groups-rings-fields

[wooden ember]

been struggling at it for a decent bit

[lethal dune]

what about ZxZ -> Z?

that's a elusive way to give an exercise I admit

[wooden ember]

hmmm i can see this works but i dont want it to

because then the last part of the exercise cant possibly make sense

[lethal dune]

can you show the exercise?

[wooden ember]

(here i mean part iii)

[lethal dune]

huh

[agile burrow]

You're supposed to be looking at endomorphisms of a module, not maps in general

[wooden ember]

well yeah

[agile burrow]

So why would R x R -> R be an issue

[wooden ember]

R is embedded in RxR

[lethal dune]

it's not from M -> M

[wooden ember]

so i can consider it as an endomorphism

right?

[agile burrow]

And when you restrict the map to R, you have a different map

[wooden ember]

oh right

yeah ofc

lol

okay that's good news

still not sure how to show sujectivity => injectivity though

[agile burrow]

Maybe try thinking about repeatedly applying f. What can you say about the kernels of each f^n

[wooden ember]

alright ill try working with that, thanks!

[tender bough]

about the second example, does a general linear lie algebra satisfy the alternating property? I don't think AA = 0 for arbitrary matrix A?

oh wait, nvm

so the multiplication in the algebra is (A, B) -> AB - BA

[tribal moss]

An n-dimensional unital algebra over a field k always embeds into the matrix ring k^{n×n}. So there's a 4×4 matrix representation of the quaternions.

But there's a celebrated theorem of Frobenius stating that this is the only example over the reals, other than R and C, up to isomorphism.

[jovial kelp]

hey real quick question over coset notation.
So say if $G = \mathbb{Z}$ and $H = \langle 4 \rangle$. So $H1 = {..., -11, -7, -3, 1, 5, 9, 13, ...}$. So $H1 = H5 = H-7 = H-11 = ... $ etc.

Would it be okay to say the coset would be something like $H1 + 4n$ for $n \in \mathbb{Z}$? Or is it sufficient enough to just use one number from the coset

[cloud walrus]

MellowDramaLlama

[chilly ocean]

You can stick to one

Also multiplicative notation is innapropiate here

[jovial kelp]

ah so something like H + 1 then? We only did the multi in class so far

[chilly ocean]

Yes. Or 1+H

[jovial kelp]

okay cool

thank you

[trail stump]

so I guess i have seen proofs of, kernel of a group homomorphism is normal. I guess I know how to prove it but i seem to lack some deep understanding of why it works like that, and just want to ask for some good intuitions here

in other words, aH=Ha for all a in G. H is subgroup of G

especially examples of non abelian, in which ah=ha is not necessary

[chilly ocean]

Image of conjugate is a conjugate.
And trivial subgroup is always normal

So this is a special case of preimages of normal subgroups being normal

But in a very special case

[trail stump]

never heard of the term conjugate subgroup before, thx for your inputs, will learn more this example

[chilly ocean]

No, I meant of elements

[trail stump]

elements?

like in complex conjugate?

[chilly ocean]

Like element of a set

Here, the group

[trail stump]

what do you mean conjugate exactly?

[chilly ocean]

A conjugate of x is gxg^-1 for some g

Homomorphism maps a conjugate to conjugate

[trail stump]

oh i see interesting

[tribal moss]

"aH=Ha for all a" is equivalent to "aha^-1 is in H for all h in H and all a in G" -- that is, "H is closed under conjugation by arbitrary elements of G".

The latter description is often a more intuitive way to think about normal subgroups.

[abstract rock]

i petition instead of normal subgroup the property should be called the abelian coset property just to spite everyone

[trail stump]

is g^(-1)xg and gxg^(-1) the same?

[chilly ocean]

Usually not

[trail stump]

oh wow

[abstract rock]

depending on your choice of left or right convention g(inv)xg maps to g(inv), and similarly gxg(inv) maps to g for the canonical homomorphism G/X to G for abelian coset subgroup X contained in G

[tribal moss]

You can make one into the other by replacing g by g^-1. So if you're talking about all conjugates of x, the difference doesn't matter.

[trail stump]

right

[tribal moss]

I don't think there's any such "canonical homomorphism" -- in many cases the only homomorphism G/N -> G is the one that sends everything to the identity.

[chilly ocean]

[abstract rock]

woope

meant the other way, evidently

[trail stump]

and my goal right now is to learn as much of those "many cases" as possible

[tribal moss]

But the canonical homomorphism G -> G/N always maps g^-1xg and gxg^-1 to the same element of G/N -- namely the coset that x lies in! -- so there's no difference between the conjugation conventions there either.

[trail stump]

can i at least get an example?

thats not in real line?

[tribal moss]

The only homomorphism Z/7Z -> Z is the trivial one.

(replace 7 by any nonzero integer too)

[abstract rock]

no difference up to isomorphism but thats why its a convention and doesnt change the outcome with the choice of left or right.

[tribal moss]

No difference up to anything.

[chilly ocean]

The identity isomorphism

[earnest jetty]

Let G be a group and suppose x ∈ G is the only non-identity element such that x^2 = ε. Then
xy = yx for all y ∈ G?

how to proof this

[void cosmos]

wha did u try

@earnest jetty

and can u post the actual problem

[earnest jetty]

The groups is multiplicative groups.

this is everything

[pastel cliff]

.

[earnest jetty]

i didnt try anything

becuase i have no idea how to start this

[void cosmos]

so u have an element x

and no other element than x satisifies x^2=e

well ig u can try a contradiction

suppose xy !=yx for some y in G

try to deduce anything

[earnest jetty]

idk how

[hollow mica]

I'm confused, can't I just let T(x) = B(x, y) and then I'm done?

[chilly ocean]

what you wrote makes no sense

is that what you meant to write or is it missing something?

[hollow mica]

for fixed x, the function B(x, y) is a linear map from V to V*, and I am saying to use that as T

wait no it's from V to V, it's just an operator

hmm

[chilly ocean]

you have the right idea

and are pretty close to the answer

think about it some more

[hollow mica]

$T: x \mapsto B(x, y)$

[cloud walrus]

mrean

[hollow mica]

this works right?

so my original answer was right

[chilly ocean]

it makes no sense as written

[hollow mica]

I just explained it poorly

huh

B(x, y) is a functional in y for fixed x

is what I'm trying to say

is that not clear

[chilly ocean]

it's clear

i'm just being nitpicky

write B(x, -) for that

[hollow mica]

ah

[chilly ocean]

well, it's clear to me, but it wasn't clear to me that it was clear to you

so i had to make sure

[hollow mica]

it is clear now

ty

[chilly ocean]

T(x) = B(x, -), basically

using the dash in bilinear or multilinear functions is pretty common notation for "the function that puts things in this component"

[hollow mica]

makes sense

wait is - or a bullet point more common

[chilly ocean]

idk

i see - more

both work

[hollow mica]

my prof writes bullet points

and he is never wrong

so now i'm in a dillema

I will flip a coin

bullet point it is

does anyone know why we can rearrange jordan blocks up to similarity

if all the eigenvalues are distinct than this is easy

but apparently it's also true when the only eigenvalue is 0

[tribal moss]

Do you recall anything about the orders of conjugate elements, such as x versus y^{-1}xy ?

[tender wharf]

I'm incredibly confused by the answer provided to this question

I can see how A would form a subgroup but

I don't really understand the latter part

Hmm nevermind I think I somewhat understand.

[sage spruce]

That's some sort of group action isn't it?

[tender wharf]

I'm not sure I really understand

It's clear that |A| = |B|

and A is in fact a group

B is however not

[sage spruce]

Yes but the addition is injective

Well it's not

But add an even number to an odd number you'll get an odd number

That's it

[tender wharf]

Yes I got that

I don't see how showing the above applies to any arbitrary subgroup though

[sage spruce]

By Zn do you mean {0,1,...,n-1} with the operation of adding mod n?

