groups-rings-fields

[winter hound]

yeah

shorter too

[delicate orchid]

write the proof using an inverse for bonus points

[winter hound]

extra credit

Okay one more, then I am off 👉 👈

I started off a proof and am unsure if i was doing it right

Prove that, if H is a subset of a group G, then H is a subgroup of G, iff a*b^-1 is in H for every a,b in H.

[delicate orchid]

oh this is a nice one

lemme think

https://tenor.com/view/dog-bored-thinking-animal-pet-gif-20642674

[winter hound]

I first said "Suppose H is a subgroup of G. Let a,b be in H. Then a*b is in H. Since every element in H has an inverse, then let b in H, so b^-1 is in H. Therefore, a * b^-1 is in H

that's the foreword direction

wants us to do this proof in "one step"

[delicate orchid]

yup

[winter hound]

okay cool

the other direction, i am ovethinking it i think

i'm overthinking my overthinking

[delicate orchid]

so the other direction is "Assume $H \subseteq G$ such that $ab^{-1} \in H$ for all $a, b \in H$ then $H \leq G$"

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

step one, use the fact that b is in H to show that e is in H

[winter hound]

we can say a=b?

[delicate orchid]

yes we can

[winter hound]

bb^-1 = e

[delicate orchid]

yup!

now, build on this to show that if b is in H, b^-1 is in H

[winter hound]

does this help with closure under operations? or just to show there's an identity in H?

[delicate orchid]

it helps with both other steps

[delicate orchid]

but the proof i have in my head does this step first, using the fact that e is in H

[winter hound]

well if we have a,b in H, then ab = a(b^-1)^-1?

[delicate orchid]

ah yes very true

[winter hound]

would that show the inverses?

[delicate orchid]

tbf that is what I've jotted down but I thought you needed to prove identitiy first

and no, that just shows closure

a, b in H => ab in H

[delicate orchid]

there's another substitution you can do for a to get b^-1 in H

[winter hound]

a=e

[delicate orchid]

exactly

and then we're done!

[winter hound]

thank you very much for the help

[delicate orchid]

no worries

good practice for when I become a TA

[winter hound]

Ah yes

in that case, you're welcome

[white nymph]

how to show that Q is not complete wrt the 2-adic norm? the hint in the book says to create a cauchy seq that converges to the cube root of 2... but i dont know how to do that

[dense pumice]

I feel like that's weird to do, idek if cbrt 2 is in the 2-adic integers

because you would have something that works mod 4, but x^3=2 mod 4 has no solutions

[delicate orchid]

No the point is that cbrt 2 isn’t in the 2adic integers and yet there’s still a sequence in the 2adics that converges to it

[white nymph]

how do i create such a seq

[delicate orchid]

I have absolutely no idea

[white nymph]

😄

[dense pumice]

if there's a sequence in the 2-adics that converges to it, by completeness, it's in it

[delicate orchid]

They’re trying to show it isnt complete

[dense pumice]

ok so like not with this one, but with numbers you have to find a solution mod p^n that works for all n and is "compatible" then take that sequence to be your number

[dense pumice]

no, they're trying to show Q isn't complete, the 2-adic numbers are certainly complete

[delicate orchid]

Oh I see I’m completely blind

[dense pumice]

oh wait it's in Q_p not Z_p

[delicate orchid]

2-adic NORM

[dense pumice]

so mayhaps you need to find cbrt 16 or something

wait no

can't divide by 2

wait yes u can

brain not work at 10 pm

[delicate orchid]

It’s 3am here

[dense pumice]

yike

[delicate orchid]

Just the one yike

[dense pumice]

yep

yea idk how to do this, instead I would just be like base 2 1, 101, 100101, 1000100101,... converges p-adically but try to show it isn't a rational number

hopefully that'll convince u

[white nymph]

im just saying cube root of 2 isnt in Q... so Q isnt complete wrt 2-adic norm if theres some cauchy sequence that converges to cube root of 2... and thats the hint in the book so idk

[next obsidian]

Can’t you grab a Cauchy sequence converging to any real number?

Just use the decimal expansion

Truncating it at the n-th digit for x_n

Or something like that idfk, what’s rhe 2-adic norm

Nvm

[woven delta]

Its just n^2xn^2

@long galleon

[pastel cliff]

chmonkey sequence

[woven delta]

The matrix L is just the block matrix with k^2T on the kth spot on the block diagonal, 0 everywhere else

Let me try to tex that actually

$ \begin{bmatrix}
T & 0 & 0 \
0 & 4T & 0 \
0 & 0 & 9T
\end{bmatrix} $

[cloud walrus]

Emma

[woven delta]

This is the 3 case

As you can see the matrix $A\oplus B$ is 2nx2n

[cloud walrus]

Emma

[woven delta]

That's $\otimes$

[cloud walrus]

Emma

[woven delta]

The Kronecker product multiplies the dimensions

Direct sum adds them

Think about what X\otimes Y looks like

Try looking at the picture of A\otimes B you had before and the picture I posted

Actually writing these matrices out helps a lot with understanding them

Maybe try actually computing some Kronecker products so you get the hang of them

I don't think Kronecker product is that weird, it's the matrix version of the tensor product

Well sure but it's a very standard tool in linear algebra

Oh huh let me see

Not quite

Just have the second one be T

But yeah otherwise that's it

[dusty sapphire]

is there unique ring homomorphism from any field to \mathbb{Z}?

[lethal dune]

unique non-zero, yes

[prisma shuttle]

how do u show this

[lethal dune]

actually

it's the other way around that I'm aware of

from field probably not

[prisma shuttle]

oh well here its obvious right because $\phi(1)$ determines everything

[cloud walrus]

JustKeepRunning

[prisma shuttle]

so yea that makes sense

for this problem i've shown that there exists a maximal ideal with respect to the property that it is disjoint with S

but ima not sure how to show this maximal ideal is prime (because while it is well known maximal --> prime, here this is maximal with an extra condition so it might not be true)

[lethal dune]

hint: what should be kernel of the homomorphism

[gritty sparrow]

Do it by contradiction: assume there are x and y such that x and y are not in P but xy is. Then P+xR contains x, and because this strictly contains P there is a in S \int P+xR, a=p+rx, so ay=py+rxy similarily a’=q+ty. we see that aa’is in P \int S but that was supposed to be empty.

[dusty sapphire]

got it kernal should be Filed.

never be just identity

[lethal dune]

no I meant kernel is either {0} of whole F

[dusty sapphire]

if field is finite then it would be problem?

for infinite field if kernal is {e} then R/ker is field, so is it possible?

if ker is {e} then it is one-one so this is not the case

[proud bear]

wait why if ay is in P then a is in P?

