groups-rings-fields

[pastel cliff]

is very important

you dont wanna get sullied

unless your name is chmonkey

[coral spindle]

True but is violence

I shan't interrupt more

[pastel cliff]

i mean i think i have it?

the same group action/reasoning as above gives us a homomorphism from S_3 to S_6

that reasoning also shows that it's an isomorphism

wait no

[coral spindle]

hmhmhhbhnhjnhjnhj an isomorphism onto its image, certainly

not the whole of S_6

[pastel cliff]

isomorphic to the image

yeah lol i caught myself there

[coral spindle]

yup

So hopefully now

this is an extension of course but

[pastel cliff]

so we have the desired isomorphism and showing that it's transitive shouldnt be too hard

[coral spindle]

hopefully now you can find a different subgroup of S_6 of order 6

[pastel cliff]

i mean it's S_3 itself isnt it

[coral spindle]

Maybe you can relate its transitivity to the transitivity of the action

[coral spindle]

Can you write it out explicitly in terms of cycles, though

It will not be S_3 as a subset of S_6, as it turns out...

[pastel cliff]

oop

i'll brb quick break is needed

[coral spindle]

I'll be going to sleep I'm afraid

[pastel cliff]

[coral spindle]

You can polish this off on your own no problem

[pastel cliff]

can i do this using the fact that elements of the symmetric group that are conjugate to each other share a cycle type?

[simple valley]

schur?

[ripe basalt]

https://math.stackexchange.com/a/3017946 can anyone explain the line of the solution starting with "N_{i+1}/N_i is minimal non trivial subgroup of G/N /N_{i}"

Actually i understand why that group would be minimal normal. But afterwards when he starts taking all of these preimages of the homomorphism i get lost

like what is he using to say $\varphi^{-1}(N_{i+1}/ N_{i})$ is the smallest normal subgroup of $\varphi^{-1}(G/ N_i)$? also the conclusion doesn't really make sense to me

[cloud walrus]

*-algebra

[simple valley]

ok yeah I like this proof

we know that an inner product is an invertible map V -> V*

and if $\phi = H_1^{-1} \circ H_2 : V \to V$ then $$\phi(x) = y \iff \forall z, H_1(y, z) = H_2(x, z)$$ and from $G$-invariance we have $\phi(g \cdot x) = g \cdot y \iff \phi(x) = y$, i.e. $\phi(g \cdot x) = g \cdot \phi(x)$. Thus by Schur's lemma we have that $\phi$ is a scaled identity

[cloud walrus]

mniip

[ripe basalt]

actually im just trying to show that if N is a minimal normal subgroup of finite G, then N is a chief factor in some chief series of G

[coral shale]

mniip doing math

[ripe basalt]

ok

[prime sundial]

i'm kind of struggling to care about permutation groups, transpositions, k-cycles, etc.

i found abstract groups way more interesting and intuitive, is there more motivation behind permutation gruops

[coral shale]

all groups are isomorphic to some permutation group?

🤔

[prime sundial]

oh really?

[coral shale]

cayleys theorem

[prime sundial]

omg i literally have that in my notes

[coral shale]

note by permutation group i dont mean Sn

[prime sundial]

ok thank you

[coral shale]

any subgroup of Sn edit: oops any subgroup of a symmetric group (underlying set doesnt have to be finite)

that aside, group actions

[prime sundial]

yep my notes say G is isomorphic to any subgroup of Sym(G)

[coral shale]

like permuting stuff is an action

[ripe basalt]

not to any subgroup

[coral shale]

and this is very relevant for various stuff

[prime sundial]

specifically the subgroups {\lambda_g | g\in G}

[coral shale]

most recent thing for me i can think of is algebraic topology

where permutation groups were definitely relevant...

for a less abstract example, theres the rubiks cube thats interesting

[prime sundial]

yeah my professor did mention the rubiks cube as an example

[ripe basalt]

since no one is biting in the help channels, does anyone know if this proof of G is nilpotent ==> G is solvable is correct?

[prime sundial]

i don't need anything particularly concrete, cayley's theorem is interesting enough to motivate it. thanks shuri

sorry algebra for hijacking the channel but i have another question
my class defined $$\text{sgn}(\sigma) = \begin{cases} 1 & \sigma\text{ is even}\ -1 & \sigma\text{ is odd}\end{cases}$$ for $\sigma\in S_n$ but then we said $$\text{sgn}(T_1\circ T_2\circ ... \circ T_n)=(-n)^n, T_i\in S_n$$

[cloud walrus]

maximo

[prime sundial]

i don't understand how we suddenly have signs that aren't 1 or -1

should it be (-1)^n instead?

[coral shale]

signs are 1 or -1. The bottom line likely typo

[prime sundial]

ok thank you

[ripe basalt]

anyone have any thoughts on my question?

it's just a verification of a proof

[sage spruce]

Looks fine to me

[ripe basalt]

@sage sprucedo you know why the G_{i+1}/H / G_{i} / H is in that center?

[sage spruce]

Well it seems logical but the proof might come from computations in the quotient

[ripe basalt]

what do you mean?

[sage spruce]

I don't think there is a theorem about that

[ripe basalt]

did you not actually check to see if it works?

this is sort of my question, i just need verification. or maybe a short explanation what needs to be improved

[sage spruce]

Or else just use the third isomorphism theorem to get simpler quotient

[ripe basalt]

so the 3rd isomorphism theorem says that quotient is isomorphic to another one, which makes my new series a central series

then, since it's a central series of a smaller order group, that quotient must be abelian

via the isomorphism from before, that means the original quotient was also abelian

?

[sage spruce]

No

You might use the third theorem twice

[ripe basalt]

you only have to use it once, since it's between the same 2 things

[sage spruce]

First to simplify the quotients and show commutativity

Then to get the entire thing

[ripe basalt]

by simplify, you mean just removing the H's?

[sage spruce]

I may be wrong though

[sage spruce]

Yes and you do what you have to do with the center

[ripe basalt]

sure

okay that says the quotient with the H's is iso to the quotient without the H's

which says it's a central series then

and then you can just say that via the inductive hypothesis, that G_(i+1)/ G_i quotient is abelian

by the same exact isomorphism

it's isomorphic to the quotient with the H's, which is abelian by the inductive hypothesis

[sage spruce]

Yes

The only difficult thing is the center

The rest follows

[ripe basalt]

is that difficult thing addressed..?

oh yeah i guess im not sure why it's in the center

[sage spruce]

That's the problem with groups intuition is very difficult to have

[ripe basalt]

do you see why it's in the center?

[sage spruce]

No

It probably uses the minimality somehow

Besides the minimal normal subgroup in the sense of inclusion is the trivial one

So perhaps a maximal one

[ripe basalt]

if it's maximal

then the 3rd iso wont work

right?

