groups-rings-fields

[cinder bone]

or is it fine to one from M->C since its a function

[next obsidian]

But who cares

It’s bijective and the inverse is obvious

a + bi goes to the matrix of the form in the photo you sent

I don’t wanna type a matrix in latex

[cinder bone]

oh okay gotcha

thanks

[next obsidian]

It’s clearly additive

Just show it’s multiplicative

[cinder bone]

i guess i should show the set of complex numbers are a group too?

p sure it only works for operations of addition and matrix addition

[past temple]

how do i show that

8 = 1 + 1 + 2 + 4

is not a valid class equation

without just citing the class equations for D8 and Q8

and saying it follows

[next obsidian]

bruh ngl

I've been thinking bout it for a while and nothing pops out to mind that isn't really involved and involves knowing a lot about the structure of groups of order 4

and maybe groups of order 8

[past temple]

well yeah if u can just classify the g roups of order 8

but like

[next obsidian]

I think the only thing I can think of

[past temple]

without having to do that

[next obsidian]

The center is order 2 right?

[past temple]

yeah

[next obsidian]

There's sort of an interplay between conjugacy classes of a group and conjugacy classes of a subgroup

I forget precisely what it is, I had to discover it for a take home midterm last year

[past temple]

i think the conjugacy classes of the subgroup like

[next obsidian]

but maybe you can try poking around going to G/Z(G) and deriving some contradiction about the class formula of the quotient?

[past temple]

divide

the conjugacy classes of the group

given the same representative

[next obsidian]

yeah

oh that's true

[past temple]

thats how u prove that A5 is simple for exampl

[next obsidian]

Like

[past temple]

idk how to use it tho

[next obsidian]

if the size of the conjugacy class of pi(g)

divides the size of the conjugacy class of g

Idk is this for a hw thing?

Or just curious?

I'm inclined to say it's kinda hard lol

[past temple]

its for homework

[next obsidian]

Yeah like tbh

I never dealth with problems of "show this isn't a valid class formula"

it was always "show no simple group has this class formula" or whatever

or is that backwards hmm

Idk, I feel like it was always with simplicity

[past temple]

show that a group is simple

given the class formula

[next obsidian]

right

[past temple]

and it usually follows from the fact that

if a normal subgroup exsist

its some union

[next obsidian]

yeah

[past temple]

of conjugacyt classes

[next obsidian]

and then you use lagrange

[past temple]

yeah

pretty lost on how to do this though

[next obsidian]

do you have to avoid classifying groups of order 8?

like explicitly?

[past temple]

the problem doesn't explicitly say so

but that's not something we've done

[next obsidian]

You technically only have to classify non-abelian groups of order 8

[past temple]

yeah

i mean im aware that D8 and Q8

are the only ones

[next obsidian]

which I've done the full classification before but I used semi-direct products

[tacit saffron]

is Zn (Z sub n) another way to say Z/nZ

[next obsidian]

so it's kinda messy

Yeah Zempro

[past temple]

yeah we havent done semi direct products yet

we just did

the classification of finitely generated abelian groups

[next obsidian]

I mean okay what do we know

So we have a normal subgroup of order 4, call this N

[sturdy marsh]

You might be able to do it with some weird counting argument

[next obsidian]

we have the center, which is the normalizer of the 4 things in the big conjugacy class

?

Wdym brofibration

[past temple]

how do u know we have a normal subgroup of order 4

oh wait

wait yeah

[sturdy marsh]

Pick representatives id = g_1, g_2, g_3, g_4 for the orbits

[past temple]

hwo do we know

[next obsidian]

cuz

[sturdy marsh]

g_1 and g_2 commute with everything

[next obsidian]

the normalizer always contains the center

[sturdy marsh]

the stabilizer of g_3 has order 4

[next obsidian]

and the normalizer is size 2 cuz of orbit stabilizler

[sturdy marsh]

g_1 and g_2 and g_3 are in the stabilizer

show that the other dude in g_3's orbit is not in the stabilizer

[next obsidian]

I meant stabilizer btw not normalizer haha

[sturdy marsh]

so the fourth dude in the stabilizer of g_3 must be one of the dudes in the last orbit

[next obsidian]

hmm

[tacit saffron]

dude

[past temple]

uhh

im lost

[sturdy marsh]

relabeling, let's just say that g_4 stabilizes g_3

[next obsidian]

he's saying g_1 = e

g_2 is the other thing in the center

g_3 is something in the size 2 conjugacy class

then the stabilizer of g_3 call this N

is size 4

N contains g_1,g_2,g_3

[past temple]

sure

[next obsidian]

show that the other thing conjugate to g_3 isn't in N

so something in the last conjugacy class of size 4 is

relabel to call that g_4

[sturdy marsh]

the stab og g_4 is g_1 and g_2

[next obsidian]

yup

[sturdy marsh]

so g_3 moves g_4

this gives us another element in stab g_3 i think

[next obsidian]

hmm

[sturdy marsh]

which is a contradiction

[past temple]

u lost me at the

[sturdy marsh]

gimme a sec

[past temple]

show the other thing conjugate to g_3 isn't in N

[next obsidian]

the thing conjugate to g_3 can't stabilize g_3

is that true?

it sounds true

[past temple]

is it?

[sturdy marsh]

yes

[next obsidian]

I believe it

[sturdy marsh]

just write it out

[past temple]

is that something that holds in general?

[next obsidian]

I just tried to

couldn't get a contradiction looking at it for like 5 secs only haha

[sturdy marsh]

oh wait

[next obsidian]

like if it did that means they commute and you get
gxgx^-1 = xgx^-1g

[sturdy marsh]

it does stabilize

rip

[past temple]

kms

[next obsidian]

hmm

I mean this is good tho

oh wait it would ahve to be

lmao the normalizer is normal

[sturdy marsh]

well yeah we know all the stabilizers

[next obsidian]

stabilizer*

if you contained g_3 you need its conjugate too

[sturdy marsh]

yup

so stab g_3 is all the dudes except the ones in the last orbit

the stab of all the dudes in the last orbit is Z(G)

[next obsidian]

Okay so

I think I have a proof using Burnside's lemma?

Doesn't burnide's lemma assert that 4 = 1/8 sum(size of conjugacy classes^2)

But I calculated the right side and got 2.75

but I think I might be fucking up Burnside's lemma

[past temple]

we haven't covered burnside's lemma

although im aware of it

[next obsidian]

oh wait

nvm

hold up it's the size of stabilizers

lemme recalculate I bet this is literally always an equality

I thought it was super weird

[past temple]

is it the stabilizers

im looking at wikipedia

the right hand side is like

[next obsidian]

okay yeah

it is

yes this is always an equality

You can never get a contradiction to a class formula using burnside's lemma

I think

[sturdy marsh]

oh wait we have a contradiction, g_4 stabilizes itself

right?

so stabilizer is size 3

at least

[next obsidian]

uh

wait is it that easy?

