groups-rings-fields

[oblique river]

yes

if you're cyclic then you have a generator

which generates the ideal

[solar wyvern]

so this maybe sorta dumb, but have you ever worked with anything where your ideals are (order theoretic) ideals as well, and it actually matters?

[oblique river]

(sry i'll get back to you in like 10min)

so

i dont know what an order theoretic ideal is

which I guess means the answer is no haha

[solar wyvern]

they're dual to filters

[oblique river]

uhhh

dont remember that either haha

yeah i don't engage with set theory in that way ever

[solar wyvern]

nonempty directed and downwards closed

yeah didn't think it'd be that common lol

apparently there are like boolean rings (where the two ideal notions coincide I think?)

[oblique river]

the thing is, most people doing number theory really don't care about orders

and like our rings aren't ordered

which makes things an issue

if you wanna talk about order theoretic ideals

[solar wyvern]

TIL number theorists have their own rings

[oblique river]

what do you mean?

[solar wyvern]

nothing just joking

[oblique river]

oh lol my b haha

[solar wyvern]

(whats your favorite ring/what's a neat ring u work over)

[oblique river]

hmmm

I feel like my rings tend to come in two flavors. there's rings of integers of number fields, like Z[i], Z[zeta] where zeta is some root of unity, things like that, along with their local versions, like the p-adic integers and its extensions

and rings related to representations and group theory, things like Z[G] where G is some finite (usually abelian) group

or F_p[G]

I spent a long time trying to understand the ring F_p[Z/pZ x Z/pZ] which is honestly an absolutely disgusting ring hahaha

it's a finite ring but it's module theory is just nasty

Idk if I can pick a single favorite ring tho

it would probably be something like Z[i] just because the arithmetic of Z[i] was what first got me into number theory

I find it interesting that things that are all rings can still feel so different. like I really don't think of Z[zeta] and Z[G] in the same way even though they're both rings that are admittedly not too dissimilar from each other. If G = Z/pZ and zeta is a pth root of unity, if you tensor with the rationals, you get Q(zeta) and Q[G], which decomposes as a product of rings into Q(zeta) and Q

but the fact that you don't have such a decomposition integrally is very interesting and a pretty important fact!

do you have a favorite ring? @solar wyvern

[solar wyvern]

by Z/pZ do you mean like as a group, that is C_p?

[oblique river]

yeah

sorry sometimes I conflate the ring and the underlying additive group

but some people use the convention that C_p is the group and Z/pZ is the ring

[solar wyvern]

i get confuse easily so I use separate notation for the group and ring/field

also easier to write than like (Z/pZ)^\ast

hmmm

[oblique river]

what do you mean?

(Z/pZ)* is a different group

[solar wyvern]

oh nvm

that's Aut(C_p)

[oblique river]

yeah

for me, I kinda just write Z/pZ for it, and if I need to specify or emphasize that it's a field, I would write F_p

[solar wyvern]

monoid/group rings seem interesting, it's a little surprising at surface level how little is known about them

[oblique river]

yeah

[solar wyvern]

was interested in seeing what rings you could get as endomorphisms/automorphisms of (finite) groups

actually someone linked me a paper on the basic stuff which I should look at

[oblique river]

I think that's known?

or maybe not

I think "automorphisms of p groups" are vaguely understood

[solar wyvern]

for stuff like Aut(C_p \times C_q) p,q coprime primes

i think those are ez actually

[oblique river]

oh, that's just Aut(Cp) x Aut(Cq)

[solar wyvern]

yeah 😄

[oblique river]

basically cuz they're coprime and commute they can't interact with each other in any way

[solar wyvern]

if it's like a semidirect product might not be as obvious?

[oblique river]

yeah

[solar wyvern]

or if not coprime

[oblique river]

well if they're not coprime

you can always rewrite it as a sum of coprime things

[solar wyvern]

although I think stuff might be known for like Aut(C_p x C_p) like you mentioned earlier?

[oblique river]

oh yeah I forgot about that group

I mean that's just GL_2(F_p)

[solar wyvern]

oh yeah

[oblique river]

automorphisms of Cp as a group are gonna end up being the same as automorphisms of Fp as a ring

wait no that's literally dumb

let me try again:

automorphisms of Cp as a group are gonna end up being the same as automorphisms of Fp as a 1-dim vector space over itself

there :)

[solar wyvern]

😃

hmm

[oblique river]

in the same way that "abelian groups are Z-modules", "abelian groups of prime exponent are Fp-modules" is true

[solar wyvern]

I'm still fairly new to the sheaf game but trying to understand stuff like ring of smooth functions on a open set

[oblique river]

so in that case you can just use what linear algebra you know

here is my favorite weird automorphism fact:

Aut(S_n) = S_n... unless...

