groups-rings-fields

[cloud walrus]

keeks

[zinc heath]

Could someone help me with this? It's probably really simple, but I'm not being able to get anywhere with it.

[tribal moss]

Suppose not. Then I will contain both something in J\K and something in K\J. Add those somethings.

[zinc heath]

aah

oh ok

got it

Thanks so much!!

[pliant forge]

Im confused about normal extensions because if we have E = Q(2^(1/4)), K = Q(2^(1/2)), how is E a normal extension of K? x^4 - 2 is irreducible in F while also having a root in K but does not split into linear factors in K?

or am i blind

[rustic crown]

what's F

[quiet pelican]

It’s minimal polynomial over K is x^2 - sqrt(2), which clearly splits

It = qtrt(2)

[pliant forge]

but the statement in the screenshot confuses me

is it not irreducible in K?

[pliant forge]

i meant K

[rustic crown]

yea, that's not irred over K. it factors as (x^2 - sqrt2)(x^2+sqrt2)

[pliant forge]

ohhhh

right

my bad

thx

[next obsidian]

Clearly they don’t this theorem

[rustic crown]

[chilly ocean]

It was mentioned here before so I mentioned it again as a remainder

Also the proof of this should be very similar

[chilly ocean]

also to myself

[solar glacier]

how do you show p(x)= (x-1)(x-2) \cdot ... \cdot (x-n) -1 is irreducible over Z i know if its reducible and it equals a(x)b(x) where a(x),b(x) are non units, then for k between and equaling 1,n that p(k)=-1=a(k)b(k) so one needs to be 1, and the other -1.

[rustic crown]

so a and b are polynomials of degree < n

[solar glacier]

at most n/2?

[rustic crown]

one is at most n/2

can't say for both

it could be degree 1 times degree n-1 for all you know

[solar glacier]

ah

[coral shale]

think about the constant term

[rustic crown]

(anyway, use that a polynomial of degree d is determined by d+1 values uwu)

[coral shale]

wut

oh nvm me

o wait no, constant term works

o wait no it doesnt

oh wait it maybe does with a little more thought

[livid rose]

Hello everyone, I have a question about the group center,
If we have a homomorphism between two groups let's say H and G, would the homomorphism send the center of H to the center of G?

[chilly ocean]

have you tried to prove it?

[kind temple]

see what happens first if f is surjective

[livid rose]

Yeah, but I got stuck
I am physics student taking a category theory course

[chilly ocean]

where did you get stuck?

[kind temple]

there isnt much to show tbf. follow your nose through the definitions

[chilly ocean]

what stopped you from proving it, and do you think that says anything about the original question?

[livid rose]

Okay I will try that

[livid rose]

I will try to prove it again

but before I continue would it help me answer this:

Is there are functor from the category of all groups to the category of Abelian
groups, which take group G to its center Z(G)?

My idea is if I can show that the homomorphism don't send centers to centers then there is no functor from the category of groups to the category of ab groups such F takes G to Z(G)

[coral spindle]

You also need to specify how the functor behaves on morphisms

[kind temple]

you need to give an example. try looking at C_2 and S_3

[livid rose]

Okay

[livid rose]

S_3 is a set of premutation, right? what about C_2?

[kind temple]

cyclic group with two elements

Z/2Z

[livid rose]

Ohhh that's true
Okay I think I got it now

[south patrol]

This is actually sorta general ring theory - what do you know about the ideal (π)

[rustic crown]

ferb

[delicate bloom]

I'd suppose for the sake of contradiction it factors as p=x^2+2y^2 and ||try to boil it down to a statement of quadratic reciprocity...||

hopefully that's not too explicit of a hint

spoilered the second half I guess for good measure

[rustic crown]

(ferb wanted something for second part of (i))

[delicate bloom]

unless someone else has a different approach in mind

oh

I didn't check and assumed this was problem 1 and not 14 😭 lol

[rustic crown]

[summer path]

why is det's emoji so cute

[delicate bloom]

AI generated it specifically to maximize cuteness levels

[rustic crown]

[summer path]

[rustic crown]

[summer path]

[rustic crown]

[summer path]

[rustic crown]

[chilly ocean]

[summer path]

[rustic crown]

(i'll go back to #1055183210444763205 before a mod comes here )

[primal tusk]

the polynomials over any field are never a field because there is not always an multiplicative inverse right?

[rustic crown]

yee, positive degree stuff can't be inverted

[primal tusk]

so im trying to prove "If F is a field, the ring F[x] is not a field."

i guess how do i just say that it doesnt exist

[coral shale]

you can think about x^-1

i suppose

oh right - iirc, think degree is the way

[primal tusk]

oh so i can just say the inverse and say it isnt an element of the ring?

[coral shale]

well u cant just assert that

for all you know x^-1 = x^3

or something

tbh I have some misgivings about the Q

hmmmmmmmmmmmmmmmmm

ok nvm im being silly =...=

But you do want to prove basically x^-1 cannot be equal to any polynomial

[coral shale]

.

[primal tusk]

so use the property that when you multiply two polynomials of nonzero degree m and n the resulting polynomial is degree m+n ?

[coral shale]

yh

[primal tusk]

Uh okay but i dont see how to show i cant be equal to any polynomial

like clearly to get an inverse for a degree m polynomial i need a degree -m polynomial

[rustic crown]

yee and degree of any non-zero polynomial is a non-negative integer

and 0 is clearly not an inverse

[summer path]

so like suppose you could have an inverse for x in F[x], you can write it out as a polynomial in F[x] and reach a contradiction

[primal tusk]

makes sense

[coral shale]

to be clear use the definition of inverse

x . x^-1 = 1

deg(x) + deg(x^-1) = deg(1) = 0

1 + deg(x^-1) = 0

[primal tusk]

that makes sense

[coral shale]

And remember this result is proven by observing the leading term

(the 'difficulty' is proving the coefficient isn't 0)

[primal tusk]

are fields integral domains

[coral shale]

yes

[primal tusk]

ohh sick

[coral shale]

half those you won't use

[primal tusk]

wait but if its a degree n polynomial the nth degree term has to have a nonzero coefficient

[coral shale]

yes, but theres no guarantee thats the case for the product

[coral shale]

crux of the issue

you will have say abx^(n+m)

and you need to check ab is non-zero

Say F was a ring, not a field, and you had like Z_6[x]

you could have (2x + 1) . (3x + 1) = 0x^2 + ...

[solar glacier]

question

[primal tusk]

oh damn

[solar glacier]

in showing the image of R \to R3 via t \to (t,t^2,t^3) is an algebraic set

we want a zero set out of it

[rustic crown]

yep

[solar glacier]

so

can i take

x,y,z \in R3

and look at V(y-x^2,z-x^3)

[rustic crown]

can you prove why that works?

[solar glacier]

thats what im having a tough time with

is it a set containment

[rustic crown]

you have two subsets of R^3. so yep to show they're equal, you would need to verify both containments

[solar glacier]

and is my subset of R^3[x,y,z] going to be the functions y-x^2 and z-x^3

that im taking the zero set of

[coral shale]

what branch is this, looks interesting

[rustic crown]

what does that mean?

