groups-rings-fields

[daring sage]

[bleak abyss]

Why would you do it on Q then R rather than just do power series?

Or just define log as \int_1^x 1/t

[inner acorn]

I think a lot of texts define exp(z) as a power series?

[bleak abyss]

And show that guy has an inverse. Doesn't matter which way you do it but yeah

[daring sage]

i'm definitely taking this exercise too far

i appreciate the help

[inner acorn]

xD use the fact it's strictly increasing to show injective

[daring sage]

It's weird cause i don't know what a prof would like to see in this case

[marble wagon]

cuz power series is bad

[bleak abyss]

Fam use common sense, if someone's self-teaching group theory and wondering how pedantic they're gonna be about e^x they're not looking forward to say that a development is ill-suited to later learning p-adics

[daring sage]

...since i've never taken a math class

[bleak abyss]

"cuz power series is bad" the hottest of takes

[marble wagon]

constructing it in Q has strictly increasing directly

so when you extend to R you get log for free

[solar wyvern]

if it's an algebra exercise, assume existence of exp and log, then simply show that it's a group hom and you're done

[daring sage]

gotcha

appreciate it

i guess i dont need to reconstruct the reals for every problem involving them

[bleak abyss]

Yeah just black box calculus as calculus for now

You won't even really have to do much of that

[inner acorn]

xD

aww

constructing the reals is fun ❤

[daring sage]

I'm doing topology next sem before analysis

wish me F

[inner acorn]

topology before analysis

sounds like physics alright xD

[bleak abyss]

The main times I remember using analysis facts in algebra were... some in linear algebra like business surrounding the spectral theorem, also the proof I know of Cayley-Hamilton

[solar wyvern]

dami how does galois theory work

[inner acorn]

tells u 'bout polys

[bleak abyss]

And in ring theory I guess some business about C[0,1]

[solar wyvern]

how do cubics/quarts work

[bleak abyss]

S_3 and S_4 are solvable

So those are ez

[inner acorn]

their galois group iso to subgroup of S_3 or S_4

which are solvable :3

[solar wyvern]

how do I know which irr polys have which galois group

[bleak abyss]

Sasha get mega sniped

[inner acorn]

shhh

xD

[bleak abyss]

You have to play around a bit

[inner acorn]

you calculate their Galois group

find the roots

[marble wagon]

start with the discriminant, buncho explained that part last time

[inner acorn]

automorphism which preserve Q

[marble wagon]

that splits the possible groups in half

[inner acorn]

see how the roots permute etc...

[solar wyvern]

see how the roots permute etc...

@marble wagon do I have to screw around with resultants to narrow down to a precise galois group

[bleak abyss]

So if you have an irreducible cubic

Even adding one root gives degree 3

So the Galois group is A_3 or S_3

[marble wagon]

not resultants, I wouldn't say

[solar wyvern]

what do you mean by "adding one root"

[bleak abyss]

As in, adjoining one root to Q

[marble wagon]

cubics are trivial as you saw last time

[bleak abyss]

Is a degree 3 extension

[marble wagon]

discriminant distinguishes between S3 and C3

[solar wyvern]

oh ya since A3=C3

[bleak abyss]

Yeah whether it's a square or something

[marble wagon]

for quartics, you reduce to an associated cubic

and that cubic, along with the quartic discriminant, gives you most info

https://planetmath.org/galoisgroupofaquarticpolynomial

the resolvent

[solar wyvern]

ah, that cubic is the resolvant

and yeah, resolvant, not resultant

[marble wagon]

there's an assortment of other techniques to use for other cases

D&F walks you through some cases

like examples of degree 7 polys that you can compute

[inner acorn]

I mean, if you have an automorphism f of Q(√2), which fixes elements in Q
then the quadratic x^2 - 2
will be mapped to f(x)^2 - 2
Moreover f(r) = r, where r is rational and
f(√2) = √2 OR f(√2) = -√2

so you have 2 different automorphisms

the identity, and a conjugate like map

which is iso to S2 I think? (because you can observe how f just permutes √2)

[solar wyvern]

@marble wagon as long as the GG is solvable you can compute it, right?

[marble wagon]

you can compute it even when it isn't solvable

but it's probably annoying I dunno the general techniques

I've only done it for some examples

[solar wyvern]

they look annoying tbh

even for quartics

[marble wagon]

quartics are easy tho

[solar wyvern]

[marble wagon]

see the link

it's very explicit and quick

[bleak abyss]

I'm an expert on Galois groups of linear polys

[marble wagon]

nice

[bleak abyss]

At some point soon I've gotta review my Galois theory

It's been a while and it was somewhat meh-ily taught the first time

[solar wyvern]

thanks a lot pear

[inner acorn]

that's how I feel with most of undergrad dami xD

[inner acorn]

fun fun

[marble wagon]

quartics are the usual capstone example on this for galois theory anyway

[bleak abyss]

Yeah same with a bunch of things in principle

[marble wagon]

usual problem to put in a final

or a pset

[bleak abyss]

But Galois theory is somewhat high priority

[inner acorn]

o: we always got an easy quintic

[bleak abyss]

Given that I'm prob gonna do arithmetic geo

[marble wagon]

you're taking core algebra though right?

[inner acorn]

to show it isn't solvable by radicals

hence proving there is no quintic formula

I'mma go back to anime x3 cya

[bleak abyss]

I'm still deciding

A large block of core algebra's just gonna be review

[woven delta]

There is a quintic formula

It's just not very nice

[bleak abyss]

That's first semester

[marble wagon]

seems nice

[bleak abyss]

Stuff up to Sylow theory I've got, I didn't really do much solvable/nilpotent groups

[marble wagon]

the material might sound familiar / boring but it's very fast paced

or it should be anyway

[bleak abyss]

Also maybe the categorical perspective will be new

Non-commutative rings I'm iffy on since we mostly ignored that in my class

Rep theory I've probably got down for the most part

Unless he's doing the stuff like Artin-Wedderburn

[marble wagon]

weddenburn's thm is pretty fundamental

at least for finite group algebras

[bleak abyss]

Yeah we never really did that in my rep theory class since we sorta religiously stuck to reps as homs G->GL(V)

[marble wagon]

then you couldn't have done a lot of character theory right?

[bleak abyss]

Uh, depends on what you mean by a lot, we did prove the orthogonality relations

[marble wagon]

odd, I thought you need some consequences of weddenburn for that

like how the sum of squared dimensions is the order of the group

I think you should take it anyway

review is nice

means less work

[bleak abyss]

Nah you don't Wedderburn, trying to remember how we did it. We did talk a bit about C[G] as an algebra in that proof but there was no semisimplicity floating around

[marble wagon]

maybe there was, implicitly

I mean you don't need the full statement of weddenburn

a lot of simplifications are made

division algebras over C are just C itself, etc

[inner acorn]

@woven delta I meant to say, there's not a quintic formula in terms of radicals o:

[woven delta]

I know lmao

[bleak abyss]

Okay so, there's showing that the irreducible characters are themselves orthonormal and there's showing that they span class functions

I think the second bit might've involved toying with the group algebra?

[inner acorn]

qwq

[marble wagon]

the way I learned it is by heavily using corollaries of weddenburn's thm

cuz that's easier

[bleak abyss]

That makes sense

[marble wagon]

I had to prove the full weddenburn's thm in a pset

which was v. annoying

essentially a lot of different characterizations of semisimplicity

[bleak abyss]

But yeah I'll see, I might consider auditing core algebra even if I pass the qual exam

[marble wagon]

as stated in D&F

[bleak abyss]

People say it's bad to take more than 3 classes at a time with the TA load

[marble wagon]

probably

[bleak abyss]

And while I've probably done harder than that before it also put me in a situation where I barely figure out enough to solve the problems in front of my face

Which is never a satisfactory state of affairs

[marble wagon]

yeah it's not nice to be pressured for time in that way

I like taking my time to read more stuff and learn what I'm doing well

[bleak abyss]

So yeah right now the options are intro topology, intro algebra, AG, and rep theory

If I'm gonna audit one it's probably intro algebra

[marble wagon]

for some reason

[daring sage]

this is just lagranges theorem yeh?

