groups-rings-fields

[potent briar]

its coming soon

sirect sum, dual of direct sum, and then quotient space

[opal osprey]

Nice, good luck with your studies

[potent briar]

thank you and thanks for the help again

it's hard to get into serious maths as a hobbyist

[wooden ember]

lmao seems overkill to conclude that the totient function is multiplicative from the CRT

[sharp sonnet]

is it? CRT is super simple to prove and i know one proof of this that avoids that and it's kinda weird

[rustic crown]

yea CRT proof is a lot nicer... not just that, the groups of units functor is right adjoint to the functor which sends a group G to the group-ring Z[G], and so the fact that it preserves products or any limits is immediate.

[wooden ember]

fair it just feels weird i guess if this was done in isolation: like if someone randomly asked you about the totient function would you start teaching them about rings or give them a non CRT dependent proof

though ig you could prove it via the special case of the CRT in Z...

and so make no reference to rings

[sharp sonnet]

you should teach CRT anyways, its important
also for reference the non CRT proof i know considers a matrix like picture and then argues that in each column either every or no element is coprime to m and phi(n) element are coprime to n

[wooden ember]

okay ive just realized the only proof i had previously seen that the totient function was multiplicative also involved CRT 🤦‍♂️

so forget what i said lol

[cloud walrus]

det

[wooden ember]

ew

[rustic crown]

[wooden ember]

i dont remember convolutions much but im pretty sure ive seen this proof actually

[golden pasture]

i feel like jacobson algebra used this proof but implicitly

[sharp sonnet]

when you are an analytical number theorist and you have to do an elementary NT class

[chilly ocean]

how do polynomials behave over quaternions?

like when they have quaternion coefficients

[robust patrol]

<@&286206848099549185>

[robust patrol]

They are not constrained at all

[viscid pewter]

i mean, for a start, quaternions are non-commutative, right?

so you can wave goodbye to the fundamental theorem of algebra

[plucky flicker]

In an arbitrary group G we consider the commutator of two elements: [x, y] = x^(-1)y^(-1)xy, and i simply want to give an example of three elements, let's say a, b, c as a counterexample to prove that the commutator operation is not associative. I guess it's easy because if a = b and c is a third element, then [[a, b], c] = 1 but [a, [b, c]] is not 1. I just need a confirmation, if it's a good solution or not

[viscid pewter]

that should work

[plucky flicker]

Thanks

[cloud walrus]

Yeetus

[rustic crown]

say f = sum of a_i * x^i

derivative is sum of i * a_i * x^(i-1)

which needs to be 0, so we have i * a_i = 0

what can you say when p doesn't divide i?

so Fp is a field right?

and p doesn't divide i means that i is not 0 inside F_p

does this say something about a_i?

i * a_i = 0 in F_p, but i isn't 0

any thoughts?

[rustic crown]

yep!

so if p doesn't divide i then, we know the coefficient a_i = 0

so recall that f = sum of a_i * x^i

only the multiples of p matter

f = sum_{j} a_{pj} x^{pj}

so you can take g(x) = sum_j a_{pj} x^j, this way f(x) = g(x^p)

[rustic crown]

it factorizes as 3(X^4 + 2X+2) in Z[x] as the primitive polynomial X^4 + 2X + 2 is irreducible by eisenstein, and by gauss it's is irreducible over Q as well

well in Q[x], 3 is a unit so usually you don't care about it

but in Z[x], that 3 is a different factor

so its correct to say that 3X^4 + 6X + 6 is irreducible over Q, but it's not irreducible over Z

3 is a unit in Q[x] right... 3 * 1/3 = 1

but in Z[x], 3 is another irreducible factor

is it not called unit elsewhere?

i see

oh okie

right

better to write it like 2 * 2 * (x^2+2) in Z[x]?

2 is irreducible in Z[x]

like if your problem was to factor it completely... that's how one would do it

yee

yea you can... but do those factors matter given that they are units?

you saying something like what if you factorize 6 as 2 * 3 and (-1) * (-1) * 2 * 3

that -1 does no real work here, same with that 4 in 4(x^4+2) if you're looking it in Q[x]

yea, that polynomial is irreducible in Q[x], so there isn't any way to factorize it non-trivially

these things just matter upto conventions, like people prefer monic irreducibles if they wanted to choose among all irreducibles which are same upto a factor of a unit

like in Z, you only think of positive primes as prime, -2 and -3 and other things are also irreducible, but we just pick on nice candidate which we like

yep

so 2 and -2 are both irreducibles, but people usually choose the postive one if they really have to pick one

okie

[rustic crown]

looks like it

if you go modulo 3, then it factors as (x^3 - x - 1)*x

so if this polynomial was reducible, then it would have a linear term and a cubic term

but the linear term would be something like x-3k, so just check if it has a root of the form 3k

okie it does actually

-3 is a root

so that gives you a factorization

(x+3)(x^3 - 3x^2 +2x - 1)

imma sleep now

[plucky flicker]

If A, B and C are normal subgroups, then [AB, C] = [A, C] [B,C]. Can someone give me some hint regarding the solution of this? I tried to prove by two sided inclusion but I didn't get anything

[unique berry]

wtf is this question even describing

[next obsidian]

It’s kinda like the thing describes

If you’ve seen this for topological spaces, it’s like gluing along a homeomorphic subspace

So like you’re taking the product and identifying the parts of each group which are the same, this is kind like if you take say… the top of a sphere and the bottom of a sphere

Then the “rim”s are isomorphic so we can glue the two halves of the sphere along that, you can take the disjoint union then identity the parts which are the same

This is sorta like that kind of thing except we’re taking the product and then quotienting out because that’s what’s more natural to do here

Part (a) shows that you can embed A and B into this central product since there’s an isomorphic copy of them

But whereas in a normal product A x B the copy of A and B are disjoint except at the identity (as the subgroups A x {e} and {e} x B), because we’re quotienting out by stuff they now meet non-trivially

And it’s “along Z1 and Z2” that we’ve glued them

[past temple]

this might sound really stupid

but given a ring A

is it true that A[x][x] = A[x]

[next obsidian]

No

It’s A[x,y]

A polynomial ring in two variables is (sometimes defined to be) a polynomial ring over a polynomial ring

It’s also impossible for them to be isomorphic (assuming A is commutative) anytime A has finite Krull dimension

Because adjoining a variable strictly increases Krull dimension

(I mean unless it’s infinite but… yeah)

[past temple]

ah i see

how does one define a monic polynomial

in A[x,y]

[next obsidian]

Depends

Or…

Yeah it kinda depends, the easiest way is to consider it just in one variable

You can consider a polynomial f(x,y) as being written as

f_n(x)y^n + f_{n-1}(x)y^{n-1} + … + f_1(x)y + f_0(x)

Or you can swap x and y

Then you can say it’s “monic in y” or “monic in x” respectively

In general though you don’t really have a good notion of monic because you don’t have a unique monomial of highest total degree

See like x^2 + 3xy + 1/2y^2 or something

[past temple]

i see

im reading this example on wikipedia and it makes sense

im working with integral closures so this deifnition is appropriate (i think?)

[next obsidian]

Well it depends

But yeah if you’re looking at things and asking if it’s integral over R[x]

Then yes you take the definition of monic as outlined here

[next obsidian]

And what I outlined here

[past temple]

gotcha

[strong moth]

could someone check my proofs

I would like proofs checks on two theorems

(1) Prove for any set A is isomorphic to itself

For there to be an isomorphism between sets X and Y we need functions

f: X -> Y and g: Y -> X

that satisfy the following compositions:

(g o f) = id_X

and

(f o g) = id_Y

we will check these compositions are satisfied.

let g=id_A: A -> A

we see that by substitution

(id_A o id_A) = id_A

let f=id_A: A -> A

and we see that by substitution

(id_A o id_A) = id_A

thus the compositions for a isomorphism between f:A->A are satisified and hence A is isomorphic to itself

(2) Prove for any sets A, B id A is isomorphic to B, then B is isomorphic to A

Proof:

Assume A is isomorphic to B then there are functions

f: A->B

g: B->A

Such that

(g o f) = id_A

(f o g) = id_B

B is isomorphic to A if we have the compositions:

(f o g) = id_B

(g o f) = id_A

But due to A being isomorphic to B, we already have such compositions satisfied

Therefore B is isomorphic to A

[tribal pasture]

For the first one, You need to define f and g before the first " we see that by substitution" since you need some value of f to plug it in.
Secondly, You also need to show why id_A o id_A = id_A .

For the second, it would be more clear to rewrite the compositions with different letters (say h and k) and then state that if you pick h to be f and k to be g, you obtain your desired isomorphisms.

[strong moth]

few questions @tribal pasture what do you mean by defining what f and g are and showing why (id_A o id_A) = id_A?

like for all a in A there exist a b in B such that f(a)=b and for all b in B there exist an a in A such that g(b)=a and for all a in A id(a) = a?

