groups-rings-fields

[thorn jay]

ig

so eral

real

[karmic moat]

hmm i see

[thorn jay]

(in particular for any signature there's only a set's worth of varieties, even though varieties themselves are classes)

[karmic moat]

do y'all call them galois correspondences

[thorn jay]

ye

[karmic moat]

W

[thorn jay]

I mean adjoint functors doesn't seem fitting when you're really not thinking about stuff categorically

[karmic moat]

fair

[thorn jay]

ikr

it's funny how they pop up everywhere when you do anything related to order stuff

[karmic moat]

oh galois connections lol

i was confused i was thinking about the field theoretic galois correspondence

[thorn jay]

lma

[karmic moat]

wikipedia says ppl use "galois correspondence" for bijective galois connections sometimes

okay makes sense

yeah that's neat tho

[tough raven]

Oh, THAT'S why they're called varieties?

[thorn jay]

yeah, I believe so

equational classes is common too

[tough raven]

Yes, but that sounds way too old-fashioned.

[thorn jay]

Oh yeah but there is a Galois correspondence between these things

[thorn jay]

e.g. a Galois connection is between subsets of k^n and ideals of k[X1, ... Xn], with a Galois correspondence between algebraic sets and radical ideals

[thorn jay]

exactly

algebro-geometric terms are modern and hip

[karmic moat]

ah yeah

[thorn jay]

(it's also why you say signature, not species)

(lol)

[tough raven]

By which we mean 100 years old, I suppose.

[thorn jay]

younger than UA :3

Raghuram have you ever heard about tame congruence theory?

[tough raven]

I know tame, congruence and theory. 😎

In three different contexts. (But at least two of them are both in number theory.)

[thorn jay]

haha @karmic moat I was looking for solvable groups in AG online and came across this paper https://www.sciencedirect.com/science/article/pii/S0021869310003169/pdf?md5=8ac0062943ab29510005b7c485ebb73c&pid=1-s2.0-S0021869310003169-main.pdf which is about the algebraic geometry of groups

[quiet pelican]

None of those words are in the bible

[thorn jay]

that's funny, I didn't know that was still done

[karmic moat]

this stupid cloudflare thing wont let me get in

[thorn jay]

this, along with the AG of Lie algebras is the motivation for universal algebraic geometry~~, in which I've done some cool stuff~~

[karmic moat]

im no damn robot i played that capatcha game

[thorn jay]

neither is "hot say gex" but it's good nonetheless

[thorn jay]

I can send you the pdf in dms if you're interested even though it doesn't really answer your question

lol

[karmic moat]

i'll prob open it when i get home onto my pc

must be a browser thing

too lazy to download chrome

[thorn jay]

lmao that is fair

[karmic moat]

rn i need to read this damn book

or books i guess

[thorn jay]

right, good luck

what book(s)?

[karmic moat]

the main one is Flag Varieties by Lakshmibai and Gonciulea

but i'm supplementing with Humphrey's Linear Algebraic Groups bc the former skips a lot of proofs for this intro setup stuff

i have to present wednesday

i think semisimple/reductive groups, then prob like abstract root systems and stuff

[thorn jay]

oof good luck with that

[karmic moat]

cheers

[tough raven]

Oooh fun

[karmic moat]

not too worried about it it's just me, another student, and a prof

presentation is whatever i just want to get a good understanding of it on my own too

[thorn jay]

ugh I'm not sure if this says anything interesting or new

hate those kinds of papers >.>

well, it is interesting if you don't know the AG for groups yet I suppose

[thorn jay]

basically it's a weird ass theory that allows you to study "locally" finite algebras in a black magic ass way

you assign to each algebra a set of types, and based on what types don't occur you can magically make Malçev-type terms appear

(terms with certain properties)

[tough raven]

Hmmm.

[thorn jay]

in trying to write down the proofs myself I accidentally discovered an alternative way to approach this lmao

relying more on monoid theory

[tough raven]

fg subalgebras are finite sets?

[tough raven]

type = the model theory type?

[thorn jay]

no I meant that you study finite algbras "locally" by considering so-called "localisations" to "nice" subsets of your algebra

it honestly took me a while to get my head around all the concepts

The original idea is looking at the ranges of nonconstant unary polynomials f : A -> A, where A is finite and simple. Then, if there is a minimal such set U, then you can prove that U must be the range of an idempotent polynomial (one where f^2 = f), and in fact any minimal such set must be the range of an idempotent and they are all isomorphic in some sense.

This concept then gets generalized to arbitrary algebras in some wacky ass way

I've accidentally stumbled upon a way to view this as the study of certain right-ideals of the monoid of unary polynomials lmao

[earnest delta]

<1>=01,2,3,4,5

<2>=0,2,4

<3>=0,3,6

<6>=0,6

What should I do next

[late python]

How do I get started to prove this? I thought the hint implied induction but that doesn't work

[kind temple]

following the hint, if there is no such n, show that \sum_j=1^n 1 = \sum_k=1^m 1 implies n = m for all natural numbers n, m

i would proceed by ruling out cases n < m and n > m. and then using the hypothesis again somehow

what is actually going on is that you are showing that the unique homomorphism Z —> (the additive group of F) determined by taking 1 to 1 is not injective

[idle girder]

What is N*? Is that positive integers or primes?

If positive integers, that hint seems overly complicated. Not sure how to explain that in a way that is not itself a hint.

[kind temple]

6 is the same as 0 in Z/6Z, so some of your ideals listed have 1 too many elements.

this problem is just definition checking. where are u stuck?

[rapid cave]

See that <2> is maximal because no other ideal other then <1> contains it

[kind temple]

its not terribly complicated, no? maybe not the simplest proof, tho.

[late python]

yes it is the set {1,2,...} all positive integers

[late python]

so I need to show a contradiction in all these cases?

[idle girder]

Yeah, it suggests a proof by contradiction, while you can just get the result directly by using that ZZ -> FF is not injective.

[kind temple]

yes

[kind temple]

yea, this is the other thing i suggested

[idle girder]

Ah, I read that as relating to the existing hint, instead of infinite -> finite, so non-injective.

[kind temple]

yea, both proofs are kind of doing the same thing

showing non-injectivity

[idle girder]

Right. I guess it's just a way to get you to consider sum(1 for j = 1, ..., m) = sum(1 for j = 1, ..., m') for m != m'. Okay, I'll shut up now to not give the game away entirely.

[sour rain]

I used to think that vector spaces could be thought of as fields acting on abelian groups
But is this definition even complete? How would we get distributivity in this definition

[cedar vault]

I think any ring action on an abelian group G will include distributivity

Since End G will have an additive structure given by pointwise addition and a multiplicative one given by composition

So more generally you can think of R-modules as abelian groups with an action of R, and the vector spaces will just be F-modules

[swift tundra]

I really like this characterization of a vector space (as a function from the field into the endomorphism ring) because it give a bit more reason imo why we have the 4 other vector space axioms.

[sour rain]

ok wait it took me a while to understand what u meant by this

do u mean like an R-module can be thought of as a unital ring homomorphisms from ring R to End G?

cause then the axioms are pretty clear and thats really neat

[delicate orchid]

yeah that's right

[sour rain]

o thats pretty sweet

[glad osprey]

That's pretty much the definition of "field acting on an abelian group" isn't it? If R is a field and G abelian

[cedar vault]

Well it looks like they defined it by listing the axioms instead

Ofcourse they are equivalent

[earnest delta]

how is this true?

nz will not have unit element

1

??

[rapid cave]

nZ is not Zn

nZ is multiples of n

Zn is integers modulo n

[earnest delta]

you meant=n is 1 then it is z right?

n=2 then 2z

ohh it fails for 2z

no identity

please be quick

[idle girder]

The claim is that (nℤ, +, ×) is a ring without unity. Why are you looking for a unit element, when none is needed for a non-unital ring?

[astral ivy]

What's the motivation for introducing stabilizer and orbits? Isn't stabilizer basically equivalent to identity element and orbits are just all other elements?

[earnest delta]

i feel like stablizer is set of all orbits?

all possible path which is made by orbits?

[delicate orchid]

The set of all orbits is the original set

[rapid cave]

union*

Hi wew

[rapid cave]

I don't know the exact motivation, but look at the case of where a group acts on itself by conjugation. Here orbits are conjugacy classes and stabilizers are centrelizers

[earnest delta]

https://youtu.be/JHNkDTJCnpw?si=3VFfWQ_wqHBAQBV1

this video is helpful in it

[delicate orchid]

I don’t really know what you mean by this. It’s true that every orbit of x is isomorphic to G/Stab(x) as G-sets

So in that sense the stabiliser can be thought of as an identity element

[earnest delta]

orbit is like when all the elements acts on a fixed point

group is G and X is set

[dull ginkgo]

shategory theory

[delicate orchid]

Orbit is where it goes. Stabiliser is what doesn’t move it

[delicate orchid]

This ain’t theory no more this a cold hard fact

[dull ginkgo]

Is G-set just the category of sets a group acts on

[earnest delta]

doesn't move it?

[delicate orchid]

No it’s the category of sets with a G-action

[earnest delta]

it is identity element no?

[dull ginkgo]

Every shit gets cold and hard on the concrete floor bucko

[delicate orchid]

Yes… the identity is in every stabiliser because they’re subgroups

[earnest delta]

how is this true?

[delicate orchid]

Im gonna be honest you’re not really making any sense

[earnest delta]

z has inverse element is z?

commutative yes

[rapid cave]

CRO?

