groups-rings-fields

[tardy hedge]

I dont really like how it did proof by contradiction cuz i dont really see why the statement is true still

[south patrol]

[K(u):K(u^2)] divides [K(u):K] which is odd, and the former is either 1 or 2 so it must be 1

I think this is the most basic way to do it

since the question is framed such that you can do this proof basically

[tardy hedge]

Why is the former 1 or 2

Oh

The proof above shows its either 1 or 2

I think i just wanna see more basically why u is in K(u^2) i guess

[south patrol]

Hm idk how you can make it more basic than that (which is a one-liner)

There won't really be a nice constructive proof because the hypothesis you've been given is pretty weak

[rustic crown]

if f(x) is the minimal poly of u, then write f(x) = g(x^2) + x h(x^2), now h(u^2) isn't 0 as degree of h(x^2) < deg f(x).
so u = -g(u^2)/h(u^2)

[south patrol]

oh lol i stand correected

well

[rustic crown]

where did i use f has odd degree tho

ah, because h can't be the zero poly

if h was the zero poly then f(x) = g(x^2), which is a contra

[ebon harness]

Hi , i'm stuck finding the order of this permutation
(1235)(24567)

[delicate orchid]

,w lcm of 4 and 5

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by wewladstbh

[south patrol]

hm i'm not sure 5 is prime though

[delicate orchid]

it's 1 mod 4 so the debate is still out

[south patrol]

lol

[ebon harness]

But some numbers are in both cycles

[south patrol]

hm why do you say that wew tho

[delicate orchid]

gaussian integers

[south patrol]

you can just rewrite it as a product of disjoint cycles

to make it easier

[ebon harness]

The answer is 12 (says the book ) but i dont know how

[delicate orchid]

(124)(3567)
it's equal to this

I must warn you that I multiply permutations backwards though

doing it the other way should get you (145)(2367)

[ebon harness]

From right to left ?

[dim widget]

AB is conjugate to BA so the order is the same

[delicate orchid]

yur but I thought I should mention it irregardless

[dim widget]

Some people like to write it like composition of functions (which is natural) other people are deranged

[delicate orchid]

oh I write composition of functions backwards as well, the normal way is not natural

$fg$ should become $X \overset{f}{\rightarrow} Y \overset{g}{\rightarrow} Z$ sorry libs

[cloud walrus]

W;3w Lads Tbh

[dim widget]

anyway @ebon harness the point is that some people consider (1235)(24567) to be (24567) first then (1235) second as a permutation of the numbers from 1-7, and some other people are wrong

[delicate orchid]

"gf -> f then g"
boustrophedon lookin...

[summer path]

Oh I thought wew meant actually going right to left

[delicate orchid]

oh no that is deranged hahaha

[delicate orchid]

furthermore, reading it in the "incorrect" order gives us 1 -> 2, 2 -> 3, 3 -> 5, 5 -> 1, 2 -> 4, 4 -> 5, ... there's no nasty skip from 6 -> 7, 1 -> 2 where your eyes jump across the entire permutation

I feel incredibly strongly about this because I legitimately think the convention is backwards. It's like electrons having negative charge in physics, it's backwards and everyone knows it but they just COPE

[summer path]

I'm not sure what your eyes are doing

[ebon harness]

I'm doing it form right to left the the way this f book is doing it

Thank you @delicate orchid @dim widget

[summer path]

Your eyes had to jump across the entire permutation to get to 1->2 to start with though?

[dim widget]

Then the product is (124)(3567) just to be clear 🙂 but the order doesn’t depend on how you multiply

[delicate orchid]

me when a^-1aba = ba

[dim widget]

it’s one of the most fun things about math that AB ~ BA for A or B invertible but they always have the same characteristic polynomial even if they’re not invertible

as a random aside

[delicate orchid]

I will invert them anyway

[cloud solar]

A finite ring, a+x invertible for every x invertible. Show a nilpotent

I took a the function from U(A)->U(A), f(x)=a+x

The function is injective

A finite so U(A) finite

So f surjective so there is x0 invertible s.t. a=1-x0

a is clearly a divisor of zero

Any hint

[slim kayak]

localization chad

[slim kayak]

why is it a zero divisor?

[cloud solar]

In a finite ring an element is invertible or a zero divisor

[delicate orchid]

feel that? that's true.

[cloud solar]

And if a is invertible because f is surjective there is x invertible s.t. f(x)=a so a+x=a so x=0 false

[slim kayak]

Oh, but in that case you could have done, by contradiction:
if a is invertible, then a-a must be invertible

Have you tried checking if a must be in every prime ideal of A?

[dire siren]

don't we need commutativity for this idea to work?

[slim kayak]

all rings are commutative

yeah

[south patrol]

Yes

[summer path]

Non commutative rings don't exist

[stark helm]

I am considering f(x)=2*(x^2+2), and it seems that we have a unit element 2 in C, then why it is reducible in C?

I mean 2 can have multiplicative inverse 1/2 in R[x], so it is irreducible. but it doesn't have 1/2 in C[x]??

[tardy hedge]

It has a root in C so its reducible

[stark helm]

You mean 2x^2+4=0 as we have solution x in C?

I can find x=sqrt(2)i and -sqrt(2)i

[delicate orchid]

Yeah, so it factors over C

[south patrol]

every poly of degree > 1 is reducible over C due to the roots too

[delicate orchid]

Almost like it’s algebraically closed

[south patrol]

SAUR TRUE

[tardy hedge]

Since we’re talking about polynomials can i have hint for q6

Idek how to think of it tbh

This stuff hard for me

[stark helm]

even reducible over R, since we must have at least one root?

[south patrol]

try to characterise "f is irreducible" in terms fo degrees

[tardy hedge]

I dont really know what i should be thinking about

Ok thanks ill try

[south patrol]

For odd degree polynomials, yes

But e.g. x^2 + 1

is irreducible over R

[tardy hedge]

Potato y r u so good

[south patrol]

i'm not

[tardy hedge]

Lol

[delicate orchid]

Haha it’s kiand’s question but for [C:R]

[south patrol]

ye i think so lol

[south patrol]

anyway the real answer in this context is that the same ideas come up a lot

once you do this sort of thing once the problems all end up being very similar

but the first time is harder of course

[delicate orchid]

It’s a similar idea right? You’re quotienting out by a degree n polynomial so the only splitting that can occur in the bigger field is in the “n-part” of the degree. Not rigourous but I hope you get the vibe

[stark helm]

Then is it possible to say f is reducible over a set even if it has not root. Like example of 2x^2+4 is reducible in Z, but it seems we can not figure out a root in field Z?

[south patrol]

Ye

[delicate orchid]

It’s reducible over Z because 2(x^2+2) is a factorisation into non-units

[south patrol]

Yes

Basically "has a root" = "has a linear factor"

But you can be reducible in other ways thn having a linear factor

[stark helm]

I think I have understand it, really appreciate that

[south patrol]

np

[glossy crag]

When does the Galois group determine the extension?