[tribal pasture]

By the defining part, I mean write "let f=id_A: A -> A" and "let g=id_A: A -> A" before the substitution line. By the second question, I mean, write why do you think id_A o id_A = id_A is true

[strong moth]

for h,k thing would you assume that

h: A -> B and k: B -> A

and are there any assumptions made whether they are distinct mappings from f and g?

[tribal pasture]

For the h,k thing, I mean write "B is isomorphic to A if there exists h: B -> A and k:A -> B such that they satisfy the composition laws .... " And then find such an h,k.

[strong moth]

i went back to write it before the substitution line and was a little confused isnt it already before that line?

what do you mean instead? and why would i do the h,k thing in the second proof but not in the lines below in this proof?

“we will check these compositions are satisfied.

let g=id_A: A -> A

we see that by substitution

(id_A o id_A) = id_A

let f=id_A: A -> A

and we see that by substitution

(id_A o id_A) = id_A”

[tribal pasture]

I mean
"let g=id_A: A -> A
let f=id_A: A -> A
we see that by substitution
(id_A o id_A) = id_A
"
(Since your composition law consists of both f and g, you should specify what value for f and g (both) are you substituting)

[strong moth]

so you’re combining both steps at once?

and how can i show why (id_A o id_A) = id_A? isn’t that by definition and compositions of the definitions?

[tribal pasture]

No, so see your composition law is (f o g) = id_A.
So it requires both f and g. If you only state g = id_A and substitute that, the resulting equation is
(f o id_A) = id_A

[strong moth]

ah, i see. it’s like a computer program. calling functions before they’re defined in lines of code

[tribal pasture]

Yes, but you should show that. More precisely, write specifically what definitions are you appealing to and what laws are you appealing to arrive at this identity.

[strong moth]

@tribal pasture

ok. i think i have enough to finish that one off tomorrow when i wake up

in regards to (2) im kind of confused by h,k business

let’s say i define

h: A -> B
k: B -> A

1. are these h,k functions different than my f,g functions?

f: A -> B
g: B -> A

then my argument would be something like for B to be isomorphic to A we need to show the following compositions hold:

(equations 1)

(h o k) = id_B

(k o h) = id_A

and

let h=f and substitute f into equation 1 for h

f o k = id_B

k o f = id_A

similarly in equation 1 let k=g and substitute g into equation 1 for k

h o g = id_B

g o h = id_A

is that what you meant?

[uneven folio]

[prisma ibex]

the advice you've been given about h and k is really bad, it overcomplicates an already overcomplicated proof

the original proof you have is valid, it's just incredibly long winded

show that A is isomorphic to A
the identity function id_A:A->A is its own inverse, hence an isomorphism.
show that if A is isomorphic to B, then B is isomorphic to A
if A is isomorphic to B then by definition we have a function f:A->B with inverse g:B->A, that is gof=id_A and fog=id_B. But by symmetry this means we have a function g:B->A with inverse f:A->B, that is fog=id_B and gof=id_A, hence B is isomorphic to A.

[tribal pasture]

I think this symmetry argument is good if you are already familiar with the proof techniques, but for someone, who is starting out new with such proofs, I think it is more useful to be perfectly clear with the usual definitions and strategy. That's the usual strategy involves "We will find maps such that the compositions are identity." And in this case, finding those maps is basically f,g. However, I presume this is merely a stylistic difference, and whatever helps the OP. I felt to clarify because the advice was termed as "really bad".

[prisma ibex]

I mean sure you can rephrase it in any way you choose, the point is that there is absolutely no need to overcomplicate things by introducing h and k like this, and there is no need to belabor what is already a really simple proof

[tribal pasture]

Yeah, I agree. There are proofs that are simple and there are proofs that are useful in building one's familiarity in writing proofs. I do agree that your proof is simple, my intention was to make the person familiar in following through in writing a proof that works generally in problems like this.

[tribal pasture]

Yes, the reason I asked you to define h,k separately is because sometimes h,k would be different from f,g (but luckily in this case h,k are equal to f,g).
Secondly, you can also combine the steps of substituting h = f, g = k, once you get comfortable with such substitutions.

[hidden haven]

That sounds very unnecessary. "We need to show that there is x such that P(x). Define x = something, and then P(x), so there does exist such an x" is a common proof phrasing and there is really nothing wrong with this even at a foundational level (even the lean prover uses phrasing like this). Introducing h,k is also not good from an understanding perspective because you're just increasing the amount of things to keep in mind

"Define y = something, check P(y), set x = y" is not really needed lol

[tribal pasture]

I believe I was suggesting the same. "We need to show that there is h,k such that compositions are identities. Define h = f, k=g, then compositions are indeed identities and so there does exist such map".

[hidden haven]

(1) Prove for any set A is isomorphic to itself

For there to be an isomorphism between sets X and Y we need functions

f: X -> Y and g: Y -> X

that satisfy the following compositions:

(g o f) = id_X

and

(f o g) = id_Y

we will check these compositions are satisfied.

let g=id_A: A -> A

we see that by substitution

(id_A o id_A) = id_A

let f=id_A: A -> A

and we see that by substitution

(id_A o id_A) = id_A

thus the compositions for a isomorphism between f:A->A are satisified and hence A is isomorphic to itself
apart from moving the "let f = ..." line 2 lines up, I see nothing wrong with this

or are you talking about some other proof

[tribal pasture]

That's what I stated as well, apart from stating that maybe the person should also clarify why they claim id_A o id_A = id_A

[hidden haven]

isn't that the definition of identity morphisms

[tribal pasture]

I mean yes, but if this was a homework exercise, graders would deduct marks, I believe, if this was an introductory class. So just to be safe writing that out would have been better.

[hidden haven]

I am sorry for what your graders have done to you lol

But deducting marks for this is dumb

[prisma ibex]

this absolutely does not need to be clarified

[tribal pasture]

My bad then.

[chilly ocean]

there are some pretty bad graders out there

better safe than sorry

[robust pollen]

I have a question about the snapshot from this paper https://arxiv.org/abs/0805.3806 .

Namely, the associativity seems wrong to me. To me, it seems like that should be $\chi(a a') = \chi( a (\eta \circ \chi)(a'))$, no?

[cloud walrus]

expectTheUnexpected

[robust pollen]

Like, if $r \cdot a = r \chi(a)$ is the right $A$-action on $R$, then $r \cdot ab = r \chi( (\eta \circ \chi)(a) b)$ by (ii), but $r \cdot a \cdot b = r \chi( a (\eta \circ \chi)(b))$ by (i).

[cloud walrus]

expectTheUnexpected

[hidden haven]

Algebroid

[robust pollen]

Btw, the solution to my problem is that I didn't read on 🙈 My naive action isn't the action that's being used, but it's rather defined as $r \cdot a = \chi( \eta(r) a)$, and then it indeed checks out

[cloud walrus]

expectTheUnexpected

[weary terrace]

Hello, brace yourselves, this is a long one:
say we have a 1-dimensional representation $\rho$ of a group $G$. Denote $H=ker \rho$.
Let $\sigma_{i}^{j}$ be irreducible representations of $G$ for all integers $0\leq i \leq n-1$ and $0\leq j \leq l-1$.
In addition, $Res_H \sigma_{i}^{j} = Res_H \sigma_{0}^{j}$ for each $0\leq j \leq l-1$.
Moreover, $Ind_{H}^{G} (Res_H \sigma_{0}^{j}) = \sigma_{0}^{j} \oplus \dots \oplus \sigma_{n}^{j}$ for each $0\leq j \leq l-1$.
Prove that $Res_H \sigma_{0}^{j}$ for each $0\leq j \leq l-1$ are irreducible representations of $H$ and that they are the only ones.

[cloud walrus]

Gromov

[weary terrace]

All I got so far is that $Ind_{H}^{G} (Res_H \sigma_{0}^{0} \oplus \dots \oplus Res_H \sigma_{l-1}^{0}) = \mathbb C [G]$ i.e. the regular representation of $G$ which means that $Res_H \sigma_{0}^{0} \oplus \dots \oplus Res_H \sigma_{l-1}^{0} = \mathbb C [H]$

[cloud walrus]

Gromov

[weary terrace]

Forgot to mention that we're over the field of complex numbers

[robust pollen]

Something is off with the indices in your first post (i.e. when you write Ind Res sig^j _0). Also, do you mean to say that \sig^j_i form a set of representatives of isoclasses of simple reps?

