groups-rings-fields

[maiden ocean]

What do you mean by "numbers only work"

Does R^3 not count lol

[pliant forge]

this is my true motivation

[celest mantle]

maybe your question's related to the cayley dickson construction @chilly ocean

[chilly ocean]

i meant numbers like complex and quaternions, elements of R^3 aren't numbers

[maiden ocean]

How are you defining a number here

[chilly ocean]

I'm not sure but I'd say elements of fields and rings like C and H

[chilly ocean]

like why aren't there numbers like T = a + bi + cj

[chilly ocean]

i think so

[prisma ibex]

these are just the ones that arise from the Cayley-Dickson construction, which doubles the dimension every time you apply it. I also wouldn't consider the sedenions to be a good example; past the octonions the construction produces things that have absolutely no good arithmetic properties

[maiden ocean]

You can get a lot of other vector spaces which happen to be rings or fields

including ones of odd dimension

[chilly ocean]

I've heard of "applying the cayley dickson construction" to produce algebras, but what exactly is the cayley dickson construction and why does it only produce 2^n dimensional numbers?

[prisma ibex]

I mean it produces algebras which are 2^n dimensional

[viscid pewter]

https://en.wikipedia.org/wiki/Cayley–Dickson_construction

[prisma ibex]

over the field you start the construction with

[maiden ocean]

well it ends up being 2^n since each application of cayley dickinson doubles the dimension lol

[viscid pewter]

here it is

it's a little dense

[prisma ibex]

since every time you do the construction on A you're taking certain 2x2 matrices in A and defining a certain multiplication on these

the space of 2x2 matrices in A that are considered here has twice the dimension of A

[chilly ocean]

I see

[prisma ibex]

also just be aware that after the octonions these things have very awful properties

the octonions lose associativity but are very weakly associative

[maiden ocean]

to be clear while these algebras do have nice properties for n = 1, 2, 3 theyre not like. uniquely number-esque

[prisma ibex]

after that it's a complete mess

[tame grove]

Is proofwiki a good review website? It seems very incomplete, but it doesnt seem wrong though.

[hidden haven]

It is good to recall proofs when you forget them, you probably shouldn't cite it as a reference if you are writing something

if you think it is not well written or is incomplete

well idk im just talking shit

[tame grove]

Maybe incomplete is the wrong word, its just hard to navigate

I was just using it to review for my Modern Alg 1 exam, nothing that fancy but I need to know what stuff means

My professor sent a list of topics on the exam and it was literally a 4 page long list of 103 topics. I like modern alg and all, but how do I even study that much reasonably?

This class is 3 credit hours and it 100% should be more than that

At least 5 tbh...

any tips about how I can not fail? I feel like even if I get all the concepts fairly well, the formal definitions will dock me points

[thorn delta]

as a general rule, if you can do exercises, then you probably know the vast majority of what you need to know tbh.

[oak grove]

i wanted to ask about this one:

in a section about products

the only note we have about isomorphisms other than those of groups with order 4 is this

which doesnt really seem sufficient or enlightening

getting a couple of these is easy but im confused about all four

like $\bZ _ 1 \oplus \bZ _{12}$ and $\bZ _2 \oplus \bZ _6$

[cloud walrus]

jan Niku

[next obsidian]

You have 2 groups which come from direct sums of cyclic groups

Then you need 2 more

12 is an even number which means you can also take the… something group

[oak grove]

hmm

yea i wasted a bunch of time checking if 3 4 would work

but good practice

lmc

[next obsidian]

And then for the last one

Maybe think about what 2•12 is

(There’s actually a 5th group of this order but… idk how you’d find it unless you just ran a program or knew these things exist lol)

[oak grove]

something group 👀

oh you can have an alternating

oh the last hint was for alternating?

im not sure what you mean about even then

dihedral

can i ask another silly question why is the dihedral group of order 12 a hexagon? unless some of the flips collapse into rotations or visa versa but i dont see how that happens

[thorn delta]

a hexagon has 6 distinct reflection symmetries and 6 distinct rotational symmetries so intuitively, the symmetry group of the hexagon is at least order 12.

oh are you saying that you think there are more than 12 elements in the symmetry group of the hexagon?

[oak grove]

yea

well i know there arent but its cumbersome to see why not

maybe i just need to take a break

if all 6 rotations and all 6 flips are not all distinct elements, how do you decide which to exclude, assuming the choice is (mostly) arbitrary

[thorn delta]

hm this is a good question. Maybe the best way is to derive the relations that generate D_12, and show that it generates a group of order 12

[oak grove]

hmm

so maybe some operation that cycles all elements, the identity, one that swaps 2 pairs, and one that swaps 3 pairs

although i have doubts you need the last one

im not sure thats what you were suggesting

[sharp sonnet]

label the vertices of the hexagon, clearly the rotations are pairwise distinct by looking at where a fixed vertex ends up
now the rotations "flip" the order of the vertices: say you go counter clockwise, it turns 1, 2, ..., 6, 1, .... around, so all reflections are distinct from rotations
if you again fix a vertex and look where reflections send it you see that they are distinct

alternatively, look where an edge ends up

a symmetry is uniquely determined where an edge ends up and in what order the vertices are labeled

there are 6 places to send an edge and each has two orientations to be in

[thorn delta]

im saying using a labeling scheme like loch is suggesting, show that if r is rotation by 30 degrees and s is a flip along some axis, then (rs)^2 = 1 and r^6 = 1. Then show that r,s generate a group of order 12.

i mean, these are basically just the relations that generate any dihedral group

[sharp sonnet]

kxrider is suggesting the algebra-brain solution
i invite you to embrace geometry brain

[oak grove]

im thinking of between the two approaches

[sharp sonnet]

i think the edge way is better

[oak grove]

it seems like youre saying that if you look at any one edge, and examine where it ends up and how its oriented, that can be associated to a single permutation

or some equivalent permutation

[sharp sonnet]

yes

[oak grove]

which if i believe sounds way easier 😄

[viscid pewter]

??

