groups-rings-fields

[supple vortex]

Z6 x Z8

(5a,5b)=0

but 6a=0, 8b=0

so its

wh

[coral spindle]

Are you aware of Lagrange's theorem?

[supple vortex]

yes

[coral spindle]

So let's say (a,b) has order 5

What's the size of the group <(a,b)>?

[supple vortex]

5

[coral spindle]

That's right

And what's the size of the group Z6 x Z8?

[supple vortex]

6*8

[coral spindle]

Now I leave you to put this information together.

[supple vortex]

ok what about order 4

which (a,b) have order 4

[coral spindle]

Hint: see if you can work out the order of (a,b) from the order of a and the order of b.

[rustic crown]

maybe a lil easier to just do like we were doing before

[supple vortex]

ok what about this approach

[rustic crown]

4a=6a=0, 4b = 8b = 0

[supple vortex]

Isnt U(n) the orders of elements in Zn

[supple vortex]

oh

ok

so

if (a,b) has order 4

then

(4a,4b)=(e,e)

[rustic crown]

yep

[supple vortex]

ie

4a=6a, 4b=8b

so

[rustic crown]

yep

[supple vortex]

a=lcm(4,6), b=lcm(4,8)?

[rustic crown]

nu

4a=6a=0 means 2a is also 0

so you're interested in elements (a, b) such that 2a and 4b are 0

since you want the order to be 4, you shouldn't have that 2b is also 0, else 2(a,b) would already be (0,0)

now just write down all elements like that

2a = 0 in Z/6
and 4b = 0 in Z/8

[supple vortex]

isnt a = 4^(-1)

[rustic crown]

4 isn't invertible in either of the two rings, so idk what you mean

[supple vortex]

ok but how can i solve

2a = 0 (mod 6)
4b = 0 (mod 8)

[rustic crown]

6 | 2a means 3 | a

8 | 4b means 2 | b

[supple vortex]

so a=3, b=2

[rustic crown]

divides

[supple vortex]

or a=3k, b=2m

[rustic crown]

so in Z/6, a can be 0 or 3

and in Z/8, b can be 0,2,4,6

[supple vortex]

how did you reduce 4a=0 to 2a=0 again?

[rustic crown]

2a = 0 mod 6 means 2a = 6 * k which means a = 3 * k

🙈

[supple vortex]

i dont know where you got 2a=0 from

dont we want 4a=0

[rustic crown]

but recall that we didn't want 2b to be 0, so b should only be 2 or 6

[rustic crown]

we have both 4a and 6a are 0

so their difference 2a is also 0

[supple vortex]

ok

so you took the difference

and you did the same for 4a=8a=0

b

sorry

and got 4b=0

[rustic crown]

yea, but it wasn't much helpful in b

since we already knew 4b = 0.

therefore having the extra 8b = 0, told us nothing.

[untold basalt]

Let E/F be a field extension. Is it true that if E is the splitting fileld of a non-separable poynomial f in F[x] then Gal(E/F) is trivial?

[vestal snow]

Let I be a homogenous ideal of A=k[x_0,...,x_n] of dimension r. How do I show that I + (x0,...,x{r-1}) is not the entire ring?

[rocky cloak]

So if you have some seperable polynomial f(x), then f(x)^2 is not seperable, but has the same splitting field.

If you add the restriction that the polynomial be irreducible, then it's still not true. The automorphisms of the splitting field of an irreducible polynomial acts transitively on the roots, and it's determined by its action on the roots. So the automorphism group is trivial iff the polynomial has a unique root. In such cases E is called purely inseparable.

[south patrol]

Hm I think this depends on your definition of separable right lol

In Galois theory it seems the common definition would say f^2 is separable if f is since we say a poly is separable if all of its irreducible factors are

[rocky cloak]

Haven't seen people define it that way, but doesn't really make a difference since you could just consider the product of a seperable and non-seperable polynomial.

Anyway the intersecting part is when the polynomial is irreducible

[rocky cloak]

I guess this is just Krulls principal ideal theorem

[rocky cloak]

Or I guess much easier, I is contained in the trivial ideal, so adding things from the trivial ideal never gets you more than that

[vestal snow]

I is in the trivial ideal?

[rocky cloak]

It's homogenous

[vestal snow]

Oh I use irrelevant ideal for that

[south patrol]

Yes exactly

[vestal snow]

For me trivial ideal either means 0 or everything

[rocky cloak]

Yeah, I'm not an AG person, so not up to speed on all that lingo

[vestal snow]

Oh wait I messed up.

I want show that I + (x_0,...,x_{r-1}) is contained in some homogenous prime not containing the irrelevant ideal

[rocky cloak]

Yeah, that makes more sense. Then I guess you reach for Krulls theorem

[vestal snow]

I'm trying to show V(I) intersect V(x_0,..,x_r-1) is not empty

[rocky cloak]

Though, shouldn't it be r-2. Like isn't V(xr, ..., xn) r-dimensional?

[vestal snow]

V(x_0,...,x_{r-1}) is codimension r

[rocky cloak]

Yeah, exactly

[vestal snow]

The theorem is that an r dimensional thing intersects a codim r linear space

How does it follow from Krull's theorem?

[rocky cloak]

But isn't I = (xr, ..., xn) r-dimensional?

[vestal snow]

I don't think so

Wait

Sorry

Yeah

It is

and (x_0,...,x_{r-1}) is n-r dimensional

i.e. codim r

[rocky cloak]

n+1 - r I guess, but sure

[vestal snow]

Yeah indexing is hard

Anyway, ignoring indexing, how does the result follow from Krull's theorem?

[cloud walrus]

Sweet Tea 🧋

[rocky cloak]

I guess dimensions get shifted by 1, since you're in projective space.

So it says that a minimal prime containing an ideal generated by n elements has height at most n.

So the smallest prime containing I + (x0, .., xr-1) has height at most (codimension of I) + r

[merry dawn]

those small results feel very much out of context I'd say

[coral shale]

So one of the early goals is the 1st isomorphism theorem

The most important concepts in an intro course are group actions and group quotients imo

[merry dawn]

where are they used then?
(I'm not asking abt the real life 😂 )

[coral shale]

As for your more specific q's idk what to say.

u can lookup on mathexchange tbh

whats the point of this

why do we need this

intuition for this

it usually has alright answers

[delicate orchid]

Where aren’t they used

[coral shale]

But it is expected in most courses your average student to see their motivation

over time

and the lecturer hopefully gives some insight

[coral shale]

If youre struggling to answer this, I wonder if you dont quite have the intuition for these definitions

and their consequences down

Like you can just memorize the set theoretical defns and it wont get u very far

===
An aside - i didnt really have intuition for either of these when i first learn them

quotients made a lot of sense after topology

[merry dawn]

maybe that's exactly why I'm asking abt their usefulness 🙂

[coral shale]

actions, they would after rep theory, although i didnt take this

[merry dawn]

and by the way, basically all the theorems we have studied so far that mention group actions can be easily reformulated without them

[delicate orchid]

Topology? That’s a strange one. Most commonly you see them used in modular arithmetic first

[coral shale]

yh well the entire idea of a quotient didnt clock for me

til then

like properly clock

[coral shale]

one case study i like is the rubiks cube

you can look at actions and quotients involving this

gowers has a nice blog post on actions

[merry dawn]

so that's it? we study such general constructions to basically easier prove results abt mod arithmetic?🙈

I mean, if it were so, then it'd be a massive overkill imo

[delicate orchid]

You’re defining a new object from the previous one using some nice equivalence relation on the previous one. It just so happens that for groups that the equivalence relation corresponds with sub objects

[coral shale]

no way, modular arithmetic is a small specific case

he was replying to me

[delicate orchid]

no obviously not. This is a vast generalisation of that concept

[coral shale]

Yeah so, the idea of an equivalence relation and the point of being interested in them

is a good starter

for understanding quotients

In fact, a lot of courses dont teach what a set quotient is first

before a group quotient

[delicate orchid]

Yeah which is a bit weird imo

[coral shale]

which i think is a bad idea

that was part my problem.