[rustic tendon]

hey

i am a year 7 student and i know how to do simultaneous equasions

anyone need help

[ripe basalt]

@rustic tendonwrong channel

[rustic tendon]

oh

which channel

hey

which channel

[simple valley]

maybe read the channel descriptions

if you're a year 7 student you probably don't belong in #advanced

[rustic tendon]

but

i know year 9 and 10 stuff

such as simultaneous equasios

i can do one if u like

[simple valley]

that's still not advanced territory

[rustic tendon]

k

[ripe basalt]

@sage spruce still here?

[coral spindle]

I don't think that's helpful. You just need to do some computation at this point.

[south patrol]

Basically like

[dusty jacinth]

Ok guys if a group G is abelian than does it follow the exponential rules which are applied over real numbers?

[south patrol]

#❓how-to-get-help @willow geyser

Not sure I asked, Joe.

[coral spindle]

Any group follows the usual exponential rules, if you're referring to a^x a^y = a^(x+y).

This is not the correct channel, Joe

[blissful crystal]

this gives me Basic Algebra I - Jacobson vibes lmfao

i remember reading a review for that book of a grandpa who got Basic Algebra I + II for his year 7/8 (i don't remember) grandson and got really upset that it wasn't really that basic lmfao

[dusty jacinth]

@coral spindle

[coral spindle]

Are you referring to (ab)^n = a^n b^n?

[dusty jacinth]

Yes

That's where I'm confused

[coral spindle]

Indeed this will only hold in Abelian groups in general

[blissful crystal]

you're technically right about abelian groups yeah

if the group is cyclic it is abelian by definition

[dusty jacinth]

Hmm makes sense

Thanks for the help guys 😁

[blissful crystal]

we have Algebra I and Algebra II as grad courses here lmfao

no no

math grad students

they're like

commutative algebra lmfao

yeah lmfao

mathematicians really want to make it seem that what they're doing is trivial in the grand scheme of math

also when i tell my parents who ask like "so what do you do at uni?" responding with "just some algebra" really makes it seem im doing nothing here

or that I "proved that 1 > 0" in class today

[sly storm]

Hello. If $A$ is a finite subset of $\mathbb{C}$ and $r, s \in \mathbb{C}$ are algebraically independent over $A$, then is there an automorphism of $\sigma$ of $\mathbb{C}$ such that $\sigma |_A$ is the identity and $\sigma (r) = s$?

[cloud walrus]

MathPhysics

[tender wharf]

Didn't you post this yesterday?

[elder wave]

And we’ll probably be seeing it again tomorrow

[chilly ocean]

I'm not sure what it means to be algebraically independent over some subset of C. Could you clarify?

[wooden ember]

does anyone see where reflexivity and non degeneracy is required in this exercise?

i must have done it wrong cause im not using these properties at all

[chilly ocean]

I imagine people'd rather check your work than solve it themselves

[tender bough]

help, I don't understand the description of the syplectic Lie algebra on my lecture notes

[next obsidian]

What part do you not understand

[tender bough]

aren't elements in gl(n) nxn matrices? why does it say 2nx2n matrices?

[next obsidian]

Yeah that’s a typo

I just glanced over that

[tender bough]

so it should be gl(2n)?

[next obsidian]

Yup

And the identity matrices in the block matrix are n by n

[tender bough]

smh my head 🤦‍♂️

gotta email the instructor about these outrageous typos

[formal ermine]

"Let G be a group and g \in G an element of it. The transformation ... is an isomorphism, because ..."

don't quite understand how in the first thing they went from $g \circ (a \circ b) \circ g^{-1}$ to $(g \circ a \circ g^{-1}) \circ (g \circ b \circ g^{-1})$

[cloud walrus]

illuminator3 (#eric4honorable)

[chilly ocean]

the g^{-1} o g in the middle cancel out

use associativity to rearrange the brackets if you need to

[formal ermine]

ohhh

I see

[coral spindle]

I assume you saw how

[formal ermine]

no

I was first gonna ask about the second thing instead of the third lmao

[sly storm]

https://en.wikipedia.org/wiki/Algebraic_independence

[coral spindle]

Go ahead then

[formal ermine]

I don't understand what we're showing in the second and third one

like f(a) = f(b) => a = f^-1(f(a)) = f^-1(f(b)) = b

how does that show injectivity

[coral spindle]

Well that's the definition of injectivity

if f(a) = f(b) then a = b

[formal ermine]

yeah

but how are we showing that

oh wait

[coral spindle]

By manipulating both sides

[formal ermine]

so iff f^-1(f(a)) = f^-1(f(b)) and a = b then f is injective?

[coral spindle]

Yes but you're focusing on this f^-1 a lot, really you see if we have an f^-1 in the first place then we're done

Maybe look at this simply in terms of applying g and g^-1 appropriately

[formal ermine]

wait

what are we trying to show in the first place

[coral spindle]

To clarify what I mean by this, if a function has an inverse, it is necessarily bijective.

[formal ermine]

what does it mean for it to be injective

[formal ermine]

f(a) = f(b) => a = b

right?

[coral spindle]

A function f : A -> B is injective if for all a,b in A, if f(a) = f(b) then a = b.

[formal ermine]

yes

[coral spindle]

You are familiar with this I hope

[formal ermine]

yes I remember the definition

[coral spindle]

Now we begin with assuming that gag^-1 = gbg^-1

And we want to achieve a = b

We do this by composing with g and g^-1 appropriately, as above

[formal ermine]

ahhhh

ok I get it now

what about the third one

[coral spindle]

Similarly for surjectivity

[formal ermine]

what do we have to show for surjectivity

[coral spindle]

Well do you remember the definition?

[formal ermine]

no

I only know what it means

[coral spindle]

Ok let me remind you

[formal ermine]

every input is mapped to at least 1 output

[coral spindle]

Mm no

that's not right

[formal ermine]

oh right

the other way around lmao

otherwise it's not a function

every output is achieved by at least 1 input

[coral spindle]

That's right

[formal ermine]

yeah

[coral spindle]

Can you write that in more formal terms?

[formal ermine]

uh

let me think

[coral spindle]

So let f : A -> B be a function; when is it surjective?

[formal ermine]

,, \forall b \in B \exists a \in A : f(a) = b

[cloud walrus]

illuminator3 (#eric4honorable)

[coral spindle]

Perfect

[formal ermine]

like that?

[coral spindle]

good job

So hopefully now you can see how this fits in the third part

[formal ermine]

ahh

[coral spindle]

I really must urge you to inscribe these definitions into your forehead

[formal ermine]

so we have to show that for every output we can find an input

[coral spindle]

They're annoyingly relevant to everything

[formal ermine]

yeah this is the first time I've ever seen the surjective formal definition

[coral spindle]

I remember coming up with a little justification for the sur- and inj- which helped me remember which way round they are

[formal ermine]

let me write them on my whiteboard so I don't forget

[coral spindle]

That's a little worrisome as I would expect people to have seen this before doing group theory—are you a self-learner?