[solar wyvern]

2,6

[oblique river]

n = 2, which is obvious because Aut(S2) = Aut(C2) = 1

or 6

because of reasons

:(

u got me

[solar wyvern]

6 is more interesting tho

[oblique river]

Aut(S6) isn't too much bigger than S6 though

the outer automorphism group is Z/2Z

[solar wyvern]

it's like twice as big

i think

[oblique river]

so basically you have the automorphisms you expect from a regular S_n

and then one more

yeah

[solar wyvern]

yeah, that's a neat one

oh

there's an automorphism tower thm which I glanced over in rotman but didn't prove/read proof

[oblique river]

like, automorphism tower as in Aut(Aut(Aut(...(G))))

[solar wyvern]

yeah

think it stablizes eventually a lot of the time

[oblique river]

it always stabilizes

transfinitely at least

If you define "stabilizes" in the right way, then D_8 is an example of a group that doesn't stabilize until the omega+1 step

[solar wyvern]

oh yeah

[oblique river]

http://jdh.hamkins.org/everygroup/

[solar wyvern]

Wielandt proved, for every finite centerless group G, that this tower is constant from some point on. Since the last term of an automorphism tower is a complete group, it follows that every finite centerless group can be imbedded is a complete group.

so maybe not "a lot of the time" lol, but for this in finite steps

[oblique river]

what do you mean?

it always stabilizes transfinitely but i'm not sure if it's expected to actually happen finitely

often

[solar wyvern]

that's why I amended my statement

since there are groups which are not finite or centerless

[oblique river]

haha yes

[solar wyvern]

it's not obvious that it'd terminate finitely even in that case tho right?

[oblique river]

uhhh, not sure

[solar wyvern]

are the endo/autos of a top group a topological ring--is that a thing?

[oblique river]

if X and Y are topological spaces then you can always put a topology on C(X,Y)

so that should work for topological groups

you'd probably use the compact-open topology

[solar wyvern]

guess you could probably ask similar q about algebraic groups

(of which I know nothing of)

[oblique river]

if endos of an algebraic group was an algebraic group?

I'm not sure about that one

[solar wyvern]

@bleak abyss

[bleak abyss]

Fam why do you always ping me?

I'm kinda busy

Also lol Bananas is far more likely to know about this than me

[oblique river]

maybe I was a bit ambiguous

when I said "I'm not sure about that" I didnt mean it like "I don't know", it was more of a "yeah I dont think that's gonna be true"

like I just don't know really any interesting nontrivial instances where homs between varieties is still a variety

[solar wyvern]

convolution product is
$$(f \ast g)(x) := \int_{\mathbf R} f(x-y)g(y)dy$$
or (according to lang) for $G$ a locally compact (abelian?) group with haar measure $\mu$
$$(f \ast g)(x) := \int_{G} f(xy^{-1})g(y)d{\mu}(y)$$

@gentle pendant

[cloud walrus]

flimflam:

[solar wyvern]

i replied as soon as i saw ur message

[gentle pendant]

Am well aware of what a convolution is lol, just you were talking about smooth functions be invertible w.r.t it. How can you expect this to make sense given that there is no smooth identity for this operation? (Dirac delta is the identity if you expand your space to distributions).

[solar wyvern]

oh darn 😦

[gentle pendant]

Talk about the identity before you talk about invertibility.

[solar wyvern]

umm, so you can have identity if you expand to distributions?

but not over like L^1 or something?

tbh still not entirely sure what convolution does

[gentle pendant]

Yes but then it gets subtle to make sense of what convolving means when you are working with distributions. For which pairs of distributions can we take their convolution? Etc, such questions require some work.

[solar wyvern]

are they questions worth asking at some point tho

[gentle pendant]

There are explanations on stackexchange with pretty pictures. When you are working with functions you can think of it as a weighted average.

Depends on what you want to do.

[solar wyvern]

also do you need your locally compact group to be abelian here? I'd imagine so but lang doesn't explicitly mention

[gentle pendant]

Do you need abelian-ness for what?

[solar wyvern]

for the latter def of convolution

or will it just not be commutative then

[gentle pendant]

I don't see why. What part of the definition do you think does not make sense for things that are not abelian?

[solar wyvern]

🤷

[gentle pendant]

Then try to figure out for yourself what properties you would lose if you make this definition for non-abelian G.

[solar wyvern]

not sure if being tricked into doing analysis

[gentle pendant]

The definition itself in no way depends on G being abelian.

I mean don't do it if you aren't interested in the answer, but since you asked me I figured you were.

[solar wyvern]

i will check it out

[gentle pendant]

I just get the sense that your "answering your own questions" / "asking further questions" ratio is lower than it should be.

have to build these things up otherwise yeah its natural that you will be groping in the dark with the higher concepts.

[high bobcat]

can anyone solve?

i have a hard time with this

[stoic rose]

1.10 or 1.4 ?

[high bobcat]

ok thx

[stoic rose]

What?

[high bobcat]

wdy mean

[stoic rose]

Do you need help for problem 1.10 or problem 1.4?

