groups-rings-fields

[solemn dew]

but this polynominal is irreducible right?

[rocky cloak]

No, that would give a cover contradiction

[solemn dew]

it has to be irreducible in alpha^2 or am i thinking wrong?

wait what

[rocky cloak]

We're proving F(alpha^2) = F(alpha)

If that's the case, then clearly the roots are contained in F(alpha^2)

(Since the roots are ±alpha)

So [F(alpha) : F(alpha^2) ] is at most 2, but it can't be 2, so it must be 1.

[coral spindle]

The mistake that you made here was assuming that x^2 - alpha^2 was irreducible over F(alpha^2).

If it were, then yes certainly the degree of the extension would be that.

But you didn't prove that.

[solemn dew]

oh okay

thanks brotha

[vale locust]

I actually didn't lol

I should finish that paper

[coral shale]

mf

[chilly ocean]

Could you share the name of the paper? It sounds interesting.

[coral shale]

when i say paper, more like 3 page snippet of lecture-note-textbook-thingy

oh dq said paper

tryna make himself important 🙄

[chilly ocean]

I'm trying to show that the degree of the splitting field of x^6-4 is 6 using the tower law. I've split the extension as Q( (2)^(1/3) , z ) : Q(z) : Q where z is a primitive cube root of unity.

[ Q(z):Q] = 2 using the degree of the 2nd cyclotomic polynomial.

x^3-2 appears irreducible over Q(z), but I'm having trouble thinking of a proof other than trying a brute force strategy of trying to solve (2)^(1/3) = c_0 + c_1z+c_2z^2.

Oh, never mind. I figured it out.

[small bramble]

In a UFD is the gcd still a linear combination or just in a PID?

Oh bet https://math.stackexchange.com/questions/1725612/example-of-gcd-in-ufd-that-cant-be-expressed-as-linear-combination

[woven obsidian]

Lemma 1.2.22 is wrong, right? Along with the statement at the bottom of the page

Shouldn't e.g. $K = F_2(X^2,Y^2), L= F_2(X,Y)$ contradict the lemma

[cloud walrus]

AoiKunie

[woven obsidian]

And for the bottom of the page instead setting L = F_2(X^2,Y) contradicts the conclusion right?

[solemn dew]

if phi is a map that sends F to A, and F is a commutative ring with 1, then phi is a isomorphism right?

phi : F -> A, F a commutative ring with 1

my argument is that all elements of F can be described as phi(a)+phi(b) = phi(a+b), so then it must be invertable ?

or is it wrong?

[wraith cargo]

it's wrong

[solemn dew]

damn

[coral spindle]

What is A?

[wraith cargo]

yeah that's another good question

[solemn dew]

a commutative ring

[coral spindle]

You argument does not make sense to me at all. What does it mean to 'describe an element as phi(a)+phi(b)=phi(a+b)'?

[delicate orchid]

Then you’re very very off the mark

[coral spindle]

Did you mean to assume that phi was a homomorphism?

Anyway, if you meant to say:

Is every homomorphism of unital rings an isomorphism?
Then the answer is no.

[solemn dew]

phi is a homomorphism between two commutative rings with the identity

sending 1 to 1

oh okay

[delicate orchid]

For one, F and A will need to have the same cardinality to be isomorphismic

Also just think like

F = F_2 and A = Z

How can they possibly be isomorphic

[coral spindle]

Maybe the other way around for that example wew

[delicate orchid]

What

[spice whale]

oh yeah i don't think there are any homomorphisms from F_2 to Z

[coral spindle]

There's no map F_2 → Z, but there is one Z → F_2

[delicate orchid]

Oh wow yeah that definitely doesn’t support my point even more

It’s a map out of a field for a start

Z is not a field

[spice whale]

no the point is there are no maps out

so it's vacuous

[delicate orchid]

So if there’s no maps out… then there can’t be an isomorphism…

So they aren’t isomorphic…

[spice whale]

they're saying any hom is an iso

[coral spindle]

But that wasn't the question. Not that it really matters though, the answer has been given

[solemn dew]

yeah thankis haha

[delicate orchid]

Another instance where it feels like you lot just decided collectively to fuck with me

[spice whale]

here

[coral spindle]

• here, after my prompting.

[spice whale]

he's saying every hom is invertible

[delicate orchid]

Ebebebebepbepebpebep

[south patrol]

Bleak

[coral spindle]

wew I'm afraid we are going to send you to timbuktu

[delicate orchid]

Have to be careful in these channels

[south patrol]

I mean just consider the inclusion of Q in C

Fun times

[delicate orchid]

Mfs take any opportunity to rip your throat out

[coral spindle]

There are so many examples

We love u wew 😭

[south patrol]

Lol harder to find nonexamples

[delicate orchid]

IM A IDIOT. WE GET IT you don’t have to keep banging on

Fucks sake why do I bother

[coral spindle]

Nooooooooooo

[spice whale]

🫂

I'm sorry

[solemn dew]

if you are a idiot... what am i

🥺

but i am not a idiot hence contradiction! so you cannot be one

[rocky cloak]

Yeah, the first step in the proof is very dubious. They seem to imply that you have a primitive element, which isn't true in general unless the extension is seperable.

For the bottom conclusion, I guess it depends exactly how you interpret it, but I would say that's also wrong yes

[woven obsidian]

Hmm yeah I am assuming they made a mistake

Seems like it's true over function fields at least, which should be enough for what the book is doing

[dim widget]

Lemma 1.2.22 is wrong but the statement at the bottom of the page is correct

As long as L is assumed finite which I assume it is from the way it’s written

[rocky cloak]

Depends how you interpret the statement though. Like it's true that for every element x of L x^p^s is in Lsep. But not every element in Lsep has a p^s root in L right

So it would make more sense to write it as an inclusion than an equality

[dim widget]

It depends on how it's defined. The notion (L^{sep})^{1/p^n} doesn't make sense except as an abstract field (in particular it is not "equal" to anything) so I assumed it was defined "inside of L" so to speak

[rocky cloak]

Sure, but then you don't use the lemma, so why mention it

You can just define it as the splitting field of {x^p^n - l}, but that's only defined up to isomorphism of course

[dim widget]

I'm not sure, the lemma is so wrong that it's not even clear what it was meant to say to me.

[rocky cloak]

It seems to me that the bottom statement is trying to say that L contains x^p^s - l for every l in Lsep (which would be wrong). If they were just saying that x^p^s is in Lsep, then that's just the same statement as the exercise

[dim widget]

Perhaps, if L^{1/p^s} is defined inside of some algebraic closure of K then that is certainly what its saying yes

[woven obsidian]

They use it later to say that x in K implies p^s:th root in L, so they are probably taking it in an algebraic closure of K

[chilly ocean]

Hi! Does anybody know the identity extension principle for rings? Idk what it is jeje

[coral spindle]

I haven't heard of that before. In what context are you seeing it?

[chilly ocean]

Conmutative rings

[coral spindle]

Well yes of course

I'm asking you to share where you've seen this term so perhaps we can figure out what it's meant to mean

[chilly ocean]

Oh sorry I just have the concept name, I'm looking for what it means (searching in google did not help)

[coral spindle]

So where did you hear the name?

[chilly ocean]

In the program of a basic ring course

[coral spindle]

OK. Then that's that I guess.

[night onyx]

Sounds like it could mean taking a non-unital ring and adjoining 1 to it. Lol that's a complete guess though, I've never heard of it

[teal vessel]

am I missing something here? not being a group automatically disqualifies H from being a subgroup of G, right, so any subset H like {r} in D_2n, right?

