groups-rings-fields

[terse crystal]

What you said is the most direct way I think, f(x)=g(x^p^r) for some g.

They always use this, like “E is an extension of F, S is the separate closure of E/F. take a x from E, such that x^p from S…”

[cloud walrus]

Shadow08

[olive monolith]

any help is appreciated

[coral shale]

So we need to show inverses 🤔

[coral shale]

'e is the a right neutral element'

Can you fix that phrasing, it's confusing for me

[delicate orchid]

and that e is also the left neutral element

[olive monolith]

okay

[coral shale]

In any case, the first step is to write down everything you know.

I assume you meant 'e is a right neutral element'

[olive monolith]

yup

[coral shale]

In which case, you can write down what this means

in addition to the given statement

[olive monolith]

let $E$ be a monoid such that $\exists e\in E$, $e$ is a right neutral element and $(\forall x\in E)(\exists x' \in E): xx' = e$. Prove that $E$ is group

[cloud walrus]

Shadow08

[coral shale]

Yes?

You can write down this symbolically

[olive monolith]

Yes understood

I am trying to figure it out

[coral shale]

👌

[olive monolith]

Showing inverses is easy, however, I cannot show it without assuming the existence of a neutral element (which can't be done).

[coral shale]

Show work

[olive monolith]

let y be right inverse of x': you have y x' = e y x' = x x' y x' = e so y is a left inverse of x' . x is also a left inverse; you have to have x = y.

[coral shale]

So we want to show e is the identity

[delicate orchid]

ok that's the hard part done

[coral shale]

$$(\exists e\in E)(\forall x\in E)(xe = x)$$

[coral shale]

We have this

Now perhaps we can consider that E is a monoid

[coral shale]

And make a smart choice here

[olive monolith]

a solution was reached, thank you all

[echo orbit]

I've got a commutative ring $R$ and $R$-modules $M, N$. I'm trying to show $\text{Hom}_R(M, N)$ has $R$-module structure.

[cloud walrus]

gristopology

[echo orbit]

Pretty clearly for any $\varphi \in \text{Hom}_R(M, N)$ we should have for any $m_1, m_2 \in M$ and $r_1, r_2 \in R$ the equation $r_1\varphi(m_1) + r_2\varphi(m_2)$

[cloud walrus]

gristopology

[echo orbit]

What I'm wondering is how I can endow $\text{Hom}_R(M, N)$ itself with $R$-module structure; firstly I need a commutative operation on the hom-set

[cloud walrus]

gristopology

[echo orbit]

but it's not obvious to me what that operation ought to be

i guess the identity should be the trivial homomorphism?

[devout crow]

i think the comm. operation on homs should be pointwise addition, and the scalar multiplication from R should be $r.\phi(x) = \phi(r.x)$

[cloud walrus]

∧res

[devout crow]

this gives a module structure by R-linearity? I think?

[cloud walrus]

gristopology

[echo orbit]

hmm... pointwise addition

so let me check the types real quick

b/c that seems to really be the fricky part here

we are really doing

[cloud walrus]

gristopology

[echo orbit]

so like

[cloud walrus]

gristopology

[devout crow]

yeah

[echo orbit]

ah, this seems really natural

and obviously works with the identity being the zero map

very nice, thanks

[devout crow]

np

[devout crow]

the scalar multiplication from R should be r.\phi(x) = \phi(r.x)
this is nonsense btw, I meant something like (r.\phi)(x) = \phi(r.x) for all x \in M

[lavish nexus]

there's no definition saying maximal ideal can't be 0 right?

[south patrol]

ye it can be 0, e.g. in fields

[lavish nexus]

idk what the point of the problem is then for comm alg
like every inclusion of Z in a field works

[south patrol]

That's the example I'd give too, ig it's more just to prove a point than anything

(as in because preimage of prime ideal is prime for example)

[sharp turret]

can anyone give a hint on this problem? not even really sure where to get started with it

[lavish nexus]

a is a zero divisor

[scarlet estuary]

Here's a good strategy for when you don't know where to start: Start by trying to think of some examples of commutative unital rings that aren't fields (why can't they be fields?) and see if they fail the condition (and if so, why?)

[echo orbit]

s/a/the

[rapid bramble]

If F is a field, then F[x] is never a field right?
Cause x will never have a multiplicative inverse since 1/x is not in F[x]

[sharp turret]

thought about that but a has to be a zero divisor for every element and i couldnt figure out an infinite ring that with an element like that

[lavish nexus]

||direct product||

[sharp turret]

oh so like R = ZxZ and then (a,0)x = 0 for all (0,b)?

for x = (0,b) i mean

[lavish nexus]

it works

[sharp turret]

god bless

ive been sleep deprived banging my head into the wall on this problem for like an hr and didnt think about direct products

[echo orbit]

@rapid bramble correct

it is a ring though by hilbert basis theorem

in fact it's a very nice ring

PID and all that

[rapid bramble]

Yeah the prof took too long to get to the fact it's a ring

so we're doing that tmr

[echo orbit]

🐲

[rapid bramble]

but yeah, like any p in F[x]-{constants} works as a counter

[coral shale]

F field -> F[x] Euclidean Domain

v. useful 👀

[rapid bramble]

We're only at Integral domains, and the lack of calculus is upsetting

[lavish nexus]

lol calculus

as long as you know how to differentiate a polynomial you have enough calculus for this stuff

[south patrol]

Another silly way to think about it is that if you have nonconstant f(x) in k[x] then it has a root in K[x] where K is the algebraic closure

and so it can't possibly have an inverse in k[x]

[next obsidian]

Bro… it’s always a ring

You don’t need the basis theorem to show that lol

[echo orbit]

:^)

o yea duh

hilberts just noetherian rings

that was the worst thing i have said today

[next obsidian]

Yeah

[hybrid island]

does my proof work?

[chilly ocean]

damn thats long

[terse crystal]

If you were showing if T=diag{d_1,…,d_n}. N(T-λI)=span<e_j: λ=d_j} R(T-λI)=span{e_j: λ doesn’t equal d_j} so their intersection is always zero, then it’s correct, example in your proof is completely unnecessary, we don’t prove something by giving examples, we only disprove something by giving a counterexample

[hybrid island]

ok

but it works tho right

[terse crystal]

Yeah

[rapid bramble]

Basis in the same sense of vector space bases?

