groups-rings-fields

[dull ginkgo]

We care about that

And that group toys with the underlying set

It acts on it to say

be careful with distinguishing the group elements from what they do to the set elements

don’t mix up the hammer with the nail

[astral ivy]

yep I see this now

[astral ivy]

crazy analogy

[dull ginkgo]

One of the main reasons groups pop up everywhere is because they can act on other things as automorphisms

[south patrol]

Real.

[astral ivy]

so can anyone explain how to do the part with inverses?

[dull ginkgo]

Are you trying to show that the subgroup contains inverses

[astral ivy]

yes

that's the other property I need

[dull ginkgo]

So

You can kind of split up what each permutation on your subgroup does

We have the fact that it doesn’t touch 7, it either fixes 2 and 3 or swaps them, and the rest can go however, but can only go to eachother

Show how you can invert each of those

:3

[glad osprey]

One of the main reasons we care about groups is because hammers act on nails through their Carpenter groups

[astral ivy]

so what you're saying basically is that 7 doesn't change at all and is thrown in as some sort of decoy to confuse me?

[dull ginkgo]

No,

[thorn jay]

well, it's nice to know that 7 is fixed

[dull ginkgo]

It’s actually very important

In fact we have a subgroup of permutations that leave 7 unchanged (the stabilizer subgroup, they stabilize 7 and leave it unchanged)

But the great thing about permutations is that we can break them up to what they do on subsets

[quiet pelican]

You could do this exercise on S_{1, …, 6, 8} (ie ignore 7) and it would work fine
But also stabiliser subgroups are important so like, it’s useful to see an echo of it now

isotropy groups are also why we can’t have nice things in geometry though

[dull ginkgo]

I believe this is an important thing to digest. So imagine we have a permutation p on a set S right

We can look at what p does in subsets of S, because if p permutes elements in a subset, then we can just invert it in that subset

In this case

Let’s say we have the set {1…8}

We have the subsets {1, 4, 5, 6, 8}, {2, 3} and {7}

Label these subsets A_1, A_2, A_3

If we have a permutation in that subgroup, why does that permutation send elements in each of those subsets to that same subset (i.e the permutation doesn’t make elements go between those subsets)?

[quiet pelican]

{1, 4, 5, 6, 8}, not {1, 4, 5, 6}

[dull ginkgo]

I said 1…7

Oh is the original group S_8

[quiet pelican]

Yes

[dull ginkgo]

@astral ivy then show that if we can “reverse” the permutation in each of these subsets, then putting it together just reverses the whole permutation :3 (effectively you’ve described the whole permutation)

I can provide an example if you like

[astral ivy]

either because of closure or because that's the property of permutation operation

[dull ginkgo]

No why does our subgroup do that

[astral ivy]

idk

[dull ginkgo]

So it sends 7 to 7, so {7} doesn’t change

pi(2) gets sent to 2 or 3, pi(3) gets sent to 2,3, so elements in {2,3} never leave {2,3} under pi

What about the other ones?

Why can’t they be sent into {7} or {2,3}?

[astral ivy]

why? I can't think of any other explanation other than it being closure of subgroup or permutation property

[dull ginkgo]

Inversion :3

Let’s say 6 gets sent to 2, why is this a contradiction?

[astral ivy]

what about inversion tho? isn't the function bijective and invertible so everything sent can be sent back

[dull ginkgo]

Basically yeah

Because some element of {2,3} would have to leave the set, contradicting pi

[astral ivy]

huh so the order (or cardinality) is fixed

[dull ginkgo]

Yes

[astral ivy]

I see

[dull ginkgo]

It’s a bijection

so the idea to prove invertibility is to just, take pi, notice how it permutes “inside” of disjoint {7}, {2,3} and {1, 4, 5, 6, 8}

Then just, reverse it inside of those subsets, and together that describes the inverse of pi

and it is in the subgroup, because it fixes 7, sends {2,3} to itself

Do you see what I mean

In fact, when you “fully” break down a permutation into the smallest subsets that are sent to themselves, it actually decomposes the permutations into cycles. You’ll see this soon

It’s important to also look at what permutations do to subsets, not just elements :3

[glad osprey]

Yep 👍 usually substructures map to substructures - subgroups, subrings, submodules etc.

But the image of an ideal under a ring hom is not necessarily an ideal (it is if the map is surjective)

I bet enpeace is gonna come and say something smart and incomprehensible about universal algebra now

Yep, that's just the first isomorphism theorem AFAICT

Yep

[dull ginkgo]

Modules are like abelian groups

So yes

[thorn jay]

ah well I guess its a general theorem

[dull ginkgo]

They just have the added map of R, R^op, or R (x)_Z R^op into End(M)

[thorn jay]

H \cap ker\phi is just the kernel of the composition of the inclusion with \phi

[glad osprey]

phew, that was much more comprehensible than I feared

[thorn jay]

and from there on it is indeed the first isorphism theorem

[glad osprey]

I guess ideals aren't strictly substructures, so it's still true that substructures map to substructures

[thorn jay]

ideals are indeed not substructures

[prime quail]

https://drive.google.com/drive/folders/1Y3Y01g0Tfe3XTLGHGSF_TYA5JNRbytIv?usp=drive_link

here are every cycle structure of the rubik's cube

[thorn jay]

depends

you've got the category-theoretical or model-theoretical definition

in category theory it's an equivalence class of monomorphisms into some object

in model theory, a substructure of a model M for some theory T is a subset N < M such that it is closed under the operations of M, and it satisfies T, when all the relations are restricted to N

in universal algebra, where T consists of only equations, and there is no relational structure, this is equivalent to N being closed under the operations of M

in both contexts, when viewing R as a ring, ideals are not substructures

[thorn jay]

basically what im saying is it depends on context, but if you view R as a ring then in no context are ideals substructures of R

youve gotta view R as a module over itself R for that

[dull ginkgo]

Unless you consider rings without unity of which you are stinky and the important properties of ideals are "lost"

[thorn jay]

those arent "rings"

[glad osprey]

they are stinky, just like you would be if you considered them

[tardy hedge]

You asked .. you shall receive …

[thorn jay]

rngs

rIngs without Identity

[glad osprey]

in stinky poopoo-land analysis

[thorn jay]

i do consider them but not as rings

theyre a cool congruence representable class of algebras but not much more to me

in that regard lie algebras are cooler because they dont have a non-linear analogue (rings have monoids and rngs have semigroups)

[glad osprey]

I don't know much about analysis, but apparently in operator algebras and stuff. There was incidentally a thread asking this question on reddit yesterday: https://www.reddit.com/r/math/comments/1nfy9uf/nonunital_rings_where_do_they_come_up/

[thorn jay]

operator theory

C*-algebras n stuff

[karmic moat]

Hodge theory

[topaz solar]

Representations, Lie groups, harmonic analysis, operator algebras, continuous model theory

[topaz solar]

Well, you don’t usually demand N |= T, that’s a submodel, but substructures need not model T

Unless T is universal

[thorn jay]

right

well that shows I'm not a model theorist 💔

[topaz solar]

There’s like purely algebraic parts, but there’s some stuff of like, matrix spectrum stuff, and fancier groups like SL(2, R) get into analysis land

[thorn jay]

at least there's no difference for me 😎

[topaz solar]

Also ideals are rngs, but what’s more important is they’re submodules of R as a module over itself

It’s an area of research on its own

But later algebra, operator algebra analysis stuff, some harmonic analysis, fancy number theory

For what

Idk learn it and find out

[thorn jay]

representation theory is important in physics

so the boring type

that's honestly a skill issue

I'm able to do what's asked of me pretty easily

[elfin wraith]

Do we?

[thorn jay]

truthfully, I don't think you can apply it directly

the most abstract algebra I know relating to machine learning and data analysis is homological algebra for persistent homology

but I think dismissing a branch of math just because it isn't directly gonna make you money is kinda stupid tbh

what you get out of studying anything is wayy more than just the raw knowledge

[topaz solar]

There’s group rep stuff in quantum computing

[elfin wraith]

I mean we follow the same books so I don’t see how that could possibly be true lol, I did Tao 1 and 2 for my analysis courses

[maiden drift]

I have a non-abelian group G of order pq for some primes p and q such that p < q. I’m pretty sure I’m able to show that there are p elements in Z(G) and then (q-1) noncentral conjugacy classes of size p. I’m trying to show that G has a nonnormal subgroup of index q. im pretty sure I can do that using the fact G has an element of order p. But how can I find a g in G such that <g> is not the center of G?

[earnest delta]

Which part of group theory has topic related to commutator, nilpotent subgroup, derived subgroup?

[thorn jay]

group theory above elementary group theory

[tidal schooner]

Have you seen the classification theorem for groups of order pq? If so, then the existence of a nonnormal subgroup of index q follows pretty immediately from that

[cedar vault]

If there were p elements in Z(G), the quotient G/Z(G) would be cyclic meaning G is abelian so I don't think it's possible that |Z(G)|=p

Can you explain how you got |Z(G)|=p?

