groups-rings-fields

[void cosmos]

if u can write (12) as powers of these two together then im done

cuz we know (1234) and (12) generate S_4

(1234) is obv in <1234,1243>

just need (12)

but how

i couldn't write it out

[delicate orchid]

I can get (14), (23), (24), and (13)

the rank of S_4 as a reflection group is 3

[void cosmos]

yeah

i see

ok bro thank u for confirmign

[delicate orchid]

np

[tardy hedge]

“This correspondence commutes with the processes of taking sums and intersections” what does that mean exactly?

[nimble folio]

I asked this question in discrete math but they're talking alot about graph theory and other stuff

If $f:A \rightarrow B$ is a function, when is it true that there is a bijection $A \rightarrow \text{im}f$.

[cloud walrus]

clubsoda14

[warm ember]

when f is injective

assuming you mean "when is it true that f: A-> Im f is a bijection"

[nimble folio]

Yes

Thank you!

This was just a clarifying question

[sly crescent]

What is the rank of D6 as a reflection group?

Is it 2 (because D6 is G2) or 3 (because D6 is A1A2)?

[topaz solar]

If you sum submodules or intersect them, it passes outside applying this isomorphism

Like f(a+b) = f(a)+f(b)

So f and + commute

[wicked patio]

Whoa, what program is this?

[mighty kiln]

Chatgpt 6.0

[somber bluff]

Can someone explain part 1.8.3

i get 1.8.2 but somethings off with 1.8.3

[topaz solar]

it turns it into a relation based on subsets of where x_i takes values

since for any $A\subset X_i$ we have the relation $\forall_{x_i\in A_i} .\rho$

[cloud walrus]

Sharp, secret keeper

[somber bluff]

i dont understand how the function is defined. x_1, ..., A_i, ..., x_n isnt an element of Rel(X_1, ..., X_n) right

oh i get it

[topaz solar]

yeah that's a bit odd, but that's the relation rule

[somber bluff]

the defining part is defining the output

[topaz solar]

it's a bit scuffed

yes

[somber bluff]

yeah notation issues

i see

normally it is input |-> output right?

but this case is like input of output |-> output of output right?

[topaz solar]

yeah its the rule of the thing spat out

I think

[trail cave]

I'm having a little trouble with problem 3 and problem 4b here, if someone can give guidance.
In problem 3, I know |G| = n for finite n. I thought about maybe treating G as a permutation group somehow and use k-cycles?
In problem 4b, I have my work on the right. When trying to find the cycle for 4, it seems like it maps back to 1, but I already have a cycle containing 1 that maps 8 to it

[somber bluff]

wait then how is the actual all^i function defined. As in what maps input from Rel(X_1, ..., X_n) to Rel(X_1, .., PX_i, ..., X_n)

[topaz solar]

It sends rho to the thing that acts as they describe

[somber bluff]

oh yeah im stupid

[warm ember]

for p3, consider the elements e, a, a^2, ..., a^n where n=|G|

[trail cave]

I'm thinking there's some way to prove that an infinite order element must map back to an element it contains as a factor somehow?

since G is finite

[warm ember]

two of these are equal

and youre done

[trail cave]

does it use pigeonhole somehow?

[topaz solar]

yes

how many points are in that list

[warm ember]

n+1 values, at most n distinct values

[topaz solar]

dont just give the answer though

[warm ember]

mb

[crystal vale]

What are the characterizations of all group homomorphisms from R to R? One is the form of ax but what about others ?

[trail cave]

okay so, e, a, a^2, ..., a^n is a list of n+1 elements in G, but G is cardinality n. Hence there exists some a^i = a^j for distinct i, j. WLOG j > i, s.t. a^i = a^h * a^i for h = j - i. Then a^h is an identity element (since argument works from both sides), so a is an identity element. Is this the right track?

[warm ember]

a^h is identity already finishes

[trail cave]

oh right

because then a power of a maps to an identity element

and a is arbitrary so it holds for all elements of G, thank you

[warm ember]

only ones

[crystal vale]

Yes but choice of h is dependent on a

[warm ember]

if R->R is homomorphism then R/ker is in R

[crystal vale]

But how can we characterize it ?

[crystal vale]

Yes

[trail cave]

Is there anything someone might say about the 4b problem? that's the last one I need and I'm not sure I see why I'm getting a repeat of 1

[warm ember]

if ker is nontrivial then cyclic group is in R, absurd

so ker is trivial

ie homomorphism is injection

or

[topaz solar]

Ok so

[warm ember]

why did i do that just consider image of 1

[crystal vale]

R/ker is cyclic, how ?

[topaz solar]

@crystal vale

[warm ember]

^

1->a

[topaz solar]

Is this R, the reals, as an abelian group?

[crystal vale]

R is real

[warm ember]

o

ok what am i doing i thought it said Z

[topaz solar]

Is this as an abelian group, as a ring, or what?

[crystal vale]

Just abelian group

[warm ember]

if its R then there are many homomorphisms?

[crystal vale]

Yes

But characterization

[warm ember]

R is a Q-vector space

[topaz solar]

So, uhhh

You don’t

[crystal vale]

Yes

[warm ember]

thats cauchy functional equation

[crystal vale]

Yes

And f(q) = qf(1) for all q in Q

[topaz solar]

R is a Q-vector space, so take a basis

This will be uncountably many points

[crystal vale]

Yes

[topaz solar]

And you can assign any of them to any arbitrary values

[crystal vale]

R is not finite dimensional over Q

[crystal vale]

Yes

[topaz solar]

So there is no nice characterization

[crystal vale]

I see

[topaz solar]

As long as you accept enough choice to get that basis anyway

Basically, imagine taking x |-> ax and shuffling it like crazy based on a choice of basis and then permuting them

[crystal vale]

Oh

[topaz solar]

So now you have stuff indexed by Sym(2^omega)

Which is uh

Very large

Not to mention choice of basis or wtv but that’s a bit beside the point

And this won’t describe all the possible maps either

[crystal vale]

Thank you

[crystal vale]

If G is a group and H_1 and H_2 are subgroups then any right coset of H_1 intersection H_2 can be written as intersection of right Cosets of H_1 and H_2, right?

Let r(H_1 intersection H_2) then I can write r(H_1 and H_2 ) = rH_1 and rH_2, right?

Let G be a group, and A a normal abelian subgroup. Show that G/A operates on A by Conjugation.

I don't want a hint, but I am not sure about what it means by conjugation here because if I define gH•a = gag^(-1) then it is group action, right?

[warm ember]

yes

[warm ember]

yes, but replace H with A

[crystal vale]

Yes

[dull ginkgo]

Yep pretty much.

If it helps, if we have a group G acting on a set S, we can also let G act on S’s power set, by sending a subset S to its orbit xS.

This is pretty nice, because in order to consider the action restricted to subsets, we’d need elements such that xA =A (to maintain closure on subset A), luckily it allows us to assert that the subgroup Stab(A) (of the power set action) acts on A :)

Similar idea can be used here, you just need to “quotient out the redundancy” of the action, I.e elements of the group where g • x = x for any x, I.e it stabilizes the whole group

[cloud lynx]

when is a field extension with characteristic p>0 separable?

its kinda confusing

[dull ginkgo]

Usually separable is in the context of an extension

[cloud lynx]

yes right sry I meant field extension

[dull ginkgo]

Well it’s inherited from seperable polynomials, do you understand what that means

[dull ginkgo]

Here’s a subproblem. Assume f(x) is a polynomial over F where F is a field of characteristic p, and has 3 roots over field H containing F. How many roots does f(x^p) have over H?

Or better yet, how would we determine how many roots it has

[rocky cloak]

Well the definition is just that the minimal polynomial of any element is seperable.

But an equivalent characterization is that for any element in the base field that has a pth root, that root is already in the base field.

[topaz solar]

Feels very pure-ish

[winged void]

@rocky cloak Can you check my solution please

because i solved it using another method but im not so sure

[rocky cloak]

What's the problem/solution?

[white oxide]

0?

