groups-rings-fields

[lusty marlin]

And yes, it is 140.

[tardy hedge]

Not sure what the OEIS is or what landaus function is

But thanks!

[chilly ocean]

google is your friend

[tardy hedge]

No way

Ppl loveeee to say that

Its like ppl come out of the shadows just to say it

[lusty marlin]

Look them up then

[delicate orchid]

yeah it's 140

[lusty marlin]

As they should.

[tardy hedge]

Will do

True

It is a certified classic

[lusty marlin]

Could have set the upper limit on j as 15-i

[delicate orchid]

landau's function is LAME we only know the asymtoptic behaviour how boring

[delicate orchid]

oohh look at me I can run in (n-1)n/2 time instead of n^2 guess what nerd they're still O(n^2) :blazed2the9s:

[lusty marlin]

Yeah my remark was more about calming my nerves than pointing out a flaw.

[delicate orchid]

dw you are absolutely right

it would remove the "and k > 0" bit too

lets see what the speed up actually is

[lusty marlin]

Are there any interesting open problems involving it?

[delicate orchid]

dunno

[lusty marlin]

(I didn't bother to check OEIS, so much for practising what I preach)

[delicate orchid]

probably a few

[abstract rock]

such a thing as asymptotically defined groups?

[delicate orchid]

you can take Landau's for any integer

but yeah you can take the limit of groups, or rings, or modules or... etc.
they're not analytic though

[lusty marlin]

Yes, it is a function on ℕ. You don't necessarily need to think in the context of groups while talking about it.

[lusty marlin]

What do you mean by 'take the limit of...' in this context?

[delicate orchid]

limit/colimit

and I suppose to keep the analysis analogy alive, w.r.t a poset category

[abstract rock]

i meant groups defined by a set of asymptotic relations but alright

[rotund aurora]

the Riemann hypothesis

see the wikipedia article

https://hal.science/hal-02177338/

[crystal turtle]

Wtf

Wacky

[rotund aurora]

superchampion numbers

[tardy hedge]

For sigma a cycle and sigma^n a cycle, is supp(sigma^n) always the same as supp(sigma)

[delicate orchid]

supp(-) is the set of points that aren't fixed right

[tardy hedge]

Ya

[delicate orchid]

then yes this should be true

sigma and sigma^n would be cycles of the same length, and thus conjugate in the symmetric group, so they fix the same number of elements, so |supp(sigma)| = |supp(sigma^n)|
and we have supp(sigma^n) \subseteq supp(sigma)

[south patrol]

Why should they be cycles of the same length?

Well we would need extra conditions i think

[delicate orchid]

you can't multiply a cycle by itself to get a cycle of a different length

[south patrol]

sigma^n not being the 1 cycle for starters

[delicate orchid]

you can get a product of cycles

[south patrol]

but maybe that is the only exception aha

[delicate orchid]

the identity isn't a 1-cycle it's a product of 1-cycles

[south patrol]

hm

okay sure

[delicate orchid]

I did think about sigma^n = Id

[south patrol]

i didn't realise that was the convention on cycles mb

[delicate orchid]

and that's the conclusion I came to

yeah it's weird

[south patrol]

has to be exactly one element with orbit of size > 1

[delicate orchid]

yeah good point, because if we consider 1-cycles as like, valid
then in S_n we can only have n-length cycles

so perhaps this notion isn't the way to go

lets just assume we can replace "\sigma^n also a cycle" with "(n, o(\sigma)) = 1"

[rocky cloak]

I think I'd say the identity is the product of no cycles. And then just have 1-cycles not be a thing

[delicate orchid]

Fairs

[rocky cloak]

Might change that definition depending on context

Anyway if gcd(n, o(sigma)) = d, then sigma^n should be the product of d disjoint o(sigma)/d-cycles right?

[delicate orchid]

yes, exactly why I replaced the condition with what I did

[torn warren]

I have a question on this proof

it only shows $(a+H)(b+H)\subset ab+H$, but how to show $ab+H\subset (a+H)(b+H)$?
I try to write $ab+h=(a+0)(b+a^{-1}h)$, but there is no guarantee for the multiplicative inverse exists

[cloud walrus]

WT

[night onyx]

It doesn't really make sense to ask if ab + H is contained in (a + H)(b + H), because the point of the theorem is to show the definition of (a + H)(b + H) as ab + H is even well defined. Like (a + H)(b + H) isn't a "thing" until you define it and show that that definition is well defined, meaning it doesn't depend on the representative of a + H or b + H

defining (a + H)(b + H) as ab + H could be meaningless, but the theorem shows that if H satisfies the properties of the theorem (which is called being an ideal btw) then that definition actually works

[torn warren]

if define $AB={ab| a\in A, b\in B}$, here A=a+H, B=b+H, and H is the ideal, can we show $ab+H\subset (a+H)(b+H)$?

[cloud walrus]

WT

[night onyx]

oh I see what you mean...

lol doing some scribbles it doesn't seem obvious

[rustic crown]

A = B = 2 + 4Z is a simple counterexample

AB contains thinsg that are 4 mod 8

but ab+H would be 4Z

[torn warren]

thank you!

[torn warren]

yes, I try to factorize it, but not work

[torn warren]

so when the proof says "well define", it means: if $a+H=x+H, b+H=y+H$, then we need to show $(a+H)(b+H)=(x+H)(y+H)$ by using the definition $(a+H)(b+H)=ab+H$, do I understand correct?

[cloud walrus]

WT

[night onyx]

yeah exactly

in that case you need to show ab + H and xy + H are the same coset

same thing as with group cosets, except the property of H being a normal subgroup is replaced with H being an ideal

a good example when something is not well defined would be like defining a map f: R / H -> R by f(r + H) = r

[torn warren]

yes, this depends on the rep

[night onyx]

exactly yeah

and theoretically the map f: (R / H) x (R / H) -> R / H defined by f(a + H, b + H) = ab + H also depends on the representatives

unless H is an ideal, in which case it doesn't!

[torn warren]

thank you for clarify this, just now I was confused by the definition, since I thought the set product is same as $AB={ab| a\in A, b\in B}$, but actually it is not (and shouldn't be)

[cloud walrus]

WT

[night onyx]

lol no thats actually a great question, I've never really thought about it!

like if a + H and b + H are cosets is the set product (a + H)(b + H) also a coset, and is it equal to ab + H

lol thats a smart question

[torn warren]

in the group theory, we have $(aH)(bH)=(ab)H$, where H is normal, and this definition is consistent to the natural way $AB={ab| a\in A, b\in B}$

[cloud walrus]

WT

[torn warren]

because the "set product" in group theory is actually the "set addition" in ring theory, right?

so we have to give another definition for the set multiplication rule in the ring theory

[coral shale]

the name for the operation is just a name

although we call it multiplication, its just whatever operation it is

theres 1 operation in a group

theres 2 in a ring

With regards to groups, we often refer to it as addition if its abelian

[coral shale]

So with regards to this, a ring is an abelian group (the addition) with a monoid structure (the multiplication) that is compatible (distributivity)

[coral shale]

we happen to write shorthands like AB = {ab} for convenience in some cases but... this is more notation than anything else.

Like us often writing AB = {ab : a in A, b in B} is unrelated to how we need to define multiplication of cosets <-- this is determined by how the natural group structure on the quotient set needs to be defined.

[torn warren]

it is on purpose to make homomorphism to be a homomorhisim, right?

so they define the product rule in this way?

[coral shale]

so you have your quotient set

whenever we can define a group operation on that which is compatible with the original group, we have a quotient group

S set
~ subset S x S, equivalence relation
[a] := {x in S : a ~ x} equivalence class represented by a
S/~ := {[a] : a in S}, quotient set

Then, we need
* : S x S -> S, group operation on S

# : S/~ x S/~ -> S/~, group operation on S/~ which satisfies:
[a] # [b] = [a * b]

[a] # [b] = [a * b] this key property makes this triangle commute. And indeed makes those maps all homomorphisms

The equivalence class represented by a is the map
[.] : S -> S/~

which is pi in that commutative diagram

[chilly ocean]

Help me solve ii) please

<@&286206848099549185>

[abstract rock]

have you worked on an example

[chilly ocean]

Yes

Can't make a general argument

[abstract rock]

well for one, split up and G and H, since you've shown they're both cyclic by (i) then what do you know about cyclic groups

[chilly ocean]

there's a generator

[abstract rock]

right and what can you say about the subgroups of cyclic groups

[chilly ocean]

they're also cyclic

[abstract rock]

work backwards from there to G times H

[chilly ocean]

uhh

I'm working on this rn

By 6.1 they mean this

@abstract rock im dying help ;-;

😭

[crystal turtle]

I think by Theorem 6.1 they mean Theorem 6.1 in the text, not Problem 6.1

That might help.

