groups-rings-fields

[solemn rain]

why

[hot lake]

in any group G, any element g of G satisfies g^(size of G) = 1

[solemn rain]

okay

lmao yea

'so that doesnt say shit

[hot lake]

so saying that you have found an element xy with (xy)^(pq) = 1 is not a strong statement

yeah

[solemn rain]

what can i do then

[hot lake]

you need to explain why (xy)^n is not 1 if 1 <= n < pq

[solemn rain]

yea

so ic na show that that its order

right?

it cant tho?

[hot lake]

if you show that the order of xy is exactly pq and not any of its proper divisors, then it should be fine

[solemn rain]

if |xy| is n where 1<= n <pq

n must divide pq

[hot lake]

yes

so it would be either 1,p, or q

[solemn rain]

yea

wait

why XD

yea yea

nvm

p and q primes forgot

if n = 1 then its done

if n = p

[hot lake]

what does cauchy theorem say exactly ?

[solemn rain]

does it have to be unique?

idk

if it is then im done

it says that

if there is a prime dividng the order of G

the nthere exists an element of that order

doesnt pq divide pq

so im done?

[hot lake]

so it says more than x^p = 1 and y^q = 1

[solemn rain]

pq isnt prime yea lmao

nvm

if n = p

then y must equal to e

which cant be true

cuz y has order greater than q

[hot lake]

why must y be equal to e then ?

[solemn rain]

cuz x^p = 1

(xy)^p= x^p

x^py^p=x^p

okay lets recap

[fading wagon]

I think I'm done

[hot lake]

oO ?

[solemn rain]

now we want to show that there cant exist n <pq such that (xy)^n = 1

[fading wagon]

I'm trying to solve it too lol

[stone fulcrum]

That's ominous

[solemn rain]

now n must dividie pq

[stone fulcrum]

Joins the conversation
"It's finished"

[solemn rain]

so n = p or q

if n = p

[hot lake]

or n = 1

[solemn rain]

if n = 1

[fading wagon]

I shall spoiler out the sketch.
||Suppose not cyclic then all elements must have order 1, p or q. Sylow stuff, we are done.||

[solemn rain]

||sylow stuff is next section for me||

[fading wagon]

I did go overkill didn't I

[solemn rain]

if n = 1

then x and y are inverses

hence same order

which is false

[hot lake]

right

[solemn rain]

finally lmao

now

if n=p

[fading wagon]

Ah yeah that works too

[solemn rain]

then (xy)^p = x^p

x^py^p=x^p

then y has order p which is false

[fading wagon]

looks done.

[hot lake]

uuuh you deduce that y^p = 1

[fading wagon]

but since p not equals q

[hot lake]

and you know y has order q

[solemn rain]

so p must divide q which is false

yea yea

bad

[hot lake]

right

[fading wagon]

and the reverse is similar

[solemn rain]

^ yea

[hot lake]

well no

[solemn rain]

XD

[hot lake]

q has to divide p

[solemn rain]

wait

did i mix them?

[hot lake]

it's just that it depends on ho you reason about them

you know y has order q and y^p = 1

[solemn rain]

if y^p =1 and |y| = q

then q must divide p

yea my bad

so n must = xy

cool

[hot lake]

yeah now it all works

[solemn rain]

ugh i cant do any problem on my own XD

[hot lake]

eh you managed some of it there

[solemn rain]

is that bad?

[stone fulcrum]

These problems are hard and take considerable time

[solemn rain]

they are

or are u just being kind

lmao

[brisk granite]

I need this result for one of the problems I'm doing: call the splitting field of $f \in F[X]$ over $F$ $K$. Given $f$ has no repeated roots over $K$, $K/F$ is Galois.

[cloud walrus]

Field Extensions & Things....:

[brisk granite]

Could I have a hint for how to prove this?

Please let me know if the result is false btw

[somber bramble]

it’s true but I don’t remember the proof. we showed it in class. polynomial with that property is called separable and by googling I found a few stackexchange threads that outline the proof or answer questions about it

such as this one https://math.stackexchange.com/questions/3124819/splitting-field-of-separable-polynomial-is-galois-extension or this one https://math.stackexchange.com/questions/962898/on-a-proof-that-the-splitting-field-of-a-separable-polynomial-is-galois

@brisk granite

[wind steeple]

|Aut(K/F)|<=|End(K/F)|<=|Hom(K/F->Omega/F)| where Omega is the algebraic closure of F

the first = comes bc it is separable, the second bc it's a splitting field

hence it is galois

[brisk granite]

@somber bramble I was kinda hoping for a hint

Cuz I think this lemma is a big part of the question I'm doing

[somber bramble]

yea as said I can’t give you a hint cause I don’t remember the proof myself :P

but I figured since I already did some googling I’d save you the time if you wanted to look it up

or look stuff up such as properties of separable polynomials

[brisk granite]

Aight, thx

[somber bramble]

knowing the name can help

[brisk granite]

I'll try a bit harder before I look at the stack exchange Qs

Btw, is an extension finite if I'm working with finite fields or if the extension has finite basis?

@wind steeple sorry but what is end(K/F)

[wind steeple]

endomorphisms

F-homorphisms from K to K

[shrewd wasp]

can anyone here help with my problem in questions gamma?

Im doing a 2nd year abstract algebra course and I don't understand the proofs very well

[somber bramble]

Btw, is an extension finite if I'm working with finite fields or if the extension has finite basis?
the extension has finite dimension when seen as a vector space over the base field

e.g. ℚ(√2) is a finite extension over ℚ (of dimension 2)

also, we have a dedicated galois theory channel

[potent briar]

hi i have an algebra question in #help-2 :c

[fringe dawn]

How would one go about determining the Hilbert-pointcare function for K[x1,...,xm]/(x_1^d1,...,x_m^dm)

Since K[X1,...,xm] is N-graded and so is the ideal (I_n is just the homogeneous part of degree n of I) we naturally get K[...]/(...) = direct sum K[x1,...,xm]_n / (...)_n

Correct?

[long robin]

Hi mussolini

[timid storm]

Just to be sure, the group of invertible congruence classes mod 14 is the set in integers coprime to 14 right?