[gritty sparrow]

P is prime and y is not in P so ay in P implies a in P

[proud bear]

i thought justkeeprunning wanted to show that P was prime

[gritty sparrow]

ah shit I confused myself in the notation

do the same thing for P+Ry and multiply the two a's

that fixes things I believe

[lapis trail]

Do I gain insight viewing C as R[x]/<x^2+1>?

[next obsidian]

Yes

For example this lets you compute C (x)_R C

Just explicitly, you can write this as
C (x)_R R[x]/(x^2 + 1) = C[x]/(x^2 + 1)

This is just by general properties of the tensor product

From here:
C[x]/(x^2+1) = C[x]/((x-i)(x+i)) = C[x]/(x + i) x C[x]/(x - i) = C x C

Where the second equality is using the Chinese remainder theorem

[rustic crown]

another nice thing is, it tells you that there's an automorphism of C swapping i and -i. And the algebra alone can't distinguish between the two.

[next obsidian]

Chmonkey

Hello det

[rustic crown]

Hello Chmonkey

[next obsidian]

What year are you in det?

[rustic crown]

third year of ug

[next obsidian]

I see. Best of luck next year

[rustic crown]

arigato 🧡

[next obsidian]

Idk what the landscape for masters are, but I assume it’s competitive

[lapis trail]

Neat

[chilly ocean]

yo guys can u help me on 3c ill send my part a and b

[dusty sapphire]

did you read Lagrange theorem?

so what are the possible order of element in group of order 33?

[chilly ocean]

1

oh wait

i wrote those

in part a

[dusty sapphire]

yep

[chilly ocean]

1,3,11,33

are the possible orders

i need help on part c bro @dusty sapphire

i know how to do a and b

[dusty sapphire]

alright

did you know Sylow theorem?

[chilly ocean]

no

not yet

only lagrange and cosets

for this hw

[dusty sapphire]

33=3*11 right so pick p=3 and q=11 then use part b

[chilly ocean]

so we say

im not sure how b works

i mean how it works in this context

[proud bear]

wait par is this your alt

[chilly ocean]

no bro

wait wtf

this dude is in my classs

lol

[dusty sapphire]

lol😂😂

[proud bear]

ah lmao

[chilly ocean]

i guess hes just as lost as me

a fellow brotha

anyways how do i do c

wait i literally said

all the possible orders of the elements

why is it guranteed that 3 is one of them tho?

[dusty sapphire]

There is one theorem state that, G has subgroup of order prime p if p divide order of G.

[chilly ocean]

yeah

but its asking about an element

in the group

[dusty sapphire]

so subgroup of G has order 3 and 11 right

one more theorem, group of prime order is cyclic i.e. there exists a element of which have order same as order of group.

[chilly ocean]

oh i have that theroem

that a group with prime order is cyclic

[dusty sapphire]

so subgroup of order 3 and 11 are cyclic right

[chilly ocean]

cause they are subgroups right

so they would be cyclic as welll

?

[dusty sapphire]

no order is prime

[chilly ocean]

oh okay

yeah

[dusty sapphire]

so now what you get from this?

[chilly ocean]

idk how them being cyclic makes a diff

maybe im missing a connection

[dusty sapphire]

a group is cyclic if it generated by single element

[chilly ocean]

yes

[dusty sapphire]

so what is the order of that generator?

take example G = Z_3

[dusty sapphire]

what are the generator of G?

[chilly ocean]

under addition right

[dusty sapphire]

yes

[chilly ocean]

1

and 2

[dusty sapphire]

alright then what's order of 1 and 2 in Z_3 under addition

[chilly ocean]

hmmm

1 and 2

respectivelty

order of 1 is 1

order of 2 is 2

[dusty sapphire]

what

1 you add 1 times then you get identity huh?

[chilly ocean]

i mean u add it 0 times

oh

identity is 0

in this case

3

and

3

[dusty sapphire]

good

so order of 1 and 2 are 3 right

[chilly ocean]

yes

[dusty sapphire]

and what is the order of group Z_3?

[chilly ocean]

3

!

[dusty sapphire]

so what you get from this

[chilly ocean]

thats a law?

so all elements of a prime cyclic group

have the order of the group

?

[dusty sapphire]

yes that's what i said, in every cyclic group there is element which have order equal to order of group and that element are generator of the group

so in your group of order 33 you get subgroup of order 3 right

[chilly ocean]

yes

[dusty sapphire]

since subgroup is also group and order is prime so what you get from this?

[chilly ocean]

that its also cyclic

[dusty sapphire]

right

so now you get your answer right

[chilly ocean]

okay let me confirm

[dusty sapphire]

ok

[chilly ocean]

before i write this down lol

We know that the possible orders of the subgroups are 1,3,11 ,13, from lagranges theorem.

so since there is a subgroup of order 3 which is odd implies that it is cyclic.

In every cyclic group there must be an element which has an order equal to the order of the group

[dusty sapphire]

this is by Cauchy's theorem

[chilly ocean]

and that element is always the generator?

what is by cauchys theorem

[dusty sapphire]

yes

[chilly ocean]

the order 3 thing?

[dusty sapphire]

yes

if p is prime and p divide order of group then there is subgroup of order p.

similarly you can say there is element of order p

[chilly ocean]

yeah

[dusty sapphire]

if you read this theorem that's good otherwise you need to approach it differently

[chilly ocean]

i dont think we learned cauchy

all we learned was that

order of subgroup divides group

and elements order divides group

[dusty sapphire]

then i think you need to approach it differently

[chilly ocean]

hmm okay

[dusty sapphire]

or you can show that there is subgroup of order 3 then it is good

[chilly ocean]

we already know a subgroup of order 3 exists tho

from part a

[dusty sapphire]

then it is clear

[dusty sapphire]

you can use this

[chilly ocean]

ohh since i proved its a subgrou

its cyclic

also right

so can i just say its cyclic as well

[chilly ocean]

and say this

[dusty sapphire]

use this

[dusty sapphire]

yes

[chilly ocean]

good?

it makes sense to me

but want to make sure if it correct

[dusty sapphire]

you can use this but still you need to show that there is subgroup of order 3 because in the part a, this is necessary condition not sufficient

[chilly ocean]

but i proved it in part a

i can say it in this class

[chilly ocean]

look

[dusty sapphire]

this is possible order are 1,3,11 and 33 but this is not say that always exist this order subgroup

this say that if exist then order of subgroup will be from this

[chilly ocean]

how do i know

if im positive it exists

so lagrange only shows possible subgroup orders?

