groups-rings-fields

[hidden haven]

I wanted to become a writer

[daring ibex]

i have this crackpot thing

[north widget]

atleast physics has a story behind every fucking discovery

[hidden haven]

You just crushed my dreams

[north widget]

same with chem and bio

math is for noobs wtf

crackpot@!

[daring ibex]

where i try to brute force visual intuition for algebra

sometimes it works

[hidden haven]

And there go my mathematician dreams too

[daring ibex]

you should've given up a long time ago loser

[hidden haven]

[north widget]

ill take that advice

im literally stupid

[hidden haven]

You don't even have active role mirza

[north widget]

and mathematically

[daring ibex]

tbh

[hidden haven]

You don't get to talk back

[daring ibex]

it isn't that bad

im pretty stupid too lmao

[north widget]

no

im stupider

in math ofc

i am a noob

[solemn rain]

yo moldilocks

you good with func anal?

[hidden haven]

yo

Not really but I have taken a course

[daring ibex]

[solemn rain]

cna u check out #advanced-analysis

[hidden haven]

Don't remember much tho

[daring ibex]

lollol

[solemn rain]

yo and/or mirza ofc

if he/she would like to help

[hidden haven]

That was a long conversation that I did not keep track of

What part should I see?

[solemn rain]

the problemn

and my 'proposed sol'

[hidden haven]

Yeah can you repost those, discord isnt even properly loading that far up

[solemn rain]

@north widget u know func anal?

[daring ibex]

idk analysis

[north widget]

un pocito

qallil

[solemn rain]

qallil

?

[north widget]

smudgin

arabic

[solemn rain]

olayel?

[north widget]

what

[solemn rain]

arabic

[north widget]

what are you trying to say

transcription isnt perfect

[steady axle]

Let a and b be two elements of S_76 . If a and b both have order 146 and ab = ba, what
are the possible orders of the product ab?

the answer is 73 right?

[solemn rain]

ab = ba implies (ab)^n = a^nb^n

[steady axle]

yes and that tells us that ord (ab) | 146

[carmine fossil]

a and b have to be a product of a 2 cycle and a 73 cycle

[steady axle]

yes indeed and as a and b commute these cycles have to be same for both a and b right?

[north widget]

if the cycles were the same wouldnt a=b?

[steady axle]

yes. but t seems to be the only possibility. also the problem doesn't mention a cannot be same as b

so idk

[rustic crown]

a could also be b^-1

[hidden haven]

It could have any power of the order 73 cycle

[rustic crown]

(yea i was just saying that a need not be b)

[warm holly]

Yo, if $G,H$ are groups, how do you show properly that $Z(G\times H)\cong Z(G) \times Z(H)$? Literally Z(G\times H)=Z(G) \times Z(H)$.. Can we simply use an identity application to show that they are isomorphic?

[cloud walrus]

Dаniil
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[hidden haven]

Yes

[warm holly]

Where Z(*) denotes a center of *

[rustic crown]

yep you can.

[hidden haven]

What's identity application?

[rustic crown]

identity map from Z(GxH) --> Z(G)xZ(H)

[warm holly]

But should I verify something? Or simply using an identity application it's done?

[warm holly]

Thank you!

[hidden haven]

I'm confused lol how would you use that? And isn't that already assuming one inclusion?

[rustic crown]

like i think they're asking to show Z(G) x Z(H) is isomorphic to Z(G x H)... can't we just show they're equal.

[hidden haven]

Yeah

You're assuming one inclusion, then using identity map and showing it is surjective?

It seems much easier than that

[rustic crown]

yea show both inclusions if (g, h) in Z(G x H) then it is also in Z(G) x Z(H) and vice versa

[rustic crown]

i assumed that that they proved the two things are equal and then asking identity works as an isomorphism. kinda like "am i missing something" question

[hidden haven]

Multiplication in the product is defined componentwise, so an element in the product commutes with everything else iff its components commute with everything in their respective groups

Ah

I see

[rustic crown]

how do you define submodule 😶

[wispy urchin]

closure doesnt 'touch' the axioms

[chilly ocean]

wau first time i see blo here

[wispy urchin]

i thought id make myself useful

for two whole days

[chilly ocean]

CTRL-F from:blo in:abstract-algebra

[rustic crown]

i hope you mean closed under multiplication by elements of the ring. But from this can you verify all the axioms for W to be an R-Mod?

[warm holly]

Why to show that $Z(S_n)={e}$ it is sufficient to construct a transpotition which doesn't commute with $\sigma \in S_n$? I know that $S_n$ is generated by transpositions but still I don't really see why if it doesn't work with one of them then it certainly won't work with other transpositions

[cloud walrus]

Dаniil

[warm holly]

might be stupid question but still don't get it 😦

[hidden haven]

For any sigma, you have found an element of S_n that it doesn't commute with

[hidden haven]

So that sigma can't be in the centre

I think you're thinking from the other direction, ie that the transposition is not in the centre

[warm holly]

Oh, okey

[warm holly]

Yes thats what I was thinking about

thank you!

[cloud walrus]

bim

[chilly ocean]

is there a general formula to calculate Tchebychev polynomial of grade n?

[latent anvil]

(to the previous question) I think you can apply nakayama's lemma

[chilly ocean]

There is one for Chebyshev's polynomials .

[latent anvil]

Take X = (a), this is a finite generated module

I claim MX = X

Since X is principal, this just says MX contains a

but we know ma | a, so a is in (ma) = MX

[chilly ocean]

i cant find

This is the definition right?

this is recurrence

to get the 10th, i need the 9th

This is the general formula.

but ye, thats how they are defined

where did u find it?

ive been looking for a while

i found eveything but that

Wikipedia.

...

https://en.wikipedia.org/wiki/Chebyshev_polynomials

can u tell me where exactly?

[minor badger]

so this implies that if $ma\mid a$ only if $a=0$, i dont see it

[cloud walrus]

bim

[chilly ocean]

No.

oki

[latent anvil]

Do you know nakayama's lemma?

[minor badger]

No

[latent anvil]

Ah okay, it's a more general statement that I was using to prove your question

I'll think about how to do it directly

[minor badger]

Because its just a small detail of an exercise and my brain dont wanna do math today.

[latent anvil]

So if ma | a then a = max for some x in R. Then a(1-mx) = 0

Yeah?

What can you say about the element 1 - mx?

[minor badger]

well we are in a domain so mx=1

or a= 0

[latent anvil]

Where does it say we're in a domain?

[minor badger]

Oh i didnt write it, but in my assignment we are in a domain

[latent anvil]

Oh

Weird

That's not necessary to assume here

[minor badger]

Hm okey let me think

[latent anvil]

But sure, either a = 0 or 1 = mx

We need to prove that 1 = mx is impossible

[minor badger]

well if we aren't in a domain couldnt just a(1-mx)=0 without mx=1 or a=0

[latent anvil]

Yup, that's true

But if your assignment says it you might as well assume it

And continue the proof you were thinking of

I'll tell you the idea I had after

[minor badger]

Okey so mx=1 is impossible since then mR wouldnt be a proper ideal

so there exists no such x

[latent anvil]

Yup, that's right

Here's why you don't need R to be a domain:

[minor badger]

ah yeah i see it

i think

[latent anvil]

Oh okay, what is it?