[sturdy marsh]

ggg^-1 = g lmaooooo

[next obsidian]

wait wtf

[sturdy marsh]

BAHAHAHAAHAHHAHAHAHAHAHAHA

[next obsidian]

that's so stupid

[sturdy marsh]

ajahahsadkhakjsdfhjklashdgjkhwtuoniourcnuieojrntvqiat

[next obsidian]

lmfao man

I didn't think of that at all either

[past temple]

wiat a second

[sturdy marsh]

im losing my shit

[past temple]

oh shit

[next obsidian]

take an element in the last conjugacy class

[past temple]

thats mad smartttt

[next obsidian]

bwaha

[past temple]

yeah yeah i got it

the stabilizer has to be size 2

[next obsidian]

yeh

[past temple]

so its the center

plus the item itself

lmao thats retarded

[next obsidian]

yeah

tfw

[sturdy marsh]

im about to explode

[next obsidian]

this is trash

I'm mad

[past temple]

gg

time to quit math

[sturdy marsh]

gg

[next obsidian]

gg = e for all g

then ur iso to a direct product of Z/2Z

😎

remember that buckos

or direct sum I guess

since it might be infinite

[past temple]

nice lmao

[tacit saffron]

rings make me cri

[sturdy marsh]

upgrade: algebras

upgrade again: dga's

[past temple]

ive got another question im working on

[sturdy marsh]

go on, my procrastination knows no limits

[past temple]

so basically

ive shown that there's an isomorphism between

GLn(Fp)

and (V, +)

where V is the n-dim vector spacce

over the field Fp

p bring a prime

i mean Aut(V, +)

my b

[tacit saffron]

in 2 hours ima have a slew of ring questions so let your procrastination know B)

[past temple]

so now i hvae to show that

there's an injective hom

from GLn(Fp) -> Aut(V)

but that map cannot be surjective

given

p = 4

and n = 2

[sturdy marsh]

p = 4 is blasphemy

[past temple]

haha oaky

GL2(F4) -> Aut(V)

[sturdy marsh]

wait you want an injective map that is not surjective?

[past temple]

yeah

well

an injective homomorphism

that isn't surjective

[sturdy marsh]

any injective map between finite sets of the same order is a bijection

[past temple]

they're not necessarily the same order though

are they

[sturdy marsh]

you showed that there's an iso

[past temple]

i thought they're only the same order if the field is of prime order

i only showed iso for p prime

[sturdy marsh]

if V is an n-dim k-vector space, then we have an iso Gln(k) --> Aut(V) no?

yeah we do

[past temple]

we do?

apparently not tho

[sturdy marsh]

are we talking group automorphisms or linear

still those should be the same for finite fields i think

[past temple]

group automorphisms

only considering the operation of addition

for V

[sturdy marsh]

ok, maybe possible then

we just need to show that there is a group hom that isnt linear

from V to V

group aut sorry not just hom

[past temple]

yeah

hmm

how is that possible though

isnt a group hom necessarily linear

[sturdy marsh]

it is definitely Z-linear

for abelian groups

for any group hom f from V to V we definitely have f(nx) = nf(x)

[past temple]

right

[sturdy marsh]

What is the size of Gl2(F_4)

[past temple]

there's a formula i proved for F_p

but idk if it works for F_4

[sturdy marsh]

is it 180?

15 choices for 1st column

and 16-4 choices for 2nd

yeah so 180

if we count Aut(V)

(which should be larger if the problem is correct)

then the obvious map works

(pick a basis)

[past temple]

how do we count Aut(V)

[sturdy marsh]

V is a 2 dim vect over F

so it is F \oplus F as an abelian group

which is really Z/2 x Z/2 x Z/2 x Z/2

so group homs between them are in one to one correspondence with |Hom(Z/2, Z/2)|^16

which is 2^16 so this is promising, there might be a lot of automorphisms

wait what is the size of Gl_4(F_2)

[past temple]

im confused

how do we know this: "so group homs between them are in one to one correspondence with |Hom(Z/2, Z/2)|^16"

[sturdy marsh]

by the property of direct sums and products

but nvm that was just a heuristic

what is (2^4 - 1)(2^4 - 2)(2^4 - 2^2)(2^4 - 2^3)

whatever it is it is more that 180

okay yeah it's true

here's the argument

Gl_2(F_4) has 180 elements

Now we want to count Aut(V)

V is iso to Z/2 x Z/2 x Z/2 x Z/2 as an abalien group

which is a 4 dim Z/2 vector space

the automorphism group if V is therefore iso to Gl_4(F_2) by the prime case that you showed

which has more than 180 elements

so we're done

pick any one of the usual morphisms from Gl_n(F) to GL(V)

[past temple]

ahh okay

wait

this is kinda stupid

but how do we show again

that F4 = Z2 x Z2

[sturdy marsh]

there are only two groups of order 4

one is Z/4Z

other is Z/2 x Z/2

Z/4Z cannot be made into a field

but you dont even need that

F_4 is a 2 dim vector space over F_2

so V is a 4 = 2 x 2 vector space over F_2

and then you just use the prime case

yeah that's cleaner

[past temple]

okay

yeah i just used that z/4 cant be made into a field

sicne 2 has no mult inverse

[sturdy marsh]

well you might also need to show that there is no other multiplication that can be put on it lol

I assume you don't really need to prove it, depends on the class ig

contructing the field of 4 elements from just the axioms is a pretty classic problem

and super annoying

there's a zillion cases

[past temple]

jesus okay

this is algebra 1

so probably not

[next obsidian]

upgrade again: dga's
@sturdy marsh what's a dga?

[bleak abyss]

Differential graded algebra would be my guess

[sturdy marsh]

yup

[past temple]

jesus fuck

finally turned in my pset

fuck algebra

[sturdy marsh]

😦

[past temple]

dam its kinda wild

the more math u learn

the harder it gets

so u never actually get good at it

mad sad

[quaint ivy]

that's kind of the fun in math

[snow cliff]

also the pain

speaking of which

how is $s \cdot 1H = sH = 1H$ true?

[cloud walrus]

camolot457656:

[quaint ivy]

the first equality is by the natural left action of <s> over the left cosets of H, and the second is because H=<s>

[snow cliff]

What is the natural left action? My book does not define such a term

[stone fulcrum]

Your book should define what a group action is. A group action allows you to pair a group with a set, and define a "group-set" multiplication

In this case, the set is the cosets

And the multiplication is simply multiplying the coset by s to obtain a different coset

[tacit saffron]

In my problem, im told to consider the ring Z/nZ = {0, 1, ..., n − 1}, where addition and multiplication are both mod n, then prove that the element a in Z/nZ is a zero-divisor if and only if gcd(a,n)>1 (or rather if a and n are not relatively prime.
To solve would I say a and n are relatively prime, and there is an b such that ab is a multiple of n. So for some x and y, xa + yn = 1 gives me xab + ynb = b, so b is a multiple of n, so b=0 and a would not be a zero divisor. So to be a zero divisor, a and n would have to not be relatively prime?

or am i not being rigorous enough for my "if and only if"

[snow cliff]

Okay. I went back to group actions in my book. I would still say I am confused by the above equation. I think I understand the first equality because it is the identity times s. But I still do not understand how the second equality is true.