[rocky cloak]

I'm guessing they mean that H need not be a subset of it's normalizer maybe.

Kind of a weird thing to consider anyway

[teal vessel]

like maybe? but it seems like, as written, any subset of G that is not a subgroup of G is certainly not a subgroup of its normalizer

which would satisfy the "not necessarily true"

if they're saying H needn't be a subset of its normalizer, that's a harder thing to show example of

(because I'm mildly dumb)

[rocky cloak]

Yeah, I don't know why you'd want to think about the "normalizer" of H if it's not a subgroup. But that's at least a true statement that you could find an example of

[teal vessel]

I'm sure I can, I just need to start thinking which is hard for humans

in fairness to the text, the normalizer and centralizer have been defined in terms of subsets of G, not subgroups, so it makes some sense

[night onyx]

does something like H = { (1 2), (1 3) } in S_n work as a counterexample? since (1 2)H = { e, (1 3 2) } while H(1 2) = {e, (1 2 3) }?

[rocky cloak]

Hint: ||you can find an example where H has just two elements||

[night onyx]

lol unless I made a calculation mistake

[teal vessel]

no, that one works, yeah.

at least, I think. Shouldn't it be the conjugation of H with (1 2) though? since (1 2) is self inverse, it should be (1 2) H (1 2) which... oh, wait, that gives {e, (2 3)} nvm, I'm silly.

yeh, that works

[solemn dew]

guys is there an explaination of why

Z_2xZ_3xZ_3 and Z_2xZ_9 are abelian, but not

Z_6xZ_9 for example?

[coral spindle]

All of the things you have listed are Abelian groups

[solemn dew]

really?

[coral spindle]

Yes.

I have no inkling as to why you think Z_6 x Z_9 is not Abelian.

Can you name two elements that do not commute in that group?

[solemn dew]

not that i can think of

[coral spindle]

So why do you think it's not Abelian

[solemn dew]

because of this example:

[crystal turtle]

Any product of abelian groups is abelian

[solemn dew]

o(G) = 72, hence our possible abelian groups are:

Z8 × Z9, Z2 × Z4 × Z9, Z2 × Z2 × Z2 × Z9
or Z8 × Z3 × Z3, Z2 × Z4 × Z3 × Z3, Z2 × Z2 × Z2 × Z3 × Z3

but I see no Z6 x Z6 x Z2 for example

[coral spindle]

Z6 is isomorphic to Z2 x Z3

So in fact it IS listed there. Find it.

[solemn dew]

then it has to be

Z2 × Z2 × Z2 × Z3 × Z3

[coral spindle]

Did you remember to read the theorem that explains why you can list out the Abelain groups like that?

[solemn dew]

yea any product of cyclic groups is abelian

[coral spindle]

No

What you say is true, but that's not the reason

The theorem you're looking for is probably called the classification of finite Abelian groups in your textbook/course

[solemn dew]

this one?

[coral spindle]

Yes.

[solemn dew]

alright i think i got it then

Zn x Zm is isomorphic to Znm iff gcd(n,m) = 1

[coral spindle]

iff too, btw

[solemn dew]

so we do not write those Znm

mb

[coral spindle]

We can write them however we like

In this case though, the theorem writes the primes completely seperately

[solemn dew]

indeed

u have been helpful thank you brother

[small bramble]

I just managed to recall a small result I learned years ago without looking it up or referencing the book . I know it doesn't mean much but after years of not doing math it feels nice to even be getting back into the groove of things

Well it's more like I rederived it rather than recalled it but I guess that's actually better

It's just the fact that if (m, n) =1 then m^phi(n) cong 1 (mod n)

Don't really have anybody to be excited about it with so I came here I guess lmao

It's also just an improvement to past me's attitude cause past me would've just googled the answer due to time constraints related to school (and the school I went to was shit so the exams weren't challenging enough that looking stuff up for homework penalized me tbh)

[lilac slate]

does it suffice to show that $\overline{a}$ + (\overline{b} + \overline{c})$ = (\overline{a} + \overline{b}) + \overline{c}$ using the property $\overline{a} + \overline{b} = \overline{a + b}$

[cloud walrus]

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

[lilac slate]

So, $\overline{a} + (\overline{b} + \overline{c}) = \overline{a} + \overline{b + c} = \overline{a + b + c}$ and $$(\overline{a} + \overline{b}) + \overline{c} = \overline{a + b} + \overline{c} = \overline{a + b + c}$$?

[cloud walrus]

Teemo

[lilac slate]

With how addition is defined on congruence classes of Z/nZ

[rocky cloak]

Yeah that works

[coral shale]

latex is pretty for bars

you might prefer [a] notation to denote it

[chilly ocean]

ah I see

interesting

anyway

can anyone help with this

struggling with doing closure

[spice whale]

basically the idea is to prove the fractional part of any real number is in [0,1)

[chilly ocean]

how do i prove that

it's pretty obviouss

[spice whale]

contradiction is probably easiest

[coral shale]

and right after we see [ . ] used for something else
But its rare we care about the greatest integer function lol

[dim widget]

It is hard to find closure

[spice whale]

trying to think of a non-finitely generated subgroup of the free group on 2 elems

[rocky cloak]

Maybe something like the group generated by || a^n b^n ||

[chilly ocean]

need help

[south patrol]

just show the RHS is an inverse

which is easy to do - what is $a_1 \dots a_n a_n^{-1} \dots a_1^{-1}$ for example?

You can also do induction on

[chilly ocean]

how do I show it's an inverse

[cloud walrus]

potato

[chilly ocean]

well

that's just e

[south patrol]

yes

then the other way round

$a_n^{-1} \dots a_1^{-1} a_1 \dots a_n$

[cloud walrus]

potato

[chilly ocean]

thats e

[south patrol]

Yes

So it's an inverse!

[chilly ocean]

but that's not rigorous enough is it?

[south patrol]

I'd say it's rigorous enough, but if you want more detail you can justify why these terms collapse down to e

Say by induction lol

[chilly ocean]

So

base case: n = 2

[south patrol]

I'd take n=1 but sure

[chilly ocean]

(a1 a2)^-1 = a2^-1 a1^-1

[south patrol]

There is another way to do induction more directly but yes

Sure

that one is straightforward enough as I'm sure you've seen ^

[chilly ocean]

so we say base case is trivial

now assume

this prop is true for some $n = k$

[cloud walrus]

zzzzz

[chilly ocean]

in other words

$a_1 a_2 \hdots a_k a_k^{-1} \hdots a_2^{-1} a_1^{-1} = e$

[cloud walrus]

zzzzz

[chilly ocean]

so for n = k we have $(a_1 a_2 \hdots a_k)^{-1} = a_k^{-1} \hdots a_2^{-1} a_1^{-1}$

[cloud walrus]

zzzzz

[south patrol]

Well bear in mind we need two-sided inverses

[chilly ocean]

do i multiply on the left by a_k+1^-1

[south patrol]

But you can just take this as your inductive hypothesis

[chilly ocean]

right

what about the inductive step

[south patrol]

Well try grouping the terms cleverly

So what's another way to write a_1 ... a_{k+1} only involving k terms

[chilly ocean]

hmm

a_k+1 a_k+1^{-1}

[south patrol]

that's just e

[chilly ocean]

oops

🙁

[south patrol]