[minor badger]

well so we have a=amx, so mx=1 but that dosnt work

so we must have a=0

[latent anvil]

But you're still using that R is a domain here

That's what lets you cancel from both sides

[minor badger]

Do i use that its a domain, like if a=ak then either a=0 or k=1

because k works as an identity

so in this case mx=1

[latent anvil]

No, it's possibly to have a = ab and a ≠ 0 and b ≠ 1. In Z/6Z we have 2*4 = 2

[minor badger]

okey, so what am i missing?

[latent anvil]

So in any local ring, if y is in the maximal ideal then 1 - xy is a unit for all x

In our case we had a(1-xm) = 0, and this tells us that 1-xm is a unit, so we can divide by it and get a = 0

[minor badger]

Ah

Im not familiar with local rings so thats why I didnt see it, thanks

[latent anvil]

It's a good exercise, try to prove it!

[minor badger]

Hm maybe later, if you feel bored would you like to help me with my exercise ?

[latent anvil]

Sure

[minor badger]

Okey so $R$ is a noetherian local domain, $\mathfrak{m}$ maximal ideal. Im suppose to show that $\mathfrak{m}$ is principal if and only if $R$ is a discrete valuation ring.

[cloud walrus]

bim

[latent anvil]

What's your definition of a dvr (discrete valuation ring)?

[minor badger]

That $R={0}\cup v^{-1}(\mathbb{Z}_{\geq 0})$

[cloud walrus]

bim

[latent anvil]

I'm not sure that makes sense, the valuation on R won't a priori be valued in Z

Ah okay, ignore my comment

[cloud walrus]

bim

[latent anvil]

Is (ra^i, sa^j) denoting a fraction here?

[minor badger]

yeah, so K* is the field of fraction of R

[latent anvil]

Yup

[minor badger]

or K is

and K* = K \ {0}

and i have checked, and v defined as above should behave as i want it

[latent anvil]

You're requiring that r, s be units (ie not divisible by a) right?

[cloud walrus]

bim

[minor badger]

yeah

so basically every a that is in an element x in R i put in a^j, so x=ra^j s.t. a does not divide r

[latent anvil]

yup

Hm so if x = a^i r/a^j s and y = a^k p/a^l q

Sorry brb

[minor badger]

no worries

[latent anvil]

Alright, so x + y = (a^(i+l)rq + a^(j+k)ps)/(a^(j+l)sq)

[minor badger]

this is how far i get on the v(x+y)

[latent anvil]

I think you might want to recheck the definition of a valuation, v(xy) = v(x) + v(y), not v(x)v(y)

[minor badger]

ah yes, typo

but the thing i get stuck with is basically v(a+1)

[latent anvil]

Right

So is that divisible by a?

[minor badger]

no

[latent anvil]

Then it has value 0, right?

It's a^0/a^0

[minor badger]

huh

ah

yeah since a+1 is not an element in the ideal i can write

v(a+1) = v(ra^0/a^0) where r=a+1

I have been stressing so much over this, i wish i could give you a cookie

[latent anvil]

🍪

[minor badger]

https://tenor.com/view/pusheen-pusheen-cat-pusheen-the-cat-cookie-munching-gif-5414301

Okey, but i have no clue for the other implication

[chilly ocean]

can someone read this

https://www.researchgate.net/publication/264422777_Image_Reconstruction_from_Legendre_Moments

and tell me, on (4), what is f(xi, yj) please?

i am missing something

[old lava]

this does not seem like abstract algebra

[chilly ocean]

Doesnt legendre belong here?

yesterday i asked about legendre here 🙂

[old lava]

I don't know about yesterday

what I do know is that specific research paper does not seem to have literally anything to do with abstract algebra

[chilly ocean]

then tell me where

[old lava]

not here

is best I can do

[chilly ocean]

that doesnt help so please, dont flood this chat just in case someone can help me, thanks 😄

[restive star]

How would I go about proving that i isnt in a certain field?

I'm genuinely not sure.

the field in question is Q((5)^1/3,a), where a is the primitive third root of unity

[delicate bloom]

I guess gross way is write i = linear combination of basis elements, then square it, to get -1, then show that the equations you get in Q for the coefficients on that basis are not all solvable

[restive star]

yeah that's how I'm gonna try it, but I was making sure there wasn't a simpler way

[delicate bloom]

there probably is using like field norm or trace

also alternatively since 2a+1 = i*sqrt(3) if you show there's no sqrt(3) in your field that would also prove there's no i

[delicate bloom]

oh, what's the degree of the field extension?

[restive star]

well it's the splitting field of x^3-5, so I think 3

[delicate bloom]

right, and what's the degree of an extension containing i?

[restive star]

oh it'll always be even?

[delicate bloom]

yeah

[restive star]

haha that's brilliant

I would have spent so freaking long on this problem lmao

thanks

[delicate bloom]

haha you're welcome

[restive star]

oh wait it has degree 6

[delicate bloom]

yeah you're right damn lol

[restive star]

that woulda been such a cool solution haha

[delicate bloom]

I think we can use a similar approach though

like show [Q(i, a):Q]=4 and 4 doesn't divide 6

[restive star]

Is that true though? intuitively it seems strange

[delicate bloom]

it at least brings down the number of basis elements to work with

I'm reasoning off the multiplicativity of the degree

[old lava]

I mean if i were in Q(5^(1/3), a), then Q \subseteq Q(i, a) \subseteq Q(5^(1/3), a), so mero's approach does work I think

and you can then prove that [Q(i, a):Q]=4, since it can only either be 2 or 4

if it were 2, it would imply i in Q(a)

which you can disprove

[restive star]

but wouldn't proving it isn't in Q(a) be similarly difficult itself?

[old lava]

easier than proving it isn't in Q(5^(1/3), a)

[delicate bloom]

well it's a smaller basis to work with than Q(cbrt(5),a)

[old lava]

we would need to have i = q + b a, since that's a basis, for some q, b \in Q

and I suppose you can find a contradiction from there

maybe by multiplying by the complex conjugate on both sides?

[delicate bloom]

6=[Q(cbrt(5), a):Q] = [Q(cbrt(5), a, i):Q] = [Q(cbrt(5), a, i):Q(a, i)][Q(a, i):Q] = [Q(cbrt(5), a, i):Q(a, i)]*4

writing it out in full so it's maybe a bit clearer

[restive star]

Ah ok I see what you mean

[old lava]

ya, multiplying by the conjugate on both sides gives -1 = q^2 + b^2

which is like not possible

over rationals

[restive star]

you multiplied by the conjugate of q+ba?

