groups-rings-fields

[chilly ocean]

And the other i,j i not equal to j form another orbit

Now I believe intuitively we can describe a permutation on these 9 elements

Just by how the diagonal elements are permuted among themselves

But I am not able to like find a nice notational way to write it

[sly frost]

this can also be seen from the fact that in an abelian group Gσ(a) = σGaσ⁻¹ = σσ⁻¹Ga = Ga, now because G is transitive we then have Ga = Gb for all a, b ∈ A, so because the intersection of all of them is 1 they have to all be 1

[chilly ocean]

Ah it's nice i appreciate it , but jagr solution was quite well motivated for me. I think the underlying idea is the same

Conjugate to reach all elements

[sly frost]

just assign numbers to each pair

I'd just do 1 - (1, 1), 2 - (1, 2), 3 - (1, 3), 4 - (2, 1), and so on

S3 has six elements, so just describe the cycle decomposition of all of them in terms of how they act on those 9 pairs

you know what a cycle decomposition is?

[chilly ocean]

No i think I am into sth with how the diagonal is permuted among itself can lead to the cycle of how other non diagonal elements permute

In S_3 we can write every permutation as a cycle decomposition

Now this is just how the cycle decomposition of diagonal looks like

Then uhh

[sly frost]

I'm not sure what you mean by "cycle decomposition of a diagonal"

[chilly ocean]

Oh the way I am thinking of this is like a 3*3 grid

[sly frost]

you have more than just the diagonal elements here

[chilly ocean]

Yeah

[sly frost]

so you have to write out all 9 of them in a cycle decomposition

[chilly ocean]

But notice that the diagonal only permuted among itself only

And like everything else permutes among itself

Also by the cycle structure of diagonal I mean like sps p has structure (1,2) the the diagonals permuted will have structure ((1,1)(2,2))

Oh also notice that these disjoint cycles determine the sub matrices on which the permution acts independently

So a permutation on the whole matrix can be broken down to permutation on submatrices

So we just have to figure out how a cycle would actually on a sub matrix

I think doing se we can also generalise the problem to a this kind of S_n action on a nxn matrix

[rotund aurora]

Let G be a non-cyclic p-group. Must it have a quotient isomorphic to (Z/pZ)^2 ?

[coral spindle]

Yes

There's this nice lemma

it's used in the proof of the Sylow theorems, btw

that states that any p-group has a chain of normal subgroups (n.b. normal subgroups of the whole group, not just the next subgroup) with size p^0, p^1, etc all the way up to p^n = |P|

Take the one of size p^n-2 and quotient.

[mighty kiln]

What if it becomes cyclic

[coral spindle]

Wym

Oh

I misread the frikking question

[rocky cloak]

Yes, this is the burnside basis theorem

For every p-group there is an Fp vector space, such that a subset is a minimal generating set iff the image is a basis

[rotund aurora]

I think we could quotient by the Frattini subgroup, the result is elementary abelian (i.e., abelian and all elements have order p). The Frattini subgroup should have index at least p^2 then. So we should show that the number of index p-subgroups is >1

[rotund aurora]

I guess to show this last claim, one can proceed as follows.

• Every proper normal subgroup is contained in an index p-group (quotient, use that p-groups have index p-subgroups, use correspondence theorem)
• Thus if G has just one index p subgroup P all normal subgroups are contained in P. But take x in G setminus P. Then, since G is not cyclic, x is contained in a proper group H. But normalizers "grow" (i.e., H<N(H)), and thus every proper subgroup of a p-group is contained in a normal subgroup

[rocky cloak]

Yeah, so take a maximal subgroup and take an element not in there.

Then since the group is not cyclic, the subgroup generated by this element is proper, and hence contained in a maximal subgroup

[rotund aurora]

oh right I guess this is easier, since maximal is not necessarily index p

not a priori at least

[rocky cloak]

Well, I guess you would use that in proving that you get an elementary abelian group anyway, but whatever

[rotund aurora]

right, yeah

[chilly radish]

I'm losing it. I'm trying to compute the decomposition of \R[Z/3], but groupprops says there's a 2-dim irrep, but how is that possible if the R-dimension of R[Z/3] is 3??

[rotund aurora]

well yeah I showed above that proper subgroups are contained in normal subgroups, and that normal subgroups are contained in index p subgroups

[coral spindle]

I'm not sure what contradiction you're seeing. There is a 2-dim irrep and the trivial rep. This accounts for 1+2=3 dimensions :P

Maybe you're used to the fact that a complex G-rep of dim n occurs in CG with multiplicity n. This fails for fields that aren't algebraically closed.

[chilly radish]

The sum of the squares of the dimensions of the irreps should be equal to the order of the group, no?

This should be true over a general field

[coral spindle]

No, this isn't true over a general field.

[chilly radish]

Why doesn't it follow from wedderburn-artin?

[coral spindle]

It's true over splitting fields only

The endomorphism algebras of these reps are no longer the base field.

In fact, they will be the base field iff they are absolutely irreducible, i.e., we are working over a splitting field.

[chilly radish]

Yea they're matrices over a division algebra

[coral spindle]

So you cannot say that the wedderburn components are of dim n^2

Or well

n being the dim of the rep

[chilly radish]

Yes ok I get what you mean now

[coral spindle]

In this particular case, we have three CG representations, and two of them 'fuse' in R

[chilly radish]

Yea

That makes sense

[coral spindle]

The endomorphism algebra of this R-rep will be isomorphic to C

So in particular, here it means that the wedderburn decomposition is just R x C

[chilly radish]

Yea

Oh well the place I think it breaks in practice is that the endomorphism of an irrep is gonna be matrices over a division ring, but that division ring needn't be central

So the dimensions go out of whack

That's basically what you said

[coral spindle]

Yeah that's the issue

So in fact you can compute the centre really easily

[chilly radish]

I guess it's pretty obvious that sending Z/3 to multiplicatiom by a 3rd root of unity works and there's no subreps because if you add a 3rd root of unity to R you get C.

[chilly radish]

Oh

[coral spindle]

if the RG-module M has some irreducible CG-module constituent N once you extend scalars, the centre of End_RG(M) is actually equal to the so-called character field of N

So let chi be the character of N

Then the character field is R(chi), defined to be R(chi(g) | g in G)

[chilly radish]

I see

[coral spindle]

Now does this mean you can calculate the division algebra? Alas no. But it's a pretty cool fact.

[chilly radish]

Well, you can characterise it (ba dum tss)

[coral spindle]

Waheyyyy

[chilly radish]

That's something

Usually the best we can hope for

[coral spindle]

In the case of the real numbers there is a very powerful result that you will probably see soon if you're looking at real representations

The Frobenius–Schur indicator will actually allow us to determine the division algebra (i.e., R, C, or H) from a character alone

But in general it's extremely difficult

[chilly radish]

Oh this was actually an orals practice question (kinda straying off my main topic)

But that's really cool

[coral spindle]

Oh nice!

[chilly radish]

I will eventually get back to learning rep theory proper

Oooh haar measure stuff

Nice

[south patrol]

Rep theory fun

[coral spindle]

Yeah it generalises in some wild way

But I just care about the finite groups case lmao

Charamchter... thoeremy..... mmmmdeliviosu

[rocky cloak]

Well, the endomorphism ring of an irrep is always a division ring.

But in general, the number of summands of M in kG is
dimM / dim End(M)

In the algebraically closed case, the denominator is always 1, but not so in general

[chilly radish]

Fair enough

[chilly radish]

Well yea but it's not always a central division algebra, so its dimension over the base field needn't be a square

[coral spindle]

Here's something interesting.

Let d be the dimension of this division ring. Then as I said earlier, its centre is F(chi). So what's its dimension as a central F(chi)-division algebra?

As it turns out, its dimension as such a thing is always a square. It is the square of the Schur index of chi

[chilly radish]

The dimension is always a square by W-A theory but the fact that it's the schur index is cool

[coral spindle]

Yeah lol

[cobalt heath]

Just checking, you can localize ring at non-zero-divisor element and it will still preserve identities, right?

So what is an easy way to see if g - 1 \in Z[G] is a zero divisor.

Argh, I forgor that Z[G] is a noncommutative ring

Can I somehow still introduce power series in g?

I will just adjoin right inverse..

[agile burrow]

If G is finite, g - 1 is always a zero divisor

[cobalt heath]

G is infinite in my case

[cobalt heath]

Hmm why?

Ah, right, yeah

[agile burrow]

Let S be the sum of elements in the group, then it's invariant under multiplication by any element of G

[cobalt heath]

You can also do like, g^ord(g) - 1 = 0

I guess I should just send to rep first and compute

[next obsidian]

Bro… but g = (g + 1) - 1 so all g r zero divisor?

[agile burrow]

g + 1 isn't an element of the group though

[next obsidian]

WHAT

🥺

My proof…..

[quartz wind]

how can i show existence of lub/glb?

