groups-rings-fields

[solar shore]

well let me think of it when gcd isnt 1

[pastel cliff]

if you want a tangible example subgroups of Z probably help

[solar shore]

so let $|a| = m, |b| = n.$ we assume that $|ab| = mn$. then assume $\gcd(m, n) = d$. then there are integers $x, y$ such that $mx + ny = d$. then $(ab)^d = (ab)^{mx + ny}$

[cloud walrus]

blanket

[solar shore]

am i on the right track atm?

[pastel cliff]

keep going and find out

i hate being obnoxious like that bc when i want an answer it's annoyinig

[topaz solar]

notice that d <= m, n

[solar shore]

hmm okay

[pastel cliff]

but it's probably for the best that you try to see things through to the end on your own first

[topaz solar]

so clearly d <= mn

since m, n >= 1

[solar shore]

$(ab)^{mx + ny} = a^{mx + ny}b^{mx + ny} = a^{mx}a^{ny}b^{mx}b^{ny} = a^{ny}b^{mx}$

[cloud walrus]

blanket

[solar shore]

atp i dont really know what to do with it

[topaz solar]

Well, the approach I'd take is

|a|=n, |b|=m, gcd(m, n)=d -> m = xd, n=yd yes?

I mean, it's a divisor so divide

[pastel cliff]

ok done with that other pf

i hope

also sharp how are you new to the server but already have emeritus lol

[topaz solar]

no idea

I am simply too powerful

[pastel cliff]

it's supposed to be a role for people who were active/very active but are no longer so lol

[topaz solar]

I've been here before lol

[pastel cliff]

ah

your roles probably got restored then lol

[solar shore]

so for integers $x, y$, then $m = xd$ and $n = yd$ right?

[cloud walrus]

blanket

[topaz solar]

positive natural x, y even

[solar shore]

ah

oh right yes

divisors

woops

lol

[topaz solar]

notice, mn = xyd^2

[solar shore]

yep im following

do we put that to ab?

so $(ab)^{xyd^2}$?

[cloud walrus]

blanket

[topaz solar]

(ab)^mn = e clearly

[solar shore]

right

[topaz solar]

(ab)^mn = (ab)^(xyd^2) = ((ab)^xyd)^d yes?

[solar shore]

right im follwing oyu

actually can i see where it might go from there first

[topaz solar]

What is (ab)^xyd

xyd = yn = xm yes?

oops

[solar shore]

$(ab)^{xyd} = a^{xyd}b^{xyd} = a^{xm}b^{ny} = e$

[cloud walrus]

blanket

[solar shore]

OH

D = 1

?

[topaz solar]

d is not 1

[solar shore]

o.

got excited for a sec

[topaz solar]

but xyd < mn

[solar shore]

fuk

is it strictly less than?

[topaz solar]

d > 1

and xyd > 1

Because d is explicitly not 1

[solar shore]

oh right we assumed that first

[topaz solar]

yeah

so xyd < xyd^2

=mn

[solar shore]

xyd^2 is mn yeah

[topaz solar]

so not gcd = 1 implies not |ab|=|a||b| yes

funny contraposition

[solar shore]

ohh i see

wow

[topaz solar]

Not gcd = 1 -> gcd =/= 1 -> gcd = d =/= 1

[solar shore]

which then leads to |ab| =/= |a||b|

[topaz solar]

since |ab|<=xyd

[solar shore]

thus |ab| = |a||b| -> gcd(|a|,|b|) = 1 by contraposition

pppp

[topaz solar]

now go the other way around

[solar shore]

oh i had the other direciton done

i was just

[topaz solar]

ah aight

[solar shore]

trippin up on this direction on the conditional

..

feelin kinda dumb now

[topaz solar]

this isn't constructive though I think

[solar shore]

how do you mean?

[topaz solar]

since it's like a contradiction I think

[solar shore]

is proof by contrapositive not good?

[pastel cliff]

unrelated, but does the last paragraph here look sufficient

[topaz solar]

nah it's fine unless you're looking for constructive results

[solar shore]

ah

oh wait yeah i get what you mean

i did see somewhat of a constructive proof dealing with lcm?

but

[topaz solar]

"Intuitively" would get be run outta town, so I'd want to show that

[solar shore]

there was a definition that lcm$(m,n) = \frac{mn}{\gcd(m,n)}$ which i hadnt seen before

[pastel cliff]

yeah i deleted that word lol

[cloud walrus]

blanket

[pastel cliff]

but without that is that enough justification?

like it feels obvious

[topaz solar]

It seems sufficient as long as you have that k has a natural interpretation as a vector space on R

[solar shore]

anyways

@topaz solar thank you so much

i appreciate it a ton

[pastel cliff]

okie i go back to original question onw

repost for ease

[lethal dune]

What’s the question

[pastel cliff]

oh it's to prove the Claim

[lethal dune]

Feels like not worded properly

[pastel cliff]

the claim?

i agree that it should probably explicitly state like "R-module R" or something like that

but this is what i was given

[lethal dune]

No the proof

[pastel cliff]

oh it's not done

lol

[lethal dune]

Then why’s there a QED

[pastel cliff]

[lethal dune]

Use sully for QED

[pastel cliff]

my cranky 70 year old prof will surely love it

but uhhhh

[topaz solar]

I'd rather say

Suppose a is a zero divisor

then xa = 0 for some a

Then what happens to x multiplying against any element in R

[pastel cliff]

im guessing it sends every r to 0?

[topaz solar]

I mean xar = 0r = 0

So how do you define finding rank of a module?

[pastel cliff]

the size of the basis

but im guessing we need something else here

[topaz solar]

Well, what happens to using {a} as a basis?

[pastel cliff]

oh it's no longer linearly independent right

[topaz solar]

multiplying that by x such that ax = 0 has an issue

I don't know the precise definitions y'all use for the rank and all but I mean

that looks mighty problematic

Conversely, suppose a has no such zero divisor

[pastel cliff]

is it not this?