[old lava]

since i = q + b a, conj(i) = conj(q + b a)

and ya, you can multiply both sides

oh wait, it's 1 = q^2 + b^2

[restive star]

oof lol

[old lava]

which also I think has limited solutions over the rationals however

[restive star]

wait I just realized this problem isn't prove, it's "prove or disprove"

[old lava]

I'm pretty sure it's wrong

since I do not think i is in Q(a)

[restive star]

yeah same

[delicate bloom]

oh you can look at a = 1/2 + sqrt(3)/2 i and the conjugate to get a system of equations for i in terms of a

and see that it has irrational coefficients

or I wrote it earlier

[old lava]

ya, it works for either a = 1/2 + sqrt(3)/2 or a is the other root

[delicate bloom]

1/sqrt(3) + 2/sqrt(3) a = i

[old lava]

since conjugating gives the same equation

to solve

at the end anyways

I think

[delicate bloom]

we could just solve directly without the system

since they share the basis vector 1 in common lol

[restive star]

wdym?

[delicate bloom]

well it's not enough what I said since it pushes the burden to showing sqrt(3) is irrational

I stepped out for a bit, I think I might just be saying what F[x] was saying

[restive star]

oh I mean that's pretty commonly known tho tbf haha

[delicate bloom]

I'm saying we can represent i in the basis {1, a} as 1/sqrt(3) + 2/sqrt(3) a = i, but since representations are unique, it means we can't represent i with rational coefficients in this basis

[restive star]

oh so that just immediately gives a contradiction!

nice

[delicate bloom]

we could also go through and square it and show the solutions aren't rational to the system of equations there too but

[restive star]

wait what theorem are you using that "representations are unique?" what about, for instance, 0?

[delicate bloom]

yeah 0 has a unique representation

and the proof of other things having a unique representation can be pulled through with this fact

[restive star]

1-1,2-2, etc. I think I'm misunderstanding something lol

[delicate bloom]

the coefficients x,y in x+ya are unique

0 = 0+0a

[restive star]

huh. I'll be damned

[delicate bloom]

is the unique representation

[old lava]

Q(a, i) is a vector space over Q, and in a vector space, every element has a unique representation as a linear combination of any basis

[restive star]

I guess that makes sense the more I think of it

[old lava]

mero proposes one basis

and uses that

[delicate bloom]

so if you had say, two representations that weren't the same like x+ya = x' + y'a you could subtract one to the other side to get (x-x') + (y-y')a = 0

and so the coefficients must be x-x'=0 and y-y'=0

so they are actually identical

[restive star]

ah I see

[old lava]

since the basis of 2 elements doesn't work, that means the vector space must have more than 2 elements, so Q(a, i) must be of degree more than 2

which forces it to be 4

by degree considerations

and then you can use mero's original argument

from there on

[restive star]

Nice! Thanks so much for the help!

[delicate bloom]

you're welcome

[chilly ocean]

i wonder, given a nonconstant poly in Z is there always a p such that the poly is same degree in Zp and splits into linear factors in it?

[oblique river]

yes, this is true

also you can assume the poly is monic and then drop the "same degree" assumption if you want

[chilly ocean]

if you prove only for monic is there simple way to then generalize to nonmonic?

[oblique river]

you can prove that not only is there one p, there are infinitely many p

so as long as p doesn't divide the leading coefficient you should still be fine

the monic version and the nonmonic version mod p will have the same splitting behavior

[chilly ocean]

oh ofc im dumb lol

i was thinking in Z[x]

ofc can unmonicize in Zp[x]

[oblique river]

the only proof i know of this uses some algebraic number theory

actually now that i think about it the proof only works for irreducible polynomials

err, my proof

it involves looking at Q(a) for a root a of f

[cloud walrus]

squirtlespoof

[rustic crown]

what did you try?

[cloud walrus]

squirtlespoof

[rustic crown]

how do you know that f would be a degree 1 polynomial?

(Also the assertion t = t^p for every t in F is false, especially false when t is the indeterminate in F[t])

[cloud walrus]

squirtlespoof

[rustic crown]

if each a_i is a multiple of p, then its same as saying each a_i = 0

which may not be true

consider 1+t^p

what's the derivative?

this is a non-zero polynomial with derivative 0

may not be p^n

could be any multiple of p

yep

maybe try to write it nicely now

[rustic crown]

(subrings without 1 are so sad)

[nova plank]

What do you get if you multiply 0 by any quaternion?

Yeah

So it's never 1

So it has no inverse

(this is true in any ring where 1 ≠0, so all nonzero rings)

[minor badger]

If i have a discrete valuation ring, does it always exist an element r in my ring R such that v(r)=1 ?

[sturdy marsh]

you can always normalize a valuation to make that happen

[minor badger]

What does it mean to normalize? I am unable to find a definition on google

[sturdy marsh]

pick the minimum positive valuation and divide by it

[sturdy marsh]

essentially, given any valuation v, I can define a new valuation v' = 2v and this wont have that property

[minor badger]

But i dont have division in the range

[sturdy marsh]

?

oh I think I see what youre saying

that wont be a problem

try proving it

[minor badger]

Just a guess is that every positive valuation will be divisible by this minimal positive valuation

[sturdy marsh]

yes

[carmine fossil]

How are you defining discrete valuation ring?

[cloud walrus]

bim

[carmine fossil]

v has to be surjective

[minor badger]

That isnt given in my notes

[carmine fossil]

So, there's an element in the valuation field such that v(z)=1

[sturdy marsh]

some texts dont ask for it

but the theory will work out the same

[carmine fossil]

ic

[sturdy marsh]

again, as you can renormalize

assume you have some element of minimum positive valuation q

then for any other valuation n

write n = bq + r

r < q

prove r is 0

[cloud walrus]

bim

[minor badger]

dont spoil D:

im thinking

well dosnt that follow directly since if such a r exist, then q wouldnt be the minimal positive valuation

[sturdy marsh]

you need to show that r is the valuation of something

[minor badger]

well

[cloud walrus]

bim

[sturdy marsh]

n = v(n) isnt right unless the second n has nothing to do with the first

but yeah you almost have it

it should be mult

not addition

[minor badger]

well thats the abuse of notation no?

[sturdy marsh]

oh ok

[minor badger]

ah yes

[sturdy marsh]

I would not recommend abusing notation this way

[minor badger]

well i only did it now because i dont have my snippets here, so writing latex is pain

[sturdy marsh]

you could have used any other letter

lmao

[minor badger]

to much to keep track of

[sturdy marsh]

like x would have been fine

or r

[minor badger]

not r

since already taken

[sturdy marsh]

anyway

[minor badger]

i was thinking of doing like $\tilde{n}$ but nah

[cloud walrus]

bim

[minor badger]

thank you 😄

or more correctly, $\tilde{n}$ah

[cloud walrus]

bim

[sturdy marsh]

tilde n is also a weird choice

[minor badger]

yeah, but i like tilde so i put it on everything

[sturdy marsh]

alright, so now you can divide out by that minimum positive valuation to get a new surjective valuation

that's what I meant by normalize

[minor badger]

thanks so much, asking question here is actually really useful. Usually when I study with people it feels like im cheating and they work ahead, but now I get just the right amount to figure it out myself, I love it

[warm holly]

If we have $H\le S_4$ such that $H\not\subset A_4$ and we consider $(12)A_4$, does it follow that if I take $g\in H\cap (12)A_4$ and $x \in H\cap A_4$, then $gx \in H\cap (12)A_4$? Clearly $gx$ is of sign $-1$ and it is in $(12)A_4$, but I'm not sure that it is in $H\cap (12)A_4$. And if I take $x\in H\cap (12)A_4$ and $\tau=(12)$, then $\tau x \in H\cap A_4$?

