groups-rings-fields

[white oxide]

yeah

[agile burrow]

Z_3 x Z_3 has 9 elements

[white oxide]

right

ohhh wait i see

[agile burrow]

So if this group were cyclic, then there would exist some element such that if we add it to itself, we get all 9 elements

The explanation shows why no such element can exist

[white oxide]

right but every time you add it to itself three times it equals the identity

which is not sufficient

wait

is that right

[agile burrow]

Right, because if we reach the identity before getting all 9 elements in the group, then we certainly can't hit all 9 elements in the group

[white oxide]

ohhhh, okay

[agile burrow]

The explanation is saying that if we fix some g in Z_3 x Z_3, then the multiples of g are 0, g, and 2g, so there are at most 3 distinct multiples of g

[white oxide]

oh ok thank you! that really helped

[agile burrow]

happy to help

[warm shoal]

Hey!

[agile burrow]

Sure, so one thing you might observe is that two cyclic permutations of the product of the same factors are conjugate

[warm shoal]

im afraid i dont understand

[agile burrow]

Right, so we say that two elements x and y in a group are conjugate if there exists some g in G such that y = gxg^-1

Let's do this more specifically with three elements: we want to show that xyz has the same order as yzx

I claim that xyz and yzx are conjugate. Can you find an element g in G such that g(xyz)g^-1 = yzx

[warm shoal]

thank you for this explanation

let me break it down

and try to understand what you're saying

[past temple]

if i have a group G with nontrivial center, and an irreducible representation V of that group such that elements in the center of that group fix elements in V, how do i prove that V is the trivial representation?

[lethal dune]

is that even true?

[past temple]

if G is finite and V is finite-dimensional that is

[lethal dune]

what about S3×Z2 with center {e}×Z2 and the representation be
S3×Z2 → S3 → GL(2, R) std representation of S3

[past temple]

oh

[astral stream]

How do I do this?

[rustic crown]

do you know first iso thm?

(use it )

[warm shoal]

hey @rustic crown

so with 4

we are given that

a^mt = e = a^n

Let a be an element of order n in a group G. is given

what would be your first step in this

[rustic crown]

well you should tell me your thoughts >.<

i would just use the standard fact about orders

which is pretty much the question

[astral stream]

Is the first iso theorem linked to this? (they didn't give it to us in the book but I googled it)

Is this theorem saying the union of sets aH for all a will form G?

[rustic crown]

yep, they partition G

[astral stream]

ohhh and since all of them are the same magnitude the Kernal will divide G?

but how could that link the phi(G)?

[rustic crown]

almost... the phrasing is a little bit weird

[rustic crown]

it's linked to first iso thm, lagranges thm, etc

but if you haven't seen it, doing directly is also fine

[astral stream]

ok ill try both ways

[tender wharf]

this is also lagrange

Ker phi is a normal subgroup which is why your left and right cosets are same

[rustic crown]

lagrange for normal subgroup

[tender wharf]

but lagrange already says the union of cosets forms the group

[astral stream]

ahh ok thanks

[tender wharf]

why 🙈 det

spillll

[rustic crown]

cuz

monke

cute

walter

[warm shoal]

what is this saying exactly

list all the cyclic elements of S6 where we have a specific cycle?

[agile burrow]

f is a specific element in S_6. This exercise asks you to describe all of the elements in the subgroup generated by f, meaning all of the powers of this element

[warm shoal]

did it 🙂

[chilly ocean]

it just asks u to prove by contradiction that its not cyclic

[quiet pelican]

||what is the maximum cardinality of a cyclic group?||

[warm shoal]

right i just dont know how to go about it

[quiet pelican]

This doesn’t follow the hint, but I think is one of the nicer ways to show it

[chilly ocean]

I was gonna say it too

[warm shoal]

it has to be countable right

finite

wait no that's silly idk what im saying

i was gunna say

only finite groups can be cyclic

[quiet pelican]

Not necessarily finite ($\bZ$ is cyclic)

[cloud walrus]

Micose

[warm shoal]

but that can't be true

right

[chilly ocean]

Id just do that but the proof that R is not countable involves more stuff than just doing it the way of the hint

[warm shoal]

can u guys help me with the contradictory proof

[quiet pelican]

By the time you’ve reached groups, you’ve probably seen a proof of that (esp as what they want you to do needs that ||there is a rational between every pair of reals||)

[tender wharf]

suppose k generates R*, now find a number x ||such that no n exists for x=k^n||

[chilly ocean]

yea but its kinda like... using a T-90 to kill a mosquito

[warm shoal]

ngl guys

ive taken

topology

number theory

like i should be able to do this

WHATS WRONG WITH MNE

love ❤️

thank you everyone today who helped me

much appreciated

[formal ermine]

fwiw the multiplicative group of any finite field is cyclic

corollary: Zp^\times is cyclic

[delicate orchid]

Bonkers…

[south patrol]

And since R is a finite field,...

I mean fwiw the group of units of an infinite group is never cyclic, so we needn't use ordering or cardinality arguments to prove R^x is not cyclic

[delicate orchid]

(Z/p^kZ)^\times is cyclic for all primes p and all k 😌 (2 is not prime)

[south patrol]

Fr

that Z/2^{k-1} x Z/2 stuff is a myth

[delicate orchid]

k-2 albeit

[summer path]

Why do people teach cyclic group of order p to be denoted as Z_p rather than Z/pZ or Z/p

[south patrol]

Idk it's a bit cringe doing Z_p

[lethal dune]

Z_p can also mean p adic integers

[oblique river]

Because theyre lazy

[summer path]

:c

One / was too much I guess

[cunning dust]

My TA wrote C_p because "writing \mathbb{Z} takes too many strokes"

[delicate orchid]

THIS

I’ve gotten to the point of just writing “n” for C_n now tbh

[summer path]

Why would you not make a new command for bbZ

.<

[delicate orchid]

Dude people write things in real life

[molten viper]

tbh I've never seen Z_p as a general cyclic group

only as a particularly useful example

[cunning dust]

Well idk about general, but writing that a group is isomorphic to Z_p to mean it's cyclic

[oblique river]

Imo C_p should be reserved for an abstract cyclic group

Which is not the same as Z/pZ

[delicate orchid]

Oh and n + m for C_n \times C_m and n:m for C_n semidirect C_m

[oblique river]

Which has a chosen generator

[delicate orchid]

I agree, but the problem is they’ll write Z_p instead of Z/pZ and also use Z_p for abstract cyclic

[molten viper]

The based option is using the group of nth roots of unity as your epitomic cyclic group

[delicate orchid]

This tbh

[oblique river]

NO

That has a nontrivial galois action on it

Whereas Z/pZ has the trivial galois action

[molten viper]

galois actions are not covered for a while in undergrad courses, so I can excuse it

[oblique river]

Mu_p = Z/pZ(1)

Where (1) denotes the tate twist

[molten viper]

The real issue is that most students have had at least some basic number theory but not much complex numbers

[summer path]

Ok well by that reasoning p-adics are generally not covered in undergrad, so Z_p is also excused?