[oak grove]

and i guess why not

how could something act on just one edge

[sharp sonnet]

(technically there is orbit stabilizer in this argument but it should be clear)

[viscid pewter]

oh wait i see what you mean

[oak grove]

or im not sure

maybe it could act on a subset

that doesnt make sense 😄

nvm

[viscid pewter]

basically

one of the rotations is the identity

so there's no issue

[oak grove]

yours was what i was attempting

but then i had questions about like

[viscid pewter]

they're all distinct

[oak grove]

okay the largest is obviously order 6

but then you have some of order 3

why should you believe that its not order 18

i know im bad at algebra though

[viscid pewter]

bc 3|6

[sharp sonnet]

the action on the hexagon is very rigid, you can't just shuffle the vertices wildly, the vertex with label n will always be next to the vertex n-1 and n+1 (mod 6)

[oak grove]

yea loch

[sharp sonnet]

thats why knowing where an edge ends up suffices

[oak grove]

thats very cool 😄

[sharp sonnet]

(if you know the orientation)

[oak grove]

im still sorta confused about what youre suggesting to exclude in the list

although listing elements is maybe a bad idea in general

or not the point

[sharp sonnet]

i mean this shows that all 6 rotations and all 6 reflections are unique

and make up the whole group

[oak grove]

okay 😄

thank you all 🙇‍♂️

[thorn delta]

the idea is that the group generated by r and s under the relations (rs)^2 = s^2 = r^6 = 1 is something like r^{n_1}s^{n_2}r^{n_3}...., finite products of integer powers of r and s, and you have to show that it all reduces to r^ns^k for 0 <= n <= 5 and 0 <= s <= 1. The argument for why the relations hold is ultimately geometric like loch was talking about.

[oak grove]

oh huh

ill try it out after i lay my head down for a sec 😄

[thorn delta]

solid plan

[lunar lintel]

For this homework problem I'm asked to prove that for a group G, Z(G), the center of G is a subgroup without doing any computations. I can do it with computations but im not sure how else I would go about it

[chilly ocean]

g is in the center if and only if conjugation by g is trivial. maybe you can use that to write Z(G) as the kernel of something

[wooden ember]

That gives you that it’s a normal subgroup without much of explicit computations

[lunar lintel]

im not familiar with concept of kernel besides faint memory in first linear algebra class. any alternative ideas or does this seem like best way to go?

[thorn delta]

have you learned what a homomorphism is?

[lunar lintel]

no

[chilly ocean]

i don't know how you'll do this without computations then

[lunar lintel]

hm ok I will ask prof. Thanks for advice anyway 👍

[tawny pine]

if you know linalg then you learned vector space homomorphisms w/o calling em homomorphisms

[chilly ocean]

He really did

2070 kids be like

[viscid pewter]

what is this from

[chilly ocean]

Idk someone took it and posted it on a Facebook group where I found it lol

[dull prawn]

just a sanity check: you don't need to be over a field or integral domain in order to say that an eigenvalue of a matrix satisfies its characteristic polynomial, right? This should be true over any commutative ring

[coarse stag]

Are functions that map Q(x)->Q(kx) automorphisms? Where Q(x) is the extension of the field of rationals by x.

[coarse stag]

in a group, if ab=cd does ba=dc?
Intuitively I feel like this follows from there being an isomorphism from the group of group actions with the group itself, but Im not sure.

[full panther]

im reviewing cycle notation and im getting confused lol

(1345)(234) = (135)(24) right?

[proud bear]

yes

[full panther]

Ok good

[hollow imp]

no, consider the dihedral group on a triangle, then let a = rotation, b = flip, c = ab, d = identity

[tropic spade]

"For A any set of points in the plane, (a,b) is constructible from A iff (a,0) and (0,b) are constructible from A." Would (0,0),(1,0) need to be an elt of A for this to be true?

(I'm not sure how relevant this is to algebra, it's from my algebra textbook. I can delete if it's too far from the channel topic)

[lavish nexus]

why does deg g >= r(q-1) follows?

|F|=q a field here

[unique berry]

I think these would be a lot easier if I understood all the terms about finite Abelian groups

like type and exponent

[next obsidian]

I think exponent might be the minimal number which satisfies g^n = e for all g in G

Maybe?

@unique berry

Exponent is what I said it is

For the type it’s almost surely the invariant factor decomposition from the classification of finite abelian groups

Type (n1,…,nt) almost surely means that G = Z/n1Z x … x Z/ntZ

Where n_i+1 divides n_i

[final pasture]

yeah

[final pasture]

and yeah

(I'm not sure if you were asking for confirmation but just in case.. )

[robust pollen]

No, depends on how you define composition of permutations. Most common would be right to left, in which case (1 3 4 5) (2 3 4) = (1 4 5) (2 3).

[full panther]

I must be mixing them up then, because I got (145)(23) for left to right 😅

[robust pollen]

Wait. Forget what I said, I meant common was left to right and what I wrote is the result of that

But I prefer right to left, because I don't like writing (x)f instead of f(x)

[full panther]

Alright

[robust pollen]

Who else writes f \square g for "f then g"? 🌜

[hidden haven]

I sometimes do, det does too 😌

Saketh as well 😌

[thorn flint]

What am I missing here?

[hidden haven]

Do you have any guesses for what the centre is? Any obvious elements that you know will be in the centre? Do you happen to know the centre of M_2(Z)?

The idea is to get an idea of what elements will definitely be there then to eliminate all other possibilities

[thorn flint]

So they want me to give specific elements instead of just a,b,c in Z?

[hidden haven]

No I want you to try and guess which matrices (elements of R) are in Z(R). Eg you know that the identity matrix commutes with everything therefore it's definitely there. Anything else that comes to mind?

[thorn flint]

Oh right

probably (1 1 0 2)

Or (2 2 0 2)

[hidden haven]

Why do you think they commute with all elements of R?

Here is a hint to try if you're stuck with the guessing part, but I'd recommend you try working it out with some trial and error first: ||consider the simplest elements of R, and get conditions for a matrix to commute with those. Prove that the conditions you get are also sufficient||

[thorn flint]

Okay give me a min, I'll work some out and guess

I've tried a few matrices and seems like everything works :/

[hidden haven]

That's weird, very few things actually work

Which matrices have you tried?

[thorn flint]

(1 1 0 2) ( 1 1 0 2)
or ( 1 1 0 2) ( 2 2 0 2)
or (1 1 0 2) ( 2 3 0 2)
or (0 0 0 2) ( 0 0 0 2)

Or is this not what I'm supposed to do?

[hidden haven]

Hmm interesting if those all commute, I see that the last one definitely does

But here's a tip

Try simpler matrices

Maybe make one matrix really simple and the other one have random a,b,c

And see what equations you get by the commutativity

[thorn flint]

Okay

[hidden haven]

And do this for various very simple matrices not just 1

By very simple I mean ||make some entries 0||

[thorn flint]

So like (1 0 0 2) (0 1 0 2)?

[hidden haven]

You can do simpler

[thorn flint]

(0 0 0 2) (0 1 0 2)

[hidden haven]

Sure

That's perfect

[thorn flint]

that one works

[hidden haven]

What do you mean by works?

[thorn flint]

I mean it commutes

[hidden haven]

They don't, the products are (0004) and (0204)

[thorn flint]

Oh oops yea

I was looking at the wrong matrix

[hidden haven]

lol F

[thorn flint]

sorry

haha

[hidden haven]

[hidden haven]

So now maybe try doing this for other simple matrices, and also this to get precise conditions

[thorn flint]

Okay I'll do that

So it seems like if there is some value in b, then it does not work

[hidden haven]

Yep

But you can say more

[thorn flint]

Hmmm should I keep trying simple matrices to observe that?