[cloud walrus]

Dаniil

[rustic crown]

if both g and x are in H then so is gx. and you already said that is odd.

[warm holly]

Nice, thank you!

[rustic crown]

the last assertion isn't true though

take H = <(1234)>

[warm holly]

Alright, thank you!

[cloud walrus]

squirtlespoof

squirtlespoof

[rustic crown]

could you think of one polynomial which alpha has to satisfy?

[cloud walrus]

squirtlespoof

[rustic crown]

yep!

now say f was the minimal polynomial.

f will have to divide that polynomial!

could you tell me all the divisors of the polynomial?

(I'm assuming that char = p)

yea you're right... so the divisors are (t - \alpha)^m for m = 0, ..., p

f must be one of these!

it cannot be the one with m > 1, cause those are inseparable.

so the minimal polynomial is t - \alpha

which means \alpha lives in the ground field

[rustic crown]

just knowing 0 <m < p actually works. because then (t - \alpha)^m = t^m - m\alpha t^m-1 ....
if this polynomial is over the base field, then m*\alpha in F(\alpha^p), and so must then be \alpha, as m is non-zero.

you can't define separability of alpha, if the extension isn't algebraic

we say the element alpha is separable if its minimal polynomial has no repeated roots in its splittting field.

what if there is no minimal polynomial to begin with?

[golden pasture]

am i misreading the statement or if we just take F=Q, K=Q(sqrt(2)), α=sqrt(2), p=2, isn't it false?

oh

ahhhh

[rustic crown]

(i assumed that char F = p)

else we won't be able to do this

(btw just to confirm, do you know that char F = p, or just that F is a finite field?)

(there are more than one primes 😶 )

[warm holly]

what is an exact definition of order of left cosets? For example if I take $G=Q_8$ and $H=\langle -e \rangle$, then $G/H={{i,-i},{-k,k},{-j,j}{-e,e}}$, what is the order of each coset in $G/H$? Is it simply the number of elements of a coset?

[cloud walrus]

Dаniil

[rustic crown]

This is a confusing terminology in this case... when you say order of gH in G/H, it should mean the order of the element gH in the group G/H. that is the smallest positive n such that (gH)^n = g^nH = H which is same as saying smallest positive n such that g^n in H.

[warm holly]

Hm, sorry, not sure to understand... The goal then is to show if G/H isomorph to Z/2Z x Z/2Z or to Z/4Z

What I was thinking is simply to look at the size of each coset in G/H and as Z/2Z x Z/2Z is 2 -torsion then G/H isomorph to Z/2Z x Z/2Z

[rustic crown]

how does knowing size of the coset tell you anything about G/H

[warm holly]

it tells nothing I guess that's why i asked that question x)

But I think I see what you told me before I'll try to check it, thank you

[unique juniper]

i dont know why H fixes F'

[rustic crown]

H is the image of phi.

so any element of H looks like some sigma in Gal(KF'/F') restriced to K

[unique juniper]

right

[rustic crown]

which has to fix F'

[unique juniper]

yeah

sorry

[rustic crown]

its daijobu

[unique juniper]

[rustic crown]

don't sully me

i forgot what sully meant lol

[hidden haven]

damage the purity of

[rustic crown]

[final pasture]

[minor badger]

I need some help to understand what my teacher means by his text

"Show that the sets $r+I^n$ for all $r\in R$ and $n\in \mathbb{Z}_+$ form a base of a topology of R"

[cloud walrus]

bim

[hidden haven]

seems wrong as this doesnt contain any sets of diameter smaller than 1

wait whats I^n

[minor badger]

oh, I is an ideal

[hidden haven]

R is some arbitrary ring?

[minor badger]

yeah

or the exercise above has R=K[T] and I=(T) but i dont know if it is the same R and I or if it is general now

[hidden haven]

Ah so r+I^n probably generate a topology by unions

like T = {unions of sets of the form r+I^n}

is a topology

[minor badger]

I dont understand if the sets are $B_n = { r + I^n \mid r\in R}$ or $B_{r,n}={r + I^n}$

[cloud walrus]

bim

[hidden haven]

The latter

You vary both r and n

[minor badger]

so a set for each r and each n

[hidden haven]

yeah

[minor badger]

okey thanks!

[warm holly]

Show that $\langle H \cup F\rangle=U(3,k)$ ($U(3,k) \le GL(3,k)$ a group of unipotent matrices, $k$ a field). To prove the statement I could proceed by double inclusion. I just have a question for the $\langle H \cup F\rangle\supseteq U(3,k)$ direction: to show that $h\in U(3,k)$ is in $\langle H \cup F\rangle$ as well, could I discuss about $h$ scalars and then show that there exists $k_1,k_2$ such that $h=f^{k_1}\cdot g^{k_2}$ for $f \in H, g \in F$?

oups

[cloud walrus]

Dаniil

[warm holly]

sorry for a long message I just wanted to pass my idea as clear as possible

By scalar discussion I mean if we have $h=\begin{pmatrix}1&k&w\0&1&m\0&0&1\end{pmatrix}$, I would like to show that I could express each case ($k=0 \ m \neq 0 \ w\neq 0$ and so on) in function of $f^{k_1}\cdot g^{k_2}$

[cloud walrus]

Dаniil

[oblique leaf]

How do I begin proving that D3xZ/4Z is nonisomorphic to D4xZ/3Z?

[oblique river]

you can look at orders of elements

how many elements of order 2 does each group have?

[oblique leaf]

@oblique river so I think in D3xZ/4Z there are 3 elements of order 2 and in D4xZ/3Z there are none?

[oblique river]

neither of those is correct

[oblique leaf]

omg

[oblique river]

but that's the right idea, yes

[oblique leaf]

wait i think i forgot the ones were its (_,0) so like is it 6 elements of order 2 for the first one and 4 for the latter?

[oblique river]

closer but still not correct

[oblique leaf]

then i have to take into account the (0,_) i suppose

[oblique river]

yep

[oblique leaf]

ok then i'm done right

[oblique river]

that will get your answer for the first group correct, but you're still miscounting something in the second group

[oblique leaf]

em in Z/3Z there's nothing of order 2 correct?

[oblique river]

correct

[oblique leaf]

so all elements of order 2 are of form (r^ks, 0)

[oblique river]

mhm

and one more

D4 has another element of order 2

[oblique leaf]

like (r^2,0)

[oblique river]

yep

[oblique leaf]

omg great

thank you so much!

