groups-rings-fields

[coral spindle]

I guess the 0 ring is an algebraically closed field if u are ok with that…

[night tartan]

lol

[south patrol]

Based F1 enjoyer

[elfin wraith]

I said the 0 ring is a field in my LA exam I’m sure

[rocky cloak]

It sure simplifies LA a lot

[elfin wraith]

Fields are already weird enough, just allow it

[serene dune]

why would i even need this ?

[bitter rover]

Assuming that's permutation cycle notation, it's giving you a way to turn extract or distribute transpositions from permutations.

As for why you need it, well, a Lemma is usually introduced right before it's used. See how it's used in whatever proofs follow.

[serene dune]

yeah right!

and yeah its permutation cycle

[bitter rover]

I'm assuming it's being used as part of a proof that the parity of a permutation is well-defined, i.e., the number of transpositions in any two decompositions of a permutation into transpositions might be different, but they have the same parity.

So it makes sense to talk about the parity of a permutation.

[serene dune]

oh yeah it IS being used in a proof in the next page

[bitter rover]

heyy

[serene dune]

too tired to go through this section, will look at it tomorrow

[bitter rover]

These are showing that any permutation can be decomposed into transpositions.

Since any permutation can be written as disjoint cycles, it suffices to prove that a single cycle can be written as a product of (not disjoint) transpositions.

Like, if we have a cycle (1 2 3 4 5 6 7) we can write it as (1 5)(1 2 3 4)(5 6 7) or (1 4)(1 2 3)(4 5 6 7) or...etc. etc.

But really what matters, I suppose, is that a cycle of length L > 2 can be written as the product of a transposition and two cycles of length less than L.

So, recursively, you can keep decomposing things until you get only transpositions.

Your choice of what transposition to pull out affects the transpositions that appear in the final decomposition, but not the parity.

[serene dune]

that was theorem 1.3, but sure

have to digest the parity part

makes sense

[south patrol]

This should be true essentially by definition for any good definition of sgn tbf

so i am curious as to what the definition you are using is for that

[serene dune]

kinda weird ngl

[south patrol]

oh lol

No this is fair enough

i guess they just haven't shown that this is true for any factorisation yet

[candid patrol]

Fake

[serene dune]

wdym by this ?

idk, should i be looking for alt definition?

even if i don't use, i think it would be nice to take a look

honestly i was too tired when i read this last section on permutation, would take a fresh look today

[white oxide]

If $k[x, y]/B \cong k[x]$ is it true that $B = k[y]$

[cloud walrus]

okeyokay

[coral spindle]

That's not an ideal so it's unclear what you mean by that

[white oxide]

Nvm then

[coral spindle]

And in the first place

1 is in k[y]

so if you could quotient by that you'd have k[x,y]/B = 0

But again, this doesn't make sense

[candid patrol]

What is a complete factorisation ?

Fix sigma is empty ?

[tardy hedge]

Did you mean B = (y)?

[white oxide]

Nah I got confused lol

[south patrol]

But also like yeah you will very rarely be able to deduce the ideal from the quotient being "isomorphic to" something

[tardy hedge]

here goes enpeace (just playing around lol)

[thorn jay]

the main reason why one usually doesn't look at objects up to isomorphism only

[thorn jay]

AH- I've been noticed

back to hiding I go

[tardy hedge]

hahahaha

[thorn jay]

anyhow, since I'm here anyways I wish to bestow upon you all that MOLS-pairs give rise to an interesting quotient group of $S_n^4 \ast K_4$ where $K_4 \cong D_4$ is the Klein four group, arising out of a particular group action. I shall be studying this quotient group until I can identify it and then I shall return.
... hopefully this does not take as long as the last time I detoured a bit with Latin squares. That detour was one lasting half a year and led me to learn universal algebra, and module theory

[cloud walrus]

.enpeace_music

[lusty marlin]

What are MOLS-pairs?

And what sort of group product is *?

[ivory ore]

now, some elements are involutions, let's say

can we get some condition then too!
such that a1.a2....an=e

i hope they dont call the stupid involution(identity) involution

[crystal vale]

If there is no element a≠e such that a^2 = e, so a ≠ a^-1 for every non-identity element

[ivory ore]

i have asked something else after solving that

but thanks

[crystal vale]

a_1a_2...a_n ≠ e if at least one of a_i is involution

[ivory ore]

i am asking the converse

say some elements are involution

[crystal vale]

So you mean if a_1a_2...a_n = e is given?

[ivory ore]

can we hope to meet a_1a_2...a_n = e

[crystal vale]

Converse?

You mean there is a non-identity element a^2 = e?

[ivory ore]

brain fried, lol

[crystal vale]

No if there is at least one involution then it cannot be a_1a_2...a_n = e

[ivory ore]

yeah looks like it,

can we hope for rigour ?

[crystal vale]

Yes you can

[ivory ore]

by contradiction?!!!

[crystal vale]

Yeah you can say there are a_1,..a_k involution then prove that a_1...a_n = a_1...a_k

[ivory ore]

alr i will think about it

[crystal vale]

Just think about which are not involution they will cancel out so only involution elements left

Cancel out means they will operate with its inverse and they will give identity

[frigid epoch]

What do quaternionic polynomial rings look like?

Are the monomials something like $q_0 x^{n_1} q_1 x^{n_2} q_2 \cdots q_{l-1} x^{n_l} q_l$?

[cloud walrus]

PKThoron

[proper jolt]

suppose i take all the finitely generated submodules of N and i take the union of all of them is the result N?

i think yes because you can just append elements to new submodules

actually idk if this is true

[alpine plank]

any x in N lies in the submodule generated by x

[proper jolt]

oh that's true

[thorn jay]

Free product, the coproduct in the category of groups.
The free product of G and H is in a sense the "most general" group containing both G and H, or the most general subgroup generated by G and H.

Elements of the free product look like words
g_1 h_1 g_2 h_2 .... g_n h_n
Where g_i is any element of G and h_i any element of H, and you can reduce words by removing identity elements, and combining elements from G or H

[thorn jay]

Two Latin squares L and M form a MOLS-pair, or pair of MOLS (Mutually Orthogonal Latin Squares) if, when you lay them on top of another, every pair of symbols occurs exactly once

These are two examples of Latin squares

They're grids where every row and column each symbol occurs exactly once

[lusty marlin]

Ah ok

Thanks!

[thorn jay]

Yw :)
they dont have many applications, and MOLS-pairs even less, but they're intimately related to quasigroups which do have applications in coding theory and knot/link theory, for some reason..

[crystal vale]

I showed the hint but not get it how it helps us

If a is left quasi-inverse then 1-a has right inverse

[bitter rover]

Are you asking a question?

[empty perch]

Hey folks, just a quick reminder that the Abstract Algebra Lecture series is meeting regularly again ^_^
We will meet to discuss Gram-Schmidt and orthogonal complements in about 10 minutes over in the ⁠⁠#1055201711679082516. More information can be found in our thread: ⁠⁠#1317307081535000606

[hard hearth]

here is what i have tried:
so here i think induction must be done on N. suppose the claim holds for all groups of size less than N. then i want to show that it holds for groups of size N also. suppose the group G is of the form {g1,g2,...,gN} and suppose there is an element gk such that it contradicts the claim. in terms of size, take the largest subset S that generates a proper subgroup of G. if this set has gk, we are done as then it can be written as a product of the r generators in less than N elements. if S does not contain gk, by the maximal size of S, S u {gk} generates all of G. im not sure what to do from here. i feel like there must be some numerical pigeonhole argument instead.

edit: an ai model solved it in a few seconds. i feel stupid now.

[rustic rapids]

Im struggling to apply the CRT here. I just dont really see where to start with this problem

[hard hearth]

if k is coprime to m, then k mod n is coprime to n (so is k).

[rustic rapids]

Right, so would i take m = nk. Use CRT on z/mk = z/nz X z/kz. Then project to its first coordinate? So I need to find some k that we can do this ?

[hard hearth]

you cannot take m=nk. else k wouldnt be coprime to m. m and n are fixed and given at the start.

all im saying is that if k is some number coprime to m, then k must also be coprime to n. this is a number theoretic fact. (use beizouts to write ma + kb = 1, since m=nc for some c, n(ac) + kb = 1, so k is coprime to n)

this just solves the problem as you can see. a unit of Z/mZ is also a unit of Z/nZ which is what we wanted to show, since now the natural map is a surjection

[rustic rapids]

Wait then where are we applying the CRT?

[hard hearth]

on the surface, nowhere.

[rustic rapids]

hmm is then is there a way to do it with the crt? Im just a bit confused on why its a hint

[surreal swan]

Is (c) wrong? I think gh=1 is also needed (in addition to commutativity)

^ from Dummit and Foote Abstract Algebra, section 4.3

[serene dune]

oh yes, i forgot to mention
i think i have a sketch to prove the hint, which i need to do first, however the main concern would be how it relates to the meta claim here, maybe i just need to put the effort which will lead me to digest it all but yeah...

(sorry for the ping)

[bitter rover]

??