[molten viper]

If you mean \mathbb{Z} then no tbh

[summer path]

I meant bbZ yeah

[molten viper]

I think it's fine if you specifically mean Z mod p, otherwise n o

unrelated, is there a name for whatever comes out when you do like, Z_6?

Like that structure where it's not quite a group

[oblique river]

How are you defining Z_6? Inverse limit of Z/6^nZ?

[molten viper]

Like the set of integers mod 6

with addition

[oblique river]

That is a group…..

[molten viper]

multiplication

I meant multiplication

whoops

[oblique river]

The same way you do for Z/pZ under multiplication

Z/pZ is also not a group under multiplication

[molten viper]

I must be thinking of something else

[summer path]

Im curious, what happens when you do this

[oblique river]

You get Z_2 x Z_3

[summer path]

Oh

[molten viper]

Hi class just started and my professor is introducing categories please help

[summer path]

This feels like something I've definitely asked about before

[elder wave]

What class

[delicate orchid]

Arrows!! That’s it

[molten viper]

It's a seminar course mostly dealing with Type theory

[elder wave]

[tribal moss]

Perhaps get off the internet and listen to the lecture instead?

[molten viper]

That's what I'm doing!

[tribal moss]

Sorry, should have disabled the ping on that one.

[molten viper]

All is forgiven haha

[chilly ocean]

What does it actually mean for a homomorphism to be canonical?

Does it have a formal meaning, or is it merely a loose term meaning "natural" or "obvious" or something

[rustic crown]

it does have a formal meaning

categories were introduced just for this lmao

[chilly ocean]

hmm, I see

I don't know much about categories yet. How should I approach the term when I see it?

[rustic crown]

the basic idea is this... you say G --> G/N is a natural map, because the definition of the map doesn't fully depend on what G and N are... the same recipe of the map would work if you choose a different group and different normal subgroup.

[tribal moss]

"Natural" has a formal meaning (though it is often somewhat oversold in informal contexts), but I don't think "canonical" does?

[rustic crown]

they're used interchangeably i thought

[chilly ocean]

I feel like my book uses them interchangably

[chilly ocean]

Thanks, that makes sense!

[tribal moss]

Oh, if we're talking specifically about G --> G/N, then "the canonical homomorphism" does have a technical meaning. It's the particular homomorphism that the construction of G/N gives you, talking each element of G to the coset it's in.

[chilly ocean]

So it's like, it's a homomorphism that arises simply from the structure itself

[rustic crown]

more precisely, it satisfies the following property. if f : G --> G' is a group homomorphism, and N, N' are normal subgroups such that f(N) is contained in N', then you have this nice square

[cloud walrus]

det

[rustic crown]

the commutativity of this square is what makes the quotient map natural

[chilly ocean]

I don't really see that

You're talking about G -> G/N here still?

[tribal moss]

I meant, I've never heard anyone speak of "canonical transformations" between functors ...

[chilly ocean]

because I don't really understand how the right part of the diagram 'affects' G -> G/N

[rustic crown]

informally you can think of "natural" maps as something where you don't need to make arbitrary choices while defining that map. this way the same recipe/definition would work for any other object.

[rustic crown]

it severely restricts the options for maps G --> G/N such that any such diagram commutes.

[chilly ocean]

ahh, I think I see

because you are enforcing the homomorphism condition both for G -> G' and G/N -> G'/N'

[rustic crown]

yep, changing your "objects" by a structure preserving map, shouldn't affect a natural map by a lot.

a standard non-canonical map is the isomorphism between a finite dimensional vector space and its dual

to define such an isomorphism, one first picks a basis and using this basis you define the required linear map.

but if you now replace this vector space with some other vector spaces, you would have to make a different choice of basis. and because of these arbitrary choices, a diagram "like" above won't commute.

[chilly ocean]

aaa

I see the connection to category theory now

like, for a field like category theory where you study the structures itself, it would be quite important to have some notion of... well, "constantness" of your diagrams, i.e. independent on the specific properties of the objects chosen

thanks :)

[rustic crown]

i hear the word "transformation" itself quite rarely :p

but yea same never heard of "canonical transformation"

maybe canonical is just a less formal word used when you wanna talk about nice maps which are not exactly natural transformations, but idk

[formal ermine]

what is a canonical isomorphism

[warm shoal]

what does it mean here by "up to isomorphism"

[coral shale]

There are an infinite number of groups of order 1

[warm shoal]

?

oh

[lusty marlin]

It means that isomorphic groups are not considered distinct while counting

[warm shoal]

gotcha

well interesting how

all the Zn where n is prime

there are no other examples listed

[lusty marlin]

right

[warm shoal]

does that mean something?

[lusty marlin]

you should prove this

that if n is prime then there is only one group of order n

up to isomorphism

in fact there are composite integers too that satisfy this property, and they are characterized in a paper by Keith Conrad

[warm shoal]

gotcha okay

ill try that

tho first question

in most cases is it obvious the order of a group?

like, to see how many elements in it

like if it's

a1 ... an

then obviously n elements

so |G| = n

or do you mostly have to find order of element and figure out order of whole group that way

[coral shale]

I think you need to first see Lagrange's thm before doing that

[lusty marlin]

Why assume that that they haven't already

hmm ok

it would be obvious then

[coral shale]

I don't know about 'most'

it's down to how you decide to define it

[coral shale]

that depends on context doesnt it?

[empty rose]

ime "canonical" generally means kind of like,
"obvious"?
there might be several objects of a particular kind but one of them is constructible from the context and often has nicer properties than a general thing that just happens to exist in this case

like if you consider products, generally A x B and B x A will be isomorphic
there might be several possible isomorphisms, given for instance by automorphisms of A or B or more complex maps, but (for most kinds of product) there's always one very obvious one where you just swap them around and leave everything the same

and, like what often tends to happen, this obvious construction has very nice properties, for instance the swap map from A x B -> B x A and the swap map from B x A -> A x B are inverses of each other

[coral shale]

when canonical is mentioned I would think about homomorphisms and more in general

1st iso theorem gives a canonical iso.

[delicate orchid]

it's a canonical map that's invertible :trollshiro:

[coral shale]

I think this notion is formalized in category

[warm shoal]

aren't the theorems of these the same?

like if GxH is cyclic then there exists some generator (a, b) of GxH

then there does of G and H

thus G and H are cyclic too

right @coral shale

? ❤️

@delicate orchid

[delicate orchid]

"a generator" not "the generator"

there could be more than one generator

[warm shoal]

oh okay

my thought is still valid tho right?

[delicate orchid]

yes it is

that first point is worded very oddly imo

[warm shoal]

why would they give

such similar proofs

i dont get it!

like literally copy paste right?

[delicate orchid]

2 follows very quickly from 1, yes

[warm shoal]

betr

thanks

[somber sleet]

I know this is gonna be random, but thx for helping me pass my exam guys🥰

[coral spindle]

You're the only one who took your exam. Be proud!

[next obsidian]

Swag, good job

[charred crescent]

does this proof suffice?

[coral spindle]

No.