[oblique river]

so I think you should find that teh first group has 7 elements of order 2 and the second group has 5

[oblique leaf]

yeah i think so! that solves the problem

[oblique river]

yep :)

[fresh vessel]

So I have a question, I think Im close but not sure how to finish this off. Suppose $\alpha, \beta,\gamma \in \mathbb{C}$ satisfy $\alpha + \beta + \gamma = 3$, $\alpha^2 + \beta^2 + \gamma^2 = 5$, and $\alpha^3 + \beta^3 + \gamma^3 = 12$. We want to show that $\alpha^n + \beta^n + \gamma^n \in \mathbb{Z}$ for all $n \geq 4$. Using Vieta's formula and some algebraic manipulation I was able to show that $\alpha$, $\beta$, and $\gamma$ are roots of the polynomial $f(x) = x^3 - 3x^2 + 2x + 1$ and the expression we are interested in is a symmetric polynomial. Then there is a theorem in my book (Cox's Galois Theory) that if you evaluate any symmetric polynomial at the roots of a polynomial $f \in F[x]$, then $p(r_1,r_2,...,r_n) \in F$. The issue is $\mathbb{Z}$ is not a field, so I don't think this applies. I think this only tells us that $\alpha^n + \beta^n + \gamma^n \in \mathbb{Q}$. So I guess is there anyway I can "upgrade" this result to apply to $\mathbb{Z}$, or am I just completely off?

[cloud walrus]

fubini

[chilly ocean]

monic integer coefficient polynomials in Q only have integer roots @fresh vessel

[old lava]

rational root theorem in action

[fiery berry]

Let $K \subseteq L$ and $K \subseteq L'$ be two field extensions and $a \in L, b \in L'$ two algebraic elements over $K$ I'm trying to prove that if the minimal polynomials of $a$ and $b$ are equal that there exists an isomorphism $\phi: K(a) \longrightarrow K(b)$ such that $K(a) = b$ and $\forall x \in K K(x) = x$.

now i think this looks suspicously familiar to the universal property of the polynomial ring and i also already know that $K(a) \cong K[x]\(p) \cong K(b)$ where p is the minimal polynomial so my gut tells me that this is all i really need to construct the isomorphism but i am struggling on the explicit construciton

[cloud walrus]

chrisply

[chilly ocean]

every element of those fields can be written as a linear combination of a few powers of a root f the polynomial

whats the most obvious map to try?

[fiery berry]

identity i guess ?

[chilly ocean]

its similar to an identity

[fiery berry]

yea i mean its just K(a) =b and K(x) = x if x is in K, right ? but i struggle to prove that its bijective

or rather still a homomorphism and bijective

[fresh vessel]

Doesnt that only tell us that alpha, beta, and gamma have to be irrational?

[chilly ocean]

irrational?

[fresh vessel]

Well if the polynomial had an integer root, it would have to be 1 or -1. But neither is

[chilly ocean]

Do you know that {1, r, ... , r^n} forms a basis for the vector space K(r) over K when minimal polynomial of r over K has degree n+1?

[fresh vessel]

Ahh my mistake, that polynomial should be f(x) = x^3 - 3x^2 + 2x - 1. The issue still stands though

[next obsidian]

Still hate your pfp

The eyes...

[fresh vessel]

Doesnt the rational root theorem rule that out?

[chilly ocean]

i was looking at one you wrote before nvm

[fiery berry]

yea i know of that, so would it be correct to just first map from K(a) to K[x]/(p) via $r \mapsto r + (p)$, then just map from K[x]/(p) to K(b) with $x^k + (p) \mapsto b^k$ and compose the two maps ?

[cloud walrus]

chrisply

[chilly ocean]

you can do it directly from ka to kb

[fiery berry]

ye true

so the assumption that the minimal polynomials are identical just plays a role to have the dimensions of the vector spaces K(a) and K(b) be the same, right ?

[chilly ocean]

not just that

you will find out what else.

[chilly ocean]

yea sry my answer was stupid i didnt read your question correctly

[fresh vessel]

All good

[chilly ocean]

guys i am not understanding something
When calculating the
Legendre moments of a MxN array (image in this case)
many sites say it is this way
https://gyazo.com/04423e0600a12dd9513df531eb162b30
So
for the pixel[0][0], the legendre moment is L_00?
i mean, does each pixel have its own Legendre moment?

[harsh pecan]

Can someone help me find the splitting field of sqrt(8+sqrt(15))?'s min polynomial?

I ended up with x^4-16x^2+49 for the min poly

but idk how to find the splitting field

I know it's roots will be +-sqrt(8+sqrt(15)) and +-sqrt(8-sqrt(15))

but how do I prove that Q(sqrt(8+sqrt(15)),sqrt(8-sqrt(15)) is the splitting field

[fresh vessel]

Well it definitely is the splitting field because every root is contained in that field. I guess the question is whether Q(sqrt(8+sqrt(15))) = Q(sqrt(8+sqrt(15)),sqrt(8-sqrt(15)))

[harsh pecan]

That should be enough actually. Thank you!

[delicate bloom]

you can simplify the proof of irreducibility because if it factors, it factors into a linear and quadratic, so you can check it's nonzero for 0, 1, 2 mod 3

second one looks fine but there's an easier way to do it

$\alpha^3+\alpha^2+2=0$ factor the left two terms as $(\alpha+1)\alpha^2 =-2$ now you can divide to get $\frac{-\alpha^2}{2} = (1+\alpha)^{-1}$

[cloud walrus]

Merosity

[delicate bloom]

but it's good you know the euclidean algo, yup

you're welcome

yeah pretty much, I didn't look at your proofs too closely, just skimmed them but they look right

yeah you're welcome

[chilly ocean]

Is there some quick argument why $Q(\omega_3 \cdot \sqrt[3]{3}) \neq Q(\omega_3^2 \cdot \sqrt[3]{3})$?

[cloud walrus]

Ledog

[chilly ocean]

Was thinking about setting up some equations of basis elements but maybe there's something quicker?

[delicate bloom]

hmm if it were, then it'd contain both elements so you'd have their ratio w^2 cbrt(3)/ w cbrt(3) = w but since this generates a degree 2 extension of Q but 2 doesn't divide 3, we have a contradiction

[chilly ocean]

oh yeah that makes sense, thanks.

[oblique leaf]

can someone explain how i could prove that <15,x-7> is an ideal in Z[x]? My professor just assumed it and i was wondering whether it's that trivial

[oblique river]

That is the ideal generated by 15 and x - 7

It is an ideal by definition

[oblique leaf]

Our definition for ideal was: for all r in R and for all i in I, I is a left ideal of R iff ri is in I

analogous for right ideal, and then the ideal is defined as it is both a left and a right ideal

[oblique river]

Yes, but what you wrote is by definition “the smallest ideal containing 15 and x-7”

So it is an ideal

[oblique leaf]

so are you saying that you can multiply any polynomial (a+bx+cx^2+....) with something generated by 15 and x-7 and you'd still be inside <15,x-7>

because that was the nonobvious part to me

[oblique river]

Yes that is the definition of <15,x-7>

[oblique leaf]

isn't the definition of <15,x-7> that any element in it can be written in the form 15a+b(x-7)^c

[oblique river]

You dont need the c

And yes that is correct

Where a and b are polynomials

a and b are allowed to be any polynomial. You dont need the c.