[earnest delta]

CRU

commutative ring with unity

[rapid cave]

oh

ring doesn't require inverses exists for multiplication

[earnest delta]

ohhhhhhh

got it

unity exists yes

thanks

because matrices are not commutative right?

[rapid cave]

yes

[astral ivy]

I see thank you

[astral ivy]

I'm just learning about them right now so I asked out of curiosity

[delicate orchid]

If you think about a group acting on itself by multiplication, then you’re right in that every stabiliser is just the identity element and the unique orbit is the entire group

[earnest delta]

can anyone simplify this statement

i am looking for unit elements in ring theory

if I take z_6

=,01,2,3,4,5

[rapid cave]

simplify?

[earnest delta]

make it simple to understand

[delicate orchid]

U(6) is the set of elements coprime to 6. There isn’t really much to simplify

[earnest delta]

by the definition i gotr 1,5

[delicate orchid]

That’s correct

[earnest delta]

how it is unit elements

[delicate orchid]

Because 1 is obviously a unit and 5^2 =1

[earnest delta]

does it not have any connectiion with other elements

[astral ivy]

Can you give me an example where there's more than one stabilizer?

[earnest delta]

5.2mod6 =4

[delicate orchid]

Let G act on itself by conjugation. Then the stabiliser of an element x is the centraliser C_G(x)

[earnest delta]

i gave u a link

which is very useful

[delicate orchid]

And the orbits are the conjugacy classes

[delicate orchid]

Ok… what’s your point?

If you multiply a unit by a non unit you get a non-unit

[earnest delta]

in z6 i took a=2,b=5

[delicate orchid]

Yes if there exists. Not for all

[earnest delta]

ohh symbol reading mistake

thanks

[delicate orchid]

No worries

[earnest delta]

yes there exists

5 into 5

[thorn jay]

this is an important example in itself

[knotty badger]

i think the presence of K

if the direct sum was 0 then you'd need K = M

[thorn jay]

because K is given, not asked

for any K there must be a complement of K in M

[earnest delta]

it is integral domain

a.b=0

opps wait

[knotty badger]

I’ll leave this to you enpeace

[thorn jay]

yeah

the real kicker (I am not 100% sure if this holds for general rings) is that the converse holds; if any submodule has a complement, then M must be a direct sum of simple modules

..?

no that's the conclusion

well it's easily seen that a module is a sum of simple modules iff it is a direct sum of simple module

s

because, if A and B are simple submodules of M, then either A = B or their intersection is (0)

so if M is a sum of simple submodules, then if you remove all the copies in the sum you get a direct sum

[twin phoenix]

is the group of 90 degree rotations isomorphic to addition modulo 4?

[proud vigil]

yes, you can map the generator 1 to the generator 90 degrees

[rapid cave]

Both are cyclic of order 4 so yes

[delicate orchid]

cyclic group? call that the tour de france

[twilit wraith]

i have to prove that if H is normal in G and it has prime index p, then for any subgroup K in G, either K \leq H or HK = G and |K: K \cap H| = p

the |K:K \cap H| = p part follows immediately from the second isomorphism theorem

but idk how to show that HK = G

[twilit wraith]

i know intuitively that H having prime index means its "right under G" in the subgroup lattice and so since K is not a subgroup of H, HK will be bigger than H but also a group and so it must be G

[thorn jay]

the usual definition for R[X] that follows the same universal property as for when R is commutative, has X not commuting with the elements of R

[twilit wraith]

but i dont know how to prove that H is right under G in the subgroup lattice

Yes

Because H is normal

[proud vigil]

G/H is cyclic, and since the size is prime, any coset of H that isn't H will be a generator

since K is not a subgroup of H, we can find k in K not in H, so kH != H, meaning kH generates G

i think this works

[twilit wraith]

Ah I had that first part

For the second I just take the union of all such k in K right

And thats equal to KH but also G

[proud vigil]

mhm yeah

or i guess i thought of it differently but i think it's the same idea

(kH)^r = k^r H

[twilit wraith]

Why is it that kH generates G and not some other subgroup tho

[proud vigil]

and every element of G belongs to some (kH)^r

[twilit wraith]

Yeah I get it now

I see why its all of G

[proud vigil]

ya just standard coset partitioning

[twilit wraith]

The part that was tripping me up is knowing that theres some k in K that allows kH to generate G/H

But since its prime any representative not in H will generate G/H

[proud vigil]

mhm

[twilit wraith]

Given a group G of order p^a m where p doesnt divide m, a subgroup P of G with order p^a, and a normal subgroup N of G with order p^b n where p does not divide n, i have to show that |P \cap N| = p^b

ive managed to show that the order is a power of p but idk how to show that it must be a p^b exactly

[unique flint]

Does it help that P is a Sylow p-subgroup?

[twilit wraith]

cant use any sylow theory

best i have are the four isomorphism theorems, coset theorems including lagrange

[unique flint]

Maybe you can show that the subgroup of N generated by elements with orders that are powers of p is a subgroup of P

Well, maybe not, actually.

[twilit wraith]

so far i just have that the if |P \cap N| = p^k then k is \leq both a and b

i guess just at this point i have to prove it cant be less than but the reason why that is isnt coming to me

[unique flint]

Maybe you can use the second isomorphism theorem

[twilit wraith]

it is relevant to the chapter

that might be the move

but im not sure how exactly i should go about it

the problem is that the next part is finding |PN/N|

which is trivial given the second iso thm

but that also requires that i know |P \cap N| so itd be circular

[proud vigil]

i think i found a solution

it feels a little too easy

by second iso

|PN| / |N| = |P| / |P intersect N|

since |P| = p^a, |N| = p^b n, showing that |P cap N| = p^b is equivalent to showing that |PN| = p^a n

P and N are both subgroups of PN, so p^a divides |PN| and p^b n divides |PN|, meaning p^a n divides |PN|

nvm this doesn't show that they're equal :/

[unique flint]

So you would need to show |PN| = p^a n... Yeah I see

[twilit wraith]

we know that p^a dividing |PN| means its the largest power of p that divides |PN|

[proud vigil]

yeah... just need to show |PN| <= p^a n to finish this but i havent figured that bit out yet

[twilit wraith]

since PN \leq G

[twilit wraith]

the other problem with this is that it would basically cause you to find |PN/N| before |P cap N|

but the problem in D&F is worded in such a way that expects you to find it in the other order

[proud vigil]

oh

[twilit wraith]

maybe theres some particular theorem i need to use that im forgetting

[unique flint]

Probably something about N being normal

[twilit wraith]

yeah

im wondering if i could do something with G/(P cap N) and the third iso thm

ooh

wait no G/P isnt a group

wait neither is G/(P cap N)

lol now im just making shots in the dark this isnt good

ok i got it but i did look up help which feels bad

we do start with PN, so |PN| = p^k t

we know that k = a tho

so |PN| = p^a t

importantly n divides t since N is a subgroup of PN

then |PN/N| = p^{a-b} s

where s is the quotient of t and n

then by second iso we have |P/P cap N| = |PN/N| = p^{a-b}s

[unique flint]

Ah okay, I see, you can show s = 1 because P cap N is a p-group

[twilit wraith]

but as |P| has to divide |P/P cap N| we must have |P/P cap N| = p^{a-b}

wait

isnt that not right

theres no situation in which p^a divides p^{a-b} s

[proud vigil]

yeah it's the other way around? |P / P cap N| divides |P|

[unique flint]

P cap N only contains elements of order a power of p, so it must have order p^k

[twilit wraith]

wait why is this true

[proud vigil]

er just because we're dividing a number by one of its factors

[twilit wraith]

oh duh

[unique flint]

|P| = |P / P can N| |P cap N|

[twilit wraith]

yeah idk why i had that train of thought thats obviously true

good thing this is the last problem on the hw bc i think im nearing needing a math break

[proud vigil]

this also just follows from lagrange on P cap N wrt P, im confused why this adds new information we havent already deduced though

[twilit wraith]

anyways from here the rest follows pretty directly

so i think i got it

[unique flint]

It's just a way to show s = 1 above

[proud vigil]

hm i dont think i get it
s is such that
|PN/N| = p^{a-b} s
right

[unique flint]

Yep and s has no powers of p

Since it divides n

[proud vigil]

right

how do u get to s = 1 from |P cap N| being a power of p

oh i see

we just use the theorem again

[unique flint]

|PN/N| = |P/(P cap N)| by second iso thm

[proud vigil]

yeah

okay cool

[unique flint]

lol I feel so out of practice

[twilit wraith]

ugh i hate looking up solutions but i was so lost

feels bad

fortunately most of the problems on this homework i was able to do independently so at least theres that

[proud vigil]

i think you're still gaining value out of it, you have a perspective to apply to other problems in the future

[twilit wraith]

very true

ive already done algebra once before so this time around its serving as a way to open up my perspective

and it definitely wont be the last time i do intro algebra

though i should only need to do it once more

@proud vigil thank you for your help

actually wait i need help phrasing something

i had to prove that if |G| = pq for primes p,q then either G is abelian or Z(G) is trivial

since Z(G) is a subgroup it has order 1,p,q,pq

how would i work the case where Z(G) = p or q?

you end up getting that G is abelian but it implies that Z(G) = pq

so it wasnt p or q to begin with

do i approach this as a contradiction despite having arrived at one of the sought-out conditions before arriving at the contradiction?