[delicate bloom]

the classes aren't the entire set of integers coprime to 14

but you can pick a representative in the range 0<n<14 for them

[timid storm]

Well not integers but congruence classes?

[chilly ocean]

How do I show the multiplicative group modulo 8, {1,3,5,7} is isomorphic to the group of symmetries of a non-square rectangle by identifying an isomorphism between them?

[timid storm]

Like {[1],[3],[5],[9],[11],13}?

[chilly ocean]

do I just draw out the Z//8Z* and K4 cayley tables and call it a day?

[timid storm]

All mod 14 ?

[latent anvil]

@chilly ocean that's probably easiest since the groups are so small

you can also write down an isomorphism between them

the groups are very symmetric, so you can actually choose any bijecton between the nonidentity elements

[leaden schooner]

Im learning about constructible numbers from pinters abstract algebra, and he says that we associate every constructible point P(p_1, p_2) with a field extension Q(p_1, p_2)

Can someone explain to me why this makes sense?

I dont really see the connection between polynomials and constructing the point in a finite number of steps

[chilly ocean]

@latent anvil is there a more elegant way to do this?

like define a function that sends Z/8Z* to K4?

what would the function look like if there is one?

[shrewd halo]

flip on one axis, the other axis, both axes

would be 3,5,7

[chilly ocean]

Does this prove anything?

[scarlet estuary]

,rotate

[scarlet estuary]

uhhh

are you trying to prove Z/8Z is isomorphic to the klein group?

im confused

[latent anvil]

they're trying to prove (Z/8Z)^* is iso to the klein four group

but presenting the klein four group as rotations of a nonsquare rectanlge

[scarlet estuary]

oh wow im blind

but uh i dont think you proved it

in generality

[latent anvil]

@chilly ocean any bijection that sends 1 to the identity will work

because these groups are extremely symmetric

[scarlet estuary]

yeah

but im assuming they dont have theorems on that if they're just learning isomorphisms

[latent anvil]

oh yeah I don't think so either

[scarlet estuary]

and they havent proved, say, that $\phi(3 \cdot 3) = \phi(3)\phi(3)$

[latent anvil]

but it will work

[cloud walrus]

Namington:

[scarlet estuary]

i mean this is trivial exhaustion of cases

and utterly tedious

[latent anvil]

yeah

[scarlet estuary]

but if youre gonna use exhaustion by cases, you have to cover all the cases

¯_(ツ)_/¯

[latent anvil]

I would write out the tables

probably label things by a,b,c and x,y,z and forget about the actual elements

@chilly ocean I don't suppose you know about internal direct sums/products?

that would make this easier

[chilly ocean]

@latent anvil we did

do that

[latent anvil]

oh

nice!

use that

[chilly ocean]

like idk how this is supposed to look

can i label 1,3,5,7 as a1,a2,a3,a4?

or what

[shrewd halo]

Side note: this discussion makes me wonder if 4 is as big as you can get a group so that its automorphism group is S_G-1. Feel like the answer has got to be yes

[latent anvil]

@chilly ocean what do you know about internal direct sums?

[chilly ocean]

i did learn about inner automorphisms?

[chilly ocean]

Is that The same thing

[vestal snow]

I have a question about free modules on a set

If F is a free module over A, does that mean every non-zero element of F has unique r_1,...,r_n in R and a_1,...,a_n in A such that x = r_1a_1 + ... + r_na_n or is it something like unique r_1,...,r_n in R and a_1,...,a_n in A such that x = r_1 f(a_1) + ... + r_nf(a_n) where f is an injective map from A to F?

[woven delta]

No @chilly ocean

[latent anvil]

Wait then what did you mean by "we did"?

@vestal snow what are R and A?

Like, how are they related?

Oh is A the set on which F is a free module? And R the ring is a module over?

[vestal snow]

R is the ring of which F is a module of

[latent anvil]

Sorry, I tend to use A for a ring

So it really comes down to definitions

[vestal snow]

Yeah I guess

[latent anvil]

Like, you might say F is free on a chosen basis for F

In which case it's the first

Or you might define it as a set of formal sums

In which it's also the first

[vestal snow]

Though I really like dummit and foote, they don't tend to formalize important stuff like the notion of equality of a given set

Which is what ends up confusing me a lot

So here's a question for you

For corollary 7, how exactly would you prove it?

I did something like let (F_1, +, *) and (F_2, +', *') be two free modules of set A over the ring R.

Define g:F_1 to F_2 as
g(summation r_i * a_i) = summation' r_i *' a_i

[latent anvil]

How did they define "free on A"?

Using a universal property?

[vestal snow]

I used summation and summation'to distinguish between the addition of F_1 and F_2

The universal property is the second part of the corollary

[latent anvil]

Oh lol

[vestal snow]

They defined F to be a free module over A if for all non zero x in F there exists unique r_i in R and a_i in A such that x = summation r_i a_i

[latent anvil]

Ah okay

Then yeah, your proof is what I'd do

[vestal snow]

But this is D&F we're talking about so the definitions tend to be more liberally used upto isomorphism if that makes any sense

[bleak abyss]

Did someone say D&F

😡

[vestal snow]

I'm guessing you're not a fan?

Is a free module alternatively defined using the universal property?

[latent anvil]

Is A a subset of F?

[vestal snow]

Yes

[latent anvil]

then d&f's definition is fine

If F is a free module over A, does that mean every non-zero element of F has unique r_1,...,r_n in R and a_1,...,a_n in A such that x = r_1a_1 + ... + r_na_n or is it something like unique r_1,...,r_n in R and a_1,...,a_n in A such that x = r_1 f(a_1) + ... + r_nf(a_n) where f is an injective map from A to F?
@vestal snow
The second definition isn't right

you can think about a pair (F, f) being free on A and use the second definition though

[bleak abyss]

D&F feels like it's written by someone who despises algebra but had to write a book on it anyway

[vestal snow]

@latent anvil do you need the definition to have A as a subset?

Because every element of a of A is 1•a

[latent anvil]

What does 1.a mean?

Are you thinking of F(A)?

[vestal snow]

Module operation

In F

[latent anvil]

If A isn't a subset of F, there's no module structure on it

[vestal snow]

Oh yeah you're right

@bleak abyss I don't think it's that bad lol

[latent anvil]

What's with the usernames?