You never prove that an arbitrary element of H can be generated by a^n, and in fact this is false in general

For example choose 2Z, and h = 4. Then 4 does not generate 2Z.

[charred crescent]

well im stupid

yeah i never showed a was in H

[coral spindle]

a is not in H in general.

[charred crescent]

so if every element h of H is in G, then why can't h be represented as a to some power

[coral spindle]

h can be represented as a to some power

[charred crescent]

if the same element can be expressed that way in G

[coral spindle]

h does not in general generate H.

[charred crescent]

okay yeah im confused

a generates h but a isn't necessarily in H?

[coral spindle]

Yes

In fact, if a were in H, then H = G.

[somber sleet]

Thx for being patient with me, and helping me through everything, I'm really grateful

[charred crescent]

okay so for any arbitrary 'h' in H, we know that there is some integer 'n' s.t. h = a^{n}. Can we say that a^{n} in H generates H?

or not quite

[coral spindle]

.

Hint: look for minimal numbers n

[charred crescent]

if h = a^{n} doesn't generate H in general then how can we even use a^{n} at all

[coral spindle]

See the hint

[primal tusk]

If i know that the characteristic of a commutative ring is prime, will the ring always be a field?

and if so how would you show that?

[coral spindle]

It will almost never be a field

[primal tusk]

oh okay

[coral spindle]

Try to find a single example for each prime p of a non-field with characteristic p

BIG hint: ||product ring||

[primal tusk]

i havnt learned about product rings yet i dont think

[coral spindle]

OK

[primal tusk]

oh wait nvm thats just a cartesian product

yes i have

[coral spindle]

I would've liked you to give it a shot before looking at the hint, oh well

[primal tusk]

i have been trying I only could think of integers mod p

okay yea im realizing product rings are good for thinking about if things always hold true

[void cosmos]

yo

F[x,y]/(y^2-x) is iso to F[x]

with this observation that y^2-x = 0 --> y^2=x

so F[x,y]/(y^2-x) = F[x,x^2]/(0) = F[x]

is this "formal" enough

i feel this is super fishy

can someone solidfy this

😦

[delicate orchid]

that looks fine to me

[void cosmos]

idk man

feels super weird

but ig yea

[delicate orchid]

actually I'd say that F[x,y]/(y^2-x) is iso to F[y^2, y] but that's nitpicking

[quiet pelican]

Use the first isomorphism theorem - map F[X, Y] into F[X] by X to X, Y to X^2 and it follows (using the generalisation of the factor theorem to find the kernel)

[void cosmos]

^

perfect

tysm

[next obsidian]

Hurb

[rustic crown]

i talked to moldi today

after sooooo long

was fun :3

[delicate orchid]

Burb

[next obsidian]

Swag

What he up 2

[rustic crown]

writing his masters thesis and learning too much cat and top

[odd forge]

Hey I'm not understanding something about adjoining elements, can someone help explain this to me?

[rustic crown]

you can use CRT maybe

ig maybe not :p

[odd forge]

maybe can't use CRT, or maybe no better way to think of it?

[rustic crown]

the above is quite concrete, you can think of it as various things like space of certain polynomial functions and stuff

but if you want, CRT still lets you embed this into a nice product

Z[x] --> Z[x]/(x^2+1) x Z[x]/(x^2-2)

the kernel is the intersection (x^2+1) n (x^2-2) which by UFD-ness of Z[x] is same as the ideal generated by (x^2+1)(x^2-2)

so your ring R has an injective map to Z[i] x Z[sqrt2]

usually for CRT you have that the ideal I and J are coprime, i.e. I + J = (1) which lets you prove that this map is surjective as well.

[odd forge]

Okay thank you I saved this and will look back when I learn about product rings and what UFD is

[void cosmos]

is 7 just the set of all polynomials that divide both 2 and x^3+1?

[coral shale]

No?

You are referring to the quotient ring right? It's not

Even if you're referring to what you're quotienting by - it's not.

[void cosmos]

yes im reffering to the quotient ring

why doens't this follow from third iso theorem?

[coral shale]

Well I mean no - for starters the polynomial 1 is in the ring

and neither 2 or the cubic divide 1

[void cosmos]

ideals that contain (2,x^3+1)

yo

i meant the ideals

those are the ideals

[coral shale]

oh alright

[coral shale]

.

[void cosmos]

yeea i meant problem 7

my bad

doesnt matter

is it correct?

[coral shale]

I don't follow

[void cosmos]

np

[coral shale]

It asks you for all the ideals right

[void cosmos]

yes

[coral shale]

the set of all polynomials that divide both 2 and x^3+1

How does this describe some ideals

What do you mean

[void cosmos]

so i need ideals that contain (2,x^3+1)

[coral shale]

What do you mean by that

(2, x^3+1) is in the original ring

[void cosmos]

yes

[coral shale]

the equivalence class that represents in the quotient is the identity

[void cosmos]

no

the ideals of the quotient ring is in correspondonce with ideals that contain (2,x^3+1)

[coral shale]

oh ok sure

yes, I follow

[void cosmos]

so now everythings correct?

[coral shale]

As far as I know, yes this is correct

[void cosmos]

ok cool

tytyty

[coral shale]

And yes 3rd iso

[void cosmos]

yea

ciol

col

[coral shale]

https://i.imgur.com/tDpcJnP.png

[sonic coral]

so as far as i can understand, we assume that the possible numbers of subgroups are 16.

2^4, but where do the 5-1 elements come from?

i think we were trying to show a contradiction to show that a group of order 80 is not simple by counting the number of elements

i just don’t get the idea behind this counting process

[rustic crown]

so assume that this group was simple

and look at the sylow-5 subgroups

how many of them would be there?

it has to be 1 mod 5

and has to divide 16

which gives two possibilities

1 or 16

it can't be 1 since we assumed it's simple

(as that implies that the sylow subgroup is normal)

so 16

[sonic coral]

yes. i got to that point by myself

[rustic crown]

now given two distinct sylow 5-subgroups

they can only intersect at {1}

[sonic coral]

so we remove the non identity element

[rustic crown]

yep

[sonic coral]

wait

we remove the identity element*

it’s at this point i don’t understand any further

counting the number of elements of a specific order

[rustic crown]

yee

so each sylow 5 subgroup will give you (5-1) order 5 elements

that gives you the 64 elements of order 5

[sonic coral]

okay, so for the sylow 2 subgroup case, we have 5(2-1) elements of order 2?

[rustic crown]

the sylow 2-subgroup has order 16

[sonic coral]

oh yes 2^4

[rustic crown]

which is precisely the space left for one such subgroup

80-64 elements = 16 elements

hence this would be normal :3

[sonic coral]

how would i get those 16 elements using the same argument? 2^4(2-1)?

[rustic crown]

nah, that would be more complicated

because two sylow-2 subgroups can intersect non-trivially

but not here

because we have space for only one such subgroup left

the remaining non-order-5 elements has to be the ones in the sylow 2-subgroup

[sonic coral]

since we assumed they were distinct right?

or is it not even an assumption, they are distinct?

