groups-rings-fields

[tough raven]

Ah, I see.

[velvet flame]

Hlo ppl I have a q for u

Let's see if u can solve it

[elfin wraith]

You cant reverse physcology me into doing your homework. I know I cant do it!

[velvet flame]

It s not hw

[thorn jay]

that's what they all say

[velvet flame]

Cmon

Pls

It from ch circle

And I am doing self study

[thorn jay]

"groups-rings-fields" lol

[velvet flame]

Cmon pls

[thorn jay]

ch circle?

[velvet flame]

Yes

🙏

[thorn jay]

no I'm asking you to clarify

wtf is that

[velvet flame]

Its a q from ch circle

[thorn jay]

what is ch circle

[elfin wraith]

Yeah, not really the channel for this, try #prealg-and-algebra

[delicate orchid]

we will never know

[thorn jay]

it seems so...

[sacred wharf]

In an artinian ring with no non-zero nilpotent ideals

there are no non minimal ideals right ?

As in no proper ideal that contains a proper minimal ideal

[delicate orchid]

I believe such a ring would be semisimple and thus have krull dimension zero but that only guarentees what you want for prime ideals

maybe someone who has thought about rings once in the past 3 years can spot the obvious thing I'm missing

[sacred wharf]

i'm assuming unity

[delicate orchid]

that's good - I was as well

[elfin wraith]

Every one sided nilpotent ideal is contained within every prime ideal so maybe theres soemthing there, but ive not fully formed a thought yet

[sacred wharf]

In this proof

[delicate orchid]

even so why can't we have 0 < (p^2) < (p) < R. I don't think the intersection of all prime ideals vanishing could detect this happening

we could try artin-wedderburn tbf but idk how it would help

[vapid vale]

you're a direct product of fields, so k x 0 x 0 ... is contained in k x k x 0 ... etc right?

[delicate orchid]

ts is true

[sacred wharf]

in this we show that in a semi simple ring any ideal can be characterized as Re where e is some idempotent

[delicate orchid]

you can just add the primitive idempotents together right

[sacred wharf]

But the proof also uses the fact that the minimal ideal in said ideal is equal to Re

[elfin wraith]

I really need to brush up on my rings before this scholarship interview

[sacred wharf]

is it martingale scholarship ?

[elfin wraith]

Yuh

[vapid vale]

ya

[delicate orchid]

I hate rings so much

[sacred wharf]

linkedin spammed it all over my feed so i saw if i could apply for it but i'm not even eligible 💔

[delicate orchid]

I just can't SEE them like I can with groups ykwim 💔

they're too big

[elfin wraith]

Yeah but they tend to have nicer presentations so you can just see them

[true bolt]

wut about E_oo rings

[delicate orchid]

better but not great

[sacred wharf]

no longer a mod

[elfin wraith]

Crazy that they let wew mod back in the day

dark times

[true bolt]

I'm on break break, I will be pink again in like 2 minutes

[delicate orchid]

strongly disagree. It tells you how the elements work but nothing structural

[elfin wraith]

What else is there to know? This is actually somewhat serious, this is I think a large reason why I prefer working with rings

[true bolt]

rings are annoying because there's more structure so you have to study more

[delicate orchid]

I can't tell you what else there is to know but the 6000 billion different entries in the chain between "domain" and "field" make my head hurt

[elfin wraith]

I feel I understand them far better because I can look at the typically much more simple presenatations and just know how they behave, even for big rings like A_n

[delicate orchid]

A_oo mention

[true bolt]

A_oo categories have such a simplie intuitive definition

[elfin wraith]

(I mean the weyl algebra, not any of that woke n-lab nonsense)

[true bolt]

A_oo cats are very concrete objects

they come up from topology

as in non schizo topology

[delicate orchid]

that's... all infinitiy categories...

oh ok non-schizo OK THEN!

[quiet pelican]

The thing that makes things nice for rings is that the quotients you care about are more likely to behave as you expect, I feel
So like generator-relation presentations are more useful

[elfin wraith]

Eyyy im all pissy here

[delicate orchid]

relations in quotients work the same for both groups and rings?

[delicate orchid]

WOW

[elfin wraith]

Apparently ive been too nice of late

[twilit wraith]

Shit role when

[true bolt]

my condolences 🌾

[true bolt]

<@&1328097118644338779>

[quiet pelican]

In the literal sense yeah
But in a more philosophical sense, I feel the ones we care about for rings tend to be nicer than the ones we care about for groups

[elfin wraith]

Yeah but like S_n is confusing

[thorn jay]

rings are nice because abelian groups are nice and monoids are relatively nice, so tensoring the two theories yields a somewhat nice theory

[elfin wraith]

I like solid concrete objects like x and a, not (123) where I need to work out which version of the notation someone using

[delicate orchid]

How it’s literally the nicest object ever

[elfin wraith]

Cycle notation confuses me

[delicate orchid]

who is out here actually multiplying permutations

[thorn jay]

not even that nice

[elfin wraith]

I am only like half joking here, S_n is of course pretty concrete but I do think in relation to like a polynomial ring, not as concrete or nice

And im including non-com rings in that, even weird quotients of them

[true bolt]

S_n is evil

I don't want to actually do computations in it

[thorn jay]

just go so abstract that the word computation doesnt even make sense anymore

reasoning with concepts atp

[tough raven]

Is there a large class of finite groups G for which in every composition series 1 = G0, ..., Gn = G, Gi is not normal in Gj whenever j-i ≥ 2?

[wraith cargo]

nobody does everybody just pretends they do

[delicate orchid]

"counting is hard" 💔

[elfin wraith]

This is based entirely just on vibes because I dont work with groups, but I would presume this is specific enough that youd want to just make a search with sage

Or your CAS of choice

[thorn jay]

yes

[rapid cave]

Music 🎵

Wanna help me on #point-set-topology?

[thorn jay]

what a coincidence! listening to music rn

[rapid cave]

Me too

[thorn jay]

(not a coincidence i am literally always listening to music)

[rapid cave]

XD

True to your name

[wraith cargo]

I barely know what 2+3 is c'mon

[south patrol]

Real

[tough raven]

Does the following proof strategy for Sylow's theorems work?

1. Consider GLn(Fp). Show that Un(Fp) (upper-triangular unipotent matrices) is a Sylow subgroup (mainly that the index is coprime to p). Show that any representation of a p-group in characteristic p maps into Un in a basis. This proves the first two theorems for GLn(Fp).
2. Prove that any group embeds into GLn(Fp) for sufficiently large n (indeed Sn embeds in GLn(Fp)). By some basic counting arguments the intersection of a Sylow subgroup is a Sylow subgroup of the subgroup. Hence using the result on representations of p-groups we get Sylow's first 2 theorems in general.
3. TODO 3rd part (just the 1 mod p bit is not very easy).

[rocky cloak]

Did you also have an argument for 1 in mind, showing that a representation lands in Un

[tough raven]

No, I realised I need to fill that in as I wrote it. But it should essentially come down to showing that in char p the only irrep of a p-group is trivial (e.g. since the group algebra is local). I'm still hoping I can find a more basic argument though.

[rocky cloak]

I think that argument is pretty good. But not super basic yeah.

For 3) you can just use that |N(P)/P| = |G/P| mod p, but that doesn't involve any rep theory

[tough raven]

I think when I had the idea I knew it's elementary (or at least just linear algebra) for cyclic groups and subconsciously jumped to p-groups

[marble hinge]

I am working on Problem 7 from Herstein:

it's quite interesting, but I think I am getting stuck

I have those considerations:

1. I don't know any "ready to use" non-abelian groups that have this property (S_3 and D_4 have subgroups that are not normal). So I started thinking that maybe this statement is true and I have to prove it.

2. I tried to prove a contrapositive: start with a non-abelian group that has a pair of elements such that ab != ba and somehow show that I can get a subgroup that is not normal, but I couldn't easily do that:

3. I tried to focus on commutators of non-abelian groups (since they are in some sense "a measure" of how non-abelian the group is). I thought that if I start with them, maybe they will automatically lead me to some non-normal subgroup. But it looks it is the opposite, in a way, i.e. I can prove that if a subgroup contains all commutators, then it is normal. For example {e, r^2} in D_4 is normal, when r^2 is a commutator. When it doesn't contain all, it's not clear, at least I can't say that it is automatically not normal.

4. Maybe if I can start with just one commutator in a group that has many, and build a cyclic group out of it then it may potentially end up with only one commutator there, so it is not automatically normal. But I couldn't find a way to prove that it will always lead to a non-normal subgroup.

5. Overall, the statement that for all subgroups H we have Ha=aH (normality) is a weaker statement then claiming that ha=ah for all possible a, h from G (abelian-ness). So I started to actually doubt the truthfulness of our statement. But on the other hand, maybe some extra power comes from the fact that it holds for all subgroups H... But I don't see how to reason over all subgroups H, i.e. how to combine their premises.

do you have any hints? 🙂

[quiet pelican]

There is such a group

There is also such a group with small order (ie, less than ||16||)

[chilly ocean]

Nvm I think i was wrong

🌾

[quiet pelican]

(Yeah I tried the same thing as you, ||reduced to the case of p groups|| then ||realised the|| answer: ||quaternion group|| is an example)

[marble hinge]

ok, thanks 🙂 I'll think more about that then

[chilly ocean]

I should really learn more group theory, it’s pretty interesting

[quiet pelican]

Group theory is fun
(I am legally required to indoctrinate people into groups)

[chilly ocean]

I’m really interested in groupoids so I’m sure learning more about groups first will undoubtedly help me learn more about groupoids too.