[next obsidian]

No, as in Hilbert's basis theorem

Hilbert's basis theorem just says that if A is a Noetherian ring, then so is A[x]

you don't need it to conclude that F[x] is a ring

[south patrol]

You need A[x] to be a ring for you to be able to prove theorems about the ring A[x]

[chilly ocean]

Hello can anyone please explain me that why are left cosets different from right cosets in a non-abelian group, while also explaining what are left and right cosets respectively? Thanks in advanced.

Any good source explaining the above message is also fine for me 👍

[delicate orchid]

in a non-abelian group you don't have ab = ba so for a non normal subgroup H you don't have gH = Hg

[chilly ocean]

Oh

Oh ya order of operations matter in non abelian group

Thank you

[delicate orchid]

of course, if H is normal we have gHg^-1 = H so gH = Hg but this isn't true in general

[rapid bramble]

I meant the term basis here being the same as basis of vector spaces

[sharp sonnet]

no, it's related to every ideal being finitely generated (thus having a basis in some sense)

[rapid bramble]

So only ideals have bases

[sharp sonnet]

i am not sure this word is still used in this context outside the name of this theorem

gröbner basis i guess

[proud bear]

Well usually you can write an element of the ideal in multiple different ways with the "basis" so it's still different to bases for vector spaces

[sharp sonnet]

yes, hence my original no

basis in LA implies linear independence

this here just means generating set

[chilly ocean]

What are modules in simple words

[delicate orchid]

vector spaces over a ring instead of a field

[chilly ocean]

Oh

Idk I saw a video saying they're vector spaces over a field and got me confused so bad

Thank you

[plucky torrent]

I'm reading Hatcher's textbook on algebraic geometry, and I've gotten to a section on page 36 where he mentions homomorphisms that act like retractions. He says such a homomorphism rho is one that maps from a group G onto a subgroup H of G, such that rho acts as the identity on H ( rho(h)=h for all h in H ). I understand this, but then he says that if H is a normal subgroup of G, G "is" (presumably he means "is isomorphic to") the direct product of H with the kernel of rho. I've been trying to prove this for a while now and can't quite do it. It seems relatively elementary, so I'm probably missing a simple trick. Does anyone know how to prove this?

[terse crystal]

K is the kernel You define f: H times K to G, mapping (h,k) to hk

H intersection K is {1}, this is obvious

So any h from H, k from K, ρ(hkh^-1k^-1)=ρ(h)ρ(k)ρ(h)^-1ρ(k)^-1=hh^-1=1 so hkh^-1k^-1 is contained in K , on the other hand, H is normal, so hkh^-1k^-1=h(kh^-1k^-1) is also contained in H, so hkh^-1k^-1=1, hk=kh

So f is well-defined homomorphism

f is clearly bijective

Surjective: any g from G, ρ(ρ(g)^-1g)=1 so g=ρ(g)(ρ(g)^-1g) is from HK

Injective: hk=1–> clearly h=k=1

QED

[plucky torrent]

Thanks!

[subtle ivy]

it should be noted that this result is simply an application of the fact that G can be decomposed into $H\oplus K$ whenever H and K are both normal, their intersection is trivial, and HK=G. it's proved early on in artin.

[cloud walrus]

shortcut

[next obsidian]

Woah wtf? Do you mean algebraic topology???? I didn’t know Hatcher has an algebraic geometry book

[weak matrix]

Could this be done using a short exact sequence like ker rho -> G -> H?

[plucky torrent]

whoops. meant topology!

[subtle ivy]

ohhh! i was wondering that too!

[subtle ivy]

well what you can show is that rho splits, which might be what you're referencing. but i'm not sure you can show the result without $H\leq Z(G)$.

[cloud walrus]

shortcut

[subtle ivy]

it holds for abelian groups, but not in general that if $0\rightarrow A\rightarrow B\xrightarrow{f} C\rightarrow 0$ is exact and splits by $g$, then $A\oplus C\cong B$. but this fails for $D_6$.

[cloud walrus]

shortcut

[subtle ivy]

(or D_3 if you are geometrically inclined lol)

sorry wait, $C$ must also be free. lol

[cloud walrus]

shortcut

[weak matrix]

Oh, I see

[subtle ivy]

yeah u run into problems with proving it is a homomorphism

[weak matrix]

The free product of groups with inclusion and projection would also be an example, right? Although it's an short exact sequence it's not isomorphic to the direct product

[subtle ivy]

what's projection on a free product look like?

[weak matrix]

Send the element $a_1 b_1 a_2 b_2 \dots a_n b_n \in A*B$ to $a_1 a_2 \dots a_n$ where $a_i\in A$ and $b_i \in B$

[cloud walrus]

Tₑ𝘖(n)

[pastel cliff]

algebra exam tm

moldi pray for me

[weak matrix]

It's not a natural projection like in the direct product, but it is an surjective homomorphism, right?

[pastel cliff]

should i think about elements of the symmetric group as actions/functions

bc the group operation is essentially function composition here

im fairly certain im correct at least but is there a better way of thinking

[delicate orchid]

sure you can think about them in terms of their natural group action on a set

you got the second thing backwards though

function composition is a group operation :troll:

[pastel cliff]

i mean like

elements of S_n are closed under that composition

idk in my head i just said "the sky is blue" and you said "no it's actually rgb(0,0,255)"

new strawb btw

[delicate orchid]

yeah I just got 8bg and then legitimately fainted

[delicate orchid]

my point was a lot more meaningful than that

I'm saying you can think of any group element in any group as a function

hell, you can think of it as a permutation acting on the group if you want to be direct

[pastel cliff]

ohhhhh 3b1b said something about this

"symmetry describes an action"

dont devastate me that man is my savior

im asking bc i was thinking about the braid relation

[delicate orchid]

sure every symmetry is some group action

[pastel cliff]

and i realized i dont actually know what an element of S_n looks like

well i do

(1)(2,3) is an element of S_3 i get that

[delicate orchid]

not an element of A_3 though , or as I like to call it, C_3

[pastel cliff]

but $\sigma_i \sigma_j \sigma_i$

[cloud walrus]

μ₂ (46/47 🪲)

[pastel cliff]

actually

should i think of each of these as like

acting on the identity

FUCK i keep thinking about matrices

[delicate orchid]

yessss let the matricies flow through you

[pastel cliff]

NO

[delicate orchid]

map each of these bastards to a permutation matrix

multiply them

[pastel cliff]

https://tenor.com/view/kinoplex-steak-allow-me-to-explain-the-guy-above-me-is-insane-talking-gif-24312956

[delicate orchid]

map em back

[pastel cliff]

you're memeing right

[delicate orchid]

look all I'm saying is be glad it's just n dimensional not n! dimensional

[delicate orchid]

yes of course I am....