[slow rampart]

my brain dies trying to figure out where i am going wrong

[serene dune]

how does proving the hint proves the main result ?

[floral crane]

Hello. Does anyone here know about irreducibility tests for modular polynomials?

[tough raven]

Prove that the function R\{0} → R\{1}: x ↦ 1-x is a bijection and maps multiplication to the operation. So if the codomain is a group, so is the domain.

[serene dune]

Would you let me know why this notion of inverse is even useful ?

[tough raven]

The one use I know of is in defining the Jacobson radical.

[ivory ore]

where am i wrong ?
is there any counter that i am missing

[next obsidian]

You technically need to use that f(b^-1) = f(b)^-1

And likewise for g

[ivory ore]

alr, coach

[ivory ore]

help

[topaz solar]

what have you considered

[ivory ore]

express an arbitrary even permutation as a product of 3-cycles.

[topaz solar]

why not 2 cycles

you know those generate right?

[ivory ore]

oh yeah

and then use the hint right ?
for both the cases

[topaz solar]

I dunno what cases you refer to

but yes use the hint

[ivory ore]

in the hint

there are two cases disjoint and not disjoint

[ivory ore]

alr

[topaz solar]

aight

yeah fair

[ivory ore]

show, for $n \ge 2$, that $S_n$ cannot be imbedded in $A_{n+1}$.

[cloud walrus]

yeshua

[thorn jay]

Any A_n cannot contain elements of order 2, as they would have to be transpositions, hence be odd

[crystal vale]

(12)(34)

[lusty marlin]

A_n must contain elements of order 2 for n≥4, since A_n is a group of even order for n≥4

[lusty marlin]

Also this is an explicit example

A product of two disjoint transpositions is an even permutation of order 2 and hence is in A_n

[thorn jay]

Im stupid

Lmao

[candid patrol]

Bro forgot about V4

[thorn jay]

I did, and worst thing is

I used that exact representation in A_4 literally yesterday when studying some group action

[candid patrol]

Let H a subgroup of An+1, then H is a subgroup of Sn+1, and so H is a subgroup of index n+1 in Sn+1.
Hence, there exists i in [|1,n+1|] such as H = { sigma € Sn+1 | sigma(i) = i }

But it works for n différent to 5

[surreal magnet]

I was toying around with automorphism groups of groups, and noticed something. Is it always true that given a commutative ring $R$ with an additive group $G$ and unit group $U$, then $\operatorname{Aut} G = U$? This is certainly true for $\mathbb{F}_p$ and $\mathbb{Z}$ and I think for any ring with a cyclic additive group.

[cloud walrus]

jp

[ivory ore]

grade this one

[thorn jay]

It is true for cyclic groups

As in, Aut Z_p ≈ (Z/pZ)*

[glad osprey]

I guess U is always at least a subgroup of the automorphism group, since multiplication by a unit is an automorphism. So I think your proposition is true when every automorphism is an inner automorphism (?). Maybe someone else can explain in more detail

[thorn jay]

The additive group is abelian, so has no nontrivial inner automorphism

[glad osprey]

Oh, right. What's the relation between Aut(R) and Aut(G)? Is Aut(R) a subgroup of Aut(G)?

[thorn jay]

Yeah, because an automorphism of an algebra must consequently be an automorphism of all its reducts

[rocky cloak]

My guess would be that it's true iff G is cyclic

[thorn jay]

That seems rather strong

[rocky cloak]

Yeah, Q would be a counterexample

[thorn jay]

Lol

[rocky cloak]

What about
Z -> R being an epimorphism

[thorn jay]

What is Aut_R G

[rocky cloak]

Aut_R(R) = U

That's true in general

[thorn jay]

Even for noncommutative R?

[rocky cloak]

Yup

[thorn jay]

Cool

[rocky cloak]

Well might be U^op

Depending on your conventions

[thorn jay]

So we're essentially asking when is Aut_R(G) = Aut_Z(G)

[rocky cloak]

Yeah, and then Z->R being epi should be sufficient I think

[thorn jay]

Sufficient, but not necessary maybe?

[thorn jay]

I've got part of the answer.

For $\tau = \langle \alpha, \beta, \gamma, \delta\rangle\in S_n^4$ denote $\tau^\dagger$ to be $\langle \beta, \alpha, \delta, \gamma\rangle$, and $\tau^T$ to be $\langle \gamma, \delta, \alpha, \beta\rangle$\par
Then, consider also the group $C_2 = \langle g\rangle$.
We construct the group:
$$\mathfrak{G}^\ast = (S_n^4 \ast C_2) / \langle \tau_1 g \cdots \tau_n g \mid \tau_1\cdots \tau_n = 1 \text{ and } \tau_1\tau_2^\dagger \cdots \tau_{n-1}^\dagger \tau_n = 1 \rangle$$
Where WLOG we can assume $n$ is odd.
We have an automorphism of $\mathfrak{G}^\ast$:
$$\tau_1 g \cdots \tau_n g \mapsto \tau_1^T g \cdots \tau_n^T g$$
Which is an involution, so we let $C_2$ act on $\mathfrak{G}^\ast$ in that way, and finally construct the group we want:
$$\mathfrak{G} := \mathfrak{G}^\ast \rtimes C_2$$

[cloud walrus]

.enpeace_music

[thorn jay]

That's the best i could do, and wouldn't surprise me if that's the best you can do in general without having to explicitly calculate stuff

I've proven it to be finite, and i know it is centerless due to the S_n

But like a proper decomposition of this group would be nice

[weak prawn]

Could someone help with this one?
I asked ChatGPT, and it said it was a homomorphism when G is abelian. That means g(xy) = gx*gy, but I really can't see why that equality holds, even when it's abelian?

And is introductory abstract algebra questions okay here, or should I go to a help-channel?

[frigid epoch]

Check what happens to x the neutral element

[weak prawn]

Could you elaborate on that?

[chilly ocean]

Take $\phi_g(x)$ for x the identity

[cloud walrus]

NAT Enthusiast

[weak prawn]

Yeah, then it's just g. But why does that matter?

[chilly ocean]

What does a homomorphism between groups send the identity to?

[weak prawn]

The identity of the other group?

Oh, g must be the identity element

or are the more "solutions"?

[rocky cloak]

That's the only solution, and being abelian is irrelevant.

[bitter rover]

To be a homomorphism, it must send the identity to the identity. That means "homomorphism implies g=e"

So there's at most one solution. And there's at least one solution because g=e makes phi the identity.

[rocky cloak]

Best not to trust our AI overlords yet

[weak prawn]

Haha yeah. I thought at first that it was when g was the identity, but when GPT said otherwise I got lost

[thorn jay]

I would recommend not using AI for math

And especially not proof based math

And especially especially not abstract algebra

[weak prawn]

Yep, makes sense. I usually don't use it, but if I do I always takes it with a grain of salt (except now I guess)

it can however be surprisingly accurate sometimes, I have noticed

Thanks anyway

[bitter rover]

Now do conjugation: $\phi_g(x) = gxg^{-1}$

[cloud walrus]

Cufflink

[weak prawn]

Then it's a homomorphism for all g in G? Since $\phi_g(xy)=g(xy)g^{-1}=gxg^{-1}gyg^{-1}=\phi_g(x)\phi_g(y)$

[cloud walrus]

Michael

[bitter rover]

Not just a homomorphism, but it's bijective, too, so an isomorphism (called an automorphism when it's from the same group to itself)

[bitter rover]

Does every group automorphism arise from conjugation?

[weak prawn]

Ah yeah, that makes sense

[weak prawn]

That I'm not sure of

[bitter rover]

That is, if $\psi: G \to G$ is an automorphism does there exist a $g \in G$ such that $\psi = \phi_g$

[cloud walrus]

Cufflink

[weak prawn]

Well I would suppose so?

Can't prove why, but still

[bitter rover]

If you answer this you will see why ChatGPT mindlessly started mentioning commutativity

In an abelian group, what does $\phi_g$ look like?

[cloud walrus]

Cufflink

[weak prawn]

Then it's just x

I suppose

or $\phi_g(x) = gxg^{-1} = gg^{-1}x = x$

[cloud walrus]

Michael

[weak prawn]

Not sure if that's what you're asking

[bitter rover]

Right, it's the identity. So can you find an abelian group with a non-identity automorphism? If so that can't correspond to conjugation by some element, because conjugation always yields the identity when the group is abelian.

[lusty marlin]

!nogpt

[flat tree]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[weak prawn]

For example $\phi: \mathbb Z \to \mathbb Z$, such that $\phi(x) = -x$

[cloud walrus]

Michael

[bitter rover]

Yep.

In general, $x \mapsto x^{-1}$ is an automorphism iff $G$ is abelian. It'll be the identity iff $g^2 = e$ for all $g \in G$

So any abelian group where there's an element of order other than 2 will have at least one non-conjugation automorphism.

Automorphisms that come from conjugation are called "inner automorphisms": https://en.m.wikipedia.org/wiki/Inner_automorphism

Let $Z(G)$ be the center of $G$, i.e., the set of all elements they commute with everything in $G$.