[dusty sapphire]

yes

[chilly ocean]

how do i prove

order 3

is a subgroup

[dusty sapphire]

that's why i said by Cauchy's theorem

@proud bear need your help😁

[chilly ocean]

damn bruh

[broken stirrup]

Do you also think algebra classes get kinds boring after groups? Rings are fun too but I can't motivate myself to start studying Modules

[proud bear]

If you are not allowed to use cauchy's theorem you can do this. First, if G is cyclic, then there are phi(33)=20 elements of order 33. There is one element of order 1 (the identity) so the 12 remaining elements have orders 11 and 3. If there is no element of order 3 then the rest all have order 11. (It's kind of hard to explain this part do hopefully it makes sense). Let a be one of the elements of order 11, then <a> is a subgroup of order 11. Every non identity element of <a> has order 11 so 10 of the 12 elements of order 11 are in the subgroup <a>. If b is one of the 2 elements of order 11 not in <a>, then <b> is also a subgroup of order 11 whose intersection with <a> is 1. This means the 10 elements of order 11 in <b> are different from those in <a>. But that would mean there are 20 elements of order 11 (10 in <a> and 10 in <b>) which can't happen. Thus there must be an element of order 3. @chilly ocean

You can do something similar for the case where you don't assume G is cyclic

[chilly ocean]

fuck it bro imma do cauchys

so cauchys theorem assumes and odd ordered group is cyclic?

On they did teach me cauchys in discussion

[dusty sapphire]

R be ring of function from [0, 1] to Real number, I ={f in R such that f vanish exactly at one point in real number(same for all)} then R/I is field? i.e. I is maximal ideal of R?

[foggy merlin]

sth is off here, if the functions in I all vanish at exactly 1 point only, it's not not even an ideal

I guess that it's probably the set of functions vanishing at some fixed point 0<= r <= 1.

in my case yes, any ideal J containing I that contains a function f that does not vanish at r allows us to choose a function g in I which is 0 at any point where f is not and nonzero wherever f is zero. Then f+g has no zeros and has therefore a multiplicative inverse. Since f+g is also in J this would imply that R=J, which proves maximality.

[prisma shuttle]

isn't this showing the inverse of the desired statement though

i thought u want to show that if x,y are in S then xy is in S

also what are q and t here?

and what does \int mean

[dusty sapphire]

Yes got it, similar if i take space of continuous function on [0, 1] then it is also same case right

[foggy merlin]

the continuous case seems slightly more tricky at first sight

[dusty sapphire]

In first case if i take R’= R/I, R’[X] is PID right

[foggy merlin]

since R/I is a field, yes

[dusty sapphire]

Wb this R[X]/I[X] is it PID?

[dusty sapphire]

But it will be maximal right if it is vanish at two fixed point then it will not remain maximal ideal

[foggy merlin]

that's just isomorphic to R/I if I understood your writing of the expression correctly

[foggy merlin]

It feels like it should be maximal for the single point case, but proving it becomes harder than in the non continuous case because you can't just construct an arbitrary function that sums to an invertible one

[dusty sapphire]

$\frac{R[X]}{I[X] }$ isomorphic to R/I?

[cloud walrus]

BLツ

[dusty sapphire]

So there i need different approach

[foggy merlin]

wait hold on, no I don't think it is I was mixing stuff up, it should be isomorphich to (R/I)[X]

[dusty sapphire]

I stuck in the $\frac{R}{I}[X]$ is not PID in non continuous case but PID in continuous case so I’m thinking that there is some notation problem

[cloud walrus]

BLツ

[foggy merlin]

maybe a similar approach works but with a lot more analysis involved

[inland otter]

right ive got a solution which is different from the solution im given and i just wanted to make sure what i did is correct because this stuff still is somewhat fuzzy in my mind\

Exercise is to show that for a prime ideal p of R, we have $R_p/pR_p \simeq Frac(R/p)$
The solution im given somewhat explicitly constructs an isomorphism between these two but i was wondering if i could consider the exact sequence of R-modules $0\to p \to R \to R/p\to 0$ and then localise at $T=R\setminus p$, using the fact that $T^{-1}$ is exact, ill only have to show that localisation at T of $R/p$ is isomorphic to $Frac(R/p)$ and localisation at T of $p$ is isomorphic to $p R_p$, its these last two statements im not quite sure about, tho they feel natural

[cloud walrus]

dadaurs

[foggy merlin]

I didn't understand that sentence

[lethal dune]

your f becomes identically zero in this case

\{ \}

[dusty sapphire]

R is set of all function from [0, 1] to $\mathbb{R}, I={f \in R | f(t)=0 , t \in [0, 1]}, \frac{R}{I}[X]$ is not PID

[cloud walrus]

BLツ

[lethal dune]

.

[dusty sapphire]

I= zero this one?

[lethal dune]

did you mean (for some t \in [0, 1])

[dusty sapphire]

Yes fixed t

[lethal dune]

no f(t) = 0, t\in [0,1] means f(t)=0 for all t making it identically zero

[dusty sapphire]

No no for some fixed t

[lethal dune]

which is not mentioned there

[dusty sapphire]

I need to use \text that’s wh

[lethal dune]

literally changes the entire question, so not worth skipping

[dusty sapphire]

R is set of all function from [0, 1] to $\mathbb{R}, I={f \in R | f(t)=0 , \text{for some} \ t \in [0, 1]}, \frac{R}{I}[X]$ is not PID

[cloud walrus]

BLツ

[lethal dune]

I think you are not writing the question properly, I is not a ideal here

R/I is not even defined

(like if you mean t is fixed then it is but that's not what the question says)

[dusty sapphire]

R is set of all function from [0, 1] to $\mathbb{R}, I={f \in R | f(c)=0 , \ c \in S}, \frac{R}{I}[X]$ is not PID where $\phi \neq S \subset [0,1]$ is singleton

[cloud walrus]

BLツ

[inland otter]

Oh wait i also need to show they are isomorphic as rings not only as modules mb

[cloud walrus]

BLツ

[dusty sapphire]

What’s different between this two?

[prisma shuttle]

um i think the theorem is $R[x]/(I)\cong (R/I)[x]$

[cloud walrus]

JustKeepRunning

[dusty sapphire]

now I'm confused

[dusty sapphire]

wb this?

[foggy merlin]

same thing, justKeepRunning has better notation tho

[dusty sapphire]

okay

[dusty sapphire]

then it is PID right

I is maximal in R then is I is also maximal in R[X]?

[dense pumice]

By I, do you mean I*R[x] in the second case?

because I is no longer an ideal of R[x]

[dusty sapphire]

I is this

thats the confusing of (R/I)[X]

Alright leave it i think some notation problem in the question. Thank you so much.

[dense pumice]

singleton means like isolated points?

[tribal moss]

A singleton generally means a set with exactly one element.

[prisma shuttle]

if you mean a singularity thats like a single point where a function is not complex differentiable

[dense pumice]

wait so cant you just say R/I is isomorphic to R (reals) since its just point evaluation and a polynomial ring over a field is a PID?

[dusty sapphire]

yes i think of it but in answer (R/I)[X] is not PID that's why, for continues function it is PID

[dense pumice]

wait so youre trying to show its not a PID?

but why isnt it?

[dense pumice]

oh wait is t not fixed?