What I had in mind was just letting like b_k = a_k a_{k+1}

[chilly ocean]

ok

that

makes sense

[south patrol]

Now see what happens

[chilly ocean]

so

oh

that's samrt

we use even the base case for this

[south patrol]

Yup

[chilly ocean]

thank you

that's very clever

[south patrol]

Np

Another way to do it would be like

to show that $a_1 \dots a_n a_{n}^{-1} \dots a_1^{-1} = e$ more rigorously i.e. by induction, you can just induct on n and say that $a_1 \dots a_n a_{n}^{-1} \dots a_1^{-1} = a_1 \dots a_{n-1} a_{n-1}^{-1} \dots a_1^{-1} = e$

[cloud walrus]

potato

[south patrol]

Like just cancel the middle two terms and then you use inductive hypothesis

[chilly ocean]

ah

[south patrol]

I think this is the sort of thing where it's clear enough how it goes btu yeah

[chilly ocean]

that's also a good way of doing it

[south patrol]

Depends on course ig lol i guess some first courses in groups would be more picky xd

Thank

[chilly ocean]

thank you again, i'll write this proof down

[south patrol]

Np

[chilly ocean]

also for this problem what i had in mind was

|x| = 1 implies x = e

so e^2 = 1

|x| = 2 implies x^2 = 1 so that's correct by assumption

that's done right

[south patrol]

Well you have the other direction too

[chilly ocean]

yep

x^2 = 1 \implies |x| = 1 or |x| = 2

so

e^2 = 1 and x^2 = 1 implies |x| = 2 by the definition of order.

[south patrol]

That's not true

Well unless you add more caveats

[chilly ocean]

hmm

how do i fix it

[south patrol]

"x^2 = 1 => |x| = 2" isn't necessarly true since you've allowed |x| = 1

[chilly ocean]

oh

righ

[south patrol]

But I would just go back to the actual definition of order

[chilly ocean]

tt

[south patrol]

What is it

[chilly ocean]

An element has an order $n$ if $n$ is the smallest positive integer such that $x^n = id$

[cloud walrus]

zzzzz

[south patrol]

Yup

so if x^2 = e, what can you say about the order

[chilly ocean]

the order is 2

[south patrol]

No, since can have x = e

[chilly ocean]

umm

[south patrol]

From the definition I'm saying

[chilly ocean]

x = x^{-1}

[south patrol]

Okay let's generalise a bit, what I mean is liek

Say x^m = e for some m >= 1

What can you say about the order of x?

[chilly ocean]

the order of x divides m

ohhh

[south patrol]

Oh wow you had Lagrange lol

I assumed you didn't

[south patrol]

But yes

In this problem it's enough to say that the order is <= m

But yeah

[chilly ocean]

since 1 | 2 and 2 | 2 |x| = 1 or |x| = 2

done

right?

[south patrol]

Well more that 1 and 2 are the only (non-negative) divisors of 2 but ye

[chilly ocean]

yessir

or ma'am

thank you very much

again for the help

i got it

and these two are just basic cancellation law questions

18 is a bit harder because its iff

[barren sierra]

Did you end up getting it?

[summer path]

You can do it via equivalent statements

[formal ermine]

what book is this

[summer path]

From the font, I'd guess d&f? But maybe fraleigh possibly

[crystal turtle]

Definitely looks like D&F

[steel light]

Love how we can all recognize D&F from font alone

Even I recognized it

[south patrol]

D&F has too many questions

[formal ermine]

the above ones feel redundant

[crystal turtle]

all of the questions in D&F are

[long nebula]

I like the D&F questions though

[thick zealot]

what is d n f

[south patrol]

Dummit and Foote

[tawny dune]

thats like chapter 1 section 1 tho

[crystal turtle]

Idk I just find they're not really interesting, at least the group theory ones. A lot of them are just calculations

[formal ermine]

deez nuts (are on) fire

[thick zealot]

is it a book or like a theorem?

ah its a book

[south patrol]

Book

V popular one for algebra

particularly like intro stuffs

[summer path]

It's a bit wordy though which may be good if you are a bit confused with some details, or bad if you're already familiar with the topics

[cloud walrus]

yatsude

[rocky cloak]

Are you familiar with Lagrange's theorem?

[safe wind]

hey uh can anyone help me with this problem:
f:M->N, g:N->M are homomorphisms of R-modules such that g(f(m))=m for all m in M, i need to prove N= f(m) x ker(g)

[rocky cloak]

Here's a hint: are you able to say anything about the composition fg?

[safe wind]

ill be real no fkin clue

[rocky cloak]

What does fgfg equal?

[safe wind]

its equal to fg

so fg is idempotent or smth

[rocky cloak]

Indeed, and do you know anything about idempotent homomorphisms?

[safe wind]

dont know any theorems that relate to that though

not in our syllabus i think

[rocky cloak]

Okay, maybe we try a more direct root:

Are you able to show that the image of f and the kernel of g don't intersect?

[chilly radish]

You don't really need Lagrange for this. This kind of stuff follows from ||division with remainder and definition of order||

[rocky cloak]

Potato tomato

[chilly radish]

Not really

One of those is usually learned befor the other

[safe wind]

if they intersect at say r, g(r) = 0 = g(f(m)) for any m but also g(f(m)) = m so they cant intersect unless M=0

[rocky cloak]

Good, so then we just need to show that every n is of the form n= f(m) + k

[safe wind]

yeah

[rocky cloak]

So, how would you think to construct an m given n?

[safe wind]

wait lol if the intersection is empty then MxN has the form M+N ?

[safe wind]

yeah i was going to do that and then like apply g so g(n) = g(f(m)) + g(k) = m bu- oh

[rocky cloak]

Alright so you have your candidate m

Then you just have to verify that k = n-f(m) actually is in the kernel

[half palm]

Hmm, okay, this is going to be a very vague question because 1) I haven't done group theory in a while (kinda abandoned the project midway), and 2) I just got reminded that this question exists, and 3) it's fundamentally an open-ended question.

Consider conjugacy classes. Elements in the same conjugacy classes are considered to be somewhat "similar" in that they share many of the same properties. b is a conjugate of a if b = gag^{-1} for some g, i.e. b can be viewed as a under "change of basis" g.

An abelian group has ab = ba for all a,b, and each conjugacy class is necessarily a singleton. In one sense, elements of the group are somewhat "all the same" as they can be composed with each other in any order (i.e. they're commutative). But, in another sense, all elements of the group are fundamentally "different." They don't belong in the same conjugacy class.

In what sense are different conjugacy classes "different" from each other? In what sense are all members of an abelian group "different"?

idk sorry for the somewhat open-ended philosophical question it just has been nagging at me for a long time and i still couldn't figure out exactly what it is that i'm trying to get at

[ivory trail]

In one sense, elements of the group are somewhat "all the same" as they can be composed with each other in any order (i.e. they're commutative).
i would disagree with this

[half palm]

ok tbh that was kinda a red herring lmao but i guess the real issue here is this: it's much easier to say why some things belong in the same class than to say why things don't belong in the same class. and that's kinda what i'm trying to figure out

like, you could imagine a notion of conjugacy class that's more restrictive or less restrictive

[paper flint]

Not sure if this is the right channel but here it goes. I have proved the forward implication (algebras being isomorphic implies they have the same structure constants with respect to some bases), but the converse is throwing me off

I'm not entirely sure what it means to have "equal" structure constants here, really. If they mean "have the same set of structure constants with respect to some basis" then I find the claim to be morally wrong (I wouldn't even be convinced the Lie algebras should have the same dimension in that case)

Or is it implicit that the bases have the same cardinality, and then they're "equal" by a bijective correspondence between the pairs of basis elements and the structure constants?