Perhaps you've seen the terms "onto" and "one-to-one" before?

[formal ermine]

not in english, no

[coral spindle]

Right that does present a barrier

Well at least you know now

[formal ermine]

uhm not really

they let me attend one uni course per semester (I'm still in hs)

last semester I did numerical analysis

[coral spindle]

Ah that does explain it

[formal ermine]

now I thought it'd be fun to do algebra

[coral spindle]

Alright, I hope you enjoy it

I look forward to seeing more questions here

[formal ermine]

yeah it's been fun so far

[formal ermine]

already got one

what does the notation

$$\bZ/k\bZ$$ mean

[cloud walrus]

illuminator3 (#eric4honorable)

[coral spindle]

Have you seen quotient groups yet?

[formal ermine]

no

this is like the first lecture

[coral spindle]

Then don't worry yet.

[formal ermine]

we've just introduced groups

[coral spindle]

Oh dear

Right so the lecturer is likely just talking about what you will see in the future

You will not be able to understand what this is yet

[formal ermine]

oke

[coral spindle]

And I don't want to teach the course here!

[formal ermine]

lmao yeah

thanks for the help c:

[coral spindle]

But put simply, this is the integers modulo k, if that means anything to you

[formal ermine]

oh

ohhhhhh

yes

thanks

[coral spindle]

Again this will become clearer later on in the course

[formal ermine]

$k\bZ$ is ${ kx | x \in \bZ}$?

[cloud walrus]

illuminator3 (#eric4honorable)

[coral spindle]

It is indeed

[formal ermine]

thanks

what does the notation $^tAA = \mathbb{1}$ mean? (orthogonal group)

[cloud walrus]

illuminator3 (#eric4honorable)

[formal ermine]

uh

here

A is an n x n matrix over K but what is the restriction

[coral spindle]

Do you know what the matrix transpose is?

[formal ermine]

yes

[coral spindle]

Brilliant. So ${}^{t}A$ is the transpose of $A$.

[cloud walrus]

Boytjie

[formal ermine]

ahhh

[coral spindle]

So it's saying that $A^{-1} = {}^tA$.

[formal ermine]

I've only ever seen people write A^T

[cloud walrus]

Boytjie

[formal ermine]

right

[coral spindle]

Yes it is unusual notation to be sure...

[formal ermine]

the definition of an orthogonal matrix

[coral spindle]

Yes.

[formal ermine]

thanks

[coral spindle]

Hence orthogonal group :)

[pastel cliff]

i got stuck on the last part last night - in the homomorphism $S_3 \to Sym(S_3) = S_6$ given by $\sigma \mapsto \alpha(\sigma, \tau)$

[cloud walrus]

stμ₂dying

[pastel cliff]

where \alpha is the group action

[coral spindle]

When you write Sym(S_3) = S_6, I mean this isn't literally true is it

[pastel cliff]

oh it's not

perhaps i should put more care into justifying things lol

[coral spindle]

I mean what're the elements of Sym(S_3) — they're functions S_3 -> S_3 in particular, right?

And elements of S_6 are functions {1,...,6} -> {1,...,6}

So while I do agree that they're isomorphic, I think you should specify the isomorphism

This will help you with the computation.

[pastel cliff]

yeah i kinda jumped to the S_6 conclusion bc i was trying to figure out the image of S_3 under this homomorphism

[pastel cliff]

should this isomorphism be obv? im having a hard time coming up with the actual map

[chilly ocean]

This doesn't say what it means for sets

As far as I can see? I only browsed it briefly

You need to specify subfield MathPhysics.

[coral spindle]

Try explaining in words why you think Sym(S_3) is the same as S_6

[pastel cliff]

Sym(S_3) is the the group of permutations on 6 "objects" since the order of S_3 is 6

[chilly ocean]

I assume he means algebraic independent over Q(A)?

[coral spindle]

Right, so we have 6 objects in S_3, and we can identify this with the numbers 1 to 6, right?

[chilly ocean]

One can only guess.

[pastel cliff]

but just putting elements of S_3 in different orders next to each other just gives other elements of S_3

[chilly ocean]

Let's not draw straws. Better to wait for a response

I'm gonna draw straws.

[pastel cliff]

GET DOWN HE'S GONNA DRAW

[coral spindle]

You agree with this @pastel cliff

?

Do you see how we might make an isomorphism like this? Maybe you can work from here

[pastel cliff]

yeah

that's where im at actually

[coral spindle]

So can you write it down more explicitly perhaps

[pastel cliff]

like if (1 2) -> 2 and (1 2 3) -> 6, (2 6) in Sym(S_3) would just be (1 2)(1 2 3) no?

[coral spindle]

I don't like that you're using equals here

they're not equal

[pastel cliff]

or well

[coral spindle]

I like the arrow better

So maybe you could give this labelling a name?

[pastel cliff]

it'd be like that then

[coral spindle]

Oh hold on I need to read this more

[coral spindle]

This is wrong.

You need to think of elements of S_3 purely as objects here. Forget their algebra completely

(2 6) is not an element of Sym(S_3). It is an element of S_6

Now there is an element of Sym(S_3) that sends (1 2) to (1 2 3) and (1 2 3) to (1 2)

We want to somehow say this corresponds to (2 6) in S_6

If we were to be horrible nasty people and write it out as a cycle, we could write ((123) (12)) but this is gross

[coral spindle]

I want to focus on this

You're assigning the label '2' to (1,2) and the label '6' to (1,2,3), that's fine

[pastel cliff]

i almost wrote this

[coral spindle]

So how do we encode a labelling mathematically?

[pastel cliff]

this was my mistake then

[coral spindle]

Indeed

Here's a thought actually

I'll wait for you to finish what you're saying though

[pastel cliff]

so we're first mapping S_3 to {1, ... , 6}

[coral spindle]

OK let's give that a name

Like there's a bijection f : S_3 -> {1,...,6}

I like that

[pastel cliff]

ok sure f

[coral spindle]

You want a different name? I don't mind

[pastel cliff]

doesnt matter

then we want to consider Sym({1, ... ,6})

[coral spindle]

Yup

[pastel cliff]

and show that it's isomorphic to S_6

i had been assuming that they were the same by defn lol

[coral spindle]

Well I would argue that S_6 is defined as Sym({1,...,6})

[pastel cliff]

oh then it is

[coral spindle]

I think rather we want to show that Sym(S_3) is isomorphic to S_6

So like, what's the isomorphism there

how can we use this function f to get an isomorphism

Just to check, are you happy with what the elements of these groups are?