[tender wharf]

Yup

In my own (failed) attempt to prove this I was able to show that no subgroup exists where everything is odd

I'm just not able to demonstrate that exactly half must be odd

[sage spruce]

Then you know the subgroups of Zn

I guess

Yes you do

Those are the Zd where d divides n

So the demonstration works fine

[tender wharf]

Oh that's what I was missing

I'll think a bit more on that

Thank you

[sage spruce]

You're welcome

To show that you just need Lagrange theorem stating that the cardinal of a subgroup divides the cardinal of the entire group

[tender wharf]

Oh um I haven't gotten that far in my book

(Gallian)

Are the exercises meant to be doable assuming no further chapters?

[sage spruce]

It depends

Sometimes they assume you know that

[tender wharf]

Right

In that case perhaps I'll do a few more chapters first and just mark this off

(I'm self-learning as a hobby so I don't have an actual curriculum)

[sage spruce]

It's unfortunate because I have a lot of good resources on group theory but they are in french

[dusty jacinth]

Guys can someone explain me that for the set R if I say that any element a has a inverse -a then it should be like a+(-a)

As for the identity element it should be it should be 0

[tender wharf]

If you use the addition notation

[tender wharf]

Why does m specifically have to be the least positive integer

I follow everything else but this choice is confusing for me

[eager willow]

The division algorithm guarantees a remainder r< m, which is where that assumption of least positive gets used.

[tender wharf]

Right but why can't m be like

a^(m+1)

That's still in H no?

[eager willow]

You're trying to prove H is cyclic

You need to find a generator

All you have is a generator for G by assumption so you must find a generator of H in terms of the generator assumed to exist

Only the smallest power is going to get you anywhere to that end

[tender wharf]

Right

That makes more sense

Yeah that makes a lot more sense thanks

[hollow mica]

For any vector space V, how can I show that the only element that is in the kernel of every functional (element of V*) is 0

I got here by trying to show that the canonical isomorphism between V and V** is injective

[little root]

try showing that if v ≠ 0, then there is a functional f such that v is not in ker(f)

[gritty skiff]

so i'm learning about ideals and factor rings and i was wondering
is Z[x]/<x^2 + 1> isomorphic to the ring of gaussian integers? since we're essentially considering x^2 + 1 = 0

[hollow mica]

got it

[glass grail]

yea

[gritty skiff]

ah so then

i've heard of the dual numbers being like
a + bj where j^2 = 0

so would Z[x]/<x^2> be isomorphic to those?

[chilly ocean]

Yes. Well, technically C[x] where C is complex numbers, or something similar

In fact, I think a+bj where j^2 = 0 forces the definition of dual numbers to be C[x]/(x^2)

just the variable x here happens to be denoted j, and instead of writing a+bj + (j^2) we happen to write a+bj

isomorphism is, in fact, just a change in notation

well, not in category theory in general, but for rings and other algebraic structures, it is

[chilly radish]

I think usually you take R[x]/x^2 instead of C, otherwise you get some kind of dual-complex hybrid

[chilly ocean]

Ah. Yes. True

[wooden ember]

any advice on finding composition series of modules for even easy examples? I think im lacking a decent amount of intuition on modules and how they behave under quotients, ive been trying to get a composition series for a quotient module of a polynomial ring on two variables and tried doing it naively by taking an ascending chain of ideals that seem to be the smallest that contain each other but the quotients dont end up being simple...

it's the existence of that second polynomial variable that just kinda messes things up

[latent anvil]

@wooden ember one thing that might be useful for you is that you can always refine a series to a composition series (assuming the module is finite length, ie has a composition series at all)

So try taking your series with non simple quotients and seeing if you can insert things between the terms with a bad quotient

[wooden ember]

that's what ive been trying to do but it feels to me like my inclusions are minimal

[latent anvil]

Also keep in mind that submodules between N and M are submodules of M/N

Can you post the example?

[wooden ember]

i cant sadly because it's an exercise i need to hand in

so dont want to go into specifics

it was in case anyone had general advice

[wooden ember]

hmm actually this might be a good perspective to think from

alright i think i see my problem thanks

it's making my comp series hella long though

[maiden heath]

How would I prove this?

(kS_n)^{S_k} is S_k invariant and S'_{n-k} is the subgroup of S_n that fixes all {1,2,...,k}, X_n are the Jucys Murphy elements

[solar glacier]

quick question for a group of order 72 are there three iso classes by the FTFGAG

are they

[cloud walrus]

MyMathYourMath

MyMathYourMath

[solar glacier]

and lastly

[cloud walrus]

MyMathYourMath

MyMathYourMath

[next obsidian]

,w factor 72

[cloud walrus]

Results provided by WolframAlpha

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

Requested by Chmonkey 2.0#9767

[next obsidian]

That seems very off

[solar glacier]

yeah and the n_i+1 needs to divie n_i

[next obsidian]

Like what about Z_2 x Z_6 x Z_6

Also this is only for abelian groups, you wrote group

[solar glacier]

thats using the second version of the fundemental theorem

i meant abelian

by FGTAG

[next obsidian]

Yeah but this isn’t isomorphic to any of the ones you wrote?

This doesn’t have any element of order > 6

It’s also definitely abelian of order 72

[solar glacier]

but the theorem states to split it into

[cloud walrus]

MyMathYourMath

[next obsidian]

Okay then just

Put Z_2 at the end lol

[solar glacier]

but you can alsp have Z_3 at the end as there are two powers of 3

[next obsidian]

What?

[solar glacier]

you can leave one behind so 3 divides the predecessor

[next obsidian]

Z_6 x Z_6 x Z_2

Is not isomorphic to any of the ones you wrote down

[solar glacier]

ahh i forgot this version

so there are 4 iso classes

[next obsidian]

No…

There’s also like

Z_18 x Z_2 x Z_2

You have missed a lot

[solar glacier]

so far I have 5 iso classes

is that all of them?

12 and 6

so theres 6 iso classes?

Z_72, Z_24 x Z_3, Z_36 x Z_2

and

Z_18 x Z_2 x Z_2 and

Z_6 x Z_6 x Z_2

and lastly

Z_12 x Z_6

[abstract rock]

what about Z_3XZ_3XZ_2XZ_2XZ_2, point is you're counting the number of ways to partition 3 factors of 2 and 2 factors of 3

[solar glacier]

but 2 doesnt divide 3

[abstract rock]

that's not what i'm saying

[solar glacier]

isnt what you wrote iso to one of mine

[abstract rock]

is there an element of order 6 in the group i just mentioned? ans: ||no||

[solar glacier]

thats what the FT says

[abstract rock]

all the FTFGAG says is that the number of abelian groups for a given order is the same as the number of ways to partition their prime factors

[next obsidian]

This is Z_6 x Z_6 x Z_2

[solar glacier]

ok so I was correct there are 6 iso classes.

[abstract rock]

whoops

[gritty skiff]

so i understand the idea that when you have an ideal = <a> you basically consider a = 0 in the factor ring
so if you have an ideal generated by two elements or more, like say <a, b, c>, would you then just think of all the generators as 0 in the factor ring?

[solar glacier]

so I was correct? theres only 6 iso classes?

[abstract rock]

sure seems that way

[chilly ocean]

Hey guys, i'm trying to draw a hierarchical graph of all ring structures, and each arrow signifies an inclusion relationship, but there's a few things i don't understand:

• How do the branches on top (that start with semiring, near simiring and non associative rings) fit in the whole picture ?
• And are Fields both division rings and euclidian domains at the same time ?

[solar glacier]

are all iso classes of group of order 432 the following 15 classes.

z_432, z_216 x z_2, z_108 x z_4, z_54 x z_8, z_144 x z_3, z_72 x z_6, z_36 x z_12, z_12 x z_12 x z_3, z_54 x z_4 x z_2, z_36 x z_6 x z_2, z_108 x z_2 x z_2, z_48 x z_3 x z_3, z_108 x z_2 x z_2, z_54 x z_4 x z_2, z_54 x z_2 x z_2 x z_2

[chilly ocean]

this is a question that google will help you more with than anyone here

[solar glacier]

lol

is there 15 classes though? and how using the prime decomp can you tell how many classes you will have

[lethal dune]

you can

try to find a formula

[lavish gull]

what do the different colours represent?