[chilly ocean]

I think imma just dip

no point obessing over one one problem

[terse crystal]

You just show that any r|mn, you can find a|m, b|n such that r=ab

This case gcd(m,n)=1, but I think it’s not needed here

Don’t know what 6.1 is in your book. But the only thing needed to prove CRT, probably is that I+J=I+K=J+K=R, then I+JK=R, where I,J,K are ideals of R

[lavish rampart]

Let G be a finite group with identity element $e$. Suppose there exists $x \neq e$ such that $x = x^{-1}$. Show that $|G|$ is even. \

How to prove this without using Lagrange theorem?

[cloud walrus]

wlmmm

[lavish rampart]

With Lagrange theorem, you can consider the subgroup H = {e, x} and it's over. But I'm not allowed to use it for this question

[molten rivet]

on this problem, not sure how to do part b.

[terse crystal]

For other elements other than order two elements, x and x^-1 always appear in pairs

[lavish rampart]

How does it prove that the number of elements y such that y = y^-1 is even?

I see that there are even elements in {z : z \neq z^(-1)}, but I don't see how it helps

[terse crystal]

Oh I am dumb. My bad. Now this:
(left) multiplication by x is a permutation of G of order 2, since this permutation can’t have fix point, so it’s disjoint union of 2-cycles

Take one x such that x=x^-1 I mean

[molten rivet]

if anyone has any hints or just a general approach ping me

[lavish rampart]

Nice, thank you

[chilly ocean]

can't use rings. We only studied direct product so far

[terse crystal]

You are proving chinese remainder theorem and you can’t use the concept of ring. Interesting…

[chilly ocean]

yeh we only know group theory so far

[terse crystal]

Okay then the idea is the same. a1,…, an, any two relatively prime, show ai and product of rest of aj are relatively prime

Like a1 and a2a3…an are relatively prime for example

[chilly ocean]

uhh

[molten rivet]

anyone?

[terse crystal]

I don’t have any smart way. By brutally considering elements of order 4, I found out that mapping (x,1), (1,i), (1,j) to ||(x,1), (1,r), (x,s)|| works

Whose inverse is ||mapping (x,1), (1,r),(1,s) to (x,1), (1,i), (x,-j)||

I listed elements of order 4 of two groups, and they should match. So I tried then it naturally emerged

Presentations of them can make it clear that what I wrote is well defined:
<x,r,s | relators of D8, x^2=r^2, x commutes with r,s>
<x,i,j | relators of Q8, x^2=i^2, x commutes with i,j>

[torn warren]

Are the two definitions different?

should it be subgroup or subring?

[crystal turtle]

They are equivalent. The condition of the second one forces it to be a subring as well.

For it to be a subring means that it's close under multiplication by elements of N. To be an ideal, you must be closed under multiplcation with any other element of R, which is a stronger condition.

[torn warren]

it requires the subgroup is with the same multiplicative operation as in R, right?

[rapid junco]

We have a group of symmetries in n space where n is bigger than 2.

Do reflections still needed to generate the group?

[crystal turtle]

The subgroup part is with respect to the addition operation.

[rapid junco]

Or can every reflection be decomposed into a product of rotations in n > 2 dimensions.

[crystal turtle]

And then this extra condition is with multiplication.

The classic examples of ideals are nZ as an ideal in Z. The fact that this is an ideal means that it is an additive subgroup of Z (simple to check), and that, whenever you take an element of nZ, say nk, and multiply it by another integer, say m, you get an element of nZ again (indeed: you get (nk)*m = n*(km) in nZ)

[torn warren]

this is what i am confused, i know the subgroup part is with respect to addition operation, but to make it become a subring, we need to define the multiplication, but this definition didn't say, so it means to use the same multiplicative operation as in R, right?

[molten rivet]

That helps alot actually, thank you!

[abstract rock]

Try to summarize this in one sentence, with the specific groups involved

[abstract rock]

reviewing most definitions of a subring they should dictate that a subring of a ring is a set that is closed under the additive operation of the ring and the multiplicative operation of the ring

[sonic coral]

So for this, i think that since N is normal in G and is contained in both A and B, then N is normal in both A and B as well. i’m not sure if that helps me but i’m stuck regardless

[rocky cloak]

So the costs of A in B look like bA, what is the condition for two bs to correspond to the same coset.

What is the condition for the cosets of A/N in B/N? Can you see how the cosets naturally correspond to one another?

[terse crystal]

Not sure I understand your question. I don’t know whether you were asking about group of isometrics of R^n. And I think it’s the other way round, like rotation factored into product of reflections. There is a result called Cartan-Dieudonne theorem, which says any orthogonal transformation (isometries this case) is product of <=n many reflections.

[warm ember]

is aut(F_p)=Z_p* in general

or uh

i mean Z_(p-1)*

?

or is it like Z_(phi(p-1))

or actually

there are no nontrivial automorphisms since 0 maps to 0 and 1 maps to 1??

[tender wharf]

Z_p* meaning group of units?

Aut(Z/nZ) is isomorphic to U(n)

[warm ember]

oh

wait whats u

U

[tender wharf]

group of units

like uh

[warm ember]

oh uh

ye

i know

whats U(n)

[tender wharf]

in this case it's just all the integers less than or equal to n, geq 1, coprime to n

$U(n) = \set{m | 1 \leq n \leq m, \gcd(m, n) = 1}$

[warm ember]

im talking about fields tho

[tender wharf]

sure, if n is prime

[warm ember]

ye

thats for groups

[tender wharf]

it holds in general anyway

[warm ember]

but F_p has no nontrivial automorphisms?

[tender wharf]

Z/pZ?

[warm ember]

ye the field

1 maps to 1

[tender wharf]

um

[warm ember]

so 2 maps to 2

[tender wharf]

oh I was thinking about them as groups

[warm ember]

yeah

[tender wharf]

I'm not totally sure if you think of them as rings

or fields whatever

[warm ember]

huh

F_p is a field

[tender wharf]

no I mean

the group of group automorphisms on Z/nZ is isomorphic to U(n)

and if n is a prime

it's still true

[warm ember]

yes

but for fields there are none?

i mean F_p where p is prime

[south patrol]

as fields yes, F_p has no non-trivial automorphisms

[warm ember]

oh ok ty

[south patrol]

Indeed any field automorphism is in particular an additive group automorphism sending 1 -> 1

and that determines where everything is sent

[warm ember]

and F_q is always galois over F_p since x^q=x is separable over F_q?

[rocky cloak]

Yes, and the Galois group is cyclic of order log_p(q)

[warm ember]

oh thank you

[terse crystal]

q=p^r, Aut(Fq)=Gal(Fq/Fp)=<f>, where f(x)=x^p, order(f)=r. So when r=1, Aut(Fp) has only the identity map.

[warm ember]

tysm

by freshman dream right

thats clever

[cobalt heath]

Uh, sounds like #real-complex-analysis to me.

(Tho I don't see what you mean by "ring of sets")

[delicate orchid]

A ring of sets is just a lattice of sets

[cobalt heath]

Ah. Ordered by inclusion, I guess

[feral junco]

Ah sorry, let me ask my questiom in real complex analysis 🙏

[untold cloud]

Hi, guys. I am thinking of if this is true: if A is a commutative ring with 1 such that dimA<infty, then A is Noetherian. I think this is not true, because one is about prime ideals, one is about ideals. Chain of prime ideals is finite may not imply chain of ideals is finite. Let p be a prime ideals, then p might be a smallest prime ideal containing infinitely many ideals.

But i cannot think of any counterexample

[void cosmos]

a ring is noetherian iff all prime ideals are f.g btw

commutative ofc

[untold cloud]

Thank you! But i cannot see how i can use this characterization

Oh

This is a counterexample of dimA=0 but not noetherian

[acoustic cargo]

Valuation rings are noetherian iff they are DVRs. However, there exist valuation rings that are not DVRs.

[acoustic cargo]

or that, yes 🙂

[untold cloud]

Thank you! That is also lots of counterexamples😊

[rocky cloak]

Another example could be an infinite product of fields. Which would be reduced and 0-dimensional, but not Noetherian.

So then you have your pick of
Connected 0-dimensional,
Reduced 0-dimensional, or
Integral domain 1-dimensional

I'm guessing connected+reduced+0-dimensional is impossible...?

[flint crater]

Bit confused. Suppose I have a PID R and an irred. element p. Is R/pR the same as R/(p)?

[tribal moss]

Yes, pR and (p) are both notations for the principal ideal generated by p.

[flint crater]

Awesome thanks. Thought someone (not here) claimed otherwise and got confused haha

[delicate orchid]

hopefully you can see why they both notate the same thing

[flint crater]

I mean. I thought pR was the set {pr | r in R} and (p) is defined exactly the same?

[delicate orchid]

exactly

[flint crater]

Yeah but I started doubting whether I remembered (p) correctly when someone said it wasn't so I just hopped on here to be sure haha

[cobalt heath]

Perhaps for some people, (p) is just a singular tuple containing p

[south patrol]

Not necessarily true when R isn't commutative

Which may be the issue others said idk

[cobalt heath]

Ah, (p) means two-sided ideal right

I forgot that.

[tribal moss]

But it was specified to be a PID.

