groups-rings-fields

[latent anvil]

saying that each entry of the left hand side is equal to the corresponding entry in the right hand side

The entries of the adjugate matrix are polynomial in the entries of the original matrix, and the entries of the product of two matrices are polynomial in the entries of the two matrices you started with

[toxic zephyr]

Okay that makes sense. The polynomial is just addition and multiplication. We could choose to "mod" the terms every time we multiply/add or just do it after all the computations are done normally. So there shouldn't be a difference.

[latent anvil]

Yeah that's exactly the proof that polynomial identities are preserved

Pretty much

So A adj(A) = det(A) I

Which means if det(A) is invertible, then A has inverse adj(A)/det(A)

This works more generally than Fp, but it does in particular answer your question

[toxic zephyr]

Alright very cool thank you.

I'm glad my linear algebra professor covered the adjoint and Cramer's rule so this stuff isn't super new. The professor whose class I tutor skipped over it :/

[chilly ocean]

Quick question:\

Let $a,b \in \mathbb{Z}^+$ and let $d=\text{gcd}(a,b)$ and $m=\text{lcm}(a,b).$ In the group $G=\mathbb{Z}_a\times\mathbb{Z}_b,$ let $M$ be the subgroup generated by $(1,1).$ Find a subgroup $H$ of $G$ with $G = M\oplus H.$

[cloud walrus]

wstayman:

[fading wagon]

why did they define gcd and lcm

@chilly ocean

they never used it

hint hint

[chilly ocean]

@fading wagon I know $\mathbb{Z}_a \times\mathbb{Z}_b \simeq \mathbb{Z}_m \times \mathbb{Z}_d$ and that ${lcm(a,b),gcd(a,b)}$ are the invariant factors for both groups, but I'm lost trying to construct the subgroup theyre asking for

[cloud walrus]

wstayman:

[fading wagon]

Well the subgroup generated by (1, 1) can only be Zm

@chilly ocean

and obviously since groups are abelian...

[chilly ocean]

@fading wagon Could I argue that since $|M| = m$ we know $M \simeq \mathbb{Z}_m.$ Then if $\varphi$ defines the isomorphism between $G$ and $\mathbb{Z}_m\times\mathbb{Z}_d,$ the subgroup $H \leqslant G$ we'd need is the image (or preimage) of $\mathbb{Z}_d$ under $\varphi$? Or is that not the right direction?

[cloud walrus]

wstayman:

[cloud walrus]

L1:

L1:

[wind steeple]

Hey, let C be a category and X in C be of finite presentation if for every filtrant inductive system a, the natural morphism lim C(X,a) -> C(X,lim a) is bijective. I was asked to check that it's a good definition in Mod(A) the category of A-modules, i.e., M is a A-module of finite presentation it the sense of modules iff in categorical sense. But I've shown that if it is the case in categorical sense, then M is projective, it is a bit weird since for instance Z/2Z isn't projective but of finite presentation. Can someone check my proof or check if it is the good definition of finite presentation object in a category ?

[toxic glade]

Zak

I don't know what category this is so can you just help me here or no

[latent anvil]

@wind steeple I'll take a look at your proof

[wind steeple]

wtf, I'm looking for help

ok thanks

[latent anvil]

lol

[wind steeple]

(I was answering to Fadi)

[latent anvil]

Yeah np

[toxic glade]

its so easy but im stupid

[latent anvil]

I haven't heard of this result before btw

But hopefully I can help proofread

[wind steeple]

ok so, if you take f:X->Y a morphism, the coker is a colimit

[latent anvil]

Yup

[wind steeple]

so if we have Coker C(M,f) -> C(M,Coker(f)) bijective, it is surjective

[latent anvil]

oh I see, C(X, —) seems to preserve colimits

but there's a restriction on what colimits it preserves. I think those are satisfied for cokernels though

[wind steeple]

filtrant colimits yes

cokernels are filtrant

so, if psi : M -> Coker(f), we have psi : M-> Y such that psi = p o phi where p : Y -> Coker(f) is the canonical projection

(by definition of the natural map and by surjectivity)

[latent anvil]

Should one of those be a phi?

You wrote psi twice

[wind steeple]

then we can generalize that since every surjective map p : X -> Y, Y can be seen as a cokernel

uh yes, phi : M -> Y

this shows that M is projective isn't it ?

[latent anvil]

So if C(M, —) preserves cokernels, M is definitely projective

[wind steeple]

yes

[latent anvil]

What's your definition of filtrant?

Maybe there's a condition that doesn't actually hold?

That's the only thing I can think of

[wind steeple]

not only preservation, the surjectivity of natural morphism suffices

uh ok

in the definition there is also for any parallel morphisms f,g : i -> j we have k and h : j -> k such that hof = hog

so it doesn't work for cokernel

[latent anvil]

okay yeah that'd be it

[wind steeple]

mh I didn't pay attention of this axiom, it is pretty weird

[latent anvil]

Sorry, I thought it was saying that you're taking the limit over a directed system

But I guess that's what the inductive part is

I'm not familiar with the terminology "inductive filtrant system"

[wind steeple]

inductive it's only the fact that the functor a is covariant, or whathever it is, you'll take the colimit of the functor

okay thanks then

[latent anvil]

Also isn't the natural map coker C(M, f) -> C(M, coker f) always injective?

[wind steeple]

I don't think so, the injectivity also looks like to some projective property

the map is just [phi] -> p o phi

if you see coker C(M,f) as C(M,Y)/Im(C(M,f))

[latent anvil]

Where p is the projection Y -> coker f?

[wind steeple]

yes

mh if M is projective then the natural map is bijective btw

so surjective => injective here lol

that's funny

[latent anvil]

Yeah okay, I think I was getting confused thinking that if a map has image contained in im f then it factors through f

But that's not true

[wind steeple]

yes

[latent anvil]

The kernel of the natural map is the class of everything with image contained in im f

[wind steeple]

that's truc if you can split of the domain

yes

uh no

wait

yes xD

yes, and you want to show that you can choose linearly an antecedent

a preimage ? idk what's the name for it

for a y such that f(y) = x

[latent anvil]

A section

Well, a section is kind of wrong but it's close

[wind steeple]

yes

here we have C(M,.) commutes with cokernels iff M is projective x)

[latent anvil]

Yup. My definition of projective is that C(M, —) is exact, and it's always left exact, so this is pretty much by definition for me

in fact you should just need C(M, —) to preserve surjectivity

[wind steeple]

yes

[toxic zephyr]

Is row reduction limited/different when the entries are in F_p?

Which I suppose becomes the question 'are elementary matrices limited/affected?'

[azure grail]

Don't mean to intrude, just gonna drop my question here as well.

Isn't this trivial by the definition of a normal subgroup, and the center of a group G? I'm worried my proof is too simple/I'm missing something.

[solemn rain]

H has two elements

one of them is the identity which commutes with everything in G

the other is an element call a

you knoww gag^-1 is H for all g in G ( normality )

H = { a^0 , a }

so its easy to just check

@azure grail good?

[azure grail]

yeah I thought it was too simple, I pretty much just did that

thanks mo2

this is a bit separate, but I'm confused why we need |H| = 2. Can't this just be generalized to any normal subgroup H in G?

[steep hull]

gHg^(-1)=H just means the elements of H are permuted. The only reason why the argument works here is because e is sent to e, forcing a to be sent to a, but with more elements, this can’t be guaranteed.

[azure grail]

ahh okay I see. When I think normal subgroup, I think of gH = Hg (rather than gHg^(-1)=H), and I forgot this doesn't mean that for every h in G, gh = hg, just that gh is mapped to some h'g in Hg

[toxic zephyr]

Is it possible to be able to reason out which elements will generate the entire group of $\mathbb{F}_p^{\times}$?