[chilly ocean]

Yep

That's the only one

[winged void]

the problem is that a ring with 10 elemnts must be commutative

@rocky cloak

[white oxide]

so the closure of an infinite subset of Prime(Z) would be Prime(Z)

[winged void]

this is the solution that i had

[white oxide]

because it's intersection would be the 0 ideal

which is a subset of every prime ideal

so the sets which are not closed are those which are infinite and those which contain 0

[winged void]

Suppose that $R$ is a ring consisting of $10$ elements. Prove that $R$ is commutative.
\begin{proof}
We will prove using the \textbf{Fundamental theorem of finite abelian groups} and the theorem \textbf{Cauchy} that it is a commutative ring. which implies that:
\begin{equation*}
ab = ba \quad \forall a,b \in R
\end{equation*}
From \textbf{Fundamental theorem of abelian groups} we know that $(R,0,+)$ must be isomorphic to $Z/5Z \times Z/2Z$. Using \textbf{Cauchy theorem} there is an $x\in R$ with $order(x)= 2$ , $order(x) = 5$. Furthermore, we know that every element in $R$ must be a divisor of the order of the size of the group and since $(R,0, +)$ has ten elements. This means that the order of each $x \in R$ can be either $2, 5, 10$. Now suppose that $ord(1) = 2$ , then this means:
[
1 + 1 = 0
]
then
[
a+a = a(1+1) = a \cdot 0 = 0 \quad \forall a \in R
]
This is a contradiction because every element in $R$ now has an order max of $2$ but that is a contradiction. Now suppose that $order(1) = 5$ then this means:
[
1 + 1 + 1 + 1 + 1 = 0
]
then
[
a+a+a+a+a = a(1+1+1+1+1) = a \cdot 0 = 0 \quad \forall a \in R
]
This leads to a contradiction because $2 \nmid order(1) = 5$. So there is no element with $order(x) = 2$ which contradicts \textbf{cauchy theorem}. This means that $order(1) must be 10$. Now we know that each element must have max order $10$. So we can certainly have $a \in R$ if $a = 1 + 1 + \cdot 1$ ​​where the number of ones is not greater than $10$. From here we can conclude that $R$ is commutatively true for every $a,b \in R$
\begin{align*}
ab &= (1+1+ \cdots + 1)b \
&= b+ b+ \cdots + b \
&= b(1+1 + \cdots + b) \
&= ba
\end{align*}
\end{proof}

[cloud walrus]

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

[winged void]

@rocky cloak this is my proof

[rocky cloak]

Looks good

[white oxide]

so the open subsets would be infinite sets, or sets containing 0?

since the closed sets are finite sets not containing 0

[chilly ocean]

There are other closed sets

[white oxide]

besides prime(z) and empty set

[chilly ocean]

There are other ones

[crystal vale]

Any hint? If | G | = p^n and H ≠ G be a subgroup of G then [ N(H) : H ] > 1.

[chilly ocean]

Uhh wait I'm sorry. The closed ones are finite ones not containing (0) and the whole space, yes

[white oxide]

nwnw

thanks!

[delicate orchid]

split it into when Z(G) is a subgroup H and when it isn't, if Z(G) isn't in H then HZ(G) normalises H and we are done because centres of p-groups are nontrivial. For the other case, try using induction on the order of G to prove it (hint: ||G/Z(G) is a strictly smaller group||)

[cloud lynx]

If Char K=p>0 and f is irreducible in Fp[X] then its separable if the degree of f is not divisible by p. But then why is x^2+x+1 over F2 is separable?

I mean the degree is divisible by 2

but the Differential is 1

!=0

is the statement wrong?

[frail summit]

Hey. There's this exercise in D&F which I don't understand: Let K be a finite extension of F. Prove that K is a splitting field over F if and only if every irreducible polynomial in F[x] that has a root in K splits completely in K[x].

By "K is a splitting field over F", do they mean K is a normal extension?

[delicate orchid]

that's a good question, your usually have a splitting field of a polynomial or set of polynomials

[frail summit]

Yeah

[rocky cloak]

It's not an if and only if.

[delicate orchid]

the result is true for normal extenstions so just prove it for those

[frail summit]

Even if the collection of polynomials is empty, or say, trivial?

[cloud lynx]

I know

[rocky cloak]

So what is the confusion?

Everything you wrote is correct yes

[topaz solar]

If it’s irreducible over F and has a root in F…

[delicate orchid]

yeah this basically

I was writing a whole paragraph

[cloud lynx]

Oh bruh now i understand it

Every irred. polynomial which is not divisible by char p>0 is separable but this does not imply that every irred polynomial is separable if the degree is divisible

My reading skills are bad lol

[frail summit]

Oh. So essentially if K is an extension over F which doesn't split any more polynomials, then it is the splitting field over F of the linear factors only and hence must be isomorphic to F by uniqueness of splitting fields?

Cause then being irreducible and having a root implies it being linear, as you indicated

Makes sense

Thanks!

[void cosmos]

hello

suppose you have the character table for a group

can u know how many elements in the conjugacy class itself?

like the cardinality of the equivalence class

[rocky cloak]

If you set up the character table with columns indexed by conjugacy classes, then the rows are orthogonal wrt to inner product weighted by the size of the conjugacy classes. So you can set up the system and solve for that.

For example the character table of S3 is
1 1 1
1 1 -1
2 -1 0
The first column is the identity, let's call the size of the other two classes a and b. Then the orthogonality of the rows says
1 + a - b = 0
2 - a = 0
-> a=2, b=3

[void cosmos]

okay so im new to this so let's like be slow.

i know that the rows are orthogonal to each other and the columns as well by schurs orthgonality

and ik that the number of irred chars is the number of conjugacy classes

so now how exactly did u get these equations ?

[rocky cloak]

It's the inner product of the first row with the other two

(sat equal to 0)

[void cosmos]

okay wait let me show u the problem

for context

now

[delicate orchid]

jagr you want column orthogonality for this

I'm guessing you're trying to get the size of the centraliser out right?

[rocky cloak]

Well, here you're already given most of the conjugacy classes. So just take the order of the group and subtract what you have.

[void cosmos]

i dont have the order of the group do i

[delicate orchid]

it's S_4

[rocky cloak]

It's the square length of the first column

[delicate orchid]

1+1+4+4+9+9

[void cosmos]

okaay i think this is like similar to the square of the degrees of irreps

is that correct

what ur syaing

[delicate orchid]

it's exactly that

[void cosmos]

just a disclimer im going to be ultra stupid so

okay so

why is the first column

[rocky cloak]

Not just similar

[void cosmos]

th edegrees

the degrees

[rocky cloak]

It's what it is

[delicate orchid]

because it's the largest number?

the character values are bounded by the degree

[void cosmos]

why

[delicate orchid]

you literally just said

[void cosmos]

yes but why the fuck is the first column the degree of the reps

oh lmao

trace(1) u mean?

or like trace(phi_e) = trace(I)

[delicate orchid]

sure

[void cosmos]

= dim(V)?

[delicate orchid]

I mean the degree of the character

I should've actually read the image posted lol it's completely unrelated to determining conjugacy class sizes from the character table

[rocky cloak]

Well, there are different ways to see it. But the size of the group is the dimension of the group algebra.

And the group algebra breaks up as a direct sum of all the irreducible modules. Then by looking at Hom(kG, V) you can see that the multiplicity of V is equal to its dimension. So dim kG = sum (dim V)^2

[void cosmos]

yes

[delicate orchid]

or you just take inner products and it's immediate

[void cosmos]

i see

[rocky cloak]

And trace of the identity is indeed the dimension of the space

[void cosmos]

yes

why would b follow then

i can do c)

[delicate orchid]

you have a non-trivial linear character with a kernel

[void cosmos]

and i can indeed prove that the tensor product of the reps are the reps that have products of characters

[delicate orchid]

this kernel is a normal subgroup, count the elements in it using the sizes of the conjugacy classes given

[rocky cloak]

Or even easier, just noticed that the image is {±1}

[delicate orchid]

ah true

[delicate orchid]

I'd also like to point out how you would find conjugacy class sizes from the table, $g \in G$ then $\sum_{\chi \in \text{Irr}(G)} \chi(g)\chi(g^{-1}) = |C_G(g)|$, and then $|g^G| = [G : C_G(g)]$.