[delicate orchid]

Mine didn’t teach that, I don’t even think they probably introduced equivalence relations

It was just “construct the set of cosets”

[delicate orchid]

The idea behind this is you’re setting your chosen subobject to act like the identity in the quotient. So in a way you’re looking at the structure of what you started with without (or modulo in old timey english) whatever you’ve quotiented by

A good example is the abelianisation. For any group G there’s a particular normal subgroup with captures “non-commutivity” in G, and when you quotient out by that normal subgroup (heuristically you’re setting the “non-commutivity” to zero) you get the abelianisation of G

[rocky cloak]

When I TAed algebra list year I made a little collection of exercise sheets with 'applications'. The students didn't seem super interested though

[tardy hedge]

oh thats cool i havent seen that

is the subgroup it the center of G or something

[delicate orchid]

No, quotienting by the centre gets you the group of inner automorphisms which is the exact opposite of what we want kinda, lol

It’s the derived subgroup [G, G]

[tardy hedge]

ya after i said that i thought the center maybe was the opposite

[merry dawn]

can u please share it? 👉🏻👈🏻🥹

maybe in DM, if u r ok with that 🙃

[tardy hedge]

I would like to see that sheet too

[tardy hedge]

What is that?

[rocky cloak]

It's here
https://wiki.math.ntnu.no/tma4150/2023v/anvendelser
But it's in Norwegian

[delicate orchid]

Subgroup generated by all elements of the form xyx^-1y^-1

[coral shale]

picking the worst order!

[delicate orchid]

I don’t care

[delicate orchid]

Exercise: prove this @tardy hedge

[rocky cloak]

I don't know if they make for very good motivation, it's just a little taste of how algebra arises in other areas of math/science

[coral shale]

"Commutator" of g and h
[g, h] := g^-1h^-1gh

It is a measure of how far off from commuting these elements are, you could say. If it is e, then they do.

For A, B subgroups of some group, say G
[A, B] := <{[a, b] : a in A, b in B}>

[tardy hedge]

ok i will try at some point. idk what inner automorphisms are atm

[coral shale]

automorphisms u good with?

[tardy hedge]

Yeah its just an isomorphism from G to G right

[coral shale]

they are automorphisms of the form
f(g) = h^-1gh
for some h

thats conjugation of g by h

[tardy hedge]

Ok thanks

[coral shale]

so inner automorphisms are conjugations 🙄

[tardy hedge]

I watched a video talking about intuition for conjugation but i kinda forget it

[coral shale]

small realisation

we have 2 words for the same thing kek

[tardy hedge]

about "shifting" ur perspective or smth

do u guys think about conjugation in that way?

[coral shale]

yeah vector space conjugation thingy is not a bad way to see it

like change of basis things

or rubiks cube also a good example

Other good one is to compute conjugations in Sn

f (1 2 3 4) f^-1 (cycle notation)
try computing what this will be

wait

[delicate orchid]

Actually continuously banging on about the Rubik’s cube. A group that complicated is not a good first example lol

[coral shale]

oh but its like if u own one

u can physically try out these things

[delicate orchid]

A good point I like to bring up is that both the trace and determinant of a matrix is invariant under conjugation

[coral shale]

yeah so see what this permutation does to the element f(x) (function notation)

thats how gowers explains conjugation in his blog

[delicate orchid]

So pretty much all of representation theory only cares about group elements up to conjugation lol

S_n is another good example though, yeah

[coral shale]

what does the trace mean again.

geometrically

[delicate orchid]

sum of the eigenvalues

[coral shale]

hmm.

[delicate orchid]

I suggest going to any other channel in the server if you want a "geometric" """""""""""""""""intuition"""""""""""""""""

[coral shale]

i think thatll do

[merry dawn]

so that I don't ask this again later: do u know of any textbooks / lecture notes / lecture recordings that miraculously try to motivate concepts / definitions along the way?

sure, I could just google all of it along the way, but having a structured resource would be much better ofc

basically looking for something similar to Abbott's book but in the world of algebra 🥲

[coral shale]

I mean, you'd hope a good book would do that job

#book-recommendations so u could try also asking here

i dont know of them myself sadly

and theres some sticky somewhere with book reccs

u could try skimming chapters of reccs ones and see if ur happy with whats said.

#books

Some youtube series are also good for this. With illustrations. Better if other people recc than me tho

Like 3b1b does a good job motivating intuition in a lot of his vids, idk if he includes the specific topics ur after/does a good job of them, but i dont see harm in checking it out. I do remember some good stuff for linear algebra

[rustic crown]

shuwi is here

[tardy hedge]

Ok bros im officially starting my self study

Class is over and i wanna try to get through the rest of the textbook

Im stuck on first question though lmao

In quotient field Q(D) show multiplication is well defined 🤦

[teal vessel]

welcome to self study: bang your head against the wall for a few hours, reread the section, and work backwards

[tardy hedge]

Looking forward to it lmao

[teal vessel]

start by assuming it's well defined, what does that mean?

[rustic crown]

don't hurt your head literally ><

[tardy hedge]

Choose different representstives of equivalence class and show multiplication gives same result

[teal vessel]

ok, so what do you need to ensure that can happen

[merry dawn]

«for objects that are equal the result is going to be the same»

I just love how that sounds 🥰

[delicate orchid]

"equality commutes with the operation"

[rustic crown]

[merry dawn]

and then the question is how do we define 'equal objects' 😏

[tardy hedge]

I need to show acb’d’ = bda’c’

But rn i have ab’cd’=ba’dc’

[teal vessel]

$\forall x,y \in F ( x=y \implies f(x)=f(y))$

[cloud walrus]

GoldenPhoenix

[rustic crown]

<

[teal vessel]

"putting parallel lines horizontally between them makes the teacher happy"

[rustic crown]

true, but f(x) makes no sense a priori, so can't write that formula

<

[teal vessel]

not without exterior context, but with the rest of the conversation, there's sufficient context to get away with it 😛

[merry dawn]

and the fun starts when the things that make the teacher happy are things that do not make happy anyone…

[rustic crown]

one needs to show that the function (S x R) x (S x R) --> S^-1R factors through (S^-1R) x (S^-1R)

[delicate orchid]

sure if you want to be a nerd about it, det

[rustic crown]

<

[teal vessel]

to be fair, pulling out quantifiers does invite nerdery

[rocky cloak]

Are you given that multiplication is commutative?

[tardy hedge]

Can i have some help pls

[tardy hedge]

I just have that D is an ID

Should i use that somehow?

[rocky cloak]

And do you know the definition of integral domain?

[tardy hedge]

Cancellation law holds, or could also say ab = 0 implies a or b is 0

[delicate orchid]

based definition tbh

[rocky cloak]

Yes, but another part of the definition is that multiplication is commutative

[tardy hedge]

Wait wut

[rustic crown]

and it's not the zero ring.

[tardy hedge]

How did i not know

Oh ur right

Bruh i spent 50 mins on this i couldve done in like 10 if i remembered ID is commutative lol

Oh well thank you

Why do we want it to be commutative ?

[rustic crown]

because non-commutative domains are called domains.

not integral domains :giggle:

[tardy hedge]

Oh ok i mean i see

Its just a thing

[rocky cloak]

Because otherwise the construction for the field of fractions doesn't (necessarily) work

[tardy hedge]

ID is used to generalize and have general results for rings like integers and polynomials which have commutative multiplication right

[tardy hedge]

Yeah cuz probably proving Q(D) is a field uses the commutativity in D

[rocky cloak]

Yeah, I mean fields are also defined to have commutative multiplication. But I mean just the fact that multiplication in Q(D) is well defined needs commutative

[tardy hedge]

Oh ok

When self studying, how many problems at end of section should i do? I mean obv “all of them” is “best” but i also want to progress at a reasonable rate

[delicate orchid]

none of them. Make up your own and then write a paper about your solution

[tardy hedge]

Bruh

[rustic crown]

ig even the fact that a/b = c/d is an equiv relation needs comm.

[tardy hedge]

It is cool how most questions u try, u do gain some new understanding of stuff after doing them

[delicate orchid]

symmetry is over-rated

that's the idea!

it's why doing exercises is the most important part of mathematics

[tardy hedge]

Yeah so true…. It forces u to learn what u dont even know u dont know

[delicate orchid]

cohomology sighting

I found an actual use for derviations the other day, they come up in Navarro's sludgebabble about p-solvable groups

[tardy hedge]

I think i need to remember this one. Sometimes im tempted to skip exercises that sound boring but even doing those , most times i take something away from it

[delicate orchid]

oh wait that's why they're called derivations. They satisfy the product rule

I feel really stupid now

oh so that's what de Rham cohomology is

hahaha it's literally the quotient rule

I'm going iNSANE

I'm LOOSING the PLOT

[tardy hedge]

Woah did u just learn something from reading one of the exercises in my book?