[sturdy marsh]

@tacit saffron If you assumed a and n not coprime, then how do you have x and y so that xa + yn = 1

[tacit saffron]

wait i meant assume they are

[sturdy marsh]

oh sorry i misread it

[tacit saffron]

no me mistype

[sturdy marsh]

yeah that direction works

[tacit saffron]

would i have to supply the reverse

[sturdy marsh]

yup

[tacit saffron]

does the reverse follow same order of logic

[sturdy marsh]

are you trying to show not zero div implies coprime

or not coprime implies zero div

[tacit saffron]

im being asked to show the latter

but i kinda just showed coprime implies not zero div

[sturdy marsh]

both of the things I wrote are the same (it's just the contrapositive)

it's up to you

[tacit saffron]

but does coprime imply not zero div imply what you said

[sturdy marsh]

no

you showed A implies B

you need to show B implies A

[tacit saffron]

but im being asked to show not A implies not B

and not B implies not A

[sturdy marsh]

that's the same thing as A iff B after relabelling

set A' = not A

[tacit saffron]

so A->B->A implies A'->B'->A'

because i started teh former in my proof but im being asked for the latter

[sturdy marsh]

A = coprime

B = not zero divisor

[tacit saffron]

ye

[sturdy marsh]

you showed A implies B

you want to show A iff B

you can do that by showing B implies A

or by showing not A imples B

[tacit saffron]

yes i understand that part but the question is asking for A' iff B'

if i can say A iff B can i say A' iff B'

[sturdy marsh]

I am lost now

just prove not zero div implies coprime

[tacit saffron]

yes i can do that

but

question asks

prove zero div implies not coprime

[sturdy marsh]

ok you already did that

you're good if they only care about one direction

[tacit saffron]

no i proved coprime implies not zero div

[sturdy marsh]

that's the same thing as zero div implies not coprime

[tacit saffron]

ok great thats what i was getting at

[sturdy marsh]

cool

[tacit saffron]

bingo bongo

i don't understand ring notation

like i understand the axioms of a ring but

if im given "Z2[x]/<x^2 + 1>" i got no idea what this mean

[sturdy marsh]

@snow cliff sH is contained in H as H is a subgroup, it is all of H as h maps to sh is surjective (as it is bijective )

if im given "Z2[x]/<x^2 + 1>" i got no idea what this mean
@tacit saffron did they not mention quotients?

[tacit saffron]

that might be the lecture i miss

[sturdy marsh]

it's just the usual thing where you quotient out by an ideal

yeah read about ideals

and quotients

it's the same idea as quotients for groups

[tacit saffron]

hm ok

i may be back

[snow cliff]

So because H is generated by s, sH is the same as H? I'm very confused by this

@sturdy marsh

[sturdy marsh]

for any x in H you have xH = H

you dont need it to generate

the definition of xH is {xh : h in H}

[snow cliff]

so the 1 is okay to put up front again for the same reasoning?

[sturdy marsh]

you can put any element in front

it is just notation

[snow cliff]

How come it changes some of them but not all of them?

so like sr is r^3

but sr^2 is r^2

[sturdy marsh]

sr^2H = r^2H you mean

[snow cliff]

Yes, I forgot the H

[sturdy marsh]

well yeah as sr^2H = r^2sH = r^2H

as sH = H

[snow cliff]

oh

because r^2 commutes here?

[sturdy marsh]

yup

[snow cliff]

Why is the last one not r^{1}H then?

[sturdy marsh]

the last one?

[snow cliff]

because shouldn't is follow that sr^{3}H=r^{-3}sH?

[sturdy marsh]

r^{-3} = r

[snow cliff]

you're right

I've been awake too long

ty this makes more sense now

but to clarify a general concept for cosets

if H containted l

or ratehr y

better letter

then yH = H?

[sturdy marsh]

yes

you should try proving it

it is a 2 liner

[snow cliff]

would I go about this doing something like permutations?

or maybe not

[sturdy marsh]

no

just unpack the definition

maybe it's a 3 liner lol

it is a 3 liner up to a finite number of lines

[snow cliff]

yH = {yh : h is an element of H} for all y in G.
I don't see how we are closed ? Or have the same set?

Because couldn't an element of G push us out of the set of H

or am I terribly wrong here?

[sturdy marsh]

I thought you said y was in H

if H containted l
@snow cliff ^

[snow cliff]

oh nvm

yup

we good then

I really need to finish this pset and sleep

[tacit saffron]

ok so an ideal is to a ring as a normal group is to a group? And Z2[x] would be all the polynomials with coefficients 0 or 1 and x^2 + 1 would be an ideal in the quotient ring Z2[x]/<x^2+1> ?

[sturdy marsh]

sorta

(x^2 + 1) is an ideal in the poly ring

so it is the zero ideal in the quotient

[tacit saffron]

ok hm

[sturdy marsh]

the same way the normal subgroup is the identity element in the quotient

[tacit saffron]

im supposed to calculate a multiplication table

[sturdy marsh]

have fun lol

luckily it is Z2 so not very big

and the poly is a quadratic

shouldnt be horrible

[tacit saffron]

would it be 0 1 x x+1

how would i show this wasn't a field

wait x+! * x+1 would give 0

so it aint a field

but if i used

x^2 + x + 1 instead

it would be field

wowie

[sturdy marsh]

yeah if you quotient out by an irreducible poly you get a field

[tacit saffron]

i gots more quotient ring question stuff

im told to describe Z[x]/<x^2 − 3, 2x + 4> idk where to go, am i using one ideal and than another or is the ideal (x^2 - 3, 2x + 4) ≤ Z[x]

[sturdy marsh]

where's the 'other' ideal?

[tacit saffron]

x^2 - 3 and 2x+4

[sturdy marsh]

you're looking at the ideal generated by those two

there's just one ideal

you could look at it as a sum of ideals too

[tacit saffron]

oh

[sturdy marsh]

all of this stuff is a matter of figuring out the definitions

which book are you using?

[tacit saffron]

artin

i find slightly hard to read

or rather hard to find information after the fact

[sturdy marsh]

aluffi is easy to read

dummit has a lot of examples

[tacit saffron]

i shall look into deez

[sturdy marsh]

I think the first chapter of atiyah macdonald also has problems related to the stuff you're doing, the problems might get hard towards the end

the first few problems are very doable tho

if you're just starting out

[steep hull]

Haha what a laughably troll problem

[tacit saffron]

que

[steep hull]

From before

[tacit saffron]

which

[steep hull]

1+1+2+4 can't be class equation for group of size 8

[next obsidian]

It also can’t be class equation for a group of any other size

😎

[steep hull]

Um ofc

[tacit saffron]

im still mega confusion what im supposed to do for "describe Z[x]/<x^2 − 3, 2x + 4>"

[sturdy marsh]

try to find a nicer generating set

for the ideal

[vestal snow]

What does $Tr_{K/F}$ mean in field theory?