[vestal snow]

But yes, I feel it does skip important things like the notion of equality when defining a set

[latent anvil]

Why are they lime?

[scarlet estuary]

see #changelog

[worn rune]

Solving advanced systems by subtraction

the book give me 4 examples

but obviously with no teacher i dont understand shit ;0

[mild laurel]

This isn't the right channel

use #prealg-and-algebra

[worn rune]

OOHO

mb

[thorn trench]

Having trouble understanding this proof of cantor-schroder-berstein theorem: https://web.williams.edu/Mathematics/lg5/CanBer.pdf. Can someone explain lemma 2* or give some hints? Thanks!

[golden pasture]

may be better to put in some set theory/logic channel

but B1=f(A), A1=g(f(A)), B2=f(g(f(A)))...

[solemn rain]

Let G = <x> a cyclic group of order n. For n = 2 , 3 ,4 ,5,and 6 write explicitly the elements of Aut(G)

solution:

at n = 2

G = {x^0 , x}

let phi be an element in Aut(G)

phi(x) = x^a

for a = 0 or a=1

at a =0 phi(x) = 1

at a = 1 phi(x) = x

hence we have 2 automorphisms in Aut(G)

at n= 3

phi(x) = x^a for a = 0 , a=1 or a= 2

( as G = {x^0 , x , x^2} )

phi(x) = 1

or phi(x) = x

or phi(x) = x^2

is this right reasoning ?

continuing with the other n values?

[stone fulcrum]

@solemn rain
The automorphisms don't have to be in the form φ(x) = x^a

However, they DO have to match generators to generators

[solemn rain]

why

isnt phi from G to G

and every element in G must be in the form x^a

for osme integer a?

[teal perch]

Yes, but not all of the elements you map x to generate G

[stone fulcrum]

In the case of n = 2, there's two possible functions:
f(0) = 0, f(1) = 1
Or
f(0) = 1, f(1) = 0

The first one is an identity, so clearly an auto. The second one fails, because it maps 1 to 0 and 0 is not a generator

I'm using mod 2 notation, to be clear

[solemn rain]

oh

yea yea

mb

i forgot |x| = |phi(x)|

[teal perch]

yeah and you need to consider the constraints on your exponent a that make x^a a generator

[stone fulcrum]

Actually perhaps the second function does work? Hol up

[solemn rain]

0 is not a generator

[stone fulcrum]

φ(0 + 0) = 1 ≠ φ(0) + φ(0)

Yes okay excuse my brain fart

[solemn rain]

i sitll dk wwhy isnt phi(x) = x^a

😦

what did i do wrong?

[teal perch]

it is!

but not all values of a give you a generator

[solemn rain]

kayenx told me no

😦

yea

[stone fulcrum]

Are they all of that form? Mb

[solemn rain]

so i had to double check

that every element

maps to an element of same order

right?

i didnt do that XD

[teal perch]

yeah but you also have to check that it sends a generator to a generator

[stone fulcrum]

Yes, it's necessary in an isomorphism that orders match up

[solemn rain]

yea yea

okay got it boys

now another problem

[stone fulcrum]

Which matching generators guarantees

[solemn rain]

let G be a group and H be a subgroup

define * GxH ---> H , g*h = ghg^-1

[teal perch]

classic

[solemn rain]

define phi_g (h) = g*h for a fixed g

g*h = ghg^-1

is an automrophism

of H

for fixed g

[stone fulcrum]

φg(x) is naturally a conjugation of x by g

[solemn rain]

define f : G---> Aut(H) , f(g) = phi_g

f is a homomoprhism

ker(f) = { g | f(g) = idmap } = { g | ghg^-1=h } = C_G(H)

so G/C_G(H) is isomorpihc to img(f)

okay

so im right with all this now right?

[latent anvil]

Yeah

[solemn rain]

okay

now

[latent anvil]

So got hang on

[solemn rain]

from this

[latent anvil]

*wait

[solemn rain]

okay

what

[stone fulcrum]

Did you want to prove that f was a homo?

[latent anvil]

No

[solemn rain]

nah fuck the details

[stone fulcrum]

Or that's given

[solemn rain]

there is another problem

its easy i think

[latent anvil]

Yes

[solemn rain]

the problem is this

[latent anvil]

Sorry I was being dumb

[stone fulcrum]

It's very easy yeah. Go on

[solemn rain]

okay okay

now

[latent anvil]

I thought you were looking at the kernel of phi_g

[solemn rain]

okay

so now

so G/C_G(H) is isomorpihc to img(f)

right?

of first isomorphism theorem

now

[latent anvil]

Yes

[solemn rain]

why is it that N_G(H)/C_G(H) is isomorphic to

a subgroup of aut(H)

?

[teal perch]

ooh that sounds like another iso theorem

[solemn rain]

i dont like other iso theorems bth

only first and thirdc

[stone fulcrum]

What's C_G(H) here?

[solemn rain]

centrlizier

[stone fulcrum]

Okay that's to be expected

[latent anvil]

C_G(H) = C_(N_G(H))(H)

right?

[solemn rain]

ig yea

[teal perch]

yeah anything that centralises is also a normalizer

i think

[solemn rain]

ye aayeaa

[teal perch]

lemme check that rq

[solemn rain]

okay

[teal perch]

yeah its right

[latent anvil]

Okay, so apply the thing you said above but considering H as a subgroup of N_G(H)

Since it's normal in N_G(H)

[solemn rain]

how cna i apply them

G/C_G(H) is isomorphic to img(f)

if G = N_G(H)

[latent anvil]

Take your new G to be N_G(H)

[solemn rain]

ye ayea

okay

now

how can i go from this

to G/Z(G) is isomorphic to a subgroup of Aut(G)?

[teal perch]

H=G

[latent anvil]

What's C_G(G)?