[topaz solar]

I've seen a few different definitions for it, use the definition from your course/text

This should be close to it though

Now, for a non-zd, what about the map R -> aR defined by λr.ar

[pastel cliff]

ok just to be sure is the following sequence of implications correct:

a generates Ra by definition -> assume a is a zero divisor -> there exists x such that xa = 0 (by definition) -> then we have that xa \cdot r = 0 (in other words, there's a nonzero r such that r times the basis element is 0) -> contradicts linear independence of {a}

[topaz solar]

I don't know why you have that r in there

it'd be more like, x and ar are nonzero and x \cdot ar = 0

[pastel cliff]

oh uhhh

[pastel cliff]

i meant that as this

[topaz solar]

aight

ar in aR

So suppose there is no such zero divisor

clearly we have the obvious map R -> aR too

[pastel cliff]

ok but it's true that a generates Ra and is therefore a basis of Ra right?

[topaz solar]

It generates Ra, but whether you consider it a basis or not is different

R generates R but clearly isn't a basis

[pastel cliff]

hmm true

ok this is where im at rn

i appreciate your patience btw im quite tired and probably not thinking at 100% rn

[topaz solar]

nonzero x in R and nonzero ar in aR

[pastel cliff]

[lethal dune]

xar is meaningless

[topaz solar]

it's associative

[pastel cliff]

comm. ring too

[lethal dune]

Oh a as an element of R

Nvm

[topaz solar]

owned

[pastel cliff]

but that's not enough right

for that direction

like i've shown that {a} cant be a basis but not why it's a contradiction

[topaz solar]

Well idk the exact definition for your rank but consider

[pastel cliff]

bc i was assuming {a} had to be a basis but that was wrong

[topaz solar]

P -> Q and -P -> -Q

but wait -P -> -Q is just Q -> P contrapositive yes?

So if we show a is zero divisor -> not rank 1, and a not zd -> rank 1, we get iff, yes?

[pastel cliff]

i know my logical implication truth tables so yeah i agree

but then is what i have enough to show it's not rank = 1?

we have a few defn's of it

[topaz solar]

what's the definition of rank 1

[pastel cliff]

of free generators

max # of linearly independent elements

if M \cong R^n, rank(M) = n

[topaz solar]

well any collection of elements S has xy = 0 for y in S

[white oxide]

this proof is flawed, correct?

Here's the question (attachment didn't upload correctly);

[topaz solar]

that uses commutativity

[white oxide]

right

[pastel cliff]

groups aren't necessarily abelian

[white oxide]

ye

[pastel cliff]

in the third like you should have (ba)^n on the right

since you're multiplying by the inverse on the left

[white oxide]

oh okay

[topaz solar]

what's (ab)^-n

and does that have any relation to the order of (ba)

[topaz solar]

This seems to be the way to show aR is not rank 1

Now, if a is not a zero div, then what can we say about it

Personally, I'd ||go with definition 3 here, I think. R->aR has (x-y) |-> a(x-y) = ax-ay =/= 0 if x-y =/= 0||

[white oxide]

well i wrote it out

we also know that it is equal to e since you can take the inverse of both sides in the equality (ab)^n = e

but when i wrote it out, for say n = 5 i got b^-1a^-1b^-1a^-1b^-1a^-1b^-1a^-1b^-1a^-1

and so i tried some things with associativity

didnt get anywhere

still working on it

maybe i can work backwards

with like (ba)^n = e

[pastel cliff]

sorry i got frustrated and went and did a different problem again lol

taking a smol break brb

[topaz solar]

Well you can get (ab)^n-1 = (ab)^-1 yes?

= b^-1 a^-1, so might be possible to do a lil shuffling in there with that?

[white oxide]

how did you get that

[topaz solar]

ab^n = e

multiply by ab^-1

oop

parenthesis as necessary

what exactly is the question it's asking btw?

[white oxide]

ah right

hm

it's not asking if G is abelian right

[pastel cliff]

reading this back i feel like this is too handwavey

but fck it we ball

[topaz solar]

I don't know I don't have the question

[white oxide]

oh my bad

i thought you were like trying to get me to understand a deeper point or smt about the question

Ok here it is:

Let $a$ and $b$ be elements of a group $G$. Show that if $ab$ has finite order $n$, then $ba$ also has order $n$.

[cloud walrus]

okeyokay

[white oxide]

idk maybe we could do something like let x in G

[topaz solar]

[white oxide]

uh

ok

nvm then

[topaz solar]

I meant about thinking I was asking a deeper question

[white oxide]

oh ok LMFAO

my b

[topaz solar]

so if we need (ba)^n = e we just need a^-1b^-1 = (ba)^n-1 hmm

[white oxide]

yup

maybe we should consider the cyclic subgroup generated by (ab)?

so we know that it's finite

and since it's a finite cyclic group of order n

it's isomorphic to Zn

which we know is abelian

so would that work?

Let $H = \langle ab \rangle$, the cyclic group generated by $ab$. Observe that $H$ is of order $n$ since $(ab)^n = e$. Then, by Theorem 6.6, $H \simeq \mathbb{Z}_n$. Since $\mathbb{Z}_n$ is a abelian group, we see that $H$ is also an abelian group; thus, for $ab \in H$, $ab = ba$ so $(ab)^n = (ba)^n = e$ and $ba$ has order $n$.

[cloud walrus]

okeyokay

[white oxide]

theorem 6.6 just says that any finite cyclic group is isomorphic to Z_n under addition btw

[topaz solar]

H has ab, not necessarily a or b

so it's a bit rough to commute those two

[white oxide]

wait but since it's a subgroup doesn't it have to have a and b

[topaz solar]

nope

integers are a subgroup of reals

[white oxide]

ah

[topaz solar]

they don't have 1/2 despite 1 = 1/2+1/2

[white oxide]

wait but i define H to be the cyclic group generated by ab

nowhere in the proof do i use a in H or b in H separately

[topaz solar]

but you commute a and b

[white oxide]

oh because it's isomorphic to Z_n

and Z_n is abelian

does that not work

[topaz solar]

but it's commutative for a, b in H

[white oxide]

i thought any group isomorphic to an abelian group is abelian

[topaz solar]

it is

but you can't commute a and b because they aren't in it

[white oxide]

hm ok

well

um

hm

this might seem a little bit stupid or hand-wavy and i'm sure there's a million counter examples as to why we can't

but can we define a, b to be in H?