[rustic crown]

we assumed there were 16 sylow-5 thingies

[sonic coral]

so the sylow-5 subgroup has order 16

[rustic crown]

nuu

.<

sylow 5 has order 5

because that's the largest power of 5 dividing 80 :3

[sonic coral]

okay, yeah i can see that with the prime factorization

[rustic crown]

think about it for a while

what we proved is that either you have a normal sylow-5 or normal sylow-2 subgroup

but we didn't prove which one it is

[sonic coral]

i’m still trying to understand the proof in the first place. i know that this is gonna lead me to a contradiction somewhere, where i have more elements than in the group. hence simple

[sonic coral]

but either way their is a normal nontrivial subgroup and so not simple

[rustic crown]

yep

[sonic coral]

i get that. but getting there isn’t clicking

[chilly ocean]

p-group has cardinality being power of p

so if Sylow-5-subgroup would have order 16

that couldn't be right

16 is not a power of 5

[sonic coral]

so is n_5 = 1 or 16 in this case, the number of these possible subgroups of order 5?

[sonic coral]

i got it (i hope). we have that there are 16 possible sylow 5 subgroups of order 5. since the order of an element divides the order of the group, we have that there are 4 elements of order 5 in each group, giving us 64 elements of order 5.

[sonic coral]

,rotate

[sonic coral]

this is the right idea?

[haughty shale]

How do I compute for the cardinality of $(\mathbb{Z}/100\mathbb{Z})^\times$

[cloud walrus]

emphatic_wax

[rustic crown]

yeep

[quiet pelican]

Do you know the Euler Totient function?

Otherwise, inclusion exclusion

(Or the general formula derivable from that)

[haughty shale]

not really

/but I can look it up. but if not using that. what do you mean by inclusion exclusion? like try some numbers

[quiet pelican]

Do you know what the inclusion exclusion principle is?
If not, we can use it informally

[haughty shale]

ahhh yes

but how do I use it for this?

[quiet pelican]

The events you want the Union of are “divisible by 2” and “divisible by 5” (then you want the complement of that set)

[haughty shale]

Ohhh okay

so is it related to the prime factorization of 100?

[quiet pelican]

Yes

A number is not coprime with 100 = 2^3 * 5^2 iff it is divisible by 2 or 5

[haughty shale]

so, for example if I replace 100 with 900, I will do 2, 3, 5?

[quiet pelican]

Yes

[haughty shale]

Gotcha! Thank you!

[vivid tiger]

the proofs of the impossibility of trisecting the angle, doubling the cube, and (given in a black box the transcendence of \pi) squaring the circle are...suprising easy? like i would have thought they'd require a fuckton of Galois theory, but they actually require only some basic field extension stuff (first 5 pages of the first section of the algebraic extensions part of Lang's Algebra for the stuff + some knowledge of the polynomial facts required to understand those 5 pages + the short proofs on wikipedia)

like, this feels like cheating

[quiet pelican]

You need a lot of machinery for them to be easy

And I don’t mean Galois

[vivid tiger]

maybe i have more machinery than i think?

[quiet pelican]

I mean all the abstract algebra to get to that point

[vivid tiger]

but like, isn't it just some polynomial stuff + some basic field stuff that honestly feels just like linear algebra

to be fair everything is obvious and simple after knowing it

and an obstruse edifice of unknown stuff before knowing it

[tribal moss]

Everything is trivial after you understand it.

[white oxide]

does anybody know where i can find a good proof on the fundamental theorem of finitely generated abelian groups? my book chooses to omit it

[rustic crown]

most books do that

[white oxide]

is it just too hard to understand

[rustic crown]

because there is a classification theorem of finitely generated modules over a pid

[white oxide]

or out of the scope of an abstract algebra one semester course

an introduction course rather

[white oxide]

wish i knew what that meant

[rustic crown]

so just redoing the proof for Z-modules = abelian groups doesn't make too much sense

[rustic crown]

oh so R-modules are like vector spaces, but the scalars come from a ring instead of a field

[white oxide]

wish i knew what a ring meant LOL

[rustic crown]

you can add and multiply uwu

[white oxide]

okie

[rustic crown]

and that's fun

modules are trying to understand these things using their actions on abelian groups

if the ring is Z

then the action is uniquely determined

[static needle]

Uh uh, R-modules are generalizations of vector spaces

[rustic crown]

because what can 3 * m mean? you're forced to define it as m+m+m

[static needle]

It's like a group, but also works with multiplication. Formally speaking, a ring is a structure (R,+,×) such that 1. It is a group under +, and 2. It is a monoid under ×

[white oxide]

what is a monoid

[static needle]

Uhhhh

[white oxide]

ah i see

[static needle]

Lemme dig thru my notes

[rustic crown]

so this is just a special case of a little more general theorem which works for certain nice class of rings called PIDs. and the ring of integers Z is one such thing, so people just prove this theorem while doing modules instead of working thorugh the proof with abelian groups

[static needle]

Ok it's a group that doesnt require inverses

[long geyser]

herstein has one

[white oxide]

wait what i thought by definition groups had to have inverses

[white oxide]

oh ok i see

[static needle]

No no groups do need inverses but monoids dont

[white oxide]

so why is it a group

[long geyser]

it isn't

[static needle]

But they have the other structures associated with groups

They arent groups per se

But they have some structure of groups

Like possessing an identity element and being associative under a binary map

[white oxide]

as in they're isomorphic to some groups?

[long geyser]

you are making this too complicated to yourself

[static needle]

No no no they just have some properties groups have but are not the same on the nose

[white oxide]

ah i see

[long geyser]

a monoid is just a set with a binary operation that misses one axiom of a group (one about inverses)

[static needle]

Only groups can be isomorphic to other groups, you dummy

Now that he's on the topic tho

Is there a map that makes a group lose structure that maps it to a monoid?

[long geyser]

I believe that's what those in the business call "forgetful functors"

not sure if it applies to this specific situation but it should?

[tribal moss]

Yeah, every group is also automatically a monoid.

And the concept of forgetful functor does apply here.

[static needle]

Ah yes

Category theory

[tribal moss]

It's a fairly nice forgetful functor too; it has a left adjoint that is actually a left inverse.

[static needle]

Noice indeed

[white oxide]

just to be clear i can think of n being the sum of how many generators there are for each Z_d where d divides n

[vivid tiger]

that is indeed what you are being asked to prove

[white oxide]

could i phrase it this way?

or like would this example apply

say i have Z_10

[vivid tiger]

what's the question?

[white oxide]

then its order should be equal to the number of positive generators of Z_10 + number of positive generators of Z_5 + number of positive generators of Z_2 + number of positive generators of Z

[vivid tiger]

no, 10 equals the order of Z_10 + Z_5 + Z_2 + Z_1

[chilly ocean]

.