[chilly ocean]

Ah litterally nothing xD i just used the first mind mapper website i could find, so ignore the 2 different kinds of arrows too

[lavish gull]

ok. i think the second part is correct. all fields are euclidean domains and are division rings.
the first part about the other structures, i'm not sure because those don't appear in my abstract algebra texts

[chilly ocean]

Ok nice, yeah for the others I've just been scouring wikipedia trying to piece everything together but sadly i can't fit them either

For groups it was easy, there's a nice diagram on Wikipedia but for rings im kinda struggling

[lavish gull]

part of the problem is that it isn't a tree
ring -> commutative ring
ring -> division ring
a ring that is both is a field, so the branches converge

[chilly ocean]

yeah, that's why i said it's a graph (since that way they can converge again)

I'm coding my algebra library to learn more, and i can easily represent the convergeance with interfaces since they support multiple inheritance with other interfaces (but that's the code side)

[lavish gull]

i don't think formalising the "is" relationships using OOP inheritence is a good approach. but that is my opinion

[chilly ocean]

hmm if you have a better approach i'm all ears 😮 i simply thought the "is a" relationship reflects well with inheritance since a field is also a commutative ring etc

[lavish gull]

maybe at some level, e.g. a high level helper interface

i think that e.g. a euclidean domain depends on the existence of a euclidean valuation
in this case it is a has (euclidean valuation) relationship that determines the is (euclidean domain)
i don't think it's possible to have this represented as static OOP structure

[chilly ocean]

ah but you can represent that with interfaces, making them force implementing classes to define a valuation function

Here's what i got until now: https://pastebin.com/pUq8PDzN
I've defined the groups, rings, lattices and modules (but the rings and modules are probably wrong for now, which is why i was asking my question here to understand better how to formalize them)

at least for rings x'] i'll ask about modules after i fix them first

[lavish gull]

ok. but imagine that the euclidean valuation is user-defined. there is no way to adjust the inheritence structure dynamically

[lavish gull]

nice work but too complicated for me to check

[chilly ocean]

I'm not sure why i would need to be able to adjust that 😮 as long as something is a euclidean domain there should be a valuation function

[chilly ocean]

yeah i know, didn't mean to ask to check, just to show if you ctrl+f "EuclidianDomain" you'll see the way i defined it i mean

[lavish gull]

ok

i still think it's too complicated to have a fixed structure like this in a program
imagine the user "builds" a structure by providing the operators and functions
how will you know whether those satisfy the needed axioms for it to be e.g. a field

[chilly ocean]

that's 100% true, it's up to the user to not define an operator that's not commutative as an abelian group (for example) yeah. I see your point

[lavish gull]

it's a good question. i hope you can find the answer to how your other structures fit in, but as i said those seem to be outside of basic abstract algebra

[chilly ocean]

Thanks for the input! Yeah anyways it's not really a general purpose library for other ppl x'] it's more for me since i usually understand better when i actually implement something

[lavish gull]

by the way, have you also considered ideals? those are similar in that you have principal ideals, maximal ideals, prime ideals

[chilly ocean]

I was going to look into that, since i haven't learnt that yet so i can't do it until i understand it =p but yeah some structures required that if i remember

[lavish gull]

ok. it's a similar situation in which an ideal is a specific kind if it satisfies axioms, meaning it has certain properties
another instance of where has leads to is

and the fact that has is a kind of dynamic property makes it hard to describe is structure statically

[chilly ocean]

yeah those are harder to implement x'] I just checked, It's basically a subgroup of the additive group of some Ring, and there's no way to check in code statically if a subgroup has closure so i'm not sure how to implement that yet (or if i would)

I would probably again leave it to the implementing user's discretion 😄

[ripe basalt]

Are the only elements of order 2 in S_n those which can be decomposed into a product of disjoint transpositions?

i think that's true...yeah

[lavish gull]

but wait, if an element has order 2 in S_n, then in cycle notation it has the form (a b), so that (a b)^-1 = (b a)

[ripe basalt]

so?

(a b) is order 2 in S_n

it's also a product of disjoint transpositions

[lavish gull]

ok, so order-2 elements and transpositions are the same

[ripe basalt]

no they arent

transpositions are cycles of length 2

order 2 elements should be those which can be decomposed into products of disjoint transpositions

but this was my question

[solar glacier]

the partitions of the exponents of the primes times eqchother?

[lavish gull]

is there an example of a order-2 element that is not a transposition? i think those are the same thing

[ripe basalt]

yes

(a b)(c d)

or if you like (1 2 3)(2 3 4)

[lavish gull]

of course, thank you, i see. you are right

[ripe basalt]

anyway im being lazy, i think it's just true but didnt feel like checking

[tribal moss]

Of course it's true.

If there's any longer cycle in the decomposition, then σ² won't map an element of that cycle to itself.

[ripe basalt]

yep

im trying to do some counting right now and it isn't working out

@tribal mosshow would you count the order of the conjugacy class of an involution?

im thinking that the things are in the same conjugacy class as the involution if they have the same cycle type

but im actually not sure what im counting there...

are the elements with the same cycle type the representatives of the conjugacy class? or are they just all grouped under the identity representative?

[tribal moss]

The conjugacy class (in the full symmetric group at least) consists of exactly the permutations of the same cycle type.

[ripe basalt]

so say an element is a transposition, then the size of it's conjugacy class is n!/(n-2)!

[tribal moss]

divided by 2 since you don't distinguish between (a b) and (b a).

[ripe basalt]

if an element is an involution then the size of the conjugacy class is n!/(n-2l)! where l is the number of disjoint 2 cycles that compose it

oh

[solar glacier]

Question, is z_36 x z_35 x z_11 iso to z_9 x z_7 x z_220 since both are cylcic as pairwise theyre coprime

and z_36 x z_11 x z_13 is NOT iso to z_9 x z_22 x z_26 as 22,26 are not coprime thus one isnt cyclic

[tribal moss]

So all in all we get something like $\dfrac{n!}{(n-2l)!l!2^l}$.

[cloud walrus]

Troposphere

[ripe basalt]

that's for the conjugacy class of an involution

[tribal moss]

Yes.

[ripe basalt]

sec let me think about that

so can we see it as we're choosing 2l elements from n

and then dividing out the ways you can rearrange those cycles?

no that's not quite right

[tribal moss]

The best way to think about this is probably the orbit-stabilizer theorem, as the group acts on itself by conjugation. The denominator is the number of permutations that don't change the original involution when you conjugate by them.

[tribal moss]

That sounds right.

[ripe basalt]

wait what

how do you get the denominator

i get the part about orbit stabilizer

[tribal moss]

You can permute the l transpositions; that gives a factor or l!. You can swap the elements in each of the transposition independently from the other ones; that gives a factor of 2^l. And finally you can permute the elements that the permutation doesn't touch, for a factor of (n-2l)!

[ripe basalt]

by swapping the elements what do you mean?

(a b) -> (b a)?

[tribal moss]

Yes.

[ripe basalt]

okay

permuting the elements that the permutation doesn't touch

what do you mean by this?

[tribal moss]

Basically we're now counting how many ways there are to fill in the blanks in
(_ _)(_ _)...(_ _)(_)(_)...(_)
such that we get a complete decomposition of the original involution into disjoint 2- and 1-cycles.

[solar glacier]

can someone answer this real quick

[solar glacier]

and this

[tribal moss]

I did so.

[solar glacier]

so im correct?

[tribal moss]

I believe so.

[tribal moss]

So the (n-2l)! factor counts the choices of order for listing the 1-cycles.

[ripe basalt]

uh

i mean aren't those identified though?

like (1 2) (3 4) is the same as (1 2) (3 4) (5) (6) and (1 2) (5) (3 4) (6)

[tribal moss]

Yes, that's the point.

We're now counting the size of the stabilizer.

No, sorry, not exactly the point.