[solar glacier]

like algebraic set is defined as

[rustic crown]

classical alg geo

[solar glacier]

the zero set of

some subset of

the polynomial ring over some field

[rustic crown]

right

[solar glacier]

and im saying is that subset

[rustic crown]

so what's the ring here

[solar glacier]

R^3

[rustic crown]

that's your space

[solar glacier]

the ring is R3 polynomial ?

R3[x,y,z]

[prisma ibex]

the points (t,t^2,t^3) correspond bijectively to the points (x,y,z) where x^3=z and x^2=y

[rustic crown]

you mean R[x, y, z]

[solar glacier]

sorry yes

lol

coefficents from R

[prisma ibex]

so this affine variety is given by V(z-x^3,y-x^2)

[rustic crown]

so which containment is troubling you?
{(t, t^2, t^3) | t in R} * V(y-x^2, z-x^3)

[solar glacier]

let me work it out

let a \in the first one

then a is of the form (t,t^2,t^3) for t \in R

then i need to verify this point satisfies zeros of those equations?

[rustic crown]

yep

[solar glacier]

but that seems trivial no?

[prisma ibex]

it is trivial!

[solar glacier]

you just plug in component for component

ok lol phew lol

[rustic crown]

yea lol, that how you constructed the polynomials :p

[solar glacier]

the bacmwards containment seems trickier?

[prisma ibex]

they're both trivial

[solar glacier]

let a \in V(y-x^2,z-x^3)

[prisma ibex]

you really should not be overthinking this

[solar glacier]

then a is a zero for both

im just making sure i get every step lol

and since its a zero for both

we get that x=x , y=x^2, z=x^3

[prisma ibex]

yes

[solar glacier]

i e. (t,t^2,t^3)

ok

sweet thanks guys!

[prisma ibex]

here's a fun variant of this problem

replace the map A^1->A^3 with P^1->P^3

[solar glacier]

oh i dont have time rn but ill try it later

[prisma ibex]

so sending [s:t] to [s^3:s^2t:st^2:t^3]

[solar glacier]

prijective space>

?

[rustic crown]

yee

[prisma ibex]

yeah

[solar glacier]

oof

[prisma ibex]

I'll say the answer since it's kinda surprising

[solar glacier]

lines through the origin

[prisma ibex]

you need three polynomials to cut out this curve

if [x:y:z:w] are homogeneous coordinates for P^3 then you can take uhhh

[rustic crown]

xw-yz is one

[solar glacier]

this is a bit beyond my scope of knowlegge i breifly have covered projective dpaces

[prisma ibex]

xz=y^2
yw=z^2
xw=yz

this is the twisted cubic in P^3, it's an example of a projective variety that isn't a complete intersection

[solar glacier]

in a course on riemann surfaces

[prisma ibex]

(so it's not cut out by the expected number of polynomials)

[solar glacier]

ahh

[solar glacier]

now i see it

welp i gtg guys, gotta blow off some steam from hw and grading

thanks for all your help!

asl always!

[rustic crown]

why does that happen tho?

det hasn't seen much actual AG

and you show it's not done by 2 or fewer by some cohomology calculations?

[prisma ibex]

if you use only two polynomials the resulting ideal isn't radical

e.g. you would get that like (yw-z^2)^2 is in the ideal but yw-z^2 isn't

[rustic crown]

oh okie

makes sense

[oblique river]

Its strange because it’s set-wise a complete intersection, but not as a variety

[prisma ibex]

early examples in AG that make you go "huh, AG is kinda hard"

[rustic crown]

lol after doing so much sheaf theory, i just wanna see some actual stuff :p

[remote dust]

how would i go about showing the free module on a finite set X has standard basis (like standard vector basis)

[vast quiver]

how did you define "the free module on a finite set X"?

[prime sundial]

i was asked to show that the map p: SU(2) x SU(2) -> SO(4) given by p(A,B)(x) = AxB^{-1} is surjective, where x is in R^4 described as the set of 2x2 matrices of the form (a & -b* \ b & a*), a and b complex
every proof of this i've found online makes use of lie algebras, clifford algebras, or some other topic we did not discuss in my class. is there some simpler way to show this? this came from a hw set on covering spaces for an AT course

[long nebula]

am I crazy or is this statement wrong? like for example if p=2, then there's no z in the group such that z^3 = -1

[oblique river]

z = -1?

[long nebula]

Wait

Idk why that didn't occur to me

Lol thanks

[oblique river]

Np lol

[long nebula]

Okay never mind

[delicate bloom]

maybe cause it seems weird to think of the map z |-> z^3 as an isomorphism of the group

what's the inverse?

[oblique river]

Well, write 1/3 as a 2-adic number…

[delicate bloom]

heh 😛

was trying to get them to discover p-adics on their own

[oblique river]

Im sorry i ruined it :c

[delicate bloom]

it's cool either way lol

[oblique river]

I didnt realize you were asking sn actual question, I thought it was rhetorical

And i was trying to be silly, acting like “oh of course this is so obvious its not strange at all” but while goving a strange answer

[delicate bloom]

I was playing innocent to lure them into the trap

[oblique river]

You played me too haha

[delicate bloom]

I was blessed today so all is good 😌

[oblique river]

Ok bed time now gn

[long nebula]

Okay figured it out, it was just some elementary number theory lol

I can't believe I had a crisis about how to find the cube root of -1

[delicate bloom]

solving cubics is no joke

[long nebula]

I was thinking about this

But I think it led me to miss the obvious argument lol

[delicate bloom]

oh lol whoops

well, once someone tells you -1, you can always "guess and check" and evaluate (-1)^3 to see if it does become -1

[long nebula]

hahahaha

[delicate bloom]

I'm guessing you did the x^3+1 =(x+1)(x^2-x+1) factorization then? heh

or some kinda thing else, idk I gotta sleep too, past my bedtime

[vapid horizon]

Let $R$ be a ring. Let $S = {1, 2, \dots, n}$. let $R \langle S \rangle$ be the all mappings from $2^S$ to $R$. let $a, b \in R \langle S \rangle$. We defines $(a \otimes b)(C) = \sum_{A \subseteq S, B \subseteq S, A \cup B = C} a(A) b(B)$.

Let $a \in R \langle S \rangle$. consider a multi-variable formal power sierie ring $R_1 = R[[x_1,x_2,\dots,x_n]]/(x_i^2-x_i)$. Because $R[[x]]/(x^2-x) \simeq R^2$, so $R[[x_1,x_2,\dots,x_n]] \simeq R^{2^n}$.

let $\phi : R \langle S \rangle \to R1$ which $\phi(a) = \sum{X \subseteq S} a(X) \prod_{e \in X} x_e$. Then we have proved $R \langle S \rangle \simeq R^{2^n}$.(Addition is add each other, Multi is $\otimes$)

Is it right and how to transform from $R \langle S \rangle$ to $R^{2^n}$?

[cloud walrus]

1048576Prog

[void cosmos]

what is R^2^n

set of all functions on the ring R?

[vapid horizon]

e

$2^S$ is the power set of $S$.

[cloud walrus]

1048576Prog

[vapid horizon]

$R^(2^n)$.

[cloud walrus]

1048576Prog

[vapid horizon]

$R^{(2^n)}$.