[pastel cliff]

yeah

[coral spindle]

OK take the stage

[pastel cliff]

not gonna list out the elements of S_6 but S_3 is easy enough

[coral spindle]

Oh you're going to list them out?

Maybe write out the way to calculate it in general first

So I mean like, given an element of Sym(S_3), how do we get an element of S_6

[pastel cliff]

if S_3 \cong {1, ... , 6}, does that imply that Sym(S_3) \cong Sym({1, ... ,6})

[coral spindle]

Remember that we now have our labelling f

[coral spindle]

This is exactly what I'm asking you to prove.

I would like you to write out what this isomorphism is explicitly

[pastel cliff]

so like an isomorphism between the maps?

i appreciate your patience

sorry if anything i say is silly lol

isomorphism btwn maps is probably silly

[coral spindle]

I do not know what you mean by an isomorphism between maps

I've said what you need to do: given an element of Sym(S_3), give me a way of finding an element of S_6, using this labelling f

[pastel cliff]

ok i'll brb then

[coral spindle]

Remember that elements of Sym(S_3) are bijections p : S_3 -> S_3 (not homomorphisms, just maps)

Similarly elements of S_6 are bijections q : {1,...,6} -> {1,...,6}.

[simple valley]

speaking of Sym

[coral spindle]

If you have a method of doing this, you probably have an isomorphism, and we'll work on proving it is after that

[coral spindle]

Symple as can be!

[simple valley]

I think I'm looking for some tweaked version of burnside lemma/polya enumeration theorem

trying to relate $\sum_{\substack{i_1, \dots, i_n\\text{all distinct}}} \lambda_{i_1} \cdots \lambda_{i_n}$ with $\sum_j \lambda_j^p$ for various $p$

[cloud walrus]

mniip

[chilly ocean]

In a cyclic group, is every element either identity or a generator?

[coral spindle]

No

[chilly ocean]

Or can an element be neither identity nor generator

[coral spindle]

E.g. in Z_4, the element 2 is neither.

[chilly ocean]

hm

[pastel cliff]

2 generates Z/2 in that case

[coral spindle]

I mean I don't think I understand the setup fully, but this is looking a lot like rep theory

[simple valley]

it is related to rep theory

[coral spindle]

Go on then

[chilly ocean]

hm ok

[simple valley]

so I have a set X and Sym(X) acting on it, as well as a map $\lambda : Y \to \bC$, and I'm computing $\sum_{\substack{i \in Y^X\\text{some condition on $i$}\\text{related to $Sym(X)$ action}}} \prod_{x \in X} \lambda(i(x))$

[cloud walrus]

mniip

[coral spindle]

I wonder if we can describe this sum as a trace?

[simple valley]

the action of Sym(X) seems important because the product with lambda is obviously constant on orbits

yeah this is exactly a trace

[coral spindle]

Right

Ach there's some result I'm forgetting here

[simple valley]

the original problem is to relate $\chi(<Sym>\nolimits^n V)$ to $\chi(V)$ of some rep $V$

[cloud walrus]

mniip

[coral spindle]

Right

[simple valley]

so my thought is that I want to calculate $S = \sum_{\substack{i \in Y^X\\text{$i$ stabilized by any non-identity of $Sym(X)$}}} \prod_{x \in X} \lambda(i(x))$

[cloud walrus]

mniip

[coral spindle]

Well that bothers me because surely i would be a constant map?

haha it's like I'm talking about myself

Presumably this is under conjugation, rather than what I'm thinking of

Or in fact would that still be the same...

[simple valley]

Then we have $$\prod_{i_1 < \dots < i_n} \lambda_{i_1} \cdots \lambda_{i_n} = \frac{1}{n!} \left(\left(\sum_j \lambda_j\right)^n - S\right)$$ $$\prod_{i_1 \le \dots \le i_n} \lambda_{i_1} \cdots \lambda_{i_n} = \frac{1}{n!} \left(\left(\sum_j \lambda_j\right)^n + (n!-1)S\right)$$

[cloud walrus]

mniip

[simple valley]

does the idea make sense?

[coral spindle]

I'm not seeing it

[simple valley]

if we're choosing some indices, then either all indices are distinct, or some coincide

S = sum of cases where some coincide

[coral spindle]

Right, yes

[simple valley]

and then "all distinct" has a multiplicity of n!

[coral spindle]

Yes that's quite nicely done, I see

So all right, this does look like an application of the hermitian inner product

I'm wondering if we can use some kind of decomposition

[simple valley]

I've computed this for n<=4 and it looks rather awful

[coral spindle]

looking at the 2nd line, on the left hand side, I'm trying to think of what rep we're looking at

[simple valley]

that's the Sym^n V

[coral spindle]

Can I make sure I'm understanding what this means

[simple valley]

1st line is \wedge^n V

[coral spindle]

Sym^n V is Sym n acting upon V with certain eigenvalues lambda_1, ..., lambda_n?

[hot lake]

I don't think the S in both lines are the same S

[simple valley]

no

[coral spindle]

OK what's the construction?

[simple valley]

suppose G acts on V

we form the symmetrized tensor product Sym^n V = V (x) ... (x) V / permutations

[coral spindle]

Right so it's the symmetric algebra, OK

[simple valley]

then G acts on it with g * (v (x) ... (x) w) = (g * v) (x) ... (x) (g * w)

[coral spindle]

OK

Just trying to think about how this behaves quickly

Alright very nice.

[simple valley]

maybe not

but like in both cases we have $\chi(Sym^n V) \approx \frac{1}{n!}(\chi V)^n \pm$ some odd bits

[cloud walrus]

mniip

[formal ermine]

kernel and image of a homomorphism

[coral spindle]

I think we can pull this Sym^n apart into some components which will help us

[simple valley]

how so?

[coral spindle]

Well what I'd like to do is use the linearity of the inner product really

I think if we find some subreps we can then separate them out and apply that

[simple valley]

what inner product

[coral spindle]

The hermitian inner product on characters (and more generally on class functions)

It would be really nice if we could restrict our attention to the n-fold tensor product but dammit we can't...

Ach I'm struggling to put my finger on the correct group to act with here

[simple valley]

we could ask what's missing in $V^{\otimes n} = (\land^n V)^{n!} \oplus {?}$

[cloud walrus]

mniip

[coral spindle]

Yeah I think I agree with that direction if not the exact one

I think we could act on the n-fold tensor product with S_n x G or something like that?

