groups-rings-fields

[formal ermine]

lol

[granite topaz]

dich auch ? 😄

[formal ermine]

huh?

[granite topaz]

you too?

[formal ermine]

oh that's a different you

[granite topaz]

As you can see, my German fails me miserably

[white oxide]

could i get a hint on the (=>) direction

I wrote sqrt(a)/sqrt(b) is in Q(sqrt(a))

and so sqrt(a)/sqrt(b) = x + sqrt(a)y for some rationals x and y

problem is trying to show that (x + sqrt(a)y)^2 is rational lol

[granite topaz]

This is too complicated, I feel

Writing sqrt(a) = x + y*sqrt(b) should be enough, no?

[dim widget]

Even writing sqrt(a) = c sqrt(b) for c nonzero is enough

[white oxide]

yea i just wrote that

wait

but how does that imply that (x/sqrt(b) + y) is rational

[white oxide]

ah yea

that's true

oops im dum

[dim widget]

So maybe the simplest: sqrt(a) - sqrt(b) = x + y sqrt(a), (1-y)sqrt(a) - sqrt(b) = x which shows that sqrt(a)sqrt(b) is rational (by squaring)

But there are many ways

[white oxide]

ye that makes sense

thankss

[formal ermine]

I hate this

[granite topaz]

I don't know why I started learning german anymore

:sadge:

[rustic crown]

should det learn german

[lavish nexus]

Es ist einfach

[rustic crown]

muri desu

.<

[formal ermine]

you already know some german!

[rustic crown]

me know zwischenkorper

[formal ermine]

nicht wenn du eine gedichts analyse zu naturlyrik aus dem expressionismus machen musst

[rustic crown]

[formal ermine]

and nullstellensatz

KRULLS HAUPTIDEALSATZ

[rustic crown]

a few days back our topology prof forgot the word "coset"

could only say nebenklasse

then other german students reminded him

[formal ermine]

lol

my alg nt prof forgot the german word for ufd

so he kept saying ufd

none of us remembered the german word either

till this one guy remembered

FAKTORIELLER RING

[rustic crown]

i expected something bizzare :p

[formal ermine]

EINDEUTIGE FAKTORISIERUNGS DOMÄNE

[lavish nexus]

Ich habe mal versucht Faust zu lesen

Die Kommentare waren eben so lang wie der Text selbst

[formal ermine]

bist du deutsch?

[lavish nexus]

Nein

[formal ermine]

mensch

[twilit notch]

could someone help me with this problem

[coral spindle]

Well remember that the euclidean proof works based on size

we say, oh well if it were reducible then it would be divisible by a smaller prime, but oho look

[twilit notch]

right

[coral spindle]

But this proof hasn't ruled out the size yet

[twilit notch]

ohh that makes sense

[coral spindle]

There's no reason to believe that the product plus one couldn't be one of the primes listed, at least not yet

[chilly ocean]

What are the subgroups of Z/nZ? This should be absolutely obvious but I still don’t get it…Can someone help, please?

[coral spindle]

What have you tried?

[solar glacier]

hint: it depends on wether or not n is prime

[coral spindle]

no it doesn't

[long nebula]

Hint: Lagrange's theorem

[solar glacier]

thats why i said it matters if n is prime or composite , the order of subgroup need divide order of group

[long nebula]

There's pretty much just one answer, you don't really have to split into cases

The case where n is prime is the easiest special case though

[formal ermine]

hint: Z/nZ is abelian

[rustic crown]

wait isn't the fact that the product + 1 being not a unit a bigger problem

[coral spindle]

True that too

good catch

[rustic crown]

because they already say that if it had an irred factor then it can't be one of the irreds you already listed

[solar glacier]

thats an interesting proof based on the infamous infinitude of primes proof, never heard it in this sense

[coral spindle]

oh yeah mb. I see what you're saying

Yeah my answer was wrong then

[long nebula]

If you want a bigger hint for the subgroups of Z/nZ problem: ||prove that any subgroup of a cyclic group is cyclic||

[coral spindle]

@twilit notch you should see what det said above ^

[chilly ocean]

I have a question: how does one get that a presentation homomorphism is surjective? is this because of some choice of the set A?

[rustic crown]

yee and notice that for negative d you don't need to worry much because if P is the prod of all primes then you can look at nP+1 for various values of n. for negative d you only have finitely many units, so life is easy.

[chilly ocean]

unfortunate lettering in the definition of \rho...

[coral spindle]

Yes

But this is a definition

[solar glacier]

quick question about showing non-primality of an ideal

consider the ideal I=(1+\sqrt{-5}) show this is not prime in Z[\sqrt{-5}] can take two elements of Z[\sqrt{-5}] that arent in I whose product is in I

[agile burrow]

YES

[solar glacier]

so like 1 \pm \sqrt{-2} would work

[rustic crown]

that not in the ring

[coral spindle]

Have you proved they aren't in I?

[solar glacier]

no

[slim kayak]

How are you defining a prime ideal anyways?

[solar glacier]

but det is right theyree not even in the ring

[chilly ocean]

but how do we know that rho always is surjective when the set A is arbitrary, since rho is unique?

[solar glacier]

(P) is prime if whenevrr ab \in P then either a \in P or b \in P

[rustic crown]

(and proper)

[coral spindle]

Define f : F(A) -> F(A)/R to be f(x) = xR. Do you agree this is surjective? Now define rho to be f composed with the isomorphism, and we're done.

[slim kayak]

My favourite prime ideal, just the entire ring

[rustic crown]

then i guess your favorite field is Frac(0) = F_1?

[slim kayak]

Yes

[chilly ocean]

Oh wtf I’m actually just dumb thanks

[slim kayak]

Products of fields are always fields too for me, I'll just pretend the category of fields is nicer than it actually is

[white oxide]

Can somebody give me a hint for this question? I’m trying to extend my induction hypothesis and I attempted to use the first isomorphism theorem but I’ve been stuck on this question for a while lol (it was show that the finite direct product of solvable groups is solvable)

[coral spindle]

what is the question

[rustic crown]

prod of solvable = solvable

[solar glacier]

finite direct prod of solvable is solvable

[coral spindle]

Have we not been through this before?

[white oxide]

yes but i'm still stuck

lol

[coral spindle]

You showed that the product of two solvable groups is solvable, right?

[solar glacier]

been there my friend been there

[coral spindle]

You happy with that proof?

[white oxide]

nah that's what i'm trying to solve

[coral spindle]

OK

[rustic crown]

(you can show G is solvable iff N and G/N are solvable )

[slim kayak]

Okay, take the homomorphism sending each (a,b) to (aN, bH) given normal subgroups N and H

What is it's kernel?

[coral spindle]

Let G and H be groups. Let H' <= H and G' <= G be normal subgroups. Prove:
(1) G' x H' is a normal subgroup of G x H.
(2) (G x H)/(G' x H') is iso to (G/G') x (H/H').

[white oxide]

tried to construct an isomorphism from Hi+1 x Ki+1 / Hi x Ki onto Hi x Ki/Hi-1 x Ki-1, because then I could use my induction hypothesis that Hi x Ki/Hi-1 x Ki-1 was abelian and hence Hi+1 x Ki+1/Hi x Ki would be abelian since they're isomorphic

[coral spindle]

This'll finish your proof that the product of two solvable groups is solvable.

[white oxide]

ok thanks guys i'll try all of these

[coral spindle]

I want die

[rustic crown]

[coral spindle]

(the RTGame character)

[chilly ocean]

i see now, tysm

[agile burrow]

Bro ngl

Higman's group goes monumentally hard

[coral spindle]

Neat

hadn't heard of it before now

[agile burrow]

It's pretty neat

The funny thing is, if you take that same presentation but with only 3 generators

You get the trivial group lol

Supposedly the group is also acyclic, meaning all the homology groups vanish

I am unable to verify this claim

[white oxide]

oh yea i think i was inquiring earlier as to if this proof would work for normal subgroups and groups, i guess it's somewhat of an analogue to be precise

[coral spindle]

yes

[slim kayak]

You didn't have to check well-definedness there, you are sending concrete elements to some element there.

[torn warren]

I have another question, how to guarantee this operation will generate all its conjugacy classes, for example, maybe there exists some $x$, and $x(123)x^{-1}=(35)$?

[cloud walrus]

Witness

[torn warren]

nice example, thank you!

