groups-rings-fields

[knotty badger]

yep yep

[thorn jay]

ooo!!

[knotty badger]

it's for mathematicians and physicists

[elfin wraith]

That’s pretty cool

The physics rep theory course at my UG was similar I think actually, but that shouldn’t have been the first time anyone in that saw groups

[thorn jay]

it's good that you're making sure everything flows nicely

[knotty badger]

oh this is definitely the first time some of my students are seeing groups

[thorn jay]

bleak

[elfin wraith]

I’m going to need to spend my next like support class teaching how to write which is fun

[vocal pebble]

wdym by check? an action of G on X is a homomorphism p from G to Sym(X), so p(e) = id

[thorn jay]

omg I have that next year it's going to be so funny

[elfin wraith]

Of the 120 homework’s I marked last week, there was like maybe 5 people who made an actual mistake, but like 10 people who got full marks because the other 110 just submitted a pile of calculations without any words

[thorn jay]

some papers have too many words too little math tbh...

99% fluff

[glad osprey]

Just to make sure I understand: when you define group actions as a map f : G -> Aut(A) you get f(e) = id "for free", but when defining it as a map f : G -> (A -> A) (that is G x A -> A) you need to specify f(e) = id. And the reason is basically that A -> A is just a monoid, so you need f(e) = id to ensure the image is actually Aut(A)?

[elfin wraith]

Yeah like it was a very computational problem, so I get it, but you can’t just start pulling integrals out of thin air

You need to give me some explanation of what’s happening

[thorn jay]

right

[knotty badger]

yes yes

[thorn jay]

do you ever take a step back and look at the math notation and think "what the fuck am I even doing bro"

[knotty badger]

when you define a group action, you "naturally" define each element of G to become a function A -> A

[elfin wraith]

Also a lot of people confused about the definition of a linear DE which is fair, it’s a slightly confusing definition

[knotty badger]

but these functions just form a monoid

it turns out that to show that it lands in Aut(A), you need to show the identity element gets sent to the identity map

otherwise, it could just get sent to some random idempotent

[knotty badger]

defining a group representation requries an extra step

you "naturally" define each element of G to become a function V -> V

you'd need to check these are linear maps

so now you're landing in End(V), the endomorphism monoid of V (whose elements are linear maps V -> V)

and then you need to check the identity gets sent to the identity map

but yeah, if i want to explain this to my students, it might be helpful for them to know what a monoid is

[elfin wraith]

I think it’s a good thing to let them know about

Monoidal categories show up a lot, and when you come to define rings you can just say, slap on a compatible monoid and you’re good to go

[knotty badger]

One of my students is complaining that the notes never define what "well-defined" means, ironically

How would you go about explaining this

I think I'd want to go via the relation route, and talk about when relations define functions or not

[rocky cloak]

I think it's also quite "natural" that the identity of the group act as the identity function, but it can probably be helpful to emphasize the reason we don't require this for homomorphism is because we have inverses, and functions don't always have inverses.

I think any more about monoids would probably be a detour, but if you have the time learning about monoids is useful anyway I suppose...

[thorn jay]

"means that, whenever two inputs are the same, the definition of the function also actually ouputs the same elements"

[knotty badger]

right so that's the relation route in my language

[thorn jay]

(e.g. when you descend a function down a projection map p : X -> X/~)

[thorn jay]

mm alr

[knotty badger]

by which i mean - the formula defines a relation from X / ~ to your codomain

and "well-definedness" amounts to showing this relation defines a function

[elfin wraith]

I remeber finding that hard at first, I think the thing that solidified it for me was seeing an example of something that wasn’t actually well defined. I.e. imposing some relation and showing some computation depends on the representative of the equivalence class

I don’t know if that’s generally helpful, but I do know that helped me when i had the same confusion

[velvet steeple]

You can probably get good counter-examples they are familiar with out of Q

[elfin wraith]

I don’t remeber what I first saw but it shouldn’t be hard to come up with something that’s wrong, I do that by accident all the time

[noble nexus]

yeah just take the "function" that assigns to each rational number its numerator

[knotty badger]

how would you explain the significance of the first iso theorem

[elfin wraith]

Significance in terms of usefulness or in some other way?

If it’s just usefulness I’d almost just say, just let them see. You use it constantly doing any form of algebra

It’s probably the main way you’ll end up showing isomorphisms

[knotty badger]

hm how about apart from usefulness

[chilly ocean]

it tells u that homomorphisms = quotienting your domain and then putting the quotient inside something bigger

[knotty badger]

right, this is the "decomposition of homomorphisms" pov

[chilly ocean]

so if u understand quotients well and extensions well, u can often understand homomorphisms well

[noble nexus]

I like to interpret it as basically saying that whenever you have a group extension, the structure of the smaller group is reflected in sort of a "fuzzy way" in the larger group

which is a bit vague but for example

you could imagine if you were looking at the general linear group but you had really shitty vision and could only see the scaling of volumes and the fchanging of orientations but couldn't make out the precise details

that the general linear group would "look" like the nonzero real numbers (or whatever field)

similarly if you were watching permutations but for whatever reason you could only see the parity, it would look like C2

so essentially the first iso theorem just says that when you take an extension of a group, the result is the same up to what makes them different, which when you say it that way sounds kinda stupid and obvious but

really the first iso theorem is actually obvious, but its one of those things that you only realize its obvious once you see it enough and internalize whats going on

(an extension of a group H by G is just a surjective map G -> H of course)

[vapid vale]

like the statement is crudely “if you get rid of the things that sent to zero, you’re left with the stuff that did get somewhere”. so spiritually it should feel believable. but now the interesting part is that there is any mathematically honesty to the statement; that something which was taught in terms of cosets is equal to honest elements in a different group. so the significance of it is really a statement about how homomorphisms behave. to me all of these identities are really equivalent to the fact that the one condition on homomorphisms makes them so well behaved

but if you understand homomorphisms, i agree with blake a good litmus test of understanding the first iso theorem is if it feels totally trivial/obvious

[thorn jay]

yeah the thing with first iso is that at some point it became like a ubiquitous law of nature for me lol

[knotty badger]

it's not always true ig

e.g. for topological spaces it's false

[thorn jay]

yes i was talking about algebraic structures :P shouldve mentioned that

[thorn jay]

(this goes for all isomorphism theorems, for that matter, the correspondence theorem perhaps the most)

[knotty badger]

famously it's not true for topological abelian groups

hence condensed math or smth

type shit

[thorn jay]

i suppose i dont think of those as purely algebraic structures

[elfin wraith]

This was exactly how I had it described to me “wouldn’t it be nice if homalg worked for topological spaces?”

[thorn jay]

the derived category of Top

[knotty badger]

are derived categories in any way related to derived groups

[elfin wraith]

I will understand if all one day, I can give Barwick a more insightful response than, “yeah that sounds pretty cool”

[thorn jay]

idk i just know theyre hom alg

[knotty badger]

derived groups come from quotienting out by the commutator subgroup

[thorn jay]

it does sound really cool though

then no

[knotty badger]

or wait

[thorn jay]

im pretty sure

[knotty badger]

was it taking commutator subgroup

[elfin wraith]

Not as far as I know but I don’t know much about derived categories beyond the vibe of what they are

[knotty badger]

it's probably taking commutator subgroup

[thorn jay]

the derived category is when category of chain complexes up to quasi isomorphism or smt

[elfin wraith]

Something something derived functors for chain complexes

[thorn jay]

(quasi isomorphism is a homomorphism which induced an isomorphism on the homology groups)

[thorn jay]

yis cuz all injective / projective resolutions are isomorphic in the derived category

[knotty badger]

yeah so derived category is a quotient but derived group is a subgroup

[thorn jay]

so that basically gives an explicit construction rather than requiring a choice of resolution

[knotty badger]

do you get these between the singular chain complex and the simplicial complex for a triangulation, for example?

[agile burrow]

I think it's just overloaded terminology, there's no relation to my knowledge

[thorn jay]

well not exactly, because you start with an abelian category A, then construct its category of chain complexes, and then localise that category to make all quasi isomorphisms invertible

[knotty badger]

right right

[thorn jay]

this localisation exists because Grothendieck is a smart fella

[knotty badger]

better a smart fella than a fart smella

[thorn jay]

exactlyy

[thorn jay]

uhhhhhhhhhhhhhh
maybe in nice cases idk i dont know topology

[knotty badger]

well i haven't done algtop but i assumed this is how you show that singular and simplicial homology are the same

that there's a quasi-isomorphism between the associated complexes

but mb i'm wrong

[thorn jay]

ive got no clue how they do that even

well there must be a quasi morphism unless the isomorphism isnt given by a chain map at all

[agile burrow]

You use the natural map from the simplicial chain complex to the singular chain complex and induct on the dimension of the complex, using the 5 lemma and stuff

[knotty badger]

is that natural map a quasi-isomorphism?

