groups-rings-fields

[hidden haven]

Ye orthogonal basis should always exist

[warm holly]

Yes but if it is not zero?

[hidden haven]

For finite dimensional V, pick a non zero vector v_1. If this doesn't already span V, the function x → <v_1, x> is linear with rank 0 or 1, so it must have a non trivial kernel. Choose non zero v_2 from the kernel, and continue this way

At every stage, the intersection of kernels of <v_i, -> is either non trivial, or the v_i will span the space

For infinite dimensional, I am guessing some Zorn's lemma argument should work

But I am not sure

[hidden haven]

Here you have to make sure that the next v_i you choose is not in the span of the previous ones

Not just that it is non zero

But this choice is always possible by counting codimensions

hmm wait

[wooden ember]

The way we learnt it in class is that this is only possible for characteristic different from 2

[hidden haven]

You do need the assumption that <v,v> neq 0

For all v, in this construction

[hidden haven]

Then ig it don't work

[wooden ember]

Oh I mean I’m in the same class as Daniil so it’s not confirmation

[hidden haven]

ah

kinda don't wanna think about it

[wooden ember]

Lmao

[charred flume]

Consider the finite field $G F(16)$.
(a) The Galois group $G=G\left(G F(16) / \mathbb{Z}{2}\right)$ is cyclic and generated by the Frobenious automorphism $\sigma{2}$. Find $G$.

what does find G mean in my assignment?
isn't it already given? which is G = $<\sigma_{2}>$
or does it mean the isomorphism in $Z_{n}$. I've found that it is isomorphic to $Z_{4}$

i'm very confused on the instructions. pls help. Thx

[cloud walrus]

Ji

[wooden ember]

Yeah that’s weird phrasing

It’s probably just the cyclic group order they’re asking

[charred flume]

$\mathrm{G}=<\sigma_{2}>=\left{1, \sigma_{2}, \sigma_{2}^{2}, \sigma_{2}^{3}\right}$

is this correct?

[cloud walrus]

Ji

[delicate orchid]

if it actually is iso to cyclic of order 4 then yeah

[charred flume]

tnx

[wise igloo]

based on the fundamental theorem of finite abelian groups, every finite abelian group is isomorphic to a direct product of cyclic groups, e.g. $$G_N\cong G_{n_1}\times\cdots\times G_{n_m}, \ \ n_i\text{ prime powers}$$ but does $N$ have to be an integer? what if it comes from a ring that is not a UFD? then does this break everything? or is this impossible to even do?

[cloud walrus]

gmod

[hidden haven]

What is N denoting here

[wise igloo]

like G/NG

e.g. $\bZ/N\bZ$

[cloud walrus]

gmod

[wise igloo]

is this a stupid idea

[hidden haven]

Then what you have stated is not the fundamental theorem of finite abelian groups

It is the Chinese remainder theorem

There is a more general version yes

For a PID R, and coprime ideals I and J in R,

[cloud walrus]

Moldilocks1337 ✓

[hidden haven]

Or stated in a simpler way, for a PID R and coprime elements a and b,

[cloud walrus]

Moldilocks1337 ✓

[hidden haven]

@wise igloo

[wise igloo]

I see

alright tysm

[pastel cliff]

what is S_A referring to at the bottom there

small chapter on group actions in DnF

is it just the group of permutations of elements of A?

[delicate orchid]

the symmetric group on A I guess

yeah

[pastel cliff]

why should are group actions useful though

im understanding that a group G acting a set A essentially just creates a permutation of A - but what information can that give us about G?

[devout crow]

it's often useful to understand groups as symmetries of some object

you probably did this with dihedral groups informally, without the language of group actions

[hidden haven]

Group actions are the reason we study group theory

We want to understand the symmetries of some object and symmetries of objects form groups so we abstract out a definition

The point of studying symmetries of an object is that often that the symmetries alone can tell us a lot about the object

For example in Galois theory you study fields by studying their automorphism groups, and that contains a lot of information about the fields themselves

[pastel cliff]

it does feel like you're knocking on the door of group actions when talking about the dihedral group

[hidden haven]

One example of information about G is the orbit stabilizer theorem

Sylow theorems are also proved using group actions

[pastel cliff]

moldi, not more than two days ago: "yeah i spend too much time answering in advanced"

[hidden haven]

Wait what definition of group actions do you have

[pastel cliff]

uhhhh

[hidden haven]

If it is defined as G x S → S satisfying some conditions then that's slightly annoying

[pastel cliff]

yeah that

i mean i kinda see it

[hidden haven]

Have you seen vector spaces over fields

[pastel cliff]

but authors proceed to be less formal about it immediately

[hidden haven]

oh ok

[hidden haven]

or modules over rings

[pastel cliff]

they even say so to make sure you know

[hidden haven]

Like both of them are examples of an object acting on another

[pastel cliff]

i mean i know how a vector space works

right

[hidden haven]

Ye so a vector space is like a field acting on an abelian group by multiplication

[pastel cliff]

ok wait does this group action talk apply to the dihedral group

[hidden haven]

It's a very analagous situation

ok here's another way to phrase the definition of group actions which most books do later but I find easier

Suppose you have an object X

X can be a set, a group, a ring, a vector space, a shape, etc

[pastel cliff]

a jar

[hidden haven]

works

Let Aut(X) be the group of symmetries of X

When X is a set, this is the group of permutations

[pastel cliff]

that makes sense

[hidden haven]

For the next 3, it is the group of automorphisms

For shape its like dihedral group thing

Now we want to do stuff with X

We could want to construct a new thing Y from it

With certain similar symmetries

Just 1 example

In this case, it makes sense to think of X not on its own

But as a pair, (X, G) where G is the subgroup of Aut(X) of "special symmetries"

These are the symmetries we care about

Sometimes an object has way too many symmetries and we don't want to think about all of them so it makes sense to restrict our attention to a subgroup

This is a group action

When X is a set, this is the group G acting on the set X

[pastel cliff]

wait wait which part

the restriction to a subgroup?

[hidden haven]

The object X and a subgroup of Aut(X)

The subgroup of Aut(X) "acts on" X

[pastel cliff]

this is funky

[hidden haven]

For now forget about earlier def

I'll tell you how they are the same

But you see how every element of Aut(X) acts on X right?

It does things to elements of X

By definition

It's a symmetry, so you can apply it

This "action" in fact satisfies the group action axioms that you have seen

Does that make sense?

[pastel cliff]

ok an element of Aut(X) is a permutation of X

[hidden haven]

Ye

[pastel cliff]

so saying an element of Aut(X) acts on X just means... permuting X..?

[hidden haven]

Yep

Every element of Aut(X) permutes X

And that's what a group action is lul

Now sometimes what happens is that you have symmetries of X and you wanna see how they act on Y, a subset of X

[pastel cliff]

wait so then that definition of a group action as a map G x X -> X... how is that rewritten as the Aut(X) thing

[hidden haven]

Ye I shall tell

[pastel cliff]

https://tenor.com/view/waiting-akward-silence-syeed-well-gif-14614626

[hidden haven]

For example if you have the set X = {1,2,3,4} with the subgroup of Aut(X) we are concerned with being the subgroup generated by (12) and (34)

You see that Y = {1,2} is stable under the action of this subgroup?

Stable in the sense that elements of Y stay in Y

When this kind of a thing happens, you can "restrict the action" of G to this stable subset

Since elements of Y stay in Y when you apply the permutations in G, this restriction permutes elements of Y among themselves, and elements of X - Y among themselves

[pastel cliff]

G being Aut(X)?

wait no

[hidden haven]

Sorry, the subgroup we took here

[pastel cliff]

[hidden haven]

So it is now like elements of G are behaving like symmetries of Y, or elements of Aut(Y)

This restriction has therefore defined a map G → Aut(Y), where an element g is mapped to the permutation of Y that it acts like

Make sense?

