groups-rings-fields

[delicate bloom]

alright cool, keep going

[maiden ocean]

yah

im thinkin about it

ig we have m | r (and also m | ar obv)

argh its probably something dumb and obvious

im assuming i have to use commutativity somehow

[delicate bloom]

I think the divisibility stuff is a bit tricky to look at

[maiden ocean]

so maybe something like (g^a)^r = 1?

and i use this somehow?

[delicate bloom]

I sort of see it as being ar=0 mod m personally since all multiples of m map g^m=1 anyways we can toss them out

I guess I'm trying to figure out what angle you're thinking about it from to help you see

[maiden ocean]

the angle of its 3 AM qwq

[delicate bloom]

haha

does the fact that ar|m make sense or is that just what the book says

[maiden ocean]

ar | m confuses me because we have g^(ar) = 1

and isnt m supposed to be the least positive integer such that g^m = 1?

[delicate bloom]

right, so you're thinking, when ar <= m then it makes sense

[maiden ocean]

oh

[delicate bloom]

otherwise what about when ar > m, is that fair to say

[maiden ocean]

hmmm

m | r and m | ar seem like the keys here

[delicate bloom]

$k^a = g^{ra} = 1 = g^m$

[cloud walrus]

Merosity:

[maiden ocean]

right

o

[delicate bloom]

brb potty break

[maiden ocean]

i guess like, it seems like im supposed to use commutativity here somehow. so maybe ra = ar and so $(g^a)^r = g^m = 1$ is useful somehow?

[cloud walrus]

hegel:

[maiden ocean]

sorry for wasting your time

[delicate bloom]

no, haha you're not wasting my time don't worry about that

just learning, sometimes it's how it goes when you get stuck

[maiden ocean]

thanks :p

[delicate bloom]

so a is the order of k, and we know that k^a = 1 and we also know g^{ra}=1

[maiden ocean]

ye

[delicate bloom]

now if I'm not too tired too, I think that means ra=m here

cause a and m are both orders so they're going to be the minimal number here

[maiden ocean]

hmm

[delicate bloom]

at the very least we know m | ra, do we agree on that?

[maiden ocean]

yah

[steep hull]

Oops one could overkill with Lagrange

[delicate bloom]

that's what I said from the start lol

but apparently not allowed

[maiden ocean]

yah not here quite yet

[delicate bloom]

trying to understand the book's way of doing it

let me go back to what I was saying before, one sec

$k^a = g^{ra} = 1 = g^m$

[cloud walrus]

Merosity:

[delicate bloom]

we know from this that

$ra \equiv 0 \mod m$

[cloud walrus]

Merosity:

[maiden ocean]

yep

[delicate bloom]

it has to be because g^{ra}=1

now is r = 0 mod m?

[maiden ocean]

i mean yah?

[delicate bloom]

well, if it was we'd have nothing to prove

[maiden ocean]

o wait

[delicate bloom]

cause then g^r = 1 right

[maiden ocean]

ye

[delicate bloom]

yikes I am second guessing myself now, I never work with mod not a power of a prime lmao

[maiden ocean]

lol

[delicate bloom]

I was going to say, well that means a =0 mod m but we can't really cancel it out unless m is prime

I think that's ok though, we just need that a | m

[maiden ocean]

ya

[delicate bloom]

since a*r divides m (since ar=0 mod m) we can say that, is that true, hold my feet to the fire if it's not

[maiden ocean]

can i hold ur feet to the fire even if it is true

[delicate bloom]

hold my face too

[maiden ocean]

spooky

[delicate bloom]

feel like I confused myself here I think I need to straighten it out on paper a bit before I say too much more

[maiden ocean]

lol

[delicate bloom]

well ok let's pretend we reasoned this out

[maiden ocean]

hahha

[delicate bloom]

we can still make sure the rest of the argument is sound

[maiden ocean]

yah

[delicate bloom]

once we have a|m and a|n then because gcd(m,n)=1 the only way a can divide both is if a=1

[maiden ocean]

from here it says then $o(k) \mid m$ and $o(k) \mid n$ and since m and n are relatively prime, $o(k) = 1$

[cloud walrus]

hegel:

[maiden ocean]

and thus $g^r = h^r = 1$ and then we have $mn = [m, n] \mid r$

[cloud walrus]

hegel:

[delicate bloom]

alright so it's just that step we have to be sure of

[maiden ocean]

ye

[steep hull]

Suppose that $g^rh^r=1$ and $r$ is minimal. Taking mth powers gives $h^{mr}=1$, so then $n|mr$, with implies $n|r$. Similarly, $m|r$ and we’re done.

[maiden ocean]

pls latex

it is 3:20 AM and i am too tired to read smol text

[delicate bloom]

this was the argument I was going to give earlier

but we were following this way the book did the lemma

[maiden ocean]

thank u tree man

but also those parens lol

[cloud walrus]

tree3:

[steep hull]

The solution you had was Lagrange without mentioning Lagrange

[maiden ocean]

?

[steep hull]

The o(k)|m thing

[maiden ocean]

the proof the book gives?

yes

ik

they are very unsubtle

btw whats the proof that the order of an element = the order of the subgroup it generates?

[steep hull]

Uh the group is literally {1,g,...,g^(k-1)}

[maiden ocean]

o wait

[steep hull]

And they’re all distinct of course

[maiden ocean]

ya

ok i rly need to sleep its 3:30 and im losing brain functionality

also i crashed out earlier today and it fucked my sleep schedule

[steep hull]

😦

[maiden ocean]

big sad

fuck ap chem

[steep hull]

I’m basically nocturnal

[maiden ocean]

lol

are you in the US?

[steep hull]

Yes

[maiden ocean]

noice

well

not all that nice but

[steep hull]

Lol

[maiden ocean]

this week was hell week QwQ

i got slaughtered by an AP chem test on monday and my grade is gonna drop by like 2 points

[steep hull]

@maiden ocean let’s move to chill

[maiden ocean]

yah migration time

[latent anvil]

lemme think about a problem on my algebra homework out loud

Let R be a possibly noncommutative ring

Let J(R) be the intersection of all maximal left ideals of R

(the Jacobson radical)

By a previous problem, this is the same as the set of x such that 1 - xy is left-invertible for all y

I want to show that it's also the intersection of all maximal right ideals of R, i.e. J(R) = J(R^op)

It seems like I want to use the other condition

So show that for a fixed x, 1 - xy is left invertible for all y iff 1 - xy is right invertible for all y (the multiplication xy should switch, but it doesn't matter by a previous problem)

[latent anvil]

Okay so for any y, there's a u such that u(1 - xy) = 1.
I want to show that for any z there's a v such that (1-xz)v = 1

[steep hull]

@latent anvil are you in grad school

[urban acorn]

Does anyone know if the concept of a generator S of a group G such that any proper subset of S generates a proper subgroup of G has a name?

I was trying to study the structure of group automorphisms and it naturally popped up.

[final gulch]

https://groupprops.subwiki.org/wiki/Minimal_generating_set
@urban acorn

[urban acorn]

Hey, that's also the name I came up with.

Definition: Let a finite generator S of a group G be minimal if any proper subset of S generates a proper subgroup of G.     

Observe that every finitely generated group has a minimal generator.
Question: Do all minimal generators have the same order?

Lemma 1: Any automorphism of G takes any minimal generator of G to another minimal generator.
TODO: Add proof.

Lemma 2: No two automorphisms take the same elements of a minimal generator to the same elements.
TOD: Add proof,

Theorem: Let there be m ≠ 0 minimal generators of G of order n. Then |Aut(G)| ≤ n!m.
TODO: Add proof.

(for the purposes of this i only cared about finitely generated groups)

[chilly ocean]

No, Z admits the two minimal sets of generators {1} and {2,3}

(unless I misunderstood what you meant)

[blissful root]

even for finite groups this doesn't work

[chilly ocean]

Essentially, this is the reason why linear algebra is nice while modules are shit

[latent anvil]

@steep hull no

Why?

[woven delta]

Linear algebra is boring

Modules are interesting

[simple agate]

why do I need to show that this is surjective?

[woven delta]

🤔

What do you mean?

To show it's an isomorphism

[simple agate]

that's not injective though is it

[woven delta]

You use the first iso theorem

Mod out by the kernel

And it's an isomorphism

[urban acorn]

@chilly ocean I think you misunderstood what I meant. The theorem establishes an upper bound on the order of the automorphism group of a finitely generated group G for each positive integer n such that G has at least one minimal generator of order n.
The way it works is basically like this: As a consequence of Lemmas 1 & 2 each automorphism corresponds to a unique (up to domain: no two such mappings exist for the same domain and same automorphism) bijections from a given minimal generator to another minimal generator.
There are m different minimal generators, and for any minimal generators there are n! bijections to it.

This isn't necessarily an equality, however, because not every possible mapping corresponds to some automorphism, and so this is only an upper bound.

[blissful root]

the point is your lemmas are wrong @urban acorn

[urban acorn]

Which lemma?

If both provide a counterexample to at least 1.

[blissful root]

oh wait

[urban acorn]

This theorem is enough to find that $Aut(\bZ) \simeq \bZ_2$.

[cloud walrus]

Intel:

[blissful root]

yeah ok the lemmas are fine I was misinterpreting them too

[urban acorn]

that's not entirely your fault; i'm being very loose in my use of language here

[blissful root]

what's as generator of order n?

a set of size n?

[urban acorn]

A set of size n such that no proper subgroup of G contains it.

[blissful root]

can you reword the theorem?

[urban acorn]

Suppose n is some positive integer such that I have a minimal generator of order n.

Let m denote the number of distinct such minimal generators of order n.

[blissful root]

you are also assuming m < infty?

[urban acorn]

Then the order of the automorphism group of G cannot be larger than n factorial times m.

[blissful root]

oh

yeah that's clear from the lemmas

[urban acorn]

yes, there's a loose sense that if m = infty than "n! * infinity = infinity and so we get a trivial bound"

[blissful root]

i thought you claimed equality for some reason

[urban acorn]

but that's very loose

[blissful root]

but that doesn't hold

[urban acorn]

what exactly doesn't hold?

the theorem?

[blissful root]

equality

[urban acorn]

oh, yeah

that's true

[latent anvil]

You might find this interesting:
Call x an inessential generator if it is not contained in any minimal generating set (equivalently, if <S union {x}> = G then <S> = G). Prove that Φ(G) = intersection of all maximal subgroups of G (where it's all of G if there are no maximal subgroups) is the set of inessential generators of G

Also fun: prove that Q has no minimal generating sets

Oh also Φ(G) = G if G has no maximal subgroups

More fun things about Φ(G) (the frattini subgroup)
• it's always nilpotent
• if G is a p-group, then Φ(G) contains G' and G/Φ(G) is elementary abelian

[urban acorn]

"Prove that the phi(G) = intersection of all maximal subgroups of G is the set of inessential generators of G."
phi(G) is a subset of G.
The set of inessential generators of G is a subset of P(G).
How does this make sense?

[latent anvil]

I defined an inessential generator to be an element which is contained in no minimal generating set

[urban acorn]

ohhh

I thought x was supposed to be a set.

okay

that makes it all make sense

[latent anvil]

Ah yeah that was ambiguous

[urban acorn]

at the beginning I was just like "wat? that's just the negation of being minimal"

[mild laurel]

I think the issue was just that the answer to your question is no

And there are easy examples

[urban acorn]

Which question?

[mild laurel]

Do all minimal generators have the same order

[urban acorn]

Can you show me an example?

[mild laurel]

He did

[urban acorn]

oh, right

{2, 3} and {1}

I wonder if it's at least true for finite groups then.

Hmm....

[mild laurel]

1read the next message

Right after the counterexample

[urban acorn]

oh lmao

i didn't realize they were referring to the "Question: do..."

[blissful root]

lol

i took a class on this stuff last year

finite generation of groups

it was uhhh

Idk I don't like finite groups

[urban acorn]

ur request to join kool kidz klub has been considered and rejected

our representatives respectfully wish that u fuck off

regards, the finite group theorists, otherwise known as the KKK (kool kidz club, totally unrelated to the racist hate group)

[maiden ocean]

hmm

i'm trying to prove that the set of left translations and the set of right translations in a monoid are the centralizer of each other

and im very lost

[mild laurel]

send pictures of these definitions

I honestly don't know what the left and right translations are

[maiden ocean]

left translations are the set of all translations $a_L : x \to ax$ for $a \in M$

[cloud walrus]

hegel:

[maiden ocean]

and right translations are the same but on the right

[urban acorn]

Aren't centralizers properties of subsets of the group?

[maiden ocean]

wut?

[hot lake]

centralizers in what

[urban acorn]

And aren't translations mappings on the groups?

monoid*

not gorup

[maiden ocean]

[urban acorn]

ffs

centralizers in a monoid

my bad

ive been doing too much group theory

i just call the object im looking at "the group" now

[maiden ocean]

the centralizer of an element is the set of all elements in a monoid (or group) that commute with that element

[urban acorn]

yes

more generally

[maiden ocean]

and the centralizer of a subset is the intersection of the centralizers of every element in that subset

[urban acorn]

yes

[hot lake]

but what monoid are you looking at that contains left/right translations of the original monoid ?

[maiden ocean]

ig you could call that a property of a subset?

[urban acorn]

yes, and when you're looking at a single element you can still reduce it to the second definition by considering its singleton set

[maiden ocean]

the left/right translations are mappings of the monoid into itself

[urban acorn]

yes

[maiden ocean]

i mean ya

[urban acorn]

mappings of the monoid into itself aren't elements

sure, they may correspond to elements

but they aren't

[maiden ocean]

they're elements of the monoid of transformations of A into itself

[urban acorn]

ohhhh

i see

[maiden ocean]

ye

[hot lake]

all permutations of A ?

[urban acorn]

not just permutations

since not all of them are bijections

[maiden ocean]

ye

[urban acorn]

since inverses don't necessarily exist in monoids

[maiden ocean]

mfw

we should simply declare all groups abelian and move on

[hot lake]

oh right

[urban acorn]

lmao

[hot lake]

so then, A^A ?

[urban acorn]

let's declare all cats smol while we're at it

[maiden ocean]

[urban acorn]

yes, a^a

A^A*

[maiden ocean]

ig im trying to think about like, what is the centralizer of a left translation

but its just

[mild laurel]

all groups are abelian

[maiden ocean]

[mild laurel]

hm

[maiden ocean]

do u have any ideas zoph?

[mild laurel]

I'm still a little confused at your question

[maiden ocean]

wot about it

[mild laurel]

So you take a single left translation and you want to show that its centralizer is some right translation?

[maiden ocean]

im trying to show that the centralizer of the set of left translations is the set of right translations

and vice versa

[mild laurel]

This doesn't make sense

The set of left translations is not a subset of your monoid

A single left translation is a subset

[maiden ocean]

centralizer in M(A), where M(A) is the monoids of transformations into itself

[mild laurel]

Oh, so you're thinking about them in the monoid M(A)

[maiden ocean]

yah

[mild laurel]

Oh

Hm

[hot lake]

does your monoid have an identity element ?

[maiden ocean]

um

yeah?

[hot lake]

then it's pretty easy

[maiden ocean]

all monoids have identity elements

[hot lake]

suppose f is a transformation that commutes with all left-translations

you want to show f is a right-translation

it pretty much has to be the right-translation by f(e)

[maiden ocean]

[hot lake]

so then you want to show that forall x, f(x) = x f(e)

[maiden ocean]

o

[hot lake]

but x is the image of e by the left translation by x

and f commutes with that

so f(x) = f(xe) = x f(e)

and done

[maiden ocean]

hmm

ok ngl ive thought about it and im really not sure

what you are trying to do here

[urban acorn]

Is $A \cross B \cong C$ equivalent to there being $B' \trianglelefteq C$ with $B' \cong B$ such that $\frac{C}{B'} \cong A$?

[cloud walrus]

Intel:

[urban acorn]

In other words - more loosely but intuitively - is $A \cross B = C$ the same as $\frac{C}{B} = A$? i.e. are quotients the inverse of direct products?

[cloud walrus]

Intel:

[urban acorn]

Because it seems obvious, but I can't show one of the directions without making some assumptions I don't know how to prove.

Actually, I don't think it's that obvious that it holds in the other direction.

[hot lake]

that's usually the definition of division

[blissful root]

you are just proving AxB/B = A

surely you see what the isomorphism is

[chilly ocean]

$\frac{A \cross B}{{e_A} \cross B} \cong A$ by the first isomorphism theorem since the homomorphism projection $(a,b) \mapsto a$ (of which ${e_A} \cross B$ is the kernel) is onto.

[cloud walrus]

Deconstructed:

[urban acorn]

@blissful root @chilly ocean No, this isn't it.
You're showing that $A \cross B \cong C$ implies $\frac{C}{B} \cong A$, but I already said I can show it in this direction.

[cloud walrus]

Intel:

[urban acorn]

It's the other direction.

[chilly ocean]

I was just replying to urboz lol

Cool

[urban acorn]

oh

[chilly ocean]

${e_A}\cross B$ is a normal subgroup in $A \cross B$ and isomorphism preserves subgroups so there should be such a $B'$ normal in $C$
I think it might just be up to the fact that subgroup structures are preserved by isomorphisms?

[cloud walrus]

Deconstructed:

[chilly ocean]

Sorry that wasn't formal at all lmao

[urban acorn]

Formal or not, it's dicernible and can be made into a rigorous argument.

How does that prove the other direction though?

discernible*

[chilly ocean]

You would just need that $\frac{C}{B} \cross B \cong C$ hmst yeah uhh D: can't think of a constructive proof rn oof

[cloud walrus]

Deconstructed:

[blissful root]

what other direction?

oh

that one's false lol

[chilly ocean]

o u got a nice counterexample? if it is false prob plenty are lemme try one

[blissful root]

think small groups

there are very easy ones

[urban acorn]

$\frac{\bZ_4}{<2>} \cong \bZ_2$ and $<2> \cong \bZ_2$ but it isn't true that $\bZ_4 = \bZ_2 \cross \bZ_2 = K_4$

[cloud walrus]

Intel:

[chilly ocean]

nice

I wonder where the partial converses are

since the full converse isn't true

[woven delta]

Did someone say that all short exact sequences split?

immediately perks up

@chilly ocean 👀

If the quotient is projective then it's true

Otherwise you're kinda fucked

Oh or the kernel is injective

[latent anvil]

Liquid no

That would be saying C is a semidirect product of C/B and B

[wind steeple]

The generalization of the fact that you can factorize a group by its quotient and its subgroup is called semi-direct product. That's possible iff you can inject canonicaly G/H in G @chilly ocean

[latent anvil]

The generalization is called an extension

A semidirect product is a special kind of extension (one where what you said happens)

[woven delta]

Fine

I only work in abelian categories

[wind steeple]

It's the semidirect product

[woven delta]

So it's all good

[latent anvil]

I have this weird belief that Grp is somehow an abelian category

[wind steeple]

An extension isn't needed to have a section

[latent anvil]

Not actually

[woven delta]

But tbh what I said was still true

It was just weaker than the actual statement

That I should have said

[latent anvil]

@wind steeple yes, I'm saying that extensions are the generalization of "you can factorize a group by its quotient and its subgroup"

[wind steeple]

Oh ok

What's an extension ?

[latent anvil]

An extension of G by H is a group E along with an embedding of H into E as a normal subgroup, such that E/H = G

Or more simply a short exact sequence 1 -> H -> E -> G -> 1

[wind steeple]

Yes

But we need the section

From G to E

[latent anvil]

Right, E is a semidirect product if there's a section

[wind steeple]

To have the reciprocal

[latent anvil]

Reciprocal?

[wind steeple]

Of the previous question

What is the partial converse

From Deconstructed

[latent anvil]

I mean the partial converse is "group extensions are hard"

And there's usually a lot of different ones

group cohomology lets you figure out what all the extensions of G by A are, where A is abelian

it turns out to be encoded in H^2(G,A)

[wind steeple]

I understand his question as "is there any conditions on H thus we can 'factorize' G by H and G/H ?"

[latent anvil]

I understood the question as "how can we recover G from H and G/H"?

[wind steeple]

Oh ok that's another point of view x)

I didn't study group cohomology yet

I'm begining algebraic topology x)

[latent anvil]

I don't really understand it, I'm hoping to do a reading course with a prof in the spring

We had some problems on it in my algebra class last year and it was cool

[wind steeple]

I can't wait teachers to teach me bc where I am is specialized in analysis lel

Unlucky

So I'm reading books

[chilly ocean]

oh god am such a noob, still don't understand short exact sequences, prof brought them up weeks ago and never again when talking about the extension problem but really provided motivation for neither the extension problem nor the definition of a short exact sequence

The extension problem seems interesting on its own at least but I'd be curious to find out the historical relevance or at least what brought people to it as an important problem as opposed to some other form of "creating groups out of other groups"

Interesting people started bringing it up when we were talking about some naive idea to do algebra on groups using quotients and direct products, up to isomorphism. Does it have to do closely with that?

[prisma thunder]

Hey everyone! I've been looking for an algebra group (heh) to be a part of.

[latent anvil]

@chilly ocean short exact sequences will make more sense and feel useful when you do stuff with modules

I think you just found out why we care about the extension problem

It would be nice if C = C/B × B

That's not always true

How much about C can we get from B and C/B?

[chilly ocean]

OHHHHHHHHHH That's so cool!!

That last question seems interesting but unmotivated out of that context, thanks!!

[prisma thunder]

I do have a couple questions regarding a project I've been working. In addition to being interpreted as a problem over algebraic geometry that deals with schemes (not an expert on that so any resources will help), this problem can also be interpreted as when a given matrix over the set of diagonal matrices D (which is isomorphic to R^n as an R-algebra) over this set of interest forces tr to be not only a additive map (i.e. a linear map), but also a multiplicative map.

[latent anvil]

Sorry, what's the problem?

[prisma thunder]

I'll link it here.

https://mathoverflow.net/questions/345359/when-does-the-product-equal-the-sum

The question is how can I determine the set of solutions that forces tr to also be a multiplicative map.

[latent anvil]

Neat question

Are you saying you want to think about what conditions on Kn force tr : M_{n×n}(R) -> R to be a multiplicative map?

I'm not sure what you mean by set of solutions

[blissful root]

seriously when people introduce S.E.S. they HAVE to make the motivating example
0->A->B->B/A->0
explicit

[latent anvil]

Yuppp

[blissful root]

new students do not realize it

[latent anvil]

And even moreso the kernel -> domain -> image

Like ik that's a special case

But showing it to students explicitly is good

[prisma thunder]

I would say determine A's in D such that tr: D -> R is a multiplicative map, where D is indicated as above.

As you noticed above, if one of the elements is in the Jacobson radical R, then we have a way of determining A.

[latent anvil]

Let A have entries a1,...,an and B entries b1,...,bn. Then tr(AB) = a1 b1 + … + an bn. So essentially you're asking when a1 b1 + … + an bn = (a1 + … + an)(b1 + … + bn)

Does that sound right?

[prisma thunder]

OOF

[chilly ocean]

wait so why do short exact sequences tell us more about B from A and B/A, I'm not sure how the maps are defined. not sure why those middle two maps are always injective and then surjective respectively

[prisma thunder]

I totally screwed the calculation up haha

[latent anvil]

@chilly ocean let B be an abelian group and A a subgroup. I claim that there's a short exact sequence 0 -> A -> B -> B/A -> 0

What is it?

The maps are "the obvious maps"

Other question: suppose you have an exact sequence 0 -> M -> N, where the map M -> N is f. Why is f injective?

You asked these questions but it's more helpful to work them out on your own

[chilly ocean]

Thanks, I'll give it a thonk

[prisma thunder]

Take A,B in D. If A+B were in the set of interest, then
tr(A+B) = a_1+...+a_n+b_1+...+b_n = a_1...a_nb_1...b_n = det(AB).

So we would have a relationship between addition in tr and multiplication in det. I think that's what I wanted.

[latent anvil]

I don't really get how matrices get us anything here

[prisma thunder]

It's an alternative description of the problem.

Not that I was sure it was going to get us anywhere Haha

[latent anvil]

Sure

[prisma thunder]

Basically, this problem has me stumped and I'm not sure how to get about it, so if anyone had any ideas here that would be great.

Luc Guyot gave the restriction with the Jacobson radical and Vic Camillo/myself did the case for n=2.

[latent anvil]

It seems hard

[prisma thunder]

So it's open for n bigger than or equal to 3.

[latent anvil]

I have no useful insight lol

I don't think you'll get a good general answer

It seems like you'd want more information about your ring in general

[prisma thunder]

Here's the following conditions that give closure to problem.

If R is a unique factorization, then we have a way of determining R by prime decomposition of product a1... an, associate the elements with certain elements of the decomposition, and just go from there. Then we would have test them to see if they satisfy the property.

If R is an ordered ring with a multiplicative norm that satisfies the triangle inequality, then we have a bound on the elements and what values they take. If R is a discrete ordered ring, then we can easily determine what elements should be which. If R is non-discrete, then...well, we would have really bad bound. Furthermore, this would fail for non-ordered rings like the Gaussian integers.

[latent anvil]

I don't see how unique factorization helps here

[cloud walrus]

ZaiD Bogard:

[prisma thunder]

Why is this popping up?

[chilly ocean]

Are the maps in exact sequences (for groups) defined as homomorphisms? Or not necessarily? They look like they're begging to be homomorphisms

Cause in that case "suppose you have an exact sequence 0 -> M -> N, where the map M -> N is f. Why is f injective?" can be answered by the kernel being trivial from the definition of exact sequences and trivial kernel implying injection for homomorphisms

got that one

[prisma thunder]

I mean if R is a UFD, then the product a1a2 all the way to an has decomposition p1p2 all the way to pn. Choose ai such that the above property is satisfied and satisfies that sum = product.

However, now that I think, we're assuming the prime decomposition could be any decomposition, which is effectively inefficient.

Hmmm, you might have a point.

[chilly ocean]

As for 0 -> A -> B -> B/A -> 0, the obvious maps are the restricted identity and the natural map, it being an exact sequence comes nicely since A is a subgroup and it's image from that first map gets mapped to itself as the identity coset in the second one, so it was the kernel of that map, and the maps to and from 0 come even more easily

aaa idk about getting B from A and B/A tho hmst

[latent anvil]

Yeah, the maps have to be homomorphisms

Your first thing is right

Usually we call the restricted identity the inclusion map

Getting B from A and B/A is impossible (consider the example A = Z/2Z and B/A = Z/2Z)

Figuring out what Bs it could be is extremely hard

[woven delta]

Short exact sequences are literally just 0 -> ker \phi -> A -> im \phi -> 0

[latent anvil]

^^^^

[woven delta]

That is what you should think of when you see one

[latent anvil]

To be clear, phi is any homomorphism going out of A

[prisma thunder]

Speaking of short exact sequences.

Suppose some setoids A,B contain 0 (i.e. setoids with identity) and with an M-monoid action on them, phi: A to B is a bijection and preserves M-multiplication (i.e. phi(mx) = m phi(x)and define ker phi = {x in A | phi(x) = 0}. Is there a definition for exact sequences in this context?

[latent anvil]

What do you mean by setoid?

Sets with an equivalence relation on them?

Also wouldn't ker phi be trivial of phi is a bijection?

[prisma thunder]

A setoid is a set with an equivalence relation on it. Oh, and yeah, I'm being dumb that is just trivial following from the definitions. However, the question still stands: is there a definition for exact sequences on setoids with identity?

[blissful root]

why

the definition of the kernel here isn't good

it doesn't have structure

[latent anvil]

I don't see how the setoid structure is being used

Does the monoid action send equivalent elements to equivalent elements? If so, shouldn't the kernel be all elements equivalent to 0?

[latent anvil]

Also you're saying "identity" but 0 isn't the identity for anything

[fickle brook]

0 is the identity for addition

[urban acorn]

My cat is the identity for addition.

My dog is the identity for multiplication.

My cat is my dog. And "dog" is my cat whereas "cat" ain't no dog, but is my "dog" without quotations.

[urban acorn]

Can someone explain to me what $H^k(P^n(\bR);\bZ_2)$ is?

[cloud walrus]

Intel:

[blissful root]

cohomology mod 2

[urban acorn]

$P^n(\bR)$ is the space of polynomials of degree at most n with coefficients in R, right?

[cloud walrus]

Intel:

[blissful root]

no

projective space

but the notation is bad

also dont write Z_p for Z/p

[chilly ocean]

@blissful root do you know what the integral cohomology of RP^n is?

[blissful root]

yea

https://topospaces.subwiki.org/wiki/Cohomology_of_real_projective_space

its a bit long to describe

[chilly ocean]

and do you know the UCT?

[upbeat burrow]

Why don't write Z_p for Z/pZ

[urban acorn]

Z/pZ*

[upbeat burrow]

Eerr, yeah

that

[urban acorn]

I guess Z/pZ highlights the fact that it's a quotient of the integers?

[mild laurel]

Z_p is p-adic integers

[urban acorn]

other than that i have no idea

ohhh, because of the ambiguity

yeah, then let's write C_n

[upbeat burrow]

Ahhh

[blissful root]

@chilly ocean yes why

[urban acorn]

not to be confused with Z_n

[cloud walrus]

Bodenhammer666:

[chilly ocean]

cause you can use that

[blissful root]

yes, you can

[urban acorn]

$C_n$, the cyclic group of order n. Not to be confused with $Z_n$, the cyclic group of order n.

[cloud walrus]

Intel:

[chilly ocean]

now let's say k is odd

what happens to Tor?

obv 0<k<n

@blissful root can you see why Tor vanishes?

[blissful root]

why are you asking me these questions?

[mild laurel]

he's not the one that asked the initial question

[chilly ocean]

oh

i didn't go that far up

then nvm

[blissful root]

anyway tor doesn't vanish

the computation gives what's in the link i posted

i guess you meant that tor vanishes in half the cases

when k is uhhh even

[simple agate]

do isomorphisms preserve integral domains? i.e. if I have by the 1st iso. thm that $A[X]/\langle x \rangle \cong A$ and I know that $A$ is an integral domain, can I also infer that $A[X]/\langle x \rangle$ is an integral domain?

[cloud walrus]

alex:

[mild laurel]

Yes

Everything is preserved under isomorphism

[simple agate]

oh that's good to know, thank you

[blissful root]

that's a slogan

but you need to check it

[mild laurel]

Yeah, it's not too hard to check directly in this case

[urban acorn]

Every structural property is preserved under isomorphism.

Even integers under addition are isomorphic to all of the integers under addition, but not all integers are even.

But most of the times you'll just create a headache by distinguishing trivially isomorphic objects.

So let $\bR$ be the set of 1x1 matrices with real entries.

[blissful root]

structural just means properties that are preserved under isomorphism

[cloud walrus]

Intel:

[urban acorn]

Well, if you define it that way - which is a good definition in most cases - than what I said does reduce to tautology.

[blissful root]

yes

[urban acorn]

But in this case, in order to give it more meaning, I'd define it as a property that can be defined in terms of the add structure here of the object. Like in terms of the operations of an algebraic structure.

It's a little loose, but it still provides insight.

[woven delta]

🤔

[blissful root]

the formalization of that is "stable under isomorphism"

[urban acorn]

you got a point

but it still provides intuition that allows you to easily see if things are preserved under isomorphism (hint: if you're doing algebra the answer is yes)

[latent anvil]

hot take: we should have an axiom saying all properties are preserved under isomorphism

yeah it's inconsistent but it would be very convenient

[urban acorn]

@latent anvil smoked so much weed he's on a higher plane of understanding

[maiden ocean]

uwu big boy channel

[woven delta]

Lmao

Okay, so what is a generating set

So you know what a basis is from linear algebra

[maiden ocean]

the intersection of all the subsets of a monoid/group that contain that set rite?

[woven delta]

You can think about a basis as a map from R^n to V

[maiden ocean]

yah

[woven delta]

And a spanning set is a surjection from some R^n to V

So generating sets are basically spanning sets

It's the same idea

[maiden ocean]

surjection just meaning a surjective map rite

[woven delta]

Yes

[bleak abyss]

R here is a ring ofc

[woven delta]

Of course

[maiden ocean]

dami i hate u

[latent anvil]

@bleak abyss you mean an affine coscheme?

[maiden ocean]

:(

[woven delta]

So you have a set of generators

[bleak abyss]

ugh you remind me I still gotta learn AG

[maiden ocean]

go awayyyy

[woven delta]

So that gives you a map from Z^n to your abelian group

But

If you know that each generator has finite order

[latent anvil]

Are you working through Hartshorne dami?

I was told by a prof on Friday to do all the problems in Hartshorne in the next 24 months

Fun stuff

[woven delta]

You can refine the abelian group you are mapping from

[bleak abyss]

Not at the moment, my class has its own notes that vaguely ish are like

A mix of Milne's AG notes and Shafarevich

For first quarter at least

[woven delta]

And instead say it is the product of the finite cyclic groups generated by the generators

[latent anvil]

Oh yeah my actual class is doing varieties in the first quarter

And then Hartshorne in like 2 months

[bleak abyss]

I found Perrin which seems to be a bit faster? Idk

[woven delta]

And that means you have a map from a finite set onto your group

[bleak abyss]

Next quarter is schemes

[maiden ocean]

[woven delta]

So your group is finite

You have to check this argument makes sense though

[latent anvil]

I'm gonna get destroyed by schemes I bet

It'll hurt

[bleak abyss]

Though we're not really working out of any one book. Hartshorne is the obvious candidate, though somehow I think Gortz/Wedhorn and Liu seem better tbh

[woven delta]

But when you see generator, you should think surjective map from Z^n/something like this

[blissful root]

liu is pretty based

I like how he covers some important commutative algebra

that few other texts cover

[maiden ocean]

mfw migrate to channel to read explanation and buried by AG

[woven delta]

What do you think of Mumford's red book?

[blissful root]

i haven't read it yet

[bleak abyss]

Yeah it's supposed to be better than Hartshorne for number theory until it's time to do sheaf cohomology at which point no

Just parroting what people say

[latent anvil]

What was the original question?

[blissful root]

hartshorne is surprisingly very good for cohomology

[latent anvil]

Oh nvm not gonna check out liu then

[woven delta]

I've been looking at this book by Warner that does sheaf cohomology for manifolds and it's kinda nice

[blissful root]

yeah a lot of this stuff is pretty general

to be fair once you see one example of homological algebra the rest is the same

[bleak abyss]

Glancing through Mumford it seems kinda

[woven delta]

Yeah I keep hearing about how the cohomology theories are just different resolutions

[bleak abyss]

Interestingly placed

It starts with varieties but not really classical stuff

[woven delta]

I think I like Mumford

Also I want to read his other book

On GIT

Once I grow up a bit

[blissful root]

yeah essentially you need to resolve your object M by "acyclic stuff" F_i, that means stuff with no (co)homology, as
0->M->F0->F1->F2->...

[bleak abyss]

It talks about prevarieties and all, like what Milne does/what we do in class

Oh GIT seems dope from what I hear

[blissful root]

and then the complex 0->F0->F1->F2->... lets you calculate the stuff about M

[bleak abyss]

But yeah after that he kinda gets into schemes but seems to do less than other books

God wait there's so many AG books I swear

[blissful root]

the morally correct way to think about it is that you are actually doing two steps

1. include your object M into the category of chain complexes as a complex 0->M->0

[woven delta]

Yeah that book is meant as kind of a crash course

Right

[latent anvil]

There's way too many fucking books

In general

[blissful root]

2. find a nice chain complex and a weak equivalence from 0->M->0 to this object

therefore M and your nice complex are the same in homology

[woven delta]

Right, I saw that idea in weibel urboz

[blissful root]

nice

yeah some books don't spell it out

[woven delta]

Fuck I should get back to weibel

[bleak abyss]

Off the top of my head: Shafarevich, Reid, Red Book, Milne's notes, Gathmann's notes, Perrin, Hartshorne, Liu, Gortz/Wedhorn

[latent anvil]

@woven delta yeah you should fucker

[bleak abyss]

Vakil ofc

[blissful root]

you're missing uhhh

[woven delta]

There's other books by Mumford

[blissful root]

geometry of schemes

what's that again

[latent anvil]

I should read vakil

In the near future

[woven delta]

Eisenbud and Harris

[blissful root]

yeah

[bleak abyss]

Oh yeah true Eisenbud-Harris

[blissful root]

that one's good

[woven delta]

Fulton

[bleak abyss]

Fulton Algebraic Curves

Fuck

[blissful root]

don't bother will fulton

you're a big guy

[bleak abyss]

Harris I guess

Lol yeah this is more just like

Wow there are so many

[woven delta]

Yeah

Which book by Reid are you talking about?

[bleak abyss]

Undergrad AG

[woven delta]

Lmao

[bleak abyss]

For now I need to know varieties so I'm looking at a bit of Perrin and Milne

[woven delta]

Oh there's another one

[blissful root]

do you?

[woven delta]

The one that focuses a lot on complex Geometry

[blissful root]

miranda?

[latent anvil]

There's that one undergrad one

[woven delta]

Principles of Algebraic Geometry

[bleak abyss]

Griffiths Harris

[latent anvil]

Ideals, varieties, and algorithms

[woven delta]

Right

[blissful root]

cox little oshea is nice

[latent anvil]

Yee

[woven delta]

Yeah

Computational stuff is good

[bleak abyss]

And yeah my first semester is doing varieties, these are the lecture notes: https://www.math.wisc.edu/wiki/images/Notes.pdf

[woven delta]

The 2 main things I've learned in grad school are computations are good and integrals are good

[bleak abyss]

Do a different part of math if you like integrals tbh

[woven delta]

Lol

Nah

[latent anvil]

I like integrals

[woven delta]

I don't like intergrals

[latent anvil]

I like the way they can be done by wolfram alpha

[woven delta]

But they are cool objects

Yeah, same

I like hyperbolic Geometry, so I appreciate integrals

I'm a simple man

[bleak abyss]

See that's your problem

[woven delta]

Also curvature is a thing you should care about

[bleak abyss]

Nope differential geometry can commit sudoku

If you need more than a smooth structure to talk about it, it's not worth talking about

[woven delta]

If you need more than a tangent sheaf

[blissful root]

the thing is like

you can say "i'm an algebra, and I only care about nice algebra, let's say projective varieties over C"

and then whoops you're automatically kahler

how can I hold all this structure?

and then next thing you know you have to care about PDEs

life's hard

[bleak abyss]

Is this the story of your life?

[blissful root]

seems to be

I didn't choose the thug life

[woven delta]

Lol some Geometer starts poking and prodding you with an anaytification functor

[bleak abyss]

Tbh I kinda wanna learn Hodge theory at some point

Sounds dank

[blissful root]

yeah it's pretty neat

I learned the classical stuff recently

need to learn more

[manic trail]

Anyone good with algebraic geometry?

And willing to help

If I can ask the question here or in some other channel?

[mild laurel]

Here or #point-set-topology is fine

Depending on the question maybe I can help

[manic trail]

Allright so the setting is

We have the curve in P^2 given by the equation

$ x_1^n = x_2 x_0^{n-1} - x_2^n$

[cloud walrus]

Ridder:

[manic trail]

And a presentation is given by the charts with x_0 != 0 and with x_2 != 0

The question is to show that

$\mathcal{O}_X(X) = k$

[cloud walrus]

Ridder:

[manic trail]

By considering we have the exact sequence

$ 0 \to \mathcal{O}_X(X) \to \mathcal{O}_X(X_1) \times \mathcal{O}_X(X_2) \to \mathcal{O}_X(X_1 \cap X_2) $

[cloud walrus]

Ridder:

[manic trail]

Hope it's clear what I'm saying: X_1 and X_2 are the intersections of the curve X with the x_0!=0 and x_2!=0 charts.

Please tell me if it is/isn't clear what the question is

I can also tell you what I tried if that helps

@mild laurel 🙂

[mild laurel]

Yeah sorry, don't know enough about this

[manic trail]

Ok thanks for your time though

Someone else know some stuff about this? Would be highly appreciated!

[bleak abyss]

@manic trail disclaimer, I'm not great at this stuff but I kinda wanna think about it

[manic trail]

Sure, thank you! Let's see how far we can get with it

[bleak abyss]

So is the idea here that we're just trying to compute other terms of the sequence?

[manic trail]

I think so

Because we know that the sequence is exact

We can identify O_X(X) exactly with the kernel of the last arrow

[bleak abyss]

Do any of these guys contain a copy of P^1?

If you contain a copy of P^1 then regular functions should be constant I think

Though that'd be a very roundabout way to do this lmao

[manic trail]

Hm

[bleak abyss]

Actually tbh the fact that this would be a bit stupid if true sorta suggests that at least X_1 and X_2 shouldn't contain a copy of P^1

[manic trail]

I don't think so because X1 and X2 are affine

[bleak abyss]

Right I'm dumb

[manic trail]

What my attempt was calculating O_X(X1) and O_X(X2)

Which I think can be done by writing them as a quotient of a polynomial ring

Since they should be closed inside affine space

[bleak abyss]

Yeah that could work aside from the worry about taking the radical

[manic trail]

But then I can't find the kernel

Oh ye that radical also

[bleak abyss]

So X_1 is given by the affine equation x_1^n = x_2 - x_2^n

[manic trail]

yep

And X_2 is given by the affine equation x_1^n = x_0^n-1 - 1

And their intersection is like

weird

😛

The difficult thing is

Even if I (attempt to-lol) explicitly calculate the O_X(X1) as a quotient of polynomial space

I'm still not getting anywhere

So yea that's where I'm stuck

[bleak abyss]

Yeah it's tricky

[woven delta]

👀

[upper pivot]

the group of all maps $\psi_y(x) = yxy^{-1}$ form a subgroup of $Aut(G)$, and hence that must be cyclic. if we let $\psi_y$ be a generator, then $\psi_y^n (x) = y^n x y^{-n}$ and that must equal to some $ \psi_g$ which implies that forall x and for all g, and some n $ y^{-n} g$ commutes with x, but since $y^{-n} G=G$, clearly G is abelian

now tell me how wrong this is lol

[woven delta]

I'll post my proof lol

[upper pivot]

also rip bot

[woven delta]

Also it's called Inn(G)

[cloud walrus]

JohnDoeSmith:

[upper pivot]

whats inn(G)?

[woven delta]

The group of inner automorphisms

Ie congugations

[upper pivot]

oh

[woven delta]

The left board lel

Z(G) is the center of G

The map G to Inn(G) is g mapsto g^-1xg

@upper pivot

[upper pivot]

mod

[woven delta]

If they are in the same coset

They are equal mod the kernel

[upper pivot]

oh i think i understand that

oh i see thats nice

wait is our proof, kinda similar

cause i also had $g= y^n k$

[cloud walrus]

JohnDoeSmith:

[woven delta]

It should be the same proof

I just know enough Algebra to make things slick

[upper pivot]

yeah probably

i spent the past hour just playing around with stuff until something stuck lol

[woven delta]

Incidentally this tells you inn is trivial

[upper pivot]

wdym trivial?

[mild laurel]

Trivial group

[upper pivot]

oh right causse if G is abelian then Inn(Z) is just {1}

[cloud walrus]

QuickMaffs:

[mild laurel]

well p | ab

[tribal pasture]

Hmm?

[mild laurel]

you see why p | ab?

[cloud walrus]

QuickMaffs:

[mild laurel]

well no, you can't really do operations like inverting p

p has no inverse in Z

[tribal pasture]

Oh right right. But yeah I dont even have to do that

[mild laurel]

but ab is in pZ

[tribal pasture]

ab = pZ <=> p | ab

[mild laurel]

yes

[tribal pasture]

I understand it follows from p|ab => p | a v p|b. But I wanted to do it completely group-theoretically. That's to go from ab ∈ pZ => a ∈ pZ v b ∈ pZ

For understanding purposes

[mild laurel]

You can't

That's the only way you can do it

[tribal pasture]

):

Yeah i guess you are right. This property more relies on the numeric properties of a prime number, than just mere group-theoretic properties of pZ. So you must necessarily rely on "number-theoretic" reasoning.

[tribal pasture]

For b) why cant I multiply both sides by [a]^{-1}?

[tribal pasture]

<@&286206848099549185>

[mild laurel]

How do you know a inverse exists

[tribal pasture]

a is in F_p\0 which is a multiplicative abelian group

[mild laurel]

Have you proven that

[tribal pasture]

Smh Nvm...... Apparently this statement comes after the theorem but it is actually a tool needed for the proof

[lament dawn]

How do we construct a Bass-Serre tree for a non-primitive block graph in which 3 maximal cliques meet at a single cut point?

for example

I imagine it would take two trees but I'm unsure

[lament dawn]

:(

[stoic rose]

I don't think this goes in #groups-rings-fields

[wind steeple]

#numerical-analysis ? #discrete-math ?

[mild laurel]

As far as I know, non primitive block graphs are a type of graph that arises from certain group actions

Well at least, that's my guess from knowing what blocks of non primitive group actions are

Also really not too hard to just google https://en.wikipedia.org/wiki/Bass–Serre_theory and see that it is group theory related

[lament dawn]

^ def group theory

I figured it out btw, so no need to answer anymore if anyone knows

[mild laurel]

Yeah nice, I don't think anyone in here has the necessary knowledge to help you out on this

Seems like pretty specific stuff

[lament dawn]

Yeah, I figured that since its a trending topic right now there might be someone in here that knew

[mild laurel]

Yeah I've heard of tits buildings a couple times and it seems that this is related?

[lament dawn]

(The answer btw is that you can divide by the index relative to the whole block. The resultant graph is isomorphic)

Yeah, its in the Artin-Tits group

[urban acorn]

69 upvotes for 69 IQ

[quiet cave]

[chilly ocean]

Semidirect products are evil.

[chilly ocean]

yes

extensions should be banned

[icy lagoon]

Guys help me with those tasks please

<@&286206848099549185>

[fading wagon]

@icy lagoon context?

what are you supposed to do with these diagrams?

are they abstract algebra?

[junior edge]

Hello hello, I'm having troubles with a math proof which involves fermat's sum of two squares. The question is: Suppose that p is a prime and p is congruent to 1 (mod 4). Show that it is possible to write p^2 as the sum of two positive squares.

I (think I) understand how to do fermat's sum of two squares in practice, but the theory still deludes me a little bit. I'm not entirely sure how to get started beyond squaring p and trying to go from there

[fading wagon]

@junior edge #elementary-number-theory

[junior edge]

My b! This is a part of my abstract algebra homework so I figured I'd ask here first lol

[fading wagon]

really? that's a bit weird

[junior edge]

It's a part of our factorization in integral domains section, we just started that and Group theory

[fading wagon]

okay, that might make more sense

what's the ring?

integers?

@junior edge

if so there's probably also a elementary number theory proof

[junior edge]

The question doesn't specify, but yes, I'm assuming it's the ring of Gaussian integers

[fading wagon]

hmm, that might be a bit harder than I originally expected...

1+4i time

wait

oh no

we can't write (1+4i)^2 as the sum of 2 positive squares

[hot lake]

tell us everything you know about the ring of gaussian integers and about sums of two squares

[junior edge]

Well, I know that a positive prime integer p can be written as the sum of two integers iff p is either 2 or congruent to 1(mod 4). As for the gaussian integers, I know that they are ring of complex numbers including both real and imaginary numbers

[hot lake]

were you given a proof of the first thing you said ?

[junior edge]

Yes, I have a bidirectional proof we did in class the other day that involved factoring an integer n as n=p*...p_kq^2*...*q_l^2

where p...p_k are primes either equal to 2 or congruent to 1 mod 4 and q,...,q_l are primes congruent to 3(mod 4)

[hot lake]

... that sounds strange given that you're only talking about prime numbers

so did you get a more general result stating the number of ways to write something as a sum of two squares given its prime factorisation ?

or you can just take the result about prime numbers and think really hard about multiplication and norms of gaussian integers

the ring of gaussian integers is {a+ib | a in Z and b in Z}

the set of complex numbers whose real parts and imaginary parts are integers

[junior edge]

Yep, for one direction of the proof, we factor as above, and then in Z[i] we take p_i = x_i * x_i', but I don't fully understand it myself which is the problem lol

Oh well, I guess I just have to think more about it

[hot lake]

you should probably study that proof a bit more

and if you got the general result then this is just applying that result to the case of p² with p=1 mod 4

[quiet ember]

Hey, I'm wondering

If we have a commutative ring with unity R
Is the factor group R/R={r+R|r in R} =R?

[golden pasture]

uh no, r1+R=r2+R=R for whatever r1,r2 in R

[wind steeple]

it's equal {R}

[quiet ember]

aaah, okay. Thx :3

[prisma halo]

Knowing that f:R--->R , defined by f(x) = 2x/(1+x^2)
How do we check if its subjective or injective?