[south patrol]

Yes, I'm saying it isn't necessarily true more generally and that may be why someone else said that (p) wasn't just pR

[flint crater]

Oh yeah I guess that would be it since I did not mention PID to the other person

[rocky cloak]

Anyone have an argument or counterexample to this?

I.e. connected reduced 0-dimensional commutative ring that is not a field

[acoustic cargo]

Is the product of infinitely many copies of a field k connected?

I just mean countably many copies, ig

[rocky cloak]

No, connected means not the product of other rings

[tiny jolt]

If a,b,c,d are linearly independent vectors in the quaternions (viewed as a vector space of R), are xa, xb, xc, xd linearly independent for nonzero x in the quaternions?

[coral spindle]

Yes

[barren sierra]

not too hard to see for yourself after expressing x in terms of a, b, c, d and multiplying (although tedious)

[tribal moss]

You can also just argue by contrapositive: if pxa+qxb+rxc+sxd=0 is a linear combination, then rewrite to xpa+xqb+xrc+xsd=0 (since the real coefficients commute with everything), and then multiply the whole thing by x^-1 from the left. That gives a linear combination of a, b, c, d.

[south patrol]

Funny answer; If you normalise x so that it is a unit, then Norm(xy) = Norm(y) for each y, so the multiplication by x map is an orthogonal map. Hence multiplying by x not only preserves linearly independent sets but also orthogonal ones lol

[lusty summit]

and
xa=ya⇒x=y

Proof:
ax=ay
⇔a⁻¹(ax)=a⁻1(ay)
⇔(a⁻¹a)x=(a⁻¹a)y
⇔e⋅x=e⋅y
⇔x=y ∎

Similarly for xa=ya⇒x=y ∎

3. In the integers modulo 4 ring Z₄ do all the non-identity elements form a group with regard to multiplication?

Solution:
Z₄={0,1,2,3}

No because {1,2,3} is not closed under multiplication modulo 4
2⋅2=4 mod 4=0∉{1,2,3}```

