groups-rings-fields

[void cosmos]

given now that this is multiplicative

and 1 is the identity

[restive birch]

yea it should be pretty intuitive

[formal ermine]

henry have you done euler's theorem yet?

[restive birch]

remind me

[formal ermine]

g^|G| = e for all g in G

[restive birch]

seems intuitive, but no, i havent seen the proof yet

[formal ermine]

oh okay

[restive birch]

if i did see the proof tho, this would imply that |g| ≤ |G| for all g ∈ G, right?

[prime sundial]

if G is finite

[formal ermine]

wouldn't make sense for non finite groups anyway lol, but ye forgot to specify that :p

[formal ermine]

yes

or stronger: the order of g divides the order of G

or more general (you can prove this via contradiction using euler's theorem, about 2-3 lines, could be a nice exercise imo): if g^n = e then the order of g divides n

[restive birch]

makes sense

[prime sundial]

a nice corollary to this is that all elements except for the identity of groups of prime order are generators

[formal ermine]

isn't that just lagrange?

[prime sundial]

i’ve seen like 3 different versions of lagrange so idrk

[formal ermine]

lol

[restive birch]

if |g| doesnt divide n, then you can kind of 'pull out' all of the g^(some number less than n) and if |g| doesnt divide n, youll be left with some g^x term that will not equal e, and e * not e = not e

not an actual proof of course but the general idea i hope

[formal ermine]

yeah kinda something like that

[prime sundial]

yeah you can use division algorithm on |g| and show the remainder must be 0

[formal ermine]

start like this: suppose the order of g is d and g^n = e. assume for the sake of contradiction that n is not a multiple of d, i.e. n = kd + r for a 0 < r < d

then use euler to get a contradiction and therefore imply that n is a multiple of d

[restive birch]

anyone have a good link to a simple explanation of the basic syntax of proofs?

and what determines if a proof is rigorous or not

[formal ermine]

#proofs-and-logic

[rotund aurora]

How would you characterize such a field, what techniques should be used? For example, if n=1 you just get the usual constructible numbers (if I'm not mistaken). If n=2, then E_2 is the smallest subfield of R closed under taking square roots (of positive elements) and cubic roots (you take the real ones)

I think this field will be uniquely determined because all subfields of R contain Q. You can make similar constructions in general for arbitrary fields. Let F be a field and K a subfield of F. Let P be a "family of polynomials". Then, define E to be the smallest field K<E<F satisfying the condition that for any p in P with coefficients in E, the roots of p that are in F are also in E

[coral shale]

well yh it will be uniquely determined because you can construct it 'recursively' starting from Q

[rotund aurora]

Sure, but in that case is there a nice characterization of E_n?. Say n=3 to look at a particular case

[vestal cairn]

Can someone help me out here? Idrk what exactly the relation here is - like ik that in general, an equivalence relation is reflexive, symmetric, and transitive.. but idk how to show it for this one

[tender wharf]

For reflexive it is trivial: the epsilon permutation should do

For symmetric, you want to show that (f,g) being in R means that (g, f) is in R. Assume (f,g) in R, now you have sigma st f = sigma circ g

For transitive, suppose that (f,g) in R and (g, h) in R. What does this imply? || We can find sigma_1 st f = sigma_1 circ g and sigma_2 st g = sigma_2 circ h ||

[vestal cairn]

well, that (f,h) is in R

oh

[vestal cairn]

what does st stand for?

[tender wharf]

such that

[tender wharf]

You might need to elaborate a bit more

But yes

[vestal cairn]

yeah idrk how to do that..

[tender wharf]

Hmm

[vestal cairn]

i lowk dont understand the relation itself

[tender wharf]

You're confused on the symmetric bit or the transitive bit?

$R$ is symmetric if for any $(f, g) \in R$, $(g, f) \in R$. Or, [ \forall f \forall g fRg \implies gRf ]

[cloud walrus]

DerpZ

[sharp dirge]

can i ask a question

[vestal cairn]

well both, like how can i prove that they are

[tender wharf]

Well, first let (f, g) be from R arbitrarily

Now you need to show why (g, f) is in R

What does (f, g) being in R mean?

[vestal cairn]

yeah

uh

[tender wharf]

There is a permutation sigma in S_m such that...

[vestal cairn]

that- yeah that

[tender wharf]

Right

Now you want to find another permutation such that...

[vestal cairn]

like $g = \sigma_2 \circ f$?

[cloud walrus]

Levens

[tender wharf]

Great

Now, can you find sigma_2?

Recall the definition of a permutation

[vestal cairn]

like.. a rearrangement of the members in a set?

[tender wharf]

Hmm, not really.

A permutation is a function that is....

[vestal cairn]

a function that performs that rearranging

[tender wharf]

Alright, so what properties should this function have?

[sharp dirge]

inject and surject

[tender wharf]

^

So can you find sigma_2 now?

[vestal cairn]

not really😔

[tender wharf]

Alright alright so

$f = \sigma_1 \circ g$

[cloud walrus]

DerpZ

[tender wharf]

So now $\sigma_1^{-1} \circ f = \sigma_1^{-1} \circ \sigma_1 \circ g$

[cloud walrus]

DerpZ

[tender wharf]

Do you know why I can simply just say $\sigma_1^{-1}$

[cloud walrus]

DerpZ

[vestal cairn]

yeah, like a productive zero

[tender wharf]

Recall that $S_m$ is a group of permutations on $m$ elements

[cloud walrus]

DerpZ

[tender wharf]

So if $\sigma_1 \in S_m$ then...

[cloud walrus]

DerpZ

[vestal cairn]

it has an inverse element

[tender wharf]

There we go

[vestal cairn]

ahhh

[tender wharf]

Can you take care of transitivity by yourself now?

[vestal cairn]

so sigma_2 is the inverse of sigma_1

[vestal cairn]

ill give it a shot

so are we trying to find $h = \sigma_3 \circ f$?

[cloud walrus]

Levens

[sharp dirge]

f ~ g and g ~ h then f ~ h

so yes

just compose the permutations

[vestal cairn]

what does composing permutations mean? mb english isnt my first language

[sharp dirge]

$(f \circ h)(x) = f(h(x))$

[cloud walrus]

pewdssssssss

[sharp dirge]

f is related to g and g is related to h

so f = sigma_1 \circ g

and g = sigma_2 \circ h

so f = (sigma_1 \circ sigma_2) \circ h

so f is related to h

[vestal cairn]

ohh alright, so there isnt really a need to define a third permutation

[sharp dirge]

sigma_1 \circ sigma_2 = some other permutation in S_m

[vestal cairn]

or that ig

yeah

ok let me try again

[sharp dirge]

does someone know what this equivalence relation is? is just matrix obtained by elementary row operations?

[true cairn]

I’d like to solve this problem but I’m kinda confused how to start it. So far, I have H={0,6} and K={0,4,8}. I tried doing the product of HK and I got {0} which does not make sense. Thank you in advance

[south patrol]

How did you get 0

[chilly ocean]

Are you multiplying the elements?

[true cairn]

Yes

That’s how I got 0

[south patrol]

Here you need to add since the group is additive

[true cairn]

Got it. So when I do the last addition, I mod it 12 right like 6+8=14 =2

[slim dragon]

yes

[sharp dirge]

do you know about my question? @south patrol

[south patrol]

I guess yeah it should be equivalence of matrices

i.e. A equivalent to B if there are invertible P and Q with A = PBQ

(Afaik for a PID to be able to prove this you do need to use operations other than elementary operations, though with a euclidean domain these would suffice)

[sharp dirge]

so in this case, how does A become B ?

[molten viper]

Hi folks, I’m trying to understand Cayley’s Theorem in the context of Dihedral Groups. The problem I’m given is to find the corresponding element of $rs \in D_4$ under the isomorphism guaranteed by Cayley’s theorem. Is it proper to say $\lambda_{rs}$, where $\lambda_{rs} (x) = rsx$?