[sly kernel]

hi friends, given F field char(F) = 3. proof that v: a -> a^3 is isomorphism

[coral spindle]

You don't need to post your question in multiple channels.

[rustic crown]

it's not, F = F_3(x) is a counterexample.

[coral spindle]

what I said in another channel

[sly kernel]

it is not isomorphism?

[coral spindle]

As I said, your claim is false

[rustic crown]

which is the other channel

[coral spindle]

linear algebra, for some reason

I didn't give the example tho. det got me there.........

[rustic crown]

do you have additional assumption like F is a finite field?

[coral spindle]

I bet Eagle is just getting isomorphisms and homomorphisms confused.

It's always a homomorphism, of course

[sly kernel]

No brother)

[coral spindle]

Well then you may indeed struggle to prove a false fact.

[rustic crown]

it's also true if you assume F is algebraic over F_3.

[sly kernel]

sorry, what is F_3 ))

?

[rustic crown]

field with 3 elements

Z/3Z

[sly kernel]

or simply Z3?

[rustic crown]

.<

[coral spindle]

Some people write Z_3, but this can also mean the 3-adic integers, so one should avoid Z_3 and write Z/3 or Z/3Z instead.

Sniped by det again wtf........

[sly kernel]

how can I proof that it is not isomorphism?

[rustic crown]

not to mention Z_3 can also mean Z[1/3] or localization of Z at (3).

[coral spindle]

It is sometimes an isomorphism, but not for all fields. Det gave an example above.

typo

[rustic crown]

field maps are always injective. so surjectivity is the only thing that can fail

F_3(x) is the field of rational functions with coefficients in F_3

[sly kernel]

is it Field?

[coral spindle]

Yes.

N.b. it is F_3(x) and not F_3[x].

[rustic crown]

what was n.b. det forgor

[sly kernel]

ok, thank you for help 🙂

[summer path]

||nonbinary ||

[coral spindle]

N.b. = nota bene = "note well" or "pay attention to the fact that..."

[south patrol]

n.b. actually standard for "Novikov-Browder"

[summer path]

potato also has an uwu in about me

[rustic crown]

[crystal turtle]

uwu!

[south patrol]

I do indeed

I forgot about that aha

Any tips for showing flatness is Zariski-local?

I forgot how to do it but don't want to cheat lol

I can see that if M is a flat R module then the same is true after localising everything at p, so im doing the converse

[rustic crown]

if localizing at each D(f_i) is exact, then original sequence was exact

[slim kayak]

You can limit yourself to the case of maximal ideals

at least what atiyah did

[south patrol]

Okay yes sure

[south patrol]

Ye

[rustic crown]

by exactness, that is same as showing N|_D(f_i) = 0 implies N = 0

[south patrol]

I've done this before ye just not sure how to go from there

Sorry I guess I mean at primes

Rather than zariski locally

Uhh I not good w Terminology

[rustic crown]

N localized at f_i is zero then N_p is zero for every prime p not containing f_i

[south patrol]

ye sure nice

[rustic crown]

since D(f_i) is a cover, these are all primes

[south patrol]

ye

okay cool thankies

[rustic crown]

[south patrol]

I will finish details

Heh

[slim kayak]

sorry how are you relating this to a ring spectrum?

flatness of some module that is

[south patrol]

Ring spectrum

[rustic crown]

say
N' --> N --> N'' is exact
we want to show
N'⊗M --> N⊗M --> N''⊗M is exact
and we're given by each M[1/f_i] is a flat A[1/f_i] module.

[south patrol]

I know what you mean but aha

[slim kayak]

yeah idk what else to associate D(f_i) with haha 😅

[rustic crown]

so we know localizing the second sequence at each f_i is exact, and that localizing at each prime is exact, so original sequence must be exact too

if you're asking why i wrote D(f_i)

[slim kayak]

if f_1 to f_n generate A, is it enough to show that localizing at all of these is flat?

[rustic crown]

yep

[slim kayak]

okay i see the value in the notation

[rustic crown]

if M[1/f_i] is zero for each f_i then M is already 0

[slim kayak]

just a bit bamboozled how it actually works algebraically. Stupid magic

yeah I guess should go through the steps

[rustic crown]

its like a partition of unity argument

if you do it directly

[slim kayak]

supposedly I can get stuff like af_k^n = 0 for all k and then get a(1)=0

with a some image

[rustic crown]

m in M, then for large enough N, f_i^N * m = 0

[south patrol]

Real

[rustic crown]

since (f_1^N, ..., f_n^N) has radical ideal A, 1 is also in that ideal

which gives m = 0

[south patrol]

Sadge

So sleepy

[slim kayak]

okay yeah that seems like enough

[next obsidian]

Prod A_fi is flat and the map on Spec is surjective by assumption that D(fi) cover Spec A so its faithfully flat

[rustic crown]

det was gonna say that ><

[slim kayak]

damn, using geometric intution for algeba for one of the first times in the wild

crazy

[rustic crown]

prove that flatness is flat-local

[slim kayak]

now i can say things like "checking flatness on points is sufficient" and be kind of correct and sound deranged

or does it correspond to check the complement of closed points?

[next obsidian]

You can check on closed points on a ring

[rustic crown]

complement?

[south patrol]

Closed points

[next obsidian]

Because all points specialize to a closed point

[south patrol]

Real

[slim kayak]

M is flat iff the localizations at maximal ideals are flat

[south patrol]

Cofinal

TIng

[next obsidian]

M

[rustic crown]

M

[white oxide]

M

[slim kayak]

my bad u right 😭

[next obsidian]

This follows from being able to check at each point

Any point has a closed point in its closure (is contained in a maximal ideal)

And the localization of flat is flat

So if you’re flat at all closed points, to get to an arbitrary point you detour through a closed one

You know you’re flat at the closed one, then you stay flat going to the actual point you wanted

[slim kayak]

actually isnt this a little bit more fitting for adv alg

[rustic crown]

det no know how to think about flat geometrically other than for ring maps where i can sort of think about fibers not jumping dimension and forget about other stuff

<

[rustic crown]

so this no make sense to me

[slim kayak]

so like, if I have localized over m I can pass onto the localization at p as long as the closure of p contains m?

Right, I can just send anything to its identical looking image

[slim kayak]

I think it just translates to that localizing at maximal ideals is looking at S^-1M with S being minimal, you then localize over some T containing S which is again exact. I am happy to be corrected though

[rustic crown]

we're going backwards tho

[slim kayak]

backwards?