And in the second post, you also have the indices switched, i.e. it should probably be $\operatorname{Res}H ( \bigoplus{j=0}^{l-1} \sigma^j_0 ) = \mathbb{C}H$,

[cloud walrus]

expectTheUnexpected

[weary terrace]

Sorry, too many indices. Correct version for the first message:
say we have a 1-dimensional representation $\rho$ of a group $G$. Denote $H=ker \rho$.
Let $\sigma_{i}^{j}$ be irreducible representations of $G$ for all integers $0\leq i \leq n-1$ and $0\leq j \leq l-1$.
In addition, $Res_H \sigma_{i}^{j} = Res_H \sigma_{0}^{j}$ for each $0\leq j \leq l-1$.
Moreover, $Ind_{H}^{G} (Res_H \sigma_{0}^{j}) = \sigma_{0}^{j} \oplus \dots \oplus \sigma_{n}^{j}$ for each $0\leq j \leq l-1$.
Prove that $Res_H \sigma_{0}^{j}$ for each $0\leq j \leq l-1$ are irreducible representations of $H$ and that they are the only ones.

And that's the second:
All I got so far is that $Ind_{H}^{G} (Res_H \sigma_{0}^{0} \oplus \dots \oplus Res_H \sigma_{0}^{l-1}) = \mathbb C [G]$ i.e. the regular representation of $G$ which means that $Res_H \sigma_{0}^{0} \oplus \dots \oplus Res_H \sigma_{0}^{l-1} = \mathbb C [H]$

[cloud walrus]

Gromov

Gromov

[strong moth]

@tribal pasture i attempted to incorporate your h,k recommendation into my proof for (2), see if this is now valid:

assume A is isomorphic to B then there are functions

f: A->B
g: B->A

such that

(g o f) = id_A
(f o g) = id_B

B is isomorphic to A if we have the following functions and function compositions:

h: A->B
k: B->A

(h o k) = id_B
(k o h) = id_A

but due to A being isomorphic to B, we already have such compositions satisfied,

we can see this if we let

h=f
k=g

and we see that

(h o k) = id_B
(k o h) = id_A

Therefore B is isomorphic to A

[lethal cipher]

This isn't quite enough. Judging by this, I am not sure you have proven that the inverse of an isomorphism is also an isomorphism (this is necessary).
You have showed bijectiveness, sure. But not necessarily that the inverse is also a homomorphism

@strong moth In case you haven't seen it yet, your argument is not enough to show B is isomorphic to A

[strong moth]

ok i just saw due to your ping

im really confused because i had another argument and someone said it's correct and then someone else said it wasnt and to use h,k stuff so i wasn't sure what they meant

[lethal cipher]

Moreover, we also know that the inverse of a bijective function is bijective, so much of your argument isn't really necessary.

What was your original argument?

[strong moth]

i'll find it hold on

[strong moth]

(2) is the original argument

[lethal cipher]

So here is the issue. All of your arguments just check to see if their is a bijective map between them.
However, you still need those maps to be homomorphic

[hidden haven]

They're trying to do this in a general category setting

[strong moth]

correct

so in the general category theory setting, is my argument correct, and if so, which version

[hidden haven]

I don't see a problem with the argument yeah

[strong moth]

either one of them, both the original and h,k one?

[lethal cipher]

Why are we ignoring the homomorphism part of an isomorphism?

[hidden haven]

In a category a morphism is just an arrow

Like you don't need to check anything

If it exists, it is a morphism

[hidden haven]

Didn't see the original, this one seems fine

[lethal cipher]

Oh, well then this is a much easier isomorphism to deal with than for groups

[strong moth]

this being the h,k one? i'll post in #category-theory next time.

sorry for the confusion

i incorrectly assumed that cat theory and abstract algebra would be the same proofs

[hidden haven]

This is a more general kind of isomorphism than for groups

In the category of groups the morphisms are exactly the homomorphisms

So even there, if some morphism exists then it's a homomorphism

[lethal cipher]

Ah, so what is a category exactly?

[strong moth]

isn't a category of a group a cat w/ one object and arrows are group homomorphism?

and if so whats the diff between a cat of a group and cat of monoid

[hidden haven]

It is a collection of objects and arrows between them, in a way that any 2 arrows with matching heads and tails may be composed, and the composition is associative and has an identity arrow at each object

[regal carbon]

I didn't understand the question

[hidden haven]

So the category of groups is the category in which the objects are groups and the arrows (morphisms) are homomorphisms

[hidden haven]

Not a single group, the above description is of the category of all (small) groups

[strong moth]

@hidden haven what is the reason i need to label h,k the way i did, as oppose to just say given the isomorphisms between A,B such compositions for the isomorphism between B, A already exist, and just write out those function compositions out expliclty and say done?

[lethal cipher]

So I have a question. I need to describe the automorphism group of Z_2xZ_2 (by showing an isomorphism with a familiar group).
I know that the set of automorphisms is just permutations of (0,1),(1,0), and (1,1), but how do I show this concisely?
Right now, it feels like my only option is to check each one is homomorphic (which is a bit tedious)

[hidden haven]

Belongs in #proofs-and-logic

[hidden haven]

If you are phrasing it as "A ≈ B therefore there are f and g such that [...]. For B ≈ A we need f and g such that [...]" then it's wrong because you're using the same symbols again for something different

[strong moth]

ahhh that makes sense

[hidden haven]

But if you phrase it as "A ≈ B therefore there are f and g such that [...]. But then g and f are such that [...] and this B ≈ A" then it's fine

Notice the reversed order of g and f in the second part just to emphasize that they exchange places when you do the check for B ≈ A

[strong moth]

ok, i see i essentially had both proofs correct

because i understood the core ideas

i needed help with phrasing things correctly

[hidden haven]

No need to check them individually, just call a permutation sigma and check it in a general form

[strong moth]

kind of rephrasing what you said phrasing it as "A ≈ B therefore there are f and g such that [...]. For B ≈ A we need f and g such that [...]" -- this is wrong because i am saying for B ≈ A then the composition of the functions need to match the order of the function evaluations in the compositions as written for A ≈ B, but thats wrong because f and g switch places in the evaluation order so instead of g and f being in this specific order (g o f) i need it to be in this order (f o g)

[hidden haven]

Another way to do this is to notice that Z2 x Z2 is exactly a 2 dimensional vector space over the field Z2 so the number of automorphisms is exactly the number of (ordered) bases, and it's easy to see that those are exactly ordered pairs of non zero vectors in this case.

[lethal cipher]

So how might I do this?

[hidden haven]

Yeah kinda

[strong moth]

maybe the word i am trying to use is there is a function composition 'template' for what an isomorphism is and im doing variable over-loading on the template

so i use new variables and say if they match the pattern of (h o k) = id_B and (k o h) = id_A then there is an iso between B,A oh wait, i already have that satisfied by the iso between A, B

where (hok) and (koh) is my 'template' that has to be satisfied

[hidden haven]

Call the permutation p. Then a ↦ p(a) is an automorphism: it's a bijection because that's what permutations are. It's a homomorphism because: handle the case for when one of the 2 summands is 0 separately (pretty easy) and for the other case notice that the sum of 2 distinct non zero things is always the third non zero thing. For the case where you have p(a+a), this is p(0) which is 0 which is p(a)+p(a)

This is still kinda brute force so way better to do the second thing using Z2 vector spaces

[hidden haven]

Yes

[strong moth]

@hidden haven i feel like these proofs are literal computer programs

thats how i make sense of them interallym

i only know python, but this compels me to learn haskell lol

btw ill post cat questions in #category-theory and will post actual algebra questions here. anyways, ty all for the help, highly appreciated

[untold cloud]

Hi, guys. Why here it emphasise integral domain rather than unital ring?

[hot lake]

it's false for general rings

for example in Z/4Z [X], (1+2x)² = 1 so 1+2x is invertible, but is not a constant polynomial

[wraith obsidian]

Weird question, but can we derive associativity of the symmetric difference Δ by noting that Δ: 𝒫(X)×𝒫(X) → 𝒫(X) has Ø as a neutral element, and every nonempty set is its own inverse?
In other words, can we rewrite a(bc)≈(ab)c using only ae≈a and aa≈e (where e is a constant)?

Hm, simpler question, can we even derive a(ab)≈b?

[untold cloud]

So the idea for integral domain is you cannot eliminate the nonconstant term by another nonzero polynomial, so you will always keep nonconstant term in product?

[hot lake]

yeah if you look at the monomials of highest degree

over an integral domain, the degree of the product of two polynomials is the sum of the degrees of the polynomials

and since 1 has degree 0, well then your invertible polynomials must have degree 0

[untold cloud]

Thank you!

[robust pollen]

I remember this rule differently. "If it exists, there is ..." wait, nevermind

[robust pollen]

Do you know this? https://en.wikipedia.org/wiki/Curry–Howard_correspondence

[strong moth]

reading

[robust pollen]

It's off topic, but it's the "proofs as programs" viewpoint.

[strong moth]

it appears to be saying there is a bidrectional correspondence between proofs and programs, i.e., you can convert one into the other and they are really the same "thing"?

like there is a system of types or logic in programs and you can encode proofs into these type/logic systems?