[cloud walrus]

Phil With Flex Tape

[molten viper]

Specifically, interpreted as an element of S_8

Pls ping me if you respond, thanks!

[tender wharf]

@molten viper here r and s are elements from D_4 correct?

[molten viper]

Yup, r with order 4 and s with order 2

[tender wharf]

Yes lambda rs will be in S8

[molten viper]

Hm ok

But that’s different than like, just interpreting rs as an element of S8 rather than S4?

[next obsidian]

Yah

Well like, sort of

[molten viper]

I seeeeeee

I think

[next obsidian]

You’re maybe conflating like

There’s a natural way to embed the dihedral group of 2n elements

Into S_n

By interpreting it as symmetries of a regular n-gon

The embedding of it into S_2n via the Cayley embedding is like

Way less natural

You’re literally seeing how an element permutes the group

[molten viper]

I see

[next obsidian]

Well okay it’s natural but in a different way haha

Like the way you do the Cayley embedding “geometrically” is to draw say 8 points

One for each element g

Fix an ordering of the 8 elements so like g_1,…,g_8

Then label each dot with g_i

[molten viper]

Ohhhhhh

[next obsidian]

Then when you want to see how g works as an element of S_8

You multiply g into g_i on say, the left

[molten viper]

Ok I think when we proved it in lecture we used that indexing

[next obsidian]

Then this sends g_i to some g_j

[molten viper]

But tbh, I got very lost

[next obsidian]

This is like say drawing an arrow from g_i to g_j and labeling it g

So you’re seeing how g shuffles up the g_i

[molten viper]

Ohhhhh ok

That’s what those visual interpretations I’ve seen are

[next obsidian]

For the natural embedding of the dihedral group into S_4

You’re considering 4 dots

And the square they form

And interpreting the dihedral group as the symmetries of that

Which you get by rotating and reflecting the square

[vestal cairn]

There's a second subquestion for this one that I can't seem to find an approach for..

ii) Now let m = n. Show that for bijective functions $f \in X$
$$
(f, g) \in R \Longleftrightarrow g \text{ is bijective}
$$
holds.

[cloud walrus]

Levens

[next obsidian]

A permutation is just a bijection

So by definition if you have f = sigma•g

[vestal cairn]

yeah

[next obsidian]

And f is a bijection and sigma is a bijection

How can you show g is a bijection

[vestal cairn]

idk😔

[next obsidian]

Do you know what a bijection is

[vestal cairn]

yeah

[next obsidian]

Can you define it for me

[vestal cairn]

an injective and surjective function

[next obsidian]

Well…

That’s true…

But there’s a more useful definition for this situation

[vestal cairn]

hm..

not quite sure what it is

[next obsidian]

Has an inverse?

I mean you can prove it from your other definition by just plugging stuff in

[vestal cairn]

oh oh

[next obsidian]

But it’s easier if you just use that a bijection is a function with an inverse

[vestal cairn]

ya i just found the other definition as u sent that

[next obsidian]

Okay

[vestal cairn]

so

[next obsidian]

So now you have f = sigma•g

If these were like numbers or something

How would u solve for g

[vestal cairn]

inverse of sigma on both sides

[next obsidian]

Right

So what do you get?

[vestal cairn]

then we'd have g = sigma^{-1} \circ f

[next obsidian]

Right

And now it’s clear g is a bijection yeah?

[vestal cairn]

well to me ya, but on paper? it feels like somethings missing

[next obsidian]

Well

I mean you can prove it one of two ways

Ultimately you want to show the composition of two bijections is a bijection

Because f and sigma^-1 are both bijections

You can either do this by showing it’s injective and surjective by hand

Or you can produce an inverse using the inverses of the two functions you’re composing

I think it may be beneficial for you to write out both proofs

[vestal cairn]

hmm okay

thanks

[sharp dirge]

example of commutative Char 0 ring but not an integral domaiin?

i cant think of anything

[next obsidian]

There’s a lot

I think u should think a bit harder tbh

[sharp dirge]

hmm

ok

[next obsidian]

Well

[sharp dirge]

nvm lol

[next obsidian]

They will work if you take it over a commutative char 0 ring which isn’t an integral domain

So it won’t really help much

[sharp dirge]

lol ye

[next obsidian]

But you could do another operation to a polynomial ring

And get an example

So how do you know to make new rings out of old ones

[sharp dirge]

my only example of rings are Z Q R C and quotients

[next obsidian]

It seems like you’re aware of polynomial rings

Any other examples?

[sharp dirge]

matrix rings but they arnt commutative

no quotients of Z Q R C can work

i think

[next obsidian]

That’s true but

Quotients

What if u tried to quotient a polynomial ring

[sharp dirge]

oh

[next obsidian]

Could you add zero divisors like that without changing the characteristic?

[sharp dirge]

hmm

errrrr

i think so

[next obsidian]

That will be Z again

But you’re on a right track

[sharp dirge]

i need to quotient by a non prime ideal

[next obsidian]

Right

So when would (f(x)) be prime?

Or rather

When would it not be prime?

[sharp dirge]

if ab in (fx) then a and b is not in it

[next obsidian]

Sure

[next obsidian]

This should be “and” though

Not “then”

[sharp dirge]

oop yes

so <x^2 + x + 2> should work

its not prime ideal

[next obsidian]

Let me think

[sharp dirge]

(X+1)(X+2) = that

[next obsidian]

I don’t think that’s quite what you want?

[sharp dirge]

oh

[next obsidian]

[sharp dirge]

lel

<x^2 + 3x + 2>

[next obsidian]

Sure

So okay

It has zero divisors

Can you show it’s characteristic 0?

[sharp dirge]

Z[X] still has Char 0

[next obsidian]

Sure

But we quotiented

And we already know we can destroy being char 0 after a quotient

[sharp dirge]

and Z[X]/<x^2 + 3x + 2> still has all of Z

[next obsidian]

Hmmm that’s true

But that’s almost just restating it

[sharp dirge]

we can embed Z into that quotient

[next obsidian]

I think you would want to evaluate why that’s true

And write a proof

Ah, okay

So can you prove that?

[molten viper]

(Thanks btw @ chmonky and @ levens)

[sharp dirge]

yeah i can prove it

[next obsidian]

How would you do that?

[sharp dirge]

homomorphism

first iso theorem

contains a copy of Z

[next obsidian]

Okay, what’s your map

[sharp dirge]

identity on Z

[next obsidian]

Well, not quite but

I get what you mean

You send n to n + <x^2 + 3x + 2>

[sharp dirge]

inclusion map

[next obsidian]

Right?

[sharp dirge]

yes

in terms of cosets yes

[next obsidian]

Okay but the point that you’re kind of glossing over

Why is this injevtive?

Proving that is equivalent to showing the ring is char 0

Which is why I think you haven’t really proven it yet

[sharp dirge]

good question

[narrow marsh]

generally its ez to prove 1 to 1 for bijection

[sharp dirge]

its because we cant divide by the ideal basically

constants

[next obsidian]

So what would it mean for n + <x^2 + 3x + 2> to be 0?

[sharp dirge]

integers

[sharp dirge]

n to be in the ideal. or n can be divided by it

[next obsidian]

Right

And so why’s that impossible?

[sharp dirge]

no integers can be divided by it

lower degree

[next obsidian]

Because?

[sharp dirge]

smaller

[next obsidian]

Right

That’s all I was looking for

You can appeal to degree

[sharp dirge]

ah

[next obsidian]

But I think u gotta say that ya know?