[rocky cloak]

Yeah, Id say the structure constants aren't just an unstructured set, but a matrix / indexed set.

So them being equal will necessarily mean there is a bijection between the bases.

[safe wind]

its more of that the elemts are 'unique'

congruency class is "same element under different perspective", and what the congruency class being a singleton means is that no matter how you look at it, you have the same element

i think what youre getting at is sort of "because each element is a unique action, it doesnt matter in what order you apply it" since theres no worry that by "looking at a different perspective" i.e. applying other elements beforehand will change the action

i.e. commutative

[vale locust]

???

it's an expository paper

it's a thing hurb

[vale locust]

yea, sry, I don't use this account on weekends, lemme fetch it

https://kconrad.math.uconn.edu/blurbs/grouptheory/wordproblem.pdf

[solemn dew]

guys i am a little confused here

F_5 x F_5 is isomorphic to F_25 right?

Thus F_5 x F_5 is a field?

[solar vessel]

no

F_5×F_5 isn't even an integral domain

[vale locust]

multiplication is defined differently

[solemn dew]

oh okay thanks

[solar vessel]

np

[somber thorn]

that moment when fundamental theorem of finitely generated abelian groups

[delicate bloom]

might also be nice to keep in mind a finite field with q elements has its multiplicative group a cyclic group with q-1 elements

(in fact every finite subgroup of the multiplicative group of a field is cyclic)

[peak rain]

Quick thing cos of a question my friend asked me.

Is the characteristic of a finite ring equal to its cardinality ?

My answer is "yes as you can form equivalence classes based on the number of repeated additions applied to the multiplicative identity.

Then you get that the set of classes has the same size as the ring as one finds duplicate classes if you try to create more classes than the characteristic of the ring.

I came here to get a second opinion to make sure this isn't complete bullshit.

[rocky cloak]

No, it could be smaller

For example the finite field of order p^k has characteristic p

Or even simpler Z/2 x Z/2 has characteristic 2, but has 4 elements

[lusty marlin]

How do we prove this?

[solar vessel]

if the subgroup has order n say

[rocky cloak]

Classification of finite abelian groups

Plus degree n polynomial has at most n roots

[solar vessel]

ye

[lusty marlin]

Could you elaborate a bit? I still don't get how it works

[coral spindle]

You can try proving that any finite subgroup of thr multiplicative group of any field is cyclic by considering the highest-order element and seeing it as a root of unity. Try it!

[lusty marlin]

Hmmm
Thanks!

[rocky cloak]

So if a finite abelian group is not cyclic, it contains Z/n x Z/n for some n. Then just verify that this is not a subgroup of the multiplicative group of a field

[coral spindle]

This boils down to the same proof, btw. Hint: consider the polynomial that nth roots of unity satisfy

[solemn dew]

the polynominal in (a) has 3 roots in F_3 since f(0) = 0, f(1) = 0, f(2) = 0 so is the splitting field itself then?

or is it this:

[rocky cloak]

So this is a bit impressise. In F3 we have 2=-1, so i and sqrt(2) I guess refer to the same element

But it is true that the splitting field is a degree 2 extension

Given by adjoining for example the square root of -1

[indigo ridge]

I haven't fully understood the usefulness of proving a homomorphism (I understand isomorphisms pretty well) is it fair to say that if we suppose for ring R and S we have a homomorphism f f:R --> S then the image of f is a subring of S, which basically means that f is isomorphic from R to a subring of S which is the image of f

[solemn dew]

isn't F_2(sqrt(2)i) a degree 4 extension?

so my solution is wrong then?

[rocky cloak]

Depends exactly what you mean by that. Like I said, the language isn't very precise

[solemn dew]

sorry i am quite new to this

[rocky cloak]

So a homomorphism need not be injective. In general if I is the kernel of f, then you can factorize f as

R -> R/I -> f(R) -> S

Where the map R/I -> f(R) is an isomorphism. And f(R) is then a subring of S, and R/I is a quotient ring of R

[rocky cloak]

So in F3 we have 2=-1. So what exactly do you mean by sqrt(2)?

Is it the root of x^2 = 2? Because that would be exactly the same as a root of x^2 = -1

Also this 6th degree polynomial should have 6 solutions, but you only list 4

[solemn dew]

i'm just thinking that because the solutions of x^6+x^4+x^2+1 = 0 are -i, i, (i+1)/sqrt(2), -(i+1)/sqrt(2), then we need to extend the field further

you are right

(x^6+x^4+x^2+1) = (x^2+1)(x^4+1)

x^2=-1 has solutions -i, i, so our solution will be in the splitting field F_2(i)

[rocky cloak]

But your talking about solutions in complex numbers?

Remember this is not a polynomial with rational coefficients. It has coefficients in F3

Complex numbers live in a completely different universe

[solemn dew]

so how do we solve x^4+1 then?

since the solution is not in F_2[x]/(x^4+1)

[rocky cloak]

So you have the polynomial factored as (x^2 + 1)(x^4 + 1), and x^2 + 1 is irreducible. Then you can adjoin a root of x^2 + 1 and try to factor it further from there

Or you can try to factor x^4 + 1 first (it's not irreducible)

[rocky cloak]

Also you're writing F2 everywhere, but it's supposed to be F3

[solemn dew]

oops

alright i'll try to do it

[solemn dew]

i think x^4+1 is irreducible

[topaz solar]

Are you sure about that one

[solemn dew]

wait a minute

the elements of F_3(i) are:
0, 1, 2, 1+i, 2+i, 1-i, 2-i,i, -i

going to investigate if the polynominal has solutions there

alright jagr i found the solutions

thank you very much

answer is F_3(i)

[topaz solar]

Also, a^2-b^2 = (a+b)(a-b) right?

[molten viper]

I think x^4 + 1 factors over F_3...

at least into quadratic terms

[solemn dew]

yes

[topaz solar]

Polynomials do be the bread and butter frfr

[molten viper]

love a good polynomial

no I don't this shit sucks lol

[solar vessel]

based

[crystal turtle]

based

[topaz solar]

based

[spice whale]

polynomials are cool

[crystal turtle]

unbased

[coral spindle]

I would like everyone to know that I accidentally super-reacted

[molten viper]

Lol

[vale locust]

I feel you

shit be embarrassing

[coral spindle]

I wonder if Wew is around hmmm

I'm getting the vibe he is

[crystal turtle]

Lmao

[solemn dew]

is x^2+x+1 irreducible in F_3?

[delicate orchid]

Yeah probs

[solemn dew]

p(x) = x^2+x+1, p(1) = 0

but

[delicate orchid]

Yeah probs

[solemn dew]

x^2+x+1 = (x-1)(x+2) = x^2 + x -2 = x^2 + x + 1

[crystal turtle]

Try to factor it

[solemn dew]

but -2 is not a solution

nevermind

[delicate orchid]

Ok then how can it possibly be irreducible

[solemn dew]

it is

[molten viper]

-2 = 1

[crystal turtle]

-2 = 1

[solemn dew]

hahah

[molten viper]

in F_3

[solemn dew]

did a oopsie

[crystal turtle]

Damn it, sniped

[molten viper]

REKT

[solemn dew]

sorcerer vs wizard

[chilly ocean]

any hints?