[solemn rain]

Z(G)

[latent anvil]

And G is normal in G

[solemn rain]

this is to omuch and getting confusing XD

[teal perch]

go back to the very first 1st iso thm part

and take ur subgroup to be H=G

bc G is a subgp of itself

and just rewrite that result

replacing C_G(G) w Z(G)

[solemn rain]

okayy

okay okay

now here is the problem

Let G be a group of order 1575. Prove that if H is a normal subgroup of order 9 in G then H <= Z(G)

help please 😦

[teal perch]

Have a look at the results you just proved

and see what they mean when H is normal

like whats the centralizer of a normal subgp

[solemn rain]

N_G(H)/C_G(H) is isomorphic to some subgroup of Aut(H)

if H is normal

then N_G(H) = H

[latent anvil]

no

@teal perch what do you mean? The centralizer can be pretty much anything afaik

[solemn rain]

why no shamrock?

[teal perch]

yeah its everything

[solemn rain]

the elements of H are all normal in G no?

N_G(H) or C_G(H)?

[latent anvil]

No, you're thinking of the normalizer

Euclid

the centralizer being everything means it's contained in the center

[teal perch]

ah shit ur right my bad

ignore me i just woke up

[latent anvil]

Anyways, the conclusion is equivalent to C_G(H) = G

[solemn rain]

uh oh why

C_G(H) = { h | ghg^-1 =g }

gh=hg*

[latent anvil]

Right

[solemn rain]

ghg^-1 = h

thats not always true for normal subgroups no?

[latent anvil]

No it's not

I'm claiming that H <= Z(G) iff C_G(H) = G

Say it outloud

The elements of H commute with everything iff the stuff which commutes with the elements of H is everything

[solemn rain]

damn im that stupid XD

yea

i got it

[latent anvil]

So you want to show G = Z(G)

Sorry

G = C_G(H)

Equivalently, G/C_G(H) is trivial

[solemn rain]

okay

[teal perch]

@latent anvil isnt that what i was saying

that means C_G(H) is everything if H is normal

[solemn rain]

N_G(H)/C_G(H) is isomorphic to some subgroup of Aut(H)

[latent anvil]

Weren't you saying that the centralizer of a normal subgroup is automatically everything?

@solemn rain N_G(H) = G iff H is normal

And we know H is normal

[solemn rain]

okaay so i gotta know stuff

about Aut(H)

how can i do that

[latent anvil]

I'm not sure actually

1575 = 3^2 * 5^2 * 7

Which isn't great

[solemn rain]

|G|/|C_G(H) must be 1

XD

[latent anvil]

Why?

[solemn rain]

cuz its trivial?

[latent anvil]

Yeah, but that's what we're trying to prove

[solemn rain]

i mean

i wanna show that

ye yea

tbh idk at all

how to do it

im stupid

and all these ficing corralrlys

are so fucking confusing lmaoa

[latent anvil]

It feels like we don't have much control over the centralizer

Oh hang on

It contains the center of H, right?

And H is actually abelian

Since groups of order p^2 are abelian

[teal perch]

H is also a Sylow 3 subgp

[solemn rain]

idk sylow yet

[teal perch]

shock horror

[solemn rain]

what is order h again?

H*

[teal perch]

9

[solemn rain]

okay

H is abelian

cuz H/Z(G) is of prime order then cyclic then H is abelian i think

if H si abelian do ik anything about Aut(H)?

[latent anvil]

H/Z(G)?

[solemn rain]

Z(H)*

XD

sory

[latent anvil]

I don't see how that applies here

Well I mean it does

[solemn rain]

no its just to show

[latent anvil]

But you need more about Z(H)

[solemn rain]

its cyclic

i think

ugh nvm about this

okay i believe u XD

H is abelian

[teal perch]

Z(H)=H if H is abelian

[solemn rain]

is Aut(H) isomorphic to anything

[latent anvil]

Do you know that p-groups have nontrivial centers?

[solemn rain]

^ yes

by class equation

[latent anvil]

Oh lol I'm stupid

Aut(H)

Not Aut(G)

[teal perch]

how does that come before Sylow???? wild

[latent anvil]

Okay we're done lol

Aut(H) can be computed explicitly

[solemn rain]

@teal perch idk

howww

[latent anvil]

@teal perch that's how my course does it

@solemn rain do you know what the groups of order 9 are?

[solemn rain]

no i dont think i can classify shit

thats next chapter

[latent anvil]

@teal perch how do you prove sylow without it?

@solemn rain do you know what an internal direct sum is?

[solemn rain]

no

thats next chapter

[teal perch]

tbh i dont remember

[latent anvil]

Direct product, sorry

[teal perch]

i did it two years ago lmao

[latent anvil]

Oof

[solemn rain]

ig aut(h) is isomorphc

to some product eh?

[latent anvil]

I run an undergrad algebra reading course

so I saw it like a month ago

Actually like 2.5 months ago

quarantine got my sense of time fucked

[solemn rain]

okay so how cna i do it

[latent anvil]

I'm not sure. You need more info about Aut(H)

[solemn rain]

should i just check solutions?

[latent anvil]

No

[teal perch]

naughty! no

[latent anvil]

you should think about it

[solemn rain]

fuck

well

by just wandering around the text

in a chapter that i dont htink i shouldd see

[latent anvil]

So you know G/C_G(H) embeds into Aut(H)

[solemn rain]

Z_m X Z_n is isomorphic to Z_gcd(m,n) XD

i think thats chinese

[latent anvil]

That's wrong

[solemn rain]

somethign somethign

[latent anvil]

lol

[solemn rain]

oh okay XD

i saw wrong then

[latent anvil]

And C_G(H) contains H

Since H is order 9

And so abelian

So |G/C_G(H)| divides 5^2 7

And it divides |Aut(H)|

If you can show those see coprime, you're done

[teal perch]

(Z_m x Z_n =Z_mn iff gcd(m,n)=1 in case u were wondering)

[solemn rain]

if is how these are coprime

then |aut(h)| must be one

hence trivial

?

just so i understand

[latent anvil]

What? Why?

[solemn rain]

So |G/C_G(H)| divides 5^2 7
why

[latent anvil]

It would tell us G/C_H(G) is trivial

Not Aut(H)

[solemn rain]

order of C_G(H) musut be bigger than 9

?

[latent anvil]

@solemn rain Lagrange's theorem and the fact that H is contained in C_G(H)

Not just bigger

It's divisible by 9

By lagrange

[solemn rain]

oh fuck

yea

[latent anvil]

So |G| = 9 * 25 * 7

And |C_G(H)| divides that

And is divisible by |H|

So |G/C_G(H)| divides |G|/|H| = 25*7

Does that make sense?