[dull marsh]

Depends on the poset

[next obsidian]

U take 2 element

And show something exists under both

And anything else under both is under that

Swag

[quartz wind]

set of all subspaces under set inclusion. heres what i got so far. i know proving boundedness isnt enough. in analysis, i guess id prove that the "closure" of any chain is also in the set, or that its "compact" somehow. not sure how it works in this context

[next obsidian]

You need candidates

Once you have the candidate it frankly isn’t hard to show they are the glb or lub

Particularly for glb

[quartz wind]

candidates as in possible glb/lub?

like can i say, "suppose G is the glb, provided it exists. then no proper subset of G or superset of G can be the glb"

because i already have the second part of that logic, i think. but idk if i can instantiate G without knowing its existence first

[quartz wind]

am i just not built for this 😭

like my argument is basically since $E\subseteq S_i$ for all $S_i$ and since no element outside of $E$ can be in all $S_i$, then all thats left to show is that $E$ is a subspace and therefore a greatest lower bound. but im not sure if that actually works

[cloud walrus]

esca (@ with reply)

[quartz wind]

updated proof

[sly frost]

not really

the way to prove this is that first you show that the intersection is a lower bound, then you show that any other lower bound is contained in that intersection

[quartz wind]

yeah i think i did that part

[sly frost]

tbh you didn't show that the glb exists, just that if it exists, it is equal to E imo

to show that glb exists, you show that every other lower bound is contained in it

[quartz wind]

ahh okay i see

thanks

[sly frost]

first you say "E ⊆ Si for all i and E is a subspace, so E is a lower bound"

then "suppose L is a lower bound, then L ⊆ E, therefore E is the greatest lower bound"

then you do the same thing on the other end

[quartz wind]

makes sense yeah

thanks 👍

[sly frost]

sure, you pretty much proved that E is a lower bound, but not that it's the glb

also, the existence of a minimal element does not prove the existence of the glb

you wrote something like that in your first proof

the minimal element is a lower bound, but that doesn't mean there's a greatest lower bound

for example {x ∈ Q | x² > 2} has a lower bound 1, but no GLB in Q

[quartz wind]

mhm yeah i just skipped proving existence so it came out looking like that lol

fairly sure this should work now

[fading field]

i think you meant to say that lower bounds don’t imply the existence of glb’s, minimal elements are glb’s when they exist

[sly frost]

sorry - in this case we had a minimal element of the whole set, and were talking about glb of a subset

minimal element is a glb of the whole set, but that doesn't mean some subset has a glb

[fading field]

i see, i’m not paying attention to the conversation enough sorry :P

[sly frost]

sure, no problem

[sly frost]

yes, I think it's more or less ok

[quartz wind]

ok thanks

[languid trellis]

I can't quite see why Tor(M) should be a submodule. In particular, if y \in Tor(M), then I'm not sure that ry \in Tor(M) for all r \in R. y being in Tor(M) means that there exists an r1 such that r1y = 0. I don't see why the "annihilation condition" should hold for ry too. Any help?

[rocky cloak]

Consider r1 * ry

[languid trellis]

I thought about this then immediately threw it out because R is not necessarily commutative.. hm

[rocky cloak]

Integral domains are commutative

[languid trellis]

oh

🤦‍♂️

thank you

[rocky cloak]

Also it's not true in the noncommutative setting. Like let R be the noncommutative polynomial ring in x and y, and M = R/Rx (+) R/Ry.

Then R is a domain, (1, 0) is annihilated by x, (0,1) is annihilated by y, but their sum (1,1) is not annihilated by anything.

[languid trellis]

Well it's no wonder I was struggling to prove it on the assumption that R is not necessarily commutative lol, thanks again

I'm not sure what you mean by (+)

[rocky cloak]

Direct sum

[languid trellis]

I'll save that example and I'll try parse it later, I appreciate it

[rocky cloak]

Maybe easier is also just let M = R/Rx, then 1 in M is annihilated by x, but y*1=y isn't annihilated by anything

[languid trellis]

Oh yeah I see

[ornate tiger]

i don't understand this

[dull marsh]

Which bit exactly?

The highlighted one?

[leaden heart]

a coset, written gN for some g in G, is defined as {gn : n in N}. The group operation on the cosets is then defined as (aN)*(bN) = (ab)N. Normality of N is precisely what is required for this definition to be well defined (this needs to be checked because there can exist g_1 =/= g_2 such that g_1N = g_2N)

[ornate tiger]

Yes

[ornate tiger]

Hmmm

[ornate tiger]

The operation is instead of + or * ?

[leaden heart]

If a_1N = a_2N, we have a_2 = a_1n_1 for some n in N, similarly b_1 = b_2N implies b_2 = b_1n_2, and thus
(a_2N)*(b_2N) = (a_2b_2)N
= (a_1n_1b_1n_2)N
= (a_1b_1n_3n_2)N (this is normality of N)
= (a_1b_1)N
= (a_1N)*(b_1N)
so the operation is well defined

[leaden heart]

i was being lazy and not writing the operation my b

[ornate tiger]

I think i need to read a lot about this because it's a little bit hard

[leaden heart]

don't worry, the fact that normalness of subgroups is important was very unintuitive for me, too

[leaden heart]

you could consider the symmetries of an equilateral triangle, a group with six elements, and the subgroup that contains only one reflection and the identity. That subgroup is not normal, it has three cosets, but the group operation doesnt define an operation on the cosets

the operation is "pick an element from each coset, and do G's operation on them, and see what coset you landed in", but different choices will land you in different ones

[lusty marlin]

You don't need induction

Also what is ℤ*?

[chilly ocean]

I meant Z+ sorry

[coral spindle]

That's not a group under addition or multiplication, so it's not clear what you mean

Also, limits for sets don't make sense ||category theorists shut||. You can already take infinite unions, so you don't even need some kind of limit.

As KnightWatch said, induction isn't the right way to think about it. Just take the (infinite) union and see what you get

[chilly ocean]

I wrote that by mistake, I meant integers under addition. Sorry once again

[coral spindle]

We just write Z for that.

[lusty marlin]

Yeah I didn't want to point out the set limit thing

Because the category theorists might pile on me otherwise

[tardy hedge]

What are we saying HomZ(Z/nZ, A) and An are isomorphic as?

[coral spindle]

Z-modules

[tardy hedge]

Thank you. I realized that rn because Hom isnt a ring if the two modules arent the same one

If function comp is the multiplication

Any element in HomZ(Z/nZ, A) is determined by where it sends 1 (a generator) in Z/nZ cause thats a cyclic group, which means that the maps have the form ka and then we showed before that this is hom iff na = 0

So then we can just identify each map with an element in An

Basically?

[coral spindle]

So construct the isomorphism and show it works

[chilly radish]

Even without category theory we can define the limit of a sequence of sets if the limsup and liminf agree

[coral spindle]

SHUT

Yeah something something pedagogy

I think iceball needs to appreciate that we don't have the same obstructions with sets as we do with the real numbers

[south patrol]

lol

[zinc plaza]

I was looking at the group of symmetries of the platonic solids and I'm confused about the order of the groups. For example Wikipedia says that a cube has 24 rotational symmetries, which I assume comes from 6 faces * 4 sides for each face = 24 rotations. But the rotation of a face is always equivalent to a reflection of another face
For example in this cube if you rotate the front face counter-clockwise by 2pi/4 radians it is exactly the same as reflecting the left face "downwards"
So if a cube is just one whole object, why do we count the symmetries of each face and not the unique symmetries of the whole object?

[coral spindle]

You are right to question this. This is a (imo nontrivial) consequence of something called the orbit-stabiliser theorem, and indeed the justification is different from the one you rightly pointed out is flawed.

[chilly radish]

I'm just being a smartass. I agree with you

[coral spindle]

Here is a more detailed argument.

Consider the set R of all rotations that fix a particular face f_0. Now for every face f_i, choose a rotation r_i sending f_0 to f_i.

Now, any rotation of the cube is of the form f_i \circ r (first doing r, then doing f_i) where r is some element of R. Furthermore, this pair (f_i, r) uniquely defines this rotation. This is a tricky thing to see, but try proving it.

There are 6 elements f_i that we chose, and R has 4 elements. So in total there are 6 x 4 = 24 pairs (f_i, r), so there are 24 rotations.

[zinc plaza]

So the reason to count it like this is so that we have "rotations" and "reflections" and not just a generic "transformation"?
Because if we were to count only the unique transformations we would have that rotations on some faces are equivalent to reflections on other faces, so we would either have to choose to talk about these transformations as rotations or reflections, or simply call them transformations and not specify the way the cube is actually being transformed
Is that right?

[coral spindle]

No it's not really anything to do with the distinction between rotations and reflections

The vital observation in my opinion is that we're not actually counting faces, but rather counting these maps f_i

Without getting into the details (because I don't know your background in group theory) it is essentially a coincidence that these are in correspondence with faces

The general orbit-stabiliser theorem elaborates on this.

[zinc plaza]

I don't know your background in group theory
Almost none. I'm studying algebra with Paolo Aluffi's book and I just started the chapter about groups. I know the definition of a group, the concept of order (both of the whole group and its elements), and some examples of groups like symmetric groups, cyclic groups and dihedral groups

[zinc plaza]

Okay. So I think I will understand all this soon, right?

[coral spindle]

I don't recall when Aluffi gets to this, but orb-stab will be covered eventually.

[zinc plaza]

Okay

[leaden heart]

The cube has 3 axes with 4fold symmetry, 4 with 3fold, and 6 with 2fold. So the total number of rotations is from 3*(4 - 1) + 4*(3 - 1) + 6*(2 -1) + 1 = 24. Remember, you can’t count the trivial rotation multiple times

Those types of axes go through the center of opposite faces, vertices, and edges, respectively

[kind temple]

are there two non-isomorphic groups of the same order with isomorphic subgroup lattices?

for example, Z/6Z and Z/15Z have isomorphic subgroup lattices (basically a square) but they are not isomorphic as groups

i was having trouble asking my question until trying to write it out here. a google search says no, but i dont quite understand it

https://math.stackexchange.com/questions/14588/does-the-order-lattice-of-subgroups-and-lattice-of-factor-groups-uniquely-det

unless i am misinterpreting the results of what was proved above

[leaden heart]

after working out a little, my guess is there no example for finite abelian groups, but there is a finite nonabelian example (i think they give one in that link)

[kind temple]

yes, the number of groups that i know and am familiar with is pretty small

and i was figuring that it wouldn't work with abelian groups

due to the decomposition theorem or whatever its called

[leaden heart]

yes, hmm, does taking the product of two groups produce a lattice that is a """product""" of their lattices, in some sense?

[kind temple]

sort of?