update: oh, by the ring identity it is meant the multiplicative identity so I should check  {0,2,3} not {1,2,3}
```Answer:

No because {0,2,3} is not closed under multiplication modulo 4:
3⋅3=9 mod 4=1∉ (not belong to) {0, 2, 3}``` <- is my answer correct?

[summer path]

yes

[still perch]

Hi,
I am stuck at understanding this problem. Can someone help me?
I do understand that the question asking for a subgroup of transformations which stabilises the line.
That is the line y=0, means (x,0)
Am I right?

[chilly radish]

Correct, the line is the set {(x,0):x\in \R}

What have you tried so far

[lusty summit]

Thank you

[still perch]

say we have a matrix of the form 2*2,
a b
c d

Now to stabilise the line y = 0, say we have a point (x,y). We consider the transformation, (ax + cy, by +dy). This is what I did, I think now according to the question, we are supposed to have the resultant transformation of the following form i.e (ax, 0).

So ax + c0, b0 + d* 0?

[delicate bloom]

double check your (ax + cy, by +dy)

[still perch]

I didnt get that

[delicate bloom]

redo it

it's wrong

[still perch]

oh, but how?

[delicate bloom]

redo it and find out

[vestal snow]

The matrix that flips row i and row j is in U(n), right?

[terse crystal]

Yeah

[lime badge]

I just had a quick question I wanted to check (moved from advanced algebra to here).

If f: R --> S is a ring homomorphism and M, N are S-modules, I need to show there is a canonical homomorphism of R-modules

M (x)_R N --> M (x)_S N

This is just simply defining a map F(r m (x) n ) = f(r) m (x) n and extending R-linearly right? So basically restricting scalars on the tensor product

[void cosmos]

yea

[spare belfry]

FUCK

[south patrol]

Standard inner product is invariant under permuting entries

Is what it comes down to ig

[delicate orchid]

permutation matrices appear in the image of the standard rep of S_n for some n, and are thus unitary :pack:

[low wyvern]

Guys, I am reading DF and whilst it's a good book, I find it has so much in it compared to, for example, 'Contemporary ab alg'. Take for example the order of an element in a group defined as, the smallest positive integer n, such that $x^n = 1$ . This does not seem to be in other books, what is its importance?

[cloud walrus]

NoName

[open sluice]

order is pretty important, what?

[low wyvern]

I know order of group, but this is defined as order of element...

Every where else it is the number of elements in a group...

[sharp sonnet]

any intro algebra book will define this term

if it doesnt seem to be in other books, you just didnt look hard enough

[low wyvern]

Maybe it's later on, DF seems to dive into things from the start, where as other books come back to it in more detail it seems

[sharp sonnet]

its the order of the subgroup generated by that element as well, so ...

DF is just very wordy, especially in the beginning

well, books generally cover more than lectures

[low wyvern]

My lectures go straight from groups to subgroups, where as here, he does types of group and actions first...

[sharp sonnet]

hm?

chapter 2 in DF is subgroups, group actions are covered in chapter 4

[low wyvern]

sec 1.7

and 1.6 is homo and isomorphisms

My lectures do that much later lol

I guess I will be jumping around a lot

I think from now on, I will just use DF as a reference...

[elfin prairie]

so automorphism is an isomorphism mapping from one group to itself

if the group is say finite

then there are n! bijections possible

so what I am trying to understand is that...

only some of these bijections are iso/automorphic right?

[lusty marlin]

Yes

[boreal cedar]

The order of an element in a group is the smallest number of times you have to use that element to get the group's identity element. It's important because it helps understand the group's structure, and plays a role in things like cryptography. Different books cover it to different extents.

[boreal inlet]

I have a ring R = {0, a, 1 - a, 1} st a^2 = a. Is this isomorphic to something known?

The only commutative rings of order 4 I know are Z/4Z, Z/2Z x Z/2Z, Z[x]/(2, x^2 + x + 1) and Z[x]/(2, x^2)

Third one is the field of 4 elements

[cobalt heath]

Z/2Z x Z/2Z ?

[boreal inlet]

But none of these seem to be isomorphic to that

[abstract rock]

looks like the sum of two Z_2 rings

[cobalt heath]

a = (1, 0), 1 - a = (0, 1) it seems like.

[boreal inlet]

Oh

My baf

[abstract rock]

just write 1-a as 1+a and its more obvious

[boreal inlet]

I did a mistake in checking the multiplication

[boreal inlet]

oh right are these the only commutative rings of order 4?

[boreal inlet]

Thanks

[rocky cloak]

Yes, they are in fact the only (unital) rings of order 4.

[boreal inlet]

Thanks for the confirmation

[tardy hedge]

If sigma is a cycle , and ord(sigma^n)=ord(sigma), can we say that sigma^n is a cycle?

[coral spindle]

Yes

[tardy hedge]

How can i show that?

I showed that orders are the same, now i just need to conclude with sigma^n must be a cycle

[coral spindle]

Hint: sigma^n decomposes as a product of disjoint cycles no matter what. Try arguing that all the cycles have the same order, and see why this gives you what you want

[tardy hedge]

All cycles in the disjoint cycle decomposition of sigma^n have same order?

[coral spindle]

That’s what I’m suggesting you try and prove.

[tardy hedge]

🫡

[tardy hedge]

I know that the order of a permutation is the lcm of the order of each cycle in its disjoint cycle decomposition

[tardy hedge]

@coral spindle so is it that having the orders be the same means that u can show that the supports are the same?

[delicate orchid]

(1234) and (5678) both have order 4 but their supports are disjoint

[tardy hedge]

Orders are the same means the size of the support is the same?

Let me think

[lusty marlin]

No

[tardy hedge]

Poo

[coral spindle]

(12)(345) and (123456) are a counterexample btw kian

[tardy hedge]

I see

Feels like this part shouldnt be too hard.

Ok so firstly, to show a permutation is a cycle, u can do that by choosing an arbitrary element in support and show that it at least takes |support| iterations to reach the same element, right?

[delicate orchid]

drop the 6, nerd

anyway

[delicate orchid]

cycles act transitively on their support, so if you can show you can get from any element to any other via your permutation it must be a cycle

for the fans in the back you get the orbits of the action by directly looking at the cycle decomposition! such fun!

[coral spindle]

wowowow

[karmic moat]

“oh thats really fun” - 🪭

[tardy hedge]

Ok makes sense but how i said it was also correct?

[tardy hedge]

😟

[delicate orchid]

yeah I think so

what's the actual problem we're trying to solve here lol

[tardy hedge]

Lol

[coral spindle]

Up here wew

[delicate orchid]

oh ok

[coral spindle]

I have already given a hint but we've veered off-course

[tardy hedge]

Showing that if sigma is a cycle and order(sigma)=order(sigma^n) then sigma^n is a cycle

[delicate orchid]

they map to the same element in the burnside ring over S_n :pack:

[tardy hedge]

This was part of a bigger proof about n and order(sigma) being coprime

I used something like that to show order(sigma)=order(sigma^n)

[delicate orchid]

kian was it you I had the discussion about what happens if |sigma| and n aren't coprime - like what cycles it decomposes into

or was that someone else

[tardy hedge]

Probably

Dont remmeber rlly

[tardy hedge]

So what about this?

Given a in support(sigma^n = tau) whats the minimum j so that tau^j(a) = a?

Proposition from my textbook says that tau^j = tau^i iff i = j mod (order tau)

Since order of tau is order of sigma = k then

tau^k(a) = a = tau^j(a) iff j = k mod k so k divides j so j cant be smaller than k

So given any element in support(tau) it takes at least order(sigma) to reach same element, and since sigma is cycle then order(sigma) = |support(sigma)|

For sure we know |support(tau)| <= |support(sigma)| , and since it took at least |support(sigma)| iterations to reach the same element, |support(tau)| = |support(sigma)| so tau is a cycle

Lol am i just saying a whole lot of nothing

[daring nova]

that's true iff tau is a cycle

[tardy hedge]

Fak

I knew something was wrong

[daring nova]

otherwise each element is on a smaller orbit and therefore has a smaller order

[daring nova]

if tau is (1 2 3) (4 5) then order(tau) = 6, but order(a) = 2 or 3 depending on the orbit of a

[tardy hedge]

Yeah

Order of whole map isnt the same as order of element

I see

[daring nova]

actually that's not quite true

[daring nova]

you did say sigma is a cycle and tau a power of sigma ?

right

[tardy hedge]

yeah what im trying to show rn is if sigma is a cycle and order(sigma)=order(sigma^n), then sigma^n is a cycle

I was just writing tau to not write sigma^n all the time

[daring nova]

then looking at a decomposition like this, you actually see that every orbit has the same length (because it's isomorphic to a certain kZ/lZ)

in which case ord(a) = ord(tau) still holds

[tardy hedge]

Fyi we only just started groups

So idk much about isomorphism and stuff

[daring nova]

don't need it

it's just a habit of always saying "by isomorphism to the group for which this structural property is most obvious to me"

[daring nova]

let l = ord(sigma) be the length of the cycle

writing s for sigma
then each element has orbit (a, s^n(a), s^(2n)(a), ..., s^(kn)(a)) before it inevitably loops

so tau will be sigma's cycle sampled evenly into subcycles

like if sigma = (2 3 4 5 6 1)
then sigma² = (3 5 1) (2 4 6) and sigma² has order 3

and every element has order 3 by sigma²

[daring nova]

so the order of the element is always the order of the permutation, iff that permutation is a product of disjoint, equal length cycles

[daring nova]

anyways, you'll always get a structure like this

where k is such that (k+1)n = 0 mod l

[tardy hedge]

Ok ok i see what ur saying but its gonna take a min for me to digest lol

[daring nova]

so you know you well get that form of permutation, and then order(s) = order(tau) forces the structure to be single cycle of length l

[daring nova]

yeah I'm using a certain insight about how the powers behave that is not quite obvious, but very good to know

[tardy hedge]

Wait so doesnt that work then

Cuz tau= sigma^n and sigma is a cycle

[daring nova]

yes then tau is a cycle

since its order is the same as sigma's

[daring nova]

because if there's m cycles, they're all of length l/m

and so tau is of order l/m

[daring nova]

but presumably it wouldn't have been

[tardy hedge]

Yeah

[daring nova]

basically what was said earlier, you may notice

[coral spindle]

Indeed

For reference, there's a really nice simple proof that all the cycles in the decomposition have the same length. They are all conjugate via some power of sigma: supposing 1 and i are in some (possibly distinct) cycles in the decomposition of sigma^n, since conjugation by sigma^(i-1) will preserve sigma^n and therefore send the cycle containing 1 to the one containing i, we must conclude that they are of the same length.

The rest is history.

[lime badge]