A group is abelian iff $G = Z(G)$, but you also have $G/Z(G) \cong \text{Inn}(G)$

[cloud walrus]

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

[frigid epoch]

I thiiiiink inversion is an automorphism iff G is abelian

For finite G at least

[weak prawn]

Is InnG the inner automorphism of G?

[bitter rover]

Oh yeah, I'm a dumdum

[cloud walrus]

Cufflink

[lusty marlin]

The statement holds for all groups, not just finite ones

[lusty marlin]

Yes, it is the inner automorphism group

[crystal vale]

yes

[bitter rover]

ChatGPT was like, "Oh, you said $\phi_g$, I know that! (Starts saying stuff related to conjugation.)"

[cloud walrus]

Cufflink

[ivory ore]

LLM

is it conscious

[bitter rover]

It's like an undergrad who memorized keywords on Wikipedia and is very good at stitching them together in a way that gets you to your head in agreement.

[rocky cloak]

I was imagining it was thinking about how G is abelian iff x |-> x^-1 is a homomorphism

Hard to say what it was "thinking" though

[weak prawn]

Ahh okay, so $\phi_g$ generally refers to the conjugation stuff?

[cloud walrus]

Michael

[weak prawn]

Haha, nice analogy

[bitter rover]

No! Haha. But that's what it associates with it, probably. Like someone who has only ever seen it used that way might walk away think that's always what it meant

[weak prawn]

Alright

Well thanks for this tiny lesson, appreciate it :)

[thorn jay]

For infinite G too

[thorn jay]

If \phi_g is understood to he an automorphism of G, and g is an element of G then yes

But i've seen $\gamma_g$ or even $\on{ad}_g$ too

[cloud walrus]

.enpeace_music

[tardy hedge]

Thanks @bitter rover for my basic group theory review of the day

Lol

[sinful fern]

This is false

Free groups and $S_6$, the symmetric group on $6$ elements, have outer automorphisms

[cloud walrus]

Ras Al-Shaytan

[frigid epoch]

I was thinking of the free groups and was like "oh yeah I'm just mapping generators to their opposites"

[thorn jay]

S_6 is so weird

[frigid epoch]

Realized that ab would get mapped to b'a' so it's an antihomomorphism still lol

[thorn jay]

Yeah

[thorn jay]

Like there doesnt even exist a MOLS-pair of order 6
???
And not of order 2
???
But all the other orders there does exist one
???

[bitter rover]

Indeed! I was just spelling out what "every automorphism arises from conjugation" means.

[surreal swan]

Can every group be extended so every automorphism becomes an inner automorphism in the extended group?

[thorn jay]

https://londmathsoc.onlinelibrary.wiley.com/doi/abs/10.1112/jlms/s1-43.1.387

Tl;dr yes

How? I gotta read the paper for that

[topaz solar]

Does G \rtimes Aut(G) work?

[slender violet]

when microsoft released Majorana processor, i decided to look in to the paper they published i couldn't understand the mathematics. so i took book "The basic mathematics for quantum computing" , i somehow found a field i can dedicate my rest of life for. Although its just a beginning and i am happy for journey am also aware that there will be haunting ladscapes i need to cross ahead. So, i want understand from you all that what keep's you motivated to learn this field. Apart from uni and exam clearing?

Pardon typos

[surreal swan]

Would it be gamma . g = gamma(g) . gamma ?
gamma . g . gamma^-1 = gamma(g) , seems promising tbh

[topaz solar]

I mean, conjugating by the Aut elements is applying the automorphism

This isn’t the same as embedding G into G’ with Aut(G’) = Inn(G’) ofc, but

[surreal swan]

Nice, TY!

[velvet goblet]

shouldnt the 2nd condition be replaced by " K-{0}⊆K* " since K* is always closed under .

because K* is definitely closed under inverses

ah wait the definition of K* isnt here

K* is the set of all invertible elements of K

[thorn jay]

Yeah it should be K-{0} lol

[velvet goblet]

ah yes or this

tysm

[thorn jay]

I fear that i may have been an idiot

Due to the nature of MOLS-pairs, the word $g \tau g$ is actually equivalent to the element $\langle \alpha, \beta, \delta, \gamma\rangle$ by some interesting skew-symmetry.
In other words, the group I'm interested in.. is way simpler then i thought. Lol
It's probably $S_n^4 \rtimes K_4$

[cloud walrus]

.enpeace_music

[thorn jay]

Ah I've

Seen my mistake now

Instead of the K_4 which i thought was acting in the MOLS pairs it's actually D_8

Which explains a lot of the asymmetry / skew symmetry :P

So, after calling myself an idiot three times, we arrive at the group
$$S_n^4 \rtimes D_8$$

[cloud walrus]

.enpeace_music

[hidden wind]

uwu

[thorn jay]

So true

💥

[tough raven]

If f(X) is a polynomial over a field F in positive characteristic p and p^k is the largest power of p such that f(X) = g(X^p^k) for some g, then is the extension generated by the p^k^th roots of the coefficients of f the smallest field extension L/F such that the splitting field of f over L is separable over L?

[carmine minnow]

How do you show that the set of nilpotents form an ideal?

[velvet hull]

same as always, closed under (internal) addition and (external) multiplication

or, you could show that it is the kernel of some ring homomorphism, both works (this one might be quite a bit harder)

[carmine minnow]

sth like let l = lcm(all powers that make them nilpotents), f(a) = a^l ?

nvm not (or hard to prove) homomorphic

[velvet hull]

that does not work

I can construct a ring with nilpotent elements with index that get arbitrarily large

so the lcm might not be well defined

it's a hard construction if you don't know about it, just do the normal way first

[carmine minnow]

for external multiplication, I'm thinking if a^n=1, something goes along the line of R, aR, aaR, aaaR... a^nR?

[velvet hull]

I don't care what something goes to

all I need to show, is that if n is nilpotent and r is in R, then rn is nilpotent as well

that's it

[carmine minnow]

that I understand

[velvet hull]

same for addition (hint: ||think about the binomial formula||)

[carmine minnow]

oh,

the problem I had is that i thought a nilpotent defined to have solution to a^n=1

[velvet hull]

should be 0, not 1

[carmine minnow]

yeah, now I see how this works

[carmine minnow]

thanks

[serene dune]

this is beautiful:

https://media.discordapp.net/attachments/1334457895563169792/1343799920901099591/973465216228683886.png?ex=67be96b3&is=67bd4533&hm=b985c5936174561640d57f1cd71eb8fe10dacb7b59ae18640044637d867c5040&

never heard of this theorem, doesn't come naturally
does it ?

[lusty marlin]

It does come naturally because a subgroup generated by a single element is cyclic

so you just take each element and consider the cyclic subgroup generated by it

we naturally get a partition of the group

[serene dune]

now after taking a look at the theorem it reveals itself

i like it

[lusty marlin]

yes it's easy to see

[serene dune]

what i meant is if you don't know about the theorem, it's probably gonna stay at blind sight, but that's dumb me
..

[lusty marlin]

Nah I had never heard of this theorem before

but when you mentioned it, it seemed very natural

[serene dune]

yeah the proof was neat however i never thought about it

https://media.discordapp.net/attachments/1334520360883978342/1343799259916664904/973465216228683886.png?ex=67be9616&is=67bd4496&hm=f704f3c5edb2ae58dc9c9248f82fcbf3a298237f1847fd6262276cda90b3e45e&

[tough raven]

That's not a ring homomorphism, so no.

[static mauve]

Im trying to show that a group homomorphism from phi: G to G' has the property that if G is a finite group, then |phi(G)| is a divisor of |G|

One approach Im trying involves using the Theorem of Lagrange (the order of a subgroup divides the order of the group.) I thought that if I can show that if phi[G] is isomorphic to a subgroup of G, then phi[G] is essentially a subgroup of G, and therefore this would prove this property.

Does this seem correct?

[bitter rover]

You don't have to prove that it's isomorphic to a subgroup, it just is a subgroup and you can prove that directly.

[static mauve]

Hmm. It is not obvious to me phi[G] is a subgroup of G. Couldnt phi map an element of G to... something not contained in G?

[tough raven]

That's almost but not quite correct: phi(G) need not be isomorphic to a subgroup of G (I'm almost sure about this, but don't have a counterexample at hand). Instead, you can find a subgroup H of G such that |phi(G)| = |G|/|H|, which also divides |G|.

[static mauve]

Could this subgroup, H, actually just be the subgroup of G that Im claiming is isomorphic to phi[G]? Im not entirely sure I follow the logic for |phi(G)| = |G|/|H|, but I feel like I can always make a map from H to phi[G] where any elements in G that arent 1 to 1 are removed until only elements remaining are 1 to 1.

For instance, if G = {1,2,3} and phi[G] = {a,b}, and lets say that (f is my map) f(1) = f(2) = a and f(3) = b, I can say that H = {1,3}. Thus f would be 1 to 1 between H and phi[G] specifically. And I think I can add the conditions of onto and homomorphic here to get my isomorphic condition?