[pastel cliff]

maybe

i gotta mull this over

[hidden haven]

Ye ok

[hidden haven]

Basically it turns out that this will always be a group homomorphism

And we want to call this an action of G on Y

So we generalize, and define

An action of a group G on a set X is a homomorphism
f: G → Aut(X)

The idea being that thus every element g of G is "acting" on X like the permutation f(g)

[pastel cliff]

ok that genral version makes a little more sense

[hidden haven]

And we are just requiring that the elements of G act together nicely, which is the requirement of f being a homomorphism

Like their individual actions should play nicely with the group multiplication

In a way that happens in the concrete situation that I described above

[hidden haven]

So now this is the definition of a group action

The action is a map f which takes an element of G, and gives a permutation of X

Suppose now you have an element of G x X

It is a pair (g, x)

[pastel cliff]

is this kinda why matrix multiplication equates to moving around basis vectors

[hidden haven]

Yes that is also an action

[pastel cliff]

this is weird

but i kinda see it

[hidden haven]

The ring of matrices acts on the vector space

Or the group of invertible matrices acts on the vector space

[hidden haven]

Notice that you can then do f(g), which gives you a permutation, and apply it to x, to get an element of X. So you get a map
G x X → X
described by
(g, x) maps to (f(g))(x)

Similarly, given a map
h: G x X → X
and an element g of G, you get a map
h_g: X → X
which fixes the first input to g, ie,
h_g(x) = h(g, x)

If this map h also satisfies the group action axioms that you know, then you can prove that h_g is a permutation (it has an inverse h_g' where g' is the inverse of g) and that this map
g → h_g
is a group homomorphism

[hidden haven]

The upshot of all of this is that

Homomorphisms from G to Aut(X) "are the same as" maps G x X to X that satisfy the group action axioms

Same in the sense that they contain the same data, you can recover one from the other, as I said above (check that the above processes for turning one thing to the other are inverses of each other)

So the definition you have seen is just an expanded out version of saying that you have a homomorphism from G to Aut(X)

[opaque pilot]

Im having trouble showing that $H_n \subseteq \langle r^2 \rangle$. So far, if I take $x \in H_n$, then $x = g^2, g \in D_n$ Since $D_n$ is generated by $r$ and $s$, then $g = r^ms^k$ for some integers $m, k$ So then $g^2 = (r^ms^k)^2 = r^ms^kr^ms^k$.

[cloud walrus]

Kurama

[opaque pilot]

I think I should use the orders of r and s, as well as the relation rs = sr^{-1}. I've tried expanding g^2 = r...rs..sr..rs..s then using that relation, but that didnt really get anywhere

[young hollow]

Ok so this is a subgroup of a dihedral group

[opaque pilot]

Yeah

[opaque pilot]

Any clue what I should do?

[strong yacht]

Looks good so far. Maybe you can specialise the m, k a bit. Really the only elements in D_n are {Id, r, r^2, ..., r^{n-1}, s, rs, r^2s, ..., r^{n-1}s}. So two cases to consider: when your k = 0 or when k = 1

[sage wagon]

Hope I'm not interrupting - I'm reading about Schur Weyl duality, but I'm just a bit confused about what it means here by 'induces a map':

Any help/suggestions would be appreciated!

[opaque pilot]

oh ok. So when k =0,l then g^2 = r^{2s} = (r^2)^s in <r^2>.

[strong yacht]

I think you mean to write your 'm' instead of 's' here but yes looks good. What happens when you square in the case where k = 1?

[cloud walrus]

Kurama

[opaque pilot]

yep.. my mistake. I was using other variables in my written work so i got confused

[cloud walrus]

Kurama

[strong yacht]

Yeah, that's correct. Another way of saying this is showing r^m.s = s.r^{-m}

Remember that r^m.s are reflections of the n-gon

[opaque pilot]

ah yes, they are

thats a good way to think about it in my head

thanks 🙂

[strong yacht]

No problem 😄

[agile burrow]

Remember that End_{CGL(V)}(V^n) is the set of GL(V)-linear endomorphisms of V^n. The preceding statement, that the actions of Sn and GL(V) on V^n commute, means precisely that that the image of the map CSn -> End_C(V^n) lies within the subset of GL(V) linear endomorphisms, hence we can redefine the map in (1) but restrict the codomain to GL(V)-linear endomorphisms. This is precisely the induced map

[sage wagon]

Ahh I see, thank you! Would you happen to know of any resources/be able tell me a bit about Schur Weyl duality? It's my understanding that there's meant to be a bijection between the irreducible representations of the symmetric and general linear groups, but the formulation in the notes I'm reading (https://tartarus.org/gareth/maths/notes/iii/Representation_Theory_2013.pdf) doesn't seem to make this clear (or I just haven't read it carefully).

[agile burrow]

Unfortunately I can't say that I understand Schur-Weyl duality myself, it's one of the things I've been meaning to learn more about. Just reading online about it from time to time, it seems there are a number of short articles/write-ups on it that you might find helpful if you just want a different perspective. I believe it is discussed in Fulton and Harris which some of my friends have recommended so that could be worth checking out, but I can't say how clean of an explanation it is

[wise igloo]

what are y'all's opinions on the Tits alternative?

[sage wagon]

Oh no worries, thanks, I've had a brief look at Fulton and Harris before, perhaps I'll go read the relevant parts in a bit more detail this time. 👍

[wise igloo]

it would be satisfactory to list the four possible cases and show why f(xy)=f(x)f(y) for each case, right?

where x and y are real numbers

[next obsidian]

Right

Alternatively if you recognizes that G is {-1,+1} under multiplication, you can probably just cite that this is the sign function which is multiplicative or whatever

[wise igloo]

true

alright ty

im stuck

I get the Z/nZ part

but not Z/(Z/nZ)

idk how to think about this

[next obsidian]

Wtf

This doesn’t make sense

Yeah this problem is pretty much uninterpretable

[wise igloo]

fantastic

lol

it was some random group theory pdf I found

[wooden ember]

I mean it’s downright just wrong lol Z doesn’t even have any subgroup isomorphic to Z/nZ do I don’t see a single way to interpret this

Probably a typo

[hidden haven]

The answer is obviously nZ

[next obsidian]

That’s a stupid problem then lol

There’s no content

[hidden haven]

Wait I was just saying stupid shit what did you interpret it as

I was saying that for numbers a/(a/b) = b

[next obsidian]

Oh

I thought you’re saying you should replace the (Z/nZ) with nZ

So it’s asking you to list the elements of Z/nZ

[hidden haven]

lol

[grand sigil]

I make this typo often

It was definitely Z/nZ

[young hollow]

Can someone confirm my understanding please:

So when it comes to dihedral groups, there are the ‘immediate/direct’ physical symmetries (which are the elements of the group) such as the reflections about the various lines of symmetry and the various rotations- all of which may themselves be represented as permutations. But there is also the perspective of viewing the immediate symmetries in terms of operations on two generator symmetries (and their powers), each of which may also be expressed as permutations.

Am I capturing the notion accurately?

[strong yacht]

It sounds like you are asking why does the typical presentation of dihedral groups agree with our geometric intuition, is that correct?

https://dougalmaths.files.wordpress.com/2020/10/dihedralgroups.pdf

You may very well want to give that a read in that case

[young hollow]

Let me take a look, because that may also be a question I have. But more so I’m just confirming that the notion of generator is really just an efficient means of expressing a group’s elements- they’re not the “actual” group elements themselves

Tricky to word, maybe I’m not conveying my thoughts well

Let me read your link

[next obsidian]

“Actual” group elements is usually not the right thing to say

You only care about these things up to isomorphism, so the actual labels you give these things don’t really matter

In one presentation of the dihedral group given relations and generators you have formal symbols which multiply in a certain way

If you instead think of the elements “as symmetries” then as a set these arent the same

But you only care about how the elements multiply, their multiplication table I guess

[young hollow]

Ok I think I see. I am familiar with isomorphism so in light of that I suppose it’s irrelevant how the structure is expressed

Although it is revealing that 2n symmetries can be expressed via a set of 2 generators

[next obsidian]

I suppose so

I don’t really know what it tells you, but minimal number of generators is surely an invariant and some people care about them, but I’ve never particularly cared haha