But yeah there’s an example

[sharp dirge]

thank you very much

[next obsidian]

Anyway so now I’ll describe a few others

[sharp dirge]

ok

[next obsidian]

First off, I would quotient by x^2

Since that’s pretty simple

And clearly gives a zero divisors

But your example is equally fine

Next up: consider the product ring R x S

If R or S is char 0 this is still char 0

But a product of rings can never be an integral domain unless R or S is 0

Namely, take (1,0)•(0,1) = (0,0)

Final example:

Consider the set of functions / continuous functions / smooth functions / whatever from say, R -> R

R the real numbers say

We can make this a ring by defining (f +g)(x) = f(x) + g(x)

And (fg)(x) = f(x)g(x)

[sharp dirge]

this is a good example

[next obsidian]

Then this has zero divisors, take day a function f which is 0 at 0, but not zero, so say f(x) = x

Let g be zero away from 0, so say g(x)=1 if x = 0, g(x)=0 otherwise

Then neither of these are zero, but their product is

(Note that g isn’t continuous / smooth, so if you want examples for continuous / smooth functions you take like bump functions where the place they’re non-zero doesn’t overlap)

[sharp dirge]

i see

hm

i also have another small question maybe you can help

i dont understand how A is equivalent to B there

do you know

[next obsidian]

I suck with matrices

I think you just like

[sharp dirge]

same

[next obsidian]

Cook up a matrix that does it tho

[sharp dirge]

do i need to explicity find them or can i just do it via elementary row operations?

[next obsidian]

I mean

[sharp dirge]

or are they the same

[next obsidian]

I think that’s the idea

They’re basically the same

[sharp dirge]

hmm k

[next obsidian]

Like via row operations WLOG the first row is nonzero

Then by Column operations WLOG a11 is nonzero

Then you find some matrix which when you conjugate by it kills off everything but a11

Or something

[sharp dirge]

i can do column operations too?

[next obsidian]

I think so right?

You can describe that by like

[sharp dirge]

i think so

[next obsidian]

A matrix

You swap the ith and jth column

It’s still gonna be invertible

Idk tho I’m bad

[sharp dirge]

hmm okay

thank you, i think i understand

[vestal cairn]

i couldnt figure this one out

[true cairn]

Suppose G - <a> is a cyclic group of order 12.
(a) Find all the generators of G. Explain your reasoning.
(b) Find all of the proper subgroups of G, and list their elements. Find all the generators of each
subgroup. Explain your reasoning. What are they asking in part b of the above question?

[smoky cypress]

Bump

[delicate bloom]

maybe I'm not explaining myself or you missed it or I'm just so obviously wrong and I'm blind here: #「ivory-tower-emeritus」 message

[smoky cypress]

Oh I straight up didn't see oops

[delicate bloom]

oh yeah I think what I said is wrong but like

the truth shouldn't be too far off

[south patrol]

smh i also answered

bleak

[smoky cypress]

Idk man just feels too good to be true

If math has taught me anything that is if anything nice is happening that's because you're still dreaming

[delicate bloom]

I think that answer potato was saying could be right, like you're taking the trace from one field to a subfield, I think that can work

like you can show it's fixed by the automorphisms it needs to be and not by the ones it shouldn't

[smoky cypress]

Ohh I didn't fully comprehend it but now I do

[south patrol]

Hm i may have mucked up though

cause like need to see how it works with generators

[cloud walrus]

potato

[south patrol]

or maybe that is true and I'm being dense lol

[south patrol]

ye so i looked it up and someone did what I did up to the bit where you say it's a generator lol

Like if you use the lemma I wrote about V^G = \sigma . V then that gives you a set of generators

i.e. sum_h h. zeta^i for all i = 1,...,p-1 all generate it

the issue is then reducing that to a single generator

and maybe we just got lucky with one of those sufficing

i can't think of any examples where you might need further linear combinations though

hm

[smoky cypress]

I don't want to do explicit calculations

[true cairn]

Can someone help me with part b?

[tender wharf]

Do you know the fundamental theorem of cyclic groups?

[true cairn]

yes

[tender wharf]

Great, then it should directly follow

[true cairn]

so the generators are 1, 5, 7, 11 right?

[tender wharf]

Are you talking about part a?

[true cairn]

yes

[tender wharf]

Yes

Follows directly by applying said theorem

[true cairn]

for part b, <1> =G are so it is not a proper subgroup right?

[tender wharf]

If 1 is the identity, then no.

<1> = {1} so it's a trivial proper subgroup

[true cairn]

0 is the identity

[tender wharf]

Ah then yes

It's the whole group so it can't be a proper subgroup

[true cairn]

got it, so when I did other generators, only 11 is a proper subgroup does that sound right to you

[tender wharf]

<11> makes the whole group

Because gcd(11, 12) = 1

[true cairn]

oh I see

[tender wharf]

So all those other generators you gave those are the whole group

Because they all have orders coprime to 12

[true cairn]

so if the generator gives the whole group, it is not a proper subgroup?

[tender wharf]

Yes, by definition of proper subgroup

If H is a proper subgroup of G then H is a subgroup of G and H != G

[true cairn]

ok, thank you

[tender wharf]

#books-old or #book-recommendations

I personally use Gallian

A friend I know used D&F for self study

What's wrong with Dummit and Foote anyway

Are you intending to self study?

Never tried nor do I know anyone who did so I can't comment

Have you actually tried it and found it "too easy"?

Let's move this discussion to #book-recommendations

[sharp peak]

Is the euclidean algorithm not equivalent to polynomial long division? and also when can you perform the euclidean algorithm?

[sharp peak]

I thought your ring had to be a euclidean domain in order for you to do the euclidean algorithm but you can long divide polynomials of two variables (\mathbb{C}[x,y])etc... and that's a ring that is a UFD and not Euclidean

[ruby sundial]

You should not be able to

[sharp peak]

https://www.kristakingmath.com/blog/predator-prey-systems-ghtcp-5e2r4-427ab

[chilly ocean]

@sharp peak This and the following conversation may be relevant.

As long as you're working over an integral domain, you can divide by any polynomial whose leading coefficient is a unit.

The precise statement is in the image I replied to.

And the relation to Euclidean domains is clarified in the conversation there.

[ruby sundial]

@chilly ocean

So you know how all integral domains have are isomorphic to some quotient by a prime

what is corresponding theorem for ufd,pid,icd,gcd

Im guessing it has to do with length of prime that gets quotiented

[chilly ocean]

Off the top of my head, no idea.

[ruby sundial]

same

[eager willow]

There isn't necessary going to be a theorem of ideals for any of those. The theorem for integral domains and fields is that a commutative ring with identity is an integral domain iff 0 is prime, and a field iff 0 is maximal. Simultaneously by zorn every commutative ring with identity can quotient onto a field, supporting the idea that these notions of primality and maximality are 'nice', as opposed to what you're searching for.

I'm rephrasing your question now as "whats a type of ideal such that a ring is a ufd, pid, etc iff 0 is of this type".

I don't think there's any nice characterization of these because the structure that is necessary in these kinds of domains is, roughly speaking, happening too deeply outside of 0. While fields also have this flavor 'finding structure (multiplicative inverse) outside of 0', I am thinking this is not so deep outside of 0, because the role of the 0 ideal is still that it contains all non-units, and is therefore clearly an important ideal to look at for determining if a ring is a field. The reformulations as maximality is a somewhat trivial consequence of that intuition. Contrast this with definition of the special domains you've considered. Does 0 have a starring role in these definitions? Not really, not as special as in the definition of integral domains and fields.

[white grotto]

can someone explain to me the minimal polynomial

what does it tell us

i know its the smallest polynome that a matrix evaluated in this polynome is 0

but whats the relation between this and the characteristic polynomial and proper values

and whats the relation between this and other matrices that evaluated at this polynome is 0

and whats does it tell us about a given matrix? if we have a minimal polynomial what info could we extract?

[cloud walrus]

grass

[hidden kite]

Well actually nevermind haha (deleted my msgs)

[fallow laurel]

This guy has some really insightful lectures online

https://www.youtube.com/channel/UCIyDqfi_cbkp-RU20aBF-MQ/playlists

found out he won the fields medal

I recommend his group theory lectures, super duper gentle

[maiden heath]

Is the tensor product of the sign representation with itself still the sign representation or is it the trivial representation?

