acorn]

This is the part of the argument when we need it to be monic: when f is monic of degree n, then given any polynomial $p = {a_m}x^m + ... + a_0$ where $m \geq n$, we can find $q = p - {a_m}x^{m-n}f$

[cloud walrus]

Ab

[urban acorn]

and q is of degree smaller than m

and q is equivalent to p mod f

so we can repeat this until eventually finding a polynomial equivalent to p mod f which is of degree smaller than n

[urban acorn]

but if for example f was not monic, like 2x^2 - 1, then we couldn't find a multiple of f that would get rid of the highest coefficient in x^5 - 1

of course, this also works if the highest coefficient in f is a unit, for example, so this works for any polynomial over a field

[urban acorn]

and this then follows from the monic case because if the highest coefficient is a unit then (f) = (f') where f' is monic

[wooden ember]

yeah makes sense

[urban acorn]

$2 \in Z[x]$ is an example for a non-monic polynomial, and $Z[x]/(2)$ is basically $[Z/2Z][x]$

[cloud walrus]

Ab

[wooden ember]

hm yeah

when you say basically is there something in the way of that?

seems isomorphic to me

[urban acorn]

no, I don't know why I said basically

[wooden ember]

aight

[urban acorn]

it's isomorphic

and I'm sure you can find some way to even maybe show the isomorphism is "natural"

[wooden ember]

wdym

[urban acorn]

lemme try to see if I can make the idea I have formal

the vague idea I have is that if we can find the right way to make the constructions of R/I [x] and R[x]/I into functors they'll be naturally isomorphic

like, maybe in some kind of category where the objects are a ring together with a selected ideal, and the morphisms (R, I) -> (S, J) are ring morphisms phi : R -> S such that phi^-1(J) contains all of I, and then from x = (R, I) we get Fx = R[x]/I and Gx = (R/I)[x]

and I think that F and G are well defined functors and naturally isomorphic

I need to think about that for a moment though

[wooden ember]

im afraid i wouldnt be able to help with that, functors are still a little confusing to me

[wooden ember]

the question

My answer

[urban acorn]

nice handwriting

[wooden ember]

not very good cause the tablet is tilted and not flat

so im writing from an awkward angle

[chilly ocean]

Sorry but i still have a question. What do you mean when you write "$x^2$"? Is it just "$x \otimes x$"? Or how do we square a vector? Or is it a polynomial? So is $Sym^2(K^2)$ the set of homogeneous polynomials of degree 2?

[hot lake]

well uh

well when you have tensor products of vector spaces you can tensor product vectors

[delicate bloom]

you should give a counterexample

[cloud walrus]

Gewisser Fler

[hot lake]

and you can do similar stuff and define symmetric product between symmetric tensor powers

and that kinda corresponds to making products of homogeneous polynomials

and all of that is functorial as in a change of basis of the original K induce change of basis between the other stuffs so that things are independent of the choice of basis

so for now you can just wing it and pretend that K is the set of homogeneous polynomials of degree 1 in x and y

then the nth symmetric power of K is the set of homogeneous polynomials of degree n in x and y

and multiplication is the usual polynomial multiplication

and you perform change of variables the uuh obvious way

but you can also build everything in terms of tensor products, where xy is now 1/2 (x tensor y + y tensor x)

and use tensor products + symmetrization as multiplication

anyway Sym²(K²) will have dimension 3

and you will have an action of GL2(K) on K² that induce an action on Sym²(K²) because Sym and stuff are functors

[wooden ember]

i mean i wasnt bothered to write any out they arent hard to find

though answering the question of when equality is true seems interesting

[delicate bloom]

sure just pointing out if I was your grader I wouldn't give you points for stating something without proof

[wooden ember]

oh of course but this isnt for class

[delicate bloom]

but yeah seems interesting to work out when this is true too, maybe there's some clean geometric argument that pops out

[urban acorn]

𝕊elf 𝕝earning 𝕚s 𝕔ool.

[wooden ember]

idk seems pretty non trivial to answer

i mean a few simple examples pop out but that's it

(at least for me)

[delicate bloom]

I guess simplest cases to solve is for arbitrary f(x) we can pick g(x)=k and that pair will always work

[wooden ember]

yeah

[delicate bloom]

are we still restricting ourselves to continuous functions

[wooden ember]

might be better to take weaker regularity first

oh i can apply the same argument from the exercise i saw yesterday

if f is linearly dependent on g it works

[delicate bloom]

I guess one way to look at it is take a continuous function with parameters that has constant integral independent of its parameters

then when multiplying with an arbitrary f(x), we can say there will be a solution by the IVT

[wooden ember]

and it only works when f is linearly dependent on g if f and g are continuous

[delicate bloom]

and then we can do any linear combination of these types of constant functions

wdym linearly dependent

[wooden ember]

f = r*g for some constant r

[delicate bloom]

not true

[wooden ember]

oop where did i mess up

[delicate bloom]

because the integral of g is constant but has parameters that can vary

[wooden ember]

oh i took a positive square root

however i do think the converse is true, namely it is necessary for f to be a multiple of g

(when continuous)

[delicate bloom]

nah

[wooden ember]

lmao what am i messing up

lemme think

oh i see what i did

nvm anything i said then

[delicate bloom]

https://www.desmos.com/calculator/qxwdhigs9l

[wooden ember]

always gut check your answers lol

[delicate bloom]

here g(x) has constant integral, it's a line through (1/2, 1/2) where I vary the slope

[wooden ember]

im a little confused by your method

wdym by constant integral

[delicate bloom]

I'm basically varying the integral f(x)g(x) while the product of the individual integrals remain constant

play around with the desmos example I gave to see, that might help, gonna grab something to eat real fast

[wooden ember]

oh i see

the integral of g is a constant function of k but the integral of gf isnt

smart

[delicate bloom]

well that might not be the best example I gave tbh

[wooden ember]

so as long as we can get the difference to be negative and positive you get a solution by IVT i see

[delicate bloom]

yeah, that's the idea

so we can imagine there are a bunch of these continuous functions with parameters that have constant integral out there

kind of lame I guess maybe

[wooden ember]

no it's a neat idea

tried to do something with integration by parts but it just gave me the "one of the functions is constant" case 😑

[delicate bloom]

yeah, thought I might try some kinda euler lagrange type thing to get an exact solution for g with a fixed f and the constraint that the integral of g is constant but didn't get anything exciting

[regal carbon]

This is a symbol of group right?