[sacred wharf]

Only if its geometric group theory

[chilly ocean]

Is geometric group theory related to interpretating group presentations as CW complexes or is there some other notion of geometric at play 🤔

[quiet pelican]

Other group theory also allowed (but the more combinatorial it is the more I’ll tell you you’re doing the wrong group theory)

[quiet pelican]

Somewhat
Although it’s more generally considering actions of groups on spaces (usually by isometries, sometimes by homeos)

That’s the answer I was guiding them to without outright saying it

[karmic moat]

Is algebraic groups a gray area then

[quiet pelican]

Alg groups isn’t really GGT
It’s more group flavoured AG

[karmic moat]

Aggt

Gagt

[chilly ocean]

Oh that makes sense, neat!

[quiet pelican]

I might end up forced to do a GRT-ish master’s thesis equivalent 😢

[candid patrol]

Sorry, I thought you were implying that the smallest counterexample was of order 16, so I deleted my message.

[quiet pelican]

Yeah I just chose 16 because the number of groups you’d have to consider jumps drastically then

Like after you rule out abelian and dihedral, there’s not that many groups of order less than 16 left
||there’s one of order 8, and some of order 12?||

[balmy python]

how do i prove soemthing is an action, do i prove that theres a homomorphism from the group to the symmetric group acting on the space? that seems quite long

[quiet pelican]

That’s one way
The other way is the whole g(hx) = (gh)x and 1x = x

[balmy python]

how do i even know what symmetric group my action corresponds to though

[candid patrol]

It’s a bit confusing, because the second counterexample actually has order 16 haha, sorry.

[balmy python]

how are they equivalent

[rocky cloak]

Some hints could be

1. ||Notice that if the sylow subgroups are normal, G is the product of its sylow subgroups, so it's enough to think about p-groups||
2. ||If the subgroup generated by a and b are both normal, then aba^-1b^-1 must be contained in their intersection.||
3. ||If you're just looking for a single example you can of course just consider the subgroup generated by two elements a and b that don't commute. What could their orders be, what could aba^-1b^-1 be? Here you can try a few things and see if something works.||

[quiet pelican]

f_g: X -> X via f_g(x) = gx gives an element of the symmetric group on X
Then prove g \mapsto f_g is a homomorphism

[balmy python]

is g a group element?

[quiet pelican]

Yes

[balmy python]

and does gx mean g acting on x

[quiet pelican]

Yes

[alpine island]

Use the class equation to show that a group of order pq, with p and q prime, contains an element of order p.
I know how to solve this generally (it's just cauchy's theorem), but how is the class equation relevant here?

[quiet pelican]

So the size of possible centralisers is 1, p, q, pq by Lagrange
So the size of possible ccls is the same list
So we get pq = apq + bp + cq + d with d > 0 (as the identity has ccl of order 1)
So we deduce a = 0 by size considerations
And we know that p divides cq + d
So either c = 0, d = p or pq (so the centre has size p or pq, so we get an order p element in the centre)
Or c > 0, in which case there is an element g with centraliser of size p
But <g> commutes with g, and has size ord(g) \in p, q, pq, so the order of g must be p

[alpine island]

Huh. yeah, that works.

[rocky cloak]

|G| = |Z(G)| + Sum |G|/|C(x)|

if G is divisible by p it means either ZG is aswell or one of the summands is not.

If ZG is, then you're dealing with an abelian group. Otherwise C(x) is a smaller group, so you can apply induction.

[quiet pelican]

And yeah I feel like the point of this is kinda just to show that you can prove Cauchy manually for simple cases

[marble hinge]

ah, holy cow, it's a || quaternion group || !

I recalled that in Pinter there were 3 sets of exercises: Survey of 6 element groups, survey of 8 element groups, and 10-element groups. The only non-abelians there were symmetrical groups, dyhedral groups and quaternion (which I forgot)

So I've built the table for quaternion, it's 8 elements, so non-trivial subgroups are either of order 2 or order 4. Order 4 is automatically normal (because it has index 2 and I already proved that it is normal). So the only one to check was the subgroup of order 2

[rocky cloak]

In some sense Q8 is basically the only such group.

[marble hinge]

(e, a^2), and checking that it turned out to be normal too, yahoo

OK, now let me read all the hints 😄

[rocky cloak]

They are exactly of the form
Q8xAxH where A is elementary 2-group and H is abelian torsion group where all elements have odd order.

[marble hinge]

Re 1: I haven't got to Sylow yet, this is just after introduction of normal subgroups and quotient groups in Herstein. Re 2 and 3 : that would be interesting to try!

[marble hinge]

mm, this doesn't make much sense to me yet 🙂

I don't know what is a torsion group

[rocky cloak]

Group where the elements have finite order

[marble hinge]

It would be interesting to learn what was Herstein's intended approach here

[rocky cloak]

I mean, they might not have a singular intended approach

[marble hinge]

I think technically, if I didn't request for hints -- I could have proceeded to groups of larger orders and tried to classify them somehow, like Pinter actually asks in his exercises

I already proved that dyhedral and symmetric didn't work for any n

[rocky cloak]

I guess you probably would have seen Q8 before at that point (?), so just trying groups you know would work

[marble hinge]

but working with those larger groups is kinda terrifying: it requires a lot of manual calculations, at least until one starts to "feel" those groups and internalise their properies on higher level

[marble hinge]

no, that's the thing!

I think he doesn't introduce quaternion groups in his book

[rocky cloak]

I see, then it's a bit of a challenge yeah

[marble hinge]

and they also are barely mentioned in Pinter (only in that exercise with the survey of all 8-element groups, that I didn't do, because Pinter has so many exercises!)

I've seen quaternion group in D&F, but they just listed the relations and that's it, it didn't spark my imagination so I promptly forgot about that 😄

[marble hinge]

let me check actually, maybe I missed it. Also, it's possible that it was mentioned in some exercise

no, I am right, in the index it comes 26 pages later 🙂

it will be introduced in an exercise a couple of chapters ahead

[quiet pelican]

Yeah knowing every group of order at most 11 (read: cyclic groups, dihedral groups, the Quaternion group, and their direct products) is generally useful for this sort of counterexample

[marble hinge]

yeah, so Pinter saved me with his surveys 🙂

[rocky cloak]

Who needs A4, amirite

[marble hinge]

I should probably look more at Carter's group atlas

or rather Group Explorer

[quiet pelican]

I can never remember the complete set of groups of order 12, but yeah remember symmetric/alternating too

[marble hinge]

just installed a GAP package for visualising groups using his template

Normal = yes in all rows 🙂

[rocky cloak]

It's just A4, dihedral, dicyclic and the abelian ones.

So I guess it's not a big ask to remember everything < 16

[south patrol]

I just remember all finite groups tbh

[rocky cloak]

Does it product html?

[marble hinge]

yep

https://nathancarter.github.io/gap-pkg-groupexplorer/

and there are usage examples here: https://github.com/nathancarter/gap-pkg-groupexplorer

[rocky cloak]

Cool, maybe I should get this in case I ever want to look to something not in groupprops

[marble hinge]

and the manual too. It generates HTML and opens it in browser

even with some cool 3D Cayley diagrams that one can rotate 🙂

[elfin wraith]

Oh thats cool

I’ve wanted to look into GAP since I read the counter example to the unit conjecture, it seems really powerful for group stuff

[tidal schooner]

it really is

[tough raven]

Wow

Too prescient? 👀

[tough raven]

How about this:

• For a cyclic group of order p (or really any power of p), the minimal polynomial divides x^p - 1 = (x-1)^p (or x^{p^n} - 1 = (x-1)^{p^n}) whose only root is 1. Therefore the element is unipotent and conjugates into U_n.
• In general, argue that there is a proper normal subgroup N with cyclic quotient (haven't yet thought of a more elementary proof than showing centre is non-trivial by class equation hence by induction all p-groups are nilpotent hence solvable hence take a suitable subgroup containing [G, G]), say generated by g in G. Find a basis where N maps into U_n. Now define V_i = { v : (x-1)^i v = 0 for all x in N }. Then g preserves V_i because it normalises N. By the previous part, we can find a basis of V_i/V_{i-1} so that g is unipotent upper-triangular; pull back those into a basis of V.

[tough raven]

Step 2 is wrong: the intersection of a Sylow subgroup with a non-normal subgroup need not be Sylow.

[vocal pebble]

how do you prove this if G is finite? id assume that first you write G as a product of its sylow subgroups, then somehow we split the 2-sylow subgroup into Q_8xA where A is of that form, and prove that the other sylow subgroups are all abelian? although i dont see exactly how to proceed with this.

ok so these are called dedekind groups but could not find a proof classification of them online. wikipedia has the classification theorem in the article but there is no reference to it after they stated it. the article as a whole has a bunch of references but honestly i cba to look through them

[sly crescent]

But then there are 14 nonisomorphic groups of order 16

[vocal pebble]

most of them will be the abelian ones which are not hard to remember by the structure theorem.

[sly crescent]

No, only 5 are Abelian

[vocal pebble]

oh order 16 my bad i read order 12, but yeah i guess for numbers below that its not hard because order 12 is easy like he said, 13 is cyclic, 14,15 are semiprimes so they are classified as semidirect products iirc. either way he said <16

[sly crescent]

Hence the “but then”

[vocal pebble]

anyways i found a proof in The Theory of Groups by Hall page 190 if anyone else also wants to look into it

[cyan skiff]

I was wondering about a question and would appreciate help, the question is prove that for a belonging to a ring R and integer n, n * (-a) = - (n*a). To prove this we have to assume that the ring has a unity but the question doesnt mention that, can it be proved for general rings as well somehow?

[vocal pebble]

Hmm, idk why you need unity, because n*(-a) + n*a = -a+-a+...+-a + a + a +...+a =0, so adding -(n*a) to both sides gives n*(-a)=-(n*a)

[crystal vale]

I think, just show n*(-a) is the additive inverse of n*a

[vocal pebble]

First equality by definitions and second one is by pairing the a's with -a's

[cyan skiff]

Thanks, I was using - (1)a = - a and distribution over individual a's

[vocal pebble]

Isn't that the same thing?