[white oxide]

oh right because r and s don't have any representatives right

or i don't know how to word it

if the inputs were cosets then i would have to check for well-definedness tho right because those are dependent on representatives

[slim kayak]

Yeah

[tribal moss]

Conjugation always preserves the cycle structure of a permutation.

In fact, a simple trick for conjugating a permutation that is already written in cycle notation is to simply apply x separately each element mentioned in that cycle notation.

So x(123)x^-1 is always the 3-cycle (x(1) x(2) x(3))

[white oxide]

also i had another question, what exactly does a canonical map mean? does it just mean that the outputs are the equivalence classes under some equivalence relation? idk how to really be precise (i know it also has to do with the most natural form in defining something)

[slim kayak]

Sometimes it is a map that is unique up to isomorphism, sometimes it's just a "natural" choice.

[coral spindle]

The word 'canonical' is not precise.

[white oxide]

huh ok

[coral spindle]

You could translate it as just meaning 'nice'.

[twilit notch]

ah thank you

[south patrol]

it's more clear what it doesn't mean like

if you show a group has order 7 then it is isomorphic to Z/7Z

but that doesn't give you some like canonical isomorphism between them

[twilit notch]

ohh

[south patrol]

or like when you have various choices involved in constructing a map

[torn warren]

but in this notation, $x(123)x^{-1}=(xx^{-1}(1), xx^{-1}(2), xx^{-1}(3))$ it will go back to (123), which is incorrect, where is the mistake?

[cloud walrus]

Witness

[empty rose]

yeah i think "canonical" generally just means, like,
sometimes there's one obvious way to do something, like the isomorphism between A x B and B x A that just swaps the things around

[coral spindle]

That's not how this notation works. Maybe look back at the definition.

[slim kayak]

It's just x not the stuff from both sides

[white oxide]

so like would the projection map: G x H --> G given by (g, h) = g be considered a canonical epimorphism?

it's quite natural to define it as such

[coral spindle]

yes

[tribal moss]

It's only (x(1) x(2) x(3)), not (xx^-1(1) xx^-1(2) xx^-1(3)).

[white oxide]

ah i see

[slim kayak]

"If you conjugate a cycle by x it's the same as the cycle obtained by the elements permuted by x" sth like that

[empty rose]

yeah (xx^-1(1) xx^-1(2) xx^-1(3)) would be if you conjugated by xx^-1, which means you're conjugating by the identity and that obviously won't do anything

[tribal moss]

For example, if we give x(1 2 3)x^-1 an input of x(2), then first x^-1 makes it into 2, then (123) gives 3, then x gives x(3) -- so the net effect of the conjugated cycle is to send x(2) to x(3).

[torn warren]

so how to show the operation $x(123)x^{-1}$ will give all permuations of $(123)$, and never gives something like (12) or $(234)$

[cloud walrus]

Witness

[empty rose]

or (234)
well that's wrong, conjugating it by (1234) or something should get you that

[empty rose]

the reason it won't get you (12) is... this

it's always going to be (x(1) x(2) x(3)) and there's no way for that to be (12)

[torn warren]

so when do conjugacy, what type of x can I use?

for example do conjugacy on (123)

then what type of x is allowed to use?

you said I can't use x=(1234)

[empty rose]

...no, i said the exact opposite

[torn warren]

[empty rose]

you said that x(123)x^-1 can never be (234)

i said it can because x might be (1234)

[torn warren]

ah i see...

but why the number can't be less than 3

I mean

x(123)x^-1=(12)(3)=(12)

[coral spindle]

we've just been through this...

[empty rose]

for what value of x is (x(1) x(2) x(3)) = (12)(3)?

[torn warren]

x change 1 to 2, change 2 to 1, and don't permute 3

[empty rose]

is (213) the same as (12)(3)?

[torn warren]

not same

[empty rose]

alright so that x doesn't work then

x(1) = 2, x(2) = 1, x(3) = 3, so (x(1) x(2) x(3)) is (213) and not (12)(3)

[torn warren]

[empty rose]

this is equivalent to "x(1) = 2, x(2) = 1, x(3) = 3", right...?

[coral spindle]

Anyone thinking of farcry right now?

[torn warren]

yes, but why we don't consider the x^-1 part

[empty rose]

what x^-1 part?
the question we're asking is "is there an x such that (x(1) x(2) x(3)) = (12)(3)"

[torn warren]

no, it is asking x(123)x^-1

[empty rose]

yes, and we decided that was equivalent to (x(1) x(2) x(3))

if you disagree on that then we can talk about that instead

[torn warren]

why it is equivalent

[slim kayak]

3?

[torn warren]

and you said I can't do it like x(x^-1(1)), x(x^-1(2)), x(x^-1(3))

[empty rose]

well, consider x(123)x^-1 applied to x(n)

...hm, how do i do this

ok let's think about x(1) as an example

x(123)x^-1 applied to x(1)

we do x^-1 first and get x^-1(x(1)) = 1

x(123) applied to 1

apply (123) to 1 and you get 2

then apply x to 2 and you get x(2)

so we put in x(1) and it was mapped to x(2)

for the same reason, x(2) will go to x(3), and x(3) will go to x(1)

which means the permutation is (x(1) x(2) x(3))

[torn warren]

x(123)x^-1 we need to do (123) x^-1 first, right?

[empty rose]

well we're doing three things: x^-1, then (123), then x
if you want to take the first two things and group them into one thing and say that's the thing we do first, then uh, ...yeah sure that's fine i guess

[torn warren]

for permuations x, y, z, the association doesn't hold, (xy)z not equal x(yz)

[empty rose]

......wait, what?

uh, do you have an example of three permutations for which associativity doesn't work?

[coral spindle]

Witness, are you confusing cycle notation (x y z) for composition xyz?

By (1 2 3) we mean the permutation that sends 1 → 2, 2 → 3, and 3 → 1.

When we write x (1 2 3) x^-1 we mean the composition of the permutation x with (1 2 3) and then this with x^-1

it is a really important fact that actually, composition is associative. xyz = (xy)z = x(yz), and note I am not using cycle notation here.

Is this clear to you?

[torn warren]

let'me write down...

[coral spindle]

If it is not clear, then just say so.

[torn warren]

the red is correct, the blue is wrong

my question is why we can't start from left

[coral spindle]

That depends on your definition

some people define the composition from left-to-right, but most define it from right-to-left (the start you indicate would then be on the right)

If you're using left-to-right composition, as you seem to be, then you would indeed start from the left.

These are completely different definitions

in fact in one, you'd write xy instead of yx to mean the same thing

This has nothing to do with associativity.

[torn warren]

do I understand correctly?

[coral spindle]

No

Wait, one moment

in fact you have; I was thrown off by the swapped order

yes, that is the idea

[torn warren]

we use the red color defintion usually right?

[coral spindle]

That is the most standard one in mathematics generally, yes

[torn warren]

thank you

[torn warren]

why this hold? x(123)x^-1 applied to x(1)
we do x^-1 first and get x^-1(x(1)) = 1

if we do x(123)x^-1 (1), we do x^-1(1), then (123)(x^-1(1))

[coral spindle]

which is...?

Oh no

oh no this is wrong

[coral spindle]

we do not do (123)(x^-1(1))

[torn warren]

I mean go back to here

[coral spindle]

we do (123) (x^-1(x(1))).

Try again.

[coral spindle]

This is correct reasoning.

[torn warren]

It should be 2 right?

[coral spindle]

What you write is correct

[torn warren]

but why you do (123) (x^-1(x(1))).? it is like to switch x and (123)

[coral spindle]

Y'all can take it from here

[torn warren]

so i don't understand this step (123) (x^-1(x(1)))., why we can do this

[tribal moss]

We're trying to establish what x (123) x^-1 is. I claim that it is a permutation that maps x(1) to x(2). So what we need to do is evaluate
x((123)(x^-1(x(1)))) and see that it equals x(2).

[torn warren]

why you can do this?

[tribal moss]

I don't understand the question. There's nothing that prevents me from doing it, and I know how to do it, so I can do it.

(I don't know why your latest images seem to have switched around x and x^-1 in the formulas).

[torn warren]

because you use x(123)x^-1, I use y^-1(123)y

so how doyou prove this?

[empty rose]

you don't

[tribal moss]

Right here from the beginning you had x(123)x^-1.