[cloud walrus]

Have a Banana Bitch:

[vestal snow]

I have always seen $Tr_{K/F}(a)$ where $a$ is an element of $K$

[cloud walrus]

Have a Banana Bitch:

[chilly ocean]

random guess, maybe the map $a \mapsto Tr_{K/F}(a)$?

[cloud walrus]

88ddda:

[vestal snow]

Hmm so it says Let Tr_{F'/F} : F' --> F

be the trace map from F' to F

[chilly ocean]

i think that makes sense yeah

[vestal snow]

What does that mean?

Like what does Tr_{F'/F} do to an element a of F'?

I'm guessing it takes a to the trace of multiplication by a as a K-linear map

[chilly ocean]

i'm not great at field theory but that sounds familiar and looks to be the definition on wikipedia

[vestal snow]

okay that makes sense

Thanks

Also, for people who have done field theory

Which definition of trace would you recommend using for doing grad level stuff in field theory?

The one using tensor product of a vector space and it's dual or the one using sum of diagonal entries of a matrix?

[sturdy marsh]

I'm guessing it takes a to the trace of multiplication by a as a K-linear map
@vestal snow yup, you look at the map from K to K (as F vector spaces) given by multiplication by a. Then Tr_(K/F)(a) is the trace of that map

similarly you also have the norm of a

if the extension K/F is separable, then the trace is just the sum of all the galois conjugates of a

and the norm is the product of all the galois conjugates

so if you're in the separable case, then you compute the trace and norm using that fact

[chilly ocean]

do you have a specific example in mind

[carmine fossil]

Definition of group operation

Product of any 2 elements of the group is in the group

Well,Without closure the operation doesn't make sense

Wdym

We don't

It could be Z_2xZ_2xZ_2

Or Z_2x Z_4

h^2 could be in <g>

h^2 has to be in <g> otherwise you get h is in <g>

h^3 will not be in <g> because if so (h^3)(h^-2)=h will be in <g>

Well, Don't think that's relevant

Honestly you can choose any 4 distinct elements

If you choose any other element,they will come out to be the same as one of the 8 elements

Think of the group operation as a function:GxG -> G

The closure condition is same as saying this function is well defined

Yes

Yes

Yes

[tacit saffron]

"Let R be a ring with unity. Prove that R contains a subring which is isomorphic either to Z or Z/nZ." I'm confused how i'd go about this proof. I know a ring with unity means for all r in R, r1=1r = r, so it has a multiplicative identity element

[chilly ocean]

maybe try adding 1 to itself a bunch

[tacit saffron]

yeah you'd get Z but is 1 a subring?

[chilly ocean]

what do you mean? i was meaning that, if R is characteristic n, then you could have the set {0, 1, 2, ..., n-1}

[tacit saffron]

how would i know R contains those other elements

[chilly ocean]

because it is closed under addition

1+1+...+1

[tacit saffron]

wait fuck i sptotopid

do i have to prove a mapping of that to Z

[chilly ocean]

i guess so

although is is fairly trivial

[tacit saffron]

yeah thats what i was thinking

idk how id really go about doing it

cuz like 0 would map to 0 and 1 would map to 1

so same properties so isomorphic?

[chilly ocean]

something like that

[carmine fossil]

Map r.1 to r

Where r.1 represents the sum (1+1+1... r times)

[tacit saffron]

what is r.1

[carmine fossil]

2.1 is (1+1) ,3.1 is (1+1+1) and so on

[tacit saffron]

im confused by how r.1 and r are different

[carmine fossil]

(1 as in the multiplicative identity of the ring)

r is not a element of the ring

[tacit saffron]

wait why not

[carmine fossil]

r is an integer

The ring may not be consist of integers

For example,Take the ring of matrices

[tacit saffron]

shid now im more confusion

[chilly ocean]

He's just saying that $1_R + \dots + 1_R = r\cdot 1_R \mapsto r$ where $r\in \mathbb{Z}$ and $1_R$ is the unity of an arbitrary ring $R$.

[cloud walrus]

miqqson:

[carmine fossil]

Sorry,that's only if characterstic is 0,if characterstic is nonzero(let's say char is a)
$r \cdot 1_R \mapsto$ r mod(a)

[cloud walrus]

DrunkenDrake:

[chilly ocean]

Yeah, you're right, I overlooked the characteristics

[carmine fossil]

This is because r$.1_R=(r+a).1_R=(r+ka).1_R$

[cloud walrus]

DrunkenDrake:

[tacit saffron]

what is characteristic

wait lemme see txtbook

[chilly ocean]

It is the least natural number $p$ s.t. $p\cdot r = 0_R$ for $r\in R$. IIRC

[cloud walrus]

miqqson:

[chilly ocean]

If it exists! Otherwise it's 0

[tacit saffron]

where r is mult identity?

[chilly ocean]

depends on author

can be arbitrary elt, some want it to be the unity

Which textbook are you using?

[tacit saffron]

"The characteristic of a ring R is the nonnegative integer n which generates the kernel of the homomorphism phi: Z -> R This means that n is the smallest positive integer such that "n times 1(R)" = 0 or, if the kernel is (0), the characteristic is zero"

artin

[chilly ocean]

Ah

[tacit saffron]

wait so whats the case if char is a

like so far this is making sense but i confusion as to relation to isomorphism with Z

(or Zn)

[tacit saffron]

figured it out

[astral galleon]

Dumb question How many permutations $\sigma$ are there in $S_5$ for which $\sigma(1)=4$ ?

[cloud walrus]

Zophike1:

[solemn rain]

so you want (14xyz)

keep permuting xyz

@astral galleon

how many ways can you rearrange xyz

x y and z*

[astral galleon]

I made a slight insight into the problem realizing that in $S_5$ it seems to come down to counting permutations of the form $(a b c)(d)$ and since we are dealing with disjoint cycles the following result seems useful

I made a slight insight into the problem realizing that in $S_5$ it seems to come down to counting permutations of the form $(a b c)(d)$ and since we are dealing with disjoint cycles the following result seems useful,

$\textbf{Result}$

If $kr \leq n$, where $1 < r \leq n$ $\alpha \in S_n$ where $\alpha$ is a product of $k$ disjoint r-cycles is

$$\frac{1}{k!} \frac{1}{r^{k}} [n(n-1) \cdot \cdot \cdot (n-kr+1)]$$

[cloud walrus]

Zophike1:

Zophike1:

[solemn rain]

i mean

[astral galleon]

@solemn rain it's just 4!