[high bobcat]

1.4

[stoic rose]

For 1.4 you just need to check the axioms of a field

[high bobcat]

oh

[solar wyvern]

skew field = field except multiplication not necessarily commutative

[stoic rose]

The most difficult is probably proving existence of multiplicative inverses, but iI don't think it's very difficult

You'll have to solve a system of 4 equations

[covert vector]

the inverse is just the conjugate divided by norm squared

that is,

$(a+(bi+cj+dk))^{-1} = \frac{a-(bi+cj+dk)}{a^2+b^2+c^2+d^2}$

[cloud walrus]

woog:

[covert vector]

wait does this work? 🤔

something like that

a²-(bi+cj+dk)²
= a²+b²+c²+d² - (bcij+bdik+cbji+cdjk+dbki+dckj)
= a²+b²+c²+d² - bc(ij+ji) - bd(ik+ki) - cd(jk+kj)
= a²+b²+c²+d²

so yeah that's the inverse

[solar wyvern]

@spark plank did u have any group questions tho
the i,j,k's above form quaternion group

[spark plank]

plz no

[chilly ocean]

Is the center of SL(3, R) {E} or {-E, E}

Because i wrote E and I'm worried af rn

One such little mistake and I ain't getting the excellent mark

Or they ask me a question about factor groups and I fail

[chilly ocean]

Nevermind

Excellent mark

😎

[random crag]

what a kewl kid

[covert vector]

{±E}

[cedar gate]

I have a few questions about Corollary 1.9 if anyone could help out

1. Why does it follow that m is contained in m_p?
2. Why is this correspondence 1-1? It seems like picking any point in Z(m) gives the same maximal ideal?

[chilly ocean]

eisenbuddy

[solar wyvern]

what m

just a max ideal?

[chilly ocean]

ye

m_p:= (x_1-a_1,...,x_n-a_n)/m (a_1,...,a_n) in Z(m)

Show that m is in m_p, m being maximal of A^n

[solar wyvern]

what's A(X)

[chilly ocean]

Algebraic set of X

ie. hol' up

Let me get my eisenbuddy

Right

Sorry

Coordinate ring

A(X) = k[x_1,...,x_n]/I(X)

Where k is field

[stone fulcrum]

Still fighting learning commutative algebra. How'd you learn Jicheal?

[chilly ocean]

I(X) = set of all polynomials in k[x...] that have all X as roots

So a polynomial in I(X) must have every x in X as a root

This is why algebraic geometry is heck

@stone fulcrum I didn't kek, bouncing between Eisenbud and the CRing project

[stone fulcrum]

Hmm. Good luck, I'm still looking at modules

[chilly ocean]

So in this case - they've decided to just mke A(X) the affine n-space over k

Or just direct product of n set of k's

So what I think Eisenbuddy et. al is trying to communicate is that you'll get a quotient ring: (x_1-a_1,...,x_n-a_n)/m and you can show that m is in this quotient.

And since maximal contained in non-maximal => non-maximal is maximal, they must be equal.

But the former step is what I'm struggling with.

[cedar gate]

Right, I get why once we have m contained in m_p they must be equal, but this containment is not clear to me

I've tried considering the homogeneous components and proceeding by induction on the degree of f (for f in m), but i couldn't work it out

[chilly ocean]

So I boiled down the following:

1. k[x.]/m_p = k (from pretext)
2. m_p_2 = m_p/I(X) (from Corr. 1.9)
3. A(X) = k[x.]/I(X) (from defn. of coordinate ring)

Applying third isomorphism theorem we get:

A(X)/m_p_2 ~= k[x.]/m_p = k

So since both are isomorphic to a field, m_p_2 must be maximal in A(X).

@cedar gate

Not quite the containment - but it demonstrates the property.

x. = x1,...,xn

The one-to-one correspondance arises from the bijection.

Good. Exercise. Thanks.

[covert vector]

yes.

both what

you didn't say a second thing

oh

yes

it's normal

it's a pretty easy to follow proof, I can give it right now if you want

ok

let f:G→H be a group homomorphism

which means for any x,y in G, f(xy)=f(x)f(y)

let ker(f) = {x \in G | f(x) = 1}
where 1 is the identity element

now suppose f(x) = 1
now take any other element g in G,

let's conjugate x by it, and see what we get

if that's also in ker(f), then ker(f) is normal

that's by definition

yeah let's use something like g' to make it smaller

g' = g^-1

and f(g') = f(g)'

inverse of f(g)

yes

yes

so g'xg is in ker(f)

for any x in ker(f), for any g in G

what

if I say extra stuff that seems confusing, then ignore it

np

[errant drum]

You can kind of see it had to be normal because the 1st isomorphism theorem is quotient group of G and ker(f) and you need normality for quotienting in the first place

[covert vector]

well, you learn about kernels before you learn about isomorphism theorems

[solar wyvern]

@chilly ocean I wonder if group homos have some property about certain things always being in the kernel

no, just a basic property

can the kernel ever be empty?

@chilly ocean

the kernel is things which map to id

[chilly ocean]

Empty kernel

EMPTY KERNEL

[solar wyvern]

what maps to id

it rhymes with id

yes

this is by definition

[chilly ocean]

Try having empty kernel: see what do. O no das no grup, proof by contrapterodactyl

[solar wyvern]

t!wiki group homomorphism

[fossil mango]

📖 | ** https://en.wikipedia.org/wiki/Group_homomorphism **

[chilly ocean]

If the id of the original group is not in the kernel then you generally lose the structure.