[vocal pebble]

This is completely valid as well. The 1 here isn't referring to the unity of the ring R. It's referring to the 1 in Z(integers)

And the n in n*a is also referring to the n in Z. n*a just means a+a+... n times

[cyan skiff]

Could be interpreted as such but the book I am reading doesn't take it as such, plus for us to use distribution we have to consider - 1 as an element of the ring rather than an integer. That was the difficulty I was struggling with, but honestly your sol is much simpler and easier to understand

[pine flint]

Does anyone have any sources they like/find are useful to grasp polynomial rings? (any good videos or textbooks they'd recommend) I'm in first year uni and doing a semester project on post-quantum cryptography so I want to understand lattice-based cryptography more in depth. I understand the basic of polynomial rings but I feel like I'm missing an intuitive understanding of moving from rings to polynomial rings (I haven't technically taken Linear Algebra yet, this is more self taught right now)

[lusty marlin]

"Abstract Algebra" by Dummit and Foote is a good resource for this.

[lusty marlin]

Seemed to do the job for me, that particular section at least

[elfin wraith]

It’s very much the opposite, some 1000 pages of yap

[lusty marlin]

Maybe they were talking about the section. But yes, Dummit and Foote is the opposite of terse

[gusty thistle]

What are you looking to learn about polynomial rings ?

I also recommend dummit & foote. I had a hard time with it when I first read it but it gets a lot better if you sit down and think about everything carefully and do the exercises ofc

If you are doing more computational things maybe ‘ideals, varieties and algorithms’ by cox little and shea is interesting to you

[pine flint]

I'm very much a noob in this area. I just don't fully comprehend what they are in comparison to regular rings. I'm looking specifically at module learning with errors (with regards to CRYSTALS-Kybe) and wanting to understand that. So it seemed like the natural jump was from understanding basic LWE to polynomial rings so I can understand how module LWE applies polynomial rings to basic LWE

[gusty thistle]

I don’t know what those things are so I cannot help you

[thorn jay]

polynomial rings are just rings too

theyre a construction much like direct products are a construction

[balmy python]

Does St(g(x)) = gSt(x)g^-1 imply that the union of all stabilisers of all elements within the same orbit is normal?

[delicate orchid]

it'll be conjugacy closed yeah, a normal subset if you will

[balmy python]

also can someone show me a small example of how i can form normal subgroups from conjugacy classes?

[balmy python]

ah

[delicate orchid]

canonical example is that $x$ is central if and only if $|x^G| = 1$, hence $\mathcal{Z}(G) = \bigsqcup_{x \in \mathcal{Z}(G)} x^G = \bigsqcup_{x \in \mathcal{Z}(G)}} {x}$

[cloud walrus]

Wew Lads Tbh
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[delicate orchid]

says compile error yet displays exactly what I wanted

is this what you're after or do you mean more generally

[balmy python]

what does x^G mean?

[delicate orchid]

the conjugacy class of x

I'll give another example anyway: the conjugacy classes of S_3 are {1}, {(123), (132)}, {(12), (13), (23)}. The union {1} U {(123), (132)} is closed under multiplication and is thus a normal subgroup

[balmy python]

ah yeah

and how do you get that conjugacy class of the 3 cycles

do you just get any element and apply conjugation to it

oh is it also that

conjugation of any permutation must produce the same cycle shape

[delicate orchid]

for S_n this is easy. Two elements are conjugate if and only if they're of the same cycle type

yeah

[balmy python]

ahhh

and normal subgroups are always unions of conjugacy classes?

[delicate orchid]

yes and it's easy to prove

[balmy python]

hmmm

let me think

so take a normal subgroup

then for all x E N, and for all g E G, gxg^-1 E N

so taking an element, keep applying conjugation to it, this will always be in N

so all possible conjugates of an element are within the subgroup

and then take elements outside these already found conjugates and repeat

but the thing is, how do I know that they are disjoint

[delicate orchid]

they're orbits under a group action they're always disjoint

[balmy python]

G x G -> G?

[delicate orchid]

in fact this proof directly generalises to show that any subset of a G-set that is closed under the action must be a union of orbits

[balmy python]

such that g(x) = gxg^-1?

[delicate orchid]

yeah

[balmy python]

hmmm so what would a formal proof be of this

[delicate orchid]

you just wrote one

[balmy python]

oh yeah 😭

that is so cool though

[balmy python]

G-set?

[delicate orchid]

a set with a G-action on it

[balmy python]

ohhh

[delicate orchid]

like R-module but less woke

[balmy python]

i don't know about modules yet 😭

starting rings in like 2 weeks time

i think modules is third year if i choose the course

[thorn jay]

im sleep deprived 💔

[delicate orchid]

they're closely related to group actions in ways people don't like telling you

[delicate orchid]

if X is a set then Aut(X) naturally has a group structure, so groups should act on sets. But if X is an abelian group then Aut(X) naturally has a ring structure, so rings should act on abelian groups. We call an abelian group with such an action a module

[delicate orchid]

WAKE UP

[thorn jay]

a presentation is used for DESCENDING A RESULT ON FREE OBJECTS DOWN TO ALL OBJECTS not something to be PLAYED WITH

[white oxide]

Can somebody please help me see why 5.19 implies this highlighted sentence

[quiet pelican]

1-v0 is not in the maximal ideal, so not in any maximal ideal

[white oxide]

Bro got it in 2 seconds

[quiet pelican]

And every non-unit is contained in a proper ideal (the one it generates) and hence in a maximal ideal

[quiet pelican]

(A) not bro
(B) it’s a standard argument with local rings

[white oxide]

Ah sorry

How do we know 1 - v_0 is not in th emaximal ideal tho

Nvm

Oh wait this claim is unrelated to equation (3)

[quiet pelican]

If it were, and x = 1 - v0, then x + v0 = 1 \in m

[white oxide]

Yeah

ty

[balmy python]

Why is the action of G x G by conjugation never transitive?

wait i get why

{e} is always its own conjugacy class

hence it's own orbit

[quiet pelican]

Of G x G on what?

[balmy python]

G

i think 😭

[quiet pelican]

Which group is acting on what and what exactly is the action, bc I don’t see it there

[balmy python]

G acting on G

g(x) = gxg^-1

[quiet pelican]

Yeah for this reason

[balmy python]

ah okay

[quiet pelican]

(There are actually (infinite) groups with only 2 conjugacy classes)

[balmy python]

really?

like what 😭

[quiet pelican]

The construction I know is kinda hard to explain
Essentially you start with G_0 = Z
Then for n > 0, fix g \in G_{n-1} set G_n to be G_{n-1}, with |G_{n-1}|-many elements attached, where the element corresponding to h, called t_h, is defined so t_h ht_h^{-1} = g (with the exception that we ignore t_g, t_e)
Then take the union of all the G_i

[white oxide]

Why does k[x] not satisfy d.c.c on ideals

Where k is a field

Oh wait

Just (x), (x^2), (x^3)... right

[noble nexus]

yeah

[tribal moss]

In order to be a chain it would need to be something like (x), (x^2), (x^4), (x^8), ...

No sorry, I'm speaking nonsense.

[quiet pelican]

No?
The ideal generated by x^2 contains xx^2 = x^3

[tribal moss]

Yeah.

[glad osprey]

You mean End(X) for X an abelian group, right? I don't think Aut(X) has a natural ring structure

[white oxide]

Shouldn't this be N?

[quiet pelican]

Yeah

[white oxide]

Why can we make this assumption? If m[x] = B[x], then the proposition gives m[x^{-1}] \neq B[x^{-1}], but how does this show that x or x^{-1} is in B?

[tough raven]

The proof after that sentence will show that m[x] ≠ B[x] ⇒ x in B or x^{-1} in B; by repeating it for x^{-1} in place of x we will be able to show that m[x^{-1}] ≠ B[x^{-1}] ⇒ x^{-1} in B or x in B. But by (5.20) either m[x] ≠ B[x] or m[x^{-1}] ≠ B[x^{-1}], and in either case we have that x in B or x^{-1} in B, so we are done.

[white oxide]

Ah okay, thanks!

[marble hinge]

Another challenging problem from Herstein.

I proved that this subgroup H can't be abelian, normal in G, or finite, because then under given condition we'll have aHa^{-1} = H.

This leaves me with non-abelian infinite groups, and I don't know many examples of those.

Basically, we need to find an element of H that can't be represented as aha^{-1} for given a and any h from H.

Or in other words, we need f(h) = aha^{-1} to be a bijective mapping that maps H to its proper subset (which is, in a way, a definition of infinite set H, if such a mapping exist).

I tried a subgroup of mappings of infinite set A to itself that moves only finite number of elements which is a non-abelian subgroup of all mapings from A to A, but then I can't see a way to construct this mapping to a subset of H...

Also tried some other non-abelian infinite groups (like group of functions f(x) = ax + b under composition of mappings), but that seems to be fine, can't find this peculiar case there.

Any hints please? 🙂

[rocky cloak]

It's perfectly possible for H to be abelian. But of course G can't be abelian, since then aHa^-1 = H

[marble hinge]

ah, right

ok, then I'll update: G can't be abelian, H can't be normal in G (because then for all a: aH=Ha), H can't be finite (because if H is finite, aHa^{-1} has at least ord(H) elements, and all of them are in H by assumption, so H doesn't have any more "space" for those elements that we are looking for, since its order is ord(H) too)

but H can be abelian, hmmm, not sure if it helps me much, because I still need to start from G, and this leaves me only with that example of mappings that move only finite number of elements 🙂

[candid patrol]

What kind of non-abelian infinite group do you know ?

[marble hinge]

some subgroups of permutations of infinite sets A, like permutation that move only finite number of elements. Or say permutations that do not move some element a_0

[knotty frigate]

whats ord[H}?