[agile burrow]

Yes, that's what the argument shows

[knotty badger]

cool

[thorn jay]

thats super nice

[south patrol]

This

Ig also like you can view singular homology of X as simplicial homology of Sing X (ok I am using simp sets instead of simplicial complexes but essentially the same) and if K is a simplicial set K have a map K -> Sing |K|. This induces a map from simplicial to singular

[vapid vale]

derived category comes from derived functors which are derived from left or right exact functors. i don’t actually know what derived subgroups are intended to be derived from when that terminology was introduced

[south patrol]

And in fact that map is a weak homotopy equivalence

[vapid vale]

beyond just the group

[south patrol]

It is sort of annoying terminology to me now lol

[vapid vale]

which

[south patrol]

Like there are notions of derived algebraic group etc which give you annoying results from derived subgroup etc

[vapid vale]

lol

[south patrol]

Commutator subgroup is nice ig

[vapid vale]

yeah

you’d like taters

[thorn jay]

it is technically derived from the original group :P

[south patrol]

Indeed

[noble belfry]

any idea how to show c)?

[rocky cloak]

There's also derivations, derivatives and differentials, and the deRham complex.

Not sure if there's actually any link there or if it's just something derived from something else...

[rocky cloak]

Do you know that elementary abelian p-groups are just isomorphic to (Z/p)^n?

[noble belfry]

no

that makes sense though

[noble belfry]

this follows from classification of finite abelian groups + the fact that it has (p^n-1) elements of order p right

[rocky cloak]

Yes

[noble belfry]

oh so basically

for any element a it lies in some Z/p

uniquely

and so it follows that if we take all the other Z/ps

and it has index p and doesnt contain a

[thorn jay]

its like normal

[rocky cloak]

Well the various uses of normal are clearly unrelated. Here I'm less sure

[karmic moat]

Overloaded terminology? In algebra? Impossible

[thorn jay]

variety is funny because in dutch manifolds share the same name

[south patrol]

Iirc this is just coincidence

Though I have heard this suggested before

[vapid vale]

math is so derivative smh

[elfin wraith]

It simply couldn’t be the case! That would be completely out of the normal!

[noble belfry]

G is isomorphic to a subgroup of S_n, if it contains an odd permutation g then g must have even order (since we know if g^k = e, g^k is the product of even permutations).

im not sure about |G:<g>| being odd though

If a group G contains an odd permutation, half of its elements are odd and half are even I believe

Oh

We only have to look at the cases of S_5 or above

[crystal vale]

Sorry I am not following your argument, g^k = e, g^k is the product of even permutation?

[noble belfry]

yeah

like

g^2k+1 can't be e

because its odd

and e is even

[noble belfry]

do you know how to show that [G:<g>] is odd though

im considering the action of G on [G:<g>] but it doesn't seem very helpful

[crystal vale]

But here what is k?

[noble belfry]

im not doing good notation lol

my argument is that if g^2k+1 for any k in N is e

its a contradiction

so it must be even

[crystal vale]

So if h is odd permutation then h has order even, but if h is even permutation then it can be odd or even

If | G : < g > | is even and g has even order, can π_g odd permutation?

It can't

Wait

I think if | G : < g > | even means if you decompose π_g into disjoint cycles then the number of cycles is even.

Therefore if g has even order and | G : < g> | is even then π_g is even permutation

[lean sail]

just a quick question. someone in here mentioned jordan-holder the other day. and called it an advanced topic. it just came up in my course. is jordan-holder and composition series and all that really considered an advanced algebra topic? maybe it depends on what program you're in?

[south patrol]

Ig depends what advanced means. But this is the sort of thing you would see in, say, a second course mentioning group theory, so as such is not particularly advanced (but also is not smth you would see immediately)

[knotty badger]

I’ve never seen or used Jordan holder

[vapid vale]

it also depends how much you do with them

we certainly defined and discussed composition series in my intro algebra course but not much further

i don’t know the context in which it was called advanced

[thorn jay]

Do you know what a composition series is?

[lean sail]

i think we have to use it to prove theorems about solvable groups.

[knotty badger]

No

[thorn jay]

Schreier refinement theorem is also useful then

[lean sail]

i see a question on my pset i haven't gotten to yet that has a chain of subgroups in it and we have to show that something is solvable.

[lean sail]

we also have to learn feit-thomson and phillip hall's theorem

it's all part of this unit we're doing about classification

[vapid vale]

that’s fun

[thorn jay]

because suppose your group has a subnormal series where each quotient is abelian (so its solvable), suppose A_1, ..., A_n. Then you can refine this to a maximal subnormal series (so a composition series) where the factor groups are the factor groups of composition series of all the A_i. These must be cyclic, so your group must be built out of cyclic groups. Conversely, if your group is built out of cyclic groups, then any decomposition series (by Jordan-Holder) will have those as factor groups, which are abelian, so it must be solvable

[thorn jay]

classifications are fun!!

[thorn jay]

a subnormal series of G is a series of subgroups
1 = A0 < A1 < A2 < ... < An = G
such that Ai is normal in Ai+1.

a composition series is a subnormal series which is maximal. Equivalently, it is a subnormal series such that each factor group (successive quotient) is simple.

[lean sail]

just getting into, so i don't know too much yet. we are learning about the holder program as well

[vapid vale]

what’s that

[lean sail]

it was an effort to classify all finite simple groups that was completed around 1980

part of it was using the feit-thompson theorem i guess

if G is a simple group of odd order, then G isomorphic to Z_p for some prime p

[vapid vale]

i never heard it called the holder program

the classification is cool

[south patrol]

Idk if it is possible to learn precisely one of composition series and Jordan–Hölder

[vapid vale]

lol

[lean sail]

no, jordan holder is built on composition series right?

[rocky cloak]

I guess it's possible to learn only about composition series

[thorn jay]

I know Jordan-Holder but no composition series

this is the way

[south patrol]

Ye maybe if they are just mentioned offhand

[thorn jay]

I've got the best idea

imagine a composition series

[vapid vale]

the only time i’ve used jordan holder is to understand what th grothendieck group of a FDA is

[thorn jay]

but for congruences

[south patrol]

I love how Jordan–Hölder is so much easier for modules than grps

[thorn jay]

rocking the UA world one step at a time

[south patrol]

Lol

[lean sail]

oh it comes up again for modules?

[south patrol]

Me when normal subgroups

[vapid vale]

imaginebalatro but for congruences

[south patrol]

Same result

[thorn jay]

yeah cuz you can just use modular lattice theory, duhh

[south patrol]

But easier proof

No Zassenhaus tears

[rocky cloak]

Is it? Isn't the proof exactly the same?

[south patrol]

Zassenhaus and stuff become almost a triviality I think right

No fiddling around worrying about normality

At least this is how i remember it

[vapid vale]

i’m assuming bc modules are abelian group

there’s something missing

[south patrol]

Like the proof has the same structure, just the hardest bit is missing iirc

[thorn jay]

^^

[vapid vale]

modular?

surely we can’t let modular have 3 definitions

[thorn jay]

this is a very old definition of modular

[vapid vale]

i c

[thorn jay]

going back to Dedekind n 1894, apparently

[lean sail]

sounds like there would be too many diagrams involved

[thorn jay]

(citation needed)

[thorn jay]

surprisingly not

[south patrol]

Sheaves

[thorn jay]

nO

we are not bringing sheaves into this

[lean sail]

does that even belong in algebra? i thought that was more related to topology or geometry or something

[thorn jay]

you've got algebraic geometry

[lean sail]

yeah that's true

[thorn jay]

that uses the formalisation of sheaves

but true I will always stand by the idea that in some sense sheaves are the "correct" geometric objects

not topological spaces

[south patrol]

I agree for much of this lol

[knotty badger]

Why are these useful

[lean sail]

in my course it has to do with classification of finite groups

[thorn jay]

I guess the evidence for this is that, when wanting to do actual geometry, the topological space is often not much more than a "carrier" of sorts: eg in algebraic geometry we really only care about polynomial maps, which do happen to be continuous in the Zariski topology and that gives nice consequences, and for manifolds we require an atlas which has certain niceness properties like smoothness or continuity or wtv

[thorn jay]

certain definitions of groups can be stated using the existence of certain subnormal series

and in general subnormal series are pretty well-behaved wrt like subgroups or extensions

and, for example, it helps prove jordan-holder (which basically says that decomposing a finite group into simple groups, like reversing a sequence of extensions, always gives the same multiset of simple groups)

[south patrol]

It's also like just generally a nice filtration you can always put on a group, with simple (lol) quotients

[thorn jay]

"simple"
the humble classification™:

[karmic moat]

holy grothendieck

[thorn jay]

maybe I am the
grothendieck

[karmic moat]

or actually i guess leray

[south patrol]

Just say topoi atp

[karmic moat]

let me look that up rq to see if i agree

[south patrol]

Ye sheaves of smooth functions

[thorn jay]

"He concealed his expertise on differential equations, fearing that its connections with applied mathematics could lead him to be asked to do war work."
this is crazy

[karmic moat]

okay W

yep

[thorn jay]

topoi are scary

[karmic moat]

someone told me he came up w spectral sequences on the walls of the prison

and he just drew rectangles and arrows in them

[south patrol]

Based

[thorn jay]

crazy man

[thorn jay]

I thought they were more logic though

[elfin wraith]

Something about pictures of an elephant comes to mind

[south patrol]

Depends on whether you use topoi for applications or not

Joking.