I mean it makes writing down the elements pretty easy lol

[young hollow]

Yeah I think I saw that mentioned too in the reading. Probably the main reason

Although, admittedly, I wouldn’t have guessed that all the symmetries could be achieved by means of compositions of a rotation snd reflection, maybe that’s why it’s more exciting to me initially than it ought to be

[delicate bloom]

yeah that's what I find appealing about generators, kind of helps digest an otherwise large and unruly group into the bare minimum essentials to get anything which is neat

[young hollow]

Yeah it’s very efficient

(And to dummies like me somewhat illuminating as to some structural info)

[next obsidian]

I think this is kinda intuitive though

If a symmetry on an n-gon preserves edge relations you just need to see where a single vertex goes then if the orientation got flipped

Namely if you single out a vertex v, and look where it goes

Then the neighbor of that needs to be one of v’s original neighbors so just go in one direction and see which one it is

Okay so now if you want to go even further there’s only one option because there’s only two vertices it could be, but you already used up v

So once you picked the first neighbor in a certain direction, you’re forced to just keep going along the polygon

[young hollow]

This is fantastic

[next obsidian]

So you have a choice of wheee you put the vertex (how much rotation) and a binary choice on which neighbor goes where (the flip)

[young hollow]

Yeah that’s the way D&F motivate it as well

[next obsidian]

O

Kekw

[young hollow]

But yeah I see what you’re saying. Pick any of the n vertices

There are n places it can go via rotation

Then subsequently a reflection is possible

[strong yacht]

Glad you enjoy

[wise igloo]

when I first started learned the definition of a group (before I formally started learning group theory), I knew of the existence of algebraic number theory

and im like how tf can that exist

the combination of two seemingly completely separate fields of math

but like there are so many instances where number theory concepts showed up in the group theory lectures (modular arithmetic, gcd/lcm, primes, etc) and it's super interesting

[delicate orchid]

a lot of both early group and early number theory feels almost combinatorics-ally so there's a lot of cross over

[wise igloo]

interesting

isn't combinatorics like permutations and shit

nCr type shit

[delicate orchid]

permutations

wow you are early in group theory

if you haven't seen S_n yet

[wise igloo]

I have dumbass

[delicate orchid]

every finite group embeds isomorphically into some group of permutations

[wise igloo]

I know

[delicate orchid]

ok so there you go!

[wise igloo]

yeah

I was making connections

[delicate orchid]

well the connections you made were great ones to see

[wise igloo]

lol

I should learn NT and combinatorics

I only know basic NT concepts

[delicate orchid]

NT very quickly becomes ring theory

[wise igloo]

lol

do u know of any good lecture series

or books

[delicate orchid]

no but maybe

[wise igloo]

(I prefer lecture series)

ah

f

I'll look some up

[delicate orchid]

eine seconde bitte

[wise igloo]

what

[delicate orchid]

https://www.youtube.com/watch?v=EzE6it9kAsI&list=PL8yHsr3EFj53L8sMbzIhhXSAOpuZ1Fov8

I knew he'd probably have one

[wise igloo]

oh ok

lol

ty

[prisma shuttle]

guys in a paper ima reading

they keep referring to a ring R'

which they call "the subring of R generated by 1"

am I wrong or is this not just the whoelr ing R itself

cuz i thought the ring generated by 1 was just the ring itself

or is this only for groups

i might be confusing myself

[delicate orchid]

uhh not always I don't think

hold on

yeah like

take $M_n(\bR)$ to be the ring of real n by n matricies

[cloud walrus]

Wew "Boötes Void" Tbh ⊗

[delicate orchid]

the identity matrix doesn't generate the whole ring

the ideal generated by 1 is always the whole ring though

[chilly ocean]

I think this is where the confusion is

[prisma shuttle]

yea becuase RAR = R right when A=1

[wise igloo]

rawr

[chilly ocean]

Yep you got it bruh

[prisma shuttle]

wait so just for clarifcation

wut does the identity matrix generate

just the set of matrices where all elements are the same right

or not elements like

entries

[delicate orchid]

it would just be all multiples of the identity matrix I think?

[prisma shuttle]

so basically wut i said right

cuz its just like a multipel fo deitnty

[delicate orchid]

well no cause you can have like, $\begin{pmatrix} 4 & 4 \ 4 & 4 \end{pmatrix}$ which is not a multiple of the identity

[cloud walrus]

Wew "Boötes Void" Tbh ⊗

[prisma shuttle]

oh oops wait

i meant like

where all entries in diagnals are $k$ for $0\leq k\leq n-1$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

and everything else is 0

[delicate orchid]

yeah but you don't need to bound k like that

[prisma shuttle]

oh right right

ok i get it now

[delicate orchid]

unless you're working in F_n

[prisma shuttle]

thx for your help

[delicate orchid]

np

[noble dust]

I'm bugging so hard right now

I can't understand well the tensor product of a right module M and a left module N over R.

If R is non-commutative that is

the following chain of equalities seem to be a contradiction

$(rs)(x \otimes y) = x(rs) \otimes y = (xr)s \otimes y = s(xr \otimes y) = s(r(x\otimes yu)) = (sr)(x \otimes y)$

[cloud walrus]

Digiteraat

[noble dust]

What am I missing here?

[next obsidian]

There’s no way to turn the tensor product into an R-module unless one of M or N is a bi-module

It is an abelian group

[noble dust]

shit

a course on representation theory I'm following used the fact the tensor product is an R module

I maybe missed something

[next obsidian]

R is probably commutative

[noble dust]

$R = \C[H]$ where $H$ is a subgroup of $G$ (which is a finite group)

[cloud walrus]

Digiteraat

[next obsidian]

Hmmm

[noble dust]

The tensor product in question is $\C G\otimes_{\C H}V$

[cloud walrus]

Digiteraat

[next obsidian]

Well as far as I know unless you have a bimodule you can’t get a module structure out of the tensor product

[noble dust]

which is how we defined the induced representation

[cloud walrus]

Moldilocks1337 ✓

[next obsidian]

Oh

[hidden haven]

This applies to the above situation

[next obsidian]

Well CG is a bimodule

[hidden haven]

Yep

[next obsidian]

CG is a (CG,CG) bimodule

[hidden haven]

It's like the middle stuff cancels

[next obsidian]

You can multiply scalars on the left and they only get to stay on the left

[hidden haven]

Here you wanna view it as (CG, CH)

So that CHs cancel

[next obsidian]

Ah

But yeah so like

[noble dust]

bimodule means you can multiply on both sides?

[next obsidian]

r(s (x) x) = rs (x) x

Yes, but it’s more than that

(R,S) bimodule means that

(rx)s = r(xs)

So it’s like, “associative”

[hidden haven]

(A,B) bimodule means you can multiply by elements of A on the left, of B on the right

[hidden haven]

Ye such that this happens

[noble dust]

Yeah ok makes sense

[next obsidian]

When R is associative

Any left module structure gives a right module structure

So everything’s a bimodule

So if you know what rx is, define xr to be rx

And when R is commutative it just works out

[hidden haven]

wait how

[noble dust]

Maybe they defined the CH structure wrong then? They said $y(x \oplus v) = xy \oplus v = x \oplus yv$

[cloud walrus]

Digiteraat

[hidden haven]

oh

[hidden haven]

commutative?

[next obsidian]

FUGG

Yes commutative

[hidden haven]

Didn't get this

[noble dust]

I used oplus but I meant otimes..

[hidden haven]

Ye then that's right

That is what it means to tensor over CH

[next obsidian]

If y is in CH

Then the y on the right can jump sides

[hidden haven]

ye

[noble dust]

Yeah but I can still do something like above no?

[next obsidian]

But if x is in CG\CH it cannot

[noble dust]

$zy(x \otimes v) = xzy \otimes v = y(xz \otimes v) = y(z(x \otimes v))$

[cloud walrus]

Digiteraat

[noble dust]

it's not even about jumping sides

[hidden haven]

What's wrong here?