[oblique leaf]

oooh ok i confused myself and though a and b were constants

tysm

i have an exam in two days and was so overwhelmed in the last part of the course

you are a lifesaver

[oblique river]

You give me too much credit

Im just bored

[oblique leaf]

you don't have anything to do?

[oblique river]

No

:P

Im a teacher but the school year is over

[oblique leaf]

are you a prof too

[oblique river]

Yes

[oblique leaf]

i feel like all my professor does is answering questions 24/7 and i don't wanna bother him more lol

i wonder when you all have time for research

[oblique river]

What if i was your prof

Jk i didnt teach algebra this term

[oblique leaf]

no prof can know I ask questions here lol, i feel like they would be offended as they always encourage us to ask endless questions

[oblique river]

Yeah i also pray that none of my students figure out that im here

Because then i would have to clean up my act

And be “professional”

[oblique leaf]

you seem pretty professional on here though

no one cares about the occasional smiley

[oblique river]

Oh i mean like in other channels i curse and make sex jokes

Lol

[oblique leaf]

haha

i mean my prof sends us drag memes

it's not sex jokes but still no one ever complained about it lol

[hidden haven]

I really doubt any student is going to mind their prof acting non clean in a casual conversation lol

[oblique river]

Oh i should send my students drag memes more often

I do say things like “what’s the tea”

I think i described the fundamental theoremmof calculus as “this is the real tea” a couple years ago lmao

[chilly ocean]

Why are splitting fields normal extensions? Just hint if possible, im dumb

What definition of normal extension are you using?

[restive star]

anyone willing to help me out with my question in question channel 9?

it's galois theory

[hidden haven]

you can post advanced questions in the advanced channels, you'd get quicker answers

[restive star]

ah ok I didn't know

[chilly ocean]

i replied in that channel, it should follow rather simply by plugging in x=y^-1

[restive star]

thank you. I assumed it was more complicated.

[hidden haven]

If the splitting field is E/F with the roots of the polynomial you split being a1,...,an, you look at automorphisms of cl(E)/F (alg closure). Any automorphism of cl(E)/F must permute the a_i's and since those generate E/F, the image of any automorphism must lie within E. Now suppose you have some x in E, which has a conjugate y in cl(E). Then there is an automorphism of cl(E)/F that sends x to y, so y must also be in E

assuming the definition is that conjugates of elements in E should lie in E

[chilly ocean]

idk what conjugates are

my definition of nornal extension is one such that every polynomial that was irreducible, if it now has a root then it splits linearly

ye

so a splitting field

[hidden haven]

y is a conjugate of x if x and y are the roots of the same minimal polynomial

so normality is equivalent to "the conjugates of any x over the base field are also there"

[chilly ocean]

why is it equivalent?

or show equivalency to this

[hidden haven]

if all irreducibles with a root split, then for x in the extension, any conjugate of x must also be there because min(x) splits

[chilly ocean]

no. this definition of normal is quantifying over all base field irreducible polynomials

[hidden haven]

if all conjugates of x are there, then min(x) splits

[chilly ocean]

why does min(x) split for other x in the extension?

[hidden haven]

because all conjugates of every element are there

[chilly ocean]

why are they there?

[hidden haven]

thats the assumption

the statement im proving is equivalent is that "for an algebraic extension E/F, if x is in E, and y in cl(E) is a conjugate of x over F, then y is in E"

so its saying conjugates of all x are there

[chilly ocean]

@chilly ocean yo did you come out?

how is that an assumption?

im confused and stupid ugh

[hidden haven]

Im proving that "E/F is normal" is equivalent to that statement

so there I have assumed the latter and proved the former

[waxen hollow]

Ideas to find all non-abelian groups of order 20?

[next obsidian]

Sylow’s theorems

And semi direct products

Actually, this is a p^2q group

[waxen hollow]

Sure, I'll try

[next obsidian]

And I think it’s not too hard to classify those

i think

[waxen hollow]

Then more generally if I have a problem for any group, I'll use Sylow's Theorems and Semi direct products?

[next obsidian]

Maybe it’s just easy to show something like no simple groups of order p^2q or something

That’s certainly where I’d start

You might have to get more creative, but often those let you begin cracking the problem

In particular, I think Sylow instantly implies you have a normal Sylow-5 group

For 20

And that lets you begin doing a lot of semi direct product stuff

Like, actually I think the two completely kill the problem

You have a normal Sylow-5, and a Sylow-2 of size 4

The product of the two must have size 20

Because they intersect trivially

This tells you G is the semidirect product of a size 5 and size 4 group

Only one group of size 5

Only two of size 4

Then you just go from there

[waxen hollow]

Understood

Thanks man

[next obsidian]

Np

Maybe should said a bit less ahaha

[chilly ocean]

these questions are clumsy for me. proposition 2.3.2 is properties of field. lets say if I want to proof addition and multiplication are associative on F[x], shouldn't I just proof normal associativity operation ?

[next obsidian]

what do you mean?

You want to show that multiplying polynomials is associative

[chilly ocean]

oh, just wait a moment, am I have to use polynomials ?

yes,

thanks

[next obsidian]

Yes, because you want to show the operations are associative and commutative on F[x]

[chilly ocean]

😄

[next obsidian]

And then you have F as a subring of F[x]

And you just want to verify that it acts like it did in the normal field

So, to be clear

[old lava]

this is extremely cringe to do

[next obsidian]

It means multiplying constant polynomials is just like multiplying them as if they were in the field

[inner acorn]

imagine proving associativity of functions 👀

[old lava]

[next obsidian]

It’s not under composition omegalol

[old lava]

but ya, polynomial multiplication being associative amounts to (discrete) convolutions being associative, and it's just cringe symbol bashing

[next obsidian]

It’s under multiplication

[inner acorn]

I am forever cursed by this role mniip bestowed upon me

[next obsidian]

Lol

It’s funny cuz det is more lewd

[old lava]

leave server -> rejoin server

[next obsidian]

Cuz their eevee pfp is eevee butt

[old lava]

chmonkey, if I ever do anything related to comm alg

I wanna let you know

it's entirely your fault

[next obsidian]

Omegalol

[hidden haven]

[inner acorn]

Oh oh, I have an abstract algebra question - well, a sanity check

it's been a lil while so I just wanna double check, given a character table of a group - idk, lemma find a fun one

[next obsidian]

Character table

It’s like sudoku

Sadly I forget the orthogonality relations sadge

[inner acorn]

I remember the orthogonality relations, they're actually really neat xD

[next obsidian]

My last hw in my algebra class in the winter last year was to compute a character table

Given partial info

It was fun, I just sat down on the floor and it felt like a sudoku problem

It was easy

Took like an hour

[inner acorn]

idk, here's a random group I picked

SL(2, F5) given I've been working with PSL(2, R) lately

just wanna double check, to find the normal subgroups, I just union the conjugate classes C_i
where ρ_i(x) = ρ_i(1) for x in C_i right?