[marble hinge]

ord(H) = number of elements in H

order of H

[candid patrol]

H has to be infinite

[marble hinge]

Yes, that's what I stated above

[knotty frigate]

why does finiteness of H mean that aHa^{-1} has ord[H] elements?
unless A is infinite wont the permutation group also be finite?

[marble hinge]

because we can build a bijection from aHa^{-1} and H, right? So when H is finite, they have the same number of elements, so number of elements of aHa^{-1} is also ord(H)

[knotty frigate]

ok maybe ill think about it later

but the second question is more important rn

[quiet pelican]

For the latter, the set of permutations of an infinite set that fix all but finitely many elements of that set is infinite

[marble hinge]

f(h) = ah^{-1}
f is a bijection because:
injection: f(h') = f(h'') => h' = h'' (by cancellation laws)
surjection: for any h', we can use a^{-1}h'a as an argument to f and have f(a^{-1}h'a) = a(a^{-1})h'a(a^{-1}) = h'

injection + surjection = bijection

well, the surjection case is actually wrong because a^{-1}h'a is not necessarily in aHa^{-1}, but actually just injection is enough there, because this means that aHa^{-1} has at least ord(H) elements. So H must be infinite

[marble hinge]

I am not sure I understand the second question 🙂

I am now actually confused who is trying to give me hints and who is just thinking about his problem...

[knotty frigate]

Permutation groups of a set A (where |A|=n) have size n! no?

[marble hinge]

yes

[knotty frigate]

so youd be dealing with an infinite set A.

[marble hinge]

what are we trying to do now?

to show that H must be infinite?

[knotty frigate]

wouldnt dealing with a set of matrices directly be more natural to work as a choice for H?

or for G even

[marble hinge]

maybe, but I don't know much linear algebra to work with that

also, Herstein marks problems that require knowledge of LA with "#"

and this problem is not marked

Ah, I think maybe there was a confusion when I talked about "permutation of set A" -- it looks like people automatically assumed that I am talking about finite A, I was not

I just meant mappings from A to itself, perhaps "permutations of A" is a wrong choice of word for that

so is the consensus that I need some LA-based examples of groups to figure this out?

(maybe then it's worth quickly reading Artin's first chapter, he should cover that 🙂

but still, this is surprising to me, if true

[knotty frigate]

i wasnt im just worried this mgiht be something thats more difficult

[knotty frigate]

does multiplying 2x2 matrices count as LA knowledge?

if you know that you may already know enough (idk maybe youre bijections of a set A approach is right or maybe you need even more LA knowledge and thats a mistake in the textbook)

[marble hinge]

nah, I know how to multiply matrices 🙂

[knotty frigate]

alright then id always try to work within 2x2 matrices. do you know what kinds of transformations 2x2 matrices can 'look' like/represent?

[marble hinge]

well, this definitely counts as LA in the book and is marked as such 🙂

you see, all the ones that deal with matrices are marked with "#"

in previous chapter

so I suspect that there is a solution to my problem without matrices

does anyone know any?

I can see that previous he introduces three types of examples: any 2x2 matrices (with non-zero determinant), matrices with det=1, and matrices of type ( (a b) (-b a) )

I suspect that the first two correspond to some kind of GL and SL, and the third one I don't know

[quiet pelican]

GL_2, SL_2 and the complex numbers

[marble hinge]

but all those examples are marked with "#"

[marble hinge]

right, maybe I should just experiment with those matrices and see what I can get

[quiet pelican]

Consider the group of permutations of the rationals, the subgroup ||that fixes everything except the integers, ||and the ||permutation given by doubling||

or just do BS(1, 2) /hj
More seriously free groups generally aren’t introduced until much later

[marble hinge]

yes, free groups are not introduced yet

this is a chapter about normal and quotient groups

[vocal pebble]

ok nvm this isnt correct

[sacred wharf]

Any module with 1 is cyclic right

[vocal pebble]

wdym by 1?

[sacred wharf]

Say the module is a ring with 1

[quiet pelican]

Not necessarily, consider k[x] as a k module

[vocal pebble]

well if you take R as an R module and R has unit then yeah

but thats not saying anything

[sacred wharf]

But we define a cyclic module as { rm | r in R} no ?

[vocal pebble]

so the cyclic module generated by 1 here will be k =/= k[x]

[quiet pelican]

I think if you try to formalise what you mean by “ring with 1” you’ll probably end up with something basically trivially equivalent to being cyclic, or that fails to k[x] (possibly quotiented out by x^n)

[sacred wharf]

Hmmm

Ok but lets say i have cosets G/A as a G module

Then g/a is cyclic if g contains one right ?

Since surely

1+A is our cyclic generator

[thorn jay]

a module does not contain a "1" element

[thorn jay]

G/A are G-sets, not G-modules

[vocal pebble]

if you mean to take the quotient ring R/A of a ring R with unity as an R module then it is cyclic (as you said with generator 1 + A), and all cyclic R modules look like this

[tough raven]

Hint: ||first find such an example with a general automorphism rather than a conjugation, i.e., find a group G, a subgroup H and an automorphism f of G such that f(H) ⊊ H. Then use a semidirect product to embed G in G' such that f comes from a conjugation in G'.||

[sacred wharf]

Sure since its a group

[rocky cloak]

It is actually a good idea to start from H.

[tough raven]

There's another fun example given by a standard example of ||an embedding of the free group on countably many generators in the free group on 2 generators||, but I don't think that solution is very motivated.

[thorn jay]

a module contains a 0 element

the additive identity

the identity of the group structure is the 0 element

[sacred wharf]

Sure but we in a sense construct a module from a ring it can contain the element that woukd act like 1 if the module was a ring

[thorn jay]

well I guess that's just the definition of a cyclic module

and then it's isomorphic to R/I

[rocky cloak]

I think the most accessible example is probably ||affine functions i.e. f(x) = ax + b with a and b rational, and a=/=0||

[sacred wharf]

There is a theorem that states

Is this what people mean by fundamental theorem of pid?

Or is that smth else

[south patrol]

Ye or like fundamental theorem of finitely generated modules over a PID

[sacred wharf]

I see and it includes uniqueness right

Which ig is not mentioned above

[south patrol]

Yes, though this is sort of easy

[sacred wharf]

Easy for oxford phd

[south patrol]

No I mean this uniqueness is a like paragraph at the end of the proof

The existence is the hard part

[sacred wharf]

I see my lecture notes spends some time developing notation ad terminology for the uniqueness bit

Hence why its not in the original statement

[rocky cloak]

I usually call it "the structure theorem", or "the structure theorem of finitely generated modules over a PID"

[sacred wharf]

Man i am dying in my algebra class with these 30 step proofs

[rocky cloak]

The proof can be a bit involved. Are you doing it through Smith normal form?

[sacred wharf]

Yea

I still didnt get smith normal form proof😢

[sacred wharf]

At least proof seems more doable than smith normal form ig

It uses some linear algebra ideas. I dont have good grasp of

[rocky cloak]

Well, once you have smf the rest of the proof is basically done, no?

[sacred wharf]

They use stuff with linear basis idk

My lin alg is too trash for this

[thorn jay]

bro is doing module theory with trash linear algebra

[sacred wharf]

Eh tbh this is the first time i got stunped due to lin alg

In my 12th lecture

So it seems alr i just need to get through it

And after this and 3 more lecs no algebra for a long (potential infinite time?)

[thorn jay]

what a sad existence that must be...

[vocal pebble]

depends on the person

[sacred wharf]

Nah topology is where math gets fun

[thorn jay]

topology is fun ngl

[sacred wharf]

More fun than algebra

[rocky cloak]

Wait till you hear about algebraic topology

[sacred wharf]

I was planning on taking it next year but got convinced not to

[thorn jay]

convinced by what

general algebra experience?

[sacred wharf]

Forgot who it was

But they told me its very visual and ill get cooked by hatcher

[rocky cloak]

I think that depends a lot, but I guess hatcher likes the "just imagine the homotopy" type thing

[sacred wharf]

Yeah i am screwed if i do algebraic topology

[rocky cloak]

I'm not sure that's a good reason not to take it, but maybe you have other fun stuff you want to do

[sacred wharf]

i can't decide between wanting to branch out and take introductory courses in other fields or go hard and specialize

i think latter could help more with career

former might be more fun

[alpine plank]

The advice i've recieved is during undergrad try to get breadth in a lot of fields and depth in (atleast) one specific one

[sacred wharf]

by end of this year i would have 2 courses in algebra, 2 courses in mesure, 1 course in top

i have one more year

hmm

if i really like mt i think i'll go down mt probability route

[south patrol]

if someone says topology is fun i assume they mean algebraic topology

[prime sundial]

in a direct product of groups, every group is normal?

[rapid cave]

They are the kernel of projections

So yes

[candid patrol]

What are the conditions on U and V for G to be isomorphic to U × V ?

[marble hinge]

Got this warning on Amazon for one of Abstract algebra textbooks! 📚

It must be very energising or something ⚡️

[winter niche]

Tell me if is this right: i have the short exact sequence 0→Zn→Zn²→Zn→0 and i have to find all n such that this splits. I know about a theorem that such graphs split (for abelian groups like i have) iff there is isomorphism between Zn² and Zn x Zn such that the graph commutes. I know that Zn x Zm is isomorphic to Znm iff (n,m)=1 and therefore there isnt any n.

[wraith cargo]

this never splits

if I'm interpreting everything right

[south patrol]

n=1

[tribal moss]

n=0, though?

[south patrol]

n=0 this is 0 -> Z -> Z -> Z -> 0 which does not split

[winter niche]

I have that n≥1

[tribal moss]

Ah, I somehow thought I saw a bunch of ^ that weren't there. Ignore the n=0 quip.

[winter niche]

So, the explanation yet is ok?

[prime sundial]

but we can prove it directly?