[dusty sapphire]

just use this one

[dusty sapphire]

this is mistake

[dense pumice]

Im confused on the question

is c fixed or not?

like is it 1 point or potentially multiple points?

because if just 1, then R/I is definitely the real numbers and R/I[x] is a PID

[dusty sapphire]

yes only point

fixed c

[frank fiber]

if i have a non-cyclic abelian finite p-group G then the exponent of G is p?

[next obsidian]

What is the exponent of a group?

[frank fiber]

the lcm of all the orders of the elements in the group

[next obsidian]

Not necessarily then

Take Z/p^2Z x Z/pZ

The exponent will be p^2

You can at least say that some element has order equal to the lcm, but this is just because the orders of every element is a power of p

[frank fiber]

oh i see

[next obsidian]

In general actually

For any finite abelian group some element will achieve the exponent as its order

This follows from the structure theorem for finite abelian groups

[frank fiber]

i want to prove that a non-cyclic non generalized quaternionic p-group G has a subgroup of exponent p and order at least p^2

[frank fiber]

G has a abelian subgroup non cyclic because every p-group such that every abelian group is cyclic is cyclic or generalized quaternionic,

[barren sierra]

If G = G_1 x ... G_n and we have N is normal in G, must N = N_1 x ... x N_n where N_i is normal in G_i?

That sounds right and I think I can show that by contradiction

But maybe there's some edge case I'm not thinking of

[next obsidian]

No this isn’t true

[hidden haven]

{(x,x)} in Z x Z

[next obsidian]

I WAS GONNA SAY THAT

Grrrrrr

[hidden haven]

[barren sierra]

Hm fuck

Wait how is that a counter example?

[chilly radish]

It's normal since ZxZ is abelian but it's evidently not a product

[barren sierra]

Ah

What if I add the restriction that the groups are non-abelian and simple lol

Cause that's how I was gonna solve this

[next obsidian]

Use the correspondence theorem relating normal subgroups containing a subgroup H and those of G/H

I think

[barren sierra]

(not a dox cause that's his public email)

[next obsidian]

You need to reduce to the fact that each normal subgroup of G_i is either trivial or the entire group

[barren sierra]

Yea

Correspondence theorem hm

[next obsidian]

Actually

[barren sierra]

Ok

Seems overkill

[next obsidian]

Also use the fact this product is finite

That isn’t overkill Hurb

[barren sierra]

I guess

[next obsidian]

Finite = inducc

[barren sierra]

Oh true

[next obsidian]

Maybe these two alone aren’t enough to kill the problem, but it’s a good start

[barren sierra]

Ok so the book has a hint

but I don't fully understand it

however a couple notes. That G_i really should say K_i

and then that really should be ({e} x ... K_i x ... {e}) cap N

I just don't get how showing ({e} x ... K_i x ... {e}) is a subgroup of N helps

and more importantly I don't get how that's always true

because what if K_i is non-trivial but N is trivial in that coordinate

definition of G_I if needed again from above

oh wait is it because a_i =/= {e}?

[barren sierra]

Yea idk

I don't know how showing this intersection

Gives me that ({e} x ... x K_i x ... {e}) where K_i is non trivial is a subgroup of N

[next obsidian]

The i is an existential variable, you have to show that there is some i so this point won’t come up. Specifically, you’re assuming already that there’s an element in N which has a non-trivial i-th component

[next obsidian]

Once you have this you can mod out by K_i and then by induction you’re finished

[next obsidian]

N\cap K_i is a normal subgroup of K_i, so now you have that N\cap K_i is non-trivial so is all of K_i

Of course here I’m identifying K_i with the subgroup {1} x … {1} x K_i x {1} x … x {1} of G

[barren sierra]

Yea I figured it out

[next obsidian]

[barren sierra]

Sorry 💀

[next obsidian]

That’s good

[barren sierra]

Ye

[next obsidian]

:petthespamakin:

[barren sierra]

Now one final small part and I gotta type this up

But at least I'm done early

Rather than frantically doing this midnight sunday

[prisma shuttle]

can someone help me understand why the ideal $(3,x)$ of $\mathbb{Z}[x]$ is prime?

[cloud walrus]

JustKeepRunning

[prisma shuttle]

Z/3Z?

[dense pumice]

which is an integral domain

[prisma shuttle]

oh oops

oh yea i see

[terse crystal]

There is a morphism f from Spec(Z[x]) to spec(Z)

Induced by natural inclusion map Z—>Z[x]

Then you use a property in fibered product which is spec(k(p) tensor product Z[x]) is homeomorphic to f^-1(p)

So you are just classifying prime ideals q of Z[x] by its image p=f(q) in spec (Z)

depends on p=(0) or (P) where P is a prime number , k(p) is Q or Z/pZ respectively

k(p) tensor product Z[x] is isomorphic to Q[x] or (Z/pZ)[x] respectively and they are pid

For example,there exists a one-to-one correspondence between prime ideals q of Z[x], such that the intersection of q and Z is (3) in Z, and the image of spec(Z/3Z[x])

x is irreducible so (x) is a prime ideal in Z/3Z[x], whose image in spec(Z[x]) is (3,x)

Anyway prime ideals of Z[x] therefore are either (g(x)) where g is irreducible or (p, f(x)) where f is irreducible in Z/pZ[x]

[chilly ocean]

(in an endearing way)

[lethal dune]

lol

[chilly ocean]

if $R$ is a ring, $A$ an $R$-module, and $I$ an injective $Z$-module, then apparently $\text{Hom}_R(A,\text{Hom}_Z(R,I))$ is naturally isomorphic to $\text{Hom}_Z(A,I)$ but I really don't see this

[cloud walrus]

wren

[chilly ocean]

by the way $Z$ is $\mathbb{Z}$ but I was too lazy to write it out

[cloud walrus]

wren

[chilly ocean]

I can see one way to take an R-module map from A to Hom_Z(R,I) and get a Z-module map from A to I, but in order to go in the reverse direction I find you can get a map from A to Hom_Z(R',I) where R' is the integer multiples of 1 in R, and you can extend each of these to a map from R to I since I is injective, but I see no nice way to do this in general

I'm just very confused

[rustic crown]

tensor hom adjunction ?

in more details, think of R as a (Z, R)-bimodule, then

[cloud walrus]

det

\begin{align*}
    \operatorname{Hom}_{R}(A, \operatorname{Hom}_{\mathbb{Z}}(R, I))  & \cong
    \operatorname{Hom}_{\mathbb{Z}} (R \otimes_{R} A, I) \\
                                                                      & \cong
    \operatorname{Hom}_{\mathbb{Z}}(A, I)
\end{align*}

[rustic crown]

@.mniip uwu

[chilly ocean]

oh yeah smart, I didn't see that cause I'm not used to doing tensor product that way with 2 different base rings

[terse crystal]

A \otimes R not R \otimes A I think

R is a R-Z bimodule , not a right R module I think

[rustic crown]

yea i'm using the (Z,R)-bimodule structure

[terse crystal]

Other direction I think

R-Z

[rustic crown]

you're looking at Hom_Z(R, I) so it has to be Z-R, we want the result to become a left R-module

(i'm assuming R is non-commutative, if everything is commutative, then probably all this doesn't matter at all)

[terse crystal]

R is Z-R A is R-R I is Z-Z?