[simple valley]

we could

[coral spindle]

Then there's an appropriate quotient, I'm sure

[simple valley]

average over S_n?

maybe not

[coral spindle]

Well we'd have to do some magic

The average over n is too simple

But the hermitian inner product would probably fit the bill

Let me get some actual paper lol

We're happy to work over C, right? I assume we are

[simple valley]

yeah

[coral spindle]

Good lol

Would be terrifying if we weren't

[pastel cliff]

gonna interject rq, but using the stuff we talked about back here, f: S_3 -> {1, ... , 6} is a bijection so we can use the homomorphism from the group action and then use something like g: {1, ... , 6} -> Sym({1, ... , 6}) to get g \circ f : S_3 -> Sym({1, ... , 6})

[coral spindle]

I think you're almost there stu, but I think the g thing is off

at least I don't see what that is

[pastel cliff]

we have the group action of S_3 on itself

[coral spindle]

I'm gonna take 5 minutes so give me a moment

[pastel cliff]

all good i really appreciate the patience and help

[simple valley]

the book where I took the exercise from seems to suggest a completely different approach

[coral spindle]

Yeah this S_n x G idea is too complex I think

I'm gonna make a thread for stu

[simple valley]

theory of symmetric polynomials

[coral spindle]

gonna interject rq but using the stuff

Ah, some Newton stuff huh

[coral spindle]

[simple valley]

seems like in the end it does boil down to partition combinatorics

[coral spindle]

As expected I guess...

Gross though

[simple valley]

which is the same kind of combo you'd get by iterating over cycle types in S_n or whatever

I think I'll skip this exercise because it doesn't seem to have any rep theory content left in it, just combinatorics

[hot lake]

the formulas I know are of the kind chi(sym²(rho))(g) = 1/2 (chi(rho)(g) + chi(rho)(g²))

[simple valley]

this only works for 2

[hot lake]

for 3 I expect it's a linear combination of chi(g)^3 chi(g)chi(g²) and chi(g³)

[formal ermine]

#1029070214459048056

[tough raven]

This and the ideas mentioned about it in chat are really interesting, thanks!

I should learn how partitions correspond to irreducible representations of S_n…

[agile burrow]

If I have an exact sequence of abelian groups, I can consider the sequences induced by looking at the p-torsion of each group. It seems like these sequences are still exact. Does this correspond to the fact that Z localized at p is a flat Z-module, hence tensoring by this preserves exactness? There's probably a simpler explanation using fundamental theorem of f.g. abelian groups as well

[tough raven]

Oh hmm if I take a partition, take the corresponding isotypic component of V^{\otimes n}, ask for the character of g with eigenvalues λ_1,…,λ_m, do you get all sums of degree-n products of λ's where the exponents form exactly that partition or something?

[next obsidian]

p-torsion is measured by Ext groups, it’s specially Ext^1(Z/p,A) and use the LES and that Ext^2 is always 0

[tough raven]

Is this exact?
I was thinking of Z -> Z -> Z/pZ -> 0 with the map mult by p at the start.

[agile burrow]

hmmm you're right

[next obsidian]

So yeah you can fail to be exact on the right

[tough raven]

It is left-exact though.

[agile burrow]

this computation is killing me!

[tough raven]

Maybe it's given by a Hom-functor?

[next obsidian]

Br.uh

I just said it’s given by Ext

😭

wait

I meant Tor

TOR TOR NOT WXT

[tough raven]

I don't know what Ext means

[next obsidian]

MOT EXT I REPEAT NOT EXT

[tough raven]

Or Tor

[coral spindle]

[agile burrow]

I do be tensoring

[coral spindle]

You want a tensor product indeed

[next obsidian]

Yeah but it’s tensor with Z/p

Not with Z_p you want to look at

[agile burrow]

hmm ok so maybe I'm thinking about this wrong

[tough raven]

Wait p-torsion is given by tensoring with Z/pZ?

[agile burrow]

why is Z_p not the right thing to look at

[next obsidian]

Like, localizing at p doesn’t really give anything about torsion

[next obsidian]

No, it’s Tor^1(Z/p,A)

[agile burrow]

I guess so

[next obsidian]

Tor is the derived functor of tensor

[agile burrow]

Yeah ok that makes sense, localizing shouldn't do anything to the quotients

[next obsidian]

Like, even in a non-integral domain

Or err

Yeah idk what I want to say

I mean what I was gonna say is, localizing at p

[agile burrow]

I'm working with the exact sequence 0 -> Z/2 -> Z/4 + Z/6 -> H(SL_2(Z)) -> 0 and author was like, look at the 2- and 3-torsion to figure it out!

[next obsidian]

The map A -> A_p has kernel things which die to elements OUTSIDE of p

[chilly ocean]

Generalization of Hom functor

[agile burrow]

and it's making me

[simple valley]

wouldn't call it a generalization

it is derived hom yes

[agile burrow]

oh also, once I know the 2- and 3-torsion I can just direct sum these by fundamental theorem of abelian groups, right?

[next obsidian]

Your mother was a whore

Are they fg?

[agile burrow]

grow up

look at the sequence bozo

[next obsidian]

Honestly

Idk why I said that

[next obsidian]

Grow up

[formal ermine]

this is the first thing I saw when I clicked on the channel

[next obsidian]

I actually okay so why I said it

[simple valley]

just math server things

[next obsidian]

I was thinking about the Harry Potter puppet pals

“The elder swear”

Or whatever

And it’s like your mother is a beeeeep

And then a bunch of other stuff

[simple valley]

chmonkey please this is abs alg

[next obsidian]

And then I just typed that

[coral shale]

its an honor to have mniip in abs alg

[untold basin]

Hello

[coral spindle]

Welcome

[untold basin]

I have t in L, a transcendant number on K. Let P(X) in K[X], deg(P) >=1. (K a subfield of L)
Show that P(t) is transcendant

[coral shale]

what have u tried

[coral spindle]

You looking for a hint?

[untold basin]

I am writing what I did

I have put the defs :
t is a transcendant number on K <=> For all Q in K[X], Q(t) = 0 <=> Q = 0 (1)
P(t) is a transcendant number on K <=> For all T in K[X], T(P(t)) = 0 <=> T = 0 (2)

Are these right to begin?

[coral spindle]

Yup that is the definition

[untold basin]

We learnt in term of Ker of the function evaluation and I developed it

Okay so :
Let T in K[X].
<= is trivial.
=> : I assume that T(P(t)) = 0. Now, (The only guess I have is to use (1) so I particularise Q = T o P and I have T o P = 0)

[coral shale]

(T o P)(t) = 0

but yh, u have the right idea

[untold basin]

That's what I assumed right ?

[coral shale]

yes, so you are going for a contradiction, really

[untold basin]

mmh no ?

[coral shale]

or not ig

[untold basin]

Where do you think I did that

[coral shale]

sorry misread your statements

yeah, continue then, u have the right idea

[untold basin]

I'm sure it is also trivial but I can't continue xd

I will search a bit, now that I know I am on the right path

[coral spindle]

Well if T were nonzero, Surely T o P would be nonzero too, right?

Try proving that.

[untold basin]

I thought of doing that but I sounded weird for me

I will dig in

Thx

[untold basin]

I might think a bit too far but, is the set of polynomials a domain ?