[hidden haven]

Yes, R has 3 "simple" matrices

Which are non zero in exactly one entry

Try multiplying those with a generic a,b,0,c matrix

You'll get conditions from all 3

Prove that a matrix satisfying those conditions commutes with everything

For this last step there's a neat way of doing this in a general case when you want to figure out if an element is in the centre of a matrix ring or not, I'd suggest looking at this after you've solved it
||Any matrix can be written as summation_{ij} e_{ij}, where e_ij is the matrix with 1 on slot i,j and 0 elsewhere (though here you will want e_22 to be (0002)). Use this to prove that a matrix M commutes with all other matrices iff it commutes with each e_ij||

[thorn flint]

Okay I have to go but thanks for the help! I'll look at multiplying that way and if I get stuck again, can I dm you? or should I just ask on here again?

[hidden haven]

Better here, you can just tag me

[thorn flint]

Sounds good

ty

[chilly ocean]

Let $G$ be a group, $R(G)$ the representation ring. How does a congucay class $\gamma$ in $G$ give a prime ideal $\mathfrak{p}$ in $R(G)$

[cloud walrus]

lime_soup

[hot lake]

if g is in G, you get a ring morphism from R(G) to C by taking a representation rho and looking at trace of rho(g). The map only depends on the conjugacy class of g and its kernel is a prime ideal of R(G)

[chilly ocean]

C is the field we take are taking out representatiosn over

then we take any representation?

oh no sorry

we have a map $R(G)\rightarrow \mathbb C$ defined by $\rho \in R(G)$ is mapped to $\text{Trace}(\rho(\gamma))$. This is a ring morphism

[cloud walrus]

lime_soup

[chilly ocean]

By property of the trace it doesn't depend on the choice of representative for the conjucacy class

[hot lake]

well you extend it to a ring morphism since not everything in R(G) comes from a representation

[chilly ocean]

so the kernel should be an ideal in R(G)

[hot lake]

C is an integral domain so the kernel is prime

[chilly ocean]

oh

I thought R(G) was by definition

representations of G over whatever field we pick

can you recommend some place to read about this

[hot lake]

that wouldn't have additive inverses

[chilly ocean]

i'm looking to understand equivariant k theory

[hot lake]

I know fulton harris talks about it a little

[chilly ocean]

representation theory a first course

okay thanks ill check it out

[hot lake]

also there are obvious size issues for whoever wants to be petty enough to point them out

[hidden haven]

umm that's a proper class tho?

[hot lake]

I think they had in mind finite dimensional representations only

which actually I could like go back to the beginning of the book and see

[hidden haven]

Is the field some particular field?

Because otherwise I think you'd still have too many reps

[hot lake]

C

[hidden haven]

ah ok

[hot lake]

and an isomorphism class of representations is probably a class but who cares

[oak grove]

can someone help me figure out what a quotient group is

looking at G/H

its just Z4 + Z4 again, right?

or does it do something special

[rustic crown]

what's the size of G/H?

let's narrow it down first

[oak grove]

16?

[rustic crown]

that's the size of G, right?

[oak grove]

[rustic crown]

Z4 has 4 elements, so G should have 4*4

[oak grove]

yea, but does H actually shrink it

[rustic crown]

yea, so do you know a theorem relating size of G, H and G/H?

[oak grove]

id have to search my notes 1s

[rustic crown]

okie

[oak grove]

oh is it just 4

but why 4

it doesnt seem like you should be losing any

[rustic crown]

Lagrange's theorem! It says that for a finite group |G/H| = [G:H] = |G|/|H|

[oak grove]

I guess im not seeing mechanically why you'd lose them I mean

[rustic crown]

um... one reason would be... why would you call it G/H is it had literally nothing to do with division 😛

[next obsidian]

You’re taking elements and making them the same so you lose stuff cuz what was once two elements is now one

[oak grove]

im having a hard time wrapping my head around these so im not sure what questions to ask, im getting a little closer each day though

[next obsidian]

Except two is the size of H

[oak grove]

two is the size of H?

[rustic crown]

maybe think it like this... elements of G/H are cosets... i'll use multiplicative notation so elements are gH where g varies over G. So it might seem that there are total |G| elements... but these gH are not distinct!!!

think it like this... when are aH and bH equal?

[next obsidian]

I mean you replace the number two with |H| in general

So in this case 4 elements become one

[oak grove]

when a and b are equal?

when they have the same elements

[rustic crown]

okie, so maybe let's go a step back. if h was an element of H, then what is hH?

[oak grove]

H

[rustic crown]

exactly!

see all those |H| elements of G are compressed into one single element of G/H... namely eH

this should motivate why sizes should go down

does that make sense?

[oak grove]

sort of? so youre saying since theres 4 elements of H its just each h in H operated on by G but then i guess uhh

i mean thats just the definition of it if that reduces the size why would it land at 4, arent a bunch of the elements the same then

like (0,2)G and (2,0)G are the same

so it seems like if it does go down to 4 it should go down further

[rustic crown]

so what i'm saying is that if aH = bH, then this should mean that any element on the left is on the right and vice versa. Check that this is equivalent to having a^-1*b in H

[oak grove]

but then all of our cosets are equal

[rustic crown]

what we want to do is look at distinct cosets of H

[hot lake]

there are 4 elements in every coset

and G has 16 elements

you should write them all down

and write all the cosets

[oak grove]

Ive wrote a couple, you just get G again but in a different order right?

but still all the same elements

[hot lake]

no ?

G is not a coset

[oak grove]

i mean if you have those 16 elements and tick one element of them forward by 2 it doesnt make you lose or gain any elements you just shuffle them around

right but every element in G is in (0,2) + G unless im totally misunderstanding what a coset is which is possible

[rustic crown]

okie lets take a smaller example? consider the group G = Z_6 = {0, 1, 2, 3, 4, 5} and the subgroup H = {0, 2, 4}.
there are two cosets of H...
0 + H = 2 + H = 4 + H = {0, 2, 4}
1 + H = 3 + H = 5 + H = {1, 3, 5}

[hot lake]

we are talking about cosets of H not of G

cosets are sets of them form xH

xH = {xh | h in H}

and G/H is the set of those cosets

[oak grove]

hmm so really i could just divide the two orders if i had to write it out

then just stop once you have 4 distinct sets

thats slow but faster than what i have been doing

sorry i just had the orders reversed because idk what im doing

that was a dumb mistake

so you end up with
0,0 1,0 2,0 3,0
0,2 1,2 2,2 3,2
0,1 1,1 2,1 3,1
2,3 3,3 0,3 1,3

[hot lake]

what's H again ?