[cloud walrus]

1048576Prog

[void cosmos]

so ur trying to show that the set of all functions on R from R to itself is isomorphic to the ring of formal power series?

dont think this is true

[vapid horizon]

wait

$S = {1,2,\dots,n}$

[cloud walrus]

1048576Prog

[vapid horizon]

then $R \langle S \rangle$ is a mapping from (the power set of $S$) to $R$.

[cloud walrus]

1048576Prog

[vapid horizon]

$R \to R$

[void cosmos]

idk man take R = Z

let f be in an element in R<S>

suppose f is invertible

then f must be 1 or -1

cuz those are the only units in Z

but let sum(a_nx^n) where a_n is in Z be an element in Z[[x]]

oh so

any unit would have f^2 = 1

but not every unit in Z[[x]] would have this multiplicative property

so they arent isomormorphic?

does this work as a counterexample lmao

so take S={1}

hopefully this works as counterexample sorry i couldnt follow ur proof

but yea ig ik what u were thinking cuz like

its easy to think of them the same but as in like element-wise

waitt i multiplied wrong

i thought Z[[x]] would have the normal multiplication

not what u defined lmfao

yea literally nvm everything i said

[void cosmos]

what is X ? when evaluating phi(a)?

[median pawn]

a hint would be nice

i was told to check n = ((p-1)/2)!, but idk where it's coming from and seems like cheating

[quiet pelican]

It comes from Wilson’s theorem if you’ve seen it

[median pawn]

ik wilson's theorem states that p is prime iff (p-1)! is congruent to -1 mod p

but (p-1)! may not be a perfect square

[quiet pelican]

But if p = 1 mod 4… then consider ||the products of the first (p-1)/2 and second (p-1)/2||

[languid osprey]

They're trying to show that the adjoint representation matrix is given in terms of the structure constants. But how did they go from the first line to the second line?

[formal ermine]

what's the context of this

[languid osprey]

If $g \in G$ for a Lie group $G$ and $T_a$ are the generators for the corresponding Lie algebra, then the matrix elements of the adjoint representation d(g) is given by
$$gT_ag^{-1} = T_b d^b_a(g)$$

[cloud walrus]

Abrar

[languid osprey]

So they apply this for a group element very close to the identity so that $g \approx 1 + i \epsilon^aT_a$

[cloud walrus]

Abrar

[languid osprey]

So my question is how does one get $d^c_b(1 + i \epsilon^aT_a) = \delta_b^c + i \epsilon^a d^c_b(T_a)$ ?

[cloud walrus]

Abrar

[lusty marlin]

A factorial cannot be a perfect power of an integer

[coral shale]

proof?

[quiet pelican]

Bertrand’s postulate + consider prime factorisation

[formal ermine]

,w bertrand's postulate

[cloud walrus]

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[coral shale]

sad i thought it was gonna be summing like that

[lusty marlin]

Yeah, Bertrand's postulate

Also, a nice paper extending this result further:

No product of consecutive integers is a perfect power

[vapid horizon]

$\sum_{X \subseteq S} a(X) \sum_{e \in X} x_e$

[cloud walrus]

1048576Prog

[tribal moss]

Counterexample: 1! = 1³

[formal ermine]

lol

[oblique river]

Another counterexample: 3! = 6^1

[tribal moss]

I think "perfect power" usually means with an exponent of at least 2.

[oblique river]

Idk i think 1 is a perfectly good number

[coral shale]

knew it also

[coral shale]

integer usually means not 1

[kind jacinth]

for (a) and (b) its not possible right?

(c) For this one can I say that for any word x over S, if y is the inverse of it its gonna cancel out (so for example x=abc then y=c^-1b^-1a^-1

[oblique river]

I think (b) is possible

[kind jacinth]

how ?

i got cb

also is there a trick in doing this?

cz im bruteforcing it sortof

[kind jacinth]

and am i heading in the right direction with (c)?

[oblique river]

Yes, i would just be clear by what you mean by “inverse”.

Youre basically asked to prove that every element has an inverse

So saying “yes, the inverse of an element is its inverse” isnt really a proof that inverses exist

[kind jacinth]

ah i see

[oblique river]

But you are correct in writing down the inverse of abc, so i would just try to do that in more generality

For (b), there are two words in R which are just straight up inside your given word; remove them

The fact that bc^(-1) is in R means that you can replace every b with c, or vice versa

Use that to reduce to only two symbols; it should be easier to see then

[kind jacinth]

thanks! will give that a second try. just one more question about c

because i am not entirely sure how i would write it in words. an inverse of ab would be b^-1a^-1 right? not a^-1b^-1. So saying inverse implies that

[oblique river]

Im suggesting that you should avoid using the word “inverse” in your proof

And instead describe in general how to find the “inverse” of a given word

And then prove that the concatenation can be reduced to the empty word

[kind jacinth]

in the Note he said that a can be thought of as a reflection and b as a rotation. So if we can view it geometrically then any transformation can be undone

[oblique river]

That note was only for that one example

[kind jacinth]

so we can undo the acitons by reverting the process

generally

[oblique river]

Yes

That is true

[kind jacinth]

so then my proof would be something like, given the words in x, let the words in y be the reverse process of x. By concatenating the two we would get an empty word

[oblique river]

I still think that’s lacking

Idk what “reverse process” means. These are words, there is no process or action in them

Write it explicitly. Let $s_1^{\sigma_1} s_2^{\sigma_2} \cdots s_n^{\sigma_n}$ be a word. What will the “inverse” word be?

[cloud walrus]

Buncho Bananas

[kind jacinth]

then Y: sn^(sigman), sn-1^(sigman-1), ...., s1^(sigmna1). and for every si in X and s(n-i) in y, sigmai != sigman-i. By concatenating x and y we get an empty set

[oblique river]

right idea but bad notation

X is not a set

also, $\sigma_i$ is a definite element of {1, -1}

[cloud walrus]

Buncho Bananas

[oblique river]

it's not a variable

like, maybe $s_1 = a$, $s_2 = b$, $\sigma_1 = 1$, and $\sigma_2 = -1$

then $x = ab^{-1}$

so then your y would be $ba^{-1}$, or $s_2^{-\sigma_2} s_1^{-\sigma_1}$

[cloud walrus]

Buncho Bananas

[kind jacinth]

are u saying that sigma = 1 always? so then we use -sigma as thats -1?

[oblique river]

no, sigma can be 1 or -1

[kind jacinth]

oh ok i see

[oblique river]

but if $\sigma_1 = 1$ then $-\sigma_1 = -1$

[cloud walrus]

Buncho Bananas

[kind jacinth]

but how would i mention that the sigmas have to be different signs for s if i cant use for an arbitrary si in X?

I do agree x isnt a set

[oblique river]

like this

the "inverse" of $s_1^{\sigma_1} s_2^{\sigma_2}$ is $s_2^{-\sigma_2} s_1^{-\sigma_1}$

[cloud walrus]

Buncho Bananas

[kind jacinth]

but x can have a finite number of s. so i should be more general right?