[rustic crown]

we need to know flatness for all m in cl{p}, which is why i'm confused by the argument

[white oxide]

so i think I came up with a proof of $\impliedby$; If $\alpha \in E$ is transcendental over $K$, then $K[\alpha]$ is not a field. But this contradicts the assumption that every subring $R$ of $E$ containing $K$ is a field, so $E/K$ must be algebraic. was wondering if there was any other way to prove the leftward implication

[slim kayak]

yeah, which you can do once you know it is flat for the maximal ideal in cl{p}

[cloud walrus]

okeyokay

[next obsidian]

Only for one gets it for p

[south patrol]

Basically the same honestly lol

like this is the idea p much

I would do it basically the other way round like

[white oxide]

wait don't prove the converse plsss

[south patrol]

pick an element a of E, then K[a] = K(a)

no i'm doing the same thing as you dw

[white oxide]

oh okay lo

[rustic crown]

contrapositive

[south patrol]

oh wait

did i just do both directions at once

lol

uh

oops

[slim kayak]

Only for one gets it for p
i cant parse this 😭

[south patrol]

Anyway K[a] = K(a) means 1/a is a poly in a

and that gives you a min poly for a

[rustic crown]

same :3

[white oxide]

well i'm hoping the converse isn't that bad, just gonna let some a in R, use the fact that it's algebraic and try to manipulate the equation for the minimal polynomial or something to produce an ivnerse for that element

[white oxide]

oof lmao

okay okay

[next obsidian]

Haha

Flat at only one such m gets flatness at p

[slim kayak]

Oh, i thought det meant any prime ideal in cl{p} with m

[next obsidian]

I’m saying that

“Flat at all p => flat” => “flat at all m => flat”

Because via this argument you can turn “flat at all m” into “flat at all p”

And then conclude “flat”

[white oxide]

i was just curious, because it's only in the next section that he introduces K[a] and some theorems about it

which implies that maybe there's another way to do it

but yeah this is the easiest i would guess

[slim kayak]

nvm yeah

[rustic crown]

do you have this even without finite generation?

[next obsidian]

Yes

[rustic crown]

[next obsidian]

It’s literally just the statement that (M_m)_p = M_p

[slim kayak]

just localize to a p s.t. m contains p, no?

[next obsidian]

If M_m is flat then you keep it forever

[next obsidian]

To an m, but yes

[slim kayak]

with localize to a p I mean going A -> B to A_p -> B_p

A and B just happen to be A_m and B_m

[rustic crown]

idk what was i thinking now

yee makes sense ><

[slim kayak]

hey you learned a new mnenomic for this

(i couldve sworn we had an emoji called det once)

[white oxide]

i mean would this work for the converse? let $R \subset E$ be a subring containing $K$. Let $\alpha \neq0 \in R$. Since $E/K$ is algebraic, the extension $K(\alpha)/K$ is algebraic and is a field; in particular, there exists $\beta \in K(\alpha) \subset R$ with $\alpha \beta = 1$

[cloud walrus]

okeyokay

[slim kayak]

powers of alpha form a basis, it being algebraic extension just means its finite

but yeah, on the existence of beta

[rustic crown]

i would write K(alpha) = K[alpha] to argue it's contained in R

[white oxide]

this would be sufficient right? all the other conditions of being a field just follow from it being a subset of E

which is a field

like there won't be zero divisors and it's commutative

[white oxide]

ye i think potato did that

probably easier

[rustic crown]

no i mean, it's not "obvious" why K(alpha) is contained in R

[slim kayak]

this is for "if every ring containing E is a field, then K(a) is algebraic" ?

[white oxide]

trying to prove that every subring R of E containing K is a field

given that E/K is algebraic

[slim kayak]

oh

[white oxide]

oh

i see

oh so with K[a] = K(a) we can explicitly describe the elements using K[a]

which are just going to look like linear combinations of powerse of alpha

with coefficeints in K

which should be contained in R

[rustic crown]

yee

[slim kayak]

thats a characterisation of being algebraic, yeah

[rustic crown]

you can also think of K[alpha] as the smallest subring containing both K and alpha. since R contains both, this is a subset of it.

[slim kayak]

so like, if R is K[a^3], it should be able to express its own inverse?

wtf nvm, powers and linear combinations of algebraic are still algebraic

||for any b in R, R contains K[b]. But being algebraic we have K[b]=K(b) so R contains the inverse of b|| ?

[rustic crown]

non-zero b, yee.

[slim kayak]

all elements are invert- fuck, yeah okay

[rustic crown]

also yo udon't need "since b algebraic over K, R must contain K[b]"

R will contain K[b] as soon as it contains both K and b

[slim kayak]

i know

i am just eternally scatterbrained

i have the "is algebraic hypothesis" floating around freely and sometimes it bumps into stuff

[rustic crown]

lol

[slim kayak]

mfw when the smallest rings containing K and b is contained in a ring... that contains K and b

actually, how long are you planning on keeping the festive attire?

[rustic crown]

dunno, when i'm the odd one out

[rustic crown]

okie i think i was all confused because i was thinking p as a bigger point, and so it made me think that (M_p)_m = M_m

det did a dum dum

[white oxide]

for complex numbers $a, b, c$, is it true that say $\overline{\mathbb{Q}(a, b, c)} = \mathbb{Q}(\bar{a}, \bar{b}, \bar{c})$

[cloud walrus]

okeyokay

[white oxide]

i'm too lazy to prove this if it's true

[slim kayak]

so sad

i got mixed up by the notation too

sometimes your thing-y is a multiplicative set, other times its an prime ideal yada yada

[rustic crown]

bar = complex conjugation?

then yee

[white oxide]

ye

ok cool thanks

[chilly ocean]

yee?

[tardy hedge]

Q6, what about x^2-2 is irreducible in Q[x], but Q(root2, root3, root5) has degree 3 over Q, so this question says that since 2 is relatively prime to 3, x^2-2 should be irreducible over Q(root2, root3, root5)?

but cant it be (x-root2)(x+root2)

wat goin on

[delicate orchid]

I don't buy that's degree 3, it's degree 8 no?

[tardy hedge]

honestly idek i just said it

[delicate orchid]

[Q(r2, r3, r5) : Q] = [Q(r2, r3, r5) : Q(r2, r3)][Q(r2, r3) : Q(r2)][Q(r2) : Q] = 2*2*2 = 8

[tardy hedge]

fgjfgk

[delicate orchid]

I'll say that while I do not know how you'd prove this result, I absolutely believe it

[tardy hedge]

hmm yeah see the problem is i look at it and my mind is blank

like i just really dont know whats going on lmao

[delicate orchid]

the vibes tell me that you can only get degree n by appending mth roots with m|n

[tardy hedge]

i need to think about the fundamentals more carefully i think

[delicate orchid]

and if something is degree d coprime to n that factors over the extenstion then it implies some kth root has been added with k|d => k not |n

that's the vibes anyway

[south patrol]

oh also lol

i realised i was dumb in that like

flatness being local is (again!) just a consequence of being able to check exact sequence at primes lol

like uh

[slim kayak]

I thought we concluded that situation lol

[south patrol]

oh rally

i just meant like i thought people gave a more specialised argument

when really it's like

[white oxide]

here do they just take F = K(X) and F' = K(a) and phi: K[X] --> K[A] in place of id: R --> R

[south patrol]

$(M \otimes_R N){\mathfrak p} \simeq M{\mathfrak p} \otimes_{R_\mathfrak p} N_{\mathfrak p}$ and so you can just apply usual lemmas lol

[cloud walrus]

potato

[slim kayak]

There was some confusion on what the argument was but yeah. I mean, we never got to the flat at maximals -> flat step but yeah

[slim kayak]

Said step

[south patrol]

oh i meant that's what it is

like

exact at all primes iff exact iff exact at all maximals

is a standard theorem too

so just apply that and use what i said and bingo

[slim kayak]

Yeah it's like 3.10 in atiyah

The argument was about maximal -> primes directly

[south patrol]

ah okay sorry lol

[slim kayak]

Which follows from localizing being exact

[south patrol]

then yeah what chomnkey said

[slim kayak]

Like this, yeah?