NOT

[pastel cliff]

https://tenor.com/view/crying-emoji-dies-gif-21956120

[shell brook]

okay

friends

[delicate orchid]

just think about them as functions

[shell brook]

Oh you're actually doing math sry continue

[delicate orchid]

nah it's fine

[pastel cliff]

feel free to interrupt im being dumb 99% of the time

[pastel cliff]

but ok like \sigma_i

suppose it's S_3

[delicate orchid]

ok

[pastel cliff]

wait im confusing myself i know what elements of S_3 look like of course, but i cant understand it in "general"

does that make sense

like

[delicate orchid]

not really no

[pastel cliff]

epic

[delicate orchid]

they're just permutations on bigger sets

they behave identically other than they have more things to act on

[pastel cliff]

ok what does \sigma_2 look like in S_3

[delicate orchid]

well what the fuck is sigma_2

[pastel cliff]

oh wow i really just created this problem for myself then

[delicate orchid]

wait no no no, what's sigma_2

what did you mean by that

are you putting some ordering on them or something?

[pastel cliff]

no no i thi

i mean

yes...? i guess? i looked back at some notes, i forgot and in class we did like

s_1 is the element of S_n that switches the first and second element

i just forgot that this whole time

[pastel cliff]

is that not done in general...

[delicate orchid]

I have never heard that in my entire life

[pastel cliff]

LETS GOOOOO

[delicate orchid]

just write (12) what the fuck

I guess if you want a generating set like ok yeah maybe writing S_n = <s_k : k = 1, ..., n-1> is useful but like get a life?

[pastel cliff]

we learned that too but he wrote the braid relation as $s_is_js_i = s_js_is_j$ if $|i-j| = 1$

[cloud walrus]

μ₂ (46/47 🪲)

[delicate orchid]

I'm sorry the what relation?

you're not doing braid groups are you

[pastel cliff]

no he said that it's a relation that holds in S_n

[delicate orchid]

lemme think

yeah it does

[pastel cliff]

wew lads you're scaring me

[delicate orchid]

(12)(23)(12) = (13)
(23)(12)(23) = (13)

yeahhhh

I believe it

ok so if you want to prove this in general just think about the images of i, i+1 = j, and j = j+1 taking each s_n as a function mapping n -> n+1 and n+1 -> n

bit of a nightmare cause it's like.... duh?

[pastel cliff]

"get a life"

[delicate orchid]

anyway back to your confusion about s_n, or as everyone likes to call it, (n, n+1)

[pastel cliff]

yeah im calling it that from now on

makes life much easier and harder to fogor

[delicate orchid]

so has your confusion been cleared up?

[pastel cliff]

yesnt

mostly

i think

i just need time to be dumb before not being dumb

[delicate orchid]

I'm still waiting on the latter

[pastel cliff]

you're gonna be waiting an awfully long time wew lad

[delicate orchid]

I meant me not you

[pastel cliff]

oh

hell ive been waiting like 20 years

[delicate orchid]

oh god I'm gonna be 22 soon

horrifying

[pastel cliff]

what the fuck does this mean

[pastel cliff]

cringe!

oh wait actually i know what it means

i just saw the word computations and got confused because we've seen like 3 whole numbers in the class

and theyve been on a clock waiting for lecture to end

[delicate orchid]

what is a cyclic group?
what is a coset?
what is a normal subgroup?
define the construction of a quotient group?
is the set of yo bitches a group?
and so on

[shell brook]

okay this is devolving so I will ask my question

I have polynomial x^4 - 12x^2 - 253 and the question is asking me to compute the Aut group of it's splitting field

so the split field is Q(sqrt23, i sqrt(11))

I have literally no idea how to compute the group

[delicate orchid]

I'm feeling big C_2 energy with this one

[shell brook]

I've looked at like 5 diff ways of doing these type of question and they are all different

Like we don't have min polys in the course or whatever

i think he wants us to explicitly compute the elements

[delicate orchid]

wait how are you d-

oh

[shell brook]

or the generators or whatever, so we can show its iso to C_2

if it is

I don't know if it is

[delicate orchid]

yeah it was just a guess

[next obsidian]

I highly highly doubt it’s C_2

That seems ludicrously small imo

[shell brook]

I think it's D_4

personally

but that's beside the point

[next obsidian]

Honestly I’m so trash with these

I forget how I ever computed these like

[shell brook]

any idea is better than no idea

I can show you how the only example he ever did was done

[delicate orchid]

I mean we just need to find the funny automorphisms

[shell brook]

See example 2

I have another question related to this but it comes later

I feel like I have no fucking clue whats going on in this course anymore

🙂

[delicate orchid]

wait it might be K_4

[shell brook]

Okay well lets think harder cuz throwing out random groups isnt useful

[delicate orchid]

just looking at the roots it's a rectangle which has K_4 symmetry (and I'm pretty sure they're not conjugate)

[shell brook]

I think from the stuff i've read w min poly

[next obsidian]

So correct me if I’m wrong

But

Isn’t this automatically Galois?

[shell brook]

It's gotta fix like sqrt23 -> -sqrt23 and isqrt11 -> -isqrt11

idk

[next obsidian]

It’s the splitting field of something so it’s normal

[shell brook]

Wdym

[next obsidian]

I believe

[shell brook]

"We" don't know what normal is

[next obsidian]

It means any polynomial with a root in there

Splits

But I think splitting fields are always normal

And Galois = normal plus separable

But this is separable cuz char 0

[shell brook]

What are you saying is Galois, the Q(sqrt23, isqrt11)?

im gonna call it E from now on

[next obsidian]

I mean

If that’s the splitting field of an irreducible polynomial

Then I think so

[shell brook]

x^2 - 12x^2 -253 is irre over Q

so ya

[next obsidian]

Oky so

I loooed it up

Splitting fields are normal indeed

[shell brook]

Ok

[next obsidian]

So if you determine the degree this at least tells you how many automorphisms it should have

Then, write down the 4 roots

[shell brook]

what is the degree

[next obsidian]

Of the extension

I’m not sure, but I’m guessing it’s degree 4

[shell brook]

I don't know what that is but okay continue

[next obsidian]

What?

[shell brook]

We haven't learned this

[next obsidian]

You don’t know the degree of an extension is?

Wtf????

[shell brook]

Not in the course

No

[next obsidian]

???????????

[shell brook]

can you define

[next obsidian]

You’ve never seen

[K:F]

Or whatever

[shell brook]

wrt groups yeah

[next obsidian]

Holy shit

So it K is an extension of F

It becomes a vector space over F right?