[vivid tiger]

n=10

phi(Z_10) = order of Z_10 since the integers less than ten that are also coprime to 10 have multiplicative inverse mod 10

[white oxide]

wait what

isn't the order of Z_10 10

[vivid tiger]

n=10

ok

[white oxide]

order of Z_5 5

[vivid tiger]

so, i assumed you meant under multiplication

[white oxide]

oh nah my bad

[vivid tiger]

(i do usually see "Z_n" as the additive version, but appear to have assumed you meant otherwise and then i ran with it)

generators for (Z_n,+) are the same as elements of (Z^(×)_n, *) which is what i use to denote the modular integers mod n that are multiplicatively invertible mod n

due to both having the coprime requirement

but if i understood you correctly, you have only said true statements

[white oxide]

okay thanks

yeah i'm just trying to clear things up

[chilly ocean]

Let R be a ring 💍
If ab=1, then ba=1?

[vivid tiger]

false

false

left inverse does not imply existence of right inverse

sorry i should provide an example

[tribal moss]

A counterexample could be the ring of linear operators on a sequence space, with a and b be the shift-left and shift-right operators.

[vivid tiger]

wow this was surprisingly hard to come up eith

differentiation and integration (integral from 0 to x) of polynomials

[vivid tiger]

this example feels thematically similar to mine

[white oxide]

does this question have something to do with a partition of G?

like is Z_n partitioned into subgroups Z_d such that d|n

or is that just not true

oh wait there was a bijective map defined from a coset to a subgroup H

and we know cosets partition a group

so it must be true then

or am i trippin

or in order to have a partition do you need to define an equivalence relation

[vivid tiger]

partitions and equivalence relations can induce each otjer

it is useful to glue (get it! because equivalence relation!) the two in your head, as in to think of the two as two things attached with some glue and duct tape

[white oxide]

ye

[vivid tiger]

think about this for a little

partitions must have disjoint sets

[white oxide]

ohhh

so no generator can be in another

right?

[vivid tiger]

are you sure?

what must all subgroups contain?

[white oxide]

the identity but an identity must be unique right

oh wait

i'm stupid

yeah you're right

but 0 isn't a generator

[vivid tiger]

no worries

[white oxide]

wait

is that wym

[vivid tiger]

i was once again thinking of the multiplicative group by mistake

[white oxide]

oh yeah ur good

[vivid tiger]

i think that's because of the totient function. i automatically think multiplicative modular Numbers

this still surely is false

subgroup isomorphic to Z_2, subgroup isomorphic Z_4, both living inside Z_8

[white oxide]

oh right

[vivid tiger]

wait

i am incorrect.

Z_4, Z_6 in Z_12.

i am still wrong

[white oxide]

i have utterly no clue how to do this then, or how to show that Z_n of order n is equal to the sum of the number of positive generators in each Z_d such that d|n

yea no worries

[vivid tiger]

(i remember proving the result of this exercise...but before i knew any abstract algebra)

i kinda just shuffled around with the prime factorizations

[white oxide]

damn i just got mogged

jk jk lol

[vivid tiger]

there was this math contest with a round for longer harder proof based stuff and there was a year with a series of questions about this, then the mobius function and square free numbers

[white oxide]

idk what those are but damn

ur cracked

[vivid tiger]

...i immediately forgot them...i don't know number theory, and couldn't tell you what the mobius function was except that it somehow had a pretty relationship with the property of being squarefree

https://en.m.wikipedia.org/wiki/Möbius_function

(a special case of the mobius inversion formula is what i think it was. i remember it because it was one of the first times i did that kind of discovery)

oh wait no it's not that

[white oxide]

would this be a stupid idea

[vivid tiger]

anyways this is a tangent

[white oxide]

could we take phi of both sides

[white oxide]

yeah ur good

but like im not sure what any of what you're talking about is lol

like what a mobius function is and square free

but it sounds cool

[vivid tiger]

my intuition was to try to use the generators to say something about the prime factorization

[white oxide]

huh okay

abstract algebra is making me dislike number theory more and more

[vivid tiger]

totient of both sides seems messy

abstract algebra made me like number theory more. considering also that afaik number theory has abstract algebra as a core part

[white oxide]

wait...

isn't this just a linearization thing

sorry if im butchering math terms

but like

say n = 18

wait no

i'm stupid

nvm

jesus

[vivid tiger]

i think the course of using partitions may work

8 = generators of
0 1 2 3 4 5 6 7 + generators of
0 2 4 6 + generators of
0 4 + generators of 0

[white oxide]

sorry could you explain this

wouldn't it be 7 = generators of Z_7 + generators of Z_1

[vivid tiger]

oh i just expanded out the formula for Z_8

[white oxide]

oh wait

oops

wait but Z_8 has order 8

bc isn't n = 8 so Z_n = Z_8

[vivid tiger]

counting is hard

[white oxide]

LOL no worries

wait how is that a partition though cuz they're not disjoint

would it have to do with taking the union or smt

[vivid tiger]

but let's see the generators:
1,3,5,7
2,6
4
0

[white oxide]

OHH

[vivid tiger]

so: if we have an element that generates Z_d1, we want to prove it doesn't generate Z_d2

aha!!

d1 divides n

if you don't share any common factors with d1

then you must share a factor with d2

because if you don't share any common factors with d1

but d2 | (n/d1) = [Z_n : Z_d1]. a generator of d1 creates a cyclic subgroup of Z_n with the order of Z_d1 (uhh duh, that cyclic subgroup is literally Z_d1)

[white oxide]

ok let me process this

you're probably right

[white oxide]

okay

wait

yeah

sorry im just trying to get there on my own and if i get stuck ill look

[vivid tiger]

i am not actually there though...

[white oxide]

oh

my

b

[vivid tiger]

no it's mine

i just realized that

[white oxide]

wait is it true that if you have some a in Z_n and gcd(a, n) > 1 then a is not a generator right

[vivid tiger]

yes, because that's what "not coprime" means

d1 | n. d2 | n. say your generator is coprime with d1. however: a generator of Z_d1 must have an order that divides n by Lagrange's theorem. in fact n/d1 is the first element in Z_d1 after 0.
n/d1 has the same order as the generator, since n/d1 is also a generator (all generators in a cyclic group have the same order).

[sonic coral]

an element a in Z_n is a generator of Z_n iff gcd(a,n)=1

[vivid tiger]

(effectively, turn the case of any generator into the case for the first generator).

so the order of your generator equals n/d1. but we know n/d1 is coprime with d1. so d2 divides n/d1. so n/d1=(the generator we are trying to show isn't a generator of Z_d2) is not coprime with d2

therefore we have disjointness

part 2 is that we need to prove we cover Z_10.

that's more obvious: if the number (say, m) is coprime to n, we are done as it generates Z_n. Otherwise, take the gcd and look at Z_(n/gcd(n,m)).

obviously m is coprime to (n/gcd(n,m))

furthermore it is similarly clearly less than (n/gcd(n,m))

so now we have a partition

so bob's your uncle, and now that sum from earlier works

[white oxide]

damn ur a genius

okay let me process this

[vivid tiger]

i'm not?

[white oxide]

i feel bad because like you did all the heavy lifting

[vivid tiger]

i'll accept "smart". ill even accept the praise of "very smart". but in my mind, genius's simply are not me.