[south patrol]

oh not what i had in mind but ye

[slim kayak]

Idk I find it a bit more intuitive

[south patrol]

wait

sorry

that was dum

ye

[slim kayak]

When thinking about the mult sets you can just send any element to itself

[south patrol]

comm alg is cute

kinda cursed that what made me do this was like homotopy theory

[slim kayak]

No 😿

Only if it wants to be

[south patrol]

wdym

[slim kayak]

Rn working on it

And that doesn't feel cute most of the time

[south patrol]

aw

i think it's like

i love the more geometric sides/interpretations lol

but ye

what comm alg u doing

it seemed you were doing hartshorne II.1 or II.2?

[slim kayak]

Just II.2 and now stating on II.3. looked at sheaves months ago

[south patrol]

ye cool

i messed up cause i read hartshorne like up to II.4 like a year ago and haven't really progressed lol

[slim kayak]

The commalg in the geometry feels like dark magic. The geometry in comm alg is a pleasant surprise

[south patrol]

since then

[slim kayak]

Yeah, I'll see how long until I run out of internal motivation

Gluing together the fiber product is giving me a headache

Harty's proof is much appreciated

[oblique matrix]

I can't quite understand why showing that each set of relations implies the other will prove that the group presentations represent the same group

[sonic coral]

if you believe that a presentation represents a unique group then surely the relations in each group satisfying each other’s presentations will force them to be the same group

[oblique matrix]

I can replace a with s and then b with sr and then yes the relations reduce in an if and only if manner to the dihedral group relations

that doesn't explain why you can just replace a with s and b with sr

[barren sierra]

A better word for replacement is "isomorphism"

[oblique matrix]

how would I exhibit the isomorphism between the presentations

[barren sierra]

So if the presentation in that problem is G

[oblique matrix]

a maps to s and b maps to sr

[barren sierra]

Yup

That's an isomorphism G -> D_2n

[oblique matrix]

and I have to prove that well-defines everything else

ok I'll give it some more thought

thanks!

[barren sierra]

The exercise is basically giving an isomorphism

Without using the words isomorphism explicitly

[dull marsh]

Will it be more appropriate to ask about modules here or in #linear-algebra ?

Or depends on the question itself?

[slim kayak]

Modules don't belong into linear algebra, here please

[dull marsh]

I see

I don't have a question yet though, I am just thinking of reading about modules

[naive whale]

If the modules are over a field then #linear-algebra would probably be fine 😏

[cobalt heath]

What is a module over group

[rocky cloak]

An abelian group (or vector space) with a linear group action.

[cobalt heath]

Linear group action?

[rocky cloak]

g(x+y) = gx + gy

[cobalt heath]

Oh wow

[rocky cloak]

It's equivalent to a module over the group ring

[cobalt heath]

So you do not necessarily need rings to act like action on modulea

[cobalt heath]

Ah. A bit underwhelming

[cobalt heath]

The addition itself is defined as abelian group op, right? How do you get a group ring for the module here?

[rocky cloak]

Yeah, the addition is the operation of the abelian group.

The group ring consists of formal linear combinations of group elements. In the group ring addition is just addition of linear combinations

[cobalt heath]

Ah, I only knew about the group ring in the form of k[G]

I guess formal linear combination of Z[G] is also possible

[coral shale]

daily reminder Z is a field

[cobalt heath]

@coral shale huh

[south patrol]

Lol

[tardy hedge]

finite field extensions are like introducing a new factorization for some elements of the field right

at least for ones like Q(sqrt(2))

[tardy hedge]

so ur saying something like if f(x) does factor over this extension, then it implies something about the roots of f(x) that is impossible

Hm i think im slowwwly seeing what im supposed to see

[rocky cloak]

A bit of a strange way of phrasing it, but finite field extensions are given by adjoining roots of irreducible polynomials. And that will then in turn allow you to factor them further

[dull tiger]

I'm currently looking at the subgroups of $C_2 \times C_4$. The group has order 8, so by Lagrange, the subgroups can have order 1,2,4 or 8. The subgroups of order 1 and 8 are clearly the trivial subgroups. But now I'm a little bit stuck finding the rest of the subgroups. They have to contain the identity element and $xy^{-1}$ has to be in the subgroup for any $x,y$ in it. So I thought I found ${<0,0>, <1,0>}, {<0,0>, <1,2>}, {<0,0>, <0,1>}, {<0,0>, <0,2>}$ . But what about some of order 4?

[cloud walrus]

Marieeee

[rocky cloak]

You seem to have a mistake because
<0, 1> + <0,1> = <0, 2>

So your third set is not a subgroup.

[dull tiger]

ouh yes, you're right! i totally missed that

[rocky cloak]

As to how to find subgroups of order 4, it helps to know what the groups of order 4 are up to isomorphism

You can start by finding all the cyclic subgroups of order 4 for example

[dull tiger]

so like C4 and C2 x C2?

[rocky cloak]

That's right

[rocky cloak]

So C4 is generated by a single element of order 4, while C2xC2 is generated by two elements of order 2. So that tells you what to look for when finding such subgroups.

[dull tiger]

so basically my subgroups of order 4 of $C_2 \times C_4$ would be ${<0,0>,<0,1>,<0,2>,<0,3>}$ and ${<0,0>,<0,1>,<1,0>,<1,1>}$?

[cloud walrus]

Marieeee

[dull tiger]

but no, that does not work with the closure of the subgroup

[chilly ocean]

suggest me a good book on group theory

[naive whale]

complete beginner, or do you have some experience already?

[chilly ocean]

i studied some lectures but my basics arent strong enough

a beginner

hows this book

[naive whale]

Well the book I got going with was Contemporary Abstract Algebra by Gallian, I would say it's pretty accessible for a beginner

Yeah D&F is like the classic algebra book, but some parts of it can be fast paced/hard to digest if you're self studying or like me and have a hard time with dense walls of text

[rustic crown]

#book-recommendations maybe?

[rocky cloak]

Almost, so again <0, 1> doesn't have order 2.