Z/6Z = Z/2Z x Z/3Z certainly does

[leaden heart]

ah, i just corrected a mistake i made working it out for Z_4 x Z_2

its not clear to me that lattices have unique decompositions under this """product"""

[coral spindle]

If they have no isomorphic nontrivial subgroups then yes, otherwise shit gets v complicated

For example the diagonal embedding of G in GxG comes up

There’s no easy correspondence

[leaden heart]

!!! oh

[kind temple]

mhm

thats neat

[coral spindle]

If f, g are homomorphisms G -> G then {(f(x), g(x)) : x in G} is also a subgroup. I’m sure you can see how this produces a lotttttt of issues

[kind temple]

how do the lattice structures change under direct sums and quotients?

[coral spindle]

We just call them products in group land

Quotients look nice though

[kind temple]

oh okie

[coral spindle]

See the “lattice isomorphism theorem”

[kind temple]

id imagine its just kind of like contractions of the original lattice

[rotund aurora]

Don't the dihedral group of the n-gon and Z/2nZ have the same lattice of subgroups? or maybe I'm clowning

[kind temple]

no, Z/8Z is a stick, but the dihedral group of the square is not

[rotund aurora]

ok take n odd maybe?

if n is an odd prime it is true I think?

[coral spindle]

No, it's the dual of subgroups. The subgroups of the quotient are in correspondence with the groups containing the normal subgroup you quotient by. This is exactly the same as the lattice of a subgroup being those elements lying below the subgroup

[coral spindle]

The dihedral group has loads of subgroups of order 2, but a cyclic group has at most 1

[rotund aurora]

mmh right

[coral spindle]

(Even if n is odd)

[kind temple]

you mean above it?

[coral spindle]

And although obviously the lattice does not 'see' order, all these subgroups will be minimal elements

[leaden heart]

perhaps, they are australian

[coral spindle]

Above the normal subgroup? Yes.

But that is what I said; subgroups containing the normal subgroup

I was specifically saying it is dual to the lattice of a subgroup

I did not make a typo.

[kind temple]

(a bunch of subgroups containing N)
|
N
|
(rest of subgroups)

[leaden heart]

yea, all those H's above correspond to a subgroup of G/N, that is, H/N

[tardy hedge]

Not sure why elements of Z2 (+) Z2 has unique representation as n1+n2 for n1 and n2 in Z2

[kind temple]

but how is this the same as a lattice of a subgroup being those elements lying below the normal subgroup

[coral spindle]

This is bad notation, they mean (n_1, n_2).

[coral spindle]

For a subgroup you look below, for a quotient you look above. They are the same computation, only the direction is reversed.

[kind temple]

are there two different lattices we are considering? would you mind clarifying this statement a bit more?

im not clearly seeing how the direction is being reversed

[leaden heart]

hold on, there doesn't exist a "dual group" of G, that has as its lattice G's lattice flipped over, right??

[coral spindle]

Not that I know of

[coral spindle]

Above is the reverse direction of below

The lattice of G/N looks like the lattice of G above N
The lattice of H looks like the lattice of G below H

Do you see how similar these are? They are completely the same situation, with the direction reversed, one might say.

[kind temple]

i thought you were saying that they were related

[coral spindle]

The lattice of G/N and N can be completely unrelated.

[leaden heart]

This gives you immediately that the lattice of everything above N and below H is H/N's lattice

[kind temple]

like, um. maybe i just have a bad understanding of what you meant by dual

[tardy hedge]

There is something im confused about, its relating to this

[kind temple]

oh, okay that is clear

[tardy hedge]

Relation between this (n1,n2…) product, and some sum of modules N1+N2 …

Im confused in relation to that Z2 example

[coral spindle]

Can you elaborate on what is confusing you

[tardy hedge]

Yes

[coral spindle]

I hope I don't need to prompt you to do so!

[tardy hedge]

😂

Sorry one moment, im trying to figure out exactly what is confusing me here 😅

[coral spindle]

OK take your time

[zinc plaza]

That makes more sense. But how would you do reflections in a similar way?

What does it mean to reflect a 3D polyhedron?

[leaden heart]

this is a great question: all distance preserving transfomations either keep orientation or reverse (you can think of it as, the determinant is either 1 or -1). It turns out it 3d, all orientation preserving maps are simple rotations (rotate around some axis by some amount), but the orientation reversing maps are NOT all simple reflections (reflect over some plane), you get some more by reflecting, and then rotating

for example, the negative of the identity matrix is one such orientation reversing map

you can think of it as reflecting (over any plane, actually), and then rotating 180 degrees around the axis perpendicular to the plane

anyway, in any of these symmetry groups, there must be exactly as many orientation reversing maps as orientation preserving, because "taking the determinant" is a group homomorphism from the symmetry group to {1, -1} under multiplication

(or rather, either that, or that homomorphism is trivial, i.e. all your symmetries are rotations)

[zinc plaza]

Do you think that it would make sense to think of 3D reflections in a 4D space?

[leaden heart]

yes, its exactly like a 2D reflection is also a 180 rotation around the line, if you think of that plane as inside of 3D, yes

if a matrix A represents an orientation reversing transformation, then the square matrix:
[A 0]
[0 -1]
is orientation preserving, of one higher dimension

also, a tangent, in 4D you have more than just simple rotations: basically, you can rotate in the xy plane and the zw plane simultaneously, and those two angles dont need to be related

[zinc plaza]

Interesting

[zinc plaza]

Imagining such thing is very hard

[leaden heart]

it is, i will try to draw a visualization

[zinc plaza]

What I'm imagining is that it is equivalent to opening the cube like this and then closing it again but with the faces in a different orientation
And the different reflections are just different ways of opening it. But idk if that makes sense

Basically inverting the cube. And depending on which sides you choose to be on the long part the result will be different (different faces going to different places)

Is this what a reflection of a 3D cube in a 4D space would look like?

[leaden heart]

here, the cube is rotating around its bottom face (yes, its weird it rotates around a plane instead of an axis), where color represents the 4th dimension

[zinc plaza]

I see. So those two faces are being swapped by kind of "compressing" the cube (from a 3D perspective)

[leaden heart]

yes, right in the middle which i didnt draw, the cube looks "flat", it doesnt extend in the z axis at all. But keep in mind, the cube at the beginning was flat in the w axis. Its inherent shape is not changing, just its orientation

[zinc plaza]

Like if a 4D being did this you would see the cube flattening like that

And then becoming a cube again

[zinc plaza]

Yeah that makes sense

So there are only 3 ways of doing this using the faces?

3 reflections with the faces, 4 reflections with the edges and 4 reflections with the axes?

[leaden heart]

there are 6 reflections like this:

[zinc plaza]

Oh right

6, not 4

[leaden heart]

and three parallel to a face, there are rotoflections (i think thats the technical term) over planes perpendicular the longest diagonals of the cube

but not normal reflections

reflect over the red hexagon, then rotate by 1/6 of a turn, thats a symmetry

there are four such hexagons, and three rotoflections for each one (1/6, 3/6, 5/6 of a full turn), but actually all of those 3/6 of a turn end up the same, the negative identity

[zinc plaza]

So there are also symmetries that are a reflection in 4D space + a rotation in 3D space?

[rotund aurora]

quick sanity check: If M is an A-module and M-->M is an isomorphism, the induced map M_p-->M_p where p is some prime is also an isomorphism right?

[zinc plaza]

And these symmetries aren't equal to any symmetry that is only a reflection in 4D, or only a rotation in 3D?

[leaden heart]

when i said "reflection" i meant reflection in 3D, you can think of this as a rotation in 4D if you like

[coral spindle]

Yeah, this follows from exactness right? Assuming you do mean localisation here & not something I’m unfamiliar with

[zinc plaza]

Right, I meant a reflection in 3D but a rotation in 4D

[rotund aurora]

yeah right, also follows from exactness

you can like check it by hand too lol, but just wanted to make sure I wasn't missing something

[coral spindle]

I reckon you’re right

[leaden heart]

theres something easy to confuse here, when we say "rotation" sometimes we mean the orientation we ended up at and sometimes we mean what action we took to get there (which is, mathematically, a path through the space of all possible orientations)

[leaden heart]

yes, then, exactly, you cant write these combination rotation + reflection is any simpler way

(you can reflect and then rotate in 2d too! it just ends up being the same as a plain reflection, over a different axis)

[zinc plaza]

I see, that makes sense. I was confused because I thought these rotations in 4D wouldn't add up to 24, which we need to have the octahedral group

But there are also symmetries by 4D rotation (3D reflection) + 3D rotation

[leaden heart]

yes, that exact thing confused me for long time, thats why im good at explaining it

(not to toot my own horn, lol)

[zinc plaza]

lol

This is not easy to count at all. Finding these symmetries is an exercise from the book I'm reading and I thought it would be much easier than imagining all this 4D stuff

Is there a better way to count these symmetries?

[leaden heart]

oh well, for this, i dont know if thinking about the reflections as 4d rotations does any good, but, whatever

like i said, the group theory gives you there must be the same amount of reflections (in the broad sense) as rotations, so you only need to count the rotations

[zinc plaza]

But why is that the case?

I can see why in 2D but not in 3D without thinking about 4D rotations

[leaden heart]

the determinant, from our symmetry group S:
det : S -> {1, -1}
is a group homomorphism ( det(AB) = det(A)det(B) ), and ker(det) is the rotations

(sorry if this language is too technical btw) the cosets of ker(det) are all the same size, and map to different things

so either, ker(det) is everything, because everything maps to 1, or ker(det) is exactly half

(i use "reflection" to mean any orientation reversing transformation here) the order of a subgroup divides the order of the group, so the only other possibility is 1/n of the transformations are rotations for some n > 2. lets say n = 3, so theres twice as many reflections as rotations. pick a reflection r, then consider the function f from S to S given by s -> r*s. this is bijective (always true for groups), but because there are more reflections than rotations, there would have to be a reflection r_2 where r*r_2 is also a reflection. but this cant be true

[zinc plaza]

Ahh

Well, I can't understand most of this explanation/proof since I don't know what some terms mean, and I don't know about some of these facts. But it's fine, I'll take your word on it and I will most likely know about all this soon

[zinc plaza]

I have one last question though. Why rotate by 1/6 of a turn here. I can kind of see why, given the hexagon, but from what I'm seeing in the picture if you rotate it by 1/6 of a turn (by this I assume you mean 2pi/6 radians), no matter the axis, you will change the cube's orientation and it won't be a symmetry

What axis are you using to rotate the cube?