You can try substituting id for something else and see what happens

Well yeah. As a direct consequence of the definition of a homorphism

Homomomomorphism

[solar wyvern]

no?

map everything to id

this a gp hom

[chilly ocean]

Monomorphism just means an injective homo

😏

So yeah

Not every homomorphism is injective

What was your question again?

Show the kernel is a singleton?

[worthy kindle]

oh this one is a classic I believe

G and H are groups I presume?

I don't really get what you've done tbh

[chilly ocean]

Are you trying to show it's an isomorphism?

This homo that you are trying to leave kerneless?

Trying to show if it be a singleton

[worthy kindle]

To show P iff Q
you usually show that if P then Q
and if Q then P

one of them is ez

[chilly ocean]

Ask away

Aight

[stone fulcrum]

Wow that's a large meal

[chilly ocean]

Making your GI tract a torsion group

huehuehue

[solar wyvern]

@chilly ocean
Monomorphism just means an injective homo
imagine restricting yourself to Set

[chilly ocean]

For all intents and purposes

I live in a world constructed by sets

What else can you really do?

Play with topoi?

ie. set substitutes

[stone fulcrum]

We talking about iii) above?

[chilly ocean]

injective boi

Think injective

1-2-1

1 identity

maps 2

1 identity

f(e) = f(e)

Correct?

[stone fulcrum]

A monomorphism is an injective homomorphism. If the kernel had more than one element in it, then there are multiple that map to the identity, flying in the face of everything that injectivity stood for

[chilly ocean]

^^^

Shitting on the flag of injectivity

[stone fulcrum]

Injectivity served for us, damnit. How could we?!

ie by the definition of injective, there's only one element in the kernel

[chilly ocean]

We've done

one-ident => singleton kernel

Now we do

singleton kernel => one-ident

We take the ident

We map it with a monomorphism

This makes kernel a singleton

Because its injective

So

one-ident w/ monomorphism => singleton kernel

singleton kernel w/ monomorphism => kernel = one-ident

[solar wyvern]

seriously tho, mono usually reserved for left-cancellative maps, just use injective/1-1

so what's that say about your kernel @chilly ocean

and the quotient group when you quotient by such a kernel

[stone fulcrum]

I don't know what you mean by the x = eg part. Where in the proof is that?

[chilly ocean]

At the end

They show that if the kernel is a singleton

Then it must contained the identity element

[stone fulcrum]

Why is the proof split into a, b, c that's dum

[chilly ocean]

@solar wyvern Pls define these things before confusing yes with them

[stone fulcrum]

Oh that's three different proofs?

x = eg proves that the identity is in the kernel

Wait no. We assumed x was ANY element in the kernel, then proved x = eg

Ergo, the kernel is only eg

[solar wyvern]

f is mono, or left-cancellative if f(g(x)) = f(h(x)) for all g,h

f is injective, or 1-1, if for each y in im(f) there exists a unique x such that f(x)=y

the first doesn't use elements, while the second doesn't use other functions

anyways what happen when you quotient by one element group @chilly ocean

[chilly ocean]

g,h being homomorphisms?

[solar wyvern]

wow you have a lot of pings

@chilly ocean didn't know if you were sure yourself 🤷

[chilly ocean]

g,h being homomorphisms

[solar wyvern]

umm so then use your 1st iso stuff

which says you have a decomposition of a morphism into surjective, iso and inj maps resp

but your surjective map also iso since it dun do nuthin

then what's iso, iso, inj give you?

it rhythms with injection

show G to G/ker(f) a surjective map

and im(f) to H injective

[chilly ocean]

Can we just construct the quotient of a group over a singleton

And be done with this

[solar wyvern]

bad teacher.

[chilly ocean]

He assumed it was a monomorphism

[solar wyvern]

check milne tbh

it's like p16

[chilly ocean]

Heh

Bijective = Injective + Surjective correct?

Cool

The point is - the assume it is a monomorphism

[solar wyvern]

@spark plank should be there, help @chilly ocean 4me

[chilly ocean]

Show it is surjective to show it is an isomorphism

[spark plank]

"sees ker and im"

[chilly ocean]

At this stage

Literally showing G/{e} = G

Recall that

,$ \dfrac{G}{\text{Ker}(\varphi)} = \text{Im}(\varphi)

[cloud walrus]

Jichael Mackson:

[chilly ocean]

So if |Ker(phi)| = 1

Ker(phi) can only be one thing

Construct like you've been taught

Alright then

Is {e} a normal subgroup?

Yep

So now

Let's find cosets of {e}

What are they gonna be?

Ye

So we're going to have an equivalence class for each element in G right?

Cool.

And the structure is the same because it inherits the operation right?

If ab=c in the group

Then a{e}b{e} = c{e} in the quotient group

Alright then

So without throwing around any terms like injective or surjective

We can see this is already isomorphic to the original group

Literally just map each a -> a{e}

You can check it's homomorphic

And then you can check its bijective

It should be abundantly clear that its atleast injective

[solar wyvern]

also group ops same 2

[chilly ocean]

I mean, an isomorphism is best thought of as just "relabelling elements in a set"

How is a -> a{e} not injective?