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

in case it's relevant to you in the future

[void cosmos]

i saee

[void cosmos]

oh yes fuck

why would it have two cosets tho im sorry

im just being as stupid as i can but like bare with me

[rocky cloak]

Well it's a group homomorphism onto {±1}

[void cosmos]

oh yeah lmao

ok

[void cosmos]

wait wdym exactly by like

a non-trivial character with a kernel

like any character that isn't X_0?

i mean there are lke

4

right?

wait

i should only consider X_1 correct?

[rocky cloak]

Gotta be 1d

[void cosmos]

like X_1?

[rocky cloak]

Yeah

[void cosmos]

and X_1 is only onto like

like X_1 is literally G-->{-1,1}

[rocky cloak]

Yup

[void cosmos]

okay i got it

sorry for being so stupid tysm

[delicate orchid]

there's only one with a non-trivial kernel. So unless you want to prove the trivial subgroup is normal I suggest looking at that one lol

[void cosmos]

yeah

ty

[arctic trail]

man I love finite group representation theory

[arctic trail]

ah my bad misinterpreted

Since the irreducibles form a basis of the functions ConjClassesOfG -> C, I assume you can associate a function to each conjugate class (1 if it's in it 0 if otherwise). By linear algebra obtain a decomposition in terms of the character table, and through the Inner product formula you obtain the size

[arctic trail]

it's like so clean

and doesn't deal with the all the yicky topology

(I love abstract harmonic analysis though, don't worry)

[delicate orchid]

yes you can decompose the indicator functions but I gave a cleaner way of doing it

[delicate orchid]

_ _

[arctic trail]

It's not like it's that hard to do either

let me see what you wrote

[arctic trail]

ah yeah

I forgot this formula existed

this is the correct way to do so

@delicate orchid why didn't you just write |xi(g)|^2

[delicate orchid]

I would never write that

here in particular it's to keep the connection to column orthogonality clear

[arctic trail]

weird

I think it's clearer

writing it like this

After all the inner product is given by <xi,zeta> = 1/|G| Sum(C conjugacy class) |C| xi(C) zetaConj(C)

Since the rows are orthogonal, the matrix is unitary and so the columns are as well

This doesn't involve g^-1 at all

unless you always write xi(g^-1) as opposed to its conjugate

[delicate orchid]

chi(g^-1) is the complex conjugate of chi

[arctic trail]

I know

I just think it's clearer to not write xi(g^-1) cause in the more general compact case it corresponds to the regular L^2 inner product

[delicate orchid]

also if you're going to write the inner product like that, replace |C|/|G| with the 1/|C_G(x)| and just sum over representatives it's much nicer

it took me 3 times to read the notation you're trying to convince me is clearer correctly

[arctic trail]

it's not LaTeX

$\sum_{\xi \in \text{Irr}} |\xi(C)|^2$

[cloud walrus]

Trivial Lemma

[arctic trail]

I'd write this

for a fixed conjugacy class C

or just replace C with g

plus you can write

$\xi(g) \overline{\zeta(g)} = (\xi \otimes \bar{\zeta})(g)$

[cloud walrus]

Trivial Lemma

[delicate orchid]

I would never write that either

[arctic trail]

That's not what I mean

The thing is that the object is also a character of a representation

[delicate orchid]

there's no tensor product here you're multiplying functions

I know it's the tensor product of a representation

but I don't care

[arctic trail]

:/

don't care in peace I guess

[delicate orchid]

\eta\bar{\zeta}(g) is fine if you want

[delicate orchid]

we've literally spent the past 354 hours debating minutia and that's where u draw the line?

[arctic trail]

I dunno, I just feel like it doesn't make sense to write g^-1

I don't see any advantage

over writing just the conjugate

[delicate orchid]

I had to type fewer characters

pun intended

[arctic trail]

if I wanted to share something with someone that's not familiar with the subject I'd prioritize better ways to write / think of something over saving characters

From the way I wrote it the identity

[delicate orchid]

they were not asked to perform this calculation, and "by column orthogonality" was already all of the required data

[arctic trail]

$\langle \xi, \eta \rangle = \langle \xi \otimes \bar\eta, 1\rangle$ becomes clear for example

[cloud walrus]

Trivial Lemma

[delicate orchid]

nah

also stop writing \otimes

[arctic trail]

no

[delicate orchid]

yup

[arctic trail]

I won't

[delicate orchid]

do it

[arctic trail]

no

[delicate orchid]

yes

[arctic trail]

no

[delicate orchid]

pretty please

[arctic trail]

let me see if Fulton harris uses the notation

if it doesn't I'll stop

[delicate orchid]

fulton whorris?

[prisma ibex]

can someone please summarize what we are bitching about I'm not reading this yapfest

[delicate orchid]

also that book was written 234 years ago

[spice whale]

i think there's an argument as to whether to denote the product of two characters as cdot or otimes

[delicate orchid]

James Liebeck, Navarro, Isaac's don't use it

[cobalt heath]

Does a character involve tensor product

[prisma ibex]

both are fine, it's also fine to just concatenate them with no symbol between

[prisma ibex]

yes in a really trivial way

[arctic trail]

no wait, it doesn't but a professor of mine uses it so Imma keep it

[delicate orchid]

ok I was being facetious with how much I disagree with using \otimes but using \cdot is deranged

[prisma ibex]

yeah \cdot is probably the least standard

I don't think I've ever seen it to be honest

[cloud walrus]

Absta

[spice whale]

ok well by this i meant just

normal multiplication

absta what

[prisma ibex]

I think it's generally best to write this as a concatenation, and to reserve the tensor product for something like an exterior tensor product

[delicate orchid]

where \simeq denotes Morita equivalence, I'm sure

[arctic trail]

in abelian* groups I've written + for the product of characters

[delicate orchid]

my face just dropped from my goofy grin down to the most serious stare I've ever pulled

you do what

[arctic trail]

there's a good reason

[delicate orchid]

the ring of characters is still a ring even if those characters are of an abelian group

[cobalt heath]

I forgor if this is true

[elder wave]

bro locked in

[arctic trail]

cause it forms a module essentially

[delicate orchid]

I understand

that an abelian group

is iso to Irr(A)

[coral spindle]

I've also seen \boxtimes for the exterior one, just in case there's confusion

Me likey

[spice whale]

yes?

[delicate orchid]

but that isomorphim embedds it as a multiplicative subgroup, of R(A)

[prisma ibex]

yeah

[spice whale]

1 times 1 is indeed 1

[delicate orchid]

not an additive one

[spice whale]

actually maybe not

[cloud walrus]

Absta

[arctic trail]

?

[delicate orchid]

squensor product...

[delicate orchid]

R(A) being the ring of characters of A

[cobalt heath]

What is concatenation, is it just

$\eta \xi$

[arctic trail]

Yeah, but what I'm saying is unrelated to the ring of characters

[cloud walrus]

Absta

[prisma ibex]

yeah

just write them next to each other

[cobalt heath]

Convenient and correct

[spice whale]

according to terrence howard $\bC \otimes_\bC \bC \cong \bC^2$

[cloud walrus]

hot girl ({ω|-ω})

[arctic trail]

It gives a non-trivial module structure to the irreducible characters of an abelian group

[delicate orchid]

as an A-module?

[prisma ibex]

like idk suppose I have a representation of GL_2 x GL_2 which is an exterior tensor product of some representations of GL_2, each twisted by a different character

[arctic trail]

what is A?