[empty rose]

i never said anything about applying x(123)x^-1 to 1

i also never really said anything about (123)(x^-1(x(1)))

i was talking about applying x(123)x^-1 to x(1), in other words x((123)(x^-1(x(1)))) (i might have the wrong number of closing brackets there)

[torn warren]

do you mean this?

[empty rose]

well the left-hand side is still irrelevant but the right-hand side there is the expression i was talking about

[tribal moss]

I think at some point Boytjie was indeed talking about applying (123)x^-1 to 1, as part of writing down (123)x^-1 in tabular form, since Witness appeared to want to do it the hard way and compute the compositions one by one.

[torn warren]

ah I know you mean now...

I didn't see x(1)

you act the whole thing on x(1)

I thought you act on 1...

so this conjugacy map (123) to (x(1), x(2), x(3)) , they always have 3 elements in the ( )

[tribal moss]

Yes.

[torn warren]

Fraleigh's book didn't cover this

about this conjugacy

and what is the use for this tool?

by introducing the conjugacy

one thing I can imagine is to use the orbit stablizer theorem right?

all x such that x(123)x^-1=(123) forms a subgroup H, then |conjugacy of (123)|=[G:H] ?

[tribal moss]

Most fundamentally, conjugation by a fixed element is a self-isomophism of the group, so conjugate elements "behave identically" as far as the group structure is concerned. Having some intuition about conjugation is also important for understanding many concretely appearing group actions. Not to mention that kernels of group homomorphisms are always closed under conjugation, so being able to reason about conjugate elements is a great help when trying to design homomorphisms with specific properties.

[torn warren]

but for me, I just know this concept today/yesterday...

[long nebula]

if you know linear algebra, conjugation is just a change of basis

conjugation of permutations is very similar

you switch around the numbers

for example (1 2) is conjugate to (1 3) via a permutation which switches 2 and 3, but it's not conjugate to (1 2 3)

because you can't get from (1 2) to (1 2 3) just by switching around numbers

[torn warren]

for linear algebra, we change basis and can diagonalize the matrix , such as in the quadratic form

[long nebula]

Yeah so there you're writing a matrix as PAP^-1 = D which is a conjugation of A by P

it's the same underlying structure, but the basis vectors are changed

similarly, if you conjugate a permutation, you get another permutation with the same underlying structure, but you change around the numbers

[torn warren]

so eigenvalue is like the group H={x| x(123)x^-1=(123)}, right?

[long nebula]

Uhhh that's more like the family of matrices which are similar to a matrix

Er wait no

Ignore me

[torn warren]

I think it is like Ax=x

[long nebula]

That's the group of permutations which commute with (1 2 3)

[torn warren]

means it is invariant

[long nebula]

So it's like the group of matrices which commute with another matrix

[torn warren]

x(123)x^-1=(123) also means is invariant under conjugacy

[long nebula]

This is called the centralizer of (123)

[long nebula]

(1 2 3) is invariant under conjugation by x, yes

in other words, (1 2 3) commutes with x

x(1 2 3) = (1 2 3)x

[torn warren]

what i try to make an analogy is , Av=v, so v is invariant under A.

x(123)x^-1=(123), so (123) is invariant under conjugation x, so (123) is analogy to the eigenvector v

but eigenvalue can be any other values, so for other eigenvalues, seems no analogy to the conjugation...

[long nebula]

This is not the right analogy to be making

If you want to make an analogy to linear algebra, you could think of (1 2 3) as being analogous to a matrix A and H being analogous the set of matrices that commute with A

[lilac mango]

What's a good place to start learning abstract algebra from scratch?

[summer path]

This is a question for #book-recommendations , but you can start by looking at the book review here: #book-recommendations message

[lilac mango]

Oh thanks

[torn warren]

I got another question

so, I need to find all g, such that gx=x, right?

[lusty marlin]

If by gx you mean the element g acting on x, then yes.

[torn warren]

but I can only find one, which is trivial one, how to find others?

[white oxide]

how is this true? if p = 7 and n = 4, then (-1)^7^4 = (-1)^4 = 1 (or am i trippin)

[empty rose]

$a^{b^c}$ means $a^{(b^c)}$, not $\left(a^b\right)^c$

[cloud walrus]

bee

[white oxide]

tfw you forget the algebra needed to do algebra

[empty rose]

so since $7^4 = 2401$ is odd, $(-1)^{2401} = -1$

[cloud walrus]

bee

[white oxide]

ah got it thx

[lusty marlin]

Group actions are not the same as composing two elements with the group operation.

[torn warren]

can you give me an example? I suppose g is from S_13, right?

then gx=x

[lusty marlin]

Do you know what group actions are?

If no, then you should read about them before trying to attempt exercises involving stabilizers and other related terms.

[torn warren]

yes, I know

I know stablizer

[lusty marlin]

Then it is strange how you are confusing it with the group operation.

[torn warren]

and orbit stablizer theorem

but I don't know what is this problem talking about

[lusty marlin]

What is the group action defined in the exercise you sent? I'm assuming it is conjugation since that is when Z is used to denote the stabilizer.

[torn warren]

ah, so it means all g, such that gxg^-1=x, right?, where x is (123)(456)(78910)(11)(12)(13)

[lusty marlin]

Yes

[torn warren]

thank you, because this problem is from online, I don't have the textbook where it comes from

but I think I am clear about this now

thank you

[white oxide]

how do the elements of E being zeros of x^p^n - x imply that f(x) is a factor of x^p^n - x in Zp[x]?

oh it has to do something with the first isomorphism theorem i believe

[elfin prairie]

Need some help understanding this proof

I know that it is an inductive proof but I can't properly grasp its structure (base case, inductive hypothesis)

[empty rose]

the base case is... i guess n <= 2
the inductive hypothesis is the "suppose that we have defined the product of r elements when r <= n - 1, and that it is the unique product satisfying (iii)"

[elfin prairie]

hmm

it doesn't use r or (r + 1) after though

[empty rose]

they don't explicitly write "r" but they use products of lengths that are <= n - 1 all over the place

the fact that [a_1 ... a_i] exists is using the inductive hypothesis with r = i

[elfin prairie]

oh I see

[white oxide]

anyone know?

<@&286206848099549185>

[lethal dune]

your extension E is finite so E^* is a cyclic multiplicative group with pⁿ-1 elements

x^n = 1 for all x of finite group G of order n

[thorn monolith]

сan someone help me how to find out if Q/2Z is a finite group and if it is cyclic?

[lethal dune]

can you show Q/2Z iso Q/Z

[slim kayak]

It probably helps to visualize what is going on. Like, each number [0,2) corresponds to an unique coset. Furthermore, can any representative element in [0,2) produce all others as Z-scalar multiples of itself?

[thorn monolith]

oh yeah Q/2Z iso Q/Z from this i can conclude that the group is infinite, but is it possible to say something about cyclicality based on this? or i should use something else

[lethal dune]

it is

say your Q/Z could be generated by 1 element

say [a/b]

can you produce an element that cannot be written as some n*a/b?

that'll show it's not cyclic

[south patrol]

Take a root a of f in E, then f is min poly of a

[median pawn]

By McCoy's theorem, we know that two matrices A,B are simultaneously triangularizable if and only if p(A,B)(AB-BA) is nilpotent for every noncommuting polynomial p(A,B). Here's what I'm trying to do - I want to find matrices A,B such that AB-BA is nilpotent, but A,B are not simultaneously triangularizable. We can certainly use McCoy's theorem here, i.e., find appropriate A,B such that AB-BA is nilpotent, and find a polynomial p(A,B) such that p(A,B)(AB-BA) is not nilpotent.

How would you go about this? Thanks!

For simplicity, maybe the 2x2 case should be okay (Idk if it'll work)

[chilly ocean]

We have a ring A with 4 elements. Show that A is a field if and only if the equation x^2+x+1=0 has a root in A

[empty rose]

there might be a better way, but,
there aren't that many rings with 4 elements, you could probably just check all of them

[delicate orchid]

if A is a field with 4 elements it must be an extension of F_2 with degree 2

[coral spindle]

If A is a field, then A\{0} is a group of order 3.

(wink wink nudge nudge)

[formal ermine]

witchcraft!