[ripe basalt]

okay so the (n-2l)! is because "we lose access to the elements in those transpositions"

like that is how many choices we have to place the 1 cycles

[tribal moss]

Conjugating doesn't change the position of the parentheses, so it doesn't make (- -)(- -)(-)(-) int (- -)(-)(- -)(-).

But (1 2)(3 4)(5)(6) is the same as (1 2)(3 4)(6)(5), corresponding to the fact that (5 6) is in the stabilizer.

[ripe basalt]

so this condition is it's in the stabilizer if it only moves the position of a 1 cycle?

im only talking about the n-2l factor now

[tribal moss]

No, it's in the stabilizer if the result of conjugating by it is the original involution.

[ripe basalt]

no no

that's what the stabilizer IS

im asking about this subset

of the stabilizer

the (n-2l)! subset

[tribal moss]

It's not really a subset of the stabilizer, but a factor. But yes, if I understand you correctly.

[ripe basalt]

so this subset is essentially "how can we order the numbers which don't appear in the transpositions"

[tribal moss]

Yes.

[ripe basalt]

ok

ty

the 2^l doesn't just cover swaps WITHIN a transposition either

so that's nice

it covers the case (a b)(c d) goes to (a c)(b d) i think

oh wait

no it doesnt

each spot in each transposition only has 2 choices

itself or the other thing

got it

[tribal moss]

The 2^l really applies after you've selected an order to list the pairs in. Then there's still a choice to make inside each pair.

It's really just elementary combinatorics:

A reastaurant has l tables for two and n-2l single tables. If l couples and n-2l single guests arrive, how many ways are there to seat everyone?

[ripe basalt]

yeah

i hate counting

[tribal moss]

Sorry about that. But at least we can keep that separate from the algebra.

[ripe basalt]

so the order of the orbit is going to be the order of the conjugacy class

can't two orbits fall into the same conjugacy class though?

[tribal moss]

A conjugacy class is the same as an orbit under conjugation, by definition.

[ripe basalt]

i thought the conjugacy class has an equivalence relation on top of it

that an orbit might not have

oh i guess

yeah it's the same hm

hm

i dont know if this is right

@tribal moss if n = 6 and l = 3 then the order of the conjugacy class of a transposition should be equal to the order of the involution composed of 3 transpositions

[tribal moss]

Yes.

[ripe basalt]

oh it does

nice

[tribal moss]

There should be 15 of them.

Wait a moment.

[ripe basalt]

yes?

[tribal moss]

Ah, this is a bit confusing. n=6, l=3 correspond to the conjugacy class containing (1 2)(3 4)(5 6), and there are indeed 15 of those.
But "the conjugacy class of a transposition" sounds like the class containing (1 2), which is different but also happens to have 15 elements, for (as far as I can see) unrelated reasons.

[ripe basalt]

@tribal moss apparently this has something to do with Aut(S_6) having a subgroup of index at most 2 which sends transpositions to transpositions

[tribal moss]

Ah, right, S6 is the only symmetric group that has a non-inner automorphism.
The "exceptional" automorphism maps (- -)(- -)(- -) to single transpositions and vice versa.

[ripe basalt]

im trying to think of why what i said is true hm

[tribal moss]

I don't think it is obviously true.

[ripe basalt]

you do what

oh

[tribal moss]

But Wikipedia says so, so it must hold. :-)

[ripe basalt]

this means like only half of the automorphisms will preserve tranpositionness

oh wiki had it?

[tribal moss]

https://en.wikipedia.org/wiki/Automorphisms_of_the_symmetric_and_alternating_groups#The_exceptional_outer_automorphism_of_S6

[ripe basalt]

i dont get it

Well because of that conjugacy class stuff, i guess you can say the automorphisms that map 2 cycles to 2 cycles are just those with three 2-cycles?

[tribal moss]

The automorphisms that preserve the cycle structure are exactly the inner automorphisms -- that is, the ones defined by conjugation by a fixed element.

[ripe basalt]

automorphisms that preserve the cycle structure are exactly the inner automorphism
this is by definition?

[tribal moss]

(It is clear that the inner automorphisms preserve cycle structure, but it's less immediately obvious they are the only ones).

[ripe basalt]

or it needs proving

oh yeah

[tribal moss]

Proving the other direction might in fact amount to understanding the exceptional outer automorphism of S6 and why it's exceptional.
(I think I've once managed to grasp one of the constructions, but I don't remember how it goes. 😆).

[ripe basalt]

well im hoping the work we just did

can help

S_6 is also the only time that those conjugacy classes have the same size

[tribal moss]

Are you solving a problem about S6 specifically, or was that just a random example you picked?

[ripe basalt]

i think you use it to identify the inner automorphisms with the non-transposition involutions

im doing this part specifically for S_6

[tribal moss]

Okay -- I don't think I've seen the particular goal you're working towards, so I'm just making conversation :-)

[ripe basalt]

well im trying to show this subgroup thing

like Aut(S_6) has a subgroup of index at most 2 which maps transpositions to transpositions

@tribal mossdo you know

i cant connect it at all, i have no intuition for this

[tribal moss]

Hmm, so the subgroup we're looking for is clearly the group of inner automorphisms.

[ripe basalt]

yes so i have the count of the conjugates of a transposition now

which is 15

[tribal moss]

The tricky part is showing it has index 2. (or at least showing it has index at most 2).

[ripe basalt]

i think at most index 2 is easier

why is it the group of inner automorphisms btw?

i get how theyd be a subset

[tribal moss]

Because we know the inner automorphisms map transpositions to transpositions.

[ripe basalt]

thats only subset

[tribal moss]

Oh.

Yeah, but if we believe what Wikipedia tells us, they're the only ones.

https://en.wikipedia.org/wiki/Automorphisms_of_the_symmetric_and_alternating_groups#No_other_outer_automorphisms_of_S6 seems to sketch a proof, but I haven't read it in detail.

[ripe basalt]

ooo

yes an automorphism has to preserve conjugacy

sec

so wiki doesnt say why this group is the inner aut group

it leaves it for the reader

but it does say a hint is that an automorphism must send each conjugacy class to another conjugacy class

oh i think it has to do with cycle type preserving

wtf

@tribal moss can i say in the concrete case that an automorphism can map C (the conjugacy class of transpositions) to itself or C' (the other conj class of order 15)

but if it sends C to C', it wouldnt be sending transpositions to transpositions?

[tribal moss]

Yeah.

[ripe basalt]

okay so that means this subgroup is Inn(S_6)

[tribal moss]

Ah, so we don't even need to know that the automorphisms that send transpositions to transpositions are inner!

[ripe basalt]

in general i was wondering about that

but

i think it goes by cases

i couldnt come up with a general proof

[tribal moss]

We can just say that the the automorphisms (outer, inner, whatever) that happen to send C to itself form a subgroup.

And this subgroup is the kernel of the action of Aut(S6) on {C, C'}.

[ripe basalt]

whoa what

that looks very close to getting that this thing is index 2

but {C, C'} isn't a group

[tribal moss]

No matter, a group action just needs a set to act on.

[ripe basalt]

yes but im trying to get this to index 2

[tribal moss]

The set of permutations of the two-element set {C, C'} is a group, though.

[ripe basalt]

it'd be nice to say it's iso to a 2 element group

[tribal moss]

So the action of Aut(S6) on {C,C'} is a homomorphism Aut(S6) -> S2.

And the kernel of this homomorphism is the automorphisms that preserve C.

The index of a kernel of a homomorphism is the order of its image.

[ripe basalt]

how does aut(s_6) act on that set?

oh i guess you take

g in Aut(G)

[tribal moss]

Each automorphism either swaps the two conjugacy classes or leaves them alone.

[ripe basalt]

and then do g(C)

and that's = C or C'

[tribal moss]

Yes.

[ripe basalt]

the image is...C_2?

[tribal moss]

Yes.

But for this proof we don't even need to know for sure that that is the image.

[ripe basalt]

wouldn't we?

[tribal moss]

Because we're just asked to prove that the index is at most 2.

And the image is certainly a subgroup of S_2.