[delicate orchid]

I feel like I've made this connection before but I have memory issues lol

[tardy hedge]

Hahaha

Thats pretty epic though

Its fun how authors do this, slyly introduce some advanced topic

[delicate orchid]

like it is obvious in de rham that the "derivations" are literally differentials (its by definition)

[tardy hedge]

Without telling you

[delicate orchid]

yeah, it's cool

[tardy hedge]

Sneaky

[delicate orchid]

one day I will understand homology properly

[rocky cloak]

Next exercise show that a derivation on k[x] satisfies

D(f(x)) = f'(x)D(x)

[delicate orchid]

yeah this one is expected though

[rocky cloak]

I'm wondering if you can make a similar statement about derivations on C^infty(R) or something like that

If you just add in enough adjectives, to avoid cursed stuff

[delicate orchid]

lemme see, so we have $D(f)g+gD(f)$ and it's additive, so lets break it up
[D(f(x)) = D\left(\sum_{i=1}^n a_ix^i\right) = \sum_{i=1}^n D(a_ix^i) = \sum_{i=1}^n a_iD(x^i)+x^iD(a_i)]
$D(a_i)$ satisfies $D(a_i/1) = D(a_i)-aD(1) \Rightarrow D(1) = 0$, hence $D(f) = 0$ for any constant $f$. Thus $D(f(x)) = \sum_{i=1}^n a_iD(x^i)$.
Now to just show the product rule

[cloud walrus]

W;3w Lads Tbh

[delicate orchid]

sorry, power rule

not product rule

hmmm

oh and also I'm implicity restricting the derivation to the field k for the a_i/1 bit

[tardy hedge]

Has anybody tried using gpt4 for help in math?

I wonder how useful it is now

[delicate orchid]

$D(x^0) = 0 = f'(x)$ with $f(x) = 1$. Assume the power rule has been proven for $j < i$. $D(x^i) = D(xx^{i-1}) = D(x)x^{i-1}+xD(x^{i-1}) = x^{i-1}+x(i-1)x^{i-2} = ix^{i-1}$, thus the claim is shown

[cloud walrus]

W;3w Lads Tbh

[rocky cloak]

Had a student use gpt for linear algebra, and they also had some kinda Wolfram alpha + python thing hooked to it.

Gave good answers whenever I looked over what he got, but idk

[delicate orchid]

I'm a luddite so I hate all technology

fun exercise btw @rocky cloak

took my mind off of this awful character theory I'm doing atm

[rocky cloak]

The upside to representations of groups is that you have characters, the downside is that you have to do character theory

[delicate orchid]

I mean you always have trace functions right

the problem is I'm doing some kind of cracked up supercharacter theory where we DO NOT HAVE LINEAR INDEPENDENCE

[delicate orchid]

problem with trace functions in general is the uhhh entire langlands program from what I've heard about it

[teal vessel]

sanity check: for positive integers n (e^(ix))^n=e^inx, yes? I'm driving myself crazy trying to prove it to be the case but I'm gonna have a ways to go with this expansion.

[south patrol]

Yes

[delicate orchid]

that's just how exponents work

[south patrol]

Eh depends how you define e^x lol

[delicate orchid]

no, it's a fundemental property of associative operations LMFAOOO

[south patrol]

I think you basically have to use power series to define smth like this

Which is what I'd do anyway

But yes this is true, just prove more generally that e^a e^b = e^(a+b)

[delicate orchid]

S^1 is a group, "m"ate

[south patrol]

And to do that, first prove that the derivative of e^t is e^t again

and then uh

[teal vessel]

this isn't strictly guaranteed for a and b in the complex numbers, hence why I had to double check

[south patrol]

Yes it is

[delicate orchid]

it is when a is real

[south patrol]

when a is complex too

[delicate orchid]

which is the one that fails

one of them fails

[south patrol]

Basically everything with e works, the issue is that logarithms can only be defined coherently on some subsets of C

[delicate orchid]

RAARRGGHHH just pass to the lie algebra of S^1 and do your computations there RAAAARRRRRRGGGGGGHHHHHHHHHHHHH

[south patrol]

Anyway for e^a e^b = e^(a+b), I would do uh

lemme remembr how this works lol

Well you can just do it directly

But there is a trick i'm trying to remember lol

[delicate orchid]

pass to SO(2)

[south patrol]

I don't really get what you're saying lol

probs memeing

[delicate orchid]

I'm being funny by quoting ways to do it that only work when you already know it's a group under that operation

[teal vessel]

OH I'm being overly paranoid, the thing that set me off was about matrix exponentiation but i forgot where I saw it (and thus that it applied to matrices), so I had a red flag go off without the conscious reason why.

[south patrol]

Yes the issue is that matrix multiplication isn't a commutative operation

[delicate orchid]

can we get a fact check on this one?

[coral shale]

II captain

[delicate orchid]

just use your half baked campbell formula

II

[tardy hedge]

For Q4. The D a field -> isomorphism part. Having [ab^-1,1] = [a,b] shows its surjective right?

[delicate orchid]

Q4 send my head spinning

I was like "Q4 is iso to V4 what are you talking about"

[delicate orchid]

yus

[teal vessel]

ANYWHO I will now take my leave and return to proving that certain nth roots of unity form a group

[south patrol]

Oh yeah here's the proof I had in mind

I can do it dw

[tardy hedge]

This reminded me of how the quotient field is trying to create the notion of “ab^-1” for any ID

But if D is already a field, u already have that idea works

[delicate orchid]

that's the idea behind why this isomorphism should even be expected to exist

I sense great things in ur future...

[tardy hedge]

:0 that is pretty cool to hear tbh

Trying my best and im grateful for you people here to help

Super helpful

[delicate orchid]

reminds me of that time jagr and I computed the automorphism group of the p-Prufer group. Good times

[south patrol]

Consider the function $f(z) = e^{a+z} e^{-z}$ (from $\mathbb C \to \mathbb C$) and then $f'(z) = e^{a+z} e^{-z} - e^{a+z} e^{-z} = 0$, so $f$ is constant. But $f(0) = e^a$, so $e^{z+a} e^{-z} = e^a$ for all $a,z \in \mathbb C$. So in particular $e^{-z} = 1/e^{z}$, and $e^{z+a} = e^{a} e^z$ which is what we want

[cloud walrus]

ẞ

[south patrol]

But you can also just do like

[delicate orchid]

ok now this is based

[south patrol]

$e^a e^b = \Big( \sum_{k\ge 0} \frac{a^k}{k!} \Big)\Big( \sum_{\ell \ge 0} \frac{b^\ell}{\ell!}\Big) = \sum_{k, \ell \ge 0} \frac{a^k b^\ell}{k! \ell!} = \sum_{n \ge 0} \sum_{k + \ell = n} \frac{a^k b^\ell}{k! \ell!} = \sum_{n \ge 0} \Big( \sum_{k+\ell = n} \frac{1}{n!} {n \choose k} a^k b^\ell \Big)= \sum_{n \ge 0} \frac{1}{n!} (a+b)^n = e^{a+b}$

[cloud walrus]

ẞ

[delicate orchid]

Kunneth theorem

[south patrol]

LOL

wait why are we in #groups-rings-fields lmao

oh i guess the original question is fine

[delicate orchid]

Kunneth theorem

[teal vessel]

I'm partway through proving (e^ix)^t becomes multiplication, which has been interesting, but I might save the rest of it for later because it's giving me a headache. a four-term expanded summation is quite the endeavor.

[delicate orchid]

I feel like you're not really going about this in the right way then

like, at all

[teal vessel]

nah, it's fine. At this point it's for fun

[south patrol]

hm

[delicate orchid]

if you've shown that e^ae^b = e^(a+b) then this should be completely unneeded

[teal vessel]

there's a particular symmetry going on that seems intriguing

[delicate orchid]

and you already knew that roots of unity multiply to other roots of unity via a simple norm argument

[south patrol]

is t an arbitrary complex number? if so then this is essentially true by definition

[delicate orchid]

I imagine it's an integer no

all groups are Z-modules :pack:

[south patrol]

lol

[teal vessel]

it was an integer for the original point of the question, which I've since answered, but the trig expansion here seems to be demonstrating some nice facts about trig identities that looks like fun to explore.

it's not groundbreaking by any means, but just an interesting diversion

[south patrol]

tbf yeah there is an issue here in that there isn't one choice of putting stuff to the power of t

e.g. for t = 1/2 there are two choices of square roots

whereas for integer t this is fully well-defined

[teal vessel]

autodidactry moments of getting distracted by shiny new knowledge about things I already mostly understood

[icy bear]

are they all just the cyclics of prime order?