[median pawn]

why so

[median pawn]

also, bump

[delicate orchid]

finite fields of the same order are isomorphic

[coral spindle]

Ok that's a much more complex fact...

[delicate orchid]

gets the job done

[coral spindle]

but it's easier than that. Any field has a 'prime subfield' which is generated by the elements 0, 1, 1+1, etc

[south patrol]

Lol wew

If you can assume that then this is immediate

[coral spindle]

This is a quotient of Z in the case of a field of order 4, so it's gonna be Z/pZ for some prime p

[delicate orchid]

and?

[coral spindle]

It can't be a non-prime since then it wouldn't be a field.

Now then our field A is a vector space over Z/pZ, so it's of order p^n for some n

since p^n = 4 we have p = 2 and n = 2.

This is an elementary way of deriving this fact @median pawn

[delicate orchid]

oh yeah this sounds way easier.... NOT

[coral spindle]

easier to understand, 100%

I'm a big proponent of elementary explanations for things

[south patrol]

yh

I think what I'd be tempted to do honestly is like

Actually lol have we done the other direction

I was gonna give that one lol

Okay I mean I'd be tempted to say like => follows from group of units being cyclic (as order 3) immediately

as has been pointed out

for <=, you can take any root y of x^2 + x + 1 and then y has order 3 in the group of units and is not 1

so that {1,y,y^2,0} is the entire ring

[median pawn]

thanks!

[south patrol]

and clearly every non-zero element has an inverse

(and it's commutative)

[formal ermine]

FINITE DIVISION RINGS ARE COMMUTATIVE

[solemn garden]

why?

[formal ermine]

wedderburn's little theorem

[delicate orchid]

That’s a pretty complex fact…

[coral spindle]

truly is

quite hard to prove

[solemn garden]

is there a more elementary way to see why finite division rings are commutative

[lethal dune]

proof is already pretty elementary

granted you know somethings about cyclotomic polynomials

they are easy facts to pick up

[south patrol]

Lol

[lethal dune]

lol Boyt said it's hard to prove

[median pawn]

for this problem, if x^2 + x + 1 = 0 has a root in A, one possibility is that 1 + 1 + 1 = 0, i.e., A has characteristic 3

what about that possibility?

if the root is not equal to 1 then I can show that A is a field

[delicate orchid]

jumping off of boytjie's exposition of the prime subfield

if A has characteristic 3 then the prime subfield has 3 elements, 0, 1, 1+1

see if you can use this to show that A cannot have both order 4 and characteristic 3

[wooden ember]

Our rep theory teacher gave us a small introduction to quantum groups lecture for the end of term and made me realize once again how vast every field in math can get

Seems like interesting stuff but I didn’t really get anything lol

[lusty marlin]

The order of a finite field must be a power of the order of its prime subfield, which would be 3 if A has characteristic 3.

[coral spindle]

I find it hard to prove okay

[rustic crown]

https://media.discordapp.net/attachments/796976320067272704/818150000482582528/612933479730249728.gif

[delicate orchid]

damn spoilers.....

[lusty marlin]

Apologies, although I think it would be weird to delete the message now.

[delicate orchid]

it's alright boss

[median pawn]

ahhh is it just that the prime subfield is a subgroup of the ring (as an additive group), and 3 doesn't divide 4, so that's a contradiction?

[coral spindle]

see here.

[delicate orchid]

that's a really good way of seeing it

[south patrol]

nerd

[formal ermine]

noob!1!11 get him!!,,

I can't believe you still find wlt hard to prove at (insert age here). I proved it myself in my mother's womb,,,,

[delicate orchid]

silence, cringer

[plucky topaz]

I created An intriguing collection of rings where R(α,β)={a+bi: a,b∈R & i^2=α+βi}. R(-1,0) is equivalent to the regular complex numbers so i^4=1. But in R(-1,-1), i^3=1. I am trying to find out if i^n=1 for some α and β

[echo gull]

Could you clarify what you mean by some $\alpha$ and some $\beta$?

[cloud walrus]

kodiak

[echo gull]

Notationally this seems a bit confused

[delicate orchid]

these are just quotients R[x]/(x^2-βx-α), are you asking if you can find a root of unity of arbitrary high order in these rings?

[plucky topaz]

α and β are real numbers. so for every α and β, R(α,β) is a ring. if we evaluate R(-1,0) we get that R(-1,0)={a+bi: a,b∈R & i^2=-1+0i}=C

[delicate orchid]

that's like, everything except what we wanted you to clarify

[south patrol]

Uh is this like R[x]/(x^2 - a - bx)

Lol

[delicate orchid]

ty potaot

[south patrol]

Oh

LOL

[oblique river]

maybe I can help clarify: this ring is just $\mathbb{R}[x]/(x^2 - \beta x - \alpha)$

[plucky topaz]

Oh. I didn't know it was a quotient. but yes, I am trying to show if there are always a solution for nth roots of unities

[oblique river]

it depends on alpha and beta

[delicate orchid]

well every root of unity exists in C but I'm guessing you mean specifically the thing you're extending by is the root of unity

hmm

[oblique river]

if beta^2 + 4*alpha > 0

then your ring is just isomorphic to R x R

i'ma ssuming you don't want trivial solutions to x^n = 1

namely x = 1

[delicate orchid]

ah yeah I forgot the ol discriminant meme

[oblique river]

if beta^2 + 4*alpha < 0

then x^n = 1 will have a primitive solution for all n

because the ring youg et is isomorphic to C

to be clear: your rings R(alpha, beta) are either isomorphic to R or to C

oh sorry wait not true

[delicate orchid]

or a more sinister third option

[oblique river]

R + R or C

oh wait yes the third option

R + R, or C, or R[t]/t^2

[delicate orchid]

the third option is when the discriminant is zero in case it wasn't clear, dubdub

[oblique river]

let D = beta^2 + 4*alpha. if D < 0 then your ring is just the complex numbers again in which case you have primitive solutions to x^n = 1 for all n. If D > 0 then your ring is isomorphic to R x R, and there are primitive solutions to x^n = 1 only for n = 1, 2 and for no other n. If D = 0 then your ring is slightly more complicated but you get the same answer as the previous case

[plucky topaz]

I was able to find this inequality, and realize when these rings are also fields. but I failed to notice the isomorphism. thanks

[oblique river]

the isomorphism isn't immediately obvious

but it comes from the chinese remainder theorem

at least, that's how i like to think about it

(well, at least in the case D > 0)

[plucky topaz]

are desmos links valid

[oblique river]

?

What do you mean

[plucky topaz]

are they allowed, is there rules against it

[summer path]

why would there be

(there aren't)

[plucky topaz]

https://www.desmos.com/calculator/tovtwyoidn I found another pattern, almost all solutiond happen at alpha=-1

[oblique river]

Why dont you read the rules yourself

Solutions to what

[plucky topaz]

for i^n=1, I was fixing n, and solving for α and β, not the other way around. and in that case α=-1 execpt for R(1,0) with n=2

[thorn monolith]

How many ways to color a necklace of 6 red and 6 blue beads (located in the vertices of a regular 12-gon), if any two necklaces that differ only in rotation or axial symmetry are considered equal? It seems that such problem is somehow solved by Burnside's lemma, but I don't really understand how to use it

[formal ermine]

I thought this was gonna be a probability question

[south patrol]

So basically a specific group acts on the set of such necklaces and the number of different necklaces is the number of orbits

[formal ermine]

had me in the first half ngl

[south patrol]

Burnside tells you how to compute # of orbits by breaking it up

Consider Fix(g) for different g in the group

[oblique river]

oh i see, you specifically want i^n = 1, not just x^n = 1 for some x

in that case, yes, the only solutions will be when alpha = -1

your i is a root of the polynomial x^2 - beta*x - alpha, and so the question is "when are the roots of that polynomial roots of unity?"

and that's only going to happen when alpha = -1

[valid night]

My professor just proved a theorem today that says that every abelian group is isomorphic to $\mathbb{Z}/n_{1}\mathbb{Z} \times \mathbb{Z}/n_{2}\mathbb{Z} \times ... \times \mathbb{Z}/n_{k}\mathbb{Z}$ where each $n_{i}$ is a whole number that devides $n_{i+1}$

He also showed what those $n_i$ are for $(\mathbb{Z}/16\mathbb{Z})^\times$

[cloud walrus]

Kroros

[valid night]

But is there some kind of procedure to find those $n_i$s?