Hey guys, just a quick question. If $f: R \rightarrow S$ is a ring homomorphism and we have a homomorphism of S-modules $\phi: N_1 \rightarrow N_2$, I need to show that $\phi$ is an iso iff the corresponding map given by restriction of scalars to R is an iso. One direction is obvious, the other direction is where I was confused

[cloud walrus]

Eternal Way

[rocky cloak]

Are you able to show that a homomorphism of modules is an isomorphism iff it is a bijection?

[lime badge]

Well that's precisely what I was confused by, since it seems I get that anyway by assuming the map restricted to R is an iso. I feel like I have to make some argument involving the ring S

[rocky cloak]

I'm unsure what you're confused about? You are able to show that a bijective homomorphism is an isomorphism yes? Once you have that it's pretty obvious

[lime badge]

Yes, I know a bijective homomorphism is an iso

[rocky cloak]

Right, but then phi is bijective as a function and badaboom

[lime badge]

I guess it felt weird at first that injectivity and surjectivity doesn't really depend on the ring R or S

[rocky cloak]

Modules are abelian groups first and foremost

[lime badge]

Ah true. Ok, that's clear now. Thanks!

[rocky cloak]

Everything about injectivity and exactness and all that stuff is just about the underlying abelian groups really

[karmic moat]

abelian groups are Z-modules too 😎😎😎

[lime badge]

Yeah, its obvious now that I think about it this way

Thanks as usual 🙂

And I guess a similar logic imples the restriction functor is faithfully exact, right?

[rocky cloak]

Indeed it is

[lime badge]

Perfect

[lime badge]

Last question: if i: k --> F is a finite field extension of degree d and W is an F-vector space of dimension n, its clear that W when restricted to k has dimension dn

Similarly, if V is a k-vector space of dimension m, the extension of V given by tensoring by F has dimension m as an F-vector space

There's a third extension given by taking V --> Hom_k(F,V). I'm a bit confused how to compute the dimension as an F-vector space of this last one

[rocky cloak]

So F = k^d and V = k^m, so Hom_k(F, V) can be seen as the set of mxd matrices

[lime badge]

Man

That should've been really obvious

Oh but this would be matrices with coefficients in k

[rocky cloak]

Yes, but you can deduce the dimension over F from that

[lime badge]

Yeah fair, its m

That's pretty interesting

The fact that the last two have the same dimension as F vector spaces is just the hom-tensor duality in action, right

[rocky cloak]

Yeah, you have D(F(x)V) = Hom(F, DV)

And since V and DV have the same dimension, things work out

It's still true if V is infinite dimensional though. In which case I guess you can use that Hom(F, V) = (DF)(x)V because F is finitely generated projective.

[lime badge]

I'll get to projective and injective modules in a bit

That's the last two sections in this chapter of Aluffi

[white oxide]

here, by assigning each ordered set to it's "opposite" ordered set, do we just reverse the order of the relations or smt? presumably this is because D is not the opposite functor, since if it were, it would just assign each set to itself right and only reverse the morphisms

[crystal turtle]

Just the order of relations. So x <= y becomes y <= x

(also, the "functor" C --> C^op which takes f : X --> Y to f^op : Y^op --> X^op is not actually a functor btw, since it does not preserve composition)

[white oxide]

ah facts

thanks

[white oxide]

i don't understand what the purpose of D is here lol

well

i guess if I can show that D is an equivalence then the equivalence class generated by it and the identity functor constitute the two equivalence classes

and hence the automorphism class group has at least two elements

[white oxide]

i mean

isn't D an equivalence since DD is naturally isomorphic to the identity functor on C?

am i overthinking this problem

but then we have to show that the equivalence classes are disjoint right

[white oxide]

i'm dumb and my brain's not working how is a left module A over a commutative ring R a R-R bimodule if we define ra = ar

more specifically how is the associativity condition r(as) = (ra)s satisfied

[barren sierra]

ok so r, s in R, a in A

r(as) = r(sa) = (rs)a = a(rs) = (ar)s = (ra)s right?

@white oxide

I think this works

[chilly radish]

Yes

[chilly radish]

What equivalence classes?

Isn't the automorphism class group just the group of all automorphisms of the category?

[upper inlet]

up to natural isomorphism

[timber blaze]

can someone explain the part about cyclic groups please

this is theorem 10.3

[rocky cloak]

It's saying that for each divisor d of n, you have a unique subgroup of order d (the one generated by n/d), and hence a unique factor group up to isomorphism.

The homomorphism Z/n -> Z/d is not unique though. There are d of them, and phi(d) of them are surjective.

Does that answer your question?

[timber blaze]

why does a unique factor group for each divisor of n determine a homomorphic image?

[rocky cloak]

That's the first isomorphism theorem. If N is a normal subgroup of G, then there is a surjective homomorphism G -> G/N

[summer path]

[timber blaze]

wait what

[timber blaze]

is that equivalent to this statement

or is it a different result

[rocky cloak]

I mean I guess it's not really a result at all, is just the definition of a factor group

[timber blaze]

oooh i get it now

the map is f(x) = xN right

and the kernel here would be N

[tender wharf]

exactly

[tender wharf]

the first isomorphism theorem pops out nearly for free with the definition of normal subgroups

really goes to show the power of a good definition

[tardy hedge]

How can I interpret lemma 3.2.9?

some a and b in G can be deemed equivalent if their combination is part of a subgroup ? Or something like that?

[open sluice]

are you familiar with equivalence relations?

[tardy hedge]

Yeah

I know what the lemma mechanically means

And how its used in the next part to prove Langrages theorem

Hm im looking at the text right below and i think that example helps

With congruence mod n and the subgroup nZ

A subgroup of G inherently defines some equivalence with the elements of G

^something like thr

[delicate orchid]

LOL?

[tardy hedge]

Ok

[tardy hedge]

Is this sort of what lemma 3.2.9 is saying

[delicate orchid]

fuckin chatGPT posting unironically

[cursive arch]

I sometimes use my own words just so u know

I use chatgpt if it's too long.

To explain or I'm just lazy.

It's unintelligible but I'm lazy right now.

[tardy hedge]

I sent a pic so im not sure what ur talking about

Do u not see the pic

[cursive arch]

Lemme aee

See

[tardy hedge]

Ill send again

[delicate orchid]

here's how you interpret it, if ab^-1 in H then ab^-1H = H => b^-1H = a^-1H

it's basically a proof that cosets are equivalence classes

[tardy hedge]

Can u give an interpretation using words like an intuitive one is what im trying to do

Im not rlly sure what u wrote out

[delicate orchid]

they give you a very explicit example underneath, this is building up to quotient groups

[tardy hedge]

Yeah the example does help for sure

I think i got it im just being anal

[delicate orchid]

well i'm not seeing what you're not getting lol

[tardy hedge]

Lol yeah im just trying to get a good intuitive understanding of it all, lagrange thm

I do “understand” it

[delicate orchid]

the proof or the statement of lagrange

[tardy hedge]

I understand the proof and statement of lagrange in the sense of, each step makes sense logically and whatever

[delicate orchid]

cause if it's the proof I can draw some pictures

[tardy hedge]

But im just tryna see some of the “big picture” ideas a little better

[tardy hedge]

If you want to i think thatd be great

[delicate orchid]

so the main idea is that, because ab^-1 \in H is an equivalence relation (or more to the point, a ~ b if and only if aH = bH is an equivalence relation, which makes it more clear why we care) we have that the cosets aH, bH are equivalence classes - hopefully you can see why |aH| = |bH|?

anyway |aH| = |H| for all a because multiplying by any element in a group is a bijection, cause they're invertible, so |aH| = |H| = |bH|
and since the cosets are equivalence classes they're also disjoint sets

so we start with this picture, with our subgroup H inside of our big group G

then because all of our cosets are the same size and disjoint, our cosets will look like squares of the same size:

and since every element of G is in some coset (g is in gH), we can divide up our big rectangle into rectangles of size |H| like this

so, since we've evenly divided up |G| into some (here, six) copies of |H|, we have to have that |H| divides |G|

this is lagrange's theorem

[tardy hedge]

Awesome, ill look at this in a bit

Thx tho

[tardy hedge]

I like group theory so much more than analysis

[delicate orchid]

based

[white oxide]

category theory moment

[south patrol]

?

[chilly ocean]

any ideas?

[coral spindle]

This is very annoying to read because it's an image, and you have made it very unclear what problem it is you're attacking

Anyway, yes in essence this is number theory. Hint: think about what the lcm does to powers of p

[chilly ocean]

hm can you give me another small hint?

[chilly ocean]

sorry

[coral spindle]

Hint: lcm(5, 25, 125) = 125.

[chilly ocean]

this says n=p^c * q with c the maximum exponent of p in the factorizations of the orders

and again c must be >=k

i got it

i think

ok so we have x an element of order p^c * q with q not divisible by p and x>=k

if we look at the element x^(p^(r-k)*q)

the order of this element is the ord(x)/gcd(ord(x),p^(r-k)*q)

and we get that the order of this element is p^k

it is ok?

[coral spindle]

Yup

[white oxide]

reverse the arrows and you get a new definition

heh heh

nvm

why is a magma called a magma

[coral spindle]

The earliest reference I can find that uses the word is Serre's book on Lie algebras from '65. He does not explain the terminology.

[abstract rock]

and here i thought it came from condensed matter for whatever reason

[white oxide]

man must have been hot terminology back then

[open sluice]

it’s named after the magma CAS

[abstract rock]

it definitely spread like fire

[alpine island]

It's called a magma because it magma brain hurt

[minor fulcrum]

Ok I have thought about this in past and have insane take

So the other term around is groupoid yeah? But that conflicts with categorical groupoids so magmas started getting called "Ore groupoids" after their inventor

So it's a magma because it's an ore groupoid? Maybe? The word magma in French can also mean something like a chaotic mixture iirc, pardon my French but something like "propositions constituent un magma incohérente"

So it could be both of these

It's a bourbakism tho, serre didn't coin it

[tardy hedge]

@coral spindle

I just said “Two nondisjoint cycles sigma and tau commute iff one is a power of the other”

Is that still correct?

[gleaming shell]

Im trying to solve this solve this problem using a direct proof. I seem to be stuck in the work i posted. I was hoping it would simplify all down to b*a. can someone tell me the answere or give me a hint?