[lapis latch]

Pretty much

[icy bear]

cus like, if it doesn't have a proper subgroup, it's cyclic
and cyclics of prime order famously have no proper subgroups
I just don't know if all the cyclics whith no proper subgroup are necesseraly of prime order

oh actually

I forgot the infinite ones

[south patrol]

Well lol

I thought normally a proper subgroup is any subgroup that isn't the whole thing lol

But ig this is proper nontrivial

[icy bear]

ye

it's said before that proper subgroup means it's neither trivial nor the whole group

[icy bear]

well they'll be cyclic too

I just don't know the converses

[rocky cloak]

There is a pretty straightforward description of all the subgroups of a cyclic group, in terms of the order.

[icy bear]

huh

don't tell me I'll think about it later

I found it lol thanks

if it isn't prime I can do a subgroup that's like
a², (a²)², (a²)³...

so ok prime cyclics for finite groups

[daring nova]

That infinite cyclic => no proper subgroup ?

It's very wrong

[icy bear]

just realized that infinite cyclic doesn't make sense I think

[daring nova]

Z

[icy bear]

oh yeah

ohhh yeah

[daring nova]

And only Z, up to isomorphism

[icy bear]

I forgot negative stuff counted as cyclic groups

negative "exponent"

[daring nova]

"cyclic monoid"

Is a group, nvm

Assuming finite

[icy bear]

wait actually for the infinite ones you can always find a proper subgroup

[prime sundial]

i don’t think this is true

[icy bear]

just skipping every other term since no exponent will get you back to the identity

sooo

yeah all prime order cyclics

and only them

[icy bear]

(and consequently never get you back to anything in general)

[rocky cloak]

Z is the only infinite cyclic group. All cyclic groups are of the form Z/n for nonegative n.

Cyclic groups are completely determined by their order

[prime sundial]

ah i kept thinking of just infinite

[rocky cloak]

Yeah, lots of infinite groups out there

[tardy hedge]

I love how cyclic groups are so easily understood

[coral spindle]

Once I was in a reading group where people were talking about finite semigroups generated by a single element – so "cyclic" in some sense of the word. Their cayley graphs have some kind of tail lead-up and then they have a loop after a certain point. Apparently semigroup theorists call this a saucepan shape, but it tickled me that the person talking called it a noose!!

[tardy hedge]

So for Q5, i suppose the point is that if we have a (1-1) ring hom between IDs we can create a corresponding homo between their quotient fields by “extending” the original map somehow

At this point tho, im still not sure why if the original map isnt 1-1 this fails

[rocky cloak]

Considering something like
Z -> Z/2

Why doesn't this extend to a map
Q -> F2?

[south patrol]

You can use the "universal property" of fraction fields too if you want (though i'm not sure if that'll have been covered)

[languid trellis]

What's the motivation for the quotient group being called the "quotient" group? Is it in anyway related to my "normal" understanding of a quotient like dividing numbers?

[tardy hedge]

ye

[coral shale]

it is

to explain to a primary school kid 12/4

you might draw a 3*4 grid of dots

and then draw circles

So if there are 3 rows, you draw a circle around every row of 4 dots

the exact same picture applies for C12/C4

[coral shale]

this picture would give you the underlying quotient set. The circles are the cosets

[glossy crag]

"Quotient" is an adjective that shows up all over algebra and usually has more to do with equivalence relations than outright fractions. This has to do with the fact that the set of equivalence classes of an equivalence relation is called a "quotient set" and equivalence relations abound in algebra. Of course the term "quotient set" itself probably came about originally because this has its roots in divisibility and the "quotient group" you speak of is closely related to this, e.g. the order of a quotient group is literally a quotient (of two known quantities). And besides, the whole idea of quotient groups comes from Z/nZ, which has everything to do with divisibility. My personal historical perspective on this is Z/nZ (which is literally divisibility) -> people try to generalise this and we get both equivalence relations and quotient objects.

[languid trellis]

I see

thanks to both of you

[tardy hedge]

Field of quotients of Z/2Z would just be Z/2Z?

Is that true?

[rotund aurora]

Z/2Z is a field

[tardy hedge]

Yes

Thanks

A field to its “field of quotients” is an isomorphism

I just did a question that was about that

[tardy hedge]

Somehow the map doesnt even work

Dividing by 0!!

[cobalt heath]

Field can lack a sign?

Oh wait one can have -1 = 1

[celest furnace]

Well finite fields arent ordered (if 0 < 1 then 0 < 1 + 1 + ... + 1 (p times) = 0 so 0 < 0 which isnt true)

So yes but they still have additive inverses

[cobalt heath]

I mean, did you consider what horror -1 = 1 can bring

[celest furnace]

?

[south patrol]

Positive characteristic is the best

[cobalt heath]

Not even comparable to being ordered

[rocky cloak]

A very direct connection to division of numbers is that

| G/H | = |G| / |H|

(When |G| is finite)

[cobalt heath]

In a normal field, "x = -x" implies x = 0

But..

[coral spindle]

What is a 'normal field'

There are many, many fields in which 1 = -1.

[cobalt heath]

Oh no

[coral spindle]

It is not exactly a usual property, but it is not by any means a crazy one.

[rocky cloak]

Life is so much easier, when sign errors become impossible

[cobalt heath]

Do you like lack of alternating forms

[coral spindle]

Alternating forms exist in characteristic 2... in some sense. They're just much easier to work with :)

[rocky cloak]

I mean you can still have alternating forms, they're just more symmetrical that way

[cobalt heath]

Just.. oh no

[rocky cloak]

You keep writing "oh no".

You know it's spelled "Ah yeah!" right?

[tardy hedge]

Idk im not really satisfied with how im thinking of it

[rocky cloak]

Alright, how are you thinking of it?

[tardy hedge]

So if the original map is not one-to-one then it maps nonzero things to 0. If you want to make a map from Q(D1) to Q(D2) by extending the original map, then …

I got that for the condition in a) to hold then the map must be [a,b] to [theta(a),theta(b)]

[rocky cloak]

Indeed

[tardy hedge]

If theta sends nonzero things to 0 then that map isnt defined on all of Q(D1)

[rocky cloak]

Exactly

[tardy hedge]

Im not satisfied with that though

I feel like i dont “reallyyyy” see it

I just know thats right cause i managed to do some algebra to find it

[rocky cloak]

I mean you are doing algebra, that's kinda the channel we're in

[tardy hedge]

But do you know what i mean when i dont feel satisfied?

Im just not seeing through all the layers

[rocky cloak]

I guess another perspective is that a ring homomorphism, must map units to units (hence doesnt map units to 0).

So if we make more things units we'll have fewer homomorphisms

[tribal moss]

Basically going from Z to Q you say "I want everything nonzero to have inverses".
When you then apply a non-injective homomorphsim to Z, some of the "everything nonzero" you just gave inverses turn out to map to 0, but if you want to extend the map to a homomorphism from Q, the homomorphism condition claims that the inverses still work! So the codomain of the homomorphism can only be something where 0 has a multiplicative inverse -- which is the zero ring!

[rocky cloak]

So basically
f(1/x) = 1/f(x) and if f(x)=0 you're in trouble

[south patrol]

||Well...||

[tardy hedge]

My prof

Lol

[tardy hedge]

Thank u

I will read and consider this answrr

Old man doing ^_^ is funny

[delicate orchid]

this guy seems based

[rustic crown]

^w^

det always ends up putting emojis in emails

my most recent email to a prof

[tardy hedge]

hahaha

[tardy hedge]

true

[graceful elm]

how would you go about counting group homomorphisms say from S6 to S5?
i know S6 can be generated by (234561) and (21) so i focused on where these could be mapped
they could both map to the identity, so thats 1
S5 has elements of order 6 we can map the 6cycle to (namely elements like (23154) which are a 3cycle and a swap, of which i believe there are 20)
elements of order 2 fall into two conjugacy classes- single transpose and double transpose. there are 10 single transposes and 15 double transposes.
20 choices for (234561) and 25 choices for (21) gives 500 nontrivial, and the 1 trivial means there should be 501, right?

unless we can map the generators to the identity independently? but wouldnt that mean there are actually 5 conjugacy classes?

if a is a 6 cycle and b is a transposition. and e is the identity ofc. then the conjugacy classes would be
f(a) = e, f(b) = e
f(a) = e, f(b) = (12)
f(a) = e, f(b) = (12)(34)
f(a) = (123)(45), f(b) = e
f(a) = (123)(45), f(b) = (12)
f(a) = (123)(45), f(b) = (12)(34)

with 1 element, 10 elements, 15 elements, 20 elements, 20*10 elements, and 20*15 elements
so now its 546
are all of these valid homomorphisms?