[next obsidian]

You don’t put zy on the right of x

It has to go on the left

[noble dust]

well I get $(zy - yz)(x \otimes v) = 0$

[cloud walrus]

Digiteraat

[noble dust]

which seems pretty fishy to me

[hidden haven]

Ah ye zy goes on the left

[noble dust]

okay

that explains shit

[hidden haven]

The multiplication in the middle gets all forgotten

[noble dust]

the definition we were given in the course was on the right

so the right CH-module structure is $y(x \otimes v) = yx \otimes v$?

[cloud walrus]

Digiteraat

[noble dust]

just to check

[hidden haven]

There is no right CH module structure

Once you tensor

There is only left CG structure

[noble dust]

left CH-module structure no?

[hidden haven]

This is what happens

The Bs in the middle disappear

There is no nice way to define a B module structure on either side on the tensor product

[noble dust]

Yeah I guess the thing is $\C G$ is a $(\C H, \C H)$ module

[cloud walrus]

Digiteraat

[noble dust]

so it works out

[hidden haven]

Here you want the result to be a left CG module

So you have a CH module V

(CH, Z) bimodule

You want to induce a CG module from it

So you tensor this on the left with the (CG, CH) bimodule CG

[noble dust]

Okay yeah I want a CG module of course

it's clearer now

[hidden haven]

And you get a (CG, Z) bimodule, ie a left CG module

[noble dust]

Ok thanks

why do you specify Z here?

I'm only intereseted in the fact V is a left CH module right?

[hidden haven]

Just wanted to phrase everything in terms of bimodules, since I stated the original result for bimodules

[noble dust]

I mean it doesn't carry any information right?

[hidden haven]

But ye you can ignore the Z part

[hidden haven]

Ye

[noble dust]

Okk thanks

I think it cleared up a bit tensor products in general

[next obsidian]

When the module is over Z

😫

[agile burrow]

just to make sure i'm not going crazy, the homology at the middle term is Z/2 \oplus Z/2, right?

[hidden haven]

ye

[agile burrow]

bet, thanks

been trying to construct an SES of complexes such that the i-th homology functor is neither left nor right exact, but i suck at making counter examples

ended up skipping the exercise until i learned about cones and i think I finally figured out how to do it

[prisma shuttle]

what does it mean when a polynomial is said to be normed

specifcally, a sentence in a paper ima reading says "Assume by contradiction, that $p(x)\in R[x]$ is a normed null-polynomial of degree $3$"

[cloud walrus]

JustKeepRunning

[prisma shuttle]

and they don't define wut normed means beforehand

so ima assuming that means it should be a common term that is well-defined

but i haven't seen it before

[north sand]

coefficient of x^n is 1

[devout crow]

what does null-polynomial mean

[prisma shuttle]

isn't this called monic

[north sand]

yes

[prisma shuttle]

or am i misunderstanding u

[north sand]

I've heard both

[prisma shuttle]

then why are they caling it normed her

oh u mean like they are two diferents ways of saying the same thing

[prisma shuttle]

a vanishing polynomial

[chilly ocean]

sometimes two different words can mean the same thing

[north sand]

or do you have a norm on your polynomial ring?

[prisma shuttle]

idt it means it in that way

cuz theres no well defined norm

at least not given in the context of the paper

[north sand]

vanishing of degree 3?

[prisma shuttle]

so i think wut u said is correct

[devout crow]

vanishing like, has a root?

[prisma shuttle]

yea vanishing and monic ig

[prisma shuttle]

vanishing, like when u plug in any element of the ring u get 0

[north sand]

is your polynomial over F_3 or something?

[devout crow]

oh I see

[prisma shuttle]

wut makes this interesting is that vanishing polynomila is not neceessarily the zero polynomila

[prisma shuttle]

no

[devout crow]

nonzero vanishing polynomials necessarily have to be over finite fields, right?

[prisma shuttle]

its over the ring $\rho$ with four elements ${0,1,a,1+a}$ with $1+1=0$ and $a^2=0$

[cloud walrus]

JustKeepRunning

[prisma shuttle]

not necessarily

the only important property of fields is that every function is a polyfunction

which follows from lagrange interpolation

[north sand]

of finite fields

[prisma shuttle]

and the converse is also established in a paper by Redei and Sleze

anyways i think wut u said is correct

@north sand

thx for your help

[opaque pilot]

https://math.stackexchange.com/questions/1132946/show-that-a-group-that-has-only-a-finite-number-of-subgroups-must-be-a-finite-gr

In this proof, the person used that G is the union of finite cyclic group <a> where a in G. But aren't we trying to prove that G is finite? So why can they say this isnt an infinite union, if that makes sense

[chilly ocean]

there are finitely many subgroups so the union has to be a finite union

it is a union of subgroups

[opaque pilot]

ahh ok, that makes sense

ty

[chilly ocean]

Round 2:
Let A be the subset of Z×Z, containing all ordered-pairs (a,a)
Consider a ring homomorphism f from (A,+,.) to (Z,+,.), which in terms of Binary Relations is defined as:
f = {((a,a),b) | (a,a) in A and b in Z, b = a}

It's kernel is {(0,0)}. From what I have researched is that the cokernel of this homomorphism is Z/im(f).

im(f) = {im(x) | for all x in A and im(x) = f(x)}
Basically the whole range of f which is Z.
Z is a trivial subgroup of Z, and all subgroups of Z are normal

Hence, the cokernel of f will be:
cokernel(f) = Z/im(f) = {a + im(f) | a in Z}

.
.
.
a = 0: {...-2,-1,0,1,2...}
a = 1: {...-2,-1,0,1,2...}
.
.
.

Hence Z/im(f) = {{...-2,-1,0,1,2...}}
Hence Z/im(f) = {Z}, which is trivial.
The kernel is also trivial.
The mapping is also bijective (only happens when both kernel and cokernel are trivial)

So this must be correct now right?

Interestingly, the cokernel here is also isomorphic to the image of the kernel of f, right?

[hidden haven]

Lol please describe f in terms of inputs and outputs instead

The f(x) = y thing

[chilly ocean]

Ah okay

[hidden haven]

Because the thing you have written is not a well defined function

[chilly ocean]

f((a,a)) = a

[hidden haven]

But this is then not a function on ZxZ

It is a function on a subset of it

[chilly ocean]

Oh I see

Ah wait let me reframe it

[chilly ocean]

Done

[hidden haven]

It is correct except I didn't get the last line

cokernel is isomorphic to image of kernel

Like they are both trivial so yeah?

[chilly ocean]

The cokernel has one element
Image of kernel has one element (the additive identity of Z)

idZ + idZ = idZ

G/G = {G} could be also defined as G+G = G

So a group homomorphism G/G -> {idZ} is a group isomorphism

And both are trivial

So G/G is isomorphic to {idZ} or the image of kernel

[hidden haven]

Yeah, but the kernel itself is also isomorphic to the cokernel

Because the kernel is trivial too

[chilly ocean]

Oh

Oh yes I see

Thank you very much for verifying my answer o7

[agile burrow]

to show homotopy of morphisms of complexes is reflexive, you can just set h = 0, right?

[wooden ember]

This is more #point-set-topology I believe

[agile burrow]

walter

[chilly radish]

Is it? If i'm understanding correctly he's asking about chain homotopy which while arising from algebraic topology is an abstract algebraic concept

Anyways, what is h in this case?

[wooden ember]

lol im far from an expert in any of this if you say it belongs here it probably does

[chilly radish]

It can honestly go either way

Like it's a concept at the very start of Homological algebra which is still strongly related to classical algebraic topology

So neither channel is wrong per se

It's more about context

Of which there wasn't, really

[agile burrow]

Fwiw, this came from a section on homological algebra

[agile burrow]

I meant h^i = 0 for all i, I was just lazy to type it out

[chilly radish]

I'm still not sure who is h in this case? Do you mean the like, prism morphism?

[agile burrow]

Where h^i: A^i -> B^i-1 in the definition of a chain homotopy

[chilly radish]

Yea

Then yes I think that's right cuz it reduces to the definition of a chain map

[agile burrow]

Bet thanks

Do you know a good bit of homological algebra?