[ivory trail]

induction?

[chilly ocean]

don't think too hard

[delicate orchid]

Write out a+b x’s and then rebracket

[chilly ocean]

$(x \cdot x \cdot \hdots \cdot x)$ a times multiplied by $(x \cdot x \cdot \hdots \cdot x)$ b times

[cloud walrus]

zzzzz

[chilly ocean]

using the generalized associative law?

[delicate orchid]

Yeah so put both of those mofos next to each other and what you got

[chilly ocean]

$(x \cdot x \cdot \hdots \cdot x) \cdot (x \cdot x \cdot \hdots \cdot x)$

[cloud walrus]

zzzzz

[chilly ocean]

$= (x \cdot x \cdot \hdots \cdot x \cdot x \cdot x \cdot \hdots \cdot x)$

[cloud walrus]

zzzzz

[crystal turtle]

Just count them

[chilly ocean]

what

a and b are arbitrary though no?

[delicate orchid]

… so??

[ivory trail]

so you use induction

[delicate orchid]

They’re two positive numbers and this works for all positive numbers

[chilly ocean]

okay, so let's use induction then

on b?

[delicate orchid]

You’ve got two degrees of freedom here, I wish you luck with that

But why the hell do we need induction?? We’ve literally just shown it

[ivory trail]

actually no

not formally

[delicate orchid]

a copies of x concatenated with b copies of x is a+b copies of x

Yes formally

[ivory trail]

ok cool

[delicate orchid]

These two expressions do not depend on bracketing by associativity

[chilly ocean]

can I say x^a x^b consists of a factors of x multiplied by b factors of x for a total of a + b factors of x, thus x^(a+b) = x^a x^b

[ivory trail]

the point being is that naturals exist for the purpose of induction

[delicate orchid]

Yeah, just mention associativity somewhere

[delicate orchid]

This is N^2 boss

[ivory trail]

ok?

[coral spindle]

It's really not a great idea to spend your time trying to prove that associativity generalises like this. A complete proof, of the sort that a proof assistant would accept, is a pretty big pain.

[delicate orchid]

You can reduce it down using some symmetry argument but again this is way too complicated

[coral spindle]

Also yeah as wew alludes to, doing induction in two variables is a massive pain too

[ivory trail]

well if you're doing intro proofs level stuff

[chilly ocean]

I'd like to prove it rigorously

[ivory trail]

I don't think this is as careful as you could be

[delicate orchid]

My rebracketing thing is rigorous

[chilly ocean]

by the generalized associativity law?

[delicate orchid]

The only subalgbera of N^2 is not N^2 so induction doesn’t work a priori

[delicate orchid]

Yup

[coral spindle]

I must strongly suggest you do not, again. If you really want to then that's up to you but it will not reveal any insight into how these things work.

[delicate orchid]

I take issue with the fact that you’re insinuating that rebracketing is not rigorous

I can’t even believe this is a conversation

[ivory trail]

subalgebra of 1+X^2?

[delicate orchid]

What

[chilly ocean]

what about b)?

for b) I was thinking multiplying by x^-a

[coral spindle]

I'm not, but if you want to prove that rebracketing works then this is just an immense pain in the arse

[ivory trail]

you're not being clear at all

[delicate orchid]

I really don’t care

[ivory trail]

then don't be so abrasive

[delicate orchid]

I’ll be as abrasive as I wish

[delicate orchid]

Anyway, yeah give this a go!

[ivory trail]

lol ok

[delicate orchid]

Remember that you haven’t proven that x^ax^b = x^(a+b) for negative a, b though so you can’t use that fact here

[chilly ocean]

So my thinking for (b) was, since $(x^a)^{-1}$ is the inverse of $x^a$ by definition, we have $x^a (x^a)^{-1} = id$, multiplying by $x^{-a}$ on the left, we obtain that $(x^a)^{-1} = x^{-a}$

[cloud walrus]

zzzzz

[delicate orchid]

Ah but how do we know that x^-ax^a = Id

[chilly ocean]

hmm

we don't

[delicate orchid]

I suggest expanding out x^a and x^-a like we did for the first question

And here you actually can use induction if you want

[chilly ocean]

ohhhh

[ivory trail]

to do induction on N^2, you just do it one variable at a time

you don't bring out F-algebras for whatever reason

[chilly ocean]

WAIT @delicate orchid THAT MAKES SENSE ACTUALLY

[delicate orchid]

I’m sure that’s very enlightening

[delicate orchid]

Yurrrrrrr

[chilly ocean]

because we know $(x \cdot x \cdot \hdots \cdot x)^{-1} = x^{-1} \cdot x^{-1} \cdot \hdots \cdot x^{-1}$

[cloud walrus]

zzzzz

[ivory trail]

this looks like homework, not a monastery

[delicate orchid]

Each copy of x^-1 in x^-a cancels with exactly one x in x^a

[chilly ocean]

exactly

from that we conclude that x^-a is the inverse of x^a which is represented as (x^a)^{-1}

[delicate orchid]

Oh and look you didn’t need any induction clogging up your brain to realise this fundamental aspect of the proof

Almost as if that’s not the focus of this exercise

[ivory trail]

I already made this clear

[ivory trail]

maybe we simply treat induction with differing weights

I think it is best to get very familiar with it rather quickly

[delicate orchid]

In all the texts I’ve read in the past year they’ve completely hand waved any inductive argument

I agree doing it for x^-a would be an ok exercise

[ivory trail]

well I guess you're not doing intro to proofs, or what immediately follows

[delicate orchid]

But doing it for N^2 is just such a pain

[ivory trail]

there's something nice about being able to formalize an argument

so at least understanding the presence of induction is important

it depends on the class for the level of rigor you actually put into writing

I don't think there's any good reason to react the way you did

[ivory trail]

if you had the "just associate" idea but didn't know how it came about via induction, you would be at risk of not doing math rigorously enough

[delicate orchid]

I still stand by saying that the result holds for all a and b a priori is completely rigorous though

[ivory trail]

that sort of thinking can easily get off the rails in the future

[delicate orchid]

You assumed nothing about them and showed the result holds

[chilly ocean]

@delicate orchid is this okay?

[delicate orchid]

Hold on I’m going through a tunnel

[open sluice]

the classic challenge of formality at the cost of understanding

[ivory trail]

I don't care about your particular definition of "rigorous," since I think the way I used it was pretty clear

[delicate orchid]

Sorry about that lol

[chilly ocean]

are you in a car?

[delicate orchid]

Yur

[delicate orchid]

Train

[chilly ocean]

ty much appreciated

don't math and drive kids

[ivory trail]

well I'm describing a way they're not opposed to each other here

[delicate orchid]

In between the carriages in my little hole

[delicate orchid]

Bogging down an elegant idea with a 2 dimensional induction absolutely opposes understanding lmfao

[open sluice]

when I look at the original problem, it’s like
yeah if you really want to be formal induction is the way to go, more power to you
but I really don’t think it’s something worth getting all worked over

[ivory trail]

the elegant idea is "just count and use coherence of associativity"

[ivory trail]

I only mentioned it because it looks really early on in abstract math

congrats if you understand induction

[ivory trail]

I didn't advocate for "bogging down"

it's just a supplement

you seem to have assumed I said your idea was not worth using

[delicate orchid]

Yeah that’s how I interpreted it

[ivory trail]

I missed that somebody could possibly look at this and wonder if it was true

[chilly ocean]

is this okay for this problem?

do I need to consider x having infinite order

wait no i just proved that x^-1 has order at most n

[delicate orchid]

Yeah you will need to consider that case too

[solemn dew]

guys i can't factor x^5-x in F_3 please help me

p(x) = x^5-x,
p(0) = 0
p(1) = 0
p(2) = 0

[delicate orchid]

Yeah

[chilly ocean]

so what should i do

[solemn dew]

so we have roots 0,1,2
thus we can write x^5-x = x(x-1)(x-2)(x^2+ax+b), but i see no solutions for a and b. what do you do from there?