We've removed nothing tho

The cosets of {e} are literally just all the group elements g times {e}

The quotient group of G quotiented by {e}

Is the group of cosets of {e} in G

So when you take the quotient group you can just think of it is as "relabelling" each g in G with g{e}

Ie. the cosets

The relabelling is the isomorphism

You only have 1 representative

That's the point

Well

Try it out

What do you get?

Heh

Have you ever looked at the more general idea of quotient sets?

Yes

Well no

Hold up

So I have a set S

And I then I define an equivalece relationship

~

Nop nop

So two elements in my set

Are related by an equivalence relationship

So a ~ a is true of the relation for any object in my set

If a ~ b then b ~ a

And if a~b and b~c then a~c

So over integers we'll define a relation to be

x-y is even

So a ~ b if a-b is even

So we check point 1:

a ~ a, well a-a = 0 and 0 is even (yes it's even, go check)

a ~ b => a-b is even, does this mean that b-a is even so b~a?

Yes, yes it does.

a ~ b => a-b is even, b ~ c => b-c is even

So does this mean that a-c is even?

You bet your ass it does.

So a ~ c

So this "relation" is an equivalence relation

x,y are related

If

x - y is even

a is related to 0

If a-0 is even

Am I making sense?

The relation is

x and y are related

if x-y is even

The squiggle

~

Is a shorthand for "is related to"

It is also the relation itself

We'll see shortly when we get onto quotient sets

Alright

Your relation is just this property

x and y are related if x-y is even

Das it

So now

If we take the integers

We can build an "equivalence class"

{x in Z : x ~ 0}

What will be in the equivalence class?

Right

Which will be

In this particular example

x~y if x-y=2n

Just means even my guy

2 is even

2 is not a multiple of 4

So I write 2n

Instead of 4n

So yeahh

Original question

{x in Z : x ~ 0}

Is what

What x

Are related to 0

[errant drum]

Pick some values of x and y

See if you spot a pattern

[chilly ocean]

x-0 must be even

For x ~ 0

No tricks - this hammers out what we really mean by quotients in set-based objects

So

For what x in Z is x-0 even?

~ is shorthand for "is related to"

It provides a rule to define a relation

[errant drum]

Don't be so unconfident (?) in yourself

[chilly ocean]

But yeah, you got it right

It is the even numbers

Since an even number - 0 is even

[errant drum]

[chilly ocean]

An even number - 0 isn't even?

Well

We're using my relation

So

[errant drum]

Then they're not related

[chilly ocean]

x is related y if x-y is even

That is my relation

So

If I make a set

{x in Z : x is related to 0}

Then x is related to 0 if x-0 is even

By the relation ~

It provides a rule to tell you if something is related

x~y := True if x-y is even

x~y := False if x-y is not even

{x in Z: x~0 = True}

Does this clear things up?

I just defined it

It is True when x-y is even

It is False when x-y is not even

Well let's try it

Give me two numbers

Alright

4~3?

Well

4-3 = 1

1 is not even

So

4~3 = False

IF x-y is even

This is the key

You compute x-y

If it is even

Then our output is True

It is False otherwise

[solar wyvern]

for groups x~y iff xy^{-1} in your normal subgroup

[chilly ocean]

x - y

Is not

x ~ y

[solar wyvern]

actually for abelian groups it is

[chilly ocean]

Can you see the difference between

~

and

x ~ y

IF

x

y

is even

Yes

I have stated it 10 times

You might be getting confused

[solar wyvern]

tbh ~-look similar on my phone too

[chilly ocean]

Because I tried to run you through the properties of an equivalence relationship quickly

Alright alright alright

New notation for relation

x R y

Better?

Okay

{x in Z : x R 0}

x R 0, iff x - y is even

It makes sense this would be the even numbers?

Alright

Let's iterate over the naturals then

{x in Z : x R 1}

etc.

Okay

Let's kick it up a notch

{{x in Z : x R a} for all a in Z}

Okay

So what's the cardinality of the set?

How many elements are there in my jumbo set?

Remember that sets in sets are just elements

Ye

Okay

You just made quotient set

Congrats

No no no

Quotient set

Not quotient group

Different animal

Different color

The equivalence classes are elements of the quotient set

The quotient set is constructed in the manner above - find all elements that are related to a given element in our set and making unique equivalence classes.

In our case

We call this Z/R

R being the relation

Z/~

If we stuck to the original thingy migigger

set*

Quotient set

It's important

You can take a quotient of a group and not get a group back

That's why all the weird extra conditions

To stop you newbs breaking math

Well yeah, the equivalence relation has to define cosets of a normal subgroup

Mmm

The set getting quotiented must be a group

The equivalence relation must be bunching together a normal subgroup and its cosets

The reason for this is that the nice relationship between the original group operation and the coset multiplication doesn't hold otherwise

You can try breaking this if you want

Eyup

Which is why we do away with quotient sets altogether

And just say "break it up into equal size cosets"

In defining a quotient group

They are

I fudged

So ye

With this in mind

So let's try a concrete example.