[hidden haven]

Yeah a program (or a function) of a certain type corresponds to a proof of that type viewed as a statement

[strong moth]

where can i learn type theory 101 at

i want to start w/ haskell

to learn its type systems

can you prove things using haskell?

[robust pollen]

look up Lean and waste your entire life on that (or become kenny lau and get incredibly good at lean and maths at the sam time)

[hidden haven]

Haskell can probably be learned from some random online tutorials

For CH correspondence you should probably look at a book or lecture notes

[strong moth]

i found these lectures from hutton (associated with his book on haskell) https://www.youtube.com/c/GrahamHuttonNotts/playlists

i'll start with these

[hidden haven]

oh neat

[strong moth]

i heard cat theory related to haskell

type systes

i can learn both

[hidden haven]

Yeah but it is not super useful from what I know

[strong moth]

i suspected that lol

so where is category theory actually used in a natural way in cs that makes sense

[hidden haven]

You can view certain operations as functors on the category of types in Haskell

[simple valley]

when all you have is a hammer everything looks like a nail

[simple valley]

recursive data structures are related to algebras on functors

structurally recursive buildup/teardown algorithms are also related to this

cartesian closed categories capture the essence of (pure) computation

lambda calculus is the internal language of a cartesian closed category

this connects to the church-turing thesis

then for effects you have various ways of encoding/representing them

the most popular one is monads and monad transformers (functors on the category of monads)

data aggregation is related to semigroups and monoids

then there's profunctor optics

using various kinds of profunctor to represent what is essentially a pure functional reference

[strong moth]

@simple valley this is insanely interesting

ty

i decided i’ll learn haskell now

[oak grove]

so i have a feeling x=4

im stumbling through proving it

well thats an overstatement im not sure how to start

maybe just enough to say "its 4" and nothing else

[weak oriole]

Have u studied the first isomorphism theorem?

[oak grove]

I think we saw it briefly

let me review

hmm im not really sure how id use that here

so $\phi : \frac{\bZ _{24}}{\langle 4 \rangle} \to \psi( \bZ _{24} )$ by $g \langle 4 \rangle \mapsto \psi(g)$ is an isomorphism?

[cloud walrus]

jan Niku

[weak oriole]

Yup

[oak grove]

[viscid pewter]

well

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

[weak oriole]

And $\psi(\bZ_{24})$ sits inside $H$

[cloud walrus]

Noob666

[viscid pewter]

ok but even more simply

[viscid pewter]

.

[oak grove]

I'm not sure I follow what you mean

so you have to pick some x to ensure that its inside H?

[weak oriole]

If you've understood the theorem, it should be obvious
Otherwise I suggest reviewing

[weak oriole]

Ya

[oak grove]

i guess i dont really understand the pieces of the theorem

i have a lot of reviewing lol

[weak oriole]

Can u say give a goodish classification of all homomorphisms $\bZ_n \to \bZ_m$

[cloud walrus]

Noob666

[viscid pewter]

to be clear, phi(Z_24) is precisely H, not just inside/a subgroup

surely

[weak oriole]

Aren't u assuming that the homomorphism is surjective

[viscid pewter]

oh

[oak grove]

hmm i know what a homomorphism is im not sure what statements you could make that generally

[weak oriole]

$\bZ_n$ are rather nice groups, so you can say a lot actually

[cloud walrus]

Noob666

[oak grove]

are you saying about the mapping or n and m

[weak oriole]

mapping

n,m are arbitrary (in the thing I asked)

[oak grove]

im sorry im really not sure

we only saw homomorphisms a few days ago

well wednesday

[weak oriole]

Oh, like have u done cyclic groups?

[oak grove]

yea

[weak oriole]

Ya that's all u need

And maybe some elementary number theory

[viscid pewter]

i mean if it's not surjective aren't there... multiple answers, at all?

[weak oriole]

Ya, there are
|| all multiples of 4 ||

[oak grove]

i mean you can make statements about m and n but im not sure what you mean the mapping

[viscid pewter]

yeah thought so

[oak grove]

unless you mean generators

[viscid pewter]

here's another way to think about it: G/ker(phi) = im(phi)

so what group actually is im(phi)

[weak oriole]

You can characterize the whole homomorphism by the image of a generator (say 1)

[oak grove]

sure

youre saying this relates to the first isomorphism theorem?

[weak oriole]

No, I'm saying that this helps characterize all homomorphisms between those two cyclic groups

and gives a way to solve ur problem without the isomorphism theorem

[robust pollen]

Why do people always call modules algebras tho :(

[simple valley]

are they modules?

@robust pollen explain

[robust pollen]

So where I'm from, the canonical example of a monad is given by tensoring with an algebra, in which case the "algebras" for the monad are exactly modules for the algebra.
Another point of view is that "algebras" over a monad M are actually right M-modules when considered as constant endofunctors (i.e. an object gives a constant endofunctor, and that functor is a left module over the algebra M in End(C))

But of course it doesn't matter, I along with many people I know seem to prefer the terminology "module" though.

[simple valley]

these are not algebras over a monad

these are F-algebras

[robust pollen]

Well, people say "the algebras for this monad are.." though

[simple valley]

yea but I'm not talking about those?

[robust pollen]

Wait what

Oh

[simple valley]

algebras of a monad M are certainly F-algebras for F=M, but the kind of F-algebras you usually encounter when talking about recursive datatypes are not algebras of a monad

[robust pollen]

I see, didn't know those
I someone says functor and algebra, my brain goes monadic, mb

[simple valley]

consider the (1+) functor on some (cocartesian) category C

it certainly is a monad (factoring into an adjunction between C and 1 \downarrow C)

but the monad algebras here would be maps 1+X -> X, that fix the X part

i.e. an object with a chosen point

F-algebras on the other hand are any maps 1+X -> X

and we are interested in the category of all such F-algebras

because we want to claim existence of an initial object there and make use of its universal property

[robust pollen]

• = \sqcup?

[simple valley]

coproduct yea

such an initial algebra would be the natural numbers object in C

[lethal cipher]

So I am having a bit of trouble with this: Let G be a finite abelian group where over half of the elements are their own inverse. Show that all elements are their own inverse.

I can't seem to find anything notable that will force that. I could use some help figuring out how to approach this.

[next obsidian]

It’s probably pigeonhole

My suggestion is, for every element which is its own inverse g

Look at the set of all g + h

Do some weird shenanigans with the sizes of sets

Something might pop out

[simple valley]

it really is just that

[lethal cipher]

I'm not sure what weird shenanigans we are talking about here

[next obsidian]

Like

There ought to be a clash

Between some g + h = g’ + h’

[simple valley]

suppose A is the set of elements that are their own inverses and g is in G\A

[next obsidian]

Where g,g’ are distinct

[simple valley]

what can you say about g+A

[lethal cipher]

Well, I can say that gA=Ag, but I'm not sure that's very helpful here

[simple valley]

can g+A intersect A?

[lethal cipher]

No

[simple valley]

and how big is g+A?

[lethal cipher]

Ohhhhhhh

gA is the same size as A

I feel silly for not thinking of this

[hot lake]

can't you show that the set of elements who are their own inverses is a subgroup, then use lagrange ?

[simple valley]

true

[split flicker]

Local Algebra

[lethal cipher]

Thank you for the help mniip!

[oak grove]

are you saying find what some generator would have to get mapped to here then figure out where everything else would get mapped to then show you end up with Z_4?

[oak grove]

maybe ill try to start further back and work there

[weak oriole]

Figuring out image of just one generator suffices
But yes. Also x=any multiple of 4 works

[oak grove]

any multiple okay

[lethal cipher]

Okay, final question. I need to show that Aut(Z_2xZ_2) is isomorphic to S_3.
The automorphism group is literally cycles of non-identity elements, so it is not hard to see why this is.
However, I am having a hard time figuring out how to do this concisely, instead of case by case (which is a bit much)

[hot lake]

what do you mean by "The automorphism groups are literally cycles of non-identity elements"

[lethal cipher]

For example, ((0,1),(1,0)) is an automorphism of Z_2xZ_2.
So (0,1) maps to (1,0) and (1,0) maps to (0,1). It really is just a cycle

[stoic rose]

A neat way to do it is to notice that its automorphisms as a group are the same as its automorphisms as an F2 vector space

Which are just given by invertible matrices

[hot lake]

so you are saying that automorphisms permute the three non-identity elements ?

[lethal cipher]

So we can construct a function so that ((0,1),(1,0)) maps to the cycle (12).

[hot lake]

that gives you a map from Aut(Z2xZ2) into S3

you need to explain why it's surjective I suppose ?

[lethal cipher]

No, that is a bit obvious. It is more about defining such a function. I'm having a hard time with it

[hot lake]

:/

[stoic rose]

Which function?