So here:
ρ_1 gives SL(2, F5) normal
ρ_4 gives C_1 ∪ C_2 is normal (which is just the cyclic group of order 2)
any obviously the trivial group is normal

[chilly ocean]

when Z_p isn't a field, how come K is a subfield of Z_p?

Z_p doesn't have multiplicative inverse

[final pasture]

is Z_p integers modulo p with p prime ?

[chilly ocean]

yes

[final pasture]

it is always a field, then

but anyway, you can speak about subfields of a ring

For example, $\bQ$ is a subfield of the ring $\bQ[X]$

[cloud walrus]

Shika-Blyat

[final pasture]

but $\bQ[X]$ isn't a field

[cloud walrus]

Shika-Blyat

[final pasture]

[chilly ocean]

[chilly ocean]

can I ask why ?

[final pasture]

Do you know some condition on $n$ and $k$ so that there exists some $q$ such that $kq = 1 \pmod n$ ?

@chilly ocean

[cloud walrus]

Shika-Blyat

[chilly ocean]

we can have an element in a set with mod n that is invertable but not all of them

[old lava]

If p is prime

Everything but 0

Is invertible

Which is exactly what you need for a field

[final pasture]

that's not quite helping, he's asking why

[old lava]

In general it depends in coprimeness

[final pasture]

[chilly ocean]

let's say p=3, which number is invertible in that set ?

[restive star]

Alright I've made no progress on the problem I asked earlier, is anyone here good with galois theory?

[final pasture]

1 and 2, 0 never being invertible

[chilly ocean]

only 1 is invertible. 2 isn't

[final pasture]

it is, though

2*2 = 1 mod 3

[chilly ocean]

got it

thanks

[old lava]

by euler's theorem, $a^{\varphi(n)} \equiv 1 \mod{n}$ for every $a$ coprime to $n$

[cloud walrus]

F[x]-module

[old lava]

for n = p a prime, every a from 1 to p-1 works

as a number coprime to p

so you can easily see using euler's theorem why every nonzero element is invertible

the inverse of any nonzero $a$ is just $a^{\varphi(p) - 1}$ (and since $\varphi(p) = p - 1$ for $p$ prime, the inverse is just $a^{p-2}$)

[cloud walrus]

F[x]-module

[chilly ocean]

I didn't consider 2*2=1 under modulo 3. thanks for your answer

[final pasture]

Another way:
Saying that for some $p$ (just an integer, not necessarily a prime) there exists some $q$ such that for $pq = 1 \pmod n$ is equivalent to say that there exists some $k$ such that $pq = kn + 1$, or $pq - kn = 1$.\
By Bezout's theorem, this is only possible iff $gcd(p, n) = 1$.\

[cloud walrus]

Shika-Blyat

[old lava]

ya, that also works well

for a more understandable reason

euler's theorem does give a nice constructive way to find the inverse however

which is why I like that approach

[final pasture]

I mean, bezout is also constructive ?

[old lava]

I mean true

[final pasture]

it's just euclid's algorithm ?

[old lava]

but not as immediately obvious

as just a^{p-2}

[chilly ocean]

cool, thanks

[final pasture]

hmm yeah, fair enough

This approach works in any finite group, btw (I mean yours, not mine)

[restive star]

Could someone help me with this: we have that for every root y of f(x), y^-1 is also a root, and that the polynomial is 4th degree. We have to show that the size of the galois group divides 8

Im pretty lost

I know it must divide 24, but I'm not sure how to get that it must divide 8

(we have the specific polynomial too, I just wasn't sure if it's relevant since we used it to get the thing about y^-1)

[old lava]

why not give the specific polynomial too

[restive star]

wanted to simplify as much as possible haha

[hidden haven]

Try showing that there are no elements of order 3 in the Galois group

what happens if you have a 3 cycle of roots?

[restive star]

I've been trying that but haven't been able to find a contradiction

[hidden haven]

Notice that the last root will have to remain fixed under this isomorphism

[restive star]

what do you mean by the last root?

[hidden haven]

the root that is not being cycled by the 3 cycle

[restive star]

Ah ok so really think of this in terms of S4?

[hidden haven]

yeah you would pretty much always think of the gal group as acting on the roots

[restive star]

but then what's the issue with the final root being fixed?

[hidden haven]

what happens to the inverse of that root?

[restive star]

so say the roots are r,s,r^-1,s^-1

r->s,s->r^-1, and r^-1-> r

wouldn't this be valid?

[hidden haven]

Thats a 4 cycle

(rss^{-1}r^{-1})

[restive star]

ah ok I changed it

[hidden haven]

in that, s^{-1} is fixed

[restive star]

I guess what I'm missing is why that turns into a contradiction

[hidden haven]

but you have the equation ss^{-1} = 1

[chilly ocean]

if r maps to s, then how can r^-1 map to r?

r^-1 should map to s^-1 then

[restive star]

ah I see! So basically I do this for all the possibilities?

[hidden haven]

you can do it in one go

if r goes to x, r^-1 has to go to x^-1, so its clear that you cant have 3 cycles

alternatively what i did was show that if s is fixed by an automorphism then so is s^-1

[restive star]

Ah ok. Thanks so much!

[oblique leaf]

if a group A has some element of order 2, why does that automatically mean that Aut(A) is of even order?

[final pasture]

because Aut(A) also has an element of order 2, then @oblique leaf

[old lava]

Z_2 has an element of order 2 but |Aut(Z_2)| = 1 isn't even?

[oblique leaf]

I understand Aut(A) also has an element of order 2 but I still don't see it ...

[final pasture]

Lagrange's theorem alyosha

wait

[old lava]

how can you prove something

that isn't true

what are y'all on

[final pasture]

Yeah wait

[old lava]

literally the most trivial example of a group with an element of order 2 fails

[oblique leaf]

no like we don't consider the trivial group and Z/2Z

it states it in the problem that those are the exceptions

[final pasture]

yeah that doesn't work only with these cases

like if the group has atleast three elements

my proof works

[oblique leaf]

so using lagrange works right? i think it does

[final pasture]

yes

(and it doesn't work with Z/2Z because there's no reason for the said element of "order 2" (the interior automorphism induced by the element of order 2) to not be the identity, but if there's atleast one element different from the identity and the element of order 2, it can't be the case )

[oblique leaf]

yea that makes total sense

tysm!

[final pasture]

np (and thanks F[x] for correcting my mistake )

and thanks @chilly ocean for clapping your hands ! 👏

[next obsidian]

Shamrock is a psyop

[chilly ocean]

Actually, what is the order 2 automorphism? (Order 2, or just even order?)