[thorn jay]

https://ncatlab.org/nlab/show/Sketches+of+an+Elephant

how am I ever gonna fucking learn all this 😭

why is there so much math

[elfin wraith]

See I often find myself feeling that, yet I somehow keep knowing the stuff that seemed so far away before

But then I just realise that was still so much farther away from the frontier than I even knew

[thorn jay]

just work in a dead field

you'll be at the frontier in no time

🔥

guys my research will properly revive UA into the mainstream I swear!!!!

[elfin wraith]

I will be the worlds best general topologist

Ignore the fact that metric spaces and general topology are by quite a considerable margin by worst results in uni

[thorn jay]

looking at my current grades in math I am not good at math

the average is like a 6/10

[elfin wraith]

I think this is a common occurance for people who already know a lot because their mind is just elsewhere

[thorn jay]

this is because my only grade is a 6/10, for graph theory, because we didn't have enough time

and because it was too cs related, I'm not abt to learn the algorithm to calculate the max flow under a certain capacity 💔

[elfin wraith]

Also you just wont get the same grades in uni as in school

[elfin wraith]

Yeah my first couple graph theory homeworks were uber cooked and they only got easier and shorter because the tutors complained about the ammount of marking so I feel this lol

Tbf they werent that hard, but they were silly long

[thorn jay]

this is how the exam felt

not hard, just 3 questions and each had like 7 subquestions

linear algebra homework consisted of proving some stuff for general vector spaces, where each algebraic step we did we had to mention the specific axiom used

[rapid cave]

noooo enpeace changed pfp

[south patrol]

By the axiom of foundation, ...

[thorn jay]

people kept confusing me with the guy in the pfp

[rapid cave]

By the axiom of UA

[thorn jay]

I am NOT The Caracal Project

[karmic moat]

Floppa

[rapid cave]

put an anime girl pfp

[thorn jay]

omg yes

maybe I'll even get gender euphoria from that who knows

[fiery dirge]

Can anybody help me understand proving isomorphisms?

I know the theory, 1. Show homomorphism, 2. Show bijection

But in practice I am just so confused.
Here's the first example:

Let $G = { \begin{pmatrix}
1+a & -a \
a & 1-a \
\end{pmatrix}, a\in Z}$. Prove that $(G, *)\cong (Z, +)$. I don't know how to do this at all.

[cloud walrus]

HMD

[rapid cave]

Btw my lecture today was weird @thorn jay

[thorn jay]

G is that one matrix indeed

[rapid cave]

we did sheafification, and then we defined a ringed space, and then co-limits and filtered co-limits😵‍💫

[thorn jay]

I swear there was an emoji called clopencry

[thorn jay]

maybe try writing out the general formula for the multiplication of two such matrices

[fiery dirge]

what?

[thorn jay]

then, perhaps, a natural bijection will show itself

[thorn jay]

in that order?

[rapid cave]

yes

[thorn jay]

how are you defining sheafification without colimits or filtered colimits

sections of the etale space?

wait no thats direct limits

[rapid cave]

universal property

[thorn jay]

wut

I see

so not even a construction 💔

[fiery dirge]

I know, the result is something like $ \begin{pmatrix}
1+a & -a \
a & 1-a \
\end{pmatrix} \cdot \begin{pmatrix}
1+b & -b \
b & 1-b \
\end{pmatrix} = \begin{pmatrix}
1+a+b & -a-b \
a+b & 1-a-b \
\end{pmatrix}$

[rapid cave]

we did do a construction of regular functions on stalks

[thorn jay]

because the construction uses a limit over refinements of covers

[rapid cave]

to show existence

[thorn jay]

or that

failed to render 💔

[rapid cave]

we also did the g++ construction

[thorn jay]

but geee, you may ask, why are there two +'s??

[rapid cave]

because the first one only makes the presheaf g into a mono-presheaf

[cloud walrus]

HMD

[thorn jay]

well, now you can see there is an obvious mapping, right? You want the mapping to satisfy f(A * B) = f(A) + f(B)

is there an obvious way to get such a matrix and get an integer?

[thorn jay]

yus

[fiery dirge]

yes I see it's obvious as everything in abstract algebra but I dont know how to prove it with technical steps

[fiery dirge]

determinant?

[thorn jay]

no

[rapid cave]

dw not everything is obvious in abstract algebra

[thorn jay]

the matrix (1 + a & -a \ a & 1 - a) gets sent to a

then you just have to prove that this is a homomorphism, and surjectivity and injectivity are kinda by definition

[fiery dirge]

i know the definition but dont know to apply it that's my problem

[thorn jay]

well you just have to show that this mapping is a homomorphism

i.e. that, if you have two matrices A and B, then f(AB) = f(A) + f(B)

[thorn jay]

you have the formula for AB here, so you just have to match sides

[fiery dirge]

Okay so $ f(A + B) = \begin{pmatrix}
1+a+b & -a-b \
a+b & 1-a-b \
\end{pmatrix} = \begin{pmatrix}
1+a & -a \
a & 1-a \
\end{pmatrix} \cdot \begin{pmatrix}
1+b & -b \
b & 1-b \
\end{pmatrix} = f(A * B)$??

[cloud walrus]

HMD

[fiery dirge]

that's the only line I have to write regarding homomorphism?

[thorn jay]

well yeah

okay no

youve notated it wrong

f is applied to the matrices, not to the integers

or i guess you can go the other way but then you should end with f(A) * f(B), not f(A * B)

[rapid cave]

btw @thorn jay this co-cones thing kinda melted my mind

too much commutative diagrams

[thorn jay]

basically, think of it as a sink

everything goes to the cocone and its well defined

so no matter what path you take to the sink, its the same

[fiery dirge]

bijection

if a = b then f(a) = f(b).

So we take a = b then $ \begin{pmatrix}
1+a & -a \
a & 1-a \
\end{pmatrix} = \begin{pmatrix}
1+b & -b \
b & 1-b \
\end{pmatrix}$?

not sure how to show surjection though. Matrix consists of $a \in Z$ so clearly it should map to entire $\mathbb{Z}$

[cloud walrus]

HMD

[rapid cave]

the map a --> (1 + a, -a, a, 1-a) is obviously surjective

[fiery dirge]

I don't get it, can you show me the correct example?

[fiery dirge]

so many things is obvious I know fr

[lean sail]

i think you need to define a function between the g \in G (using the variable a in the matrix), to the elements z \in Z

then, show it's a homomorphism. then show it's a bijection.

[fiery dirge]

yeah but I don't know how to think through that process, I've never done it before

it's a notation hell

[lean sail]

did you define a map yet?

[fiery dirge]

no

[lean sail]

you need something that takes a g \in G and maps it to a z \in Z.

i'm assuming that Z is the integers? not the cyclic group?

[rapid cave]

Z are integers with addition, which makes it a cyclic group of infinite order

[lean sail]

ah, duh

off the top of my head, could be wrong here, but just take the $a$ out of that matrix and send it to the corresponding integer $a \in Z$

[cloud walrus]

proofman

[rapid cave]

yeah thats the map

and you have an inverse a -- > (1+a, -a, ...)

[lean sail]

ok, then compute f(a)f(b), should get f(ab)

[thorn jay]

cant be cyclic, { 2, 3 } is a minimal generating set and we all know that abelian groups have invariant basis property

[lean sail]

im confused, Z is definitely cyclic

[rapid cave]

are you making ||23|| the new ||67||

[rapid cave]

its a joke

[fiery dirge]

how do i do that tho?

[thorn jay]

rest assured, 99% of what i say is covered in layers of irony

[lean sail]

f(g) = a, where a is the coefficient in the matrix

okay, then declare two matrices, g, h \in G with coefficients a, b \in Z. then show f(g)f(h) = f(gh)

[lean sail]

i can't understand 99% of what you've said so far, so i guess that doesn't help me much 🙂

[lean sail]

usually you have to build the map using the "stuff that's lying around" is how it was explained to me...

[fiery dirge]

Yeah... I don't recall the last time I did it so I'm really really rusty

I only have the "useless" theory

[spark veldt]

How much Galois theory do you need to be able to follow Sharpe's steps into commutative algebra, or even Matsumura?

Will the Galois theory section in Herstein's Topics in Algebra field theory chapter suffice?

namely this

[rocky cloak]

So I don't know these books at all, but typically the Galois theory needed for commutative algebra is like characterization of Galois extensions as seperable and normal, the correspondence theorem and basics of field extensions.