Infact this tells you Z(G) is trivial

[south patrol]

(Can also use the funny trick: if the centre has order p then the centraliser of any element x not in the centre contains x and the centre, and is therefore the whole thing)

[maiden drift]

My thinking is that by the class equation pq = |G| = |Z(G)| + the sum of r indexes of the centralizers of the representatives of the non central conjugacy classes. My thought is that if |Z(G) = 1 then the sum of indexes must be pq-1. And since the summands are indexes, they must each divide pq. Which means they can only be 1 each and pq-1 of them. But that’s only true if G is abelian right?

[cedar vault]

Why must they be 1 each? For example take p=2,q=3 then 2*3 - 1 = 2 + 3

If some things divide pq, their sum need not divide pq

If some things are divisible by pq, then sure, their sum is divisible by pq

[coral spindle]

Why not look at an example? Dih(2p) where p is a prime, the Dihedral group. What's its centre?

[maiden drift]

Ok but if you have a group of order pq with p=2 and q=3 and a trivial center then 6=1+?

[cedar vault]

6=1+2+3

[cedar vault]

So here you want to look at dih(6)

[maiden drift]

Thanks. I’m tripping over myself

[elfin wraith]

Dih

[merry summit]

for this question, i applied the sylow theorems and the possibility for A = {1, 3} and B = {1, 4} but i don't know how to proceed further. It's not given that it is a normal subgroup or not else eliminating 1 would be easy.

they want me to find what A and B are btw

[wraith cargo]

what's the size of these sylow subgroups?
Are they normal?

that's the question you should be asking

[tardy hedge]

if I = rad(I), then I = intersection of minimal prime ideals that contain I?

[wraith cargo]

yes

[tardy hedge]

ty

[south patrol]

rad(I) = intersection of minimal primes containing I, so yes

[wraith cargo]

prove this hehe

[south patrol]

or define it as such

:chad:

[wraith cargo]

wow

didn't know big algebra got to you too

sadge

[merry summit]

for sylow 2 subgroup the order is 8, idk if its normal.
for sylow 3 subgroup the order is 3, again idk if its normal

[quiet pelican]

To tell if they’re normal, how many element of order 3, 9, 27, … (resp. elements of order 2, 4, 8, …) are there in S_4?

[merry summit]

im unaware of smth like that

is there a theorem?

[quiet pelican]

We’re essentially gonna do a counting argument to show that if you only had one sylow subgroup, then it wouldn’t be able to contain all the elements of order a power of 3 (resp 2), so contradiction by sylow 2

[quiet pelican]

It’s just some combinatorics, after you figure out which cycle types have the relevant orders

[rapid cave]

this is a classic trick

[merry summit]

do you have a resource? I'll give it a read. i never came across this trick

[quiet pelican]

It’s not really something I can figure out a resource idrt

The cycle types you want to consider are errr
1, 1, 1, 1 has order 1
2, 1, 1 has order 2
2, 2 has order 2
4 has order 4
3, 1 has order 3
And nothing else has order a power of 2 or 3

In fact, there are no other elements of any other order in S_4

[thorn jay]

very simple lattice theory (as long as you define rad(I) to be the intersection of all primes containing I)

[cyan skiff]

hello everyone, can someone give some reviews on this article that i wrote https://medium.com/@taahatariq2005/from-groups-to-homomorphisms-making-algebraic-structures-click-ebc93bb14de5

[tulip otter]

what should i do to prove that the map a|->σ_a is surjective? I mean isnt it like surjective by definition

i mean take g=σ_a, a in (Z/NZ)* then g(w)=w^a for all w in G and here is the a appearing right there

[quiet pelican]

Prove that g as defined there is an automorphism too

If you’re assuming you’ve already done that part then yeah that’s enough

[tulip otter]

ohhh so that is sufficient for surjectivity

alright i will write a full proof of the whole problem to make sure that i am not missing anything, if you can check it then i will be grateful.

[glad osprey]

It's somewhat non-trivial that this map is surjective - how do you know every automorphism of G is of the form g |-> g^a for some a?

btw, this is not the same as saying the map g |-> g^a is surjective

[tulip otter]

let $f:(\mathbb{Z}/N\mathbb{Z})^\to Aut(G)$ be the map defined by $f(a)=\sigma_a$ and let $a,b\in (\mathbb{Z}/N\mathbb{Z})^$, then given $w\in G, f(ab)(w)=\sigma_{ab}(w)=w^{ab}=(w^b)^a=\sigma_a(w^b)=\sigma_a\sigma_b(w)$ so that $f$ is a homomorphism.\ \Next, $\ker f={a\in (\mathbb{Z}/N\mathbb{Z})^\mid f(a)=id}={a\in (\mathbb{Z}/N\mathbb{Z})^\mid\sigma_a(w)=w\ \forall w\in G}={a\in (\mathbb{Z}/N\mathbb{Z})^\mid a\cong 1\pmod N}={1}$ where $id:G\to G$ is the identity map of $G$ so that $f$ is injective.\ \Finally, let $G=\langle w\rangle$. Then given $g\in Aut(G), g(w)=y$ for some $y\in G$ but $y=w^a$ for some $a\in (\mathbb{Z}/N\mathbb{Z})^$ since $G=\langle w\rangle$ is a cyclic group of order $N$ and generator $w$. Hence $g(w)=w^a$ and $g(x)=x^a\ \forall x\in G$ so that $g=\sigma_a$ and $f$ is surjective.

[cloud walrus]

yassine

[tulip otter]

ah yes i have to show that every automorphism of G is of this form, i was doing something else. After i read your message i tried to show that but idk if i succeeded or no

[velvet hull]

for the last part, can you tell me why a has to be in (Z/nZ)*?

[tulip otter]

ok so |G|=N and G=<w> so G={0,w,...,w^{N-1}}. So for any element of G of the form w^b with b in Z, you can reduce b mod N to get the element of G that it represents

[velvet hull]

okay, sure, what next?

because (Z/nZ)* is not the same as (Z/nZ)

hint - ||sending w to w^a does not always induce a group automorphism for all 1 <= a <= n, you need something more about a||

[tulip otter]

ah wait, for some reason i was thinking about Z/NZ and not (Z/NZ)* when i was coming up with the proof

ok let me think about it for a bit

if i dont get anything for a while i will check the hint

ohhh wait

if (a,N) neq 1 then |w|<N

but G=<w> and |G|=N so |w|=N?

is that correct or am i saying non-sense

i see, tysm for pointing this out so that i dont skip it unknowingly

[tulip otter]

are there any other notes/mistakes/... about this other than your previous note and the weird congruent sign

i intended to write $\equiv$ but thought that $\cong$ gives the usual sign

[cloud walrus]

yassine

[glad osprey]

You're on the right path, but I think you can be a bit more precise. What is the relation between a and |w|?

[tulip otter]

in terms of why a\in (Z/NZ)*?

[glad osprey]

yeah, specifically how do you go from (a, N) != 1 to |w| < N?

Remember that a isn't the order of w, it's just the exponent in the map g(w) = w^a. I think the conclusion |w| < N is on the wrong path actually, you need to think about another assumption you haven't used

[tulip otter]

hmmm maybe i see what you are trying to tell me. Actually i didnt use the fact that g is an automorphism. Since g is an automorphism then ker g={e}. Now if (a,N) != 1, then g(w^(a,N))=(w^(a,N))^a=w^N=e but w^(a,N) neq e since |w|=N and (a,N)<N which would contradict the injectivity of g

is that it?

[glad osprey]

yep, you got it 👍 but can you explain why (w^(a,N))^a = w^N?

-# sorry for nitpicking, I'm not trying to be pedantic, I'm just stupid and need help seeing all the steps

[tulip otter]

dw, that doesnt bother me at all. In fact your notes always point out mistakes/things that i didnt give much thought because i thought they were kinda obvious but that always turns to not be the case lol

[tulip otter]

maybe thats false

i reached this result because i had (a,N)a=N in mind

but take a=5 and N=15 for example, then (5,15)5 neq 15

[glad osprey]

Yeah, (w^(a, N))^a = w^N is not true. I showed that (a, N) = 1 by using the fact that g is surjective, but I'm sure there's a way to use the fact g is injective too

[tulip otter]

ohhh nice

then i will try to use surjectivity first

so consider an integer b in [0,N-1] such that (a,b)=1. Then there doesnt exist y in G such that g(y)=w^b since if there exists such a y then g(y)=g(w^k)=w^{ka}=w^b. I think it is sufficient to consider k where ka in [0,N-1], by doing this you should have ka=b for w^{ka}=w^b to hold. But (a,b)=1 so ka neq b which contradicts the surjectivity of g (?)

i think that i did use injectivity here too no? (assuming that what i wrote is correct to begin with)

[glad osprey]

Yeah, I think that works, you're saying that if (a, N) != 1 and (a, b) = 1 then w^{ka} = w^b is impossible. You can simplify a bit by considering b = 1:
Since g is surjective, we must have g(w^k) = w^(ka) = w for some k. Then w^(ka - 1) = e, so ka = 1 (mod N), since w has order N

[tulip otter]

ohhh i see, thats a nice solution. Doesnt this in fact also use injctivity of g

[glad osprey]

hmm, I don't think so?

we're just assuming that there exists a k such that g(w^k) = w

[tulip otter]

doesnt w^(ka-1)=e\implies ka=1 (mod N) in fact use it?