[ripe basalt]

by S_2 do you mean C_2?

[tribal moss]

So the only sizes it can have are 1 and 2.

S_2 and C_2 are (up to isomorphism) the same group. In this context it arose as the group of permutations of {C, C'}.

[ripe basalt]

im trying to see what the image is actually

to see how it's a subgroup

mixing too many data types

[tribal moss]

The image of a homomorphism is always a subgroup.

[ripe basalt]

yes but

if g in Aut(S_6) then g({C, C^{-1}) feels like {C, C^{-1}}

which isnt a group

so how do you see that it is

[tribal moss]

In order to know that the image is actually the entire group, we need to know that the exceptional outer automorphism exists, which looks hairy -- and the reason the exercise says "at most two" must be to let you avoid that hairiness.

[ripe basalt]

yes im not asking that rn

im just asking about this group action

[tribal moss]

The other conjugacy class is not C^{-1}.

[ripe basalt]

that was a typo

C'

i mean

okay Aut(S_6) is acting on {C, C'}

[tribal moss]

You either have g(C) = C and g(C') = C', or you have g(C)=C' and g(C')=C.

[ripe basalt]

we can't call that a homomorphism

because Aut(S_6) x {C, C'} isn't a group?

[tribal moss]

It's a general fact that if the group G acts on a set A, then the action can be viewed as a homomorphism G -> S(A).

Where S(A) is the group of permutations of A.

[ripe basalt]

oh

what are the permutations of {C, C'}

i guess either e or (C C')

ok

well wait

(C C') feels like what g would be

[tribal moss]

Depends on which g we're talking about.

[ripe basalt]

the automorphism

[tribal moss]

Yes, but there are many different automorphisms.

If g is an inner automorphism, then g(C)=C, so it maps to the identity in S({C, C'}).

[ripe basalt]

whatever S({C, C'}) is

[tribal moss]

The group of permutations of {C, C'}.

[ripe basalt]

what is that

is it just (e) and (C C') or what

[tribal moss]

The group whose elements are e and (C C').

[ripe basalt]

ok

[tribal moss]

S(A) is isomorphic to S_2 (= C_2) whenever |A|=2.

[ripe basalt]

what maps to the identity in S({C, C'})? the inner automorphism g?

jesus

group theory blows

all this mixing of data type

[tribal moss]

Yes, but we don't need to know that.

g maps to the idenity if g(C)=C.

g maps to the other element of S({C,C'}) if g(C)=C'

[ripe basalt]

ok

i guess i have to get used to this "group action is a homomorphism" thing

[tribal moss]

Yes, it is pretty crucial.

[ripe basalt]

okay ty, i need to go walk around or eat or something

[trail stump]

so in complex analysis, the conjugate of a conjugate of complex number is the original complex number. But I guess in group theory, the conjugate of the conjugate of an element h in a normal subgroup H of G does not have to be the original element h right?

[tribal moss]

Right, the two usages of the word "conjugate" are not directly related.

[trail stump]

gotya

so lets say for a normal subgroup H, it has say 8 elements, and then a random g in G, the conjugate of h1 can be like it goes to h2 and h3 and maybe h5 and then back to h1 and then repeat right?

does it have to go through every element in H?

probably not? since if g is e, then it is trivial?

[tribal moss]

Right.

But it can definitely cycle through several elements.

[trail stump]

yeah idk, maybe it has something to do with some property of g

[tribal moss]

For example we can let G = H = S_5 and g = (1 2 3 4 5). Then if we start with h1 = (1 2) then it moves through (2 3) and (3 4) and (4 5) and (5 1) before it returns to h1.

(Or the other way around, depending on whether you've defined conjugation as g^-1hg or ghg^-1).

[trail stump]

G=H?

[tribal moss]

Yeah, just to fit with your notation above where you wanted H to be a normal subgroup.

[trail stump]

ok

[tribal moss]

Since G is a normal subgroup of itself, I was choosing that to the H for simplicitly :-)

[trail stump]

yeah i guess so

i was not thinking about if g is also in H

[tribal moss]

It doesn't make much of a difference for the situation.

[trail stump]

it probably would simplify the situation

[tribal moss]

If you want a g outside the subgroup, we could have G = S_5, but H = A_5 and g = (2 3 4 5). Then (1 2 3) would move through (1 3 4) and (1 4 5) and (1 5 2) before returning to (1 2 3).

[trail stump]

holy shit, how did you calculate everything so quick?

[tribal moss]

There's a trick for conjugating with cycle notation.

For example, if h = (a b c)(d e), then ghg^-1 is (g(a) g(b) g(c))(g(d) g(e)) -- you simply apply g to each element of each cycle.

[trail stump]

oh

still looks complicated

[tribal moss]

Only at fist sight.

        (  a    b    c  )(  d    e )
becomes (g(a) g(b) g(c)))(g(d) g(e))

[trail stump]

thx for the example, i will take time think about it

[tribal moss]

There are still some intriguing connections between conjugation in the group sense and complex conjugation. For example, if we embed the complex numbers in the quaternions, then j(a+bi)j^-1 = a-bi.

[trail stump]

“the group conjugate of a complex number is also its complex conjugate"

[trail stump]

i guess so

[tribal moss]

Only in very particular senses such as the quaternion embedding above. In an abelian group, conjugation never does anything, so within C itself there's no hope of the complex conjugate also being a group-theoretic conjugate.

[trail stump]

right

[tough raven]

\newcommand*{\ext}{\operatorname{Ext}}
\newcommand*{\res}{\operatorname{Res}}
\newcommand*{\Mod}[1]{(#1)-(\mathsf{Mod})}
If
[ \begin{tikzcd}
B \arrow[r] & C \
A \arrow[u] \arrow[r] & D \arrow[u]
\end{tikzcd} ]
is a commutative diagram of rings in which $\ker(B \to C) \subset \operatorname{im}(A \to B)$, is it true that
[ \res_C^B \circ \ext_D^C = \ext_A^B \circ \res_D^A, ]
where $\ext_A^B$ is extension of scalars from \Mod{A} to \Mod{B}, $\res_A^B$ is restriction of scalars from \Mod{B} to \Mod{A}, etc.?

[cloud walrus]

Raghuram

[tough raven]

(More general conditions on the commutative square under which this happens also appreciated.)

As some motivation, this came up in the context of group representation theory, with $A = k[H], B = k[G], C = k[G/N], D = k[H']$, where $H$ is a subgroup of $G$, $G/N$ a quotient group of $G$ and $H'$ is the image of $H$ under the quotient map, i.e., the image of the composite map $H \to G \to G/N$.

It is easy to see that in general, $k[G]$ is a $[G \colon H]$-dimensional $k[H]$ module (pick representatives of cosets), so similarly $k[G/N]$ is a $[G/N \colon H']$-dimensional $k[H']$ module, so just by comparing dimensions over $k$ (say, starting with the trivial representation of $H'$ as a module of dimension $1$ over $k$), it is necessary that $[G/N \colon H'] = [G \colon HN]$ and $[G \colon H]$ be equal, which is easiest to satisfy by having $H = HN$, i.e., $N \subset H$. This holds iff $\ker(k[G] \to k[G/N]) \subset k[H]$ (since for every $n \in N$, $n - 1$ is in that kernel and $1 \in k[H]$).
Hence the question.

[cloud walrus]

Raghuram

[flint cave]

When can we naturally identify quotient groups as subgroups (like in vector spaces R^2/R can be naturally identified with R x {0}). Or am I mistaken that this isomorphism is anymore natural than others?

Or in some sense is this always possible when we think of a group as the quotient group of the free group modulo presentation and then use the third isomorphism theorem? (this sounds like gibberish?😅)

Ah this is clearly not possible always

Mb

I guess my question is, when does a quotient group have a section.

[tough raven]

So if N is the normal subgroup of G and pi the quotient map, a section would be a subgroup K such that pi restricted to K is a bijection.

Equivalently, K and N intersect trivially and KN = G.

When this happens one says it splits

[flint cave]

Oh, when it is a semi direct product?