[solemn rain]

idk it seems simpler to me

yea so thats a number

[astral galleon]

oh so it's 24

[solemn rain]

keep playing around forms

of (14)(xyz)

ig its just 24 i mean

maybe im mising somethign obvious?

idk

[astral galleon]

oh so that's the answer

[solemn rain]

yea?

idk if im right

check me on this

[astral galleon]

maybe someone else in chat can help

[solemn rain]

i just saw your reply

and i thought dman thats alot

[sharp sonnet]

if your permutation in S_5 fixes one element, it's just a permutation in S_4

[astral galleon]

I failed on exam and i'm looking through what I got wrong

[sharp sonnet]

so the elements of S_5 that fix an element are in bijection with the elements of S_4

[solemn rain]

i mean fuck cycle notation

just draw dots and arrows

1 goes to 4

how many other bijections

nothing wrong right/

[sharp sonnet]

well, in this case not fix

[solemn rain]

not fix but assign

yea i gotu

[sharp sonnet]

but it's the same

[astral galleon]

@sharp sonnet so just count $S_4$ but the problem says look in $S_5$ through

[cloud walrus]

Zophike1:

[astral galleon]

@sharp sonnet is the answer 24 ?

[sharp sonnet]

well, then just send 4 to 1

and see how many choices you have left

or 1 to 4, wtv

[astral galleon]

I think i got it earlier it's 24 right ?

[solemn rain]

just permute the rest 3

3*

(14xyz)

[astral galleon]

now i'm confused again

[sharp sonnet]

honestly cycle notation is suboptimal for this

[solemn rain]

bro a cycle that assigns 1 to 4 looks like(14xyz)

[sharp sonnet]

just map 1 to 4

[solemn rain]

or (14)(xy)(z) or whatever

permute the rest

[sharp sonnet]

how many choices are there to map 2?

[solemn rain]

u get different permutations

[carmine fossil]

It's not (1 4)

[solemn rain]

why

[astral galleon]

wait before I get to that nobody answered my eariler question is the answer 24 ?

[sharp sonnet]

yes

[astral galleon]

oh then why didn't you say it eariler

[carmine fossil]

That would mean 4 gets mapped to 1

[solemn rain]

@astral galleon

i told you

[sharp sonnet]

because i wanted you to understand it yourself

[solemn rain]

@carmine fossil and?

he wants maps that map 1 to 4

doesnt matter anything else really right?

[sharp sonnet]

but probably the easiest way is map 1 to 4

then you have 4 choices to map 2, namely 1, 2, 3, 5

and so on

[carmine fossil]

(1 4) would mean 1 gets mapped to 4 and 4 gets mapped to 1 simultaneously

[solemn rain]

whats wrong with that

he wants permutations that map 1 to 4

thats it

why should we care where 4 goes

[sharp sonnet]

exactly

[astral galleon]

I still can't believe I failed the exam I feel so ashamed 😢

[solemn rain]

was the exam hard?

easy:?

i would have failed if i were you too

without looking at the exam

[carmine fossil]

Pretty sure hard

[astral galleon]

The professor said I should read more carefully before I turn something in I was able to get some of the questions correctly

I have to rebuild my grade from the ground up i'm at like a 70 now

I should have gotten those problems I did well on the homework

[solemn rain]

wanna share the problems

so i fail with you aswell?

[astral galleon]

I was able to resolve some of them myself I'll have to look at the rest tommrow

Why is it that when I do well in a math class I always fail the first test 😢

[obsidian path]

the matrix is throwing me off.. how do you define the function here?

[smoky cypress]

Try to think of each complex number a rotation and scaling of the complex plane

Then try to make that into a matrix

[obsidian path]

any hints to make it into a matrix?

[smoky cypress]

Remember the rotation by $\theta$ clockwise is $\begin{bmatrix}\cos\theta&-\sin\theta\\sin\theta&\cos\theta\end{bmatrix}$

[cloud walrus]

Whoever:

[thorn delta]

*counterclockwise but yea

[smoky cypress]

yes

counterclockwise

but that gives you an isomorphism as well

[slate forum]

Hey fellas, I'm trying to show the tensor product of an $(S,R)$-bimodule and and left $R$-module can be given the structure of a left $S$-module, but I'm having trouble defining the scalar multiplication, does anyone know how to properly do this? I've tried defining an $R$-balanced map $S\times (A\times B)\rightarrow A\otimes_R B$, but I'm not sure what this gives me, will it give me a map from $S\times (A\otimes_R B)$ or a map $S\otimes_R (A\otimes_R B)$?

[cloud walrus]

HelixKirby:

[sturdy marsh]

just multiply on the left

[slate forum]

Well, yes, but I want to make sure that's a legit thing to do, like show that it's well-defined and whatever

[sturdy marsh]

Okay so you just want a map from S to End(A\otimes B)

an element of End (A \otimes B) is the same as bilinear maps from AxB to A \otimes B

So you just need to define a map from S to Bil(AxB, A \otimes B)

show that the obvious thing satisfies the required bilinearity condition

i.e. the map that sends (a,b) to sa\otimes b

so check that (a, rb) and (ar, b) go to the same thing

(which uses the bimodule structure on S)

and check additivity

and check if it is a homomorphism

[slate forum]

Ok, yeah, that's what I'm doing now

Checking it's a homomorphism, though, do I need to compare the explicit construction of the maps\ from r+r' and the sum of the two maps from r and r'?

[sturdy marsh]

you also need to check multiplicativity

[slate forum]

Like, perform the construction on the sum and then compare it to the sum of constructions and perform the construction on the product and then compare the composition of the constructions?

[sturdy marsh]

yup

[slate forum]

Is it enough to say they perform the same way on the cartesian product?

[sturdy marsh]

?

it's the usual process of checking if a map is a homomorphism

nothing weird

by doing so you're just checking if s(s')a = (ss')a

and the similar statement for sums

[slate forum]

mmmk

[vestal snow]

Can someone explain where the s comes from?

Shouldn't $-a_{r-1} = Tr_{M/L}(x)$ because $a_{r-1}$ is the sum of all roots of $\phi$ so it is the sum of all $\sigma_i(x)$

[cloud walrus]

Have a Banana Bitch:

[sturdy marsh]

trace is the coefficient corresponding to the characteristic poly, not the minimal

the char poly is the min poly raised to s

[vestal snow]

So why do we multiply by s?