[glad osprey]

no, that's just because the order of w must divide (ka - 1)

[toxic sapphire]

I'm not really sure where to go with this. we're given the hint to think about centralizers of an element and the center of the group

I've found some basic facts, like that becuase alpha is a homomorphism it preserves centralizers

could Lagrange's theorem be helpful?

[toxic sapphire]

hm. if I can show that I is a subset of C(G) then it's done by Lagrange, but I'm really struggling to show that

[velvet hull]

in any group, does sending x to x-1 always define a group homomorphism?

[toxic sapphire]

what is x-1 in general?

I'm not sure what that would be for the reflection in D_n for instance

oh wait do you mean x^{-1}?

nvm sorry I'll think about that for a second

no, right? this is a homomorphism precisely when the group is Abelian?

[velvet hull]

can you tell me why?

[toxic sapphire]

that map sends ab to b^-1a^-1, which is a^-1b^-1 only when a and b commute

[velvet hull]

right

now think about what that means if an isomorphism sends a to a^-1

[toxic sapphire]

if an isomorphism sends all group elements to their inverse, clearly G is abelian. but I'm not sure what to do in the case that ab is not sent to (ab)^-1, which is possible if |I| < |G|

since the map is homomorphic, ab goes to a^-1 alpha(b)

I don't think there's anything else we can say about where ab is sent without assuming b or ab is in I

[earnest delta]

100

If I took n=2

0,2,4...
1,3,5

so i got 2

C option

[toxic sapphire]

I think that's correct. the cosets are nZ, 1 + nZ, 2 + nZ, ..., (n-1) + nZ, where k + nZ is all integers equivalent to k modulo n

[velvet hull]

hmmm, maybe think about it like this, is there anything you can say about the subset of all elements that gets sent to their inverse?

[merry harness]

notational q. is there a difference between
[ (\bZ/30\bZ)^* \text{ and } (\bZ/30\bZ)^\times ]

man wat

...

[toxic sapphire]

lol try ^{*} and ^{\times}

[cloud walrus]

qiu

[merry harness]

it wasn't math mode though 😔

[toxic sapphire]

oh right. in that context, no, those are both the multiplicative group of units mod 30

[merry harness]

my prof uses the former but dummit and foote the latter

[merry harness]

Any contexts where they are different?

[toxic sapphire]

as in the two symbols? I'm not sure, maybe, but that is just two common ways to notate the same group

[merry harness]

Yeah the symbols \times vs * in superscript

[toxic sapphire]

not that I'm aware of

it would probably be kind of confusing, so I assume most authors would avoid using them both in similar ways to mean different things

[toxic sapphire]

this is the set I, and all elements in I commute with each other

[velvet hull]

anything more?

[toxic sapphire]

hm I'll think for a moment

[merry harness]

math peeps otw to find yet another application for the symbol [ \mathcal{L} ]

[cloud walrus]

qiu

[toxic sapphire]

of course, we assume I constitutes more than 3/4 of G? I'm not thinking of anything else

[velvet hull]

is it a subgroup?

[toxic sapphire]

oh yeah that's an obvious question to ask. I'll check

actually it's not obvious to me that it's closed. that depends on it being commutative, which I've claimed, but I don't think I've proven

[toxic sapphire]

I don't think I defined this way is always commutative, and thus I don't think it's always a subgroup, but it may be under the assumption that it constitutes the majority of G

so I guess, I think I is a subgroup because I think I is a subset of the center of G, which is exactly where we started, I'm still not sure how to show that I is in the center of G though

[crystal vale]

S3 and D3 are isomorphic

[crystal vale]

What have you tried?

[earnest delta]

D6 has 12 elements

e,a,b...b^5...ab,ab^5

and center elements would be identity and?

I am stuck here

12/?

[rapid cave]

ab^3

remember what realtion defines Dn

bab=a

so b^3ab^3=a --> ab^3 = b^-3a

but b^-3=b^3 so a and b^3 commute

[twilit wraith]

yo, i have to prove that given an abelian subgroup H of G, the group generated by $H \cup Z(G)$ is abelian

[cloud walrus]

hiidostuff

[twilit wraith]

its pretty obvious were the group to just be H \cup Z(G) itself

but im having trouble proving that, if the two sets arent equal, then the element not in H or Z(G) commutes with the elements of H

obviously it commutes with the stuff in Z(G)

[kind temple]

if an element x is generated by H U Z(G), you can always write x as hz for some element h in H and z in Z(G) by collecting all terms in Z(G) to the right to form z

[twilit wraith]

why is that

[kind temple]

an element x generated by a subset S of G is just a product of elements of S and their inverses, right?

[twilit wraith]

i was not aware of that

augh but now that i am looking at the textbook im seeing it now

i guess i just didnt register that section for some reason

then of course its clear now why the subgroup generated by H and the centralizer of H is not necessarily abelian

because the elements of the centralizer may not commute among each other

[limber sequoia]

There are almost always two-three definitions of "generated thing"

1. (top down definition) "smallest thing that contains the generators"
2. (top down definition, concrete variant) "intersection of all things that contain generators"
3. (bottom up definition) A concrete description of the set, usually as the set of all possible operations on the generators.

[kind temple]

imo 1 subsumes 2

and usually it’s called bottom up lol

[limber sequoia]

yeah I was translating in my head 😆

[kind temple]

from what language

[limber sequoia]

hebrew, we say "from up to down" literally

[kind temple]

oh cool

[limber sequoia]

One way in which groups are nice is that the bottom-up definition is very simple.

[twilit wraith]

i forgot about the third definition completely

i was thinking about it in the intersection sense and thus didnt have much info to go off of

unless i thought about it further i guess

[kind temple]

free constructions and presentations are cool

[cedar vault]

Well even if you think about it in the intersection sense you will get that the elements will have form hz with h in H and z in Z(G)

[twilit wraith]

true because theyre equivalent

i guess its just not obvious from that definition

[kind temple]

any nice way to see this?

[twilit wraith]

at least in my mind i cant come up with a reason why a stray element couldnt come about

[limber sequoia]

Here's a different proof:
H is abelian, Z is abelian. The set HxZ therefore must be abelian, and there is a surjective group homomorphism from it to HZ inside G, so HZ must be abelian.

I think im missing a step here and shoving something under the carpet.

[twilit wraith]

of course showing that the two definitions are equivalent does the trick but that somewhat presupposes their equivalence

[cedar vault]

It must be closed since you are taking an intersection of subgroups, so you will end up with a product of the form ...h1(product of stuff in Z)h2(product of stuff in Z)... And Z(G) commutes with everything, so..

[twilit wraith]

the only thing shoved under the carpet is showing that the group generated by H and Z is HZ but i guess thats a clear step

[twilit wraith]

i dont see why the first part immediately implies the second

[limber sequoia]

no, the function from HxZ to HZ is a group homomorphism if the image is abelian. So this suppouses the outcome.

[twilit wraith]

ah, i missed that

[cedar vault]

Because I should be able to take any some elements of H and any some elements of Z(G) and multiply them together in any order I want and it should end up still in the subgroup generated by H U Z(G)

I.e it's closed

[twilit wraith]

yes i understand that all words of elements from H or Z(G) is a subset

[kind temple]

but that’s just the first definition lol

[twilit wraith]

but the issue is why are they equal

[limber sequoia]

On the one hand, the generated set must contain HZ.

On the other hand, HZ is a group (because H or Z are abelian), so it must contain the generated set (as the generated set (as the generated set is the intersection of all containing groups)

The same argument applies, but replace HZ with the third definition.

[twilit wraith]

wait lol i just realized

i have to give an explicit example of a abelian subgroup H where the group generated by H and its centralizer is not abelian

couldnt i just let H be trivial in a nonabelian group lol

[cedar vault]

Well if something is not of that form then it can't be in the intersection, because I can take the subgroup of words made by H and Z(G) and it won't be in the subgroup, so it can't be in the intersection of all subgroups

[twilit wraith]

oh thats actually a very good point

since the word group would be one of the groups we consider in the intersection

[cedar vault]

That's basically also the proof of the equivalence

[twilit wraith]

that makes it abundantly more clear, thank you

[kind temple]

maybe this is what you are bumping into: H and Z(G) are in Ab, but G is in Grp. H \coprod Z(G) gives you a homomorphism into G respecting the natural inclusions, and by construction, the image of this homomorphism is exactly words in H U Z(G) interpreted in G, which is precisely the group generated by H U Z(G). you may alternatively be able to analyze the kernel of the homomorphism H \coprod Z(G) —> G, but that feels a bit more difficult

[limber sequoia]

There's an "almost counterexample" to this construction I like to think about.