[delicate orchid]

an abelian group I've used it like 4 times already

[prisma ibex]

the most sensible way to write this is probably like

$\chi_1\pi_1\boxtimes\chi_2\pi_2$

[delicate orchid]

I didn't think so, so as a module over what ring

[cloud walrus]

nGroupoid

[prisma ibex]

replace \boxtimes with \otimes if you like

[arctic trail]

wait I'm confused about what you mean

[prisma ibex]

but writing something like

[arctic trail]

but I have a terrible stomach ache

[delicate orchid]

the irreducible characters are a module over what ring

[prisma ibex]

$(\chi_1\otimes\pi_1)\boxtimes(\chi_2\otimes\pi_2)$

is kind of annoying to write

[cloud walrus]

nGroupoid

[coral spindle]

Also I want to write things like the action of Lin(G) on Irr(G) nicely

just by concatenation

[delicate orchid]

this isn't loaded btw this is genuinely new to me unless the ring is Z

[arctic trail]

ehttps://groupprops.subwiki.org/wiki/Supercharacter_theory_of_an_algebra_group err it's related to this but I gotta brb

[delicate orchid]

lets have a peep

[prisma ibex]

I think between two sections of my thesis I forgot to pick a consistent convention of writing character twists on the left or right, although it doesn't matter a whole ton

[coral spindle]

Incomprehensible; may g-d have mercy on your wretched soul

[delicate orchid]

I've encoutnered supercharacter theories before

first time I've ever seen them actually be used though

[cobalt heath]

,nlab exterior tensor product

[cloud walrus]

No results found at:
https://ncatlab.org/nlab/search?query=exterior+tensor+product

[delicate orchid]

they're like blocks but with much weaker structure

[coral spindle]

If you have a rep of G and a rep of H, their tensor product is a rep of G x H

[cobalt heath]

But what is exterior tensor product?

[coral spindle]

But if you have two reps of G, you can also see their tensor product as another rep of G, just G.

So you want to distinguish between the two

[delicate orchid]

(via the diagonal action)

[cobalt heath]

Oh

[coral spindle]

There you go

So the first one is exterior

the second is interior

[cobalt heath]

Ahh, I see. So it is to distinguish two representations?

[coral spindle]

Yes

[cobalt heath]

I wish I would be learning these stuffs in my rep theory class

[coral spindle]

Which we've agreed to write $\chi \boxtimes \vartheta$ vs $\chi\vartheta$ in the case of characters

[delicate orchid]

idk what else you could be doing this is 101 stuff

[cloud walrus]

Boytjie

[cobalt heath]

Instead it is confusing idempotents handling and projective covers

[delicate orchid]

wtf is taht font

[delicate orchid]

oh are your fields arbitary characteristic

[coral spindle]

It's a clone of Palatino. I think the packages are newpxtext and newpxmath

[cobalt heath]

It is not my field, it is just a class

[delicate orchid]

doesn't matter actually. What we're saying is still true

brother

in christ

you take vector spaces to be over a field

what field are they over

[coral spindle]

h

Koerper

Corps

[cobalt heath]

Ah, likely arbitrary

[delicate orchid]

yeah that makes sense you need gross machinery

except you DON'T sucker!

it's still all character theory!

[cobalt heath]

Quiver is not character theory tho

[delicate orchid]

q-q-q--q-q-q-q--q-q-q-q--q-q-q-GULPq--q-q-q-q-q-quive-r-r-r-rr??!!??!

[coral spindle]

So you're not doing rep theory of groups

OK

[cobalt heath]

Damn, I forgor

[delicate orchid]

A1 reps

[cobalt heath]

Yeah, it was not a class about representation of group theory

[coral spindle]

Quivering in my groups boots rn

[delicate orchid]

Benson 1 has a great exposition on quiver reps. Chapter 4 iirc

[cobalt heath]

I heard the class will handle quantum representations

[coral spindle]

Quiver reps seem cool. I haven't looked into them that much I admit.

[arctic trail]

they have been used in like describing characters of unitriangular groups

[delicate orchid]

that sounds like a buzzword

[coral spindle]

Quantum groups: not groups, and not really quantum

[prisma ibex]

quantum groups are pretty standard

[delicate orchid]

I didn't doubt they had a use but it's nice to actually see one

[cobalt heath]

I mean (I heard) quantum rep is well-defined tho

[delicate orchid]

are they infinite

[prisma ibex]

[coral spindle]

They're not even groups wew

[arctic trail]

I haven't really seen quivers

are they like automaton?

[delicate orchid]

well now isn't this just woke

[coral spindle]

They're literally just digraphs

quivers are, I mean

[delicate orchid]

read the channel name :midelfintger:

well yeah quivers can stay

[prisma ibex]

quantum groups are defined through their categories of representations

[delicate orchid]

because uhhh Denzel lusting theory or whatever

[prisma ibex]

in a sense the quantum group is its category of representations

[cobalt heath]

This will be a suffering, I feel

[delicate orchid]

https://tenor.com/view/heath-ledger-and-here-we-cope-joker-gif-9467009

[coral spindle]

Woah this is cool

[arctic trail]

so many quivers

[prisma ibex]

it is yeah

[delicate orchid]

this is true of any category by Yoneda

[prisma ibex]

I took literally this course minus quantum groups and it's uhhh

have fun!

[delicate orchid]

(up to equivalence?)

[cobalt heath]

Wait what

[arctic trail]

ah bruh

[cobalt heath]

I guess I am right that it is insanely fast

[arctic trail]

quivers are quite literally multidigraphs

[prisma ibex]

honestly Auslander-Reiten stuff is the only kinda painful part

the earlier stuff is not bad at all

[cobalt heath]

Also it is my skill issue but welp

[delicate orchid]

quivers are literally some little arrows I draw sometimes

[cobalt heath]

Dunno, I am struggling with even at projective cover

[coral spindle]

Auslander--Reiten looks so sick though!!!!!

[prisma ibex]

it is really sick yeah

[arctic trail]

quiver is literally where I draw my commutative diagrams

[coral spindle]

Where's jagr to tell us how cool it is

[delicate orchid]

projective covers are kind of stinky

[delicate orchid]

give me my associated projective characters instead please!

[prisma ibex]

I also kinda stopped paying attention at that point in the course and computing basic examples of AR theory is a bit painful

quantum groups are fun though

[arctic trail]

man drawing commutative diagrams with tikz is awful

💀

[delicate orchid]

https://q.uiver.app/

[coral spindle]

Yooo Wew did you see that Chris Bowman is gonna be doing a MAGIC course on Lusztig's conjectures? Do you even care?

[arctic trail]

@delicate orchid

[prisma ibex]

people will really be like "I need to open quiver to draw a simple commutative square"

[delicate orchid]

oh I thought you said your commutative diagrams WERE quivers, which is true

[delicate orchid]

hmm... NO. No I don't think I do

I've secretly opened MAGIC in a browser tab

[coral spindle]

It's in the autumn semester

[arctic trail]

I had to learn tikz for drawing graphs it was awful 💀

doing analytical geometry

[cobalt heath]

@prisma ibex should I stop math if this feels challenging

[delicate orchid]

hahah good thing I'm not INCREDIBLY BUSY until novemeber

[arctic trail]

just to have nice spacing

[delicate orchid]

everything is hard until you've learnt it

[cobalt heath]

I mean I also put 2 hour each week into this class other than class time

But still, I am getting skill issued

[arctic trail]

Anyway, I'm off to bed.

Gotta love me some Nullstellensatz

semester starts monday, I'm cooked

[delicate orchid]

how explicitly do you need to work with projective covers here

[coral spindle]

Why are you asking someone else to tell you to quit math, I don't get this behaviour

[delicate orchid]

they want affirmation that it's ok for them to stay

and that struggling is normal

hence why I said what I said (I also said it because it's true)

[coral spindle]

Right, right ok

[delicate orchid]

anyway what the fuck is an injective hull

rhetorical question don't answer that

resetting my MAGIC password at 1am

truly unhinged behaviour

[coral spindle]

Aw shit it's 1am

[delicate orchid]

it's not there

[coral spindle]

Shit I meant the spring semester

It's right at the bottom of the list

He links a preprint of his book on diagram algebras, pretty amazing tbh

[delicate orchid]

the one above it looks more appealing

I think I need to take this one as well... been meaning to learn this stuff

[rotund aurora]

Character theory is nice in particular

[delicate orchid]

so I've heard!

[coral spindle]

This course sucks actually

Well I'm being mean. It's just very introductory. They do very little.

[delicate orchid]

my thesis be like

[arctic trail]

Its sooooo nice

[cobalt heath]

I am sorry

[cobalt heath]

Maybe not that explicitly, I just expect it to jumpscare time to time

[vapid vale]

its so cool!