[topaz solar]

not by that definition of H

[white oxide]

ok i'll think about it a little more

[pastel cliff]

still havent gone back to original question but

im gonna guess that basis of M is union of basis of N and M/N?

[lethal dune]

Not necessarily basis tho

[pastel cliff]

hmmm yes

then generating set

[lethal dune]

yes but can't just take their "union"

[pastel cliff]

yeah that would be too easy

so i assumed it wasnt enough

[lethal dune]

[m] ∈ M/N then take a lift of these and then take their union

[pastel cliff]

lift = preimage of natural projection map?

[lethal dune]

yes

[pastel cliff]

https://tenor.com/view/lifting-workout-bonding-me-and-my-dog-doyouevenlift-gif-16764439

[pastel cliff]

stuck here atm

[topaz solar]

take a representative from preimage of each element of B?

[pastel cliff]

got a suggestion from google but this is what i arrived at

end feels a bit scuffed tho

[topaz solar]

I mean, is p^-1(B) finite tho

[hot lake]

that looks suspicious

[novel parrot]

the determinant can be computed for each fixed i ?

[cloud walrus]

Eso

[tawny pine]

@novel parrot yes, basically you can cofactor expand along any row. the formula even works for any column. in practice we try to expand on rows/cols with many 0s

[simple valley]

Any tips for the character table of SL_2(Z/3)? I figured out the 4 rows that lift from A_4 and that the remaining three irreps of dimension 2 map {{0,-1},{1,0}} to some sort of symplectic unit matrix (e.g. itself)

[kind jacinth]

is it fine the way I tried proving identity?:
G is a group so for any g there must also exist an inverse g1. Same applies for H.
Take arbitrary g1, h1 and the inverses g-1, h-1 then
(g1, h1)(g-1,h-1) = (gg-1, hh-1) = (eG, eH)
thus both identities are contained in this group composition

[south patrol]

I'm not really sure what you've proven there

what exactly are you trying to prove?

[covert cove]

Assume G is finite and cyclic and V is finite dimensional

I feel like this boils down to choosing the right basis and considering the dual basis of that

But I haven't been able to find one

[delicate orchid]

what have you tried so far?

[covert cove]

I've tried most tricks I know (considering v + gv + g^2v + ... g^{k-1}v which is g-invariant)

[novel parrot]

I suppose that I_A is maximal

If A empyt, then I_A is whole ring

if A had two elements, say A = {a,b} then I_A is contained in I_{a}

not maximal

So A only have 1 element

This works for one way right?

[delicate orchid]

it's certainly convinced me yeah that's good for the forward direction

[novel parrot]

sarcasm or its actually corect ?

[delicate orchid]

it's actually correct

[novel parrot]

nice

let me tell u my other way

for other direction

i use contraposoitve

I_A not maximal => A not single element

Essentially same argument works

do you agree ?

i see solution and its very different to what i come up with

[delicate orchid]

yeah my solution was also different

[novel parrot]

is mine correct ?

both direction

[delicate orchid]

my only problem is, how do you know there's not some other ideal that isn't of the form I_A that contains I_{a}

if you've shown that then it works I think

[novel parrot]

which direction

second direction?

[novel parrot]

this

?

[delicate orchid]

backwards yeah

oh wait no sorry I'm still thinking A singleton => Maximal

[novel parrot]

ok

[delicate orchid]

although this could be a problem regardless

[novel parrot]

this is backwards right

i mean

you aggree with Maximal => singleton right?

Just my singleton => maximal you are asking about

[delicate orchid]

Yes

[novel parrot]

So for Single => maximal

i prove instead not maximal => not singleton

So I_A <= J and J not = R

oof

yeah

i see

it doesnt work for this case

[delicate orchid]

Yeah, there’s an easier way to do it

[novel parrot]

hmmmmmmm

hmmmmm

[delicate orchid]

What else do you know about maximal ideals

[novel parrot]

er

divide by maximal to get a field

R/M = F

[delicate orchid]

So see if R/I_A is a field

When A is a singleton

[novel parrot]

so for all f in I_{a}

f(x) = (x-a)g(x)

since root

a

so like

R/I_{a} = R/(x-a)

and x-a is irreducible/prime => maximal

?

[north sand]

you dont have polynomials here

[novel parrot]

right

oh

[north sand]

iso theorem

[novel parrot]

how ?

i need to show that R/(x-a) is a field right

[north sand]

R/I_{a}

[novel parrot]

but R/I_{a} = R/(x-a) anyway?

[north sand]

I dont like that notation

screams too much polynomials

[novel parrot]

i use little x 😄

[north sand]

and we already have the notation I_{a}

[novel parrot]

ok so we need to show R/I_{a} is field

[north sand]

what does the iso theorem say

[novel parrot]

R / ker = image

you mean first iso them right

like consturction a map from R -> R with kernel I_{a}

that is surjective

[north sand]

well depends on what you mean with the letter R here

the first one is the ring in question but the second should preferably be a field

[novel parrot]

oh

but what field ?

[north sand]

well that's the question. well or one of the questions

[novel parrot]

okay so, for function that isnt 0 on a, we need to find inverse

[north sand]

so I_{a} = {f in R: f(a)=0} and we want that this looks like a kernel {f in R: phi(f)=0} for some homomorphism phi

[covert cove]

reposting this because it got buried

[novel parrot]

dont get it 😅

i have no idea what the image field could be

[north sand]

do you know a famous homomorphism on a ring of functions

[novel parrot]

no

there exists a homomprism from Z -> R for any Ring R

you mean that

[north sand]

no

[novel parrot]

fix a value from real number, then f -> f(t) ?