[rustic crown]

R is Z-R
A is R-Z = (left R-mod)
I is Z-Z = (left Z-mod)

[rustic crown]

also the second line here is abusing the notation... technically i mean restriction of A as a left Z-module

[terse crystal]

What is Hom_Z(R,I)?

What module structure does it have?

[rustic crown]

so if M is (A,B)-bimod and N is (A, C)-bimod then the set of left-A-linear maps Hom_A(M, N) naturally acquires a (B, C)-bimod structure.

so that should become an R-Z bimodule = left R-mdoule

i'm sure the details work out... it's been a while since i last verified any adjunctions 🙈

[terse crystal]

Oh thank you

Got it now

[tall jay]

If G is isomorphic to H, then is H isomorphic to G?

[rustic crown]

sure is

[tall jay]

Hmm, I guess I asked the wrong question

If I am asked to prove that G is isomorphic to H, is it equivalent to prove H is isomorphic to G?

Wait, it is

Bc if H is isomorphic to G, then G is isomorphic to H

[rustic crown]

that's same as what you asked in the start

[tall jay]

Yeah, I realized it lol. Sorry.

But one more thing

Im trying to prove $g: \mathbb{Z}_9 \rightarrow \mathbb{Z}_9$ defined by $g(x) = 2x$ is isomorphic, and I'm having a little trouble proving it to be onto. I could just compute the values for all 9 elements in the domain, but is there a way to just find the function's inverse?

[cloud walrus]

beeswax

[tall jay]

Bc like x=y/2 wont work

[rustic crown]

what's g(5x)?

🙈

[tall jay]

10x ?

[rustic crown]

and we're modulo 9

[tall jay]

x

[rustic crown]

🥳

[tall jay]

Aha

[rustic crown]

in general from Z/nZ --> Z/nZ you can find inverse of multiplication by a if and only if a and n are coprime

[rustic crown]

the reason is exactly this, but like you said 1/2 isn't available to us directly

in particular we were looking for a solution of 2x = 1 (mod 9)

[tall jay]

Ohhh I see now

Modular arithmetic....

[next obsidian]

There’s a useful trick here

For a map between two finite sets of the same size, it’s bijective iff it’s injective iff it’s surjective

This follows by the pigeonhole principle

So you only need to show that the map is injective

And this is kinda just some number theory, it follows because 2 is coprime to 9

I mean surjectivity also follows from it being coprime but

It seems like you had already shown it was injective so

You can save yourself some trouble

[tall jay]

So, for finite sets, onto implies injective

finite sets of equal size

and injective implies onto

[next obsidian]

Yes

[languid osprey]

Isn't it obvious since the identity is an even permutation and so is not contained in the set of all odd permutations?

Or is there more to that?

[prisma shuttle]

yes this is very useful its used a lot

like considering the map $x\to ax$ for some element $a$ in the set

[cloud walrus]

JustKeepRunning

[languid osprey]

They're saying this is the isomorphism between the permutation group n and it's subgroup n-1. But is this bijective? Only elements with n-1 permutations gets mapped to the n-1 permutation subgroup. What about elements with n permutations?

[prisma shuttle]

what is the next sentence that starts with "but an isomorphism is ..."

[languid osprey]

But an isomorphism is not always so trivial

[iron vessel]

Can someone tell me why I’m getting -6 instead of 2 as the coefficient of s_2^2? I can’t seem to figure out what I’m doing wrong. Any help is appreciated

[languid osprey]

nvm I got it

[pastel cliff]

@hidden haven you're gonna love this one

my crank professor is now requiring that we do homeworks in groups of 3-4

this fucking professor

[rotund aurora]

The first point is accurate though

[delicate orchid]

@pastel cliff actually submit a complaint

[pastel cliff]

?????????????

i made perfectly sure literally everything was legible to a fucking toddler

this is the only thing i missed

AND IT'S RESTATING THE QUESTION

[delicate orchid]

overly enforcing the structure of proofs.... fucking zringe

[delicate orchid]

he literally disobey's his own rule in that claim

he started the claim with an alpha

[rigid cave]

lmfao what is this

[delicate orchid]

this is a prime example of too much attention in the wrong places of a course

[pastel cliff]

unfortunately the claim is what i wrote

that's what i lost a point for

which is still fucking stupid

[delicate orchid]

omfg WHAT

[pastel cliff]

literally everything is correctly and excessively explained

i feel bad for whoever lost points for this

on their algebra hw...

[delicate orchid]

nietzba I'd legit submit a complaint lol

[pastel cliff]

so yeah basically uhhhh

[hidden haven]

[pastel cliff]

https://tenor.com/view/heres-pipe-bomb-santa-pipe-bomb-gif-24143658

[hidden haven]

[hidden haven]

[pastel cliff]

i am LIVID

[delicate orchid]

wtf a reference to hit song Kaczynski claus by wew lads tbh

[delicate orchid]

go tell HIM TO HIS FACE BIATCH! Sort em out

[pastel cliff]

nono the best thing i can do now is get a fucking A in the class and wipe my ass with the final in front of him

[delicate orchid]

let the rage fuel you... an excellent maneuverer

I legit spell maneuverer by typing "man" then spamming vowels and then an r, autocorrect sorts the rest out

what a stupid word

[pastel cliff]

queue

[delicate orchid]

queueing

[pastel cliff]

wednesday

[rigid cave]

q

[delicate orchid]

let q be a polynomial in D[n]

[hidden haven]

This reminds me, Clerk was teaching me french so that I can read godement's book

[rigid cave]

wtf

[delicate orchid]

is there not a translation?

[rigid cave]

what is that book about?

how did "q" remind you of this?

[hidden haven]

Clerk translated the first 60 pages but then ditched it

[hidden haven]

Quimst'd've speak French

[pastel cliff]

baguette

[hidden haven]

Sheaves simplicial sets spectral sequences homological algebra

[delicate orchid]

⚠️ 🚨 france detected 🚨 ☣️

[rigid cave]

😵‍💫

[pastel cliff]

good thing france isnt real

[rigid cave]

daim sounds nice tbh

did you read the other book you planned reading?

simplicial stuff

[hidden haven]

Clerk sometimes joins when saketh and I study stuff together and it's

[hidden haven]

A little

I am planning to speed run that

[rigid cave]

moldi go zooooom

[hidden haven]

Once I'm done writing something

[delicate orchid]

I'm speedrunning galois theory atm

[hidden haven]

I'm writing a pdf on string diagrams 😌

[pastel cliff]

im speedrunning this fucking bleach god im so pissed

[rigid cave]

oooh daim

[delicate orchid]

me when I see the word "field-representation"

[pastel cliff]

but topology time

[rigid cave]

still haven't read about those

[hidden haven]

Field-representation?