[errant drum]

[chilly ocean]

Cyclic group of order 8, quotiented by equivalence relation x R y are related if m^2x = n^2y

Well just that m^2x = n^2y for some m and n in G

G/R we're finding this

Then we'll find this G/N

Okay

Remember this is addition

What you're doing here

Ah you're getting confused

I should've written it in abelian form

My goof, my goof

z^0 = 0 correct?

Right

Let's keep this simple

Addition mod 8

The relationship stated in addition

x R y if 2m+x = 2n+y

Alright

So of these

Which is the set of a normal subgroup?

Alright then.

So now C8/[0]

What's up?

?

We attach the original operator to [0] my guy

({0,2,4,6}, +)

Quickly define some coset addition

a+[b] = {a+b,a+c,...,a+d}

= [a+b]

And now we can take the quotient group

[0] in C8/R

Nah

{x in C8 : x R 0}

Remember?

x R y iff 2m+x=2n+y

nah

Cuz

2m + x = 2n + 0

x = 2(n-m)

n-m can be anything

So

[0] = {0,2,4,6}

Cuz

0 R 2, 0 R 4, 0 R 6 and 0 R 0

Does though

in this case

These are all elements of the for 2m

Which must have even order

O

ye

ayy lmao

even elements

You read the relationship right

I goofed on the definition of order

Heh

x R y iff 2m+x = 2n+y

for some m, n in C8

Stick to the relation defined

Eyup

So now

Let's have it so that

x R y iff 2c+x = y+2c

Retract that

I'm going to get a glass of water

Math stoomina is running low

The basic idea is to show that these equivalence relations define normality on the cosets.

Once you've done that - the quotient multiplication/addition works and the quotient group is formed.

The quotient set can be taken at any point

But only equivalence relations which define normality on the cosets can make groups out of their cosets

The ease of use to quotient group notation

Is to leverage lagrange theorem and normality

And let that spit out cosets which form a nice quotient group.

Because once you know the normal subgroup - you pretty much have the corresponding quotient group itching to be defined.

Equivalence relation is normality, normality requires subgroups, subgroup's cosets divide the group evenly, these equivalence relations then can be moved between by chosing a representative constructive the quotient group and using the usual group operator

Answer flimflam's question for best insight tbh

show that for subgroup H in G that aH~bH is an equivalence relation on left cosets of G by H

Then similarly show for right cosets

Basically show

aH = aH

aH = bH => bH = aH

aH = bH and bH = cH => aH = cH

And repet for right cosets

[simple valley]

thonk

rly hard

and it appear

[stone fulcrum]

Always

Sometimes

Usually

[solar wyvern]

@chilly ocean so this only nets you a quotient set on your cosets {aH: a \in G}, not a quotient group

yeah, equivalence of cosets

I guess it's not quite obvious why you'd want to endow your quotient set with group structure tho

um

how would you define the group operation on your cosets

yes

since you just showed there's an equivalence relation, i.e., the cosets partition the group even if H isn't normal

that's the part u need to show :<

so you know aH = {ah:h \in H} right

so say you have two cosets aH, bH not necessarily equal or anything, and you want to consider all the ahbh', where h,h' in H

if H is normal is there anything sneaky u can do

(this isn't difficult, but it's not obvious either)

be a bit more explicit?

and what coset is that in

okay

now let's say you wanted to do this with aH, a⁻¹H

and eH=H, aH

sorta see where I'm going with this?

if H is normal you can define (aH)(bH) = (ab)H

and this gives you a group structure on the cosets of G over H which we write G/H

so here's the cool part

we have a way of getting normal subgroups right?

hint: group homomorphisms

(you can and should show that each normal subgroup arises as the kernel of some group hom)

(now even)

you can look at it that way

(once you show that each normal subgroup is in fact the kernel of some group morphism)

tbh I'm not sure

did ur prof mention universal props or some shit before teaching 1st iso thm

i feel bad for their students

saying a map is canonical without showing it's canonical

really makes this cat think

...is it possible you're actually doing comparatively well in your class

O:

gj

i mean like some of your assigned problems are fairly sophisticated, so idk

yeah for this sorta stuff sometimes it's better to assume students know literally nothing

rip

brb seeing if chalk arrived

it arrive!

[random crag]

i suppose if you take h(x)=xN that’s canonical

yeah

[runic bear]

Hi guys, is this the correct place to ask about linear algebra? 😮

[thorny slate]

sure

[runic bear]

Is the answer 2 correct?

Because they are two non-pivot columns, so they would be assigned an arbitrary parameter to those variables

I converted it into row-echelon form first

[thorny slate]

yes

[runic bear]

Thank you! @thorny slate

[errant drum]

[chilly ocean]

What r u doin

Understand what's going on?

Kk

MILNE

Yas

Da wae

Yes. Freedom smells gud

[solar wyvern]

@chilly ocean how far u get in milne

[chilly ocean]

I think your course is just the first 30 pages

No, I actually think is.