[proud vigil]

yeah that's fine, we do a similar thing when proving groups of size p^2 are abelian

[twilit wraith]

should i just in parentheses state that it technically implies that Z(G) could not be p or q

well that just seems flimsy then

[proud vigil]

like you showed that

Z(G) = p -> Z(G) = pq

and that is a contradiction, so Z(G) != p

that's totally valid

[twilit wraith]

cool

[novel spruce]

I am currently proofreading a book, and at some point the authors claim that S_4 is isomorphic to a subgroup of GL_2(R), without any further explanation. I somehow find this hard to believe. Is there anyone who happens to know either an embedding of S_4 into GL_2(R) or a proof that this does not exist?

[quiet pelican]

It can’t happen
Consider where the transpositions map to
By considering the eigenvalues, they must map to reflections in some lines
Call the lines corresponding to (12), (23), (34) respectively l_1, l_2, l_3
The fact that (12)(23) has order 3 implies the angle between l_1 and l_2 is some multiple of pi/3
Similarly the angle between l_3 and l_2 is some multiple of pi/3
As (12) and (34) commute and are distinct, the angle between l_1 and l_3 is pi/2 or 3pi/2
But this is a contradiction, as we get that some integer multiple of pi/3 is pi/2 or 3pi/2

[knotty badger]

Another way to see this is that (1234) has to map to a matrix A with A^4 = 1

This has simple roots over C, so is diagonalisable over C, with possible eigenvalues 1, -1, i, -i

In order to have order 4, the eigenvalues have to be i and -i, meaning that in some basis (1234) maps to a 90 degree rotation

But then (1234)^2 = (13)(24) maps to a 180 degree rotation, i.e. -id, which commutes with every matrix

This would imply (13)(24) commutes with every element of S_4, which it does not

[novel spruce]

Thank you for your answers!

[thorn jay]

this is a really good argument

[knotty badger]

:D

It’s rare that i can make a good argument in algebra

[tough raven]

Another argument using more machinery: GL_2(ℝ) ⊆ GL_2(ℂ), so S_4 maps into the latter as well. By the representation theory of symmetric groups (specifically, because S_4 has no 2-dimensional irreducible representations), up to conjugation, the only possibilities are the trivial representation s ↦ id, the "sign" representation s ↦ sign(s) id, and the direct sum of both s ↦ diag(1, sign(s)). None of these are injective - they have kernel S_4 or A_4. The same is then true for any group homomorphism from S_4 to GL_2(ℝ) as well.

[knotty badger]

yep

equivalently, a basis for a module is a chosen isomorphism with a free module

oh uh

for me every ring has unity

[thorn jay]

well yes

infinite sums are not algebraic things

they are analytic

[knotty badger]

yep, this is also how a free module on an infinite set works

it's a specific way to turn a set into a module

[south patrol]

Eh I mean free module does mean this lol

A free module is indeed a module with a basis

[knotty badger]

interesting

for me i prefer free module referring to the specific construction via functions with finite support etc

[knotty badger]

since then this makes more sense

[south patrol]

This is standard

Multiple equivalent ways to define it

Probably easiest is that the only submodule of M containing that set is M

Yes, every element can be written as a linear combination of elements of the set

This is equivalent

That's what linear combination means

Yes, same as in lin alg

Oh I guess you are worried cause of non-unitality lol

I cannot speak about them

Lol

I have been assuming unital

Shouldn't matter cause you can just absorb into the linear comb

[kind kindle]

question. What is an algebra exactly?

Like what's the definition of an algebra

[rapid cave]

an A-algebra is a ring B with an A-action

[kind kindle]

an action is just like generalized multiplication right

[rapid cave]

you can think of this as a way to multiply by an element of A

[kind temple]

alternatively, an R-algebra is a ring with an R-module structure

[kind kindle]

alright sounds reasonable

also i've asked similar questions before, but I know elitpic curve operations basically use a rational function as a group action over a finite field in 2d

is there a way to construct rational functions like this that represent some sort of group over a finite field?

sorry if this is trivial/annoying

i know finite groups of lie type are a thing

i'd be interested in some sort of low dimensional way to represent one of those lie type groups using rational functions over a finite field on some tupple of finite field values i guess

eliptic curves feel like a trivial case of this where the group being represented is a direct product of finite cyclic groups

if we find a sufficiently good non abelian group/way to represent elements of that group in a way where conjugacy is hard, we should get post quntum crypto for free

Matrix representations have 2 issues:

1. Linear algebra attacks on conjugacy
2. really big

But if we have this group, the public key cryptography scheme i came up with is as follows:(Could be generalized if we have an otherwise desirable group where conjugation is easy to some other hard non abelian group problem)

Alice wants to send a message to bob. They agree on the following public parameters:

A public finitely presented non abelian group.(With efficiently solvable word/inverse problem and where taking powers is also efficient and some normal form)(finite group prefered)
a public element of a such that ord(a) >= 2^b where b is the bit security

bob has his private key k. Which is a private element of G. His public key p is p=k*a*k^-1.

If alice wants to send a private message M to bob using his public key, she does the following.

Chose a random number between 1 and ord(a)

2. Compute a^r and hash it using some hash function H. compute c = H(a^r)

3. Alice uses a symmetric algorithm to encrypt her message m using c(we denote this E(c, m))

4. Alice then computes p^r and sends (p^r, E(c, m)) to bob.

5. Using his private element k, p^r = k*a^r*k^-1 , so bob computes k^-1pk = a^r. From a^r bob computes H(a^r)=c and decrypts E(c, m) and recovers m

If eve is interloping on this conversation, and gets (p^r, E(c, m)). Eve would have to both break the conjugacy problem over G in order to recover the key. if She can compute discrete logs over G efficiently, she can also break the encryption however. She could compute r from p^r than recompute a^r to recover the key.

[noble nexus]

It should include anything you can get to from the set by applying ring operations

if ur ring isn't unital then saying linear combinations is maybe a bit sloppy

[karmic moat]

Not sure about your confusion

What do you think it should be a subset of?

Well K is a submodule of F which is isomorphic to R^n

Pi is a map from K to R as defined

If pi =0 then it sends everything in K to 0

Ie the kernel is K

As desired

By definition of pi

It’s not all possible n-1 tuples to be precise

It’s all n tuples with 0 in the first coordinate

No the domain of pi is K

There’s no M in your screenshot lol

Ah i see what youre saying then

It probably means ker pi as a subset of K

Not all n tuples w 0 in first coordinate

Maybe. I’ll reread

Yeah should just mean ker pi as a subset of K, which ends up being all of K by definition

the result follows because K = ker pi is a submodule of R^{n-1}

which we know is free by induction on n

[earnest delta]

9z is maximal ideal of 3z but it is not for z why?

[rapid cave]

9Z is a subset of 3Z which both are ideals in Z. So 9Z is not maximal in Z by definition of being maximal

[unique flint]

This is very interesting. Have you looked into Anshel-Anshel-Goldfeld key exchange? It's not exactly the same, but it's similar in also using conjugation in a non-abelian group.

I'm also working on some group-based cryptography, though what I'm working on isn't anywhere near as secure, yet I'm still having trouble coming up with an algorithm that can crack it.

[kind kindle]

Yeah I have. My goal though wasn't just a one time key exchange but a non interactive public key cryptosystem(all alice needs to send bob a message is a public key. no other communication is required). That was also relatively compact and able to work over arbitrary groups. Someone has probably come up with this before, but i'm suprised it's not on wikipedia making me think that maybe it has some trivial flaw i'm overlooking?

But if we found a suitable group, this could be far far faster than lattice based cryptography an on par with ecc in terms of speed assuming a suitable group and (possibly nonlinear) representation can be found(by nonlinear representation i mean for example the way ecc points are reprsented despite technically being a cylcic group)

i.e, the group operation does not resemble a bilinear map over a vector space in this presentation

general linear groups over a finite field would probably work for this. But that feels suboptimal in terms of keysize

My thought is some weird nonlinear automorphism from some finite simple group of lie type to F^n under which element inversion/exponentiation re computable efficiently

sporadic grpups may or may not work but even the monster group(which is too structured and computationall intensive) only has security of liek 153 bits assuming no better conjugacy aglorithm than brute force

but for the monster group, there probably is a way better algorithm

my hunches are weird representations of symmetric/alternating groups and lie type groups

that are relatively low dimensional

like technically via polynomial interpolation you ccould make a nonlinear representation of any group over any field but like it'd be so absurdly big as to not be worth it

we want to find these representations that are simple enough for practical use.

eliptic curves are special because they give an automorphism for a cyclic group over which discrete log is hard for that automorphism

So essentially, it's a one way automoprhism that's hard to convert back into a simple form

[unique flint]

I feel like maybe the reason it's not commonly used is that most groups that are complex enough to be useful for cryptographic security are hard to do computations in

[delicate orchid]

big matrix. the classic. el classico.

[quiet pelican]

A map M -> M is composed of
(A) a map M_1 -> M_1
(B) a map M_1 -> M_2
(C) a map M_2 -> M_1
(D) a map M_2 -> M_2
If you figure out how so, your problem is essentially done

Yeah

[delicate orchid]

it's a direct sum because the matrix is diagonal

that's why the matrix is diagonal

yeah that's cause the matrix is diagonal

actual answer is this argument doesn't work. You can only add morphisms in the same Hom-set

you'd have to embed them inside End(M_1 (+) M_2) implicity and that defeats the point

[tardy hedge]

Wow u know Hom and stuff now

Just the other day u were confused by the definition

My bad G

[earnest delta]

Prime ideal and maximal ideal are the same thing?
If I have to find these ideals for z_12
What will be they?

2z,3z
Both are maximal and prime?via definition maximal says that it shouldn't be equal to z_6 and less than it

[mint seal]

maximal and prime ideals are not the same in general

[earnest delta]

In this example they are same?