[solemn rain]

yea

[latent anvil]

But it's also isomorphic to a subgroup of Aut(H)

So it divides the order of |Aut(H)|

If we can show that |Aut(H)| isn't divisible by 5 or 7, we're done

I happen to know |Aut(H)| is either 8 or (3^2 - 1)(3^2 - 3) = 8 * 6 = 16 * 3

[solemn rain]

why

[latent anvil]

But it relies on the classification of groups of order 9 and some linear algebra

[solemn rain]

okay

okay so we are done

in both cases

5 or 7 do not divide

okaay

[latent anvil]

Well I'm done

You're not

You don't know how to prove the thing I said

[solemn rain]

yea

well ig its a badly placed exercise

[latent anvil]

Probably not

I assume there's a clever way to do it

[solemn rain]

i really think that

|Aut(H)| is just isomorphic to some shitty group

that idk about

that has the same orders u just asid

said*

[latent anvil]

Why does that make it badly placed?

Just because I'm using advanced knowledge doesn't mean you have to

[solemn rain]

maybe the 'shitty group' is some produt

product

mao therea re like 3 exercises

same idea just diff orders

thanks for helping thjo

and if ur the same guy who helped me like prove the first thing i ever did ( epimorphisms are surjective in set ) then ty again

time to finally learn about sylow theorems

[latent anvil]

I was

[solemn rain]

yea tysm

im still stupider but learening other stuff lmao

[elder valley]

i'm trying to solve this nonlinear recurrence in $\mathbb{F}p[x]$ given by $f_1 = x^2-1$ and $f_n = (f{n-1}-2^{n-1}xf_1 \cdots f_{n-2})(f_{n-1}+2^{n-1}xf_1 \cdots f_{n-2})$. any ideas?

[cloud walrus]

Auvera:

[elder valley]

p is very very large

[jolly crypt]

y'all I have yet another question to ask

So the question I was originally trying to answer was determining the value $[\mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{3}:\mathbb{Q}]$.

[cloud walrus]

4TL:

[jolly crypt]

So I fired up mathematica and what do you know?

So if anyone has insight about a) the question above or b) why this polynomial is irreducible I'd really appreciate it.

[latent anvil]

How do you know that that's a primitive element for the field extension?

[jolly crypt]

I don't. But with that mathematica calculation I do.

since the degree of the field extension is at most 9, so with that mathematica calculation it is exactly equal to 9 (9 linearly independent Q(cbrt(3), cbrt(2)) vectors)

[latent anvil]

It's easy to get a bound of 9 on the degree

[Q(cbrt(2),cbrt(3)) : Q] = [Q(cbrt(2),cbrt(3)) : Q(cbrr(2))][Q(cbrt(2)) : Q]

That first degree is bounded by 3 since cbrt(3) satisfies x^3 - 3 over Q(cbrt(2))

So if the x^3 - 3 failed to be irreducible over Q(cbrt(2)), it would have a root

By degree considerations

That root would have to be real, because Q(cbrt(2)) is contained in R

So either the degree is 9 or it's 3

I'm not sure how to do this without messy calculations

[chilly ocean]

can someone help me with this

[latent anvil]

take em away boys

By which I mean post in #prealg-and-algebra

[scarlet estuary]

what field are you working over

and how are you defining an ordering on it?

[latent anvil]

C and by magic

[scarlet estuary]

the axiom of choice is fairly indistinguishable from magic, so i feel ya

[latent anvil]

tfw you need choice to define a total ordering on something of cardinality |R|

[scarlet estuary]

hush

[chilly ocean]

You can define an ordering from model theory

Without choice

[elder valley]

how can i compute roots of unity in algebraic form? meaning i want to write the 8th roots of unity as sqrt(2)/2 + sqrt(2)/2 i and not ~ 0.707 + 0.707 i. how can i do this for say 10th roots, 16th roots, etc?

[latent anvil]

@elder valley ζ_n = e^(2π i/n) = cos(2π/n) + i sin(2π/n)

So, like, use trig identities

I'm not sure if it's possible in general

For n a power of 2 you can use the half angle formulas repeatedly

For 5 I think there's some connection with the golden ratio

[elder valley]

oh nice, didn't think of that

what about 3?

seems there's a 1/3rd angle formula but it's quite ugly

this answers my question though, thanks!

[toxic zephyr]

Is it possible to define S3 with any two elements (not including the identity)?

[latent anvil]

Maybe you can use the sum formulas inductively

Like 2π = 2π/n + … + 2π/n

Then work backwards

@toxic zephyr could you clarify?

Are you asking if it's generated by any pair of distinct nonidentity elements?

[toxic zephyr]

Yes

[latent anvil]

No

There's exactly one counterexample

Can you think of what it might be?

[toxic zephyr]

I'm going to work on it a bit and see if I can figure it out

[latent anvil]

Good idea

[toxic zephyr]

I messed around with it for a while but I couldn't figure it out :( @latent anvil

[latent anvil]

Well what possible pairs are there?

flip, flip

flip, three-cycle

three-cycle, three-cycle

Right?

@toxic zephyr

[toxic zephyr]

Yes @latent anvil

[latent anvil]

so let's go by cases

Do two flips generate the whole group? If so, why?

[merry pollen]

hi could someone explain why the extension Q(third root(2))/Q is not galois

[latent anvil]

@merry pollen what's your definition of galois?

size of automorphism group equals degree?

[merry pollen]

yeah

[latent anvil]

@merry pollen okay, so let's show that the identity is the only automorphism

Suppose α is an automorphism of Q(cbrt(2))

What can we say about α(cbrt(2))?

What we want to say is α(cbrt(2)) = cbrt(2), because that will imply that α is the identity, right?

[toxic zephyr]

@latent anvil I would think two flips would generate the whole group. I don't think I'm advanced enough to articulate quite why.

[latent anvil]

@toxic zephyr do you know Lagrange's theorem?

That's all you need to do this problem (well you could do it directly, but eww)

I guess it's not so much eww because there's only 10 pairs

[toxic zephyr]

The order of a subgroup divides the order of the group, right? I'm still really new to the subject so I don't really understand all of the implications of that. @latent anvil

[mild laurel]

(can't you use the fact that any automorphism must sent cbrt(2) to some other root of x^3 - 2?)