[cloud walrus]

Kroros

Kroros

[echo gull]

This is just some version of the fundamental theorem of finite abelian groups, no?

[valid night]

Correct, I misspoke. My apologies

[echo gull]

Oh I misread

[valid night]

But I have to find these direct products and I don't know how

Hmmm

I found that $V_{4}$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^2$

[cloud walrus]

Kroros

[valid night]

Is that correct?

Excuse my countless edits

[oblique river]

yes but isn't that how V_4 is defined to begin with?

[valid night]

Not sure

[summer path]

how can you find that that it's isomorphic to something you don't even know the definition of though

it's correct though

[valid night]

I know V_4 as the definition that is also on wikipedia

This one

[summer path]

i see

[valid night]

Though there the primary definition given is ideed Z/2Z ^2

[delicate orchid]

this one here?

[echo gull]

There are two abelian groups of order 4, isomorphic Z4 or Z2 x Z2

V_4 is significant in that it has all nonidentity elements as self inverses which is Z2 x Z2 by definition

[valid night]

Ah

[oblique river]

that is true for (Z/2Z)^n for any n, though

I don't think that's really all that special about V_4

[delicate orchid]

the only reason we even give C_2^2 a special name is because it was a very early example of a group

[summer path]

tbh i don't think i learned klein 4 group as having the notation V_4; only found out about that from this server

it was always just referred to a Z/2 x Z/2

[echo gull]

It's from the German

[formal ermine]

KLEINSCHE VIERERGRUPPE

[summer path]

oh no the germans

[echo gull]

lovely language

[echo gull]

not that this behavior in general is special, it is just a way of stating V_4 as the minimal abelian group that is not cyclic

[coral spindle]

it's also just the smallest non-cyclic group period, let alone Abelian

that is a pretty special property

[echo gull]

my wording was redundant but ofc all groups up to order 4 are abelian

[oblique river]

it may not be a cyclic group but it is cyclic as a module over the ring Z/2Z x Z/2Z

[coral spindle]

Buncho you're a meanie

[oblique river]

i didnt take abstract algebra 2 for nothing

[coral spindle]

if abstract algebra is so good, why isn't there a—PPFT

[lethal dune]

can anyone help me understand the induced character formula?
I get that we can write induced representation as
⊕gV for g each g representing a coset of H in G. So in the character computation of s ∈G, the only component of the direct sum matters that is fixed by s ∈ G but how is the translates to the formula we see? I feel it's trivial but I'm not seeing it atm

[south patrol]

Ye fam 1 second

So like the point is that okay say you have some s, well as you say we only care about the bits fixed by s. If s sends gV -> gV then sg and g must represent the same coset of H in G, so then g^-1 sg is in H and s acts on gV as g^-1 sg acts on V

[torn warren]

what is this sentence meaning? I don't understand, what are the 6 conjugacy classes does this refer to?

[coral spindle]

Cycle type (3) is not odd

(123) = (12)(13) is even.

[torn warren]

I know it is even, I am asking the last sentence

there are four conjugacy classes which are even, right? (1), (123), (12345), (12)(34)

why the OP says conjugacy classes split? what do they mean split

a class is just a class, why split?

[south patrol]

When they talk about split they mean like

Suppose C is a conjugacy class of S_n which is contained in A_n. Then it "splits" in A_n in that it is actually a union of two conjugacy classes in A_n

Like, conjugacy in S_n doesn't imply conjugacy in A_n

[lethal dune]

that's the part I don't get is why s acts as conjugation by g

[south patrol]

It doesn't act as conjugation by g, it acts as g^-1 sg acts i mean

[lethal dune]

or is the definition of the action?

[south patrol]

As for why that is the case uhhh i think it depends on your exact construction but the point is you can transfer the action of any element of H on the g stuffs to the action on V

[lethal dune]

surely the definition naturally should arrive from kG ⊗_kH V

[south patrol]

Though I fear you may be using a different construction of the induced rep to me lel

yeah okay sure so the point here is like we are decomposing as the subspaces like

gkH (x) V right

[torn warren]

can you give an exmaple? such as S_5

[torn warren]

what it is not working in A_5

[lethal dune]

well yeah I don't see why the action on Induced should be the one coming from tensoring

Rather I would like to see the equivalence

[south patrol]

Wait lol how are you defining it then i am confused

s.(g (x) v) = sg (x) v = gg^{-1}sg (x) v = g (x) g^{-1}sg.v basically is the point

[lethal dune]

I've see 2 definitions

[delicate orchid]

if I see the goofy submodule definition I'm flipping out

[south patrol]

i believe lol

[lethal dune]

one using the direct sum I described and another using tensors but I don't see why tensor one gives the same action

[warped fable]

Draw the matrix

[delicate orchid]

yesssss

[south patrol]

i mean that's the bit ryu understands right lol

I think the key bit is how you can move over the H action to V

Bimodule structure ting

or whatever

[delicate orchid]

the key bit is viewing the goober matrix as a tensor product

[warped fable]

exactly

or, try writing an element of kG o_kH V explicitly

[south patrol]

i have a reps exams next week and this has reminded me to look under the hood a bit lol

[warped fable]

it necessarily looks like g o v

[delicate orchid]

it's literaly the tensor of the rep with C[G] taken to be a H rep

they are... the same 😌

[warped fable]

where H linearly holds when you multiply… however

then break it up into cosets

[south patrol]

Aaaa

Yeah the way we defined the induced rep was slightly interesting lol like

[delicate orchid]

_ _

[lethal dune]

yeah so does kH acts on the kG by left mult?

[warped fable]

who cares man fix a direction, i only remember this block matrix

[lethal dune]

then we transter it to the right by inverse multiplication?

[delicate orchid]

oh now I see why you're bringing up the bimodule stuff potato

yeah

[south patrol]

Take an $H$-rep $V$, form the tensor product $\mathbb k[G] \otimes_k V$ (of vector spaces), both $kG$ and $kH$ act on it by $g \cdot (x \otimes w) = gx \otimes w$ and $h \star (x \otimes w) = xh^{-1} \otimes hw$, then form $kH$-coinvariants

[cloud walrus]

potato

[south patrol]

Boils down to the same as non-commutative tensor product

[delicate orchid]

lower case mathbbk my beloved

[warped fable]

Its just the upper star operator of V considered as a sheaf over Spec kH to Spec kG

[south patrol]

Anyway with the tensor product formulation the key thing is that like gh (x) v = g (x) h.v

[delicate orchid]

time to spill the beans

[lethal dune]

so h•g= gh^-1?

[south patrol]

SHIFT THAT BOY OVER

Uh I'm not sure wht you mean by that

why is h acting on g

Sidenote ryu, love the name lol

[lethal dune]

kG as kH mod?

I wrote it in reverse mb

[delicate orchid]

define for $X \subseteq \bC[G]$ the subspace $X(\bC[G]) = \langle xg \colon x \in X, g \in G\rangle$. Let $V$ be a $\bC[H]$-submodule, then $V \uparrow G$ is $V(\bC[G])$.

[cloud walrus]

Wew (

[delicate orchid]

this is how I learnt about induction thanks reps and characters of groups by James and Liebecks you HACKS

[lethal dune]

(it's just notations ahem)

[south patrol]

I don't see what's sexy about the arrow notation lol well

Okay actually fair enough

I was gonna say I prefer usually when people use like uh

$\chi^G$ or $\chi_H$ but then it makes sense for general tings to avaoid confusion with $V^G$ and stuff

[cloud walrus]

potato

[torn warren]

can you give an exmaple? such as S_5

[south patrol]

Oh sorry

[delicate orchid]

non arrow users be like

[south patrol]

So for example pretty sure (12345) is not conjugate to its square in A_5

[chilly radish]

WTF is this definition

[south patrol]

Lol

[chilly radish]

Isn't induction just an extension of scalars

[warped fable]

This seems the most sensible to me, am I wrong

[delicate orchid]

indeed

[south patrol]

Yes I agree Ibsen lol

[delicate orchid]

ohh mr wew lads mr wew lads why is it taking you so long to learn block theory idk maybe because every statement looks like this?

[south patrol]

Probably avoided out of laziness to not have to inroduce the non-comm tensor product right

kek

[torn warren]

I don't understand, what is the defintion of "split"?