[leaden heart]

view that plane head on and the cube looks like this:

[zinc plaza]

Oh I see it now lol

Wait, this is only 3 fold symmetry

Why can you rotate by 2pi/6?

[leaden heart]

if you reflect over the plane we are looking at it looks like:

so the reflection isnt a symmetry, but rotate it by n/6 of a turn for an ODD n after, and the picture looks exactly like you started

[zinc plaza]

Oh

Yeah that makes sense

It really does

Thanks!

I think I have no more questions now

[leaden heart]

np, this was fun, ask me anything about this stuff anytime lol

did you know there are five platonic solids in 3D, and theres like a 4D version of each one... but there is also exactly ONE extra regular 4D shape? and its cool

[zinc plaza]

I didn't

What is this extra one?

[leaden heart]

ok so, take a cube, with the coordinates of the vertices at (±1, ±1, ±1), if you add some more vertices at (±2, 0, 0), (0, ±2, 0), (0, 0, ±2), you get a cool but not completely regular shape:

(look up rhombic dodecahedron for a better picture lol)

but the nutty thing is, do the same thing in 4D, and it IS completely regular. in particular, the points (±1, ±1, ±1, ±1) and (±2, 0, 0, 0) are the same distance from the origin, the center of the shape (they are both 2 away!)

i am like, legitimately upset i dont have 4D eyes to look at it, because im sure my nerdy self would find it very beautiful

[zinc plaza]

lol

Yeah, 3 dimensions are too few

It would be much more fun if we had 196,883 dimensions

[leaden heart]

it would be phenomonally easy to get lost in the woods, though

[zinc plaza]

Imagine 196,883 dimensions in hyperbolic space

[leaden heart]

hmmm, no thanks

[zinc plaza]

Also I saw that orbits (the ones in physics, not the group theory term) aren't stable above 3 dimensions so the universe would be just full of black holes because everything nearby collapsed

[leaden heart]

oh yes, i was gonna say so, though, maybe there is a completely bizarre and alien physics in high dimensions that could have complex life in it

[zinc plaza]

I have to go now, bye

[leaden heart]

same here actually bye!

[tardy hedge]

But theyre talking about uniqueness of the sum n1+n2

Before it was talking about if every element of M can be written uniquely as n1+n2, then the external direct sum (direct product) is the same as internal direct sum

I just do not understand the example with Z2

This is my understanding of what was being talked about previously in the book:

My understanding so far is this, i want to try to be as thorough as possible with details: for R-modules M1 … Mk, the external direct sum is those ordered pairs (m1, … , mk), which is again an R-module with action defined component wise.

The question after that is, when is an R-module M equal to an external direct sum of some of its submodules? Well, if every element of M can be written as a sum of elements in the submodule, M = N1+N2+…, then in fact M is equal to the external direct product of these submodules if the Proposition 5 is satisfied, i.e every element of M can be written UNIQUELY as a sum of elements in the submodules, in this case the internal direct sum is the same as the external direct sum, so use same notation (M1 (+) M2 (+) … same notation as using cross M1 x M2 …)

For the Z2 example, the elements of N1+N2 (Z2+Z2, which is just Z2) do not have unique representation. 0 = 0+0 = 1+1

Which is why im confused when it says Z2 (+) Z2 has unique representation of the form n1+n2 for n1 , n2 in Z2

[sly crescent]

Seven, actually

There are ten Schläfli-Hess polytopes but only four Kepler-Poinsot solids

[chilly radish]

What is "seven" in reference to

If you define platonic solid as a regular convex polyhedron there are only 5

[quartz wind]

is it asking me to prove existence of $\text{proj}_S$?

[cloud walrus]

esca (@ with reply)

[quartz wind]

or is there a simpler way i can do it

[chilly radish]

There's no need to do that

[quartz wind]

i feel like considering the content given in the book so far, projections might be too high tech. id rather use a more direct argument like "every limit point of S is also in S" rather than doing the complement hint given

[chilly radish]

And this is true of general normed spaces without a need for orthogonal projections to exist

[quartz wind]

hmm yeah makes sense

alright thanks

[chilly radish]

That would work too

[lilac sluice]

hey guys, I really hope someone can help me...

the Galois group $Gal( w, \sqrt^3 2018 )$ where $w=e^{2\pi i/3}$ is isomorphic to what? And what are the automorphisms? I think the automorphisms are $\sigma (\sqrt^3 2018)=\sqrt^3 2018 , \sigma (w)=w^2$ and $\tau (\sqrt^3 2018)=w \sqrt^3 2018 , \tau (w)=w$ but I just know that that dont commute and $\sigma$ has order $3$

[cloud walrus]

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

[lilac sluice]

it was cubic root of 2018 😦

[quartz wind]

$\sqrt[3]{2019}$

[cloud walrus]

esca (@ with reply)

[lilac sluice]

the Galois group $Gal( w, \sqrt[3]{2018} )$ where $w=e^{2\pi i/3}$ is isomorphic to what? And what are the automorphisms? I think the automorphisms are $\sigma (\sqrt[3]{2018})=\sqrt[3]{2018} , \sigma (w)=w^2$ and $\tau (\sqrt[3]{2018})=w \sqrt[3]{2018} , \tau (w)=w$ but I just know that they dont commute and $\sigma$ and $\tau$ have order $3$

[cloud walrus]

andrei_doronin

[lilac sluice]

ok I found my mistake
you have to ignore me
thank you all

[sly crescent]

but there is also exactly ONE extra regular 4D shape?
Without reference to convexity

[naive lance]

if we want to construckt a quotient ring of a ring ( commutative with 1) such as Spec(R/I) = MaxSpec(R) I is the Jacobsen radical right?

[tardy hedge]

Is the fact that any free module F with basis A is isomorphic to that "universal property" free module on A construction F(A) actually ever useful?

I'm talking about defining the free R-module with basis A as that F(A) = set functions f: A -> R with f(a) = 0 for all but finitely many a in A

[naive lance]

Yes there are proof where you need finiteness

And these functions

I can understand why it seems to be very unhandy in comparison to express the Module as isomorphic to the direct sum over some R quotient rings

But having already given functions can help a lot if you want to construct morphisms with codomain M

[median ether]

hope this channel is correct place for such questions from elementary cyclic groups topic. is showing that -a is generating the same subgroup as a, but in reverse order is a right way to show <a>=<-a>? also, is k>0, k<0 section is right?

[eager mango]

Hello, I'm working on ECDLP, I need help to define intermediate group between two group.

If anyone interested please DM me.

[glad osprey]

I don't know why you went from talking about <a> to <a^k>. a is by definition a generator of <a>. I think the quickest way to prove this is using the characterization of <a> as the smallest group containing a

[languid trellis]

I'm struggling to get started with the second part of this exercise. I've reviewed all the definitions involved and I can't get anywhere. My best thought is that all maps in Hom(Z/nZ, A) are of the type given in the first part of the exercise, and then we would have a 1-1 correspondence between An and Hom(Z/nZ,A). Also, An is a submodule of A considered as a Z-module so maybe I should apply some isomorphism theorem? I'm not entirely sure how to proceed. Any help will be appreciated (:

[mighty kiln]

Consider where 1 gets mapped to

[languid trellis]

by 1 you mean the identity in Hom(Z/nZ, A)?

[mighty kiln]

1 ∈ Z/n

Yea the correspondence is correct

[languid trellis]

I imagine it'll get mapped to a \in A just ebcause of the given phi

[mighty kiln]

Yup

[languid trellis]

oh and 1 getting mapped to a guarantees that na =0

which is what we need for Z/nZ -> A to be a z-module hom

so the a to which 1 gets mapped to determines our z-module hom

and this a is then in An

and thats a z-module isomorphism

[mighty kiln]

Yea

[languid trellis]

oh this makes sense because 1 generates Z/nZ so the image of 1 is enough information to fully determine the homomorphism

[mighty kiln]

Yup

This is a good way to think of homomorphisms if you know the generators and relations

A homomorphism R → S is just a choice of an element in S for each generator of R that is compatible with the relations of R

Same for groups

[languid trellis]

ya

[chilly ocean]

I was able to get through the first bit. I am stuck on the second one

I know that as O_i s behave as blocks G_a \le G_O_1

So |G_O_1||O_1|=|H| right?

Now if this is true then why is |H:G_O_1|=|H:H intersection G_a|

[tardy hedge]

bruhh i just did that exercise

are you also working through Ch10 dummit and foote?

[languid trellis]

Yep

[tardy hedge]

Nice!

10.4 on tensors looks like a lot

im wrapping up section 10.3

[languid trellis]

I'm on 10.2. the section on tensors and exact sequences looks like a lot of work

but it's exciting, lots of applications to commalg and of course canonical forms

very much looking forward to it

[tardy hedge]

Me too!

[small bramble]

bruh section 10.5 is so long

[tardy hedge]

i thought 10.4 was so long

tensor stuff looks like it goes on forever

[small bramble]

yeah that part is p long too

[tardy hedge]

are you also studying ch10?

[small bramble]

I've seen it before but I'm reviewing baby stuff rn since it's been a while

[tardy hedge]

Noice

algebra epic

[small bramble]

Ye

[tardy hedge]

Ye

analysis poo

[small bramble]

aw I like analysis too

I don't have much intuition for it but I like that

[tardy hedge]

im so bad at it and i also dont care

so its not a good combo

[small bramble]

rip

[ornate tiger]

why e is in every H_i

[coral spindle]

Every subgroup, by definition, contains the identity. I hope I am not revealing new information to you by saying this

[coral spindle]

Like I said, they are using bad notation. They really mean (n1, 0) + (0, n2).