[oblique river]

yes

[kind jacinth]

so is saying let y be s1(sigma1)s2(sigma2)....si(sigmai) is si(-sigmai)....s1(-sigma1) sufficient?

where 0<i<n

[oblique river]

what is n?

i think you only need one variable for the length of the word

[kind jacinth]

they used n here

[oblique river]

okay, then you should use n as well; the i is not necessary

[kind jacinth]

oh alright. I was using i instead of n to imply that the word doesnt need to contain n letters

but n already implies that

also just one more question about a separate exercise. was just wondering if i understood the question.
Exercise 1.4.4 Show that addition and multiplication mod n are well defined operations. That is, show that the operations do not depend on the choice of the representative from the equivalence classes mod n.
Are they saying that no matter which elements we take in the equivalence class, we would still get the same remainder?

[oblique river]

yes. more precisely, if a = b mod n, and if c = d mod n, then a + c = b + d mod n, and ac = bd mod n

[kind jacinth]

u mean a congruent b and c congruent d then a+c is congruent to b+d right?

[oblique river]

how ios that different from what i wrote

[kind jacinth]

was asking

[lusty marlin]

I thought it was apparent that perfect power means exponent≥2

[oblique river]

i dont have a triple bar equals sign on my key board

[kind jacinth]

yeah got it. was just wondering if i understood it right

ok thanks!

[lusty marlin]

And base ≥2

[oblique river]

ok np haha sorry i was just confused for a sec

[kind jacinth]

i managed to reduce it to cb and then to cc but dont see a pattern from here

[oblique river]

what steps did you take to get to that point

[kind jacinth]

first i got rid of baa and bc-1 and we remain with cbca then i get cc-1 at the end so (cbcacc-1) this gets rid of ac and thus (cbcc-1) and then cb

[oblique river]

from cbca you can use bc^-1 to switch b to c and vice versa

and you get bcba which is an element of R

like i said earlier my strategy is to recognize that since bc^{-1} is in R we can freely interchange b and c

so let's just rewrite all c's as b's so we only have two lettesr

now our word is bbaabbabb^{-1} which we just reduce to bbaabba. now our new set R is {ab, baa, bbba}. now I also recognize that since ab is in R, we can replace a with b^{-1} and b with a^{-1}, so this lets us convert everything down to a single letter

bbaabba becomes a^(-1)a^(-1)aaa^(-1)a^(-1)a which just simplifies to a, and then we feel like we get stuck... but looking back at R our two remaining relations (elements of R) are a^{-1}aa and a^{-1}a^{-1}a^{-1}a, which just simplify to a and aa

therefore we can replace a with e and we are done

(e = empyt word)

[kind jacinth]

i am confused on how u got bcba. I dont see where i can add the empty words to get there (i think the only empty words we can add are bb^-1 cc^-1 etc right?)

or wait. we can actually add any word in R

this is why i was confused

thanks! this makes sense now

but how would we realize if its not possible to get an empty word?

[oblique river]

it's hard

like, proven that there is no algorithm which can answer this question for a general word with a general set R

ther eare some cases you can do by hand though

like for part (a)

recognize that no matter what substitution you do, you will always keep the same parity of number of a's

like, if you start with an even number of a's then no matter what you do you will always have an even number of a's

so if you start with an odd number of a's, it's impossible to get it down to 0

[kind jacinth]

i see

thanks!

[prime sundial]

if P and Q are two sylow subgroups of G of different orders, is PQ a subgroup of G? is it a normal subgroup of G? id like to say yes but im not confident

[agile burrow]

P and Q would have trivial intersection, but you would also need P and Q to commute for PQ to form a subgroup

[prime sundial]

im trying to show that a group of order 385 is not necessarily abelian. my idea was to use the sylow subgroups P_5 and P_11
if (P_5)(P_11) do not commute, then we're done. if they do, then (P_5)(P_11) is a subgroup of order 5*11. since 11 \equiv 1 mod 5, then there are 2 isomoprhism classes, one of which is a non-abelian group (showed this in class)
so there necessarily exist noncommutative elements

im pretty weak when it comes to sylow subgroups so any pointers would be appreciated

[agile burrow]

sure, so 385 = 5 * 7 * 11. Can you count the number of Sylow subgroups?

[prime sundial]

3

well

the 7 and 11 subgroups are unique

i haven't said anything about the 5 subgroups yet

[agile burrow]

Right, P_7 and P_11 are unique. This implies they are normal, right?

[prime sundial]

yes

[agile burrow]

So in particular, the product P_7 P_11 is a normal subgroup of G, call it N

[prime sundial]

makes sense

[agile burrow]

Now let P_5 be any 5-Sylow subgroup. By order considerations, P_5 has trivial intersection with N

Furthermore, P_5 N = G

So we can deduce that G is a semidirect product of P_5 and N

[prime sundial]

haha i think this may be beyond what im asked to do. i don't recall going over semidirect products

[agile burrow]

Ah

That's unfortunate

[prime sundial]

yeah i actually found a proof of this online using semidirect products

[agile burrow]

Ok I'll think about a different way to show it then. This is a fun problem though

[prime sundial]

does my "proof" above make sense?

taking the "if |G| = pq, p = 1 mod q, then there is 2 isomorphism classes" statement for granted

[agile burrow]

ah i see

That seems correct to me

Cool proof

Really the semidirect product proof is the same underlying idea - it comes down to the fact that there'll be a non-trivial homomorphism from Z/5 -> Aut(Z/11) = Z/10

But if you haven't learned semidirect products then there's no need to worry

[prime sundial]

i do recall going over automorphisms and i imagine we'll talk about semidirect products soon enough

thank you anyways!

[agile burrow]

They're fun, gives you new ways of constructing groups

Happy to help 🙂

[formal ermine]

when is the algebraic closure of a field equal to its separable closure

[agile burrow]

The separable closure is the subfield of the algebraic closure containing the separable elements. We say a field is perfect if every algebraic element over the field is separable

Fields of characteristic 0 and finite fields are perfect

[formal ermine]

I don't see the connection between a perfect field and this

[agile burrow]

The separable closure is algebraically closed when the field is perfect

[rustic crown]

a field is perfect if every algebraic extension is separable by definition. this is equivalent to "char= 0" or "char=p and frobenius being an iso"

[solar glacier]

hey im supposed to show the coordinate axis in R3 form an algebraic set

what are the coordinate axis in R3 is is (x,0,0),(0,y,0),(0,0,z)

[rustic crown]

ig

[rapid relic]

tryna show this is or is not a subgroup of S_n

[solar glacier]

can u check it using subgroup criteria

[rapid relic]

I said yes cuz the identity exists, inverses always exist because the inverse of the identity is the identity, and I don’t think you can make a non-identity permutation out of identity permutations but I’m second guessing

[chilly ocean]

there is a tiny bit more to being a subgroup than containing the identity and inverses

you're missing one condition to check

[solar glacier]

theirs closure

[rustic crown]

(you're trying to prove P or not P, which is always true )

[solar glacier]

check closure

and existence of inverse

it is indeed a subgroup

[rapid relic]

you cant make a non-identity permutation just operating with identity permutations

so it should be closed

right?