[bitter rover]

Oh of G

[static mauve]

Or maybe my fault here is that Im saying the property of being injective is related to the set the map is being acted on rather that the actual domain and range of the map?

[velvet hull]

how about this: do you know of the First Isomorphism Theorem yet?

[static mauve]

I havent learned that quite yet 😄

[bitter rover]

Well, this is a prelude to the first isomorphism theorem, because what you wind up having is $G/\mathrm{ker}(\phi) \cong \mathrm{im}(\phi)$

[velvet hull]

hmmm, how about this: do you recall the equivalence relation used in the proof of Lagrange's theorem?

a very similar equivalence relation is present here

[thorn jay]

I prefer using \ker and \operatorname{im}

[cloud walrus]

Cufflink

[bitter rover]

I have no clue what the latex environment is, so I try to do the dumbest thing.

[thorn jay]

Haha it's alright, if no one has to see your LaTex doc you can do the stupidest shit

[velvet hull]

so, here's the secret: although it may not seem like it at first glace, but group homomorphisms are actually really restricted.
Specifically, what I am thinking of here, is that phi(a) = phi(b) if and only if a,b differ up to some element in the kernel of phi

or, in other words, the kernel partitions the group (this is just the statement of Lagrange's theorem, but I am restating it here because it is an important fact to notice for this proof)

so, if phi: G -> G' is some group homomorphism, phi^(-1)(x) (which is not necessarily a well-defined group homomorphism!) always must look like some copy of the kernel of phi, in G

now, what can we conclude about phi(G)

[ivory ore]

A has order 4
B has order 4 as well
and A^2 = B^2
AB \neq BA

[candid patrol]

Im(phi) is always a subgroup

[ivory ore]

oops, sorry to drop at the mid of convo

[candid patrol]

It’s okay

[ivory ore]

i can see the existence of order two subgroup here

A^2 = B^2 = - I

so the group would be {I, -I}

[candid patrol]

What is ur question ?

[ivory ore]

any hint for the first part

[ivory ore]

this

i have an urge to see the whole group sturcture and find the isomorphism to some group

[hidden wind]

you need to show there are exactly eight elements in the group, for this just take all products

[ivory ore]

but i'm resisting myself to do that for there can be some slick argument

[ivory ore]

awight

thanks, would

[candid patrol]

Tips : ABA = B and for k € {0,3}, A^(4-k)B = BA^k

[ivory ore]

two g3enrators for order 8

we know what it is

[candid patrol]

?

[ivory ore]

or Q8

its Q8

[hidden wind]

yep

[candid patrol]

Let G be a group of exponent 2. Show that the set of subgroups of index 2 in G has cardinality 2^k - 1 for some k in ℕ.

Really cool exercice 👍

[candid patrol]

Try it jagr

[rocky cloak]

G = (Z/2)^k so there are 2^k homomorphisms to Z/2, and exactly one of them is not surjective

[candid patrol]

Right

You are de too strong

[rocky cloak]

A similar statement could be that if G is a p-group, then there are (p^k - 1)/(p-1) subgroups of index p for some k.

[candid patrol]

There is a bijection between thé dual of G {trivial morphism} and the set of subgroups of index 2 un G

[rocky cloak]

And fun fact, k is the minimal size of a generating set for G

[candid patrol]

Didn’t know, how do you prove it ?

I’ll try by my own nvm

[candid patrol]

But I’d like some tips for this

[crystal vale]

How can I get this group is abelian?

[rocky cloak]

This is the burnside basis theorem, and it's like a Nakayamas lemma for groups.

Basically you have this Frattini subgroup, which is the smallest normal subgroup such that G/F is (C_p)^n. Then you want to show that any element of F can always be dropped from any generating set.

[crystal vale]

Oh group of exponent 2, I got it

[candid patrol]

Alr thanks you, i’ll try

[rocky cloak]

Using that p-groups are nilpotent + induction can be a useful approach

[candid patrol]

Try to show that if G is a groupe of exponent 2, then |G| = 2^k - 1 if G is finite, it’s a classic one

[round portal]

hello guys, hope this is the right channel to be posting this question... but I'm struggling with a basic proof from group theory, here's the exercise

Let $(G, \cdot)$ be a group with $|G| = 2n$ for $n \in \bZ_{>0}$ show that: \

$\exists g \in G, g \neq e$: $g = g^{-1}$

[cloud walrus]

texaspb

[round portal]

how do I go about this one?

found a post on stackexchange but it didn't help me at all 😦

[crystal vale]

exponent 2 means x^2 = e for x in G, right? Then G is abliean and also only prime divisor of | G | is 2 so how | G| = 2^k -1?

[crystal vale]

since it has even order

[round portal]

i don't understand how it having even order helps

😭

[rocky cloak]

Think about the number of elements g such that g is not equal to g^-1

What can you say about this number?

[crystal vale]

can you prove that if | G | is even then there exist an element of order 2?

[round portal]

my guess is that it would be odd

[crystal vale]

i mean | G | = 2n so cardinality of G is even

[round portal]

I saw this with some trial and error

[round portal]

I get it but idk how this helps

how to use this fact

[rocky cloak]

Maybe check it for some small examples, to see if you're guess makes sense

[crystal vale]

contradiction, assume that there are no non-identity element such that g = g^-1, that means for each non-identity g, g and g^-1 are different, now do some pairing

[round portal]

okay... that helps!!!!

ok so I've noticed with some small examples that if |G| = 3... then it doesn't hold the g=g^-1 thing...

but with |G| = 2 or |G|=4 it does...

[thorn jay]

Fun fact, this exercise is actually a generalisation of a property of Latin squares!
But if you applied it you'd only be able to prove it for abelian groups, so not that useful >.>

[candid patrol]

Fastest way (without using Cauchy), is to say that if g is not trivial, then o(g) is even, let's say 2k, then g^{k} € <g> has for order 2

[candid patrol]

My bad haha, |G| = 2^k sorry *

[round portal]

ok I think i finally got it! is it okay if i send a proof here

[round portal]

is this aight?

i get the counting part but i feel like something is still missing

[candid patrol]

It's good but...
You should explains why the number of non identity elements would have to me odd with some maths, not only with a sentence.

[crystal vale]

yeah then it is trivial

[candid patrol]

You've already done it ?

[crystal vale]

i mean i can show | G | = 2^k

[candid patrol]

Show me then : )

[crystal vale]

since G is abliean and if any odd prime p divide G, then there will be an element of order p but every element has order 2, so only 2 is a prime divisor of |G|

[candid patrol]

Fact, Cauchy is the fastest way again

But we don't need G abelian

[crystal vale]

yeah Sylow, same thing

yes we don't need that

[candid patrol]

Try with a recurrence also

[crystal vale]

recurrence, you mean induction?

[round portal]

thanks for the help guys 🙂

i get the intuition now

groups are weird

[candid patrol]

I lied, the fastest way is to say that we can equip G with a Z/2Z-vector space structure

[candid patrol]

yes my bad haha, recurrence is a french word

[round portal]

🦧

[rocky cloak]

Did you mean to say that it must be even?

The number of nonidentity elements is 2n-1 which is odd

[somber sleet]

Is R noetherian topological space?

[round portal]

yes!

[rocky cloak]

Does it have an increasing sequence of open subsets?

[hidden wind]

oooh

[somber sleet]

Increase the radius of the unit ball?

i thought every field is noetherian, but I guess regarding them as ropological spaces is different

[rocky cloak]

Being a Noetherian ring and being a Noetherian topological space are different things yeah

[somber sleet]

What’s the difference then?

[rocky cloak]

The definitions

A Noetherian ring is a ring without increasing sequences of ideals.

A Noetherian space is a space without increasing sequences of open sets

[somber sleet]

Okay fair that makes sense, so a decreasing sequence of closed subsets(?)

[rocky cloak]

Yeah, the complement of open sets gives closed sets and vica versa

[somber sleet]

So in the zariski topology this corresponds to subvarieties right?

[rocky cloak]

I'm not sure what "this" refers to.

But in the Zariski topology open sets (or closed if you will) correspond to radical ideals of the coordinate ring. So if the coordinate ring is Noetherian, then so will be variety be.

[crystal vale]

jagr can you verify one thing for me?

[rocky cloak]

That would probably depend on the thing

[pearl fog]

Let K be comm field, is there other way to write K[x]/(x^2-2x+1) other than maybe say x^2=2x-1? Since deg((x-1)^2)=2 extension degree is 2 probably means every ax+b a and b in field but what else? What is it different from K[x]/((x-2)^2)? (I know if I don't stack power together chinese remainder kicks in)

[kind temple]

do you mean a commutative ring? fields are always commutative…

[pearl fog]

Not quaternions

Oh quaternion isn't exactly field right then field

[rocky cloak]

It just depends on your definition.

In the English speaking world people usually say "skew field" or "division ring" if they don't want to assume commutativity.