The external direct sum A (+) B, as a set, is the Cartesian product A x B. It is the internal direct sum of A x {0} and {0} x B, which are submodules of A (+) B.

[ornate tiger]

yeah but is e in H_1 the same as H_2 ?

like in i) there is e i think it's (1)(2)(3)(4) but what about the one in ii) is it (1)(2)(3)(4) ? or something different

[coral spindle]

They are all subgroups of A_4. The identity is the identity of A_4.

[ornate tiger]

the identity of A_4 is the same as S_4 right ?

[coral spindle]

A_4 is a subgroup of S_4.

[ornate tiger]

yeah

[tardy hedge]

I think there is some sort of disconnect between what youre saying and dummit and foote

[coral spindle]

I assure you there is not

[tardy hedge]

Yeah probably not but something is still weird to me

[tardy hedge]

and then this is the proposition they mention

KK wait I may understand now what happened

[tardy hedge]

Oh .. so every external direct sum is an internal direct sum of some of its submodules, in that way…

[coral spindle]

In fact, the internal and external direct sums are isomorphic.

[tardy hedge]

I guess im getting confused by that whole Proposition above, about when an external direct sum IS the internal direct sum ???

[coral spindle]

They are always the same, up to isomorphism.

[tardy hedge]

I think im also getting confused by the N1+N2… notation

[tardy hedge]

The proposition makes me think that they are not always the same, im trying to figure out what im missing here

Honestly whatever lmao

[chilly radish]

One can form the external direct sum even when the submodules do not form an internal direct sum

[chilly ocean]

hello im looking for papers on ring theory anyone have a link or a tip where i could find them?

not anything in particular it the field just grabbed my attention by pure chance

however thank you for recommending the site

thank you very much

[coral spindle]

They are always the same. Moreover, they are the same with respect to the way that A and B embed.

A submodule A of C is really a module A and an injective map A → C

Let's say C is the interal direct sum of submodules A and B. Then there are injective maps A → A x B and B → A x B. It should be clear what these are.

Then this diagram commutes:

(& similarly for B)

Furthermore the map A x B → C defined by (a, b) |-> a + b is an isomorphism.

Idk why I'm saying this really, this is all in the proposition.

Every direct sum of two modules A, B is isomorphic as an overmodule of A and B

Another perspective: every two dimensional vector space over a field (i.e., every v.s. which is a direct sum of subspaces which are isomorphic to k) is isomorphic to every other such space.

[cobalt heath]

Never seen this diagram, what is this for?

[south patrol]

I guess the perspective I have is well, given two submodules A, B of X there is a canonical addition map A (+) B -> X. The image is by definition A + B. To say the sum is direct is to say the canonical map is an iso onto its image

Very similar to how given a collection of elements v1,...,vn of a vector space V over a fjeld k, there's a canonical map k^n -> V sending like "formal" addition to addition of the x_i within V. The image is, by definition, the span of the x_i, and the x_i are said to be linearly independent if this map is injective . In particular the x_i form a basis iff this map is an isomorphism

I guess what I mean is that like, external direct sum is a natural thing you can just do to any two things, and then an internal direct sum is essentially what you get when that can be "modelled" "within" a big module

[versed light]

Can someone tell me if Z[x] is isomorphic or homomorphic to Z[i]?

[rocky cloak]

They're not isomorphic, but there is a surjective homomorphism Z[x] -> Z[i]. Don't know if that's what you mean by 'homomorphic'

[quartz wind]

is that polynomials with integer coefficients?

[versed light]

Yes thats what i meant, thanks

[versed light]

Yes

[versed light]

Im having trouble finding kernel of the evaluation function after defining it, i defined it to be $$\varphi (f(x)) = f(a+bi) , \varphi : \mathbb{Z} [x] \rightarrow \mathbb{Z}[i]$$
then kernel would be \ \
Ker$\varphi = { f(x)\in \mathbb{Z}[x] : \varphi (f(x)) =f(a+bi)=0 }$
\
\
So if a function is in the kernel it should get divided by x - ( a+bi) but the ideal (x-a-bi) isnt an element of Z

[cloud walrus]

Cyrenux

[rocky cloak]

That's right, so the answer would depend on whether b is 0 or not.

A possible hint could be that if x^* is the complex conjugate of x, then f(x^*) = f(x)^*

[versed light]

So you are trying to tell me complex conjugate is the root too, and by that x - (a-bi) also divides the functions that are in the kernek

Since every function in kernel gets divided by both (x-(a-bi)) and (x-(a+bi)) they must get divided by (x² - (a² +b²) ) which is in set if integers

Thanks

[versed light]

What is the optimal way to show that factor field Q[x]/(x²+1) is isomorphic to Q[x,y]/(x²+1,y-1)

[coral spindle]

First iso

[south patrol]

Well the latter is the first where you add another variable y and then you set y = 1

So hopefully that makes it clear which map you should consider between the two

[marsh scaffold]

What is the statement in the a bit even saying

[mighty kiln]

Let S be the set of left K-cosets

[marsh scaffold]

right?

[mighty kiln]

There is a group action of H on S given by h(xK) = (hx)K

[marsh scaffold]

Shouldn't it be left?

Oh okay

[mighty kiln]

Yea sorry

An orbit of this action is a subset T ⊂ S of K-cosets

Consider the union of all the K-cosets in T

This is an H,K double coset

[marsh scaffold]

Ah

Now it's a lot clear

Thanks @mighty kiln

[versed light]

I believe this suggests me to define the evaluation function from Q[x,y] to Q[x]/(x²+1) but i have no idea how to express image of an element

[versed light]

Nvm it would be the converse? $$ \phi : \mathbb {Q} [x] \rightarrow \mathbb{Q}[x,y]/(x^2+1,y-1)$$
$$\phi(f(x)) = f(x,1) + (x^2+1, y-1)$$

[cloud walrus]

Cyrenux

[ornate tiger]

why for e.g (13)A_4 is not a coset ?

[tribal moss]

It's the same as (1 2) A_4.

[ornate tiger]

so we can also say it's a coset not just these 2

[south patrol]

Well it's a coset by definition

[ornate tiger]

when its not a coset

[tribal moss]

We can say it's a coset and it is one of these two. Namely, it is (1 2) A_4.

Everything of the form [some element]·[some subgroup] is by definition a left coset of the subgroup.

[ornate tiger]

ok thx

[fossil shore]

Hey, I'm following Dummit and Foote, and I need to read up on finite fields in chapter 14. I've only read chapter 13.1 on field extensions. Could someone give an idea of which sectins from chapter 13 and 14 finite fields depend on?

[burnt sandal]

Anybody have any good recommendations for video lectures for an intermediate algebra course? I've taken undergrad abstract 1, but we used Hungerford's book so it was more focused on ring and field theory with the basics of groups at the end. I'm looking for a playlist or channel that would be equivalent to an undergrad abstract 2 course or grad level intermediate algebra course.

[marsh scaffold]

Hey, I have been struggling with this excercise

Also as G is not assumed to be finite here I am guessing we have to work with cosets

Also why is the relatively prime part important

[tardy hedge]

Is one reason why its called “free” module, is that for any other module over the same ring, there is always a module homomorphism between them?

Like a nontrivial one or whateve

And u have like free control over it

Send the basis elements of the free module to whatever u want in the other module and then the homorphism extends from there

So ur free in the choosing

[marsh scaffold]

Hey, as far as I know that's mostly the universal property of free objects

Like even free groups

[tardy hedge]

I see yeah

So thats just what it means to be free in general

?

[marsh scaffold]

Yeah

[tardy hedge]

I see. Ty

[coral spindle]

Think about the orders of A and B in terms of their prime factors. Remember also that AB will have subgroups A and B hence its order is divisibile by the orders of A and B

[coral spindle]

I think of it as free as in unconstrained. You have a set of generators, and you produce the most unconstrained object that contains things with those names. Same thing with free group etc.

The universal property is a better way to think about these things, though, so perhaps your internal justification is better.

I don't know the historical motivation for the name but I would put money on it being one of the two.

[marsh scaffold]

Ah but there's a problem through we don't know that A,B are finite?

[coral spindle]

Good point, I overlooked that. Nonetheless a similar idea holds. Look at the index of AB in G instead.

You may know some relationship between the indices of A and B in G and the index of AB in G

Not an enormously strong one, but...

[marsh scaffold]

Suppose I didn't...say is there a way to directly do the problem via group actions?

I am not very good at remembering most formulas tbh

[coral spindle]

I don't see a direct way to use group actions here.

Hint: let A <= B <= C be groups. Then every coset of A in C is contained in exactly one coset of B. So how many cosets of A are there if we know the number of cosets of B in C and the number of cosets of A in B?

[leaden heart]

I think you understand this already now, but certainly some extra condition is needed, otherwise just take A = B =/= G

[marsh scaffold]

I get that |G:AB|<= |G:A||G:B|

[coral spindle]

That's too strong

Or rather

lol

That's just not what we need

think about divisibility, not size.

If you see the word "coprime" in a question, you should think about ordering by divisibility, not by the usual ordering on the integers.

[marsh scaffold]

Okay....

Hmm

[coral spindle]

You remember the definition of coprimality I hope? It states that the only natural common divisor of two numbers is 1. I.e., if p, q are coprime and n | p and n | q, then n = 1 (assuming all naturals). You can probably imagine what number n we need to use here...

[leaden heart]

Actually, why is |G:AB| an integer? Because you don’t know (until the proof is finished) that AB is a subgroup

[marsh scaffold]

Ah fair

[coral spindle]

Oh wow that is a problem huh

[leaden heart]

Yeah it’s annoying, that proof you were alluding to seems slick and natural

[coral spindle]

I had just assumed that we could argue it was a subgroup by some easy argument but they're not giving us much to work with

So I mean, <A, B> = G definitely

By my argument lol

But now let me think...