[wooden ember]

Yes the star is the binary operation associated to the group structure

[regal carbon]

What if it has this symbol

[wooden ember]

(At least I think this is what your book means, it’s standard notation but you’ve only showed a very zoomed in portion)

[wooden ember]

The symbol doesn’t matter

As long as what the operation does is defined

[regal carbon]

Direct sum or delts

Delta doesn't matter

Okay

[wooden ember]

(G, 👌) could be a group as long as I define 👌 as an operation

(And that it satisfies all the relevant group properties)

[regal carbon]

Subscribed to your channel

[south patrol]

Let 🖖 be a vector space over 💯 . Then a map 👌: 🖖 x 🖖 -> 💯 is called an inner product if

[wooden ember]

sure why not

[regal carbon]

In example 1 I didn't understood the second line

a+e-1

And the question a*b=a+b-1

[wooden ember]

the operation $\oplus$ is defined by $a\oplus b = a+b-1$. An identity $e$ must satisfy $a\oplus e = a \Leftrightarrow a+e-1 = a$

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

(which is true iff e = 1)

[nova plank]

The only group is ( , )

[wooden ember]

this is the most interesting of the finite simple groups i agree

[nova plank]

Based narwhal

[wooden ember]

i really dont wanna multiply stuff out eternally 😭

[cursive temple]

doesnt it suffice to find a subring of M_2(C) which is isomorphic to H

[terse crystal]

Oh it’s not complicated

[cursive temple]

which seems a bit more doable

[wooden ember]

sure but it's still gross

id rather just skip it lmao

[terse crystal]

You remember that C can be viewed as subring of M^2(R) right?

Mapping a+bi to

[wooden ember]

sure that was the previous ex anyways

(a,b;-b,a)

[terse crystal]

( a b

Yeah and in H

a+bi+cj+dk =(a+bi)+(c+di)j

Kind of double

[wooden ember]

i really am not bothered to think about how that translates to a 4x4 matrix over R sorry

[wooden ember]

when the dimensions exceed 3 i just say no cause it's ugly

[terse crystal]

H—>M^2(C)=M^(R)

[wooden ember]

as pappa said if it had asked M_2(C) i wouldve done it but bleh

fair it doesnt say construct the explicit isomorphism

so i guess i could just say that

yeah lol im a bit stupid thanks

[wooden ember]

we dont know what you mean

and please dont multipost

[terse crystal]

a+bi+cj+dk —>(a+bi c+di ; -c-di a+bi)—>( a b c d ; -b a -d c ; -c -d a b ; d -c -b a)

?

[wooden ember]

4x4 matrix 🤮

[terse crystal]

Told you not complicated at all just do this thing twice : a+bi—>(a b ; -b a)

Like C=R+Ri, we have that H=C+Cj

[wooden ember]

my point is i know im being unreasonable but writing out the 4x4 matrix is ugly

[wooden ember]

C + China

autocorrect go brrr

[terse crystal]

Lol

[wooden ember]

honestly im so polarly opposite in the exercises i want to do or not lmao

on one end there's this one which is easily doable and frankly not even all that ugly that i dont want to do and on the other end there's "show there are no non-abelian simple groups of odd order <10000" which i wanted to do

[robust pollen]

Is there a joke about modules over quantum groups hidden in this?

[wooden ember]

i tried reading about quantum groups once

i was quickly lost

[robust pollen]

Learn Hopf algebras first, and then never learn quantum groups. It's better, trust me

[wooden ember]

lmao okay

[hidden haven]

Learn cat theory first

Don't you wanna know what this means

[wooden ember]

i sort of do but not fully

[robust pollen]

That succs

[wooden ember]

scrap that

i just dont

[hidden haven]

You'd rather find all finite simple groups

[wooden ember]

lmao no

[hidden haven]

[wooden ember]

though doing that on my own would be one hell of an achievement

an impossible one

[hidden haven]

Why impossible

If you get stuck somewhere

Post on this channel

[wooden ember]

lmao

[hidden haven]

We'll make fun of you and you'll get motivated

[wooden ember]

[robust pollen]

Senpai Moldy best senpai

[hidden haven]

Don't 🤡s get motivated from making others laugh

[wooden ember]

bitch please

[hidden haven]

[wooden ember]

pretty stuck on this one, a small hint on the right manipulation track would be appreciated

second part is easy but im struggling to show the zero divisor part

i concluded that phi wasnt surjective cause no element of R could be mapped to 1_S and that phi(g) is a zero divisor for every other g than 1_R not in the kernel

wait im a moron

g can be 1_R

🤦

[gusty halo]

i feel like its kinda hard to give a hint without giving it away

[wooden ember]

no it's aight i realized my mistake

[gusty halo]

if phi(1) = a, what can you say about powers of a?

[wooden ember]

i dont need help on this anymore 👍

[hidden haven]

[rustic crown]

moldi senpai

[hidden haven]

.

[rustic crown]

[wooden ember]

is a new emoji on the server

since we had a short mention of constructive mathematics yesterday this makes me wonder: is there a way to handle such questions without reasoning by contradiction?

cause here i showed this was true by contradiction but im wondering if there's a more "constructive" way to do it

[gusty halo]

I mean given i,j in the intersection and r in R, then clearly r-s and ri are in each ideal, hence in the intersection?