[shell brook]

Sure

[next obsidian]

Like you can literally interpret scalar multiplication by elements in F

As if you were multiplying an element in K with an element in K

Cuz like F < K

Does that make sense?

[shell brook]

Okay

Right yeah we mentioned this part

[next obsidian]

So the degree of an extension is the dimension of K as an F-vector space

[shell brook]

Ah okay

[next obsidian]

So it has a basis of [K:F] elements

Galois is also the same as saying that Aut(K/F) = [K:F]

Or like whatever the fuq your notation for like

Automorphisms of K fixing F is

[shell brook]

what

[next obsidian]

So you have “the right number” of automorphisms

[shell brook]

|Aut(K/F)|?

[next obsidian]

Is equivalent to Galois

Yeah

[shell brook]

oh okay

[next obsidian]

Whoops

[shell brook]

yeah sure

[next obsidian]

Lmao

Anyway

Assuming that you wrote the field correct

The one you called E

I think it is degree 4

So there should be 4 automorphisms

I’m about to board a flight

Lowkey

@latent anvil

Help Jesse pls

🥺

[shell brook]

any algebrahs in chat

[shell brook]

Here's the problem 🙂

[delicate orchid]

jesse don't mean to be rude but what the fuck is your course

[shell brook]

ring n fields :)

can i just

like do this

wait

I dont feel like typing it out

is this not like. reasonable

whats wrong with this

is it cuz I don't knwo that there are only 4 elements or something idk

[delicate orchid]

chmonkey said there are only 4

[shell brook]

he just said the degree is 4

idk if thats the same claim

[delicate orchid]

_ _

[shell brook]

oh

poggers

[delicate orchid]

so it's either C_4 or K_4

[shell brook]

Hm

idk do you get cyclic vibes

[delicate orchid]

these are my thoughts on the matter

[shell brook]

ah

[delicate orchid]

alternatively you could just like

[shell brook]

How do i mak ehtis epxlicit tho

[delicate orchid]

show it's not cyclic lol

[shell brook]

yeah

[delicate orchid]

cayley table

[shell brook]

god

[delicate orchid]

it's 4 elements it's viable

[shell brook]

maybe I'll just say "by a trivial checking of the cayley table"

[delicate orchid]

in general though I'd try to construct an isomorphism to whatever group you think it is

[pastel cliff]

what is A_3

[latent anvil]

Sorry Jesse

Didn't see ping

[shell brook]

It's okay sham

idk what I'm doing still

[pastel cliff]

got maidens

[latent anvil]

😎

I just beat the ulcerated tree spirit twice

The worst boss fight in any game I've ever played

[pastel cliff]

you havent played hollow knight

dont wanna derail convo tho

[latent anvil]

I have

[shell brook]

Here's the question Sham

[latent anvil]

Oh are you still working on it?

Sorry I thought someone had helped

[shell brook]

No

I just guessed a solution

but it's not relaly a solution

[latent anvil]

So it's a degree 4 field, right?

You got it via two quadratics

[shell brook]

I did?

Oh well yeah the polynomial factors into two quadratics

[latent anvil]

?

Yep

[shell brook]

like it's (x^2 + 23)(x^2 - isqrt(11)) or something I think idr

[latent anvil]

You know the field is galois, right?

[shell brook]

No

[latent anvil]

(x^2 - 23)(z^2 + 11)

[shell brook]

This is not a notion we have learned

[latent anvil]

oh

Okay that makes this harder

[shell brook]

Here's what I did

[latent anvil]

Okay, can you think of an upper bound on the number of automorphisms?

Oh looking now

[shell brook]

I just watched a youtube video and copied the exact technique

well it's supposed to be n! isn't it

lol

[latent anvil]

No, not at all

[shell brook]

well it's a very loose upper bound

[latent anvil]

Oh I see

Sorry I didn't realize that was the upper bound

I thought you were saying "well the number of auts is n!"

[shell brook]

So my course doesn't have the notions of degrees or min polys

[latent anvil]

Ah

[shell brook]

or Galois/normal/seperable fields

[latent anvil]

well, you don't really need min poly

[shell brook]

Okay

tight

[latent anvil]

But it's a useful way to think about things

The idea is like

You have 2 generators of the field, right?

So any automorphism is actually determined by where it sends each generator

Yeah?

[shell brook]

Yes

[latent anvil]

So it suffices to determine what the image of a generator under some automorphism is

If α is an aut, what can you say about α(sqrt(23))?

[shell brook]

It's gotta fix it or send it to -sqrt(23) right

[latent anvil]

Yup

Can you prove that?

[shell brook]

with min polys I can

[latent anvil]

Well, it's essentially that

You just don't need to say "min poly"

What's your proof?

[shell brook]

Well the min poly of sqrt(23) over Q is x^2 - 23 I think. and any aut must send sqrt(23) to one of the roots of the min poly, so it could only be sqrt 23 or -sqrt23

I think that's the proof

[latent anvil]

Why does any aut send it to a root?

:)

[shell brook]

Uhh if a is a root and P is an aut, f(P(a)) = P(f(a)) = P(0) = 0? idk

sorry

im nto sure

[latent anvil]

Yes!

So write that

[shell brook]

Oh nice okay

[latent anvil]

Specific to sqrt(23) and sqrt(-11)

You don't need minimality or anything, right?

[shell brook]

wdym minimilarity?

[latent anvil]

Like we don't need to know P is the minimal polynomial

Just that it's some polynomial that a satisfies

[shell brook]

I don't think so

I can't see why you would

[latent anvil]

Yep, I agree

[shell brook]

Okay so

[shell brook]

Is this just clarifying that this isn't wrong

I'm confused about what this is for

[latent anvil]

What you wrote is right

It's enumerating every possibility for where the roots could go, and saying these define automorphisms

Now that's not actually right

You don't know that there is an automorphism for each possibility

Like how we exclude swapping roots of different polys

[shell brook]

wdym

[latent anvil]

So we get the n! bound by saying each root has to be sent to another root, and this permutation determines the whole map

But you can't just write down all possible permissions and say that there's an automorphism for each one

[shell brook]

Right

[latent anvil]

Well there could be more constraints than just being roots of different irreducible factors

[patent crescent]

what does this mean if P is a group? Is this asking to prove that theta is an automorphism?

[latent anvil]

So you can't just say "there's an automorphism which sends sqrt(23) to -sqrt(23) and sqrt(-11) to -sqrt(-11)"

It might be that no automorphisms permute the roots in that specific way

In this case there is such an automorphism, but you need to prove that

Does that make sense @shell brook?