[lethal cipher]

Moreover, it is the automorphism group is all such possible permutations

The isomorphism between Z_2xZ_2 and S_3.
It really is basically just denoting (0,1) by 1, (1,0) by 2, and (1,1) by 3.

[hot lake]

yes

[deep nova]

hey im having trouble understanding symmetric groups and cosets. So I know how to do cosets, but there's a questions that has me stomped

,,D_4 in S_4

[cloud walrus]

Zoro

[deep nova]

what does this imply?

d4 is just a square with all the respective rotations and reflections as elements right?

and if im not mistaken those are permutations, so they are functions

but then S4 is a symmetric group with 4 objects

that's what my textbook says

but im not sure what that means? how do you do a coset of this?

so an idea i have is that since S4 is a group, then i can represent all its elements with the cayley table right?

where the binary operation is composition of functions?

[sharp sonnet]

you can interpret D_4 as a permutation group on the 4 corners of a square and as such you can view it as a subgroup of S_4

then you just do cosets like you do in any group

[weak oriole]

Oh, does it suffices to show for one transposition
It should I think?

Or maybe two transpositions

[uneven sparrow]

Does anyone know any rings formed on the unit complex numbers? Especially with complex multiplication as the additive operation? Basically, the circle group would be the additive group of this ring

[lethal cipher]

Okay, so I am near the finish line with this question. Now how do I show that phi is homomorphic w/o going element by element?

So if $\alpha,\beta\in A$, I need to show that $\phi(\alpha\beta)=\phi(\alpha)\circ \phi(\beta)$.

[cloud walrus]

dackid

[void cosmos]

can you shjow

the original problem

[wooden ember]

i think the original problem was to show Aut(Z_2 x Z_2) = S_3

[void cosmos]

i did it before

in linear algebra

can u use linear algebra

[wooden ember]

i dont directly see a way to neatly show phi is a homomorphism dackid but that might be cause im just tired (there's probably just a straight definitional way to do it)

[lethal cipher]

I'd rather not. The ideas y'all were trying to do with linear algebra were a bit above my head.

[wooden ember]

if i may i would have approached this problem by considering how the automorphisms act on the generators of Z_2 x Z_2

[lethal cipher]

I mean, I already have the automorphism group. Also, we haven't discussed generating sets yet.

[wooden ember]

ah i see

youve discussed order though right

[lethal cipher]

Pretty sure that is all there is to it, but I am having a hard time figuring out what to say.

[void cosmos]

this is better

[wooden ember]

honestly id just say "phi is trivially a homomorphism"

[void cosmos]

and easier

[wooden ember]

because you only have 4 elements to consider

[void cosmos]

yea

[wooden ember]

i wouldnt really bother

[void cosmos]

thats better

[lethal cipher]

Well, 6 elements

[wooden ember]

i mean from Z_2 x Z_2 and as a matter of fact 3 since (0,0) is out of the game

[lethal cipher]

Oh, I see what you mean. My bad

[wooden ember]

the homomorphism just follows from your identification

[void cosmos]

6

but idk how

[wooden ember]

i mean you gave a labelling to each element of Z_2 X Z_2 and each automorphism just permutes these

so it's a permutation of 3 elements and hence S_3

[void cosmos]

i think you still

should argue

why its not

Z_6

@narwahl

[lethal cipher]

That isn't the question. But to answer it, order

[wooden ember]

cant exhibit an element of order 6

[void cosmos]

yea

[wooden ember]

i mean a clean way to go about it is to know all isomorphism types of groups of order 6 and then just argue by the properties you have from your automorphism group

but clearly not the intended direction

[hidden haven]

Why sully this

[wooden ember]

because this is probably not what's expected of dackid

in the general case i agree it's better

[rustic crown]

hello moldi

[hidden haven]

Hello det

[wooden ember]

gets you the automorphism group of every elementary abelian group

bruh im tired today

[rustic crown]

did you try that problem from above? it was asking if you could define a multiplication on the additive group R/Z such that it forms a ring.

[hidden haven]

What's an elementary abelian group 😵‍💫

[wooden ember]

(Z_p) ^ n

[hidden haven]

No

[hidden haven]

Oh ok though there's it's slightly harder to count all bases

[wooden ember]

can always be viewed as an n dimensional vector space over F_p

[hidden haven]

Ye

[wooden ember]

and then automorphism group is just GL_n(F_p)

[hidden haven]

Ok maybe it's not hard actually should be fine

Dackid are you not allowed to use linear algebra?

[wooden ember]

what's wrong with regular multiplication and taking fractional parts?

[rustic crown]

cuz Z is not an ideal, so doesn't work

[wooden ember]

right nvm was just not finding a counterexample

[lethal cipher]

I would like to know how to use LA here for future knowledge. But I guarantee I would not have come up with it, so I do not wish to use it in my assignment.

[hidden haven]

The number of automorphisms of a vector space are exactly the number of bases that the space has (but all the bases have to be ordered by the same indexing set I, or in the case of finite dimensional spaces you can just say you count bases as ordered tuples, so (e,f) ≠ (f,e)). This is because given a basis, you get an automorphism by mapping some fixed basis to that, and conversely an automorphism must map a basis to a basis

[lethal cipher]

Can you remind me what a basis is?

[hidden haven]

Linearly independent spanning set

[lethal cipher]

Moreover, how do we know that first claim?

[hidden haven]

First claim?

[lethal cipher]

"The number of automorphisms of a vector space are exactly the number of bases that the space has"

[hidden haven]

Oh I explain it later in the message lol

But if you aren't familiar with bases yet then I'm guessing you've also not seen vector spaces over arbitrary fields? Like Z2 x Z2 is a vector space over Z2

[lethal cipher]

No, definitely not.
To put it in perspective, we have only talked about groups. We have not discussed rings/fields yet.

[hidden haven]

I see, alright

[lethal cipher]

We'll get there, but we have a minute

[potent briar]

the integers with usual addition and product are a ring right?

can you have an algebraic structure similar to a vector space

where the scalars are a ring and not a field?

im asking because im trying to think what would be a basis for the "vector space" of positive rational numbers with positive integer scalars

like, can you get any positive rational number by scaling some rational number in [0, 1] by some positive integer?

[hidden haven]

Yes, look up modules. They are exactly vector spaces over rings (invertibility is not mentioned anywhere in the vector space axioms)

But not all modules have bases, only the free modules do (all vector spaces are free modules hence they all have bases). Q is not a free Z module, so it doesn't have a basis. You can still talk about minimal generating sets though

It's not a free Z module because ||any 2 elements of Q are linearly dependent over Z: if you have a/b and c/d then they are dependent because bc(a/b) - da(c/d) = 0||

[potent briar]

nice

i like it when amorphous thoughts in my head actually turn out to be real things

but [0,1] is not a basis becuase bases have to be finite right?

that's the only difference between bases and minimal generating sets?

[hidden haven]

No bases can be infinite, we then call the vector space infinite dimensional

[potent briar]

because [0,1] in Q is enumerable

[hidden haven]

[0,1] a basis for what?

Oh intersected with Q

It's not a basis because it's not a linearly independent set

[potent briar]

oooh

i see right

[rustic crown]

how do you even talk about linear independence when it's positive scalars?

are we looking at modules over half-rings?

[hidden haven]

Why only positive scalars?

We are taking Q as a Z module

[rustic crown]

im asking because im trying to think what would be a basis for the "vector space" of positive rational numbers with positive integer scalars
like, can you get any positive rational number by scaling some rational number in [0, 1] by some positive integer?

[hidden haven]

Oh I see

Yeah that doesn't work out as nicely I guess

[deep nova]

what im struggling to understand is S4. D4 has 8 elements, and I think S4 must have 64 elements right? Because after all S4 is a group which by definition is the binary operation (composite function) applied on a set D4

but since its S4 it must be 4 objects not 64

and the problem doesnt specify the elements so i dont know what to use for my cosets

[hollow imp]

S4 has 4! elements

and it's the set of bijections between a set of cardinality 4 and itself

[deep nova]

oh i thought it was 64 because i thought you do a cayley table on D4 elements

[deep nova]

what does that mean exactly?

[hollow imp]

you know what a bijection is right?

[deep nova]

yes

[hollow imp]

you can just think of S4 as the set of bijective functions f: {1, 2, 3, 4} -> {1, 2, 3, 4}, with the group operation being composition

[deep nova]

ah so s4 is a number then, it has nothing to do with all the reflections and rotations of d4?

sorry i meant

[hollow imp]

no, S4 is a group

[deep nova]

s4 has number elements

[hollow imp]

the elements of S4 are functions

and the elements of D4 can also be interpreted as functions

[deep nova]

i see now

yeah makes more sense now

thanks

[deep nova]

just to make sure im on the right track, let's say if im doing the left coset, would this be correct? (1234)[u1]; where (1234) is an element of S4 and u1 is an element of D4 and it's a reflection

[viscid pewter]

??