[chilly radish]

Wait nvm

It comes out to 0=0 in the definition

I misremembered the defn

Still works

[chilly radish]

The slightest bit. I'm taking a reading course on it this summer

[agile burrow]

Ah nice

Just started learning it too

[chilly radish]

What r u using

[agile burrow]

Last chapter of Aluffi's book

[chilly radish]

Ah

I stan rotman but weibel is a standard resource

[agile burrow]

Been working through this for 2 years so I really wanna see it through to the end

[chilly radish]

Damn

[agile burrow]

Yeah I figure I can learn more of the details from Rotman or Weibel later

[lament dawn]

What does it mean for fundamental domains to be adjacent

If $D$ is the diameter of the fundamental domain (finite group) a then maybe two fundamental domains are "adjacent" if $\sup_{a\in A, b\in B}{d(a,b)} < D$?

oops

[cloud walrus]

Migillope

[opaque pilot]

How did they get (a b) (a c) = (a b c)? Should it be (a c b)?

[next obsidian]

Are these permutations?

Written in cycle notation?

If so, there’s actually competing conventions for how you multiply them

One is how functions work so you go right to left, so a goes to c

The other (what I prefer) goes left to right

So a goes to b, done

b goes to a, goes to c, done

[opaque pilot]

Ah ok so it seems this video is done in the convention that you mentioned second

[next obsidian]

Yeah I believe so

[opaque pilot]

Then that makes sense

[next obsidian]

If you’re drawing from multiple sources, they probably won’t say which they use

[opaque pilot]

Yeah, its just some youtube video I found

https://www.youtube.com/watch?v=28tC6E6wdUo

[next obsidian]

So just keep an eye out for it until you can see what convention they’re using

Like here, you now know they go left to right

All that matters is that you pick one, then stay consistent

If they flip-flop though, then

RIP

[opaque pilot]

oh man that would be confusing

why would any sane person do that

[next obsidian]

I mean probably as a mistake

Lmfao

[opaque pilot]

fair

[next obsidian]

But there’s a couple things where it’s conventions like this

you can go between, but you need to just keep an eye out for what they use

For example, conjugate g by h, does this mean h^-1gh

Or hgh^-1?

It’s whatever you pick

Later if you see semi direct products, there’s also a left-right choice you have to make

So just be aware of it

[opaque pilot]

yeah, ill keep that in mind

thanks

[next obsidian]

[native sand]

are there any fields people care about that aren't isomorphic to some subfield of the complex numbers?

[next obsidian]

Yeh

Finite fields

Although model theoretically I think they’re really similar

[native sand]

for those with a prime number of elements at least (say p), they're isomorphic to Z/pZ I believe, which is a subfield of C. I don't know the other ones, can you name an explicit example which isn't isomorphic to a subfield of C?

[next obsidian]

Noooooo

[agile burrow]

Z/pZ is not a subfield of C as they have different characteristic

[next obsidian]

A positive characteristic thing can’t go into C

Since p ≠ 0 inside of C for any p

[grand sigil]

what about function fields?

[next obsidian]

If you’re a function field over a number field you’re still iso to a subfield of C

[golden pasture]

Hi! I'm currently trying this qn but I'm kind of reallly stuck at trying to relate Z(G)\cap H_1 and Z(G)\cap H_2 to something that (at least im aware off) in rep theory

could anyone help give a bit of hint in the right direction

[wooden ember]

when they say H_i invariant vectors do they mean V^H_i is invariant under H_i or every element of V_H_i is invariant under H_i

[hidden haven]

The latter

[wooden ember]

i think i see what to do?

[hidden haven]

Can always rely on resident finite group theory expert

[wooden ember]

the fact that im not using the fact that the dimensions are equal is sus lmao

Take an element $g\in Z(G)\cap H_1$ and consider its action on any vector $v\in V^{H_2}$. $\forall h\in H_2$, $g(v)=gh(v)=hg(v)$ which means $h$ fixes $g(v)$ so $g(v)\in V^{H_2}$.

[hidden haven]

Isn't that the only hypothesis on H_i

[wooden ember]

this was my argument for one inclusion and then the other was symmetric so something is wrong lol

[cloud walrus]

𝓛ittle ℕarwhal ✓

[wooden ember]

probably the fact that i dont know any rep theory and am likely misunderstanding the question

[hidden haven]

How are you concluding that g is in H_2?

That's what you have to conclude right?

[wooden ember]

oh i see it: just cause g fixes V^H_2 doesnt mean its in H_2 yeah

[hidden haven]

Ye

[wooden ember]

youd need H_1 and H_2 to be the largest such subgroups

which might be where dimension comes in idk

[hidden haven]

g isn't fixing V^H_2 pointwise anyway

H_2 does

F

[wooden ember]

okay next time ill think before i write

[hidden haven]

But the solution should probably be very similar lol

Like this seems like a lemma you would prove about elements of those intersections

[wooden ember]

While ive got you here ive got a question of my own and then ill put ari's question back to apologize for burying it

is it true that polynomials (over multiple variables) over F_2^n all reduce over some extension of F_2^n?

[hidden haven]

lol idk anything about multivariable polynomials

[wooden ember]

i swear i hate 2 it ruins everything

[hidden haven]

2 is the nicest prime in alg top lol

No signs and orientations to worry about, and graded commutativity becomes commutativity

[wooden ember]

because i watching a video on finite geometry and at one point they say that the number of points on irreducible conics in PG(2,q) is q+1. I can show this for q odd by using an argument of quadratic residues in F_q but im struggling to show it for q even and starting to think these conics dont exist

so i guess the proper question is does every degree 2 homogeneous polynomial over F_2^n reduce in some extension

[delicate bloom]

yeah I think they're all lines cause like x^2+y^2=(x+y)^2 or somethin

[wooden ember]

yeah for F_2 i agree

[delicate bloom]

every element of F_{2^n} can be written as a power of 2 of another element for the sake of factoring out at least if that's the problem, idk I have to think about it more

[wooden ember]

alright thanks, ill put ari's question back now and think about this more

[wooden ember]

somebody answer this

[golden pasture]

like the trivial ideas jus look at all your simple modules of C[G] but really the only time i see Z(G) coming in is like when doing a dimension count or something

nothing too much that helps sadge

o wait

V^{H_1} would just be some direct sum of C with the trivial H_1 action

so the restriction of the rep to H_1 is like

[C^dim] \oplus [idk]

i wonder if this helps

maybe say restrict to H_1\cap H_2 then perhaps ind back to H_1 for smt interesting

It is true in trivial cases when G is abelian or Z(G)={e} but

now how to prove for the intermediate cases

[delicate bloom]

I went to make some food and was thinking about your response, I wanted to clarify a bit that because in $\mathbb{F}{2^n}$ we have all elements $a_i=a_i^{2^n}$ then we can take our diagonal forms $$\sum{i=1}^k a_ix_i^2 = \sum_{i=1}^k a_i^{2^n}x_i^2 = \sum_{i=1}^k (a_i^{2^{n-1}}x_i)^2 = \left(\sum_{i=1}^k a_i^{2^{n-1}}x_i\right)^2$$

[cloud walrus]

Merosity

[delicate bloom]

the real problem is finding a cute factorization for the non diagonal forms

which is probably not too bad since multiplying the two affine planes together is gonna give you k^2 variables to solve for k(k+1)/2 variables so it should be doable

like speaking about the coefficients here

[wooden ember]

yeah i realized that as well after i said i only agreed in F_2

[wooden ember]

im not sure what you mean here

[delicate bloom]

a nondiagonal homogeneous quadratic form has k(k+1)/2 coefficients

an affine plane has k coefficients

[wooden ember]

oh right okay

[delicate bloom]

so if we take two separate polynomials for two separate affine planes and multiply them together and set it equal to the polynomial for the homogneous

yeah

basically just equating coefficients and creating a big system of equations to solve, idk like probably this can be proved by induction or some other well known argument

[wooden ember]

tbf even before thinking about >2 variables, non diagonal forms in 2 variables kinda stump me

because upon creating a system for the coefficients i end up getting that it's solvable iff ax^2+cy^2=bxy has a root which is redundant lmao

[delicate bloom]

well like (ax+by)(cx+dy)=Ax^2+Bxy+Cy^2 is what I'm saying

solve for the lowercase in terms of the upper case

[wooden ember]

yeah but like you end up winding back to Ax^2+Bxy+Cy^2 has a root (at least i did)

cause ac = A, bd = C, ad+bc = B and multiplying the last equation by cd gives Ad^2+Cc^2=Bcd

[slate mortar]

Hello guys, can someone explain me (or find any reference explaining) why the wedge product on free modules is a perfect pairing?