What kind of Galois theory are you studying

[delicate orchid]

it's not a representation of a field

you just smash a group into the multiplicative group of a field

[hidden haven]

Smash as in some sort of smash product or you mean like take homomorphic image?

[delicate orchid]

homomorphic image

[hidden haven]

Lmao smh

[delicate orchid]

"clang" might've been better than smash tbf

[hidden haven]

Clang product

[delicate orchid]

I'm sure they're a very useful construction but I cannot remember why

[hidden haven]

I don't think clang product is a thing

[delicate orchid]

no not that one

the field reps

[hidden haven]

Oh 😌

Multiplicative group of field is just GL_1 F

So you are doing 1 dimensional rep theory when you do that

[wraith obsidian]

Do lie algebras belong here or in #diff-geo-diff-top ? I'd say here, right?

[hidden haven]

Probably here

[latent anvil]

I think either is probably fine

[rigid cave]

moldi is lying

get it?

[hidden haven]

[rigid cave]

lmao okay gotta stop procrastinating

[wraith obsidian]

I'm wondering whether there is a specific name to the "non-direct sum" algebra consisting of matrices
[
\begin{pmatrix} A_1 & \ast \ 0 & A_2 \end{pmatrix}
] where $A_i\in L_i$ matrix lie algebras

[cloud walrus]

lux

[hidden haven]

Is that the extension of L_2 by L_1 kinda thing

[wraith obsidian]

(also, I spent half an hour asking myself whether so(3)⊕so(3) stays (semi-)simple lmao)

[wraith obsidian]

I mean we do have a lie-hom to L₁⊕L₂ and the right hand corner is in its kernel

[hidden haven]

I meant that this would be a non split extension maybe

[wraith obsidian]

Well it's not an extension of L₁ by L₂ for dimension reasons

or vice versa

[hidden haven]

Right yeah

[wraith obsidian]

The interesting thing about this is that if L₁ and L₂ are perfect, then this often continues to be perfect (as long as you get all matrices by applying col operations to L₁ and taking the span of that)

so eg when L₁=L₂=so(3) this should be perfect, but not simple

[rotund aurora]

What is the reason for group actions to be shortened by f(g,x)=gx ? seems odd to me.

[tribal moss]

Convenience.

[rotund aurora]

come on

oh ok

I read "consensus" sorry

[delicate orchid]

same reason we write ab for group multiplication instead of *(a, b)

[tribal moss]

Once an action has been fixed, repeating a name for the mapping function over and over would be the least interesting part of formulas, but fairly visually distracting.

All notation is a compromise between "being able to see the forest for trees" and "not being so ambiguous that the point fails to get through".

[rotund aurora]

I see

thanks

[simple mulch]

Guys, I don't see why corollary 1 says that

For V the regular representation we have V = w1 W1 + w2 W2 + ... + wi Wi where wi is the multiplicity and Wi the irreducible representations?

So taking the traces of both sides we have that sum in yellow?

[hot lake]

taking the characters evaluated at s

[simple mulch]

since every irreducible representation is contained then the sum of the characters at s equals the character of the regular representation at s

that's what it means?

[hot lake]

Yeah with the correct multiplicities

Every irreducible representation is contained in the regular representation

And corollary 1 says how many

And we know the character of the regular representation

[pastel cliff]

actually @delicate orchid the damn prof actually accepted the "subgroup generated by set" bullshit

[delicate orchid]

remind me of the specifics of this bullshit

[pastel cliff]

which wasnt wrong but i wouldnt have been as enraged if he had taken points off for that as opposed to the other stuff

[delicate orchid]

wait didn't I say this

[pastel cliff]

yes

[delicate orchid]

[pastel cliff]

i actually understand it i promise (believe me or not this was my first line of reasoning the other night i just dismissed it bc i didnt think i was rigorous enough - i reevaluated my opinion at around 4 in the morning)

[hidden haven]

Bro what

What is this prof on

[delicate orchid]

idk but I want some 😋

[languid meteor]

Hi all, I'm starting a course on Galois Theory and I don't quite understand the motivation for separability

so far it just seems to be a 'nice-ness' condition but surely there's something deeper going on here

[next obsidian]

Honestly… there’s definitely deep stuff going on but I think just treating it as a technical “niceness” condition for the moment is fine

I think you get more out of understanding separability by exploring what happens when you have some inseparability and see how some theorems fail. The examples I can think of are mainly really deep results though

It may also help to look into the seprable degree, very briefly you can measure the degree of the separable part of a field extension in terms of the number of embedding into an algebraic closure which fix the base field

https://stacks.math.columbia.edu/tag/09HA

In a sense a separable extension has “the right number” of these

There’s also some issue when you start using derivations. A derivation is a generalization of the derivative, and it has the property that d(a^n) = na^n-1

If you have a non-separable extension L\K then there’s an element a in L such that a^p is in K but a is not, and then things become weird because d(a^p) = pa^p-1 = 0

In short, all sorts of things become really screwed up in weird ways but I think it’s hard to really get at why that is until you start doing some advanced stuff. Maybe someone else has better motivation for it

[languid meteor]

thats a great answer thank you so much! I really appreciate it

[next obsidian]

Oh there’s one final thing to mention

This requires knowing what the tensor product is but

If you have L a separable extension of k, then for any finite extension K of k you will have that L (x)_k K is a reduced ring

This becomes useful for technical reasons as well, reduced here means that no element is nilpotent

[languid meteor]

ah cool, im assuming this is an important construction? I see reduced rings are important in alg geometry

[next obsidian]

Yeah

For a lot of technical reasons again hahaha

It says among other things that this is a (non-irreducible) variety

A lot of these things become important deep into algebraic geometry for a lot of hard commutative algebra that underpins things like smoothness and non-singularity

For example this stuff is (probably) used in the process of showing that for classical varieties that smoothness is equivalent to non-singularity

This statement holds when you’re over a perfect field because over a perfect field every extension is separable and so smoothness becomes equivalent to something being a regular local ring which is what non-singularity means

But if you have an inseparable extension the two notions no longer agree

But this is the sort of stuff that’s contained really deep inside of EGA IV or in the end of a book on commutative algebra like Matsumura

[languid meteor]

thats so cool, I can't wait to see some proper algebraic geometry is seems so sick

it's quite incredible that separability enables so much

[next obsidian]

Its pretty remarkable. If you want an even better example of a seemingly innocuous condition enabling so much, you should hear about Noetherian

It’s actually ludicrous how much things change just from having the ascending chain condition