And there are more elements of order 4, so you should expect another cyclic subgroup

[chilly ocean]

do u have this in pdf

[rocky cloak]

Yar yar philadidee 😉

[dull tiger]

Okay, so i get one subgroup like this ${<0,0>, <0,1>, <0,2>, <0,3>}$. Then there is a subgroup with elements of order 2 looking maybe like this ${<0,0>, <0,2>,<1,0>, <1,2>}$ and then maybe one like this ${<0,0>,<0,2>,<1,1>,<1,3>}$?

[cloud walrus]

Marieeee

[rocky cloak]

You got it!

[dull tiger]

Thanks a lot for your help!

[tardy hedge]

Ok yeah. Im stuck on q6.

Spent too long on it now

I was thinking of trying to use that degree of field extension formula when u have a tower of field extensions

[rustic crown]

draw diamonds

[summer path]

🔶

[tardy hedge]

Diamonds?

Like a field diamond thing

[rustic crown]

yee

[tardy hedge]

Should i work with a specific example?

Yea like i was trying to do things like this

I felt like i was close

[rustic crown]

take \alpha to be a root of f

[cloud walrus]

det

[rustic crown]

now write all degrees

show that [F(a):F] = deg f, that proves f is also irred over F.

[rustic crown]

yea, that can help sometimes

try K = Q, F = Q(sqrt(2)) and f = x^3 - 5 in Q[x].

[tardy hedge]

Ehh ok

Im trying not to be too hard on myself but i just feel dumb

[tardy hedge]

Like im assuming u use that degree formula of a sequence of finite extensions on both sides of that diamond

[tardy hedge]

And im not entirely sure why this is true. I sort of get it but its still messy to me

[rustic crown]

okie one thing you should note is that the element alpha always satisfies f which can be considered as a polynomial in F[x].
so [F(alpha) : F] <= deg f.

this didn't require any coprimality stuff

the coprimality puts heavy restrictions on what [F(alpha) : K] could be.

do you see why this is the case ><

[chilly ocean]

If I am working with torsors and I want to turn them into a tensor space of the N-dimension, is this just a matter of defining the origin? With an abstract or undefined origin representing a torsors space and the newly defined origin defining the rank-N tensor space as a Euclidean space of the N-dimension?

I understand this is a non-standard function but I'm more curious on a hypothetical standpoint and do not know how to approach finding this out.

To express this as a theorem instead of a question I would try it like this:

Let T be the torsors under a group G with a free and transitive action. Suppose there exists a vector space V associated with G and a tensor space T^N(V) constructed fromV. Under certain conditions and mapping, one could define a representation of T in T^n(V) such that the actions and structural properties of T are preserved and reflected within the tensor or Euclidean framework.

Would this theorem be true?

[oblique matrix]

Ok I want to prove that A = <a,b|a^2=b^2=(ab)^n = 1> is isomorphic to B = <r,s|r^n=s^2=1, rs = sr^{-1}>

I want to choose an isomorphism where phi(a) = s and phi(b) = sr

I don't know how I can extend this to the rest of A in a well defined way and I don't know how to prove the homomorphism property along with the injective/onto property

[clear fiber]

Hi all, I’m struggling a lot with the last part of this biconditional proof. I don’t know how to show that S = (e)

One thing that bothers me is that I feel that my whole proof so far is valid in any ring, and I haven’t used any properties of Z/nZ yet. (Which I’m guessing I should, since otherwise why would they make the problem statement specifically about Z/nZ?)

[cobalt heath]

(Missed something)
Consider what a subring of Z/nZ satisfies.

[slim kayak]

Have you heard of zero divisors?

Consider the product of two rings R x S. Then (1,0)(1,0)=(1,0), but its ideal is just R x {0}

[tardy hedge]

Haha this is random but i was reading definition of topology on a set and i finally saw a glimpse of what the point of it is

Im taking topology for first time next semester

[cobalt heath]

There is #point-set-topology btw

[delicate orchid]

hahah nobody cares about that definition (in here). We just use them for invariants

seethe more analysists

[hidden kite]

Just to check, my proof that the additive group of rationals isn't isomorphic to the multiplicative group of rationals should be alright right?

If you have any comments/suggestions on anything other than the validity of the proof (e.g. phrasing), feel free to share it as well! :D

[rose prism]

its pretty overcomplicated and written in a convoluted way, and the last conclusion is not quite right (but almost)

you can just observe that 1 and -1 have order 2 in (Q, x)

ok on second look its not that overcomplicated

but i would suggest avoiding run-on sentences

[hidden kite]

Oh right oops had a brain loading moment

Yeah yeah the cardinality isn't necessarily 1 😅

[hidden kite]

Mind elaborating on this?

[rose prism]

so many commas

so few periods

[hidden kite]

Isn't that quite common in math writing?

Is the issue it being a bit of a chore to read because of long sentences?

[rose prism]

“furthermore, since …, suggests that” is not correct grammar

[hidden kite]

Isn't there some grammatical liberties one can take in math writing though

[rose prism]

lol

[hidden kite]

Tbh I don't see the issue with grammar at that part but that's besides the point: either ways, isn't it pretty readable? Well, to me it is at least, which is why I'm asking.

[hidden kite]

why the "lol"? Seems like you're mocking me or smt tbh, in which case, not very nice man.

[cobalt heath]

Is that that many commas

Dang, I need to reduce my comma consumption

[rose prism]

i found it quite annoying to read

but if you already made your mind up then why ask

itd probably be more readable if you put the math on its own line

[cobalt heath]

Is "Since ..., one has ..., so ..." too many of commas*?

[rose prism]

instead of treating it like words

[rose prism]

idk

depends on the sentence

[cobalt heath]

Hmm

Since G is finite, its element has finite order, which generates a cyclic subgroup of G.

[hidden kite]

I didn't/haven't 'made up my mind'. I was just trying to figure out why my phrasing might not be ideal, as you seem to be suggesting. So that I can improve.
(I tried to emphasize this by stating "Well, to me it is at least, which is why I'm asking")

[hidden kite]

How do you decide whether to write inline math or not?

What's wrong/hard to read about my inline math 1=eta(0)=... btw? I can follow it just fine and putting it in a new line seems to only improve the readability a bit? No?

Again, just trying to understand what's less than ideal about my phrasing.

[rose prism]

idk how to explain why its hard to read

it just is

im not actively monitoring whatever my visual cortex is doing

when i look at the words on my computer screen

https://kconrad.math.uconn.edu/blurbs/proofs/writingtips.pdf

[hidden kite]

Hm I see

[prisma ibex]

so many undergrads never got this message

[rose prism]

using these is like that I Q bell curve meme

well \forall and \exists

the others are inexcusable

[cobalt heath]

But but short memo-

[rose prism]

math is an oral tradition

so its ok to use symbols that you use in chalk talks

[cobalt heath]

At least /\ and \/ are not much shorter than 'and', 'or'.

[rose prism]

they also have mathematical meaning

in many many contexts

so are really never acceptable to use in place of and/or

[cobalt heath]

Indeed.