[delicate orchid]

(12345) is conjugate to (13245) in S_5, but it isn't in A_5

[lethal dune]

is it the correct one? that's what I'm confused about

[delicate orchid]

we say this conjugacy class splits upon restriction to A_5

[lethal dune]

ok too much clutter

[warped fable]

it seems the decomposition you want works out with this one

big rep theory thread

make it

[south patrol]

Ughhh

Is there a nice way to see when a representation is induced from another one lol

[warped fable]

mckay lol

[torn warren]

can you give me an example, what is the x and y, such that xyx^-1=what?

[south patrol]

So like I am doing something with restriction to normal subgroups rn

No I mean like prove that smth is induced hm

[warped fable]

oh induced no idea

[south patrol]

Actually it shouldn't be bad from universal properties

Like show that it extends in a unique fashion

[delicate orchid]

always banging on about universal properties

[south patrol]

from that ting

Yes

Specifically what I'm doing is like

[delicate orchid]

ur rep theory is so much more sophisticated than mine I just bang characters together

[lethal dune]

someday

[torn warren]

who have time now?

[delicate orchid]

it should be fairly easy to see that (23) conjugates that element to the other one

which is not in A_5

[south patrol]

Basically it's uhhh

[torn warren]

what is x and y?

[delicate orchid]

wow I really do need to spoon feed this

x = (23) y = (12345)

[south patrol]

If V is an irrep of G and N a normal subgroup, then Res_N V is either isotypic or induced from an irrep of a proper subgroup H containing N

And ye

[warped fable]

nobody understands characters one just computes with them

[south patrol]

lol

[delicate orchid]

I understand characters way more than I understand this module nonsense

[south patrol]

Yeah this course basically morphed from a more non-comm-alg kinda focused thing

[delicate orchid]

cringe, banned

[south patrol]

Ok

[torn warren]

xyx^-1=(23)(12345)(23)=(13245) they don't split

[south patrol]

x isn't in A_5

[torn warren]

I don't know the definition of split

[south patrol]

I thought we explained it like

[delicate orchid]

yeah I have just told you this

[south patrol]

If $C \subseteq S_n$ is a conjugacy class, then $C$ is a union of conjugacy classes in $A_n$. If it's a union of more than one conjugacy class in $A_n$, it "splits" in $A_n$

[cloud walrus]

potato

[echo gull]

Conjugacy classes in the symmetric groups order n are by cycle type

[delicate orchid]

even better, potato

[echo gull]

In the alternating group, these conjugacy classes can split

[torn warren]

I don't understand, can you give an exmaple of split?

[delicate orchid]

I

just

did

(12345) is conjugate to (13245) by (23) in S_5

(23) is not in A_5

so the conjugacy class of (12345) splits into the classes (12345) and (13245)

[torn warren]

ah ,i see

[echo gull]

Also note iirc this splitting behavior is relegated to elements with cycle decomposition of k-cycles of distinct odd lengths

[delicate orchid]

it is, those are the only ones that split

[south patrol]

Hmm

I need to review uhhh

Actually no

[delicate orchid]

I need to review uhhh ermmm uhhh

[south patrol]

But character table of S_n being just integers is pretty cool

[delicate orchid]

Fulton Harris chapter 4(?) is all about S_n if you want to read that

[south patrol]

But not too bad iirc using GALOIS THEORY and ALGEBRAIC INTEGERS

[delicate orchid]

it has the funny tetris pieces

[south patrol]

Lol

Well I think this is much more elementary

[delicate orchid]

alternatively you use the funny tetris pieces

[south patrol]

In some sense

[echo gull]

S_n is beautiful

[torn warren]

wait, what about this phenomenon? for example, let x=(241), a=(1234), y=(3,4) then we have $$xax^{-1}=(2431)=(1243)=yay^{-1}$$, there exists two elemnets x and y, where x and y are different types, such that $$xax^{-1}=yay^{-1}$$

[cloud walrus]

Witness

[delicate orchid]

yeah you just use the fact it's an action of a symmetric group iirc

[south patrol]

Yh yh

Yh yh take a splitting field for the group &c. whitney and induct

Actually wew you probably know too much about p-groups right

nice pfp kodiak

[coral spindle]

can you please explain simple and intuitive-like what "whitney and induct" means

[torn warren]

your arguemnt says, since (12) is not in A5, but maybe other elements in A5 can do this conjugacy

[delicate orchid]

(241)(1234)(241) = (1324) no? -no, it isn't

[echo gull]

Note that xax^-1 in general is just the permutation represented by x acting on the elements in the cycle decomp of a

[delicate orchid]

conjugation is a bijection

it's invertible

[echo gull]

For clarity, if a = (1 2 3) then xax^-1 would be (x(1) x(2) x(3))

[delicate orchid]

ah wait yes I see what you mean now

[torn warren]

(241)(1234)(142) = (1243)

[coral spindle]

oh no it's happening again

[delicate orchid]

c_x = c_y if and only if x and y are in the same coset of G/Z(G) = Inn(G)

good thing the centre of S_n is trivial so it all works out

good spot

[delicate orchid]

go on

[torn warren]

here y is not in A5, but x is in A5, just now your argument only says y is not A5, but you can't exclude other elements such as x, and x can do this job in A5

[south patrol]

Uuuh no I don't think I will.

[delicate orchid]

xax^-1 = (1324)
yay^-1 = (1243)

used wolfram alpha to calculate these there's no room for error

[delicate orchid]

the conjugation maps of two elements can only be equal if x = yz for some z in the centre of the group

hmm

[echo gull]

wolfram alpha for some reason can misinterpret cycle notation while in natural language (source: my own trauma) for some reason but I anticipate these are correct regardless

[delicate orchid]

but why must this imply they're different pointwise

[delicate orchid]

I put "permutation" in front of it so it had better not

[south patrol]

Oh Wew it was like

uhhh

[delicate orchid]

if c_x(a) = c_y(a) => c_{y^-1x}(a) = a => y^-1x \in C_G(a), weird

[south patrol]

Proving that p-groups are monomial lol do you happen to know how to do this

/ where to look lol

Hmmm

Kinda cool though

Actually

[delicate orchid]

yeah lemme google the definition of a monomial group first though

[south patrol]

lol

if P is a p-group then every (complex) irrep is induced from a linear rep of a subgroup

So really actually I guess you've just got to say it's induced from some rep of a proper subgroup if it's not linear

Which should be ok i imagine

[delicate orchid]

I am so bad at thinking about induction

[south patrol]

Hm

i will think anyway dw lol

apparently this is true for nilpotent groups in general tho

[torn warren]

No, your first line is incorrect, $xax^-1\neq (1324)$

[cloud walrus]

Witness

[torn warren]

I verify it, this is incorrect

[south patrol]

Oh nvm I know how to do it Wew lol

[delicate orchid]

ok sure don't care anymore

[south patrol]

Piecing together a few things

[delicate orchid]

it's a nilpotency thing I bet

[torn warren]

could you explain this

[delicate orchid]

witness I really don't care about this conversation anymore

[south patrol]

Yes Wew look

Wolfram is using a different piece of notation to some

So yeah probably wrong order

[delicate orchid]

I do it backwards as well

[south patrol]

Oh lol

Ye fair

[delicate orchid]

anyway tell me about ur magical nilpotency adventures

[south patrol]

Okay this is funny because I have read more of Serre's book on linear reps and then looked at a past paper and was like OOH

now I understand

lol

Anyway no so I mean dw I just basically think I pieced together stuff so uhhh

let P be a p-group, obviously if it's abelian we're done

[delicate orchid]

don't give me that dw I wanna KNOW

[south patrol]

So suppose it's non-abelian

It has an abelian normal, non-central subgroup Q

[warped fable]

love over gold, knowledge over certainty - wew, probably

[south patrol]

So let V be an irrep of P

If it's not faithful then we are also done by inducting after quotienting out right lol

So we may as well assume it is faithful

Now you can prove that if $W$ is a faithful irrep of a group $G$ and $N$ a normal, abelian subgroup, then if $\mathrm{Res}^{G}_{N} W$ is isotypic, $N$ is central

[cloud walrus]

potato

[torn warren]

for example, let x=(241), a=(1234), y=(3,4) then we have $$xax^{-1}=(2431)=(1243)=yay^{-1}$$, there exists two elemnets x and y, where x and y are different types, such that $$xax^{-1}=yay^{-1}$$, surely, $y\notin A_4$, but it is not sufficient to conclude (1234) is not conjugate to (1243), because you can find another element $x\in A_4$, such that $x(1234)x^{-1}=(1243)$, can anyone explain this?