You probably won't need anything about transcendence of e, compass straight edge construction or solvability by radicals.

And I'm guessing 5.8 is about constructing specific groups as Galois groups over Q, which doesn't seem relevant to CA either.

I might even suggest just starting on CA and then going back to Galois as you need it

[elfin wraith]

I have done a pretty reasonable ammount of CA and i am only just now learning galois theory (not because I needed it for CA)

[south patrol]

Agreed lol

(I guess Galois is most pressingly needed in algebraic number theory things initially)

[elfin wraith]

A few ideas popped up in my alggeo course but these were things that were either easy to look up, obvious, or some combination there in

[rocky cloak]

If you care about groups acting on rings there are some things with primes lying over fixed primes etc that goes into Galois theory territory

[south patrol]

When you say this characterisation, which definition is your preferred one, jagr?

I guess probably like L^{Aut(L/K)} = K

[rocky cloak]

Well I'm saying you should be familiar with several and why they're equivalent.

But if I am to pick a favorite, then sure that one

[south patrol]

Yee makes sense

[elfin wraith]

I have honestly not found my galois course to be all that enlightening so far, I dont know if thats because ive done quite a bit of other algebra before now or if the course just doesnt cover much ground

That being said I have only read so far as to actually define the galois group so there could be more coming (though im about 80% of the way through the notes I think)

[south patrol]

i think it is more enlightening when you actually use it for stuff like number theory tbh

though the fundamental theorem is v cute in itself

[elfin wraith]

The fundamental theorem is the main reason im taking it lol, in my first algtop class the correspodence between covering spaces and (deck transformations? i dont actually remember) was pointed out to just be the galois correspondence and that seemed to make everyone else quite happy

So i guess thats something I should understand well

[south patrol]

i wouldn't say "the" but rather "a" galois correspondence, like formally the same but yeah

(though there is more that can be made precise lol with étale maps in algebraic geometry)

[elfin wraith]

Yeah sorry, something something Galois categories

[south patrol]

Yee

[elfin wraith]

What is the correspondence anyway? I feel like it was something to do with covering spaces and like subgroups of the deck transformation or something

I honestly didnt super follow the stuff on deck transformations, its something I should review

[rocky cloak]

Covering spaces correspond to subgroups of the fundamental group, and then for normal covers the deck transformation group is the quotient of the fundamental group by the corresponding normal subgroup.

[elfin wraith]

Ahh yeah that sounds right, I remember becoming quite lost around that part of the course lol

I need to learn more homotopy theory soon anyway so I guess thats a good time to review it

[thorn jay]

Ive discovered a book on Galois theorie_s_ which seems super cool

in the uni library

of course it was grothendieck who did a lot of work on the stuff covered there

[knotty badger]

the proof of young diagrams classifying irreps of S_n is quite pretty actually

[austere shadow]

hi groupers i've been directed here since not getting much of a reply in help channels

am doing some basic galois theory at uni, got given this question

as mentioned in the screenshot im currently stuck on the 4th one - finding the splitting field of Q for the polynomial X^3 - 3X + 1. We're told that 2cos(2pi/9) is a root, but I don't really see how that helps us there, since we have no idea if the other two roots are generated by that root as well.

I'm assuming that the answer is that those other two roots are in fact in Q(2cos(2pi/9)), but I am lost on how to show that

and this is annoying me greatly since the other 4 parts of this question are borderline trivial 😭

[velvet hull]

you don't have to prove that the other two roots are generated by it, it is enough to show that the polynomial is irreducible

because ||we are working over an extension of Q, so there are no concerns about separability||
and to that end it is enough to show that ||the polynomial has no roots in Q, because a cubic polynomial is not irreducible iff it has at least one linear factor||

[rocky cloak]

So I'm assuming part of this hint lies in how you showed 2cos(2pi/9) is the root of this.

What comes to me is that if you let z be a primitive 9th root of unity, then Q(z) is a Galois extension with Galois group (Z/9)^* = Z/6, and 2cos(2pi/9) = z + z^-1.

So Q(2cos(2pi/9)) is a Galois extension with Galois group a quotient of Z/6

[rocky cloak]

The exercise is to find the splitting field, so it's relevant what the other roots are.

For example x^3 - 2 has a splitting field of order 6, while in this example the order is 3

[velvet hull]

oh, find the splitting field ok

[rocky cloak]

Another thing that's useful is the discriminant
https://en.wikipedia.org/wiki/Discriminant

An irreducible cubic has splitting field of degree 3 iff it's discriminant has a square root, otherwise it's degree 6.

And in this case it's 81 = 9^2

[austere shadow]

Ooh ok

[tough raven]

I remembered this much better once I learnt the formulation in terms of G-sets (the one part of May I really like).

[austere shadow]

Yeah it was exactly that.

[elfin wraith]

Oh, noted, ill take a look there!

[tough raven]

Don't go beyond the covering spaces part (i.e. chapter 4 or 5 I forget), unless you want to be... changed.

[thorn jay]

changed?

[vapid vale]

horrifying

[elfin wraith]

I just looked at the contents page for the book because I’ve always heard it’s an experience

Good lord is the word “concise” ever doing some heavy lifting

[tough raven]

The most ominous word that didn't seem inaccurate I could think of on short notice.

Actually maybe that's also inaccurate

[south patrol]

Lol

[south patrol]

Tbh I need to have a look again lol and I am curious what the opinion on it is nowadays

[tough raven]

Oh, well.

[elfin wraith]

Everything I’ve seen people say is that it’s good but learning from it yourself looks impossible, but honestly it looks like it could be a good thing for me to go through. I’ve worked through quite a lot of Hatcher now, but my knowledge is still pretty patchy and I’ve not seen much of the categorical stuff, because Hatcher, so it seems like a nice book for review

I wonder if I’ll feel the same way about that book as I do about Rudin

But yeah starting from defining homotopy and building all of Hatcher, and more category theory and an introduction to K theory in like 230 pages is just

[south patrol]

Hmm

Idk I think it is kinda different from Rudin

Ig yeah I think it is a nice resource but not to learn from

Like putting cofibrations and stuff so early seems a bit much – I like the way Hatcher does stuff with homology etc which has less machinery

Whereas idk how you would get someone to just care about cofibrations of topological spaces before more stuff

I found some of the "topics" bits near the end (e.g. with manifolds or K-theory) quite useful tho

[south patrol]

(And lol this is coming from a homotopy theorist)

[elfin wraith]

Yeah my first course kinda made a point about cofibrations when we did the HEP and also with mapping cylinders, but we didn’t expand on that, I know they’re useful and important but not how or why

[knotty badger]

I’m trying to understand how you can reconstruct a representation given its character

Would the following approach work:

Take the regular representation

Use the character to make a projector onto the component corresponding to an irrep

Then… I guess you need to block-diagonalise the matrices simultaneously

[thorn jay]

CG is semisimple so every module can be written as a direct sum of simple modules, i.e. the irreducible representations.

you then take the inner product with the irreducible characters, and these will be the amount of times the respective irreducible rep appears in the decomposition

[knotty badger]

Yes I’m aware, this isn’t quite what I’m asking though

If I have a known group G, and a character corresponding to an irrep, what’s the fastest way to create a representation that realises this character?

[thorn jay]

ah okay, I'm sorry

yeah take the regular representation, and then you can retrieve the irreducible representation by construction the corresponding projection from the regular rep onto it

[merry summit]

i want to know how many group homomorphisms are there from S3 to Z6? i tried understanding from chatGPT but it uses commutator subgroups (???) quotient s3/A3 and whole lot very confusing things that i don't get. can someone tell me an easier and intuitive way

[elfin wraith]

ChatGPT is just generally going to be a bad idea for maths you’re doing at this level.

One thing I would suggest you think about is the order of elements in S3 and those in Z6 and see where things could be sent

[merry summit]

so S3 has 3 elements of order 2 and 2 elements of order 3. While Z6 has only one element of order 2 and two elements order 3. so these elements from s3 must go to their respective order elements in z6?

[elfin wraith]

<@&268886789983436800>

[elfin wraith]

Yeah that’s the idea! You need to refine it a little, one idea is to notice that Z/6Z is abelian and S3 isn’t, that should help you!

[merry summit]

hmm i thought of abelian-ness too, but didn't get anywhere. what element would even go to order 6 element cuz s3 doesnt have any order 6 element. i feel like im almost there but missing smth

[elfin wraith]

You can also think about the fact that the kernel of a homomorphism is a normal subgroup, and look at the different normal subgroups of S_3 but these ideas are pretty similar

[elfin wraith]

Possibly this second idea I just mentioned will be more helpful to you then, because that is the right idea

[tribal moss]

Especially in the case of spambots, it would be useful to write something to that effect in your modping, or make it a reply to the spam.
When a spambot is banned, all its recent posts are automatically deleted so we don't need to hunt them down one by one -- but that means that there will be a lot of orphaned modpings in channels other than the one we first found it in, and the quicker we can ascertain "oh, this was just for the spambot I just banned" rather than some interpersonal problem that needs further action, the easier for us.