[cloud walrus]

nix:

[toxic zephyr]

An exercise said to find generators for all the groups with p<20. So far I've been able to do it with 2 or 3. 2 had order less than p-1 for p=7,17 so far.

[mild laurel]

No

[toxic zephyr]

I see. A task for a computer I suppose for large p.

Thanks

[solemn rain]

whats F_p

[toxic zephyr]

Prime field. $\mathbb{F}_p=\bZ/p\bZ$

[cloud walrus]

nix:

[solemn rain]

multiplicative group?

[toxic zephyr]

Yes

[toxic zephyr]

Would it be satisfactory to prove $a^{p-1}\equiv 1 \mod p ; \forall a\in \mathbb{F}_p^{\times}$ with
$$a^{|\mathbb{F}_p^{\times}|}=1$$ since $\mathbb{F}_p^{\times}$ is a cyclic group.
$$|\mathbb{F}_p^{\times}|=p-1 \implies a^{p-1}=1$$

[cloud walrus]

nix:

[mild laurel]

Hint, lagranges theorem

[solemn rain]

if ur talking about fermats little theorem

a^(p-1) = 1 ---> a^p = a

so yea

but ur missing a case

@toxic zephyr

oh i didnt notice

well

[toxic zephyr]

Right. So a^(order of the group) has to be 1. Since there are p-1 elements, a^(p-1)=1? @mild laurel

[solemn rain]

yes

this is mod p ofc

but if ur trying it prove fermats little theorem

0 is not in your group

so you have to do this as a case on its own

its easy tho obv

[toxic zephyr]

Yeah the exercise was to prove FLT using the fact that Fp is a group.

[solemn rain]

0 is not in F_p

[toxic zephyr]

Absolutely

[solemn rain]

0 is an integer tho

so 0 should satisify FLT

so prove that

[toxic zephyr]

It said integers which are not divisible by p, which 0 is.

[solemn rain]

Fermat's little theorem states that if p is a prime number, then for any integer a, the number a^p − a is an integer multiple of p. In the notation of modular arithmetic, this is expressed as a^p = a mod p

[woven delta]

Lmao FLT

[solemn rain]

?

[woven delta]

But yeah if the order of F_p^x is p-1, then what does that mean about the order of the elements

That's how you prove Fermat's little theorem as a corrolary of Lagranges theorem

In fact this argument immediately proves Euler's theorem

[lone niche]

is this possible to prove from defn of psotive operator and that T* = -T

where T* is adjoint

[fading wagon]

-(T^2)v=T*Tv and I think you are done

@lone niche

[lone niche]

is that from applying T with some v in V

to T* = -T

[fading wagon]

but $-T^2=T^*T$ so I think from there you are done

[cloud walrus]

Element118:

[fading wagon]

@lone niche

[lone niche]

ye

thanks for clarfication

[lone niche]

how can we prove that T applied to that generalized eigen space is contained in the eigen space itself

[lone niche]

ok thanks so that shows T preserves the subspace G(lmabda, S)

[mild laurel]

No if you read it closer, it shows that its in the eigenspace itself

[brisk granite]

Are there finitely many polynomials in $\mathbb{F_{p}}[X]$?

[cloud walrus]

PolyBeanDip:

[brisk granite]

because x^p = x mod p by fermat's little theorem and coefficients can only be 0,1,2,...,p-1

[quiet cave]

It would be kinda weird if a finite field corresponded to an infinite polynomial field

[brisk granite]

Follow up: if f(x) divides g(x^p) in Z, does that mean f(x) divides g(x) in F_p?

Or more simply, does g(x^p) = g(x) in F_{p}[X]

[quiet cave]

I would say so, yeah

[brisk granite]

ok cool

[quiet cave]

Fermat's little theorem again, right?

[oblique river]

sorry to barge in but that's just false

[brisk granite]

oh ok

[oblique river]

x^p and x are different polynomials in F_p[x]

F_p[x] is infinite

because it contains x^n for all n

and those are all distinct

it's not true that g(x) = g(x^p)

but it is true that g(x^p) = g(x)^p

in F_p[x]

[quiet cave]

Hmm

[brisk granite]

why are they distinct tho? isn't it true that x^p = x mod p?

[oblique river]

no

it's true that a^p = a for all a in F_p

but x is not in F_p

[brisk granite]

oh, I see

it's supposed to a transcendental over F_p, right?

[oblique river]

yeah you can think of it like that

[quiet cave]

Thanks bananas

[brisk granite]

how do you know g(x^p) = (g(x))^p?

[mild laurel]

it comes from the fact that a^p + b^p = (a + b)^p in F_p

and the fact that a^p = a in F_p as well

[brisk granite]

oh, thanks

[oblique river]

zoph it's more than that

[brisk granite]

it comes from the fact that a^p + b^p = (a + b)^p in F_p
This part is just binomial theorem, right?

[oblique river]

it's not jsut that a^p + b^p = (a+b)^p for all a and b in F_p

it's that that statement is true in any ring of characterstic p

yes, by the binomial theorem

it's true in F_p[x] for example

[mild laurel]

okay yeah sure, I was just noting that this is the fact you need to prove the thing he stated

[brisk granite]

it's that that statement is true in any ring of characterstic p
Does every extension of a field of characteristic p also have characteristic p?

[oblique river]

yes

[brisk granite]

oh yea, that's obv lol

[maiden ocean]

ok may be im stupid but this doesnt make any sense to me :|

this book says that the ideal generated by a set $(a_1, \ldots, a_n)$ in an arbitrary ring is the elements of the form

[cloud walrus]

Sloth:

[maiden ocean]

$\sum_{i_1} x_{1 i_1} a_1 y_{1 i_1} + \ldots + \sum_{i_n} x_{n i_n} a_n y_{n i_n}$

[cloud walrus]

Sloth:

[maiden ocean]

but he says that if we let our ring be commutative, this formula just simplifies to

the set of elements of the form

$\sum_{i = 1}^n x_i a_i$

maybe im just dumb but

[cloud walrus]

Sloth:

[vocal river]

you can use the commutativity to rearrange $x_{1 i_1} a_1 y_{1_i1}$ to $x_{1 i_1} y_{1 i1} a_1$ and then denote $x_1 = x_{1 i_1} y_{1 i1}$ because the product of these elements is always a member of the ring, and do this process for all the indices to get the sum of the required form

[cloud walrus]

Jesse?:

[maiden ocean]

I understand that, but shouldnt that get you something like, $\sum_{i_1} x_{i_1} a_1 + \ldots + \sum_{i_n} x_{i_n} a_n$

[cloud walrus]

Sloth:

[mild laurel]

just use distributive property to combine the coefficients into one

[maiden ocean]

so we let $x_1 = \sum_{i_1} x_{1 i_1} y_{1 i_1}$?

[cloud walrus]

Sloth:

[maiden ocean]

ohhhh yeah that should also be in R

okay thanks

i get it

[eternal furnace]

How is (I,x) = R ?

[wind steeple]

because I is maximal

[eternal furnace]

why does that make it true

[woeful flint]

how can i prove that $sqrt(5)\notin\mathbb{Q}(5^{1/3})$

how can i prove that $\sqrt(5)\notin\mathbb{Q}(5^{1/3})$

[cloud walrus]

Kraft Macaroni:

[wind steeple]

because (I,x) is an ideal @eternal furnace

containing I

and not equal to I

since I is maximal, (I,x) must be R

[woeful flint]

One thing I know for sure is that $\mathbb{Q}(5^{1/3}) \cong \mathbb{Q}[x]/(x^3 - 5)$

[cloud walrus]

Kraft Macaroni:

[chilly ocean]

doubt that helps I think

[woeful flint]

damn

[chilly ocean]

I think its doable by playing with degree of the extension, like showing Q(5^(1/3), 5^(1/2)) is different from Q(5^(1/3))

[woeful flint]

ah that could be an idea

is it the case tho that Q(5^(1/3), 5^(1/2)) = Q[5^(1/3), 5^(1/2)]

[chilly ocean]

whats the difference?