IIRC at least lattices use these as operators

[rose prism]

\forall and \exists are ok but only very sparingly, and not for undergrads

[summer path]

Do whatever up to reason in talks/ lectures where you may have time constraints

[rose prism]

i meant in papers

in talks its fine

[summer path]

Yeah

[rose prism]

to use \forall and \exists

[cobalt heath]

How about my personal notes?

[rose prism]

personal

who cares

[summer path]

But if you're writing stuff for others to read

It better be readable

[rose prism]

some fields have a custom of using these symbols a lot

probability for one

well

many people use them

[summer path]

You can choose to do so for personal notes to get into a habit of doing so but if you don't intend on having other people read it, it doesn't really matter

[rose prism]

but nobody uses them a lot in a single paper

[cobalt heath]

Indeed ofc, like who would write these symbols for shorthand in a paper

[rose prism]

probabilists

[cobalt heath]

Huh

Straight to jail

[rose prism]

depending on what youre doing

it often can help readability

because in probability (and certainly elsewhere) you often need to define sets using fairly complicated quantifiers

in math mode its not such a big deal if the first symbol after the { is \exists

or whtaever

[cobalt heath]

Ah, I see.

How about ⇔, by the way?

[rose prism]

i think its horrible

but idk i am sure many disagree

[cobalt heath]

Ouch, was going to use it (in a paper in the future)

[rose prism]

i use it all the time in notes

there are no rules

just write whats readable

[cobalt heath]

Thanks, helped a lot!

[stark helm]

For sentence" f(x) is primitive because we can divide both f(x) and g(x) by the content of f(x)." , I am wondering why we can just assume that, it seems that f(x) is reducible ovrer Q can not imply f must be primitive?

[next obsidian]

If f(x)/content(f(x)) = p•q [where p,q are in Z[x]] then f(x) = p•q•content(f(x))

So f(x)/content(f(x)) is reducible over Z => f is reducible over Z

[stark helm]

so is it adequate to say f is reducible over Z if we can decompose f into p*q where p and q is in Z[x]?

[next obsidian]

That’s the definition

Well, assuming p,q aren’t invertible

[stark helm]

Therefore, reducibility over a field implies that factorization of p and q over a field right?

[stark helm]

It seems that if we have roots over a field, we can also say f is reducible over a field( does it imply zero?)

[next obsidian]

Well yes

But that’s a thing about division with remainder

[clear fiber]

Hi all, I’m struggling a lot with the

[balmy belfry]

My book says that in a finite cyclic group G with order n and generator <a>, that the number of elements with order k is phi(k) (Euler phi function) and then it says those elements of order k is a subset of of the only subset with order k

Did i restate it correctly?

I’m not understanding the proof, he shortens the order of the group randomly

[rocky cloak]

One crucial part your missing is that k must divide n. But other than that yes.

For every k that divides n, there is a unique subgroup of order k, which contains all elements of order k, and there are phi(k) of them

[balmy belfry]

I see

Thanks

[balmy belfry]

Every cyclic group is isomorphic to some subgroup of Z right

And isomorphic to Z_n if finite?

[coral spindle]

No, as every subgroup of Z is either trivial or isomorphic to Z.

Rather, every cyclic group is isomorphic to a quotient of Z.

Quotients of Z are Z and Z_n for n>0.

[balmy belfry]

So every cyclic group is isomorphic to Z or Z_n

[south patrol]

Yes

Note a group G being cyclic with generator g can be rephrased as saying that the map Z -> G determined by sending 1 -> g is surjective

And now use first iso

[coral shale]

show no cyclic groups are uncountable

[dim widget]

they have one generator

[south patrol]

Well they admit a surjection from a countable set

See above

[coral shale]

hmm so are finitely generated free groups and their quotients

wait.

countably generated

[coral spindle]

All finitely generated groups are countable, yes

[south patrol]

You can see this by noting like

[coral spindle]

(because every f.g. group is a quotient of an f.g. free group)

[south patrol]

You can model fg free groups as consisting of finite words in a finite alphabet (modulo obvious relations)

[coral shale]

looks like id misremembered this

[south patrol]

There are only finitely many words of given length

Now take a countable union

[coral spindle]

We really are stating the same result like, three different ways

[south patrol]

Wdym

[coral spindle]

Nah I'm just musing. These are all the same statement with slight variation

Alright alright tteg spit it out

[south patrol]

UwU

[coral spindle]

But the point being, f.g. free groups are exactly those finite words. Once you take quotients you can choose representatives, which is what you're doing in the argument just above. This is really exactly the same argument but unwrapped slightly.

I just think that's neat

insert marge simpson

[rustic crown]

topologically cyclic

[south patrol]

I don't see how that is any different

I'm confused lol I thought i was the only one providing a proof

[coral spindle]

This was the remark I made which is the same as your explanation

[coral shale]

whats an example

[coral spindle]

R/Z

[south patrol]

I was only talkain about free groups

[coral spindle]

Oh mb

[south patrol]

i meant only to complement your explanation aha

but yes the same argument works in general ofc

if you add in more relations

[rustic crown]

absolute galois group of F_p

[south patrol]

ig when I say model I mean that like

[coral spindle]

mb lol

[south patrol]

i care about free groups as a universal construction rather than having a particular incarnation

[rustic crown]

i.e. inverse limit of Z/nZ

[coral shale]

I get the r/z ig

[coral spindle]

Showing R/Z is top cyclic is not trivial if I'm not mistaken (it has been a while...) It's essentially an ergodic result, right? Perhaps someone more knowledgeable than I can comment

[coral shale]

its the integer points of sin right

thingy

iirc its not yh

{sin n} dense in [-1, 1]

[coral spindle]

Whereas showing the p-adic integers are topologically cyclic is more straightforwardly by definition

[coral spindle]

Yeah exactly

And ofc Q/Z is not topologically cyclic lol; lmao even

[rustic crown]

you can choose your irrational carefully to make the proof easy ig

one in my head is the Louville's constant

[dim widget]

it's an ergodic result to show equidistribution but it's pretty trivial to show density

which is all you need to show that it's topologically cyclic

[south patrol]

ergodic

[coral spindle]

Er god, ick

[south patrol]

just to check like 1 is the intended generator of Z_p right lol

[coral spindle]

Yeah I think so

[south patrol]

just cause it generates all of the tings in da limit diagram

[dim widget]

yes since Z_p is a completion of Z Z is dense

[south patrol]

yh

[coral spindle]

I say this to my dynamicist friend whenever he mentions ergodic theory; I hope to someday drive him to madness <3

[south patrol]

if he's doing dynamics he's already been driven to madness

joking, it seems a v cool field

[coral spindle]

True, maybe that's why he always chuckles

[south patrol]

Okay time to finish another phd appa

[coral spindle]

Good luck 'tater

[south patrol]

Thank

[coral spindle]

Welc

[rustic crown]

do people give name to injective limits?

like similar to procyclic

[coral spindle]

I honestly don't know, sorry

[dim widget]

what do you mean injective limits?