[deep nova]

though im using abcd for my vertices, should i use (abcd) instead of (1234)

[viscid pewter]

bit weird to take the left coset of a set with only one element

[deep nova]

it's not one element, it's just an example

[viscid pewter]

oh

[deep nova]

just a sample of the many that are to come

[viscid pewter]

usually the set you're taking cosets of is capitalised

so it would be like

gH

or (1234)H

[deep nova]

yeah i was trying to be specific here

but i meant gH

[viscid pewter]

ok

anyway yeah so it doesn't really matter whether you use abcd or 1234 although numbers are more standard

but i don't see how that's a reflection

[deep nova]

so u1 is the reflection where A goes to B and D goes to C;...... where A is bottom left point of square, B bottom right of square, D upper left of square and C upper right, so it's a reflection where the line splits a square in half

[viscid pewter]

so yeah

ok wait

ah so fine

that's u1, ok

[viscid pewter]

right, i thought you meant (1234) was a reflection

but yeah it doesn't really matter what u1 is

[deep nova]

no thats from s4

ok

[viscid pewter]

i mean yeah there was an element and a set and you took a left coset

[deep nova]

i just thought maybe in order to compute the gH, they all needed to be abcd so i was making sure

[viscid pewter]

i mean yeah

the notation should be consistent

they should be in the same group

[deep nova]

gotcha

[coarse stag]

Perhaps this is meaningless, but if S_n is the set of bijective functions from a set of n elements into itself, is of order n!, then is the group of bijective functions from R to R of order |R|!

Which I assume is roughly R^R?

[viscid pewter]

the actual set is called R^R

|R|! is meaningless

can't take factorials of infinities

[coarse stag]

what does R^R even mean?

[viscid pewter]

exactly what you said

the group of functions from R to R

[coarse stag]

the set of all functions from R to R?

[viscid pewter]

yeah

set

ok so

how did it go

right so

any specific list of three numbers in R^3

can be thought of a function from {1, 2, 3} to R

[coarse stag]

wait

[round jay]

If there is a homomorphism from group A to group B, is that equivalent to A being isomorphic to a subgroup of B?

[coarse stag]

is this why the exponential object is notated like that?

[viscid pewter]

yes

[coarse stag]

wow

[viscid pewter]

cool, right?

[coarse stag]

yeah

[viscid pewter]

no, for example consider the homomorphism from G to the trivial group

[round jay]

So for any given two groups, there is always a homomorphism from one to the other

[viscid pewter]

yes

the trivial one

if it's an injective homomorphism then it's trivially true

[coarse stag]

@viscid pewter Wait, can you go over your {1,2,3}->R example?

[viscid pewter]

so f(1) = 8

f(2) = 4

f(3) = 2

then f corresponds to (8, 4, 2)

[coarse stag]

sure

oh

the image is {R, R, R}

[viscid pewter]

image of what?

f is like

[coarse stag]

not image

but like range of where it can map to

[viscid pewter]

yeah it's a little hard to describe

[viscid pewter]

basically

[round jay]

thanks

[viscid pewter]

np

[past temple]

im confused by the solution to (i)

it only shows that the roots of f,g are integral over A in the splitting field F

so how does it conclude that those roots are in the integral closure of A in B?

when B is strictly contained in F

<@&286206848099549185>

[golden pasture]

coefficients of f,g are integral over C and they are contained in B, but since the integral closure of C in B is C, you're done

[past temple]

oh jesus yeah im stupid lmao

thanks

[deep nova]

would a permutation (1) just be 1 mapped to 1?

[viscid pewter]

permutation (1) is used to mean the identity permutation i think?

[deep nova]

and if it's in a subgroup that has elements {(1), (123), (132)}, would it also imply that 2 maps to 2 and 3 to 3?

[deep nova]

oh yeah that is right

isn;t that usually represented by id?

[viscid pewter]

sure

[south patrol]

meh, id, (), e, various things

[thorn delta]

meh is my favorite

[deep nova]

im gonna start writing meh from now on and be forced to explain to my prof where i invented this new object

[chilly ocean]

what are some resources that are good for understanding the krull topology on galois extensions?

like if someone knows a book with a good section on it, just for reference

[coarse stag]

Actuate example of a homomorphism into the trivial group:
https://youtu.be/AuEY5JYbWlI

[terse crystal]

Don’t know whether it helps, Jacobson gave some discussion about it on charter 8 volume 2 of basic algebra

[chilly ocean]

thanks, I have that

[trim grove]

How can i show a subset of $\mathbb{Z}[i]$ , such that $I={a+ib /}$ where a and b are multiple of 3 is a maximal ideal,

I have proved that this an ideal, next is how to show this is maximal ideal ( without using homo, or isomorphism)

I consider some J such that

$I \subset J \subset Z[i]$ ,now by contradiction i want to show that this is wrong , please help

[cloud walrus]

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

[rustic crown]

are you allowed to assume that every ideal in Z[i] is generated by a single element? That would make it very easy, as I = (3) and say J = (a). since I is contained in J, we get 3 in (a) implying that there is b such that 3 = a * b. If you know 3 is irreducible, then what does this tell you about a? else just take norms and see what the norm of a can be.

[tropic spade]

Would showing Z[i]/I is not a field work?

Because it seems like (3+I)(u+I)=1+I would give 3u-1 divisible by 3, which seems contradictory?

Maybe I'm misunderstanding the q tho

[rustic crown]

well, i see them writing "without using homo, or isomorphism"

[tropic spade]

Oh the stuff I'm talking about relies on that right?

[rustic crown]

kind of?

like the way i like to think about m is maximal iff R/m is a field is via correspondence theorem. ideals of R/m correspond to ideal of R containing m. If m was maximal then it says R/m has only 2 ideals. so is a field.

[tropic spade]

I see what you mean.

[rustic crown]

and if i were proving that it's a field, then i would have probably used that Z[i] is isomorphic to Z[x]/(x^2+1), then quotienting by (3) would be isomorphic to Z[x]/(x^2+1, 3) which is isomorphic to F3[x]/(x^2+1). But x^2+1 has no roots over F3, making it irreducible and F3[x]/(x^2+1) a field.

[tropic spade]

That makes sense. I would never have even thought to prove it along those lines.

[rustic crown]

what would you have done?

[tropic spade]

No idea, wing it and try and figure something out lol.

[rustic crown]

oh lol

[tropic spade]

I'm not particularly experienced with algebra since I am learning about this stuff in one of my classes rn.

[trim grove]

This is some interesting fact

[plucky flicker]

Hi guys! How can I prove that [a, c1][b, c2] = [a1b1, c3], for some normal subgroups A, B, C?

where [a, b] = a^(-1)b^(-1)ab is the commutator

and a, a1 is in A, b, b1 in B and c1, c1, c3 in C

[simple valley]

so you're asking whether [A, C][B, C] is a subset of [AB, C]

[bronze jay]

are the abelian groups here isomorphic to eachother?

[simple valley]

no I think none of these are isomorphic

[bronze jay]

why not?

[simple valley]

which of these groups have an element of order 4, and which have an element of order 9?

[bronze jay]

Z60xZ3 has a element of order 4, namely (15, 0) ?

[simple valley]

good, so does Z180

but not the other two groups

[bronze jay]

how can you see that?

[simple valley]

if g in G has order n, and h in H has order m, then (g, h) in GxH will have order lcm(n, m)

(notably not n*m)

[plucky flicker]

yes

[plucky flicker]

i proved that [AB, C] is a subset of [A, C][B, C] but I don't know how to prove the other inclusion

[simple valley]

yeah maybe the way I wrote it is a little misleading

when taking [A, C] you want the two instances of a to be the same

[plucky flicker]

Well, we know something: for all a,b,c [ab, c] = b^(-1)[a, c]b[b, c], but this is not enough for the inclusion we want to prove

I used this to prove the first inclusion

[bronze jay]

I don't quite see how this is used in the above example

[simple valley]

hmm that doesn't prove anything

if a divides x and b divides y, then lcm(a, b) divides lcm(x, y)

[bronze jay]

mhm

[simple valley]

elements of Z30 have orders that divide 30

elements of Z6 have orders that divide 6

therefore elements of Z30xZ6 will have orders that divide lcm(30, 6) = 30

[bronze jay]

oh and 4 doesn't divide that

there exists no such a in Z30 and b in Z6, such that lcm(a,b)=4, since 4 doesn't go up in lcm(30,6)?

[final oasis]

Is it euclidean @rustic crown

[viscid pewter]

think so

[final oasis]

If so then it would be a PID, and the answer would be yes

[simple valley]

@bronze jay you know CRT right?

[bronze jay]

what?