I dont remember the notation

I meant []

[eternal furnace]

Ohhhh @wind steeple thanks I get it now

[woeful flint]

one is a ring the other is a field

[chilly ocean]

its a field right

pretty sure

idk

[woeful flint]

like [] means the smallest ring containing 5^(1/3), 5^(1/2)

[chilly ocean]

and () means smallest field?

[woeful flint]

yeah

[woeful flint]

@chilly ocean Just in case you were interested we can prove it like this. For $L \subset K$ fields we write $[K:L]$ dimension of K as an L vector space. We know $[\mathbb{Q}(\sqrt{5}):\mathbb{Q}] = 2$ and $[\mathbb{Q}(5^{1/3}):\mathbb{Q}] = 3$. If $\mathbb{Q}(\sqrt{5})\subset\mathbb{Q}(5^{1/3})$ letting $n$ be the dimension we would have that $3 = 2n$ for an integer $n$; contradiction.

[cloud walrus]

Kraft Macaroni:

[chilly ocean]

yeah thats what I was thinking

[serene cliff]

Hi guys quick question. Is the answer to this question no because 3 and 2 are in the ideal but 3 - 2 = 1 is not in the ideal?

[smoky cypress]

Yeah

[serene cliff]

Thanks!

[solemn rain]

prove that A_n is the commutator subgroup for S_n for all n>=5

i always struggle with finding explicit things for given groups

like iw oujldnt know how to find the commutator subgroup or like the center of a group especially a permutation group

any help

[bleak abyss]

Commutator subgroup is normal

Only normal subgroups of S_n for n\ge 5 are 1, A_n, and S_n

S_n/1 isn't abelian

S_n/A_n is abelian, so A_n is the commutator subgroup

[solemn rain]

oh

okay

ty

got it

so for these kind of problems im not supposed to explicity compute stuff right/

[bleak abyss]

I mean, can't say you're not supposed to explicitly compute stuff

It happens to be the case that knowing the normal subgroups of S_n for n\ge 5 makes life very easy

But like, in general S_n is generated by (12) and (12...n)

Also S_n is generated by transpositions, in fact adjacent transpositions

So in principle you could just exhibit elements of a nice generating set as commutators

[solemn rain]

cool

ty

just checking proof

prove that every non abelian simple group is perfect

proof: suppose G is a non abelian simple group

since H = <[x,y]| x is in G , y is in G> is normal in G

H must equal G

is that right?

[chilly ocean]

Is {e,(12)} a normal subgroup of S3? Is {e,(123),(132)}
a normal subgroup of S3?

is there a better way to do this than just showing for all elements s in S3 sH = Hs?

so H is the subgroup given and then I show sh1 = h2s for every element h in H?

[solemn rain]

show that ghg^-1 is in H for all g in G for all h in H

conjugating in S_n isnt too hard

phi tao phi^-1 is just phi(elements in cycle of tao)

dk how to saay that ^ xd

@chilly ocean

and u can use some theorems too

like for example for the second subgroup that you have

you can notice something

consider [G:H] where H is the second group and G is your group

i struggle too with computing stuff but im learning ig]

[stone fulcrum]

Two elements are conjugate in Sn, if they have the same cycle type

So if you can show that the subgroup of {e, (12)} has an element that isn't just a transposition, then it's not normal

[solemn rain]

[stone fulcrum]

Wait I've got something mixed up here

1sec

Haha yeah I did the dumb

[latent anvil]

if you can show there's a transposition which isn't in there

normal subgroups are unions of conjugacy classes

[stone fulcrum]

Two elements are conjugate iff they have the same cycle type

So (23) can be made by conjugation, but that's not in the subgroup

[solemn rain]

to find the yccle type you msut have

distinct cycles right?

[stone fulcrum]

Yeah something like (13)(23) doesn't count

Write it as (132) instead

[solemn rain]

so this has cycle type 3

[stone fulcrum]

And is conjugate to (123) but is not conjugate to (12)

[solemn rain]

yea

so yea

(123)(12)(321) isnt in H so its done

@chilly oceans

[chilly ocean]

thank you for your advice

i used the first possible input (123)(12)(132) to show the 1st sub group was not normal

and for the 2nd subgroup showed [s3:H] = 2 and H <= S3

so it was a normal subgroup of S3

[solemn rain]

yes

gj

[woeful flint]

a bit stuck on this. I have to prove that $\overline{\mathbb{Q}}$ is countable and that therefore transcendental elements exist. Defining $\overline{\mathbb{Q}}n = {\alpha \in \overline{\mathbb{Q}} : \deg m\alpha \leq n }$

[cloud walrus]

Kraft Macaroni:

[woeful flint]

i have shown that $\overline{\mathbb{Q}}_n$ is countable for all $n\in\mathbb{N}$

[cloud walrus]

Kraft Macaroni:

[woeful flint]

im unsure how to extend this to $\overline{\mathbb{Q}}$

[cloud walrus]

Kraft Macaroni:

[stoic rose]

$\overline{Q}$ is the union of the $\overline{Q}_n$, thus the union of a countable family of countable sets

[cloud walrus]

radiateur-man:

[stoic rose]

@woeful flint

[chilly ocean]

All abelian groups of order $n$ are isomorphic $\Leftrightarrow$ the integer $n$ is square-free.

[cloud walrus]

wstayman:

[latent anvil]

Is this a question?

Or a statement

[chilly ocean]

Have to prove the statement, my bad for not clarifying. I was able to show that if the integer $n$ is NOT square free, then we get 2 non-isomorphic abelian groups of order n, which in turn implies that all abelian groups of order $n$ cannot be isomorphic.

[cloud walrus]

wstayman:

[latent anvil]

Do you know cauchy's theorem and internal direct products?

[chilly ocean]

I know Cauchy's theorem, but not internal direct products..

[latent anvil]

hmmm

[chilly ocean]

So I have the $\Rightarrow$ direction, just need to show that if $n$ IS square-free, then all abelian groups of order $n$ are isomorphic.

[latent anvil]

Right

You know one game of order n right?

[cloud walrus]

wstayman:

[chilly ocean]

I'm not sure I know what you mean. One game of n?

[latent anvil]

Sorry, group

Not game

You know one abelian group of order n, right?

[chilly ocean]

yes, Z_n

[latent anvil]

So you want to show any other abelian group is iso to this

Here's my thought process

let n = p1…pk for distinct primes p1,…,pk

Note that Z/nZ ≈ Z/p1Z × … × Z/pkZ

So my instinct it to try and "pull off" a prime factor at a time

And use induction

Does the general idea make sense?

[chilly ocean]

Couldn't we just say that since gcd$(p_1, \ldots, p_k) = 1, \ \mathbb{Z}{p_1} \times \cdots \mathbb{Z}{p_k} = \mathbb{Z}_n,$ which is cyclic. And then since "all abelian groups of order $n$ are isomorphic" $\Leftrightarrow$ "all abelian groups of order $n$ are cyclic, we can conclude that $n$ is square-free?

[cloud walrus]

wstayman:

[latent anvil]

Don't you already know that if all abelian groups of order n are isomorphic then n is squarefree?

I thought you proved the contrapositive of that above

[chilly ocean]

oh yeah duh

I meant we can conclude that all abelian groups of order $n$ are isomorphic. (If we assume $n = p_1p_2\cdots p_k,$ where each $p_i$ is a distinct prime)

[cloud walrus]

wstayman:

[chilly ocean]

I'll write it up nicely and post it here to see if you agree with it or not

[latent anvil]

Why can we conclude that?