[agile burrow]

david i think you misunderstand what the set is. If you view S_n as the group of permutations of {1, .., n}, this is asking you to consider the set of permutations that fix n, the element in the set. It's not asking about the permutations that fix every element of {1, .., n}, in which case you'd be correct that the only element is the identity

[rapid relic]

oh that makes it so much clearer

thank you very much

[solar glacier]

oh i thoght it was that fixes every n

good point watler

[formal ermine]

what's an example of a non algebraically closed field with char p where frobenius isn't an iso?

[agile burrow]

Happy to help, I can see why you might have been confused

[rustic crown]

standard example is F_p(x)

[solar glacier]

youre right reading the problem again its those n that get fixed not all n

[agile burrow]

Thanks for your support

[formal ermine]

walter said finite fields are perfect though? or am I misunderstanding

[rustic crown]

that's not finite

[solar glacier]

but the set that fixes every. n is also a subgroup

[formal ermine]

oh lmao I didn't see the (x)

[rustic crown]

[solar glacier]

so then by this what is meant by the set

[formal ermine]

how does frobenius fail to be an iso here? well definedness or bijectivity?

[solar glacier]

is it ${(x,0,0),(0,y,0),(0,0,z): x,y,z \in \Bbb{R}}$

[chilly ocean]

the set of points in R^3 which are in one of the coordinate axes

[rustic crown]

the image would be F_p(x^p)

[cloud walrus]

MyMathYourMath

[delicate orchid]

yus

[solar glacier]

ok cool

[formal ermine]

ah I see

so surjectivity

[solar glacier]

thats what i thought just wanted to make sure i have the right definition before attemtping the problem lol

[formal ermine]

cuz x doesn't necessarily have a pth root

[rustic crown]

necessarily

[feral agate]

Hey all, I'm trying to understand the futurama theorem through an exercise. However, I am unsure about my proof in c because I did not use the fact that all cycles can be written as product of disjoint cycles, which is given as a hint in the exercise. Could someone check out my solution and perhaps point out where this fact should be used?

\subsection*{(c)}
Define $\tau := (yz)\rho$. This gives $$\tau\pi^* = id,$$ as $$\tau\pi^* = \begin{bmatrix}
1 & \ldots & k & \ldots & n & y &z\
1 & \ldots & k & \ldots & n & y &z\
\end{bmatrix}$$ Note that, $(yz)$ is a distinct transposition and that the transpositions used to define $\rho$ are also all distinct. Furthermore, all transpositions in $\tau$ either contain $x$ or $z$. Hence, we have proven the theorem.

[delicate bloom]

can you say "look at xyz=0"

[cloud walrus]

FrankF

[pastel cliff]

the futurama thm?? what

[feral agate]

See the screenshot attached, it contains a description of the theorem

[formal ermine]

it's a tv series

in one of the episodes the scientist does some group theory on a blackboard

[solar glacier]

for this

the x axis is the zero set V(y,z), for y axis its thezero set V(x,z) and for z axis its V(x,y) right

[chilly ocean]

yes

[solar glacier]

so by coordinate axis they mean the subspace of R3:

[cloud walrus]

MyMathYourMath

[chilly ocean]

bad set builder notation but yes

the union of the x-axis, the y-axis, and the z-axis

[solar glacier]

how would you set build that

ah

${(x,0,0): x \in \Bbb{R}} \cup {(0,y,0): y\in \Bbb{R}} \cup {(0,0,z): z \in \Bbb{R}}$

[cloud walrus]

MyMathYourMath

[solar glacier]

is the algebraic set of the union the union of the algebraic sets? lol

[chilly ocean]

i can't understand this sentence

"the algebraic set of the union"?

[solar glacier]

is $V(\bigcup_{i=1}^n S_i)= \bigcup_{i=1}^n V(S_i)$

[cloud walrus]

MyMathYourMath

[chilly ocean]

nope

elements in the left are sent to zero by every single element of each S_i, but elements in the right are only sent to zero by every single element in one of the S_i's

the correct equation is V(union S_i) = intersection V(S_i) (and this holds for arbitrarily many S_i)

the finite union of algebraic sets V(S_i) is indeed another algebraic set, but i think you should think a bit more on how to show that this is the case

(it might be helpful to assume that the S_i's are ideals in the polynomial ring)

[solar glacier]

ahhhhhhh

[coral spindle]

If you have more restrictions, would you expect more points to satisfy them?

[solar glacier]

no

[coral spindle]

Well this is what the formula says!

[chilly ocean]

something something inclusion reversing

[solar glacier]

so V(union)=intersection(V)

[chilly ocean]

as i wrote

[chilly ocean]

https://youtu.be/t1CRjNriCR8

possibly relevant

[coral spindle]

I hate these text-to-speech 'how to pronounce' videos, they're too often wrong

[chilly ocean]

i fucking love this video

[formal ermine]

lmao

[formal ermine]

no it's 120% accurate

I can confirm

[coral shale]

its pronounced the N theorem

[solar glacier]

how would i compute $V(y,z) \cap V(x,z) \cap V(x,y)$ is it all sums of all variables

[cloud walrus]

MyMathYourMath

[delicate bloom]

I'd visualize it

[coral shale]

@uneven folio do u also pronounce it nullstullensutz

[delicate orchid]

he'd better not

[coral shale]

i bet he would.

[prime sundial]

let R be a ring with an infinite sequence of ideals I_1, I_2, I_3,... where each I_k \subseteq I_{k+1}. show the union of all I_k is an ideal of R
is this as trivial as it looks?

[solar glacier]

im having a tough time

is it

the origin lol

[delicate orchid]

if you're struggling with visualising maybe try writing out the actual definition of each variety using set notation

but I'm pretty sure it's the origin lol

[solar glacier]

ive done that

and i get the origin when intersect

[coral shale]

tldr me how V(...) is defined

[chilly ocean]

is this a new question or your original problem?

[coral shale]

V(y, z) = {points st y = 0, z = 0} ?

[delicate orchid]

V(I) = {x \in C^n such that f(x) = 0 for all f in I}

[solar glacier]

yes

[delicate orchid]

I'm assuming we're over C here

[coral shale]

sick

[solar glacier]

all points in R3

[coral shale]

then yh surely x=y=z=0 is intersection?

[solar glacier]

same problem showing coordinate axis form algebraic set

[chilly ocean]

then you've definitely gone wrong somewhere

[solar glacier]

so im bascially computing the intersection of the V(S)

[chilly ocean]

no

the union

[solar glacier]

im given a union

[chilly ocean]

.

[solar glacier]

and im taking V of that union

[chilly ocean]

no

you're taking the union of the V's

[chilly ocean]

.

[solar glacier]

ohhh ur right

im taking a union of the V(S)

[chilly ocean]

V(y, z) cup V(x, z) cup V(x, y)

[solar glacier]

yes yes

cause each of those is the axis

and the coordinate axis is thier union

your right

[prime sundial]

are ideals nontrivial by convention?

or can they be trivial?