[delicate orchid]

Show it can’t have a lower order - maybe by contradiction?

[chilly ocean]

oh okay

[delicate orchid]

If you want another hint: || assume a has order n and a^-1 has order m < n, then we can write Id = a^n(a^-1)^m||

[spice whale]

polynomial long division?

[solemn dew]

does it work in F_3?

i will just get some coefficients that are not in F_3 then

i think

[spice whale]

yes...

[solar vessel]

it works over any field

[crystal turtle]

Euclidean algorithm go brrrr

[spice whale]

I've got an answer

[solemn dew]

please share it

[spice whale]

hint: factor out x and look at the form of the other bit

x(x⁴-1)

[solemn dew]

...

x(x^4-1)=x(x^2-1)(x^2+1)

[spice whale]

yep

[solemn dew]

i am so stupid

thank you

❤️

[spice whale]

tbh i did long division

[solemn dew]

XD

[spice whale]

but yeah you can obviously factor x²-1 as well

and you get x(x+1)(x-1)(x²+1)

[solemn dew]

yes ofc

i was too blind due to that p(2) is a solution, but as @rocky cloak stated, p(-1)=p(2) in F_3, so the factorization works since (x-2)=(x+1)

thank you!

[spice whale]

it's better to think about F3 as 0,1,-1

[solemn dew]

i will remember this^

[paper flint]

Thanks, my response didn't go through before for some reason

[chilly ocean]

So, for this problem, I did the following:

Let $n = 2r+1$. Then $|x| = 2r + 1 \implies x^{2r+1} = 1$.

[cloud walrus]

zzzzz

[chilly ocean]

Multiplying by $x$, we have $xx^{2r+1} = x1 \implies x^{2r+2} = x$. This can be re-written as $(x^2)^{r+1} = x$. Let $r + 1 = k$. Then $(x^2)^k = x$, as desired.

[cloud walrus]

zzzzz

[chilly ocean]

Is that correct?

[formal ermine]

yeah

[chilly ocean]

thanks

btw forgot to reply - yes this is d&f @formal ermine

[formal ermine]

ah

[chilly ocean]

too many problems but overall a great book

for this problem

My solution is

Suppose $|x| = n \implies x^n = 1$. Then multiply by $g^{-n}$ on the left and $g^n$ on the right to obtain
$$g^{-n} x^n g^n = g^{-n} g^n = g^{n-n} = \text{id}_G$$. So $(g^{-1} x g)^n = 1$.

[cloud walrus]

zzzzz

[chilly ocean]

And $|g^{-1} x g| = 1$.

[cloud walrus]

zzzzz

[chilly ocean]

Is this correct?

[rocky cloak]

(abc)^n does not equal a^n b^n c^n

[chilly ocean]

I think you're assuming that the operation is commutative.

oh

right

I forgot that g may not be commutative

no but wait

g^{-1} x g g^{-1} x g n times becomes g^-1 x^n g

and x^n = 1

and g^{-1} g = 1

right?

[rocky cloak]

Yeah, that's right

[chilly ocean]

so does that complete the proof

[rocky cloak]

Pretty much yeah

[chilly ocean]

no wait

i multiplied by g^-n and g^n on the left and right respectively

im confused now

I think you want to instead just do (g^(-1)xg)^n.

Suppose $x^n = 1$, then multiply by $g^{-n}$ on the left and $g^n$ on the right to obtain $g^{-n} x^n g^n = 1$

[cloud walrus]

zzzzz

[chilly ocean]

This doesn't get us anywhere

does it?

unless G is abelian

When you multiply it out, g and g^(-1) will cancel, leaving x^n in the middle.

[delicate orchid]

^

look closer at (g^-1xg)^n

[chilly ocean]

no yes I understand that

(g^-1 x g)^n = g^-1 x^n g

= id_G

[delicate orchid]

ok then...?

[chilly ocean]

wait

am i stupid

[delicate orchid]

sorry am I misunderstanding what the exercise was

[chilly ocean]

It follows shortly that the order of g^(-1)xg is n.

[delicate orchid]

oh right we've only shown x^g has an order that divides n

but yeah, showing that it can't be smaller than n is easy

[chilly ocean]

when i multiply it out i have g^-1 g^-1 g^-1 ... g^-1 x x x .. x g g ... g

[delicate orchid]

no you don't

you have (g^-1xg)(g^-1xg)(g^-1xg)(g^-1xg)...(g^-1xg)

[chilly ocean]

i do for g^-n x^n g^n

that's (g^-1 x g)^n

not g^-n x^n g^n

no?

[delicate orchid]

what does g^-nx^ng^n have to do with anything?

[chilly ocean]

or am i missing something

[chilly ocean]

This is the expression you should be working with

oh

sorry guys

[delicate orchid]

np boss

[chilly ocean]

(g^-1 x g)^n = g^-1 x^n g

which simplifies to id_G

and that shows that |x| = |g^-1 x g|

right?

Not exactly, it just shows that the |g^-1 x g| divides n

But as previously mentioned, showing that it can't be less than n shouldn't be too bad

how do I show it is equal to n

I think the simplest way I can think of is by contradiction.

[delicate orchid]

yeah

so assume (g^-1xg)^m = 1 with m < n

[chilly ocean]

assumed.

now

wait but doesn't m have to divide n

Sure, this proof would prove a more general statement.

(for any m<n)

[chilly ocean]

how should I proceed from here (sorry, this is the part of the proof I always have trouble with)

Work with the left side of the equation as before

okay

so

g^-1 x^m g = id

then x^m = 1

and g^-1 g = 1

See an issue?

[delicate orchid]

well, yes - which is the contradiction

[chilly ocean]

why is that a contradiction

[delicate orchid]

because the order of x is n

[chilly ocean]

What's the order of x?

isn't that kind of obvious though

[delicate orchid]

sorry username I'm kinda spoiling your stuff here

[chilly ocean]

Ok I will write a formal proof

[chilly ocean]

It shouldn't be too hard of a proof.

Let $|x| = n \implies x^n = 1$. Consider $(g^{-1} x g)^n$. This reduces to $g^{-1} x^n g = g^{-1} 1 g = \text{id}_G$. So the order of $g^{-1} x g$ divides $n$. Now suppose $m < n$ and $(g^{-1} x g)^m$. Then $g^{-1} x^m g$ implies this is equal to $1$. Contradiction, since $|x| = n$.