[solar wyvern]

@chilly ocean have you actually opened a book (any book) and (tried) going through basic defs

since this would be p helpful

what's the map?

why is it canonical

in ur own words

okay

so how is f:G ->G/N defined

which is?

more concretely

in terms of N

yes

so do you know how the iso works?

um

the iso between G/N and umm

im(f)

ok

so what are the elements of G/N

and what are the elements of im(f)

are they both groups

explicitly they're aN,bN,...

cosets

they're classes, but here they're also cosets

and then if N is normal, you can do stuff

okay, sorta easier, but is im(f) a group and how

i.e., show it's a group by providing its id, multiplication and inverses

might be easier to just show them individually but w/e floats ur boat 🤷

start with identity

yes

multiplication pls

sure

🆗 🆗 🆗

so let's try making a group homomophism first

u want id of G/N to map to id of im(f)

invereses to inverses and that multiplication thingy

ok ez question

what's the id in G/N

and how do inverses/multpilication work again for good measure

i think we went over this but what are our classes here

what are those tho

how does that help you put a group structure on them

like what's [a]*[b] or [a]^{-1}

so u haven't really showed this is well defined, and besides that no idea how this is related to N still

do u remember cosets boi

if [a] = [b] what does that say about a,b

okay

what's the congruence relation for G/N

yes

:<

so we have a surjective map G →G/N right

[stone fulcrum]

This is a ton of fun with examples. We should go over D4/{e, r²}. That one is always good

D4 being symmetries of the square

[solar wyvern]

we don't need examples yet tho

so call this map umm

phi

im(phi)= G/N right

since it's surjective

(if you're not sure it's surjective, now is the time to verify this)

and figure out the kernel of phi

remember, elements of G/N = {aN}

and if N normal, aN=Na

r u even sure it's surjective

okay go verify this

also verify this will be the case for any subgroup, not just for normal subgroups

@spark plank has shown this, even if they haven't realized it

so this like basicx

basics important tbh

[spark plank]

[solar wyvern]

sure if you have an equivalence relation

but you haven't demonstrated this

okay gotta eat brb, will read replies, but phone screen bad for detailed replies

show me ur stuff

[spark plank]

[stone fulcrum]

@chilly ocean
Hay. Let's do example and we can make everything more concrete

You seem to have a pretty good handle on it though

Let's look at D4/{e, r²}
D4 is the symmetries of the square and has 8 elements.
{e, r²} is a normal subgroup of D4

First, what are the cosets?

Start by multiplying any element that isn't in {e, r²} into {e, r²}

This will give the coset containing that element

Actually though, before even finding the cosets, how many should there be?

Group of 8 elements, and it's a 2 element subgroup. Every coset has 2 elements

So it's like taking the 8 element group and splitting it into bins, where each bin gets 2 elements. How many bins should there be?

Dere you go!

The cosets partition the group. So, it's like breaking the group into equal pieces. 8/2 = 4

{e, r²}, the normal subgroup itself, is one of the cosets. What's another one? Try multiplying by r

Nooo. I mean {r, r³}

[spark plank]

[stone fulcrum]

But indeed, that is two cosets right there.
{e, r²} and {r, r³}.

Got any others? You'll need to include some flips

{s², s²r²} = {e, r²}
Because s² = e

So indeed, we already have that one

{s, sr²} is a brand new coset though

Take s × {r, r³} and we have {sr, sr³}

That's four cosets, like we predicted!
{e, r²}
{r, r³}
{s, sr²}
{sr, sr³}

They are a partition of D4. That is, every element of D4 goes into one coset exactly

Now for the interesting question, since {e, r²} is a normal subgroup, these cosets form a group under coset multiplication. What group is this isomorphic to?

First question, how many elements does this group have?

Perfect. There are two groups of order four

Those are both the same one. You've got the cyclic group on 4 elements, and you've got the kline four group

Might have spelled it wrong. I mean Z2×Z2

But you've got the right idea! You want to find the order of the cosets and match them to the order of those groups

So before anything, you do know how to do coset multiplication?

Kk. I trust you

A lot of people struggle with that part but if you've got it

[solar wyvern]

@chilly ocean the subgroup is a coset too

:(

[chilly ocean]

As many of my friends know, I have an extremely high IQ, so don't bother answering this question if you're too dim-witted. How would the Y Tangent become the X axis if the dividend is 9?

[solar wyvern]

um

@chilly ocean do u know how quotients for abelian groups work

so

figure out those

they're the ez case

like say you have C_8, what are its subgroups, and what are the possible quotient groups by its subgroups

since honestly if you're unsure about the abelian case, nonabelian is a lost cause

also try to construct group homomorphisms such that the kernel is your normal subgroups

@spark plank u do this 2 while i struggle through ordinal indexed things

[spark plank]

[solar wyvern]

nani

[spark plank]

wut we doin'

wuts C

[solar wyvern]

cyclic group

so C_8 means a^8 =e

[spark plank]

o

ok

but wut we doin'

[solar wyvern]

cooking up subgroups

[spark plank]

aren't we always doing that

[solar wyvern]

also to be clear, we should try out endomorphisms over C_8 first

and think about what those are

[spark plank]

[solar wyvern]

@chilly ocean list em

like

[spark plank]

=pup define endomorphism

[mellow vapor]

Do more with Wolfram|Alpha pro

Query made by

@spark plank

[solar wyvern]

group homomorphisms of C_8 into itself here

[spark plank]

o rite

[solar wyvern]

not necessarily invertible, but still
mapping e to e
doing the f(a)f(b) = f(ab) thing,
and f(a⁻¹) = (f(a))⁻¹

[spark plank]

dat ting

So maps from C_8 to C_8

nani

[solar wyvern]

so u should check this, but letting everything map to e will be a valid group homomorphism

@spark plank checkem

[spark plank]

tfw confused wuts going on

[solar wyvern]

so

not using generators

[spark plank]

So we're looking at endomorphisms of C_8 that behave like group operators...?