Sometimes the thing you generate is closed to multiple operations, and you need to close it by steps, that is, close it under, say, +, then close it under *, then again under +, then again... each time gaining new objects. Then sometimes this process stops or sometimes you need to take the limit of it. But here in groups you do one step and you're done which is very nice.

[limber sequoia]

one moment not sure I understood.

[kind temple]

https://ncatlab.org/nlab/show/transfinite+construction+of+free+algebras

this is a general process to obtain free objects

they are fixed points to these closure operations

[limber sequoia]

I agree with everything, but I didn't understand the "alternatively" part

[kind temple]

the image should be H \coprod Z(G) modulo the kernel by the first iso theorem

but i’m finding that describing the kernel is tricky

[limber sequoia]

yes, but why do I want to analyze that is what im missing

[limber sequoia]

it's the normal subgroup generated by all sentences of the form g_1...g_n=e which are true in G.

Very helpful I know.

[kind temple]

lol

[limber sequoia]

Do you mean, in order to show that the image is abelian?

[kind temple]

yea, like maybe knowing the kernel gives you some nice relations to tell you that the image is abelian

but with the original way, it’s clear to argue element-wise why the image is abelian

[limber sequoia]

You can show that the coproduct is abelian, which is repeating the argument element-wise but on the coproduct.

and then any image is abelian and you're in the clear, but... extra work in order to think in a categorical way.

[kind temple]

the coproduct in Grp shouldn’t be abelian unless one of the groups is trivial

[limber sequoia]

oh right.
what I was thinking about is doing what you were saying

[tardy hedge]

for a squarefree monomial ideal, it's not actually obvious that its associated primes are monomial prime ideals right? (just ideals generated by single variables like (x1, x2) )

i guess my question is why are the associated primes for squarefree monomial ideals necessarily also monomial ideals

monomial prime ideals being of the form (x1, x2, ... xn) makes sense

[tardy hedge]

I dont think this is obvious, Herzog and Hibi have a whole algorithm for computing the primary decomposition here

[tulip otter]

ohhh i see. tysm for your help yesterday and sorry for not replying until now but yesterday i had to do something so i left the laptop and didnt use discord until i slept.

tysm HChan for your help too

have a great day/night both of you

[minor gazelle]

Guys in the context of monoid, and we define a property on monoid homomorphism called special and f is special if whenever fg = fh implies g=h and how can we derive that every map with special property in the context of monoid is injective

fg means f \circ g

[limber sequoia]

I think you can argue in contradiction here effectively.

Say we have an arbitrary monoid homomorphism A -f> B which is not injective.

The main challenge is finding two functions B -> (something) which are equal under composition but not equal in themselves.

[minor gazelle]

I shall work on that! 🥰

[limber sequoia]

gl

[minor gazelle]

I actually think I over complicated things after your hint, contradiction is gonna work 🥰 hopefully

[rustic patrol]

Hey, can someone please give me some hints or direction on how I should be thinking?

[limber sequoia]

Can you describe where you are stuck?

[rustic patrol]

I'm clueless. How should I use the associative binary composition to meet the required proof?

[limber sequoia]

ok so let's unpack the goal

We want to show that T is closed under composition.
That means, that taking arbitary objects a,b in T, we want to show that a \circ b is in T

by \circ I mean your operation. let's use * instead.
That is, we want to show that for all x,
( a * b )* x = x * (a * b)

[rustic patrol]

I saw a solution in a YouTube video. It showed some horrible things.

[limber sequoia]

What do you mean by horrible?
Anyway, can you spot the error in the picture?

[thorn jay]

do you know about free monoids?

[rustic patrol]

Yeah, he simply took closed under S means closed under T too.

Am I right?

[limber sequoia]

Yes that's right

Also, her/his proof is a bit confused in other ways

he says that ax = xa, by T's definition, therefore "ax, xa in S"
But ax,xa being in S is entirely unrelated to ax = xa, so this is very strange.

[thorn jay]

youve got it the wrong way around lol

the question asks to show monomorphism => injective

[limber sequoia]

I always confuse which way around is which 😆

[thorn jay]

here youre asking noninjective => not epimorphism

which is false

[thorn jay]

haha can happen

just think of it like this: if you know the composition of a surjection with a mystery function, then, as the surjection gives you all values of the codonain, you can deduce the mystery function from the composition

[thorn jay]

you actually dont need contradiction here! Do you know what maps from the additive monoid of natural numbers N to some arbitrary monoid M look like?

[limber sequoia]

If you're stuck I suggest the use of two methods:

1. rolling down hill
2. whack the mole

rolling down-hill: If you don't know how to progress in a proof like this, you can write down the start of how a proof should look. In this case, we want to show closed under T, so the proof should look something like, "let a,b in T, we want to show that ...."
Many times ideas form in this way.

whack the mole: after rolling down the hill a bit, you may reach an algebraic expression you need to prove and you're not sure how. Then you can look at all the things that were given to you in the prepositions and try to use them one after the other till you get the result or an easier form of the problem. In this question you only have 4, maybe 5 things to try, so sooner or later it will work out.

[limber sequoia]

in reply to this

[minor gazelle]

But with that the proof will be too technical

[minor gazelle]

I will follow this route

I am just trying and I am very patient these days 🫣

Or maybe I should just do some research on free monoid 🫣 if that’s more efficient

Though admittedly even I am patient, trying 100 times isn’t a nice experience but fortunately I have some leads now in my head

[rustic patrol]

Thank you,
I will re-try the problem.

[kind temple]

it isn’t extremely technical or difficult. i think this is the “categorical/algebraic” way of thinking of it.

given an element x of a monoid M, there is a monoid homomorphism g : N -> M from the natural numbers N (with addition as the monoid operation) such that g(1) = x; you define g(0) = e, g(n) = x * x * … * x (n times).

given two elements x and y of M, you get two monoid homomorphisms g,h : N -> M with g(1) = x and h(1) = y.
now, if f(x) = f(y), then fg = fh since x and y completely determine g and h.

the proof is almost complete from here, so ill let you fill in the details.

[thorn jay]

this, indeed

the beauty of this proof is that it works with any representable functor

i.e. if F is representable then any monomorphism f gets sent to an injection

in some cases (like for forgetful functors of algebraic structures) F reflects monomorphisms too so if F(f) is injective then f is a monomorphisms

i think this happens when F is faithful? (edit: nlab agrees)

too much category theory for this channel

[topaz solar]

How do I extend a monoid to a group? (And what conditions other than cancellative do I need?)

[delicate orchid]

non-commutative I presume

[rapid cave]

why would commutativity prevent this?

[delicate orchid]

commutativity makes it trivial

[topaz solar]

Indeed

[kind temple]

pretty sure its the same construction as the free group construction, except that you already have a multiplication and identity, so you just need to throw in the inverses.

[topaz solar]

You definitely need it to be cancellative

[thorn jay]

hmm, there's something called a calculus of right/left fractions

[topaz solar]

But is that enough?

[thorn jay]

that's tangentially related I think

[delicate orchid]

https://mathoverflow.net/questions/193400/non-abelian-grothendieck-group

[rapid cave]

just to understanding what you are trying to do, you are trying to add elements to a monoid to make it a group?

[kind temple]

i don't understand this comment, sorry

[thorn jay]

if your monoid is embeddable into a group then we must have that xy = zy => x = z and xy = xz => y = z for all x, y, z

[topaz solar]

If xy = y for all x, we have an issue

Like, we can do that free group thing and get some kind of map M -> (the group)

But it sure ain’t injective, so what exactly do we need for that

Is cancellative enough?