[solar glacier]

they can be trivial

(0)

[coral shale]

()

jk, depends on defn iirc

[solar glacier]

as in the ideal w 0 only

lol

I was always taught by convention to denote the trivial ideal as (0)

[coral shale]

hmm hmm

ok, looks like it usually is one

there are so many ifs and donts in ring thry tho

[solar glacier]

if you contain 1 youre the whole ring

[coral shale]

just because rng commutativity or not

[prime sundial]

thank you both

[coral shale]

i dont feel like u ever consider this ideal

for starters u cant even quotient can u

[prime sundial]

well it could have been a counterexample for one of my hw problems

[coral shale]

or can u

[formal ermine]

ideals are subrings

[prime sundial]

but there was another condition that took care of it

[coral shale]

do u get the whole ring

[coral shale]

shut

yh u get the whole ring

[prime sundial]

another not strictly alebgra question.
does the union of intervals (1/n, 1] for n in N equal [0,1] or (0,1]?

[solar glacier]

stronger than subbrings they absorb everything from the ring

[prime sundial]

it must be (0,1] right?

[quiet pelican]

(0, 1]

[formal ermine]

(0, 1]

[prime sundial]

thank you

[solar glacier]

(0,1]

[lapis trench]

(0,1] because 0 is not in any of those intervals.

[solar glacier]

yes

[quiet pelican]

Which interval would 0 be in?

[coral shale]

isnt it equiv to subring if u consider rng or something silly

[solar glacier]

good point, i think so since you have no 1 in a rng

[prime sundial]

so if we say $B_i \subseteq B_{i+1}$, and A = $\bigcup_{n\in\mathbb{N}} B_n$, then $A=B_n$ for some $n\in\mathbb{N}$, right?

[cloud walrus]

maximo

[solar glacier]

this seems a lot like the ACC condition

oh wait no thats a single one gets stablized

not the union

[coral shale]

why would this be true

are these random sets

or

[prime sundial]

you're right i don't think it's true

let me rephrase

if x in A, then we can find a B_n such that x in B_n

[quiet pelican]

Yes

[coral shale]

sure.

[prime sundial]

great, thank you

[kind jacinth]

unsure what they want in part b. is saying f(a, b) = a-b enough?

[coral shale]

suppose
(f o g) o h \neq f o (g o h)
in general

then figure out how f^n could be defined

[kind jacinth]

defined such that these 2 arent equal right?

[coral shale]

no........

do you understand the first sentences

Why f^n is well defined

[kind jacinth]

because if id plug in f(a, b) lets say or f(b, a) id get same output

[coral shale]

no?

where do you see it being stated f is a function taking in 2 parameters

f : X -> X

f^n is well defined because function composition is associative

(f o f) o f = f o (f o f)

in general it doesnt matter how we bracket f o f o ... o f

it will be the same function

[kind jacinth]

i see

[coral shale]

If they werent, f^n wouldnt be well defined because we wouldnt know what bracketing is meant

in which case you would need to specify

[glossy crag]

What's an example of a local ring that is not a DVR? Surprisingly, not much comes up when googling.

[next obsidian]

Uh

Localize any ring at a height > 1 prime ideal

[delicate bloom]

elements x with |x|<=1 of the algebraic closure of Q_p

local, but no longer discrete valuation

[uneven folio]

Null shtell en zatz

[kind jacinth]

So if we take this as an example and we would instead define f: C->D g: B->C and h: D->B
(fog)oh would be defined as B->B
fo(goh) would be defined as D->D
right?
but for the exercise its just f and X->X so im not sure how that would be defined in this case. I would still get X->X when composing

[next obsidian]

Z_p is totally a DVR

Oh

Alg closure

Oog

[coral shale]

forget function composition and take real exponentiation

tldr, it wants you to decide what 3^^3 is for example

3^(3^3) vs (3^3)^3

it wants you to come up with a definition that leaves no room for ambiguity

for 3^^n = 3^...^3 n times

You define possible ways on how the bracketing should happen

[delicate bloom]

we can be a bit sharper and only looking at only infinitely many ramified extensions, and also I think this kind of trick in general will work for any local field, like complete on the X-adic topology on R[[X]] by adding all the nth roots of X

[kind jacinth]

ahh i see.. so if we were to number the functions so lets say f1of2of3o...ofn (so first function, second, third, up until nth) we can say that we can place the brackets such that f1(f2(f3(...(fn))) or (f1o(f2o(f3o(o(...(fn)))))

so in this case the exp example would always be 3^(3^(3..(.^3)))

[pastel raft]

can someone help me with this

[warm wyvern]

This is not abstract algebra

[pastel raft]

where would this belong under

its economic theory

[warm wyvern]

I'm not really sure :/

This is a math server lel

Maybe #prealg-and-algebra ?

[coral shale]

it belongs in help channels at best possibly

but without having taking economics u cant answer these

[pastel raft]

k

[coral spindle]

Maybe an economics question ought to be asked in an economics server, not a math server.

[novel plover]

Prove that a cyclic group with only one generator can have at most 2 elements

[tribal moss]

No! You can't make me!

[coral spindle]

Sorry I'm busy proving that if it has two generators, then it has 3 elements xoxox

maybe even 4

[tribal moss]

I got 6, what am I doing wrong?

[coral spindle]

more research is needed

[novel plover]

I'm supposing G has 3 elements, e a b, and hope to find a contradiction

[formal ermine]

what

[coral spindle]

I don't think that'll get you very far

[novel plover]

Oh you think it's easier to do it directly with cases

[formal ermine]

your question is missing context

[coral spindle]

No, the question is complete

[formal ermine]

ah lol I misread the question

[novel plover]

Well clearly obviously G has at least 1 element because all groups do

[formal ermine]

I thought it meant like "G is cyclic and generated by 1 element, prove that it has order 2" Lol

[novel plover]

Ohhh haha, lol.

[coral shale]

im positively confused

[formal ermine]

look at the multiplicative group of Z/nZ

[coral shale]

oh nvm figured it

[coral shale]

i couldnt stop thinking this

[formal ermine]

ie all elements that can generate Z/nZ

for what n does that group have size 1

[novel plover]

Is Z/nZ just Z_n?

[formal ermine]

yes

[coral shale]

its the other way round really lol

Z_n is usually defined to be Z/nZ

Theres a special function that counts the number of generators a cyclic group has, but im guessing u havent seen

[formal ermine]

this is overkill

[coral shale]

[novel plover]

Oh yeah, I'm just using bare basics no theorems developed yet really

[coral shale]

its the general answer

Can also be proven, first principles

[novel plover]

Well I mean Z/2Z has 2 elements, Z/3Z has 3 elements, so I guess Z would have 1 element but that's obviously not right

[formal ermine]

Z/1Z

is not Z

[coral shale]

Take Z/8Z, list us the generators it has

[novel plover]

Okay thats easy, I can do that hang on

[coral shale]

Then do the same for Z/6Z

And see if you can guess why most cyclics have more than 1

[formal ermine]

anyway, then ||notice that it consists of all elements coprime to n|| after that ||you can just use the fact that any number > 2 will have 2 numbers coprime to it|| so ||n has to be 2||

||proof of the fact: for n > 2 we have that -1 \neq 1 in Z/nZ but (-1)^2 = 1 therefore -1 is an element of order 2 in Z/nZ^\times, by lagrange |Z/nZ^\times| = phi(n) will therefore be even. thus we get that phi(n) can't be 1||

[tribal moss]

The second of those spoiler blocks is basically the fact to be proved.