[solar wyvern]

group homomorphisms

[spark plank]

[solar wyvern]

(usually that's what we'll call endomorphisms really, maps from an object into itself preserving whatever structure that object has)

[spark plank]

So C_8 is defined inside a bigger group?

[solar wyvern]

what u mean

[spark plank]

C_8 = {a : a^8 = e}

but like

[solar wyvern]

@stone fulcrum no spoilers pls

[spark plank]

wut kinda stuffs is a

[solar wyvern]

it's just a thing for now 🤷

[stone fulcrum]

Did we abandon my q? If so I can just show the answer

[chilly ocean]

Dwai

Just a thing

[spark plank]

[chilly ocean]

Just gotta have a^8 = e

And satisfy group axioms

That actually locks you into one kind of structure

Which you can prove

[spark plank]

So we got some multiplication thingy

[solar wyvern]

so the map which sends everything to the identity in $C_8$ is generated by $a \mapsto a^8$

[spark plank]

And some stuff

[cloud walrus]

flimflam:

[solar wyvern]

also should check the identity map in $Endo(C_8)$ is a group homomorphism, that is, $a \mapsto a$

[cloud walrus]

flimflam:

[stone fulcrum]

Okay, I'll just answer it real quick. These are the orders.
{e, r²} - 1
{r, r³} - 2
{sr, sr³} - 2
{s, sr²} - 2

Note there's no element of order 4. Ergo, it can't be Z4, so it must be Z2×Z2

[spark plank]

so...

[solar wyvern]

$a$ is a generator for the cyclic group $C_8$, that is $\langle a|a^8=e\rangle \cong C_8$

[cloud walrus]

flimflam:

[stone fulcrum]

So, D4/{e, r²} = D4/Z2 = Z2×Z2

[spark plank]

we wanna show that $f:C_8\mapsto C_8$ satisfies certain properties?

[cloud walrus]

Simple_Art:

[solar wyvern]

$f:C_8 \to C_8$ but yes

[cloud walrus]

flimflam:

[spark plank]

[solar wyvern]

like we don't care about $f:e \mapsto a \neq e$, this a bad boi

[cloud walrus]

flimflam:

[spark plank]

wut's the difference between what I wrote and what you wrote?

[stone fulcrum]

Finally, this implies there's a homomorphism φ: D4 → Z2×Z2 such that {e, r²} is the kernel.

Now I'm done.

[spark plank]

wait

homomorphisms are like

group axiom transitive stuff?

that is f(e) = e, f(ab) = f(a)f(b), and f(a⁻¹) = (f(a))⁻¹?

[solar wyvern]

@spark plank here they're preserving group properties, but if it was between like metric spaces you'd want them to preserve continuity or whatever

[spark plank]

This defines a homomorphism?

And what do we want to show about these things?

[solar wyvern]

umm

we wanna show there are some

[spark plank]

[solar wyvern]

which ones are always there

[spark plank]

"some"

wdym

[solar wyvern]

and for small ez cases like this, just list them all out

[spark plank]

obviously mapping things to themselves gives you that endomorphisms exist

also Endo(G) is the set of endomorphisms?

[solar wyvern]

structure preserving endos tho?

ya

[spark plank]

wdym structure preserving endos?

[solar wyvern]

that is f(e) = e, f(ab) = f(a)f(b), and f(a⁻¹) = (f(a))⁻¹?

that do this

[chilly ocean]

^

[spark plank]

isn't that by definition

[solar wyvern]

existence isn't by def

[spark plank]

am confuzzled again kek

ya but identity function => existence

[solar wyvern]

ok

[spark plank]

wait

[solar wyvern]

so you got one

[spark plank]

[solar wyvern]

now try finding others

[spark plank]

[solar wyvern]

I gave u 2

also show how many there are for cyclic groups, if you're being ambitious

[spark plank]

I feel like I should verify things

[solar wyvern]

u really should

verify those 2 if you haven't

then try to find others

[spark plank]

like why should ab even be in C_8 for a, b in C_8

wait didn't we do that already

[solar wyvern]

what's b

[spark plank]

pretty sure that was like day 1 stuff with woog

[solar wyvern]

well humor me then

what are the endomorphisms of $Endo(C_8)$, or in general $Endo(C_n)$, if you're stuck, what about $Endo(C_p)$

[spark plank]

it was that one proof I did where u and woog were confuzzled by my contradiction

[cloud walrus]

flimflam:

[spark plank]

and where I complained that ordering > divisibility

wew boi already 740 messages in here

[solar wyvern]

how do u even check that how many do i have

[spark plank]

u just search

C_8 has unknown order?

[chilly ocean]

It has known order

[spark plank]

nani