[delicate bloom]

yeah I get what you're saying

[golden pasture]

Suppose WLOG $g\in Z(G)\cap H_1$ and $g\notin Z(G)\cap H_2$.

if somehow you can define some "minimal" $\mbb C[G]$ module $V$ such that $V^g=0$ which gives us $V^{H_1}=0$ and somehow? show that $V^{H_2}\neq0$

[cloud walrus]

ari 亲

[golden pasture]

obv cant be too minimal else V=0

[noble dust]

Hmm, this might be a bit of a dumb question, but how can one show a representation is not completely reducible? For example consider the natural representation of $S_3$ on $\mathbb{F}_3^3$, it seems that it's not completely reducible, but I'm missing the right argument.

[cloud walrus]

Digiteraat

[golden pasture]

find a submodule that is not a direct sum basically

oh right easier way is

check if the jacobson radical is trivial

[golden pasture]

the result is trivial for Z(G) i wonder if you could lift the Z(G) module up hmm

[golden pasture]

actually

consider

if $g\in Z(G)\cap H_1$ and $g\notin Z(G)\cap H_2$

[V=\text{Ind}{H_2}^G\mathbb C{H_2}]

[cloud walrus]

ari 亲

[golden pasture]

then clearly the element $1\otimes g$ is not an invariant under $H_1$, but $V^{H_2}=V$

[cloud walrus]

ari 亲

[golden pasture]

i think this works?

[noble dust]

Hmmm, I'm not sure I understand...

Also, in this case $Z(G)$ is the trivial subgroup if I'm not mistaken

[cloud walrus]

Digiteraat

[golden pasture]

basically like
radical of a module is trivial iff the module is semisimple

furthermore here we have rad(M)=J_AM

[ripe glacier]

hey what is R[x,y,z] / (x^2 + 1, y^2 + 1, z^2 +1, xyz + 1)?

[delicate orchid]

quaternions

[pure mulch]

[spice whale]

yeah i think that's the quaternions

[prisma shuttle]

yep it is the quaternions

[obsidian sleet]

then why is it commutative

[prisma shuttle]

wdym

[delicate bloom]

xy=yx

[prisma shuttle]

oh shoot wait

then it might not be theq uaternions?

is there any other ring that satisfies i^2=j^2=k^2=-1 and ijk=-1

cuz i know thats one of the fundamental relatoins of the quaternions

but like also the ideal doesn't seem to guarantee that ij=k, ji = -k, jk = i, kj=-i, ki = j, ik = -j

so i think it might not be "strong" enough to be the quaternions

[delicate bloom]

@prisma shuttle is it possible to have a group homomorphism that maps two elements that commute to two elements that don't commute?

[prisma shuttle]

pretty sure no?

cuz if ab=ba

then \phi(ab)=\phi(ba)

or \phi(a)\phi(b)=\phi(b)\phi(a)

so the images commute as well

[chilly ocean]

xy = (-1)z^2xy = z
So 1 = x^2y^2 = z^2 = -1.
It follows that 2 = 0 in this ring
If R is field of characteristic other than 2, then your ring is the trivial ring

[ripe glacier]

ah well R is supposed to be the reals, so guess that's the trivial ring

[chilly ocean]

yep

[opaque pilot]

So possible lengths of element in S4 written as a product of disjoint cycles are: 4, 3, 2 + 2, 2, identity
so then all possible orders are the lcms, so 4, 3, 2, 1?

Is this what this question is asking?

[chilly ocean]

Hey, could anyone share what kind of prerequisites there would be for studying Abstract Algebra?

[next obsidian]

👍

[chilly ocean]

Pls ping me if and when you have an answer! Ty

[noble dust]

@chilly ocean high school mathematics I guess... there's not any prerequisites per se. Just start with group theory and work your way up from there. A natural progression would be studying groups, rings, fields and then modules and then there's many paths to take from there

[chilly ocean]

oh, I see, would you recommend Artin to start?

[noble dust]

I haven't studied with it, but I think it's good

From what I've seen it's certainly a book you can pick up with very little prerequisites

But just know that it's a big book, and math is not your evening lecture 😅 you actually have to work your way through the exercises and such

It's also material that is usually covered in several years in uni

[young hollow]

Nowadays with all the resources available and places such as this I feel like formal tracks aren’t necessarily as strict. It can be kind of fun to just ‘drop yourself in the pit.’ When something comes up that’s unclear or needs prerequisite, you just take that detour; you may return to the original track or chase that detour more, that’s another beautiful flexibility of it.

This is within reason, of course. Probably don’t want to start ‘too high up’

[noble dust]

I'd still say it's logical to start with groups at least

get familiar with basic algebraic structures

since most advanced material assumes you have good knowledge of these

[young hollow]

Well yeah I’d agree, within a discipline you want to follow a coherent or traditional presentation

[lapis trail]

If I'm working with permutation groups and I want to prove that transposing element r with element s, r<s, then that is the product of 2(s-r)-1 transpositions of adjacent elements, would I proceed by induction?

[next obsidian]

That sounds pretty reasonable

[lapis trail]

Technically the problem is interchanging row r and row s of a matrix, but I think using a permutation group would be easier

[next obsidian]

It’s actually the same thing

Transposition matrices form a group isomorphic to S_n

For nxn matrices

[lapis trail]

The base case is trivial. Assume that it takes r-s adjacent transpositions to move s to the initial r position. Then it takes r-s-1 adjacent transpositions to get r to the initial s position. The total adjacent transpositions are 2(r-s)-1. Now, for our inductive step, we want to interchange r with s+1. It takes r-s adjacent transpositions to move s+1 to the initial r+1 position by assumption. It takes 1 more adjacent transposition to move s+1 to the initial r position. It takes r-s-1 adjacent transpositions to move r to the initial s position, by assumption. It takes 1 more adjacent transposition to mov r to the initial s+1 position. For a total of 2(s-r+1)-1 adjacent transpositions.

[fickle brook]

apparently every group algebra can be made into a coalgebra, and it is possible to compute the coalgebra operations (comultiplication, counit and antipode) but i am for some reason struggling to conceptualize what this would look like and why

[fickle brook]

been over an hour now. might as well try pinging <@&286206848099549185> if only to see how little that does anymore.

[languid moss]

Have you check out stock exchange

[next obsidian]

Lmfao, I assume you mean stack exchange?

[languid moss]

Yeah 💀

[fickle brook]

stonks

anyway, no.

i don't even really know what to look up tbh

[chilly ocean]

can anyone help me in the help channel

help-17

[hot tinsel]

Suppose $T:G\to End(V)$ and $S:G\to End(W)$ are finite dimensional representations (all linear spaces over $\mathbb{C}$). How can I show that $$dim(Hom_G(V\to W))=dim(Hom_G(W\to V))$$

[cloud walrus]

K零ꓘ

[hot tinsel]

I was thinking if I take a $A:V\to W$ which is $G-$linear then $A^t$ will be in $Hom_G(W\to V))$

[cloud walrus]

K零ꓘ

[red imp]

Hi, I have a free abelian group $\mathbb{Z}^3=\langle a,b,c\rangle$ and I want to determine what the quotient group $\langle a,b,c\rangle/\langle a-b+c\rangle$ is. Are there any linear-algebra-esque ways of doing this?