[open sluice]

consider expanding the equation in the problem statement

[gleaming shell]

What do you mean?

[gleaming shell]

This if I do this, can I say that the only way the bottom equation is true is if the two arbitrary elements a, b commute (the two elements underlined in red) this means G is abelian. is this a valid way to prove it?

[cobalt heath]

Depends if G is a group or not.

[gleaming shell]

G is given to be a group by the problem! @cobalt heath

[cobalt heath]

So you need to use properties of groups

(I wish ppl learned just the def of monoids so that they know what is unique with groups)

[gleaming shell]

I understand how to do it now

[cobalt heath]

Oh, my bad.

[cobalt heath]

(Yep, inverted element is important in groups)

[gleaming shell]

Ive been doing math for too long today. What should be obvious isn't anymore lol

[cobalt heath]

Ah, that feeling =P

[tender wharf]

[cobalt heath]

Nooo

why sully

[errant shadow]

Can you formally prove that any r^i =/= s for any i in D_2n? Like, can we do it without any pictures?

[cobalt heath]

Depends on definition, but you don’t need any pictures

It’s just that pictures demonstrate them well.

[errant shadow]

I mean like, the relations came from pictures right

Oh wait I think I was asking the wrong question

Like is it possible to prove that r^I =/= s for any D_2n

[cobalt heath]

Geometric symmetries forming relations, which is demonstrated by pictures.

[cobalt heath]

Yes, I believe

You can construct D_2n with transformation matrices.

[errant shadow]

Ahh so use a different formulation?

[cobalt heath]

Well I think it is possible with any formulation.

Which is why I am asking: which formulation did you learn it with?

[errant shadow]

Like we were introduced to the group with this as a relation

[errant shadow]

In terms of rotations and flipping of an n-gon?

We weren't taught any other formulations but I just found it on the wiki page

[cobalt heath]

So symmetry of n-gon, right?

You can rigorously formulate the n-gon.

[boreal inlet]

I presume here I can ask questions about Modules right?

So I'm trying to characterize the submodules of the Un(F) module F^n. Here, F is a field, and Un(F) is the ring of upper triangular matrices over F.

What do I do here? I only know that for a particular left module structure, there exists a ring homomorphism between the ring and the Endomorphism ring of the module.

So, here there exists a ring homomorphism between Un(F) and End(F^n) which is NOT (my hope was ruined) isomorphic to GL(n, F) but rather Mn(F)

But I need to find submodules

[cloud walrus]

Riku

[delicate orchid]

GL(n, F) is Aut(F^n), M_n(F) is End(F^n)

[boreal inlet]

Oof

my bad yeah

Ohh wait

Holy shit so

The pi is actually just a map from Un(F) to Mn(F). But yeah does it add anything

Btw I do see the trivial module is a submodule, and also the whole F^n is one.

If I consider a non-trivial proper submodule, it must contain a non-zero element.

Now, any non-zero vector v in F^n can be realized as a product of a square matrix and a column vector. If I can make sure the matrix can also be upper triangular, we have essentially proved this module is simple.

[delicate orchid]

F^n isn’t simple, (0,….,0,1) should be a sub module

[boreal inlet]

You mean the module generated by that vector?

[delicate orchid]

Hard to do the matrix multiplication in my head

[delicate orchid]

Nvm, I was wrong

[boreal inlet]

No it isn't closed

At all

[delicate orchid]

(1,0,0…,0)

Yeah that’s the one

[boreal inlet]

So if we can show for any vector v in F^n there exists an upper triangular matrix A and another vector u such that Au = v then it's simple, so if not true counterexample must exist

Oh wait lenme see

[delicate orchid]

In fact

(x_1, x_2, …., x_k, 0, …., 0) should be a sub module for all k

[boreal inlet]

Oh right

All of these are finitely generated too

[south patrol]

Indeed upper triangular matrices as like those that preserve the "standard flag"

[delicate orchid]

I just ignored the fact that they weren’t 1s on the diagonal and treated the entire thing like Aff_{n-1}(F) ngl

[boreal inlet]

[delicate orchid]

Then thought about orbits under the action of that on F^n

[boreal inlet]

Okay wait so these are the only submodules right?

If we have zero on the first element, i think the generated set doesn't stay multiplicatively closed anymore

[delicate orchid]

It won’t, cause of the element at (2,1) in ur matrix

And obviously since this is a field we’re acting on we can’t have a subspace of (x_1,…,0) without just setting one of the x_is to 0

I say obviously but maybe it’s not so clear to you (consider just the diagonal matrices acting on (x_1,…,0))

[boreal inlet]

Oh right, zero divisors don't exist

And other sums don't contribute

Yeahh

Thankss

[delicate orchid]

It’s more that everything non-zero has an inverse but close enough!

[still perch]

hi I am having trouble exhibiting a homomorphism f: G ----> Aut(G). I am watching these lecture series by Harvard extension school.

The attached picture explains the homomorphism. I do not understand how he came up with such a map.

[tribal moss]

Do you agree that this does in fact define a homomorphism?

[still perch]

From what I understood this is sort of a natural homomorphism bw a given group and the Aut(G).

I mean, I think I could prove that this is infact a homomorphism.

[tribal moss]

OK, great.

So your trouble is more to understand why that homomorphism is interesting?

[still perch]

The map should be something that takes an element g, in G to the function in the Aut(G) ? Is that right?

[tribal moss]

Yes.

[still perch]

Honestly, yes you are right. But, I read bits about this from here and there and I am confused about this whole thing.

now in this ss, he has written that f(g)(h) = ghg^-1

So he is talking about the functions in the Aut(G) right?

Aut is isomorphism from a set G to itself, so he is talking about this isomorphism which takes an element h in Aut(G) to h itself?

[delicate orchid]

perhaps if we write it as $g \overset{f}{\mapsto} (h \overset{f(g)}{\mapsto} ghg^{-1})$ it's a bit clearer?

[tribal moss]

f is the homomorphism G -> Aut(G).
That means f(g) must be an automorphism, that is, an isomorphism G->G.
f(g)(h) is what that isomorphism maps h to.
Does that make the notation clearer?

[cloud walrus]

Wew Lads Tbh

[delicate orchid]

it sends g to (the map which sends h to ghg^-1 for any h in G)

[still perch]

So this h to ghg^-1 , is this the isomorphism that we are talking about?

[tribal moss]

Yes.

There's one of those for each choice of g.

[still perch]

Ohhhh

[still perch]

I think I need to re-watch it again by keeping this in mind. Thanks tho, I somewhat understood that

@delicate orchid thanku

[cobalt heath]

Where does discussion of homological algebra with modules belong to?

[delicate orchid]

#advanced-algebra

[cobalt heath]

Thanks!

[errant shadow]

Yep, eventually got it.

[chilly canyon]

Heyo! Does anyone know anything about generators for the group $\mathrm{SL}(3,{\bf Z}[t])$?

[cloud walrus]

Matplotlib

[dim widget]

The natural map coming from setting t to zero is a group homomorphism since the determinant is still +/-1

I think it might be nice to describe the structure of this group by describing the structure of the kernel of that map, however I’m not sure if you’ll get a minimal set of generators

[abstract rock]

if you make a restriction on the degree polynomials you'd care to know about atleast you'll be able to have a system of equalities that will be easily solvable with computer (and that should grow cubicly for n)

but there's nothing outright nice about that space (outside the aforementioned natural map)

[oblique matrix]

for polynomials in one variable, there is a "factor theorem" stating that if P(a) = 0 then (x-a) is a factor of the polynomial. This follows from the polynomial division algorithm.

One may also factor things like P(a,b,c) = (a+b+c)^3-(a^3+b^3+c^3) by noting that P(a,-a,c) = 0 so (b-a) is a factor. This works because you can consider the multivariable polynomial P(a,b,c) as a single variable polynomial in b with a and c fixed, and plug in -a and notice it's a root, and then apply the factor theorem and see the resulting equation holds for all choices of a,b,c.

Repeating this reasoning 2 more times, we see that

P(a,b,c) = Q*(a+b)
P(a,b,c) = R*(b+c)
P(a,b,c) = S*(a+c)

Why is it possible to combine these three equations and deduce that

P(a,b,c) = K(a+b)(b+c)(a+c)?

I tried looking up a multivariable version of the factor theorem online, but I ended up stumbling upon Grobner bases, which I don't actually understand and I don't know how it would help formulate the factor theorem for multivariable polynomials without having to go through the motions of reinterpreting polynomials as single variable polynomials with all other variables fixed

[abstract rock]

Starting from one of your equalities, say P=Q*(a+b) from the other two you know that since (a+b) is a monomial that isnt divisible by the other two monomials, Q is divisible by those monomials

if you see that you can follow similarly from R and S, then there is some GCD between the three

Grobner bases formalize that in the language of ideals in ring theory, which can be more succinctly defined in terms of varieties

or something like that, its been a bit

[chilly canyon]

Mmmh yes ofc, but that doesn't tell me much about the group itself, does it?
Anyways, that was a random question I asked myself earlier. A shower thought if you wish, I don't really have a precise question anyways!

Thanks for your quick output

[oblique matrix]

so if A|BC and not A|B then A|C is what you're saying with this part? If A|C and A is a monomial then any factorization of C into primes will involve a term that is A?

[dim widget]

Yes it does! Because you can explicitly describe the structure of the kernel

[chilly canyon]

I mean, that's just one evaluation map out of the Z worth of it there are?

[dim widget]

Yes any of them would do

[abstract rock]

for some reason an alarm bell in my mind tells me to remind you this should be true for nice enough polynomials (coefficients from a field or PID, isnt an infinite polynomial), but yes since you've already shown that P has monomial factors as shown then the other factors must factor out those monomials as well

[abstract rock]

was there a motivation for this group?

[chilly canyon]

Not particularly, as I said it was more of a shower thought

[abstract rock]

lol ok

[chilly canyon]

I'm still trying to understand why it should be easier to describe the kernel of this evaluation map

I understand that if I know this kernel, I can understand the whole group. But that's like saying if you know perfectly the pure braid group you can fit the general braid group in a SES...

[abstract rock]

i dunno the kernel doesnt seem all too interesting tbh

[cloud walrus]

Matplotlib

[abstract rock]

but all the members of the kernel are just elements P of SL(3,Z[t]) such that P=I_3+t(R) for some R in SL(3,Z[t])

well, for any such R we know there is certainly a P in that kernel

[chilly canyon]