[sturdy marsh]

well what is the coesfficient of the term of degree rs-1 when you raise the poly to s

[vestal snow]

Okay I got it

I thought it was the coefficient of r-1 in char

Thanks

[golden pasture]

alternatively cuz $Tr_{M/L}=Tr_{M/L(x)}Tr_{L(x)/M}$ and $Tr_{L(x)/M}$ is $-a_{r-1}$ and $Tr_{M/L(x)}$ of anything in $L(x)$ is $[M:L(x)]$ times of it

[cloud walrus]

ariana:

[leaden finch]

can someone help a sistah out

[chilly ocean]

Divide through, until you can't then divide remainder with divisor polynomial and so on

[chilly canyon]

Hi ! Trying to give a presentation for a group, I'm struggling with some methodology ^^" So I considered $\psi:F_2\to G$, where $G$ is generated by two elements and relations $R$ hold. I get that $\psi$ is onto, and that $\ker(\psi)$ contains the normal closure of the relations : $<<R>>\subset\ker(\psi)$

[cloud walrus]

Matplotlib:

[chilly canyon]

To prove the converse inclusion, that is $\ker(\psi)\subset<<R>>$, since I don't have cardinality, how to proceed ?

[cloud walrus]

Matplotlib:

[chilly canyon]

Elements of $<<R>>$ are the products of conjugates of elements of $R$, so am I supposed to show that $w\in\ker(\psi)$ writes as $g_1x_1g_1^{-1}...g_nx_ng_n^{-1}$ with $x_i\in R$ and $g_i\in F_2$ some word ?

[cloud walrus]

Matplotlib:

[thorn delta]

can't u just say F_2/ker(psi) is iso to G so a word w in G is 1 (w is in R) iff its image under the canonical epimorphism is 1, i.e. w is in the kernel of psi?

i have a question real quick. How do we know that S4 has 3 sylow 2 subgroups, and not 5, or 7, etc??

[civic linden]

yo how can i prove $\phi : E(n)=\mathbb{R}^n \rtimes_\delta O_n \rightarrow GL_{n+1}(\mathbb{R})$ where $(u,A) \mapsto \begin{pmatrix}
A & u\
0 & 1
\end{pmatrix}$ is a group homomorphism. I'm a little rusty, do I just do phi(u,A)*(v,B)) = phi(u,A) * phi(v,B) and then write some expression for the map from (u,A) to the matrix?

[cloud walrus]

!R. ¬:

[civic linden]

where E(n) is the euclidean grou

group

[thorn delta]

(i figured out the answer to my question. the number of sylow p subgroups divides the order of a group G for each prime p)

[covert otter]

@leaden finch ahh I'm afraid your question asked you to find the GCD using Euclidean algorithm so you should write that in the form of a linear combination and then have the GCD.. but ahh these things are kinda boring to deal with tbh.

[leaden finch]

can you show me

lol

[civic linden]

<@&286206848099549185>

[compact needle]

@civic linden You already have the map between groups. You're left with checking the property of homomorphisms. Namely you need to take two elements $(u,A)$ and $(v,B)$ and compute $(u,A)(v,B)$, $\phi(u,A)(v,B)$, and $\phi(u,A)\phi(v,B)$. If $\phi(u,A)(v,B)=\phi(u,A)\phi(v,B)$ for arbitrary $(u,A)$ and $(v,B)$, then $\phi$ is a homomorphism

[cloud walrus]

Turgul:

[civic linden]

yeah i understand, can we use the determinent to show this then?

$\begin{pmatrix}
A & u\
0 & 1
\end{pmatrix}$
and
$\begin{pmatrix}
B & v\
0 & 1
\end{pmatrix}$

$\begin{pmatrix}
AB & Av+u\
0 & 1
\end{pmatrix}$

taking the dets of these we will get AB = A*B

[cloud walrus]

!R. ¬:

[compact needle]

Just because matrices have the same determinant does not mean they are equal

[civic linden]

no im just showing det(AB)=det(A)det(B) in this case

and the same would be under a homomorphism?

[compact needle]

It is, in principle, possible to have a composition of maps of groups $G \rightarrow G' \rightarrow G''$ where the first map is not a homomorphism but the composition is. What you would be showing is that $\phi$ followed by the determinant is a homomorphism.

[cloud walrus]

Turgul:

[compact needle]

But in general, multiplication of block matrices is exactly the formula you'd hope it would be. You can use this to show the homomorphism property directly

[civic linden]

so I can use this?

[compact needle]

All you need is the fact that for properly dimensioned matrices $A,B,C,D,A',B',C',D'$, you have
$\begin{pmatrix}
A & B\
C & D
\end{pmatrix}
\begin{pmatrix}
A' & B'\
C' & D'
\end{pmatrix}
=\begin{pmatrix}
AA'+BC' & AB'+BD'\
CA'+DC' & CB'+DD'
\end{pmatrix}$

[cloud walrus]

Turgul:

[civic linden]

right and what about this part? show $\phi (O\rhd_\delta \cdot):\mathbb{R}^n \rightarrow \text{the "column
matrix" spanned by translations in } GL_n$ is an algebra intertwiner for every $O\in O(n)$

[cloud walrus]

!R. ¬:

[compact needle]

I don't remember what the definition of an intertwiner is

[civic linden]

A linear map that commutes with the action is called an intertwiner

so im guessing phi just preserving the semidirect product?

but how do we show its preserved?

[compact needle]

So the question would appear to be that matrices
$\begin{pmatrix}
A & 0\
0 & 1
\end{pmatrix}$
and
$\begin{pmatrix}
0 & v\
0 & 1
\end{pmatrix}$
commute.

[cloud walrus]

Turgul:

[compact needle]

I'm not sure why my matrices aren't rending correctly

[chilly ocean]

Maybe double backslash for new line?

Also I'm not sure you need dollar sign for matrix environment

[compact needle]

@chilly ocean You're probably right about the double backslash, that's normal latex (though I copied this from what !R. ¬ wrote which seemed to work for them)

@civic linden I'm not sure I'm correctly interpreting what they mean by "column
matrix" spanned by translations

[civic linden]

probably Av+u and AB on the other side?

[compact needle]

Yeah, I'm not sure. In general translation and orthogonal actions on a vector space shouldn't commute (that's why you end up with the semidirect product in the first place)

Anyway, I have to go. Hopefully you get it sorted out

[civic linden]

ty for your help

[vocal depot]

Perhaps I am overthinking this one but I'd like to get some constructive input on this problem I'm working on.

Let $t,n\in\mathbb{Z}^{+}$ such that $t|n$. By the fundamental
theorem of cyclic groups, there is a unique subgroup $H$ of order $t$
in $\mathbb{Z}{n}$. Prove that for every $x\in H$, its order $|x|$ in
$\mathbb{Z}{n}$ divides $t$ if and only if $x\in H$.

[cloud walrus]

HisMajestytheSquid:

[vocal depot]

I can't just assume x is in H right? Because that would require that I then prove that it divides t in some other direction

Where it isn't in H?

I'm not sure.

I feel like I can just make some argument about how the order of a subgroup generated by the element x (and therefore the order of the element itself) in H must divide the order of H simply by the fundamental theorem of cyclic groups.

If I don't reply, just ping me I guess. I've gotta get some sleep.