Like, I thought of doing T o P = 0 => T = 0 or P = 0

I might have used the wrong words to use that

[pastel cliff]

can a simple group be itself normal

[coral spindle]

Well it is under multiplication, but you'd have to prove it the other way.

[coral spindle]

Yes, trivially N is normal in N x G for any group G.

[pastel cliff]

havent gone back to that problem from earlier, doing other ones now

[coral spindle]

Just wanted to add to this: the set of polynomials does not form a ring under composition, as the composition will not distribute over addition in general.

So it is not correct to call it a "domain" in this sense.

[untold basin]

Yes, I realised that after saying it

Can you give a hint ?

Do you develop the def of a polynomial to do that ?

[coral spindle]

The proof of this is really quite straightforward. Try it.

[untold basin]

I know :/

[chilly radish]

Consider the term of highest degree

And recall that K is a domain

You just have to sbow it's nonzero

[coral spindle]

Spoilerrrssssssssss

[untold basin]

Even with that, I can't 💀

[chilly radish]

It seems they weren't getting anywhere, I wanted to give another hint

[untold basin]

Okay so I have :
I assume that T =/= 0 i.e., there exists a in K such that T(a) =/= 0

[coral spindle]

This isn't a good start

[untold basin]

lol

That's why I'm struggling ?

[coral spindle]

It is best to think of polynomials as symbols rather than functions. In general there is a HUGE difference

So rather say T =/= 0, so T(x) = a_nx^n + ...

And maybe you can continue from that idea.

[untold basin]

I did that

[coral shale]

small lemma u can convince/prove to yourself for this Q is deg(f o g) = deg(f)deg(g)

[coral spindle]

To demonstrate what I mean, consider the field F_2. The polynomial x^2 + x in F_2[x] is constantly 0, but is not 0 as a polynomial.

[coral shale]

in fields of char 0, oops ignore this lol

hmmm

[coral spindle]

It's true in any ring without zero divisors

[coral shale]

yh, cba to think lmao

[pastel cliff]

is this a valid proof

im 99% sure it's good just double checking

[coral spindle]

Looks good to me

[formal ermine]

does $S({1,\ldots,n})$ where $S(X) = {f : X \to X : f \text{ bijective} }$ have $n!$ elements?

[cloud walrus]

illuminator3 (#eric4honorable)

[coral spindle]

Do you think so? Can you tell me why?

[formal ermine]

if f must be bijective then it's like having n boxes and n different numbered cookies to place in the boxes

for the first box we have n possibilities, for the second n - 1, etc

so just factorial basically

[coral spindle]

I agree that's a good intuition

Indeed it is true that there are n! elements of Sn

If you wanted to show this thoroughly I think induction would be helpful

[coral shale]

can be done via induction

[coral spindle]

jinx

[formal ermine]

why is $S_3$ not abelian

[cloud walrus]

illuminator3 (#eric4honorable)

[formal ermine]

how is the commutativity broken?

[chilly radish]

It is also not true (Over a general ring) that a nonzero polynomial will give a nonzero function. It IS true eg over an infinite field

oh you said it

[formal ermine]

ah wait

I think I got it

wait nvm

lemme try something

[coral spindle]

Choose pretty much any two cycles and see what you get

[formal ermine]

what are two cycles

[coral spindle]

E.g. (12) and (123)

Ok

(12) is the function that sends 1 to 2, 2 to 1, and 3 to 3.

(123) is the function that sends 1 to 2, 2 to 3, and 3 to 1.

Check if these commute.

[formal ermine]

(abc) means a->b,b->c,c->a?

and (ab) is a->b,b->a,c->c?

[coral spindle]

That's right.

[formal ermine]

what do you mean with 'to commute' (language barrier probably)?

[untold basin]

I tried this but I get a horrible thing with T o P

[chilly radish]

that's why you wanna focus on the term of highest degree

again

[coral shale]

if ab = ba, then a and b commute

[formal ermine]

ah

thanks

[pastel cliff]

im being asked to find the number of Sylow 2-subgroups of a group of order 144 (no info on what G looks like)

[formal ermine]

yeah if you plug in 1 to (123) then (12) you get 1, but if you first put it into (12) then (123) you get 3

[coral spindle]

Good job

[formal ermine]

so they're not commuting

thanks

this was simpler than I thought

[pastel cliff]

what i've got right now is that by Sylow 1, we know that a Sylow 2-subgroup exists ofc. the highest power of 2 that divides 144 is 16, so it's order must be 16

and by Sylow 3, we know that the number of Sylow 2-subgroups is 1 mod 2 (so it's an odd number) and that number divides 144

where can i go from there? my next idea is to just look at the prime decomposition of 144 but that feels silly bc it wouldn't scale well for really large groups

[coral spindle]

You need to look at the prime decomposition of 144.

[pastel cliff]

oh damn alright lol

144 = 2^4 * 3 * 3 so it must be 3 or 9 right

there's no way to decide which without more information about the group right?

[formal ermine]

,, (g^{-1})^{-1} = (g^{-1})^{-1} \circ e = (g^{-1})^{-1} \circ (g^{-1} \circ g) = ((g^{-1})^{-1} \circ g^{-1}) \circ g = e \circ g = g

[cloud walrus]

illuminator3 (#eric4honorable)

[formal ermine]

oh it doesn't fit

is this a valid proof of $(g^{-1})^{-1} = g$

[cloud walrus]

illuminator3 (#eric4honorable)

[next obsidian]

I think it’s more useful to show that inverses are unique

[coral spindle]

It is valid though, even so

[next obsidian]

So if gh = e and gk = e then h = k

[south patrol]

interesting use of circ

[formal ermine]

how would that help us here?

[pastel cliff]

just wanna confirm pls

[next obsidian]

well g^-1•g = e

[coral spindle]

I think orbit-stabiliser would help you here

[next obsidian]

But also g^-1•(g^-1)^-1 = e

By definition of inverse

[pastel cliff]

i mean the question said that the ans should be a list of possible values

[coral spindle]

(Also known as the 3rd part of the sylow theorem)

[south patrol]

(and the other way round ofc ^)

[formal ermine]

ahhhhh

I see

[pastel cliff]

that sylow p-subgroups are in the same orbit under conjugation?

[formal ermine]

thanks

[coral spindle]

I was wrong, I don't think orb-stab helps

I think the question does just want you to write 1, 3, or 9 as the list of values.

[pastel cliff]

yeah i tried using the conjugate thing but i got nowhere

wanted to make sure

merci

[next obsidian]

If you want to show that this doesn’t contain any extra

[coral spindle]

de rien

[next obsidian]

You should be able to produce groups with that many Sylows

But tbh I would just say fugg it

Seems like massively ass

[coral spindle]

Not the easiest thing ever to do so

[next obsidian]

Yeah

[untold basin]

I'm done

It's been < 2 hours xd

[coral spindle]

OK try again later

[untold basin]

done mentally

It's 11pm here

[coral spindle]

You can do this, you're just not doing it rn

[untold basin]

Can I tell you what I have

[coral spindle]

OK that's nice

[next obsidian]

Boytjie are you a grad student in some algebraic subject?