[novel plover]

The generators of Z/8Z are 1,2,3,5 and 7

[formal ermine]

2 isn't a generator

[coral shale]

2

[novel plover]

Oh yeah, it's not

[coral shale]

Ill also rewrite your list. -3, -1, 1 ,3

then try for Z_6

[solar glacier]

so back to #groups-rings-fields message

how is $V(y,z) \cup V(x,z) \cup V(x,y)=V((x^2+y^2)(y^2+z^2)(x^2+z^2))$

[cloud walrus]

MyMathYourMath

[chilly ocean]

for example, x^2 + y^2 = 0 if and only if...

(this only works over R - hint!)

[solar glacier]

both are 0

[coral shale]

oh thats cool

[solar glacier]

but how do you go from both are zero to discovering to write it at x^2+y^2

[coral shale]

cool way to do union and intersects

[novel plover]

What I'm seeing is that 1 and n-1 seem to always be generators for Z/nZ

[coral shale]

^ its a 'trick'

thinking how to and/or equations

into a single equation

to or, you multiply and use zero product property

[tribal moss]

Yes.

[coral shale]

to and, you do the squaring and sum trick

[coral shale]

dont write n-1

think of it as -1

(because it is, and its simpler)

[coral shale]

can you prove this fact?

[tribal moss]

In the context of cyclic groups I'd say, think of it as g^-1.

[novel plover]

By induction maybe?

[coral shale]

no, consider their order

[coral shale]

think this

if order(g) = n

what is order(g^-1)

[novel plover]

I mean, if I could prove g and g^-1 are always generators for any group of order n>=3 then I think the result would follow immediately

[coral shale]

this generalises for all n

[novel plover]

But G = {e} has only one generator

[coral shale]

it actually has none

depending on your definition of generator though

in any case e^-1 would also be a generator, if u take e to be one

===
I suppose e is a generator yes, just checked

[coral shale]

Anyways, to show what you want, you should consider this

[novel plover]

What is order(g)? Like ... for Z/6Z the order(2) = 3 because 2 generates the set {0, 2, 4} ?

[solar glacier]

whats tour question

order of 2 in Z_6 plus is 3 yes

[solar glacier]

also in Z_6 times

but z is cyclic

and has 1 generator

and has infinite number of elements..

[novel plover]

Is -1 a generator too under addition?

In any case I guess this is only for finite groups

It should say that though in the statement of the problem

[coral shale]

review what the definition of order is

for an element

[novel plover]

It's the order of the cyclic subgroup <g>

If the group is finite

[summer path]

say order(g) = n, what does that actually mean?

[novel plover]

Well like we showed for Z/6Z order(2) is just the order of the subgroup generated by 2 which is {0, 2, 4} so order(2) = 3

[prime sundial]

it has 2 generators

[prime sundial]

what the other people are hinting at is what happens if you take g^n

[novel plover]

Take it where, out to dinner?

[coral shale]

^

[coral shale]

its not this

thats not the definition.

non-generator elements also have an 'order'

[knotty frigate]

well <a>=<-a> so in general if we have another one like <b>=<a> then we know b=a^n

but also we know that every element of <a> is an integer power of b

so b^q=a^2

and b^p=a^3

so we have a^pn=a^2 and a^qn=a^3

so we know that p=2 and n=1 or p=1 and n=2

but then if n=2 then q isnt an integer

so therefore n≠2

giving us that n=1

ig i did forget about the negative options

which would allow for the other option

which is n=-1

[white oxide]

how would i begin here? i know that if m|n, then m generates a cyclic subgroup containing n/m elements, but im not entirely sure where to start on proving that x^m = e has exactly m solutions. does it have something to do with the cardinality of <m>?

[void cosmos]

wdym <m>

m is an integer

begin with experimenting

look at Z_whatever

and do stuff

@white oxide

[white oxide]

nvm I got it

[void cosmos]

okay gj

[toxic zephyr]

is a group action of G onto itself by conjugation transitive if and only if G is simple?

it seems to me that there can't be a normal subgroup

[agile burrow]

the action is transitive iff G is trivial

[toxic zephyr]

is it because of the identity element?

[agile burrow]

The issue is that the identity is always in its own orbit

yes, exactly

[toxic zephyr]

ah okay

yeah i was wondering because conjugation of the identity element always yields the identity so i didn't see how it was possible. but now i see that that if it's trivial of course it works

thank you @agile burrow 🙂

[agile burrow]

happy to help 🙂

[languid tusk]

Can someone explain the Euclidean algorithm to me? Maybe I'm just tired, or maybe I'm stupid. I don't understand where the numbers are coming from.

How do you get q?

[glass flint]

Finding the gcd of two positive integers

[languid tusk]

The objective of the Euclidean algorithm is finding the gcd, yeah

But where did they get 5 from?

[void cosmos]

this mod that

[languid tusk]

Where does q come from?

[void cosmos]

whole point of it is that gcd(a,b) = gcd(a mod b , b)

and u keep going

[languid tusk]

I don't know modular arithmetic

I just got an abstract algebra textbook and started reading it

I'm on page 5

So... assume I know nothing

[dapper sluice]

I have a question about the adjoint representation of simple Lie groups. Suppose you have a simple Lie group and two fixed generators x and y. Is it always possible to a group element g such that g^-1 x g = y?

[glass flint]

Not always possible

the set of all group elements that conjugate x to y is not necessarily non-empty . there may not exist a g that satisfies the equation g^-1 x g = y

[glass flint]

,

[dapper sluice]

under what conditions is it possible to find a g?

we obviously can't have x and y belong to different sub-algebras, but what else?

[glass flint]

1. x and y are part of a Cartan subalgebra of the Lie algebra of the group.
2. x and y are part of a set of root vectors of the Lie algebra.

For pt 1, it’s cus the elements of a cartan subalgebra are simultaneously diagonalizable, and hence can be put in a common diagonal form by conjugation

For pt 2, it’s cus the elements of a root space are connected by the action of the weyl group, which is a discrete group of symmetries of the Lie algebra

[dapper sluice]

ah right this is a good point. we clearly need x and y to be similar

[dapper sluice]

but I don't understand how points 1 and 2 aren't contradictory. the cartan elements have zero weight, but in part two you are saying they have non-zero weight?

maybe there is a difference in terminology - I study physics not math so I apologize in advance

[glass flint]

Just giving 2 conditions , not connected

It depends on the structure of the Lie algebra and the choice of generators

Cus there r other conditions / techniques that apply to specific Lie groups or Lie algebras

[dapper sluice]

when you say x and y are part of the set of root vectors of the Lie algebra, presumably you don't mean non-zero root, right?

that is what is confusing me. cartan elements have zero root vector

[glass flint]

Yes u r right

[dapper sluice]

ok, maybe I should ask my direct question

I am studying the conformal group, and I am curious if it possible to conjugate a dilation into a translation

if D = dilation and P = translation, can we find a g such that g D g^-1 = P

[glass flint]

No

As a dilation scales distances and angles and a translation preserves distances and angles, not possible to conjugate a dilation into a translation by an element of the Euclidean group

But

[dapper sluice]

dilation preserves angles i believe

[glass flint]

it is possible to conjugate a dilation into a translation by an element of a larger group such as the group of affine transformations or the group of projective transformations

[wooden ember]

That’s why they asked for the conformal group

[dapper sluice]

I think there is a miscommunication

yeah I'm not talking about the Euclidean group but the conformal group

[glass flint]

Oh welp then yes

Soz

[dapper sluice]

no problem

how can we see that it is possible to find a conjugation in the conformal group?