[quiet pelican]

It’s the trivial rep, as the trace of the product with itself is the square of it, and (-1)^2 = 1^2 = 1

[pastel cliff]

how does localization connecr to UFDs

[maiden heath]

Amazing thanks

So characters of tensors just multiply?

[quiet pelican]

Yes

[nocturne bone]

I'm currently trying to do e. I've proved that ord(Z) = p or p^2 (so it is not trivial). I know why G is abelian if ord(Z) = p^2 (because then G = Z and boom)
But why is G abelian if ord(Z) = p?

[quiet pelican]

Consider the quotient group G/Z(G)

[void cosmos]

^

[nocturne bone]

Alright, so the quotient group would have p elements if ord(Z(G)) = p. How does that imply cyclicity?

[quiet pelican]

Because what can the orders of non identity elements possibly be?

[nocturne bone]

Ah! Only p

And then that implies G being abelian because the center is cyclic

Thank you!

[formal ermine]

I'm having a hard time showing that $F : R \to R, x \mapsto x^p$ for $R$ a commutative integral domain and $\operatorname{char}(R) = p > 0$ is a ring homomorphism

[cloud walrus]

illuminator3 (I/you)

[formal ermine]

specifically the $F(a + b) = F(a) + F(b)$ part

[cloud walrus]

illuminator3 (I/you)

[formal ermine]

do I need to use the binomial theorem or something?

[chilly ocean]

Have you tried using the binomial theorem?

Do that, and then carefully look at the binomial coefficients you expect to vanish.

Argue that they do.

(You must use the fact that p is prime in doing so.)

[south patrol]

Isn't sigma f basically the whole point of it? You want a hom K -> M which induces a map K[x] -> M[x] sending f to smth which splits

[thorn delta]

if F/K and F'/K are two splitting fields for a polynomial f in K[x], then take sigma : K --> F'

that should be it

[restive birch]

struggling with this problem. It seems intuitive, but I can't find a way to formalize it.

any insight?

[formal ermine]

what's the context

[little root]

i assume youre trying to prove rs = sr^{-1}? If you want to prove it algebraically you have to write r and s as permutations of {1, ..., n}, like the book says. The rotation r would just be shifting everything either left or right, so either (n, 1, 2, ..., n - 1) or (2, 3, ..., n, 1) depending on how you label vertices and whether the rotation is CW or CCW. If you draw a picture then you should also be able to figure out which permutation s corresponds to. Since permutations are just maps of {1, ..., n} you should be able to work out explicitly what rs and sr^{-1} represent

[south patrol]

wdym lol

uh

[little root]

(i) basically says that if f splits in a field extension M / K, then the splitting field of K should be a subfield of M.

But you can't say this exactly since L isn't necessarily a subfield of M, but rather something that is isomorphic to L. So what I said above would be more correctly stated as: if L / K is a splitting field of f and M / K is a field extension such that f splits in M, then there is a field homomorphism tau: L -> M extending the identity K -> K. Then we have M / tau(L) / K, where tau(L) is the isomorphic copy of L.

Even more generally, you want to be able to consider when K is only isomorphic to a subfield of M. The way you do this is to consider maps sigma : K -> M

[south patrol]

Isn't this one of the cornerstones of galois theory

Maybe what Frank said is what u were concerned about

remember all field homomorphisms are injective and so these are basically just embeddings of fields

[little root]

basically you just want to consider things up to isomorphism and while being precise about it

[south patrol]

Also for this like

if F is some field and we form some extension K of F, then we can view polynomials in F as polynomials in K by applying sigma to the coefficients

Of course, if F is actually a subfield of K then this is just the obvious inclusion of F[x] in K[x]

Very often in field theory shenanigans it's convenient to not have to worry about actual set theoretic inclusions so you can form extensions more easily

Well the in particular is just taking L and M to be splitting fields

Because i) shows the map K -> M extends to a map L -> M and ii) shows that this extension is an iso

[formal ermine]

ahhh yeh got it

thanks

another question

is this argument true?

Let $R$ be a PID, then for any $x \in R \setminus \Set{0}$ the ideal generated by $x$ will be maximal iff $x$ is irreducible. \
\
Proof:
[
\begin{align}
(x) \text{ maximal} &\iff (x) \text{ is not contained in any } (d) \neq (x) \
&\iff x \text{ has no divisor } d \neq x \
&\iff x \text{ is irreducible}
\end{align}
]

[cloud walrus]

illuminator3 (I/you)
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[tender bough]

Hello guys, lie algebra question again.
So by wikipedia, a Cartan subalgebra over C can be defined as a maximal abelian and ad-diagonalizable subalgebra.
i.e. there exists no containing subalgebra which is both abelian and ad-diagonalizable.
However, my lecture notes defines a Cartan subalgebra over C only as a maximal ad-diagonalizable subalgebra.
Additionally, my lecture notes also points to the fact that every ad-diagonalizable subalgebra is abelian.
This makes me wonder whether the two definitions are equivalent.
Indeed, with the fact, (maximal ad-diagonalizable) entails (maximal ad-diagonalizable and abelian). Can be seen by proof by contradiction.
But how about the other way around?

[chilly ocean]

Is showing that the coset operation for groups is well defined as simple as ghN=(gN)(hN)=(g'N)(h'N)=g'h'N for gN=g'N and hN=h'N?

[south patrol]

Seems fine ye, though ofc the first and last equality are justified by normality of N as a subgroup

[willow mason]

It’s not really clear how you proved that x is irred

In a PID, there is no distinction between prime and irreducible elements

you can use that if you like

[chilly ocean]

ghN = (gN)(hN) ? Is that not just the definition of ghN ?

[south patrol]

No, it's not

[chilly ocean]

I guess not, yea

[south patrol]

ghN is simply (gh)N like the coset

You need N to be normal here

e.g. if N is normal then you can do gNhN = ghNN =ghN

which feels abusive but does work when you go through it all lol

[chilly ocean]

What is ghNN ? Where do the parenthesis go?

Is it subset multiplication (ghN)N?

[south patrol]

Well okay maybe that is abusive but when i write ABCD i mean the set of products abcd where a in A, b in B etc

if you use this convention then the parentheses actually don't matter

[formal ermine]

yeah that's what I used

I probably forgot to specify that d isn't a unit

[real scarab]

is there a standard notation for the set of join prime elements of a lattice?

[iron vessel]

I want to show that if we have G a finite group, H a subgroup, and R the complete set of representative of the left H-cosets, then the multiplication map $\pi:\mathbb{C}[R] \times \mathbb{C}[H] \to \mathbb{C}[G]$ given by $(a,b) \mapsto ab$ induces a linear isomorphism $\mathbb{C}[R] \otimes_{\mathbb{C}} \mathbb{C} [H] \to \mathbb{C}[G]$.

[cloud walrus]

Évariste Galois

[iron vessel]

It should be trivial to see that $\mathbb{C}[R]$ is a $\mathbb{C}$-module. So, using the universal property, we can find a unique linear map $\alpha: \mathbb{C}[R] \otimes_{\mathbb{C}} \mathbb{C} [H] \to \mathbb{C}[G]$ such that $\pi = \alpha \circ \gamma$; where $\gamma$ is the universal complex bilinear map associated to the tensor product above.

[cloud walrus]

Évariste Galois

[iron vessel]

I struggle to see a way to show it is an isomorphism.

[rustic crown]

show surjective, and use rank-nullity to get injective (by showing both sides have same C-dimension)

[iron vessel]

Yeah I ended up figuring it out.

Thanks alot!!!

[tender bough]

If anyone is interested. I have determined that those two definitions are equivalent. The reverse direction relies on a proposition that an ad-diagonalizable subalgebra is always abelian. An interesting note is that, the less wordy definition without the "abelian" is a logically stricter requirement. Due to how quantifiers work.

[sweet echo]

Homology of R-modules: the nth homology module of a chain complex is Z_n/B_n where Z_n is the kernel of the out map and B_n is the image of the in map. This quotient only makes sense if B_n is a normal subgroup of Z_n right? My textbook doesn't mention anything about existence of these things so maybe it's easy to see and I'm just missing why that's true?