[elfin wraith]

Noted, will do!

[knotty badger]

Well you get a big direct sum of the same irreducible

[candid patrol]

Hello! The smallest group making the fusion of two elements $x$ and $y$ of a group $G$ possible is indeed given by $H = \langle G, t \mid t x t^{-1} = y \rangle$, right?

[cloud walrus]

UGOAT

[candid patrol]

It seems obvious, but I'm not familiar with categories.

[delicate orchid]

if the order of Out(G) isn't coprime to 2, yes this is the smallest possible

you take the semidirect product of a C_2 < Out(G) with the obvious action on G and this group realises the fusion you're after

this group might not have that exact presentation though

if it isn't coprime to 2 then you just take the smallest prime dividing the order - not sure why I specified 2

[candid patrol]

Ok thx, I'd like a proof, do you have any link plz ?

[marble hinge]

ha, I just recently did this exercise from Pinter: finding all normal subgroups of S_3 and D_4. I think I missed two subgroups in D_4! Those Klein groups. Probably because I was too eager to call it a day 🙂

it was somewhat tedious, because I couldn't find any better way than just finding all subgroups first and then checking which ones are normal

Lagrange theorem certainly helps, and having the group operation table ready - to speed up the checks, but after all that tedium I quickly installed GAP computer algebra system 😄

hmm, now I wonder if the kernel(f) fully defines the homomorphism f: G - > H? probably no...

[thorn jay]

no, there are two different homomorphisms Z → Z/2 x Z/2 with kernel 2Z

[marble hinge]

then how @elfin wraith 's suggestion is going to work?

[thorn jay]

or just any automorphism :P

[marble hinge]

there might be two different homomorphisms with the same kernel

so @merry summit might miss one of them

[elfin wraith]

Thats true, but having looked at the kernerls, you just have to think about what maps could give you that kernel

The point for S_3 being that it cant be trivial (because S3 isnt abelian and youd get an isomorphism in this case) so its either all of S_3 or its A_3, and you can reason these out easily

S_3 is also trivial, A_2 gets you most of the way there by looking at the size of S_3/A_3

[marble hinge]

ok

[honest sierra]

I don't know if I have the full context but it seems that you want to find all normal subgroups of S_3. Why don't you try Sylow?

[marble hinge]

haven't got to Sylow yet, it's in later chapters

I mean, it's not like I need to find them by any possible means in the quickest fashion!

[honest sierra]

Well listing out subgroups is nice way. Because there aren't much subgroups here.

And S_3 is very easy to calculate and work with.

[marble hinge]

for D_4 there were some, and there are 8 elements, so I managed to miss Klein's!

S_3 yes -- I know that one too well by now 🙂

[rocky cloak]

You can start by finding normal subgroups generated by a single element by looking at the subgroup generated by an element and all it's conjugates.

Then the normal subgroups will just be products of these.

You can also think about what the order of a subgroup is and then find appropriate homomorphisms instead.

[marble hinge]

I don't quite like when authors use arbitrary letters for its elements, using cycle notation seems much better to me: (12), (123), etc.

but Herstein uses phi, psi, etc. And Pinter uses kappa 🙂

[rocky cloak]

r for rotation and s for speilung (reflection) is sensible I guess

[south patrol]

Basiert

[delicate orchid]

D_8 subgroup lattice wr 13 seconds

put me on speedrun.com

[marble hinge]

ah, I meant that they use phi, psi, etc. for S_3 sub-elements

using s and r for D_4 generators is perfect (and the way it's done in D&F)

[delicate orchid]

I use r for reflection and t for totation

[marble hinge]

I like "f" for flip

[delicate orchid]

I'm not sure where I picked this up from

[marble hinge]

and "r" for rotate

[marble hinge]

is this like a graph of subgroups?

[delicate orchid]

yeah it is and I've drawn squiggly lines for the conjugate ones

[marble hinge]

i.e. each vertex is a group and if there is an edge from A to B this means that A is a subgroup of B?

[delicate orchid]

exactly yeah, it's the poset of subgroups under inclusion

fun exercise is to draw the one for S_3

[honest sierra]

I guess you could try to find all subgroups which might be normal in D_4 and try to argue that they are the only normal subgroups by looking at normal closures.

[delicate orchid]

at this point just find the conjugacy classes and see which unions of them are closed under multiplication

[marble hinge]

hm, interesting

[elfin wraith]

Bro got the crayons out

[delicate orchid]

it's just a few automorphisms away from being the inner fusion system on D_8 but drawing them would be time loss 💔

[elfin wraith]

Do it for S5, no balls

[delicate orchid]

I think I did at one point

no that was S_4

[honest sierra]

For the arguement, I would try to do it like this: If a normal subgroup contains a particular element, it will force the subgroup to be a particular one (basically trying to find the normal closure).

[elfin wraith]

My galois class had that as a problem on the most recent problem sheet, and like, im good

(For S_4 and S_5)

[delicate orchid]

the WHOLE lattice? not even up to conjugacy?

[elfin wraith]

Thankfully, not assessed, no requirement for me to do it lol

[delicate orchid]

for S_5 that would have well over 100 vertices

[honest sierra]

It was nice to find all of them.

[elfin wraith]

Perhaps for some. Not the kinda thing I enjoy.

[honest sierra]

I have done S_4, and some of A_5 tho. Not S_5 wholly.

[delicate orchid]

S_4 is reasonable

[honest sierra]

A_5 is also nice to work out (group of order 60).

[elfin wraith]

I dont like to think about symmetric groups, cycle notation confuses me

[delicate orchid]

I just got mansplained the order of A_5 💔

[elfin wraith]

Bringing you back to earth after too many bifunctors

Real maths that real men do, not that woke nonsense

[honest sierra]

I didn't mean that. I meant that not too high of a number.

[delicate orchid]

they're biset functors I only know of one biset bifunctor

[elfin wraith]

I do appoligise, please forgive me

[delicate orchid]

two actually

actually no that's the same one twice

[elfin wraith]

Perfect, make a family of them, you group theory types seem to enjoy that

[delicate orchid]

it's literally just Hom(-, -) boss

but I won't tell you what category it's in

[elfin wraith]

I dont know wtf a biset is boss

[delicate orchid]

bimodule but group actions instead of ring actions

[honest sierra]

(-, -) in Hom(-, -) looks like my face rn. /s

[elfin wraith]

This I get

Feels nice, ive spent about 3 says failling to understand about 5 pages of hatcher

[delicate orchid]

someone needs to write an alg top textbook that isn't A) as dense as Benson or B) dog water

[elfin wraith]

Ive been saying this

[delicate orchid]

and calling Benson II a "textbook" is like calling the ATLAS a representation theory textbook

[honest sierra]

How do you feel about Spanier?

[delicate orchid]

haven't read it

[thorn jay]

what is a bifunctor even

[elfin wraith]

I just dont get Hatcher, bro spend about 15 pages talking about subdivisions in 4 different ways, then gets to an example he says is really important for cohomology and just throws you in the deep end

I dont know whats going on bro

[honest sierra]

I feel that Hatcher talks too much. I can't handle too much of it.

[elfin wraith]

He really does, but seemingly not in the right places

[rocky cloak]

A functor with a very open sexuality

[delicate orchid]

just a functor from a product category, at least in the sense I'm using it

[thorn jay]

oh Right

[delicate orchid]

it can also mean a functor between bicategories if you are woke

[thorn jay]

i knew that

💀

[delicate orchid]

it is quite funny how many things I study have the bi- prefix

[rocky cloak]

The duality of wew

[thorn jay]

theyre so me fr

[thorn jay]

wew is a category of algebraic structures hence has an isbell duality

[thorn jay]

Nope coming out!!!

this was not on my 2025 bingo card

[elfin wraith]

Ive been openly bi lol

[thorn jay]

i didnt know!!

good to know though

[honest sierra]

Haven't read bifunctors so this might sound crazy. Can they have different variance simultaneously?

[tardy hedge]

im not bi but i like partying w them folk

they know how to have fun

[delicate orchid]

wdym by variance

[honest sierra]

One begin co and another contra

being*

[delicate orchid]

yeah

Hom, for example

[honest sierra]

Oh yeah...

[delicate orchid]

in fact the ones people care about are usually out of an opposite category x a category

it's rather strange to see (co, co)-variant

[thorn jay]

(e.g. Hom)

[delicate orchid]

Hom, internal hom

[honest sierra]

Thats nice.