[shell brook]

Im reading

Right

[latent anvil]

The computation you did gives an upper bound of 4

[shell brook]

Okay

Yes I understand

Hm okay

How does one prove this automorphism exists

[latent anvil]

So the nice thing is that you can define maps on vector spaces like this

Right?

On a basis

So you would need to check that the linear map is a field homomorphism

[shell brook]

you lost me

[latent anvil]

1, sqrt(23), sqrt(-11), and sqrt(-253) form a basis for Q(sqrt(-11),sqrt(23)) over Q

Does that make sense?

[shell brook]

Okay

[latent anvil]

And a linear map can be defined just by giving the image of this basis

[shell brook]

why do you need sqrt(-253)?

[latent anvil]

sqrt(23)*sqrt(-11)

[shell brook]

👍

[latent anvil]

Does this make sense?

[shell brook]

No I have no idea how this works

[latent anvil]

oh

[shell brook]

LOL

sorry

[latent anvil]

So you have a field k = Q

And an extension F/k, F = Q(sqrt(23), sqrt(-11))

[shell brook]

Yes

[latent anvil]

You want to define a field isomorphism F -> F which is k-linear

k-linear says α(cx) = c α(x), and taking x = 1 we see it says α(c) = c

Does that make sense?

[shell brook]

Okay

Why

[latent anvil]

Sorry I realized you just wanted field automorphisms

[shell brook]

Yes

[latent anvil]

Not automorphisms of extensions

Sorry

Okay so

If you have a field F

And it's like, characteristic 0

Then F is a Q-vector space

F has a subfield which looks like Q

[shell brook]

Ok

[latent anvil]

Because it has an element 1 which you can keep adding and subtracting to get the integers, and then multiply and divide to get Q

So we can think of F as a Q-vector space

n*x = x + ... + x

yeah?

[shell brook]

Hm okay

[latent anvil]

And a field automorphism α : F -> F is Q linear

α(n x) = α(x +... + x) = α(x) +... + α(x) = n α(x)

[shell brook]

Yes

[latent anvil]

Well then, defining field automorphisms can be done by defining them on a basis for F

[shell brook]

this feels like overkill

[latent anvil]

and then you can check that the resulting linear map is multiplicative, sends 1 to 1, and is a linear isomorphism

[shell brook]

Okay I'm following though

[latent anvil]

All I'm really saying is that every element of your field is uniquely of the form a + b sqrt(23) + c sqrt(-11) + d sqrt(-252) for a, b, c, d in Q

And so you can define functions by these representations

[shell brook]

Ahh

I see

[latent anvil]

Define then on a, b, c, d

This is better stated in terms of vector spaces though

[shell brook]

right so then thats how you guarantee the AM exists

[latent anvil]

Right exactly

[shell brook]

Okay surely I can blackbox that part

[latent anvil]

It's more natural in terms of field extensions

Bc then to define stuff in Aut(F/k) you define k-linear maps F -> F

It's just that k = Q so we can leave it implicit

Q is the minimum field in any char 0 field, everything uniquely extends it

[latent anvil]

Honestly the more important thing is that you understand why we can't just write this and claim they exist

[shell brook]

I do not understand that point still

Why could they not exist

[latent anvil]

Why can't I write down α(sqrt(23)) = sqrt(-11), α(sqrt(-11)) = sqrt(23)? I'm sending each root of the polynomial x^4 - 12x^2 - 253 to another root of that polynomial

That doesn't define an isomorphism α, right?

[shell brook]

cause that doesn't respect the min poly

[latent anvil]

Right

Maybe there's some 2 variable polynomial this doesn't preserve

The σ, τ you wrote down

[shell brook]

Ah

I see

[latent anvil]

Bi quadratic extensions are just really nice

[shell brook]

God this is painful I feel like I only barely understand anything

[latent anvil]

You will learn better ways to think about it

[shell brook]

Okay well I will read this over and try to write something comprehensible

I have one more uqesiton

I think it's easier

[latent anvil]

Yeah

[shell brook]

what is (i)? It's weird cause F already has one thing adjoined

like we did this in the notes already but we were splitting over just Q

[latent anvil]

Right, but that one thing doesn't let you write down sqrt(2)

x^3 - 2 is still irreducible

[shell brook]

cube root 2?

[latent anvil]

Oh lol

Sorry

Fixed

[shell brook]

Right so is it not litearlly just the same field as this

I don't see why this would be different

[latent anvil]

Yeah, it is that

[shell brook]

wtf

[latent anvil]

But it's also F(cbrt(2))

And that is not Q(cbrt(2))

[shell brook]

which is Q(w, cbrt(2))?

[latent anvil]

That's what interesting about it

[shell brook]

Ohh ok

hm

What's the actual difference though

well okay nvm I guess the difference is obvious

lmao

I just meant what do you think the intention of this question is

[latent anvil]

To point out that splitting fields can look different as you adjoin roots of other polynomials

[shell brook]

Wait okay I'm a little confused

is Q(w, cbrt(2)) the same as Q(a, wa, w^2 a) as in the second screenshot?

They are both splitting fields of x^3 - 2

Oh splitting fields are unique right

[latent anvil]

They are equal fields

Yeah

[shell brook]

equal or iso

[latent anvil]

Equal

You know what, just ignore me

I don't think I'm explaining this well

[shell brook]

LOL

it's okay

you've helped more than anyone else has

Galois theory is awful

tyvm

[latent anvil]

I really liked this stuff :P

[shell brook]

I think the Gallian way of teaching it is just awful

so it transfers over

[latent anvil]

This time two years ago I was taking a final and doing weird galois shit in a coworking style place

And I thought it was weird that this was the last possible day to registry

*register

And about 3 days later there was a shelter in place

[shell brook]

oh shit its march

[latent anvil]

Ah yeah I don't like that book

[shell brook]

yeah ur right thats fucked up

[latent anvil]

Good problems

[latent anvil]

Prove that θ is a surjection of P onto itself. If you're finite this is the same as an automorphism

[patent crescent]

Also in other words, you can say θ(P) = {θ(p) for all p in P} right?

[shell brook]

these look awful sham

[latent anvil]

They were fun

[latent anvil]

Yep, f(X) usually means { f(x) : x in X }

[patent crescent]

alright thanks, I haven't seen that notation before for some reason

[shell brook]

@latent anvil sham are you sure that it's Q(w, cbrt2) lol, cause it looks like it's pretty much completely worked out in our notes

[shell brook]

ex 2 here

sry 2 question u. future Brown student or whatever

[latent anvil]

What?