[cloud walrus]

Witness

[south patrol]

Anyway, so using that abelian normal, non-central subgroup, we find that Res^P_Q W is not isotypic

Then [theorem in Serre] says that it's induced from an irrep of a proper subgroup

That's not too hard to prove by considering the isotypic decomposition and just picking anything in that decomposition

[delicate orchid]

right and the proper subgroup is abelian here, hence you're inducing a linear

[south patrol]

Well sure actually yes

But it wouldn't even matter because we could just use induction on size of group

But yes

Kinda cool how everything fits together

[delicate orchid]

this is the only bit I'm not 100% on

[south patrol]

Sure hm

[delicate orchid]

assuming we're quotenting by the kernel of the rep

[south patrol]

like

Okay yeah there's one thing I need to check but it's meant to be like

$V$ an irrep of $P$, sps kernel $K$ is non-trivial so our rep is inflated from a rep $W$ of $P/K$. Then again by induction on size of the group, $W$ is induced by a linear character of a subgroup of $P/K$. This is of the form $Q/K$ of course for some subgroup $Q$. Then I'm pretty sure like inflating this to $Q$ and induction gives us back $V$

[cloud walrus]

potato

[delicate orchid]

oh right yeah duh you just

yeah all you had to say was "lift from the quotient by the kernel wew lads you muppet"

[south patrol]

Lol

Only thing is uhh would need to check that like

lifting and induction are compatible in this way lol

But I'm p sure they are

[delicate orchid]

you can probably do that using the character formulas

[coral spindle]

I still wanna know >:(

what does whitney and induct mean @summer path why mock me so

is it a meme, is this what's happening

[delicate orchid]

OK FINE I FORGOT SOME DETIALS

JESUS WEPT

[torn warren]

how do you exclude other elements z in A5, such that z(12345)z^-1=(13245)

[chilly ocean]

some dude on the physics server was discussing a differential topology problem and their solution involved "applying whitney's embedding theorem and inducting on the dimension"

[coral spindle]

Oh right it was a meme then

[chilly ocean]

the dude suggesting it was most likely 100% serious

[coral spindle]

well what I mean is now it's a meme

[torn warren]

my question is it is possible right?

[delicate orchid]

$x(12345)x^{-1}= y(12345)y^{-1} \iff xy^{-1} \in C = C_{S_n}((12345)$. $|C| = 5$ and so the only things that can centralise $5$ cycles must have an order that divides $5$ by lagranges theorem therefore $xy^{-1}$ must have order $5$ and therefore it must be a $5$ cycle, so either $x, y$ are both even or are both odd as $(12345)$ is even. Therefore if $x(12345)x^{-1}= y(12345)y^{-1}$ either $x, y \in A_5$ or neither of them are, thus my statement holds as $(23)$ or whatever it was is not in $A_5$.

I feel light headed

[cloud walrus]

Wew (

[delicate orchid]

time to sort the latex out

only thing that remains to be seen is that the order of the centraliser is 5 in which case I suggest you google the formula

[echo gull]

good explanation

[delicate orchid]

thanks I made it up as I went along

[coral spindle]

If there's even a single element in S_n outside of A_n that centralises a given element, then that conjugacy class splits

[south patrol]

Whitney and induct

[coral spindle]

this is because of orbit-stabiliser

[coral spindle]

can't fool me now, potato

[delicate orchid]

I'll orb ur stab in naught but a moment

I legitimately did not think to do that

[coral spindle]

It's how I learned this

but my patience wears thin for some

as it seems yours does too

[delicate orchid]

no it's fine, I was more pissed at myself for not being able to answer it

apologies if I came off brash, witness, it was misdirected anger

[torn warren]

I see, I think I can understand now, but only one thing: why |C|=5, why only 5 elements in this centralized group C?

[echo gull]

^

[torn warren]

okay, you have already known what I am going to ask...

[echo gull]

For a particular answer on this, recall that conjugacy acts on a cycle decomposition as if the permutation you were conjugating by acted on each element in the cycle decomposition of the conjugated element, that is $\tau \rho \tau^{-1}$ with $\rho = (1 2 3)$ is exactly $(\tau(1) \tau(2) \tau(3))$.

With this in mind, remark obviously (2 3) is not in the alternating group on order 5. Suppose then there is some element in $A_5$ which conjugates (1 2 3 4 5) into (1 3 2 4 5). As all elements of the cycle are enumerated, we can see clearly this permutation is (2 3), but there is no such permutation in $A_5$.

I’m on my phone so I cannot browse effectively for the formula Wew is referencing

[cloud walrus]

kodiak

[echo gull]

This explanation works though because you have cycles which are aloof on elements in the cycle decomp, like how (1 3 2) is conjugate (1 2 3) by both (3 2) and (3 2)(4 5) etc.

[torn warren]

the flaw is: since (13245)=(32451), if we consider x, such that $x(12345)x^{-1}=(32451)$, then we get $x=(1542)$, surely, x is not in A5. But we get some troubles, since you only exclude $\tau=(23)$ as you selected, but we also need to exclude $x$ as I selected. Therefore, we have many false candidates need to be excluded. Therefore, we need a general proof, i.e. the proof provided by @delicate orchid finally.

[echo gull]

Oh yeah that’s totally right lol

It’s been a while

[cloud walrus]

Witness

[torn warren]

I think I know why |C|=5, since for any g in A5, g(12345)g^-1=(12345)=(23451)=(34512)=(45123)=(51234)

and g(12345)g^-1=(g(1), g(2),...,g(5)) , so there are five cases for this 5-tuple to be assigned.

[torn warren]

I understand my previous example, $$(241)(1234)(241)^{-1}=(34)(1234)(34)^{-1}$$, let $x=(241), y=(34)$, so $xy^{-1}\in C_{(1234)}$, and $|C_{(1234)}|=4$, so the order of $xy^{-1}$ can be 1, 2, 4. If $|xy^{-1}|=4$, then $xy^{-1}$ must be a 4-cycle, which is odd permutation, and odd permutation is the product of odd*even, that's why x=(241) is even and y=(34) is odd, and both of them leads to the same conjugation of (1234).

[cloud walrus]

Witness

[torn warren]

but this can't happen for the example provided by @delicate orchid , because he did is (12345)

can anyone confirm this reasoning?

why you all disappear at the same time

[torn warren]

oh,no, where are you?!

[long nebula]

No one here is obligated to answer, anyone helping is doing so in their own free time out of niceness

!volunteers

[flat tree]

Helpers are just people volunteering their time to help you. Be polite.

[long nebula]

Be patient lol

[torn warren]

This is false, counter-example: consider $S_5$, let $a=(12)(34)$, $x=(12)\notin A_5$, we have $xax^{-1}=a$, hence $x\notin A_5$ can centralize $a$, but the conjugacy class of $a$ do NOT split.

[cloud walrus]

Witness

[warped fable]

yeah this is a famous one

[warped fable]

remembered this example a while back while teaching a kid the class equation for S_4 vs A_4

[echo gull]

that is a particularly pathologic pathological counterexample

[torn warren]

maybe he means something else? could you confirm it? @coral spindle

Does this claim hold in the single cycle case, for example, $a=(123)$?

[cloud walrus]

Witness

[torn warren]

no, it doens't

$(45)(123)(45)=(123)$, but conjugacy class of (123) do not split

[cloud walrus]

Witness

[torn warren]

and (45) is not in A5

so this claim is indeed false...

[torn warren]

two hours ago they were all here, but they disappeared at the same time, which makes me feel scary

[agile burrow]

Lol I think meant Boytjie meant if there's a single element outside A_n which centralizes an element, then the conjugacy class does NOT split

Because then the stabilizer in A_n is an index two subgroup of the stabilizer in S_n and you get the same orbit/same conjugacy class

[slow wave]

i didnt get it

[long nebula]

I think you might be in the wrong channel; this channel is for group theory, ring theory, etc

[slow wave]

sorry i am new

[long nebula]

If this is a high school algebra question, then #prealg-and-algebra might be a better fit

[torn warren]

do you mean this?