[thorn jay]

no

if only

https://www.cambridge.org/core/journals/mathematical-proceedings-of-the-cambridge-philosophical-society/article/abs/on-embedding-categories-in-groupoids/00AE2B773EDD9C1D725F723177E15EA3

this one might be helpful

[topaz solar]

The test case I wanted was something like Z x N, where (0, 1)(1,0) = (n,1)

[thorn jay]

lmao Malçev pops up

the goat

[topaz solar]

So like a semidirect thing Z \rtimes N

Does this extend? Idk

[topaz solar]

Oh boy, Johnstone

[thorn jay]

😭 is that bad

[topaz solar]

No but not who I expected fr

[thorn jay]

https://www.cambridge.org/core/journals/canadian-journal-of-mathematics/article/immersibility-of-a-semigroup-into-a-group/18B024785167521B73A0EE068013BBEB

this is a preceding one which is more down to earth

i.e. no pure category theory

[topaz solar]

Gotta love when there’s no super easy thing to check

[thorn jay]

that's monoid theory for ya

[topaz solar]

This is probably not impossible to check wrt the conditions but it’s still like vruh

[tardy hedge]

Hey, Algebros! Keep up the Algebra’ing, and ill see y’all later! 👋

[topaz solar]

Some general thing like G \rtimes N in a similar fashion for (injective) endomorphisms sounds quite troublesome to do

When that endomorphism isn’t an automorphism

[thorn jay]

that's a trouble when your quotients aren't decided by a single fiber haha

[topaz solar]

Yeah real

You’d think there’s some that aren’t automorphisms where this works though?

[delicate orchid]

can you not just throw in (0,-1)(1,0) = (-n, -1)? Am I stupid?

[topaz solar]

Well, if it sends 1 to n, you’d probably want a 1/n in there?

So you can probably extend it to like, Z[1/n] or smth since multiply/divide by n is an automorphism?

But nastier endomorphisms of like Z^2 or something are probably cooked

[thorn jay]

the only endomorphisms I like are projections

[twin phoenix]

this may sound stupid but what is the point of this property of a function

[rapid cave]

homomorphism means a function preserving the structure

in the case of groups its preserving the group structure

[twin phoenix]

so it keeps being a group with the same operation after the homomorphism is applied?

[rapid cave]

not exactly

we cannot multiply elements of H with the operation of G

[twin phoenix]

then what properties are preserved?

[rapid cave]

maybe preserved is not the best word for this, but most group properties of one group will translate through a homomophism to a similar property of the othe group

for example the image of any subgroup is a subgroup

the inverse image of any normal subgroup is a normal subgroup etc

[twin phoenix]

the bit i don't understand is why the product of the outputs for all members of G having the same value as the output of the product of the members making it a homomorphism

[rapid cave]

that is the definition

[twin phoenix]

but what bit about it makes it preserve properties

[rapid cave]

this property

of course it doesn't preserve everything

[twin phoenix]

ah i see this makes sense

[rapid cave]

but this is what we are looking for to talk about maps between groups

[twin phoenix]

that's why i was a bit confused how you can have a bijection from one group to another

but i guess they're just sets in a sense so of course you can

[rapid cave]

yes

if there is an isomorphism we can think of both groups as being the "same"

a better phrasing of what I said earlier is probably that the function is compatible with the operations

[thorn jay]

a homomorphism can be seen as some form of interpretation of the group G in the group H

for example, the group of integers has interpretations into any group by choosing an element g and taking the homomorphism f(n) = g^n

that is what this "preservation of operation" actually is; it tells you that your function gives an interpretation of G in H

of course, like any transfer of information, group homomorphisms may not faithfully interpret the group structure, this corresponds to the fact that group homomorphisms might not be injective

why care about group homomorphisms you may ask? well, you take groups that you understand well (like groups of matrices), and by seeing how some arbitrary group can be interpreted in those, say, matrix groups, you can study your group this way

this is what representation theory is

[kind temple]

uhhh, how are you going to show that G/ker phi is a group?

[thorn jay]

show that ker ϕ is normal, and construct an isomorphism between G/ker ϕ and im ϕ

[twin phoenix]

ah i see this is pretty cool since homomorphisms will work for all groups

[twin phoenix]

ye but to me it feels like "it's a homomorphism so this property is true" instead of "this is property is true so it's a homomorphism" (which makes more sense usually)

[karmic moat]

you should care about group homomorphisms because rings are group homomorphisms and a little more duh

[kind temple]

yea, but you are going to use normality to show that the collection of cosets is a group;
G/H is a group (with the usual product) if and only if H is normal.

[twin phoenix]

like the property seems to have no relation to why it gives an interpretation of g in h

[kind temple]

okeh

[thorn jay]

i mean it is the definition of a homomorphism

so i dont quite get what you mean by this

ah okay i see

hm

okay, a function is an interpretation of some set X into another set Y, can you see this?

[twin phoenix]

yeah

[kind temple]

non-invective set functions interpret many elements of X as the same in Y

[thorn jay]

then we add a group structure to these sets. if we want this function to be an interpretation of the group structures, we expect it to be a change of perspective, that is, the function should take the group structure of G and put it in H, right?

[twin phoenix]

ye that bit makes sense

[thorn jay]

so, the change of perspective is seeing the element g as the element f(g). In terms of representation theory, this means representing a group element as some matrix.

in order for the structure to be carried over, we expect that, if a * b = c in the first group, then changing our viewpoint should not make this equation become untrue, else we wouldnt have the same group structure. this must mean that f(a) * f(b) = f(c). does this make sense still?

[minor gazelle]

Thanks so much 🥰🥰🥰

[twin phoenix]

ahhh i see

ok that does make sense a lot

thank you

the definition is the bit that represents the properties

[thorn jay]

and then setting c to a * b we get the usual definition f(a)f(b) = f(ab) (leaving out the multiplications)

[thorn jay]

exactly

you see this for general (universal) algebraic structures too

hell, even at the high level of category theory

[karmic moat]

oh lord here we go

can't go one breath without mentioning UA

like an addict

[kind temple]

nah, nah, im with him lmao

[karmic moat]

i only jest

[thorn jay]

be afraid

[karmic moat]

just shit myself

[crisp siren]

hello

[knotty badger]

I’m curious, what do you mean by this?

[thorn jay]

🙏 🙏

[twin phoenix]

like the definition is equivalent to saying "any two elements of the group, how the operation on them behaves in the group has an analogue in the group after the homomorphism has been applied"

[knotty badger]

Mhm that’s true - but, wasn’t that what the original definition was saying too?

f(ab) = f(a)f(b)

[thorn jay]

in an abstracted way

that abstraction can be hard to parse at first; i know it was for me

[knotty badger]

In the sense that it uses letters instead of words?

[twin phoenix]

i don't think i appreciated what the function actually did when i first saw the definition and didn't see why it mattered why f(ab)=f(a)(f(b)

[knotty badger]

I’m not trying to be facetious to be clear

[kind temple]

i just think it gives some perspective to the symbolic definition of homomorphism, a way to interpret it, if you will

[knotty badger]

Hm I see - so the equation on its own was unenlightening, but actually unpackaging what the equation meant helped?

[thorn jay]

yeah i guess, its like learning the definition of a number again

or addition

[twin phoenix]

well it's more why it works i guess

[knotty badger]

Why it works?

[twin phoenix]

why group homomorphisms have that effect on groups

[knotty badger]

I see - in your own words, why do they have that effect?

[twin phoenix]

with the effect and the definition on its own, it didn't initially make sense how they were linked

but now it makes sense, if the function preserves how the elements of the group interact with each other before the function just in terms of how it is after the function, it preserves the group in a sense

[knotty badger]

Interesting - if I understand you correctly, what you’re saying is the following

What you want group homomorphisms to do is preserve equations, right?

Stuff like - if a^2 b = c d c^2, then f(a)^2 f(b) = f(c) f(d) f(c)^2

[twin phoenix]

no that's not what i meant

[knotty badger]

Hm, then continue

[twin phoenix]

i mean like if u have the reals

u can express them in terms of how they work before you apply the function

and after you apply the function

[knotty badger]

Right - the first is x + y

The second is f(x) f(y)

[twin phoenix]

the operation u apply on the reals is preserved somehow

it's not quite that

[knotty badger]

I suppose that’s the equation f(x + y) = f(x) f(y)

[twin phoenix]

the operation of the 2nd group (post function applied) is just a way of expressing the operation of the 1st group

[thorn jay]

thats a consequence of it

[knotty badger]

Hm, right…

I guess I still don’t quite understand what you mean, but maybe that’s my fault

[twin phoenix]

nah it's most likely my fault

i'm learning this for the first time so i'm not sure the best way to express it formally

[knotty badger]

To be clear I’m not saying your understanding is bad

I’m just really interested in how you went from not understanding to understanding

[thorn jay]

but the initial idea is to want the equality a * b = c to be preserved

[knotty badger]

For context I’m gonna be a tutor for a group theory course next term

[twin phoenix]

like when i understood it, it didn't make sense why i didn't understand it before

[knotty badger]

So I want to have a good understanding of the difficulties people have with the material

[knotty badger]

We have your message here, right?

[twin phoenix]

i think because i saw f(g1) as just being an "output"

not as like after a mapping

even though those are similar concepts

but i didn't realise that the function changes the way the elements of G behave (although i guess it wouldn't if it was an isomorphism)

[twin phoenix]

like this was the perfect way to explain it

[thorn jay]

i dont remember where i first heard it

but it stuck with me

[knotty badger]

Hm ok

I’ll keep that in mind

[knotty badger]

I don’t quite understand this, but maybe I don’t have to for now

Maybe it’s viewing homomorphisms as a change of perspective…?