[chilly ocean]

If $|G| = n$ and prime factorization of $n = p_1p_2 \cdots p_k$ doesn't include any prime powers greater than 1 (i.e., n is square-free), then we know $G \simeq \mathbb{Z}{p_1}\times\mathbb{Z}{p_2}\times\cdots\times\mathbb{Z}_{p_k} = \mathbb{Z}_n$ (since gcd($p_1,p_2,\ldots,p_k) = 1). $ So then $G \simeq \mathbb{Z}_n$ which is cyclic, and we know that if this is true for every abelian group $G$ of order $n,$ then all these abelian groups of order $n$ are isomorphic, (by something I proved earlier in the assignment)

[cloud walrus]

wstayman:

[latent anvil]

Why we we know G is iso to that product because the gcds are 1?

[chilly ocean]

we know Zn is iso to the product since the gcds are 1.

[latent anvil]

Right

But why is G iso to it?

[chilly ocean]

Isn't every abelian group a direct product of it's Sylow p-subgroups? (I may be misunderstanding something, I'm very new to the Sylow Theorems)

[latent anvil]

oh

Yeah that's true

I didn't realize you knew that, sorry

[chilly ocean]

and those sylow p-subgroups are isomorphic to Zp_i, etc.

[latent anvil]

Then yeah it's immediate

[chilly ocean]

Perfect, thanks for the help!

[latent anvil]

Np

How did you prove that theorem about the direct products?

Basically the proof I has in mind was to show that result

But it involves internal direct products

[chilly ocean]

Do you mean just the theorem of how every finite abelian group is a direct product of its Sylow p-subgroups? We

[latent anvil]

Yes

[topaz solar]

i mean, couldn't you also just show that product of Z/p is the same as Z/ prod p directly w/o throwing sylow at it?

[chilly ocean]

proved it on an earlier assignment and can make use of previous theorems after we've proved them

[latent anvil]

No darkrifts this is for an arbitrary G

Yeah I was just wondering how you did it on the previous assignment

[topaz solar]

oh the G iso to the prod yeah you could throw sylow at it

[latent anvil]

Sorry, I don't mean to pry

[topaz solar]

i was thinking it was for cyclic iso to prod, ignore me am rart

[chilly ocean]

All good, here's a picture of the problem in question if anyone is interested:

[unkempt ridge]

I have two variables:

1. Number
2. Speed
   Each iteration speed is multiplied by speed coefficient which is 0.90 and adding to the number.
   (Sorry for my poor english)

What i know:

1. Iteration count which is 11 in example down below
2. Sum which is 152.48356222222225 in example down below

7.747649999999999 + 8.6085 + 9.565290000000001 + 10.6281 + 11.809800000000001 + 13.122 + 14.58 + 16.2 + 18 + 20 + 22.22222222222222
sum = 152.48356222222225

you can see here what im trying to do is finding first speed(7.7476499999999) or last speed(22.22222222222222)

[sharp sonnet]

this is neither abstract algebra nor analysis

[scarlet estuary]

hey please help with my algebra problem

d/dx (x^2) = 2x, solve for d

[sharp sonnet]

,w solve d/dx (x^2) = 2x for d

[cloud walrus]

Display results online and refine query

Requested by Lochverstärker#5585

[sharp sonnet]

interesting

also, you're welcome

[solemn rain]

whats your definitions?

[dawn gorge]

Well by definition, the identity element is in H and every element has an inverse

Ah yes associativity is obvious too

By definition yes

How do u define the subgroup of H

I mean G

Well do u get why it is associative?

I’m not sure of the top of my head how to prove that it has the identity element because that’s just the definition I learnt

[chilly ocean]

the operation is associative, and its the same operation so of course its associativity is inherited

[dawn gorge]

Yes

[chilly ocean]

and to show that it contains the identity, use closure under multiplication

if x and x^-1 are in H,

[chilly ocean]

Quick question:\

Suppose that $H, N \leqslant G$ with $N$ a normal subgroup of $G.$ Prove that if $H, N$ are solvable groups, then so is $HN.$

[cloud walrus]

wstayman:

[latent anvil]

@chilly ocean do you know anything about the group HN?

[chilly ocean]

I think I've worked it out. I will post a proof here

Since $H \cap N$ is a normal subgroup of $H$ and $H$ is assumed to be solvable, we know $H/(H\cap N)$ is also solvable. Since homomorphisms preserve ``solvability," it follows that $H/(H\cap N) \simeq HN/N$ is solvable. We have that $N$ is solvable and $HN/N$ is solvable, so it follows that $HN$ is solvable.

[cloud walrus]

wstayman:

[latent anvil]

yeah, that's what I had in mind

[chilly ocean]

Took me a while, I was trying to work with the "chain" definition of solvability for a while and didn't get anywhere

[solemn rain]

i had something too with solvable groups

[chilly ocean]

Yeah I have a few more questions involving them as well

[solemn rain]

let G be a finite group such that if for all divisors ,n, of order |G| where (n,|G|/n) = 1 there exists a unique subgroup of order n then

G is solvable

[chilly ocean]

I don't quite understand. Does n divide the order of G? If so, (n,|G|) = n (assuming you're denoting GCD). Although, I think I'm just totally misunderstanding you

[solemn rain]

oh sorry

(n,|G|/n) = 1

not |G| lmao

proving easy shit is hard for me too

quotient groups of solvable groups are solvable

is hard abit for me

[chilly ocean]

So n divides |G| and GCD(n, |G|/n) = 1 implies G is solvable is the question?

And yeah same, last semester of algebra for me. I'm an analysis person

[solemn rain]

for each divisor of |G| , call it n, such that (n,|G|/n) = 1 there exists a unique subgroup of order n

if this happens G is solvable

lmao its even an iff statement

iirc no need for uniqueness

let me just get it cuz i buthcered it

The finite group G is solvable if and only if for every divisor n of IGI such that (n,|G|/n) = 1, G has a subgroup of order n.

[latent anvil]

Weird

So n and |G|/n are coprime iff n is a product of maximal prime power divisors dividing |G|

Which makes me think about sylows

[chilly ocean]

Ohhh I was actually just looking at this theorem. Was going to use it to prove something else. Something along the lines of "If $[G:H]$ and $[G:K]$ are relatively prime, then $G = HK.$ I believe this is equivalent to what youre stating

[cloud walrus]

wstayman:

[solemn rain]

lmao sorry for butchering it then wstayman

im new to this shit XD

weirdly this theorem isnt proved in the text

so dk

[latent anvil]

So for the reverse direction

We can go by induction on the order of |G|

If G is a p-group, it's solvable

Otherwise pick a prime p and write |G| = p^k m for p not dividing m

[chilly ocean]

all good, I actually misread it the first time, wasn't your fault

[latent anvil]

Then m and |G|/m are coprime, so there's a subgroup H of order m. H is solvable by induction

Not sure where to go from here though, there's no reason for H to be normal

[solemn rain]

okay so its too hard for me id say

to prove it

[latent anvil]

Why?

[solemn rain]

cuz its not proven in the text

maybe the proof is too advanced for me

like burnsides lemma

iw ouldnt know proving this would require rep stuff

and i would have just wasted horus tryign XD

hours*

i think this is the same case here

[latent anvil]

You'd probably think of some interesting strategies when trying

[solemn rain]

the only thing ik about solvable stuff is prob feit thompson

which i dnt know how to prove

[latent anvil]

Oh wait no, I think you're probably right. Take |G| = p^a q^b. Then G satisfies the given condition by existence of sylows, so this problem implies it's solvable

So this is strictly stronger than burnsides

[solemn rain]

yea see

fuck that shit

u maybe can do it

gl with it and hf

rep theory is liek chap 15

im struggling with semidirect products overhere

what is the heisenberg group?