[knotty badger]

They’re the same for finite rings

[mint seal]

also apparently for commutative Artinian rings (with 1)

[earnest delta]

Is there any difference between equivalence class and equivalence relation?

[knotty badger]

Yeah, equivalence classes are element-specific

[earnest delta]

[2]

Which means 2,4,6...like that?

And the relation is bound with reflexive,transitivity and symmetry

[knotty badger]

Depending on the equivalence relation, yeah

You may be thinking of interconverting between equivalence relations and partitions?

[earnest delta]

Class of element should satisfy equivalence relation properties?

[knotty badger]

Given an equivalence relation, the equivalence classes partition the set

And given a partition of a set, you can define an equivalence relation by saying a ~ b if they’re in the same subset of the partition

[knotty badger]

The equivalence class of an element is just a subset of the parent set, so things like “reflexivity” and “transitivity” don’t really apply

[knotty badger]

(This is cause finite integral domains are fields)

[earnest delta]

I didn't understand the equivalence class mean here?

@knotty badger

What are they trying to ask me?

[knotty badger]

It means which of the following subsets of A is of the form [a] for some a in A

[earnest delta]

But how I do I make [a] here?

For specific [2] ={0,2,4,6...}

[knotty badger]

Ok wait R isn’t even an equivalence relation

[earnest delta]

But in this question totally lost

[knotty badger]

I don’t see where you got this from

[earnest delta]

I didn't check

[earnest delta]

I am just explaining

How I see [] class

But struggling how do I make here?

[knotty badger]

The definition is $[a] = {x \in A \mid a \sim x}$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[earnest delta]

(2,1) is not there

(2,5)(5,1) is there

So equivalence not there

So the question is wrong?

[knotty badger]

Yeah

[thorn jay]

wtf

yeah

[knotty badger]

I mean I guess you could consider the equivalence relation generated by R

But at this level I think that would just be more confusing

[thorn jay]

"Let 1 be a negative number" ahh

[thorn jay]

yeah

[placid ginkgo]

Would it be right to think of this as a "recursive" proof where if G/N doesn't contain an element yN of order p and therefore p does not divide |y|, then we consider (G/N)/⟨y⟩ which has a smaller order than G/N, and then repeat this until we reach the condition that p divides a quotient group? Sorry if this is a really basic question, I've never come across this type of inductive proof before.

[rapid cave]

Its just induction:
p divides ord(x) or it doesn't divide ord(x)
in the first case there is an element of order p.
Otherwise you reduce to a quotient group and by the induction hypothesis there we have an element of order p.
Then we lift to an element of order p in G.

[placid ginkgo]

Oh makes sense

[tardy hedge]

lol

[placid ginkgo]

Ig I was overcomplicating it lol

Thanks

[tardy hedge]

Kind of basic questions, but the proof that every vector space has a basis relies on Zorn's lemma? Where does the property of k being a field come into play?

I just wanted to get clear in my head where exactly the ring being a field matters

[knotty badger]

i believe it's the steinitz exchange lemma

afaik this doesn't work for arbitrary rings

[tardy hedge]

yea the "replacement" lemma thing? honestly its kinda my bad cause anytime i see that proof it looks annoying so i just dont think about it much

[thorn jay]

well the simple modules are precisely the quotients by a maximal, and if F is a field then F is the only simple module, which all modules are built out of. But because it is also projective every (fg) module is a direct sum of copies of F, i.e. free. I'm sure you can extend that argument in some way using Zorn

ah yeah F is semisimple so every algebra is a direct sum of simple algebras

[kind temple]

you end up dividing out by a certain scalar in the proof of the steinitz exchange lemma

you aren't guaranteed to be able to do this in a ring

[tardy hedge]

i mean i would like to understand why you said all this but i think initially im not sure why you mentioned simple modules

If M is a simple R-module then M is isomorphic to R/m as an R-module?

[thorn jay]

yes

[tardy hedge]

i think i remember doing that exercise

[thorn jay]

the other answers are better and more direct

[thorn jay]

this is in my opinion the more underlying reason ig

and if every module has a basis, then every simple module must be isomorphic to R, but this means that R can have no nontrivial (left) ideals, as those are submodules. In the commutative case this already implies a field, and in the noncommutative case (for left-modules) we have a divison ring.

[rocky cloak]

Something you can prove in general using Zorn is that every R-module has a maximal linearly independent set.

Then when R is a field you can show that this is a basis.

Exactly where k being a field comes up in your proof would depend on which proof you have in mind I guess

[noble nexus]

If you have a maximal linearly independent set S that does not span, there is a vector v not in the span of S. We claim that S together with v is linearly independent. Suppose we have a linear relation that involves v $av + b_1s_1+\dots+ b_ns_n=0$. If a is nonzero, we can invert it and solve for v and find that v is in the span of S, which is a contradiction. Thus a=0, then each b is zero by linear independence of S

[cloud walrus]

Blake

[noble nexus]

the place where you use the field axiom is inverting the coefficient a if it is nonzero

[fiery dirge]

can anyone please clarify if this is what elementary and normal forms are?

Let's say we have Z10 x Z20 x Z30.

Then elementary form would be Z2 x Z5 x Z4 x Z5 x Z2 x Z3 x Z5 or rewritten from smallest to largest: Z2 x Z2 x Z4 x Z3 x Z5 x Z5 x Z5.

Now, normal form is we take the largest (powers) first and combine them. So Z4 x Z3 x Z5 = Z60, Z2 x Z5 = Z10, Z2 x Z5 = Z10. So the normal form will look like Z60 x Z10 x Z10?

Is this what I wrote correct?

(it is unclear to me if we are allowed to merge the same ones so instead of Z10 x Z10 we'd have Z100. I assume this is not allowed)

[rapid cave]

your forms are correct

The general rule is that
$\bZ_n \times \bZ_m \cong \bZ_{nm}$ if $n,m$ are co-prime

[cloud walrus]

ExpertEsquieESQUIE

[fiery dirge]

Okay thank you!

Can you tell me how do I examine if 2 groups are isomorphic? Unrelated to the previous example

[rapid cave]

this really depends on what 2 groups

[fiery dirge]

Okay, let's say Z12 and D3 x Z2

their orders are the same check

Z12 is abelian but D3 in the product isn't so that's where they dont match right?

they're all cyclic except D3

[rapid cave]

D3 is of order 6

[fiery dirge]

my bad

fixed it

but what else is there to check? Highest order of the element, group order, abelian, cyclic?

[rapid cave]

there are a ton of things

[fiery dirge]

thats why I dont know lol

but this is introductory course

so at least 5-6 most important things

[rapid cave]

the subgroup structure, normal subgroups, number of p-sylow subgroups

[fiery dirge]

i didnt learn sylow so we stop there

[rapid cave]

you will learn it sometime probably

[fiery dirge]

what do you mean by subgroup structure and normal ones? How do I examine that in the groups?

[subtle crypt]

Yes, if one group is abelian and the other isn't, they're not isomorphic.

[fiery dirge]

like how to see if D3 has a subgroup to begin with

[rapid cave]

D3 has 2 non-trivial subgroups, one is of order 3, and one is of order 2

[fiery dirge]

but if only 1 is non-abelian or non-cyclic then the product with that group is automatically not either, correct?

[rapid cave]

yes

[fiery dirge]

of order 2 being D2 right? I don't know which one is of order 3 since dihedral groups have 2n orders usually

[rapid cave]

D3 is a semi-direct product of C2 and C3

C2 is reflection

C3 is rotation

[fiery dirge]

ohhh I didn't actually get introduced to C denoted groups/subgroups, what are they?

[rapid cave]

C2 is Z2

just in an abstract way

C2 would mean "cyclic of order 2"

[velvet hull]

as long as a group is not prime cyclic it has to have at least one nontrivial proper subgroup, try and see why

[fiery dirge]

oh that makes sense I see the correlation to Z2

[fiery dirge]

figuring out stuff on my own is not my strength 😭 I would give up before even starting

[tribal moss]

Unfortunately there are two different naming conventions for dihedral groups. In one of them the subscript is the number of elements in the group; in the other it's the order of a primitive rotation.

[velvet hull]

I would still ask you to give it an honest attempt, at least

it's okay if you can't figure anything out just try

[rapid cave]

D3 will not make sense in the first naming conventioon

[tribal moss]

Right, so if the subscript is odd, you know it has to be the second convention (which it sounded like HMD was not aware of).

[subtle crypt]

Right. I think you can write them as tuples. For example, with D2 x Z3, you can write (element of D2, element of Z3). You can then look at (x, 0) + (y, 0), and if x and y aren't commutative, (x, 0) and (y, 0) won't be commutative, so the combined group won't be commutative.

[tribal moss]

Unfortunately even subscripts can be either one or the other convention...

[fiery dirge]

at D2 I mistakenly confused it with S2 and calculated the order as 2! instead of 2n (in which case the order is 4)

[rapid cave]

I always assume the second convention

[tribal moss]

Actually there are three different (but conjugate) subgroups of order 2: each of the three reflections together with the identity.

[rapid cave]

true

[fiery dirge]

what are possible orders of all subgroups? Do all of them have to divide the order of a group?

or is it contrary

they have to be coprime with order of a group

[rapid cave]

yes. This is lagrange's theorem

H <= G --> |H| | |G|

[fiery dirge]

so Z8 for example will have Z1 (e), Z3, Z5, Z7 and Z8 (Z1 and Z8 being trivial)

[fiery dirge]

yee

[proud vigil]

they cant be coprime, they have to be factors

additionally, not all factors correspond to subgroups

[fiery dirge]

alr so its Z1 Z2 Z4 Z8

got it

idk where being ccprime came from i mixed smth

[proud vigil]

in cyclic groups they do (like Z8), but not in all groups

[rapid cave]

hello

[astral ivy]

Hii

I have a question

[rapid cave]

I have an answer 🤑

[astral ivy]

When checking if 2 groups are isomorphic, one of the conditions is to make sure they have elements of the same order (or orders?). My question is how to find the order of various elements?