[shell brook]

It's weird that they would ask a question thats fully worked out

Cause the next question asks to show its a cyclic of order 3

but they did that

[latent anvil]

Yep, you can even prove it using the result in the notes directly

[shell brook]

Weird

okay

sry

[latent anvil]

Let k = Q, F = Q(ω), E = Q(ω, α)

Then we have k < F < E

We know E = Q(α, ωα, ω^2α) right?

Do you see how each generator of one is in the other?

We know E is the splitting field of x^3 - 2 over Q from the notes

[shell brook]

Right

[shell brook]

wait I dont see this

[latent anvil]

Okay, which generators might not be in which field?

[shell brook]

can you define what a generator is in this case LOL

I have not heard it wrt fields

[latent anvil]

Q(ω,α) has generators ω and α

[shell brook]

Oh okay

nvm yes I do lol

[latent anvil]

If K/F is such that x^3 - 2 splits in K, then K actually extends Q as well because F contains Q. So K contains the splitting field E of x^3 - 2 over Q. So E is the smallest extension of F where x^3 - 2 splits

[shell brook]

Right

[latent anvil]

(ie the splitting field)

[shell brook]

Ahh

okay I got it

Okay thx

[subtle ivy]

just read about identities and proving linear algebra results over a general ring.

that stuff is so satisfying & cool :)

[hidden haven]

They should make more math

[terse crystal]

That’s what category theory does I think, make more math

[subtle ivy]

so true

[subtle ivy]

im scarily close to understanding this. what have i become. i used to only like analysis and combinatorics

wait but maybe i dont understand what an algebra is. i thought an algebra over a ring A was a ring B with a homomorphism A\rightarrow B...

hm

[iron vessel]

Also it should map 1_A to 1_B

But in this case B is an A-algebra

[hidden haven]

An algebra of an endofunctor T is an object a along with a morphism Ta → a

[subtle ivy]

that's part of the definition of a ring homomorphism

[subtle ivy]

hm so for an algebra (in my sense) $B$ of a ring $A$ is there always an endofunctor T on Ring such that $A=TB$?

i think yes

[cloud walrus]

shortcut

[subtle ivy]

actually im tripping this question is weird

[chilly ocean]

Have I proved this correctly

[sharp sonnet]

what does it mean for two groups to be homomorphic?

[chilly ocean]

For two groups to be homomorphic, they must preserve similar algebraic structure of elements/objects such that one element from one group could map to the other (through a function) and be similar?

Like here the evens can map to 0, and odds to 1 and behave exactly similar like the elements of G1 under addition

I hope my intuition is correct-

[sharp sonnet]

so existence of a homomorphism?

i mean your idea is correct, but its written down very non-rigorously

[delicate orchid]

It looks like you assumed it was a homomorphism in step 3

Never mind you’ve just ordered it strangely

[subtle ivy]

also... any two groups are "homomorphic" via the 0 map, trivially lol

[delicate orchid]

Lol, I assume they’re looking for something injective though

[sharp sonnet]

then they will have a hard time

[subtle ivy]

um, then definitely not? because Z is infinite and Z_2 is finite.

[delicate orchid]

What?

They’re mapping Z_2 into Z which can be done injectively very easily

[subtle ivy]

lol. show me an imbedding of Z_2 into Z.

[delicate orchid]

The one they’ve written lmfao

[sharp sonnet]

[subtle ivy]

Z_2 is a cyclic group of two elements my friend.

[delicate orchid]

Yup

[subtle ivy]

there is no element of order 2 in Z.

[delicate orchid]

And 0+2Z -> 0 and 1+2Z -> 1 is an injective mapping into Z

[subtle ivy]

i mean it is a mapping.... but certainly no homomorphism.

[delicate orchid]

When did I claim it was

[subtle ivy]

[delicate orchid]

New exercise: prove that there can’t be a non-trivial homomorphism

[subtle ivy]

you make me feel insane. best of luck to you.

[hidden haven]

Ye, if you want to study R-algebras, define T to be the constant R endofunctor, which means that all rings map to R and all ring homomorphisms map to the identity map on R. Then a T-algebra is exactly a ring A along with a map R → A since TA = R and this is exactly what an R-algebra is (at least for commutative R)

[chilly ocean]

Can you please link down some resources if you can related to proving the existence of homomorphism

Or just homomorphism

[tribal moss]

As Shortcut pointed out, for every two groups G and H, there is a homomorphism from G to H, namely the map that sends everything to the identity in H.

And it is unclear to us what you're actually trying to do, since "homomorphic to" is not a concept in common use.

[chilly ocean]

Oh

[tribal moss]

_Presumably you're looking for a homomorphism with some specific properties in addition to being a homomorphism.

[chilly ocean]

Oh so the group operation also matters

And under that we check if there exists a homomorphism between the two groups

Right?

[tribal moss]

Again, there always does, so that is not a very interesting thing to ask.

[chilly ocean]

Ohh okay

[tribal moss]

Are you confusing "homomorphic" and "isomorphic", perhaps?

[chilly ocean]

No no

I was studying homomorphism yesterday

So I tried to frame a random question and prove it

[tribal moss]

Actually I just note something is even worse off. You write $G_1=(\mathbb Z^+,{+})$, but that is not a group at all -- it doesn't have inverses!

[cloud walrus]

Troposphere

[chilly ocean]

Wait a second

Wait yeah

I didn't notice that, thanks for pointing that out

[tribal moss]

Okay, that is important information. The sort-of default assumption is that you're trying to solve an exercise set by someone who (for lack of a better word) knows what they are doing. So we'll tie ourselves into knots trying to reinterpret the work you show in a way that could be a meaningful exercise.

[hidden haven]

You're talking about groups

The only maps between groups are the homomorphisms

[chilly ocean]

Okay

[tribal moss]

When we know it's just something you came up with for yourself we can give much better answers :-)

So first, "homomorphic to" is not a thing. There is always a homomorphism going between any two random groups you can name.

[chilly ocean]

Oh

[tribal moss]

If we change your groups to $G_1 = (\mathbb Z,{+})$ and $G_2 = (\mathbb Z/2\mathbb Z,{+})$, though, so they are actually groups, then there are \emph{exactly two} different homomorphisms $G_1 \to G_2$.

[delicate orchid]

It’s called a little trolling we do a little trolling

[cloud walrus]

Troposphere

[chilly ocean]

Ohh

[tribal moss]

One of them is the trivial homomorphism that sends everything to 0.
The other one is the one you were trying to define, which sends every even number to 0 and every odd number to 1.