[thorn jay]

yeah

[knotty badger]

I suppose this makes sense to me because homomorphisms are like functors

Sometimes I do like to think of them as putting on a pair of tinted glasses

Where if you view a group element g through the glasses, it looks like f(g)

And then if applying g then h looks like another element x without the glasses, then putting on the glasses you’d have f(g) then f(h) being f(x)

[thorn jay]

yeah

thats why functors are so important too because there are a lot of times where youd like to interpret some area of math into another to simplify certain problems (e.g. tbe multitude of functors from topological/geometric categories to algebraic ones)

[knotty badger]

I’ll keep this in mind, then

Because I would definitely say that functors generally are a change in perspective

[thorn jay]

a particular fun one being the functor from the groupoid of tame knot diagrams and homotopies to the category of quandles

:>

[knotty badger]

Actually enpeace do you mind if I ask a Q

[thorn jay]

sure

[knotty badger]

I’m gonna be tutoring for a groups and reps course next term

What sorts of confusions do students tend to have

[thorn jay]

hmm, im honestly not sure

ive never TAd myself, of course, and my method of learning groups was.. unique

[knotty badger]

Do you remember what confused you when you started off?

[thorn jay]

it was rather painful to have to come up with visualisations of the quotient stuff myself

ill give you a couple images though

[knotty badger]

Mhm, I see - I’ll keep that in mind

[twin phoenix]

it's like the operation has an equivalent in the 2nd group

[knotty badger]

One way I like to view this is wanting the group operation to be “polymorphic” or “uniform”

[thorn jay]

these are generally very helpful

[knotty badger]

It’s like how we use the same symbol, “+”, to denote addition of naturals and of integers and of rationals and of real numbers and of complex numbers

[delicate orchid]

this is really good

[knotty badger]

If you add 5 to 3, the result is 8 no matter if you consider these as naturals, or integers, or rationals, or reals, or complex numbers

[thorn jay]

and of course my beloved

[karmic moat]

did you draw these

[twin phoenix]

ye sometimes u unconsciously read the group operation signs as being actual multiplication/addition

[karmic moat]

these are quite nice

[thorn jay]

the universal algebraists actually gave you something useful

[twin phoenix]

i think if u're looking for a reason why i was confused, it might actually have been just that

it isn't a product, it's the operation

[thorn jay]

no, theyre from A Course in Universal Algebra

[karmic moat]

i see

ruined..

[knotty badger]

It’s like how there’s a sense in which + for naturals is a different function than + for integers

[thorn jay]

i think these pictures should be standard in any algebra textbook

[knotty badger]

But if a + b = c as natural numbers, then a + b = c as integers too

[karmic moat]

tbh i never really thought about visualizing quotient groups

[knotty badger]

Right yeah

Often the group operation is some kind of “composition”

In that the group elements can be interpreted as functions, and the operation as function composition

[twin phoenix]

well aren't the natural numbers with addition a subgroup of the integers with addition?

[thorn jay]

theres a madman in universal algebra who studied congruences (the stuff you quotient by) in a geometric manner

[knotty badger]

A submonoid actually

Since you don’t have inverses

[karmic moat]

geometry prevails once again

[thorn jay]

like the equivalence classes were treated as lines

[twin phoenix]

ah ye ofc u can't do subtraction since group wouldn't be closed

[karmic moat]

ohhh that's interesting

[knotty badger]

What you can do is like

If $\iota : \mathbb{N} \to \mathbb{Z}$ is the inclusion map

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

Then $\iota(a) + \iota(b) = \iota(a + b)$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

On the LHS you’re adding integers, on the RHS you’re adding naturals

[glad osprey]

@knotty badger do you think it helps pedagogically to formulate the homomorphism condition as a commutative diagram? I guess the issue then is you need to answer why do we want this diagram to commute

[delicate orchid]

we're like 2 steps from Lawvere theories here so I approve

[knotty badger]

Well I’m not sure - I didn’t want to interrupt enpeace when they were explaining

[thorn jay]

approving Lawvere's stuff but hating on UA i see how it is..

everyone here is a hypocrite i am starting my villain arc

[glad osprey]

when I learned about homomorphism, I didn't really understand why we want f(ab) = f(a)f(b), but I didn't really question it either. I guess I just accepted it as something that might be useful later

[delicate orchid]

I like Lawvere's stuff because it's UA through a categorical lens

[thorn jay]

eh, thats fair

[knotty badger]

Hm maybe one way to say what you were saying earlier is

If you want to compute f(a) f(b) in the new group, you can instead do ab in the old group, and apply f to that?

So the new operation is “determined” in terms of the old operation

[thorn jay]

i wonder, do congruences exist for models of lawvere theories?

[delicate orchid]

wdym by congruences? the shit you can quotient out by?

[thorn jay]

i dont know what the meta is for them

[thorn jay]

yeah

[knotty badger]

There’s a categorical notion of congruence

It’s an internal equivalence relation

[thorn jay]

its basically the central object of study of UA

[delicate orchid]

I know about them for posets (other than the obvious derivatives of groups and rings) but that's about it

[thorn jay]

i know i guess i meant more like are they as important for the study of lawvere theories

[knotty badger]

This I’m not sure

[thorn jay]

because basically anything you might care about for UA boils down to studying some congruences

hey if i get into lawvere theories i might even get a position in academia sometime

[delicate orchid]

why would you want that

[karmic moat]

lawvere theories or a position in academia?

[delicate orchid]

the latter

[karmic moat]

real shit

[thorn jay]

money

i wanna live

[karmic moat]

>position in academia
>money

[thorn jay]

its better than not having a job at all

[delicate orchid]

there are better ways to make money. Like fraud

[thorn jay]

im gonna steal the moon

[karmic moat]

osrs flipping makes me a lot of money on osrs

[karmic moat]

this guy will never steal the moon

[thorn jay]

🌑

🤌

i have stolen the moon

[karmic moat]

man spitting out cereal.png

too lazy to find it

[thorn jay]

i feel that

[topaz solar]

Also btw necessary and sufficient for G \rtimes N extending is being an injective endomorphism I think

For nastier structures idk tho

[thorn jay]

how to prove a theorem

step 1: assume everything is nice

step 2: trivial

[south patrol]

Hey I mean it often works

...If you get that far

[karmic moat]

1% of the time it works 100% of the time or however the saying goes

[south patrol]

Lol I guess I mean like

permanent position in academia can often be good like in US ig

non-permanent at least here is a bit different

[knotty badger]

this sounds like the joke of mathematicians always rigging their problems to be solvable

[karmic moat]

tough job market in the US too

[thorn jay]

it does feel that way sometimes

[karmic moat]

pay's not very good either

[thorn jay]

in response to this

[karmic moat]

talked to a prof and he was like "yeah you shouldnt do math for the money, there's not much of it here"

[thorn jay]

doing it for the love of the game >>>

[karmic moat]

to be precise, academia

forgot to include that

not for math

well depends heavily on what field you teach and where you are

if you do the right field and at the right school you can make a lot

it still counts as academia

[topaz solar]

Real, I’m trying to assume things are not and somehow derive that it’s nice anyway

[karmic moat]

i mean even teaching at smaller schools, if you teach the right thing then the pay is big

[thorn jay]

foolish

[topaz solar]

Clearly

[karmic moat]

math is particularly bad because a lot of research in it seems useless to the real world

[topaz solar]

But yeah if we took like, an n-tuple \bar e of injective endomorphisms of G, I dunno but maybe there’s always a way to extend G \rtimes_{\bar e} F_n^+ or some similar thing like G \rtimes <\bar e> where the F_n^+ is free monoid or <> to denote sub-monoid in End(G)

[karmic moat]

sure there's still a lot of it which is relevant but there's a lot of it which doesn't seem relevant

[topaz solar]

If it does extend you’d probably get like G \rtimes F_n or similar based on extendibility of that monoid <\bar e> to a group

[karmic moat]

im not saying applied math is easier or harder lol

[thorn jay]

hmm

[karmic moat]

i'm not in a position to comment on whether it is or isn't

[thorn jay]

I'll give it a thunk maybe

[karmic moat]

difficulty is subjective

[thorn jay]

if the monoids carry extra structure then this question might become easier

[karmic moat]

because it's more immediately relevant to the real world than algebra or something like that

[thorn jay]

for example, what if we think about rings

[topaz solar]

I think you’ll be able to do it if <\bar e> extends to a group, and in which case it’s gonna be a semidirect, the issue is you can’t do it as G \rtimes F_n

You need to like, include pre images

So you need to like, expand G some

[thorn jay]

what is G here?

group?

[topaz solar]

Yeah

By like, taking a direct limit based on e_i G < G

And similarly for each e_i in order, which should work out fine if it extends to a group?