[delicate orchid]

you need to check that these still satisfy the relators your transposition and 6-cycle do in S6

cause then you'll know that f(xy) = f(x)f(y)

you could potentially map a to elements of order 3 or 2 as well because you didn't mention anything about injectivity (which would make the question much easier to answer lol)

but here might be an easier way of looking at it

if we had a homomorphism f : S_6 -> S_5, this implies by the first iso theorem that S_6/ker(f) = H < S_5

but S_6 only has 3 normal subgroups, S_6, A_6, and 1

so ker(f) has to be one of them, and it obviously can't be 1 as |H| <= |S_5| < |S_6|

assuming f isn't the identity map sending everything from S_6 to 1 in S_5, ker(f) = A_6, meaning H is a group of order 2

I think these f's will look like the sign function followed by mapping -1 to some order 2-element in S_5

[graceful elm]

im processing this here gimme one sec

[tardy hedge]

@delicate orchid do you like topology at all? im gonna take it next term

[delicate orchid]

I quite liked it when I took it and it's important for algebra

[tardy hedge]

thats cool. I only have a vague idea about what its about rn

[delicate orchid]

the most general form of analysis possible

[tardy hedge]

thats cool

[delicate orchid]

but it has a completely different vibe

it's strange

[tardy hedge]

thats cool cause i think i am actually terrible at estimating

inequality things i actually am genuinely shit at

[delicate orchid]

well yeah we stop working over R lol

[tardy hedge]

I have a hard time with comparing sizes

[delicate orchid]

we stopped working over R years ago

we replace \leq with \subseteq or whatever

[tardy hedge]

yeah maybe id enjoy that aspect more

the funny part is how i still havent had any formal class in analysis lmao

i did an honours calc class in first year that did epsilon delta stuff

and complex analysis

but convergence of functions and all this stuff i see i actually havent learned yet

isnt that silly

[graceful elm]

did you mean S_5 at the very end here?

[delicate orchid]

I did yeah

[graceful elm]

i think i understand then

[tardy hedge]

Its also interesting just how DIFFERENT algebra feels to learn and do than something like complex analysis for example

[graceful elm]

we have to pick a normal subgroup to be the kernel. it obviously cant just be the identity, bc we have to shrink the group if we want to map it to S5. the only other normal subgroups are S6 and A6. S6 as the kernel means we send everything to the identity giving 1 trivial map. choosing A6 as the kernel means the even elements all get sent to the identity, and we just need an element of order 2 to send the odd elements to. and that can be any of the 10 transpositions, or any of the 15 double transpositions. for a total of 26 possible homomorphisms across 3 different conjugacy classes. right?

[slim kayak]

is the course called point-set / general topology?

[tardy hedge]

yeah i asked him and we are covering the first x chapters of Munkres

first 4-5 or smth?

[slim kayak]

yeah thats a little bit more on the geometric side of topology, sorta

But very cool and interesting. The arguments and intuitions used are extremely useful in almost any field of math

[tardy hedge]

nice hopefully id like it

[delicate orchid]

if those numbers are right on the size of the conjugacy classes (i forgor) then yeah

[celest furnace]

Do you guys know of a website that shows you the lattice of subgroups of a group for most groups?

[delicate orchid]

there's one for cyclic groups that I know of

that's it though

even for small stuff like S_4 it's very unwieldy

[celest furnace]

Cyclic group lattice must look nice

[delicate orchid]

they do!

[rustic crown]

anything beyond order 20 and its weird

[celest furnace]

Just one subgroup for each divisor right

[delicate orchid]

https://demonstrations.wolfram.com/SubgroupLatticesOfFiniteCyclicGroups/

yup

[delicate orchid]

I mean even groups of order 16 are nuts

subgroup lattice for SD_16 my behated

[rustic crown]

okie anything beyond 12 is something i don't care about

unless it's nice in some other way

[delicate orchid]

my favourite part of these is if you put in a cyclic group of order p_1p_2...p_n you can see the n-cube in it's subgroup lattice

[celest furnace]

😮

I gotta try this

[delicate orchid]

try order 30 for the cube

order 210 for the hypercube

there's some distortion but it's there

[celest furnace]

Thats awesome

[rustic crown]

lattice of subsets of {2, 3, 5}

[delicate orchid]

(the idea is that cyclic groups C_p_1...p_n are isomorphic to C_p_1 x C_p_2 x ... x C_p_n, the subgroup lattices of each of those are just a line so you get an n-cube by direct producting them)

[rustic crown]

[celest furnace]

360 has so much

[rustic crown]

and for arbitrary factorization n = prod p_i^e_i, you have a product of linear posets {0, 1, ..., e_i}

[delicate orchid]

yur, but you get a nice cube when e_i = e_j for all (i, j)

and it's easiest to see the cube when you have e_i = 1

check this out as well
https://juliapoo.github.io/Cayley-Graph-Plotting/

my favourite one here is A_5. You can see the natural action on the icosahedron and the stabliser subgroups of each vertex because the cayley graph is just an icosahedron lol

[celest furnace]

Holy shit that’s awesome

[rustic crown]

what was a cayley graph again?

[celest furnace]

Idk either but I do know that it is awesome

[rustic crown]

i can make sense of it if i had a presentation of the group

but we're not using all elements as generators right, that would just give the complete graph

[delicate orchid]

assign each element of ur group a vertex, and assign each element of some fixed generating set for ur group a colour

draw a directed edge of colour "g" between two elements x, y if x = yg

[rustic crown]

right makes sense

what's this website using for edges?

like a minimal set of generators?

[delicate orchid]

oh I don't think you can access the presentation, I assume it's either minimal or the one that gives the prettiest picture

[rustic crown]

makes sense

[ivory trail]

yeah they don't say, but you can usually tell by looking

[delicate orchid]

it's a shame you can't see the tetrahedron in A_4

[ivory trail]

unfortunately it's not powerful enough to let me see M_11

[delicate orchid]

good

[ivory trail]

i want to see it though

[delicate orchid]

some things were not meant to be witnessed by mortals

[ivory trail]

i guess i just have to think about some random stuff about multiply transitive permutations instead

[delicate orchid]

something something octet action something something

[icy bear]

what's the problem ;-;

is it that it is assuming a or b exist or something?

maybe actually

since without aCa it's not clear any elements are related to any element

[charred iris]

it is an issue with assuming existence, though not with aCa (given that aCa existing is a conclusion)

[icy bear]

I know like

without tha aCa property

aCb can't necesseraly be stated

[charred iris]

?

< isn't reflexive, but I can state 1 < 2?

[icy bear]

I mean

I mean not necesseraly

you can't state there are a and b such that aCb

can't assume

[charred iris]

yes

[celest furnace]

Ah thats crazy

[brave agate]

Show that a group of order 210 with normal subgroup of size 15 has a normal subgroup of order 5.

Sylow isn't enough here from what I can see, you can only conclude the normal subgroup itself has a normal subgroup of order 5, and normality is not transitive

[celest furnace]

You're almost there

Hint: A subgroup of order 5 of a subgroup of order 15 is a Sylow 5-subgroup. Second hint: Characteristic subgroups.

[brave agate]

oh i didnt know this. that's it then?

because every characteristic subgroup is normal in the entire group

[celest furnace]

Yes. Why is that fact true?

[brave agate]

If M is characteristic in H, a normal subgroup of G, then automorphisms of H send M to M, but these automorphisms can also be realized as conjugation (inner automorphism)

[celest furnace]

That’s true but why is every (normal)* sulow subgroup characteristic?