*how to find all orders of elements

[rapid cave]

what two groups are you thinking of?

[astral ivy]

Well let's start with something basic like Zn

To be more concrete let's go with Z8 and Z15 for example

[velvet hull]

in general that's a pretty hard / impossible thing

[tribal moss]

It depends critically on which kind of information you have about the two groups you start with.

[velvet hull]

but for groups with simple structure like abelian groups, dihedreal or symmetric groups it is doable

[rapid cave]

of course they won't be isomorphic because they have different orders

[astral ivy]

No no they're examples for finding orders of their elements but okay let's take Z8 and Z4 x Z2 then

[rapid cave]

every non-trivial element of Z8 has order 8

but every element of Z4xZ2 has order 1,2 or 4

[tribal moss]

"Finding the orders of elements" is not a general procedure that you can just apply to a random example. It's one of several possible strategies for proving that two groups are not isomorphic, but it doesn't always work.

[astral ivy]

Those are the ones I'm interested about as well as the Euler group Fn (x in F if gcd(x, n) = 1) might be the most complicated one for me

[rapid cave]

they are more interesting yes

[astral ivy]

Yep I got that (because we can take n = 8 and apply it to 1 aka 1+1+...+1 = 8

[tribal moss]

That's for a rather unusual sense of "non-trivial element", I think. There are also elements of Z8 that have order 2 and 4, and those are not really "trivial".

[rapid cave]

oh yeah lol

my brain though it was Zp

[astral ivy]

But because of that they're not isomorphic

Is this some term I'm not familar with? Why is this suggested 😭🙏

[rapid cave]

you typed that sometime

[astral ivy]

No way

I don't have typing history

[astral ivy]

Okay so regarding them how would it work?

[rapid cave]

what would you like to know about them?

[twilit wraith]

interesting, i didnt know that was called an euler group

[astral ivy]

Let's take F8 for an example.

F8 = {1, 3, 5, 7} right? gcd(n, 8) = 1 for every n from F8.

How would we check order of each element here? I don't understand the group operation well enough

[astral ivy]

We call it like that here, I don't think it's used internationally though

[twilit wraith]

i see

[tribal moss]

The group operation would just be multiplication modulo 8.

[twilit wraith]

maybe not in the whole US, but at my uni in the US we just call it (Z/nZ)^\times

[rapid cave]

or (Z/nZ)*

[tribal moss]

(And it turns out the orders in this particular case are very easy to find just by brute force computation of powers of each element until you see the identity come up).

[twilit wraith]

say youre trying to figure out which elements of (Z/nZ)* are generating elements

[astral ivy]

Yeah but take Z8+, to get to element 4 we do 1+1+1+1 ie apply group operation to 1 four times hence the order of element 4 is 4. But here it doesn't make sense

[rapid cave]

in general this group won't be cyclic

[rapid cave]

you do 1 * 1 * 1 * 1

[twilit wraith]

yes, but you can optimize the brute force computations when checking if it is

[astral ivy]

True but I already wrote what group operation was

[slow bison]

hi

[twilit wraith]

you only need to multiply as much as half the order of the group to see if you have a generator

[tribal moss]

No, the order of 4 in (Z/8Z, +) is 2, because when you add two copies of 4 together, you get 0 -- that is, the identity of the group.

[twilit wraith]

if you get -1 mod n earlier than half the group order's worth of multiplying some element by itself, it cant be a generator

if you do get it at the halfway point then it is

but only if thats the first time of it showing up

[astral ivy]

Ahhh yeahh I see

[astral ivy]

So can you show me how it'd work in Z/8Z *

[twilit wraith]

checking will demonstrate that (Z/8Z)* isnt cyclic

[rapid cave]

one more thing is that
$(\bZ/n\bZ)^\times \times (\bZ/m\bZ)^\times \cong (\bZ/nm\bZ)^\times$ by the CRT

[cloud walrus]

ExpertEsquieESQUIE

[twilit wraith]

which tells you what more standard group its isomorphic to

[tribal moss]

(Z/nZ)^× is only cyclic if n is a prime or twice a prime.

[slow bison]

don't n and m have to be coprime?

[rapid cave]

yeah

didn't write that

[tribal moss]

For example, 5 has order 2 because 5×5 = 25 == 1 (mod 8) and 1 is the identity in the multiplicative group. So the order is at most 2; on the other hand the order is not 1 because 5 itself (that is, the "product" of a single copy of our element) is not the identity.

[rapid cave]

isn't it a prime power

[velvet hull]

Odd Prime power, twice an odd prime power and 2 and 4 iirc

[tribal moss]

(Z/4Z)^× is cyclic of order 2, right?

[twilit wraith]

it has to be

[tribal moss]

Hmm, yeah, I think odd prime powers (and twice an odd prime power) will work too.

(But anything with two distinct odd prime factors, or multiples of 4 with any odd prime factor, will have too many elements of order 2 to be cyclic).

[coral spindle]

(Z/p^nZ)^x is always cyclic for odd primes p (which you can prove, if you like, by some funny p-adic arguments, but is not easy to see)

And indeed if a, b are coprime then (Z/ab)^x is (Z/a)^x x (Z/b)^x

so if you get two distinct odd primes you get a copy of C_2 x C_2 so can't be cyclic

It's kinda easier to see if you take this perspective of applying the CRT first and remembering the group-of-units functor respects products... so oft forgotten...

(of course it's an adjoint, so)

[rapid cave]

isn't it the galois group of a finite field extension or something?

I rememeber something like this

[coral spindle]

Yes it's the galois group of a cyclotomic field

but this doesn't really help to see it's cyclic lol

At least, I don't see why it would help

[rapid cave]

oh it was that the multiplictive group of a finite field is cyclic

mb

[coral spindle]

Right yeah that would make it simpler

I was hoping this would be what the p-adic proof looked like when I first heard of it

alas it isn't. but nonetheless it is a very nice proof. Let me find it

See 2.7 here https://kconrad.math.uconn.edu/math5020f11/padicinterpolation.pdf

[rapid cave]

this actually uses this result

[coral spindle]

indeed!

[astral ivy]

So in a sense a set for Euler's group F_8 is technically a subset (probably also a subgroup) of Z_8 (because Euler's group has gcd (a, n) = 1 condition it has to satisfy as well)

[rapid cave]

not a subgroup because its not the same operation

[tribal moss]

Definitly not a subgroup, because the group operations are completely different.

[astral ivy]

But the operation is still multiplication under mod 8?

[tribal moss]

When you view Z_8 as a group, the group operation is not multiplication modulo 8. (But instead addition modulo 8).

[astral ivy]

But what you did with 5x5 = 25 == 1 (mod 8) was multiplication

Even though Z_8 under multiplication is not a group indeed

[tribal moss]

When I said 5×5 = 25 etc, I was speaking about order in the multiplicative group Z8^× rather than in Z8.

[coral spindle]

To be crystal clear, Z8^× as a set does not contain the same elements as Z8

[astral ivy]

Z8 = {1, 2, 3, 4, 5, 6, 7, 8} whereas Z8* = {1, 3, 5, 7}?

[rapid cave]

Z8 = {0, 1, 2, 3, 4, 5, 6, 7} more conventionally

[dire turret]

idk which channel i should be asking this so ill ask here:

i am looking for some sort of algorithm which determines whether a solution to a system of linear equations modulo m where m is not necessarily prime exists (in particular im concerned with m=12 but the chinese remainder theorem says 4 is sufficient since 3 is prime). im not having a huge amount of luck (best I've found is to compute a hermite normal form or something which looks expensive) so i just thought id ask here to see if anyone knew a good algorithm for this?

[tough raven]

So the question is an algorithm to determine for m ∈ ℤ_+ and A, b k⨯n and k⨯1 matrices whether Ax = b has a solution for the n⨯1 matrix x (all with entries in ℤ/mℤ)?

By some change of variables (i.e. column operations on A) and linear combinations of equations (i.e. row operations on A) which are invertible over ℤ (and therefore over ℤ/mℤ as well), A can be brought to Smith normal form, i.e. a matrix with its only non-zero entries on the diagonal and the diagonal entries dividing each other in sequence. In other words, for some a_1 ∣ ... ∣ a_{min(k,n)}, A (x_1, ..., x_n) = (a_1 x_1, ..., a_{min(k,n)} x_{min(k, n)}, 0, ..., 0). Thus there is a solution for x iff a_i ∣ b_i for 1 ≤ i ≤ min(k, n) and b_i = 0 (mod m) for n < i ≤ k if n < k.

There is a formula for the a_i's without the full Smith normal form algorithm (namely, if x_i is the gcd of all i⨯i minors of A, then x_i = a_1 ... a_i, so a_i = x_i/x_{i-1}), so there are probably somewhat efficient algorithms for it, but I'm not sure how you would calculate the b_i (which are supposed to be the modified ones after doing the same row operations to b that were done to bring A into Smith normal form) without doing the row operations.