[latent anvil]

that's what I was trying to lead them to zoph

@toxic zephyr yeah, so the order of the subgroup generated by two flips has to be divisible by 2 and divide 6, right?

Since it contains an element of order 2

[mild laurel]

oh wait sorry, I got confused cause you were helping two people

I was wondering why Lagrange's theorem mattered

[latent anvil]

Oh lol

[toxic zephyr]

@latent anvil Yes. So it either misses two elements or generates the whole set.

Right?

[latent anvil]

What would the order be if it misses two elements?

[toxic zephyr]

4

[latent anvil]

Does that divide 6?

[toxic zephyr]

No it does not.

I see.

[latent anvil]

And the order isn't 2 because we know three distinct elements in it

[toxic zephyr]

So two flips does generate the whole set?

[latent anvil]

Yup

[toxic zephyr]

Okay I see.

[latent anvil]

So there's two more cases

2,3 and 3,3

[toxic zephyr]

I would think 3,3 would generate the set. Is it because 3+3 is divisible by 3 and divides 6? 😬

[latent anvil]

I'm not sure what your logic with 3+3 is there

[toxic zephyr]

Rip yea sorry

The order of the subgroup of 3,3 must be divisible by 3 and divide 6

Which works if it does get 6 and generate the set, right?

[latent anvil]

It's true that 6 divides 6 and is divisible by 3

But it's also true that 3 divides 6 and is divisible by 3

[toxic zephyr]

Right

How can we know that 3,3 only generates 3 elements total?

[latent anvil]

Are you sure it does?

I said there was one counterexample

Let's think more specifically about the three cycles

What are they? How are they related?

[toxic zephyr]

Could you say one is clockwise and one is counterclockwise? They go in the opposite order?

[latent anvil]

Yeah, that's a good way to put it

Have you seen cycle notation?

(if you think of S3 as the rotations of an equilateral triangle, it's exactly clockwise and counterclockwise rotation by 2π/3)

[toxic zephyr]

Yes that's the notation I'm most comfortable with

(1 2 3) and (1 3 2)

[latent anvil]

Cool

[toxic zephyr]

Oh. They appear to be inverses.

[latent anvil]

Yup!

And what does that say about the subgroup they generate?

[toxic zephyr]

The subgroup must just contain the two elements and the identity. No combination of 1, a, and a inverse will get anything else. It's order is 3.

[latent anvil]

Yup

And then the final case?

[toxic zephyr]

2 and 3. The only number less than or equal to 6 which is divisible by both 2 and 3 is 6 so they must generate the whole group.

[latent anvil]

Exactly

Lagrange's theorem is super powerful

[toxic zephyr]

It really is.

So to recap,
2,2 must generate the group because we know we have at least 3 elements, and the order of the subgroup must be even and divide 6 (so it cant be 4 because of the second condition)
2,3 must generate it because 6 is their least common multiple less than or equal to 6.
3,3 cannot generate the set because they are inverses

[woeful marten]

Excuse me

[toxic zephyr]

Is knowing they are inverses necessary, or is there another observation that could lead me to that conclusion?

[woeful marten]

What if I don’t know what math I’m doing

[chilly ocean]

@woeful marten what grade are you in?

[woeful marten]

8th

[latent anvil]

@toxic zephyr I think you need to know more concrete stuff

[toxic zephyr]

,rotate

[woeful marten]

I’m doing rewriting radical expressions

[chilly ocean]

What can I change in my proof to show that Inn(G) is a group to make it better?

[toxic zephyr]

haha

[chilly ocean]

Consider you are a professor

Let this be the proof in front of you

[long robin]

,rotate

[chilly ocean]

Suggest possible edits

[long robin]

Jesuz

[toxic zephyr]

@latent anvil Alright thank you so much for the help! 🙂

[latent anvil]

lmao

[chilly ocean]

For a perfect score

[latent anvil]

@toxic zephyr no problem! Let me know if you have more questions!

pebble, my advice is to use latex

You should also maybe say at the start what you're proving

Except I guess I would know that if I were your prof

b^(-1)a^(-1) = (ba)^(-1) and (ba)^(-1) = (ab)^(-1) are false unless G is abelian

But also, don't insert subproofs like that. Prove your lemma before starting the main problem

Say "Lemma n. Let G be an abelian group. Then (ab)^(-1) = (ba)^(-1) for all a, b in G."

[chilly ocean]

Does g have to be abelian?

[latent anvil]

no

[woeful marten]

can you pls help me

[latent anvil]

You made a mistake

You said b^(-1)a^(-1) = (ba)^(-1), right?

[chilly ocean]

Yea

[latent anvil]

@woeful marten you haven't said what you want help with

And if you're in eighth grade this is probably the wrong channel

[woeful marten]

ok

[latent anvil]

You might want #prealg-and-algebra

[woeful marten]

what channel

thx

[latent anvil]

@chilly ocean that's false in general

In fact, it implies ab = ba

Take a = (12) and b = (123) in S3. We have b^(-1)a^(-1) = (132)(12) = (13) but ba = (123)(12) = (23) and so (ba)^(-1) = (23)

If you don't know what cycle notation is, let a be a reflection of a triangle and b a rotation in D3

[chilly ocean]

I understand that

Shit

[latent anvil]

Yee

Have you heard of the socks-shoes theorem?

[chilly ocean]

No

[latent anvil]

it says that (ab)^(-1) = b^(-1) a^(-1)

The proof is just to multiple b^(-1)a^(-1) by ab

You b^(-1)a^(-1) ab = b^(-1) b = 1

Right?

[chilly ocean]

Yeah

[latent anvil]

The way to remember it is to consider the case a = put socks on, b = put shoes on

How do you undo ab?