[thorn jay]

monoidal stuff

[delicate orchid]

they're called profunctors in general and they're a generalisation of BISETS to categories

[thorn jay]

or did you mean contravariant x contravariant

[delicate orchid]

thats right buddies you can't escape that easily

[delicate orchid]

nah I mean (contra, co)

[thorn jay]

yes yes

[delicate orchid]

(contra, contra) is just (co, co) really

[thorn jay]

a (co, co)-functor immediately has some monoidal feeling tbh

[delicate orchid]

yeah they're like monodial products but without any of the coherence conditions

cause it's morally just a binary product

andddd I think we've just reinvented single sorted Lawvere theories

BUUURRRRPPPPPPP

[thorn jay]

COHERENT??
CONDITION????

COHERENT CONDITION???????!!!?

[delicate orchid]

what is your PROBLEM

[tribal moss]

Hush, you two.

[thorn jay]

brainrotted by my research

[thorn jay]

dont tell me what to do grrrrr

[delicate orchid]

he's a senior mod his job is kinda to tell you what to do lol

[delicate orchid]

oh? what coherence conditions do u work with

[rocky cloak]

I guess we need to make an alge-unchill channel

[elfin wraith]

Alge-ments

[delicate orchid]

also I just meant the associator pentagon it's not that deep

or the unit traingles

[thorn jay]

a generalisation of prime spectra suitable for any variety of algebras, which happens to have cool ties to both universal algebraic geometry and logic

[fading acorn]

[thorn jay]

yes lol ik

[delicate orchid]

this is the woke kind of spectra isn't it

[thorn jay]

(just recently published a paper abt it wink wink)

[south patrol]

Published = arxiv

Jk

[thorn jay]

its nothing super complicated ngl

[south patrol]

Valid

[delicate orchid]

unless you mean like, Morava K-theories by "prime spectra" idk what else that could mean

potato what else could that mean

[thorn jay]

shushh let me have smt

[south patrol]

published

[delicate orchid]

this but unironically

[south patrol]

I was joking tho dw

[delicate orchid]

I've been thinking about how small of a result I can put on arxiv and get away with it

[elfin wraith]

pt8 maybe pt7 id guess

[delicate orchid]

we \footnotesize in this mf

[south patrol]

As small as you want

[thorn jay]

a coherent condition assigns to every algebra a set of distinguished congruences such that the preimage of a distinguished congruence is distinguished, and the quotient of a distinguished congruence is distinguished

[elfin wraith]

An entire paper in a footnote would be amusing

Start doing that rather than citations

[delicate orchid]

like I have a really nice little combinatorial result that doesn't fit into my thesis but it really is just "apply this known formula" but the application of the formula is highly non-trivial

[thorn jay]

that sounds like it could be a nice little paper

[delicate orchid]

I see, don't really know what a distingusished thingy is icl

like what would be a distinguished congurance for groups

[thorn jay]

the set of congruences are the distinguished congruences of that coherent condition

[delicate orchid]

I need an example rather than more adjectives/nouns I don't know

[thorn jay]

so like there is a coherent condition for which the distinguished ideals (as those are equivalent to congruences for comm rings) are the prime ideals

[delicate orchid]

can't believe potato "this" reacted me, admitting to their diabolical plan to scoop my result

[south patrol]

I will scoop yours

[delicate orchid]

ah ok

[fading acorn]

https://tenor.com/view/gang-roingus-gang-roingus-chungus-meme-gif-19827832

roingus ideals

[thorn jay]

but theres also one where the distinguished ideals are radical ideals

[south patrol]

Eepus peepus

[delicate orchid]

so it just lets you pick out certain types of quotients by looking at what you're quotienting out by?

or perhaps you get the congruences from a specific condition on the quotient?

that sounds more right to me

[thorn jay]

yes, to both, in some sense

it turns out coherent conditions on a variety V correspond to subclasses of V closed under embeddings

[delicate orchid]

interesting

variety in the logic sense, I presume

[dull ginkgo]

Semishrimple modules

[thorn jay]

of course

[fading acorn]

what's your favorite definition of variety

[thorn jay]

then you can take the radical of a coherent condition (which corresponds to taking all congruences which are some intersection of distinguished congruences), and those correspond to subclasses of V closed under products and embeddings

anyways, neat little thing

[south patrol]

Maybe like separated and integral scheme of finite type over a field

Algebraically closed field if that is ur taste

[thorn jay]

category of representations of a clone in Set

[south patrol]

Could also make it not quasicompact

[delicate orchid]

the spice of life

[merry summit]

so the normal subgroups could be {e}, A3 and S3. which means those are my options for the kernel. now if the kernel is {e} it would mean an isomorphism which aint possible so im left with 2 kernels, can i conclude from just that there are 2 homomorphisms? is knowing the possible no. of kernels enough for such questions?

[delicate orchid]

the identity map from S_3 to itself has trivial kernel but I don't really know what you're doing

[elfin wraith]

It isnt enough no, because as was mentioned somewhere above you could have different maps with the same kernel. You need to look at what maps could give you A3 and S3 as a kernel and then you can conclude

[dull ginkgo]

So uh

[elfin wraith]

There are 2, but your reasoning isnt quite right

[delicate orchid]

Ah Hom(S_3, C_6) ok

[elfin wraith]

Yur

[dull ginkgo]

I’m taking a third year engineering quantum course and the prof this morning just casually tossed around the ideas of Lie algebras in full abstraction to a group of crayon eating engineering students

Catastrophic

I tried providing an idea here in the group chat for the course but we’re cooled

[merry summit]

hmm i see, ok so how do i check if there are more maps when the kernel is A3 🤔

[delicate orchid]

honestly why are we messing around with kernels? just see where the generators of S_3 can go

[fading acorn]

mfw yapping about (\mathfrak{sl}(2,\mathbb{R})) representations

[merry summit]

but S3 isnt cyclic how would it have generators

[delicate orchid]

... all groups have generators

(123), (12) generate S_3

[dull ginkgo]

Cyclic is if it has one generator brother

There can be many generators. Typically the number of generators is the size of the smallest family of generators

Which isn’t unique

[fading acorn]

they might not be always finitely generated or generated by one element(cyclic) however we can always take a generator for groups

[elfin wraith]

I had also mentioned this as an idea to start with, along with looking at the order of elements, but the kernel thing is another way to go

[dull ginkgo]

Technically the whole group is a set of generators.

A set of generators is basically where the smallest subgroup containing that set is the whole group

[merry summit]

this is confusing im sure I've seen cyclic groups with multiple generators...

[dull ginkgo]

(The generated subgroup of a set is the smallest subgroup containing the set)

[merry summit]

and not one

[elfin wraith]

A group is cyclic if it has 1 generator, so those dont exist

[dull ginkgo]

You’re fighting ghosts

[delicate orchid]

the cyclic group of order n has phi(n) generators where phi is the euler totient function

[elfin wraith]

Every group looks like <g_1,...g_n|r_1,...,r_m> for some set of generators g and relations r, but this is probably going to come up a little later in your course

[delicate orchid]

but the entire group is generated by any SINGLE one of them

[dull ginkgo]

Euler toilet function

[merry summit]

doesnt Z8 has generators {1, 3, 5,7}

[fading acorn]

https://tenor.com/view/leonhard-euler-euler-time-to-shred-euler-shred-shredder-gif-24591321

[delicate orchid]

yes but any SINGLE ONE OF THEM generates the whole group

as I just said

[merry summit]

yes!

but you said one 😭

[quiet pelican]

murderer

[elfin wraith]

There are phi(n) choices of generators, but you only need one to generate the group

[merry summit]

yes

[fading acorn]

i mean there can be different set of generators
i assume you should have taken linear algebra, and as you know there can be different basis for them
and a cyclic group is a group where there exists a generating set with one element.

[dull ginkgo]

A generator is described as a part of a generating set which generates the group. The minimum number of generators is the size of the smallest generating set(s), which are not unique (there is often a fuckload).

A group is cyclic if the minimal generating sets are singletons, so the group can be generated by one of many “lone” generstors

[merry summit]

i think i dont know what a generator of the form <a,b> means, or how it works. do i multiply ab repeatedly?

[thorn jay]

well in my experience a minimal generating set is "locally" minimal in the sense that removing any element makes it not a generating set

so for example { 2, 3 } is a minimal generating set of Z

[dull ginkgo]

Well that’s just minimality of a family of sets.

[dull ginkgo]

Subgroup generated by a,b

[delicate orchid]

<a,b> is all words you can make out of {a, b, a^-1, b^-1}, like a^2, aba, ab^-1, aba^2ba^3ba^4b...

[dull ginkgo]

Or in other words, the smallest subgroup containing {a,b} which you can show is the group of words

[thorn jay]

yes, a minimal generating set wrt inclusion

minimal overall is minimal wrt cardinality

[dull ginkgo]

Well the size of the minimum generating sets

yeah

[fading acorn]

<S> is just left adjoint of the forgetful functor F:Grp\to Set

[delicate orchid]

when write write, for example <a, b | a^2 = 1, b^2 = 1> we're setting a^2 = 1 , b^2 = 1 in all of the words - so the words I've written become 1, aba, ab^-1, ab, for instance

[dull ginkgo]

Free falling

Adjoints:

Forgetful/Free or Tensor/Hom

Every other one is alg geo/top witchcraft

[thorn jay]

galois connections:

[dull ginkgo]

True

[delicate orchid]

Tensor/Hom is forgetful-free

[merry summit]

alr so back to the question on homomorphisms from s3 to z6...💀

[delicate orchid]

just see where you can send (123) and (12)

[dull ginkgo]

Algebraic structures are great because {x : f(x) = g(x)} is a proper sub object

Unlike fuckin topo spaces where you need hausdorffness

[delicate orchid]

just take the homotopy equaliser

[fading acorn]

topoi

[merry summit]

how to know where they can go

earlier i just figured out by the orders

[thorn jay]

ah yes equalisers thats the only reason why algebraic structures are nice

[delicate orchid]

just think about it,: writing g for the generator of C_6 and f for our homomorphism,(123) is either sent to 1, g^2, g^4. Because (12)(132)(12) = (123) we need f((123)) = f((12))f((132))f((12)) = f((132)) so (123) must go to 1. Then (12) is either sent to g^3 or 1 giving us 2 morphisms

[dull ginkgo]

you're mother

[thorn jay]

i am earthbound

[dull ginkgo]

Explosion

[thorn jay]

*deltarune explosion sfx

[marble hinge]

have you found at least one homomorphism?

[kind kindle]

would questions about nonassocative rings go here?

and also loops

[thorn jay]

doesnt really matter ngl

[quiet pelican]

Could easily go here or #advanced-algebra tbh

Probably slighrly less likely to get swamped out by more elementary qs over there

[kind kindle]

okay good because i'm trying to work on a cryptosystem based on octonions over finite fields, and as far as I can tell this is not anywhere near as well studied as cryptography over nonabelian groups(this would work over a finite moufang loop instead of a group)

alr to #advanced-algebra

[lean sail]

struggling to show that if G is a solvable group and N \trianglelefteq G, then G/N is solvable.
i was able to show that i have a subnormal chain, now i'm trying to show that the composition factors are abelian. i'm struggling a little bit with how to define the homomorphism. i know that once i have the homomorphism defined, i'll probably invoke one of the isomorphism theorems (i'm guessing)... any thoughts?

[candid patrol]

Is a transversal (on the right or on the left) the same thing as a system of representatives? In group theory, relative to G/H

[knotty badger]

A nonabelian group can’t be abelian right

Or can it

[rapid cave]

huh

[tribal moss]

I think "nonabelian group" usually means "group that specifically has some elements that don't commute".
A group that can be either would be just "group" or "(not necessarily abelian) group" if it needs emphasis.

[knotty badger]

Ok cool

Because things like “noncommutative ring” often mean “not necessarily commutative ring” right

[rapid cave]

confusing yeah

[knotty badger]

Or nonassociative algebra

[tribal moss]

Yeah, a bit inconsistent.

[rapid cave]

Maybe the abelian case in whatever you are doing is trivial?

[thorn jay]

theres an nlab page on this lol

[south patrol]

probably red heering principle there lol

This is why often people say like "associative algebra" ig lol

[sacred wharf]

Dat confusing why not just call it ring

[sacred wharf]

Wym🥔

[south patrol]

rings are often assumed commutative unless otherwise stated in some fields

[vapid vale]

fields

[sacred wharf]

Bro had to use the word fields out all of em

[south patrol]

e.g. generally in algebraic geometry, and much of alg top tbh

[sacred wharf]

Is dat what we call commutative algebra

[south patrol]

ig that too

[elfin wraith]

I had a whole arc the other day of realising cohomology rings aren’t commutative and thinking there’d be some really cool overlap in the stuff I’m interested in

Then I realised that it’s jus graded commutative and that i definitely already knew that from my intro to geometry course lol

[south patrol]

Lol

[tardy hedge]

If (q) is maximal wrt to principal ideals containing it, is q an irreducible element?

I recall something like this but i forget the statement

[south patrol]

This should be an equivalent definition (outside 0 and 1)

[tardy hedge]

If q is irreducible then no principal ideal contains (q) ?

[south patrol]

Besides (1) lol

But yeah I mean it's like

[south patrol]

if u have this condition and q not a unit ofc and q = rt then WLOG r is not a unit. (q) is contained in (r) so (q) = (r) i.e. q,r are associates

Then this proof works the other way too I guess

like if q is irreducible and (q) strictly contained in (r) then q = ru for some u which is a contradiction unless r is a unit

[tardy hedge]

Which means t is a unit so q irred right

[south patrol]

It doesn't mean t is a unit in general

Well depends on whether ur assuming the ring to be an integral domain for example

If so then yes

[tardy hedge]

Talking about irreducible when ur not in a domain is weird isnt it

[south patrol]

Though there is a more general notion of irreducible where you say like if a = bc and c is not a unit then (a) = (b) or smth

[south patrol]

But yeah I think so

[tardy hedge]

Also, this is just the same as saying no principal ideal (excluding (1)) contains (q) isnt it

[south patrol]

Idk what else you would mean by maximal here

Though specifying "containing it" is not needed

Like just "it is maximal among proper principal ideals"

[tardy hedge]

Yea the context was i reading something just showing why pid has dimension 1, and he said for (q) since q is prime … no other prinicipal ideal contains it

But that is like using prime —> irreducible right

Or ig i was thinking u just need q to be irreducible to use that argument

[south patrol]

Ye

Well for PID everything coincides

[tardy hedge]

Oh yea

[south patrol]

(At most 1)

[tardy hedge]

Because fields?

[south patrol]

Indeed

Maybe sounds pedantic but often important to bear in mind lol

[tardy hedge]

Yeah true

[tardy hedge]

Where are these concepts even important? I know they obv are for number theory but idk where else

[south patrol]

Which concepts?

This stuff is important for some AG for example cause dimensions of ring are essentially the notion of dimension for varieties/schemes, and things like unique factorisation correspond to geometric properties too

Though personally I have never rly worried about irreducibility other than showing something is not a PID or smth lol

[tough raven]

Why not just exclude fields?

[south patrol]

From the original ting? Sure

[raw anvil]

$LaTeX$

oops

$/LaTeX$

oops

$\LaTeX$

⭐

[elfin wraith]

#latex-testing

[rapid cave]

This is not #latex-testing

[elfin wraith]

This is an algtop question but I thought id ask here since my question is really, how should I reason about Hom(Q,G) where G is a free abelian group? I want to say this has to be trivial but im not sure how exactly to argue that

Like for Z/mZ you can argue that free abelian groups are torsion free so its trivial, but Q is also torsion free. Its also a lot more "rich" though, since you can divide

So I think it should be 0, I just dont know exactly why

[quiet pelican]

No element of a free abelian group is n-divisible for all n
Cyclically reduce the word, then the nth power has length n times the original
So there's a lower bound of n on the length of an nth power cyclically reduced word

[rocky cloak]

Just think about what the image of a non-zero q can be? It will have to map to something such that n f(q/n) = f(q). So 0 is the only option.

Or subgroup of free group is free, so the image would need to be a direct summand and Q is indecomposable (and not free), so image is 0

[elfin wraith]

Ah these are all very nice thank you!

I like this one a lot though, im terrible for remembering that subgroups of free groups are free because ive had the modules thing drilled into me so well lol

[thorn jay]

oh this is really good

[rocky cloak]

I mean it's way more machinery than needed, but it works

[elfin wraith]

Im a big fan of a sledgehammer proof though lol

If our fancy machinary cant help with the small problems whats even the point

[rocky cloak]

... and proving that Q is indecomposable would be about the same amount of work as just the direct proof I guess

[thorn jay]

well divisible groups cant be free

unless theyre 0, right?

[rocky cloak]

That's true, but the proof is essentially the original statement

[thorn jay]

rahh

1984

[tardy hedge]

is that because if u assume u have a basis then there would always be relations among the generators since G divisible ?

[thorn jay]

yes

[tardy hedge]

https://math.stackexchange.com/questions/129337/how-can-we-show-that-mathbb-q-is-not-a-free-mathbb-z-module

so that makes the top answer here even easier doesnt it

dont need to go through showing if Q were free its rank 1 and then also Q cant be cyclic

[quiet pelican]

But combinatorial group theory

[elfin wraith]

These are not words that bring me joy

[quiet pelican]

[thorn jay]

i will have to agree with Nope here

[quiet pelican]

You like UA so you have no opinion here /j

[thorn jay]

UA is super respectable idk what youre talking about

[quiet pelican]

Unfortunately only group theorists are allowed opinions

[tough raven]

Group theorists, ring theorists and field theorists*

[quiet pelican]

No, just group

[tough raven]

What if they are all the same

[quiet pelican]

They are not
Proof: me

[tough raven]

Field theorists are Galois theorists, which are group theorists. Group theorists are representation theorists, i.e., module theorists, as are ring theorists.