[iron finch]

when dealing with groups is it a valid operation to bring everything to the power of -1
for example a^-1 = xay a^-1y^-1=ay a^-1y^-1a^-1=y a^-1y^-1a^-1x^-1=e ayax=e^-1 ayax=e yax=a^-1

[oblique river]

not unless the group is abelian

in general, (xy)^(-1) = y^(-1)x^(-1)

so if you wrote xy = z

it's not true in general that x^(-1) y^(-1) = z^(-1)

what would be true is that y^(-1) x^(-1) = z^(-1)

[iron finch]

thanks 😄

is it valid to do other exponents?

like xy=z
x^4y^4=z^4?

or (xy)^4 = z^4

[stone fulcrum]

You can raise everything to the power 4:
(xy)⁴ = z⁴

But unless the group is abelian,
(xy)⁴ ≠ x⁴y⁴

[cerulean crag]

i have 3 questions I need help answering ASAP

anyone know anything about the union and intersection of intervals?

[mild laurel]

This doesn't seem like the right channel

[cerulean crag]

which would be the right one

[mild laurel]

probably #proofs-and-logic

[chilly ocean]

Quick question\

Show that if $G$ is a group with $|G| = p^nq$ where $p > q$ and $p,q$ prime, then $G$ is solvable.

[cloud walrus]

wstayman:

[chilly ocean]

It's easy to see that G has a normal Sylow p-subgroup of order p^n, but I haven't made much progress outside of that

[solemn rain]

induct on n

[chilly ocean]

anyway you could be more specific? don't mean to be rude, brain is just fried at this point

[solemn rain]

same lmao

but i mean

just induct on n

assume the result for all values less than or equal to k

show that its true for k

i had this exercise too

and i couldnt do it

and thats how i learnt to it

and apparently its from am uch stronger result

[chilly ocean]

So suppose G with |G| = p^kq is solvable for all k < n, then use that hypothesis to prove its true for n?

[solemn rain]

groups of of order p^aq^b are solvable

yea

thats how strong induction works

but u ahve to prove

the base case first

namely n=0

[chilly ocean]

okay the base case was handled in another exercise. if I write a proof up quick would you take a look at it

[solemn rain]

yea sure

i hope im good enough tho

[latent anvil]

@chilly ocean p-groups are solvable

So take a sylow

It's solvable

And as you mentioned it's normal

[chilly ocean]

I think my confusion lies in forming the chain that proves the group is solvable

[latent anvil]

Don't think of it in terms of chains

Use the result that G is solvable iff G/N and N are solvable

For any normal subgroup N of G

[steep hull]

Yeah

Idk if I should spoil tho

[solemn rain]

u did tho

👀

[chilly ocean]

Okay so if the Sylow p-subgroup (call it P) is normal in this case, I would want to show G/P is solvable and that P is solvable?

[latent anvil]

Yeah, exactly

One of those is pretty easy

[chilly ocean]

Okay that does make more sense. Where does induction come into play?

[latent anvil]

Well let's see what happens

Why is G/P solvable?

[chilly ocean]

Because |G/P| = q is prime so G/P is cyclic and therefore abelian

[latent anvil]

Yup

Now P is a little harder

P is just an arbitrary p-group

So we've reduced your original problem to "If P is a p-group, then P is solvable"

Right?

[chilly ocean]

Right, I was looking at a proof of it earlier, and could mostly make sense of it, but the induction step confused me

[latent anvil]

Oh okay

[chilly ocean]

This link in particular: http://mathonline.wikidot.com/every-p-group-is-solvable

[latent anvil]

Well, let's try to prove it

Don't look at the link

Suppose |P| = p^k

And let's induct on k

The base case is easy, right?

[chilly ocean]

yes, because it'd just be cyclic right?

[latent anvil]

Right

Well, I was thinking of k=0 rather than k=1 but it doesn't really matter

So now assume all groups of order p^j for j < k are solvable

How can we show that P is solvable?

[chilly ocean]

This is exactly where my confusion lies. I don't understand how the induction hypothesis is implemented

[latent anvil]

Sure, so if we want to apply it we need some smaller p-group

In particular, any proper subgroup or quotient of P by a nontrivial subgroup will work

So that suggests to me that we'll want to use the N & P/N condition

If we can find a normal subgroup N such that N ≠ 1 and N ≠ P, we're actually done

[chilly ocean]

Could we just say by Cauchy's Theorem that p^k divides p^n so there exists an element of order p^k in G. Does it form a normal subgroup though? Or am I not looking at this right

[latent anvil]

I think your statement of cauchy's theorem is a little off

Cauchy's theorem only gives you an element of order p

[chilly ocean]

Oh yeah you're right my bad

right

[latent anvil]

Otherwise all p-groups would be cyclic

[chilly ocean]

So you use Sylow p-subgroup of order p^k

[latent anvil]

It's a good idea though. Unfortunately the subgroup such an element generates might not be cyclic

Well the sylow subgroup is just the whole group, right?

[chilly ocean]

oh right, i read your first statement wrong. So we want to find a normal subgroup N in G where |G| = p^k such that N != 1 and N != G?

[latent anvil]

Yup

[chilly ocean]

Is the answer obvious? I might've overlooked something because I'm drawing a blank here. I know it has something to do with the center of G, right?

[latent anvil]

What can we say about the center of G?

[chilly ocean]

Its a normal subgroup of G and nontrivial

[latent anvil]

Yes

And we're looking for a normal subgroup of G which is nontrivial and not the whole group

So what happens if the center is the whole group?

[chilly ocean]

oh right, that makes sense. How do we apply the N & P/N condition once we're here?

[latent anvil]

Well we don't know that the center isn't the whole group

That could happen

[chilly ocean]

True, but then it'd be abelian, so it'd be solvable anyway wouldn't it?

[latent anvil]

Yes, exactly

Okay, so we can assume our p-group has a nontrivial normal subgroup N = Z(P) which is not all of P

And we want to apply the induction hypothesis somehow and use that to prove P is solvable

[chilly ocean]

We know Z(P) is abelian and therefore solvable, but do we know that P/Z(P) is solvable?

Or is that what we're assuming in the induction hypothesis?

[latent anvil]

Good question

What's the induction hypothesis?

[chilly ocean]

That a group G with |G| = p^k is solvable for all k < n, right? or no?

[latent anvil]

Yeah

[chilly ocean]

So our induction hypothesis that G is solvable forces P/Z(G) to be solvable? Is that what we're going for?

[latent anvil]

Right

[chilly ocean]

Due to the iff statement

[latent anvil]

Why does our induction hypothesis apply?

[chilly ocean]

Still not quite sure on that. My understanding is we're assuming |P| = p^k is solvable for all k < n. Then we can guarantee a nontrivial normal subgroup, Z(P), that is solvable, then by the induction hypothesis, we must also have that P/Z(P) is solvable because of the equivalence relation.

[solemn rain]

consider the order of P/Z(P)

@chilly ocean

[chilly ocean]

its some power of p, less than or equal to p^(k-1) wouldn't it be?

[solemn rain]

does that 'lie' in your induction hypothesis

or lay

or whatever i have bad english

[chilly ocean]

I'm not sure.. could you explain

[latent anvil]

Your induction hypothesis is on groups with order p^j for j < k

k-1 < k

[chilly ocean]

ok yes it does i see. OHHHHHHHHHHHHHHH

I see now!!