Do we need double cosets or something

I don't know shit about double cosets

Yeah OK I think a double coset argument works. A double coset in A\G/B is something of the form AgB. Since A\G and G/B are finite, we can argue that |A\G/B| is also finite. And I think we can argue again some divisibility thing. No spoilers lmao.

I might imagine that the context is about double cosets, if not I don't know.

[marsh scaffold]

Well this was in the subgroups section

[coral spindle]

@leaden heart what do you think, does that sound feasible?

[leaden heart]

ah, distinct double cosets are disjoint, so it does work I think

[coral spindle]

I think the trick is the divisibility part and I haven't checked the details very thoroughly

[marsh scaffold]

I would like to know in on the details 😅

[coral spindle]

OK fine, the trick is we just show that |A\G/B| divides |G:A| and |G:B|

Then it must be 1

My idea is that every double coset AgB contains the same number of cosets Ah and h'B

Ah sorry what I mean is

The number of cosets of A in AgB and Ag'B are equal

ans similarly for cosets of B

[marsh scaffold]

There are cosets of double cosets

I'm sorry what are cosets of A in AgB?

[leaden heart]

in, as in a subset of

[marsh scaffold]

Oh ok

Okay i would like to suggest that isn't the double cosets just the orbit of a certain group action

Like action of A on the cosets of B recognised as a.cB = acB

[coral spindle]

Yes they are

They are an action of A x B on G

[marsh scaffold]

Ah

[coral spindle]

orbits of* an action

I think there may be some way to argue the divisibility condition via group actions, I'm not sure. I can't see it immediately without writing things down.

[marsh scaffold]

I would like to suggest How about we consider the natural map from G:A intersection B to G: A x G:B

gAnB mapsto (gA , gB)

?

I believe it's well defined?

[leaden heart]

Bs action is backwards (composition is backwards), so it’s like, A x B^(dual) (however you write that)

If you consider it as a single action, it’s neither a left nor right action, it does one of each on each component

[marsh scaffold]

I see

[coral spindle]

N.b. you can force it to be forwards by working by inverses on B. So for example (a, b) acts on g sending it to agb^-1, and this will work

[marsh scaffold]

Interesting

[leaden heart]

oh that looks natural anyway, as an generalization of conjugation

[marsh scaffold]

I have noticed that this is a injection too?

And the sets in consideration are finite

So it's automatically a bijection?

[leaden heart]

since there can be gA =/= hA where gA n B = hA n B, I don’t think it is well defined

But I’m not actually sure what the domain of your function is

[marsh scaffold]

It's just the cosets of A n B
For well definedness
if aAnB = bAnB then ab^-1 in both A and B

So aA and bA are same?

Similarly bB aB

[leaden heart]

oh I was thinking (gA ) n B not g(A n B), my bad

[marsh scaffold]

It's in similar vein to G acting diagonally on the set G:A x G:B whose satibilizer coincidentally is AnB i suppose

[leaden heart]

“stabilizer of x” is the subgroup that maps x to itself, is there another word for the things that stabilize every element?

my algebra vocabulary is rusty and cringe

[marsh scaffold]

I meant the one that stabilizes (eA,eB)

[marsh scaffold]

I want to primarily consider this mapping

[marsh scaffold]

I was thinking that since |G:A| and |G:B| are coprime their LCM is their product?

I forgot to turn off the ping sorry

Also as boytjie said we would like to use divisibility

Is it the case that G:A divides G:AnB

I think so yes

[leaden heart]

You are doing it “backwards” from how we suggested

G:(A x B) divides G:A, and also divides G:B

[marsh scaffold]

I'm sorry to bother By G:(AxB) we mean ?

[leaden heart]

It’s swirling not fully formed in my head, but, | {AgB} |

[marsh scaffold]

Size of that double coset...okay...then by our argument were saying that this size is 1?

[leaden heart]

Yes, which means it only contains one element, which you can write AeB, and the cosets partition G

So AeB = G

[marsh scaffold]

I see

[marsh scaffold]

Okay now how is this done then ? (Consequence of some kind of orbit stabilizer?)

[leaden heart]

No wait hold on, not the size of the coset, the amount of double cosets there are! My bad for ambiguity

[marsh scaffold]

How exactly are we showing the divisibility part

[leaden heart]

Ag is contained in AgB, for sure, and… wait

I don’t see why other right cosets of A partition AgB, maybe this doesn’t work

[marsh scaffold]

Okay

[leaden heart]

As far as I can see my argument doesn’t work (unless you have one of the groups normal)

Oof

[marsh scaffold]

[leaden heart]

no no no this works, suppose Ah intersects AgB, then a_1h = a_2gb, so Ah is contained in AgB, because ah = aa_1^-1a_2gb

[marsh scaffold]

I see

Then pls proceed

[leaden heart]

your pfp is too distracting

I thought that was the only part you didn’t understand, so now we’re done

to be fair, you have to believe every double coset AgB is partitioned into the same number of cosets of the form Ah

[marsh scaffold]

Hmm okay

But I need more familiarity with double cosets i guess

[leaden heart]

I… really hope that doesn’t turn out to break

[marsh scaffold]

Also maybe this map leads to sth cause it seems very strong

But that's for later purposes

Thanks @leaden heart and @coral spindle

[wraith swan]

Anyone got a hint on how to prove monomorphism from a trivial kernel.

Let ker φ = e and let φα = φβ
=> ker α = ker β
How to prove α = β?

[tardy hedge]

I havent started working out stuff for this Q yet but … im not sure how quotienting with maximum ideal and using vector space stuff helps

Im not sure of the intuition here

Or im just not sure of the bigger picture of how this solution is gonna work

[safe pendant]

what book/resource do you people recommend for free, projective and injective modules? i want something full of examples

[leaden heart]

I assume because the quotient of R by a maximal ideal is a field? and (R/I)^n = R^n/I^n

[rocky cloak]

I feel like the hint for this exercise is just the full solution. Like there isn't really more for you to do.

[tardy hedge]

Oh ok lol, i mean im at the point tho where i dont yet see how the hint is gonna be used to solve it yk? Im noob

But i will just try to do it and see what happens

[quartz wind]

under what operations can $\mathbb{Z}_2$ be a field? i tried xor as addition and xnor as multiplication and its already unclear to me whether or not that distributes

[cloud walrus]

esca (@ with reply)

[quartz wind]

unless it means $\mathbb{Z}/2\mathbb{Z}$?

[cloud walrus]

esca (@ with reply)

[quartz wind]

oh

and thats a field since 2 is prime right

oh yeah that would make a lot of sense

thanks

[sly rain]

Maybe the general theme is translating a problem into a nicer environment where you can assume certain things that you could not have assumed before

[rocky cloak]

It's indeed Z/2Z which is the same as XOR for addition and AND for multiplication.

[quartz wind]

i was just about to say AND doesnt have an inverse and then felt really stupid

[rocky cloak]

Yeah division by 0 is though

[quartz wind]

yeah

anyway thanks for pointing that out

[glad osprey]

Use the homomorphism properties of φ. ker α = ker β doesn't mean anything, since α and β are elements

[wraith swan]

No α,β are two arbitrary morphisms

[glad osprey]

Ah, φα is composition?

[sly rain]

Are you pressed on using trivial kernel? Maybe you can use injectivity

[wraith swan]

Yes it is, sorry

[chilly radish]

I don't remember the pages, I think it's all in the same chapter in rotman's introduction to homological algebra. It's very comprehensive

[wraith swan]

Well that's what I want to try. So I don't want to prove injectivity -> monomorphism

[safe pendant]

thanks.

[chilly radish]

A lot of the examples are in the exercises

[sly rain]

You want to show that the 3 are equivalent?

[quartz wind]

what about proof by contrapositive

[wraith swan]

Correct

[glad osprey]

I think you can prove trivial kernel => monomorphism basically the same way as trivial kernel => injective. Just start with (φ o α)(x) = (φ o β)(x)

[wraith swan]

Ok, i will try some more

[tender wharf]

we have injectivity <=> monomorphism

in Set

[twin sand]

Can someone help me with Galois Theory?

[delicate bloom]

better to just ask your question than ask if someone exists

[twin sand]

I am having trouble with item d)

I dont know where i have to use the hint

[rocky cloak]

zeta_n = cos(2pi/n) + i sin(2pi/n)

So if you can prove cos(2pi/n) is in the left hand side, you're done

[twin sand]

Yes yes

But i dont know how to prove it

[wraith swan]

it is so easy... I am just stupid

[twin sand]

Someone know?

[barren sierra]

only thing I have left for this is showing that $\ker [a, b] \subseteq \text{im} \begin{bmatrix} b \ -a \end{bmatrix}$. So basically I want to show that if $ax + by = 0$ then there exists $r$ such that $ \begin{bmatrix} br \ -ar \end{bmatrix} = \begin{bmatrix} x \ y \end{bmatrix}$ and I'm not sure how to get that

[cloud walrus]

Spamakin🎷

[leaden heart]

just solve ax + by = 0 for y to get y = -a/b * x, which is exactly what you want (reparametrizing by letting r = -x/a, if you like)

[barren sierra]

But how can I divide by b

It may not be a unit

(R is a commutative unitial ring)

[leaden heart]

ah, a is not a zero divisor

yes im dumb and didnt read

hmm, in a UFD, the fact ax = -by has a unique factorization implies ax = -by = abr (a and b are coprime because b not a zero divisor in R/(a) ), i dont know if this line of thinking can be modified to work here

[south patrol]

Hm well if ax + by = 0 then reducing mod a we have by = 0 mod a

So y = 0 mod a

so y = az and z is unique since a is not a zero divisor

[leaden heart]

y is a zero divisor, not necessarily 0 mod a? but might still work

no im stupid lol

[south patrol]

b is not a zero divisor mod a

Which is why y must be 0 mod a

[leaden heart]

yea, thats the stuff

[south patrol]

This is also a fun example of the "Koszul complex" btw

Like this isn't smth random lol

If you have a sequence of central elements x_1,...,x_n in a ring A with like x_k not a zero divisor in A/(x1,...,x{k-1} then this is called a regular sequence in A.