[next obsidian]

your statement isa bout x in H

so you get to assume x is in H

Like |x| doesn't change based on if what group you view it in. I think what you wanted to say is that for all x in Z_n that |x| divides t if and only if x in H which is true but not as trivial

[sturdy marsh]

never knew that result was called the fundamental theorem of cyclic groups

[next obsidian]

I didn't either

I don't think it's very widely used, maybe whatever textbook they're using decided to call it that ¯_(ツ)_/¯

[covert otter]

@leaden finch Should look something like that, and I don't really know if your question strictly belongs to this channel

[next obsidian]

I wouldn't disagree with the name, although TBH if you said fundamental theorem of cyclic groups the first thing that comes to mind is the characterization of Hom(G,H) where G is acyclic group

[golden pasture]

@covert otter you really only need the remainder for gcd, somewhat faster to only implement modulo P as computing the quotient takes some time
@leaden finch it should end up being gcd(x^2-x,-x^2+1) which has a immediate common factor of x-1 (-1=2)

[vocal depot]

@next obsidian I didn't write the statement so I can't really make that kind of change to it. That's how the problem is worded from my professor.

But I get what you're saying

[chilly ocean]

Hey guys, just wondering. Is this always true?

a^(n) = e

[carmine fossil]

What is n and what is a?

If I understand you correctly,consider (Z,+)

[chilly ocean]

yeah

[carmine fossil]

There is no n such that (a)^n=e

[chilly ocean]

ever?

[carmine fossil]

Tell me what n makes n(2)=0(normal multiplication)

Except for n=0

[chilly ocean]

I agree, I dont understand that either. Thats my thought.
Im just trying to understand my textbook that says it.

[carmine fossil]

Can you show the statement?

I think he specifically means finite groups

[chilly ocean]

This is the full section.

[carmine fossil]

There is an if part

[carmine fossil]

So,You are assuming x^n=e for some n

[astral galleon]

How do you show that each $g_i\in G$ lies in the coset $gH$ is using the coset defintion enough ?

[cloud walrus]

Zophike1:

[astral galleon]

$g_i$ is the elements of $G$ ? @carmine fossil i'm asking this in general

[cloud walrus]

Zophike1:

[carmine fossil]

Consider subsets, $g_1H,g_2H...g_nH$ where you cover all elements in G

[cloud walrus]

DrunkenDrake:

[carmine fossil]

Coset definition is that except we remove duplicates

[astral galleon]

ahhh okay

@carmine fossil would computing $|g_1H|,|g_2H|...|g_nH|$ be enough to say that $\forall i g_i \in g_{i}H$

[cloud walrus]

Zophike1:

[carmine fossil]

I mean,I would just say 1 is an element of H so g.1 is in gH

[pallid ember]

you dont need to compute anything

g=ge

thats all u need

[astral galleon]

oh but is what I did correct ?

[carmine fossil]

I don't think you will have sufficient info to do it your way

[astral galleon]

What do you mean ?

[carmine fossil]

What is your plan of action?

[astral galleon]

Fix $g_i \in G$ now we will have to show that $\forall i , g_i \in g_i H$, to do this using a result we have that $|g_i H| = |H| , \forall i$ this implies that $g_iH = g_i \forall i$ @carmine fossil

[cloud walrus]

Zophike1:

[carmine fossil]

Wdym $g_iH=g_i$

[cloud walrus]

DrunkenDrake:

[pallid ember]

[astral galleon]

oh that all elements of $g_i$ lie in $g_iH$

[cloud walrus]

Zophike1:

[astral galleon]

sorry I was in a rush

[carmine fossil]

Are you using the fact that g_iH partition G?

[astral galleon]

yes

[carmine fossil]

Well then how do you know that without assuming every g is in some g_iH?

[astral galleon]

In the problem i'm working on we can assume it

In my book it says Prove that each $g \in G$ lies in the coset $gH$

[cloud walrus]

Zophike1:

[carmine fossil]

How did you show g_iH partition G?

For your case,it's just g.e=g

[astral galleon]

oh I have to show it isn't it just a Thm or result

@carmine fossil do I have to show that $g_IH$ partition $G$ because that was already proven in the book i'm using

[cloud walrus]

Zophike1:

[pallid ember]

why insist on complicating it tho

it literally just follows from the definition of a coset

[astral galleon]

oh I actually did it via definition the first time around

Here's the solution Fix $g_i \in G$ now we have to show that $\forall i , g_i \in g_iH$ via defintion we have that, $g_iH = {g_{i} * H: h \in H }$

[cloud walrus]

Zophike1:

[astral galleon]

@burnt glen ^

[pallid ember]

nice ping

[astral galleon]

My bad but is what I wrote correct @pallid ember

Did I prove it

@carmine fossil is the solution correct ?

[carmine fossil]

I mean,it should be g_i * h

And you haven't shown that g_i is in g_iH

[astral galleon]

I thought that was enough

Do I also have to show that $g_i = g_iH$ ?

[cloud walrus]

Zophike1:

[carmine fossil]

You finish with $g_i =g_i \cdot 1=g_i \cdot h$ implying $g_i \in g_iH$

[astral galleon]

Now I don't understand I thought using the defintion was enough

why play with the identity element

[carmine fossil]

Yea,This is the Definition

[astral galleon]

ahhh okay

[cloud walrus]

DrunkenDrake:

[astral galleon]

oh ok I see the error I made never mind I had the proof written down I mistyped

[bronze trench]

Ok kinda just a comment, don't want any answers or anything since that would be cheating. I have a group theory homework question asking to prove that out of all groups of order below 100, only A_5 is both non abelian and simple.

I just want to state that Sylow's theorems are so overpowered here

I'm plowing through the orders proving there's no simple group of that order, Sylow helping me a lot!

[next obsidian]

Totally. Sylow let’s you hit huge classes of groups at the same time, based off the prime factorization

Once you prove a few general ones you probably only have to manually do like 3 or 4

[bronze trench]

Yeah I'm just left with one case left, most were essentially trivial, a few more required 3 or 4 line proofs and this last one is resisting my attacks... not for long hopefully 😄

[next obsidian]

Which order if you don't mind sharing?

I had to do something like this once for a midterm lol

[sturdy marsh]

I just want to state that Sylow's theorems are so overpowered here
@bronze trench hahaha

Feit-Thompson - "am I a joke to you"

and Burnside

[next obsidian]

Classification of finite solvable groups when?

[golden pasture]

[snow cliff]

is a 2, 3 cycle structure the same as a 3, 2 cycle structure?

[thorn delta]

define structure?

[snow cliff]

@token yes

so the cycle type does not have to be strictly increasing?

so long as the same cycle types are included two cycle structures are equivalent?