[hot lake]

{(0,0) ; (1,0) ; (2,0) ; (3,0)} isn't an H-coset

[oak grove]

how do you know

[hot lake]

the coset containing (0,0) should be containing (0,0) + (0,2) because (0,2) is in H

[oak grove]

oh i drew my lines the wrong way

ok
0,0 0,2 2,0 2,2
1,0 1,2 3,0 3,2
0,1 0,3 2,1 2,3
1,1 1,3 3,1 3,3

[hot lake]

yeah

[oak grove]

then { (0,0)+H, (1,0)+H, (0,1)+H, and (1,1)+H}

ive asked before but how youd know its gonna look like that without writing anything down?

[hot lake]

well we know because we wrote it down

so is it isomorphic to Z4 or Z2 + Z2

[oak grove]

it looks cyclic?

i guess you just get a better feeling for it over time

im not sure about the isomorphism part id have to think about it more

i was just trying to figure out what it was

[hot lake]

so then you should do the addition table of G/H

[oak grove]

oh since some coset would have to generate all the other cosets and youre not talking about orders of each element anymore

[hot lake]

do you know what the group law on G/H is ?

how it is defined I mean

[oak grove]

like the operation?

[hot lake]

yes

[oak grove]

its just inherited from G right

component wise addition mod 4

[hot lake]

G's operation is component wise addition mod 4 yeah

[oak grove]

so its not cyclic

[hot lake]

G isn't cyclic

[oak grove]

well i guess you have weirder collapsing in the table

its more like mod 2

table of the cosets i mean

[hot lake]

well G/H is G mod H

[oak grove]

sorry this is so rough

[hot lake]

so it's like mod H

[oak grove]

oh

so it just happens that these numbers are iso to H and its not really mod 2 any more than H is

bleh i ordered this wrong

ill have to keep working after class

[hot lake]

yeah so G/H is isomorphic to Z/2Z * Z/2Z

[oak grove]

I will get there 😄 I appreciate your time 🙇‍♂️

[simple mulch]

,rotate

[simple mulch]

(5) the subset is {a^m : m in N} or?

[viscid pewter]

yes

[cloud walrus]

det

[strong valve]

$0 \notin \mathbb{N}$

[cloud walrus]

Master|Y|

[rustic crown]

I'll use my first sully on you

[brisk crescent]

I think I solved this isomorphism proof but wanted to get it double checked.

Prove that <Q, +> is not isomorphic to <Q+, *>

Suppose towards a contradiction that an isomorphism phi(q) exists where q in Q and phi(q) in Q+ . Choose phi(p) = 2.

Then, phi(p) = phi(p/2 + p/2) = phi(p/2) * phi(p/2) = phi(p/2)^2

This implies 2 = phi(p/2)^2 but this contradicts phi(p/2) being a positive rational number.

[stoic grotto]

Looks fine to me

[acoustic pine]

If R is a UFD, and r is an irreducible element, there is no ideal smaller than <r> smaller containing r right?

[rustic crown]

do you require UFD-ness?

[acoustic pine]

We don't require it but its a nice feature of the ring

[rustic crown]

if an ideal I contains r, then (r) is contained in I

[acoustic pine]

Oh its that easy huh lol

[rustic crown]

(r) may not be maximal

[acoustic pine]

r is irreducible

[rustic crown]

yea even in that case

consider k[x, y] and r = x

[strong valve]

bro like just say yes or no

[acoustic pine]

They're asking questions to make me think, don't bash them for helping, unless youre bashing me

[rustic crown]

this is pretty much what you're looking for ig...

[acoustic pine]

which is fine

[acoustic pine]

don't think we explicitly learned this, but it is sufficient

[strong valve]

I'm basing them not you

[rustic crown]

it follows straight from the definition...

r in I, then the ideal I absorbs multiplication by elements of the ring

[acoustic pine]

[rustic crown]

so rR = (r) is contained in I

[acoustic pine]

So we can at least say it is not contained in any SMALLER ideal, but a larger one, sure

which is sufficient for what I need

[rustic crown]

[acoustic pine]

I'm just trying to prove that a given ideal is the smallest ideal containing some affine subset over R

[strong valve]

How does it define transitive?

[plush quartz]

oh sorry hold on

the exact same thing as the bottom photo

nvm I'm a bit off

[strong valve]

uwu?

[rustic crown]

why did you summon me

why did you unsully me moldi

[strong valve]

Moldi has this god like presence which is only observed through emotes

[rustic crown]

my sully is back

[past temple]

is the fixed field of a finite galois group always algebraic?

if so, why

[thorn delta]

It’s true if G = Gal(F/K) is the Galois group of a finite galois extension. This is because if H is a subgroup of G then [H’:K] = [H’:G’] = [G:H] (and finite implies algebraic)

If F/K is not galois, then you could have G’ strictly bigger than K in which case [H’:G’] <= [H’:K] so idk

[tired granite]

What does the |Z C A} do in regards to this question? It doesn't really restrict what A can be right?

[chilly ocean]

it restricts A to those sets of rationals containing all the integers

[shell brook]

can every eleemnt of SU(2) be rewritten as $\begin{bmatrix} x & y \ {-\bar{y}} & \bar{x} \end{bmatrix}$ where $|x|^2 + |y|^2 = 1$?

[cloud walrus]

jesse

[shell brook]

Seems insane !

like i figure u need to do some funky stuff composing multiple of these or something?

[uncut girder]

Yes

[shell brook]

[uncut girder]

Wait

SU(1,1) i thought

[shell brook]

idk what SU(1,1) is

[uncut girder]

SU(1,1) is a bit different . Its exactly the same thing but the bottom left corner is y-bar instead of negative y-bar, and the condition is |x|^2 - |y|^2 =1 instead of |x|^2 + |y|^2 = 1

[shell brook]

sry this isn't a t/f question it is claimed to be true and i must prove it

i just wanted to make sure that he didnt make an error or something cuz lol

ah i see

[uncut girder]

I have a proof that the subgroup of SL(2,C) which preserves the unit disk is SU(1,1)

[shell brook]

this is for intro group theory

[uncut girder]

But maybe thats not relevant to what you're trying to do

[shell brook]

ys

i just dont really see how to show this

[uncut girder]

Have you tried multiplying the matrix with its conjugate transpose and seeing what relations you get?

Wait

[shell brook]

that should be identity no?

[uncut girder]

When you write |x| + |y|=1 do you mean |x|^2 +|y|^2=1

[shell brook]

yeah

sorry

whoops

[shell brook]

well. should be obvious from context pty

[uncut girder]

Take a general matrix (a b ; c d) and multiply it by its conjugate transpose

And see what relations you get by setting that equal to the identity

[shell brook]

ok

these do not seem ghelpful

[uncut girder]

What if you did the same thing with the inverse of (a b; c d)?