[final pasture]

I said it lol

[next obsidian]

Wtf

This isn’t sham

[final pasture]

take a an element of order 2, consider axa^{-1}

[hidden haven]

a the

[chilly ocean]

Ah, right

[hidden haven]

a an

[final pasture]

its order divides 2, and if the group isn't trivial or Z/2Z, the order can't be one

wait

actually is it true that the order can't be one

that sounded so obvious in my head

But now I'm doubting

[hidden haven]

It's not

a could be in center

[final pasture]

yeah right

[hidden haven]

[final pasture]

so huh

@oblique leaf actually what I said doesn't work, there's still some work ig

[oblique leaf]

wait why not

we are working with finite abelian groups fyi

[chilly ocean]

Wow, you should have said this from the start

[oblique leaf]

sorry

[final pasture]

[hidden haven]

In abelian groups, all inner automorphisms are identity

[final pasture]

yeah, so what I said doesn't work at all in that case, oopsie doopsie

[oblique leaf]

so, i have another question: why is the division ring (\mathbb{R}\0) a ring at all? (Sorry this is prob super dumb but it's 4 am and my exam is in two days and i just started now bc family crisis)

and i'm screwed

[hidden haven]

What ring is that?

Like R minus 0?

[oblique leaf]

yea like the reals minus 0

[inner acorn]

pretty sure that wouldn't be a ring

no additive identity

[hidden haven]

Yeah

[oblique leaf]

it's called the division ring? yea exactly i was confused about that

how is it a ring without 0

[inner acorn]

Are you sure you don't mean R

[hidden haven]

R is a division ring

[oblique leaf]

oh so with the zero in it is a division ring

[inner acorn]

Mhmm, it's a field

[oblique leaf]

ok everyone who helped me today, tysm!!

[inner acorn]

[hidden haven]

For the abelian group problem, first try to prove that that is the case for Z/2^nZ for any n

Then use the structure theorem for finite abelian groups

[final pasture]

huh

I feel like there should be a simpler answer

[hidden haven]

Perhaps

[oblique leaf]

yea i got so far i think, like all finite abelian groups where all elements except 0 are of order 2 are all of the form 2Zx2Zx...

and then for all other groups you can also find elements of order 2 in aut(G)

i think you guyses proof should work!!

[final pasture]

hmm

no

[hidden haven]

I didn't understand what you did there

[final pasture]

Z/2Z x Z/2Z isn't isomorphic to Z/2²Z = Z/4Z

[oblique leaf]

no like that's not what i mean

[hidden haven]

That has an order 4 element

[final pasture]

Oh sorry, I misread what he said

[oblique leaf]

*she

[hidden haven]

But I don't get how you used that claim

[final pasture]

she*, sorry

[oblique leaf]

i'm sorry i just like bro K @final pasture lol

[final pasture]

(Thought she was speaking about all abelian group of even order, that's why I said that, don't mind me )

[oblique leaf]

@hidden haven i think that if you have a group where not every element is of order 2 then you can still find an automorphism a-->-a of order 2 always

and then when every element is order 2 you can also construct such an automorphism of order 2

[hidden haven]

That is what you want to prove

[final pasture]

Well that is possible, since the exercise claim it is 🤷‍♂️

[chilly ocean]

-(-a) = a

[oblique leaf]

^^

[final pasture]

I feel dumb

[oblique leaf]

I'm confused what everyone is confused about

i feel somewhat confident the proof works?

why is it wrong

[final pasture]

Carla just answered the question

x |-> -x is an automorphism of order 2

[hidden haven]

Where does an element of order 2 come in?

[final pasture]

wait

huh

yeah

[oblique leaf]

oooh sorry i should have written f(a)=-a instead of a-->-a lolol

[final pasture]

ok I'm confused

[hidden haven]

Lol this is just true for all abelian groups then?

[oblique leaf]

all finite abelian groups

[final pasture]

I should go sleep

well looks like it yeah lol

[hidden haven]

Other than the 2 exceptions

[hidden haven]

Infinite as wwll

Same automorphism works

[final pasture]

I mean

in the infinite case

it's not clear Aut(A) is finite

so speaking about even order doesn't quite make sense

[hidden haven]

oh right

[oblique leaf]

no but if it's infinite you can't assure that 2Zx2Zx ... are the only groups where all nonzero elements are order 2 right

so the argument falls apart?

[final pasture]

no but I mean you don't need all that machinery

[hidden haven]

How?

No

[final pasture]

it is of order 2

except the 2 exceptions

[hidden haven]

Not if every element is its own inverse

[oblique leaf]

wooow i'm so confused first of all why isn't Aut(A) finite? A is finite after all

[final pasture]

oh wait

yeah

[hidden haven]

So you handle that (Z/2Z)^n case separately

[final pasture]

yeah

god I can't think

[hidden haven]

By an exchange of coordinates automorphism

[final pasture]

[hidden haven]

For infinite A

[oblique leaf]

@final pasture true but the argument requires that 2Zx2Zx... are the only groups with all order 2 els cuz finitely generated abelian groups right

[final pasture]

forget what I said

Seems like I'm too tired to think tonight, I'll let moldilocks help you

[hidden haven]

[oblique leaf]

i think we can safely conclude that what you came up with initially was correct given finite and Abelian @final pasture

and goodnight!

[final pasture]

you meant the axa^{-1} thingy ?

[oblique leaf]

lmao what??

[final pasture]

that's the order 2 automorphism I was initially thinking about

[hidden haven]

Tragic

[final pasture]

'cause if the group is abelian, it literally always is the identity

so well, that's not quite of order 2

close enough, I guess

"Aut(A)'s order is close enough to an even order"

[oblique leaf]

omg what

define "close enough to even"

[final pasture]

I'm joking

(for the "close enough" part)

[oblique leaf]

at this point i can't tell anymore lol

my whole brain is a joke

[final pasture]

I mean, I just concluded while working on some proof that $\varepsilon < 0 < \varepsilon$

[cloud walrus]

Shika-Blyat

[final pasture]

I think I'm a funnier joke than your brain

[oblique leaf]

lmao

you know sometimes you just need one day in a year where the brain isn't a joke

in two days for me

but i'm pretty sure that most days you don't have this issue lol

[final pasture]

Anyway so if I sum up:

• If every element of G is its own inverse, handle the case separately (by using that G ~= (Z/2Z)^n for some n, and using an exchange of coordinates automorphism as moldilocks said)
• Else, take f: x |-> -x. It's not the identity (because there exists a such that a != -a), and its order is 2.

In both case, that gives you an order 2 automorphism, and lagrange's theorem allow you to conclude that Aut(G)'s order is even

I think I didn't say anything wrong here ?

[hidden haven]

Is good

[oblique leaf]

right so exhange of coordinates like the first and the 2nd one right? cause if you take n\geq 2

[hidden haven]

Yeah

[oblique leaf]

and yea i think all that looks great

[final pasture]

yes, that's why you want n >= 2, 'cause else you can't quite exchange coordinates

Anyway now I should definitely go to sleep, good night alyosha and moldilocks

[oblique leaf]

goodnight! and thanks @hidden haven @final pasture

[hidden haven]

Good night

[candid pier]

Is every subgroup of the permutation group S_4 normal?

[hidden haven]

No, order 2 elements aren't in the center, so their conjugates are not themselves

[restive star]

@hidden haven I just realized Im not sure it actually divides 24. RIP lol

I need the poly to be irreducible

[hidden haven]

doesnt that proof work regardless of whether polynomial is irreducible or not?