[hidden haven]

no

[gusty halo]

on an unrelated note, does constructive always entail that you are working with finite (or finitely presented) objects?

[wooden ember]

doesnt the clearly here work by contradiction though

[gusty halo]

why?

[wooden ember]

cause you cant talk about "each" ideal since there are uncountably many right?

[gusty halo]

r,s is in one of the ideals

so r-s is also in that ideal, by ideal axiom

do this for every ideal

[wooden ember]

the do this for every ideal part fails because you have uncountably many

with countably many you could argue for each of them by induction

at least i think

[gusty halo]

hmmm

[wooden ember]

so im pretty sure contradiction is the only way to go about it

which is kind of unsatisfying

let's hear what others have to say about this cause im not at all confident in my analysis of the problem here

[gusty halo]

so by your logic, whenever i would want to prove something for an uncountable collection of objects, the underlying logic would have to be via contradiction?

[wooden ember]

i believe? im not sure

[gusty halo]

im also not familiar with this formal proof theory stuff

maybe ask in #foundations

[wooden ember]

yeah same

[wooden ember]

yeah no this doesnt sound right this is exactly why we have the for all quantifier

[wooden ember]

how the fuck does one nicely handwrite the nilradical of R

[viscid pewter]

cursive

[sharp sonnet]

isn't it "just" fraktur?

the handwritten analogue of that is called Kurrent if you want to learn a new way to write letters, alternatively i usually just underline my fraktur letters

[cloud walrus]

det

[wooden ember]

so you just write the analogue in the regular alphabet and underline it?

underlining looks weird in math...

but much better than my poor excuse for a capital N in fraktur

[sharp sonnet]

yes

real fraktur requires a calligraphy pen (it was not designed to be written by normal people)
you can try imitating Kurrent (https://en.wikipedia.org/wiki/Kurrent#/media/File:Deutsche_Kurrentschrift.svg) if you want

[simple valley]

for which rings R, all abelian groups can be extended to R-modules?

obviously it's true for Z and false for something like \mathbb{R}

seems to be related to existence of ideals of various orders?

[hidden haven]

Whenever Z is an R-algebra you will be able to

I feel like the converse should be true too

Oh yeah if you can extend any abelian group to an R-module then you can extend Z to an R-module and that is the same as existence of a homomorphism R → End(Z) = Z

So Z is an R-algebra in some way

[urban acorn]

imagine not being able to type ℤ with your keyboard

[hidden haven]

ℤ

[urban acorn]

you copy pasted it

[hidden haven]

I just choose not to

Bruh

[delicate bloom]

$\bZ$

[hidden haven]

I have gboard tex dictionary

[cloud walrus]

Meroseous

[delicate bloom]

whee

[simple valley]

so to put it another way

iff R has an ideal I such that R/I = Z

I was close, I was thinking about a family of ideals I_p such that R/I_p =Z/pZ

[hidden haven]

Oh are all ring homomorphisms into Z surjective

Yes it should be that

[urban acorn]

yes

1 ∈ ℤ must be in the image since morphisms are identity-preserving, and then since the image is an additive subgroup it has to be all of ℤ

[simple valley]

hmmm caveat

if we consider {0} as a ring with 1=0

but that's the only exception

[hidden haven]

We won't have a ring homomorphism in that case

[simple valley]

oh true

the only 0-module is 0

if we consider that

ok

[urban acorn]

lol if there was a map 0 → ℤ then there would be a map R → ℤ for all rings R

[simple valley]

I mean in the category of unitless rings it's true

there's a zero object and whatnot

[urban acorn]

ew unitless rings

I'm gonna go wash my eyes

[wooden ember]

do you mean a ring with identity

[urban acorn]

the biggest mistake dummit and foote made was not defining all rings to have identity

I mean not that there's anything wrong with looking at rings without identity - but they shouldn't be introduced at the beginning

also I personally wouldn't wanna think of rings without identity at all

[wooden ember]

lmao

yeah then all the non zero morphisms into Z are surjective

[urban acorn]

also @wooden ember don't let d&f lie to you: ring morphisms always preserve the 1!!!

[wooden ember]

haha

[simple valley]

well think about ideals

they're kernels in the category of rings without identity

[urban acorn]

okay so I'm not saying no one should ever look at rngs

[regal carbon]

Any YouTube video playlist you recommend who teaches you to solve problems?

Of different kind of proofs?

[wooden ember]

you can only learn to solve problems by solving problems

[hidden haven]

Rings should always be unital because the same way that the groups arise as symmetric groups, rings arise as endomorphism rings of abelian groups (so they should really always be non commutative unital)

[urban acorn]

okay so rings have multiplicative monoid structure, so their morphisms need to be monoid morphisms

[regal carbon]

No, i got no menor

Mentor*

[urban acorn]

and monoid morphisms need to preserve 1

why, you ask? well for once, we want to say monoids are 1-object categories

[regal carbon]

Even in abstract algebra?

[urban acorn]

and we want to say that all of their morphisms are functors

and functors preserve the 1

[simple valley]

semigroupoids

[urban acorn]

boom

[regal carbon]

...

[hidden haven]

wtf

Is it like non unital rings arise from endomorphisms of semigroupoids?