Oh everything is abelian

[white grotto]

When is the algebraic multiplicity greater than the geometric multiplicity for an eigen vector
what goes wrong

[rotund aurora]

If you have m linearly independent vectors with eigenvalue Lambda, you can extend this set to form a basis for the whole space

Then, the characteristic polynomial is det(A-Ix) where A is the matrix representation of your linear operator with respect to your basis

and A has the top mxm upper-left-corner (after a suitable rearrangement), a diagonal matrix with Lambda as its entries. Those correspond to your linearly independent vectors of eigenvalue Lambda

[south patrol]

One perspective on this: if $V$ is a vector space over an (algebraically closed) field $F$ and $T: V \to V$ a linear transformation then we can factorise the minimal polynomial $m_T$ into irreducibles as $m_T = \prod_{i=1}^{n}(x-\lambda_i)^{r_i}$ and correspondingly decompose $V$ into $T$-invariant subspaces by $V = \bigoplus_{i=1}^{n} \ker((T-\lambda_i I)^{r_i})$. Now the snag here is that that the algebraic multiplicity of $\lambda_i$ is $r_i$, the dimension of $\ker((T-\lambda_i I)^{r_i})$, whilst the geometric multiplicity is the dimension of $\ker(T-\lambda_i I)$. This means the algebraic multiplicity $\ge$ geometric multiplicity and yet inequality may not hold in general because of the possibility of the $r_i$ being $> 1$

[cloud walrus]

potato

[formal ermine]

how do I show that if f and g are endomorphisms on G then f(x) + g(x) is also an endomorphism

this seems stupidly simple

[south patrol]

So ultimately, in a sense, it comes down to the fact that End(V) may have nilpotent elements, which perhaps motivates studying nilpotent operators e.g. in jordan normal form

[rotund aurora]

definitions

[formal ermine]

but I can't figure it out

[rotund aurora]

read the definitions

[south patrol]

Try to show that f + g satisfies all the necessary properties to be an endomorphism

[formal ermine]

f(x) + g(x) is G -> G because + : G -> G and both f and g are G -> G?

[rotund aurora]

• is nto G->G, its GxG->G

That's why that expressoin makes sense

[formal ermine]

same thing

[south patrol]

Besides, being an endomorphism is not the same as "being G -> G"

[rotund aurora]

just read the definitions

[south patrol]

The important thing here is being an endomorphism, not just a function G -> G

[formal ermine]

wut

that's how my homework defined an endomorphism

oh wait

[rotund aurora]

no it didn't

[south patrol]

I'm sure it didn't

[formal ermine]

group homomorphism

[rotund aurora]

read the definitions

[formal ermine]

I forgor

[white grotto]

How do we make sure that the generalized eigen vectors are independent?

and form a basis

[south patrol]

Not sure what you mean

Oh like do you mean like how do you prove the assertion I made that you have that decomposition?

[white grotto]

ker(A-lambda I)^n gives us generalized eigen vectors

[south patrol]

Indeed

[white grotto]

Which form a basis

so i am asking why?

[south patrol]

Of ker (A - lambda I)^n?

[white grotto]

Of the whole space

[south patrol]

Well that is just the decomposition I gave above

:)

[white grotto]

Yeah i know i am asking why are they independent

why do they form a basis

[south patrol]

So yeah isn't that basically just asking why the decomposition I gave holds

[white grotto]

yeah

no

no i get that each subspace is disjoint from the others

what i dont get is how we make sure to get vectors that form a basis to each subspace

[south patrol]

I'm still not sure what exactly you mean by that like

We can form a basis of each generalised eigenspace like

Since it is a vector space we can just pick a basis

Then whenever you have a direct sum decomposition, you can string together bases of each direct summand to form a basis of the whole space

Linear independence follows immediately from the definition of direct sums

[south patrol]

Or do you want me to justify this

[formal ermine]

$\operatorname{End}(\bZ_2) = \Set{\operatorname{Id}, n \mapsto n + 1, n \mapsto 0, n \mapsto 1, 0 \mapsto 0, 0 \mapsto 1, 1 \mapsto 0, 1 \mapsto 1}$?

are those all

[cloud walrus]

illuminator3 (I/you)

[simple mulch]

If we have a group of order $p^2$ what can we tell about the order of its elements?

[cloud walrus]

mns

[simple mulch]

p is prime tho

[formal ermine]

well the order of any element must divide the order of its group

by lagrange

[simple mulch]

but does it guarantee us that there's an element of order p^2 since the division would yield 1?

[formal ermine]

it doesn't

if there exists an element of order p^2 then the group is cyclic

[simple mulch]

p being prime doesn't affect anything?

[formal ermine]

it means the order of the elements can only be 1, p, or p^2

also some additional info: the group must either be $\bZ_{p^2}$ or $\bZ_p \times \bZ_p$ (cartesian product)

[simple mulch]

okk thanks

[cloud walrus]

illuminator3 (I/you)

[restive birch]

struggling with the idea of generators and relations- does a value being in the generator set mean that its inverse is also used for the generation?

so $\mathbb{Z} = \langle{1}\rangle$ and not $\langle{1, -1}\rangle?$

i know the carat symbols are supposed to be different, i dont know the code for them

[rotund aurora]

\rangle 1 \langle

$\rangle 1\langle$

[restive birch]

thanks

[cloud walrus]

Croqueta

[rotund aurora]

ok, I am 5 years old, but you get the point

cannot distinguish left and right xd

[rotund aurora]

In group theory you usually include the inverses yes

so Z=<1> and Z=<1,-1> are both correct

[cloud walrus]

henryduke

[restive birch]

okay, and theres no relation needed?

[rotund aurora]

no

[formal ermine]

@restive birch $\gen{g} = \Set{g^n | n \in \bZ}$, you can see that it includes inverses and the neutral element too

[cloud walrus]

illuminator3 (I/you)

[restive birch]

i'm still having trouble with what $\mathbb{Z}/n\mathbb{Z}$ means exactly. Is it a set of sets of numbers, grouped by the number mod n, or is it just a set of all of the numbers 0≤a<n?

[cloud walrus]

henryduke

henryduke
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[formal ermine]

it's Z but in mod n

so the numbers 0,1,2,...,n - 1 (assuming addition as operation)

and if you do a + b you gotta mod n at the end

so 2 + 3 = 5 = 1 (mod 4) in Z/4Z

[restive birch]

so its just a group of n numbers under addition, with an additional rule

[formal ermine]

yes, kind of, but not really

have you done cosets already?

[restive birch]

maybe? what are those again

[formal ermine]

$x \circ G = \Set{x \circ g | g \in G}$ is a left coset of G

[cloud walrus]

illuminator3 (I/you)

[restive birch]

okay, so like, all of the possible outputs of an operation x when applied to elements of a set G?

[formal ermine]

yeah, something like that

G/H means all possible left cosets of H where the x is in G

$G/H = \Set{g \circ H | g \in G }$

[restive birch]

im confused

[cloud walrus]

illuminator3 (I/you)

[restive birch]

where $\circ$ is just some binary operation?

[cloud walrus]

henryduke

[formal ermine]

yes, the operation of G

note that H must be a subgroup of G

[restive birch]

wait, are we talking about sets or groups?

[formal ermine]

groups

but remember that a group is just a set with an operation and some conditions

[restive birch]

so cosets are defined by groups? thats probably why i havent come across them yet, im at the beginning of the section on groups

[formal ermine]

cosets come right after subgroups, usually

[restive birch]

yea, those havent come up yet

subgroups are chapter 2

(working through Dummit and Foote, per the server's recommendation)

[formal ermine]

ye then just go with this 'definition' of Z/nZ for now

[formal ermine]

nice

I'm in an algebra 1 course at uni right now

[restive birch]

so its just a special form of addition on a set {0, 1, 2, ... n-1}

does that work for now?