[white oxide]

ok very smart person

[vivid tiger]

so, i accidentally did the heavy lifting?

mostly i got carried away

remember that i've done this before

albeit proven differently, what i just did feels vaguely what i did back in...early 9th grade?

but this time it's better

[white oxide]

lmfao

[vivid tiger]

because group theory

[white oxide]

i was doing geometry in 9th grade

[vivid tiger]

it makes it much cleaner with group theory

anyways.

[white oxide]

with sin cos tan

anyways

[vivid tiger]

there's a central concept here: the example

i thought "they can't possibly be disjoint, lemme come up with the obvious counterexamp...oh shit"

then, i thought "ok, maybe they are disjoint, lemme try to prove it"

then i thought "ok, that means a generator of Z_d1 must not generate Z_d2"

then i just gradually inserted in "coprime" and "divides n" and came up with that

and then the cleaned up two paragraphs (1. for disjointness 2. for covering) was hopefully straightforward

i knew from the start that there would be some sort of partition of the factors.

i did not know that there'd literally just be a partition of Z_n. i only started trying that after seeing you say it and being unable to prove you wrong

is this post mortem helpful or am i beating a dead horse without a medical student to carefully notes, instead just splattering blood over the student while the student stares at me in horror?

[white oxide]

uh lmao

yeah that's helpful

thank you but yeah i see your point

it means a lot

i'm still a bit confused on disjointness

[vivid tiger]

oh! gotcha

uhh, wait, which part?

the rephrasing is that if you are coprime with divisor 1, you must not be coprime with divisor 2.

the reasoning is that the order of your generator equals n/d1

[white oxide]

i'm a little bit confused as to why n/d1 is a generator - is it because it's coprime to d1? how would we prove that?

[vivid tiger]

i thought of this by first thinking only of the first generator.

[white oxide]

or is that just a fact of number theory that i haven't encountered yet

or is it just trivial and i'm dumb lmao

[vivid tiger]

so, i thought like this:
0 4. 4 generates this. 4 = 8/2
0 2 4 6. 2 generated this. 2 =8/4

[white oxide]

wait are you thinking multiplicatively again

[vivid tiger]

...i don't think so?

[white oxide]

wait cuz they're not disjoint

[vivid tiger]

but they are?

[white oxide]

wait but 4 is in both

[vivid tiger]

4 doesn't generate the second

[white oxide]

oh we're talking about generators

sorry my brain is fried

i've been on this problem for 2 hours

don't know if that's normal or not

but regardless okay i see your point

[vivid tiger]

mine's less good than it normally is

[white oxide]

shieeeet ok

[vivid tiger]

definitely is

[white oxide]

okay that's good to know

i come from the realm of computational math

so this is sort of new

[vivid tiger]

honestly better to break up the 2 hours

[white oxide]

yeah ur right

i have to go soon anyways

[vivid tiger]

it'll feel like little time for me, because if i'm stuck and really stuck and don't know what to do and google doesn't help/i don't want to google, then i just come back to it in the morning/checking the mail/waiting in line/something else

[white oxide]

can't think anymore even though you did all the thinking lol

[white oxide]

ohhh ok fair

[vivid tiger]

if i get somewhere, it may take 2 hours

where every so often i make progress

it may also take 2 hours to understand a single sentence in a textbook

hopefully you enjoy it.

[sonic coral]

the great thing about it though, is regardless of how long you spend on a problem, once you get it then it becomes trivial 😂

[vivid tiger]

for me, it makes the early algebra stuff less boring, because for me, algebra starts with a bunch of things i personally don't care as much about.

well, it starts with a bunch of stuff i do (like proving grade school algebra properties! proving anything! also groups, so cool and deep!)

[white oxide]

ah that makes sense, given the seemingly limitless stream of pedantic and condescending assholes I’ve encountered in this channel LOL

[vivid tiger]

even the non condescending assholes feel that

i feel that

[sonic coral]

i havnt encountered anyone like that

[vivid tiger]

like instantly after being confused by this problem i feel like it's super obvious

[sonic coral]

and i’m in here quite a lot now

[vivid tiger]

math is a cycle of going "what the actual fucking fuck fuck this makes no sense and is meaningless and there are a million things required to understand this of which i know -5 truth is fake and reality is an illusion" to "huh, ok, that makes sense, oh my god it makes sense it's the most beautiful thing in the entire universe i'm now religious and pray to this theorem and chant euler's name in cathedrals shaped like platonic solids!" to "Proof: Trivial - go ask a nearby 5-year old."

[white oxide]

LOLLL

Proof trivial

from what I’ve seen so far that’s been true

Well regardless thank you so much for your help

It really means a lot!!

[vivid tiger]

why thank you. but it is weird that my late night rambles are called the most helpful

[sonic coral]

could i get a hint on how to show that a group G of order 2p^n where is an odd prime is not simple?

[tough raven]

Do you know Sylow's Theorem?

[tacit hemlock]

for a Dihedral group, what does it mean that the rotations "don't reverse order?"

[tender wharf]

wdym reverse order

what

[north sand]

Well do you know the geometric interpretation of D_n? If you draw the polygon and check the vertices in clockwise order, that order will be the same before and after rotation, it just starts at some other point

[coral shale]

flip sign of permutation, likely

[tacit hemlock]

ok I thought so too

Can someone please give me an intuitive way of understanding what a subgroup is?

[coral shale]

a group in a group...?

[tacit hemlock]

So for e.g. here

Are they all subgroups?

The notation Cn, Dn, Sn represent the Cyclic, Dihedral, and Symmetric groups respectively

[tender wharf]

well yeah sure

C_n is isomorphic to a subgroup of D_n specifically

and likewise

[tacit hemlock]

This is the definition of a subgroup given by our lecturer

[coral shale]

Whats the issue

[tacit hemlock]

I am currently interpreting this something like

"if H is a group, it is also a subgroup"

[tender wharf]

sure

it's a subgroup of itself

[coral shale]

If a subset is a group under the same operation, then its a subgroup

[tacit hemlock]

Thank you

[coral shale]

The 'closed' is redundant

[tacit hemlock]

How so?

[coral shale]

Take n vertices equally spaced on a circle.

• The group of permutations is isomorphic to Sn
• The group of reflection and rotations is isomorphic to Dn
• The group of rotations is isomorphic to Cn

[tacit hemlock]

Have not gotten to that part yet : )

maybe next part of the lecture

will keep watchign

[coral shale]

Its fairly easy to see the set containments, you just need to justify they are groups

[coral shale]

im replying to what that screenshot says

[tacit hemlock]

not sure what morphisms refers to in context of group theory

i.e.

[coral shale]

equal

'is isomorphic to' essentially means equal

in the sense that 2 groups have the same structure

They are the same except we have relabelled some elements

[tacit hemlock]

I see, thank you!

Took me forever to digest

[tacit hemlock]

I thought ∅ denotes the empty set?

Does it make sense to write H = ∅?

H is a group and ∅ is a set

[coral shale]

it does

its an abuse of notation used everywhere

Where G is a group, in writing G can denote either the group or underlying set depending on context

Besides, you are already using \in set containment symbol for elements 'in' a group

[cunning dust]

Well it doesn't make sense to write that a group is equal to \varnothing since groups are non empty

I think that's what they meant

[coral shale]

Apriori H is not a group

it is a set.