[prime sundial]

U inter V ={e}, G=UV and U and V commute

[wraith cargo]

https://tenor.com/view/dog-cooking-perro-dog-dangerrfox-gif-23889683

[south patrol]

Though tbf it would be interesting if you asked about 0 -> Z/nZ -> Z/nZ x Z/nZ -> Z/nZ -> 0 but where the maps aren't the standard ones

But then splitting lemma

[wraith cargo]

abelian groups are too easy 😔

[wraith cargo]

but yeah this explanation makes sense

[winter niche]

This is the only case

[south patrol]

Exactly

[wraith cargo]

just ig you have to account for the trivial n=1 case

[rocky cloak]

Mfw Z/n is selfinjective

[winter niche]

Nice, thanks 🙏🏿

[south patrol]

Gotem

[marble hinge]

Hmm, I haven’t got to semiproducts yet. But I can try to use the first part of your hint

[rapid cave]

Yes?

[marble hinge]

Well, I tried to look at those basic examples of 2x2 matrices, i.e taking GL_2 as a group and SL_2 as a subgroup but then it’s normal

[rapid cave]

If G = U x V then V is the kernel of the homomorphism
U x V -> U

[marble hinge]

So not getting anywhere with matrices, I think I don’t have enough intuition about them to come up with something clever

[marble hinge]

Free groups haven’t been introduced yet

[marble hinge]

Right, I looked at ax+b functions as a group, but didn’t consider rational coefficients! Let me see what changes then…

[marble hinge]

Hm, this is interesting, I’ll try. For now I still don’t fully grasp why rationals should be involved in those examples!

Thank you all for the hints!

[quiet pelican]

Rationals is just a nice way to describe…

[quiet pelican]

This sort of construction

[quiet pelican]

Of which this example is a special case (the BS(1, 2) one)

[marble hinge]

What is BS(1, 2)?

Haven’t seen this before

[quiet pelican]

If you know what this notation means: <a, b | bab^-1 = a^2>
If you don’t, then it’s pretty hard to explain

[marble hinge]

Yeah, generators and relations

[quiet pelican]

Then that’s your G, H is <a>, and the element you’re conjugating by is b

It should be intuitively obvious why that example works (it’s basically by construction that b<a>b^-1 = <a^2> which is a strict subset of <a>)

[marble hinge]

Cool. So BS(1, 2) is a general notation to describe that group with generators and relations?

2 Because of a^2?

[prime sundial]

okay tnx

[marble hinge]

Right, found that!

[quiet pelican]

Yeah

[quiet pelican]

They’re very useful for counterexamples to all your dreams

[marble hinge]

I wonder if someone should write a book “Counterexamples in group theory”

Like similar books for topology and real analysis

[delicate orchid]

oh these are cool. They're like C_n \rtimes C_m but woke

[quiet pelican]

They’re the HNN extensions of Z over itself

[delicate orchid]

HNN?

[quiet pelican]

Higman-Neumann-Neumann
Everyone calls them HNN extensions

[delicate orchid]

I mean I can see how they're kind of Z \rtimes Z with different actions

[quiet pelican]

They’re extensions which are essentially “add an extra element making two specified isomorphic subgroups with a specified isomorphism conjugate”

[delicate orchid]

oh now you're speakin my language

FUSING subgroups you say?

[knotty badger]

Is this the right channel to ask about topological groups

[rapid cave]

Probably

[knotty badger]

So I’ve been trying to figure out what special properties topological groups have compared to arbitrary topological spaces

It seems a big one is that you have a family of automorphisms for free, given by left or right multiplication by a group element

[noble nexus]

connected components are particularly nice for topological groups I believe

since the connected component of the identity is a subgroup, and then all of the other components will just be cosets

so in particular I guess you could say all the connected components are homeomorphic

[knotty badger]

Yeah it seems that there’s a lot of homogeneity present

[noble nexus]

and yeah the automorphism groups acts transitively

[knotty badger]

Lemme think about why the connected component of the identity is a subgroup

[noble nexus]

its a very nice argument

[knotty badger]

Hm so first

Inversion is a continuous map

So it sends connected spaces to connected spaces

So the connected component of the identity is closed under inverses

[thorn jay]

the fundamental grouo is also abelian

[noble nexus]

another fun argument is showing that every discrete normal subgroup of a connected topological group is in the center

[knotty badger]

Ok maybe you can do

The map (a, b) -> a b^-1 is continuous

If C is the connected component of the identity, then C x C is connected

So the image under that map is also connected

[noble nexus]

yeah

[knotty badger]

And contains the identity

Hence is a subset of C

Is that enough to show C is a subgroup?

[noble nexus]

yeah since its closed under multiplication and inversion

[knotty badger]

Mhm

Cool

[tough raven]

Any topological group is uniformisable and therefore completely regular (a point can be separated from a closed set not containing it by a continuous function to the real numbers).

[noble nexus]

oh yeah the uniform structure is important

[knotty badger]

The uniform structure?

[noble nexus]

it allows you define things like uniform continuity

[tough raven]

I have a hunch this may be as good as it gets, i.e., any (T0) completely regular topological space is a topological group, but I don't know for sure at all.

[noble nexus]

uniform structure essentially gives you a way of comparing open sets at different points roughly

and a top group you can do this by translation

[knotty badger]

Ooh

[noble nexus]

tbh the definition of a uniform space is totally unhinged and I still don't really know what it is

but the way it manifests in topological groups/vector spaces is quite intuitive I find

[tough raven]

Oh yes.

[tough raven]

This is probably absurd; ignore it.

[noble nexus]

yeah u need a lot more qualities

[thorn jay]

assuming choice any nonempty discrete space is a topological group

😎

[rocky cloak]

The empty space would like a word

Reminds me of a fun exercise I made for a group theory course:

Define a group as a set with an associative binary operation and inverse such that
b^-1 b a = a = a b b^-1

Part 1: show that this is equivalent to the usual definition.
Part 2: show that you actually made a mistake in part1, there is a "group" satisfying this definition and not the usual one.

[thorn jay]

reminds me of heaps

[knotty badger]

I want to check if I’ve proven this property of topological groups correctly

Let $G$ be a topological group, $N$ a normal subgroup

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

Consider the quotient group $G / N$ given the quotient topology

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

Then the quotient map $q : G \to G / N$ is open

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

I think this works because, if $U$ is an open subset of $G$, $q^{-1}(q(U)) = UN = \bigcup_{n \in N} Un$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[noble nexus]

yep

[knotty badger]

Then since right multiplication by any group element is a homeomorphism, Un is open for every n

[noble nexus]

Also holds if the subgroup isn't normal

[knotty badger]

Oh right, yeah

[noble nexus]

of course then G/H is just a topological space

But still comes up a lot in the theory

[knotty badger]

Hence q^-1(q(U)) is an open subset of G

So q(U) is open, as reauited

[noble nexus]

Yep exactly

[knotty badger]

Yeah I’m currently trying to show that G/N is a topological group

And want to figure out an efficient way to do so

[south patrol]

Hm maybe it'd be good to show that product of open maps is open

[knotty badger]

see #point-set-topology lol

[knotty badger]

just posted a proof attemtp in #1388211506314870824 if you want to check it out

[south patrol]

Ye I was memeing cause I said yes there before coming here and understanding aha

[noble belfry]

If R is a commutative ring and $f \in R[x]$ i want to show that if f is a zero divisor, there's some a in R s.t. af = 0

[cloud walrus]

sudo

[noble belfry]

my current thoughts are that:

if g(x)*f(x) = 0, we know the leading coefficient of g, c gives us
f'(x) = cf(x) where deg f' < deg f

and f' is a zero divisor as well or its zero (in which case we're done)

so c^n f(x) = 0 for some c

but i'm not sure how we can avoid the problem of c possibly being nilpotent

anyone have any hints?

[marble hinge]

Yay, verified that this one indeed is a good example for my problem. aHa^-1 is a subset of H (because it still moves only integers), but H still has some elements that are not representable in aHa^{-1} form, for example a permutation (12) can’t be represented, because any aHa^-1 form would send 1 to 0.5 and then it’s fixed and then multiplied by 2, so we always have 1 -> 1

[quiet pelican]

Yeah
(Fwiw this was an exercise in my first year groups course that I didn’t get, and that either is, or is very similar to, the example I was given)

[marble hinge]

I see. For me this exercise is quite tough, probably the most difficult from all the exercises from Herstein that I tackled so far (and it’s like 30 or so…). Not sure why it doesn’t even have a star…

[noble belfry]

oh nvm i figured it out

[marble hinge]

Now I need to connect the dots to that BS(1, 2) stuff. I can’t see yet how I can generate all permutations of rational numbers from just two generators

[quiet pelican]

You can’t
It’s a subgroup of the permutations of the rationals

[marble hinge]

Right, but what part BS(1, 2) plays in this construction that you suggested? It’s not G, and it doesn’t look like H either…

[quiet pelican]

It doesn’t
You can get BS(1, 2) as a “minimal subgroup” of (a slight edit of) my example

Where by “minimal” I mean something more like “minimal subject to satisfying the condition”

[marble hinge]

I see, got it. So basically instead of using “mappings that move only integers” I could use this BS(1, 2) for some generators satisfying that relation and still get everything working for my example, right?

[quiet pelican]

Yeah

[marble hinge]

Cool, thank you!

[arctic lance]

Can someone help me solve 7. And 8. Exercise ?

[noble nexus]

for 8, it's mostly routine except for division

For division, you want to think about how you can simplify a fraction with radicals

[arctic lance]

Thank you , but I don’t know how to solve it by myself

[noble nexus]

7 it's a bit unclear what is meant by analyze to be honest

[arctic lance]

Just do binary operation on it

Like prove it’s closed, associative , inverse and identity

Element

[noble nexus]

as in are you trying to show it is or isn't a group? Ah ok

[arctic lance]

Just to prove whether or not it’s a group

[noble nexus]

For most of those you just need to write out the definitions and calculate

Though I suspect it's not actually a group

actually no I think it should be fine

but associativity takes a bit of thinking at one step

[rocky cloak]

So I'm not entirely sure what they're looking for in 7.