Its a group so its inverse must be in the group as well

And you get more relations between the coefficients from that

[shell brook]

i have lots of relations

now

[uncut girder]

At the end of the day you wanna conclude d = a-bar

and c = -b-bar

[shell brook]

what is -bar

[uncut girder]

Conjugate

[shell brook]

oh ok

ok

i see

[uncut girder]

Note d = d*1 = d(|a|^2 + |b|^2) = d( a•a-bar + b•b-bar) = ad•a-bar + bd•b-bar -bc•a-bar +bc•a-bar = (ad-bc)•a-bar + b(d•b-bar - c•a-bar) = a-bar

[shell brook]

oh wow

thats gnarly

[pallid ore]

Hi guys, i got a question, can someone explain how i show that both equations are equal?

(i think that i know how to resolve properly, but can't be that easy)

[chilly ocean]

so first you assume that ab = ba and try to prove (ab)^2 = a^2b^2

then you assume that (ab)^2 = a^2b^2

and prove that ab = ba

[shell brook]

is there a way to show this is a group hom without compooting

S(H) is group of quaternion z s.t. |z| = 1 under multiplciation

[chilly ocean]

if working in a ring ( not specified to have unit) and you have an expression ab + b, can you right factor out b? what does the second term become?

[shell brook]

maybe its literally not a group hom

its definitely an isomorphism but

[unreal portal]

I don't think so, being able to factor would imply that you could distribute back to get the original expression, but there's no guarantee to exist such an element e such that eb = b

[chilly ocean]

thats what im thinking too

so i guess you just can't undo the distribution there

[past temple]

i have the easy containment

but im completely lost as to how to show that for f(x,y) in the kernel

f is divisible by y^2 - x^2 - x^3

is there some kind of trick i need to use?

[next obsidian]

So

Generally I think the best way is to take an in the kernel then in K[x,y]/(y^2 - x^2 - x^3) show that its 0

This implies f is in the ideal we want to show is the kernel

The reason this is beneficial is we can replace any y^2 with x^2 + x^3

So you can write f like
f = yf1(x) + f2(x)

Because when you replace every y^2 with a polynomial in x the only power of y you can’t just rewrite in terms of x is the y^1 term

From here we know this still maps to 0 because we’ve only messed with f by something in (y^2 - x^2 - x^3)

Which we know maps to 0

So now you want to do some shenanigans to show this is divisible by y^2 - x^2 - x^3

It should be something like, note that under phi yf1(x), if it’s not 0, maps to something which has an odd power of T

Meanwhile f2(x) maps to something which has even powers of T

So for them to cancel both yf1(x) and f2(x) have to be 0

Or something like that

@past temple

This is generally how I do these problems involving polynomial rings and showing the kernel is some specific ideal

[past temple]

okay i think this makes sense

i get how in the quotient, any polynomial divisible by y^2

can be rewritten with x^2 + x^3

so u have the decomposition into f1 and f2 as u wrote

[next obsidian]

Yup

[past temple]

im a little confused by this line of reasoning though

[next obsidian]

Well

Y maps to T^3 - T

And every X maps to T^2 - 1

So when you have a polynomial only in X

The result is a polynomial in (T^2 - 1) and after you expand that out you only have even powers of T

So take that multiply by T^3 - T

Now every power of T is odd

Sorry, I should have wrote “has odd powers of T”

As in when you write it like a0 + a1T + … + anT^n

If k is even then ak = 0

[past temple]

ahhhh okay thats very clever

[next obsidian]

This is like

The only way I know how to solve these

The other strategy might be that some term is like degree < n

But for some reason the other side is >= n

So again, they can’t cancel each other out unless they’re both 0

Stuff like that, you just wanna see how you can rule out them cancelling each other

Cuz then it forces everything to be 0 to start with

[rustic crown]

You can use resultants to make stuff simpler ig... Like call that polynomial f and say g was something in the kernel, then you can kill y and get fu+gv=h where h is a polynomial purely of x and deg v < deg f... h must be in the kernel as well, but h clearly can't die unless it's 0. Since f is irreducible f divides gv means f divides g.

I learnt about resultants recently and they look very OP

[past temple]

wait but how can we conclude that yf1 and f2 are zero from the fact that their images under phi are zero?

[next obsidian]

Well cuz

Like

Just show that this is true

Like in particular look just at f2 rn

It’s a polynomial in x

Then look at the fact that if it’s degree n

Then phi(f2) has a single term anT^{2n}

Like the highest degree term will map to the highest degree term

And there’s no way for it to be cancelled

Do a similar thing for yf1(x)

The highest degree term is anT^{2n + 3} I think

And nothing can cancel that out

[past temple]

i get how theres no way it can be cancelled in their images under phi

but im still a bit confused on how this shows that they cant be cancelled

just as polynomials in x,y

[next obsidian]

Wdym?

The point is you conclude f1 = f2 = 0

Like because the image of yf1 and f2 can’t cancel each other out if the images are nonzero tells you that yf1 and f2 map to zero

But because f1 and f2 are only polynomial in x

you can show f1 and f2 have to be 0

You just look at what the highest degree term of the image or yf1 and f2 are

It will have the coefficient, corresponding to the highest degree term of f1 and f2

So if those are non-zero, then yf1 and f2 can’t map to 0

[past temple]

ahh okay that makes sense

wait

is it just like

trivial

that if the coefficient of the highest order term of a polynomial is zero

then the polynomial is zero?

[next obsidian]

yes

I mean if it isn't 0 there's a minimal one where it's non-zero

[sleek violet]

Would this be the correct channel for questions on mechanics?

[chilly ocean]

#groups-rings-fields? probably not

[upper cape]

If I have field extensions, L / K, and S / L. Does an embedding from K to S induce some embedding from L to S?

[chilly ocean]

My guess is no in general

[upper cape]

Or would it be possible to write an embedding from K to S, as some restriction of an embedding from L to S?

[next obsidian]

The answer is no

Take L/ K transcendental

And S/L algebraic

Something like R(x) / R vs C / R

[upper cape]

ahh

[next obsidian]

Oh wait

Once again the things are wrong

Umm

I misread it

But I think transcendence might be an issue time to look at the letters again kek

[upper cape]

If it makes it easier, restricting to algebraic extensions would be fine

[next obsidian]

Wait so…

Don’t you already have S / L?

[upper cape]

yup

[next obsidian]

So I guess you’re wondering if there’s an embedding of it which factors the one S / K

[upper cape]

yeah

[next obsidian]

I think it’s true for algebraic extensions

Like… I thought I remember there being something about being able to lift isomorphisms like… up a ladder

The reason is we have this kinda vertical ladder

S
L
K

And then this other embedding of K into S gives you something like this

S S
L
K -> K

Where the right is representing your new embedding of K into S

And the arrow K -> K is just an isomorphism

Then I swear there’s some theorem in Dummit and Foote which lets you like find a copy of L in between the S and K on the right

I feel it’s somewhere around where they prove the primitive element theorem or something

I think you needed algebraicity for this but you said you’re happy to assume that

[upper cape]

I'll take a look for it, thanks!