[simple valley]

chinese remainder theorem

[bronze jay]

uh not really

[simple valley]

it's the statement that Zn x Zm = Znm, when n and m are coprime

[bronze jay]

ah

okay

what page is that even in dummit and foote

[simple valley]

I have no idea

[bronze jay]

but what about CRT?

[simple valley]

well, it comes with a bit of intuition for why it fails when n and m are not coprime

it's very easy to see the difference between Zp x Zp and Zp^2

Zp x Zp doesn't have an element of order p^2

[bronze jay]

alright

I think it is this?

[simple valley]

so like, it's kinda obvious what orders a group like this does and doesn't have

yeah

[bronze jay]

Another question, elementary divisor decomposition and invariant factor decomposition are two ways of uniquely determining what a finite abelian group is up to isomorphism?

[simple valley]

I'll leave that to someone else

[plucky flicker]

Well every finite Abelian group is isomorphic to a direct sum of finite cyclic groups of prime power order

This is one of the most famous characterization theorem in group theory

Or isn't it your question? 😄

[viscid pewter]

false

consider Z4

[plucky flicker]

Z2xZ2

[viscid pewter]

no

Z4 has an element of order 4

Z2xZ2 has none such

every finite Abelian group is isomorphic to a direct sum of finite cyclic groups of prime-power order

[plucky flicker]

woops

[viscid pewter]

get dunked on

[plucky flicker]

i meant to write that

sorry

[viscid pewter]

:P

[plucky flicker]

So any ideas regarding my problem?

[plucky flicker]

..

[simple valley]

we just talked about how they're different

[plucky flicker]

Yeah, I didn't follow that

[bronze jay]

but is this not correct?

[plucky flicker]

in my opinion you are right

you can see it that way

[bronze jay]

do we also have that G is isomorphic to Z2xZ9xZ5?

[viscid pewter]

depression

[bronze jay]

what

?

[viscid pewter]

.

[bronze jay]

I don't get why we can just factor 6 and 15 here?

what theorem are we referring to

[deep nova]

is there a theorem that states the left cosets of H in G must be equal to the right cosets of H in G? Specifically for permutation groups. I've seen examples where this is not true, but is there a fast way to determine if the right and left cosets are equal instead of having to do all the right ones again

[viscid pewter]

normality

[deep nova]

im not sure if we've done such concept in class

[sharp sonnet]

surely you have covered normal subgroups?

[deep nova]

nope, hasn't been in the syllabi so far

it's the next chapter after cosets and lagrange's theorem

really dont feel like doing 8 more right cosets right now lol

[sharp sonnet]

well, one way to easily deduce gH = Hg is g commuting with every element in H, in particular if your group G is abelian you get gH = Hg for all g and subgroups H

in general proving normality is hard

[deep nova]

i see

[plucky flicker]

is that true if I take A, B, C, D normal subgroups in G such that, A < B and C < D, then AC < BD?

[rustic crown]

Oh... I wasn't really asking a question.. the question above asked to not use isomorphisms, so was wondering if we're allowed to use Z[i] is a PID.

[simple valley]

AB < BD?

[plucky flicker]

sorry

AC < BD

[simple valley]

"<" just means subgroup?

[plucky flicker]

yes

[simple valley]

then sure

AC < BC < BD

[hot lake]

is AC the subgroup generated by A and C or is it {ac ; a in A and c in C} ?

[simple valley]

they're normal so

[plucky flicker]

yeah, they are normal

AC is not the generated

[hot lake]

ah yeah then it's the same

[plucky flicker]

thanks

I want to prove that if A, B are nilpotent normal subgroups than AB is also nilpotent, and I'm guessing that I need that inequality here, but not sure

[hot lake]

not sure how your lemma is going to help there

[plucky flicker]

well actually it doesn't

[sly crescent]

The simple group PSL(3,4) gives rise to ten almost simple groups, including PSL(3,4) itself, PGL(3,4), PΣL(3,4), and PΓL(3,4). Do any of the others have names like these?

[simple mulch]

Hey guys, I followed the hint given the question but I don't understand why showing that both functions are equal we're answering the problem

[kind temple]

manoid lol

[rustic crown]

To show it's a monoid, you need to show there is an associative operation and an identity element for that operation. What you have done is shown that the usual composition operation restricted to the set of such mappings is closed. That if you take two things in the set, and compose them, you get another thing in the set. You just now need to verify if identity function is also here.

[kind temple]

also that it’s associative

[rustic crown]

right

[simple mulch]

what about ahah

[kind temple]

you have a typo

[simple mulch]

where??

[kind temple]

“… set of all such mappings is a manoid under the circle …”

[simple mulch]

its how it is in the book o.O

Question 2 tho

.rotate

[rustic crown]

the actual word is "monoid" with an "o" and not "manoid" with an "a"

[simple mulch]

I am trying to understand how is this the set

oh

lol

wtf

sorry

[viscid pewter]

,rotate

[rustic crown]

woah

Texit can do this?

[viscid pewter]

ye

quite handy

[rustic crown]

[viscid pewter]

,rotate 45

[simple mulch]

thanks!

[viscid pewter]

,rotate 315

[simple mulch]

nice xd

[rustic crown]

Woah

[viscid pewter]

quite handy

[rustic crown]

right

[viscid pewter]

anyway yeah what was the q

2

i mean just show it satisfies the monoid axioms?

[rustic crown]

yea lol

[rustic crown]

a function f : R --> R is in that set if you can show there are a, b such that f(x) = a + bx = f_{a, b}(x)

[simple mulch]

Yeah, I was thinking how I could define the set

So that I could take the elements to prove closure, associativity and that there exists e in the set which is the identity element

[rustic crown]

S = {f_{a, b} | a, b in R}

[simple mulch]

uh, so taking s in S we have some a,b in R such that f_{a,b} = s. So s = a + bx?

[rustic crown]

like think of it as abstracting stuff out

functions are big objects

but now you zoom so far out, that these functions look like elements to you

elements of S are functions from R --> R which look like that

so s = f_{a, b} which means s(x) = a+bx

[simple mulch]

humm

I see

[simple mulch]

oh for associativity I need to show that $f_{a,b} \circ (f_{c,d} \circ f_{e,f}) = (f_{a,b} \circ f_{c,d}) \circ f_{e,f}$ ?

[cloud walrus]

mns

[simple mulch]

and for the identity element I need $f_{e_1,e_2}$ such that $f_{e_1,e_2} \circ f_{a,b} = f_{a,b} = f_{a,b} \circ f_{e_1,e_2}$

[cloud walrus]

mns

[strong moth]

would it be correct or incorrect to think of a ring as being an commutative group under addition and a monoid under multiplication along with the left/right distributivity property?

[nova plank]

That's correct

That's literally the definition on wikipedia iirc

[strong moth]

oh sweet!

i was trying to condense the definition as much as possible

[nova plank]

A ring is a field without units where 1 may be equal to 0

[strong moth]

is the additive identity part of this definition a typo:

[nova plank]

Yes

= a

[coarse stag]

Does anyone know what P^n (R) could mean?

in this context

[prisma ibex]

Projective space

[chilly ocean]

[coarse stag]

yeah its from that

lol

[lethal cipher]

Damn! Cosets are powerful

[coarse stag]

its says that projective space is the space invariant under all general linear homogeneous transformations

Whats the difference beetween a linear homogeneous transformation and just a general linear transformation?

I assume its just a homomorphism from one vectorspace to another vectorspace

[lethal cipher]

So I am a bit stumped on something. I need to show that the order of U(n) is even when n>2.
We will end up using the fact that the order of an element divides the order of the group

Nevermind. n-1 is always relatively prime to n. And (n-1)^2=n^2-2n+1 which is congruent to 1 mod n

There is our answer

I could use some help on 25. I am not even sure where to begin

[chilly ocean]

have u tried an example

that'd be somewhere to begin

[lethal cipher]

No, but worth exploring. Z_5={0,1,2,3,4}
The sum of all of those is 10, which is congeuent to 0mod5

One thing I do know for certain is that all elements have odd order.

But I am not sure how that helps tbh

[rustic crown]

notice that you can write the sum as 0 + (1+4) + (2+3)

can you do such stuff everytime?

[maiden ocean]

so as a light hint: ||if G has an odd number of elements, it has an even number of non-identity elements||

[lethal cipher]

Yea, I already caught that thembossing

[maiden ocean]

ok so we're taking g_1 g_2 ... g_2n of the non-identity elements in G

[lethal cipher]

So you are just pairing an element with its inverse

Of course, this is assuming that an element is not its own inverse

[chilly ocean]

maybe there's a silly way you can do this using the fundamental theorem of finite abelian groups

[rustic crown]

and what are the elements which are their own inverses?

[lethal cipher]

Ooo, but if that were the case, you'd have to have an even number of such elements

Otherwise you couldnt pair every other element with its inverse (pigeonhole principle)

[rustic crown]

um... almost... it's necessary that we need even to be able to do this... but is it sufficient? what if there were elements which couldn't be paired

[lethal cipher]

Well, if they couldnt be paired, it would be its own inverse

[rustic crown]

right... but what are those elements? is identity the only one?