[brave agate]

since it is the unique subgroup of order 5, an automorphism must map the group to itself

[celest furnace]

Yep great

[graceful elm]

what does this notation mean? (in the context of sylow subgroups)

[prime sundial]

do you have the context?

it could be the special linear group over the field with 7 elements

[graceful elm]

the order of sylow 7-subgroup of SL_2(7)
the number n_7 of sylow 7-subgroups of SL_2(7)

i dont remember going over special linear groups but admittedly i didnt pay a ton of attention

[prime sundial]

i imagine it is that then

SL_2(7) probably refers to the space of 2x2 matrices with determinant 1, with entries coming from the field of order 7

[graceful elm]

geez, ok

[graceful elm]

i just calculated that that group would have 336 elements, is that correct?
49 choices for first row minus the 0 vector leaves 48
49 choices for second row minus the 7 multiples of row 1 leaves 42
divide by the 6 possible determinants (since we only want to allow matrices of det 1)
48*42/6=336

[icy bear]

any hint on c? (idk if it fits here)

made basically no progress

I'm just trying to think of a necessary condition for an equivalence class to contain S but I'm not finding anything

and I initially assumed this T would be S' or related but I can't prove it and have no actual reason to believe it

I initially thought like
xTx+1
x+1Tx+2
so
xTx+2 but I have no reason to think x+1Tx+2 since x+1 can be greater than two

[tribal moss]

I think you're probably looking for too nice a condition -- it's not going to be particularly pretty.

[icy bear]

probably ;=;

[chilly ocean]

Is there a category where the polynomial ring over R is an initial object? I am trying to better understand its universal property among R-algebras.

[coral spindle]

I mean sure, if you look at the category of R[x]-algebras lol

[icy bear]

I think I may try examples

like, trying to do some kind of sub equivalence relation of S'

which contains S

[chilly ocean]

Is this even true?

[tribal moss]

For example consider
U = { (x,y) | x=y or both of x,y are > 0 }
This is an equivalence relation that contains S. So what you're looking for is a subset of both S and U.

[coral spindle]

Yes

[chilly ocean]

It seems likely you can be an R-algebra in more than one way.

[coral spindle]

Who said you can't?

[chilly ocean]

Wouldn't that make it not initial?

[coral spindle]

I think if you look at 'pointed' R-algebras — pairs (A, x) where A is an R-algebra and x is in A, and you require algebra homomorphisms to preserve these distinguished points — then (R[x], x) is initial in this category. This encapsulates the universal property.

[tribal moss]

Those different ways are then different objects.

[chilly ocean]

Ahhh I see, this is a helpful perspective, thank you

[chilly ocean]

ahh I see, because the R-algebra includes the morphism

[icy bear]

huh

thx

ok I was very wrong in what I was thinking

what type of thing I was thinking at least

[tribal moss]

Perhaps that was too roundabout a hint.

[icy bear]

what is roundabout

[tribal moss]

This.

[icy bear]

(I searched for the translation of this word for my language and one of the translations was merry-go-round)

[tribal moss]

The general idea of "smallest equivalence relation that contains A" is that you need to relate any two points that can be connected by a finite chain of points where each link in the chain is in A either forwards or backwards.

Ah. 🎡 No that wasn't the meaning I had in mind. 😆

[icy bear]

the intersection thing and smallest equivalence relation that contains A is the same thing right

[tribal moss]

Yeah. Sorry, I'm so used to considering them the same that I didn't check which of the wordings your problem used.

[icy bear]

lmao

the more I think about this problem the weirder it gets I'm going crazy

I can't think of choosing any point to relate to anything

I mean, number

[tribal moss]

Try being concrete and choosing, say, the point 0.1762. What does that need to relate to?

[icy bear]

1.1762 and 2.1762

[tribal moss]

Yes.

How about 3.91?

[icy bear]

3.91

[tribal moss]

Right.

[icy bear]

I'm trying to think if it should relate to anything between 0 and 2

wait actually

stop everything you're writing I thought of somethign

I mean not everything just don't tell me anything

wait...

is it just

S' union all points just related to themselves?

with no other connection?

[tribal moss]

Not entirely quite, because S doesn't contain (0.1762,2.1762).

[icy bear]

oh right

but like

ig the idea is right

[tribal moss]

It sounds like you have a good intuitive idea of which pairs should be in T and are just looking for a succinct way to describe the set.

[icy bear]

for all x between 0 and 2, xTx+1 and the rest is just (x,x)

[tribal moss]

That still doesn't say that 0.1762 T 2.1762.

[icy bear]

well 0.1762 is between 0 and 2

oh should I like

describe it specifically in pairs

(x,y) such that

[tribal moss]

[Also this should probably have been in #discrete-math or perhaps #proofs-and-logic]

[icy bear]

(ty I didn't know where to put it so I thought "well I learned about equivalence classes the first time in group theory")

T = {(x,y) | x = y or y = 1 + x for 0 < x < 2}

?

[tribal moss]

Still doesn't give you a set that has (0.1762, 2.1762) as an element.

[icy bear]

it does but I realized I'm wrong

[tribal moss]

What?

[icy bear]

OHHHH

I see sorry

[tribal moss]

x=0.1762 and y=2.1762 satisfies neither "x=y" nor "y = 1+x and 0<x<2".

[icy bear]

aaa guaranteeing the transitive property is hell

is describing a set algorithmically like, wrong

cus I'm thinking of adding 1 to x until it's a number greater than 2

I have no idea how to describe that without it being a process

I thought of something

[tribal moss]

My solution would involve (as one component) something like S' intersect (0,3)²
where (0,3) is the open interval, not the pair.

[icy bear]

maybe just something like y = x + k such that y < 3

and k >= 1

an integer

[icy bear]

welll nowhere in the exercise it says that the set should be written prettyly

but that's cool

[tribal moss]

You can spell it out like that too, but youl'll need bounds on both x and y.

And k>=1 is not what you want because T also needs to cotain (2.7, 0.7)

[icy bear]

the thing I was thinking is just write out the same thing but changing x for y

[tribal moss]

Hmm, I suppose you could do that, by why not just k be any integer regardless of sign?

[icy bear]

fr troposphere you're a genius

I think I'll write it like y - x = k

cus it's prettier

[tribal moss]

And makes it look like the definition of S'

[icy bear]

indeed

ig just bounding then beteen 0 and 3 works

soooo

T = {(x,y) | y = x or y-x = k for k in Z and 0 < x,y < 3}

1- thank you a lot troposphere you made me realize a lot of errors and wrong ways of thinking I had no idea
2- lmao I made a joke up there with the 3.91 and it actually made sense

I mean, it actually lead me to the answer

good exercise great learning experience

[tribal moss]

Right.

I would write it with parentheses around "y-x = k for k in Z and 0 < x,y < 3" just to avoid misunderstandings.

[icy bear]

good idea

[languid trellis]

I asked a question in my thread to not clog up the channel I would appreciate it if anyone took a peek

[stark helm]

Please Help me check one question: et p be a prime. Show that if H is a subgroup of a group of order 2p that is not normal, then H has order 2. My solution is to suppose there is an element a in H so that |a|=p, by lagrange we know that |H|=p or 2. Then we have a subgroup H with order p, namely <a>=H. Therefore, we must have an element b in G that is not in H, so we get bH in G. since b is not in H, so bH intersect H is identity. since H is not normal, so bH doesn't equal to Hb, so bH intersect Hb is trivial. Since Hb intersect H is also trivial, then we have bH, H, and Hb in G, then the order of group G is at least 3p, which contradicts the statement that G has order of 2p.

Is it a valid proof for this question?

[delicate orchid]

this looks good to me

it's like the proof that index 2 subgroups are normal but backwards

[stark helm]

OK, really appreciate

[tardy hedge]

That was a question on my final exam

Index 2 subgroups are normal

[delicate orchid]

it's a good one

harder version, if p is the smallest prime that divides |G|, then a subgroup of index p is normal

[stark helm]

I think it is a sylow p-subgroup, so we need to prove it is the only p-subgroup in G?

[delicate orchid]

no it's not a sylow p-subgroup

sylow subgroups always have index coprime to p by definition

consider the action of G on the cosets of an index p subgroup and the homomorphism into a particular symmetric group that gives you

[tardy hedge]

In my book it covers a lot more group theory in chapter 7

Havent gotten to it yet

Conjugacy, group actions, sylow stuff etc hasnt been covered Ye t

Need to go thru chapter 6 on fields and then ch7

[delicate orchid]

conjugacy is surprising lol

[tardy hedge]

Ya im looking forward to getting back to groups

Although i did enjoy the rings and fields diversion

[icy bear]

is the product of two transpositions always a 3-cycle?

[celest furnace]

No

Take the same transposition

Or disjoint transpositions

[icy bear]

oh yeah

[stark helm]

How can I prove that Z/2Z is the only group that has exactly two conjugacy classes?

[barren sierra]

So how do you prove Z/2Z is the only group with 2 elements?