[simple mulch]

What do you think of the book Groups, Rings and Fields? I feel like the proofs use the conclusion to be made

[scarlet estuary]

author?

[upper cape]

If I take an embedding from a number field K into Q closure, and then compose it with an automorphism in Gal((Q closure)/Q ) would that still be an embedding from K into Q closure?

I think it is obvious, but I'm pretty new to Galois/field theory

[thorn delta]

every field homomorphism is an embedding so yes

[rustic crown]

yea it's correct... try to ask such stuff #point-set-topology next time

[simple mulch]

D.A.R Wallace

[void knot]

Is it true that if $C\subseteq X\times Y$ then there are $A\subset X$ and $B\subset Y$ such that $C=A\times B?$

[cloud walrus]

RaD0N

[viscid pewter]

no

for example consider Z2xZ2

ie. (0, 0), (0, 1), (1, 0), (1, 1)

consider the subgroup {(0, 0), (1, 1)}

[void knot]

Yes but my question is about the set not in sense of group

[viscid pewter]

oh

well it still works

[void knot]

Z2={0,1}

[viscid pewter]

yes

[void knot]

Ohh I see.

Thanks

I had that question because of an exercise of topology about disjoint union and the definition of subdirect product

[mortal elbow]

Last second line

@anyone?

[chilly ocean]

it's a group (H is, to be clear) so it contains the inverses of its elements

[mortal elbow]

Here in 4th part a belongs to Integer so -a is also an integer. But in above case how did you conclude -a and -b are inverse of a and b ?

[viscid pewter]

??

a + -a = -a + a = 0

[chilly ocean]

-a literally means the inverse of a

it's the notation being used in the first image for the inverse of an element of an arbitrary group

[mortal elbow]

Ok.

[lunar crater]

What is the best method for defining/notating a group

I know a bit on how they are supposed to be written, but I don't think I fully understand it.

[sharp sonnet]

a group is a groupoid with a single object

[lunar crater]

Hm?

When writing/defining/notating a group should I write
G = {contents}﹡
Or
Gₓ﹡
Maybe G = {contents}ₓ﹡
Or perhaps I should write out the table itself

[sharp sonnet]

wtf is ﹡

[lunar crater]

I was taught to use the ﹡ for the operation the group is under

[sharp sonnet]

[lunar crater]

If it is not specified

[sharp sonnet]

we just write "Let G be a group"

[hidden haven]

Usually you write (G, *) where G is the underlying set and * is the operation

[sharp sonnet]

if the group operation is not clear, then what moldi said

[lunar crater]

Alright

[simple mulch]

Ok so I have a question

Consider D_4, the set of all symmetries of a square

Now let Z_2 = {0,1}

why is it that the homomorphism g : D_4 -> Z_2 is g(x+y) = g(x) + g(y) for all x,y in D_4? It is because multiplication multiplication doesn't make {0,1} a group. right?

unless we consider the set {-1,1}, then g(xy) = g(x)g(y) because multiplication would be an operation on {-1,1} ?

[viscid pewter]

why is it that the homomorphism g : D_4 -> Z_2 is g(x+y) = g(x) + g(y) for all x,y in D_4?
you shouldn't use + in both cases, to avoid confusion

let's say the operation in D_4 is idk "

[small bison]

• 🤢

[viscid pewter]

then any homomorphism g from D_4 to Z_2 must satisfy g(x"y) = g(x) + g(y)

[hidden haven]

" 🤮

[viscid pewter]

look, at least it's symmetric

[viscid pewter]

i've used $ in the past

and £

[hidden haven]

did asterisks hurt you

[viscid pewter]

i don't like having to backslash them to avoid italicism

[rustic crown]

hey moldi

had a quick question

[viscid pewter]

basically yes

Z_2's operation is +

not *

[rustic crown]

if R is a ring, then "how many" non-isomorphic simple R-modules are there?

[viscid pewter]

5

[rustic crown]

a sets worth or need proper classes?

[hidden haven]

They should all be quotients of R right?

[simple mulch]

I see

[rustic crown]

F lol

right

[hidden haven]

because they are cyclic

[rustic crown]

also |2^A| = |2^B| doesn't easily imply that |A| = |B| right

[viscid pewter]

... i feel like it should

[hidden haven]

nope

[viscid pewter]

rip

[hidden haven]

it is independent of ZFC

See Martin's axiom

[rustic crown]

is there some intuitive way to understand that?

[hidden haven]

There may be cardinals between omega and 2^omega, whose power set is 2^omega too

ZFC+GCH proves it though

[viscid pewter]

oh what

[hidden haven]

MA says that this is the case (so you usually assume MA with not GCH)

[viscid pewter]

oh wait i was misreading it

'there may be cardinals between omega and 2^omega such that these cardinals have power set 2^omega'

that's what you meant, ok

[hidden haven]

yeah

[sinful mirage]

why are left and right actions equivalent?

[hidden haven]

Left action of G is the same as right action of G^op, where G^op is G but with opposite multiplication

[sinful mirage]

this is what we wrote in the lecture too

but i can't see why

[hidden haven]

xg := g^-1 x

I guess that works too

Because it reverses multiplication order in the group

G^op is like
a (•^op) b = b • a

Try and see why defining xg := gx fails

That won't be a group action necessarily

[sinful mirage]

yes the property of (gh)x=g(hx) will fail

[hidden haven]

And then it should be clear why reversing multiplication fixes things

Right

[sinful mirage]

cause of composiiton

[sinful mirage]

but still,i don't see why this would turn a leftt action into a right aciton

lets say gx is my left action

then how is g^{-1}x a right action?

or do we actually mean, we DEFINE a new right action, in terms of the old lef taciton?

xg:=g^{-1}x, where g^{-1}x is a left action

so every left action gives rise to another right action?

my issue is that a left action can't be a right action at the same itme so i'm pretty confused with terminology

[sinful mirage]

an action is either a left action or a right action,no?

can't be both

[hidden haven]

Yes

Yeah it's not both

It's a bijective correspondence

[sinful mirage]

right,so every left action produces a right aciton and vice versa

this makes sense

[hidden haven]

Yes

[sinful mirage]

here does the g have to be fixed?

or not necessarily

i.e. what I mean is, do we look at a family of maps,or the action itself? or can look at boht?

both*

GxX->X is not the same as gxX->X is what I mean

is there a correspondence both on the level of actions and on the level of |G| many induced maps?