Z(P) is a P-group so our induction hypothesis applies to it as well, I was overlooking this

[solemn rain]

so are ud one?

cuz i have a problem u can help me with

[chilly ocean]

Yeah was just double checking I understood it all. It's all good. Thanks again @latent anvil @solemn rain

I can help with your question

try to at least

[solemn rain]

nvm i got it anyways

i am not getting semidirectp roducts and am jusut going to sleep on it

and work for them tmrw

ty tho

[chilly ocean]

no problem, I haven't looked much at those either.. good luck!

[solemn rain]

thanks u2

[latent anvil]

@chilly ocean you're not applying it to Z(P)

Well you could, but Z(P) is abelian and thus solvable

You're applying it to P/Z(P)

[chilly ocean]

Yeah that was a typo, meant to say P/Z(P) is also a p-group with order that falls within our induction hypothesis

[ocean magnet]

pls help

[mild laurel]

This is not the right channel

[latent anvil]

Try looking at H^2 of the keys?

[stone fulcrum]

I mean yeah nobody's got it, it's a terrible question

What is a "triple"? Do they mean three key rings interlocked? Is any possible pattern allowed?

What is "left" and "right"?

[ocean magnet]

@mild laurel what's the best channel for this?

it looks like algebraic topology to me, telling knotted rings apart

[latent anvil]

lmao

[mild laurel]

lmao

[golden pasture]

lmao

[mild laurel]

honestly the problem description is just so bad and so vague that its impossible to tell

[last ether]

Can anyone explain in English what prop 2.9 is?

[mild laurel]

do you know what it means for a sequence to be exact

[sinful zephyr]

I'm asked to show that C[[x]] is a unique factorization domain. Does this just follow from the fact that every non-zero element is a unit?

[hot lake]

if every non-zero element is a unit then it's a field !

[sinful zephyr]

All fields are UFDs?

[hot lake]

trivially so

[sinful zephyr]

tyty

[hot lake]

by the way

[sinful zephyr]

just making sure I'm understanding the definitions correctly

[hot lake]

C[[x]] is not a field !

x isn't a unit

[sinful zephyr]

oh wah

[tribal pasture]

In this notation, is i the column number? Because usually I have seen the product to be written as sum_j a_ij e_j (which translates here as sum_i a_ji e_i)

[delicate bloom]

I assume you're asking because e_i are column vectors or something?

it's technically irrelevant so long as you know what it's representing, rows and columns are just notational devices for putting numbers into boxes for computation

[tribal pasture]

Yeah I am trying to clarify the notation whether here i represents the column or the row index for a_ij

[delicate bloom]

so what are e_i

[tribal pasture]

They are just denoted to be basis vectors

[delicate bloom]

ok

so you have now a linear combination of them, one for each j

if you decide to place them as columns or rows, it is up to you for whatever is convenient

you just have to stay consistent with your choices once you make the choice

[tribal pasture]

I am trying to interpret this in terms of matrix multiplication.

[delicate bloom]

it's like asking where the positive x axis is in the real world

look in your book if you want to stay consistent to their convention

I can't answer that for you

[tribal pasture]

This is all that is said

[delicate bloom]

you should be able to see how to put it into a number box to line up with how you want it to be multiplied with matrix multiplication

[tribal pasture]

It seems like this notation wants me to think of matrix multiplication as multiplying a column vector in a matrix with a column vector e_i

Which seemed a bit unconventional

[delicate bloom]

do they use a_{ij} somewhere else

[tribal pasture]

This is just the start of chapter

[delicate bloom]

maybe they pick the backwards convention for some reason

[tribal pasture]

So I just wanted to clarify, it is backwards right, just wanted to be sure I wasn't misunderstanding anything

[delicate bloom]

write it as a_{ji}^T if it helps you

[tribal pasture]

Yep yep that's fine, I had this doubt that maybe I forgot the conventional notation

[delicate bloom]

cool

[solemn rain]

can some1 prove with me

if N is a normal subgroup , H is a subgroup of G and H intersects trivially with N

then HK = G

and i think HK = H x K for someee conditions also

can some1 help mew ith those theorems

[bleak abyss]

Well

Then HK is a subgroup of G

If you also know that |H||K| = |G| then you know HK = G lol

[solemn rain]

okay

how do i now |H|K| = |G|

@bleak abyss

[bleak abyss]

I'm pretty sure it was implicit somewhere in the theorem/problem statement and you missed it

[solemn rain]

it was a lemma yes

wwo ur genius

it said i think

anay element of G can be written as hk

for some h in H k in K

the profo was leeft as exercisee

and for some reason once is aw that i couldnt follow the rest

i was put off

when is the external product = internal

whne is HK ~= HxK

[bleak abyss]

If both H and K are normal

[solemn rain]

thats it?

col

now im going to try to define a semi direct product if nayone would correct me that would be amazing

[bleak abyss]

I've kinda got stuff to do so don't rely too much on me

[solemn rain]

nvm i cant wtf

can u like do a TLDR on semi direct products?

b4 u leave

i would really appreciate it

[stone fulcrum]

What's a semi-direct product? Lol

[solemn rain]

its like a generlization of a direct product

you define a map Pi H ---> Aut(N)

h--> f_h where f_h is the inner automoprhism induced by h on N i think

no onot H fuck

N is a normal subgroup of G and H is a subgroup of G

so now you use this function to define an operation

on GxH

i really butchered shit here so idk im sorry

[tribal pasture]

So my teacher writes p1 ≊ p2 in the second part instead of p1 =p2. One can prove that p1(s) o (f - λ) = (f-λ) o p2(s) if it was the case that p1 = p2. How can one do it for the case p1 ≊p2? where λ is the eigenvalue of f?

[oblique river]

I think you're overthinking it. If p1 and p2 are isomorphic you can treat them as the same

[last ether]

Can anyone explain in English what prop 2.9 means?

[scarlet estuary]

you've asked this multiple times but you've never clarified what's confusing you

do you know what it means for a sequence to be exact?

[stone fulcrum]

Oh you've actually bought AM

M,N are modules

Hom(M,N) is the set of all module homomorphisms between them. Hom(M,N) is itself a module over the same ring as M and N

Basically, if you have an exact sequence, you can construct an exact sequence using hom modules onto some unrelated module N. It goes backwards.

English for this is weird, lol

[scarlet estuary]

but note that these modules all have to be over the same ring

otherwise it doesnt make sense

[stone fulcrum]

The beauty here is that hom(M,N) isn't always easy to capture, but you can still easily describe an exact sequence using them

[scarlet estuary]

yeah, this proposition partially motivates the very definition of exact sequence

at least in the context of modules

[last ether]

The first line of 2.9 is just a sequence of transformations that eventually results in zero?

[golden pasture]

look at the definition of a exact sequence

it is stricter than eventually 0

So like you should be able to find what is Im v

[last ether]

So it’s a series of transformations where the image of the previous module is the kernel of the next module

[stone fulcrum]

The image of one, is the kernel of the next

Images and kernels really belong to the homomorphisms, not the modules themselves. But there's semantics here. Point being that the arrows are very specific

[golden pasture]

image and kernels are for functions not modules btw

yea

like a simple example
$$0\rightarrow2\mathbb Z\rightarrow\mathbb Z\rightarrow\mathbb Z/2\mathbb Z\rightarrow0$$
is exact while
$$0\rightarrow4\mathbb Z\rightarrow\mathbb Z\rightarrow\mathbb Z/2\mathbb Z\rightarrow0$$
isnt exact

[cloud walrus]

Ariana:

[last ether]

Hmm ok so sequence 4 is is exact only if sequence 4' is exact. So does 4' means that M, M', and M'' are all homomorphic to N?