[dire turret]

woah that's very cool! but (maybe i am not understanding this correctly) wouldn't it be that a solution exists if a_i x_i + 12l = b_i for i ≤ k? ie gcd(a_i, m) | b_i? thank you by the way this is very helpful

[earnest delta]

Q/Z is not quotient ring why?

a.b=1

z+2 so we don't have multiplicative inverse like
1/2

Am I right?

z has no rationals...and 0 element is also not having multiplication inverse

[twilit wraith]

Not really able to follow what ur saying here

[violet spade]

Multiplicitive identity is the element such that 1a=a1=a

[earnest delta]

I'm asking for rings

[velvet hull]

yes, and rings need to have multiplicative identities

Q/Z does not

therefore it is not a ring

[earnest delta]

Yes. So how do I check Q/Z is not having multiplicative inverse

[earnest delta]

.

[velvet hull]

having a multiplicative inverse is not the same as having a multiplicative identity

[earnest delta]

Z is abelian in +

Z is semi group in ×@velvet hull

[velvet hull]

okay sure, what's your point

[twilit wraith]

What is the multiplicative identity of Q

Is that identity in Z?

[earnest delta]

Obviously 1

[twilit wraith]

And 1 is in Z right

[earnest delta]

Of course yes

[twilit wraith]

So then Q/Z is not gonna have a mult identity

[earnest delta]

Why?

Where is the problem?

[twilit wraith]

1 goes away

[earnest delta]

1 is persent in both

[twilit wraith]

No it isnt

Q/Z is the rationals in [0,1) looping over and over kind of like modulo

[earnest delta]

I still don't see the problem

Which restricts q/z to be ring

I meant quotient ring

[velvet hull]

1 is equal to 0 in Q/Z

you cannot divide by 0

so 1 is no longer a multiplicative identity

[earnest delta]

1.0=0.1=1

Okay got it

[twilit wraith]

There is no 1

[earnest delta]

Z should not have 0

[twilit wraith]

Ok instead think about Q and Z as additive groups

If you take the quotient group Q/Z what kinds of elements do u get

[earnest delta]

Z should be left and right ideal

okay so a-b where ab belongs to z satisfied

now let me check second property

a.r where a is from z and r is from q

Clearly a.r will not be in z

So it fails here to be ideal

[twilit wraith]

Thats a good way to see why the quotient ring fails

[earnest delta]

Why do we need ideals im quotient?

[twilit wraith]

Q/Z works totally fine as a group but multiplicatively theres problems

[twilit wraith]

Its kinda like why normal subgroups are needed for G/N to be a group

We need both left and right multiplication gives you closure for operations on ring cosets to be well defined

[earnest delta]

Thanks a lot. Actually I am learning concepts so I ask such low level questions thanks for helping

[thorn jay]

Z does not absorb multiplication by Q, so does not act as a proper zero element

in a ring you expect 0a = 0, so for any element a of Q, we'd need to have aZ \subset Z

[rapid cave]

||universal algebra||

[somber sleet]

Hey guys, can someone tell me which properties must hold for G, such that the subgroups of G are in one to one correspondence with the divisors of ord(G)?

[quiet pelican]

It’s exactly if G is cyclic

[somber sleet]

does this never happen otherwise? Like for abelian groups?

[astral ivy]

Let $n \in \mathbb{N}$ and $G$ a group of finite order such that $\left(g_1 g_2\right)^n = g_1^n g_2^n$ for all $g_1, g_2 \in G$. If $H = \left{g\in G: g^n = e \right}$ and $K = \left{g^n : g\in G \right}$, show that $H$ and $K$ are normal subgroups of $G$ and that $|K| = [G:H]$.

Firstly, I am slightly confused by the definition of $H$ and $K$ (element property and belonging to a set orders are different for some reason?).

Anyways, to show they're normal subgroups of $G$, I would first show they're subgroups. To do that, I demonstrate the rules:

1. $A$ must be closed and non-empty. By definition of $H$, it has element $g^n = e$ already in it. $K$ also has an element $g^n$.
2. If $g_1, g_2 \in A$, then $g_1 g_2 \in A$. For $H$ we take some $g_1^n$ and $g_2n$ such that $g_1^n g_2^n \in H$ but I am not sure how to show $\left(g_1 g_2\right)^n \in H$ as well. Maybe since all $g^n \in H$ are equal to $e$, then $g_1^n g_2^n = e = \left(g_1 g_2\right)^n$? $K$ looks like a direct application and should be more straight forward, for any $g_1^n, g_2^n\in K$, $g_1^n g_2^n \in K$ also and so is $\left(g_1 g_2\right)^n \in K$ by definition of the operation in group $G$.
3. If $g\in A$, then there exists $g^{-1}\in A$. For $H$, the identity either has no inverse or is it's own inverse, in both cases $g_1^n, g_1^{-n \in H}$ and group operation $g_1^n g_1^{-n} = e = (g_1 g_1)^n$.

For them to be normal we will need to additionally show conjugates $gHg^{-1} = H, g\in G$ or alternatively left cosets = right cosets $gH = Hg, g\in H$. Is what I've done so far good or I have some issues? This seems so simple but that's also what makes it hard to not make a logical mistake. I feel like I made mistakes but I also don't know how else I would rewrite them.

[cloud walrus]

danilojonic

[rocky cloak]

A finite group G is cyclic if and only if it has exactly one subgroup of order n for each n that divides it's order

[quiet pelican]

We prove it for “at most one subgroup of every order”
It holds for primes trivially
By induction on the order of G, every proper subgroup must be cyclic (as they have at most one subgroup of every possible order)
Then count the number of elements by their order - you find there’s at least one left over

[rocky cloak]

I'm pretty sure it's also true that any noncyclic group will have more subgroups than divisors of the order, but that's a little more tricky

[tribal moss]

It doesn't have a well-defined multiplication at all. (Never mind whether the operation that isn't there would have an identity).
Since 1/2 = 3/2 in Q/Z, there would need to be a common result of (1/2)·(1/2) and (3/2)·(1/2). But 1/4 and 3/4 are not the same element of Q/Z.

[rocky cloak]

I guess the fun thing here is that even ignoring that the multiplication would need to be related to Q somehow, there is no possible ring structure on Q/Z at all.

[tribal moss]

Right. If it had a ring structure with p/q counting as multiplicative unit, then (p/q)+(p/q)+...+(p/q) with q terms would be 0, but distributing ((p/q)+(p/q)+...+(p/q))×(1/2q) gives 1/2, a contradiction.

[rapid cave]

Addressing your concerns:

First, $H$ are all of the elements of $G$ of order dividing $n$. While $K$ is something else entirely.
By the property of the group $G$, you know $$\varphi: g \mapsto g^n$$ is a homomorphism.
Using $\varphi$ we can describe $H,K$ as $$H=\ker(\varphi),K=\mathrm{Im}(\varphi)$$

Second, regarding your proof.
You show the 3 group axioms for both $H,K$. I would advise you to not use $"A"$ as its not clear to the reader what $"A"$ even is.

Now, when you show $G$ and $H$ are non-empty you say that $H$ is not empty because it has an element $g^n=e$.
This is what you are supposed to show! So what you should really explain is why $H$ contains the identity. Likewise with $K$.

Then you say "let $g_{i}^{n} \in H$" and this is just wrong as the elements of $H$ are not of this form, this instead would apply to $K$.

[cloud walrus]

ExpertEsquieESQUIE

[earnest delta]

{0} is prime ideal of z but not maximal ideal why?

[rapid cave]

its the furthest it can be from maximal in fact

any ideal contains (0)

so as long as Z is not a field (and it isn't) it will have a non-trivial ideal and (0) won't be maximal

[astral ivy]

I wouldn't use A, it was just a shortcut to write it as "universal" definition since we already use H.

[astral ivy]

Now, when you show $G$ and $H$ are non-empty you say that $H$ is not empty because it has an element $g^n=e$. This is what you are supposed to show! So what you should really explain is why $H$ contains the identity. Likewise with $K$. Then you say "let $g_{i}^{n} \in H$" and this is just wrong as the elements of $H$ are not of this form, this instead would apply to $K$.

How can I explain why H contains the identity?

[cloud walrus]

danilojonic

[astral ivy]

And how would I do it for K?

[rapid cave]

e^n = e

this shows it for both K and H

[astral ivy]

What about showing they're normal? Okay, kerG = H so it's automatically true but what should I do for K?

[rapid cave]

what is gKg^-1?

[thorn jay]

sleeper agent activated

[rapid cave]

enpeace music summoned

happy noises

[thorn jay]

so true

[astral ivy]

Conjugates of K

gKg^-1 needs to be equal to K

And K only has elements that are in g^n form

So I guess g k^n g^-1 = (g^-1 k g)^n?

[rapid cave]

yep

[thorn jay]

yes

yes

[rapid cave]

But rx comes from the vector space structure of R^n

Not from the ring structure

[thorn jay]

well, it's used there

but you can use the diagonal map R -> R^n

this makes R^n into an R-algebra (given that R is commutative)

genuinely cannot tell if youre being sarcastic lol

[proud vigil]

Q/Z isn't a quotient ring because Z isn't an ideal of Q, right? is that the same idea as saying there's no identity element etc in Q/Z

[thorn jay]

yeah

[earnest delta]

Any famous examples of hamiltonian group rather than quarternion group

[rocky cloak]

I wouldn't really say those two are the same idea.

Like you need Z to be an ideal for there to even be a well defined multiplication on Q/Z. I guess it doesn't even make sense to ask what the identity of Q/Z is without any defined multiplication

[proud vigil]

thats what i figured, but originally people were saying that Q/Z wasnt a ring bc there wasnt an identity so i was just curious if those were equivalent

[karmic moat]

SU(2) is isomorphic iirc

[rocky cloak]

They are classified, but I wouldn't say any are famous.