[tough raven]

Or you just don't know it yet. 😉

[thorn jay]

discrimination >:(

[fading acorn]

opinions on homotopy theorists?

[quiet pelican]

I’m a group theorist and not a rep theorist

[quiet pelican]

Opinions banned

[fading acorn]

@true bolt

[true bolt]

They need to spend less time online

[thorn jay]

just because every student is forced to learn geoup theory doesnt mean group theorists should be the only ones allowed with an opinion

smh

[true bolt]

Why do I feel like #groups-rings-fields every time I've ever opened it has just been #math-discussion if #math-discussion was actually good
(I'm not telling y'all off, I'm just observing)

[thorn jay]

no it does not make your field "better" it makes it OLDER

[rapid cave]

What are these channels

[true bolt]

the first one is groups-rings-fields the second 2 are math-discussion

[thorn jay]

never heard of her

[rapid cave]

Oh yeah I have studying role

[elfin wraith]

We are the role models of the server

[true bolt]

Then why don't you have honourable

[rapid cave]

Yellow is closer to pink

[thorn jay]

yeah why DONT we

[elfin wraith]

Hey the ones of us who know things do

[thorn jay]

this is scandalous

[true bolt]

nuh uh

[karmic moat]

I am pure of heart and that’s the highest honor I could ask for

[elfin wraith]

By those of us I mean the regulars here. So jagr and raghuram

[tough raven]

Maybe I was an appeasement appointment

[elfin wraith]

You’re not in the in group, I’m sorry

[south patrol]

:(

Maybe I'm not that much of a regular idk

I feel like one

[rapid cave]

You are pink

Like lancey

[elfin wraith]

The pink supersedes the piss

You know things potato

[true bolt]

Are you not also a busy grad student

[south patrol]

:(

Lol

[south patrol]

Maybe!

[delicate orchid]

consider ur account blocked and reported

[thorn jay]

gamma

[delicate orchid]

I got gamma, saturation, AND offset

[tough raven]

That's the problem. Where's the rho for representation?

[delicate orchid]

I'm a character theorist so I'd use chi. I have not seen a matrix in 15 years

[tough raven]

I was going to say chi next actually.

[tough raven]

Also no chi.

[elfin wraith]

Bro was doing LA along side his multiplication tables

[rocky cloak]

That's just what happens when you put cool people in a channel together I guess

[worldly canyon]

Why is the term "unique" used here? I undersand the theorem and proof, but not why the langauge unique is used

It says Q is "smallest ring" containing R, this seems to imply you can have another strictly larger ring S which contains R as a subring and has every element of D a unit in Q. But if such was the case, couldn't you repeat application of this theorem swapping S and Q to show that they are isomorphic?

Maybe I am misunderstanding, it seems the use of langauge "uniqueness" and "smallest" concurrently is contradictory

in that if you drop the sentence "smallest" its somewhat clear how would arrive at uniqueness. Whereas if you include the sentence of "smallest", it seems to imply there is sometihng larger, which seems impossible from uniqueness

[vapid vale]

i also agree the language is a little strange, partly because its getting at a point that seemingly isn't made apparent yet. uniqueness just means that anything nonisomorphic which containing R and with all elements of D units must in fact have more elements than D turned into units

[worldly canyon]

When the theorem sates, there is a commutative ring Q with 1 such that Q contains R, and every element of D isa unit in Q. The proof constructs a specific Q. Does it mean that THIS speicfic constructed Q is hte unique instance satisfying property (2)?

or on the other hand, is it saying "any comm ring q with 1 st q contains r, and every element of d is a unit in q" is a unique instance satisfying property (2)?

[vapid vale]

if you are familiar with universal properties, you can understand the "Let S be.." part of (2) as a universal property statement, and unique is to be interpreted as "unique up to unique isomorphism which satisfies that factoring through property"

[vapid vale]

well Q hasn't been constructed yet

[worldly canyon]

this is the Q they use

[vapid vale]

yes it means this specific constructed Q is the unique istance

[worldly canyon]

yeah I guess I understand the phrase "smallest" then

[worldly canyon]

yeah I have done a little bit of category theory. I guess this problem is missing the statement "there is a unique map"

it just states "there is a map"

[vapid vale]

right. presuming this is D&F and i dont remember the way they motivate things but i don't think theyre really trying to convey universal properties at this point in the book? the point theyre trying to make is the injective map implies S is bigger so Q is the smallest

[worldly canyon]

do you know how to show uniqueness without adding the statement "there is a unique injective homomorphism phi:q->s..."

[vapid vale]

you still get uniqueness without a unique map

its just not unique up to unique isomorphism

[worldly canyon]

Im trying to show P that satisfies property of being smallest, contains R, and has elements of D as units, then there is an isomorphism between P and Q

i guess I have been able to find maps from P to Q and Q to P such that they agree with identity fxn on R

but don't know how to show they agree with idenetity fxn on Q/P

[vapid vale]

the theorem states that P and Q should contain isomorphic copies of each other as subrings

this is enough

[worldly canyon]

oh yeah

😂

thank you

[worldly canyon]

wait is this true? mutual embeddings mean you have isomorphism?

like you can generalize schroder bernstein to groups/rings?

[vapid vale]

wait no thats definitely not true

lol

[vapid vale]

the uniqueness of the homomorphism is given by 1) if we have f, g: Q \to S then f(rd^{-1}) = \phi(x)\phi(d)^{-1} = g(rd^{-1})

[worldly canyon]

Oh I see, I think i was misinterpreting this point

[noble nexus]

no it doesn't hold in general for groups and rings

I believe it holds for certain nice classes of modules

it also holds for unitary representations of a group, which is fun

[twilit wraith]

ok so if I and J are isomorphic ideals

and theyre generated by i and j respectively

if I is generated by i^2 - i, then must J be generated by j^2 - j

[spice whale]

No
consider (2) and (3) in Z

[twilit wraith]

2Z and 3Z arent isomorphic

i actually asked that question trying to show the two arent isomorphic lol

but fortunately i ended up finding an easier way later

[spice whale]

wait what do you mean by isomorphic here

They're definitely isomorphic as Z-modules

[twilit wraith]

they arent isomorphic as rings

[spice whale]

ideals aren't rings though

oh wait are your rings non unital

[twilit wraith]

given our definition they are

yes

good ol dummit and foote

[spice whale]

if they're isomorphic as rings then like

[twilit wraith]

the problem lies in that f(2)f(2) = f(4) = f(2) + f(2)

[spice whale]

this should be true surely

[twilit wraith]

thats what i was thinking but idk

[spice whale]

if f is our iso

[twilit wraith]

yeah

[spice whale]

wait are we saying isomorphic as rings or as R-algebras

[twilit wraith]

rings

[spice whale]

oh ew

[twilit wraith]

yeah rings are kinda ugly

[spice whale]

i mean less that

[twilit wraith]

the homework on them this week has been terrible

[spice whale]

more now our isomorphism doesn't preserve any of the actual important structure on the ideal

i.e. the fact it's an R-module

there's got to be a counterexample here surely

oh wait (2) and (-2)

[kind temple]

why are we considering ring isomorphisms in this special case when the ideals are rings

[spice whale]

They're considered as rings (non-unital)

[spice whale]

where the isomorphism is just taking the negative

[kind temple]

oh i interpreted this as saying that they are unital

[twilit wraith]

i mean isomorphisms have to send 2 to -3 or 3

either way it doesnt work out

[spice whale]

actually this is an algebra homomorphism

[spice whale]

-3?

[twilit wraith]

-3 generates 3Z

[spice whale]

i said above sending 2 to -2 in (2)

[twilit wraith]

were looking for isomorphisms from 2Z to 3Z

[spice whale]

ok but that obviously doesn't work

if we're considering ring structure

[twilit wraith]

im confused what were talking about then

bc that was the whole issue

[spice whale]

your original problem

[twilit wraith]

oh right i forgot i said something more general

[spice whale]

if you have (2) with generators 2 and -2 in Z, these are clearly isomorphic (as algebras even!) via the identity map (or multiplication by -1), but while 2²-2 = 2 generates (2), (-2)²+2 = 6 does not

[twilit wraith]

yes

hm

well that sucks but at least i found an easier way to do it

[tough raven]

(-a)(-b) ≠ (-ab) though.

I think there are no non-zero non-identity rng homomorphisms between ideals of ℤ.

||Consider the property of an element's square equaling the result of multiplying it by n (in the sense of adding it to itself n times). This has to be preserved by any rng homomorphism and in ℤ is satisfied by (only) {0, n}. Therefore any rng homomorphism from nℤ to ℤ (or any subrng) has to map n to n or 0, so must be the identity on nℤ or 0.||

[rocky cloak]

In not really sure what the problem is trying to get at anyway. It's like very unclear and very arbitrary. What would it matter if i^2 - i also generates?