Can you take a group with 2 elements and prove that it’s isomorphic to Z/2Z

Unless I’m misunderstanding the question

[delicate orchid]

I don't see how this is relevant ngl

assume that there's a larger group that only has two conjugacy classes and recall that conjugation is an automorphism and thus preserves the order of elements

[barren sierra]

Oh conjugate classes

I can’t read

I read cosets

[icy bear]

corsets

[barren sierra]

And like elements of Z/2Z are cosets

I’m dumb

[delicate orchid]

didn't even let me quip

Z/2Z^n does have only two orbits under all automorphisms, that is true

[hot lake]

yeah

that's what I was thinking

[delicate orchid]

transitive fusion systems :pack:

[rocky cloak]

At least for finite groups I can think of an argument that uses that p-groups always have nontrivial center, but I feel like there should be an easier argument.

[delicate orchid]

there is

||assume G is a larger group that satisfies this condition, then G cannot be abelian obviously. But then G must have at least two non-identity elements with different orders||

||conjugation preserves the order of elements||

[rocky cloak]

I don't see why ||the elements would have different orders||, but another argument I could think of was to ||consider the orbit of a non-identity element to conclude that |G|-1 divides |G| ||
Still this is using that G is finite though

[delicate orchid]

oh I didn't even consider G infinite

[delicate orchid]

||true, we need to exclude p-groups, I forgot about p_+^{1+2n} lol||

||The only way for all non-identity elements to have the same order is for the group to be exponent p, and thus a p-group - so we would need to use the non-trivial centre thing||

[delicate orchid]

ah no wait. This is definitely the way to go, lol

[stark helm]

So when the group is of order power of prime, then we have nontrivial center. (but I am confused how can we compute number of conjugacy classes. The only thing I know is that if a and b are conjuage, then xax^-1=b for all x, and cl(a)=a iff a is in Z(G)

[delicate orchid]

oh so we do have that result

each element of the centre is obviously in it's own conjugacy class

[rocky cloak]

So if there is an element in the center, it gets it's own conjugacy class. So if there's only supposed to be 2 conjugacy classes you can only have 2 elements.

[rocky cloak]

Apearantly, it's not true

https://math.stackexchange.com/a/88997/306319

[delicate orchid]

something weird happening in infinite groups no way

[barren sierra]

Infinite groups are useless

Name one useful infinite group, you can't

[delicate orchid]

S_9

it's pretty big it's gotta be infinite right

[barren sierra]

O(1) = ∞ for large values of 1

[stark helm]

if the group has non-trivial center, can we conclude that it only has its own conjugacy classes?

[rocky cloak]

Every element of the center is in a conjugacy class of its own

[stark helm]

I may need to know how to justify that Z/2Z has exactly two conjugacy classes?

[lapis latch]

Maybe the class equation will help

[rocky cloak]

I mean, you just count the conjugacy classes

There are two of them

[stark helm]

is it 0 and 1?

[coral spindle]

You mean the only two elements of Z/2Z?

Yes

Yes indeed

[static glen]

Z/2Z has only 2 elements that are obviously not conjugate so the proof is kinda trivial

[stark helm]

I am kind of confused about if conjugacy classes just represent elements in a group. I mean if we say a is in Z(G), then cl(a)=a. and we write a as one conjugate class for element a.

[coral spindle]

cl(a) = {a}

Does that clear your confusion?

[static glen]

No they don't if your group gets non-commutative

[coral spindle]

Generally a conjugacy class — which I will remind you now is a set of elements — contains many elements of the group.

However, central elements live in their own conjugacy classes, all alone.

[static glen]

There are even groups (so called i.c.c. groups) such that every element (expect the identity) has infinite conjugacy class. These groups are important when constructing Type II factors (von Neumann algebras). an example would be the free group on n generators (n >= 2; the free group with 1 generator is just Z)

[rocky cloak]

Or for S3 the conjugacy classes are {identity}, {3-cycles}, {transpositions}

[stark helm]

So go back to Z/2Z, if i say two conjugate classes, and only element 0 and 1 in this group. then cl(0)={0} and cl(1)={1} represent two conjugate class.

[static glen]

they not represent more precisely they are

For some background: being conjugate is an equivalence relation and the conjugacy classes are the equivalence classes w.r.t. this relation; so it is not surprising that the conugacy classes partition the underlying set

[stark helm]

OK, then the question:how to prove Z/2Z is the only group that has exactly two conjugacy classes. Do you have any ideas for this one?

[delicate orchid]

all matrices in GL_n(R) are maximal rank, they're invertible

:letrollface:

[delicate orchid]

I'm going to steal jagr's solution for this one cause it's much cleaner than mine. Do you have the orbit-stabiliser theorem

[stark helm]

I actually don't know about this theorem

[static glen]

You're right, a dumb mistake; take different jordan canonical forms then it works

[delicate orchid]

yurrr. Multiplicty of eigenvalues is preserved and what not

[delicate orchid]

ok we'll have to go with my proof then, cause you do have that p-groups have a non-trivial centre right?

which is incredibly strange to me but you know

some professors don't know how to structure courses in a way that makes sense

[static glen]

But not the multiplicity in the minimal polynomial

would be nice if every matrix is diagonalizable but unfortunately jordan canonical form is the best one can do (at least over algebraically closed fields)

[delicate orchid]

the diagonalisable matrices are dense at least

[static glen]

Never seen this. Do you have a reference? (Just for personal interest)

[delicate orchid]

or I can just prove it

an invertible matrix isn't diagonalisable if and only if one of the eigenvalues has multiplicity greater than 1

say the matrix is n by n

then this corrisponds to a codimension >= 1 space in R^n

the complement of this space is thus obviously dense

[static glen]

this should be have different geometric and algebraic multplicitys

[delicate orchid]

I literally don't know what that means

[static glen]

algebraic multplicity= multplicity in the characteristic polynomial; geometric=dimension of corresponding eigenspace

[delicate orchid]

if two of the eigenvalues are the same, when you attempt to diagonalise it you get a jordan block instead

[rocky cloak]

I mean you just prove that the matrices with all distinct eigenvalues are dense, then it follows that the diagonalizable are dense

[delicate orchid]

that's all I've ever needed to know

[static glen]

The identity matrix has eigenvalue 1 of multiplicity n is is obv diagonalizable

[delicate orchid]

yeah that's true for any diagonal matrix with a repeated element on the diagonal

the problem is I just don't care

sorry I don't mean to be rude but I just don't

the gist of the proof holds

[stark helm]

I see that if I can suppose a larger non-abelian group, then we have at least two non-identity elements of the same order, and it satisfies condition for conjugacy class. like we say a and b are conjugate, so we have at least two conjugate class: cl(a) and cl(b).if p-groups, then have non-trivial center, so the conjugate classes will be at least 4, otherwise at least 3?

[static glen]

I think you have shown density of a subspace of R^n not in GL_n(R), or M_n(R) whatsoever

[delicate orchid]

the map is continuous

[static glen]

I guess the statment is false

[delicate orchid]

the continuous preimage of a dense set is dense

[rocky cloak]

Which statement?

[static glen]

wait i have to think about it for a second

[delicate orchid]

if you judge an entry in a matrix by a little bit the eigenvalues only change by a little bit as well
that's a heuristic as to why the map should be continuous

same logic behind why the determinant is continuous

[static glen]

the determinant is a polynomial so continuity is really easy to check

[delicate orchid]

yur

[static glen]

The Leibniz formula shows that det(A) is a polynomial in the entries of A

[delicate orchid]

if we have a larger non-abelian group we have two options, either everything not equal to the identity is the same order, or two things are different orders. If two things are different orders they cannot be conjugate and so we have at least 3 conjugacy classes. If everything is the same order the group has to be exponent p (an exercise) and thus a p-group (follows from Sylow), but then p-groups have a non-trivial centre, so we have at least |Z(G)|+1 > 2 conjugacy classes (at least one for the stuff outside of the centre, which is a non-empty set because we're assuming it's not abelian, at least one for the non-trivial element of the centre, and one for the identity)

[stark helm]

if G is abelian, what will happen, I am thinking |Z(G)|=G so if it will be too many conjugate classes?

[delicate orchid]

it's pretty trivial to see that the conjugacy classes in an abelian group are of size 1

gxg^-1 = gg^-1x = x

[stark helm]

Then if we have non-trivial center, elements in Z(G) form a distinct conjugate classes right?