[chilly ocean]

map a to b-c, b to b and c to c into the free abelian group <b, c>
kernel of this map is <a-b+c>, so we get an isomorphism

[chilly ocean]

here

[red imp]

makes sense, thankssss

[odd flame]

Hey guys i am trying to find minimal polynomial of element $\frac{/sqrt{2}-/sqrt{7}+1}{2}$ over Q

[cloud walrus]

horridharry96

[odd flame]

but im very confused

$\frac{\sqrt{2}-\sqrt{7}+1}{2}$

[cloud walrus]

horridharry96

[odd flame]

Not really sure how to start here

[delicate bloom]

I would make my life easier and find the minimal polynomial of sqrt(2)-sqrt(7) since we can always shift and scale the polynomial by a rational number afterwards

[odd flame]

Alright

[noble dust]

Did you figure it out?

[delicate bloom]

also just fyi you can check your work with wolfram alpha

,w minimal polynomial of (sqrt(2)-sqrt(7)+1)/2

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Merosity#0135

[spice whale]

I'm doing 1.3
by first iso theorem on the determinant homomorphism i got $O_n/SO_n \cong C_2$ therefore $O_n/SO_n \times SO_n \cong SO_n \times C_2$

[cloud walrus]

ally 🌈

[spice whale]

not sure how to establish that $O_n / SO_n \times SO_n \cong O_n$

[cloud walrus]

ally 🌈

[tired parcel]

you can think of $O_n/ SO_n$ as equivalent classes right

[cloud walrus]

Toto_31416

[tired parcel]

it is a tedious computation

[spice whale]

yeah

[tired parcel]

but first decompose any matrix in SO_n into a representative from the equivalence classes and an element in SO_n

perhaps the singular value decomposition helps

[latent anvil]

You should not need svd for this

That is very much overkill

[latent anvil]

Are you sure this is true?

[spice whale]

it is
but I'm going about it a different way

[young fiber]

is this even true? O_2 has infinitely elements of order 2 but SO_2 x {I, -I} just has two such elements

[latent anvil]

The second one has three such elements, right?

(I, -I), (-I, I), (-I, -I)

But also O_2 is nonabelian while SO_2 × {I, - I} is abelian

[young fiber]

oh yea i forgot about (I, -I) (i was thinking about it geometrically)

[spice whale]

ah

it's only true for odd n

i believe

[dull root]

If I have the polynomial x^n - a in Z[x], then any rational root must necessarily be an integer right

by the rational root test

[rustic crown]

for odd it's true, {I, -I} and SO_n are two normal subgroups with product O_n and trivial intersection

[rustic crown]

yep!

[slate mortar]

Hello guys, can someone explain me (or find any reference explaining) why the wedge product on free modules is a perfect pairing?

[rustic crown]

well the observation that it has no rational roots is enough for this. think about it, how many way can you factor a degree 4 polynomial over Q such that no factor is degree 1?

[rustic crown]

ah, don't factor it over C. that gives us different information than what we're looking for

we're saying it has no rational root, which is same as saying it has no linear factor if you factor it in Q[x]

so among the partitions of 4
4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1
the only partition which doesn't have a 1 in it is 2 + 2

yep

ah, then directly use something like Eisenstein's criterion

yep!

[young fiber]

is the orthogonal group ever abelian?

[delicate orchid]

O_1(F) over any field is abelian

bit of a cheat but I don't care

[chilly ocean]

SO(2) is, O(2) isn't

to give an example

in general, SO(n) and O(n) for n > 2 will be non-abelian. the n = 2 case i outlined and the n = 1 case is trivial

[young fiber]

how would you show this? can you somehow 'embed' a non-commuting pair from O(2) in O(n) to create a non-commuting pair in O(n)?

[chilly ocean]

idk

maybe

[young fiber]

apparently you can do this for SO(n)

[chilly ocean]

SO(n) is a compact connected lie group, so if it were abelian it would be a torus. but its fundamental group is finite

(n geq 3)

:)

[young fiber]

i meant the embedding thing

to show it's non-abelian

[chilly ocean]

i know what you meant, im just bringing in some extra firepower

[young fiber]

oh cool, sounds fancy

[opaque pilot]

I understand why psi is completely determined by its value on 1. But Im having trouble following the next sentence, why psi = phi iff psi(1) = phi(1)

Also, what is capital phi's relation to small phi

[latent anvil]

Yeah they're just matrices

Make the 2 by 2 matrix into an n by n block diagonal matrix with the identity as the other block

[lapis trail]

So a 24 element abelian group is isomorphic to one of
Z8×Z3
Z2×Z4×Z3
Z2×Z2×Z2×Z3

[next obsidian]

Indeed

[red imp]

or Z_24

[lapis trail]

Oh yes

[vast quiver]

Z_24 is isomorphic to Z8xZ3 btw. You’ll know you’ve covered all of your options if you look at the elementary divisors. I.e, 24=2^3 * 3^1, so look at all the partitions of the powers (3 and 1), and list them.

3=3=1+2=1+1+1
1=1

Then, doing all combinations of those partitions, you get exactly the ones you originally mentioned. These are exactly all of the (isomorphism classes of) abelian groups of order 24, no need to worry about any others. There are multiple ways to write the (isomorphism classes) of these abelian groups, e.g. Z_24 vs Z8xZ3, but that’s no problem.

[celest cairn]

Hi, I wanted to make sure I had the right answers for this system of equations.
Is (-1, 5) correct?

[delicate bloom]

that's only one, there are 3 solutions

also -1=5 in Z_6 so might help you to restrict to residues from 0 to 5, like (5,5)

[celest cairn]

Ah ok thanks 🙂

[delicate bloom]

might help if you consider the chinese remainder theorem and what the solutions are for Z_2 and Z_3

[lapis trail]

Hey thanks. But can you explain why Z8×Z3 is isomorphic to Z24 but not the others?

[vast quiver]

as for showing Z8xZ3 is isomorphic to Z24, let's do something a bit more general. If n,m are coprime, then ZnxZm=Z(nm). We claim that Znm = <(1,1)>. To prove this claim, we want to show any element (x,y) can be written as k(1,1) for some k so that ZnxZm is cyclic, and once we know it is cyclic we are done, since |ZnxZm|=nm (since the underlying set is just the cartesian product).

Thus, let (x,y) be given, and we want to find k such that k(1,1)=(x,y), i.e., k=x mod n, k=x mod m. Since n,m are coprime, this k exists by the Chinese Remainder theorem. Thus, ZnxZm is a cyclic group of order nm, so it is isomorphic to Znm.

Now, let's see why, for example, Z2xZ2 is not isomorphic to Z4, we can look at orders of elements. If Z2xZ2 were isomorphic to Z4, then there would be an isomorphism f:Z4->Z2xZ2. Then, f(1) would have to be an element of order 4, but Z2xZ2 only has elements of order 2 or 1 (i.e., (0,0) is order 1, (1,0),(0,1),(1,1) are order 2)

[lapis trail]

ohhhhhhhhhhhh lightbulbs

it makes sense

[vast quiver]

aw yus

[lapis trail]

thanks a lot

[vast quiver]

there's also a theorem that says that two finite abelian groups are isomorphic if and only if they have the same elementary divisors

so you'd need to prove this theorem, but it would tell you right away that Z2xZ2 and Z4 are not isomorphic

[cloud walrus]

SkyTwX

[slate mortar]

For reference, this has been asked here:
https://math.stackexchange.com/questions/4372611/reference-request-for-proof-that-the-multiplication-map-is-perfect-pairing-for-e

[young fiber]

I see, thanks

[lapis trail]

Can anyone explain how they get 23 in the example?

It's CRT

[sharp sonnet]

$105=3\cdot 5\cdot 7$, so the CRT gives you an isomorphism $$\varphi\colon\bZ/105\bZ \to \bZ/3\bZ\times \bZ/5\bZ \times \bZ/7\bZ$$ the question is now to compute the preimage of $(2, 3, 2)$ under this isomorphism. the easiest way to do is is by brute force and computing the $\varphi(x)$ for $x=1, 2, \dots, 104$.
you can see that $\varphi(23) = (2, 3, 2)$, so this is the result

[cloud walrus]

Lochverstärker

[delicate bloom]

I think it's easier to call it x then start with x=2+3n then reduce mod 5 and you have 2+3n=3 mod 5, so 3n=1 mod 5, and so n=2 mod 5, plug in n=2+5m and now you have x=2+3(2+5m)=8+15m now one last time reduce mod 7, 8+15m=2 mod 7 and you have m=1 mod 7 so you can just use that directly, x=8+15=23

[north sand]

you can also just use the proof of the CRT. it's constructive

[chilly ocean]

I have heard somewhere that the order of a permutation is the LCM of the orders of its disjoint cycles from the decomposition.