Now that I think about it, what about the natural map $$\mathrm{SL}(3,{\bf Z}[t])\to\bigoplus_{n\in{\bf Z}}\mathrm{SL}(3,{\bf Z})$$
given as all possible evaluations?

[cloud walrus]

Matplotlib

[dim widget]

R should actually be a traceless matrix

[dim widget]

Thats (of course!) injective

But the image is pretty annoying to describe

[chilly canyon]

Yeah I would assume so

[dim widget]

Actually hmmm

Maybe it’s not so bad

[chilly canyon]

Wait, isn't it surjective?

[dim widget]

Yes I am thinking it is

Okay so that’s quite nice!

[chilly canyon]

I'm thinking interpolation, but integers, but still surjective

[dim widget]

Although it’s not really a map of groups

[chilly canyon]

Is it not? Each individual projection is a group homomorphism

[dim widget]

In fact there is not such a map

[chilly canyon]

That's like concatenation of 3² evaluation map for each projection

[dim widget]

There is a natural map to the product

But not the direct sum

[chilly canyon]

Ah, my bad, it's true that it sometimes differs xD

[dim widget]

In this case in particular you’d be imposing impossible conditions on the polynomials

[chilly canyon]

Yeah well I'm being dumb because it very much does indeed

[dim widget]

Like having infinitely many roots

But there is a map to the direct product and the image is a bit like a direct sum in some way

[chilly canyon]

I definitely meant the product when writing this \bigoplus

Naaaaah I need a rest, this is not surjective; you can interpolate one matrix, but not infinitely many

[dim widget]

Yes like I said the image is a bit like a direct sum

[chilly canyon]

[dim widget]

Probably Lagrange interpolation gives some explicit description of the image, though not particularly nice for finding generators

[chilly canyon]

But Lagrange won't give integer polynomials though

But yeah, some kind of interpolation

[dim widget]

It will! Not so hard to see

You can just multiply out by d! Where d is the degree

[chilly canyon]

Ah, but you're cheating, you're being clever there!

You're clever! even

Another approach could be to look at the maps $$\mathrm{SL}(3,{\bf Z}[t])\to\mathrm{SL}(3,{\bf Z}[t]/I)$$
for some ideal $I$ of ${\bf Z}[t]$, a special case being reduction mod a prime with $I=\langle p,t\rangle$

[cloud walrus]

Matplotlib

[chilly canyon]

Anyway, I'll stop there because, again, #showerThought, and I'm not even sure it's an interesting one

[dim widget]

This was my original approach with I = (t)

It’s an interesting question! Definitely doable but not totally obvious

But also it’s hard to find a well-defined question, what makes a set of generators a “satisfying” answer? For instance there are probably countably many

[chilly canyon]

Of course there's going to be countably many, it would be more than surprising if there were not!

The question was very vague indeed; someone else might want to give a presentation of the group (most likely there won't be a presentation with either finitely many generators or finitely many relations anyways)

Thanks for the chit chat

[warm ember]

why is C the algebraic closure of R

or i mean why is R[i] the algebraic closure

actually nvm

[chilly radish]

This is essentially what the fundamental theorem of algebra says

[boreal inlet]

Let's say we take C[0,1] as a R[x] module with the multiplication structure pv = p(T)(v) where T is the indefinite integral operator.

How do I show this module has a trivial Annihilator?

If this is required - I have proved both Polynomials over [0,1] and C^\infty[0,1] are submodules of C[0,1] over this structure

[cloud walrus]

Riku

Riku

[cloud walrus]

Riku

[boreal inlet]

Okay uh that might not be true for any f

But well if one such f does exist - let's assume it does. For that f to satisfy this equation, we need all coefficients a_n to be zero.

Now let's say the polynomial p we started with is non-zero, hence, one of these coefficients is actually not zero. But then, that particular function we found before does not satisfy the condition anymore, then p fails to belong to the Annihilator.

Hence, finding even one such f is sufficient, right?

I think then the constant function f(x) = 1 for all x in [0,1] works

[vapid stone]

Hi! Does someone have an idea or any references regarding classification of the subgroups of $(\mathbb{C},+)$ ?

[cloud walrus]

LF

[vapid stone]

So far I've found a classification of the subgroups of (R, +), so we have those plus those rotated in the complex plane as subgroups of (C, +)

And you can have a discrete group generated by finitely many elements

[ivory trail]

what is your classification of subgroups of (R, +)?

[vapid stone]

Do you think there are other discrete subgroups or are the other subgroups dense?

[vapid stone]

either dense or generated by a single element (of the form pZ)

I dont think i need to (or can) classifiy the dense groups further

[ivory trail]

ok so you are just interested in the discrete subgroups then?

the dense subgroups seem to be headache-inducing

[vapid stone]

indeed i have given up on those

i think its more about topology than groups

but for the subgroups of (C, +) i think we can have a subgroup that is dense on a line of the complex plane

does that make sense?

[ivory trail]

yeah i see

so you're wondering if all the non-dense subgroups are either lattices or isomorphic to Z x (a subgroup of R)

[vapid stone]

yes well put

[ivory trail]

https://mathoverflow.net/questions/335657/subgroups-of-mathbbrn looks like this is what you want, if it's right

Hence G=D⊕M where D is dense in E and M is lattice

[graceful elm]

I just asked a question over in #elementary-number-theory about cyclic groups of composite order, wasn't sure if it belonged there or here, if anyone sees this and wants to take a look i would appreciate it :)

[drifting plover]

If G is a subgroup of SL(2,q) that doesn't contain just upper triangular matrices, can we always get an element of G whose trace is not 2?

Answer: yes, lol

I guess I just thought of an argument as I typed it

[coral spindle]

No I don't think that's right

If you take certain conjugate subgroups of the upper triangular matrices you will typically not just get upper triangular ones, unless this group is just too small.

Let me see if I can cook up an example really quickly.

Yeah indeed the simplest example I could think of worked

If we denote G = U(B) as the subgroup of upper unitriangular matrices, then conjugation by the element [0 -1 \ 1 0], which is always an element of SL_2(q), gives us the subgroup of lower unitriangular matrices.

You can calculate this by hand very easily.

And more importantly, they're all of trace 2.

[drifting plover]

Ah, damn

[coral spindle]

The condition on the subgroup would probably be that it's not conjugate to a subgroup of the unipotent radical of any Borel, but I don't know if you'd need some weird conditions on the stability of the Borel under the Frobenius map. I can't see that quickly.

In any case, the point is that the condition is probably not going to be nice.

That being said, there is one very nice almost-condition which is admittedly way too strong: let the order of G be coprime to q — then all the elements are semisimple, and there oughtn't be too many semisimple elements in SL_2(q) with trace 2

I'd have to work this out, I'm so bad with this stuff but oh well

[drifting plover]

I think if q divides the order of G, then it must be all of SL(2, q)

[coral spindle]

That's definitely not true

Think about the upper unitriangular matrices

There are exactly q of them

But anyway, you can quite easily see this via elementary group theory too: we know about Sylow subgroups, so there must be a subgroup of order q (the Sylow p-subgroup) which cannot be the whole of SL(2,q)

[drifting plover]

Hold on, I think that's not what I meant

[drifting plover]

How do we know this is not upper triangular?

[coral spindle]

It is...

[drifting plover]

Yeah, I mean something that's not upper triangular, and has order divisible by q

[coral spindle]

But anyway, you can also just use the example that I gave earlier to get a different Sylow p-subgroup. Just conjugate.

[drifting plover]

Yeah

[coral spindle]

So I've now gotten a subgroup of order exactly q that isn't upper triangular.

In fact, since the normaliser of this Sylow is just B – all upper triangular matrices – if you just pick any non-upper-triangular matrix and conjugate, you'll get a group of order q that contains at least one non-upper-triangular matrix :)

I don't think it's trivial to see that the normaliser of U(B) is B, but it happens to be true.

I forget the proof.

[drifting plover]

I'll get back to you in a bit, in a seminar lol, but thanks!

[delicate orchid]

something to do with them preserving the standard flag perhaps? Conjugating would change the flag meaning the resultant group isn't the same

[coral spindle]

Oh yeah that makes sense. I guess I was worried the det = 1 condition would screw things up but it just doesn't

[prime quail]

Is there a classification of infinite simple groups?

[topaz solar]

There’s a “”big”” (it’s at least got a few texts on it) conjecture that all simple groups satisfying a certain condition are of a certain type so

Wew can probably say more on how it’s doomed but

[delicate orchid]

The Janko 42069 group

[topaz solar]

Even the parts of the conjecture we know about more (“”close to finite”” ones), we still can’t say yes or no to the conjecture

And what we do know draws a lot from the classification of finite simple groups, if you believe that result

[delicate orchid]

which conjecture is this

[topaz solar]

Cherlin-Zil’ber, the “algebraicity conjecture” or smth

I think I have that first one right

[delicate orchid]

ah I see the term "omega-stable" so this is definitely your area

[topaz solar]

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

[delicate orchid]

yeah that's exactly what I'm reading rn

[topaz solar]

So, free group on 2 generators ain’t omega stable

So this is rather restrictive

They’re stable but the omega in front is a specific size bound thing, and implies totally transcendental, which implies superstable

[delicate orchid]

guessing that's to do with the Morley rank gizmo

[topaz solar]

Ye, they can’t have finite Morley rank

[delicate orchid]

and in turn I guess that's because F(2) contains F("\omega")

[topaz solar]

In fact, their Morley rank > w for all ordinals w

Yeah basically I think that would do it

[south patrol]

Morley rank

[topaz solar]

Make some infinite binary tree thing

[south patrol]

i get excited when i hear a term i've never heard of lol

in this channel of all lol

[delicate orchid]

uponthewitnessing

oh yeah duh because something in model theory is a vector space or somesuch

[topaz solar]

Like taking your generators for a F(omega) iso subgroup (x_n), make it like vertex t splitting to branches tx_n and tx_n^-1?

[delicate orchid]

I remember you talking to the wall about that at some point sharp

[topaz solar]

It’s more like it can only be split up into infinite sets that many times

[delicate orchid]

a... vertex? of what

what graph is here

[topaz solar]