[formal ermine]

yeah

it gets a bit more tricky with multiplication

because not every element will be invertible

[restive birch]

only those that are coprime with n, right?

[formal ermine]

yes

[restive birch]

so $\mathbb{Z}/n\mathbb{Z}^{\times}$ is just multiplication on the set ${a\in\mathbb{Z}|0\leq{a}\<n, (a,n) = 1}$ where $(a,b)$ is the greatest common factor of a and b

[cloud walrus]

henryduke

[restive birch]

?

[formal ermine]

multiplication mod n

[restive birch]

right

[formal ermine]

you usually write gcd(a, n) btw

[restive birch]

makes sense

[formal ermine]

also $\bZ_n = \bZ/n\bZ$

[cloud walrus]

illuminator3 (I/you)

[restive birch]

and multiplication mod n is $a \times_n b = a \times b - cn$ for some $c \in \mathbb{Z}$ and such that $ (a \times b) - cn \in \mathbb{Z}_n$

[cloud walrus]

henryduke

[south patrol]

Idk this isn't normally how I would define the operations, seems a bit confusing to do it like this

But then idk if you have done quotient groups

[restive birch]

how would you define them?

i have not done quotient groups as far as i am aware

[formal ermine]

he hasn't seen subgroups yet

[south patrol]

Oop

[restive birch]

but even though its clunky, does it work?

[chilly radish]

(a,b) is common enough notation

[chilly radish]

You can explicitly define it through.operations on equivalence classes

[formal ermine]

I've never seen it and it can be quite ambiguous imo. but yeah, it doesn't really matter what notation you use at the end of the day, as long as you define it properly

Let $R$ be a PID, then for a $x \in R \setminus \Set{0}$ we have:
$$\begin{align}(x) \text{ maximal } &\iff (x) \text{ is not contained in any } (d) \neq (x) \text{ where d is not a unit}\&\iff x \text{ has no divisor } d \neq 0 \text{ where } d \not \in R^\times\&\iff x \text{ is prime}\&\iff x \text{ is irreducible}\end{align}$$

is that correct?

[cloud walrus]

illuminator3 (I/you)
Compile Error! Click the reaction for more information.
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[willow mason]

whats ur definition of prime

@formal ermine

[formal ermine]

an element $p \in R$ is prime, if $p \neq 0, p \not \in R^\times$ and from $p | (xy)$ always follows $p | x$ or $p | y$

[cloud walrus]

illuminator3 (I/you)

[willow mason]

i think its fine

[formal ermine]

oke thank you c:

[nocturne bone]

So I'm trying to do c. If ker(f) = {e}, why does that imply G is isomorphic to f(G)?

[quiet pelican]

First isomorphism theorem

Also there’s no part e on the picture you sent

[nocturne bone]

Ah I meant c

[formal ermine]

yeh iso theorem time

[nocturne bone]

I'm not sure I follow why

[formal ermine]

if f : G -> G' is a homomorphism then G/ker f is isomorphic to the image of f

[quiet pelican]

Bc then $G/ker f \cong im f$, so that $G \cong im f = f(G)$

[formal ermine]

if you now have G' = G you get that G/ker f is isomorphic to f(G), but you know that ker f only contains the neutral element, so it simplifies to G being isomorphic to f(G)

[cloud walrus]

Micos

[winged quail]

im stuck on this dont rly know where to go from here thought about using counting formula but that doenst rly help

[nocturne bone]

oki! that makes sense. Thanks guys!

I appreciate the help

[formal ermine]

how is G_s defined?

[winged quail]

G_s is the stabliizer = {g in G | g*x=x}

[quiet pelican]

First part: consider the size of the set ||{gs = s, g in G, s in S}|| in two different ways
Second part ||apply orbit stabiliser to the LHS||

[lethal dune]

Can anyone help me see why the ker is the homology group

[winged quail]

does it suffice to show that the Union across all s in S of G_s = that set and similarly the Union across all g in G of S^g is presicely that set

[quiet pelican]

Yes

And that the intersections Sg \cap Sh and Gs1 \cap Gs2 are all empty

But that second part is trivial

[formal ermine]

how do I show that $\operatorname{gcd}(3, (1 + i\sqrt{5})) = 1$ in $\bZ[i\sqrt{5}]$?

[cloud walrus]

yes yes yes no

[formal ermine]

I can't really use euclid's algorithm because we're not in a euclidean ring

[quiet pelican]

Use the norm in the complex numbers, N(3) = 9, N(1 + i sqrt(5)) = 6, so any non unit divisor has norm 3, but it’s easy to see there’s no elements with norm 3

[formal ermine]

which norm?

we never defined what 'norm' is in my course

[south patrol]

ye that's what i was finna do ig

[formal ermine]

I only know the term from numerical analysis

[quiet pelican]

N(x + yi) = x^2 + y^2

[formal ermine]

ah

[quiet pelican]

Interesting, it’s quite standard to use that N to help with divisibility in subrings of C

[formal ermine]

yeah we randomly started using it but never defined it lol

[elder wave]

You were probably assumed to be familiar with it

[formal ermine]

what does "norm 3" mean here?

[quiet pelican]

N(z) = 3

[formal ermine]

so any non unit divisor has norm 3,
why?

[quiet pelican]

As it must divide 6 and 9 (as N is multiplicative, N(xy) = N(x)N(y)), and can’t be 1, as anything with N(z) = 1 is a unit

[formal ermine]

uh I don't quite get the line of reasoning

so we have N(3) = 9 and N(1 + isqrt(5)) = 6. why must the gcd divide both of these norms?

[quiet pelican]

Suppose x divides 3 and x divides 1 + i sqrt(5). Then 3 = xy so that N(3) = N(xy) = N(x)N(y) so that N(x) divides N(3) and similarly for 1 + i sqrt(5)

[formal ermine]

ah ok

the norm is only multiplicative if the imaginary part is 0, right?

[quiet pelican]

It’s always multiplicative

Just write it out in re^i theta form

And calculate out

[formal ermine]

N(ae^(ix) * be^(iy)) = (ae^(ix) * be^(iy))^2 = (ae^(ix))^2 * (be^(iy))^2 = N(ae^(ix))N(be^(iy)) like this?

wait nvm no

that doesn't make sense does it

[quiet pelican]

Almost, just drop the e^ix and e^iy in the middle two, using that N(a e^i theta) = a ^2

[formal ermine]

mhm

yeh

[formal ermine]

so we have N(a + bisqrt(5)) = a^2 + 5b^2 = 3

b has to be 0 because otherwise we'd already be overshooting by at least 5

so a^2 = 3

but that has no integer solutions

correct?

[quiet pelican]

Yes

[winged quail]

is it valid to say sum across s of G_s = {g in G | g""s=s, s in S} = {g""s=s | g in G, s in S } = {s in S | g"*"s=s, g in G} = sum across g of S^g

[quiet pelican]

You need sums in the 2nd and 4th terms

I assume you just forgot to write them

Oh wait I can’t read

It’s fine

[formal ermine]

how do I conclude that therefore 2(1 + isqrt(5)) and 6 must have no gcd either? is gcd(xa, xb) = gcd(a, b)?

[winged quail]

do i need to put order's around the sets cuz the sets are not identically equal or wait r they ... maybe im just not grasping what the set {g*s=s | g in G s in S} means

[winged quail]

gcd(xa,xb) = x * gcd(a,b)

[quiet pelican]

Since we’re working in a non UFD, you probably want to check all factors of each number

[quiet pelican]

I would put orders in

[quiet pelican]

Actually, this is an example where it fails, as 2 and 1 + i sqrt(5) divide both, but their product doesn’t, so there is no gcd

[quiet pelican]

As demonstrated, this doesn’t necessarily hold outside a UFD

[winged quail]

sorry for the second part how do you go from |G| / | O(s) | to |G| * n orbits in S
i get summing order of orbits across all of s gives the order of S but then you are left with (sum over S of |G|) / |S|

sum over S of |G| should just be |G| * (elemnets in S) which is |G||S| so the resulatnt sum should just be |G| not this |G| ""n

[formal ermine]

they want me to go from gcd(1+isqrt(5), 3) = 1 to gcd(2(1+isqrt(5)), 6) = 1

[quiet pelican]

How does your course define gcd?