[rustic crown]

Q/Z is the colim of roots of unity

so inj-cyclic?

[coral spindle]

Oh I read that as meaning colimits

[dim widget]

indeed, it's the pontryagin dual of Z

[dim widget]

the relevant property is that it's an injective/divisible module

[rustic crown]

right

[dim widget]

the dual terminology would be ind-cyclic

[rustic crown]

ah

[dim widget]

but I don't think that's a very useful notion

[coral spindle]

Ooh very nice

[balmy belfry]

Is this hard

Can’t you just

T: N -> G, G = <a>
T_n = a^n?

[coral shale]

its not hard

You haven't considered a^-1

[slim kayak]

Cyclic group as in the sense that it has one generator?

[coral shale]

thats up to the prover to define

[slim kayak]

||I am just trying to think of some cursed long-line analogue of the integers for sufficiently cranky definitions of "cyclic" honestly||

Hm I guess not

[tardy hedge]

Ok so since alpha is algebraic over K, it is of course algebraic over F. So [F(alpha) : F] is equal to the degree of the minimal polynomial of alpha over F. Since f is irreducible over K and has alpha as root, it is minimal polynomial over K. When we are considering the degree of minimal polynomial of alpha over extension fields, it will continue to be deg f as long as f is not reducible among those extension fields, or else it will be <= deg f

i think i just need to work through it very carefully like this in order to understand rn

[rustic crown]

Yep, try to stress out every single detail. eventually when you're comfortable with this stuff, you'll know which ones are "ignorable" 🙈

[tardy hedge]

Yeah totally. this stuff has been deceptively tricky for me lol

[rustic crown]

keep drawing diamonds

the slogan is: "moving diagonally reduces the degree"

So here [F(alpha):F] <= [K(alpha):K] = deg f

[velvet steeple]

How did you draw that?

[rustic crown]

went to quiver app, drew there, copied tex code.

[velvet steeple]

Ah, thanks

[tardy hedge]

hmmmmm

ohhh wait so if alpha was actually in K, it would be degree 1 f right

then [K(alpha) : K] is 1, K(alpha) is the same as K

Ur helping me immensely in learning rn btw so thanks lo

yeah its something like trying to really internalize that connection between polynomials and fields is taking time to digest i guess

[tardy hedge]

do we need to do something with [F(alpha) : K] = [F(a) : K(a)][K(a) : K] = [F(a) : F][F : K]

[rustic crown]

yep!

so the degree of the whole tower is divisible by both [K(a) : K] and [F : K].

[tardy hedge]

We are trying to show that [F(a) : F] is deg f right

[tardy hedge]

Im not sure what to do with this again

[rustic crown]

oops sorry for leaving

that's where we use that these two numbers are coprime.

Initially you knew that
[F(a) : K] = [F(a):F] * [F:K] <= [K(a):K] * [F:K]

but because of coprimality, since both these coprime numbers divide the degree

their product must also divide it

in particular [F(a):K] is also >= [K(a):K] * [F:K]

and so [F(a):F] = [K(a):K] = deg f

[delicate orchid]

I'll forgive you... this time...

[south patrol]

Coprime0

[tardy hedge]

[F(a) : K(a)] is the same as [F : K]?

[cloud solar]

A is a commutative ring s.t. there is n,k natural numbers ≠0 with x^k=x^(n+k) for every x in A. Show that there is a polynomial P in A[X] with deg(P)=n(n+1) s.t. P(x)=0 for every x in A. Any idea?

[coral shale]

[balmy belfry]

Nah fuck all I said

Map the nth even to the nth power of a

nth odd to -nth power of a!

[coral shale]

You could have just chosen a map from Z

Anyways.
<a> := {a^n : n in Z}

[balmy belfry]

It isn’t feel as right ok

[delicate orchid]

I buy this

[balmy belfry]

Its just not the same

Isn’t the definition of a countable set mean bijection to N

Not Z

[coral shale]

You use the minor lemma that Z is countable

...

[balmy belfry]

If you are gonna use Z then also prove that even tho it’s easy

[coral shale]

Ill let you have a think

[delicate orchid]

I no longer buy this. What if the minimal polynomial over F is different than the one over K

[balmy belfry]

Think about what

[balmy belfry]

I literally did this..

What’s a map from N to Z?

[coral shale]

If Z is countable, and you have a bijection to Z

[balmy belfry]

Evens to positive odds to negative

I did that

And then I replaced it

Where u put ur z..

[coral shale]

its trivial to achieve this

[balmy belfry]

I just wanted to do it properly

[coral shale]

[balmy belfry]

Instead of improperly

[coral shale]

No amukh, show that given Z is countable, a bijection between S and Z shows S is countable for a general S.

[balmy belfry]

I know

I know that, but I do not wanna prove Z is countable each time

[coral shale]

????????

[balmy belfry]

And it doesn’t feel good to assume that it’s already finished

[coral shale]

Thats exactly what you are doing here

as opposed to using results

[balmy belfry]

But it isn’t really a result

[coral shale]

[balmy belfry]

It’s like a baby lemma lol

Feels eh

[cloud solar]

Any idea

[balmy belfry]

No idea at all

[rustic crown]

in this case yea

but in general going diagonally up in the diamond reduces the degree

the coprimality makes the diamond super rigid

[dire siren]

multiply that relation by x^(n-k)

you get x^n=x^(2n)
so x^n=x^(2n)=x^(3n)=...=x^(n(n+1))

so you have the polynomial x^(n(n+1))-x^n

[rustic crown]

another way of saying what i said above would be,
[K(a):K] divides [F(a):K] = [F(a):F][F:K], since [K(a):K] and [F:K] are coprime, we get [K(a):K] divides [F(a):F]. So coprimality doesn't let the degree of [F(a):F] reduce to something smaller.

[dire siren]

actually, this only works for n>=k

[balmy belfry]

Basical

Y

Im try to prove that if

It’s a set that isn’t generated by an element

It can’t be subgroup

Ok

[coral shale]

restate the problem

[balmy belfry]

And obv all generated sets is a subgroup

[balmy belfry]

Prove cyclic groups have cyclic subgroups

So anahays

[coral shale]

only have right

[balmy belfry]

Let G = <a>, |G|=n

[coral shale]

i think thats a hard approach to go for

[balmy belfry]

Yuh

Shit

Ok how about

[coral shale]

so my approach

[balmy belfry]

H is an arbitrary subgroup

[delicate orchid]

hint, any element in G can be written as a^k for some k

[balmy belfry]

Show it cucle

[balmy belfry]

Ye

[coral shale]

yh sure u can do that

but i wanna point out

that i usually

try small explicit examples

when im stuck

it sounds stupid but can help u find the right approach

[balmy belfry]

I mean

[coral shale]

so for example, what are some subgroups of Z6 and why are they cyclic

[balmy belfry]

I already know it’s just the generated sets

In fact

There are k subgroups of G, if there are k factors of G

Each has that many elements

[coral shale]

[balmy belfry]

I proved that using this theorem

But I didn’t prove this theorem

[coral shale]

ok so u cant use that then

right now

[coral shale]

do this first

[balmy belfry]

Yuh

[coral shale]

{0, 1, 2, 3, 4, 5} mod 6

[balmy belfry]

<1>, <2>, …, <0>They are cyclic because that’s the definition

[coral shale]

thats not too explicit

so i meant like

[balmy belfry]

Cuz they have a generator?