[tough raven]

Yes

[flint cave]

I see! I had long suspected this . Thanks!

[tough raven]

which is why it's kind of important; you can more or less classify all G which are extensions of N and K such that there's a section as a semidirect product of N and K

[flint cave]

I see

[neat valley]

Why is C[x,y]/(x^2-y^5) not a UFD?

In particular, can someone please give me an explicit example of an element in that ring that does not have a unique factorization

[glass grail]

i think x^2-1=y^5-1 works?

[neat valley]

I don't follow

[glass grail]

(x-1)(x+1)=(y-1)(y^4+y^3+y^2+y+1)

[tender wharf]

If the order of a group element x is 1, must x be the identity?

Intuitively it seems to make sense, x^1 = e => x=e but then wouldn't that also imply x has infinite order?

[neat valley]

oh thanks!

[neat valley]

why would it imply infinite order? order is smallest n>0 such that x^n = e

[glass grail]

y^4+y^3+y^2+y+1 isnt *irreducible but it doesnt matter because y-1 isnt in the other factorization

just to note

[tender wharf]

But the order is 1

[neat valley]

What are you asking?

[tender wharf]

One moment

I don't see how an element x can be in H with order 1 and not be the identity

[wooden ember]

The |x| is 1 condition makes sure that e is in H to not have a trivial counterexample

[tender wharf]

Oh right the order of the identity is 1

[wooden ember]

What they want you to show is that in general the product of two elements of even order may not have even order

Even in an abelian group

[tender wharf]

Yeah I got that part I think

Thank you

[tender wharf]

So yeah I was a bit silly I forgot that the order is the smallest n>0

[ripe basalt]

What's a quick way to show that if A_n is simple in S_n then it's also the only normal subgroup in S_n?

[lethal dune]

If there's any other, it either lies in A_n which makes it a normal subgroup of A_n, otherwise try intersecting it with A_n.

try to complete the argument

[ripe basalt]

yes i had this already

the intersection i guess would have to be assumed to be trivial

is it possible that it can be nontrivial?

mm

[lethal dune]

can you show it'll have an even perm?

that'll show it intersects with A_n so intersection cannot be trivial

[ripe basalt]

if it's nontrivial then the intersection would be a normal subgroup of A_n no?

[lethal dune]

yes

since you are assuming A_n is simple, n>=5

anything comes to mind?

[ripe basalt]

Wait, if i have two normal subgroups of G (call them A and B)

is A \intersect B normal in A?

[lethal dune]

no not necessarily

[ripe basalt]

well you said yes

in the A_n case

okay anyways

[lethal dune]

actually yes it is

[ripe basalt]

it's normal in A?

[lethal dune]

take $x \in A\cap B$ then $axa^{-1} \in B$ as it's normal and also in $A$ because $A$ is closed

[ripe basalt]

you would have to show that axa^-1 is normal in A \cap B

is in A \cap B*

my bad

[lethal dune]

isn't that what I just showed?

[ripe basalt]

oh yeah

ok cool

okay so take a normal group N

if it intersects nontrivially with A_n we get a contradiction on A_n being simple

if it intersects trivially with A_n then..

[lethal dune]

if N is normal then it can't intersect with A_n trivially, that's the point

[ripe basalt]

why is that

[lethal dune]

you will be able to construct an even perm in N given N is normal

try to think how

[chilly radish]

It's also not a contradiction that it intersects nontrivially, it just means that N\cap A_n=A_n

From which you can derive your stated result

[ripe basalt]

yeah that implies N = A_n or G

i dont really feel like constructing an even permutation in N...hm

[chilly radish]

Spoiler: ||I think you can also argue by normaliry (which might be easier) that N can't have order 2, and so it can't intersect A_n trivially otherwise we would have a contradiction via reasoning about orders||

[ripe basalt]

i saw a proof that said that N would be contained in the center if it intersected A_n trivially

and the center is trivial

of S_n anyway

why would N be contained in the center?>

ok

@chilly radishwhy couldnt N have order 2?

like look at this https://proofwiki.org/wiki/Normal_Subgroup_of_Symmetric_Group_on_More_than_4_Letters_is_Alternating_Group

this leaves out the fact that N could be normal in S_n without being a subgroup of A_n no?

that's all i want help with actually

if N is normal in S_n then N is a subset of A_n

@chilly radish or @lethal dune

this is the fact i always get stuck on

oh wait

well my proof tries to discount the intersection being trivial

[chilly radish]

Prove it. For a normal subgroup if you have an element all elements with the same cycle structurr are also members of this subgroup

[lethal dune]

if you want a proof that why $N$ intersecting with $A_n$ would imply N is in the center, try this
$[N, S_n] \subseteq N$ and $[N, S_n] \subseteq [S_n, S_n] = A_n$ so $[N, S_n] \subseteq A\cap N = {e}$ that is N commutes with every element of $S_n$ so $N \subseteq Z(S_n)$

[ripe basalt]

did u see the proofwiki i shared

[chilly radish]

That proof seems insufficient

Ye proofwiki one

Tbe

The*

[lethal dune]

no

[ripe basalt]

@lethal dune i mean the intersection being trivial implies that it's in the center

so...

[lethal dune]

well I didn't follow the convo you had with Shin.

[ripe basalt]

im just saying

[lethal dune]

came back though I should post it,

[ripe basalt]

ur proof cant be right

because it would mean nontrivial intersection implies this

but it's trivial intersection which implies it

oh you write that it's trivial intersection

weird

[lethal dune]

I'm just saying no nontrivial normal subgroup of S_n can have trivial intersection with A_n

[ripe basalt]

well u write the opposite in your first sentence

but the rest seams fun

[lethal dune]

well not in a mood to correct that

[ripe basalt]

[N,S_n] \subset [S_n, S_n]

this is obvious right

since a commutator on the left

is a commutator on the right

N < S_n

[lethal dune]

yes

[ripe basalt]

ok

i think i understand

[lethal dune]

great

[tribal moss]

(Oops, misread the question, sorry for the ping).

[chilly ocean]

hi, could someone explain what a knit product is in groups?

i'm reading through Visual Group Theory by Nathan Carter (page 128 is the only mention so far) and the book only says that it will not be referenced later

i've tried to find information about it online but have found nothing

[tribal moss]

Googling "knit product" group immediately led to https://en.wikipedia.org/wiki/Zappa–Szép_product

[hexed mantle]

Let (E, *) be a "group" ( * is not associative) such that

Prove that * is commutative.

[tribal moss]

When you write "group" in scare quotes, do you mean that is has an identity element and inverses?

[hexed mantle]

Ah, no.

[cloud walrus]

Jonathan Phan

[tribal moss]

So just a magma satisfying those two rules and nothing else?

[hexed mantle]

Yes.

Okay, it's a groupoid, apparently. I suck at translating math terminologies from my language to English.

[tribal moss]

||xy = (xx)(yy) = ((yy)x)x = ((yx)y)x = (yx)(yx) = yx||

[chilly ocean]

Groupoid and magma means the same

People avoid the former because of collision with category theory nomenclature

[hexed mantle]

Okay, sure.

Okay, that was horribly stupid of me.

[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

[chilly ocean]

thanks

[lavish gull]

yes

[cloud walrus]

MathPhysics

[chilly radish]

It is not, since it doesn't fix 1

[chilly radish]

How are you justifying 2->3?

[gusty thistle]

What do we know about two polynomials when they have the same splitting field?

Two monic irreducible polynomials

Ideally I want to show that they share all roots

But idk if that’s even true

[abstract rock]

well think about the base field, would it have to be the same for the two polynomials to be monic?

[tribal moss]

(yy)x = (yx)y ?

[chilly radish]

But it's not (yy)x

It's (yy)(xx)

How are you justfying going from that to ((yy)x)x to use the 2nd rule

[tribal moss]

I don't have (yy)(xx) anywhere.

[chilly radish]

You have (xx)(yy)

Same thing

2nd equality

[tribal moss]

No, because we don't yet know the operation is commutative.