I'm new to this and your ability to just know this is magic to me hehe

also I should clarify

[dapper sluice]

here, D and P are generators, not the group elements

and I want to find a group element g such that g D g^-1 = P

[wooden ember]

I don’t think this is true? You have 2x2 matrices mod scaling. Dilations will look like lambda & 0 \ 0 & 1 while translations will look like 1 & mu \ 0 & 1. Unless lambda = 1 first has distinct eigenvalues while second has the same eigenvalues (and this is preserved under conjugation). When lambda = 1 then clearly you can only make it work with mu = 0 but that’s just saying the identity is both a dilation and translation

Or am I being dumb

[dapper sluice]

D and P are differential operators acting on functions

[dapper sluice]

how do you know it is possible?

[wooden ember]

Yeah I am being dumb anyways even in my interpretation of the question I mixed up the projective linear and conformal group

[glass flint]

We consider the following dilation and translation transformations:

Dilation: D(z) = kz, where k is a positive real number and z is a complex number

Translation: T(w) = w + a, where a is a complex number

To conjugate the dilation D(z) by the translation T(w), we can apply the conjugation formula

T^-1 D T(z) = T^-1(D(T(z))) = T^-1(k(T(z))) = T^-1(kz + ka) = (kz + ka - a)/(k^2) ,

where I have used the fact that the inverse of a translation T(w) is given by T^-1(w) = w - a

This transformation is the composition of a scaling by k^(-2) and a translation by the complex number -ka/(k^2)

Thus I have conjugated the dilation by the translation to obtain a composite transformation that includes both a scaling and a translation

Since the conformal group includes both dilations and translations as special cases of conformal transformations, it is possible to conjugate a dilation into a translation by an element of the conformal group.

This may result in other conformal transformations such as rotations and inversions

[wooden ember]

Why are you conjugating by a translation

[dapper sluice]

this isn't what I'm asking

I'm saying D and P are GENERATORS

D = x partial_x

P(a) = partial_x

the exponentials of these are the dilation and translation group elements

exp(a D) f(x) = f((1+a) x)

exp(a P) f(x) = f(x + a)

I am asking if it is possible to find a group element g in the conformal group that takes D to P

g^-1 D g = P

my question is if it is possible to find such a g

[wooden ember]

So I don’t know much about Lie groups (nothing in fact) and I might be saying something dumb but don’t g and D live in completely different objects?

Like isn’t D an element of a Lie algebra and g an element of the group?

What does it mean to conjugate them

[dapper sluice]

yes

it means composition of operating on f

[wooden ember]

Aight if it makes sense I’ll leave it to people who actually know about the subject then

[dapper sluice]

this is a standard notion in Lie groups. It is called the adjoint representation of the group

the Lie group acting on its own Lie algebra through conjugation

you may prove that this takes the Lie algebra to itself, is linear, and preserves group structure

therefore it is a representation

the adjoint rep

sadge, i feel like this entire discussion was just miscommunication

[wooden ember]

I think you communicated fine on your end but though I can’t speak for the other person we probably just don’t know enough about Lie groups to help

So yeah just wait for someone else or ask in #diff-geo-diff-top

[dapper sluice]

oh really? is that an appropriate channel?

[wooden ember]

I think so

[dapper sluice]

ok cool thanks

I will ask there tomorrow

[hazy flax]

Shouldn't the union of conjugacy classes of G be G itself? What am I missing here?

[upper pivot]

it doesnt say the union of all the conjugacy classes

just that H is the union of some of the conjugacy classes

[hazy flax]

Ahh okay

thanks

[hazy flax]

So according to the theorem arbitrary union of conjugacy classes would also form a normal subgroup?

[upper pivot]

well not necessarily, such a union doesnt have to be a subgroup

the theorem says if a subgroup is such a union then its normal

[hazy flax]

But the theorem says iff

[upper pivot]

right but H is a subgroup of G is assumed above right

[hazy flax]

Ohh I see

So if H is a subgroup and it can be written as a union of conjugacy classes then it's normal

Is that right?

[upper pivot]

yep

[hazy flax]

Thanks a lot

[formal ermine]

I'm having a brain fart right now

is G/trivial = G and G/G = trivial?

[lusty marlin]

If by trivial you mean the singleton set containing the identity element in G, then the both claims of yours are correct if you replace equality with isomorphism.

[coral shale]

G/1 = G, G/G = 1

[rigid cave]

G/0=G

[coral shale]

If I said I was serious would anyone believe me

smh

[chilly ocean]

why sully

1 is standard notation for the trivial group

[rustic crown]

because fractions

[coral shale]

Do any authors use - for quotient on rings

G - 0 = G

[coral spindle]

Nope

[coral shale]

[lethal dune]

you can be the first one

[coral spindle]

(please don't)

[lament dawn]

Do not do this

[coral shale]

why

[lament dawn]

This is often used for set minus

[coral shale]

yes is there any other reason

[lament dawn]

Does there need to be?

[coral shale]

good

,,R-_\mathcal{R}0=R

[coral shale]

[lament dawn]

I'm calling the police

[formal ermine]

,, R = R + 0

[coral shale]

oh wait ofc

we have oplus; just use ominus

,,R\ominus0=R

[formal ermine]

do we have $\oslash$

[coral shale]

bahaha

[rustic crown]

[next obsidian]

[coral spindle]

Do not advertise your question in irrelevant channels

[coral shale]

omg did they spam like 10 channels

[coral spindle]

🤷 I didn't bother to check

[cloud walrus]

hausdorff

[median pawn]

the ideal generated by p and i-a would have elements of the form (c+id)p + (i-a)(m+in)

the real part is pc - n - am, and the imaginary part is pd + m - an

pd + m - an = 0 says naught

or am i hallucinating lol

[terse crystal]

$G_{1}$ , $G_{2}$, T are finite groups. Does $G_{1} \oplus T \cong G_{2} \oplus T$ imply $G_{1} \cong G_{2}$ ?

[cloud walrus]

Cogwheels of the mind

[median pawn]

can't u construct the isomorphism explicitly

[terse crystal]

G_1 times T to G_2 times T not necessarily maps T to T

When they are not restricted to be finite groups, counterexample can be easily constructed that’s why I am curious about finite group case

[median pawn]

ahh, injectivity might fail

surjectivity won't

but can u choose the map in such a way that injectivity doesn't fail

[white oxide]

is seven asking for the cardinality of the cyclic subgroup generated by tao^2?

[formal ermine]

hmm

[terse crystal]

pd+m-an=0-> m=an mod p-> am=a^2n=-n mod p, pc-n-am=0 mod p, answering his question

[formal ermine]

so for finite groups oplus is equivalent to cartesian product

what if we like

take two groups a,b with the same order but different structure

and a third group c, such that the resulting group is always cyclic

[median pawn]

why are you working mod p

[formal ermine]

nvm what I thought of here wouldn't work

I was thinking like