[hollow tartan]

Is this for ug? I wish I had taken it for ug for a stronger app to grad school but the school didn't offer it I want to learn it unsure to get lorenz book btw what is character theory?

[vapid vale]

what is MAGIC

oh whoa

holy crap thats so cool

[delicate orchid]

By Boytjie’s description of that particular course, yes

[hollow tartan]

Wow it's like a website that does online courses?

[delicate orchid]

Yur

[vapid vale]

thats a rather fair price too

[hollow tartan]

I already signed up lol

[toxic sapphire]

hey so I'm trying to show that the group of rigid motions in R^3 of a dodecahedron is 60.

I think I understand one good method: any rigid motion maps a given vertex (say, 1) to one of the 20 total vertices.

then, vertex 2 will have to be one of the three adjacent vertices, giving 3 * 20 = 60 total rigid motions

it's intuitive that the motion is uniquely determined by the first two vertex mappings, but how do I show this in a proof?

the idea is presented as an extension to finding the order of dihedral groups, which are more obviously determined by the first two vertex mappings

maybe another way to find this solution is to find some proper way to label the vertices of polyhedra, which then forces a structure that would be uniquely determined, since that's what makes the 2d analogue obvious

any hints would be appreciated

[crystal vale]

I don't get it operation here, is it x^(phi(h)(x_2) ) or xphi(h(x_2)) ?

[vapid vale]

i personally think this is fine to just say - if you know how to transform one edge everything else follows

[warm ember]

we are reading the same book

its the latter

[chilly ocean]

What book?

[crystal vale]

Lang

[crystal vale]

😭

[warm ember]

which edition do you have lol

[chilly ocean]

the big one?

[warm ember]

ye

[crystal vale]

Third one

[warm ember]

i have third revised

[crystal vale]

Is there any other way?

[crystal vale]

I skipped some sections after Sylow

[cobalt heath]

Wild, so that was a typo?

[crystal vale]

Yes

Let G be a group of order p^3 such that it is not abelian.

Then if x^p = 1 for all x in G then there exists subgroup H such that H is isomorphic to Z/pZ × Z/pZ.

Now we know the existence of the subgroup H which has order p^2. Now that subgroup is abelian, by fundamental theorem of abelian group it will be Z/p^2Z or Z/pZ × Z/pZ.

But if it is Z/p^2Z then there exists an element which has order p^2 but this is not true, so it is Z/pZ × Z/pZ.

Is it correct?

[topaz solar]

I think some of the order in your writing of the message is scuffed, but I think it works to show a p^3 order group with all elements of order p (or 1) has a subgroup of that shape

[crystal vale]

Should I write again?

[topaz solar]

Nah

[crystal vale]

Okay thank you

[crystal vale]

If I want to show that a group of order p^2q is solvable, where p and q are distinct.

So we can use the result that if H is a normal subgroup of H also solvable's and G/H solvable then G is solvable.

And G is a group of order p^2q so one of the Sylow subgroups is normal.

If Sylow p-subgroup is normal then H is a subgroup of order p^2 so it is abelian implies H is solvable and G/H is a group of order q which is solvable thus G is solvable.

If Sylow q-subgroup is normal then H is a group of order q so it is solvable and G/H is a group of order p^2 so it is solvable thus G is solvable.

Is it correct?

[topaz solar]

Well, if either Sylow subgroup is normal, yes, since that’s actually saying it’s very nicely solvable even, but do make sure at least one of those two groups is normal of course

As in, prove it or have proven it for another exercise etc

[crystal vale]

I proved that, but now I am trying again

[warm ember]

you could also repeatedly use the fact that a subgroup of order |G|/p is normal where p is smallest prime factor of |G|

[crystal vale]

Yes

It works when q < p

[crystal vale]

In other cases if number of Sylow q-subgroup is p^2 then p and q are consecutive primes which are 2 and 3 so then G has order 12

[rapid junco]

what am I doing wrong here?

I never use HK \subset KH

so I am suspicious

oh wait nvm HK is not closed under products

[warm ember]

easier sol is (HK)^{-1} \subseteq (KH)^{-1}

[crystal vale]

So when HK is a subgroup then HK = KH

[rapid junco]

right ye

i was just tryna see if you didnt need to use that

[serene dune]

i think one of them has to be normal in order for it to be closed

would be nice if it all can be summed up with element analysis (i mean how B acts on Aut(A))

any help is greatly appreciated

[cloud walrus]

yeshua

yeshua

[cloud walrus]

yeshua

yeshua

[rapid junco]

if A is a set of sets, then how does one prove this function is well defined?

It seems obvious but I just want to make sure im doing this right

[serene dune]

what is K ?

[rapid junco]

subgroup

H, K subgroups of G

[serene dune]

umm i might be missing the whole picture here

[rapid junco]

it should just be really easy but im getting lost in the logical.

[serene dune]

lost in liminal

[rapid junco]

like this should work, but I have no idea how to prove formally the well defined part

if K \cap xH = K \cap yH then how do I show xH = yH?

[serene dune]

isomorphism theorems coming to mind

[rapid junco]

for the well defined part?

[serene dune]

nah i havent looked at answer

can we not just show that it is indeed a subgroup then use the ladder construction

[long obsidian]

Can someone help me understand semifields which are abelian groups (P,) with an extra structure (P,,+) which is commutative, associative, and distributive as p*(q+r)=pq+pr

It's a basic result that the underlying abelian group (P,*) is torsion free.

I'm forgetting my basic abelian group theory, does this mean the underlying abelian group is free?

[grave sedge]

If it's finitely generated, yes

In general a torsion free abelian group need not be free

Q being a counterexample

(i mean the additive group of Q, because the multiplicative group of the semifield of positive rationals is actually free)

[crystal vale]

another way is you can show that right coset of H cap K is right coset of H cap right coset of K

[rapid junco]

this works right tho?

[crystal vale]

it is not true what if K = H and take x and y such that xH\neq yH and xH\neq H and yH\neq H

[rapid junco]

I see

so then I need to fix somethings

[crystal vale]

take H = 3Z in Z H cap 3Z + 1 = H cap 3Z + 2 but 3Z+1 \neq 3Z+2

you can try my hint

[long obsidian]

Ah okay I see thank you!

[serene dune]

for each coset $g\left(K\cap H\right)$, there exists some coset representative in K and H such that $g\left(K\cap H\right) \subset g_1 K$ and $g\left(K\cap H\right) \subset g_2 H$

[cloud walrus]

yeshua

[rocky cloak]

I mean, you can just pick g1=g2=g

[serene dune]

oh no

[old hollow]

Have you considered group actions

[serene dune]

i dont seem to understand that in this situation

dam i deleted that by mistake

in anyways its 0 1 0 1 0 1 in Z2
0 1 2 0 1 2 in Z3
0 1 2 3 4 5 in Z6

[rocky cloak]

?

[serene dune]

i dont see how g1=g2=g

[rocky cloak]

I mean you can pick them differently if you like

[serene dune]

well yeah

but wait

the order and shi can mess up no ?

[rocky cloak]

Order mattess yes (if the group is not abelian)

[serene dune]

are all the cosests equivalent upto representation?

[rocky cloak]

"upto representation" ?

[serene dune]

u mean the different coset mean anything differently or they are equivalent ad partitioned by equivalence class

[rocky cloak]

I'm still not sure what you're asking. Different cosets are different yes. But a coset can have several different representatives

[serene dune]

i mean its just a different name to repe represent different coset?
or they have more difference that's not apparent just from the representation

[rocky cloak]

If gK = hK, then they are literally equal, and it's just notation that distinguished them

gK = {gk : k in K}

[serene dune]

well there us bijection bw those sets right?

now are their orders same too ?

[rocky cloak]

Different cosets are in bijection yeah, their all in bijection with K by k |-> gk

[serene dune]

that's what i meant by upto representation

well can we conclude anything further about their orders?