[agile burrow]

Yeah, that looks right

I suppose you technically have to show that the centralizer of an element in A_n is either equal to the centralizer in S_n or is an index 2 subgroup, but that isn't terribly difficult either

[torn warren]

but how to show "stabilizer in A_n is an index two subgroup of the stabilizer in S_n"

right, yes, that's my question...

why either index=1 or index=2

[agile burrow]

I'm thinking second isomorphism theorem may be helpful

[torn warren]

why all these content are not covered in Fraleigh's boook?

I think this is very useful, right?

[agile burrow]

I don't know, you'd have to ask Fraleigh

[torn warren]

ok, let me call him

[agile burrow]

Some books treat it, others don't. Symmetric groups are just a pretty familiar family of groups so some books choose to explore it in more detail

[torn warren]

I see, Fraleigh's book just do an introduction on symmetric group

[solemn garden]

What are the isomorphism theorems used for? Are they like some defining property of some more general structure, like how the open sets become more abstract from analysis to topology?

[slim kayak]

So in algebra you wanna understand the algebraic properties of various objects like groups, ring and more, right? If you have an isomorphism you know that both groups must be essentially the same thing. So knowing that two ways of presenting the same group can be useful since insight from one can carry over to the other. They will also come very handy for making other constructions later on

[solemn garden]

What other constructions can they make?

[sharp sonnet]

the first isomorphism theorem is just a very basic fact that underpins all of algebra; all the other ones are essentially a consequence

one important construction is the correspondence lemma

[torn warren]

All the 3 sentences are wrong?

[hot lake]

none of the given conjugacy classes split

so yeah blue and green are wrong

red tried to be wrong but they forgot that its cycle type is 5,1,1, so it doesn't consist of distinct odd integers

[torn warren]

right, the 1-cycle should also count, thank you

[median pawn]

Hey! Any thoughts on this linear algebra business?

I'm really asking for a construction

[teal vessel]

question: do the rigid motions of a tetrahedron in R3 count rotations through the origin? or would that only count as one in R4?

[delicate orchid]

do you mean reflections?

[teal vessel]

reflections are just rotations through higher dimensional space but yeah

[delicate orchid]

alright Brainiac lets relax a little

reflections are usually considered rigid motions

[teal vessel]

I'm working on some exercises on the 3d equivalent of dihedral groups, and they want me to show that for the group G of rigid motions of a tetrahedron, |G|=12, but that seems low if you include reflections through the origin

[delicate orchid]

right, yeah

it's S_4 if you include reflections and A_4 if you don't

weird

they're definitely rigid motions I'm sure of it

maybe they have to be orientation preserving? idk

[south patrol]

Also I assume you mean like centre-of-mass preserving or smth lol because rigid motions would include translations

So like neglecting translations and reflections makes it clear they don't really mean rigid motions lol

[analog garden]

The way I'm thinking about it is that you have 4 rotations along each axis. And I think any other reflection could just be a combination of those rotations

[delicate orchid]

good point

[south patrol]

So ye must be rotational symmetries

[delicate orchid]

unfortunately not, the symmetry group including reflections is twice the size of the purely rotational group

lets see uhh

there's the rotations around a corner and rotations around the midpoint of two edges, corresponding to cycle types (***) and (**)(**) in A_4

but there's a reflection that just swaps two corners, corresponding to the permutation (**) in S_4

[analog garden]

But its a rigid motion, so I don't think you're allowed to only swap two corners.

[delicate orchid]

what definition of rigid motion are you using

I'm just using it as a synonym for "isometry of Euclidean space" i.e. a distance preserving transformation

[analog garden]

Orientation preserving, meaning that all adjacent vertices have to remain adjacent after the transformation

LOL

neverming youre right

i knew i had this as a homework problem

just found the proof and i said 24 lol

[delicate orchid]

that's definitely still true, the adjacency graph of the tetrahedron is complete

[analog garden]

ah, its not just four rotations per axis, its four roations per face

[delicate orchid]

I did slides on this one moment

[analog garden]

Here's my proof:

Label the faces of a cube 1 to 6. Fix face $i$ as the top face. Then there are only four rigid motions which keep $i$ in place, namely four $90^{\circ}$ rotations about the center of the face. Thus, there are at least 24 rigid motions of a cube. \
Consider any other rigid motion of a cube, such as a rotation or reflection. These motions, however, map the faces of the cube to the top face, where every rigid motion has already been accounted for. Therefore, there are only 24 rigid motions of a cube.

[cloud walrus]

Roald Amundsen (Paul)

[delicate orchid]

wait the CUBE?!

[analog garden]

oh wait tetrahedron?

[south patrol]

Lol

[delicate orchid]

LMFAOOOOOO

[south patrol]

Bruh

[analog garden]

lol sorry i have no idea what im saying

[south patrol]

You said tetrahedron originally 😭

[delicate orchid]

you're right about the cube though!

[south patrol]

Okay this is funne

[delicate orchid]

it is 24 with just rotations

[south patrol]

Yes, the same argument works if you change the numbers

[delicate orchid]

the rotation group of the cube is the same as the "symmetry group" of the tetrahedron

[analog garden]

okay well then now im convinced is 12 for the tetrahedron then. Fix any point at the top and you have 3 rotations. I don't see how that doesnt give all possible orientations of the tetrahedron

unless im missing something about the symmetry group

[delicate orchid]

uh you only have 2 non-trivial ones, corresponding to different 3 cycles in A_4

[delicate orchid]

you have the rotations about an edge as well

[analog garden]

okay i think i see that. you mean like tipping it from face to the next?

[delicate orchid]

oh and to add, if you imagine a plane extending vertically from axis in the figure on the right, and reflecting about that plane, you should see that it fixes 1 and 4 (they lie in the plane) and swaps 2 and 3 - these are the missing symmetries

[analog garden]

i dont see how you could fix 1 and 4 and swap 2 and 3 while maintaining the orientation of the tetrahedron

but im starting feel like im working under different assumptions and thats why are answers keep flying past each other

[delicate orchid]

I'll see if I can find the corresponding matrix

nope, I can't

I'm pretty sure it's in SL_n though so it should be orientation preserving

if only I had a little triangle to spin around

[teal vessel]

I'm imagining a mapping like (a,b,c,d)→(a,b,d,c) which would preserve all distances, but reflect a face, or something similar. It would change the chirality of the figure.

But I'm barely started on this so idk if that's a legal move

[delicate orchid]

it would

any odd permutation would

oh yeah we were talking about the tetrahedron

[teal vessel]

To my understanding, as explained in the original restrictions on the dihedral group, these transformations have two conditions: 1. Map vertices to vertices, and 2. Preserve adjacency. With a tetrahedron, all vertices are adjacent, so a chirality change would be a legal move. I'm guessing the extension they were thinking of wasn't considering students that ignore the intuition that "you can't flip objects inside out."

[delicate orchid]

right yeah, that seems like a mistake in the quesiton

[teal vessel]

They haven't formally defined some of the notions, they're just building intuition, and it's only a little blurb. I can work with chirality preservation, just wanted to double check if my instincts were accurate given my assumptions.

When I get back to the office I may post the paragraph to see if I'm just reading weird

[warped fable]

This embedding of the tetrahedron in the cube gives the embedding of A_4 in S_4.

Which relates the two pictures pertaining to the orientation preserving isometry group of the tetrahedron vs the cube.

You can see for yourself that there are exactly two cosets, by observing that {vertices of the cube} \ {0,1,2,3} forms another embedding of the tetrahedron on the other side of the cube

And the symmetries of the cube either preserve or permute these two tetrahedra

[uncut girder]

Nice

[warped fable]

There’s an interesting variant of this

There’s five ways to embed a cube inside the dodecahedron. The corresponding permutation representation of the oriented dodecahedral group inside S_5 is an isomorphism onto A_5

[summer path]

Wow spoiling all the fun

Pretty pictures though

[warped fable]

There’s more. Consider the action of the oriented cube group on the three pairs of opposite faces of the cube. The permutation representation is the quotient map S_4 -> S_3 given by quotienting out the Klein Viergrouppen V_4 generated by the 2+2 cycles.

Likewise, there’s an action of the oriented icosahedral group on the six pairs of opposite faces of the dodecahedron. The permutation representation A_5 -> S_6 is an exceptional embedding, not conjugate to the standard one.