Given any finite sequence of central elements you can form the Koszul complex K(x1,...,xn), and it is a theorem that this is a free resolution of A/(x1,....xn) if x1,...,xn is a regular sequence

This is the case that n = 2

However fortunately this case is more elementary - usually you prove it using a little homological algebra

[leaden heart]

ahhh, I remember Koszul complexes from the thesis i never finished

[south patrol]

Oh nice

[manic cairn]

ok simple question about Integral domains and UFDs.
On Integral domains: prime => irreducible
On UFDs: irreducible => prime, thus prime <=> irreducible since UFDs are Integral domains

Now my prof says "an Integral domain is a UFD <=> every element is a unique product product of irreducible numbers". The rhs is just the definition of UFD but with irreducible instead of prime.
In the proof "<=" it gets proven that irreducible => prime.
Doesnt this just mean that prime and irreducible are equivalent on Integral domains? And wouldnt then Integral domains and UFDs be the same thing? Or does that implication only work if the rhs of the statement is true in that way?

[topaz heart]

well you still need the rhs to conclude irreducible => prime right?

so it's stronger than just being a domain since this rhs condition isn't true for any domain

in general domains you are right there are irreducible elements that are not prime

[manic cairn]

im looking at the proof to see where the rhs is invoked

ok yea without it the proof wouldnt work, so only if that statement is true the 2 are equivalent

[untold basalt]

Suppose that H,K and HK are subgroups of G. How can I prove that H and K have trivial intersection?

[coral spindle]

Well you'll struggle to prove something that's false in general

H = K = G is a counterexample

[twin sand]

Can someone help me?

[untold basalt]

oh so HK can be a subgroup without being a direct product?

Because direct product would require trivial intersection?

[topaz heart]

in an abelian group for example HK is always a subgroup of G for subgroups H and K of G

so take any two subgroups of an abelian group with nontrivial intersection for a counterexample

but in general HK will not be isomorphic to H cross K yeah

[untold basalt]

Allright, now I got it

[coral spindle]

Yes

[sly frost]

no, you can take a direct product of groups that don't have a trivial intersection

you can take product of any two groups

HK is in general all products hk for h ∈ H and k ∈ K within some group G where H, K ⊆ G

[coral spindle]

They're clearly talking about the internal direct product

[sly frost]

that's different from H x K

[coral spindle]

I also can't see why this would be true in general. I can get cos(4pi/n) in the left hand field but this doesn't seem sufficient

[coral spindle]

It is isomorphic to H x K. That is what the internal direct product means.

HK is typically not even a group

[severe linden]

Could anyone check the argument below which I am slightly confused about?\

\underline{Lemma:}
Let $f \in \mathbb{Z}[x]$ and $g,h \in \mathbb{Q}[x]$ be \textit{monic} such that $f = gh$. Then all the coefficients of $g$ and $h$ are integers.\

\begin{proof}
Suppose $g = \sum_{k=0}^n \frac{a_k}{b_k}x^k$, denote $m := \mathrm{lcm}(b_0, \dots, b_n)$, $d := \mathrm{gcd}(a_0, \dots, a_n)$, $g_0 := \frac{m}{d} g$. Then it can be checked that $g_0$ is an integer polynomial that is primitive, i.e. the gcd of its coefficients is $1$. Take $h_0$ accordingly. Because $g,h$ are monic, $g_0 = ag$, $h_0 = bh$ for $a,b \in \mathbb{N}$. So $ab f = g_0 h_0$. $f$ is monic and thus primitive, and $g_0 h_0$ is primitive by Gauss' lemma. Thus $ab = 1$, i.e. $a=b=1$, thus $g = g_0$ and $h=h_0$ and the claim follows.
\end{proof}

[cloud walrus]

Yuese

[coral spindle]

There is an exercise in Humphreys' Reflection and Coxeter groups, stating:

The height of a non-simple positive root is > 1.
Where the height of a root is the sum of the coefficients of each part in the simple system.

I think this is false. Suppose we have some root beta for which ht(beta) = k > 1. But then we can form a new root system substituting beta for beta/(k+1) — and doing the same for its negative — whence this new root has height k/k+1 < 1 since the simple system remains exactly the same.

Am I missing something here?

The actual exercise is exactly this. This succeeds the proof that a finite reflection group is generated by simple reflections.

[twin sand]

Me too, but it should be easy

[teal vessel]

so speaking of sets HK, I'm going through a proof of the second isomorphism theorem, and I'm dealing with subgroups A and B of G with the condition that A≤N_G(B). I understand that this means that AB is a subgroup of G, and that B is a normal subgroup of AB. The next part of the proof is to show that the intersection of A and B is a normal subgroup of A, and that AB/B is isomorphic to A/(A⋂B). I have a question, though: shouldn't AB/B be isomorphic to A?

AB is defined here as the cartesian product of the sets A and B (not to be confused with the direct product of groups A and B). In this case it is a subgroup of G because of the condition on A. This means that it can be thought of as the union of (left) cosets of B of the form aB with a in A, so... Nevermind. this only holds if A and B are effectively disjoint, since any element of A that is also an element of B would then be part of 1B. ok, AB/B is isomorphic to A iff A intersect B=1, otherwise the map f:A -> AB given by f(a)=aB would only be surjective, not injective. Dangit.

[coral spindle]

Firstly:

AB is defined here as the cartesian product of the sets A and B [...]
No! No it is not! We never write AB for the cartesian product!

AB = {ab | a in A, b in B}

A x B = {(a, b) | a in A, b in B}

These are EXTREMELY different things!

[woeful sage]

chat do I have to do all these "group of rigid motion of polyhedron" exercises?

[coral spindle]

Secondly:

Shouldn't AB/B be isomorphic to A?
Working in the group G = Z/3 x Z/3, let A = Z/3 x {0} and let B = A also. Then AB = A = A and AB/B is the trivial group.

So no.

[teal vessel]

defines an operator that maps (a,b) |-> ab out of spite to write AB in terms of A x B

[coral spindle]

OK

[teal vessel]

also "extremely" seems a bit of an overstatement here, but yeah, I figured it out. Had to be dumb for a moment

[coral spindle]

They are extremely different. Typically AB is much smaller than A x B, and AB may not even be a group in general.

The distinction is so vital, I cannot quite express this clearly enough

[teal vessel]

Technically, the cartesian product and direct product are distinct as well, but that's splitting hairs

since one is acting on sets while the other is acting on groups

[coral spindle]

But they are isomorphic as groups, whereas there is no correspondence here.

One might even say they are extremely different

Whereas the internal and external direct product are not very different

[teal vessel]

idk, they seem very similar in terms of the structure to create them. The groups (or sets) they create look different but they're much closer than say A x B and A/B

they're both elementwise arrays, just got a different protocol for what the meaning of the pairing is.

[coral spindle]

AB is not an array

If A = B, then in particular AB = A = B, which is certainly not an array in any reasonable sense of the word

[teal vessel]

you seem to have a very different sense of the notion of reasonableness than I do

if A=B then the array is just the multiplication table of the group A, which by definition contains all and only elements of A, hence A=B=AB

[tardy hedge]

I'm waiting for Boytjie response

[coral spindle]

I am doing work

[tardy hedge]

I mean like the similiarity youre talking about is the way you build the structure but Boytjie is talking about comparing the structure itself

I think. someothing like that

[teal vessel]

yeah, I know the resulting structures are different, and can be quite different, but that comes as very little surprise, as the procedure for building them is different, despite being similar

[tardy hedge]

i mean to be fair tho the way you talked about AB vs AxB here suggested you were saying the resulting structures are similar

not just the "procedure" for building them

[teal vessel]

one is a reduction on the other

[coral spindle]

Every finite set is a 'reduction' of every infinite set

[teal vessel]

that's true, and in some cases it's meaningful

Technically, given that a group is a set with associated operation, the subset AB of group G is just the operation of the group applied to all elements of AxB

[tardy hedge]

Tbh i think whats happening here is that youre saying they are similar in like a superfluous way or smth

[teal vessel]

I'm being a little pedantic for the bit. Idk about being superfluous, but it's an important distinction that shouldn't be confused, so pedagogically speaking it would be a bad idea to introduce HK as the image of H x K under the group operation of G

this discussion is actually helpful, though, as it has more firmly cemented what's going on in my head

[chilly radish]

There's generally not a unique way to write an element of AB as a product of an element in A and an element in B, so you can't reasonably interpret this as an array without making arbitrary choices

[teal vessel]

in the context of A and B being subgroups of G there is

which is the context being described here

[chilly radish]

That's the context in which one talks about AB. What I said is still true, like in boyt's example

And there is not

For example, if A=B=G=Z/3, then I can write
2= 0+2 = 1+1
Are both presentations of 2 as a+b

There is A presentation. It is NOT unique. That is one difference between the direct product and AB

[teal vessel]

is there some secret definition of array you guys are using that I'm not understanding?

[chilly radish]

Define an array

Seems like you want your elements to be of the form (a,b)

[south patrol]

Computer science moment

[chilly radish]

Or be representable as such

That's not possible to do uniquely in general

[teal vessel]

oh, sorry, you meant unique in the opposite direction to how I'm thinking

yeah, it's not unique, that's fine

[teal vessel]

probably, more commonly talk about them in terms of either basic arithmetic or compsci

I haven't traditionally understood arrays to need unique elements, they're not pure sets

[chilly radish]

My point is that you can't represent it as an array unless you make choices for the presentation

[teal vessel]

that seems like a non-problem

[chilly radish]

You're trying to claim that AB and AxB are very similar and i'm showing you where that fails

The differences between these two objects are very important

[teal vessel]

I know they're different, and all differences in math are important, otherwise they wouldn't be there.