[rocky cloak]

Well, I'm not sure what more one could want. |gK| = |K|, and then you can prove Lagrange: |K|*[G:K] = |G|

[serene dune]

i mean if the K ain't normal just the right and left becomes different

that's why i wad being cautious about that g1= g2

but u mentioned the abelian there

[rocky cloak]

But you were only ever taking about left cosets

[serene dune]

and then another question occurred
how about all the elements that are in bijection
can we conclude anything about their orders too

[serene dune]

oh yeah true

[rocky cloak]

There isn't really a fixed bijection, so depending on your choice every element is mapped bijectively to every other element.

[serene dune]

i mean yeah it's not really an isomorphism

as all cosets aren't group

hmm so nothing but a mapping which doesn't preserve the structure in general

[cobalt heath]

I am stuck on a proof of complete set of principal orthogonal idempotents for bounded quiver algebra.

Suppose w + I is an idempotent in KQ/I, where I is an admissable ideal. Why is w \in I?

[serene dune]

a what algebra?

nvrmind 😭

[rocky cloak]

I think you have your question mixed up.

Did you mean to ask why there exists an idempotent e in KQ such that w+J = e+J or something?

[cobalt heath]

In the proof, after the calculation, it concludes "w = w + I is an idempotent, so e = 0".

[rocky cloak]

Well, is there some context maybe

[cobalt heath]

I am outside so I cannot post the proof rn, maybe some time later

[rocky cloak]

And how are w and e related, and what are you even trying to prove

Like if w = 1, then it's certainly an idempotent and not 0. So something is missing

[cobalt heath]

To show that c.s.p.o.i. of KQ/I is {\bar{e_a} | a \in Q_0},
It is enough to show e_a KQ/I e_a is local.

Now, write an idempotent as
e = l e_a + w + I

Computation shows that l^2 = l

In case of l = 0, the proof directly concludes that w \in I.

[cobalt heath]

(This is for admissable ideal I)

@rocky cloak do you know what is going on in such a proof?

[cobalt heath]

(It is to show that e + I = 0 + I or e_a + I)

[rocky cloak]

So you're considering an idempotent in e_a KQ/I e_a.

So you would write that as e_a l e_a, no. Or what is l and w supposed to be in your decomposition?

[cobalt heath]

I think l is in K, and w is linear combination of cycles at point a.

[rocky cloak]

Ahhh, then it makes sense.

So w+I is an idempotent, but w is a sum of paths

So w^n is in I for sufficiently big n

But w^n + I = w + I, because it's an idempotent

[cobalt heath]

Huh

Oh, because R_Q^n is included in I

Damn, I am dumb

Anyway, thanks a lot! That clears it up for me.

[rocky cloak]

Just remember, quivers is love, quivers is life. Stay strong!

[hidden wind]

i feel like everything i read somehow leads me to the ADE-classification (which i’ve no idea what is) help

[serene dune]

may i ask what's the fuss there?

[rocky cloak]

Dynkin diagrams is the secret elite that controls the world.

[rocky cloak]

Quivers is a combinatorial tool to study (homological) representation theory of algebras. And it makes everything nice

[cloud lynx]

If we have Char K= p>0, then why is the polynomial x^p=x for x in prime field?

[serene dune]

do i have to face it if i take graduate homological algebra

[hidden wind]

i knew it!

[serene dune]

hope a physicist doesn't see this

[crystal vale]

In the finite field if char K = p then x^(p-1) = 1 so x^p = x.

I am not sure about the prime field but I think subring generated by 1, right?

[cloud lynx]

but why is then x^(p-1)=1?

ah x(x^p-1)

[serene dune]

multiplicative group

[cloud lynx]

so what do u mean ? because its multiplicative there exists a inverse of x so that x^p-1 -1=0 => x^p-1 * x - 1*x=0?

[serene dune]

the order of the group is (p-1) and its cyclic

[rocky cloak]

Depends

But maybe

[cloud lynx]

so?

[hidden wind]

we constructing the tensor product of modules in class

[serene dune]

so whatever element u take
if u mutiply it (p-1) times it will be 1

[cloud lynx]

know i understand it

[serene dune]

is k identity element in your alphabet ?

[serene dune]

like depends on the school and professor
enlighten me on the dependence

[arctic trail]

Lagrange theorem

the multiplicative group has p-1 elements

so x^(p-1) = 1 forall x

[cloud lynx]

Ok ok so every order of elements divides p-1

[arctic trail]

yes

in the prime field

but looking at the prime field is just looking at Z/pZ = Fp

[cloud lynx]

So x^(p-1)=x^*(ord(x)*y=p-1)

But x^ord(x) =1

So yeah

[arctic trail]

yes

[cloud lynx]

So if we look at X^p-x then the elements are the roots

And every field ist factorial so yeah

[arctic trail]

the roots correspond exactly to the elements of the prime field

[cloud lynx]

Its the elements of prime field

[arctic trail]

don't understand what you're saying here

[cloud lynx]

Öhm

ZPE?

[arctic trail]

A field is an integral domain

unique factorization domain

and principal ideal domain

and in integral domains polynomials can have at most as many roots as its degree

(if a is a root of a polynomial f(x) then f(x) must be factorable as f(x) = (x-a)q(x) for some q with degree less than f)

[rocky cloak]

Yeah, depends on those things. Like, I said it's a tool in homological representation theory, but if your course is more concerned with algebraic topology or algebraic geometry, then you'll do that instead.

And something like an intro course in homological algebra might just focus on abelian categories, the derived category, Morita equivalences and you know homological algebra. And not necessarily go to other contexts.

[serene dune]

i might take that for alg geo

[crystal vale]

By induction, right?

[arctic trail]

yes by induction

[serene dune]

if i get into the school im hoping for

[crystal vale]

If X is non-empty set such that there is bijection between X and field F then X can be field, right ?

[chilly ocean]

X can be given a field structure yes, in fact even one that makes it isomorphic to F

[crystal vale]

Yes I see

[chilly ocean]

You also don't need the requirement that X is nonempty, since every field is already nonempty

[crystal vale]

Yes

Thank you

[cloud lynx]

is the automorphisgroup of G just the symmetrical group of G?

[coral spindle]

No

[serene dune]

it's a long story, gl

what took me a long time to realise was by construction its a group of maps(structure preserving)

[chilly ocean]

What is the symmetrical group of G?

[arctic trail]

noo

[arctic trail]

I assume they mean Sym(G)

[chilly ocean]

They could have meant something else though

[cloud lynx]

yes

[arctic trail]

@cloud lynx the automorphisms are much less

one element in Sym(G) is a transposition (1 g) for some g != 1

this won't result in a automorphism as 1 has to be fixed

[cloud lynx]

Oh ok ty^^

ahh so is there a trick how to determine the order of a galois group?

[delicate orchid]

the order is the degree of the splitting field extension of whatever polynomial you have

[cloud lynx]

so |Gal(L/K)|=[L:K]?

ah ok

ty

:))

[glad osprey]

If A is an abelian group, is Aut(A) necessarily a group under pointwise addition?

[arctic trail]

yeah

[arctic trail]

abelian group written additively?

[glad osprey]

yeah

End(A) would be a group in that case, but I don't think Aut(A) would necessarily be closed under pointwise addition

[arctic trail]

so like if phi, psi in Aut(A) (phi + psi)(g) = phi(g) + psi(g) forall g in A?

[glad osprey]

yep

[arctic trail]

consider the identity morphism, when A is finite, and notice |A|id = id + id + ... + id (|A| times)

[glad osprey]

aha, you get the trivial homomorphism, which is not in Aut(A)?

[arctic trail]

yeah

[glad osprey]

nice, thanks 👍

[arctic trail]

unless A = trivial group

[arctic trail]

yw!

[pliant rivet]

I've been tasked with proving that no subgroup of S_4 is isomorphic to Q_8. I know from looking it up that the subgroups of S_4 of order 8 are all isomorphic to D_8, which means none of them are isomorphic to Q_8.

My question is if there is a better way to approach this problem than to prove that S_4 has 3 subgroups of order 8 and all of them are isomorphic to D_8. It seems incredibly computationally boring

[rocky cloak]

Instead of focusing on S4 you can focus more on Q8.