[chilly radish]

I just mean it doesn't matter which direction you start with

x and y are just names

[tribal moss]

(xx)z = (zx)x with z = yy.

[chilly radish]

Right ok

Yea you're right

I missed that

[chilly ocean]

This is correct from what I checked

[gusty thistle]

I don’t know : / and I don’t rlly know what kind of things to consider

I have been thinking abt this for a bit

[tribal moss]

For less confusion in the main steps we could also say: (as)(bc) = ((bc)a)s = ((ca)b)s = (bs)(ca), so therefore (xx)(yy) = (yx)(yx)

[gusty thistle]

I am trying to do exercise 11 and I want to show that all irreducible factors of f in F[x] share no roots (I have assumed Wolog that f(x) is monic) so that it can not be the case that an irreducible factor in K[x] divides two distinct irreducibles in F[x]

Oh it’s rotates 90 degrees sorry

[tribal moss]

,rccw

[lavish gull]

which exercise is it that you are working on?

[gusty thistle]

11

I have found a proof but it seems to me to be incorrect because it assumes f is inseperable due to F being perfect

But f is not necessarily reducable so I don’t think you can say that it’s seperable

I have shown that if two irreducible factors in F[x] share a root that then they have the same splitting field

[lavish gull]

is it possible that f(x) is irriducible in F[x] and yet has repeated factors in an extension of F[x]?
maybe that is where perfect and separable get involved

[gusty thistle]

No but f(x) is not irreducible

[chilly radish]

Prove it first for f irreducible

And then do it for the case when it's a product of irreducibles

[gusty thistle]

I have

[chilly radish]

Use the fact that the gcd of 2 polynomials remains invariant under field extensions

I.e. if f,g are polynomials in F[x], and K is some extension, the gcd of f,g over K[x] is the same as their gcd over F[x]

[gusty thistle]

I am having trouble with the proving that that two distinct irreducibles in F[x] can not have a common irreducible factor in K[x]

[chilly radish]

For that use the fact I just stated

And the fact that the gcd divides both

[gusty thistle]

I would have to prove that first

Is that a difficult thing to prove ?

[chilly radish]

No

[lavish gull]

then they would have a factor of multiplicity in K[x], despite having no common factors in F[x]
it sounds like this is prevented by being a unique factorisation domain

[chilly radish]

Call r_F their gcd over F[x] and r_K their gcd over K[x]. For one direction, use the fact that r_F is a divisor of both when considered as a polynomial in K[x]. For the other direction, use the euclidean algorithm

[chilly radish]

Yes, but you can also prove it using properties of euclidean domains, which is easier

I'm actually not sure if UFD is sufficient, but I think you could do a similar argument by factoring into irreducibles over F[x] and K[x]

[gusty thistle]

I don’t really understand this

[chilly radish]

Yea UFD is sufficient I think

[gusty thistle]

My plans is formally write down for f irred. and prove the gcd lemma

I hope that will order my thoughts a bit

[chilly radish]

Good luck, feel free to @ me if you need more help

[gusty thistle]

Thank you very much :]

[chilly radish]

Np

[gusty thistle]

I do like this problem

[chilly radish]

It's important

[gusty thistle]

Oh nice

[chilly radish]

Generally the idea of irred polynomials remaining coprime even when split

[gusty thistle]

Cool

Is it important for Galois stuff ?

[chilly radish]

Yes very

[gusty thistle]

Oooh nice 😋😋

[gusty thistle]

I want to show that r_K divides r_F using the Euclidean algorithm. How do I show that degree of r_K is less than or equal to the degree of r_F ?

[chilly radish]

I was thinking of using the fact that r_F = af+bg for some polynomials a,b in F[x]

And then showing from yhat that r_K|r_F

[gusty thistle]

Thanks

G2g now I’ll try again later

Tyvm for all the help

[chilly radish]

Np

[gusty thistle]

I did it :]

Thanks

[chilly radish]

Nice!

[gusty thistle]

It was a nice problem

And I also feel way more comfortable with working with gcd in rings now

[tough raven]

How BTW?

Of irreducible representations?

So there's a way to manufacture invariant maps between irreducible representations from linear maps by averaging

$T' = \frac{\sum_{g \in G} \sigma(g) T \rho(g^{-1})}{\lvert G \rvert}$

[cloud walrus]

Raghuram

[tough raven]

where ρ, σ are the representations on domain and codomain respectively

You can ensure T' is non-zero if T has non-zero trace (if the irred reps are isomorphic, which they are in this case), at which point it will be an isomorphism of reps

If I'm remembering correctly 💀

Imagine forgetting representation theory while taking a first course on representation theory

Pick any T whatsoever to begin with

So you can take T = [[1, 0], [0, 0]]

[tough raven]

.

They're the two irreducible representations you want to find an isomorphism between

[chilly radish]

I'm actually not sure about this again

I'll have to think about it

[tough raven]

I thought the GCD being a linear combination (i.e. being a Bezout domain) was essential
So please do tell if you come up with something

[chilly radish]

No I think you are right

[tough raven]

It's any linear map (here, 2 by 2 matrix)
You are manufacturing an invariant linear map from an arbitrary linear map by an averaging process

[chilly radish]

Yea this is false. Consider eg the gcd of 2,4 in Z vs Q

That would be a trivial counterexample

[tough raven]

To ensure the nice invariant linear map you end up isn't just the zero map

That would not be theinvariant linear map you're looking for

Since these are irreducible reps, Schur's lemma guarantees if you can cook up a non-zero invariant linear map, it'll be an isomorphism, so all you need is to get T' to come out non-zero

and IIRC tr(T) non-zero is how you ensure that
I might have gotten this part wrong though; if it doesn't work, try a few different T s until you successfully get a non-zero T' IG

[tough raven]

I mean 2 is still a GCD of 2 and 4 in Q, it's just not unique because GCDs in fields are kind of broken

[chilly radish]

Hmmm wait you're right

[tough raven]

Upto units (in the larger ring), the GCD is unchanged

[chilly radish]

Yea yea

You're right

Maybe we can find an example in Z[x] vs Q[x]

[tough raven]

This should be true in any Bezout domain (finitely generated => principal, or equivalently any two elements have a GCD which is a linear combination of them) for essentially the same reason as the usual proof

[chilly radish]

Is Z[x] bezout

Should be yea

We'd need a more pathological example then

And yea you're right

[tough raven]

Nice idea
The canonical failure for Bezout's identity in Z[X] is that gcd(X,2) = 1 but no linear combination

[tough raven]

No

[tough raven]

Unfortunately, the gcd still happens to be 1 in Q[X] because we made 2 a unit

Hmm

You could look for the isomorphism explicitly

These matrices are diagonalisable, right?

I mean i is already diagonalised

That shouldn't be an issue

The point is you want the isomorphism T to be such that
T (first rep(i)) T^{-1} = (second rep(i))

Which means T will carry eigenvectors of (first rep(i)) onto eigenvectors with the same eigenvalue of (second rep(i))

which already constrains T quite a bit

No

Why should it

Also this is terrible notation

Let's call first rep P and second Q or something

The eigenvectors of P(i) are (1,0) with eval i and (0,1) with eval -i
For Q(i) it is (1,0) with eval -i and (0,1) with eval i
So T has to send (1,0) to an i-eigenvector which must be (0,b) for some non-zero b
Similarly T(0,1) = (a,0)
So T = (0 a)
(b 0)

Yep.

Now try to narrow it down more using P(j), Q(j).

Why?

Oh I see

Does that work?

Fair enough

[chilly radish]

Yea that's the problem with taking FoF in general I reckon

[tough raven]

I'm not sure if that's the issue per se. You can try a localisation that doesn't invert 2 and then the problem is that the gcd is still 1 and you still can't represent 1 as a linear combination.

So if gcd_A(a,b) = c and you go to a bigger ring B gcd_B(a,b) can get larger, but the greatest among elements of A is still c. So B has to add some new elements (upto units), rather than collapsing some things (upto units) by making things invertible

which IG means FoF won't work

Perhaps F = Z and K = Z[i]?