[urban acorn]

yeah but then the rings that are more interesting to look at end up being commutative imo

[simple valley]

no actually I retract what I said

[urban acorn]

like, this is the same way we look at groups instead of monoids

groups can be represented with automorphisms of some set, monoids can be represented with endomorphisms of that set

[prisma ibex]

depends on your taste, there are many very interesting non-commutative rings that show up in nature so to speak

[urban acorn]

okay fair, hence I added "imo"

[simple valley]

End(X) in a semigroupoid is indeed a semigroup

but it's not clear how to Ab-ize that

[prisma ibex]

[simple valley]

Ab-enriched semigroupoids? that' more ugly than what we started with

[prisma ibex]

I mean these things simply do not have very nice categorial properties

[simple valley]

ye but kernels man

[urban acorn]

whenever a mathematician wants to enrich a category over Ab they come to me and ask me to sign a legal document

[simple valley]

ok hmm, I thought this other question would be simpler but it doesn't seem to be

basically when is U : R-mod -> Ab full or not

[urban acorn]

example: it is full for R = Q

[simple valley]

I don't think so

but the reasoning is involved

wait is it

[urban acorn]

Let A, B be Q-linear spaces and ϕ : A → B an Ab morphism. Let x ∈ A, then clearly ϕ(nx) = nϕ(x) since this holds for all Ab morphisms, so now for some n/m ∈ Q we can see that mϕ( (n/m)x ) = ϕ( m (n/m) x ) = ϕ(nx) = nϕ(x), and and therefore ϕ( (n/m) x ) = (n/m) ϕ(x), so ϕ is also a Q-linear map.

[simple valley]

right

but this clearly not true for R the reals

what gives

something something uniqueness (epicness?) of Z -> R ?

[urban acorn]

epic 😎

[urban acorn]

but yeah I have a gut feeling that's about right

[simple valley]

if it's not epic then we have f,g : R -> S

then we consider R and S as R-modules

f-g is(?) an Ab-homomorphism but not an R-mod homomorphism?

[urban acorn]

yeah, I think that logic is sound

[simple valley]

not sure how I used Z here though

[urban acorn]

so let h be the inclusion Z -> R

the maps that you chose f, g are such that f is not equal to g but fh = gh

so they are equal on Z, but not everywhere

so f-g is zero on Z, but not everywhere

[simple valley]

hmm

[urban acorn]

but this is impossible for an R-module homomorphism from R

[simple valley]

am I just using the 1 from Z

[urban acorn]

since it is determined by the image of 1, yeah

[simple valley]

I guess I am

does it work backwards too?

full => Z->R epic?

[urban acorn]

isn't this the direction we just did? we assumed Z -> R not epic and concluded U not full

[simple valley]

oops yea

[wooden ember]

wtf epic is an actual mathematical term

[urban acorn]

an epimorphism is a categorical analogue of a surjective map

to say that a map is an epimorphism is to say that it is epic

[wooden ember]

i know what an epimorphism is

[urban acorn]

then that's what epic is

[wooden ember]

i never heard the epic terminology lmao

that's great

[urban acorn]

you could say

that's epic

but is it mono?

[wooden ember]

i just realized i miswrote epimorphisms as exomorphisms when i tried to remember what the terminology was when writing some notes for my lin alg class the other day

not an epic moment

at least i remembered monomorphisms

[edgy vault]

@wooden ember imagine that these are just alien bois

like xenomorph

[urban acorn]

brb coining the term "xenomorphism"

[wooden ember]

if only xenomorphism was a thing

[edgy vault]

what if it could be

xenomorphism must be a concept in mathematics

[wooden ember]

could be the names for morphisms that arent epi or mono morphisms

would be lame but hey

cool name

better than just "morphisms"

[edgy vault]

okay..

I'm going to ask Gilbert Strang about the radical ideal proof

[wooden ember]

if anyone in this server becomes a cat theorist and finds a new interesting class of morphisms please call them xenomorphisms

[edgy vault]

alien theory

[wooden ember]

alien cats

[urban acorn]

"radical ideal" sounds political

how about "radical left ideal"

I remember where I got that from

there was a stackexchange post asking for mathemtical urban legends

this is one of the answers

When I was at the University of Oklahoma in the early '80s, we were all required to write a brief description of our research for the (rather conservative, this being Oklahoma) Board of Regents of the University. An colleague in algebra, perhaps hoping for more state support, wrote that he was studying "annihilating radical left ideals."```

[wooden ember]

lmao

[hidden haven]

this is what I imagine narwhal's yt to be like

[wooden ember]

lmao

[hidden haven]

Maybe I will visit and confirm

[wooden ember]

i have yet to make a single group theory vid

[hidden haven]

[wooden ember]

i havent made a video since ive started d&f i dont think

[hidden haven]

oof

[wooden ember]

maybe the metric space video was like a week in im not sure

[hidden haven]

yo you have so many videos

I was expecting like 3

5 mins each

[wooden ember]

all the nt ones are shit

[hidden haven]

tell me your best one

[wooden ember]

idk i like the exponential function one personally

[hidden haven]

lesgo

[wooden ember]

or pythagorean triples

or properties of counting tools yeah id say those are my three favorites

but i really need to up the quality on these

[hidden haven]

you sound exactly like some other math yt channel I watched before did I accidentally watch one of your videos before

I can't remember the channel now

[unreal portal]

have 50k views on one of my youtube videos

except it's not math related lol

[wooden ember]

a musician i see

that's still pretty dope

it's crazy how many people who enjoy music also enjoy math and vice versa

[unreal portal]

I do have 1 math video on there about knot theory

[small bison]

that's probably because the vast majority of people enjoy music

[unreal portal]

except it was before I knew what it was doing so it's shite lmao

[wooden ember]

yeah im watching it rn

[unreal portal]

[wooden ember]

but it's not bad for now

it's at least better than what i do lmao

[unreal portal]

this was for an intro to group theory class

so I didn't know any path homotopy like I do now

let alone the wirtinger relations at the crossing of a knot

[hidden haven]

yo cool video narki 😌

watched the one on exponential function one

[unreal portal]

I thought about making a video for the 3b1b contest about knot invariants because I think invariants in general are really cool

but I just didn't have the time

[wooden ember]

your knot video is 👌

you dont waste time which i do too much

[unreal portal]

talking about how you have to find a balance between an invariant that's easy to calculate, but not too easy that it doesn't distinguish objects

thank you!