[barren sierra]

aight so like

I have found how to generate elements of the form (0, b)

that is easy

that's just (0, 1)^b

however I don't have that (1, 0)^a = (a, 0)

so like (1, 0) * (1, 0) = (2, 1)

(1, 0)^4 = (4, 6)

etc

I can't actually nail down a closed form for that second component (since I can just subtract that away as needed if I have a closed form)

it doesn't actually matter, I can black box the problem away

however I would like to find a closed form

[delicate orchid]

you have (0, b) for arbitrary values of b
subtract (0, b) from whatever you get for x(1, 0) to leave you with just (x, 0)

[barren sierra]

well yea

[delicate orchid]

there we go

[barren sierra]

that's what I mean by black boxing the problem away

I was just curious if anyone knew of a closed form

[delicate orchid]

oh like a specific formula

hmm

smells "triangular number-y"

compute up to (1, 0)^10 and then put the 2nd components into the OEIS and see what comes out

[barren sierra]

oh yea that site exists

I'll do that

[delicate orchid]

$a_0 = a_1 = 1 \
a_{n+2} = a_{n+1}+a_n$ \newline
$b_n = \prod_{i=0}^n a_i$

I think

[cloud walrus]

GM Wew Lads Tbh

[barren sierra]

a_n+2 = a_n+1 * a_n

not + a_n

right?

[delicate orchid]

I don't think so

the first term is a sum

the only product is going on in the 2nd terms, which is why b is a product

[barren sierra]

no but for the product in second term

yea

[delicate orchid]

ah wait y_0 y_1 aren't always 0

[barren sierra]

yea

[delicate orchid]

I knew I missed something

[barren sierra]

also that

it seems like a pain in the ass lol

[delicate orchid]

yeah it's definitely non-trivial

hopefully oeis will come in clutch

[barren sierra]

this problem sucks tho

[delicate orchid]

idk I think it's alright

[barren sierra]

actually this is worse

no I mean like showing it's a group

particularly associativity

it's not hard, just very tedious to write out

[delicate orchid]

oh yeah associativity always sucks

[barren sierra]

finding generators is whatever

[delicate orchid]

it's literally the same thing but just... again

[barren sierra]

this is even easier

cause (1, 0, 0)^a = (a, 0, 0) and (0, 1, 0)^b = (0, b, 0) and (0, 0, 1)^c = (0, 0, c)

and (a, b, c) = (a, 0, 0) (0, b, 0) (0, 0, c)

but just a pain to write out all of the working

[delicate orchid]

at that point just construct an isomorphism to Z^3

[barren sierra]

hm

well wouldn't defining the obvious isomorphism just reduce to showing all this anyways

[delicate orchid]

maybe

actually, yes

yes it would

[barren sierra]

yea

cause I need to show it's a group

and the isomorphism is surjective, and injective

so it's the same anyways

[delicate orchid]

fuck surjective and injective just find an inverse

[barren sierra]

sure whatever

either way

[delicate orchid]

I'm team inverse

[barren sierra]

I'm team easier

[delicate orchid]

a flip flopper... gross...

[barren sierra]

growth mindset

[delicate orchid]

although your funny group has got me thinkin

it's got me thonkin

[barren sierra]

See I'm a believer in the fact that problems should teach something and focus on something useful

and like

I do not see a point at this level asking people to prove something forms a group

when the main value of the problem is finding the generators and invarient factor form

[delicate orchid]

I agree with you tbh

especially invarient factor form

[barren sierra]

like if this were an intro course

sure

but like this is a grad level class

[delicate orchid]

I can barely even understand what invarient factor form is

[barren sierra]

there's a neat algorithm for it if you're given a set of relations for the generators

[delicate orchid]

that's where you write it as the ol

Z/pZ product right?

[barren sierra]

yea

[delicate orchid]

for p prime?

[barren sierra]

wait

no

that's elementary decomposition

[delicate orchid]

no like

[barren sierra]

you write it as Z/n_1 Z x Z / n_2 Z ... x Z / n_m Z x Z^r

[delicate orchid]

ok say you've got G = Z/n_1Z x Z/n_2Z

one min lemme explain what I think it is

[barren sierra]

ok

[delicate orchid]

so yeah this follows from the structure theorem for abelian groups

and then you write all the ways you can decompose this into Z/pZ for primes p

[barren sierra]

yea but that form is invariant factor form I thought

and then further decomposition was called elementary decomposition

or I think primary decomposition

[delicate orchid]

hmm

[barren sierra]

idk it's just names

[delicate orchid]

I shall google it

[barren sierra]

point is that's the meat of the problem

not

"hi plz show this is associative plz"

cause it's so annoying, especially for that second one

[delicate orchid]

ONLY time I will accept a "associative" proof is proving the tensor product is associative via the universal property

[barren sierra]

via the universal property
again

more interesting

and more useful I presume (idk what universal property is)

but this is just computation go br

[delicate orchid]

ok so in this example they're looking at G = Z/10Z x Z/12Z x Z/30Z

I don't know how you do the ? step

but this is kinda what I had in my head

[barren sierra]

I'm sure there's a nice algorithmic way to do it

but with all the primes you just kinda

shuffle around till it works

[delicate orchid]

hmm

[barren sierra]

I think

[delicate orchid]

ah I see

the n+1th element needs to be a multiple of the nth element

good refresher

[barren sierra]

https://en.wikipedia.org/wiki/Smith_normal_form#Algorithm

you can use this

which is interesting

[delicate orchid]

🚪 🚶‍♂️

[barren sierra]

http://eg.bucknell.edu/~zls002/Math503Sp16/Notes/InvariantFactors.pdf

[barren sierra]

yea lol

had to do this on a final last semester

[delicate orchid]

oh I can definitely see how this applies

how you actually compute it, I'm not sure

but I can see the connection

[long obsidian]

Where may I ask a question about representation theory, here?

[delicate orchid]

here is fine

[long obsidian]

I might be looking too deep into the meaning of the word "representation" but is there any interpretation of the singular value decomposition in terms of representation theory?

I feel that the Fourier transform is in the family of unitary representations. Yet the singular value decomposition is also related to unitary tranaformations

Pardon me if this is an ill-defined question

[delicate orchid]

I suppose you could view the singular value decomposition as the "character" of a non-square representation

but I don't know how useful of a concept this is

especially for group representations, where the image of any group element has to be invertible

as for the fourier transform, I have no idea

[long obsidian]

Hmmm I've never heard of that term 'character' before thank you so much.

[delicate orchid]

character theory is a big thing

and a very cool thing

[pastel cliff]

what if YOU

wanted to be HAPPY and LEARN

but GOD SAID

i'll stop bitching at some point but that point is an undetermined distance away from now

[delicate orchid]

sick of this guy

[pastel cliff]

me or my prof

[delicate orchid]

I'll leave it up to the fans to decide

[pastel cliff]

understandable have a good day

[desert dome]

Hi, I remember in cyclotomic polynomial, there is something like "if prime p divides a^n - 1, p does not divide a" or something like that. Could anyone point me where to look it up or explain why it's true?