Perhaps an example will clarify

would (1 2)(3 4 5) have the same cycle structure as (4 2 1)(3 5)?

so I guess be in the same conjugacy class

or are conjugates

My book and abstract stuff on math stack exchange seem to suggest that the cycle types need to be the same

but it's unclear that the order of the cycle type matters

Completely forgot about that

ty

[carmine fossil]

S3 is a subgroup of S4

And Sn is non abelian

Yes

The convention is D_6 though

Yes

Take {1,2,3,4} now consider all permutations that fix 4

Yes

(Technically isomorphic)

[cloud walrus]

God V2:

[next obsidian]

C[x] is a Euclidean domain so any ideal is principal. If (f(x)) = (g(x)) for any g(x) which is associate to f(x), aka we can write g(x) = uf(x)

Take u to be the inverse of the leading coefficient of f, this is a unit, then g is monic

[snow cliff]

This seems too easy an answer to the proof part of the problem above (have not gotten to the deduction yet), so I wanted to double check I wasn't doing anything wrong. It feels like I am.

M is a proper maximal subgroup of G, so it may or may not contain the union of conjugacy class of G. If M does not contain this union then we know that M is not a normal subgroup of G. However, every group contains two normal groups, itself and the identity. Therefore, in this case, M is normal to itself and so we have N_G(M) = M.

In the case where M does contain the union of conjugacy classes of G, it is a normal subgroup of G. Therefore, we can say that N_G(M) = G.

[carmine fossil]

The first part is pretty obvious

Ng(M) is a subgroup containing M

So,It's M or G

[snow cliff]

wow that was rapid

ty

I'll have to give the deduction a think

[snow cliff]

I think I just got the deduction

Nope nvm

What I have found out so far...
The index of M in G is the number of cosets of M in G. However, there is a bijection from cosets to conjugacy classes, therefore |G:M| also tells us the number of conjugacy classes in M.

Have also rewritten it to |G| - |G:M| and its variations, but still trying to figure out what is important/how |M| - 1 relates (the minus one is the removal of the identity, but can't make sense of the multiplication yet).

[snow cliff]

Not sure I’m getting this one before I sleep. A nudge in the right direction or confirmation I’m going the right way would be appreciated. Thanks in advance.

[hard kestrel]

@next obsidian ok i just saw your message thanks

[carmine fossil]

Are you sure that ping was worth it?

[drowsy quest]

If x is a right divisor of ab, then x is a right divisor of either a or b. Is this true in an arbitrary ring or do we need an integral domain?

or is it not necessarily true in either

[queen vine]

4 is a divisor of 2*2

[drowsy quest]

right so we would need some generalization of prime integers right

[hard kestrel]

@carmine fossil actually i never heard of euclidean domains and learnt about them and PID and UFD after he mentioned so

[carmine fossil]

Cool

[next obsidian]

Oh lol

[chilly ocean]

This is binary right?

because it would be b*a= [b^2 +2ba + a^2] / [b+a]

[carmine fossil]

Take b=-a

(a,-a) is not well defined

[chilly ocean]

I thought you just switch them. You can set them to something?

Yeah i guess youre right, because the denom would be 0 right?

[carmine fossil]

Yes

[chilly ocean]

hmm ok, thnx ily ❤️

[chilly ocean]

what is the group on the bottom?

No way it's like 2*r for r in Q coset cause that would just be isomorphic to Q^{\times} nah?

and this quotient should noooot be trivial

[queen vine]

It might be the subgroup of Q that consists of all squares in Q^{\times}, i.e. all elements that can be written as r^2, r in Q^{\times}

is it about Kummer theory?

(if yes, then it is probably what I said above)

[astral galleon]

Is $S_5$ a subgroup of $S_n$ ?

[cloud walrus]

Zophike1:

[smoky cypress]

For n≥5, Sn always contains many subgroups isomorphic to S5

[next obsidian]

I mean S_n contains S_m for all m <= n, you can simply embed it as the group of things that only permits the first m things

[astral galleon]

Oh so yes also isomorphism means the groups are the same right ?

[next obsidian]

I mean

What does same mean to you

It doesn’t mean they exist as the same set but for anything you can say that’s group theoretic they operate the same

[astral galleon]

the groups are equivalent in terms of their opermations

[next obsidian]

I mean once again that isn’t precise but sure

If you had glasses that could only see group theoretic properties

you wouldn’t be able to tell them apart

If you put on glasses that lets you see the actual “physical” elements so to speak you’d see that they’re different

[astral galleon]

Oh there's a one to one correponsdence between the elements of groups in a way that the operations are perserved

[next obsidian]

Oh yes

[astral galleon]

Ok another dumb question what is the union of $S_5 $ and $S_4$ would it just be $S_9$ ?

[cloud walrus]

Zophike1:

[next obsidian]

No

You can even Union them like this

S_5 and S_4 don’t live in the same set

[astral galleon]

oh that's fair

[next obsidian]

You have to pick some larger group which contains them

And then there’ll be subgroups isomorphic to S_5 and S_4, but then the answer can wildly vary

The reason this isn’t a great notion is that you start to care about the actual elements, which isn’t group theoretic

Also generally unions of subgroups are not nice. You want to look at products, unions are very rarely actually subgroups

[astral galleon]

The book gives a hint so i'll have to look at it thanks for the help

[chilly ocean]

yo can someone explain why this is legal? : When mentioning properties of associates in integral domain they say $a \cong b \Leftrightarrow ab^{-1} \in D $, but b isn't guaranteed to have to have inverse. In the example they say: In $\mathbb{Z} + \mathbb{Z}i: $1+i \cong 1-i$, because $\frac{1+i}{1-i} = i \in U\left(Z+Zi \right)$. The problem is, that $\frac1{1-i}$ isn't in Z+Zi, so why can we do so? Is there some kind of D is embeddable in F argument they're using?

[next obsidian]

Sorry is this isomorphism symbol saying that a is associate to b?

Also, is D the integral domain?

[cloud walrus]

Godel:
Compile Error! Click the reaction for details. (You may edit your message)

[chilly ocean]

yes if ure talking about my stuff

and sorry I thought the channels open

[next obsidian]

These ones don’t really close

Anyway, so let me think about this for a second, but to answer your question b^-1 always exists in the field of fractions

So you can actually make sense of ab^-1 if you expand where you view this

[chilly ocean]

I know of course, they dont mention it anywhere tho, Im just guessing

[next obsidian]

But I’m not sure that ab^-1 is in d actually means they’re associate

[chilly ocean]

wait, ab^-1 in U(D)

[next obsidian]

Like I feel like this should say ab^-1 in D^*

Oh yeah this is exactly the case

So it means once you embed D into F, the field of fractions

That ab^-1 is in U(D) implies ab^-1 = u so a = ub

Where u is a unit

[leaden finch]

whats noncommunitative ring ?

im lost on the wording

[chilly ocean]

ring thats not commutitatiev

[next obsidian]

It means ab is not always ba

[chilly ocean]

wait chmonkey

[next obsidian]

Commutative means ab = ba alwyas

[chilly ocean]

im not sure what youre talkingabout

[next obsidian]

Okay so you can embed D into a field F

By letting F just be fractions