And therefore they are all homomorphic to each other

[mild laurel]

What does it mean to be "homomorphic"

[stone fulcrum]

Homomorphism isn't an equivalence relation. Two things can't be "homomorphic"

[last ether]

Well I know in topology it means you keep a similar shape like a donut is the same as a coffee cup

[mild laurel]

What

[last ether]

Oh wait thats homeomorphism my bad

[stone fulcrum]

I said homeomorphic by accident cause autocorrect ignore that

Sorry if I confused you

But definitely know what an algebraic homomorphism is. That's important stuff

[last ether]

So Hom(M,N) just means that the operations are preserved between M and N?

[latent anvil]

Abirr, what context are you looking at this in? It seems like you're missing some background knowledge

[mild laurel]

Honestly, atiyah MacDonald can be kind of hard to read if you've never seen these things before

[last ether]

Honestly I think thats what it is. I know this textbook is like 40-50 years old so its hard to understand all the semantics

[mild laurel]

Uh, I'm not sure that second thing has anything to do with it

There's a reason it's still the most recommended commutative algebra book

[stone fulcrum]

But you'd be expected to know abstract algebra before commutative algebra ofc

[golden pasture]

Have you seen like
space of linear functions from a space to another
before

aka Hom(U,V)

if so this is basically the same thing

also yea learn abstract algebra from another book first

[last ether]

Yeah I have taken abstract algebra but that was about 1.5 years ago so im a bit rusty

[golden pasture]

is this the first time you've encountered modules or like have they appeared in the alg class
not exactly sure if the book is suitable rn

[stone fulcrum]

A module homomorphism is a function between two modules
φ : M → N

that splits over addition:
φ(x + y) = φ(x) + φ(y)

And scalar multiplication:
φ(ax) = aφ(x)

The set of ALL module homomorphisms between M to N is itself a module. We call it Hom(M,N)

[shrewd halo]

Starting to do a first pass at representation theory with a teacher during quarantine. This shit is incredible. That is all

[solemn rain]

boom gl hf

[stone fulcrum]

I tried. I can try again. I know character theory gets lit

[toxic zephyr]

How could I actually go about doing this factorization?

[latent anvil]

isn't this just gaussian elimination?

Get your matrix into reduced row echelon form

Then get the new matrix into reduced column echelon form

The only matrix in both forms is one as depicted in (a)

And applying elementary column operations to a matrix in row echelon form will keep it in row echelon form

Actually that's fake but you can be careful about it and like work left to right and then it will stay in ref

[chilly ocean]

Can anybody help me out with this?

For the following problem, I must use the definition of $K/F$ being Galois if every irreducible polynomial with a root in $K$ splits completely in $K$ and is separable

[cloud walrus]

Azu:

[chilly ocean]

The problem: Let $K$ be a field with automorphism group $G$ and assume $K/F$ is Galois

[cloud walrus]

Azu:

[chilly ocean]

Part a) is to prove that $K$ is the splitting field for a separable polynomial in $F[x]$

[cloud walrus]

Azu:

[chilly ocean]

Part b) is to prove that $F$ is the fixed field of $H$ for some finite subgroup $H \leq G$

[cloud walrus]

Azu:

[chilly ocean]

I was able to do part a)

How do I go about part b)?

[stone fulcrum]

@chilly ocean
a) is the definition of Galois

[chilly ocean]

@stone fulcrum Galois has several equivalent definitions

[stone fulcrum]

Oh shoot, no it isn't

[chilly ocean]

Part a) is ok, I finished it already

[stone fulcrum]

Oh pfft you did get it. I apologize I can't read

[chilly ocean]

Part b) is the real problem - we also aren't allowed to use the fundamental theorem of Galois theory or any of its consequences either

The assignment prompt is that for this problem we aren't allowed to use any of the results from Dummit and Foote 14.2 - 14.9

[stone fulcrum]

Wait, what's a splitting field of a group?

[chilly ocean]

Need different ideas here somehow

Wait wtf I must have typed that wrong

[cloud walrus]

Azu:

[stone fulcrum]

Oh I see

You can pretty easily show that the automorphisms that preserve F is also a group, and it would naturally be a subgroup

[chilly ocean]

Yep, but how to prove that F is exactly the fixed field?

And not just a subfield of the fixed field

[mild laurel]

This feels weird, since you can take G itself to be H

[stone fulcrum]

G is the automorphisms on K, not the Galois group of K/F

[mild laurel]

oh

Yeah in that case, you want to let H be the subgroup of all automorphisms that fix F

[chilly ocean]

This problem has stumped me for 5 hours because of the artificial condition that you can't cite results from Dummit and Foote 14.2-14.9

[stone fulcrum]

Is the fact that it might also fix larger fields really a concern?

[chilly ocean]

Yes

[mild laurel]

Yes?

Because the field fixed by H might be larger than F?

[stone fulcrum]

Hol up I need to visit my definitions

[mild laurel]

Anyways, you can use the galois property to show that its exactly F, and no more

[chilly ocean]

We're not allowed to ._.

Results from D&F 14.2-14.9 aren't allowed

[mild laurel]

Well I mean, you can write it out yourself

[chilly ocean]

By Galois property, are you referring to the definition that we have to use for this problem, or the fundamental theorem of Galois theory?

[mild laurel]

the former

[chilly ocean]

Hmm

Ok so we have that every polynomial with a root in K must split completely in K

And be separable

This led to our result in a)

We must then have $|H| = [K:F]$

[cloud walrus]

Azu:

[chilly ocean]

Let $U$ be the fixed field of $H$; then $F \subseteq U$

[cloud walrus]

Azu:

[chilly ocean]

@mild laurel not sure how to proceed

[mild laurel]

take an element in U, show there's an automorphism that doesn't fix it

[chilly ocean]

U was defined to be the fixed field of H

[mild laurel]

Ah sorry, I mean ttake an element of K\F

[chilly ocean]

Aut(K/F)?

[mild laurel]

no

[chilly ocean]

Oh, an element of K not in F

[stone fulcrum]

Got it

I'll not spoil if that's how we wanna do. The hint was good Zoph

[chilly ocean]

@mild laurel thank you

I'll try and do it now

@stone fulcrum does your solution involve the polynomial construction over all of the outputs of the element under the automorphisms of H?

[stone fulcrum]

Yes I think it does

I basically just found a way to get an automorphism for any choice of K/F

That clearly doesn't work for a member of F

[chilly ocean]

ahhh

hold on

I may have the solution ready

Let $m_{\alpha,F}(x)$ be the minimal polynomial for some $\alpha \in K\F$ in $F[x]$

Must have degree >= 2 since $\alpha \not\in F$

[cloud walrus]

Azu:

[chilly ocean]

Let $\beta$ be another root of $m_{\alpha, F}(x)$

[cloud walrus]

Azu:

[chilly ocean]

Then we can define an automorphism of $K$ that leaves $F$ fixed such that $\alpha$ gets mapped to $\beta$

[cloud walrus]

Azu:

[chilly ocean]

But this has nothing to do with the Galois property... am I missing something @stone fulcrum ?

[mild laurel]

uh, how do you know \beta exists?

or like, how do you know another root is in K

[chilly ocean]

If alpha were the only root, then alpha would have to be in F, since the minimal polynomial is in F[x], and just look at the coefficients if alpha were the only root

[mild laurel]

Uh, that doesn't really make any sense

[stone fulcrum]

Why can't it be a poly with a double root at α?