In order for set H to be a group, it must be non empty

[cunning dust]

Sure. I'm not sure what their issue here is, but you're right about the abuse of notation. But if H is a group it would not make sense to say H=\varnothing not because one's a group and one's a set but because groups are non empty.

[coral shale]

'for set H to be a group' <- already, the abuse of notation thinking comes in

[coral shale]

Its part of the proof to show H is a group

[cunning dust]

I don't know what they're proving

[cunning dust]

I was responding to this

[coral shale]

As youve pointed out, H being non empty is a necessary condition to be a group

[cunning dust]

I think I missed some context

[coral shale]

Well I mean yes, inherently H = phi is always false for group H

[cunning dust]

"Does it make sense to write H=\varnothing if H is a group...?"

[coral shale]

by definition

If it makes 'sense' to write H \neq phi, then it also makes 'sense' to write equality too

[cunning dust]

Yeah yeah I thought their question was specifically about equating a group with the empty set, not sets in general

[coral shale]

[tacit hemlock]

What does it mean for a function to "respect" a (group/binary) operation?

[quiet pelican]

We have that f(x * y) = f(x) * f(y)

For all x, y

[tacit hemlock]

is that interpretable intuitively

i.e. in the case of group homomorphism from group G to H

[summer path]

whatever action * you had before is the same as whatever action you have after applying the homomorphism f

though im not sure what you mean by intuitively here

[coral shale]

combine then map

map then combine

same thing comes out

[summer path]

this reminds me of how i didn't see commutative diagrams in my group theory class, but the other section had commutative diagrams everywhere lol

[delicate orchid]

You can describe “respecting the group operation” as a commutative square if you really want

[tacit hemlock]

why is the projection map's operation additive

[delicate orchid]

See if you can prove it yourself, it’s not too difficult

[summer path]

if you are unable to, try to write out your thought process and we can go from there

[tacit hemlock]

because they want to prove group homomorphism under modular addition?

since after the map, a+b (mod n) = a modn + bmodn

[coral shale]

you misuse the notation for modular arithmetic

That's something CS students would write

5 + 8 === 1 (mod 12)

This is how its written (equiv sign)

the mod 12 indicates we consider the equation mod 12

like everything - we're working 'mod 12'

[cloud walrus]

Tubular Cat

[coral shale]

Or formally, we are working within Z_12

and considering equality of classes

[tacit hemlock]

this is what the professor wrote

I think she's just trying to show why this projection map is a homomorphism

[coral shale]

informally

[summer path]

i think your prof is trying to emphasize that p(k) is in Z/nZ

[coral shale]

yes

a, b, a+b these are classes

formally its sometimes written [a] or a with a bar over

to make this explicit

[a] would then mean 'the set of all numbers in the same class as a'

So in Z_12
[13] = {13, 1, 25, -11, 37, ...} = {13 + 12n : n in Z}

[tacit hemlock]

Why can't the group operation(s) here be multiplicative?

[coral shale]

you mean notation wise?

it can be, itd just be confusing

we're used to denoting addition this way

[tacit hemlock]

I meant like

$$\phi (a * b) = \phi (a) * \phi (b)$$

[cloud walrus]

normalAtmosphericPa=101,325

[coral shale]

when the group operation is addition you generally stick to + symbol. It would cause confusion down the line in stuff like Rings (they have 2 operations, 'addition' and 'multiplication')

[tacit hemlock]

considering the definition for homomorphism that I was given as "a function that maps one group to another, respecting the binary/group operation," why can't the group operation refer to a group multiplication

rather than the addition

[tacit hemlock]

used here

[coral shale]

There is truly no multiplication or addition

There is one operation

,,\phi(ab) = \phi(a)\phi(b)

[coral shale]

we omit the operation and use the above multiplicative notation when it is clear from context

But in the case of specific groups like Z, Z_n, you don't want to do this

Because we are already have usual multiplication as well, and it would be confusing

Its a matter of notation that avoids confusion

===
When you get to rings - they have 2 operations; addition and multiplication.
Their homomorphism satisfies

[coral shale]

Z and Z_n is also a ring under the usual addition and multiplication

So it would be confusing to use the multiplication notation for the addition operation within the context of groups

[pliant forge]

$$\phi (a *{\text{dom}\phi} b) = \phi (a) *{\text{cod}\phi} \phi (b)$$

[cloud walrus]

Eso

[coral shale]

Oh and yes, they are different *

[blissful crystal]

in general, in Z mod n, you don't necessarily have inverses when it comes to multiplication

[tender wharf]

an exception is if n is prime

(make sure to throw out 0 tho)

[blissful crystal]

even stronger, you can use any n, as long as you only keep coprime elements

[blissful crystal]

but yeah if you look at 2 in Z_6 for example, 2 * 0 = 0, 2 * 1 = 1, 2 * 2 = 4, 2 * 3 = 0, 2 * 4 = 8 (= 2), and 2 * 5 = 10 (= 4). Hence 2 is not really invertible

[tacit hemlock]

wat does this mean

"up to isomorphism"

[chilly ocean]

well

for e.g. groups of order 4

it means that all groups of order 4 are isomorphic to either C_4 or C_2 x C_2

[quiet pelican]

Every group of order 1 is isomorphic to C1
Every group of order 2 is isomorphic to C2
Every group of order 3 is isomorphic to C3
Every group of order 4 is isomorphic to either C4 or C2 x C2

[summer path]

where Cn is the cyclic group of order n (i think they have been using Z_n notation)

[cloud walrus]

Raph

[chilly ocean]

Any hints?

[rustic crown]

use tensor and duals

the left side is iso to (CX)^v ⊗ (CY) and the right is iso to (CX) ⊗(CY)

since X is a basis of CX, you can define the isomorphism CX --> (CX)^v sending a basis element to the corresponding dual basis thing

[boreal inlet]

How do I show every k in K stabilizes all elements of G

we know that every element of K stabillizes x

[coral spindle]

Think about what it means to be in the kernel of the action

[boreal inlet]

OH I think I figured it out wait

is this correct?

[formal ermine]

use latex

[boreal inlet]

ok wait

lemme edit

[formal ermine]

you can just omit * everywhere

[boreal inlet]

let $K \subseteq of G_x$ for some $x \in X$. $K \text{is normal in} G$.
now $\forall k \in K$

$k \cdot x = x$

As $k \in K$, $gkg^{-1} \in K \forall g \in G$

$\implies gkg^{-1} \cdot x = x \implies g \cdot ( k \cdot ( g^{-1} \cdot x)) = x \implies k \cdot (g^{-1} \cdot x) = g^-1 \cdot x$

So, $k \in G_{g^{-1} \cdot x} \forall g \in G$

$\implies K \subseteq G_{g^{-1} \cdot x} \forall g \in G$

Now the action $\cdot$ is transitive, hence $O_x = X \forall x \in X$

hence $g^{-1} \cdot x \in X \forall g \in G$

So, $K \subseteq G_y \forall y \in X$

Hence, any k in K fixes any y in X.

hence, $K \subseteq Ker \Psi$, where $Psi$ is the associated homomorphism of the action $\cdot$.