Well you have to take your shoes off before taking your socks off

So (ab)^(-1) = b^(-1) a^(-1)

[chilly ocean]

Okay I see

I will rewrite this

thanks 😄

[chilly ocean]

@chilly ocean I don't know if someone said this but $$(ba)^{-1} (ab) = b^{-1}b^{-1}(an)$$ is only true when you're in an abelian group

Not sure what the TeX isn't working

[vestal snow]

If $M$ is a module over $R$ and $I$ and $J$ are comaximal ideals of $R$, is it true that $(I \cap J)M = IM \cap JM$

[cloud walrus]

Have a Banana Bitch:

[vestal snow]

Where $IM$ is the set of finite sums of products of the form $im$ where $i \in I, m \in M$

[cloud walrus]

Have a Banana Bitch:

[latent anvil]

Sounds fake tbh

So the left hand side is always contained in the right hand side, right?

[vestal snow]

Yeah

[chilly ocean]

it's true

they are equal

[latent anvil]

Nice

[chilly ocean]

comaximal also means I\cap J = IJ

[vestal snow]

Wait

[latent anvil]

Are you even using that here? You always have IJ contained in I cap J

[vestal snow]

Something seems weird here

[chilly ocean]

but i m using that they are comaximal

[latent anvil]

Yeah

[chilly ocean]

to justify the existence of a+b=1

since I + J = R

im using the last statement at the very end

i didn't write it there

[vestal snow]

Yeah this seems correct

Thanks

[vestal snow]

@chilly ocean I think there might be a mistake in the proof

Specifically I(JM) = (IJ)M

How do we know that this is true?

Also, is $I(JM)$ the finite sums of elements of the form $ia$ where $i \in I a \in JM$?

[cloud walrus]

Have a Banana Bitch:

[chilly ocean]

the claim is that I(JM)+J(IM)\subset (I\cap J)M = (IJ)M where the last equality follows from I and J being comaximal

I(JM) is the collection of finite sums of the form \sum a_i (bm)_i where (bm)_i\in JM

[vestal snow]

Oh okay

[chilly ocean]

but this is the same as J(IM) and (IJ)M

since \sum a_i (bm)_i = \sum a_ib_i m_i

[vestal snow]

Yeah, I guess you can prove it by definitions

[chilly ocean]

yeah cross check just in case. i haven't done this in a while

[vestal snow]

Does this seem correct?

[chilly ocean]

yeah

[chilly ocean]

@downtown#5090

[toxic zephyr]

Can we say that taking the remainder of an integer when divided by n is a homomorphism from Z+ to Z+?

[latent anvil]

What's Z+?

And a homomorphism of what kind of objects?

[solemn rain]

a --> a mod n

is a homomorphism from Z to Z/nZ ( i hope im right )

if htats what your asking

[chilly ocean]

anyone know how to simplify a kronecker delta times a tensor?

i.e, if i have $Q_{ijj}\delta_{ij}\delta_{kl}$, what does this simplify to?

[cloud walrus]

Will-:

[latent anvil]

<@&268886789983436800>

[tulip barn]

Dealt with already.

[chilly ocean]

👀

[toxic zephyr]

Must the inverse of an element in a group be unique?

[mild laurel]

Yes, try proving it

[toxic zephyr]

Is this good? or did I only prove that the left and right inverse are the same rather than that they are unique >.>

[smoky cypress]

Part of the definition of an inverse is that they work both ways

So you are good

[toxic zephyr]

Alright great thank you 🙂

[toxic zephyr]

I'm trying to make sure I understand cosets in terms of homomorphisms. The idea is that multiplying an element by something in the kernel does not change where that homomorphism maps the element? It makes me think of homogeneous solutions from Linear and DE, though I suppose in a much more simple form.

[mild laurel]

Yes that's the idea

[weak quiver]

If a is not in the subgroup <b, c> and that subgroup’s order is 28, what does that say about the order of subgroup <a, b, c>? I think the order must be 28 * the order of <a> but I’m self-teaching so hoping someone can confirm

[solemn rain]

whats <a,b,c>

[weak quiver]

Subgroup generated by elements a, b, and c

[solemn rain]

define it in set terms

[woven delta]

No, this is wrong

@weak quiver

[random crag]

lagrange;s theorem

[woven delta]

As an example, S_n is generated by an element of order 2 and an element of order n, and has order n!

@random crag what do you mean?

[random crag]

the order of a group is equal to the order of a subgroup times the index of that subgroup over the group

[woven delta]

Yes but that doesn't help

[random crag]

i guess the index isnt exactly calculable

[woven delta]

Nope

If you mess around with finite quotients of free groups you can even easily get an infinite subgroup generated by a, b, and c

[random crag]

ahh true

still it holds as a statement over cardinal numbers

[woven delta]

What do you mean?

[latent anvil]

I think they're saying Lagrange's theorem is true for infinite groups

Interpreting the index as the cardinality of the set of cosets and the product |H|[G :H] as a product of cartels

Dunno why they're saying it though

[solemn rain]

i think i got confused a bbit can soem1 recap for me?

now we do strong induction on |G| and assume the result for the less orders

now |G| = p^a(m) where p does not dividie m

if p divides |Z(G)| then we can find a normal subgroup ( since Z(G) is abelian ) to G of order p ( by cauchy )

call it N

|G/N| = |G|/|N| = p^(a-1)(m)

now im lost here

this has a sylow subgroup cuz of induction

order p^(a-1)

now what XD

why must P bar ( the sylow subgroup of G/N ) be P/N for some P ?

[toxic zephyr]

"All groups of prime order p form one isomorphism class: The class of a cyclic group of order p."
Does this mean that if any group has prime order, it is isomorphic with a cyclic group which can be generated by any one element?

[oblique river]

yes

[solemn rain]

G = <x> , H = <y>

f: G---> H , x^k --> y^k

[serene cliff]

Anybody has any ideas how to do part b? I figured out that K is an extension field and proved it in part a. But I have no idea how to do part b.

[latent anvil]

Well you know it has a root, right? So f has to factor as t - that root times a quadratic

@serene cliff

And we need t^3 + 3t^2 + 3t + 3 = (t-x)(t^2 + at + b) = t^3 + (a - x)t^2 + (b - ax)t - xb, so a - x = 3 and b - ax = 3 and -xb = 3. Then a = 3 + x and b = -3/x = 3 + 3x + x^2. Then by the quadratic formula, the polynomial t^2 + at + b factors iff a^2 - 4b = x^2 + 3x + 9 - 12 - 12x - 4x^2 = -3x^2 - 9x - 3 = 4x^2 + 5x + 4 is a square.

hmm, this doesn't look very nice

Maybe I screwed up

[oblique river]

there's a much easier way to do this since we're dealing with a finite field

over the finite field F_q, if x is a root of an irreducible poly then so is x^q

[latent anvil]

Oh lol yeah

Gal(F_p^n/F_p) is just frobenius

I mean cyclic generated by frobenius

[ripe crest]

For some reason I've having a bit of difficulty internalizing the definition of finitely generated modules

If there exists a surjective homomorphism $R^n \rightarrow M$ where $M$ is an $R$-module, then we say $M$ is finitely generated, right?