AB is just the image of AxB under the group operation, which is not necessarily a group.

[tribal moss]

That "just" is doing a lot of heavy lifting that I think you're not paying it for.

[chilly radish]

All groups are homomorphic to the trivial group

[tribal moss]

Wut?

[teal vessel]

I mean, yeah

let f:G->e
f(hk)=e=e·e=f(h)f(k)

[teal vessel]

I wasn't, but we negotiated a new contract to pay it a fair wage so it can feed its kids "simply" and "only"

[coral spindle]

I am having some mental block with this exercise in Humphreys' book on reflection groups. If anyone has a good hint I'd be very grateful, I just need a push in the right direction.

[boreal inlet]

Does this statement mean zero homomorphisms always exist from any group to the zero?

[tardy hedge]

For any group there is a homomorphism to the trivial group yeah

[boreal inlet]

got it

[tardy hedge]

G is a normal subgroup of G for any G

[chilly radish]

I was trying to emphasise that "just" being an image doesn't mean anything

[tardy hedge]

I was enjoying that conversation because you more advanced people were verbalizing more accurately sort of what i was thinking in a merely intuitive kind of way

[tribal moss]

Ah, okay.

[tardy hedge]

anyone know what happened to wew lads tbh

[coral spindle]

Yeah where's Wew

I wonder how he's doing

[tardy hedge]

Me too

[leaden heart]

do you need help finding that alpha, or showing the contradiction once you've found it? (or both)

[coral spindle]

No no, the simple root alpha exists by assumption of the proof by contradiction! I'm having trouble figuring out what contradiction to find. For example I can, by assumption, express s_alpha as the product of other simple reflections, but from then I can't seem to find a decent way in

[leaden heart]

I literally understood that when I first read it, thought about the problem, and then forgot that fact

my brain is unparalleled

[dim widget]

what is a system $\Delta$?

[cloud walrus]

Math_Discord_Final_Girl

[coral spindle]

So I have a root system Phi with a choice of positive system Pi induced by a a base Delta. Maybe you know this as a system of ‘simple roots’? Specifically, every positive root in Pi is expressible as a unique positive linear combination of roots in Delta, and Phi is Pi union -Pi

I'm looking at this in the context of reflection groups, so we have some finite reflection group W and Phi is a set of roots for the reflections in W

[dim widget]

maybe it's easiest to show that $w: \Phi \to \Phi$ has an expression as a product of $\ell(w) := #{\delta \in \Delta|w(\delta) \in -\Delta}$

simple reflections

[cloud walrus]

Math_Discord_Final_Girl

[tardy hedge]

I have to go to work so i havent thought about this too much but

[coral spindle]

w being this imagined element sending -alpha to something in Delta? I suppose so, but I'm worrying that it would just end up telling me l(w) = 1 and w = s_alpha?

[dim widget]

just any reflection

[tardy hedge]

Is it something like take any maximal ideal and u can form a quotient and the quotient implies homomorphism that eventually implies n must be equal to m

By using vector space structure

So its something to do with, you cant ignore the vector space structure with the free module stuff ok ok

Ok im being so incredibly messy here ik

I should probably just wait until i havw time to do this properly 😂

[coral spindle]

OK, I'll give it a try. Thanks tteg

[dim widget]

hmm I guess that's true.

[tardy hedge]

So the only way R^n is iso to R^m as R modules is if n = m b.c just for R^n take maximal ideal and quotient it to find a certain homomorphic image and in this case youre dealing with vector spaces and that forces n to be equal to m

When u apply vector space theory

[coral spindle]

I was going to say that I think $\Delta\setminus\set{\alpha}$ is a simple system for the root system $\text{span}(\Delta\setminus\set{\alpha}) \cap \Phi$, but I don't think I can squeeze anything out of this

[cloud walrus]

Boytjie

[tardy hedge]

And ofc maximal ideal exists, if it is trivial case ur dealing with R a field anyway

Bc field have trivial ideals

[coral spindle]

Sorry kiand I might actually look at this after I sort this out haha

[tardy hedge]

No worries haha

[dim widget]

Well I guess this tells you what you need if you know that $W(\Phi)$ acts (transitively) on the simple roots for instance

[coral spindle]

No in fact I CAN

[cloud walrus]

Math_Discord_Final_Girl

[coral spindle]

Yes yeah exactly

I don't think this was what Humphreys was going for. I might ask someone at my instutution who knows this stuff well

[dim widget]

Yes you can quotient by any maximal ideal and this shows you that the ranks being equal is necessary

[coral spindle]

At least I know now I wasn't just losing my marbles

[tardy hedge]

Awesome!!

Thanks! Thats cool

[dim widget]

to show that it's sufficient one can just explicitly construct the map

The hard question is can there be a surjective map from $R^n \to R^m$ if $n < m$ 🤔

[cloud walrus]

Math_Discord_Final_Girl

[dim widget]

then the next level of difficulty is can it be injective if n > m

[tardy hedge]

Ohh yeah thats a good question because it shows one way how free modules differs from
Vector spaces right? If i remember correctly hom between vector space is injective iff surjective … was it?? Im kind of forgetting

If thats not true dats embarrassing and i gotta review linear algebra lmao

[coral spindle]

You should look at a theorem called the rank-nullity theorem

[tardy hedge]

Hahaha yeah its been a while

[coral spindle]

This proves what you're looking for, if you recall that the only vector space of dimension n in an n-dimensional vector space is the whole space

[tardy hedge]

I was just looking at that recently cuz i was like oh yea this is the first isomorphism theorem in some way right

Sort of was perusing it recently

[coral spindle]

Yes that's exactly right

dim(V/W) + dim(W) = dim(V). That's rank-nullity.

[rotund aurora]

my answer: ||fake|| ?

I'm assuming R is commutative btw

anyway, I based my proof in the fact that a square matrix is invertible iff its determinant is a unit

[quartz wind]

is this not a pigeonhole principle thing? |R^m| > |R^n| so uh how can any function be surjective

[coral spindle]

|R^m| = |R^n|

[chilly radish]

This isn't true tho

[coral spindle]

They are sets of equal cardinality, namely |R|.

[chilly radish]

What boytjie 🔹 said

[coral spindle]

I had to get it out fast 🤓

[chilly radish]

Unless |R| is finite

[quartz wind]

oh true ig

[coral spindle]

Ngl I did assume that R meant the real numbers lmao

[quartz wind]

if they have equal cardinality then there exists a bijection, but they arent isomorphic because this bijection cannot be a homomorphism

how would one show that

[coral spindle]

This property of a ring, that R^n and R^m are isomorphic R-modules iff n = m, is called invariant basis number. Not all rings have this property, but commutative rings do and this can be argued by quotienting by a maximal ideal and looking at the dimensions of the resulting vector spaces

So at least for commutative rings, this is inherited from vector spaces.

[quartz wind]

is this a first iso theorem thing

sorry im not exactly sure how quotienting by the maximal ideal would help

[coral spindle]

Because if we quotient a commutative ring by a maximal ideal, we get a field. And modules over a field are vector spaces, which we understand very well.

[quartz wind]

oh i see

thanks

[rotund aurora]

R=0

[coral spindle]

Krull crying rn. Finally we found a commutative ring with no maximal ideal

[clear ingot]

i feel like (10, 11) and (12, 13) have overlap

since u can show they are isomorphic to each other

like the rotational group of octahedron is isomorphic to the rotational group of a cube by considering each vertex of an octahedron being the center of the faces of a cube

[woeful sage]

So...I can't skip these exercises?

I can manage with figuring out groups of 2D polygons....but 3D polyhedra? not sure

[tender wharf]

i'm sure there are online programs that can help

if anything it's worth writing down a presentation for the group

[rotund aurora]

For the cube: ||fix a vertex. There are exactly 3 symmetries that fix it. You can bring a vertex to any other by rotations. There are 8 vertices. Thus the number of symmetries is 8 x 3=24||

You can do the icosahedron and the tetrahedron in a similar way

Well and the rest

[errant wedge]

Its the subgroup of identity plus products of 2 transpositions right? And any bijection that sends () to 1 works as an isomorphism since both are isomorphic to klein 4?

[wraith swan]

the 3 transpositions, yeah

[quartz wind]

in a module over a principal ideal domain, is it true that every submodule is cyclic?

it seems to follow from the definition, or is there a distinction between submodules and ideals that im not catching?

[topaz solar]

Yes

[topaz solar]

Well, we can have modules like R^3 which has a submodule isomorphic to R^2. But if you mean a submodule of R, that’s an ideal, yeah?

[quartz wind]

to be clear, this submodule isomorphic to R^2 is not cyclic?

[topaz solar]

Unless I’m misremembering the definition, how would it be?

How do I get to (0,1) from (1,0) after all

I mean, look at a field

A field is a PID

(Verify this, of course)

[quartz wind]

mhm yeah i see your point

ohh wait i get it

[topaz solar]

But if we look at submodules of R, how does that differ from an ideal?

[quartz wind]

an ideal is generated by elements of the ring 🤦‍♂️

mbmb that was dumb

thanks sharp

[topaz solar]

Well I was also saying submodules of R are ideals

So it’s not too far from what you were thinking, but it’s not something applying to all modules

[woeful sage]

by 3 symmetries you mean rotations along 3 independent axes?

[quartz wind]

ah yeah i see now

[quartz wind]

to my understanding, R is an integral domain does not imply that any R-module is torsion free, but im finding it hard to conceptualize how a module over an integral domain can have a torsion element. does anyone have an example?

[dim widget]

Z/m is a Z module

[quartz wind]

wait whats m

sorry im new to all this

[dim widget]

any integer

[quartz wind]

as in, Z/mZ?

[dim widget]

yes

[quartz wind]

ohh

i see okay thanks

[limber bramble]

Does there exist a general method to form an injective function from G/ker(\phi) to G ?