Because that’s the first example used of where it doesn’t exist in a lot of places

[winged quail]

i think my assumption that Sum of S of |O(x)| is wrong this looks like question looks like a backwards way to prove burnisde lemme

where X/G is set of orbits

i dont understand the reasoning tho you sum x in first the orbit and then sum across all the orbits

is that it?

[rotund aurora]

|Orb(x)| is constant for x in the the same orbit

And there are exactly |Orb(x)| elements in Orb(x)

[winged quail]

ohh makes sense

ok thx everyone for help lol

[formal ermine]

Let $R$ be an integral domain and $x_1, \ldots, x_n \in R$. An element $d \in R$ is called greatest common divisor of $x_1, \ldots, x_n$ if $d: |: x_i$ for all $1 \leq i \leq n$ and if $d' \in R$ with $d': |: x_i$ also holds true for $1 \leq i \leq n$ then $d': |: d$.

[cloud walrus]

yes yes yes no

[quiet pelican]

Yeah, then we have that 1 + isqrt(5) divides d, and 2 divides d, but there is no number with those properties which divides 6, so there is no gcd

And that is the standard way

[formal ermine]

so we have the following conditions for our gcd?
d | 1 + isqrt(5)
d | 2
d | 6

yeh

so we have d | 2 => d = 1 or d = 2 but neither of those divide into 1 + isqrt(5)

[quiet pelican]

No, the first two should be the other way around

[formal ermine]

why?

[quiet pelican]

2, 1 + isqrt(5) take the role of d’ in the definition

[formal ermine]

wait

we have d being the gcd of 2(i + isqrt(5)) and 6 => d | 2(i + sqrt(5)). but if d divides an even number then d has to be even itself, so d | 2 and d | i + sqrt(5) => d = 1 or d = 2 but neither of those divide into i + sqrt(5). is this wrong?

[quiet pelican]

Since it’s not a UFD, we don’t have that d|2(1 + isqrt(5)) => d|2, and we wouldn’t even have that in a UFD

[formal ermine]

right

[formal ermine]

I don't quite understand what we are contradicting here

[quiet pelican]

N(d) divides N(6) = 36, but also 4 =N(2) divides N(d) and 6 = N(1 + isqrt(5)) divides N(d) so 12 divides N(d) so N(d) is 12 (as can’t be 24 (as doesn’t divide 36), or 36 (as must divide 24 = N(1 + isqrt(6)))) but there is no element with norm 12

[formal ermine]

d is the gcd?

[quiet pelican]

Yes

[formal ermine]

how did we get N(2) | N(d)?

[quiet pelican]

As 2 | d

[formal ermine]

why do we have that again

oh

because 2 divides both of our x_is, right

N(d) | N(6) = 36
4 = N(2) | N(d)
6 = N(1 + isqrt(5)) | N(d)
12 | N(d)

4 | N(d) => N(d) = 4k
N(d) | 36 => N(d) in { 4, 12, 36 }
6 | N(d) => N(d) in { 12, 36 }
12 | N(d) => N(d) in { 12, 36 }

how are we eliminating 36?

[quiet pelican]

Because it must divide N(2(1 + isqrt(5))) = 24

[formal ermine]

ah

N(d) | N(6) = 36
4 = N(2) | N(d)
6 = N(1 + isqrt(5)) | N(d)
12 | N(d)
N(d) | N(2(1 + isqrt(5))) = 24

4 | N(d) => N(d) = 4k
N(d) | 36 => N(d) in { 4, 12, 36 }
6 | N(d) => N(d) in { 12, 36 }
12 | N(d) => N(d) in { 12, 36 }
N(d) | 24 => N(d) = 12

ahh

yes

thank you so much for this and the past answers!

[chilly ocean]

Let $G$ be an abelian group. Prove that the set of elements of order $2$, together with the identity element, i.e. $H = {a \in G : a^2 = e}$, is a subgroup of $G$.

[cloud walrus]

Ayanokoji

[chilly ocean]

Yes

so we're fine with closure

also we have $1 \in H$

[willow mason]

yes

[cloud walrus]

Ayanokoji

[willow mason]

so you just need that all elements have inverse

[chilly ocean]

and $a^{-1}$ is $a$ itself

[cloud walrus]

Ayanokoji

[willow mason]

yes

[chilly ocean]

hence H is a subgroup

thank you

Let $G$ be an abelian group and $H = {x \in G : x^3 = 1}$. Prove that $H$ is a subgroup of $G$.

[cloud walrus]

Ayanokoji

[chilly ocean]

for this, $H = {1, x, x^2}$. Hence we don't need to worry about closure because any element composed with another element will yield another element in the group (order of the element is equal to the order of the group). The inverse element of $x$ is $x^2$ and the inverse element of $x^2$ is $x$. Similarly, the identity is in the group, hence H is a subgroup of G.

[cloud walrus]

Ayanokoji

[coral shale]

no.

H = {1, x, x^2} makes no sense at all

Let G = integers mod 12 under addition

H = {x in G : x^3 = e} is still well defined

But H explicitly written out should not have any 'x' in it.

@chilly ocean

Maybe you're confused with set builder notation?

[chilly ocean]

perhaps, sorry for the late response

how would I go about proving this then?

[coral shale]

Do you know the example I gave

integers mod 12 under addition

this group

[chilly ocean]

yes

[coral shale]

ok, so what is H

in this case

H = {x in G : x^3 = e}

[chilly ocean]

um, x = 4

[coral shale]

only?

[chilly ocean]

no

also

8

and 12

wait

[coral shale]

(tbh i chose a bad example, woops)

well 12 is 0

so 0, 4, 8

[chilly ocean]

yeah 12 is 0

0, 4, 8

[coral shale]

I wanted an example where you dont just have 3 elements

uhhhhh

[chilly ocean]

so H is the group of elements of order 3

[coral shale]

Yeah, and that doesnt necessarily have exactly 3 elements

it could be 1

or more than 3

eg. for integers modulo 5

H = {0}

[chilly ocean]

oh, true

how would I prove generally that this is a subgroup then

[coral shale]

As for more than 3, i think we need an example other than modular arithemetic

well, you need to show the 4 axioms

so start with closure, perhaps

H = {x in G : x^3 = e}

[chilly ocean]

closure, inverse, identity

[coral shale]

So we need to show for any x, y in this H, xy also in H

so you start with x, y in H

[chilly ocean]

ok so

[coral shale]

and all you know is x^3 = y^3 = e

[chilly ocean]

take x, y in H

[coral shale]

yh yh.

then proceed to show xy in H

[coral shale]

and a piece of important info is needed to show this

[chilly ocean]

how would I do this?

[coral shale]

well what do u know about things that live in H

[chilly ocean]

they have order 3

so x and y have order 3

i.e. x^3 = y^3 = 1

[coral shale]

more precisely a in H if a^3 = e

so you need to show (xy)^3 = e

(i would avoid saying order 3 for reasons here)

[chilly ocean]

oh

right so (xy)^3 = x^3 y^3 = e e = e

[coral shale]

how does the 1st equals work

[chilly ocean]

uh

not sure

power rule?

[coral shale]

you claim xyxyxy = xxxyyy

[chilly ocean]

oh

[coral shale]

which is not necessarily true in any group

[chilly ocean]

you're right

$(xy)^3 = xyxyxy$

[cloud walrus]

Ayanokoji

[coral shale]

yh but there is something u can do, related to

[chilly ocean]

i'm not sure what that is

[coral shale]

read the question

You are given 2 pieces of information

G is abelian and the definition of H

[pine patio]

the definition isnt making sense, how can we say X1 and X2 are disjoint, and if they are disjoint then why do we need the second union in the relations of the push out

[coral shale]

g1 and g2 just need to be 'of different symbols'

[rustic crown]

Maybe it helps to understand pushouts in simpler categories first?