[mild laurel]

consider the extension Q(cube root of 2) and the minimal polynomial x^3 - 2

the only root of the minimal polynomial in the field is cube root of 2, the other two roots are complex

[stone fulcrum]

Oop that's a great example

[chilly ocean]

Any irreducible polynomial has to split in K

[mild laurel]

why

[latent anvil]

This is part of the definition of galois they're using

[mild laurel]

I mean yeah

I'm just trying to get them to realize where they used the galois property

[chilly ocean]

ohh, that's where I invoke Galois

haha thanks

[latent anvil]

oh sure, sorry

[chilly ocean]

god, I'm so stupid

[stone fulcrum]

Now you also need seperability

[chilly ocean]

$\psi: K \mapsto K$

[stone fulcrum]

A poly that just has a double root at α would be a problem

[cloud walrus]

Azu:

[chilly ocean]

Defined uniquely by $\psi|_F = id$ and $\psi(\alpha) = \beta$

[cloud walrus]

Azu:

[mild laurel]

it might not be unique

[chilly ocean]

oh, right... it would be unique for $F(\alpha)$ mapping to $F(\beta)$

[cloud walrus]

Azu:

[chilly ocean]

right?

But not necessarily K

@mild laurel is that sufficient for the proof?

[mild laurel]

I don't know what you mean by that

There could be many automorphisms that map \alpha to \beta

[chilly ocean]

I mean, have we established existence?

I think we have

@mild laurel again, thank you so much

It's frustrating that I wasn't able to do this relatively simple exercise

[leaden finch]

@oblique river

[oblique river]

i didn't "pick 3"

it has to be true for all x in G

so e*3 = 3
e*6 = 6
e*9 = 9
e*12 = 12

[leaden finch]

@static seal

[static seal]

9*9 = 6 mod 15, so 9 can't be an identity

also , please verify this, if you find an element that works, then you don't need to check other elements

by uniqueness of identity (for all groups) theorem

i.e. if you find that a*6 = a for all a in G

you dont need to check that there may be another identity-like element that exists

oh i kind of gave away the identity

[leaden finch]

@oblique river

[static seal]

also it is not necessary to verify 6*a = a for a a in G, since you get that for free by a theorem

the group axiom is for all ain G, a*e = a and e*a=a .

oh nevermind, we don't have to use proposition 2, since G is commutative. therefore ae = ea =a

[merry pollen]

can I ask a question

[oblique river]

also it is not necessary to verify 6*a = a for a a in G, since you get that for free by a theorem

sorry, how else would you show that 6 is the identity?

[static seal]

you already did

[oblique river]

oh I see

6*a vs a*6

yeah you only need to check one side

[static seal]

yes, a*6 = a is sufficient

right ;o

i think its sufficient because (G,*) is commutative

[oblique river]

yep

[static seal]

but there are situations, nonabelian groups, where it isn't so obvious that ae = ea

meethonoob go ahead

[merry pollen]

thanks

anyone got a copy of dummit and foote?

[oblique river]

you can find them online very easily

[merry pollen]

just asking in case anyone can go to the page Im on

I'll find an online screenshot

[static seal]

@oblique river also do you agree it is not necessary to check, if you find an element that works as an identity, for another identity-like element

so if you happened to start checking with 6, it is not necessary to check 9 or 3. since identities are unique

[oblique river]

yes

[merry pollen]

where does he get the x^3 + x^2 from

[hot lake]

a simple computation

[mild laurel]

Hm, I'm not sure its easy to calculate where you can get that answer

But it tells you to verify that its true

[hot lake]

actually, isn't f2 the mirror of f1

so it's just 1/x ?

[mild laurel]

x(x^3 + x^2) is -1 though

[merry pollen]

yeah

[mild laurel]

anyways yeah, I'm not sure in general there's an easy way to find this polynomial. It's finding a root of an irreducible polynomial in a finite field so

[merry pollen]

"simple computation"

hmm

[mild laurel]

Well that's what I was saying

The simple computation is about verifying this fact

Which you can do easily by plugging x^3 + x^2 into f_2(x)

The computation isn't about how you would find this polynomial

[merry pollen]

oh I see

[leaden finch]

can someone help me with this one

[solemn rain]

you can use the (imo easier) definition for D_2n

D_2n = < r,s | r^n=s^2=1 , rs=sr^-1 > = {1,r,r^2,.....,r^(n-1),sr,sr^2,.....sr^(n-1)>

any element in D_2n would be of form sr^k where k is in {1,2,3....,n-1}

using the relation given it should be easy to do operatiosn

[oblique river]

that's clearly not the intended solution

and I think really misses the point of the problem

[buoyant scarab]

plz help me

[oblique river]

plz post in the correct channel

[buoyant scarab]

huh

but i neeeed help

plz

[sharp sonnet]

read #❓how-to-get-help

[oblique river]

then post in the correct channel

[sharp sonnet]

i need help as well, but i don't spam the abstract algebra channel

[oblique river]

either #prealg-and-algebra , #precalculus , or one of the #❓how-to-get-help channels

[sharp sonnet]

honestly just cut out a triangle to do that exercise @leaden finch

[leaden finch]

got it loll

[solemn rain]

if H and K are normal subgroups where H intersects K is the trivial subgroup

is G = {hk | h is in H k is in K} or is that another assumption you would need

to say G ~= HxK ?

[obsidian zealot]

Need some help

[mild laurel]

what have you tried

[obsidian zealot]

i know that for it to be non abelian it is not commutative and, I also know that S4 is non-abelian

[mild laurel]

And what's the order of S_4?

[obsidian zealot]

S4 is the group of all permutations of 4 elements (so it has 4!=24 elements).

[mild laurel]

So what exactly are you unsure about?

[obsidian zealot]

Im not sure how to prove that group of order 24 is a group

[cobalt flume]

Noob question. It is proven that no algebraic formula can exist for finding the roots of a polynomial of degree 5 or higher. Are there proofs that such formulas are impossible if we include differential operations?

[stone fulcrum]

S4 is a group because Sn is a group

And S4 has order 24.
You win!

[obsidian zealot]

so it's that all to proving this question

[stone fulcrum]

I think you just brained right and found an easy example

Z24 comes to mind as well

Kind of a dumb question

[obsidian zealot]

Z24 = Z8 * Z3

[mild laurel]

@cobalt flume what exactly do you mean by differential operators

[cobalt flume]

So if there is a derivative defined on the field.

[upper pivot]

Z24 is abelian tho

altho u could do quaterion group *Z3 if you wanted ig

[cobalt flume]

Actually I think I’m mixing ideas from differential algebra. I’m going to do some searching then ask the question properly

[chilly ocean]

Is it ok if I ask for a sanity check on a problem I am currently doing?

Problem: Determine the Galois group $Gal(K/\mathbb{Q})$ of the polynomial $(x^3-2)(x^3-3)$ and all the subfields of $K$ that contain $\mathbb{\rho}$ where $\rho$ is a primitive cube root of unity

[cloud walrus]

Azu:

[chilly ocean]

My reasoning is that both factors of that polynomial are irreducible, so elements of the Galois group must permute their roots, hence the Galois group must be $S_3 \times S_3$

[cloud walrus]

Azu:

[chilly ocean]

Here is where I need a sanity check: I assert that a subfield $U \subseteq K$ satisfies $\mathbb{Q}(\rho) \subseteq U$ if and only if $Gal(K/U)$ contains no elements with orders that are multiples of 3

[cloud walrus]

Azu:

[chilly ocean]

This is because $\rho$ would have to be fixed by all elements of $Gal(K/U)$; the elements of $Gal(K/\mathbb{Q})$ that permute $\rho$ have orders of multiples of 3

[cloud walrus]

Azu:

[chilly ocean]

I could be wrong

[bleak abyss]

Well the Galois group permutes the roots but it doesn't necessarily apply any permutation to them

[chilly ocean]

wait, no...

[stone fulcrum]

It helps to know what the roots are, then it should be clear which can be swapped with what

[chilly ocean]

The roots of $x^3-2$ are ${^3\sqrt{2},\rho(^3\sqrt{2}), \rho^2(^3\sqrt{2}})$

[cloud walrus]

Azu: