groups-rings-fields

[rocky cloak]

"groups are like symmetries" is just asking you to imagine a representation of a group

[hidden wind]

whaaat

[coral spindle]

Gonna tell all my group theorist friends that they do their work so we can understand the reps better

[gilded sundial]

What can rings and fields be used for? I’m gonna take it to get a math degree however I’d rather get a stats degree since the rings and groups are seemingly useless. There only somewhat useful in quantum mechanics I’ve heard, but have no relation to financial math or data science (from what I’ve heard), which is what I want a career in…

[mighty kiln]

Mostly pure math

[elfin wraith]

The Brouwer fixed point theorem is useful for economics so everyone should learn some algebraic topology really

[amber burrow]

is just memoring random actions/subgroups a good problem solving technique?

like, i cant tell what is meant to teach vs an actually good idea

[gilded sundial]

Should I take on a stats degree just because I want to focus solely on applied mathematics. I mean Inknow nothing about coding in R and two of the class are a kin to internships, so I thought I would be able to just learn via practical experience the coursework composited in the stats degree, however, AI told me that stats would be nice for my career ambitions…

[amber burrow]

in dummit and foote 4.4, a common strat is to use that N_G(H)/C_G(H) is a subgroup of Aut(H)

[elfin wraith]

Developing a solid example space is an important part of understanding a subject, you should certainly be aware of common techniques and illustrative examples

[amber burrow]

and in the exercises you look at the ation of G/A on an abelian group A

[amber burrow]

yeah - i'm just wondering if these are actually common techniques

[coral spindle]

Sure, if you have to memorise this. But I must say that I don't need to think about this too much – to me this doesn't need memorisation. And yes, these are common ideas.

[velvet hull]

not random actions, but useful lemmas, which is what that is

[amber burrow]

not neessarily memorization

[elfin wraith]

Learning common tricks is definitely part of learning a subject

[coral spindle]

N_G(H) obviously acts on H by automorphisms, and we kill out the kernel of the obvious map N_G(H) -> Aut(H) -- which is C_G(H) -- to get a subgroup. This just works out.

[amber burrow]

i get why its a thing

[elfin wraith]

It’s like realising adding 0 in useful ways is essential to analysis

[amber burrow]

its inner automorphism

[coral spindle]

No they're not necessarily inner automorphisms

[amber burrow]

ooh thats an insightful analogy

[coral spindle]

But I can see what you are trying to communicate

[amber burrow]

well yeah for G it would inner automorphisms

like G/Z(G)

i guess one big thing i do is like, im not always sure what steps lead to an actually new insight

like in competition math geometry, a think i often do is to use theorems like power of a point, and then when i get stuck i try everything (like similar triangles), only to realize that yields redundant information since power of a point is just similar triangles

[elfin wraith]

Why are these the possible cycle types? is it just that all of the 4 cycles with a single fixed point can be written as the product of 2 2-cycles?

[woeful sage]

arent these just partitions of 5?

[elfin wraith]

Possibly, im horridly bad at combinatorics and it leads to me struggling with S_n lol

[woeful sage]

relatable!

I really wanna go through "A walk through combinatorics" by Miklos Bona when I find the time, I'm too bad at combinatorics

But yea I'm pretty sure cycle types in S_n are just partitions of n, I don't know how to prove that right off the top of my head though

[alpine plank]

same tbh I can't prove anything about binomial coefficients that doesn't immediately follow by writing it's definition in terms of factorials

[elfin wraith]

Well i mean the possibilities are clearly just the partitions

[woeful sage]

Right yeah

[alpine plank]

(evidence is in #discrete-math save me)

[elfin wraith]

but im asking about why we dont dont have any of type (4,1)

I think its just because we could write all of those as type (2,2,1) but im not sure

[velvet hull]

..a 4-cycle is not the same as 2 2-cycles?

[woeful sage]

we do though (1 2 3 4)

[woeful sage]

is this table complete? because as it stands there seem to be 24 + 20 + 15 + 1 = 60 elements in S_5

but there are 120 elements

[elfin wraith]

Oh ffs

The table only lists the even ones, its for A_5

[rocky cloak]

If your main interest is data science it could make sense to study statistics as that's what data science is. But I'm not sure you should tie it to the significance of rings and fields.

There's lots of math, like measure theory, dynamical systems, numerical analysis that underpins financial math and data science.

As for job prospects I think it depends on a lot of factors, and exactly which concepts are directly applicable on paper might not be the most relevant.

In many cases it might be best to choose depending on what you find interesting. Whether that be math or stats

[woeful sage]

that makes more sense

[thorn jay]

this is the realest, cool constructions in math are cool intrinsically

[woeful sage]

also isn't it kind of like a generalization of algebras over a ring?

[elfin wraith]

I was really confused about why there wouldnt be any for (4,1). In any case this highlights the fact that I am inexcusibly bad at both basic combinatorics and working with symmetric groups and definitely need to work on that

[woeful sage]

and algebras are inherently cool because operator algebras are a thing

[rocky cloak]

There's 30 of type (4, 1)

[thorn jay]

not really

[coral spindle]

algebras over a ring are a strict generalisation of group algebras lol

it's like literally the opposite way around

[rocky cloak]

It's not a generalization, it's just what it is

[thorn jay]

yeah

[woeful sage]

oh

[coral spindle]

(ofc I have no idea historically if this is where the idea came from, and I doubt it)

[gilded sundial]

I find applied mathematics the most interesting, which involves statistics. I just heard that group/ring theory aren’t particularly applicable in financial mathematics or data science neither machine learning, which are the fields I want to pursue. I’ve heard that group/ring theory aren’t particularly more attune to theoretical mathematics, cryptography, and quantum mechanics, however, I foreseeably won’t be spending my entire life on that. I mean I’m just weighing the career prospects in terms of what employers may think when they see a stats degree vs a math degree in the job application repository.

[elfin wraith]

5 choices for your first element, then 3! Ways to pick the next 3 in the cycle?

[woeful sage]

Algebraic topology is used in data analysis

[rocky cloak]

5 choices for which element is fixed, then 3! ways to cycle the rest

[woeful sage]

like Topological data analysis

and for algebraic topology you need algebra i.e group/ring theory

[woeful sage]

one day I will learn this, it seems so cool

[elfin wraith]

Persistent homology is cool but I think not super applicable because it’s so slow and complicated

[woeful sage]

I see

[elfin wraith]

Robert Ghirsts book is nice though

[woeful sage]

Robert Ghrist the GOAT

He even started a Topological data analysis series in his youtube channel

[elfin wraith]

I essentially only know about TDA because I got put on a homology/combinatorial comalg project with a stats student who knew 0 algebra so I don’t know it super well but it seems pretty cool

Discrete Morse theory and all of that seems quite interesting

[wintry sluice]

is there a term for the set of elements that are mapped to 1 in a ring homomorphism?

[woeful sage]

kernel

well

kernel maps to zero

[tardy hedge]

Ive never heard it being given a name

[hidden wind]

kernel of the map on the multiplicative structure?

[next obsidian]

The fiber of 1

[glad osprey]

isn't it just the kernel + 1? atleast for unital ring homs

[thorn jay]

It is

[tardy hedge]

Y it be called da kernel doe

[stone vault]

It's part of this extended "fiber" "germ" "sheaf" "stalk" metaphor

[elfin wraith]

I have somehow never connected the dots between fibre kernel and sheaf

Probably because I don’t really know about sheafs but I just never clocked they were the same metaphor

[glad osprey]

I want to count the number of monic irreducibles of degree p over GF(q), where p and q are primes. I've looked up the solution, I just want to make sure I understand it:

So the idea is that GF(q^p) is the splitting field of x^(q^p) - x over GF(q), and every linear polynomial and every monic irreducible polynomial of degree p is a factor of x^(q^p) - x, and those are its only factors. Therefore we can just add up the degrees to get q + p*k = q^p where k is the number of monic irreducible polynomials of degree p. I think I see why every irreducible of degree 1 and p are factors, but why are there no other factors of x^(q^p) - x? If f(x) is an irreducible factor of degree n, then GF(q)[x] / (f(x)) is an intermediate field of degree n between GF(q) and GF(q^p), but [GF(q^p) : GF(q)] = p is prime, so n = 1 or n = p, is that correct?

[tough raven]

Yes. The key is that p is prime. You could also do this for general degree n, getting ∑_{d ∣ n} d n_d = q^n, where n_d = # irreds of degree d. This can be solved by Möbius inversion (and then dividind by n) to get n_n = 1/n ∑_{d ∣ n} mu(n/d) q^d.

[glad osprey]

I see, thanks

[night tartan]

<@&268886789983436800> spam

[faint forge]

This isn't really groups, rings or fields, but this feels the right channel to ask: suppose we have a nonempty set $A$ with a binary operation $\cdot$ and an unary operation $'$ such that $(x'\cdot y)\cdot z = y$ for all $x,y,z$ (we're not assuming any kind of associativity, commutativity etc). I know that if $A$ is finite, then it must be a singleton (which isn't hard), but what's the answer without the finiteness requirement? Is there an intuitive example of a nontrivial structure that would satistfy this?

[cloud walrus]

Outsider

[rocky cloak]

Let your set be the natural numbers and define
x' = 0 for all x.
0 * x = x+1
x * y = x-1 for x not 0

You can even enlarge this by adding in more elements and just defining 0 * x = x and x*y = x for the new elements x

[faint forge]

neat, thanks! are you able to explain how you came up with this? what's the motivation?

[arctic trail]

is it really supposed to be (x'y)z = y?

(Genuine question, making sure no typos cause it's very restrictive which in some sense seems to be the point)

[faint forge]

yes, that's what's in the problem set (not mine, but I got curious)

the general context seemed vaguely adjacent to Boolean algebras

but I don't see the connection

[arctic trail]

Neat

[rocky cloak]

So f(y) = x' y would need to be injective. And because
(x' y) z = y, if x' y was equal to x' that would probably be a problem. So okay, what's a function that's injective but not surjective? Successor on N. Does that work? Yup

Why did someone ask this question? meh it works

[arctic trail]

I wonder if you can like classify the type of algebra that satisfies this

Clearly the function f: A -> P(A) given by f(y) = A'y must be injective.

[rocky cloak]

I guess it will be some kind of tree structure where ' takes you somewhere to the bottom of the tree, and then
x' * moves you "up" (possibly in different branching paths depending on x') and then *z will take you down reversing whatever x'* does

[arctic trail]

I didn't pay attention to your example, but I wonder if there are commutative semigroup examples

[amber burrow]

Is there a smart way to show that $(\mathbb{Z}/24\mathbb{Z})^\times$ is elementary abelian besides explicity writing an isomoprhism

[cloud walrus]

donut123

[amber burrow]

cuz i mean if you write out the elements its obvious

[rocky cloak]

They're very much not commutative

[arctic trail]

Is there no non-trivial commutative model?

Oh wait

nvm

y = (x'y)z = (x'z)y = z leads to a trivial model

[rocky cloak]

Yeah

[arctic trail]

assuming both

what if you only assume one (like associativity)

y = x'yz
yw = (x'(yz))w = yz
forces yw = yz.
So multiplication is trivial

[rocky cloak]

Z/24 = Z/8 * Z/3 so if you know that (Z/8)^* = C2xC2 that would be enough

[arctic trail]

so (x'y) = (x'z) forces y = z

so no non-trivial model

[coral spindle]

Z/24Z = Z/3Z x Z/8Z therefore (Z/24Z)* = (Z/3Z)* x (Z/8Z)* -- exercise if this is not obvious to you! Then you can just calculate what jagr said above. This still does require you to know (Z/8Z)* is an elementary Abelian 2-group, but you should hopefully know that (Z/pZ)* is iso to C_{p-1} when p is prime.

[elfin wraith]

|| Is it just by the CRT? It doesn't feel like enough but we have n = \prod{i=1}^t p_i^{s_i} so CRT gives Z/nZ = (+){i=1}^t Z/p_i^{s_i} Z, then units must be mapped to units. I feel like im being silly and missing a step though||

[coral spindle]

So the first part is by the CRT

But (R x S)* = R* x S* for any rings S and R

N.b. this is a true blue equality, not merely an isomorphism!

Is it possible the notation is confusing you? By R* I mean the group of units ofc

[elfin wraith]

Oh is that literally just by the definition of the direct product then? Im possibly making this harder than it needs to be haha

(its been a long day)

[coral spindle]

Yeah, try it out!

It really does just follow pretty immediately

[elfin wraith]

Yeah no I see that haha, its just by how we actually define multiplication

[coral spindle]

Note: this is the 'real reason' that Euler's totient function is multiplicative

[elfin wraith]

Is there a "fake" reason generally presented? I think ive erased ENT from my brain

[coral spindle]

I've seen some awful direct proof by counting things coprime to n or whatever

some kinda direct counting thing

It's a shite explanation

This is the real reason, in every way that I can imagine

[elfin wraith]

Wait no I remeber the proof we gave in my ENT course because it was fucking insane lol

But yes I agree just defining the totient as the size of the group of units then using the definition of the direct product is much nicer

[glad osprey]

Maybe ENT should be taught after abstract algebra

[elfin wraith]

It was

We had already taken a course on group theory and on ring theory prior to that class

Page 1 of the notes lol

[coral spindle]

This is vile

[arctic trail]

Hmmm?

Isn't there another one using convolutions?

[elfin wraith]

Yeah it got me thinking about what the proof we seen was and it made me remember thinking it was absolutely wild

[coral spindle]

I'm not at all aware of it, please feel free to share it

[coral spindle]

Like I think this is actively unhelpful lmao

The matrix is a red herring

[arctic trail]

1*totient = Id
So totient = Id * mobius

[elfin wraith]

Yeah this lecturer is not great tbh...

[arctic trail]

hence multiplicative

[coral spindle]

Ah OK moebius stuff, that's fair

[elfin wraith]

Shes lovely, and like cracked out of her mind, but a horrid lecturer

[coral spindle]

I would argue maybe a bit too high powered

and I think the CRT is still the 'real' reason

[arctic trail]

For me this was the "canonical" proof for phi being multiplicative

[coral spindle]

but still neat

[arctic trail]

Because it stays within the realm of arithmetic functions

[elfin wraith]

I would still argue worse than the horrible "matrix" proof is not requiring rings to be unital

[arctic trail]

(I've seen non-unital rings in practice)

(and modules over non-unital rings*)

The ring of upper triangular matrices can be decomposed into diagonal + Nilpotent.
Iirc I saw modules over the nilpotent part, or something inspired by that

[arctic trail]

This proof is SOOOO horrible

🤢

[tough raven]

Seems fine to me. This is essentially saying that {1, ..., nm} → {1, ..., n} ⨯ {1, ..., m} by reduction modulo n, m resp. is bijective, i.e., CRT.

[coral spindle]

So why not just state this directly instead of concealing it like this?

You've spotted how to make it conceptual, but I don't think that's made so clear as written

[fair sentinel]

heyo

im new to fields somene pointed me here from the help channels

[tough raven]

It is stated directly, just using the words "every residue exactly once" instead of "the following map is a bijection". It's largely a matter of taste, and while the latter style is clearer at a more advanced level because it distracts less from more complicated ideas that might be present, the former is clearer at a level of less familiarity where someone might not be used to these things enough to immediately get what's being said when they see/hear "bijective function".

Maybe it's a little more wordy than it needs to be, but I think this is within the bounds of adapting to the audience.

[tough raven]

This is the right channel to ask a question about fields, yes.

[fair sentinel]

ok

I wanted to understand and solve this question

[velvet hull]

what progress have you made?

[rocky cloak]

I guess K is not of characteristic 2

[elfin wraith]

See I would possibly agree but this was a 6th semester course, like in no way was this introductory. I agree that it is a proof, and I can follow it easily but I do think its a far more complicated presentation than is needed, and as Boytjie points out, the matrix is more of a red herring than anything else, so actually if anything its likely more confusing to a weaker audience because the matrix is largely superfluous

[fair sentinel]

terrible progress im just learning the topic

We know that $[L:K] = 2$.

Letting $L = K[\sqrt d]$ we know that $[L:K] = 2$ as the minimal polynomial is $(x^2-d)$

Now assume $d = c^2$ for some $c \in k$. then $L = K[\sqrt d] = K[c]$ and then as C is already in K we simply get $[K(c):K] = 1$ which is a contradiction.

[cloud walrus]

BOSS

[fair sentinel]

its wildly incorrect ive been told, just trying to understand how to approach problems like these

[velvet hull]

you can't let L = K[sqrt(d)] as that is what we are trying to show

all we know is that [L:K] = 2, that's it, you need to show that at least one such d exists first

[fair sentinel]

sure yes

So we are given [L:K] = 2

From here, im not sure how to approach this

[velvet hull]

recall the definition of the degree of a field extension, is the dimension of the field as a vector space

[fair sentinel]

[L:K] = 2 means that elements are either deg 1 or 2 right

[fair sentinel]

sorry yes

L is a vector space over K with dimention 2

[velvet hull]

(also as jagr said, I'm pretty sure you need to assume that the characterisitc is not 2 for this result to be true)

[fair sentinel]

My class didnt really cover characterisitc so ill take it as truth and note it down to prove later

[velvet hull]

it just means that 2 is not 0 in this field

[fair sentinel]

ah ok

[velvet hull]

you'll see why that's necessary later

[fair sentinel]

ok so

L is a vector space over K with dimention 2

[fair sentinel]

also does this hold true

[velvet hull]

yes, but that's not necessarily useful here

[fair sentinel]

ok

[velvet hull]

algebraic extensions need not be finite

i.e. just because every element in the extension has degree 2, does not mean that L=K[sqrt(d)]

[fair sentinel]

ah

ok i see the mistake

[velvet hull]

the local property does not pass into a global property

go back to the dimension thing

[fair sentinel]

sure

L is a vector space over K with dimention 2

so the set is something like

${a l_1 + b l_2 \mid a,b \in K}$ where the basis is ${l_1,l_2}$

[cloud walrus]

BOSS

[velvet hull]

you can assume that l1 is equal to 1 (why?)

[fair sentinel]

we need to get elements of K in L?

[velvet hull]

because any linearly independent subset of a vector space can always be extended into a basis

[fair sentinel]

can you expand sorry

[velvet hull]

so, this tells us that L = K[l1,l2]

[fair sentinel]

yes

[velvet hull]

but that's still annoying to deal with, ideally we want a single element l such that L = K[l]

and then maybe we can figure something out from there

[fair sentinel]

ok

[velvet hull]

point is, if you can get rid of a variable theres no reason not to do it

[fair sentinel]

we can get rid of 1 because its in K?

it is K i mean 1 times any element in K gives us K

ok

[velvet hull]

K[k] = K for all k in K

[glad osprey]

{1} is a linearly independent set, which can be extended to a basis {1, l_2}

[fair sentinel]

ah

[fair sentinel]

ohh

ok that makes sense*

[velvet hull]

in field theory you're going to be digging up a lot of the old LA facts

[fair sentinel]

Yeah im figuring

adding getting better to LA to my list

ok now continuting

L = K[l]

we know that

$[K[l] : K]=2$

[cloud walrus]

BOSS

[fair sentinel]

that means the minimal polynomail for l has to be order 2 in K[x]?

[velvet hull]

has degree 2, yes

[fair sentinel]

sorry, l has to be oder 2

ok that means, there exists some solution in K[x] where l^2-d = 0

theb we know

l = sqrtd

?

[velvet hull]

alternatively here's the way I like to phrase it:

we know that {1,l} is a basis for the vector space L over K.
then consider the vector l^2.
now as the set {1,l,l^2} has more vectors than dimension, it must be linearly dependent; i.e. there must exist scalars a,b,c such that al^2 + bl + c = 0

you are deriving essentially the same thing, but I like this proof more because it's a good showcase of the ping-pong between abstract algebra and linear algebra that field theory is about

[fair sentinel]

i like ur explanation much better

this just feeds into l being d tho right

sqrtd

[velvet hull]

no, it does not have to be d

for instance Q[sqrt2] = Q[2+sqrt2]

but it does have to have a component of d.... try and figure out how that works

[fair sentinel]

al^2 + bl + c = 0

is it based on these linearly dependent vectors

[rocky cloak]

There is a magical formula known as ||the quadratic formula||

[fair sentinel]

whats that

jkjk

ok so we are just factoring out al^2 + bl + c = 0

dont we just get back teh quadratic formula here

because a=a,b=b,c=c

[rocky cloak]

The quadratic formula gives you a formula for l involving a square root

[fair sentinel]

ah

ok so we get some $l = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

one sec im on my phone

[cloud walrus]

BOSS

[rocky cloak]

That's right, so what d can you use here?

[fair sentinel]

wdym

sorry

oh its a there exists

hmm

im not sure

mainly what does d need to be

[velvet hull]

that's a good question

[velvet hull]

also, this is why you need to assume that 2 is not equal to 0

[fair sentinel]

yeah i see that now

[rocky cloak]

Well it needs to be something where we can write l (and everything else in L for that matter) as
p + qsqrt(d) for p and q in K.

If only there was some expression like that nearby...

[fair sentinel]

oh wont it just be the formula itsself then

like p+ql?

im lost

it needs to be something where we can write l (and everything else in L for that matter) as
p + qsqrt(d) for p and q in K.

uhh its a basis for L

[rocky cloak]

Yes, now have we manage to write l as something akin to that?

[fair sentinel]

yed

yes

[rocky cloak]

And what is d in that case?

[fair sentinel]

interesting we just found our basis explicitly basically

i^2

oh sorry

other way

sqare root of I

[glad osprey]

For this, is it enough to observe that Gal(N / K(a)) is a normal subgroup since the Galois group is abelian, then by the Galois correspondence K <= K(a) is a normal extension, thus every root of p(x) is in K(a) since it's irreducible and one of its roots are in K(a)?

[rocky cloak]

So I'm not sure what I is here, but recall that you managed to find a formula for l that contained a square root in the expression

What was in that square root?

[fair sentinel]

b^-4ac?

[glad osprey]

getting a thumbs up from jagr is the best feeling in the world thanks

[fair sentinel]

i strive to one day now

[fair sentinel]

ok im confused on whats going on

we got l

and we are trying to find dome

[K(l):K]=2 right

[rocky cloak]

Yes, we saw sqrt(b^2 - 4ac) appear.

So it's natural to think that d=b^2 - 4ac

[glad osprey]

good luck 💪 don't get addicted

[fair sentinel]

ok

[rocky cloak]

Since
sqrt(d) = 2al + b
it's in L

[fair sentinel]

but we can only get this by saying b=0,a=1 right

ohhhhh the ret of it

is in Q

wait

no

K

[rocky cloak]

Well you're not the one deciding what a and b is

[fair sentinel]

yeah...

but how can we just

take itout of the equation

[rocky cloak]

Take what out of the equation?

[fair sentinel]

sqrt(b^2 - 4ac) from $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

[cloud walrus]

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

[velvet hull]

you can't, but remember in what context we're considering this equation in:

[rocky cloak]

So the equation is

l = (-b + sqrt(d))/2a

if you multiply both sides by 2a and add b you get
2al + b = sqrt(d)

[fair sentinel]

Ok i think i need to trace my steps and understand how i got here

to have it make sense

[velvet hull]

we are claiming that L[-b+-sqrt(...)/2a] = L[+-sqrt(...)]. why can we do this?

[fair sentinel]

Ok so we know that $[L:K] = 2$, thus we have it so L is a vector space of two dimentions over K

[cloud walrus]

BOSS

[fair sentinel]

so we can say the basis {1,l} are linarly independent

thus basis {1,l,l^2} are going to be linearly dependent, giving us the equation

a+bl+cl^2 = 0

so we can easily solve for l using the quadratic formula and we get $\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

[cloud walrus]

BOSS

[fair sentinel]

right?

Additionally, we know that L = K(1,l) = K(l)

[fair sentinel]

because thats legit the difinition of what is going on here

so now we have it that K(l):K = 2 and l = the quadratic formula

oh

is l possible combinations of the quadratic formula?

no right

[fair sentinel]

im not sure

[hard hearth]

what is 1 here(underlined in red)? identity map or trivial map? i am asking because in both cases ~ is becoming equivalence relation

[kind temple]

the identity

[hard hearth]

thanks

[woeful sage]

D&F detected

[chilly ocean]

thanks for saving my time

[thorn jay]

If it were the trivial map then every map would be trivial

as from (i) we would have
ρij = ρjj • ρij = 0 • ρij = 0

[vocal pebble]

Yes

[woeful sage]

We know that the subgroup operation $uN \cdot vN = (uv)N$ is well-defined i.e $u_1N = u_2N \text{ and } v_1N = v_2N \implies (u_1u_2)N = (v_1v_2)N$ iff $N$ is normal

[cloud walrus]

Neamesis

[woeful sage]

what would be a counter-example to this? what is a abnormal subgroup and a group for which this isn't well defined?

[woeful sage]

Not "subgroup operation" I mean the quotient group operation

[rocky cloak]

C2 as a subgroup of S3 would be the smallest example

[woeful sage]

ah so any 2-cycle subgroup

[rocky cloak]

Yup

[woeful sage]

in just S_3 or any S_n?

[rocky cloak]

Works for any n>2

[vocal pebble]

If it isnt normal in S3 then it wont be normal in Sn, n>=3

[woeful sage]

Right since there's an isomorphic copy of S_m in S_n iff m <= n

[hard hearth]

so still an equivalence relation

[rocky cloak]

It is, but it would just be a~b for all a and b

[hard hearth]

it is boring though

yeah

[rocky cloak]

So all that extra data would be useless

[thorn jay]

"Yes let me attach a sheaf to this space" and it consists of singletons

[hard hearth]

[glad osprey]

How do I do the normality part? My first idea was to show that K(a, b) is a splitting field over K(a), but I'm not sure that works

[coral spindle]

If f(x) is a min poly with root alpha and g(x) for beta, surely f(x)g(x)'s splitting field is F(alpha, beta)?

Is that right

[glad osprey]

lol, I think that works, I was overcomplicating it in my head 😅

thanks

[coral spindle]

No worries

[woeful sage]

Supernatural projections when?

I guess this is the algebra version of adding a zero in analysis

expanding identity into gg^{-1}

[weary frost]

nvm

[amber burrow]

Prove that if $p$ is a prime and $P $is a subgroup of $S_p$ of order $p$, then $|N_s(P)| = p(p- 1)$

Wanna check if I'm doing this right. Since $|P| = p$, $P$ is cyclic. Suppose $P = \langle g \rangle$. $g$ must have order $p$, so it is a $p$-cycle. Thus, $P$ contains $p-1$ $p$-cycles. Consider the action of $G$ on $\mathscr{P}(G)$ by conjugation, and specifically the orbit of $P$. Conjugation in $S_p$ is just about cycle type, so every conjugate of $P$ has $p-1$ $n$-cycles. There are $(p-1)!$ $p$-cycles in total, so the size of the orbit is $(p-1)!/(p-1) = (p-2)!$. By orbit stabilizer, $p!/|N_S(P)| = |S_p:N_S(P)| = (p-2)! \implies N_S(P) = p(p-1)$

[cloud walrus]

donut123

[amber burrow]

oh if i randomly say $G$, i mean $S_P$

[cloud walrus]

donut123

[woeful sage]

the order of the normalizer of P in S_P is p(p - 1)?

very interesting

[woeful sage]

So we have normal subgroups

if N is normal then G/N has a group structure

do people study subgroups H such that G/H has not necessarily a group structure but a weaker one like a monoid or a semi-group structure?

perhaps we can name these "subnormal groups" or semi-normal, or pseudo-normal would work too

[thorn jay]

Subnormal groups unfortunately already exist

Although, I'm pretty sure that the well-definedness of the group action implies normality already, anyways

Would have to verify that

[woeful sage]

but do subgroups like these exist?

[thorn jay]

.

[woeful sage]

the group action of left multiplication? i.e G acting on H by left multiplication?

[thorn jay]

suppose g in N and for any h that ghN = hN
=>N = h^-1 g h N
=> h^-1 g h in N for all h
Thus N is normal

So even just requiring that the induced multiplication on the left cosets be well defined implies normality

[woeful sage]

crazy

that's pretty cool

[thorn jay]

Groups are very nice objects, all things considered

[woeful sage]

So nice!

I guess it's because function composition of bijections is so nice

[boreal inlet]

Let A be a commutative unital ring, and Spec(A) be its prime spectrum. For a subset F \subset Spec(A), define I(F) to be the intersection of all elements of F.

I have proved I(F) is radical, but I also want to show that V(I(F)) = F. How do I do that? F being contained in V(I(F)) is trivial, I'm not sure how to show the other side.

[thorn jay]

Because it's not true in general

It's only true if F is the set of prime ideals which contain some fixed set S in R

[boreal inlet]

Ooof

Yeah I should have mentioned that F is closed in the topology

[rocky cloak]

The easiest example might be if F consists of just a single prime

[rocky cloak]

Which you define as F being equal to V(I) for some ideal?

If so you might try to prove that rad(I) = I(F)

[boreal inlet]

Yeah, I already did prove that I(F) is radical

[rocky cloak]

Right, so then you're halfway there I guess

[boreal inlet]

Oh okay I think I got it

Let F = V(J) for some ideal J in A.

Then J \subset I(F) \subset q for any q in V(I(F))

Hence q is in V(J) = F

[amber burrow]

If $p$ is a prime and $P$ is a subgroup of $S_p$ of order $p$, prove $N_{S_p}(P)/C_{S_p}(P) \cong \text{Aut}(P)$

$N_{S_p}(P)/C_{S_p}(P)$ is isomorphic to a subgroup of $\text{Aut}(P)$. Since $|P|=p$, its cyclic generated by a $p$-cycles. This will commute with powers of itself (no disjoint cycles), so $C_{S_p}(P) = P$. Thus, $|N_{S_p}(P)/C_{S_p}(P)| = p(p-1)/p = p-1$. However, $\text{Aut}$ has order $p-1$ too, since $P$ is cyclic. Done

[cloud walrus]

donut123

[boreal inlet]

So this should show F -> I(F) is the inverse of the map I -> V(I) and hence V forms an one-to-one correspondence between radical ideals and closed subsets.

[amber burrow]

Also, general question, but what do you write when you do like group theory problems

like, i always just have to do it all in my head

[boreal inlet]

I personally just jot down literally everything that goes on in my head, it works most of the time

Like "G is a group of order pq. Hence it is cyclic."

Everything you can infer and which seems worthwhile just jot down. It lets you think for the next steps needed

[woeful sage]

This is actually insane

What do you need to prove this?

Sylow theorems?

[rocky cloak]

With the sylow theorems you can prove statements like these.

But it's not completely true as written

[woeful sage]

What additional hypothesis is required on p and q?

[amber burrow]

Isnt it line if p doesn’t divide q-1

[woeful sage]

I see

[amber burrow]

And I dont think you need sylow, just Cauchy

[woeful sage]

Oh great!

I will try this then

[amber burrow]

The counter example to cyclic always is unique, I think it’s order 12

[woeful sage]

12? that's not a product of two primes though

[amber burrow]

Oh oops

[rocky cloak]

Well you need something to guarantee normality as well, but it doesn't need to be the sylow theorems.

[rocky cloak]

What is "cyclic always is unique"?

[amber burrow]

i remember seeing somewhere (not actally done/proven yet) that if p divides q-1, theres a unique non-abelian group

[amber burrow]

for the p divides q-1 ase, i think you need semidiret produts

[woeful sage]

damn

D&F only covers semi-direct products in chapter 5

but I recently read that semi-direct product of G and H is like a direct product G x H but now you introduce a map phi : G --> Aut(G)

reminds me of the difference between product spaces and fiber bundles

[rocky cloak]

Yes, that should be true

[amber burrow]

yeah, im still on chapter 4 lol

finishing up 4.4 exercises

[woeful sage]

I will catch up with you soon enough

I'm starting 3.1 exercises right now

[barren sierra]

Let $\Gamma$ be a finite group acting on $\mathbb{C}[\overline{x}]$ some polynomial ring.
Let $I$ be an ideal fixed by this action.
It's true that $(\mathbb{C}[\overline{x}] / I)^\Gamma \simeq \mathbb{C}[\overline{x}]^\Gamma / I$ right?
Really, what I want to know is that if $f + I \in (\mathbb{C}[\overline{x}] / I)^\Gamma$ then $f \in \mathbb{C}[\overline{x}]^\Gamma$.
To show this isomorphism, consider the quotient $\mathbb{C}[\overline{x}]^\Gamma \to (\mathbb{C}[\overline{x}] / I)^\Gamma;\ f \mapsto f + I$.
Then this has kernel $I$ yea?
I mean it all seems to work but idk, I'm having doubts for some reason

[cloud walrus]

Spamakin🎷

[woeful sage]

Why is there a bar on x?

[barren sierra]

some number of variables

$\mathbb{C}[\overline{x}]$ is common shorthand for $\mathbb{C}[x_1, \ldots, x_n]$ for some unspecified $n$.

[cloud walrus]

Spamakin🎷

[woeful sage]

Oh Interesting I didn't know that

[barren sierra]

I guess my main doubt is showing surjectivity. Say $f + I \in (\mathbb{C}[\overline{x}] / I)^\Gamma$, then why is $f \in \mathbb{C}[\overline{x}]^\Gamma$.

[cloud walrus]

Spamakin🎷

[thorn jay]

Sometimes even for an infinite number of variables indexed by some set

[south patrol]

Oh I would've used an underline

[thorn jay]

Or \vec

[rocky cloak]

So this is not a finite group, but if Gamma is the infinite cyclic group acting on C[x, y] by
x |-> x+y
and y fixed.

Then C[x, y]/(y) is fixed by the action, but the fixed ring of C[x, y] is just C[y]

So if it's true you need to leverage finiteness

I guess you can just take the average of the Gamma orbits of f

Yeah
f' = Sum_g g(f) / |Gamma| is clearly fixed
and it's image is f+I

[tough raven]

This may be true. However, I think the difference between trivial and general fibre bundles is also analogous to the difference between semi-direct products and general extensions.

[thorn jay]

One is doable and the other horrible? Lol

[tough raven]

Do you know a counterexample with a finite group in positive characteristic?

I'm guessing involving the Frobenius endomorphism.

[elfin wraith]

Can I just double check this proof? We can write $A \cong \bigoplus_{i=1}^k C_{{p_i}^{n_i}}$ for some primes $p_i$ and integers $n_i$. We see that for each $C_{{p_i}^{n_i}}$ we have $C_{p_i}\triangleleft C_{p_i^2} \triangleleft\cdots \triangleleft C_{p_i^{n_i-1}}\triangleleft C_{p_i^{n_i}}$ and $C_{p_i^{n_i-k}}/C_{p_i^{n_i - k -1}} \cong C_{p_i}$. Hence, each $C_{p_i^{n_i}}$ has a composition series, with the factor $p_i$ $n_i$ times. Then, since $A$ is Abelian, all of its subgroups are normal, so in particular, $C_{p^{k_i}_i}\triangleleft A$ and by the following lemma, we get the desired result.

[cloud walrus]

Nope

[elfin wraith]

The lemma is:

[tough raven]

The question and your proof use n_i with different meanings (in particular, your proof must allow p_i's to repeat), but otherwise this is fine.

Also, you don't really need C_{p^k} to be normal in A. A composition series only requires each subgroup to be normal in the one immediately after it.

[elfin wraith]

Dont I need the C_p^k to be normal to apply the lemma (i.e. so I can quotient by it)?

[elfin wraith]

Also yes I do apologise my notation is disastrous there, Ive used k in multiple different ways and switched between using n and k pretty freely, oops

[tough raven]

Oh, just the summands. Then yes.

I thought you were talking about the intermediate subgroups too.

[elfin wraith]

Yes apologies my quick back of the napkin notation here was less than ideal, I was just meaning the summands

[rocky cloak]

The example I gave for the infinite cyclic group should work for the cyclic group of order the characteristic

[tough raven]

But over F_p, x^p - x would be fixed.

It is true that R is integral over R^G for finite G regardless of characteristic.

IG it's still not surjective.

[rocky cloak]

Yeah I'm saying I don't think that will surjects onto Fp[x, y]/(y)

[tough raven]

F_p[x] vs F_p[x^p - x]

Sorry, my brain wasn't running apparently

[woeful sage]

Oh I see I haven't learnt group extensions yet but I'll keep that in mind

[hard hearth]

whats the reason for this name(germ)?

[coral spindle]

I'm not sure if this is the origin, but it is consistent with sheaf terminology. The germ is a part of wheat.

This is the ongoing lexical field right, the sheaves, the stalks, the germs...

All very pretty

[frigid shard]

As far as etymology goes I believe it has to do with “germ of an idea”

Or maybe not

The sheaf metaphor is prolly more accurate

[coral spindle]

I think the "germ" of an idea has the same etymology as the germ of a seed and so on

Let's just be clear though -- this of course has nothing to do with bacteria

[hard hearth]

i dont know about this analogy maybe i will have to read more to get an idea of why its called so

[coral spindle]

It's just a name.

[rocky cloak]

Well, I think the word "germ" for bacteria also has the same etymology

[coral spindle]

C.f. germination

[hard hearth]

also \rho_U is surjective because given any equivalence class \bar{f} \in A (equivalence class of a function f), say f has domain V, we can consider a continuous extension or restriction of f from domain V to domain U (whichever is appropriate) and this extension or restriction gets mapped to \bar{f} by \rho_U?

and \rho_U is not injective because there are many continuous extensions of a function and many functions whose restriction matches a given function on a domain

this seems simple but i want to verify because its slightly abstract for me

[hard hearth]

(question is regarding this exercise)

btw the equivalence classes are just the continuous functions which match each other in some neigbhourhood of p, right?

[tame sierra]

basically. it’s really equivalence classes of pairs (f,U) where U is the domain of some continuous function f that contains p. then it’s basically what you said: (f,U) is equivalent to (g,V) if there exists a neighborhood W of p contained in both U and V such that f and g agree on W

[balmy wraith]

Can someone please help me understand the 1-1 step: g * (g’)^-1 in ker(theta) -> gker(theta) = g’ker(theta)

[tardy hedge]

Are u writing gtheta to mean theta(g)?

[elfin wraith]

Your notation confuses me to be honest, but the proof that it’s injective is just the same as the proof that it’s well defined but backwards

If that helps at all

I can write out how to do it if you need but perhaps you’ve seen how to show the map theta is actually well defined

[balmy wraith]

Yes

[tardy hedge]

Why do you write it like that?

[balmy wraith]

My professor uses this notation where he applies the function on the right side. He’s from the UK if that means anything 😭

I have no other reason for it other than that’s what I’ve seen from him

I just adopted it

[balmy wraith]

Sorry about the confusing notation, it’s how my professor writes it. I would like some help for showing how the function is well-defined then

[paper burrow]

Who need they subgroup normalized

Nah but its an interesting class

Very interesting material overall

[glad osprey]

So the map psi : G/ker(phi) -> im(phi) is defined by mapping g*ker(phi) to phi(g), to show that it's well-defined you need to show that psi gives the same value no matter which representative of g*ker(phi) you choose. Do you see how to do that?

[paper burrow]

one thing that irks me though, at least with the conventions that the author used for our textbooks is the notation suckedd

[tardy hedge]

Ong i sometimes completely forget analysis is a part of math

[tough raven]

Or do you sometimes forget that it isn't?

[drifting mauve]

hi everyone, I was hoping someone could give me a hand with this question. I need to find all homomorphisms from $f: \mathbb{Z}^3 \rightarrow \mathbb{Z}$

[cloud walrus]

theaveragejoe6029

[drifting mauve]

ive got that if x,y,z = f(i), f(j), f(k) respectively, then xy = xz = yz = 0 and x^2 = x, y^2 = y and z^2 = z, but I don't really know where to go from there

also f((a,b,c)) = ax+by+cz

[next obsidian]

Isn’t it pretty clear that this makes the only viable map the 0 map?

Err

Sorry not quite but close to it

[crystal vale]

U(p^n ) is isomorphic to Z/(p^n - p^(n-1) )Z, when p is odd prime, any hint how can I show that?

[next obsidian]

What is U

[next obsidian]

I would say think about what x^2 = x means when x in Z

[crystal vale]

Sorry, U(n) = { m in Z | gcd(m,n) = 1, 1≤m≤n } under multiplication modulo n

[next obsidian]

You just count them

[crystal vale]

Yes its order is p^n - p^{n-1}

But how it is cyclic group?

[next obsidian]

Oh

This is annoying

If I remember correctly the right way to do this is via induction

I think you show manually for Z/p^2Z though?

But uh, I think it’s something like if u mod p^n-1 is a generator you show that u itself is a generator or something

Sorry I kinda forget

[crystal vale]

No problem

[drifting mauve]

sorry, I dont really see this

oops, didn't mean to send that

No, but I can't quite see it tho

[alpine plank]

I think you need a couple of lemmas, there is a proof in Ireland and Rosen. Page 43 theorem 2 and 2'.

[drifting mauve]

ah, okay. So I got how the zero map, works. but what do you mean by not quite. I can't think of anything else that would work

[woeful sage]

algebraists be like: "What the hell is convergence 🗣️🔥🔥🔥"

[next obsidian]

Try harder

[drifting mauve]

my bad lol

oh haha, lol.

got it. just had a blonde moment

[crystal vale]

thanks

[topaz solar]

Rip completions of rings

I mean making completions of rings is analysis tbhtbh

[hard hearth]

i dont really see how they can be tuples... we are ordering the intervals containing $p$ by reverse inclusion, and defining $A_U=$ ring of continuous functions defined on $U$ and taking $A = \varinjlim A_U$ so the equivalence classes of $A$ can only be functions... i mean the function contains its domain information, so specifying $(f,U)$ is redundant anyways, right?

[cloud walrus]

fastrack_and_backtrack

[rocky cloak]

Yes. It's just a bit convenient to specify it this way because then you can use the same name for f and f restricted to some subset of its domain

[woeful sage]

Can I get a hint for this?

I know that in order to show that N is normal

I need to show $gNg^{-1} = N \ \forall g \in G$ which is implied by $gNg^{-1} \subseteq N \ \forall g \in G$ which in turn is implied by $gx^{-1}y^{-1}xyg^{-1} \in N \ \forall x, y, g \in G$

[cloud walrus]

Neamesis

[woeful sage]

since you know? You only have to check that a subgroup's generators are normalized by every element in G order to show that the subgroup is normal in G

[mighty kiln]

(gx'g')(gy'g')(gxg')(gyg')

[rocky cloak]

Yeah you can actually prove the stronger thing that any automorphism f: G -> G has f(N) = N

[woeful sage]

oh interesting

and in this case it's just the inner automorphism

[woeful sage]

I thought about this but I'm not sure how this would help

[mighty kiln]

This is one of the generators

[woeful sage]

you are so right

how did I not think of that

[mighty kiln]

This is called a characteristic subgroup

[woeful sage]

a subgroup fixed by any automorphism is called a characteristic subgroup? or is that another name for the commutator subgroup?

[mighty kiln]

Intuitively they occur as subgroups that can be "defined without arbitrary choice of elements"

Fixed by any automorphism

[woeful sage]

Thanks Jagr and Arki

[crystal vale]

U(p^n ) is kernel, how?

[chilly ocean]

Oh sorry i missed something wait

Let me edit

So it must be cyclic

[rocky cloak]

I think the easiest way to see it is because it surjects onto U(p) which is cyclic of order p-1 it has an element of order p-1.

Since p-1 and p^n-1 are relatively prime you just need to show the existence of an element of order p^n-1.

1+p is such an element, which can be seen by binomial theorem and some counting of powers of p in the binomial coefficients.

[chilly ocean]

i dont know why i am typing some wrong things maybe still in hangover forget to erase the wrong thing and copies it as it is

all the elements you need to find , that are relatively prime to p so U(p^n) is outside of the kernel of the map Z/p^N to Z/pz using canonical projection a+<p^n> goes to a+<p> so, restriction on U(p^n) is injective

[rocky cloak]

Are you saying that the map U(p^n) -> U(p) is injective?

1 and 1+p both map to 1

[chilly ocean]

no

[rocky cloak]

Then idk what restriction on U(p^n) means

[chilly ocean]

oh i think you are looking at it as group homomorphisms

but i forget to say it as ring homomorphism

[rocky cloak]

Then idk what you're talking about

[chilly ocean]

still not injective

[vivid kestrel]

if a semisimple algebra is written as a direct product of simple subalgebras, does that direct product contain instances of every isomorphism type of simple subalgebras

[coral spindle]

Assuming by algebra you mean unital, associative algebra over a field, the answer is no: take the semisimple Artinian algebra M_2(k) and consider the simple subalgebra of scalar matrices.

This semisimple algebra is in fact simple and indecomposable

[round portal]

hey guys good morning

[late matrix]

good meowning

[round portal]

I need to show that the Klein 4-group V_4 is the only subgroup from A_4 (alternating group) of order 4

then I have to conclude that V_4 is normal to A_4 thus A_4 is not simple

I have no idea how to go about this

😢

groups of permutations are so confusing to me fr

for reference V_4 = {id, (12)(34), (13)(24), (14)(23)}

[vivid kestrel]

When you've shown uniqueness normality is easy: consider that the conjugate of a subgroup is a subgroup of the same order

[round portal]

yup i guess normality should be easier but i don't know how to start showing the uniqueness

[velvet hull]

since it has order 4 it has to contain only elements of order 2 or 4

does A4 have elements of order 4?

[round portal]

yes

plenty

[velvet hull]

sure, give me an example

[round portal]

let's see, im actually not sure

ok

no

A_4 does not have elements of order 4

[velvet hull]

how many elements of order 2 are there in A4?

[round portal]

i would say

all of them lol

[velvet hull]

does (123) have order 2?

[round portal]

from my calculations

wait wait

ok i get it

the elements of order 2 from A_4 are precisely the elements in V_4

right?

so this does it for the proof?

the proof of V_4 being the only subgroup of order 4

[velvet hull]

yes

[round portal]

ok!!

hey thanks that was really helpful

i struggle a lot with groups of permutations

[velvet hull]

as a quick remark, V4 in A4 is what is known as a Sylow 2-subgroup

what you have shown is that if all the elements with order the power of some prime forms a subgroup, then that group is necessarily normal

[round portal]

this is by the first sylow theorem right

[velvet hull]

you'll learn more about them when you study group actions

[round portal]

since V_4 is in Syl_2(A_4)

great

[round portal]

ok yeye

we're starting to see the sylow theorems

[velvet hull]

not sure what you are referring to here, but Sylow's 1st theorem only guarantees the existance of a Sylow p-subgroup

what I'm describing here is stronger; namely that if you can show that there is a unique Sylow p-subgroup, then that subgroup is normal

[round portal]

i am referring to the second theorem my bad

[velvet hull]

and that subgroup will consist of exactly the set of all elements of order p^k

[round portal]

this theorem?

to show the normality

[velvet hull]

yes, it follows as a corollary

[round portal]

thanks!

[topaz solar]

Meowing for groups

[topaz solar]

Yes

Which this is a lot nicer than just the normality iff unique, since it also says something about a nice action of G on them ofc

not that remember what theorem name/number is which matters much, of course

[sick spear]

do you guys consider group theory as a necessary unit for one to study?

[velvet hull]

for what?

[sick spear]

well im a electrical engineering and applied math student

interested in AI/ML

the units i take include: linear algebra, computational lin alg, cal 1 to 3, complex analysis, real analysis, computational mathematics,probability, pdes, odes, network mathematics (graph theory). for my last elective i can choose between optimisation research (mathematical programming) or group theory. im not sure what i should choose.

[velvet hull]

for your interests they are not so useful

[fair sentinel]

Heyo

[sick spear]

@sick spear im not sure if i choose GT for the sake of knowing it exists even tho it may not be related to my field, or if i do a optimisation research unit

[fair sentinel]

I missed this problem on a group theroy test I just took

was wondering how you would actually solve it

[sick spear]

yeah that was my main concern

but do you think just knowing it exists is useful? is that a good enough reason to take a unit?

and GT is pretty fundamental if im not wrong?

[velvet hull]

well it is massively useful and foundational for pure math

but optimization research will be more practical for your interests

[sick spear]

okay tq tq

i agree

@velvet hull is this more group theory or is this advanced algebra

[velvet hull]

yeah thats group theory

[sick spear]

wait my uni has 2 group theory units

so why does it have algebra at the start?

[velvet hull]

part 2 is not group theory

[sick spear]

hmm

but the name of this channel is: groups-rings-fields

[velvet hull]

it's still introductory abstract algebra, but the objects of study are rings and fields instead of groups

groups, rings and fields are the first 3 objects studied in abstract algebra

[sick spear]

okay sorry if im asking silly questions. my knowledge in pure math is rlly bad right now

[velvet hull]

well, it doesnt have to be

thats why this channel exists

[elfin wraith]

For this, we can work out that $A \cong C_2\times C_3 \times C_{30}$ by Lagrange and FTFGAG's. Then in each component we have 2, 6 and 6 elements such that $g^6 = e$ (not necessarily order, but like a multiple of the order), but I dont think we can just say there are $2\cdot 6^2$ elements of order 6, but im not sure how to fix the amount of like "over counting"

[cloud walrus]

Nope

[elfin wraith]

It feels very lcm-y but im not quite seeing of what

[rocky cloak]

C2xC3xC30 has only 180 elements.

Anyway, it's easier to count the order of elements if you factor it into cyclic p-groups. So C2xC3xC5 instead of C30 for example.

Either way the order of an element in a product is the lcm of the order of each coordinate

[elfin wraith]

Oops I mean C2xC6xC30

[maiden crater]

quick question, really silly, but (1 2)(1 2) would be an example of the identity function , right

[elfin wraith]

Yeah

2 cycles are self inverse

[maiden crater]

cool! Thanks!

[round zodiac]

Hey!, could this be true without asking H normal in G?

$Let ( G ) be a finite group, ( H \triangleleft G ), ( p ) a prime divisor of ( |G| ), and ( P ) a Sylow-( p )-subgroup of ( G ). Then ( H \cap P ) is a Sylow-( p )-subgroup of ( H ).$

Asking because in proving this I didn´t use it so is surely wrong but, idk. (I did it observing that $H\cap P$ is a p-subgroup of H and then taking Q a p-sylow of H. Then I know $Q \subseteq gPg^{-1}$ for some g in G. Then $|Q| \leq |H \cap P|$, so H \cap P must be a p-sylow of H, right?)

[cloud walrus]

ArgR4N
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[coral spindle]

No it would be pretty far from being true if H were not normal. For example take H to be a Sylow subgroup of G other than P

[elfin wraith]

Thinking about this more, I think I just need to remove the elements I’ve counted which have order 2 and order 3, which I should be able to do as a basic counting argument

But I don’t think that’s what you’re saying, could I ask what you had in mind?

[rocky cloak]

"Then |Q| < |H cap P|" is incorrect

[round zodiac]

OOOO is see

[rocky cloak]

If you factor everything into cyclic p-groups then there's just prime factors to deal with. Then you just need to count the number of stuff with both a factor of 2 and 3

[round zodiac]

You need normality for doing H\cap gPg^g{-1} = g(H\cap P)g^{-1} to conclude that, rigth?

[rocky cloak]

For example
(C2)^m x (C3)^n
has
(2^m - 1)(3^n - 1) elements of order 6

[rocky cloak]

Yeah
g (H \cap P) g^-1 = gHg^-1 \cap gPg^-1

[round zodiac]

thx

[maiden crater]

How would I show that there's no 3 or 4 cycle of order 2

in S_4

I mean I could brute force it, for S_4, but that doesn't seem very fun

[elfin wraith]

Ah ok yes this is much nicer

I did my counting argument and got 56, but that’s far more annoying than saying A=(C2)^3 x (C3)^2 x(C5) so have 7*8 elements of order 6

[rocky cloak]

An n-cycle has order n

I'm not sure how you're defining n-cycle without that being immediate

But either way you can try to prove that

[maiden crater]

Unless I'm missing something, artin hasn;t defined n-cycles that way

I could have just missed it though, for I can't find a defn on a n-cycle

[elfin wraith]

He doesnt spell it out but it should be pretty instantaneous

[kindred spindle]

Let $\rho$ and $\rho$ be two (not necessarily equivalent) irreducible representations. Let $R$ be the matrix of our representation $\rho$ and $R'$ the matrix of $\rho'$.
Serre now defined a blinear form
$\ \langle r_{i_2,j_2,} , r_{j_1,i_1} \rangle = \frac{1}{|G|} \sum_{g \in G} r_{i_2,j_2,} (g^{-1}) r_{j_1,i_1}(g) \$

And said, that his would be an inner product of our Vector space $\mathbb{C}^G$ (this contains all functions $f: G \to \mathbb{C}$ iff $\rho$ and $\rho'$ are unitary representations. Then he wrote, that we can always get a to $\rho$ isomorphic representations which is unitary regarding the inner product $\langle \psi, \phi \rangle = \frac{1}{|G|} \sum_{g \in G} \psi(g) \overline{\phi(g)}$ with $\psi, \phi \in \mathbb{C}^G$. And i dont understand why we can have such unitary representations

[maiden crater]

I thought it's just a cycle with n distinct elements

[kindred spindle]

damn

[cloud walrus]

galev

[kindred spindle]

Oops. i Talk about representations of finite groups.

[elfin wraith]

I suggest trying to prove it, suppose you had an n cycle with order less than n and see that it doesn’t work

Look at the action on 1

[rocky cloak]

I can't quite parse your sentence. Are you asking why there is always a unitary representation isomorphic to a given representation rho?

[kindred spindle]

yes

[maiden crater]

hmm, okay

[kindred spindle]

I googled and the internet talks about 'Weyl's unitary trick' which i dont get, because they define a new scalar product, where our representation is unitary regarding the new scalar product

[kindred spindle]

But couldn't it be, that the new scalar product isn't as defined as this one here?

[rocky cloak]

So your representation is G acting on C^n.

Define an inner product by
<v, w>_G = 1/|G| Sum <gv, gw>

Notice that the action of G is unitary with respect to this inner product.

Now pick an orthonormal basis, then the change of basis from standard to the new basis gives you an isomorphism of representations.

Or closer to your notation, if $\rho$ is an n-dimensional representation you define inner product on $\mathbb C^n$ by

$\langle v, w\rangle_G = \frac1{|G|} \sum_g \langle \rho(g)v, \rho(g)w\rangle$

If P is a change of basis from standard basis to an orthonormal basis wrt the new inner product, then

$new-\rho(g) = P \rho(g) P^{-1}$ is unitary.

[cloud walrus]

jagr2808

[kindred spindle]

oooh, i think i get it

Thanks

[round zodiac]

I don´t understand a thing of this is incredible, can I ask what´s rho?

[kindred spindle]

rho is defined as rho: G \to GL(V). In this case V is C^n and G is a finite group

(Because im too dumb for representation theory on compact groups)

[rocky cloak]

It works the same for compact groups just replacing sum by integral

[kindred spindle]

Oh, good to know.

I just don't like serre's introduction to representation theory

[round zodiac]

And what makes the role for 1/|G|?

[rocky cloak]

So really you replace
1/|G| \sum
By integral wrt the normalized Haar measure

So if you will |G| is replaced by the volume of your group

$|G| \coloneqq \int_{g\in G} 1$

[cloud walrus]

jagr2808

[round zodiac]

Oh, nice

I don´t know nothing about Haar measure but it seems interesting

[rocky cloak]

It's the unique (up to constant) measure on G that is preserved by left multiplication by elements g in G

[stone elbow]

Hi all. Does there exist a section of a quotient map $\mathbb{Z}(p^{r+1})^{\times} \to \mathbb{Z}(p)^{\times}$, where $p$ is an odd prime? The fact that $\mathbb{Z}(p^r)^{\times}$ is cyclic seems to imply that $\mathbb{Z}(p^r)^{\times} \cong \mathbb{Z}(p)^{\times} \times (1 + (p))$ because $(p - 1, p^r) = 1$, so the short exact sequence $1 \to 1 + (p) \to \mathbb{Z}(p^r)^{\times} \to \mathbb{Z}(p)^{\times} \to 1$ should split, but so far I failed to construct the section of the quotient map.

[cloud walrus]

a-square

[round zodiac]

How difficult is it from 100 to 1000 to understand existence and uniqueness? 😅

[rocky cloak]

I think it's not too difficult. Let's say 450.

Wikipedia describes the construction
https://en.m.wikipedia.org/wiki/Haar_measure

[thorn jay]

From 100 to 1000 is a unique scale

[rocky cloak]

It does exist, just from the classification of finite abelian groups. But I'm not sure how easy it is to give an explicit construction of a section.

Actually,
x |-> x^p^r should work

[stone elbow]

I'm working on an exercise in Lang to prove that $\mathbb{Z}(p^r)^{\times}$ is cyclic. The reason I'm interested in this particular section is because that would conclude my proof by induction (I already proved that $1 + p$ generates $1 + (p)$).

[cloud walrus]

a-square

[stone elbow]

Thanks, I'll try it

[rocky cloak]

I don't see how this step would help anything in proving the group is cyclic.

If you've proven that 1+p has order p^r, then your done anyway

[stone elbow]

BTW should I be concerned by the fact that this exercise is difficult for me? The previous exercises about rings were pretty easy.

[stone elbow]

damn, you're right 😅

[rocky cloak]

Idk, it's somewhat hard. I don't know what other exercises you've been doing, but some are harder than others of course

[stone elbow]

Okay, last question: why am I getting a deja vu from this exercise? The structure of $\mathbb{Z}(p^r)$ looks a whole lot like that of $K[C^{p^r}]$. Is it just because it's a local ring?

[cloud walrus]

a-square

[rocky cloak]

I guess so.

Assuming K has characteristic p they're pretty similar

[stone elbow]

Even in characteristic zero, $K[x]/(x^{p^r})^{\times} \cong K^{\times} \times (1 + (x))$, for example

[cloud walrus]

a-square

[stone elbow]

There's a proof in Bass that I researched for another hobby project

[rocky cloak]

Yes, K[x]/(x^n) would be a similar ring in general

K[C_p^r] is very different in characteristic not p though

[vocal pebble]

1 need not necessarily be in the cycle, i suppose just look at the action on any number in the cycle? Or do you mean on 1 as in identity permutation

[elfin wraith]

I was being imprecise, it can be any element of the cycle. My point was if you have sigma = (a_0,...,a_n) and k<n then sigma^k(a_0) = a_k \neq a_0

[vocal pebble]

One could say wlog 1 is in the cycle, yeah (but if you were able to see that you should immediately also know why order is n)

[elfin wraith]

Where in this case a_0 = 1 I suppose

[somber badge]

why can't we follow the same proof (using zorns lemma) as showing all vector spaces have a basis for modules?

im referring to the one here: https://en.wikipedia.org/wiki/Zorn's_lemma#Every_vector_space_has_a_basis

[tribal moss]

For a module you might get into a situation where you have a linearly dependent set that doesn't span, and yet if you add any of the elements missing from the span, you get a linearly dependent set.

So all Zorn's lemma will tell you in that case it that there exists a non-extensible linearly independent set, but it's specific to vector spaces that such a set will span the entire vector space.

For example, consider Z as a Z-module. The set {2} is linearly independent, but its span consists only of the even numbers.

I we try to add one of the missing numbers -- say 17 -- we get {2,17} which is not linearly independent because 17·2 + (-2)·17 = 0.

[white oxide]

Dumb question but is rank Z_2 = 0 since there is no linearly independent set in Z_2

[velvet hull]

in linear algebra, the key to "linear independence -> spanning" is the fact that you can always divide by any scalars to get to a set of vectors with "minimal magnitude", per say.
in modules, as Tropo pointed out, it's possible to have a maximal LD set that has "too big of a magnitude" to reach some of the "smaller" vectors in the module

[velvet hull]

yep, it's because the torsion forces any nonempty collection of vectors to always have a nontrivial homohgenous solution, i.e. be LD with the 0 vector (assuming you're talking about it as a Z-module)

[white oxide]

Got it thanks

[tribal moss]

Which kind of rank? As a Z-module it is 0; as a Z_2-module it is 1.

[white oxide]

As a Z-module

[elfin wraith]

Can anyone give me any pointers for how to do part b of this? Also, pretty sure a) is false for p = 3, but that doesnt matter

The only examples weve had in workshops for composition series have been for like S_4 or C_n so have been rather easy

[south patrol]

Without stating the obvious, try to find a nice normal subgroup

[elfin wraith]

Ok so N is normal and isomorphic to F_n

[south patrol]

Oh wait yeah I forgot they gave you that lol OK nice

[south patrol]

What do you mean by Fn

[elfin wraith]

I genuinely could not possibly hope to tell you

[south patrol]

I guess Fp lol

[elfin wraith]

Yes sorry lol

Ok so I think we have N \cong F_89 and G/N \cong F_89^\times

[south patrol]

If you replace like a^-1 by a this is scaling and translation but I misread

[south patrol]

Yeah exactly

[elfin wraith]

Ah wait but then thats a group of order 88, and I can do that

then just stitch them together

[south patrol]

Also you can just quote what it is

Like there is a theorem on what Fp^x is as a group

[velvet hull]

how about this: can you identify what G/G' is?

look at the matrix components

[south patrol]

Well that is a) aha

Nope has just done that approach heh

[elfin wraith]

C_p-1

[south patrol]

Yeah exactly

So you are gucci

[elfin wraith]

Yeah ok, I was kinda forgetting how I did a lol

[south patrol]

In fact I mean this is a group of order p(p-1) and there aren't too many options

I wonder a) can be done without much computation

[elfin wraith]

I did exactly no computations

[south patrol]

Very good

I guess 2) and 1) show they are all commutators and then you check the quotient is abelian?

[elfin wraith]

Yeah I mean I didnt write out an actual isomorphism I just kinda looked at it and realised it looked like F_p lol

[south patrol]

Hm wdym

Like I mean how do you show that that is the commutator subgroup

[elfin wraith]

I just kinda hand waved part a) and done it by vibes, I think you can say something about F_p having no proper subgroups and the commutator is non trivial

[south patrol]

I think if you combine 1) and 2) you get like the thing with a^2b - a as a commutator but yee

But that still uses minor computation ig

c) is interesting

[elfin wraith]

I need to head home now but ill try it when I get home, im guessing its just a bunch of case checking

[south patrol]

Pog nice

Head home now at this time 🥀

Jk

[elfin wraith]

Im a chronic last bus catcher, I’ve been grinding

And it’s not paid off this has been my worst semester of exams ever

[green fox]

What are the ideals in the ring $M_n(\mathbb{C})$? I've spent a while and can't figure it out

[cloud walrus]

Luke

[mortal lagoon]

I am confused about how to do this

So I understand that the ideals of Z2 are {0} and itself and the ideals of Z4 are itself, 2Z2 and {0}. So, the ideals of Z2xZ4 are all the pairs of ideals.

[velvet hull]

well then you're basically done; recall that an ideal I is maximal in R iff R/I is a field, and I is prime in R iff R/I is an integral domain

[next obsidian]

You need to specify left or right

But when you multiply by elementary matrices you can start to realize their structure, its related to ideals in C

(I guess this isn’t very helpful when you have a field, but it’s basically the same process to identify ideals of M_n(R) for any ring R)

[crystal vale]

I have to show if H and K are subgroups of G, HK is a group if and only if HK = KH.

But I can only conclude that if HK is a group then KH \subset HK, how can I show HK \subset KH?

[green fox]

for left is it just sets of matrices with only non-zero entries in a particular column

[woeful sage]

Well we want to show that any product of the form hk, can be written as k'h' right? We know HK is a subgroup, so given any $hk \in HK$ let $hk = a^{-1}$ for some $a \in HK$ so that means $a = h_1k_1$ now I think you can finish the argument

[cloud walrus]

Neamesis

[woeful sage]

(This is proposition 14. in D&F btw, I didn't come up with this )

[crystal vale]

I got it, thank you ❤️

[woeful sage]

Thank our lord and saviour D&F:

🙏

[rocky cloak]

Those are left ideals, but there are more.

Anyway, if this is an exercise then I'm assuming you might be meant to find the twosided ideals.

They have a bit of a simpler structure (and there are much fewer of them)

[elfin wraith]

Is this where Morita equivalence comes in?

I’ve only ever looked at the case of 2 sided ideals but I vaguely remember reading the description of a project on Morita equivalence and I thought it mentioned something like that

[rocky cloak]

Morita equivalence would be more helpful for finding the left ideals

[near gazelle]

These are commutative rings. Surely the left ideals are equal to the right ideals in that case?

[rocky cloak]

Matrix multiplication is very much not commutative

[near gazelle]

oh wait nvm. I missunderstood your answer

[near gazelle]

Yes. I get the same answer. In fact $\mathbb{Z}_2$ is a field so there are only the trivial ideals. Note that $2\mathbb{Z}/4\mathbb{Z} \subset \mathbb{Z}_4$ is a prime ideal. We can lift it up to $\mathbb{Z}_2 \times \mathbb{Z}_4$ to create prime ideals for the product.

[cloud walrus]

Ante0417

[near gazelle]

Moreover, $\mathbb{Z}_2$ and $\mathbb{Z}_4$ are principal ideal domains. I then claim that it follows that $\mathbb{Z}_2 \times \mathbb{Z}_4$ is P.I.D. In principal ideal domains the notions of prime and maximal are equivalent.

[cloud walrus]

Ante0417

[green fox]

Yes, the right is the same except for the rows, and the two sided ideals are the trivia ideals

[rocky cloak]

The twosided ideals are indeed the trivial ones. And you get a bijection between left ideals and right ideals by transposing. I guess that's what you mean by "the same"(?)

[near gazelle]

Let $K$ be a field. Suppose I want to characterize the units in the non-integral domain $K[X,Y]/(XY^2)$. Is the following a valid argument?

$\bar{g} \bar{\epsilon} = \bar{1} \Leftrightarrow g\epsilon \in \bar{1} \Leftrightarrow g\epsilon = 1 + pXY^2$ for some $p \in K[X,Y]$. But clearly, $K[X,Y]$ is an integral domain. Hence, it must be that $g\epsilon = (r + a(XY^2))(r^{-1} + b(XY^2))$ for $r \in K$. Then, $\bar{\epsilon} = r$. Hence, we conclude that the units are exactly K.

[cloud walrus]

Ante0417

[rocky cloak]

I don't know how you're concluding the "hence it must be that" part, but have you considered
(1+XY)(1-XY)

[near gazelle]

No you are right. This proof is incorrect and I think its precisely the line referring to.

Maybe we can just add another case to it

We can maybe consider another case $g \epsilon = (r + p)(r^{-1} - p)$ where $p^2 | XY^2$

[cloud walrus]

Ante0417

[rocky cloak]

I think it would probably be better to write out a general product of polynomials and see which conditions are needed for them to multiply to 1 in the quotient ring

[elfin wraith]

Why more so for left ideals? Is it not a symmetric notion? I do perhaps just need to actually read and understand this so telling me to do that is also a valid response lol

[thorn jay]

I can't imagine an equivalence between categories of left modules implying an equivalence between categories of right modules

[elfin wraith]

Yeah I get that left and right modules are generally quite different, I guess my thought was more that it would be odd if equivalnce of left modules gave you more information than equivalnce of right

But youre not wrong in what you say, that would be a surprising implication I suppose

[rocky cloak]

I meant as opposed to twosided ideals

Morita equivalence gives you information about modules and left/right ideals are left/right modules

[thorn jay]

Yes, that was also what I meant lol

[elfin wraith]

Yes no sorry my question was about one sided ideals to begin with lol

The proof is straightforward for 2 sided

[rocky cloak]

Though it's an interesting question whether Morita equivalence of left modules implies a Morita equivalence of right modules

My first thought is no, but also that a counterexample would need to be ugly...

[elfin wraith]

I feel like an answer either way would be interesting none the less

[rocky cloak]

It can be shown that the left module categories R-Mod and S-Mod are equivalent if and only if the right module categories Mod-R and Mod-S are equivalent.

Actually, it's not too hard come to think of it.

Hom(-, R) is a duality between fg projective left modules and fg projective right modules, so would send a progenerator to a progenerator

[elfin wraith]

Ah yeah that makes sense

I really want to learn more about Morita equivalence an all of that, my noncom lecturer usually supervises dissertations on it and it seems pretty cool. She had suggested a mini-project about rings which are morita equivalent to the first weyl algebra but that seemed like too much work

[rocky cloak]

Possibly fun exercise: Morita equivalent rings have isomorphic centers. -> define the center of an abelian category such that Z(Mod R) = Z(R)

[thorn jay]

I saw some videos about this by a channel called K-Theory

[thorn jay]

Oh, well of course

Modules are nice objects, I should've known

[elfin wraith]

I have my group theory exam in an hour but I will think about this when I have a chance lol

I think this will come down to the same reason that Morita equivalence is only interesting for noncom rings though

[rocky cloak]

Yeah, commutative Morita equivalent rings are isomorphic, but you can still have equivalences between one commutative and one non-commutative

Which can be interesting

[thorn jay]

And then that commutative ring is necessarily isomorphic to the center of the non-commutative one :0

How much does that actually say about the noncommutative ring?

[elfin wraith]

Oh that’s pretty nice

[rocky cloak]

What the center is? Not a lot. Tells you if it's connected or not

[thorn jay]

No I meant whether it's Morita equivalent to it's center

Does that make it "pseudo commutative"? Or anything

[elfin wraith]

All is well, nothing to see here

[rocky cloak]

I guess yeah, idk if there's anything that particularly stands out about the rep theory of commutative rings, but you would have that the category of left modules is equivalent to the category of right modules, that's something

[thorn jay]

Hmm

Morita equivalence is surprisingly well defined and says more about the ring itself then I'd imagine

[thorn jay]

Men in black type shit

[thorn jay]

Well maybe you can say something about the ideals of the center and left ideals of your ring

[rocky cloak]

Actually, since it equals End(P) for a progenerator and you have this duality you should get that it's isomorphic to its opposite

[thorn jay]

Interesting

[rocky cloak]

No, scratch this. Was thinking Hom(-,R) would be the identity for commutative R, but thats not true 😢

[thorn jay]

That sucks, would have been a strong necessary condition

At least, I believe

[rocky cloak]

I mean, it's not weak I guess, but there are loads of such algebras.

[cobalt heath]

Is it normal that I forgot group and field theory as I do not use it in practice?

I forgot the Sylow theorems completely, except for that it uses actions to analyze structure

[kind temple]

i think that’s normal. i too forgot all of that same stuff. i do not use it in the math that i am interested in (god knows it will come back. the stuff you never thought you needed again always comes back…)

[cobalt heath]

Btw what are you majoring in?

[cobalt heath]

Also thanks for reassurance

[kind temple]

i recently graduated. i majored in math and cs

wbu?

[cobalt heath]

I am doing PhD in cryptography

[cobalt heath]

Graduated uni?

[kind temple]

yea, undergrad

[cobalt heath]

May I ask what you are planning to do?

Ah, sorry if you are not invlined to reply

[kind temple]

searching for swe, quant, and hardware verification roles. really just something in cs

planning to go back to school later

[cobalt heath]

Ahh, you going for CS?

[kind temple]

well, i need an income, so i am hoping to use cs for that. but i want to go back to school for math

[cobalt heath]

I see, sounds like a good plan!

As long as you have honed skills, you would settle pretty well

[kind temple]

thanks! wbu? are you planning to go into academia or industry after your phd?

[cobalt heath]

I am not sure, likely research position in industry

[kind temple]

i see, that sounds like fun!

[cobalt heath]

Yea, thanks!

[south patrol]

gl you both heh

[green fox]

Yeh

[crystal vale]

Hello Absta

[tough raven]

I think this is normal, and there is a silver lining that the understanding has more gotten lost than disappeared and you will be able to pick it up again if and when you try much more easily than the first time.

[tardy hedge]

z(y-1) is not a zero divisor on k[x,y,z]/(x(y-1), y))? But x is nonzero in k[x,y,z]/(x(y-1), y)) and z(y-1)*x = 0?

i was just looking at an example of a regular sequence

[rocky cloak]

What you wrote is the definition of being a zero divisor, so that would make it a zero divisor

[tardy hedge]

so x(y-1), y, z(y-1) is not a regular sequence?

this example in matsumuura says its a regular sequence

[rocky cloak]

x is not nonzero in that quotient ring is the mistake

Since (x(y-1), y) = (x, y)

[tardy hedge]

ohh shoot

[tardy hedge]

If A is noetherian, then the length of a maximal M-sequence contained in an ideal I is well-defined because of the ACC ?

[elfin wraith]

That group theory exam was the thing dreams are made of

I may have just had the worst exam period of my life but I ended it on an insanely high note Jesus Christ

All I needed was some unadulterated algebra, none of that topological or geometric fluff

[tardy hedge]

This is what dreaaaams are made of

Anyone else remember lizzie mcguire orrr

[elfin wraith]

And combined with getting an 80 on my diss yesterday I should actually be on for a first despite my horrific performance in the others

Now time to sit in the sun and drink all the maths back out of my head

[next obsidian]

F yeah

[next obsidian]

Your diss? Were you like “Riemann, imma show you sum, your math so whack it attract the cranks”

[elfin wraith]

Exactly what I did, wasn’t an honours dissertation

/ UG thesis depending on your proclivities

[south patrol]

Well done hehe

[stone elbow]

@kind temple sup bro

[kind temple]

sup

[woeful sage]

inf

[chilly ocean]

glb

[kind temple]

lub

[worn tinsel]

Hey, I got interested in understanding algebra a little more in depth, more than the usual introductory (first courses of an undergraduate) that involves stuff like matrices, sets, relations, functions, vector spaces, beginner to number theory stuff like congruence, polynomial concept, induction.

Specifically my interest is more about the abstract part, never got to learn in depth stuff like groups, rings, classes.

The thing is that the books I got so far are hard to understand the general idea because I guess most of them are used as reading material for courses (even if the preface says you can study it on your own).

Anyways, I'm looking for recommendations in books that somehow bridge that gap between undergraduate algebra to abstract algebra. My interests are mainly discrete mathematics, computer science, formal logic.

I post here because I don't know if I'm blind but didn't see an algebra channel, ping me if I'm wrong!

The TLDR is I want to bridge gap between algebra and abstract algebra, but books I've seen so far are too basic (gets up to undergraduate algebra) or simply too hard because it seems like the "second exposition" kind of resources.

[lusty marlin]

Undergraduate algebra is abstract algebra

Read Dummit and Foote

[worn tinsel]

Is that kind of "read in the order you need" kind of book? I'm looking at the contents and it looks like so

[lusty marlin]

You should start with groups in my opinion, but some later parts are sort of interchangeable in terms of order.

[woeful sage]

Maybe. They will tell you in the text at what point you can skip around, until then keep moving forward in a linear fashion

[lusty marlin]

A good understanding of groups is essentially necessary for algebra in general.

[woeful sage]

If you know groups, you will find rings easy

if you know rings, you can do anything 🔥

[woeful sage]

This is section 3.3

[lusty marlin]

Yes these little notes on what parts are now accessible are one thing I really like about the book.

[worn tinsel]

I'm just interested in general topics, I'm not studying mathematics formally yet, just computer science, my favourite part of CS was the discrete math, number theory, axiomatic logic part, and I also found out a real grasp on abstract algebra is mandatory for not getting lost in more advance topics.

[woeful sage]

Another thing I love about D&F is the 10 trillion examples they provide every section

[worn tinsel]

The book you mentioned looks like a big chunk

[woeful sage]

You don't have to do the entire book

[lusty marlin]

There is no abstract algebra without groups. You need to study them.

[woeful sage]

you can do like chapters 1, 2, 3, 4, 7, 8

chapter 5 as well, that's also important

it has the fundamental theorem of finite abelian groups

[lusty marlin]

Start with groups, make some progress, then decide which parts you don't need as per your requirements.

[worn tinsel]

What do you think about this one? is cheaper on my country
https://www.amazon.com/Contemporary-Abstract-Algebra-Joseph-Gallian/dp/1133599702

[woeful sage]

That one is good too

it's quite colorful and there's a lot of pictures

and there are fun little quotes at the end of every section

and at the end of every chapter there's a biography about a mathematician

quite a fun book my friend has it

But I myself prefer D&F as I love everything about this book

[worn tinsel]

Hard to judge, I just tend to read a big source and use other sources as complementary material, but always returning to the single source, I don't really know the technique to study math in a real setting, I guess my approach doesn't work with the university times

[woeful sage]

it's not just university times

learning math is like a ladder

without the bottom rungs, you cannot get to the top

[worn tinsel]

Not used to just skip straight to other chapters, that is why it is hard to find the material

[lusty marlin]

The 'technique to study math in a real setting' is to work through a well-chosen book.

[woeful sage]

Yes it's usually a bad idea to skip straight to a chapter without have the prerequisties to study that chapter

[worn tinsel]

Yeah that is my point but you always need something from the other field to learn the field you are at, like you could start from naive set theory and I guess you will still need some first order logic at minimum

[woeful sage]

I don't see what the confusion is

[woeful sage]

No?