[lethal cipher]

Not always.

[maiden ocean]

||recall that the order of an element divides the order of the group||

[lethal cipher]

U(n) is a fine example. n-1 is its own inverse when n>2

Oh good point. So this actually cant happen

Every non identity element has to have a distinct inverse since the order of all elements are odd

[rustic crown]

[maiden ocean]

[lethal cipher]

Lagrange's theorem coming in clutch

Okay, one last thing. I need to determine all finite subgroups of C^*.
I know a class of subgroups is cos(theta)+isin(theta) where theta is period multiples of 2pi/n (n positive integer).
But how do I show this is the only possible subgroups (if it is)

[rustic crown]

this is actually true not just for C but other "fields" as well, that pretty much means a structure where you can add, multiply, subtract and divide all nicely.

so let's try to first write down precisely what we want to prove

So the examples of subgroups you gave are of the form <zeta_n> where zeta_n = cos(2 i pi/n) + i sin(2 i pi/n)

[lethal cipher]

Yes

[rustic crown]

so if i just gave you the subgroup... can you give some description of that zeta_n?

try to draw the elements of this group on the complex plane, and think of if you just knew all the points in that subgroup... how would you describe the point given by zeta_n

[lethal cipher]

Well, the subgroups are points on the unit disk centered at 0 each having an argument of 2pi/n between each other

I took complex analysis. I know this all too well :p

[rustic crown]

hehe nice

so looking at the picture... how would you distinguish that one zeta_n with all other points?

[lethal cipher]

By the argument of 2pi/n? Not sure what you mean here

[rustic crown]

so what i'm getting at is... given an n, you have this picture which gives you a finite subgroup of C*. we want a way to reverse this procedure, if I give you some finite subgroup of C*, how would we determine n? or if it has a generator or not?

[lethal cipher]

Umm, I am not entirely sure tbh

[rustic crown]

we wanna show that the subgroup is generated by a single element, right... how should we go and find that element?

the subgroup apriori could look very wacky... we need a simple criterion that will pin down that generator

so why can't something like this happen. Consider the finite group Z_2 x Z_2? it's not generated by a single element, you need at least 2.

[lethal cipher]

Well, in that situation, all non identity elements have the same order

[rustic crown]

right, but why can't that appear as a subgroup of C*?

[lethal cipher]

Well, for starters, -1 is the only element of order 2 in C^*

But, how does that restrict our possibilities? -1 is not gonna be in every <zeta_n>

[rustic crown]

okie, say i gave you this subgroup... can you point out one generator?

[lethal cipher]

Sure. It's the point just anti clockwise of 1

[rustic crown]

exactly!

[rustic crown]

this is one description of that generator that doesn't depend on our knowledge whether all subgroups look like <zeta_n> or not

phrased differently, what we can do is look at the non-identity element of the finite subgroup which makes the smallest angle with the positive real axis

so does this element generate that subgroup?

[lethal cipher]

Absolutely

[rustic crown]

also this is where we use the finiteness hypothesis... if the group was infinite taking minimum would be hard, like there is no minimum of all the positive rationals

[lethal cipher]

Yea, I am with you

[rustic crown]

so let's give some names, H is a finite subgroup of C*, and let g be the element which has the smallest angle with positive real axis

if h was another element, then we're claiming that h = g^r for some r

[lethal cipher]

Yes, that is the claim

[rustic crown]

if this wasn't the case then we can find an r such that, h * g^(-r) would have a smaller angle than g!

[lethal cipher]

Why is this exactly?

[rustic crown]

um, say g makes an angle of alpha and h makes an angle beta. So we can take r = floor(beta/alpha) this way 0 <= beta - r * alpha < alpha

[lethal cipher]

I follow everything except the <alpha bit

[rustic crown]

oh so find the integer r such that r <= beta/alpha < r+1
multiplying by alpha gives r * alpha <= beta < r * alpha + alpha
and now subtracting r * alpha gives that inequality

[lethal cipher]

Ah, I see now.

I hate to cut it short my friend, but I got to head off to class. To be continued?

[rustic crown]

okie!

this was one proof, there are plenty others...

i maybe chose this one because it doesn't use any group theory lol

you can make use of Lagrange's theorem and what not to give much shorter proofs

[lethal cipher]

Well, I am taking abstract algebra. Maybe using group theory isnt the worst thing in the world :p

[lethal cipher]

Especially this!

[rustic crown]

okie i'll just say the idea, you can fill in the details... not sure how much longer i'll stay up.
take the finite subgroup H, and say it has size n. the all the elements of the subgroup H satisfy x^n = 1... but the equation in C has exactly n roots, namely the n-th roots of unity!

[lethal cipher]

Ohhh. Because all elements have order that divides n. So x^n=1 for all elements

[rustic crown]

yep

[lethal cipher]

That's beautiful!

[rustic crown]

elements of H are distinct roots of x^n -1, but there are only n roots of that polynomial, so H better be the set of roots of x^n-1

[lethal cipher]

I see. Wow, lagrange's theorem is coming in full force

It's such a simple result, but it is incredibly useful

[rustic crown]

[elfin patrol]

For this question I’m asked to find nilpotent element greater than 2 of Z_n where n=p^a, p prime. I’m not sure how to find it generally

I’ve tried playing around but can’t seem to find anything. Any help is appreciated

[wooden ember]

Lagrange’s theorem is the fundamental driving force of a lot of finite group theory

[wooden ember]

Ultimately it all relies on equivalence relations

[rustic crown]

right

to remind, the reason is that in Q[X], 3 is a unit, so factoring it out is same as doing nothing

or equivalently, you can just say 3X^4 + 6X+6 is irreducible in Q[X]

[elfin patrol]

Any suggestions for my question above

[rustic crown]

what exactly do you wanna do there?

isn't anything in pZ_n nilpotent?

what do you mean by "greater than"... don't see a way to order elements of Z_n

[elfin patrol]

Sorry maybe I worded my question poorly. I’m trying to find a nilpotent element in Zn where n=p^a for p prime and a>/ 2. Is everything nilpotent?

If it is I don’t follow

[rustic crown]

no, not everything

but everything in the ideal generated by p, i.e. (p) = pZ_n is

because (p*k)^a = p^a * k^a = 0 * k^a = 0

stuff outside that ideal are units! so they can't be nilpotent

(as our ring is not 0)

[elfin patrol]

Oh that is way simpler than I thought. Thanks, I feel like I should have seen that right away

[rustic crown]

i don't know what you meant by greater than 2 tho... for n=4, the nilpotents are {0, 2} so what do we do?

[elfin patrol]

The question just defines n=p^a but isn’t that fine. 2^2=0 so isn’t that what we want

[rustic crown]

so i was referring to the part where you said "nilpotent greater than 2"

[elfin patrol]

Oh I see, I made a typo there, my bad

[rustic crown]

oh okie

[deep nova]

what does <(1234)> mean? How can you generate a group where the generator is a permutation group?

[void cosmos]

like you usually do

what does <2> mean for example in the group Z_8

additive

you just keep repeating the operation

till yoou get back

thats the definition

right?

@deep nova

[deep nova]

yeah

so <(1234)> would just be permuting and repeating it?

[void cosmos]

<(1234)> = {(1234)^n | n in Z}

where exponentiation here is in context of the operation of the group

so if i were to be in an additive group for example Z/8Z

2^2 = 2+2 = 4 mod 8 =4

not 2^3 = 8 mod 8 = 1

etc

2^3 = 2+2+2 =6

not 8

so (1234)^2 for example would be

composition ( as in the context of S_4 group )

(1234)(1234)

[deep nova]

ah makes sense

thanks

[void cosmos]

np

[prime sundial]

is the order of an element x in Z/nZ LCM(x,n)/x?

[zealous mortar]

Q[x]/x^3-7 is a field?

[hidden haven]

ye

[zealous mortar]

thx

[prime sundial]

i realize now that this is more like number theory D :

[chilly ocean]

still fits here

[hidden haven]

do you mean order in the additive group?

because multiplicatively it is not a group

[prime sundial]

additive, yeah

[hidden haven]

ah ok I just confused myself then 😌

yes that seems correct

[prime sundial]

neato burrito

time to prove

[zealous mortar]

Any good reference on deriving Jacobson radical?

non-commutative.

[hidden haven]

Deriving as in?

[chilly ocean]

Why do numbers only work when the dimensions are powers of 2? like real numbers are 1D numbers, complex numbers and 2D numbers, quaternions are 4D, octonions are 8D, sedenions are 16D

why is that?

why can't we just have something like T = a + bi + cj

or just any other other than powers of 2

[maiden ocean]

Um