[cloud walrus]

silent flower:

[serene cliff]

@latent anvil Thank you I think I kind of understand it. g(x) would be a root and then t-g(x) is a quadratic

where g(x) = x in this case

[oblique river]

@ripe crest yes

it's finitely generated if it can be generated by finitely many elements

and those finitely many elements are exactly the images of (1,0,..,0), (0,1,0,...,0), ..., (0,0,...,0,1)

in R^n

[ripe crest]

Hmm

I've read that as well

But, how does surjectivity imply that those finitely many elements are the images of (1,0,..,0) and so on

[oblique river]

no

being finitely generated doesn't mean that there is a unique set of generators

[ripe crest]

Also, what is the action of $\mathbb{Z}$ in $\mathbb{Z}/n\mathbb{Z}$? Is it just $3 \cdot \bar 4 = \overline{3 \cdot 4}$

[oblique river]

it just means that some finite set of generators exists

the action of Z on any abelian group is just multiplication

You want \overline

but also with the braces

[ripe crest]

ah ty

[oblique river]

being finitely generated means that there is some finite set of generators {a1, ..., an}

and you should check

[cloud walrus]

silent flower:

[oblique river]

that {a1, ..., an} generates the R-module A if and only if the map R^n --> A defined by sending e_i to a_i is surjective

where e_i = (0,...,0,1,0,...,0) where the 1 is in the ith spot

[ripe crest]

hmmm

Is that map completely determined by where the e_i go?

[oblique river]

yes

[ripe crest]

Ah, so $3\cdot \bar 4 = \bar 4 + \bar 4 + \bar 4 = (\bar 1 + \bar 1 + \bar 1) \cdot \bar 4 = \bar 3 \cdot \bar 4 = \overline{3 \cdot 4} = \overline{12}$

[oblique river]

yes those are all equal to each other

[cloud walrus]

silent flower:

[ripe crest]

nice

okay, ill think about it for a bit

[latent anvil]

@serene cliff I'm not sure what you mean by g(x)?

I was thinking x itself is a root of f

But as Buncho said, if α is a root of an irreducible polynomial f(t) with coefficients in Z/7Z, then α^7 and α^49 are also roots

[toxic zephyr]

I'm trying to prove that scalar multiples of the identity matrix form the center of the general linear group. It's easy to show that they commute, but is it possible to show that no other element could form the center? Is there a way to find the center of a group directly, or is it just intuition?

[latent anvil]

It is possible to show no other element could form the center, otherwise we wouldn't know it was true

If your matrix isn't a scalar multiple of the identity, it either has two different diagonal entries or it has a nonzero entry off of the diagonal, right?

Try to find a matrix which doesn't commute with it in either case

It might be useful to think of the matrices as linear transformations

So given your (invertible) matrix A which isn't a scalar matrix, try to find a(n invertible) matrix B and a vector v such that ABv ≠ BAv

[toxic zephyr]

@latent anvil Interesting. So do centers of non-abelian groups usually have only a few forms?
Also most matrices aren't commutative, even if they are invertible, so I'm not entirely sure what I'm looking for.

[latent anvil]

I'm not sure what you mean by either of those two statements

I'm trying to give some hints as to how you'd show that the center of GL is scalar matrices

[solemn rain]

do i need big stuff to prove that groups of order p^aq^b for p and q primes and positive integers a and b are solvable?

[latent anvil]

yes

[solemn rain]

fuck

ok

[bleak abyss]

Did someone (implicitly) say representation theory?

[solemn rain]

is it me?

[latent anvil]

Yes

I didn't really understand the proof of burnsides theorem

It was really long

Maybe I should read it again

[solemn rain]

can some1 help em with proving that groups of order 200

are not simple?

the problem is asking me to show a group of order 200 has a normal sylow-5 subgroup

[latent anvil]

200 = 2^3 * 5^2 right?

[bleak abyss]

I have an excellent writeup

[solemn rain]

yes @latent anvil so a sylow-2 subgroup would be of order 8

[latent anvil]

I'd take a look at it

[solemn rain]

now i think i am confused with the whole third part thing

cuz thats how everything is proven

[latent anvil]

So the first thing to do is to try and get as much information as possible

[solemn rain]

okay

[bleak abyss]

Just look at that beautiful formatting

[latent anvil]

Let n2 count the number of sylow 2 subgroups and n5 similarly

oh my

[bleak abyss]

Yeah I had a phase where I insistently wrote like that because I was salty about having to print psets for a class

[latent anvil]

lol I do that in my manifolds course

[bleak abyss]

So I was like okay I'm gonna save paper at the cost of your sanity for not accepting emailed psets

And then it stuck

[latent anvil]

But only on problems that are like "partition of unity and do the obvious thing" because they're annoying

[solemn rain]

okay

[latent anvil]

So n2 divides 5^2, right?

[solemn rain]

idk why

^]

this is true but idk why

[latent anvil]

Well, what do the sylow theorems say?

[solemn rain]

yea the fucker just t yped that out between the letters

that n_p divides m if |G| = p^a(m)

where p does not divide m

[latent anvil]

Are you asking why that theorem is true?

[solemn rain]

i have the proof ig

i think its because n_p = |G|/N_G(P)|

where P is any representative

[latent anvil]

Yeah, exactly

Since the conjugates of P are exactly the sylows

[solemn rain]

yea

( which is cool btw )

okay

now turns out

im trash with modular arithmetic and i dont understand shit

np is congrunt to 1 mod p

[latent anvil]

So unfortunately that doesn't help with n2

[solemn rain]

why

[latent anvil]

Because each of 1, 5, 25 are 1 mod 2

Those are the possibilities for n2 since it divides 25