[north sand]

yes

[novel parrot]

oh

[north sand]

evaluation homomorphism

[novel parrot]

ooh

hm

okay and show kernel is I_A now

we fix the value a

[north sand]

yes

[novel parrot]

Yeah so evaluation homomorphism on a, is surjective with kernel of I_a

so R/I_a is isomorphic to R (field)

so I_a maximal

i see

@north sand you meant this correct?

the solution had a different solution

[north sand]

I often find using iso theorem nicer than "manually" showing an ideal is maximal

but doing it manually is of course still doable

I assume that's what they did?

[novel parrot]

they do this

[north sand]

yes

[novel parrot]

but i dont understand why their i is inside I_{a}

[north sand]

well what is i(a)

[novel parrot]

oh

nvm 😄

thanks for your help!

[covert cove]

re-reposting because it got buried again

[novel parrot]

Why did we require that f has to be continous function ?

doesnt seem like we need continous condition ?

[north sand]

yeah I don't think we do unless I'm missing a detail rn

[kind jacinth]

it was to show identity is contained but I figured it out now

I need to prove these. I am trying to do part 1 but I am stuck.
I am trying to do this by induction so:

g^n = gog.....og (n times)
what we can do is take n=2 then g^2 = gog = g^(1+1) which holds.
same for n=3 (g^3 = g^2og = g^(2+1))

do I now say assume true for 0<=i<=n-1 and g^mg^(n-1) = g^(m+n-1)?

[summer path]

i really don't think you need induction here

each part should be pretty immediate just by writing out what you have on both sides

[kind jacinth]

so like (gog....og) m times and (go....og) n times?

[summer path]

yes

[kind jacinth]

yeah then id have gog....og m+n times and I think this is sufficient?

[delicate orchid]

gog

[solar shore]

gawg

[kind jacinth]

but does this count as a valid proof?

because in that case 2 would be very similar

[summer path]

yes

the point of this theorem is to show that exponents behave how you'd expect them to

[kind jacinth]

then for 2 id just do (gog...og m times)o(gog....og m times)o....(gogo.....og m times) n times so ignoring the brackets its g^(m*n)

for 3 ill think about it

[summer path]

ye

[kind jacinth]

but what if m and n can take negative values?

i mean it can

[summer path]

$g^{-m} = (g^{-1})^{m}$

[cloud walrus]

Tubular Cat

[summer path]

then you do the same thing

[kind jacinth]

ok and instead of gog...og id write it as g^-1

[summer path]

yeah, but you can kind of just abuse notation and say it's just -m times

[kind jacinth]

alright thanks!
For 3 I rewrote h^-1g^-1 as (hg)^-1 and then from there i can get (hg)^n using property 2

[coral shale]

you prove (g^-1)^n = (g^n)^-1

hence g^-n is well defined

[charred crescent]

if we are asked to show that either G is cyclic or ab != ba, does it suffice to show that G is cyclic if and only if G is abelian?

because if G is cyclic, then !(ab != ba) => (ab = ba) and if ab != ba, then G is not cyclic

[coral shale]

.

[summer path]

abelian does not imply cyclic

[topaz solar]

Presumably you know more about the group, but if exactly one of those cases are true (non abelian vs cyclic) then yes

[charred crescent]

yeah its a group of order 6 with some a,b in G s.t. o(a) = 3 and o(b) = 2

[topaz solar]

If you don’t know more, then ofc consider ZxZ the abelian but non cyclic

Fair

[charred crescent]

you're right, so then would it suffice to show that if G is cyclic then G is abelian?

[topaz solar]

Then yeah showing cyclic iff abelian suffices

[topaz solar]

Other way around

Cyclic is obviously abelian, you just need abelian -> cyclic

[charred crescent]

so the if and only if works with the given information about the group

[summer path]

yes

[charred crescent]

but in this case it is necessary to show both conditions

okay fair, thank you both

[summer path]

though as sharp said previously, cyclic => abelian is obvious

so there isn't really anything to show per say for that direction

[topaz solar]

Though I mean

If you need a proof for it you need a proof for it after all

[charred crescent]

i might just explain something brief

[summer path]

well regardless, it's like a one-liner for that direction lol

[charred crescent]

okay, thanks y'all

what have you tried thus far?

maybe you don't have to

can you express a in terms of b or vise versa?

yeah but through algebraic manipulation

the rule is (a^2)(x^3)(b^2) = x^2 for all x right?

[topaz solar]

Do you have anything like identities or inverses?

[charred crescent]

well then (a^2)(x^3) = (x^2)(b^2)^(-1) right?

[topaz solar]

Are a, b fixed for all x, or are they dependent on x

[charred crescent]

i think for a given x there are some a and b that satisfy the rule

so they probably depend on x

[topaz solar]

I don’t have the problem statement so I can’t help you on whether a, b are dependent or not

But probably yeah

I think semigroup has identity?

[delicate orchid]

it does not

[charred crescent]

nah just closure and assoc.

[topaz solar]

I hate these naming conventions fr

Ain’t no way I memorize them

[delicate orchid]

no reason to, barely anyone uses anything other than monoids and groups

[charred crescent]

foreal, like who cares what a magma is

[topaz solar]

Anyway, a2 x3 b2 = x2, so what about unfolding x3 = x2 x1

No you don’t

[delicate orchid]

semigroup but the operation need not be associative

[charred crescent]

yeah just closure is satisfied

[topaz solar]

realistically you just add parenthesis to the symbol list & guarantee they close

[charred crescent]

so you can kind of think of a semigroup as an associative magma, and a monoid as a semigroup with identity, and a group as a monoid with inverses

[charred crescent]

so you can't say a2x3 = x2b^(-2) ?

[delicate orchid]

there might not be an inverse

[topaz solar]

No

[charred crescent]

darn

[delicate orchid]

semigroups aren't even cancellable in general - we've got basically nothing to work with

[topaz solar]

Yeah that’s an odd statement to prove

[delicate orchid]

we can shuffle around the a, x, b as much as we want at least

[formal ermine]

fuck semigroups

all my homies hate semigroups

[topaz solar]

$x^2=a^2 x^3 b^2 = x^2 x a^2 b^2$ yes?