Is this true? If yes can someone please cite a source showing this proof?

[prisma shuttle]

for this problem does $R_f$ mean the localization of $R$ by the ideal generated by f?

[cloud walrus]

JustKeepRunning

[prisma shuttle]

cuz idt u can just localize at an arbirary element, the thing u are localizing by has to be an ideal

and in fact in some cases i think it has to be prime

and also wut does it mean for a localization to equal 0 again

[junior harbor]

Can someone explain to me the concept of an invariant subgroup and quotient groups? The book I'm reading just isn't quite getting it through for me

[chilly ocean]

Do you understand the notion of cosets?

[rustic crown]

you localize at multiplicative subsets, not ideals.
ideals would always contain 0, and if you wish to invert it, the whole ring would collapse to the zero ring.

localizing at f refers to localizing at the multiplicative subset {1, f, f^2, ...}

[chilly ocean]

it would help if you shared what your book says so no one here ends up repeating exactly what you've already read

[junior harbor]

I think so, having a set A and another B, AB is the set with all ab with a in A and b in B

I think that's it? Unless I misread

[junior harbor]

Lectures in Abstract Algebra by Nathan Jacobson, Basic Concepts

[chilly ocean]

That's not really the definition

Okay let's go to subgroups first

[junior harbor]

Well then there's our first issue lol, lemme see if I can find it here

[chilly ocean]

Do you know what is a subgroup

Not invariant ones

As in general, what are they

[junior harbor]

iirc like a subset, but it fulfills the requirements for a group? And it is a subset of a group

[chilly ocean]

More or less

It is a subset of a group G

[junior harbor]

I have some trouble getting the concepts fixated sorry

[chilly ocean]

Which is also a group under the same binary operation of G

It's alright I also went through that

For example

Say a group of Integers

Which acts under a Binary operation +

Addition

So we define a group (G,+)

"A group G under addition"

[junior harbor]

Which means the positive integers are a subgroup?

[chilly ocean]

Actually no

[junior harbor]

I mean, with all the positive integers in a group I+ (includes negatives). It's closed under addition. The positive integers would be a subgroup no?

It is a subset of I, group under the same operation, and also a group by itself

[chilly ocean]

If you only consider positive integers

[chilly ocean]

It violates this

It is not a group

Since a group must contain an inverse element

[junior harbor]

Oh right

[chilly ocean]

And identity element as a result

[junior harbor]

Has to have an inverse, closed under the operation and identity

[chilly ocean]

And associative

[junior harbor]

Hm, ight

[chilly ocean]

These 4 are the axioms of a group

Under a binary operation

[junior harbor]

Ight

[chilly ocean]

And every subgroup is a group

Under the binary operation defined for the group

This is called Binary restriction

Basically you narrowed the domain of the operation that it can accept.

Okay?

[junior harbor]

Ight

[chilly ocean]

Now subgroup H of a group G under a binary operation + is defined as H < G

Confusing notation

But you read it as "H is a subgroup of G"

Okay?

[junior harbor]

Gotcha

[chilly ocean]

Let's move forward

Any group

Always has two trivial subgroups

Do you know what are they?

[junior harbor]

Itself and empty?

[chilly ocean]

Not really empty

Itself and the Identity

[junior harbor]

Right, I was gonna correct myself but discord died

[chilly ocean]

Oof

Alright so let's move to real definition of cosets now

So you have a subgroup H

Of a group G

Or in other words

H<G

Under the operation, say +

Not all operations are commutative

That is why we have the notion of left and right cosets

Okay?

[junior harbor]

Ight

[chilly ocean]

So first of all

We use cosets to divide the subgroup into equal disjoint sets

equal meaning of same size

disjoint meaning an element does not overlap between one or two cosets

[junior harbor]

Ok, but what would a coset really be?

[chilly ocean]

A set

It's really a set

Remember cosets are not groups

We divide the underlying set of a group into equal disjoint sets called cosets

a group is a set under an operation right?

We do that to the underlying set of that group

Okay?

Take ur time to grasp these concepts, when ready u can tell me to move forward

I think their discord died lol

[junior harbor]

Sorry for the delay, had to take my phone away for a moment

I get it I think

[chilly ocean]

Alright

Since not all operations are commutative

Like Subtraction and division

We have left and right cosets

Say a subgroup H of a group G

Under addition

Then the left cosets of H are defined as a + H, a in G

[chilly ocean]

Read from here again, made minor corrections

[chilly ocean]

$a+H = {a + h | h \in H}$

[cloud walrus]

Pencil/Idris

[hot lake]

wdym with "equal disjoint sets"

equal how ?

[chilly ocean]

Equal means of same size or cardinality

Afaik

[hot lake]

ah okay

[south patrol]

what about the trivial group

jk

[chilly ocean]

A trivial group (G,+) containing only identity e also has two trivial subgroups, subgroup containing e and the subgroup containing e

[south patrol]

lol

I have three friends called me myself and I

[thorny knoll]

Three friends, never claimed they were pairwise disjoint. ;)

[rustic crown]

hi moldi

[hidden haven]

het

[south patrol]

gay

[hidden haven]

So I can't even wave back without being called gay

[south patrol]

Why are you taking it as an insult

Anyway let's get back on topic

[junior harbor]

I am back, currently in somewhere people might just call me to talk so I'd have to go momentarily

But ok, that's a coset

[chilly ocean]

It's alright 👍
Take ur time

A left coset btw

A right coset is defined as $H+a = {h + a | h \in H}$

[cloud walrus]

Pencil/Idris

[chilly ocean]

For a subgroup under operation +

[junior harbor]

Got it

[chilly ocean]

Now if a subgroup has right and left cosets same (or equal), we define that subgroup as the invariant subgroup or a normal Subgroup or a normal divisor

[junior harbor]

So basically an invariant subgroup is like associative but for cosets?

[chilly ocean]

So if a + H = H + a for a subgroup (H,+), H is a normal subgroup of G

[chilly ocean]

Associative?

[junior harbor]

Like

a + b = b + a

Commutative

Is what I mean

[chilly ocean]

That's called commutativity

[junior harbor]

I think

[chilly ocean]

Yes

Associativity is different

[junior harbor]

Yeah sorry lol

[chilly ocean]

It is a+(b+c) = (a+b)+c for a binary operation +

Okay

So I hope you got what is an invariant subgroup right

[junior harbor]

Yep, commutative law but with cosets. The book makes it mighty hard to get tho

Need at least 50+ more iq points for their definition

[chilly ocean]

Equivalently we can also say if $xax^{-1} \in H$ for every $x \in G$ and $a \in H$ where $H ≤ G$

Btw

H<G means "H is a proper subgroup of G", meaning H≠G

[cloud walrus]

Pencil/Idris

[chilly ocean]

And H≤G "H is a subgroup of G" where H can be equal to G

[junior harbor]

Gotcha

[chilly ocean]

Remember that for a group of Integers under addition, every subgroup of that group is a normal Subgroup

Because Z is abelian (Z is group of Integers here)

Abelian group is a group that is commutative under the defined operation for G

Integers are commutative

[chilly ocean]

Also I would like you to yourself find out why this is true as an excercise

[junior harbor]

Well, I'd think that's because if we have a group Z of integers, say {-a, -a + 1, ... , 0 , ... , a - 1, a}, a subgroup of that would be "shaving" off the ends of it, so for b <= a, the subgroup S would be {-b, -b + 1, ... , 0, ... , b - 1, b}. Now, that is still kind of the same thing as the group Z in that it is abelian (I think that'd be it?)

So this means it is normal