In french they use
field vs commutative field

[lime junco]

i mean its different in the sense different elements are 0

[pearl fog]

But [K[x]/(x-1)^2:K]=2?

[lime junco]

well

K[x]/(x - 1)^2 isnt a field outright

becuase (x-1) =/= 0, and (x - 1)^2 = 0

same for K[x]/(x - 2)^2

[pearl fog]

It should not be a field since not irred, but like what is in K[x]/(x-1)^2 but not in K[x]/(x-2)^2, cuz since degree is both 2 I just assume that {1, x} is a basis

[rocky cloak]

If your literal about the definition they have no elements in common, since the elements of a quotient R/J is the cosets of J.

The cosets represented by 1 and x give bases in both cases, but
1 + ((x-1)^2) and 1 + ((x-2)^2) are different cosets

[lime junco]

they might be the same as vector spaces, but this doesnt mean theyre the same as rings

[pearl fog]

So the extensions are just what they are, no prettier presentation than maybe x^2=2x-1, and although {1, x} is indeed a basis they're bound to different arithmetic rules, and that's it

[lime junco]

i mean this isnt an extensions

when you have a field extensions you have the duality between like

modding out by minimal polynomials

and the K(a)

[rocky cloak]

Still an extension of rings I guess

[pearl fog]

I'm just treating x as "extension"

[lime junco]

sure, but i dont think thats the intention

[lime junco]

no yeah, but like

if a is some element i wanna "Add" to the field

like lets say i wanna add sqrt(2) to Q

[pearl fog]

I start from a bigger structure, like polynomial ring

And I define stuff on it

[lime junco]

Then Q(sqrt(2)) = R[x]/(x^2 - 2)

[pearl fog]

So it shrinks down just right that x behaves like the root

[lime junco]

in some sense here, yeah

[pearl fog]

Q[x]

[lime junco]

yes Q[x] sorry

lol

because the isomorphism just evaluates polynomials

[pearl fog]

I'm just annoyed by the fact I can't chinese remainder away these stacked powers

So I have to do a ton of junior high polynomial manipulation

[rocky cloak]

I guess it's worth noting that both of these rings are isomorphic to K[x]/(x^2), which might be easier to deal with if that's what you're trying to do

[pearl fog]

So they are indeed isomorphic

[rocky cloak]

They are isomorphic yes.

Just sending x-1 to x-2

[pearl fog]

oh right cuz K[x] is ED

just divide

[frigid epoch]

Z[sqrt(-19)] is PID but not ED, right?

[crystal vale]

Yes

[narrow shuttle]

if x is primitive in a finite field then is 1-x primitive too?

[crystal vale]

primitive?

[narrow shuttle]

an element which generates the multiplicative group

[crystal vale]

can i say 1 generates the multiplicative group?

no no

[narrow shuttle]

no ofc

like we know that the multiplicative group if cyclic, therefore there is a generating element
and actually there's usually more than one

[crystal vale]

but if i take Z/2Z then no

[narrow shuttle]

true

[rocky cloak]

Z/3 also gives you trouble

[narrow shuttle]

no

[crystal vale]

yes

[narrow shuttle]

1-2=-1=2

[crystal vale]

2 is generating element but not 1

oh

[rocky cloak]

I read x-1 oops

But F3[x]/(x^2 + x - 1) works

In the sense that x generates the multiplicative group, but 1-x does not

[white oxide]

Why is C[x, y]/(x^3 - y^2) not isomorphic to C[x]

[coral spindle]

Well in C[x], is x^3 a square? In the quotient it is equal to the square y^2.

But I guess this is assuming that we're identifying x in both rings

Maybe you could show this by arguing that in the former there is a nonzero element which has a square root, a cube root, but no 6th root. But no such thing exists in the latter.

[white oxide]

Hmm ok

[south patrol]

It is also natural to identify the first with the subring C[t^2, t^3] of C[t]

which makes it a lil clearer in what way this is "weird"

For example, C[t^2, t^3] isn't even a UFD, let alone a PID or euclidean domain

[coral spindle]

Much nicer way to distinguish the two ^

How'd you see that so quickly Potato?

Or did you just happen to know it

[south patrol]

Another efficient way: an isomorphism between C[t] and A:= C[x, y]/(x^3 - y^2) would send t somewhere. Now you need to consider where it can go

[south patrol]

This is a very common example in algebraic geometry, for example

Basically the point is like

if you sketch the curve y^2 = x^3, it has a cusp

[coral spindle]

Ah ofc

[south patrol]

(obviously this is only real points, but still)

[coral spindle]

Yes, singularities

Makes sense

I should have known this

[lusty marlin]

Intuitively it is about 'parametrising' x³=y²

[white oxide]

Is it because it can only be x or y? And it can't be both, but they're both generators

[coral spindle]

Wait, I thought I knew what you meant here but I don't. Can you elaborate?

[south patrol]

Well you can send it anywhere, but it'll never be enough to hit everything

[coral spindle]

So you can argue it's not a monogenic C-algebra

OK, nice

[lusty marlin]

The vanishing locus of x³=y² is given by points of the form (t²,t³)

[south patrol]

An iso between C[t] and C[x,y]/(x^3 - y^2) would basically say this curve is equivalent to the usual affine line

[coral spindle]

Ah, I see

[south patrol]

which is the same as parametirsing the curve basically

[coral spindle]

So there is a natural map C[x, y]/J -> C[t]. Neat.

[south patrol]

Oh yeah, this also came up in a funny thing lol

Well like

You can call a ring A seminormal if the natural map A -> { (a,b) | a^3 = b^2} sending (a,b) | -> (a^2, b^3) is a bijection lol

[white oxide]

Is C[x] viewed in this way a vector space over C? Because if we're talking about generators, don't the coefficients usually come from Z in a ring

[south patrol]

Which is funny

[south patrol]

Well no i mean like

[coral spindle]

C[x] is being viewed as a C-algebra

[south patrol]

Okay tbf I have made a mistake there that I am assume it is a map of C-algebras

[coral spindle]

not merely a C-vs

Oh good point

[south patrol]

Which will certainly be true in your case i.e. in AG but isn't necessarily the same

[white oxide]

Oh right VS with multiplication

[south patrol]

Though you can probably argue that like the image of C is a copy of C inside C[t,t^2] which contains 1 etc

[south patrol]

Well, just a ring with a map from C, I mean

[coral spindle]

I guess this truly is the best way to see it

[south patrol]

but equivalent here

[south patrol]

I think you can also make stronger statements lol like

[white oxide]

So it would be enough to show that C[x^2, x^3] \cong C[x, y]/(x^3 - y^2) \cong C[x], but show that C[x^2, x^3] is none of those things Potato mentioned

[south patrol]

Well the second \cong isn't true ofc but sure for the sake of cotntradiction

[white oxide]

Ye I meant for contradiction

Anyways thanks a lot

[sonic coral]

literally went through this example yesterday in my comm algebra class

[south patrol]

Lol nice

[toxic zephyr]

instead of writing

$((rm+n)+I_1M, \ldots, (rm+n)+I_kM)$

can i just write like

$\bigoplus_1^k (rm+n)+I_jM$?

[cloud walrus]

eigentaylor (STfFGMOaPID)

[toxic zephyr]

im so lazy is there some like shorthand 😆

in the context of R-modules. I_j are ideals of R

[coral spindle]

People usually write tuples like $\Big( (rm+n) + I_i \Big)_i$

[cloud walrus]

$\mathbf{Boytjie}$

[coral spindle]

It's not correct to use the \bigoplus notation

[velvet hull]

$((rm + n) + I_jM)_{i=1}^k$

[cloud walrus]

HChan

[toxic zephyr]

oh this is bodacious i like it

[toxic zephyr]

this is good too

okay if we have a homomorphism from M to the direct product of M/I_jM
do we need to show that it's well-defined? or because it's just on M and not a quotient then is it not even worth mentioning?

[coral spindle]

I don't see what issue there would be in defining this map

Why wouldn't it be well-defined, can you explain

[toxic zephyr]

right that's sort of my point. like you only ever need to worry about well-defined if your domain is a quotient set right?

[coral spindle]

Sure

You hopefully understand why we need to worry about well-definedness in general

And I just don't see why there would be an issue here if the map is using the formula above

[next obsidian]

This is… nebulous

You worry about well definedness when there could be an issue with well definedness

[south patrol]

Yeah lol

[next obsidian]

Like if you defined a map from Z into Z which was f(x) = roll a random die and then take that output

This isn’t well definedness

And saying “quotient set” is not really well defined because any set can be expressed as a quotient set of some other random ass set

[coral spindle]

I think this example is maybe bringing too many philosophical questions...

[next obsidian]

It isn’t an intrinsic property

[coral spindle]

The point is that we are defining a relation and then asking if it's a function or not

Whereas your example isn't even defining a relation, arguably

[next obsidian]

So what you’re telling me is

It’s not well defined

[coral spindle]

Well sure lol but when people talk about well-defined functions, they really mean what I said above

[next obsidian]

I jest

Im a joker

A chmoker

[coral spindle]

A CLOWN?!?!?!?!

For example if f : A -> B is a homomorphism of groups and we're doing the first iso, really we define a relation consisting of the pairs (a ker(f), f(a))

And then we prove this is a function

that's precisely what the proof of well-definedness is

But yes ur joke was funny

I feel I haven't made that clear enough...

[next obsidian]

You’re just saying that to make me feel better

[coral spindle]

I never say things I don't mean

[next obsidian]

I never say things I don’t mode

[coral spindle]

I never say things I don't average?

Shit I should have said median

Joke ruined

[toxic zephyr]

yeah ofc. i think i'm just like... in the mindset where i see "show this is a [something] homomorphism" and step 1 in my head is "show it's well defined".
but that seems stupid to do here.

[next obsidian]

Look also I’m gonna say I’ve never in my life seen a dataset that I’ve been trying to analyze where the mode was useful at all

Most useless ass statistic ever

[coral spindle]

Idk it's just unclear to me where the question of well-definedness was coming from so I thought I'd ask for clarification

[toxic zephyr]

absolutely

i think im just paranoid that the prof would be like "ERM WHERE DID YOU PROVE IT'S WELL DEFINED"

[coral spindle]

In general, alarm bells should ring whenever you assume that the argument of your function is in a specific form

So e.g. we are choosing a specific x when we take in xN in the first isomorphism theorem

Or perhaps when we are defining a function on the rationals, we choose a form a/b

And so on

The question is then whether or not equivalent forms lead to equivalent results

In imprecise terms of course

[thorn jay]

Well-definedness can often be the hard part

[tardy hedge]

My life sure as hell ain’t well-defined

[thorn jay]

[tardy hedge]

just tryna survive out here bruh

[toxic zephyr]

if we have like $c_1m_1+c_2m_2\in M$ as an $R$ module, and $c_i\equiv c'_i\bmod{I}$ for $I$ an ideal of $R$, is it correct to write something like
$$c_1m_1+c_2m_2\equiv c'_1m_1+c'_2m_2\bmod{IM}$$

[cloud walrus]

eigentaylor (STfFGMOaPID)

[tardy hedge]

That looks right

Cause u can just directly check that their difference is in IM

[minor fulcrum]

What is the name of the subgroup of the 3d euclidean group that maps integer points to integer points

It's like, S_4 x Z_2 x Z³ ?

[toxic zephyr]

yeah true, i didn't think of that. the proof is pretty easy now i think about it.
but yeah it's super nice to be able to bring congruence in I to congruence in IM

[next obsidian]

Carl, I think

[minor fulcrum]

clever

[tardy hedge]

Yeah the more algebra I study the more of this ring-module interplay is revealed to me, but at the same time its still hard for me to pinpoint exactly just what this interplay is and how its happening i guess, idk

I guess an R-module is just some “ring representation” R -> End(M)

But modules say something about groups too right, like how Z can’t be made into a Q-module

[sonic coral]

when talking about an algebraic set, does the set of polynomials need to be finite?

we talked about this in class but he never mentioned if the set of polynomials was finite or not

[tough raven]

It doesn't matter because the zero set of a set S of polynomials is equal to the zero set of a finite subset of S, by Hilbert's Basis Theorem.

[sonic coral]

oh i see, thanks

[acoustic igloo]

hello, i'm wondering about my approach to this problem

I think I need to show 3 things\
$(a_i b_j) K$ is a coset of $K$ in $G$ for all $i,j$\
$(a_i b_j) K = (a_m b_n) K$ implies $i=m$ and $j=n$\
For all $g\in G$ we have $g\in (a_i b_j) K$ for some $i,j$

[cloud walrus]

Axe

[acoustic igloo]

does that sound right?

[crystal vale]

yes

[minor fulcrum]

is there a proof that every group of order p^k has nontrivial center without going through conjugacy class wrangling

the proof that I know is really inelegant and I wish I knew a better one

[crystal vale]

induction?

[minor fulcrum]

not really sure how you'd do it inductively

do you know of such a proof?

[velvet hull]

what proof do you have?

the "standard" proof is through group actions

[acoustic igloo]

thanks 🙏

actually it's obvious that abK is a coset of K in G

[crystal vale]

My bad, I thought there is

[minor fulcrum]

yeah that's basically wrangling conjugacy classes but in different words

I think I have one for p^2 tho

maybe it can be inductive too idk

[ivory ore]

Let $S = \mathbb{Q} \setminus {0}$ under multiplication. Then $SS = S$, but $S$ is not a subgroup of $\mathbb{R}^\times$ since it is not a subgroup of $\mathbb{R}$ under multiplication.

[cloud walrus]

longboard kayak

[ivory ore]

this is interesting

as it seem i still can't go through the proof

any help would be greatly appreciated

[tough raven]

S is a subgroup of ℝ^⨯, actually.

[ivory ore]

🤦‍♂️

ok, what would be such a counterexample for a infite group

[crystal vale]

In Z, NN = N but N is not a subgroup of Z

[acoustic igloo]

this should show distinctness

[alpine island]

Is there an example of a ring where every element can be written nontrivially as the product of two other elements? specifically for all x in R, there exist a,b neither equal to x such that ab = x

[rocky cloak]

Q for example, x = x/2 * 2 or x/3 * 3

You can throw in a nilpotent element of you also want 0 to be covered

[alpine island]

what nilpotent element exactly? I do want 0 to be covered, that's the hard part

I thought Q[e]/e^2 but then e=ab is hard

wait

nevermind

-e*-1

alright I was overcomplicating this

[rocky cloak]

Everything except 0 can be covered by something like e/2 * 2 or indeed -e * -1

[alpine island]

Hmm now I wonder if there's an example in characteristic 2

(Z/2Z)[a,e]/(a^2+a+1, e^2)

works

any field k other than Z/2Z, you have k[e]/(e^2) works

[crystal vale]

I think we can also use fundamental of Abelian groups

[ivory ore]

Let $S, T \leq G$, where $G$ is a finite group, and suppose that $G$ is the disjoint union
$$G = \bigsqcup_{i=1}^n Sg_i T.$$
Prove that $[G: T] = \sum_{i=1}^n [S: S \cap g_i T g_i^{-1}]$. (Note that Lagrange's theorem is the special case of this when $T = 1$.)

[cloud walrus]

longboard kayak

[ivory ore]

without using isomorphism how to think about it

for each si break them up and then i feel like we need to show the isomorphism with the conjugacy part

oh how stupid i am isomorphism was used by the author already

this sounds? pls let me know if there is some other way to look at it

[brave remnant]

so can somebody explain to me why sometimes we write $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$?????????????????????

[cloud walrus]

redoftwored

[brave remnant]

because it is my knowledge that direct sums are defined when $\cap$ blah $={0}$ and this is clearly not the case for two copies of $\mathbb{Z}/2\mathbb{Z}$

[cloud walrus]

redoftwored

[frigid epoch]

This is called an "external direct sum"

And its a bit like saying "what if the two Z/2Zs were different", say the first one is the set {0, 1} and the second one {0', 1'}

Then you get the set consisting of (0, 0'), (1, 0'), (0, 1') and (1, 1')

[brave remnant]

its weird because there are arguments that rely on the trivial intersection of two sums while at the same time writing stuff that doesn't have trivial intersections......

that makes sense because then $\mathbb{Z} \oplus \mathbb{Z}\cong \mathbb{Z} \times \mathbb{Z}$??????

[cloud walrus]

redoftwored

[frigid epoch]

The stuff with the trivial intersection is called the "internal direct sum"

[frigid epoch]

Yeah

[brave remnant]

so is there no distinction in symbols

because texts use them interchangably

[frigid epoch]

Correct

Cause the internal and external direct sums turn out to be naturally isomorphic

Aka they're the same thing

[brave remnant]

or actually is it just that we classify using an adjective that some direct sum is considered internal when intersection is 0 and othewise it is called external

is it a naming thing?

[thorn jay]

Internal direct sum is when you've got two subgroups/modules of an ambient abelian group/module

So N is the internal direct sum of M_1 and M_2 if M_1 \cap M_2 = 0 and M_1 + M_2 = N

But the direct sum of two arbitrary not necessarily distinct modules N and M is an abstract object N(+)M and we can canonically identify N and M with submodules N' and M' of N(+)M, such that N(+)M is the internal direct sum of N' and M'

[thorn jay]

The difference between direct sum and direct product is only when you're direct summing / producting infinite collections

[frigid epoch]

Much better answer

Thank you

[thorn jay]

Because the direct sum of an infinite collection of modules M_i is the submodule of the direct product of all M_i where every element has only finitely nonzero components

In a sense it's the collection of finite sums of elements from M_i

[brave remnant]

to clarify do we define the direct product as a direct sum that ends up forming some sort of structure

wait i typed that so wrong

[thorn jay]

The direct product is different from the direct sum

[brave remnant]

i meant do we define the direct product as a cartesian prodcut that ends up forming some sort of structure

[thorn jay]

Yes, exactly

(Has to do with a certain property of algebraic structures namely an adjoint relation with sets, but thats not important)

[brave remnant]

so can we use the rectangle square comparison analogy for direct products and cartesian products

or are there exceptions

[thorn jay]

Eh, only for direct product of finite sets

Arbitrary direct product of a family of structures A_i indexed by a set I has underlying set the sequences (a_i) indexed by I where a_i is an element of A_i

[brave remnant]

and that is not necessarily a cartesian product

[thorn jay]

I will give you this as i have to go to class now

[brave remnant]

and this is basically saying I can take a nontrivial intersection and restructure it using an isomorphism to structures that do have trivial intersection?

[thorn jay]

No that is the definition of cartesian product

[frigid epoch]

Yes!

[brave remnant]

ty

[frigid epoch]

It means the sum will also be bigger now

[brave remnant]

oh rlly

[brave remnant]

manmade horrors beyond my comprehension

[frigid epoch]

Yeah, for C2 (+) C2, the sum is isomorphic to C2 × C2

[brave remnant]

what is C2?

[thorn jay]

You haven't seen category theory...

[frigid epoch]

Whereas in ambient C2, the sum would just be isomorphic to C2 itself

[thorn jay]

Z_2

[frigid epoch]

Shorter name for Z/2Z

[thorn jay]

Cyclic group of order 2

[brave remnant]

oh ic

[thorn jay]

You can see the direct sum of modules as the most obvious subdirect product :p

[frigid epoch]

Basically you just "forget" the two summands have nontrivial intersection

[brave remnant]

so external/internal direct sums are two sides of the same coin

[frigid epoch]

But that also gives you new sums that were "collapsing" before

[frigid epoch]

Yeah

[thorn jay]

External better though

Because thats categorically defined

[brave remnant]

oh wait like span 01 and span 10 internal direct sums to R^2 but if we have span 12 and span 24 it external direct sums to R??????

as vector spaces*

[frigid epoch]

In the C2 example, 1+1 was previously 0, but in the direct sum, it becomes a formal sum which may be (and is) distinct from 0

[frigid epoch]

External direct sum is R2

[brave remnant]

isnt internal trivialintersection?

[frigid epoch]

Internal sum is R, internal direct sum can't be defined

[brave remnant]

why not

[frigid epoch]

Cause <(1,2)> and <(2,4)> don't have trivial intersection

[brave remnant]

oh yea so that makes it external right

but for R^2 01 and 10 are orthogonal

or lin indep

then span only has trivial intersection?

so its internal?

[frigid epoch]

But as enpeace said, there's an external direct sum M(+)N with identifications <(1,2)> iso M and <(2,4)> iso N

Such that M(+)N is the internal direct sum of M and N

[brave remnant]

yes

does that apply for 01 and 10?

[frigid epoch]

Nah it just makes it a non-direct sum

[frigid epoch]

Yes

I can't really explain it any better rn

[brave remnant]

so if we have

polynomial space

subspaces are odd and even functions then their direct sum is the whole space

which type of sum is that

[frigid epoch]

Internal

For internal, you need the ambient module and the intersection condition

For external, the direct sum is a "new" object (for lack of an ambient object)

It turns out that if you take the space of odd funx and the space of even funx

And took the external sum

The new object you'd get is isomorphic to the space of all polynomials anyway

[limpid river]

Is there a deeper reason as to why <(1 2), (1 2 3 4 ... n)> = S_n but not <(1 3), (1 2 3 4 ... n)> ? Thanks

[candid patrol]

(2 3) notin <(1 3), (1 2 ... n) >

[crystal vale]

i think he wants deeper reason so maybe he is looking for generalization of generators of these types of elements

[limpid river]

yes, to rephrase my question what makes a generator a generator in the case of S_n. May be there is some hidden reason as to why (1 2) and not (1 3) is one of the generator

[limpid river]

i am sorry, but could you prove it please?

[crystal vale]

https://math.stackexchange.com/questions/2505426/for-which-k-can-s-n-be-generated-by-1-k-1-2-ldots-n?rq=1

[limpid river]

Thanks @crystal vale

[candid patrol]

Oh sorry

[velvet goblet]

a subdivision ring is a subset of a division ring which is a division ring itself?

[crystal vale]

what is subdivison ring? if its elements already has inverse in that space then yes

[coral spindle]

I can only assume so, but you should maybe post the context just in case it is some weird specialist terminology

[crystal vale]

hi, Boytjie, don't you like analysis?i never seen you on other channel

[coral spindle]

My research is in algebra

I'm sticking in my lane

[thorn jay]

The obviously superior branch of math

(Im kidding pls dont kill me)

[crystal vale]

i see

[velvet goblet]

well i asked here to verify this

[velvet goblet]

sadly i dont have any context i just heard the term

[coral spindle]

Well then I could only assume it is a subring that is also a division ring

[velvet goblet]

that is what i assumed too

ig nothing more than guessing can be done without having context

tysm both of you

if i come across some context with subdivision rings i will make sure to inform you

[tardy hedge]

If you have ideals J subset I with I finitely generated, is J finitely generated?

[coral spindle]

Well R is an ideal of any ring

And there are certainly ideals of rings which are not finitely generated...

R = <1> however...

[south patrol]

All maths ends up being algebra

except combinatorics

because algebra ends up being combinatorics

[frigid epoch]

I lowk think all math is combinatorics + compatability + x

Not sure what x is though

[coral spindle]

x = everything else

[ivory ore]

lmao

[ivory ore]

catego....💤

[frigid epoch]

Tbh category theory is still algebra imo

[thorn jay]

And algebra ends up being logic

[frigid epoch]

And logic... combinatorics and computability 😬

Oh and philosophy

[thorn jay]

We had a philosopher in the r/mathmemes discord who claimed to have discovered a numerically stable way of integration and solved the Riemann hypothesis with it

[frigid epoch]

Shucks that that didn't work

[thorn jay]

yeah

I guess

he used AI btw

like, to write his papers

[frigid epoch]

So is the RH a numerically stable method of integration away from being solved? 🤔

[thorn jay]

apparently

yes

[toxic zephyr]

if t is a torsion element of an R mod M with rt=0 and r≠0, then can we still guarantee a homomorphism from M to R/(r) via
g(t)=1+(r)?

R is a PID

M is finitely generated

if that's not enough maybe if (alpha) generates like the torsion stuff for M then maybe doing R/(alpha) instead?? idk but the main idea of my question is if we can define a homomorphism such that t maps to 1 in some quotient of R

[toxic zephyr]

ex. if the structure theorem torsion parts are R/(a1)...R/(an)
with a1|...|an
then maybe picking R/(a1)?

[tough raven]

This does not define a homomorphism from M to R/(r), it defines a homomorphism from R/(r) to M (whose range is Rt). If r is the generator of Ann(t), then this will be injective and show Rt is isomorphic to R/(r); maybe this is what you are looking for?

[toxic zephyr]

i just need some homomorphism g from M to some module N=R/(?) where g(t) is nonzero. not sure if that's possible

[tough raven]

You want a homomorphism from M to some cyclic R/(r) such that t ↦ 1, given that t is torsion? This is not always possible: let M = ℤ/4ℤ and t = 2 + 4ℤ, then t will never go to 1. (Intuitively, t is a multiple of 2 and we cannot make 2 invertible without killing M.)

[tough raven]

Ah. Well, take a direct sum decomposition of M. If t ≠ 0, one of its components has to be non-zero and you can project to that component. If t is torsion, one of its components in the non-free summands (i.e., one of the R/(a)'s with a non-zero) must be non-zero (or it wouldn't be torsion). So you can choose that factor to project to.

[toxic zephyr]

oh so you're saying we can just project to any of the components in the structure theorem product. pick a torsion one (ex. R/(a1)), and then we just pick our t to be the element of M corresponding to having 0 in every element except 1+(a1) in the R/(a1) element?

[tough raven]

I thought t is given and you have to find the projection...

If you have a non-zero linear map M → R/(r), just pick a non-zero element of its range and any t mapping to that.

[tough raven]

i.e. what you have done here.

[toxic zephyr]

i just know that one exists. i was kind of hoping i could construct something more direct for any given nonzero t. but yeah

thank you!

[acoustic igloo]

doesn't this still work if G is infinite?

thanks

[minor fulcrum]

if I have a short exact sequence 0 -> N -> G -> G/N -> 0 of (not necessarily abelian) groups, what are conditions that it splits i.e. G \cong N \oplus G/N

[tough raven]

Usually splits in this context means that G is a semidirect product.

I believe G is a semidirect product iff G → G/N admits a section and is a direct product iff N → G admits a retraction.

More concretely, G is a semidirect product iff there is a subgroup H such that H ∩ N = 0 and HN = G, and is a direct product iff H is additionally normal.