They are all of the form Q8 times elementary 2-groups times odd abelian group.

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

[karmic moat]

How do you write SU(2) as such a product?

[earnest delta]

Are they all order 8?

[rocky cloak]

SU(2) certainly have subgroups that are not normal, so perhaps you have a different definition of hamiltonian in mind than me

[rocky cloak]

No. They're Q8 times something, so they will have order a multiple of 8

[karmic moat]

Tbh I didnt have a definition of Hamiltonian in mind (so prob shouldnt have said anything) but it is weird that a hamiltonian group can be iso to a non-hamiltonian group

[rocky cloak]

It cannot

[karmic moat]

Oh wait SU(2) is unit quarternions

[rocky cloak]

SU(2) is isomorphic to the group of unit quaternions under multiplication, but the quaternion group usually refers to
{±1, ±i, ±j, ±k}

[karmic moat]

Ahh yeah my bad

Is there a linear group isomorphic to the quaternion group

[quiet pelican]

By the Quaternion group you mean Q_8?

If so, yes, by cayley’s theorem, it embeds in S_8, and S_8 embeds in GL(k, 8) by permuting the basis vectors

[karmic moat]

Nice

[rocky cloak]

It is a subgroup of SU(2), that would be the minimal faithful representation.

(if your field is the complex numbers anyway)

[karmic moat]

I see

I only care about complex numbers (for now) anyway

[rocky cloak]

The minimal faithful representation should be 2d if and only if -1 is the sum of two squares. Otherwise it's 4d

(This is for fields not characteristic 2. If you have characteristic 2 I think 8d is the smallest)

[karmic moat]

Oh that's interesting, didn't know that

[fiery dirge]

Can anybody explain Burnside's lemma to me? I can't understand it well... (and group actions in general)

An example: Find the number of ways you can paint points of an 18-gon with 4 different colors if two colorings are the same when the 18-gon is rotated.

[rapid cave]

About group actions, the most intuitive examples are the geometric descripition of common groups

for example S4 can act on {1,2,3,4} by permuting the elements

[proud vigil]

i imagine expert will explain this better so ill just send what i wrote up

the set we're acting on is the set of colored polygons, not checking rotations. there's 4^18 different colorings this way, and they all belong to this set S.

take the action of G = C18 on this set, where an element g^n in G rotates a given polygon by 10n degrees. Then the orbit of a polygon (the set of polygons it can reach via actions from the group) corresponds to 1 unique coloring, and we just want the number of orbits

according to Burnside, the number of orbits is the average number of set elements fixed by each element of our group. For instance, g^1 only fixes a polygon if all of the vertices are colored the same, g^2 only fixes a polygon if every other vertex is colored the same, etc, and if we count each of these and average them, we will get the total number of orbits

[rapid cave]

and Dn acts on a regular n-gon by rotation/reflection

[tribal moss]

Do you have a statement of the lemma? Is your trouble to grasp what it is the lemma claims, or how to apply it in your example situation, or how to convince yourself that the claim is true?

[fiery dirge]

Is your trouble to grasp what it is the lemma claims, or how to apply it in your example situation, or how to convince yourself that the claim is true?
All of that

i need an analogy for group action

is it like a regular group with operation but expanded onto some set?

Or is it more similar to a function from group to a set

[rapid cave]

think about this example

[rapid cave]

and this one is more geometric

[tribal moss]

Hmm, do you know vector spaces? Scalar multiplication might be the most commonly encountered example of an operation that goes as A × B -> B.
(It's even a group operation if you ignore the zero scalar and consider the multiplicative group of the scalar field).

[fiery dirge]

but {1, 2, 3, 4} is kinda the part of the group no? It's the set

[rapid cave]

S4 is more of an abstract structure

you could replace {1,2,3,4} with {a,b,c,d}

[fiery dirge]

okay

[fiery dirge]

I had linear algebra last year so kinda yeah but only the trivial stuff about them and how they're graphically represented

[tribal moss]

This is very basic stuff about abstract vector spaces.

[fiery dirge]

I like your approach but don't know what C18 is and I don't quite understand the last part.

[rapid cave]

C18 - cyclic of order 18

[tribal moss]

For a vector space you need to have some set V of things you call vectors, and an addition operation which I'm going to ignore, and a scalar multiplication R × V -> V which satisfies some axioms such as a·(b·v) = (ab)·v.
That axiom gives that scalar multiplication is an action on V of the group of nonzero reals under multiplication.

[fiery dirge]

like integers or dihedral with rotation only?

[proud vigil]

think of Cn like Z/nZ

if you take 1 and keep adding it to itself, you reach all numbers before getting to 0

so its cyclic

[fiery dirge]

yeah I got that

[proud vigil]

rotating a polygon vertex-by-vertex is a very similar idea

after 18 ticks, you return to the start, and you'll have cycled through all rotations, so this is a natural group to use here

[fiery dirge]

So, orbits are sets of elements we can get by applying group action onto a single element at a time? And stabilizers are elements that, when action is applied, go back to themselves i.e. they don't change (acting kinda like identity in a group).

[fiery dirge]

got it

[tribal moss]

Group actions can be thought of in one of two ways:

• either as a function G × X -> X, satisfying a few axioms,
• or by considering the group of all bijections X -> X and then saying "a group action by G is a group homomorphism into that group of bijections".
  Both views are important; you get the best understanding by being able to switch back and forth between them as necessary.

[tribal moss]

You're right about orbits here, but a stabilizer doesn't consist of elements of X -- its a subgroup of G, consiting of all the group elements that do nothing to whatever it is they stabilize.

[fiery dirge]

yeeee that's why I asked if it's like a function from group (or group product) into a set

[proud vigil]

you're basically describing the set of points that are invariant under a specific group element's action (these are the sets we're interested in in burnside)

the set of group elements that don't move a specific point in our set are the stabilizer of that point

so they're very similar ideas but sort of opposite

[fiery dirge]

ahh I see

[tribal moss]

That's what I was trying to answer yes. I'm trying to say that sometimes we think of a group action as a special function from G into the group of bijections of X, but that is not always how they're thought of.

[fiery dirge]

so orbit(element from domain) = {set of all elements in codomain} and stabilizer = elements from group that act like a kernel kinda

[fiery dirge]

got it

[tribal moss]

It's not super useful (nor really clear) to speak of "domain" and "codomain" here.

[fiery dirge]

Okay, can you explain the Burnside's lemma now when we got the ground set?

X/G looks similar to that set of cosets in Lagrange's

[tribal moss]

I think X/G might not be a super enlightening notation here.
(Though it turns out that in the special case where X is group and G is a normal subgroup of X, and G acts on X by e.g. left multiplication, then the orbits are just the cosets of G).

How I think intuitively of orbits is: Start with some element of X and keep picking random elements of G to operate on the element of X in your hand with. The "orbit" is all the elements of X you can possibly end up with.
It turns out that the orbit is also the set of all things that can be reached in a single step, but my default mental image allows a whole chain of G-elements that I let act one at a time.

[proud vigil]

i think the notation is a little useful from a coset interpretation

e.g. Gx is like letting the group elements of G act on x, where the orbit is sort of like a coset, and Gx = Gy if and only if x and y belong to the same orbit

[tribal moss]

The risk here is that one may end up thinking that two different orbits have to "look alike" like cosets do -- which is not the case; different orbits of the same operations can have different cardinalities, for example.

[proud vigil]

that's fair, i do think the partitioning intuition is valuable though but yeah its generally good to be careful

[fiery dirge]

in this case elements of G are 4 colors and elements of X are 18 points on an 18-gon?

[tribal moss]

Partitioning intuition: Yes, definitely, we need that.

[tribal moss]

No, G is the group of all 18 possible ways to rotate the polygon.
X is the set of all 4^18 possible ways to color the polygon without starting to rotate it.

Unless all points get the same color, turning the polygon generally makes one coloring into another coloring -- that's an action of the rotation group on the set of coloring.

So here each orbit consists of colorings that can all be turned into each other by rotating the polygon appropriately.

And when the problem says:

if two colorings are the same when the 18-gon is rotated
what it means is exactly that it wants to count those orbits instead of the colorings themselves.

[thorny cobalt]

two colorings would be distinct under reflection right?

Because G is not equal to D_18

[tribal moss]

Correct, that's how I read the problem.

(We could also use Burnside's lemma to count colorings where we do consider reflected colorings equal, of course -- it would be a pretty similar computation, but of course slightly different details).

[rocky cloak]

If you make the 18-gons out of paper you can't really reflect them without coloring the other side as well. So it's reasonable to not allow reflections.

Of course you could make them out of wiremesh...

But then I would need to go to the craft store

[rapid cave]

my 18-gons defy physics

[tribal moss]

If we want four different physical colors I think I might need a trip to the craft store anyway.

[rapid cave]

yep I only have black pens

and pencil which is gray

[fiery dirge]

okay so how do I utilize it in the burnside's formula?

[tribal moss]

With this identification, the left hand side of the formula is (by definition) the number we're after.

[fiery dirge]

|G| = 18 so 1/18 is the part before the sum

but what does the sum do?

[tribal moss]

And the right-hand side tells us to sum up just 18 subcounts (one for each group element) and divide by 18.

[fiery dirge]

it goes through every element of G

and X is 4 right?

4^18 is a bit too much

because we have g powers in the sum

[tribal moss]

X is a set with 4^18 elements.

X^g is not really a power, it's notation for "the subset of X that contains all elements that are unchanged when the single element g acts of them".

That's explained in the lines before the formula in Wikipedia:

[thorny cobalt]

is that the stabilizer?