[thorn jay]

wdym by this?

[topaz solar]

Like we have an embedding of G into G from the G -> e_i G \leq G

Since injective endo

Lay this into itself infinitely many times, then e_i is an automorphism on this limit

The issue is handling the other endomorphisms

Since if you have two of them, you get a terrible tree thing instead of a straight line when trying to have both extend to an automorphism of G at once

[thorn jay]

haha but

what if you assume they commute

😎

fixed

[topaz solar]

Truly

The issue should maybe probably be fine if these extend to a group, which is a necessary condition at least

[thorn jay]

you mean the <e_i>?

[topaz solar]

Yeah

[lime magnet]

hot take: using algebra to prove number theory facts is a lot more fun than having to use number theory facts to prove things in algebra

(e.g. Euler's theorem, Wilson's theorem)

[earnest delta]

If H is the subgroup of a group G then

Ha=Hb if and only if ba^-1 belongs to H

Is this true?

[lusty marlin]

Yes

[earnest delta]

ab^-1?

[proud vigil]

ab^-1 in H and ba^-1 in H are equivalent

you also can just write it as

Hb = Ha

and then it's immediately true from symmetry

[cedar vault]

Subgroups are closed under inversion.

[twilit wraith]

Not gonna lie I personally find the whole ab^-1 one-step proof thing to just be kinda stupid

But yes

[limber sequoia]

You can think of it in terms of the G-set G/H if it helps.

[lime magnet]

i'm sure some - or many - would disagree: people have been studying the integers for centuries before groups, rings, and fields were conceptualized, so in a sense, it's more natural to study the integers first.

i guess what i'm thinking (and i'm not sure if this is right), is that math pedagogy where we learn more general/abstract ideas later is somewhat questionable. is it a waste of time to study the riemann integral for so long when it could be added as a special case of lebesgue integral? to an extreme: why not learn category theory or (at least naive) set theory first and then study specific areas later?

perhaps it's because the more abstract ideas aren't immediately applicable, and at most schools, sequences of courses need to be prepared to suit those who want to go into more depth as well as those who only want to learn, e.g., calculus.

[lime magnet]

in my class, $ba^{-1}\in H$ is defined first as an equivalence relation on $G$, then cosets are defined as equivalence classes, and then we prove the equivalence that you've mentioned.

[cloud walrus]

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

[lime magnet]

hmm, 🫤 latex bot

[rocky cloak]

it doesnt like mixing markdown and latex

first

[lime magnet]

the equivalence is clearer with the intermediate $\exists h\in H$ such that $1_Ha=a=hb$

[cloud walrus]

ransom

[lime magnet]

i did a thing!

[elfin wraith]

I think the specific difference here is that the number theoretic proofs you see in elementary number theory are like a half step away from the more general proofs you can write with some very simple group theory, and it’s pretty natural to do these things together

In general it’s certainly worth studying the less general less abstract objects first, and tbh in this case you still have, you spend all of your school days learning about the integers. You need to understand the context and the need for the generality, and it’s quite frankly usually easier to visualise. It’s far easier to draw some rectangles under a graph and explain integration to someone than have to start talking about measure spaces, and it would still feel pretty unnoticed even if you could convince them. Having the context that not all functions are Reimann integrable, even ones which we kinda expect to integrate, motives the lesbeque integral

Yes you can be kinda vague about measures and introduce the lesbeque integral as a bit of a special case, but I’ve always found this to be harder than it’s worth

[knotty badger]

Does it actually motivate the lebesgue integral

[elfin wraith]

To me it does, or at the very least it motivates the need for a more carefully thought out definition

[knotty badger]

The Riemann integral seems pretty well thought-out though

[quiet pelican]

It’s kinda your first indication that the Riemann integral doesn’t play nice with limits
And the fact that a lot of basic interchange-limits-and-integrals theorems are much nicer from the lebesgue PoV is to me the big motivation for it

[knotty badger]

Yeah this is more convincing to me

[south patrol]

You need to be able to multiply by R

[quiet pelican]

We want the module generated by m to be an R-module

[south patrol]

And hence you get (at least) Rm

And one then checks this is like the smallest ting

[elfin wraith]

I think it’s both of these things, but in any case I’m making a more general point

[karmic moat]

The only motivation I had for lebesgue measure and stuff was probability class where every proof the prof presented was either “true/false bc it has/doesnt have lebesgue measure 0” or some weird convergence in measure thing

[tardy hedge]

last year i thought i was gonna take measure theory so i was studying it a bit over the summer

My God. I hated it

[elfin wraith]

Yeah I was thinking about taking some probability classes next semester for employment reasons until I looked at the course notes and remembered how much I do not enjoy analysis

[cloud walrus]

mq

[karmic moat]

when context is known, authors tend to drop the subscript

So there is likely no difference but you should check, the authors usually remark this kinda thing

Well I’d need full context then

Can you send the whole page(s)?

[quiet pelican]

The latter is probablt abelian group homs

Yeah

[karmic moat]

Yeah

[quiet pelican]

The latter doesn’t have to respect the R-module structure

All R-module endos, yeah

It’s my usual way to say yes in emoji form

[white oxide]

Does this work for iv? iv. Suppose that $M_{\mathfrak{p}} = \bigl(\bigoplus M_i\bigl){\mathfrak{p}} \neq 0$. Let $\overline{x} \in M{\mathfrak{p}}$. If $(M_i){\mathfrak{p}} = 0$ for all $i$, then $x = m{\alpha_1} + \dots + m_{\alpha_n}$, so that $\overline{x} \in (M_{\alpha_1} + \dots + M_{\alpha_n}){\mathfrak{p}} = (M{\alpha_1}){\mathfrak{p}} + \dots + (M{\alpha_n}){\mathfrak{p}} = 0$. Therefore, $M{\mathfrak{p}} = 0$, a contradiction. If $(M_j){\mathfrak{p}} \neq 0$ for some $j$, then $M{\mathfrak{p}}\neq 0$, for it contains the nonzero image of $(M_j)_{\mathfrak{p}}$.

[cloud walrus]

okeyokay

[south patrol]

It seems what you are saying is equivalent to saying localisation commutes with direct sums - indeed you use this implicitly

[thorn jay]

because R might not be the integers

[south patrol]

Or 0

[thorn jay]

I rarely consider 0 to be a ring

[south patrol]

Sad!

[thorn jay]

in ring theory the only purpose it serves is as an annoying special case

[vapid vale]

sure looks like one

[thorn jay]

I should rephrase: I rarely consider 0 when thinking about rings

[vapid vale]

it was a shape joke

[thorn jay]

aguh

too algebra brained atm

spent the past 3 or so hours butting my head with a fuckass badly written book which is basically the only resource about the topic out there

[tidal schooner]

Spiritual successor to #algebraic-geometry message

[wide spade]

what book/topic ?

[thorn jay]

On The Structure of Finite Algebras by Ralph McKenzie and David Hobby

it's a book on some universal algebra and all the math in there is amazing, it's just so stinking hard that the book goes wayy too fast

and barely gives motivation for anything

[wide spade]

i see

[rapid cave]

You can help me in #algebraic-geometry instead xD

[lime badge]

I have a silly question.

Consider the polynomial ring R = k[x,y]/I where k is a field. Let c = f(x) and d = g(y) be univariate polynomials in R

What conditions on I would I need to ensure that (c) \cap (d) is contained in (cd)?

My intuition is that the ideal I should not "mix" x and y, i.e. it should be of the form I = (p(x)) + (q(y)). Is that correct? If so, how can I formalize that?

I know there's the vanishing of Tor_1(R/I, R/J), but that's just restating the fact that I want

[thorn jay]

a general sufficient condition for I ∩ J = IJ is I + J = R

iirc

[lime badge]

But it isn't necessary right?

[south patrol]

More generally like (I \cap J)(I + J) = IJ i think

[thorn jay]

what

[south patrol]

Uh maybe after taking radicals

[thorn jay]

oh i see

[karmic moat]

this is only an inclusion i think

[thorn jay]

lol im slow

[south patrol]

Ah no yeah it is an inclusion lel

or smth

but true for Z, famously lol

I think more generally true for Dedekind domains (by imitating the proof for Z but using unique factorisation of ideals)

[karmic moat]

for a general ring you should have $\text{rad}((I \cap J)(I+J)) = \text{rad}(I \cap J) \cap \text{rad}(I + J) = \text{rad}(IJ) \cap \text{rad}(\text{rad}(I) + \text{rad}(J))$

i hate you texit

[south patrol]

(nice proof is that if I+J=R then write 1 = a + b for a in I, b in J. Clearly IJ is contained in I \cap J, and conversely if c is in I \cap J then c = ca + cb and each of ca, cb is in IJ lol)

[cloud walrus]

anamono

[thorn jay]

that is very cool

i dont think i appreciate coprime ideals enough

[quiet pelican]

Yup that works

[spark veldt]

Hi I think I'm just really confused with the wording more than anything.