[coral spindle]

I am yeah

[next obsidian]

Like rep theory or something

Idk I took a random guess

[coral spindle]

Bingo

[next obsidian]

Lmfao

[coral spindle]

yeah rep theory of algebraic groups

[next obsidian]

Oh cool

[coral spindle]

which I would like to tell you more about if I actually knew...

[next obsidian]

It do be like that

[agile burrow]

sounds cool though

[coral spindle]

They are pretty neat

You get nice representations from the coordinate ring

it's very swanky

[next obsidian]

Do you get to like

[agile burrow]

hopf algebra!

[next obsidian]

Mainly pretend stuff are varieties

Or do you have to be pretty well-versed with schemes

[coral spindle]

Yeah well I'm nowhere near group schemes

[next obsidian]

I know Waterhouse deals with schemes

But they define them funny

Lol

[agile burrow]

i wanted to get more into hopf algebra rep theory for a while

[coral spindle]

My supervisor wants me to give this particular problem he selected a shot before I get my head in the clouds

so it's just kinda, getting up to speed for that yknow

[next obsidian]

Is group schemes having your head in the clouds…

[coral spindle]

I'd say so

But yeah a lot of learning to do

[next obsidian]

Idk

I mean it’s nice to have an actual group

Keeping it classical

[coral spindle]

Exactly

[next obsidian]

Haha

Idk I mean

[coral spindle]

Plus like, as I said he's got a particular project for me to do with these things

[next obsidian]

You get groups with group schemes you just have to Yoneda

[coral spindle]

Aiyah

[next obsidian]

Yeah idk I just find it stunning to call scheme stuff head in the clouds when like

To me stacks aren’t things that qualify as head in the clouds

[coral spindle]

Hahaha

Well the point is there's still loads of work to be done with just plain ol' algebraic groups

[next obsidian]

Yeah

I mean group schemes do get pretty nasty

And then you have to read SGA

[coral spindle]

Pain

[next obsidian]

French

[coral spindle]

Ceci n'est pas du pain

Il faut... il faut que je te dise....

il faut que je te dise nuts

C'mon that's hilarious

[agile burrow]

I did enjoy it

[pastel cliff]

what kind of taco is that

[coral spindle]

You don't wanna find out...

I don't want to harass you into doing this question buttttttttt

have you found that isomorphism between Sym(S_3) and S_6 yettttttt

[pastel cliff]

im doing Sylow stuff for now lol

it's corrections for homeworks so im jumping around between two of them

[coral spindle]

Fine fineeeee

[pastel cliff]

i'll do it soon i cant avoid it forever

especially with a petrifying exam on wednesday

[next obsidian]

?????

What????

That’s not even possible by size no???

Or is Sym the automorphism group?

[coral spindle]

Sym's just the group of bijections haha

[next obsidian]

Chmonka

Lmfao

[pastel cliff]

[next obsidian]

I considered it but…

Writing S_3…

[pastel cliff]

:bleakchmonkey: when

[coral spindle]

This is part of an exercise to find a way that S_3 acts on {1,...,6} transitively

And ofc it uses Cayley

[next obsidian]

Wait what lmfao

[coral spindle]

Haha maybe you should do it too!

[pastel cliff]

here one sec

[next obsidian]

Okay that makes sense

No I thought about it harder

[pastel cliff]

A subgroup of $S_n$ is called transitive if its natural action on ${1, \dots, n}$ is transitive. Find a subgroups of $S_6$ which is transitive and isomorphic to $S_3$. (Hint: look at the proof of Cayley's theorem).

[cloud walrus]

stμ₂dying

[next obsidian]

I already know how to do this

Is what I meant

[coral spindle]

Well the computation is proving a stumbling block

[pastel cliff]

im not a bitch i wont take offense to that lol

Proof: left to chmonkey

[coral spindle]

Mate you'd be surprised how many people base their self-worth on mathematics...

It's sad but what can you do

[pastel cliff]

i mean 90% of my class is grad students

i may be incredibly behind but at least im gonna get graded slightly more leniently

[next obsidian]

In my experience UGs get graded harder because grad student grades don’t matter

Either that or it’s equal

[pastel cliff]

dont say that

shut the hell your mouth

[next obsidian]

🤐

[pastel cliff]

i just need to make it through this week

[next obsidian]

Me talking to myself every week

[next obsidian]

[pastel cliff]

my life is cyclic in structure

[coral spindle]

generated by one element: you!

:)

[untold basin]

@coral spindle My friend did it in 5 lines ! xddddddddddddddddddddd

I am gonna do a backflip ^^

[coral spindle]

Ok I'm glad your friend managed it

Did you look at the proof, do you understand it

[untold basin]

He directly went for absurd

Not of all you recommended me

[coral spindle]

Good for him

[untold basin]

I am pissed off tbh

[coral spindle]

Okay

Have you looked at the proof and understood it?

[untold basin]

Yes I'm pretty sure his proof is correct

[coral spindle]

That's great

[untold basin]

Thanks for your help btw

[solar glacier]

i have a question

[agile burrow]

ask away friend

[solar glacier]

of proof verfication

ok so heres my proof solution to group of order 48 having unique normal subgroup

48 is 2^4 times 3

im gonna consider 3-subgroups, i.e., n_3

[coral spindle]

Hold on hold on

That's just not true in general

[solar glacier]

n_3 = 1 or 16 since its congruent mod 1 to 3

[coral spindle]

You need more restrictions on this group. At the very least, it needs to be non-Abelian

[solar glacier]

really

i thought all groups of order 48 are not simple

i.e., contain non triviial normal subgroup

[next obsidian]

But you said unique

[solar glacier]

i end up showing n_3=1 your done

and if n_3=16 then you get than n_2=1

[next obsidian]

So you’re showing the existence of a non-trivial normal subgroup

[solar glacier]

yes

[coral spindle]

So in fact, this word "unique" was completely extraneous

[next obsidian]

You are not showing that groups of order 48 have a unique normal subgroup

[solar glacier]

a normal subgroup

sorry

[coral spindle]

explain this please, I don't see why this is true

[solar glacier]

if n_3 = 16 there are 16 times 2 = 32 elements of order 2

[coral spindle]

You've also forgotten the n_3 = 4 case

[solar glacier]

thus 24 elements of not order 3 remain

oof ok

so was this case ok

cause i could make a similar argument probably

[coral spindle]

I don't see how that follows

[solar glacier]

and those 16 remaining elements are s 2 subgroup

cause they share the identity

[coral spindle]

So why would they be of order 2

[solar glacier]

yes

times 16 since theres 16 of them

and theyre disjoint by langrange

[coral spindle]

No I'm asking why

[solar glacier]

sorry of NOT order 2

[coral spindle]

You're looking at sylow 3-groups, and you're talking about elements of order 2?