Like a map Q8 -> S4 is the same as a group action of Q8 on a set with 4 elements. Then you can ask yourself what are the stabilizers of the different elements. Since 8 > 4 each stabilizer is a nontrivial subgroup, and every nontrivial subgroup of Q8 contains -1. So -1 is in the kernel of the map Q8 -> S4, and the map cannot be injective.

This also has the advantage of working for Q8 -> Sn for every n<8

[pliant rivet]

thank you thank you

[cloud lynx]

If we look at the galois group of Q(sqrt(2),sqrt(5))/Q why can only sqrt(2) map to sqrt(2) or -sqrt(2)? and not also to sqrt(5)?

[coral spindle]

Suppose f is an automorphism. Remember that it has the property f(xy) = f(x)f(y)

Now note that f(sqrt 2)^2 must then be f((sqrt 2)^2) = f(2) = 2.

[cloud lynx]

mmh

but then sqrt(2)->-sqrt(2) is also not a automorphism

f(sqrt(2))=-sqrt(2)

[coral spindle]

What I have written does not imply that, no.

f(sqrt 2) = - sqrt 2 is compatible with what I have written.

[cloud lynx]

OH RIGHT I thought f(sqrt(2)^2)=-2 but 2 is element of Q so it map to 2

right

[coral spindle]

OK

[cloud lynx]

ahh ok if sqrt(2) maps to sqrt(5)

then (f(sqrt(2))^2=5 != 2= f(sqrt(2)^2)

so this is not a automorphismn even a homomorphism

an*

[coral spindle]

Indeed

[cloud lynx]

is that the only reason? or is it also because they have two different minimal polynomials?

[coral spindle]

You could probably find several reasons.

[cloud lynx]

Thank you!

[chilly ocean]

Those are pretty much the same reason. If x^n+ax^n-1+...=c then you can apply f to it and conclude that f(x) satisfies the same minimal polynomial (assuming f fixes the constants you use in the polynomial)

[cloud lynx]

ahh okk tyy

[cloud lynx]

ufff there are also polynomials with roots, which cant map to every root of the polynomial

for example X^4-1

it has i,-i,-1,1 as roots

but i cant map to 1

f(i*i)=-1 != 1 = f(i)*f(i)

[rapid junco]

If H and K are subgroups but not necessarily normal then is [G : H cap K] = [KH : K]?

This holds if normal by second isomorphism

But I don't see how the same bijection is well defined

[coral spindle]

I'm not sure that makes sense considering KH may not even be a group

I think maybe you also meant [H : H cap K], but that's just a minor typo

I'm not sure I can put my finger on what exactly the additional assumptions you might need here are

[stone sky]

Maybe someone could help with me with some understanding here, but my professor claimed that “If two group’s (he said things but i assume he meant groups) are isomorphic, then their centers should be the same”, I’m not sure why that happens, and he emphasized should, so when does this not happen?

[coral spindle]

He was trying to appeal to your sense of what ought to be true

In fact, their centres are isomorphic.

[stone sky]

I think I’m going to attempt to prove it. That way I get a better understanding

[knotty badger]

yeah that sounds like a good idea

[stone sky]

ok i proved it. so the intuition is if i have an element that commutes with everything in one group, its isomorphism should also commute with everything in the second group, and vise versa.

thank you guys

[knotty badger]

yep yep

there’s like an informal theorem that anything you can state “in the language of group theory” must be invariant under iso

you might actually be able to make this precise with model theory? im not sure

but at least, the approach I took was verifying this myself for lots of individual properties like “abelian”, “center”, “conjugate” etc and eventually convincing myself that it was true

[stone sky]

yeah, i’m simply having trouble with connecting all the ideas. this is a kind of math that i’ve never seen before. but i also really like it, just challenging

not sure if it’s considered pure math, but if it is, it’ll be the first pure math i’ve looked at (aside from possibly set theory and logic?)

[rocky cloak]

Assuming HK is a group, then
h(H cap K) |-> hK
should be a bijection between H/(H cap K) and HK/K

[cloud lynx]

if we have the polynomial X^5-1 over Q then the splitting field is Q(w)/Q where as w is the fifth unit root

why is the degree of this field extension 4?

and how do I determine the minimalpolynom of it lol

[rapid junco]

yeah I see now

HK is not always a group though.

So I just used instead the injection H/(H \cap K) into H/K

if gHg^{-1} = xHx^{-1} then nessecarily must xH = gH?

[rocky cloak]

Consider for example abelian groups

[rapid junco]

right okay

how do I prove that {gHg^{-1} : g \in G} is of less cardinality than G/H?

like I dont immediately see a well defined map here

[rocky cloak]

Well if gH = xH then certainly gHg^-1 = xHx^-1, so you could define a map from G/H to that set

[rapid junco]

Yes

[rocky cloak]

So there you go

[rapid junco]

so then we get G/H is of less size than the set I care about

right because G/H injects into this set

[rocky cloak]

No, if you have a surjective map from one set to another, then the domain is bigger

[rapid junco]

Oh I see.

And the existence of a surjection from A to B implies the existence of an injection from B to A?

[topaz solar]

Yes

[knotty badger]

interesting

[rocky cloak]

It does, that is in fact an equivalent characterization of the axiom of choice

[rapid junco]

hmm interesting okayt

[topaz solar]

$Z(G) = \phi[G]$ where $\phi(x) \equiv \forall y. xy=yx$

[cloud walrus]

Sharp, the forgotten

[rapid junco]

thanks

[topaz solar]

As with any isomorphism, $\phi(f(x))\iff \phi(x)$

[cloud walrus]

Sharp, the forgotten

[topaz solar]

So anything definable will be “isomorphic” with the appropriate object in the other one

This holds also for like, any other logical thing preserved under isomorphism

I say “isomorphic” since you can have some things maybe not defining subgroups e.g., like “elements of finite order”

L_\infty,\infty is more than strong enough for sane people purposes anyway, and L_\infty,\omega is already pretty nice

[knotty badger]

Ok now you’ve lost me

[topaz solar]

Yes if you can write it in the language of groups you can do nice things

[knotty badger]

Lol

[topaz solar]

But not just first order things

[arctic trail]

Isomorphisms preserve all first order properties

You can define the center of a group G as

$Z = {g \in G: gh = hg \quad \forall h \in G}$

[cloud walrus]

Trivial Lemma

[topaz solar]

Wow I wonder where I’ve seen that

[chilly ocean]

They even preserve second order properties and infinitary logic properties

[arctic trail]

I didn't read the whole conversation when I opened the thing I just saw that message and replied

[arctic trail]

sure

[long obsidian]

The last exercise is to show that a semifield homomorphism doesn't have a well defined kernel. Can anybody tell what they mean by this?

I feel like I can define ker=f^{-1}(1) and since (P,•) is an abelian group if f(p)=f(q) for p,q in P then f(pq^{-1})=1 so if the kernel is a singleton ker=f^{-1}(1)={1} then p=q since inverse are unique.

I don't see a problem with defining a kernel

[topaz solar]

The addition stuff doesn’t really have an inverse tho

[grave sedge]

This is actually open, isnt it?

It's obviously equivalent to choice if you assume the injection you get is a right inverse of the surjection

[topaz solar]

I believe that is open

Though set theoretic stuff isn’t my main MO

[arctic trail]

wait really?

[long obsidian]

How is this any different than the case of rings? There is an underlying abelian group and elements of a ring aren't necessarily units so that don't always have inverses. I thought kernels being well defined for rings would have the same proof as semi fields having well defined kernels

[topaz solar]

Well your distributivity has changed sides so something could change

[long obsidian]

I thought the existence of kernels for rings followed directly from the existence of kernels of abelian groups? Do you use distributivity to show kernels exist for rings?

[topaz solar]

Well you do want things to be nice

[chilly ocean]

Did they define "kernel" before?

There is a definition in terms of category theory, but that's probably not what they want

[long obsidian]

This was at the beginning of the document. i don't know If they defined it previously somewhere else.

I really just want a way to tell if a semifield hom is injective. I can't tell how to do this without a kernel.