But in the particular case of finite RM simple groups, they should be algebraic groups (linked on the page)

[topaz solar]

I was linking via reply to an infinite binary tree remark

[delicate orchid]

yeah and what does that tree corrispond to

I cannot emphasise enough how much I feel like a first year ug walking into a lecture on cohomology

[topaz solar]

If you can build out some binary tree based on formulas which ramifies like trees (with the different branches being incompatible), it has RM \infty

Basically we just crunch any hope of nice ramifying

This

Anyhow we make a bunch of subgroups

[delicate orchid]

I do not know what a formula is, or what ramification means here, or what incompatible means here

[topaz solar]

So the ones to the left are larger

And as you go to the right, two adjacent branches have trivial intersection?

Or rather than subgroups we’re doing coset nonsense

So we have an infinitely descending chain of proper subgroups

So it’s very infinite

[crystal turtle]

Me whenever Sharp rants

[delicate orchid]

hell I barely even know what ramfiication means in algebra it's just the funny group :smoked:
https://en.wikipedia.org/wiki/Ramification_group

[summer path]

Me whenever anyone here rants

[topaz solar]

By ramification I mean splitting lmao, not a fancy formal term

[delicate orchid]

anyway back to doing REAL work like finding out what the f*lp the sylow p-subgroups of G_2(p) are

[delicate orchid]

and splitting here means...?

[crystal turtle]

f*lp

[topaz solar]

We get disjoint cosets of our descending chain of subgroups

[delicate orchid]

sharp everything you've been saying has been a fancy formal term and if this was my first interaction with you I would be complaining to the mods about a new crank in the advanced channels

and I'd expect you to fully do the same for me

[delicate orchid]

right, ok I can actually understand this

[topaz solar]

So we can draw this graph out in terms of cosets, left->right being that the left contains the right

[delicate orchid]

that's a pretty strong condition no?

[topaz solar]

And each layer is like G_n/G_n+1

Where G > G_n > G_n+1….

[topaz solar]

Yeah lmfao

We can still have things of infinite Morley rank that this doesn’t happen to

[delicate orchid]

I might do a wellness check on this Morely fella...

[topaz solar]

If it’s “totally transcendental” (RM isn’t \infty), then we can replace an intersect of arbitrarily many definable subgroups with an intersection of finitely many of em

Since no descending chains

[topaz solar]

It should be noted, these chains have to be definable subgroups, so there can be nonsense that isn’t definable

[delicate orchid]

I could define them

[topaz solar]

But that’s basically saying you need some stronger principles (choice, for example) to make them, or that they’re not naturally reachable with our operations

So finite RM in particular is even stronger

[delicate orchid]

yeah so lemme just recall

[topaz solar]

You can’t split it up into infinitely many sets infinitely often

[delicate orchid]

definable means you can create the group using a first order(?) formula in the model(?) of groups?

it was something like that anyway

[topaz solar]

Yes

Model of group = a specific group

[delicate orchid]

yur

but like you can access the elements and stuff is what I mean

the example I recall from before is that the centre of a group is definable?

[topaz solar]

You can’t just pick out particular x in G too easily

Yes

[delicate orchid]

although I can't remember if the formula you used was "fixed under conjugation" or "maximal subgroup with C_G(H) = G"

[topaz solar]

I used xy = yx for all y

So commutativity

[delicate orchid]

right so that's accessing the elements

but in a general way

[topaz solar]

Well, here’s the trick

We don’t pick out particular x, y

[delicate orchid]

you could "assign" each element to it's cyclic subgroup ig

[topaz solar]

x is our predicate variable, the input, and y is bound by \forall

[topaz solar]

That’s hard to say “y is in <x>”

[delicate orchid]

no need

<x><y> = <y><x> is what I was thinking about

[topaz solar]

Since x^n = y for some n has n range over Z

[delicate orchid]

right nevermind, ignore my cyclic bullshit

we have \phi(x) : xy = yx \forall y

[topaz solar]

Yeah lmao

We put \forall in front though

[delicate orchid]

no. no I don't think I will

[topaz solar]

L

But yeah saying y is in a subgroup generated by x is harder I think

If you know that all your orders are, say, bounded by n, then it’s easy

Wait a minute, don’t Sylow subgroups have niceness

[delicate orchid]

well that's ok because all groups are finite

[topaz solar]

[delicate orchid]

depends what you mean by niceness

[topaz solar]

So yeah this helps out I think

[delicate orchid]

F_p-SWAG spaces ykwim

"ermmm aktually... the correct terminology would be F_p-comple-" RAAHAHHHHHHH

[topaz solar]

Well actually we might have some unbounded order nonsense ig but

Elements order p^k but k unbounded?

Anyhow, definable subgroups are very nice in finite RM since they’re kinda finite-y

[topaz solar]

“omega stable” groups don’t allow this to happen either btw, but it’s stronger than simply asking it not happen since we’re bounding some size of “we have only so many kinds of elements” stuff strongly

They can have infinite ordinal (but not the RM=\infty unbounded kind) of Morley rank

(This stuff should look like krull dimension stuff?)

[topaz solar]

:3 also yeah I definitely jargonize this stuff but it’s not too bad if you learn it from some more sane source

[delicate orchid]

so glad the mathematics I do is sane and resonable, anyway time to try and find the pattern in these numbers!

[topaz solar]

Reminds me of the cranks which spent years looking at OEIS

[abstract rock]

i was unironically in this situation when a computer scientist gave me a somewhat nice problem

beware of computer scientists, they are the devil's shepherd

[delicate orchid]

I'm not actually just staring at these numbers - I don't really care about more than the first 2 cycles of them

[drifting plover]

Yeah so I think what I wanted is a subgroup that doesn’t contain any conjugate of an upper triangular subgroup, rather than just not containing an upper triangular subgroup

In this case your argument for conjugation fails

I don’t know any rep theory but essentially I want a rep that’s isomorphic/conjugate to one that’s upper triangular

[delicate orchid]

so you want a subgroup of GL that doesn't contain gBg^-1 for any g

if so, the trivial subgroup satisfies this

[drifting plover]

This is the statement about the SL: Let H ⊆ GL(2,Fl) be a subgroup with l | #H. Then either H contains SL(2, Fl), or H is contained in some Borel subgroup.

[coral spindle]

Then indeed, since all Sylow p-subgroups are conjugate, this is clear

[drifting plover]

Wth, I mistyped

[white wraith]

Just done Sylow p-subgroups and heard the concept of conjugates with relation to orbits etc. Taking this as a spoiler for my next lecture 😆 (/j)

[drifting plover]

I’m saying it can’t also be the trivial subgroup, because that’s contained in Borel

[drifting plover]

The different containments are confusing me, but this is what I was trying to ask, but removing the l | #H condition

When I say upper triangular I don’t mean the Borel subgroup (which was why I tried to avoid saying that), I just mean anything without a non-upper triangular matrix in it, or any conjugate

[delicate orchid]

ok I admit it. I have no idea what you want answered. Post the full problem statement please?

do you want a proof that B covers SL in the subgroup lattice?

the existence of a subgroup of GL that intersects trivially with the upper triangular matrices and all of their conjugates?

what is it

[delicate orchid]

disregard this suggestion, the inclusions were confusing me as well

[drifting plover]

Yeah very sorry, it’s a mess trying to ask a question during a seminar lol

[delicate orchid]

no worries

[drifting plover]

I only care about subgroups up to conjugacy. I have a subgroup G of SL that’s not contained in the Borel subgroup B or any conjugate of it, and I want to prove that it has an element of trace not 2. It does not necessarily intersect trivially with B nor any of its conjugates.

[delicate orchid]

the borel subgroup of SL I presume?

[drifting plover]

Yeah

[delicate orchid]

ah ok I think that was a source of confusion, I was thinking borel of GL

it's a good thing I have the character table of SL_2(q) ready haha

[coral spindle]

ME TOO BABES

[delicate orchid]

ok really stupid question but borel of SL_2 is unitriangular matrices right

[drifting plover]

Could have q-1 on the diagonal too

Borel is always maximal right?

[delicate orchid]

good observation

[coral spindle]

Upper triangular. The unipotent radical of the borel is unitriangular

[drifting plover]

I can prove what I want assuming some hardcore results in number theory + an open conjecture, but obviously this should be an elementary question

[delicate orchid]

this does not seem elementary to me at all lol

there are many many different conjugates of the borel subgroup here

and even once I do characterise them, I have absolutely no idea what the trace has to do with anything

[delicate orchid]

everything inside the borel subgroup has trace +/- 2, so anything conjugate to something inside a borel subgroup must certainly have trace \pm 2. So if we somehow show that everything with trace \pm 2 is conjugate to something inside the borel subgroup then we're done, because your G has something outside of any conjugate of the borel subgroup and thus cannot have trace \pm 2 - how to show from there the trace must be positive I do not know

for q = p I believe this result because the borel subgroup is very large inside SL but idk if it's true for q = p^k

[drifting plover]

Ohh nice, I think I just need q = p

But general q is cool too

[delicate orchid]

oh thank heavens

[drifting plover]

It’s not elementary in the sense that it’s group theory that I don’t know, but it’s elementary in the sense that it should be a purely group theory proof

[delicate orchid]

right then yeah I think we can do it pure group theory, maybe
there are p+1 conjugates to the borel subgroup, meaning we have a highest bound on the number of elements conjugate to something inside the borel subgroup at (2p-1)(p+1)

[sly crescent]

Are the Borel subgroups of a compact Lie group just its maximal tori?

[wet zodiac]

how should i go about proving that the torsion subgroup is closed in an abelian group

[rocky cloak]

How about this?

M has trace 2 means the characteristic polynomial is x^2 - 2x + 1, so M is conjugate (in GL) to an upper triangular matrix. Hence M has order dividing q. So if G is a group consisting only of matrices with trace 2, then it is a p-group, hence contained in the p-sylow subgroup

[topaz solar]

If x^n = 1, and y^m = 1, what about x*y?

[drifting plover]

Lol I thought they meant closed as in a closed subgroup

[wet zodiac]

(xy)^(nm)=1

but where does the abelianness come in

oh wwait

nvm i see now

[topaz solar]

mfw xyxy=x^2 y^2 is abelian

[wet zodiac]

it has to be abelian or you can't rearrange