[oak grove]

I think i might be misunderstanding something

or maybe the question is easy

it should just be [n-1], right?

[hidden haven]

1 maps to 2 is not an automorphism of Z2

[oak grove]

oh, right

surjectivity

err

no yea thats why like Z->nZ is never at automorphism unless n is 1

i think

[hidden haven]

1 maps to 2 is an automorphism of Z3

[oak grove]

wait but you say 1 maps to 2 is not an automorphism of Z2

oh i guess the only possibility would be 1

so why not just [n-1]\0

oh

[hidden haven]

replace Z2 with Z4

[oak grove]

wait

do i need totient

[hidden haven]

yes

[oak grove]

okay

since you have to preserve order

i get it 😄

the only way to do it is to map 1 to a number thats coprime with n

[hidden haven]

yep

[oak grove]

okay another part on this same section im not sure about

phi here would be the automorphism itself right

[viscid pewter]

no

totient function

wait

yeah, totient

[chilly ocean]

solution kinda jan Niku.

the correspondence is: x coprime to n <-> f \in Aut(Z_n) such that f(1)=n

[oak grove]

👀

i dont really get how that helps with the problem

[chilly ocean]

phi is totient function right

[oak grove]

yea

[chilly ocean]

so phi(n) is number of numbers coprime to n

[oak grove]

wondering now if my answer to the last part was right

[hidden haven]

It was

[oak grove]

no this one

I got $|G| / \phi (n-1)$

[cloud walrus]

jan Niku

[oak grove]

seems fine right

[hidden haven]

phi(n) right?

[oak grove]

no

[hidden haven]

[oak grove]

wait

idk how youd tell

i guess theyre saying n didnt change?

so G is still Aut Zn

[hidden haven]

Yeah I'd assume so

[viscid pewter]

it's just G right?

[hidden haven]

And n-1 in Zn should be fixed by exactly the things that 1 is fixed by

[oak grove]

let me rewrite this part

then i can look at 3 again

[viscid pewter]

oops i pinged

[oak grove]

i dont understand that moldilocks

well i dont really understand the function of a lot of these objects lol

or thats the feeling i get

[hidden haven]

So same as asking stab 1, which has cardinality |G|/|orb 1|

[hidden haven]

This one?

If an automorphism maps 1 to 1, it must map -1 to -1, and vice versa

[viscid pewter]

n-1 is coprime to n

wait that's not quite relevant? hmm

[hidden haven]

You don't need to do this, I just find it easier to think about 1 than about other generators

[oak grove]

hmm the n-1 is throwing me off now

[hidden haven]

[oak grove]

but shouldnt |stab(n-1)||orb(n-1)|=|G|

oh that only works for n = 2

no wait

it works since <1> = <n-1>

i dont understand why you feel my answer is wrong

it should be |G|/phi(n-1)

[hidden haven]

This is always correct

[oak grove]

not phi(n)

[hidden haven]

I'm saying that |orb(n-1)| should probably be phi(n) because n-1 can be mapped to exactly the things which are co prime to n

Not coprime to n-1

[oak grove]

then why is orb(1) phi(n)

oh

orb(1) = orb(n-1)

[hidden haven]

orb(1) = orb(n-1)

[oak grove]

well

yea

[hidden haven]

Yes

[oak grove]

bleh

i hate this day lmao

okay

so then |G|/phi(n) = |stab(n-1)|

fair

that helps with 3 right

actually lemme just try it

gimme like 15 min

bein lazy for no reason

[hidden haven]

[oak grove]

yea i dont like this

does it imply that |stab(n-1)| is always 1?

i dont believe that at all

but idk like

i guess maybe

is it just because only the identity?

otherwise youd need n

but you only get up to n-1

am i close

im just trying the base case

is |stab(i)|=1 for all i in Z_n

@hidden haven sorry to ping

[viscid pewter]

if g exists in stab(i) in Z_n, g+i = i

[oak grove]

right

having trouble thinking now when |stab(i)| would ever not be 1

[hidden haven]

Very close, but it is not 1 for 2 in Z_4

[oak grove]

🤔

why not

[oak grove]

? only 0

[hidden haven]

1 maps to 3 is an automorphism that stabilizes 2

[oak grove]

👀

i dont understand at all

thats not an element

[hidden haven]

What isn't

[oak grove]

isnt stab(2) a subset of elements in Z_4

[hidden haven]

Stab(2) should be a subgroup of aut Z4

[oak grove]

👀

[hidden haven]

The group Aut Zn is acting on the set Zn

[oak grove]

why Aut z4

[hidden haven]

So you talk about orbits and stabilizers of elements of Zn, and these orbits are subsets of Zn, and stabilizers are subgroups of Aut Zn

Wait did I misinterpret the question

I thought this was the set up lol let me see again

[oak grove]

Idk are we talking about the question i posted? or just in general

I'm just talking about Z_n not like

Aut Z_n

Well I guess i dont really understand what a stabilizer would even mean in that context

[hidden haven]

So here it seems the group is Aut Zn

And they just ask for the stabilizer of

Without mentioning what the group is acting on

So I think it is implicit that the action is the natural action of Aut Zn on Zn

[oak grove]

action of aut zn on zn

[hidden haven]

Where f acts on s to give f(s)

Which definition of group action have you seen?