[cloud walrus]

zzzzz

[rocky cloak]

Guess you should justify why g^- x^m g = 1 implies x^m = 1

[chilly ocean]

oh okay

thank you

[delicate orchid]

hopefully you shouldn't have too much trouble with the next part of the exercise

[chilly ocean]

yeah I got it

you just conjugate it

and it simplifies to |ab| = |ba|

thanks @chilly ocean and @delicate orchid, much appreciated

[solemn dew]

isn't the intermediate field Q(i)?

x^8-1=(x-1)(x+1)(x^2+1)(x^4+1)

so every solution will be in Q(sqrt(2)*i)

but x^2+1 has a solution in Q(i)

so we get:
Q<Q(i)<Q(sqrt(2)*i), thus Q(i) must be the intermediate field?

answer says Q(sqrt(2))

[chilly ocean]

Both look like intermediate fields to me

[solemn dew]

Q<Q(sqrt(2))<Q(sqrt(2)i)

is that what you mean?

but which one do you choose haha

[chilly ocean]

It asks to determine all of them.

[solemn dew]

ooh you are right

so the answer is Q(sqrt(2)) and Q(i) then

thank you!

[glossy crag]

There ought to be one more.

[solemn dew]

really?

i am on it now

can i get a hint?

x^2+1 has roots +-i
x^4+1 has roots cos(pi/4*n)+i sin(pi/4 .n)

[glossy crag]

These are called roots in English.

[solemn dew]

right haha

[glossy crag]

I don't really have one atm, need to think about it, but since E/Q is Galois, there is a correspondence between intermediate fields and subgroups of Gal(E/Q), which has 3 proper subgroups, so there are 3 intermediate fields, one of them being Q(i).

[solemn dew]

o(Gal(E/Q)) = [E:Q] = [E:Q(i)][Q(i):Q] = 4

and I think Gal(E/Q) is isomorphic to Z_2 x Z_2

[glossy crag]

For that you'd have to show E:Q(i)=2, idk if it's immediately obvious.

[glossy crag]

That is so, by theory of cyclotomic fields.

[solemn dew]

whoa

i have never heard of that

will check it

[glossy crag]

Whatever book you're reading will likely have a chapter on it.

The splitting field of x^n-1 over Q is Galois with group (Z/nZ)*

[chilly ocean]

I think a simple argument is that the roots of x^4+1 are expressions in sqrt(2) and i, Q(i) does not have a square root of 2, so quotienting out by x^2-2 which is irreducible over Q(i) gets you Q(sqrt(2),i) (something like that).

[solemn dew]

does the start "*" mean something?

[glossy crag]

The units of Z/nZ

[glossy crag]

Right, so the 3rd subfield is Q(i*sqrt(2)), idk how i missed that.

and that's indeed all of them because you know there are 3 of them.

[solemn dew]

Q[x]/(x^n-1) is isomorphic to (Z/nZ)* then
and in our example
Q[x]/(x^8-1) is isomorphic to (Z/nZ)* = {1,3,5,7} ?

[glossy crag]

although idk how you would prove that w/o Galois theory.

[glossy crag]

No, it is not, you're confusing different things together.

[solemn dew]

oh

[glossy crag]

a) Q(a)\cong Q[x]/(f), where f is the minimal polynomial of a, indeed, but that has nothing to do with the group Gal(Q(a)/Q).

b) The minimal polynomial of a primitive 8th root of unity is not x^8-1.

[chilly ocean]

That does seem difficult. Let's see, so it has to be degree 2 by the tower law which means that it has be generated by a degree 2 root and be a root of x^8-1, then maybe do some case work. Would something like that work?

[solemn dew]

alright thanks @glossy crag

[glossy crag]

I don't see how it has to be a root of x^8-1, sqrt2 is not.

[solemn dew]

i'll try to find the subject in the book!!

[chilly ocean]

oh yea, you're right

[glossy crag]

You can use Steinitz's/Artin's (don't remember whose name it was) theorem to show that E/Q has finitely many intermediate fields, but I don't remember if it gives a precise bound.

K[a]/K has only finitely many subfields, that is known.

[solemn dew]

nvm i thought of something else

those are solutions for x^4+1=0

so intermediate field is Q(sqrt(2))

[glossy crag]

So? We know already Q(sqrt2) is an intermediate field, that's easy

[chilly ocean]

Don't you need i?

[solemn dew]

yea you do

[glossy crag]

If z is a primitive 8th root, then (z+z^-1)^2=2, so Q(sqrt2)\subset E=Q(z)

[chilly ocean]

Fellas, done with part 1 showing they commute - not sure how to deduce that o(a, b) = lcm(|a|, |b|)

[chilly ocean]

Anything less would not make a or b equal to e, by the group law for the product

how do I make it rigorous

[chilly ocean]

Maybe that's something for you to try

true

Well, suppose a^n = 1 and b^k = 1

oh

[glossy crag]

Use the fact that if an element g has order n, then g^m=1 if and only if n divides m.

[chilly ocean]

then (a, b)^{lcm(n, k)} = (id_A, id_B) in a group A x B

[glossy crag]

And use the definition of an lcm: m,n divide lcm(m,n) and if m,n divide t, then lcm(m,n) divides t.

[chilly ocean]

it's kind of obvious no?

[glossy crag]

This only shows the order of (a,b) divides lcm, you need the reverse too

[chilly ocean]

isn't this a corollary of Lagrange's thm

[glossy crag]

Not exactly, although I suppose it may be proved like that.

[chilly ocean]

I see

so

[chilly ocean]

this will help

so suppose a^n = 1 and b^m = 1

then the order of (a, b) is lcm(n, m), that's what I have to prove

The lcm of two numbers is the smallest number divisible by both numbers. The order of an element is the smallest positive integer such that g^r = 1. Since lcm(m, n) is the smallest multiple of m and n, this is the smallest possible positive integer such that (a, b)^r = 1

will that do

[glossy crag]

Too wordy and vague for my tastes, I'd say no.

let |a|=m, |b|=n, |(a,b)|=t, lcm(|m|,|n|)=k

m and n divide k => a^k=b^k=1 => (a,b)^k=(a^k,b^k)=1 => t divides k.

1=(a,b)^t=(a^t,b^t) => a^t=b^t=1 => m and n divide t => k divides t.

=> t=k.

[chilly ocean]

does this proof work too?

it's a bit different style than yours

but it uses more familiar techniques to me

[glossy crag]

It's pretty much what I said.

Just instead of showing k divides t and t divides k => k=t, it shows t divides k => t less equal to k, show t<k not possible.

I find that stilted and unnatural.

[chilly ocean]

i see

also

another question

would it be incorrect to use lagrange's theorem to show that the order of an element divides the order of the group

given that the book hasnt taught it yet

(implying that |a|, |b|, |c| has order 2)

and thus using a previous result that I proved (Let G be a group such that g^2 = 1 for all g \in G. Then G is abelian)

to prove that it's abelian

[glossy crag]

It would not be incorrect, but the exercise clearly wants you to prove it w/o Lagrange.

[chilly ocean]

yeah likel

would it be bad-natured

or

I would try and see if you can just rule out that 3 is a possible order. The group is small enough that this seems simple.

[glossy crag]

idk what you mean by bad-natured.

if you want to do it, do it, if you don't, don't.

[chilly ocean]

I think I'll just use Lagrange's theorem

seems easier that way

[glossy crag]

That's what it wants you to do, exactly.

[chilly ocean]

how would I do that

contradiction?

gotcha

Suppose a^3 = 1

uhh

lemme think

then a^2 = a^{-1}?

this question's basically asking me to classify all groups of order 4

quite weird that they introduce that in the first subsection of the first chapter

[chilly ocean]

Often the motivation for such problems is to show how difficult some problems are without more theory.

I see

can I get a hint

for showing that a, b, c cannot have order 3

[severe linden]

Might I ask what book this is from?