[cloud walrus]

Sharp

[charred crescent]

yeah that's what im trying to use

an abelian semi group huh

and you're sure "semi" didn't get added when translating? lol

[delicate orchid]

yeah, but we don't know if it's cancellative - if we did then that would actually be useful

[summer path]

if we write down all 210 orderings, surely we'll get somewhere

[topaz solar]

But then you can x2 -> x3 a2 b2 hmmm

[delicate orchid]

ah I see I see

x^2 = a^2b^2x^4a^2b^2

[topaz solar]

Yeah idk I’m not immediately seeing a solution

[charred crescent]

if cancellation is allowed then it would be easy to prove

[delicate orchid]

true

[charred crescent]

if x = xab does that imply that x = x in this case ?

[summer path]

uh x = x?

[charred crescent]

im trying some backward resoning

start with x3 = x2 and try to work it backwards

clearly x = x lol

[summer path]

yeah lol

[charred crescent]

nvm im dumb it's not getting me anywhere

[topaz solar]

It’s not cancellative as far as I know, and you’re not getting x^3 alone there unless you do x x^2

[summer path]

if there were cancellation, this problem would be very easy lol

[chilly ocean]

this statement is really weird for me

because the property is the same as to say that for any x there is a with a^2x^3 = x^2

and the assumption of at least 2 elements adds basically nothing

could you give us the exact problem?

no, show me it

[astral stream]

What does the last sentence in the answer mean?

nvm makes sense now

[chilly ocean]

are you sure they don't mean that there are elements a, b in A such that for all x in A the equation a^2x^3b^2 = x^2 holds

oh, I see I guess

[topaz solar]

Maybe could follow a^2, b^2 along that to reduce terms?

[chilly ocean]

I've solved it 🤔

Let c = ab, c^2x^3 = x^2

Then c^5 = c^2, and putting cx instead of x we get c^2x^3 = c^2x^2. Then x^2 = c^2x^3 = (c^2x^2)x = x^3

@south arrow

but yeah, the two assumptions weren't useful anywhere and its odd they included them

[novel parrot]

How to show that every finitely generated ideal is principle ?

no idea where to start

[south patrol]

By induction etc it suffices to assume you have an ideal generated by two elements and then have a think about that

[novel parrot]

ok

so we need to show that (f_1, f_2) = (f)

for some f: Z -> R

@south patrol what should i do ?

(x_1f_1 + x_2f_2)(a) = x in real

[toxic zephyr]

okay so this last paragraph loses me. this is the proof for the third part of sylow's theorem.
i don't see how what comes before implies this.

i see that Q*Q is obviously an orbit of size 1, and i see we proved that if |P*Q|=1, then P=Q. but then how does that translate to conjugation partitioning S? or is that something separate?
and conjugation partitions the set, we know that.
but then i also don't see how partitioning and there only being one orbit of size one implies n_p=1 mod p.

wait is it counting the orbits like
1 orbit of size 1
some number a_p of orbits of size p
some number a_p^2 of orbits of size p^2
etc.
so |S|=1+a_p*p+a_p^2 p^2+... = 1 mod p?

as in |S| mod p is just the number of orbits of size 1, since the number or orbits of size p^k are divisible by p so they go away?

[charred crescent]

real quick, does this proof suffice ?

[south patrol]

Yes, that's good

The <- direction can be simplified though

Well

Have you shown that all those 6 words are distinct

Like what is ruling out, say, a^2b=1

[charred crescent]

how could a^2b be equal to 1?

[south patrol]

That isn't the issue

You have to justify that the 6 are distinct

[charred crescent]

if they're irreducible then are they necessarily distinct ?

no they're not

[south patrol]

So for example, if o(a)=4 and b = a^2 then o(b)=2

And a^2b = 1

You need to use the orders being 3 and 2

You may be confusing this w group presentations or smth idk

[charred crescent]

probably confusing it with something

that's always a safe bet

[south patrol]

Incidentally, your proof is correct

Just incomplete

[wet zodiac]

4.3 made me think of something interesting

[south patrol]

Ye

[wet zodiac]

nvm it isnt as interesting as i thought

[south patrol]

There are nice interpretations for it

Oh rip lol

[charred crescent]

im still not sure as to why the words are not implicitly distinct by my imposed definition of them

[south patrol]

I mean, what about the example I gave

With 4 instead of 3

[charred crescent]

well that works for the example, but does the fact i supposed o(a) = 3 and o(b) = 2 not suffice?

[south patrol]

Well what reasoning do you have

It does but you don't seem to have a reason why

[wet zodiac]

i was thinking if a and b are infinitely cyclic its still possible for ab to be cyclic (assuming ab=1)

[south patrol]

An element isn't cyclic in itself

I'm not sure what you mean

Groups are cyclic (sometimes)

[wet zodiac]

oh no i was talking about a cyclic group generated from a and b

[south patrol]

Sure

[wet zodiac]

yeah it isnt as interesting as i originally thought it was

[toxic zephyr]

why is this the case?

[south patrol]

Every group of prime order is cyclic

[formal ermine]

there's an overkill solution to 4.3 lol

[south patrol]

lol

[wet zodiac]

i just used least common multiple stuff

[toxic zephyr]

ah i see

[south patrol]

Hm how

[formal ermine]

||conjugation is an automorphism||

[south patrol]

That doesn't seem relevant to me

[wet zodiac]

let a and b have order c and d respectively

[south patrol]

Illum how is that overkill, isn't that the intended solution basically

That's how i'd have done it

[formal ermine]

last time I recommended that solution I was yelled at by boyjtije

telling me it's too overkill

[south patrol]

lol

[coral spindle]

I would never yell smh

[south patrol]

[charred crescent]

cuz if a word is irreducible and satisfies the supposition that a^3 = e and b^2 = e, then two words x,y in G are equal if and only if (ab)^k = (ab)^j for some integers k,j, implies k = j (mod |G|)

[south patrol]

I just worry cat bread that you're about to appeal to a group being abelian

[coral spindle]

I wouldn't call it overkill, I think it's the best solution

[wet zodiac]

then for (ab)^n=1 we would need to have that (a^n)(b^n)=1

[south patrol]

why?