[oak grove]

i dont recall hearing that before

i mean weve seen group operations

i dont think weve ever used the term group action

[hidden haven]

Wait then how is stabilizer defined

😵‍💫

Maybe group operation = group action

[oak grove]

$\text{ stab } _G (i) = {g \in G : g i = i }$

[cloud walrus]

jan Niku

[hidden haven]

And where's i

[oak grove]

so maybe you see why i think its always 1 👀

i's in G

[hidden haven]

oh

bruh

[oak grove]

i mean theres only one identity

how can there be anything else

but this is the definition we have

so IDK

[hidden haven]

Ok yeah this is the case of a group acting on itself by left multiplication 😵‍💫

[oak grove]

but this works fine for our problem i think

well it works until Aut is introduced

[hidden haven]

Lol I'll also have to go through the previous parts again let me do that quickly

[oak grove]

then idk theres some additional fuckiness

[hidden haven]

Wait so 1 is the identity automorphism in G when we write orb_G(1)?

But then asking for orb(1) is silly

Because that will be G itself

[oak grove]

👀

[hidden haven]

are you defining orb similarly? orb_G(i) = {gi}?

[oak grove]

yea

so it doesnt matter here that were in Aut zn right

youre gonna get the same answer as just zn

[hidden haven]

So for any element g of G, g = g1, so g ∈ orb(1)

Um no I'm saying that for any group G, orb_G(1) = G

[oak grove]

👀

[hidden haven]

Are you sure that these are the definitions of orb and stab

[oak grove]

idk

[hidden haven]

Like what kind of group operations have you seen?

[oak grove]

its not really clear what the operation even is in this context

not a lot

weve looked at Zn, klein groups

and symmetric groups

[hidden haven]

Yeah that's true, but I think Aut Zn acting on Zn is a very natural one

[oak grove]

so is the group operation here the automorphism or modular addition

[hidden haven]

Could you give the definition of a group operation that you've seen?

[oak grove]

hrm

[hidden haven]

Like vaguely

[oak grove]

i mean klein 4 is easy like

but idk if thats helpful

[hidden haven]

Wait by group operations I don't mean the + in (Z, +)

[oak grove]

oh thats what i mean

[hidden haven]

oh ok I see

[oak grove]

then what are you talking about 👀

[hidden haven]

Given a field, you can talk about an "action" of that field on an abelian group, and this gives you vector spaces and linear algebra

Given a group, there's a natural notion of that acting on a set to give you group actions

Similar to how any field F acts on itself by its own multiplication to give a 1 dimensional vector space, any group can be made to act on itself to give a 1 orbit G-set

But that's why I feel that it's silly to ask for orbits in this situation, because it seems that you've only seen a group act on itself by left multiplication? But then there's always only one orbit, the entirety of G

[oak grove]

I'm not sure weve covered fields either

well we talked about them briefly in real analysis but never did anything with them

[hidden haven]

oh whenever I said fields you can replace that with R or C

You might have seen vector spaces with real or complex coordinates already

[oak grove]

👀

im not sure, late last week we were talking about isomorphisms

then earlier this week she introduced automorphisms

im not really sure if thats what youre talking about

like the mapping of some element to another element being the group action

[hidden haven]

with this interpretation, orbits should be the whole group, stabilizers always {1}, the singleton containing the identity

Ohh

I think I misunderstood you

Ok so you have 2 groups H and G = Aut H

So you're saying that for an element h of H, orb(h) = {h' | g ∈ G, h' = g(h)}?

Is this correct?

[oak grove]

im not sure what what you wrote means

what is the prime

[hidden haven]

And stab(h) = {g ∈ G | g(h) = h}

h' is just a variable lol

[oak grove]

oh wait

bleh

im gonna disengage from this convo I'm not gonna be any use

i appreciate your effort i wanna be able to reciprocate

[hidden haven]

Lol it's alright, sorry for confusing you further

[oak grove]

sorry for my pessimism this assignment has left me really discouraged

tomorrow is a new day

[wooden ember]

i showed that the augmentation ideal is included in the nilradical by using the previous exercise but im struggling to show the reverse inclusion, namely that every nilpotent element is in the augmentation ideal

i gotta have food will think again about it afterwards

[hidden haven]

Nilpotents must map to nilpotents under homomorphisms

[wooden ember]

yeah i thought i might be able to use that by looking at a polynomial ring and using previous results about those and when that didnt work i gave up. Ill give the general approach another thought though without polynomials thanks

i should probably be using the augmentation morphism 🤦

yeah that works obviously grrr

[wooden ember]

tried doing one direction by contradiction: assume that there is a non-unit in R and consider its principal left ideal, then this must contain a unit and so our non-unit has a left inverse but I can't show that it has a right inverse to reach a proper contradiction 🤔

[hidden haven]

You can do better, its principal left ideal contains 1

[wooden ember]

sure

[hidden haven]

And you don't need to do it by contradiction

[wooden ember]

but that still give me a left inverse

[hidden haven]

Oh right hold on lol

[wooden ember]

to clarify im trying to show the if direction

the only if direction is easy

[hidden haven]

So only 2 left ideals → division ring part right?

[wooden ember]

yeah

[hidden haven]

Given a non zero x, we know that x has a left inverse y. y is also non zero, so has a left inverse, and we already know that it has a right inverse x

Show that y is a unit

Then x must be its 2 sided inverse

[wooden ember]

ah i see

thanks 👍

[plucky flicker]

Hi guys! I would like to prove that a Tarski monster is simple group. So suppose the contrary: let N be a proper normal subgroup of G, and pick another subgroup of G let's say M. Both M and N have order of a fixed prime p. So if we take the product MN it's another subgroup of G since N is normal. And if MN has order p^2 than it is a contradiction, but can i say that it has order p^2? I mean we have the formula |MN| = (|M| * |N|) / |the intersection of M and N|, so i would need that MN is a semidirect product, which is not necessarily true

[plucky flicker]

I found the proof, it says that MN would have order p^2, which is indeed a contradiction, but i'm not convinced that |MN| is p^2, because of that formula given for |MN|. I mean, i have to prove that i can always choose another subgroup M such that the intersection of M and N is trivial (??), maybe Zorn's lemma provides this (??)