[mild laurel]

This isn't abstract algebra

[prisma halo]

what is it

thats legit what the module is called in our uni france

[mild laurel]

This is like #proofs-and-logic

[prisma halo]

oh true its logic

[mild laurel]

You're getting to abstract algebra, but this isn't really it yet

[prisma halo]

Thank you so much, im beyond confused

[wind steeple]

#precalculus

[barren delta]

I need to determine if this is a homomorphism and I don't know where to start

$\varphi: G\rightarrow G,; \varphi(g) = g^{-1},;g\in G$

[cloud walrus]

kickpuncher:

[barren delta]

I'm not sure how to check if $\varphi^{-1}[{e'}]$ is not a normal subgroup

[cloud walrus]

kickpuncher:

[barren delta]

G is any group, btw

[barren delta]

<@&286206848099549185>

[latent anvil]

Well φ^(-1)(e) is just e

So it will be a normal subgroup

But that's not sufficient to tell you whether φ is a homomorphism

What's the definition of a homomorphism?

[barren delta]

preservation of operation

[latent anvil]

Sure, so let's write that down here

[barren delta]

$\varphi(xy) = (xy)^{-1} = y^{-1}x^{-1} = \varphi(y)\varphi(x)$

[cloud walrus]

kickpuncher:

[latent anvil]

Sure. What's (xy)^(-1)?

[barren delta]

^

[latent anvil]

Oh yeah lol sorry

I misread

[barren delta]

np lol

[latent anvil]

Okay, so what you wrote isn't the homomorphism condition then

[barren delta]

right, because of the reordering

[latent anvil]

So what has to be true if φ is a homomorphism?

[barren delta]

is that enough to claim it is therefore not a homomorphism?

[latent anvil]

Nope

Sometimes it will be

[barren delta]

hmmmmmm

I'm not sure what else has to be true... I thought it was simply showing that the operation is preserved

[latent anvil]

It is

I'm asking you to write down what that would mean for this specific φ

[barren delta]

iiiiiii'm not sure I follow. I've shown that $\varphi(xy) \neq \varphi(x)\varphi(y)$

[cloud walrus]

kickpuncher:

[latent anvil]

No you haven't

You've shown that φ(xy) = φ(y)φ(x)

How do you know that's different from φ(x) φ(y)?

[barren delta]

I mean I can't assume G is abelian

[latent anvil]

Sorry, what's the original question? This proves that it will be a homomorphism when φ is abelian

But in general it might not be

[barren delta]

ah sorry yeah it asks for any group G

[latent anvil]

There's not a uniform answer independent of whether G is abelian

Oh, then no. You need to find an explicit counterexample

Or figure out what is implied by φ being a homomorphism

Like, do you think G has to be abelian for φ to be a homomorphism?

[barren delta]

from what I can see G is abelian iff phi is a homomorphism

[latent anvil]

That's exactly right

Can you prove it?

(and then this tells you φ isn't always a homomorphism because there are nonabelian groups)

[barren delta]

yeah I guess I could do a proof from both directions... or could I use contradiction right off the bat?

[latent anvil]

I don't think contradiction is that helpful

Just write out the definition of "φ is a homomorphism" and see what happens

[potent lynx]

👀

[barren delta]

Let $\varphi : G \rightarrow G'$ be a group homomorphism. Show that if $|G'|$ is finite, then $|\varphi[G]|$ is finite and is a divisor of $|G'|.$

[cloud walrus]

kickpuncher:

[wind steeple]

what's the prob ?

[latent anvil]

do you know lagrange's theorem?

Because this is just a clunky statement of that

[barren delta]

uggggggh yeah lol lemme look it up

okay so if H is a subgroup of some finite G then |H| divides |G|

Then there's also that if phi is a hom. G → G' and H is a subgroup, then phi [H] is a subgroup of G'

[latent anvil]

Yup

[barren delta]

🙄

[latent anvil]

and G is a subgroup of G

[barren delta]

yeah

ugh lame

[latent anvil]

?

I'm not sure what you expected

[barren delta]

haha it's just a lame problem

[latent anvil]

Oh sure

Is it homework or something?

Because if so your homework is bad lol

[barren delta]

it is HW lol

¯_(ツ)_/¯

[mild laurel]

I mean it's not that bad

It's easy sure

But it requires you to figure out that the image of a homomorphism is a subgroup

[visual turret]

So I have just been learning about quotient groups, and I do not understand why, if for an element g of a group G with H normal subgroup of G, if o(g)=n, then why o(Hg) is not necessarily n. Does anyone have a simple example of a case where the o(Hg) is not o(g) (or intuition as to why it is this way)

[blissful root]

H = G

less trivial example, G = Z/4, H = Z/2 generated by 2

[stone fulcrum]

Why would you think it is? Multiplication almost never preserves order

[visual turret]

I was having trouble thinking of a case where (Hg)^m=H for m != o(g)

Specifically with m<o(g) of course

[stone fulcrum]

g may not have an order m after taking mod H

It will likely drop

What's your question again? Your statement is odd

[blissful root]

the smallest two examples are right up there

[visual turret]

Yes, thank you, they helped

Would it be true that the order of <(2,2)>+(5,8), where <(2,2)> is a normal subgroup of Z/12 X Z/12 is just the lcm(5,8)?

[blissful root]

dont think so

in particular <(2,2)> + (5,8) = <(2,2)>+(1,4)

so no formula like that will work

you know the order divides 144/6 = 24

[visual turret]

Ah, good point, thanks

[blissful root]

you want 5n = 8n (mod 12)

this isnt enough, but at least puts you in the diagonal <(1,1)>

from here it's either the smallest n, or 2n

depending on where you land on <(d,d)> for even or odd d

[charred dew]

if I may butt in for a quick second here; what does this notation mean? <(2,2)>

[blissful root]

subgroup generated by (2,2)

[charred dew]

okie. Will have to look into that :+1:

[blissful root]

why

[charred dew]

haven't learned it yet 🤷

[visual turret]

Since the order divides 2^3*3, and the order is not 2 nor 3, as neither 2 nor 3 is a solution to 5n = 8n (mod 12), then it has to be 4, as 4 is the smallest number for which n(<2,2>+(5,8)) = <2,2>

Is this correct?

Thanks for all the help

[blissful root]

well you ruled out 2 and 3, it could still be 4, 6, 8, 12, 24 in principle

but yeah the fact that 4 gives n(<2,2>+(5,8)) = <2,2> proves it

along with no smaller solutions existing

[potent lynx]

a very bad question

whats the point of automorphisms

[mild laurel]

Automorphisms of a group?

[potent lynx]

yes

[mild laurel]

The automorphisms of a group is also a group

And examining this group can give you lots of information about your original group

[potent lynx]

oh

spoiler alert but ok

tt

ty

[thick fractal]

Sometimes I think Algebra in general is just abstract 😂

It doesn’t make senseeee

😂

[urban acorn]

it does make sense

but if it didn't, it wouldn't make it abstract

it'd make it nonsensical

[thick fractal]

English please I’m a freshmen 😂

In Algebra 2:1 sooo