[coral shale]

{0, 2, 4} is a subgroup. And this is cyclic as its generated by 2

[balmy belfry]

The factors of 6 = 1,2,3,6

[coral shale]

<2> = {0, 2, 4}

[balmy belfry]

So there r 4 subgroups I think

One subgroup is

[coral shale]

im trying to get u to do this because i think it hints at the general approach

[delicate orchid]

hint: gcd

[balmy belfry]

<6>, another is <1> another is <3> and another is <2>

[balmy belfry]

Huuuh

[coral shale]

to be honest, this approach works with the GCD

theres quite a few

ways

[balmy belfry]

GCD of what

[coral shale]

anyways, i would rewrite this in a more general way

[balmy belfry]

<a> = {0, a, a^2,…}

[coral shale]

yes

well

{a^0, a^1, a^2}

[balmy belfry]

Yes wait

Hmmm

Well I mean

By the previous theorem..

I already know it only has 3 elements lol

[coral shale]

lets go with your original approach and wews hint

[balmy belfry]

Ok but

Also

[coral shale]

it can work out

[balmy belfry]

Wait ok

[coral shale]

Assume the subgroup is generated by 2+ elements

[balmy belfry]

So let the subset be S

[balmy belfry]

Why

[coral shale]

thats your approach?

It’s a set that isn’t generated by an element
It can’t be subgroup
Ok

[balmy belfry]

And since is it a subset some a^k is in S for some k

[balmy belfry]

Yeah but I didn’t think to use

2+ generators

I was doing it raw

[coral shale]

[balmy belfry]

It’s cheating if I use ur idea too easy

[balmy belfry]

Also any multiples of k is in it

Ummmm

[coral shale]

I think slow down

[balmy belfry]

So anyway

[coral shale]

Start this

Let S be your generating set

[balmy belfry]

If there is no other elements it is cyclic, if there is call it a^j, then a^k+j is in S

[coral shale]

and we are interested in the subgroup <S>

[balmy belfry]

Ok

[coral shale]

too messy amukh your thoughts are
for me to follow at least

its ok

[balmy belfry]

Ye ye I realized it went noware

[coral shale]

Let G be a cyclic group.
Let S be a subset.
Show <S> can always be generated by one element

====

Does this approach make sense

[balmy belfry]

Yes

[coral shale]

So if |S| = 1 this is trivial

so consider >1

[balmy belfry]

Hmm

[coral shale]

and I would try this with an explicit G first

for example G = Z_12

[balmy belfry]

What is C

Oh

[coral shale]

then pick various examples of S

[balmy belfry]

12 😨

[coral shale]

and then try to show this always works

[balmy belfry]

<3,2>

[coral shale]

ok

[balmy belfry]

This is basically

The set of elements of the form 3^k 2^j

For some k,j

[coral shale]

(you can visualize this on a clock)

if u want

[balmy belfry]

Huh

[coral shale]

yeah i guess, but just compute a few elements

and you should be able to see which group this is

the hours on a clock is Z_12

[balmy belfry]

Ic

I guess that’s why everyone use it

{0,3,6, 2, 4, 8, 10, 12,…}

[coral shale]

so <S> = {2j + 3k (mod 12) : k, j in Z}

[balmy belfry]

Yes

[coral shale]

theres more interesting elements.

[balmy belfry]

256

[coral shale]

[balmy belfry]

It’s just all evens and all multiples of 3

[coral shale]

no, thats already written, this is mod 12

you're missing elements

[balmy belfry]

In the equiv class

[coral shale]

256 = 4, and you already wrote 4

[balmy belfry]

Damn it

[coral shale]

you're missing elements.

[balmy belfry]

How many

[tardy hedge]

Oh bruh that makes more sense to me. Thank you so much for your help this problem was loads more complicated to me than i thought

[coral shale]

most

[tardy hedge]

Now i have to relook at the whole problem and understand it all again lol

[balmy belfry]

… is most

[coral shale]

most of the important ones.

[balmy belfry]

Can I have an example

Oooh wait

Shuri

[coral shale]

i mean cmon amukh u only wrote the multiples of 2 and 3 here

[balmy belfry]

What if

Wait nvm

What if

[coral shale]

thats not everything in it is it?

[balmy belfry]

I just show one of the elements

Has order of the subgroup

[balmy belfry]

And 6

Also that’s all it is

[coral shale]

that doesnt make sense to me when you dont know what the subgroup is

[balmy belfry]

Plus some others

14

[coral shale]

Im asking you to compute it

[coral shale]

this is a multiple of 2

[coral shale]

so is this

[balmy belfry]

{0,3,6, 2, 4, 8, 10, 12,…}

15?

15

Then 15 + 2

15 + 3

[coral shale]

Is there anything that ISNT a multiple of 2. Or a multiple or 3. That should be in there

[balmy belfry]

Ohh wait

[balmy belfry]

15

[coral shale]

bro thats a mult of 3

[balmy belfry]

Ok

I know

I’m thinking

It doesn’t make any sense what would be there that isn’t

[coral shale]

should 5 be in there.

[balmy belfry]

Ye

Oh shit

This is adding

[coral shale]

there are only 12 possible numbers in there

[balmy belfry]

Is it <5>

[coral shale]

as well, like... theres not too much to consider

well actually yes.... but

if u tell me that

[balmy belfry]

Is this always work

[coral shale]

id ask you to compute what <5> is

what is its order

[balmy belfry]

Nah I’ll just computer the order

So

5 = 1^5

Wait

I’m confused one sec

[coral shale]

The order of <5> not 5

[balmy belfry]

Not <5>, 5

[coral shale]

^

im asking you what the order of <5> is

[balmy belfry]

It should = the order of 5

[coral shale]

or list its elements in roster form

[balmy belfry]

Which I will find out rn

12/gcd(5,12)

[coral shale]

youre overcomplicating, no theorems are required here

[balmy belfry]

Uhhhh

12?

Is it 12

[coral shale]

you should just write for me

<5> = {5, 5+5, 5+5+5, ...}

[balmy belfry]

Ye

[coral shale]

= {5, 10, 3, 8, 1, 6, 11, 4, 9, 2, 7, 0}

[balmy belfry]

But the 3,8 take a long time for me to figure iut

[coral shale]

which took me 30 seconds

[balmy belfry]

How ru go so fast