[wooden ember]

if M and N are distinct and have prime order their intersection is trivial

so indeed |MN| has order p^2 and MN = N semi M

assuming the previous steps were correct (im not familiar with the tarski monster so idk if you can always pick an M and an N as you did) your proof is fine

[plucky flicker]

yes, they are distinct and both of them have the same order p

[wooden ember]

yeah just looked up the definition of a tarski monster your proof is correct

[plucky flicker]

thanks 😄

[wooden ember]

So i just bashed through this but could I have used the fact that x-1 divides x^3-2x+1 to simplify some of the reductions? It feels like I should be able to do so but im not sure..

does this conclusion mean that if i am working with a quotient of a finite commutative ring and I show that it's an integral domain, then I automatically get that it's a field??

that sounds powerful

nevermind that just follows form the fact that every finite integral domain is a field

[hidden haven]

bash as in long division?

[plucky flicker]

yes, in general if R is not commutative maximal ideals are prime ideals, but in the commutative case these two things coincide

[wooden ember]

as in i just kept applying x^3 = 2x-1

[hidden haven]

divide the polynomial by x^3-2x+1 and the remainder will be the answer

[wooden ember]

sure that also works but that's not my point

[hidden haven]

wait prime ideals aren't maximal in all commutative rings

[wooden ember]

well rather that's not my question

[hidden haven]

hmm idk if you can do much simpler than long division

[wooden ember]

i just felt that the fact that x-1 divided x^3 -2x + 1 could maybe simplify reducing things like (x-1)^4

[plucky flicker]

i mean if R is finite, with identity

[hidden haven]

right

[plucky flicker]

in general maximal => prime => primary, radical

[wooden ember]

forget what i just wrote lmao

[hidden haven]

couldn't even read it 😵‍💫

[wooden ember]

too fast for you

[hidden haven]

what embarrassing high school story did you post

[wooden ember]

it's like pressing the enter key on my computer gives me a small burst of insight everytime

half of the time i answer my questions the moment i send them

i should just spam the enter key and see if i solve the riemann hypothesis

[plucky flicker]

One more question about the Tarski monster: the minimum number of generators. So i guess it is two. Because if a have any subgroup of G, then it has prime order, which is automatically cyclic, so it is generated by some element g_1. And if I take the subgroup <g_1, g>, for an arbitrary g in G \ <g_1> then i get the whole Tarski, am i right?

[wooden ember]

sounds right

at least the reasoning makes sense to me but i mean i didnt even know what a tarski monster was until 30m ago so dont take my word for it

[molten silo]

Need help with only if direction

[wooden ember]

applying the subgroup criteria, youll get a multiplication of the form h_1 h'_1 * h_2 h'_2. Apply the identity HH' = H'H somewhere to this expression to conclude

nvm you said you needed help with the only if direction oop

i thought you just did poor grammar 😂

[molten silo]

assuming HH' is subgroup

[wooden ember]

use inversion

[molten silo]

need to show HH'=H'H

Go on

[wooden ember]

(hh')^-1 = h'^-1 * h^-1

there's another argument to add but this should put you in the right direction

[molten silo]

hmmm

thanks

This is all i got mate

@wooden ember

[wooden ember]

perfect and the other inclusion follows symmetrically

giving you equality

[molten silo]

Doesnt work, because we can assume HH' is a subgroup not H'H

[wooden ember]

fair enough

i mean

[plucky flicker]

do the same trick for (h^-1)^-1 or smth like that

[molten silo]

true

[plucky flicker]

just a sec i will try to write this down

[wooden ember]

the way i would have written it from the start is just (HH')^-1 = H'^-1H^-1 = H'H which hides these details

[molten silo]

i think your onto something

[wooden ember]

but yeah something like what Boti said would do the job

[molten silo]

i see

you did it in one line

[wooden ember]

the way to complete your argument is to note that the inverse is a bijection on H and H'

so you dont just get inclusion but equality

the best is to work straight with the subgroups and to show that (AB)^-1 = (B^-1A^-1) and then conclude as I did

[molten silo]

yeah

[plucky flicker]

like this

[molten silo]

holy

yes

thanks

[wooden ember]

trying to do b) by showing that we arent quotienting by a prime ideal but im having some trouble

nvm i just got what they meant by preceding exercise 🤦

[dire bramble]

so I'm doing this exercise

it's straightforward showing it is in fact a chain complex

my idea for the second part is that the $H_n$ are basically extraneous information, so we define the chain homomorphism $f_n : V_n \to V_n \oplus H_n \oplus V_{n-1}$ by $v \mapsto (v, 0, \dd v)$

[cloud walrus]

bacono

[dire bramble]

this commutes since d^2 v = 0

so it is in fact a chain map, but to my understanding we need each f_n to be an isomorphism for f to be an iso in Ch(R-mod)

but even though the H_n are extraneous in the sense of chain complexes, the f_n aren't isomorphisms for each vector space

[plucky flicker]

Can someone clarify a proof? I have this problem: If H is a non-trivial normal subgroup in the finite p-group G, then the intersection of H and Z(G) is non-trivial. So i found this proof on stackexchange, and i don't see why H must contain other classes with one element.