Yeah exactly.

[solar glacier]

sorry

[coral spindle]

Maybe you should write out your proof in full instead of going through it message-by-message, because I keep stopping you to point out mistakes.

[solar glacier]

i meant to say the rest of the elements are of order NOT 3

[coral spindle]

Write it out first, then check it yourself before posting here.

It's very hard to keep track of what's going on right now.

[solar glacier]

does my argument somewhat make sense to you

[coral spindle]

I haven't seen your whole argument

[celest cairn]

How do I solve 3x = 12?
(Wheel Theory)
I need help understanding the step by step process.

[pastel cliff]

what do we get from dividing the order of a group by the order of one of its elements?

im trying to find all Sylow 2-subgroups of S_5

by Sylow 3 there must be an odd number of them and it must divide 120, so choices are 1,3,5, or 15

[next obsidian]

A natural number

[pastel cliff]

[next obsidian]

Wdym find “all Sylow 2”

Like are you trying to literally write them down element by element?

Or are you just trying to determine the number?

[pastel cliff]

no im assuming it means determine the number and what they "look like"

[next obsidian]

Bleka

[pastel cliff]

[next obsidian]

,w factor 120

[cloud walrus]

Results provided by WolframAlpha

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

Requested by Chmonkey 2.0#9767

[agile burrow]

well they look conjugate to each other

[next obsidian]

That’s true

And conjugation is easy to describe

Also they’re only 8 elements big

[agile burrow]

real

[next obsidian]

I would just find a subgroup of order 8

Then use that to find the others

It won’t really tell you the number of them but…

[agile burrow]

iirc there's a fun trick to finding the number

[next obsidian]

[pastel cliff]

for finding the number of them, i was wondering if 120/8 = 15 indicated anything

or if it's just a coincidenc here that order of S_n/order of Sylow 2-subgroup = number of Sylow 2-subgroup

[next obsidian]

Sort of coincidence

By orbit stabilizer # of Sylow 2 = index of the normalizer of a Sylow 2

This tells you that the index of the normalizer is the index of the Sylow 2

But normalizer contains Sylow 2, so they’re equal

[agile burrow]

in fact

methinks equality holds iff p = 2

but i forgor why

[next obsidian]

Cuz

Joe mama

[pastel cliff]

proof by

[agile burrow]

chmonkey i couldn't get it to work

i couldn't compute 2-torsion of homology of SL(2, Z)

devastated

[next obsidian]

Rip

Time 2 dead

[agile burrow]

yes

[pastel cliff]

D_8 :O

[next obsidian]

:O

[agile burrow]

:O

[next obsidian]

You can prove the existence of a D_8 using group actions

Which is coo

[pastel cliff]

wha

i was just gonna be like

[next obsidian]

D_8 acts faithfully on a square

[pastel cliff]

observe D_8 \leq S_5

[next obsidian]

So you get an injevtive map into Sym([4]) = S_4

Then S_4 embeds into S_5

[pastel cliff]

it's a sylow 2-subgroup and they're all conjugate to one another

[next obsidian]

Yeah but how do you show D_8 < S_5

You could find one by hand

Or you can show it abstractly

Actually my proof gives you 5 of them

Cuz you can embed S_4 in 5 different ways (maybe more, but they won’t be the “obvious one”)

[pastel cliff]

shoutout to user169852

[next obsidian]

Oh yeah

That makes sense

[celest cairn]

How do I solve 3x = 12?
(Wheel Theory)
I need help understanding the step by step process.

[next obsidian]

Literally nobody on earth knows wheel theory

[pastel cliff]

what the fuck is wheel theory

it spins

what more do you need

[agile burrow]

It would be wheely cool if someone could help you

[next obsidian]

A wheel is some really memey algebraic structure, the main point of it is that you can “divide by 0”

[sage spruce]

Rings with division?

[next obsidian]

Idk what that means

[pastel cliff]

https://tenor.com/view/rubber-tyre-tire-gif-12706826

[glass grail]

no need to retheorize the wheel

[next obsidian]

But like, nobody knows anything about them, it’s like asking about semigroup theory

Like, nobody actually cares about those lmao

[celest cairn]

lmao ok

so.. no one here knows a solution?

[tribal moss]

It seems likely that x=4 would be a solution, but you'd have to check that yourself against the definitions of your "wheel theory".

[celest cairn]

x is indeed 4, but I just wanted learn how to compute it using wheel logic.

It's just a different way of solving for x. A very rigorous computation.

[pastel cliff]

idk why but "wheel logic" is the funniest shit i've seen all day

[celest cairn]

lmao

[pastel cliff]

im a terrible math major

[prisma ibex]

nobody studies wheel theory lmao

[celest cairn]

Exactly the reason why I want to study it lmao

[prisma ibex]

maybe there's a good reason why nobody studies it

[solar inlet]

wheels are not studied precisely because they aren't structurally that different from rings

[celest cairn]

Probably, but I dont care 😂

[prisma ibex]

also, just some advice, studying something for the reason that it's obscure is a very bad move

[solar inlet]

especially when you know nothing about it

[prisma ibex]

you won't get any help here for such endeavors

[pastel cliff]

ive got two wheels right here that you're free to study

[solar inlet]

also I'm willing to bet the couple people who have studied wheel theory don't care that much about the arithmetic of it's elements

[celest cairn]

https://www2.math.su.se/reports/2001/11/2001-11.pdf

page 31

Example 4.19

[celest cairn]

I know, I was joking. But I’m just interested in it.

[toxic zephyr]

in Q[x], the ideals generated by x+2 and 2x+4 are the same ideals correct? since 1/2*(2x+4)=x+2

[proud fox]

yeah

[pastel cliff]

counting the number of Sylow 3-subgroups of S_5 - how does this exclude 40?

[agile burrow]

if there were 40 Sylow 3-subgroups, then this would account for 40 * 2 = 80 distinct 3 cycles. A_5 only has 60 elements

[pastel cliff]

gotcha

For $p = 2,3,5$, find all Sylow $p$-subgroups of $\mathbb{Z}/15 \times \mathbb{Z}/10$.

can i get a nudge for this pls

S_5 was easier

ugh no preamble

[cloud walrus]

stμ₂dying

[pastel cliff]

well actually ok

sylow 2-subgroup of this must be of order 2 cuz 150 = 2 x 5 x 3 x 5

so it has to be a 2 cycle

is this argument common? i used it to "find" all Sylow 3 and 5 subgroups of S_5 as well