[tribal moss]

No, it's not a stabilizer. I don't think it has a nice name.

(Or if it does, I don't remember what it is).

[thorny cobalt]

oops nvm i mixed them up

[fiery dirge]

Ahh I see, so how do I find the fixed elements?

they shouldn't exist because of different colorings and rotations tho?

[tribal moss]

We'll need to go through the 18 group elements one by one -- but many of them turn out to so similar that we can handle them pretty quickly.

First, and easiest: the identity element in G is "rotate by 0°". Which of the 4^18 colorings stay the same when we rotate the polygon by zero degrees?

[fiery dirge]

all of them

[tribal moss]

Yes, so the first term of the sum is 4^18.

[thorny cobalt]

yeah you can js do casework on gcd(r, 360) i think where r is the rotaion angle

[tribal moss]

Next, "rotate by 20° clockwise". The condition for staying the same under such a rotation is that every corner has the same color as the next one. How many such colorings are there?

[fiery dirge]

the rotation angle is 20 degrees right? (360/18)

[tribal moss]

Yes.

[thorny cobalt]

yeah

[fiery dirge]

none?

because of the rotation we don't count them as unique

[thorny cobalt]

what if all vertices are colored the asme color

then the figure would be preserved under a 20 degree rotaiton

[fiery dirge]

there's only 4 of them then

(1 for each color)

[tribal moss]

The elements of X is a sense of coloring where we do count them as separate no matter what happens under rotations. Perhaps we should speak of them as "raw colorings" or "precolorings" to avoid confusion.

[tribal moss]

Right, so the second term of the sum is 4.

What comes next?

[fiery dirge]

various permutations Im not sure

[proud vigil]

what's the next group element?

[fiery dirge]

rotation by 40

[proud vigil]

you can do any of them really, we just need to get through all 18

yeah that works

rotating by 40 means what in terms of the vertices

[fiery dirge]

they all would rotate for 40 degrees meaning we wouldn't get any new permutations

because they could've been gotten from the previous 2

[tribal moss]

We're not trying to make "new" colorings here.

[proud vigil]

can you define for us what X and G represent right now

i think youre a little confused

[slow bison]

i think its just called the fixed points

[tribal moss]

The formula just asks to count how many elements there are in the set called X^g, no matter whether they already appeared for a different g or not.

This doesn't look like it has much to do with counting orbits, sure. The claim of "Burnside's lemma" is that we'll end up with the right result anyway.

[fiery dirge]

G is a dihedral group consisting of 18 rotations without translations. X is a set of all permutations for the colors so 4^18

[tribal moss]

Minor trouble here:

1. That's not what "dihedral group" means but you're thinking of the right group (even though it's not dihedral).

[fiery dirge]

there's 18*4 of them (all painted with 4 different colors so they're stabilizers in that case)

[tribal moss]

2. The set X does not consist of "permutations", but just the set of functions from {1,2,3....,18} to {red,green,blue,yellow}.

[tribal moss]

We're here talking about the precolorings that are fixed by a 40° rotation (that is, two steps of the polygon), right?
18×4 is not the correct count here. How did you arrive at that?

[fiery dirge]

no I was talking about all possible stabilizers

[tribal moss]

Why? There's not even any stabilizer in the statement of the lemma?

[fiery dirge]

I think I confused it with X^g

[thorny cobalt]

dihedral group would also contain the 18 reflections i think

[fiery dirge]

I said 'without translations' yeah

[thorny cobalt]

translations?

[fiery dirge]

but I meant reflections

[thorny cobalt]

oh alr

[thorny cobalt]

||for a 20n degree rotation it would just be 4^(gcd(n,18)) right?||

[tribal moss]

Yeah.

[fiery dirge]

is the number of elements for rotation 40 maybe gcd(2, 18) = 2 so |X^g| = 4^2 = 16?

[thorny cobalt]

yeah

[fiery dirge]

mf

[tribal moss]

Yes.

[fiery dirge]

he could've gave me the formula instead of spinning me around bruhhh

[thorny cobalt]

formula?

i mean

[fiery dirge]

I dont have time for intuitive understanding...

[thorny cobalt]

alr

you want to develop

[fiery dirge]

technically yeah

[thorny cobalt]

intuitivie understanding

[tribal moss]

You don't have time to skip intuitive understanding.

[thorny cobalt]

fr

[fiery dirge]

its way too abstract 😭

[thorny cobalt]

thats why intuition is insaely important

otherwise you'll js get lost in the sauce with notation and whatnot

[tribal moss]

Bulldozing through one problem at a time without having intuitive understanding is way, way, way, too slow.

However if you got to the gcd formula without looking at Vivdax's hint, you should now be able to fill out a table with all the 18 terms quickly.

[fiery dirge]

but I dont have it, why is it the gcd(rotation count, number of vertexes)

[fiery dirge]

I found it looking at a different source

[thorny cobalt]

ok lemme try to explain it

basically, consider rotation count = 5 (100 degree rotation)

the whole concept of the gcd formula relies on the residues of nk (mod 18), where 20n is hte rotation amount

so for n=5, take the modular inverse of 5 modulo 18, which is 11.

Basically, since the coloring must be identical to itself after a rotation by 5 (100 degrees), we see that the co.oring msut be identical to itself after a rotation by 55

however, 55 = 1 (mod 18), so rotating it by 55 (55*20 degrees) is equivalent to rotating it 20 degrees, and thus we get that the coloring must be identical to itself after a 20 degree rotation

which is 4^1 = 4^gcd(5,18) as shown sbove

the same idea generalizes for other gcd values

like if you take n=4 (80 degree rotation)

you see that the residues of 4k (mod 18) are just 0,2,4,8,.., 16

so the coloring must be identical to itself after a rotatin by 2 (40 degrees), which gives 4^2 = 4^gcd(4,18) preserved colorings

the key idea is that the smallest non-zero residues of nk (mod 18) is gcd(n,18) which you can try proving if you want

and then that gives way to the answer

[fiery dirge]

what would happen if the task had reflexion also included as well as rotations being unique colorings?

[thorny cobalt]

that would js be applying burnsides lemma on D_18

so all the rotation stuff stays the same, you js have to check for what colorings are presrved under certain rotations

i think all rotations would have an identical number of colorings that they preserve right?

its js 4^10 for each

[proud vigil]

you can do this from a really intuitive counting approach

if i want the shape to be the same after rotating each vertex forward by 2 vertices, then vertex 1 = vertex 3 = vertex 5.... = vertex 17 = vertex 1, and vertex 2 = vertex 4 = .... = vertex 18 = vertex 2

this gives us 2 subsets of vertices, and each subset needs 1 color for all its vertices, so we have 4^2 colorings

just try solving the problem directly, don't rely on fancier constructions, they'll be a crutch if you don't get why they work

[tribal moss]

4^10 for half of the reflections, 4^9 for the rest -- depending on whether the reflection axis connects two vertices or two side midpoints.

[thorny cobalt]

oh true i forgot about those

[white oxide]

Why don't we have f finite type => finite? For if f is of finite type, then every element of B can be written as a finite sum \sum f(a_i) x_i. If we consider B as an A-module via a x b := f(a) x b, then doesn't this show that B is finitely generated as an A-module, in other words f is finite?

Oh wait never mind the x_i would appear in increasing powers

Still couldn't we take {x_0, x_1, x_2^2, x_3^3, ... x_n^n} as a finite generating set then?

[rocky cloak]

Finite type just means finitely generated as an algebra, so for example k[x] is finite type over k.

[white oxide]

What does finite mean then? According to AM it's if B is f.g. as an A-module (assuming a ring homomorphism f: A -> B)

Isn't that the same thing as being f.g. as an (A)-algebra?

Or is it that B = A[x_1, \dots, x_n] (where A is really f(A) here, but if we view B as an A-algebra, then we can just say f(A) = A), and by Corollary 5.2 (here we use the integral assumption), A[x_1, \dots, x_n] = B is f.g. as an A-module, which is the definition of being finite

[rapid cave]

when you move from being generated as a A-algebra to being generated as a A-module you remove the ability of the generated elements to be multiplied with each other

[white oxide]

Ah so {x_0, x_1, x_2^2, x_3^3, ... x_n^n} as a generating set wouldn't make sense

In any case I think I got it

[fiery dirge]

Determine which of these are isomorphic. Can anybody help?

Фn = Z/nZ* where gcd(x, n) = 1

[rapid cave]

check orders

[fiery dirge]

What is the order of the first product? Is it 180 or is it much smaller (because Z/10Z* doesn't have 10 elements)

[rapid cave]

it has 4

[fiery dirge]

Yeah

So would the order be 18*4 = 72 then?

[rapid cave]

the size of (Z/nZ)* is phi(n) the euler phi function

[fiery dirge]

Which is phi(10) = phi(2) phi(5) = 4

[rapid cave]

yes

so it turns out they all have the same order

of 72

[fiery dirge]

Wait what's the A one

I don't know what they are

[rapid cave]

alternating symmetric group

the subgroup of Sn containing all even permutations

[fiery dirge]

How do I find its order?

[rapid cave]

it has size n!/2

[fiery dirge]

Ahh I see

[rapid cave]

so as I said they all have the same order

[fiery dirge]

Yep

Next we can check if some are abelian/cyclic?

[rapid cave]

so we can't disprove that any two are isomorphic with this

abelian is good

its easy to check

is the first abelian?

[fiery dirge]

Yes

Z/nZ* is always abelian

2nd one isn't tho, Symmetric isn't abelian

(and I assume A isn't as well since it's a subgroup)