[hidden haven]

|G| = |Z(G)| + some p^i's

[wooden ember]

the order of H is divisible by p

[hidden haven]

Intersecting with H,
|H| = |Z(G) ∩ H| + some of the above p^i's

[wooden ember]

so we need to have at least p-1 other elements with trivial conjugacy class

[hidden haven]

If the first term on the right were 1 then mod p you get 0 = 1 which is contradiction

[plucky flicker]

ohh, i see

[plucky flicker]

because then, 0 = 0 + 0 mod p

i mean

yeah

[wooden ember]

lmao why are yall hating on what i said

[plucky flicker]

i actually like it

[hidden haven]

I was there first therefore I was more correct

And the yall is really just me

[final oasis]

Let p be the smallest prime dividing the order of a group G and H a subgroup of index p then show that H is normal.
Since by Sylow's theorem there exist an element g of order p. One idea is to quotient out by <g> (but how do I know if <g> is normal)

maybe I should look at the orbit of K=<g> under conjugation by G .

[oak grove]

I have a pretty easy question but I wanted to ask for help on the specifics

So its safe to assume we have some element in the order 3 group of order 3 and some element order 5 in the group of order 5

im struggling with the mechanics of getting from that (or maybe thats not important) into an element in G has order 15

MSO is saying to use lagranges theorem, I'm not seeing how its helpful here

[final oasis]

is it iso to $Z/5Z \cross Z/3Z$

[cloud walrus]

bchaotic

[oak grove]

I'm not sure, I saw this notation in a video but wasn't sure what it meant

[final oasis]

if it is always iso then (1,1) works as generator

this is pretty much identical to a problem I am looking at

[oak grove]

no joke

I don't really see how you'd involve cosets in this

but i dont understand cosets either

[oak grove]

does this have another name than product?

I'm trying to find something related in our class notes but we haven't talked about these

[final oasis]

each subgroup is cyclic

[oak grove]

correct

[final oasis]

but are they normal

[oak grove]

why would that matter

[final oasis]

conjugation should work

[south patrol]

Basically Z/nZ is isomorphic to Z_n (or whatever notation you want for cyclic groups order n)

(Here they're quotient groups anyway)

But I think worrying about normal subgroups and conjugation is unnecessary for this problem anyway

[final oasis]

$x^{-1}Hx$ for x in G is always a group

[cloud walrus]

bchaotic

[final oasis]

of same order as H

[south patrol]

yeah nvm I agree w your method now lol, I assume you mean like

[final oasis]

since only one such H exit then H is normal

how you get that they commute

[oak grove]

so <g,h> would have to be order 15

thats good enough right

[south patrol]

Eh yeah sorry, my method would be that if x,y generate the groups order 3 and 5 respectively, then <x> and <y> are normal (by hypothesis of the subgroups order 3,5 are unique), so xyx^-1 y^-1 is in both <x> and <y> and hence is equal to e

[oak grove]

well maybe that doesnt make sense

[south patrol]

i.e. xy = yx

[oak grove]

there'd have to be an element of order 15

[south patrol]

Nah, <g,h> can be defined as the smallest subgroup of G containing g and h

<g,h> is necessarily order 15 since it has <g> and <h> as subgroups, but we don't know it's cyclic w/o more work

[oak grove]

isnt having an element of the order of a group enough to say its cyclic

i think thats the definition weve been using

[south patrol]

yes, but how do you know there is such an element yet?

[oak grove]

oh

yea i guess we dont

thats just the upper bound

[south patrol]

in my method, you can use the fact xy is such an element i guess

because (xy)^k = x^k y^k and so (xy)^0,...,(xy)^14 are different

[oak grove]

I'm having trouble understanding what you wrote

[south patrol]

Sorry yes I can explain a bit further lol

[oak grove]

okay x and y generate the subgroups

[south patrol]

Ok, so again let x generate the subgroup order 3, y the subgroup order 5

[oak grove]

im not sure what you mean we get that <x> and <y> are normal but maybe thats just some given theorem

[south patrol]

yeah, <x> is normal because for all g in G, g<x>g^-1 is order 3 and so it must be equal to <x>

and same for <y>

Now xyx^-1 is in <y> as it's normal, so xyx^-1y^-1 is also in <y>

similarly xyx^-1 y^-1 is in <x>

but the intersection of <x> and <y> is a subgroup of both <x> and <y> and must be order 1 by Lagrange's theorem

i.e. xyx^-1 y^-1 = e and xy = yx

[oak grove]

can you recommend a topic if im not following this

does it have a name

my class notes are useless 😓

oh if <x> = H with H is the subgroup of order 3 then we'd want g<x> = <x>g so g<x>g^-1 = <x>?

but then how are you sure that its order 3

[final oasis]

if g is of order k then $xgx^{-1}$ is of order k

[cloud walrus]

bchaotic

[oak grove]

but g is in G?

[final oasis]

yeah

[oak grove]

then how can we know anything about the order of g other than its 1, 3, 5 or 15

i mean wouldnt you hope its possible for it to be 15

[final oasis]

note the if

and we know g of order 5 and 3 exist from the problem statement

[oak grove]

i guess i mean im trying to understand this

"for all g in G,.... is order 3" doesnt seem to follow from what you posted

i guess yea we can be sure its not true unless i just have brain damage

[final oasis]

there exist g of order 3

[oak grove]

right

but not all of them

oh wait im just dumb

im gonna take a break

so both <x> and <y> are normal ✔️

[final oasis]

as for topics{ quotient groups, normal subgroups , kernel of maps}

[final oasis]

@oak grove u can forget the normal group and quotient stuff for now, (those came to mind first because it's similar to something I was looking at )
Just use the uniqueness of the subgroups and that 3 and 5 are primes and the order of each element most divide the order of the group and that should be enough.

[oak grove]

yea im trying to understand what a normal subgroup is atm

i dont think weve seen the other two topics since they're not mentioned in my teachers notes

half thinking of going into office hours tomorrow and asking if i take a zero on the assignment if shell explain the questions to me

[final oasis]

not needed for this problem

[final oasis]

this has everything you need

[oak grove]

im not sure how you get there through this