[next obsidian]

You would get that p divides 1

[chilly ocean]

to go even further, write (a^n-1) + a(a^{n-1}) = 1 and pull out their common factor of p and you would have pk = 1 and that's pretty sussy

[next obsidian]

Isn’t that what the definition of p dividing 1 is

[chilly ocean]

yeah just writing it explicitly

[next obsidian]

I see

[desert dome]

Sorry I'm still confused. Why (a^n-1) + a(a^{n-1}) = 1? and why p divides 1? doesn't that mean p = 1

[next obsidian]

Reduce mod p

a^n - 1 = 0 mod p

But also a^n = 0 mod p

So you get that -1 = 0 mod p which means p divides -1

Which is the same as p dividing 1

[desert dome]

oh I see!

Thank you

[next obsidian]

[desert dome]

[desert dome]

Hi, I am working on this problem. I proved the first half, but had a hard time figuring out the later half. I assume a^m = 1 mod p for contradiction, so a^n - a^m = 0 mod p. Then p | a^m or p | (a^{n-m} - 1). p|a^m is not possible because if so, p|a^n. I dunno how to show p|(a^{n-m} - 1) is not possible and how to use (p,n) = 1. Any hint would be appreciated

[barren sierra]

as is tradition

I have done the easy part of this problem

and am stuck on finding the invarient factor form

I need the elements of finite order I guess?

i.e. find G_tors

I got that (1, 0) and (0, 1) generate the whole group but does that necessarily mean that Z^2 is the invariant factor form?

my textbook doesn't talk about how to compute this really for infinite groups

at least I can't find it

and my prof's notes only consider finite groups

[hot lake]

Well you could check if those two generators have finite order or not

Though that wouldn't be enough

[barren sierra]

have what?

[hot lake]

Finite

[barren sierra]

ah

they do not

that's easily shown

but yea that's not enough I figured

[hot lake]

Then either show that there is no element of finite order

[barren sierra]

I know that G is isomorphic to G_tors x Z^r

[hot lake]

Or find an isomorphism to Z^2

[barren sierra]

aight

I might do the latter I guess

[hot lake]

Is pretty easy to show that if x has finite order then it's (0,0)

[barren sierra]

Yea exactly

[hot lake]

Also you would need to show is not isomorphic to Z

[barren sierra]

Hm true

[tall jay]

So,I’m trying to prove this statement and I’m having trouble choosing the x* for the surjective proof. Need some help. Also, is the proof okay so far?

[barren sierra]

Why choose x*

why not construct it?

injectivity proof looks good tho

(don't forget to show that h o g or is indeed an isomorphism, not just a bijection)

[tall jay]

Any advice on how I can construct?

Yeah, I think I'll do it for my last part

[barren sierra]

use the fact that h is surjective

to determine that there is some x such that h(x) = y

and then go from there

[tall jay]

Nice, thank you

[oblique leaf]

In the universal property of products, does it mean that there exists a unique f given a family of fi and pi or that there exists a unique f for each individual fi-pi combination?

[hidden haven]

pi are fixed, and there is a unique f for each family of f_i

pi are part of what a product is

[marsh fossil]

i'm really stuck on part d, proving the new group law is associative, i've tried writing O as the O' * (O' * O) and just general manipulating but i'm not seeing the trick

[barren sierra]

well you know what the end result you want for associativity is

so take that

and work backwards

and meet in the middle with what you have currently

[chilly ocean]

guessing game

this isn't me asking for help in disguise

i have in mind a nontrivial non simple group that's its own only homomorphic image

I actually can't think of 2

this is the only one I can think of

so the guessing game is guess the group

[marsh fossil]

Thanks for the suggestion @barren sierra but I’ve been just sitting here playing with symbols for about 4 hours and can’t get it so if you have any other hints I’d be appreciative

[chilly ocean]

What does Inn(G) mean

[hidden haven]

inner automorphisms of G

[next obsidian]

Sorry, this doesn’t make sense to me really?

If G is a group, you can just grab like, a product of Z so large that G x Z^alpha isn’t isomorphic to G, and then G is the image of the projection map G x Z^alpha -> G

If you mean that G is such that the image of G under any homomorphism is either G or is trivial, then any simple group satisfies this simply because its image must be G/N for N a normal subgroup, and the only options for N are then G or {e}

[simple mulch]

So we're all acquainted with the finite dimensional vector space L^2(G) for G a finite group

[cloud walrus]

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

[hidden haven]

Sorry I'm not acquainted with that vector space

[lethal dune]

yes but deltas are not the basis tho

[simple mulch]

hum

$L^2(G) = {f : G \rightarrow C}$

[cloud walrus]

mns

[simple mulch]

where f is any complex valued function on G

Let me correct the notation to $\delta_t(s)$

[cloud walrus]

mns

[simple mulch]

So $f(s) = \sum_{t \in G}\delta_t(s)f(t)$

right?

yes

[cloud walrus]

mns

[simple mulch]

+1

and it has a basis and an inner product as well as a norm

So, I have some questions

That makes L^2(G) a finite dimensional vector space with basis ${\delta_t(s)}_{s \in G}$

[cloud walrus]

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

[simple mulch]

Lets say we want to prove the inverse fourier transform

We could do all the calculations on the right side but it is tedious

So we want to show the right hand side is linear on L^2(G)

And that's the case if we can write it as linear combination of elements in the base

namely the delta functions

So if we make f(s) = delta_t(s) we must show that delta_t(s) equals the right side

I mean, in fact we already know the right side is linear since everything in there is linear

[simple mulch]

I have doubts in here. Why can we make this equal? Shouldn't it be $f(s) = \sum_{t \in G} f(t)\delta_t(s)$?

[cloud walrus]

mns

[lethal dune]

is L² not the square integrable function on open set G?

[lethal dune]

well I tried scrolling up

[simple mulch]

I guess?

why we can make f(s) = delta_t(s)

here

why

we can make f(s) = delta_t(s)

[cloud walrus]

Todd

Todd

[simple mulch]

ok

cause the right side is linear?

[cloud walrus]

Todd

[simple mulch]

yes

L^2(G) is linear

I mean, it has linearity property for any function in it

what you wrote just now, isn't it linearity?

yes

Proceed please, sorry

[cloud walrus]

Todd

[simple mulch]

yep

[cloud walrus]

Todd

[simple mulch]

hum

I agree with the last statement

but

[simple mulch]

this

It is like we're proving for Pdelta_t(s) ?

I am not understanding.. I might be overcomplicating things sorry

yes

yes, so we want to prove the formula for delta

oh

I see

and indeed, for t != s we have 0 on both sides

[cloud walrus]

Todd