[cloud walrus]

rikusp2002

[boreal inlet]

is this ok?

[coral spindle]

You could shorten this considerably by simply recalling that:

[cloud walrus]

Boytjie

[boreal inlet]

ohhhh shit I forgot about that

[coral spindle]

Try using this fact to shorten your proof

But it should be said that your proof is perfectly good

[boreal inlet]

thank you

ohhh wait

$K \subseteq G_x$

[cloud walrus]

rikusp2002

[boreal inlet]

$\implies gKg^{-1} \subseteq gG_xg^{-1}$

[cloud walrus]

rikusp2002

[boreal inlet]

as K is normal in G, we have

$K \subset G_{g \cdot x}$

[cloud walrus]

rikusp2002

[boreal inlet]

yeah got it

[obsidian loom]

I guess I understand what the set is but is there any better way to think about it?

[south patrol]

Yes, in fact this isn't how I'd define it

$\mathbb Z/n\mathbb Z$ has multiplication and addition (defined in the normal way). Some of the elements $[a]$ have multiplicative inverses (elements $[b]$ such that $[a][b]=[1]$). Such elements are called units. If you then go and compute what these are, you'll see the units form that set above (indeed, the U stands for unit)

[cloud walrus]

potato

[obsidian loom]

that makes more sense

So for example U_9 = {[1], [2], [4], [5], [7], [8]} right?

I'm asked to find out the order of each element of a few of these unit sets

I started doing: x = 2 (mod 9) so x^6 = 2^6 = 1 (mod 9) so ord([2]) = 6

but it quickly becomes difficult (and long) to do it that way

what do you think is the intended method?

[white yoke]

If R is a local ring with maximal ideal m, how can I prove that a ring homomorphism f:M_n(R) --> M_n(R/m) is surjective on units? I.e. for a unit A in M_n(R/m) there exists a unit A' in M_n(R) such that f(A')=A

I asked in the help channels but no one seems to be able to help with my questions as of recent

Also if a is a unit in a local ring, and b is a non unit, when why is a+b a unit?

[delicate orchid]

I think that is the intended method it’s just there’s an easier way to do it

Instead of doing x^6 and then mod 9 you can mod 9 at each step, so 2*2 = 4 mod 9, 4*2 = 8 mod 9, 8*2 = 7 mod 9, 7*2 = 5 mod 9, and 5*2 = 1 mod 9 - 5 steps so 2*2^5 = 2^6 = 1 mod 9

Not particularly needed for x = 2 but for larger x it’s useful to keep the calculations nice and small

[obsidian loom]

oh wow that sounds so much quicker

each day i learn something new about modular arithmetic

[delicate orchid]

If you want there is a way you can skip right to the orders of everything from just that sequence as well - note how every element of U_9 appeared at some point in that sequence

[obsidian loom]

i dont see how it shows the order of each element

[delicate orchid]

||2 is a generator for U_9|| I may have assumed you knew more about cyclic groups than you do, sorry

[obsidian loom]

alright, thanks a lot

[delicate bloom]

one way is since you showed U_9 has 6 elements you know by Lagrange's theorem the order of an element must divide 6. Good thing is, that means you just need to check if they square or cube to 1 mod 9, since if they don't - then they must be order 6.

I guess cubing anything gets you +1 or -1 so maybe that's just the way to go lol

[chilly ocean]

I don't see why the checking that this defines a ring relies on the ring properties of S. It seems to me that it instead relies on the ring properties of R, no?

[cloud walrus]

alcithoe

[agile burrow]

We need that S is a ring to ensure that the kernel I is an ideal, which let's us define the quotient structure

[delicate orchid]

Oh so that’s what a fibre is

[agile burrow]

It's analogous to how normal subgroups are precisely the subgroups that arise as kernels of group homomorphisms

[chilly ocean]

I see that they're about to introduce ideals, so I guess I might understand better if I read on a bit.

[agile burrow]

Oh yeah, once you read that all will be elucidated

[delicate orchid]

It seems a bit backwards you’re doing quotient rings before ideals

[agile burrow]

Just to answer your question explicitly at this point, to verify that the multiplication is, say well-defined, we require that if $r_1 + I = r_2 + I$, then $r_1 s + I = r_2 s + I$. To show this, we have $r_1 - r_2 = x$ for some $x \in I$. Then $r_1 s - r_2 s = xs$ which lies in $I$. The last part follows because $\varphi(xs) = \varphi(x) \varphi(s) = 0$ in $S$

So here we are actually using the ring structure of S (alongside the fact that we can only define a ring homomorphism if the codomain is in fact a ring)

[cloud walrus]

walter

[past temple]

is there an easy way to ascertain whether a representation of a group is irreducible from just looking at the properties of the matrix representations of the generators?

for example, if i have a 2-D representation in R2, and i see that the matrices of the generators only have complex eigenvalues, i can ascertain that the representation is irreducible because no vector is going to be sent to its own R-span

[west folio]

Euclidean algorithms is correct in that example ? :
a = 898754 b = -15654 898754 = -15654 x -57 + 6476 -15654 = 6476 x -2 + -2702 6476 = -2702 x -2 + 1072 -2702 = 1072 x -2 + -558 1072 = -558 x -1 + 514 -558 = 514 x -1 + -44 514 = -44 x -11 + 30 -44 = 30 x -1 + -14 30 = -14 x -2 + 2

[delicate bloom]

looks about right to me

[west folio]

i add a new test tell me again ...
a = -156588884
b = -898754
-898754 = -156588884 x 0 + -898754 -156588884 = -898754 x 174 + -205688 -898754 = -205688 x 4 + -76002 -205688 = -76002 x 2 + -53684 -76002 = -53684 x 1 + -22318 -53684 = -22318 x 2 + -9048 -22318 = -9048 x 2 + -4222 -9048 = -4222 x 2 + -604 -4222 = -604 x 6 + -598 -604 = -598 x 1 + -6 -598 = -6 x 99 + -4 -6 = -4 x 1 + -2

[west folio]

-156588884 % -898754 is negative ? Are you sure

a mod by 2 numbers can be negative ?

[meager jolt]

This should go in #elementary-number-theory

[west folio]

yes sorry

[meager jolt]

Here's a question:

When we have an ideal that's a subring of a ring of algebraic integers of some number field, can we pick an element of the ideal that doesn't generate it?

This might be a simple q

[delicate bloom]

yeah it's allowed

[west folio]

ok

thx

[meager jolt]

The reason why I'm asking is because I don't know what happens if r=1 in this proof

nvm I've realised that any such ideal has an integer in it

[past temple]

^

[past temple]

ok my understanding of checking irreducibility in general over C is

lets say you have a representation V of G

if you have the character table of G

then u take the inner product of the characters of V with the character table

to get the multiplicities of each of the irreducible repns in V

and if u get a multiplicity of 1 for one of them

and 0 for all the others

then your representation is irreducible

is this what the point of character theory is?