[coral shale]

i believe

i think they just want the disjoint union down there

[pine patio]

but then what's the purpose of the second union in the relations

phi1 and phi2 won't ever be same then

[rustic crown]

it's identifying the copy of G0 in G1 and G2 respectively

[coral shale]

hmm this feels familiar. algebraic topology

[pine patio]

oh so there's an underlying isomorphism

[cloud walrus]

det

[rustic crown]

you want this diagram to commute, and the object you put at the bottom right should be universal with respect to this requirement

if G_0 was the trivial group, you're basically asking what is the universal group which contains both G1 and G2

(in the category of sets, this would correspond to the disjoint union)

but here, we take presentations of both groups and put them together

[pine patio]

i havent done category theory yet, but i guess i understand what they might mean

[rustic crown]

ah okie, just ignore that word then :p

If G_0 is a non-trivial group, you would also like to identify the copies of G_0 you get in the free product

that's exactly what the extra set of relations do

[balmy python]

for group theory when do i use the star sign and the circle sign for a binary operation

i just started it today

[pine patio]

but X1 and X2 are disjoint so why would there be any g with phi1(g) = phi2(g)

[rustic crown]

ah i see your question

so like when you write a presentation
< x, y | x = y>

[coral shale]

G0 -> G1
G0 -> G2

You can have G1, G2 disjoint but these homomorphisms work?

[rustic crown]

this means that you look at the free group generated by x, y and then quotient by the normal subgroup generated x^-1y

so you want to force them to be equal

[coral shale]

Like G1 and G2 may even be isomorphic copies, but that doesnt stop them being disjoint
if G1 === G2 then you'll find {phi1(g) = phi2(g) | g in G0} = G0 (assuming the 'same' homomorphism)

[rustic crown]

ofcoure there is no g such that those things are already equal, they live in different groups afterall

[pine patio]

oh so they mean phi1(g) phi2(g) ^-1

[rustic crown]

yee!

you want to force this element to become identity in the pushout

so you add it as a relation

but writing relations by using "=" is a normal abuse of notation

the standard presentation for D_n is like < r, f | r^n = f^2 = e, frf = r^-1>

[pine patio]

yeah but i wasnt expecting them to use both of those notations together lol

[rustic crown]

which ig properly should be written like <r, f|r^n, f^2, frfr>

[pine patio]

yeah that makes sense now, thanks!

[rustic crown]

[coral shale]

do you really have to brute force this???

[rustic crown]

so say we want to brute force the existence of the additive identity, in the worst case we have to check all 4 elements to really find it

to check one element would require you to compute a * e and e * a for every a and check whether this is indeed a

that's 4 * 4 * 2 ig?

same number for multiplicative identity

[coral shale]

this is just a waste of time bruh

unless ur writing a program

[rustic crown]

lol

i think its more of a combinatorics exercise than really asking someone to verify it by bruteforce

[dusky canyon]

k

[rustic crown]

why did ya delete the question >.<

[dusky canyon]

cause i think i just got it but yea...

[civic tapir]

Given the free group Z<a,b,c>, when we look at the subgroup Z<a-b+c, a-b-c> and we apply a "change of basis" by subtracting the two generators to get the subgroup Z<a-b+c,2c>, what exactly is going on?
More specifically, how can we guarantee that we haven't refined our subgroup any? Surely, one can just check and prove for this case that we're good, but I'm moreso interested in the Linear Algebra mechanics at play, since it's definitely a bit more nuanced than when we're allowed to have R as coefficients.

[rustic crown]

as long as you do operations which are reversible, you're good

here, you subtract again to get back one of the old genrators, (a-b+c)-2c

[coral shale]

you can 'guarantee' by checking both inclusions ig

[civic tapir]

And I'm assuming the set of operations that consist of adding or subtracting the generators, when iterated, span the "reversible" operations?

[coral shale]

think u can do 'times' and 'divide' too

oh nvm lmao

[rustic crown]

i think for abelian groups, yep, but when you have a more complicated ring which isn't a PID, then you also would have other operations like multiplying a set of generators by an invertible matrix

[rustic crown]

yee, by units!

basically in the end i think you want to see whether the matrix map that sends one sets of generators to other is invertible or not

in your case the matrix was

[1  0]
[1 -1]```

and the question that you're asking is same as "is every invertible matrix over R, same as a product of elementary matrices?"

for euclidean domains this is definitely true

it might fail for PIDs as well, i'm not super sure

[civic tapir]

Ah, there it is! thank you!!

[lethal dune]

question got buried so asking again

[plush wasp]

By Cauchy's theorem, any even order group would not be simple right? excluding o(G) = 2. Because then there would be an element of o(x) = 2 and you can make a normal subgroup {e, x} that way which isn't trivial or G itself.

then why did I just read that o(G) = 60, 168, 360, etc. are exceptions

am I missing something maybe?

[next obsidian]

No

Simple means no nontrivial normal subgroup

There’s no reason the subgroup you make there has to be normal

[plush wasp]

ooo mb, that's why it felt wrong (@_@;) I kept wondering why gag^{-1} just didn't necessarily feel like it'd belong to {e, a} unless it's from Z(G)

[south patrol]

(Also note this argument would actually show no finite group is simple besides the prime order onesif it worked)

[obtuse bear]

worth mentioning that in S_n, all 2-cycles are conjugate to each other, so the subgroups of order 2 are very far from being normal

same is true in any group of order 2k where k is odd, if it contains more than one subgroup of order 2. In that case, the subgroups of order 2 are sylow subgroups, which again are all conjugate to each other

[south patrol]

Groups are weird

[nocturne bone]

Why does ord(sigma) divide k here?

[rustic crown]

k is in the kernel if and only if sigma^k = id

you wanna show that k would be divisible by ord(sigma)

this happens if and only if ord(sigma) divides k

more abstractly, if g is an element of a group G, then g^n = e is true if and only if n is divisible by the order of g.

(if you haven't seen the proof of this, argue as follows. assume g^n = e, and since we need to show it's divisible by ord(g), it makes sense to look at the remainder on division by ord(g), say n = ord(g) * q + r, with 0 <= r < ord(g). Show g^r = e, and think what this really means :3)

[rustic crown]

hi ryu :3

[lethal dune]

Hi det

[gusty cave]

A thought popped into my head, I noticed that the first isomorphism theorem rather useful to show various isomorphisms. Im wondering whether every G/N isomorphic K will have some surjective homomorphism from G to K with the kernel being N?

[coral spindle]

The natural map G -> G/N has kernel N, so yes.

[gusty cave]

ooo ic thats pretty cool

[white grotto]

Is there some fundemantel reason other than the proof on why if f is nilpotent then x,f(x),f^2(x),...,f^n-1(x) are independent

i mean what does f exactly do to a vector

[coral spindle]

if they weren't you'd get into a cycle, as it were

some vector would be repeated and would occur from then on

[white grotto]

why

[south patrol]

What do you mean by "other than the proof" - which proof are you using

I guess one way is to see that if T is an endomorphism of an F vector space V, then the polynomials p(t) in F[t] with p(T)(x) = 0 form an ideal J. This is generated by t^n, where n is minimal such that T^n(x) = 0

(J certainly contains t^n and doesn't contain any proper divisor of t^n by hypothesis)

Hence any polynomial p with p(T)(x) = 0 must be divisible by t^n and thus x, Tx, ..., T^n-1 x are all independent

Or you can do the standard way of multiplying it by T until you kill off all the terms but one, lol

[white grotto]

not this one, the one where we compose lambdax +lambda f(x).....+lambda f^n-1(x) by f to show that each lambda =0

[south patrol]

ye so this ig

[white grotto]

wait why are they hence independent

[south patrol]

Any polynomial p with p(T)(x) = 0 must be of degree >= n (in particular divisible by t^n)

[white grotto]

yeah