I don't get what you mean

[coral spindle]

but for someone who doesn't know what conjugation is, it's overkill

[topaz solar]

well the trick is proving that

[charred crescent]

idk im dumb

[south patrol]

I can't really read what you said

No you're not dum

[south patrol]

Yeah lol you appealed to abelian ness

(ab)^n needn't be a^n b^n

In general

:(

[wet zodiac]

curse you nonabelian groups

[charred crescent]

well i don't immediately see an explanation as to why the words are distinct, other than the fact that they certainly feel distinct

[south patrol]

Yeah I get you dw

Would you like some hints as to how to get started?

[charred crescent]

i wanna think through it a bit more first

see if i can't take the intuition and formalize it

[south patrol]

Yeah sure

Gl!

This is important stuff to have intuition for so probably worth thinking about owo but feel free to like @ me if you need any more help or anything

!

[wet zodiac]

what am i supposed to do then

[south patrol]

Okay so like let's be a meme and like

Start with the basics

If ab = 1, why is ba = 1

[wet zodiac]

because b is an inverse of a

[south patrol]

ye sure

how about abab=1

why is baba = 1

[formal ermine]

bababa ba ba bababa

[south patrol]

So true...

[formal ermine]

I am speaking facts and nothing but facts

[delicate orchid]

this, albeit

[topaz solar]

; however,

[delicate orchid]

nevertheless

[charred crescent]

okay here goes, let x,y be in G. Then x = (ab)^k and y = (ab)^j for some integers k,j.
so x = y <=> (ab)^k = (ab)^j if and only if i = k (mod |G|). thus for any x,y in G, if i != k (mod |G|), then x and y are distinct. if i = k (mod |G|) then x = y.

[wet zodiac]

uh

[south patrol]

Okay, so you are assumign ab has order |G|, right?

Which is what you want to prove

like

[delicate orchid]

do we have to work with words or can we just structure theorem it

[south patrol]

How do you know that if (ab)^k = (ab)^i then i = k mod |G|

There is no need for words or structure theorem

Anyway, so

[charred crescent]

well ab = ba, and (ab)^6 = e

[south patrol]

That doesn't mean that, say, (ab)^3 isn't e

[delicate orchid]

oh right you're given the orders

[south patrol]

That's what you have to do

[charred crescent]

so o(a)o(b) = o(ab) because gcd(o(a),o(b)) = 1

[south patrol]

Oh lol

Do you have that as a theorem already LOL

Then there's nothing to prove

[delicate orchid]

lol

[south patrol]

Just immediate that ab has order 6

[south patrol]

Yeah I was gonna just lead you to proving this

[wet zodiac]

its just given to me with 2.4.3 but

[charred crescent]

well i shouldn't use it because the theorem is introduced after the exercise

[south patrol]

Oh okay well

Fair sure I mean

[charred crescent]

but im looking at the proof for it and there's no way i could've come up with it

[south patrol]

Anyway, so hint is that you wanna show that if (ab)^i = 1 then 6| i

And think about how to do that

What is another way to write "(ab)^i =1"?

[delicate orchid]

yur it's like a specific case of the theorem

[charred crescent]

by "1" do you mean identity?

just want to be sure

[south patrol]

Yes sorry, some say e and some say 1

[charred crescent]

figured

just wanted to be absolutely sure

[wet zodiac]

idk how to explain this

[charred crescent]

okay so if (ab)^i = e, then a^i b^i = e, thus because o(a) = 3 and o(b) = 2, 3 | i and 2 | i, thus 6|i because gcd(2,3) = 1

?

[delicate orchid]

hint, lagrange's theorem

[wet zodiac]

lagrange theorem isnt introduced until 4 sections

[delicate orchid]

so many newbs...

use it anyway...

[wet zodiac]

i dont even know what it is

[south patrol]

Wait you needn't use Lagrange's theorem i think right lol

wait no

[delicate orchid]

not the whole thing no, but meh might as well go overboard

[south patrol]

Like why does a^i b^i = 1 imply a^i = b^i = 1

[charred crescent]

hold up

[wet zodiac]

but basically we can say that if (ab)^i=(ab)^k then we can also say that (ab)^k=((ab)^(k-|G|))(ab)^|G|

[pastel cliff]

what's the problem

[wet zodiac]

but then (ab)^|G|=1 and we can just repeat this until we end up with k-n|G| and this is basically just

k mod |G|

[delicate orchid]

this convinces me tbh

if you're allowed to assume (ab)^|G| = 1 that is

[pastel cliff]

https://cdn.discordapp.com/emojis/982405752297771068.gif?size=96&quality=lossless

[delicate orchid]

https://tenor.com/view/splooshysploosher-funny-dog-doge-doge-stare-dog-look-gif-26301702

[pastel cliff]

licking your abelian groups

[south patrol]

My brain hurties

[charred crescent]

you're tellin me lol

[wet zodiac]

we are assuming ab has an order |G| wew

[formal ermine]

|G| = c|H| for a subgroup H of G

c = |G/H|

[south patrol]

That is

[formal ermine]

if that means anything to you

[south patrol]

the most cursed way to state Lagrange

[delicate orchid]

then sure :pack:

[south patrol]

jk it okay

just using c

[formal ermine]

idk if cat bread knows |G/H|

[wet zodiac]

nope

[south patrol]

Fair nuff

[pastel cliff]

[south patrol]

Basically the key ting is that like

[delicate orchid]

not on quotients yet
oh no no no

[south patrol]

if u have a subgroup H of a group G (needn't be normal or anything cringe)

Then

#H divides #G

[charred crescent]

okay im stuck, idk how to answer this

[south patrol]

Where the # symbol is a twitter reference

[south patrol]

Okay, so rewriting

a^i = b^-i

[wet zodiac]

how does this help with the problem though

[south patrol]

In fact, I claim that is a^i = b^j for ANY j and i then both sides are just equal to 1

Have a think about that

[charred crescent]

but how did you think to write it like that

[south patrol]

Hint: what is the order of a^i? what is the order of b^j?