[chilly ocean]

In general kernels being trivial is not equivalent to injectivity

For example f:N->[0,+oo] given by f(0)=0 and f(n)=+oo for every n>0 is a monoid homomorphism with trivial kernel which is not injective

[knotty badger]

wait that’s a monoid hom?

ah i see

[chilly ocean]

f(a+b)=f(a)+f(b) and f(0)=0

[knotty badger]

yeah you mentioned something about kernel pairs before

[long obsidian]

A standard semifield is trop(y1,...,ym) which is the free (multiplicative) abelian group generated by yi with the auxiliary addition + given by (a1y1+...+anyn)+(b1y1+...+bnyn)=(min(a1,b1)y1,...,min(an,bn)yn)

As an abelian group there is a Z action. Consider the map induced by 10 in Z, say f: trop(y1,...,ym)->trop(y1,...,ym) by f(h)=h^10. How can I tell this is injective?

It's obvious to me that 1=(y1)^0...(ym)^0 is the only element in f^{-1}((y1)^0...(ym)^0) and 1 is the identity on the abelian group.

I think this is an injective map? Can anybody tell?

[chilly ocean]

What does min mean?

What order structure are you putting on the free abelian group to have minimums?

[long obsidian]

The coefficients of the free abelian group are integers so I'm using the total ordering of the integers

I got scared by this exercise I saw to think this map isn't injective

[rocky cloak]

Oh yeah, I think you're right there

[topaz solar]

@delicate orchid rep theory of semidirects question for you, as a treat. How can I turn representation data of G \rtimes_f H into representation data of f(H) < Aut(G) :3

Of particular interest is wreath products ofc, but that’s a bit less “generic” and nice

[grizzled spindle]

If i have 2 isomorphic subgroups H1 H2 in G, does an isomorphism H1 to H2 induce an isomorphism G to G?

I thought maybe this could work by bijecting coset representatives

but it seems like not quite right

[crystal vale]

I think it is not true

Take 3Z to 5Z by 3n -> 5n so it is isomorphism but if you induce Z to Z then it is not isomorphism because there are only two isomorphism between Z to Z one is identity and second one is x -> -x

[grizzled spindle]

hm yeah that makes sense

i wonder if theres a condition that would make it true

maybe normal subgroups

anyway thx for this

wait no thats dumb u just gave normal subgroups

welp disproven

[topaz solar]

There are some conditions which happen to imply it based on, like logic bs, but I am not aware of any algebraically nice characterizations

[crystal vale]

What if the H_1 and H_2 generated set of G ?

[warm ember]

3Z and 5Z generate Z

[crystal vale]

Individually

Let H_1 generated set and H_2 also generated set

But then H_1 is G

[topaz solar]

Then H_1 = G

[crystal vale]

Because it is a subgroup

[topaz solar]

But it is true in that case yes

Further, if G is definable from H_1, I think you can do something

[peak marsh]

quick question, in order for an ideal I to be prime, then for the intersection of ideals J and K, at least one of J or K is contained in I. Does this mean for the ring of integers, where prime(Z) is the prime ideals of Z, (6) is prime?

[rocky cloak]

(6) is not prime as it is the intersection of (2) and (3)

[hollow tartan]

came across with this nice pic of algebraic numbers on wiki, adding the rationals (for differentiating purposes), is this how the absolute Galois group of Q would "look" like? sry for the stupid question!

[rocky cloak]

I mean this is a picture of the algebraic numbers, while the absolute Galois group is their automorphism group. So that's two different things I guess.

And the position of the numbers in the complex plane doesn't really say much about how the automorphism group acts on them. I'm guessing the color signifies degree of minimal polynomial(?), which says something, but still not a lot.

But I'm not sure there really is a very good answer to what the Galois group looks like

[hollow tartan]

What if we stack (not in the scary sense idk what it means mathematically, I mean like stacking a bunch of paper) of this galois group and do automorphisms like S_n

[long swan]

why do we care about distinguishing between internal and external direct sums?

I'm not able to find a good example of when there would be a meaningful difference yet they seem to be talked about separately everywhere I encounter them to the point that it makes me think there's some subtle importance that im not picking up on

[chilly ocean]

I don't know if this is helpful, but I think of internal direct sums as a partial function/functor

Internal direct sums are for subgroups/subspaces of a fixed group/space

I think of subgroups of a group as a different kind of structure than a group by itself (it has extra structure)

And internal direct sums reflect that extra structure by not existing sometimes (or having more information other times since external direct sums are only defined up to isomorphism but internal direct sums can give different embeddings)

Isomorphic (as groups) subgroups of a group can have different behaviour, for example C_2xC_2xC_4 has two subgroups isomorphic to C_2xC_2 that give different quotients. So the extra information of which subgroup it is exactly rather than just its isomorphism type could be useful

[knotty badger]

Strictly, the set-theoretic constructions of internal and external direct sums are different too

Two subspaces having an internal direct sum is like a true/false property

Whereas external ones aren’t

What is true is that if two subspaces have an internal direct sum, then their internal direct sum is isomorphic to their external direct sum

[chilly ocean]

In some categories it is. There are categories where not every pair of objects has an external direct sum

[knotty badger]

Mhm, I’m aware

[knotty badger]

And yeah I definitely agree with this

Subgroups are an instance where it is genuinely useful to care about a notion of “sameness” finer than isomorphism

Namely, set-theoretic equality

[chilly ocean]

You can turn it into a kind of isomorphism if you consider the uh I always forget if it's slice or coslice category

Then two subgroups are isomorphic in this category iff equal

[knotty badger]

I see

[rocky cloak]

Well, internal direct sums is really a statement about subgroups.

Like an external direct sum isn't just a group, but it also comes with extra data, namely inclusions and projections. For the internal direct sum you don't need this extra data.

For example saying that Z/6 is the internal direct sum of 2Z/6 and 3Z/6 is saying something more than merely Z/6 being isomorphic to the external direct sum of 2Z/6 and 3Z/6. It is fixing a specific isomorphism.

[toxic zephyr]

reviewing group theory. can someone give me a hint for 2? mostly I'm just struggling with what it means for the index to be 2.
does that mean that we have some x not in H such that for all g not in H that g=xh for some h? basically that the cosets are eH and xH for some x

also if all the cosets are eH and xH, how do we write that?

$ G=eH\sqcup xH$? $\oplus$?

[cloud walrus]

eigentaylor

[chilly ocean]

The former

Erm

Actually the former is false because both contain 0

And the latter makes no sense because cosets are not subgroups (except for eH)

[toxic zephyr]

ah true

so just unions I guess? idk is there a more appropriate operation?

[chilly ocean]

$G/H = {eH, xH}$

[cloud walrus]

Bequi

[chilly ocean]

G/H is the usual notation for the left cosets of H

[toxic zephyr]

fair enough

[chilly ocean]

This notation is used even if H is not a normal subgroup

[toxic zephyr]

ah. and if it's not a normal subgroup then the cosets just don't form a group?

[chilly ocean]

Right

[toxic zephyr]

wait, why does xH have 0?

if e is in xH, then x inverse is in H, then isn't x in H so xH=eH?

[chilly ocean]

I'm sorry that was stupid of me

[chilly ocean]

The former is correct

[toxic zephyr]

oh okay cool

I think I got the proof as well

my argument is basically that if g is in H, then ghg^-1 is obviously in H. otherwise g=xh1 so we get
xh2x^-1 for h2=h1hh1^-1. if it wasn't in H, then it's equal to some xh3, so x^-1=h''^-1 h3 implies x is in H contradiction

[chilly ocean]

otherwise g=xh1 so we get xh2x^-1 for some h2=h1.
I don't understand what you mean by this

[toxic zephyr]

either g is in H or g=xh1 for some h1 in H right? since those are the only two cosets

[chilly ocean]

Right

[toxic zephyr]

oh the line accidentally ended lol
should be
h2=h1hh1^-1

(xh1)h(xh1)^-1

=xh2x^-1

[chilly ocean]

x^-1=h''^-1 h3
Where does this come from?

[toxic zephyr]

shoot I didn't change all the ''s to 2

[chilly ocean]

It's okay

But I still don't see how you got that