Does this mean that given x,y,z in G, then choosing any two ensures they commute?

[quiet pelican]

Given x, y, z, there is some pair that commute

[rapid cave]

Its means there is no x,y,z where any pair do not commute

[quiet pelican]

That's a nice question

[rapid cave]

Indeed. Short & Sweet

[spark veldt]

ahh i see okay thanks!

[south patrol]

i don't think this is on you lol

[elfin wraith]

It took me a couple of read throughs lol

[glad osprey]

yeah, why did they use a double negative instead of just saying this 😭

[south patrol]

I think solving the problem takes less time than reading and understanding the question

[glad osprey]

Fixed it:
Let G be a group with the property that there do not exist three elements x, y, z in G, no two of which commute. Prove that there does not exist x, y in G such that x and y does not commute

[karmic moat]

not enough negatives, i'd hardly call that a fix

[merry falcon]

It’s like I unlock a new channel every time I learn new math (just started group theory)

[quiet pelican]

Group theory is the best maths

[elfin wraith]

“Group theory is the best maths” - woman who has not yet discovered ring theory

[quiet pelican]

I discovered ring theory then realised groups are better

[elfin wraith]

See that just doesn’t sound fun, but I’m glad you’re happy

[rapid cave]

Fields are amazing

[merry falcon]

So far so good

[quiet pelican]

Groups > Rings = Fields

[thorn jay]

"group theory is the best maths" - woman who has not yet discovered universal algebra

[quiet pelican]

Universal algebra boring logicy

[thorn jay]

only if you let it be

commutator theory and tame congruence theory are decidedly not logic

the latter is very lattice theory though

[quiet pelican]

Lattice theory scare mico

[thorn jay]

its not that scary

until it is

i have noticed a lot of theorems have a nice categorical interpretation though

[tardy hedge]

in commutator theory? ong wtf is ts sht

Tame congruence theory

like bruh

(that was your cue to tell me about commutator theory cuz im in the mood)

but ngl id prob be interested for like one sentence tho

[thorn jay]

this is the shit ive been STRUGGLING with

today

[tardy hedge]

Bruh

i should try learning it then cmon

Easy

What the hell is tame congruence theory man

I am preparing myself

[thorn jay]

commutator theory is a little easier to explain; basically you have the intuition that a "nice" algebra should be one where you can essentially cancel terms. that is, if t(x, a1, ..., an) = t(y, a1, ..., an), then x = y for any term t. For example, this holds in modules/vector spaces, and it so happens that we use these as nice algebraic objects to encode information (i.e. homology).

It just so happens that under surprisingly loose conditions, any algebra that satisfies this "niceness condition" must be a module, in some universal algebraic way. This is a pretty old and fairly classical result, and commutator aims to generalize this using the definition of a commutator of congruences, which essentially extends this "cancellation" idea and generalizes it to congruences in some way.

and it turns out that it is an extremely powerful tool

[tardy hedge]

what is a term?

[thorn jay]

something something looking "locally" using idempotent polynomial operations

[thorn jay]

youve got your basic operations, right

and a term is basically any composition of these operations

so a term for rings would be a polynomial with integer coefficients, and a term for groups would be a word, a term for vector spaces a formal linear combination etc

[tardy hedge]

oh ok i was thinking like in the polynomial x^2+2 the terms are x^2 and 2 or something lol

but ok that makes sense

[thorn jay]

ye

[tardy hedge]

Tbh that is actually kind of interesting

But i would still be hesitant to learn that stuff because of how few other ppl learn about it

[thorn jay]

thats fair, but really any modern universal algebraist knows it, its just that there arent many of these online

[tardy hedge]

what schools have a uni algebraist group.. if any?

[thorn jay]

our library seems to have plenty of books on it though 👍

[tardy hedge]

Or is it just more like individual professors do their thing

[thorn jay]

hawaii

[tardy hedge]

Oh wow hahaha

thats fun

bro gona be sippin pina coladas

gettin a nice tan

[thorn jay]

it's where McKenzie (the guy who wrote the commutator and tame congruence books) does research i think

[tardy hedge]

I wonder if my supervisor is aware of universal algebra

[thorn jay]

ask him :0

[tardy hedge]

her. but sure I will actually haha

meeting next week

ill let u know lol

[thorn jay]

oh im so sorry lmao

[elfin wraith]

https://tenor.com/view/sexism-awareness-anti-sexism-gender-equality-fighting-misogyny-confronting-sexism-gif-10672709015865364186

[tardy hedge]

Hahaha np

[thorn jay]

the inherent misogyny of mathematics

BRO SENT IT TWICE

[tardy hedge]

😂 😭

[elfin wraith]

Discord moment

[tardy hedge]

thats what made it funny to me 😂

the spam

[thorn jay]

its like alarms blaring

[tardy hedge]

i thought bro sent it 3 times

[tardy hedge]

Fr

[thorn jay]

guys i promise im not sexist i love women!

[tardy hedge]

its ok me too

[thorn jay]

i even have a woman friend

[tardy hedge]

lucky you

[thorn jay]

whom i am very respectful towards

[tardy hedge]

im out there shooting my shots

[thorn jay]

love will find its way to you

on a logical redstone discord server apparently

[tardy hedge]

dam

I did meet a friend i met from club penguin last year

online friends becoming irl friends is funny

[tardy hedge]

like. I knew her since 2013 basically

met on club penguin

club penguin is becoming unc status now

[thorn jay]

it js

i dont even know if it still exists because it got shut down once

or twice

[tardy hedge]

it got shut down once in 2017, after that has been private servers that fans have made

they usually shut down and open again

disney takes them down sometimes

[thorn jay]

speaking of taking down

have you heard of the insanity that nintendo's up to

[tardy hedge]

yea theyre getting hated on nowadays 😂

theyve fallen pretty hard

[thorn jay]

theyve got a patent on the game mechanic summoning creatures that fight for you

[tardy hedge]

oh yea i think i saw video on that

[thorn jay]

thats fucking insane

[tardy hedge]

nintendo is just silly they always do stuff like that

[thorn jay]

this will start a patent war on game mechanicd

[tardy hedge]

theyve gone way more corporate in recent years than they were before tho

[thorn jay]

and of course indie games will suffer the most

[thorn jay]

gotta love capitalism

[tardy hedge]

Yes i love it i love it 😻

(I need a job)

(financial industry)

[thorn jay]

all hail capitalism

all hail capitalism

[spark veldt]

Hi I’m not really sure how to show that something is a permutation group. Can i get some help on this?

[noble nexus]

given any set, the permutations of that set (bijections from the set to itself) form a group under composition. A permutation group is just any subgroup of the group of permutations of some set

[thorn jay]

basically, all you need to show is that R and L are closed under composition and inverses

composition being composition of functions

[spark veldt]

ahh i see, thanks blake and enpeace

[elfin wraith]

Yes

[south patrol]

But yeah I mean I suppose what you want to say is like

You can often give the abelian group R the structure of an R-module in many ways

But when people say like "view R as an R-module" or whatever, they mean the standard structure

Not sure what you mean

Only if you work with nonunital conventions

But there are still similar things in the unital case, like

view k[x] as a k[x]-module by like x acting as zero

(and k as usual)

[thorn jay]

not if you dont require that

a nonunital one

even if R and S both have unit, there may exist rng homomorphisms not sending 1 to 1

nope

thats only guaranteed if S is an integral domain

[south patrol]

Pain

[thorn jay]

no

let R = Z and S be any ring with an idempotent e bot equal to 1 or 0, i.e. e^2 = e. Then there is a unique map Z → S sending 1 to e

[south patrol]

(E.g. Z -> Z x Z sending a to (a,0))

Nice example

[thorn jay]

ye

[south patrol]

I meant your example was nice aha

[thorn jay]

no

oh, yeah

misread

im tired

yes

this is true

not going to classes today all of them are boring

but thats not a very useful fact if your range doesnt have special conditions like no idempotents or whatever

i have never seen someone use O for homomorphism

crazy convention

[south patrol]

Real

[cloud walrus]

mq

[thorn jay]

you can also use f

for _f_unction

[rapid cave]

O is for
Omomorphism

[thorn jay]

o is for omOri

[rapid cave]

Amazing game

Made me cry

[thorn jay]

havent played it yet but i will some dau

[rapid cave]

Highly recommended

[thorn jay]

i mean if I like deltarune theres a high change ill like omori

[rapid cave]

Yes they have a lot in similar

[thorn jay]

have a lot in similar is one ive never heard before

yes

[rapid cave]

... whats funny?

I don't understand

[thorn jay]

you merged the phrases "are very similar" and "have a lot in common"

[rapid cave]

Oh xD

It sounded normal to me

[thorn jay]

yeah

its a biproduct, meaning that finite products and sums coincide