[hidden haven]

It should be a correspondence between maps g x X → X that extends to left G actions and the other thing I guess

Oh right also all maps g x X → X and X x g → X ok

But then it's really just a bijection whereas in the other case it's something much nicer

It's actually an isomorphism of categories of left G actions and of right G actions

Maybe anti isomorphism

So it has a lot more structure in that sense

[sinful mirage]

yep,makes sense

thanks

also:a quick sanity check

,rotate

[viscid pewter]

tf is an anti-isomorphism

[sinful mirage]

is this ok? I wanted to clarify notation.

i hate writing dots withou being explicit

[hidden haven]

Yes looks fine

[hidden haven]

Isomorphism that reverses arrows

[viscid pewter]

yeah i'm not actually a feline afficionado

i don't know cat stuff

[hidden haven]

You can also talk about anti isomorphisms of groups

[viscid pewter]

please do

[hidden haven]

An anti isomorphism G → H is an isomorphism G^op → H

[viscid pewter]

fuck

what's op and what's ^

[hidden haven]

[hidden haven]

This

You just reverse the multiplication to get another group

[viscid pewter]

oh

[hidden haven]

So it does nothing to abelian groups

[viscid pewter]

wait woah

ok so

basically it's an isomorphism from G to G^op?

if you have G -> H and G^op -> H

[hidden haven]

G → H you don't have an isomorphism

[viscid pewter]

oh

[hidden haven]

An isomorphism G^op → H is called an anti isomorphism between G and H

[viscid pewter]

ok

so

G and G^op aren't necessarily isomorphic

i really thought they would be ngl

[hidden haven]

Ye they're just very similar in an obvious way

And that similarity is what anti isomorphisms capture

[viscid pewter]

can you give me an example where they're distinct

[hidden haven]

Of course G and G^op are antiisomorphic

[viscid pewter]

can you give me a G where G^op is not isomorphic to G

[hidden haven]

let me think

[viscid pewter]

aw is this gonna be some awful infinite bs group

[hidden haven]

Man it'd be embarassing if they're always isomorphic They're not for monoids

[viscid pewter]

oh well

monoids suck

[hidden haven]

Wait they are always isomorphic lol

[viscid pewter]

yeah thought so lol

[hidden haven]

g ↦ g inverse

Monoids are very useful lol

But you can also talk about anti isomorphisms of rings

[viscid pewter]

yes but they're so weaaak

[hidden haven]

[viscid pewter]

??

[hidden haven]

You can do the same thing to get R^op for a ring R

Reverse mult

[viscid pewter]

oh

hang on

oh because it reverses multiplication but has no effect on addition?? or

[hidden haven]

Then left R-modules become right R^op-modules

[viscid pewter]

wait no

it should still be an automorphism

[hidden haven]

Well you're defining a thing so you can just say that it doesn't affect addition

[viscid pewter]

no but addition in rings is commutative

anyway

so it couldn't

ok wait no

i feel like even for rings it's an isomorphism

[hidden haven]

Yes but we're not using that fact

We are just saying take a new ring R^op with same addition but reversed mult

[viscid pewter]

well i just meant that whether or not you were reversing both operations or just multiplication, it'd look the same

[hidden haven]

Ah yeah

[viscid pewter]

but anyway is it not also an isomorphism for rings as well

[sinful mirage]

@hidden haven actually doing the proof clarified exactly why it must be like that

[hidden haven]

Nope I don't think so

[sinful mirage]

the first axiom of right action gest violated if I don't use ^{-1} when (gh)^{-1}=h^{-1} g^{-1}

[viscid pewter]

can you give me an example

[hidden haven]

Otherwise there won't be so much stuff surrounding it 😵‍💫 there's like a wikipedia page for it

I'll have to think lol

[hidden haven]

Yep

[viscid pewter]

this is the whole wikipedia entry

[hidden haven]

See "Opposite ring"

[viscid pewter]

ok

[hidden haven]

There's also an opposite group page, but that says that G is naturally isomorphic to G^op while the ring page doesn't

[viscid pewter]

aw come on what are these examples

[hidden haven]

https://math.stackexchange.com/questions/45085/an-example-of-a-division-ring-d-that-is-not-isomorphic-to-its-opposite-rin

[viscid pewter]

what is that

[hidden haven]

Apparently there's even a division ring counterexample

Lol what's wrong with the examples

[viscid pewter]

there has to be a simpler thing

what's the core of it

[hidden haven]

https://mathoverflow.net/questions/64370/simplest-examples-of-rings-that-are-not-isomorphic-to-their-opposites

[viscid pewter]

look you can't just spring something on me when it has things like 'relative norm map' and 'cyclotomic field' and 'primitive ninth root of unity'

[hidden haven]

You need a non commutative ring The simplest examples I know are matrix rings, operator algebras, quaternions and free algebras

[viscid pewter]

i could chug through a matrix ring, ig

ok well that's a fun time

ty

[sinful mirage]

what does the diagram mean?

i'm confused by it,since we never defined 'the canonical projection' for example

i don't see how orbist are naturally identified with cosets

[viscid pewter]

i think cosets are like. the image of the subgroup under the group action

ie. a coset is like the slices of the orbits of the elements of the subgroup? so that's what springs to mind

canonical projection sends all the elements to their respective coset

[tawny pine]

if H is a subgroup of G then we can define a group action of G on H via the group operation

[sinful mirage]

wait,to clarify notation here, the orbit (image) of o_x is Gx:={gx|g in G}?

[tawny pine]

upon doing so, orbits are precisely H-cosets

[sinful mirage]

where do we have a subgroup here?

[viscid pewter]

??

you have a subgroup H in G

i think in your pic it's Gx??

[tawny pine]

it's given "H<G"

[viscid pewter]

Gx is a subgroup

[sinful mirage]

we have a point x in X and its orbit o_x:=gx

[viscid pewter]

wait

no i'm a fool

[sinful mirage]

is Gx:=gx|g in G?

[viscid pewter]

wait

no yes

it's a subgroup

ok

[viscid pewter]

i presume so

[sinful mirage]

how can this be a subgroup of G? would no tmake any sense

[viscid pewter]

since you're talking about orbits

[viscid pewter]

why not

[sinful mirage]

g applied to x is an element in X

by the def of left action

[viscid pewter]

what's X

[sinful mirage]

a set

let G be a group X a set

[viscid pewter]

the fundamental set?