[ornate hound]

This isn't so much a hint, but a suggestion in how to proceed in problem-solving this question: try and write down some multiplication tables for this group of order 4. The cancellation rule implies that the rows and colums can't have repeats. (Most textbooks have an example showing this property early on, so I'm assuming you've seen it before).

[chilly ocean]

Dummit and Foote

Chapter 1 Section 1.1

[chilly ocean]

Ah, I see

thank you

[severe linden]

Thank you very much

[chilly ocean]

no worries!

will this suffice for this exercise?

[delicate orchid]

sure, but you can see its abelian directly from the fact that the cayley table is symmetric

[chilly ocean]

oh ya, i proved that before

[chilly ocean]

If anyone's interested in joining an online reading group of D&F (dummit and foote), please let me know in DMs. I will provide more detail soon or in DMs.

[little shadow]

doesnt the group order (together with sylows theorem) force the number of 9-groups to 1 and therefore it cant be simple? the solution was very different so i wonder if i have misunderstood something about sylow's theorem

[wraith cargo]

no?

you need
number of groups = 1 (mod 3)

[delicate orchid]

so we have that either n_3 = 1 mod 3 and n_3 divides 8
so the only possible value is n_3 = 1

[little shadow]

oh right, i took it mod 9 so that explains why

[wraith cargo]

why would n_3 have to divide 8?

[delicate orchid]

sylows theorem

the number of sylow subgroups divides the index

[wraith cargo]

ah

[delicate orchid]

also, yeah you're right I got them backwards, it's immediate from sylow's theorem

[wraith cargo]

okay I forgor that one lol

[delicate orchid]

has to be 1

[wraith cargo]

couldn't it be 4?

[delicate orchid]

hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

[chilly ocean]

8 mod 3 = 2?

[south patrol]

||only two primes divide 72 so we're done by Burnside QED||

[wraith cargo]

burnside has a million theorems which one is the good one 💀

[delicate orchid]

yeah and if I could remember the proof of burnside's I wouldn't be having this problem

well, I know the character theory one I think

or is that the p^aq^b one

[south patrol]

ye i mean p^a q^b

[delicate orchid]

yurrrrrrrr

[wraith cargo]

ahh that one

[delicate orchid]

ok so lets proceed by assuming n_3 = 4

there's some number of order 8 Sylow 2-groups in this thang as well

n_2 = 1, 3, 9

n_2 = 1 we've found a normal subgroup

n_2 = 9 implies the whole got dang group is full of 2-elements which is a contradiction as we know there are 3-subgroups

[little shadow]

n_2 = 9 implies that there are 9 elements of order 2 no?

[delicate orchid]

2-elements

not elements of order 2

[south patrol]

wdym

[little shadow]

what

[south patrol]

9 subgroups of order 2 means 9 elements of order 2

[delicate orchid]

I didn't make a single claim about elements of order 2

[south patrol]

oh

[wraith cargo]

wait doesn't burnside just say that such a group is solvable?

[south patrol]

but it's 8

lol

i got confused by last message

[delicate orchid]

involutions be damned

[little shadow]

what is a 2-element

[south patrol]

the proof constructs a non-trivial proper normal subgroup

[wraith cargo]

man the only time you need to remember the proof you forget :(

[south patrol]

anyway solvable + simple would just mean like

prime cyclic right

well

like isn't it impossible

[wraith cargo]

no?

[delicate orchid]

prime cyclics are definitely simple

I presume you mean solvable + simple?

yeah

[wraith cargo]

okay you changed it lol

[south patrol]

yeah i meant to negate and did a dumb dumb

lol

because we were already showing not simple

but yes as written is what i mean

[delicate orchid]

do we have that order pq implies solvable here

gonna assume not and keep thinking

[south patrol]

i like how i literally do not know how to solve it besides the sledgehammer rip

[sonic coral]

do you mean without burnside?

[delicate orchid]

no I'm not getting a contradiction from assuming n_2 = 3

not immediately anyway

[south patrol]

Yeah

Idk how you're meant to do sylow questions other than basic counting

Occasional tricks ig

[sonic coral]

this is something i actually know how to do

i went from 1-101

[south patrol]

lol nice

[little shadow]

the solution:

[south patrol]

nice

[delicate orchid]

ah that's sneaky

[little shadow]

i would 100% not come up with that lol

[delicate orchid]

yeah same

[south patrol]

sort of shows how reps are useful for finding subgroup tho

[delicate orchid]

get magma to generate the character tables

[sonic coral]

that’s not how i did it wait

[delicate orchid]

yeah hahahahaha

[sonic coral]

i didnt learn anything about group actions

[delicate orchid]

do you know about group actions now

[sonic coral]

nope but i hope to learn about them in my grad algebra course in the fall

[delicate orchid]

why the fuck are they in a grad course

[little shadow]

i stg group actions have given me severe hair-loss

[delicate orchid]

wacko ass university

[sonic coral]

i dont know if they are typically, i just didnt learn them in undergrad

and im not sure if they will be in my course also

[delicate orchid]

they most definitely are not typically a post grad thing

how did you prove the sylow theorems without group actions?

[sonic coral]

im not sure if he proved them id have to look through the notes

[little shadow]

it was revealed to him in a dream

[sonic coral]

i dont have them on hand though but when i get them ill look

[south patrol]

some people are american it's ok

[sonic coral]

i dont remember him proving them though but i had a whole assignment on showing groups werent simple and 72 gave me a headache

[delicate orchid]

this is clever though

[south patrol]

ah yes considering normalisers is a nice trick i remember seeing

thank you for reminding me lol

[sonic coral]

48 and 36 i did also

multiples of 12 seemed to be the pattern for that but im not sure why

maybe just bc its got so many divisors

[delicate orchid]

2^2*3 and then you tack on more multiples

yeah exactly

gross!!!

[dim widget]

Here is a quite cute proof

[last spoke]

What are the sylow theorems lol

(I have also done undergrad algebra, group actions where done briefly but not really)

[delicate orchid]

the full statements easily googlable but it boils down to:

every group of order p^km with m coprime to p has subgroups of order p^k

these subgroups are all conjugate

[granite topaz]

existence and uniqueness of sylow p-subgroups, the fact that they are all conjugate(?, ehh, can't remember if it's conjugate or isomorphic)

[delicate orchid]

and the number of them is 1 mod p and divides m

[granite topaz]

and the statement about the number of such subgroups

[dim widget]

Suppose there are 4 3-sylow subgroups, then the action on the 3-sylow subgroups induces a homomorphism from the group to S_4, which must have a kernel because 72 is larger than 24. This contradicts the group being simple

[delicate orchid]

conjugate implies isomorphic

[upper pivot]

group actions are like, the main part of a course in group theory. Always wierd when people say their classes didnt cover it oof

[delicate orchid]

||it is cute, but that was the answer given by the book ||

[granite topaz]

yes, but I didn't remember if it's only isomorphic or smth stronger

[delicate orchid]

and I will be keeping it in mind! It's so cool

[dim widget]

Oh no

[little shadow]

this was the solution on the exam, its quite neat

[delicate orchid]

stronger than iso? woahhh

[last spoke]

We did groups in 6 weeks. Group action was week 6

Or part of it

[upper pivot]

what was the first 5 weeks oof

[delicate orchid]

yeah this sounds like a first year course maybe?

I did something similar in first year anyway

[last spoke]

Senior year

[delicate orchid]

dunno what that means I'm not american

[last spoke]

I'm not either lol