Oh I mean like

idk exepreicne probably

Like kinda the key thing is you're comparing a with b and this lets u do that

[delicate orchid]

o(g) = |<g>| which is a subgroup of G

[wet zodiac]

whats o(g)

[delicate orchid]

order of g

[south patrol]

Basically

the equation wew wrote is the definition of o(g)

[wet zodiac]

oh

[south patrol]

😎

Or at least equivalent to it depending on ur conventions

[wet zodiac]

i never used that convention before

[south patrol]

Yeah i avoid using o(g) lol

f(x) = x^2 + o(g) :trollshiro:

unaware

Am I impersonating wew correctly

[delicate orchid]

skinwalker energy

[wet zodiac]

ok so where do i go with this information

[delicate orchid]

I thought you were finished with your proof lol

[wet zodiac]

oh no i have to prove that if a and b are elements of group G then the order of ab is the same as order of ba

[delicate orchid]

wtf...

[south patrol]

Lol

[delicate orchid]

perhaps writing out (ab)^n = e as aba....bab = e will help here

[south patrol]

that's what iw as doing

roasted

Just deal w small n for concreteness if you want

[delicate orchid]

u know what I don't come here for this type of vile chitchat to be hurled at me

[south patrol]

like if you can show why abab = 1 => baba =1 then you're cone

[charred crescent]

does this come with any implicit assumptions about a or b, or is this just a general statement?

[south patrol]

This is where you use the fact o(a)=2 and o(b)=3

(or the other way round lol i can't remember)

[charred crescent]

yeah other way, but okay let me think about that

[delicate orchid]

wait I thought shook was done as well

too many chatters

[wet zodiac]

#groups-rings-fields -2

[south patrol]

Hi sebbb

[pastel cliff]

abstract chillgebra

potato

[charred crescent]

god im so dumb, i've lost sight of what we still need to show

[south patrol]

Ripped mathematicians be like lol my abs aren't merely abstract

Nah shook ur good dw

It is confusing cause as you said this is kinda intuitively true

[pastel cliff]

me af

[delicate orchid]

BY CLASSIFICATION OF GROUPS OF ORDER 6 THE RESULT HOLDS!!!!!

[south patrol]

Stfu wew

[charred crescent]

we need to show that every word constructed from a and b is distinct, correct?

[pastel cliff]

he's not wrong

[south patrol]

Yes but uh

Well

All we really need to show is that ab has order 6

And then we'll be done (agreed?)

[charred crescent]

yes

[charred crescent]

im curious about this

[south patrol]

And to do that, suppose (ab)^i = 1 for some i < 6 and show i = 0

[delicate orchid]

bros...

[south patrol]

Now I've claimed that in fact a^i = b^j => a^i = b^j = 1

[pastel cliff]

finite groups are classified

like we know all of them

[south patrol]

And that clearly implies this as we've discussed

So like

[delicate orchid]

this is ABSOLUTELY not true

[south patrol]

Lmao

I assumed sebbb was joking right

[delicate orchid]

spreading lies and slander on the line

[pastel cliff]

oh am i misremembering the thing

[pastel cliff]

what's the thing

[south patrol]

Finite simple groups

[wet zodiac]

i thought this problem was trivial

[delicate orchid]

finite simple groups

[charred crescent]

what?

[wet zodiac]

[pastel cliff]

a wee bit of trolling

[south patrol]

Anyway the point is that

[pastel cliff]

[south patrol]

One can write a list of all of the groups of order 6

[delicate orchid]

you basically did it lol, I don't really know what potato's talking about

unless 6 divides 0 now

[south patrol]

Seeing as there are only two, we are done

No I mean like wew

The point was they said they hadn't proven taht result yet right

Well

There was a gap

[south patrol]

Bruh what

lol

[delicate orchid]

well no they proved a specific case though

[charred crescent]

but he brought up an interesting point that how do we know (a^i = e and b^i = e) if (a^i b^i = e)

like i don't have an answer for that

unless it's trivial and im just re

[delicate orchid]

if it wasn't the case then b^i would be equal to a^-i which contradicts our statement about their orders

I think

[topaz solar]

a very abstract way of "If you have a tree and decorate its nodes by elements of your relevant type of objects you can in fact lift it to a tree decorated by these objects"

[charred crescent]

i think that makes sense

if this was an exam question i would be screwed

there's no way i can spend this long on one problem

[south patrol]

Yes that's what I was helping them through wew 😭

Bruh aha

[wet zodiac]

i cant see where to go
the group not being abelian is severely throwing me off

[south patrol]

Bleak

[delicate orchid]

I am an armstrong away from vomiting answers

[wet zodiac]

[south patrol]

Okay so

let's do the abab thing

sps abab=1

why is baba=1

think about dat and then it should generalise

owo

[delicate orchid]

HINT!!! conjugation....

[south patrol]

that's overkilll

[delicate orchid]

what

[charred crescent]

do we know if the group is cyclic or not?

[delicate orchid]

cyclic implies abelian

[charred crescent]

hahaha

[delicate orchid]

the group isn't abelian

[wet zodiac]

4.3

[topaz solar]

I think it’s pretty much not overkill here

[south patrol]

It was a joke lol

[delicate orchid]

you get faster
||o(a) = 2, o(b) = 3, (ab)^i = a^ib^i = 1 => a^i = 1, b^i = 1 => 6|i|| is made up of implications I would accept on the spot

[south patrol]

reference to a prior part fo this convo

[delicate orchid]

guys i ewas just pretending to be satupid hugyhdugsyguysg

[south patrol]

Lol

i think the implication you've omitted a proof of it the core to the problem though

[topaz solar]

I am very much so still slow

[delicate orchid]

uh dur durr hurr left as an exercise to ur mum

[south patrol]

Okay

[charred crescent]

im slow and not just in terms of speed

[south patrol]

btw i don't actually hate wew this is just funny to berate him

[delicate orchid]

catbread!!! Dont u... forget about me....

[delicate orchid]

likewise

[charred crescent]

rough out here smh

[wet zodiac]

i dont think i covered conjugates

[delicate orchid]

have u covered the definition of a group yet

[wet zodiac]

yes

[delicate orchid]

ok that's a good start

conjugation is just when you do the ol x -> g^-1xg it ain't no thang

[wet zodiac]

wait thats true?