[tough raven]

OK so I got somewhere with K = Z/11Z, even though F = Z is not exactly a subring of that.
Consider gcd(X^2-X+2,2X-3). In Z[X] we can show this is 1 by taking cases on the degree. But in Z/11Z[X], they have the common factor (X+4) because -4 is a common root in Z/11Z.

[chilly radish]

Yea this maybe shows taking quotients doesn't preserve gcd whenever it's non-zero but I don't think it really helps our case

[simple valley]

struggling with what seems like a simple linear algebra problem

I'm asked to prove that on an irrep of G, a G-invariant inner product is unique up to scalars

I can see that in the language of C[G]-modules, an irrep means it's a factor-module of C[G], meaning an epi alpha : C[G] -> V. Then by right-exactness of tensor product we have that alpha (x) alpha is epi and thus V(x)V is also spanned by a single vector and thus a C[G]-linear function on V(x)V is defined by a single value

but this argument seems a little too opaque and I'm looking for something more direct

is there an elementary way to show that if V is spanned by a single vector then so is V(x)V ?

wait I may be getting confused

V (x) V the tensor product of reps of G is not the same as V (x) V the tensor product of C[G] modules?

[weak matrix]

At least not trivially, since the action of G is not defined in the same way

With the assumption of finite dimension of the representation, an inner product induces an isomorphism with the dual space

Composing the map induced by one of the inner products with the inverse of the induced by the other inner product gives an automorphism of the vector space, and maybe G-invariance of the inner products implies that this automorphism is G-invariant

[weak matrix]

Another way that can work without finite dimension sometimes is to prove that the radical of an invariant bilinear form is a subrep

Taking the invariant bilinear form given by the difference of the inner products, this form is either non degenerated or identically 0

If it's non degenerated, just choose a smart scalar to multiply in the difference to make it degenerated

[celest cairn]

Does (i + j) / k = j + i?

[coral spindle]

I assume you're talking about the quarternions. You have made an error: a/b is ambiguous in the quaternions. It could mean either b^-1 * a or a * b^-1, which may produce different answers.

Lo and behold, you get a different answer here.

[celest cairn]

Ok

Thanks

[steel root]

For this prepostition, say I'm given a subset (a,b) that belongs to a field F where a=b. How do I go about proving that lambda and mu also belong to said field without any other information?

[coral spindle]

Huh?

Wym by "a subset (a,b) [...] where a = b"?

How is that related at all to proving that lambda, mu belong to a field?

I do not see how this is related at all to the proposition you've posted

[steel root]

I was trying to explain it without just taking a picture of my homework sorry

[coral spindle]

We cannot read your mind.

[steel root]

Wasn't sure if it was allowed or not

[coral spindle]

There we go.

How do I go about proving that lambda and mu also belong to said field [...]
You don't have to prove that lambda, mu are in F_q—that's assumed.

[steel root]

Am I correct in assuming that (a,b) is just x,y then?

[coral spindle]

No.

x and y are arbitrary elements of C.

Which means that, for example, x = (a,b) for some elements a,b in F_5 where a = b.

Similarly for y, although you must note that y will not necessarily be equal to the same pair (a,b).

[steel root]

I'm not sure where I should be starting with this then

Since a=b, would C only contain (0,0),(1,1),(2,2),(3,3),(4,4)?

[coral spindle]

That is true, but you do not need this information to prove what you want.

Spefically you do not need a complete list of elements. The answer is the same in any field.

[steel root]

So how would I go about proving that this subset is a subspace if not by saying that this list of elements can be found within the field?

[coral spindle]

That's nonsense, this list is not within the field.

I don't know what you mean by this. You just need to show the definition stated above.

[steel root]

You mean showing that x and y are in C?

Or if they aren't?

[coral spindle]

No. You assume that x and y are in C.

You then are required to show that mu x + lambda y is in C also

Have you done proofs before?

[steel root]

I have but idk why you are being such a condescending prick just bc you're knowledgeable in this field. I hope you never go into teaching so people don't ever have to be subjected to you.

[coral spindle]

Well I'm sorry you feel insulted, I don't especially think I've been rude. If you have specific feedback about any particular message I'd appreciate hearing it.

[chilly ocean]

this wasn't necessary

it just comes off as condescending and adds nothing to the conversation

[coral spindle]

I can see how it can come off as condescending now, and certainly I should have clarified it

but I think you can also see how I might infer that YellowPapi has not experienced proofs before and wanted to check if this was true

it was in fact very important for the conversation for that reason

Unfortunately YellowPapi seems not to care anymore

[chilly ocean]

skill issue for you

[coral spindle]

Seems to go two ways

[pastel cliff]

just wanna make sure im on the right track cuz im a lil stuck

can i start this by observing that the only subgroups of order 6 of S_6 are either the cyclic group or the group generated by (1 2)(3 4 5)

[coral spindle]

Well that's a very top-down approach but I think the hint is the most straightforward way to start this

I'm not sure what you mean by distinguishing the cyclic group and the one generated by a single element—the latter is cyclic after all

[pastel cliff]

oh oops

[coral spindle]

But unfortunately those are not the only subgroups of order 6 in S_6

[pastel cliff]

ok brb then

this is the proof we did in my class but looking back at it i dont see how the last line follows from the rest of the argument

[coral spindle]

Are you aware of the fact that if ker f = 0, f is an injection?

[pastel cliff]

yeah

[coral spindle]

This is an example of that

[pastel cliff]

i see why it's an injection

[coral spindle]

Right, and it is a surjection onto its image trivially; everything is

[pastel cliff]

oh that part is just omitted lol, that's obv eenough

but what i meant is how does this give an iso to H \leq S_G specifically

is it bc G acting on itself just "moves the elements of G around"

[coral spindle]

Well, the proof mentions a homomorphism. What is this homomorphism, and what is is from and to?

No as in, what is the homomorphism in question

[pastel cliff]

yeah i realized lol

oh well ig g1g2 is always just gonna give another element of G

[coral spindle]

That's right but what is the homomorphism that the proof is talking about? If you see this, you will see the proof.

[pastel cliff]

maybe im misunderstanding

\alpha((x,y)(a,b)) = \alpha((x,y))\alpha((a,b)) but this is kinda obvious because it's a group

[coral spindle]

Do you remember when we were talking about the different ways to see what a group action is?

[pastel cliff]

either a map G x X -> X or a map G -> Sym(X)

[coral spindle]

Bingo

the latter map is the homomorphism.

[pastel cliff]

oh the proof is showing that the homomorphism we get from this group action is exactly the same as the one from G into Sym(X)?

[coral spindle]

No, it is talking about that homomorphism from the start

the homomorphism maps g to a function that sends h to g.h

[pastel cliff]

since \alpha is a group action we already know that G -> Sym(X) is a homomorphism

[coral spindle]

That's right

So all we need to prove is that it's injective

Which is what we do.

Of course the proof would've done better to say all this more explicitly, as while it's true that alpha as written is a homomorphism, it's not the homomorphism they describe

[pastel cliff]

that's where i was getting mixed up

will try to connect this to the question brb

[coral spindle]

Ah hold on

In fact I was incorrect

alpha as written is not a homomorphism

[pastel cliff]

[coral spindle]

To be a homomorphism from the product group GxG, it would require something else

[pastel cliff]

i kinda just assumed it was before this convo just bc it's a grape

[coral spindle]

But this is totally irrelevant as the argument is talking about the action homomorphism

[pastel cliff]

yeah i see that now

[coral spindle]

The point is, (a,b)(c,d) = (ac, bd) so it would only work in general if the group G was Abelian

[pastel cliff]

should still be besides the point i think

[coral spindle]

Indeed

[pastel cliff]

ooooo ok so i think the next thing i need to use is that Sym(S_3) = S_6 right

[coral spindle]

I don't want to say anything yet, I think you should try seeing this through to the end on your own

[pastel cliff]

[coral spindle]

I still have no idea what half the emojis mean on this server but I'll take it

Give me and and I'm happy