[sinful mirage]

and hten we have a left action gx

[viscid pewter]

or what's it

[sinful mirage]

what do you mean by the fundamental set?

the set which upon the group acs

[viscid pewter]

what's the name

[sinful mirage]

acts*

i don't think it has a name

[viscid pewter]

underlying set

there we go

[sinful mirage]

yes

[viscid pewter]

right

if gx is in X then gx is in G

like

we can just say informally

if an element is in X it's in G ok

[sinful mirage]

how?

it's eiher in X or G

[viscid pewter]

no

both

we can just say this

and it all works out

[sinful mirage]

how? this would contradict the definition of left action

[viscid pewter]

why

[sinful mirage]

it can't be an element of G

[viscid pewter]

??

why not

[sinful mirage]

Let X be a set and G be a group. A left action is a map GxX->X, which satisfies: 1) g(hx)=(gh)x for all g,h in G x in X
2) ex=x for all x in X.

[viscid pewter]

right

so let G now act on itself

the left action is GxG -> G

this is a common thing we like to do

[sinful mirage]

I don't think that was assumed here

[viscid pewter]

i think it is

otherwise there's just a whole bunch of random information

nothing tying it together

[sinful mirage]

it even says Gx is a subset of X

[viscid pewter]

so?

[sinful mirage]

so Gx can' have elements in G

[viscid pewter]

if G = X, then Gg is a subset of G

which checks out

[sinful mirage]

yes,but this should (somehow) be more general

[viscid pewter]

what do you mean

[sinful mirage]

I mean prof would have specified X=G if it only worked like that

[viscid pewter]

i think it's implicit

i mean it all works, right

assuming X = G

then G acts very obviously on like G/H

[hidden haven]

Are you proving that an orbit looks like the set of coset of a subgroup?

That doesn't require action of the group on itself

Can be on any set

[sinful mirage]

trying ot prove that there's a 1-1 correspondence between cosest and orbits yes

but my prof just sketched a diagram and i don't understand

he said the commutativity of the diagram is the proof itself

i'm mega confused

[hidden haven]

Ok so have you seen maps of G-sets/equivariant maps at all?

[viscid pewter]

arrows 😨

[sinful mirage]

no

this is a manifolds course

[hidden haven]

ok hmm

[sinful mirage]

and prof said we review undergrad algebra and analysis firs tweeks

[hidden haven]

So is this just a bijection between the sets of cosets and the orbit

[sinful mirage]

and he defined tensor product via universal property

I wrapped my head around it,but he's pretty cat-theoristy 😦

which is the explicit bijection?

say,I have a point x in X. fix it. then we have ist orbist

Gx, which is a subset of X

[hidden haven]

I don't think you can do better than the universal property for the tensor product, the construction is very bad

[sinful mirage]

this is the top part of the diagram

what is the bottom par?

[hidden haven]

I mean if he is just proving bijection then this is fine, but this is actually more that's why I'm confirming

It's an isomorphism of left G actions

[hidden haven]

You know how to quotient a set by an equivalence relation right?

[sinful mirage]

yes,but I have no idea where here should be an equiv relation

he didn't say it

[hidden haven]

He is just doing that, the equivalence relation is "being in the same coset of G_x"

[viscid pewter]

if they're in the same coset

[hidden haven]

Similar to group quotients

[viscid pewter]

partitions are eq. relations, the cosets are a partition

[hidden haven]

G/G_x is shorthand for G/~ where ~ is the being in the same coset relation

[sinful mirage]

G_x meaning?

the stabilizer?

[hidden haven]

Stabilizer of x

Yeah, that's a subgroup (not necessarily normal)

So this quotient won't be a group

[sinful mirage]

wait so there's wrong notaion here

[hidden haven]

But it's a set

[sinful mirage]

Gx up is not same as down

righT?

up it's the action of G on x,down it's the stabilizer

[hidden haven]

G_x and Gx oof

One is subscript and the other isn't

Gx is orbit, G_x is stabilizer

Bad notation

[sinful mirage]

yes

[hidden haven]

So G → Gx is the map g ↦ gx

G → G/G_x is g ↦ [g]

The map in the middle is induced by the first isomorphism theorem for sets

You might have seen it under different/no names

[sinful mirage]

this one?

[hidden haven]

That's the first isomorphism theorem for groups

But same idea as first isomorphism theorems elsewhere. If f: X → Y is a function, then it induces a surjection
f: X → im f
And an injection
f' : X/~ → Y
Both of these should be extremely obvious, as long as you can convert from symbols to like actual language
Combining these you get an induced bijection, f': X/~ → im f

In the first case we throw out things in the codomain that are not hit by f
In the second case we identify things in the domain that map to the same thing

[sinful mirage]

correc,this all makes sense

[hidden haven]

Perfect

So notice that the ~ in the above process

[sinful mirage]

for these sest

but I can' see how the stabilizer comes into play

[hidden haven]

Is the same as the relation you quotient by, in G/G_x

You'll have to prove that

That g and h map to the same thing iff they're in the same coset of stab

And that directly gives you the induced isomorphism between Gx and G/G_x by the above theorem

[sinful mirage]

so in this case,he canonical projection would be the injection?

[hidden haven]

The canonical projection is g maps to equivalence class of g

[sinful mirage]

the*

[hidden haven]

It's not injective unless G_x is trivial

The injection is the diagonal upward map

Quotient to image

Here things are made easier because image = codomain

But for now you can maybe just remove the left vertical arrow and think how the diagonal map is induced

The left vertical arrow and commutativity can come later

(though that's how you should eventually start thinking)

[sinful mirage]

just to make sure:quotienting here means that I get a set of equivalence classes,correct?

[hidden haven]

Yes

I did a whole section on the first isomorphism theorem but didn't post the sticker once I'm ashamed

[sinful mirage]

what's the difference beween doing quotient wrt equiv relaiton and partitioning G into cosets?

[hidden haven]

They're the same, a partition is the same as an equivalence relation

Partitions give equivalence relations by "being in the same part"

[sinful mirage]

well,the map g->g G_x is clearly injective for each x

is this what they mean?

[hidden haven]

Equivalence relation partition into a set of equiv classes

[hidden haven]

It's not though, G_x could very well be the whole set in which case gG_x = G for all g

When you have the trivial action, gx = x

Everything is a stabilizer

[sinful mirage]

ahh

[hidden haven]

So the map G → Gx is surjective, and all they're saying is, if you identify things that map to the same thing, then it's a bijection

[sinful mirage]

right,I see this par

[hidden haven]

The only thing to prove here is G/G_x identifies exactly the things that map to the same thing along the horizontal map

[sinful mirage]

I don't see how is this related at all to stabilizers tho

[hidden haven]

So you have to prove that g and h are in the same coset iff gx = hx (images of g and h along the horizontal map are equal)

[sinful mirage]

this I've done,but I don't see how is related here

this is part of Lagrange's theorem proof

when you partition G into coses

[hidden haven]

This is what links the 2

[sinful mirage]

cosets

[hidden haven]

So G → Gx
induces injection
G/~ → Gx
and we are saying
G/~ = G/G_x

[sinful mirage]

what we are saying is the consequence of first iso thm