[delicate orchid]

it's pretty trivial to see that conjugacy classes of elements in the centre are of size 1

gxg^-1 = gg^-1x = x for x in Z(G)

[stark helm]

I mean when we assume that G is non-abelian and everything not equal to identit is the same order

[coral spindle]

Woah are we gonna see the legendary wew triple whammy

[stark helm]

Because I saw you wrote in this case, Z(G) is non-trivial, so we have at least Z(G)+1 conjugate class, and it seems to me that all elements in Z(G) are distinct conjugate classes

[delicate orchid]

no lol

[delicate orchid]

yes?

I don't understand what you're asking

conjugacy classes of elements of the centre are always size one. Because gxg^-1 = gg^-1x = x for all g in G

this seems like an incredibly advanced question to ask if you dont' know orbit stabiliser and aren't comfortable with the notion of a conjugacy class yet

perhaps there's a simpler way I'm missing that doesn't use either orb-stab or p-group theory

[stark helm]

I mean in a p-group and assume G is non-abelian, p group has non-trivial Z(G), so we have Z(G)>=2. then when in abelian group, and we want to know about number of conjugate classes, by class equation, we have |Z(G|+|G/C(a)|=|Z(G)|, and Z(G)=G, so I am wondering if |Z(G)| is of size 1, I can know that cl(a)=a in this case, but there are all a in G, so I am wondering if there is infinite finite size

[delicate orchid]

why do you have the class equation but not orbit stabilser?

what eclectic fuck up desigend your course lmfaooo

anyway

I've come up with a really nice proof using burnside's fixed point lemma give me a sec

[delicate orchid]

ok. Please write out the definition of a conjugacy class

I cannot stress enough that it should be really obvious that if x commutes with everything in g, that {gxg^{-1} : g in G} should just be {x}

and what does "infinite finite size" mean here?

we have counter examples for G being an infinite group, so we have to take it to be finite

[coral spindle]

Perhaps they mean 'arbitrarily large'

beats me tho

[stark helm]

if a and b in G, then we have xax^-1=b for some x in G. And I am thinking Z(G)=G when G is abelian, so all a in Z(G) can form a class cl(a)={a}, and |cl(a)| divides |G|

cl(a)=|G/C(a)|, so it seems cl(a) is 1?

[delicate orchid]

nevermind this proof is a mess lol

[delicate orchid]

sure if you want to phrase it like that

which seems really silly

because xax^-1 is just a in an abelian group

so obviously you're not going to have anything else in the conjugacy class of a

but yes you can see it via |G/C_G(a)| which is a consequence of orbit stabiliser and I have no idea how you prove that without just proving orbit stabilser

I feel like I'm losing my mind

[stark helm]

Because I considered like all a in G, and does that mean classes of all a means number of a as a classes

like cl(a1), cl(a2)....

[delicate orchid]

is there a language barrier here

[stark helm]

I think it has, because my first language is not english

[delicate orchid]

let A be an abelian group

let a be in A

then $cl(a) = {xax^{-1} \colon x \in A} = {xx^{-1}a \colon x \in A} = {a \colon x \in A} = {a}$

[cloud walrus]

W;3w Lads Tbh

[delicate orchid]

thus the number of conjugacy classes in an abelian group is just $|A|$

[cloud walrus]

W;3w Lads Tbh

[stark helm]

Then I ask if group A is abelian, we have |A| cnjuage classes, is that true?

[delicate orchid]

I just gave a proof of that statement, so yes

[stark helm]

That's actually what I asked at the first time , but seems not that clear

really appreciate that

[delicate orchid]

yeah and I responded with this

and gave basically the same argument that I just did lol

[stark helm]

show that there are two Abelian groups of order 108 that have exactly four
subgroups of order 3. ( what does two abelian group mean, distinct two or direct product?)

[barren sierra]

2 non-isomorphic

so there exist groups G and H with 108 elements each with 4 subgroups of order 3, G and H are not isomorphic to each other, and any other group with 108 elements and 4 subgroups of order 3 will be isomorphic to G or H

structure theorem go brrrr

[stark helm]

it seems that abelian group of order 108 Z2Z2Z3Z3Z3, Z4Z3Z3Z3, Z2Z2Z9Z3, Z2Z2**Z27, Z4Z9Z3, Z4Z27

Is that related to 4-subgroup of order 3?

[rocky cloak]

Well, some of the groups you listed have 4 subgroups of order 3, some have more and some have fewer

[stark helm]

can you name which one has 4 subgroups of order 3?

[rocky cloak]

I mean I can, but won't that spoil it for you?

[stark helm]

Take example for Z2Z2Z3Z3Z3, it seems to me that there are 3 subgroups of order 3, do you think it is correct?

[rocky cloak]

I think there are a few more than that

[delicate orchid]

just in Z3^3 there are 7 I think

[rocky cloak]

Might even be ||13||

[delicate orchid]

hmmm

oh yes

of course

||I just did <a_i>, <a_ia_j>, <a_1a_2a_3> I didn't consider ones with different exponents||

I wonder what the formula is in general. For Z_n it's just phi(n)

Something to do with orbits under GL(Z/nZ, k) I recokn

ignore me btw Kingwisdom I'm rambling

[stark helm]

You mean Z3Z3Z3? and count the number of |Z3Z3Z3|=3 right?

[barren sierra]

Use \* to type * plz

Make your messages much more readable

[coral spindle]

|Z3*Z3*Z3|=3? Are you sure about that?

[stark helm]

I typed, but it will be deleted as I send it out

[stark helm]

I mean if I want to find subgroup of order 3

[stark helm]

the subgroup of this group

[barren sierra]

Yes because * makes words italics which is why half your Z's are slanted...

[barren sierra]

what do you mean "the" subgroup and what subgroup are you referring to

[stark helm]

I mean in a group like Z3 Z3 Z3, and I want to figure out number of subgroup of order 3 in this group

[rocky cloak]

Like number of subgroups of order n in (Z/n)^k or what formula are you wondering about?

[delicate orchid]

well that's just the number of elements of order n

I think I've figured it out anyway

[rocky cloak]

I think it should be
(n^k - (n - phi(n))^k ) / phi(n)

So yeah, number of elements of order n divided by phi(n)

[delicate orchid]

they correspond with tuples of length <= k where each tuple is uniquely determined by the tuple of differences between the terms

these tuples corrispond to the exponents in the generators of (Z/n)^k of the element that generates the subgroup of order n

ah no wait this only works for n prime

sad

[coral spindle]

Yeah I would expect this formula only to work for primes

[delicate orchid]

RAAAARRRRGGGGHHHHHHHHH

all integers are prime

[coral spindle]

E.g. I don't think this counts the subgroup <(2,3)> of Z6 * Z6

[delicate orchid]

yeah, that's the issue with non-prime

you can have prime order mfs multiply in cringe ways

[coral spindle]

With primes it's not too bad to count ig

[rocky cloak]

Yeah, I guess assume n = p^m , then for n a product of p^m you just compute it for each and take the product.

Then it should work I think

[stark helm]

if determine all homomorphisms from S5 to Z/5Z. How can I know that any sigma in S5 has order 2?

[prisma ibex]

what

there are lots of elements in S_5 that don't have order 2

[delicate orchid]

and there are no elements of Z/5Z that have order 2

[south patrol]

oh yeah? name one

[coral spindle]

bob

[delicate orchid]

The identity

[graceful elm]

Considering homomorphisms from a dihedral group Dn onto Z/2Z
Couldn't all rotations be mapped to the identity, and all reflections to the element of order 2?
I know that the kernel of this homomorphism would be the reflection, and that kernels have to be normal subgroups, and that the subgroup of a reflection is not normal in a dihedral group. But I can't think of why my function wouldn't satisfy the homomorphism property

[delicate orchid]

the kernel of this homomorphism is not the reflections? It's the rotations as you said.

this is the map given by the quotient D_n/<t>

or r, idk what the kids use these days for the rotation

[slim kayak]

These damn kids and their cool and hip group presentations

[delicate orchid]

anyway, we can associate hom(D_n, Z/2Z) bijectively with the set of index <= 2 normal subgroups (lol, bit redundant) of D_n

via kernels

[graceful elm]

oh ok i just dont understand kernels i guess

OH its everything that collapses to the identity, right

[delicate orchid]

yus

every kernel is a normal subgroup, and every normal subgroup arises as a kernel

[celest furnace]

t?????

What does t stand for?

[delicate orchid]

I literally just said in the god damn post you quoted it stands for the rotation

RAAAAARGHHH

[celest furnace]

No like as words

Like r is clearly for rotation

What's t? Translation ???

[round hull]

s and t are generators

also if r stands for rotation then what is reflection

r'

[cobalt heath]

refl?