But it is possible to write this group as a (semidirect) product of subgroups, and those subgroups will be isomorphic to something familiar.

[arctic lance]

Yeah, associativity …

[rocky cloak]

A good place to start is probably to describe the cyclic submodules

[arctic lance]

What about 8.?

[vapid vale]

have you tried just doing the calculations to check the axioms?

[rocky cloak]

The fact that it's a subset of R means you don't need to check things like associativity and distributivity. Just that it's closed under multiplication, addition, subtraction and division

For division you might start assuming
1/(a + bsq5) = c + dsq5 and see if there are any c and d that make that true

[arctic lance]

Thank you

[rapid cave]

||Z |x C2 ? right||

[rocky cloak]

Yes, aka ||infinite dihedral group||

[rapid cave]

fancy name

like it

[rocky cloak]

Another fancy related name is apeirogon, a polygon with infinitely many sides.

||The infinite dihedral group is the symmetry group of the regular apeirogon||

[tribal moss]

A thing that can actually exist -- and be regular -- in the hyperbolic plane.

[balmy python]

groups exam next week and i'm so stressed

i can't do the most simple stuff 😭

like with this, i know that if H is a subgroup then |H| = p

so H must be cyclic

but then how do i even count for the direct product

i've been looking at smaller examples

[rocky cloak]

I mean, it exists in the euclidean plane as well, it's just that all the angles are 180 degrees

[balmy python]

and know that if I have a generator (a,b) then it generates the same group as (ck,dk)

hmmm

[noble nexus]

I feel like its easier to just say its the symmetry group of the integers :P

[tribal moss]

But then the symmetry group would be larger.

[balmy python]

was that to me?

nvm

😭

[noble nexus]

nah I mean about infinite sided polygons

[rocky cloak]

No you still have to take vertecies to vertecies.

(so it's just isometries of Z)

[noble nexus]

so yeah you know that every nontrivial element generates a group of order p, so all you have left to do is think about the possible overlap between subgroups

[balmy python]

which is what makes me wanna cry

[noble nexus]

so what happens if you have two subgroups generated by a single element that have nontrivial overlap?

[balmy python]

the subgroups are equal

[noble nexus]

yeah so if you've shown that, then you're basically done

[tribal moss]

Hmm, that feels like a bit of an extraneous requirement compared to just "isometries that preserve this subset of the plane".

[noble nexus]

you just need to do some counting to finish the argument

[balmy python]

which is the hard part

[noble nexus]

its not too bad really, each subgroup gives you p-1 non-identity elements

[balmy python]

yeah

[noble nexus]

and then the entire group has p^2-1 non-identity elements

[balmy python]

yeah

[noble nexus]

so can you see how to finish easily from that

[balmy python]

no 😭

ohhh

wait

[rocky cloak]

It is the same requirement, but "the subset" is the set of vertecies

[balmy python]

because all cyclic subgroups are disjoint, then we have p+1 cyclic subgroups?

[noble nexus]

yep

[balmy python]

bruh

[noble nexus]

well disjoint except the identity to be precise

[balmy python]

lemme marinate that in my head

[balmy python]

yeah

[tribal moss]

But usually the subset is the polygon as drawn in the plane.

[rocky cloak]

Not really. Usually one talks about the symmetries of a polygon by how it acts on the vertices

Like both work, but if you think about the whole polygon then things both get somewhat complicated and infinite, and also doesn't generalize well

Besides if you just think about it as a line, then it's not really much of an apeirogon anymore anyway

You need the information that distinguishes vertecies from edges

[balmy python]

By lagrange's theorem we have that for a subgroup H, |H| | p^2 which implies that |H| must be p in order to be non trivial.

So in a group of prime order, no two different cyclic subgroups can have overlap because say for sake of contradiction if they did (call the generators p and q), then we have p^a = q^b for some a,b. Since our group is of prime order, then q^b=p^a must generate the same subgroup as p and q , which results in <p> = <q>

Disregarding the identity element, these subgroups are disjoint and their union must be the group. We have p-1 elements disregarding the identity in each subgroup and we have p^2 - 1 total elements in our group. So, the number of possible subgroups are (p^-1)/p-1 = p+1.

sounds good @noble nexus ?

[tribal moss]

Hence why I feel it works better in the hyperbolic plane.

[rocky cloak]

It also works in the hyperbolic plane sure

[tribal moss]

(And that also neatly disposes of the question whether it should count as a separate symmetry to mirror-flip the entire plane across the line, given that it's the identity when restricted to the figure itself -- no matter whether we're considering the line or just the vertices).

[vapid vale]

this is not a useful comment for your homework/your exam prep but perhaps an interesting perspective.

Z/p is also a field, so you can have vector spaces over it. so Z/p x Z/p is a two dimensional vector space, and a proper nontrivial subgroup here has to be a one-dimensional subspace, which is just a line passing through the origin. but such a line is uniquely determined by its slope.

then there are p possible choices of slope (if it helps, compare it to R^2, where just picking a real number m uniquely gives you the line y = mx), and an additional one for the undefined (directly vertical) slope, which gives p+1 lines.

[balmy python]

oh yeah

that is a really cool insight

thank you very much i appreciate it

[tribal moss]

(In particular, the dihedral group of a 2-gon ought to have 4 elements, so its elements cannot be distinguished by how they permute the vertices).

[balmy python]

how do you spot that a proper nontrivial subgroup here has to be a one-dimensional subspace

[noble nexus]

yep

subgroups and subspaces are the same thing over finite fields

[balmy python]

the next question looks like hell 😭 i only learnt about orbits in lectures like a week ago and it's coming up in an exam

gonna try do it rn

[noble nexus]

because multiplication by an element in the finite field is the same thing as repeated addition

[balmy python]

yep

[balmy python]

hmmm so far I got that the klein-4 group is one of the subgroups in a unique orbit

since conjugations don't change cycle shape, I would think that gHg^-1 = K for all g in S4

[vapid vale]

what do you mean by the klein 4 group

[balmy python]

{e, (12)(34), (13)(24), (14)(23)}

aha

transpositions have order 2

so

we also know that H must have 4 elements

so it's the groups that are generated by two different transpositions

i.e. {e, (12), (34), (12)(34)}, {e, (13), (24), (13)(24)}, {e, (14), (23), (14)(23)}, {e, (12)(34), (13)(24), (14)(23)}

that is 4 subgroups

and i am tyring to find the S4 orbits of these

hope you guys don't mind me talking to myself

would love some advice on other ways of thinking if there are any though

ah so i think one orbit would be {e, (12), (34), (12)(34)}, {e, (13), (24), (13)(24)}, {e, (14), (23), (14)(23)

the other orbit would be {e, (12)(34), (13)(24), (14)(23)}

[balmy python]

check the answer and i'm right but how would i know if there are any more sets H 😭

[quiet pelican]

If it contains (12), then every other element is a transposition commuting with (12) (and similar for (13), (14)), so the only other possible elements in the group are (34), (12)(34), and those form a group
If it contains no single transposition, it must consist of e and the three double transpositions, and that works

[balmy python]

ahhh okayyy

i see

[glad osprey]

Hmm, maybe I'm being slow, but doesn't this argument work for Z/p^2 too (for which the conclusion is not true)? Where are you using the fact that it's Z/p x Z/p?

[quiet pelican]

“And we have p^2 - 1 elements in our group” the assumption that all choices work is where they differ

[rocky cloak]

So Z/p x Z/p isn't cyclic, or that each element has order p is what is being used

[glad osprey]

Oh yeah, the union of all p-subgroups in Z/p x Z/p is the whole group, which is not true in Z/p^2, right?

[quiet pelican]

All the size-p subgroups

(I would typically interpret p-subgroups as subgroups which are also p-groups)

[prisma ibex]

tfw p-subgroup versus sub-p-group

[tough raven]

The composition factors of a subgroup of a finite group need not be a sub(multi)set of the composition factors of the group, right?

But this does need to hold for normal subgroups and quotient groups?

OK, obviously. IDK why I asked. Mb

[fast kelp]

Basic question, if I have a group (Z, +), is <1> a generator for the entire group, or do I need to include -1 as well?

[prisma ibex]

1 alone is a generator

[tribal moss]

1 is a generator for the entire group.

[prisma ibex]

-1 alone is also a generator

[tribal moss]

That is, the smallest subgroup that contains 1 is the entire group.

[prisma ibex]

you can include both but this is redundant

[fast kelp]

Thanks

wasn't sure because it's an infinite group.

[proud vigil]

yeah thats fair, in general <a> refers to all powers of a, positive or negative

[prisma ibex]

nothing about the group being infinite is relevant here

[proud vigil]

(or 0, getting the identity)

[fast kelp]

if you look at a generator as forwardly applying this argument to the identity, then you'd generate 0, 1, 2, 3, ... but never generate the negatives. This was my concern. If you can also use the inverse element of 1, that changes things and allows access to the negatives.

[prisma ibex]

yes but you're talking about groups

the free group on one generator is Z, the free monoid on one generator is N

[fast kelp]

yes, I understand that one has access to the inverse element now, but I didn't a moment ago.

[proud vigil]

i guess to be more precise, <a> is meant to be the smallest (sub)group containing a, which implies existence of inverses

any such group can always be thought of as the codomain of a surjective homomorphism from the free group on a one-element set {a}, which is a bunch of words to refer to the group of all positive and negative "powers" of a. the free group is like the "least informative group possible" that encodes a group structure for one element

whatever you get after applying the homomorphism will still ensure that those inverse powers exist in some form, just maybe equivalent to other existing elements of the codomain

[tough raven]

bump (I got curious about this again)

More precisely, examples of such finite groups with arbitrarily large numbers of composition factors.