[charred crescent]

the conjugate of x is g^-1xg for any g in G (i believe)

[pastel cliff]

im gonna interject w a quick thing of my own

[topaz solar]

If x reduces to e, what can you say about g^-1 x g

[south patrol]

Don't don't don't don'ttttt

Banger

[pastel cliff]

direct sum of free modules is obviously free

proving it by showing it has a basis

[wet zodiac]

wdym “reduces to e”

[pastel cliff]

is it enough to just mash the bases together

[topaz solar]

As in, is it e

The identity

[wet zodiac]

oh

then the conjugate is also the identity

[topaz solar]

Yep

[delicate orchid]

yes unless your direct sum is disgusting and then I'll have to think about it

[topaz solar]

Reduces is just silly brain separating the word from the final group element

[pastel cliff]

im choosing to remain ignorant as to my precise definition of direct sum

[wet zodiac]

holy shit i think i got it

[delicate orchid]

thern yeah BASH EM TOGEHTER

[topaz solar]

Inject each module into the sum for finite sums

[wet zodiac]

oh my god

it was that simple

[topaz solar]

$X_n \to \bigoplus_i^N X_i$

[pastel cliff]

me to my friend when my mom says we can start a guinea pig fighting ring

$\to$ :(

[delicate orchid]

\rightarrow cmon now people

[cloud walrus]

sebbb

Sharp

[delicate orchid]

grrrr

[south patrol]

Bruh

\oplus_i^N

Like that

Dis cos tan

[wet zodiac]

holy crap the conjugate made the problem 100x easier

[delicate orchid]

$X_n \hookrightarrow \bigotimes_{i=\text{I DON'T CARE}}^{\text{I. DON'T. CARE.} X_i$

[cloud walrus]

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

[south patrol]

lol

[wet zodiac]

(it actually solved it)

[delicate orchid]

yeah almost like that's how you're supposed to solve it lol

[south patrol]

That's what I said too lol

[delicate orchid]

there is no evidence of this fact

[wet zodiac]

i had 0 idea what a conjugate was

then sharp explained it to me what it was and boom

problem immediately made trivial

[delicate orchid]

I like to narrow my PhD research topic to "conjugation"

[topaz solar]

Note that for infinite direct sums I’m assuming it’s the coproduct, so you still inject in, but that’s not quite finite ofc

[wet zodiac]

thanks guys

[pastel cliff]

im just showing M \oplus N is free

for M,N free

[topaz solar]

Define oplua

And imagine the injection

Hmm

[delicate orchid]

I got spooked by hamal basis vs the other one

[pastel cliff]

the what basis

the halal basis?

[topaz solar]

Gröbner?

[delicate orchid]

googling it

[topaz solar]

Starts with sch

[delicate orchid]

basically the other one is a regular basis but can be uncountable

[pastel cliff]

oh none of that

[topaz solar]

That’s just a basis?

[delicate orchid]

it's time to learn about direct integrals my boy

[topaz solar]

Just horrible

[pastel cliff]

im good thanks

[delicate orchid]

:weed: notation

[topaz solar]

The real nasty ones are bases where you have like countably many but you have convergence

[topaz solar]

I’ve never actually learned what this means

[delicate orchid]

I've never learnt them because I just work with finite objects

• don't know measure theory

[wet zodiac]

i wanna learn measure theory soon

probably in a couple months

[delicate orchid]

learn a tiny bit of metric spaces first and you're good

as in, get comfortable with the definition of a metric space lol

[wet zodiac]

i wanna use axler for it

(it has a cat on page 44)

axler should throw more pics of his cats in his textbook

[topaz solar]

What’s the definition of direct integrals anyway

[delicate orchid]

a dumpster fire

[white yoke]

What can I use to prove that projective modules are free over the integral group ring Z[C_2 X C_2] ?

I know that projectives are free over principal ideal rings, local rings, and fields. But I'm not sure that any of these apply in this case

[coral spindle]

Seems like a tough and interesting question

Do you have a characterisation of the ideals of the ring?

[pastel cliff]

how mcuh justification do you think i need of this

i know that's subjectivee

but i never know with things that seem obvious

[delicate orchid]

can we not write Z[C_2 x C_2] as iso to Z^4 by the standard decomposition of the regular representation? Could help

[coral spindle]

Ah, but this is Z

[delicate orchid]

should still hold

[coral spindle]

Oh right bc the reps of C_2 are nice

[topaz solar]

Well, just show something like R^n (+) R^m ~= R^(n+m) maybe?

[coral spindle]

Yeah I think that's a good idea

[delicate orchid]

the space of characters is a Z-lattice spanned by the irreducibles, so the bashing of the regular character into irreps should be the same over Z as C

I think...

[coral spindle]

This is pretty weird though, considering this would make Z projective and not free...

[delicate orchid]

hmm

[coral spindle]

I think there might be something subtle that doesn't work with Z

[delicate orchid]

could be

[topaz solar]

Well projective modules are supposedly free by the problem statement

[coral spindle]

Of course

[white yoke]

How is it isomorphic? I'm not familiar with that

[white yoke]

I'm fairly sure that they are

[coral spindle]

Have you done any rep theory in the past?

[white yoke]

I have

[wet zodiac]

ok back to doing 4.4

[coral spindle]

OK well you should know that if a field k is big enough, then k[C_2 x C_2] as a module over itself decomposes as a certain sum of irreducible submodules

in this case, there are 4 irreducible submodules, all of dimension 1, so the ring is iso to k^4, where each copy of k has a different module structure ofc

Wew's idea is that this extends to Z, and I'm inclined to agree

We can try and identify the central idempotents that would give us this decomposition

[delicate orchid]

I mean, all of the character values of C_2xC_2 are in Z so it should work, but you're right in that Z poses a problem

[coral spindle]

Yeah actually, I think that the sum may not be direct

because the central idempotent that produces the ideals would be (I think) 1/2(0,1) + 1/2(0,0)

or one of them

I hope I'm making sense

[delicate orchid]

major oopsie if true