The latter is surjective, which is an interesting property.

[chilly ocean]

Oooh

[hidden haven]

Not amused 👎

[tribal moss]

I.e. it makes sense to ask if there's a surjective homomorphism between two groups you're looking at.

[chilly ocean]

Oh so if I map the other way round, I get the trivial homomorphism

[tribal moss]

The other way around, there's only the trivial homomorphism, yes.

[chilly ocean]

Oh

Can we say

If their exists a homomorphism between two groups, there always exists a trivial homomorphism between them?

[tribal moss]

Skip the "if". There always exists the trivial homomorphism, full stop.

[chilly ocean]

Ooh

I get it now, thank you so much !!

[subtle ivy]

very cool thank you!!!

oh that emote is so cute. kitty so happy

[hidden haven]

[coral shale]

I can't spot what is wrong with my argument

If anyone cares to try decipher my writing 🙏

[hidden haven]

It is correct

[coral shale]

[hidden haven]

Epic grader

[coral shale]

ikr

thanks

[lethal dune]

Lol

btw do you memorize the proposition numbers as well? (prop 1.20?)

[hidden haven]

Open book exams

[lethal dune]

that's epic level "open book" then

when you are allowed to quote a book

[hidden haven]

What open book exam would not allow you to quote the book?

[coral shale]

For our homeworks I write it in case

In our exams we're just expected to state there's a thm telling us this ig

(if its not named)

'Because this this this, we have that that that. Therefore...'
Is what I'd just state

[hidden haven]

[lethal dune]

you guys can quote propos quotes from books

https://tenor.com/view/you-guys-are-getting-paid-gif-20686662

[hidden haven]

[lethal dune]

anyway here's a basic question

is it necessary that a finite ring with no zd has unity?

[delicate orchid]

All rings have unity so yes

[lethal dune]

good enough

In all seriousness tho

bruh

nvm

[coral shale]

Is there an intuitive way to see why? (proof is a pain to read)

else ima be staring at this for a good while

ah nvm nvm
Just derivative magic

[delicate orchid]

What did you see what was in the void

[lethal dune]

[tribal moss]

By pigeonholes and the absence of zero divisors, ax=b and xa=b always have solutions when a and b are nonzero. This means that the semigroup R\{0} under multiplication is both left and right simple, and therefore it is a group. In particular it has an identity.

[lethal dune]

I need to look up the definitions

I was looking into more elementary approach, I ab able to show there's an supposed unity but unable to show it actually is the identity

[chilly ocean]

Am I allowed to ask a question related to classes here

[tribal moss]

Let a be nonzero. By pigeonholes there is an e such that ea=a. Similarly for any nonzero b, there is an x such that ax=b. But then eb=eax=ax=b, so e is a left identity for everything.

Mutatis mutandis there's an element f that's a right identity for everything, but then e=ef=f.

[chilly ocean]

Alright so if I'm correct, according to my intuition, say S = {1,2,3,4} and T = {5,6,7}, then we say class A as:
A = (S,T)?

[tribal moss]

We'll need a bit more context for that.

[lethal dune]

I see thanks

[tribal moss]

"Class" has many different meanings, and it's not clear yet which of them you're asking about.

[chilly ocean]

Like in general
A class is like a collection of sets or other algebraic structures right

[chilly ocean]

This one is a collection of two sets

So can it be called as a class

?

[tribal moss]

Right, so the set-theoretic notion of "class".

{S,T} is certainly a class, then.

If (S,T) is supposed to remember the order of the two elements, such that it is different from (T,S), it is not merely a collection anymore, though.

[chilly ocean]

Oh I see

So a class is always an unordered pair?

[tribal moss]

A class is always unordered, but not necessarily a pair.

[chilly ocean]

Oh

Thank you so much once again 👍

[tribal moss]

(In axiomatic set theory it is common to represent ordered pairs as sets, usually with the definition (s,t) = {{s},{s,t}} which is due to Kuratowski. But that's more of an implementation trick than something one should think of as an innate property of the ordered pair).

[coral shale]

If I gave a finite group G, and asked for a field extension F over Q st Gal(F:Q) is isomorphic to G

This is an inverse Galois sub-problem right?

I was wondering if there's a method for 'easy' or 'small' G (hopefully for certain classes of G)?

=======
My specific problem is that I want to try this for G an nxnxn rubiks cube group, but that might be hoping too much. I don't even know where I could begin for 2x2x2

[tribal moss]

The 2×2×2 group has order 7!3^6 = 3,674,160 if we ignore rotations of the entire cube. I don't think I'm prepared to call that "small".

[coral shale]

No, but I want to begin with the small case first

general small case I mean

And convince myself trying is just impossible or maybe theres hope

I tried looking up and don't see anything, so I have no idea how 'hard' this is even for small groups

[terse crystal]

A person asked me something, I thought I was on the right approach but couldn’t get the correct answer. He asked given a set A={a_1,a_2,…,a_n} , how many multiplication table can be made for A to make A a (abelian) group.
I thought , okay I let X={all possible multiplication tables}, then S_n acts on it: any permutation g, the corresponding permutation matrix is P, then given a multiplication table x, I define g•x to be PxP, permutation of rows of x and columns of x in the same way.Then |orbit(x)|=n!/|Stab(x)|
For example A={a_1,a_2,a_3,a_4}, there are only two groups of order 4, G_1=Z/4Z, G_2=Z/2Z times Z/2Z. So I get two multiplication table when (a_1,a_2,a_3,a_4)=(1,t,t^2,t^3) and (a_1,a_2,a_3,a_4)=(1,a,b,ab) respectively.
Then any multiplication table should be in the same orbit with at least one of those two tables. Maybe they are the same orbit. So the answer is of the form either 4!(1/k_1+1/k_2) or 4!/k right? But the answer is 16 I thought okay two orbits, but I calculated those 2 |stabilizer|, I didn’t get the correct result. Because I got k_2=4, so k_1=12/5 not an integer impossible. Also if A={a_1,a_2} clearly the answer is 2, but when I calculate it my way, there exists only one orbit, stabilizer is S_2, so I got 1 as answer, what is wrong ?

[pastel cliff]

obliterated my algebruh exam

easiest shit of my life

i do it in my sleep

that is all

[delicate orchid]

https://tenor.com/view/rip-pack-bozo-dead-gif-20309754

[terse crystal]

I believe it’s contained in some very hardcore algebraic number theory,involves many calculating

[delicate orchid]

anyway for my contributive discussion of the day, this seems quite over complicated

each abelian group is associated with a cayley table (up to permutation)

use the structure theorem to compute the number of abelian groups of order n

then multiply by the number of permutations of n (n!)