[crystal vale]

Is that result when R has no unity?

[fading acorn]

no because that's an exercise in atiyah's commutative algebra book (which assumes all rings are unital commutative)

[twilit wraith]

I thought prime is equivalent to maximal in commutative rings

[quiet pelican]

(0) in Z

[fading acorn]

one hint is that: ||show that the nilradical equals the jacobson radical (because otherwise there is a prime ideal which is not maximal)|| if i remember correctly

[twilit wraith]

Yeah my bad I sorely misremembered a theorem in d&f

Maximals are prime no matter the ring, but not vice versa

[thorn jay]

youre thinking of dimension 0 rings

[twilit wraith]

Yeah my bad

Im still new to rings

[south patrol]

Dw it's fine

One thing that is true is like many rings that appear are "dimension 1 domains", which equivalently means the zero ideal is the only non-maximal prime

[thorn jay]

nuh uh

take a small enough nonprime quotient of a dim 2 integral domain

[south patrol]

Fixed lol good catch

[thorn jay]

(for example: Z)

[crystal vale]

I proved for commutative ring with unity but now I am wondering for ring without unity

[rapid cave]

All rings are with unity

Rings without unity are Rngs

[rocky cloak]

Let p be a prime ideal and x an element that is not contained in p.

Then for all y
x * (x^n-1 y - y) = 0 is in p, means that x^n-1 y - y is in p. Hence x^n-1 is a multiplicative identity in A/p.

From there you can reason the same as the unital case

[crystal vale]

Ah great, thank you

[mental lake]

Let V4 = {1, a, b, ab}, G a group, and
B = {(g, h) ∈ G × G | g^2 = 1 = h^2, hg = gh}.
Show that the function
ψ : Hom(V4, G) → B : f → (f(a), f(b)) is a bijection (first show that the function is well-defined).

[quiet pelican]

What are you stuck at?

[mental lake]

like the meaning of Hom(V4, G) → B

[dull marsh]

ψ takes a homomorphism V4 -> G and gives you an element of B, namely it returns the pair where the components are the images of a, b in V4

[woeful sage]

Non-commutative algebra when?

[thorn jay]

thats literally 5he rest of algebra

[winter niche]

Any hint on this exercise: let 0→A→B→C→0 be a short exact sequence of free finitely generated free abelian groups. Show that rank(A)+rank(C)=rank(B)

[thorn jay]

use the fact that the sequence must split, as C is free (and hence projective)

[coral spindle]

Alternatively, show that the rank of a free Abelian group G is the same as the Q-dimension of Q (x) G, whence this is just rank-nullity

[winter niche]

Why is this true

[thorn jay]

if you have a surjection r : B → C where C is free generated by x1,...,xn then you can choose b1, ..., bn such that r(bi) = xi and this will be a section

[tardy hedge]

A section is just a left or right inverse of a map f?

[thorn jay]

specificslly a right inverse

a retraction is a left inverse

[vapid vale]

these techniques feel a little heavy handed for the level of course i'd imagine this question to be asked for, you can just take a maximal generating set for A and C, and choose their images/preimages in B, and grind out independence and maximality

[thorn jay]

right

amounts to the same thing lol

[thorn jay]

minimal generating set btw

maximal is just all of A and C

[vapid vale]

oh derp yeah

[next obsidian]

Whence

[coral spindle]

What, without asking, hither hurried whence?
And, without asking, whither hurried hence!
Another and another Cup to drown
The Memory of this Impertinence!

[tardy hedge]

Mathematicians be talkin like its 1492

[winter niche]

Yes that was the thought. But then is it necessary any info for B like the free finitely generated?

[velvet hull]

hey can someone do a quick sanity check for me, R is a flat Z module right

[next obsidian]

Yes

Z is a PID so flat is equivalent to torsion free

Or you can see that Z -> R factors through Q, and Z -> Q is flat cuz localization and Q -> R is flat cuz a field

[velvet hull]

Tyty

[low quiver]

I didn't understand the multiplication of (12)(1)=2

And (13)(1)=3

[rapid cave]

By definition (1 2) sends 1 to 2

This is what the notation means, its the transposition between 1 and 2 in S3

[twilit wraith]

Let I be the ideal in $\mathbb{Z}_2[x]$ generated by $x^4 + x^3 + x^2 + x + 1$. Prove that $\mathbb{Z}_2[x]/I$ is a field with 16 elements

[cloud walrus]

hiidostuff

[twilit wraith]

so the fact that I is maximal follows from the irreducibility of the polynomial its generated by

but im confused how this field doesnt have 32 elements

[south patrol]

Lol I was confused what you meant by R initially

[twilit wraith]

im looking ahead at my algebra final and the question im asking is stuff we havent covered yet but its fairly intuitive i feel

[twilit wraith]

i was thinking of creating a ring homomorphism from Z2[x] into Z2^5

that just spits out the 5-tuple of elements in Z2 that are the opposite of the coefficients of the 0-4 degree terms

so for example x^4 + x^2 + 1 would map to (0,1,0,1,0)

but that homomorphism would be surjective

[rapid cave]

It doesn't have kernel I

[twilit wraith]

how so

oh wait

x^5 + x^4 + ... + 1 is in the kernel

[rapid cave]

You want to divide by x^4+...+1

And use the reminder, which is a deg 3 polynomial

[twilit wraith]

wdym

divide what by x^4 + ... + 1

oh wait i think i see what you mean

youre saying that any polynomial mod x^4 + ... + 1 is a degree 3 polynomial

[rapid cave]

Yeah

And there are 16 such deg 3 polynomials

[twilit wraith]

yeah

alright fair enough

[tulip otter]

so there is more than one dual of A?

because one can consider Hom(A,Z_m) for any exponent m of A and there infinitely many such m

[quiet pelican]

Exercise: they’re all canonically isomorphic

[tulip otter]

ohhhh

ok then let me try to prove that

[thorn jay]

(hint: prove the isomorphism for an arbitrary exponent m and the smallest exponent m0. The isomorphism between two arbitrary exponents will then follow)

[tulip otter]

i have no idea on how to proceed

[quiet pelican]

So taking the smallest exponent m0, we want an isomorphism
Hom(A, Z/mZ) -> Hom(A, Z/m0 Z)
So take f \in Hom(A, Z/mZ)
What is the image of f?
More specifically, ||if x is in the image of f, what is m0 x?||

[thorn jay]

to Hom(A, Z/m0Z)

[quiet pelican]

I can’t type 🙃

[tulip otter]

i havent read the spoiled part, but perhaps the answer you are looking for is that im f is a cyclic subgroup of order >= |A| or something like that?

[thorn jay]

cant be of order larger than |A|

besides A doesnt need to be finite

[quiet pelican]

Not quite
I’m looking for “contained in a cyclic subgroup” but that’s not the order

[tulip otter]

ohhh ok

[tulip otter]

well f(m_0x)=f(0)=0

[quiet pelican]

Yup

[tulip otter]

also since m_0 is the smallest exponent of A then m_0 is in fact the order of A right?

[quiet pelican]

False

[tulip otter]

ohh lol

[thorn jay]

Z/2Z x Z/2Z is of exponent 2

[tulip otter]

ah right

yea it was stupid of me to say that lmao

[tulip otter]

ok so for any k in Z/mZ, write k=qm_0+r where 0=<r<m_0 and q>=0. Then for any f in Hom(A,Z/mZ), f(kx)=f(rx). This means that all values of f are determined by its restriction on Z/m_0 Z

but ah i dont think that really explains the fact Hom(A, Z/m_0 Z) and Hom(A,Z/mZ) are isomorphic, does it? I am probably still missing details (assuming that what i said is even true)

[thorn jay]

well we know that m0 divides m, so there is a unique subgroup of Z/mZ isomorphic to Z/m0Z. Then the only thing you have to show is that every homomorphism from A to Z/mZ must have its image in that subgroup

[tulip otter]

given f in Hom(A, Z/mZ) and x in A, m_0f(x)=f(m_0x)=0 so |f(x)|=<m_0 and f(x) in Z/m_0 Z?

Hence im f is a subset of Z/m_0 Z

is that right?

[thorn jay]

if by |f(x)| you mean the order of f(x) then yes

[tulip otter]

yes thats what i meant by it

[thorn jay]

because the unique copy of Z/m0Z in Z/mZ is { x ∈ Z/mZ | m0x = 0 }

yes then youre correct!

[tulip otter]

took me too long

bro it was there all the time

tysm .enpeace and micose, have a great day/night

[toxic rampart]

Hello. My algebraic structures professor has defined the biyection group as the group whose elements are biyections of a set X. He has defined the composition operation like this, flipping the order:

$f \cdot g = g \circ f$

which is very confusing for me since we've learned about more complex structures that flip the order of the functions. Is this normal? Would you suggest I get used to it or to write it the other way around (since it would also work)

[cloud walrus]

dp

[thorn jay]

the convention usually is that, for functions f and g we denote f ∘ g as the function sending x to f(g(x))

[toxic rampart]

yeah, he follows that convention

that's the confusing part for me, $(f\cdot g)(x) \neq (f \circ g)(x)$

[cloud walrus]

dp

[thorn jay]

yeah because he is probably old fashioned

its an annoying convention youll probably have to get used to

[toxic rampart]

Is this convention out of fashion?