[crystal vale]

<@&268886789983436800>

[quaint ivy]

they've been already muted presumably for spamming this

[tough onyx]

yeah I upgraded to a ban since new account and multiple channels

[tardy hedge]

why did i just learn now that a "free group" is not the same as a free abelian group

free abelian groups of finite rank are Z-modules so iso to Z^n I guess

"free group" means the free group on a set S, consisting of all finite words from the set S?

so its not abelian

[thorn jay]

Yes

[thorn jay]

Yeah, because a free group is "free" with respect to the category of groups and the forgetful functor from Grp to Set

A free abelian group is "free" with respect to the category of abelian groups and the forgetful functor from Ab to Set

You can also see it from a UA perspective, where in the context of abelian groups you have the added equational identity xy ≈ yx, and the free object being the set of terms quotiented by all satisfied equational identities must thus fundamentally be different for groups and abelian groups

But people arent ready for that or smt else they'd do UA

[delicate orchid]

or in english: F(X)/[F(X), F(X)] cong Z^|X|

[thorn jay]

Lmao

Augh this ties back to some really interesting math actually

If you have varieties V < W then there is a natural projection of the free algebra of W to that of V

And the kernel of that projection is a fully invariant congruences, and the lattice of fully invariant congruences of F_V(X) is precisely the interval lattice of varieties [1, V], 1 being the trivial variety

[velvet goblet]

it turns out that it is what you expected it to be

[thorn jay]

Also has to do with coherent systems which i did a while back

[velvet goblet]

this can probably be modified to make sense to talk about subdivision rings of any ring (not just of division rings) by changing the first condition into " there exists an identity element 1_K' in K' " and the third condition into " ∀x∈K'-{0_K}, ∃x^{-1}∈K' such that xx^{-1}=1_K' "

[minor fulcrum]

Can walk me through something. Let's say N = Z_p and G/N = Z_p. Then G is either Z_p x Z_p or Z_p^2. If it's the latter, do we not have a section, not have a retraction, or both?

[bitter rover]

Straight outta Shakespeare

[thorn jay]

I was there when it was written

[tough raven]

(Assuming Z_p means ℤ/pℤ and not the p-adic integers) In the latter case, there is neither a section nor a retraction (when G is abelian, both are equivalent, since the semidirect product would have to be direct anyway).

Are there reasonable conditions on commutative rings A and R with a (pro)finite group G of automorphisms of A for any ring homomorphism A^G → R to extend to A?

An obvious one is for A to be a field and R an algebraically closed field, since A/A^G is integral. Any other ideas?

[minor fulcrum]

yeah ok cool

[crystal vale]

How can I show that if N ≠ {e} normal subgroup of A_5 then N contains at least one 3 - cycle ?

[candid patrol]

Write the conjugaisons classes

[crystal vale]

I understand at least one orbit of < \sigma > has ≥ 3 elements. Also I understand if \sigma ≠ [ ikk] then \sigma must move at least two other elements of J_n.

Now I think they take \sigma = [r j i k s ] \phi, where \phi is disjoint cycles, such that then \sigma' (i) = i, right?

I think it doesn't depend on i

They just wanted to fix one more point

[crystal vale]

How \sigma'(i) = i?

Can anyone help me to go through this proof?

[lusty marlin]

σ'(i) = τ σ τ^-1 σ^-1(i)
σ^-1(i)=j, σ(j)=i, and τ fixes both i and j

All from the definitions of σ and τ

[crystal vale]

Yes I got

So here first we showed that if sigma moves at least 4 elements and its cycle decomposition (ab)(cd)...., then it will contradict.

So it has at least one orbit of length 3, and it is not [ijk ] then it will be again contradict, right?

if f = (123...n) element in S_n then can I say there is no 0< m <n such that f^m (n) = n ?

[lusty marlin]

A cycle of length k always has order k

[crystal vale]

Yes but I am thinking that what if just fixed n

[crystal vale]

No it is not possible

[round portal]

hello guys! good morning

I have a question: how do I show that (G, .) is a group, given that G is finite and . satisfies the cancelling property (on both sides)?

I get that I have to show the existence of an identity and inverse, but I'm struggling with showing the existence of identity lol

harder than I presumed

[rocky cloak]

So one subtly is that you actually have to assume G is nonempty.

Anyway, consider g in G. Can you prove that there must exist x with gx = g?

Can you then show that x must be an identity.

[round portal]

well

I guess yes because of the cancelling property

I think?

but like

how to write that is the problem hehe

gx = g => x = ...

[rocky cloak]

So you have that gx = g

Try to show that xh = h for any h. Use g and the cancellation property

[round portal]

ok buddy!

thank you 🙂

[glad osprey]

btw, to show that there exists an x with gx = x, you need to use the fact that G is finite

[ripe crest]

You mean g?

[coral spindle]

I don’t think he did, no.

[crystal vale]

What is finite free resolution?

[rocky cloak]

A free resolution with only finitely many nonzero terms

[crystal vale]

What is the relation between modules here?

[rocky cloak]

I'm not sure I understand the question

But a finite free resolution is an exact sequence
0 -> Fn -> Fn-1 -> ... -> F1 -> F0

Where the Fi are free.

[crystal vale]

I mean I have to show, let M be a projective module. Then M is stably free if and only if M admits a finite free resolution

[crystal vale]

Okay

[rocky cloak]

And if M is the cokernel of F1 -> F0, then this is a free resolution of M

So in particular you can make an exact sequence
0 -> Fn -> ... -> F0 -> M -> 0

Then using that M is projective you can manipulate this sequence

[glad osprey]

No, I meant the set G. The proof is maybe a bit difficult to come up with if you haven't seen it before, but the idea is to use the fact that if a function on a finite set is injective, then it is surjective

[crystal vale]

I don't get it how M_1 has a free resolution of length < n?

[coral spindle]

The kernel of the map E_0 -> M is the image of the map E_1 -> E_0.

So you can guess what the free resolution of it is

[crystal vale]

0-> E_n -> ...-> E_1 -> M_1 -> 0, right?

[coral spindle]

Of course

[thin stream]

i'm trying to revise basic ring definitions and i have what might be a stupid question.

a ring doesn't need to be commutative, but the wikipedia definition for a unit says it's an element u s.t. exists v where uv=1=vu. why is it necessarily two-sided?? like can u not be a left-unit, or a right-unit, and u have some rings where the only two-sided unit is 1?

i get that if a ring is commutative all left-units are right-units and vice versa tho... but like previously to show something is a unit i just show that exists v s.t. uv = 1, i dont even think about what vu might be, is this an error on my part or is there some proof that all left-units are right-units too?

[rocky cloak]

It is possible for uv = 1 while vu is not equal to 1 yes.

In which case u and v are not units

[thin stream]

okay makes sense. so if i'm checking which elements are units in a non-commutative ring, i should check both sides?

[rocky cloak]

Yup

[thin stream]

thanks a lot! <3

[rocky cloak]

I guess it's worth noting though, that for Noetherian rings it's enough to check one side.

So these rings that have uv=1 but not vu are somewhat complicated.

[thin stream]

noetherian rings are ones where you can "factorise" prime ideals?

i'm pretty new to the rings course i won't lie to u 😭

[thin stream]

i might be thinking of dedekind rings

[rocky cloak]

Noetherian rings are those where there are no infinite increasing chain of ideals

[thin stream]

is noetherian where nested ideals need to eventually be equal?

ahhh ok

[rocky cloak]

So they are like "not too large"

[thin stream]

what's an example of a ring that isn't noetherian?

[rocky cloak]

Polynomial ring in infinitely many variables

[thin stream]

oh, i guess $\mathbb{R}[x_1,...]$

[cloud walrus]

blutac

[thin stream]

yeah lol okay

hehehe i'm glad my brain worked it out as u gave the example that gives me a little mathematical buzz

[rocky cloak]

Or noncommutative polynomial ring (in more than one variable)

[thin stream]

okay makes sense

[south patrol]

No rings are Noetherian

[thin stream]

appreciate it :)

[somber bluff]

Let F_p be a finite field of order p where p is prime. then for a field K, there exists a homomorphism from K to F_p if and ony if K has characteristic p

is this true?

[rocky cloak]

Other direction

[south patrol]

No

[rocky cloak]

Fp -> K

[south patrol]

Yeah hom from Fp to K

Rip beaten to it

[somber bluff]

oh

[south patrol]

There exists a homomorphism K -> Fp iff K is iso to Fp

[somber bluff]

when does K to Fp exist

[south patrol]

Because ring maps from a field are injective

[somber bluff]

so K -> F_3 exists if and only if K is iso to F_3?

[thin stream]

what do u mean by this

[south patrol]

Lol I was joking just like

[rocky cloak]

The zero ring would like a word

[south patrol]

Many people say like they only care about Noetherian rings

Like for some applications that is good

For other areas this is not really good enough though and it seems often good to avoid assuming stuff is Noetherian

[south patrol]

Good point lol

[somber bluff]

ring map from field is injective

[thin stream]

okay makes sense

[somber bluff]

that doesnt imply K -> F_p is iso right

only subfield

[thin stream]

kind of like how rngs are unhelpful but sometimes worth considering them instead of rings

[south patrol]

But K contains a subfield of the form F_l for a prime l

And necessarily l = p if there's a map to Fp