[terse crystal]

So you are saying the answer is 2(4!)=48

But the answer is 16

[delicate orchid]

what order is this?

[terse crystal]

4

[delicate orchid]

order 4 right

there's some combinatorial thing it's gotta satisfy that I'm missing

[terse crystal]

You can’t just multiply n! Some will repeat, that’s why I used group action

[delicate orchid]

yeah I'm now realising this lol

[terse crystal]

I see it’s not working, I just can’t figure out what went wrong

I figured it out why I was wrong anyway… g•x should not be defined PxP, but P^TgxP. Where gx is replace any a_i in the multiplication table with a_g^-1(i) then permutation rows and columns. So permutation of elements of the table first then permutation of rows and columns, I only did permutation of rows and columns…

[coral shale]

https://i.imgur.com/g8TGmAs.png
https://i.imgur.com/4UH39qJ.png
https://i.imgur.com/FYaKosg.png

I wanted to use a version of 3.3 for L -> F in general, but there doesn't seem to be one in my notes

Did I have to extend tau to a K-automorphism in K bar?

Or was there some other way round this (this felt slightly more painful than it needed to)

[latent anvil]

You could just prove the general version, it's not a complex proof

[coral shale]

Seems weird the general version doesn't seem to appear

We only proved the specific version in the last hwk

[latent anvil]

It's easy too overlook this stuff when writing course notes or whatever

¯\_(ツ)_/¯

[coral shale]

Well I was wondering if they expected us to prove without extending (since it doesn't appear)

yh

[zealous flame]

I just started group theory last week in my first year of uni and we were doing lagranges theorem and I couldn't wrap my head around what a coset is as the lecturer skimmed through it, could someone explain to me what it is ?

[coral shale]

Probably best you have an example/can come up with one for every concept

To avoid getting lost

[zealous flame]

lol im already lost

[coral shale]

For any $g\in G$ and $H$ subgroup of $G$, the left coset is defined as follows
$$gH := {gh : h\in H}$$

[coral shale]

These are the left cosets of H

[zealous flame]

ah so the right coset would be Hg

[coral shale]

yes.

[zealous flame]

aight thanks

man uni maths makes wanna die

so for a group defined by Z to the base 6 = {0,1,2,3,4,5} lets say <4>={1,4,5} , am i right in saying that <4> is a coset of z to the base 6?

[frail zealot]

wdm by Z to the base 6?

like integers mod 6?

[zealous flame]

yeah

[coral shale]

under addition?

<4> = {1, 4, 5} makes no sense to me

[zealous flame]

ah

[coral shale]

You firstly need to identify a subgroup of Z/6Z

[zealous flame]

the way i saw it is that it formed a partition on integers mod 6

[coral shale]

Well ok firstly you need to clarify which group we are even working in

Z or Z/6Z (under addition, presumably)

And then find a subgroup of this

And then figure out what the left (or right) cosets are of that subgroup

[zealous flame]

ok thanks, i'm just trying to form examples to help me understand this

[coral shale]

Sure - go ahead

Can you think of a subgroup of Z under addition?

[zealous flame]

i could think of a cyclical group of Z under addition

like using 1

[delicate orchid]

Z is cyclic, so all subgroups are also cyclic

[zealous flame]

yeha group theory is not my strong point as you can tell

[delicate orchid]

nah you're just new

[coral shale]

Keep writing your thoughts

[zealous flame]

ok

[coral shale]

If we want to think about subgroups of a group

Perhaps it is best to consider 'what ifs'

What if we had a certain element in the subgroup

Then what must happen

So in this example, what if 3 is in your subgroup?

[zealous flame]

if 3 were in my subgroup then 3 is an element of a subgroup which the subgroup must meet the criteria of having an identity, being communitive and containing an inverse

[delicate orchid]

associative not commutitive

and don't forget closed

[coral shale]

Maybe state the group axioms first (4 of them)

[delicate orchid]

whoops sorry

[zealous flame]

ah right

closure: for all of x , y belongs to S then x*y belongs to S as well

[coral shale]

no no no just the names will do, but sure

[zealous flame]

ah right

associativity

identity

and inverse

[coral shale]

Right

And if we ALSO have commutativity

this group is special

and known as 'Abelian'

[zealous flame]

abelian?

oh right

yeah

[coral shale]

ab = ba is commutativity

[zealous flame]

ahh

yeag

[coral shale]

• closure
• associativity
• identity
• inverses

So we have this list.

So firstly, any subgroup is also a group

We have a subset and under the same binary operation (with restricted domain) it is a group

So what must be in the subgroup no matter what?

[zealous flame]

identity

[coral shale]

Right

And in this case we are in (Z, +)

[zealous flame]

so its 0 in multiplecation and 1 in addition

wait no

[coral shale]

no the other way round D:

[zealous flame]

lol

[coral shale]

x + 0 = 0 + x = x

[zealous flame]

yeah

[coral shale]

Ok, so 0 is in our subgroup.

Now if we know 3 is in our subgroup

try using the other axioms to figure out what else must be in

[zealous flame]

then the inverse of 3

[coral shale]

yep whats that

[zealous flame]

1/3

[coral shale]

no

[zealous flame]

-3

[delicate orchid]

-3

[coral shale]

yes, be careful

[zealous flame]

i'll do my best

[coral shale]

{0, 3, -3} is a subset of our subgroup

What else can we figure out

[zealous flame]

there is a magnitude

[coral shale]

🤔

[zealous flame]

that is divisible by the magnitude of the gorup?

[coral shale]

I have not heard the term 'magnitude' in group theory

[zealous flame]

wait no magnitude is the wrong word here

[delicate orchid]

which group axiom have you not used

[coral shale]

order of the group?

[delicate orchid]

you mean order

[zealous flame]

yeha

[coral shale]

That is lagrange

[delicate orchid]

but that's not what we're using here - although it does tell you this group has to be infinite

[zealous flame]

lagrange is for finite

[coral shale]

but you don't need it here (also lagrange works only on finite for the dividing bit)

[coral shale]

u wot

[delicate orchid]

look not now shuri

finish the construction then I'll start ranting

[zealous flame]

ah right

in that case

[coral shale]

{0, 3, -3} is a subset of the subgroup. Yes, use other axioms.

• closure
• associativity
• identity
• inverses

======