[thorn jay]

thats what i just said

the usual convention is to just use ∘ as the group operation

[toxic rampart]

ok, thanks :)

I'll get used to it and pray the next professor doesn't change it back

[karmic moat]

Wow now I feel guilty

[alpine plank]

Don't worry too much about it

[karmic moat]

I always right fg for f circ g in groups

[toxic rampart]

It just bothers me so much since last year we did it the other way around, but I'm sure I'll get used to it :)

[thorn jay]

well you begin to treat them as abstract groups and then it really doesnt matter

[prime sundial]

What is the reason of fig 5? Is it a true visualization of the homomorphism? i dont understand rly

[untold hearth]

Look into « universal covering spaces » and related concepts

[quiet pelican]

If you imagine the circle in the x-y plane, and you spiral R above it in the z direction, then yeah the homomorphism is given by projection onto the first two coords

[untold hearth]

The cool thing about the diagram is that, by choosing a small neighbourhood around a point of the circle, you can look at it’s preimage by phi. The result is visibly a countably infinite discrete copy of « a single » preimage

As in, the preimage of the neighbourhood is « all of the points that are directly above it », and what you see is a countably infinite amount of « copies » of your original neighbourhood

Here

So \mathbb{R} is shaped like a helix in the diagram to make it intuitively clear that it satisfied the necessary conditions for a covering space (i.e. the whole, preimage of neighbourhood is a discrete set of copies of the neighbourhood, up to homeomorphism)

This seemingly random covering space property turns out to be extremely powerful

[prime sundial]

tnx

[prime sundial]

wow very intresting thank u

[errant maple]

this is so epic

[tulip otter]

quick question: for each n, the natural homomorphism
f_n: G/H_n->G/H_{n-1} is the one that sends each coset xH_n to the coset xH_{n-1} for all x in G?

[quiet pelican]

Yeah

[tulip otter]

i see tysm

[tough raven]

Honestly IDK about this. I have been reading a book that was inconsistent about these things (including whether Graph(f) = {(x, f(x))} or {(f(x), x)} for f a certain automorphism) and it slowed me down a lot sometimes.

[tardy hedge]

For primary decompositions, if p contains I then p is associated/belongs with I right? Its the radical of a qi in its decomposition

Im not sure if its that or like only if p is a minimal prime containing I, then its associated to I and a minimal one

[tough raven]

Not every prime ideal containing I is an associated prime of I. For example, if I is prime, its only associated prime should be I, but I need not be maximal.

I think Vakil has a good explanation for how to think about associated primes pictorially. At least, it should make it straightforward to guess what the associated primes of an ideal (at least of a polynomial ring) are, however much work you have to do to prove that afterwards.

[white oxide]

Can somebody hoelp me see why this is true

[tough raven]

A general element in k' = B'/m' = B[x]/m' is the image of p(x) for some polynomial p with coefficients in B; check that the image of p(x) in k' is \overline{p}(x), where \overline{p} takes the coefficients of p modulo m.

[white oxide]

They're not equal as sets though right

They're isomorphic

[tough raven]

Well, the inclusion of B in B' induces a homomorphism from k = B/m to k' = B'/m' (which is an embedding since k is a field and k' is non-trivial). This is technically not "true as sets": k is not equal to a subring of k', but isomorphic to a subring of k'. But apart from this identification, it is an equality of sets: k' = k[\overline{x}] where the latter denotes the subring of k' generated by the image of k and \overline{x}. Note that you have to use the identification to make sense of the RHS in the first place.

[white oxide]

Are they just saying that the embedding from k -> k' induces an isomorphism from k[\overline{x}] -> k'

I think I need to read the proof again lol

Oh waittt

We embed k into k' (say i(k)) then we simply take i(k)[x]

[white oxide]

Are fields Artinian and Noetherian since they have no proper ideals

[vapid vale]

yes

[tough raven]

If you are interpreting k[\overline{x}] as the polynomial ring over k, then no. If it's the subring of k' generated by the image of k and \overline{x}, then yes (and this a set-theoretic equality).

[marble hinge]

funnily enough, I've looked today at J. S. Milne's lecture notes on Group Theory and he introduces this example - that I was looking for as part of Herstein's exercise - right after the definition of normal groups!

[iron arrow]

What are good resources to introduce oneself on p-adic definitions and properties, and also the general concept of the I-adic rings?

[tough raven]

I liked Gouvea's textbook for a first reading.

[marble hinge]

@knotty frigate -- was this the example that you had in mind, when you mentioned 2x2 matrices?

[knotty frigate]

i didnt have one specific one in mind i was just trying to think of one

[iron arrow]

Thanks!

And do you know any resources on specifically I-adic rings or is this book a good introduction to that too?

[tough raven]

The first 3 chapters of Local Fields by Serre (the rest go into class field theory) are also good, if more advanced and dense / technical to read.

[tough raven]

By l-adic ring do you just mean finite extensions of ℚ_l, or something else?

[iron arrow]

No, like, the generalization of p-adic rings are I-adic rings where I is an ideal right? With all that thing about the completion and relation to topology

[vapid vale]

atiyah-macdonald has a chapter on completion

[iron arrow]

Thanks

[tough raven]

Oh

I would suggest Serre

Unless your interest is more for algebraic geometry in which case maybe a commutative algebra textbook like Atiyah-MacDonald?

Like if you just want it to understand some dimension theory thing in algebraic geometry then any full book is probably overkill and part of a CA textbook would be better

Kind of basic question but: let G be a group and H a subgroup of G and K a normal subgroup of H. How are the G-sets G/K and G/H related? Obviously there's a surjective map from G/K to G/H with fibre H/K (but this is only canonical above the basepoint * = 1H). Does the normality of K give us anything more?

IG the stabilisers of the points of the fibre above any x are all the same, i.e., only depend on x, whereas in general they would be a conjugacy class of subgroups of Stab(x).

[mint seal]

interesting question, I don't know of more that can be said

is there a typo here in the hint of ex. 1.15? I think since |h| < |g|, it should be n < m, not m < n

unless maybe we can let m = 0?

like what if |h| = 20, |g| = 21 (hypothetically, as this is in fact not possible)

are we writing |g| = 2^0 * 3 * 7 and |h| = 2^2 * 5, so m = 0, n = 2, and m < n ?

seems weird

actually, I guess one of m or n must be allowed to be 0, otherwise if |h| and |g| have no primes in common then there's no way to make sense of this

still feels weird that |h| is using the larger exponent of p

[tulip otter]

well if there is no such p, then any divisor of |h| is a divisor of |g|, right?

wait no i think this is not it too. I cant think straight today

[mint seal]

I'm not quite sure where those equations are coming from

all we're assuming is |h| < |g| and |h| does not divide |g|

and it sort of looks like he's using the fundamental theorem of arithmetic

but it feels strange that |h| is the smaller one, yet it's got the bigger power of p

[tulip otter]

ah wait i have misread something thats why i was in doubt

[tulip otter]

this should be the reason

[tulip otter]

well dont forget that there are also r and s

[mint seal]

I guess if none of the primes of |h| are in |g|, then you pick any one of them and use m = 0

[tulip otter]

so if r is sufficiently smaller than s then you can probably have the power of p in |h| greater than that in |g|

[mint seal]

yeah, I can see if we have like |h| = 2^3 * 5, and |g| = 2^2 * 17, then we can take m = 2, n = 3, p = 2

I was just getting confused about if that sort of comparison isn't possible

but m = 0 is the way out

[tulip otter]

Do you mean like if p divides one of |h| but doesnt divide |g|?

if yes then what you said is exactly right!

[mint seal]

sweet, thankee

[kind temple]

suppose |h| < |g| and |h| does not divide |g|.

• if |h| has a prime factor p not appearing in the prime factorization of |g|, then take m = 0 and n to be maximal such that p^n divides |h|
• if |h| has all the same prime factors as |g| but |h| does not divide |g|, what would happen if all the prime powers of |h| had exponents at most those of |g|?

oh looks like you figured it out

[chilly ocean]

Given a group G acting on a set X, a G-invariant subset U of X is a union of disjoint orbits, just partition U with the relation x~y iff x and y have the same orbit, then use the axiom of choice to pick exactly one element from each equivalence class. These will have disjoint orbits and the union of their orbits will be U.

Is there any proof of this fact without the axiom of choice?

[tough raven]

What is your definition of orbit?

[south patrol]

To sorta strengthen what Raghuram said, what is the fact you want to prove?

[chilly ocean]

The orbit of x is the set {gx|g in G}

[south patrol]

I basically cannot see a way of interpreting that that is not tautological

Maybe some helpful points are 1) Xis irrelevant, wlog U = X and 2) an orbit is then by definition an equivalence class under this relation

[tough raven]

So you want to prove that there exists a subset S of U such that {orbit(x) : x in S} partitions U?

[tough raven]

Not by their definition, which is why I asked.

[chilly ocean]

Well specifically I want to prove there is a set of points of U, each of whose orbits are disjoint from the others, without choice. Sorry if I phrased it poorly

[south patrol]

Well like x ~ y iff they are in the same orbit by definition of ~

[tough raven]

They mean S -> orbits: x -> orbit(x) is a bijection

[chilly ocean]

Reading back I definitely phrased it poorly

[tough raven]

Or if you like a transversal for the orbit space

[south patrol]

Oh okay

Then this is equivalent to choice I guess

[chilly ocean]

yeah sorry

[tough raven]

I think in general this is equivalent to the Axiom of Choice. I haven't proved this fully (i.e. that this would imply the full Axiom of Choice), but a sample weaker version: suppose S is a set and for every x in S, A_x is a set of size 2. Then let the cyclic group of order 2 act on the union of A_x by non-trivially permuting each A_x. Your choice of elements would be equivalent to choosing an element from each A_x simultaneously.

[south patrol]

Choice is equivalent to being able to pick a set of representatives for any equivalence relation. Given an equivalence relation on a set X, consider the action which permutes elements within each equivalence class.

[tough raven]

Oh just product of symmetric group on each equivalence class?

😂

[south patrol]

Yeah

Lol

[tough raven]

Got to think bigger

[south patrol]

But okay sorry I misunderstood the question

[chilly ocean]

Ic, thank y’all!

[south patrol]

Though I guess maybe there is the amusing thing of like

[tough raven]

identity

[south patrol]

Yeah that was stuoid

[tough raven]

Don't infect group theory with your non-canonical-ness /s

[south patrol]

We can write down all the elements