[wind steeple]

y

es

[daring sage]

kk ty

[uncut girder]

Is the contraction of a maximal ideal always maximal?

[marble wagon]

no

[uncut girder]

Example

Pls

[bleak abyss]

@uncut girder contraction means what exactly, preimage?

[uncut girder]

Preimage of a ring homomorphism

[bleak abyss]

Include Z into Q

0 is maximal in Q

Its preimage is 0, but that's not maximal in Z

[marble wagon]

if you want k-algebras instead you can do the analogue k[x] into k(x)

[uncut girder]

Yeah, this was an example in the text, facepalm @bleak abyss

[bleak abyss]

Lmao, yeah this happens

[uncut girder]

help

how to show m is maximal in A[[x]] implies m^c is maximal in A
where A is a commutative ring

m^c is the contraction of m via any ring homomorphism f: A -> A[[x]]

m is an ideal

[mild laurel]

Try proving it by contradiction

[uncut girder]

I dont know how to use A[[x]]

[mild laurel]

What do you mean?

[uncut girder]

I know I have to use the fact that m is maximal in A[[x]] somewhere

[mild laurel]

Sure, but there's nothing really that special about A[[x]] here, just have to use general properties of maximal ideals

[marble wagon]

A[[x]] is pretty special

think first about the case A = k a field

where k[[x]] is a local ring

why is it local? what do ideals look like?

generalize to A[[x]]

[uncut girder]

Is this saying the preimage of m will always be the same, no matter which ring homomorphism you use?

[uncut girder]

Yo I did it

[wind cypress]

Are groups Z/4Z and {1,i,-1,-i} <= C^* isomorphic? C^* means multiplicative group. I am thinking maybe, since both have 4 elements, {0,1,2,3} which can be mapped to {1,i-1,i}. Right?

[bleak abyss]

What's your candidate isomorphism?

[wind cypress]

Hmm... some bijection between the two. 0 <->1, 1<->i, 2<->-1, 3<->-i

[bleak abyss]

0->0?

[wind cypress]

Oh sorry 😃

I fixed it

So would that be, x^n where n is 0, 1, 2 or 3

or actually, i^n

[bleak abyss]

Alright, that's definitely a bijection, then you can check it's a homomorphism

One general tip that's worth noting

So actually here's an exercise: let C be a cyclic group and let G be any group. Then a homomorphism C->G is determined by where it sends a generator of C

[wind cypress]

Hmm

[bleak abyss]

So in particular all you really needed to do was tell me where 1 went

(or 3)

And to make this problem even faster, here's a fact: If g\in G and |g| = |G|, then G is cyclic

So just showing that i has order 4 does it (I'm assuming you know that all cyclic groups of a given order are isomorphic in order to play that game, prove it if you don't know it)

[wind cypress]

Unfortunately I don't know this stuff too well...

Another, if I have Z/9Z, and (Z/3Z) x (Z/3Z), are those isomorphic? I was thinking, it is, since i can link {0,1,2,3,4,5,6,7,8} with {(0,0),(0,1), (0,2), (1,0) ... (2,2)} thus making a bijection between those. But I think there is a faster way with this too...

Since 9 is 3^2 or something....

[somber bramble]

well, a bijection won’t give you an isomorphism yet

there are many bijections

consider this:

let’s say φ is your isomorphism

what shall φ(1) be?

pick something that looks good

actually, no, first of all

what is ϕ(0)?

wthere’s only one choice here

@wind cypress

[wind cypress]

Hmm. (0,0)

[somber bramble]

yea, because the identity is always mapped to the identity with homomorphisms

now, pick any value for ϕ(1)

we’ll see if it can work

[wind cypress]

Oh... I was just thinking (0,1)

[somber bramble]

aight, let’s go with that

what’s φ(2)?

[wind cypress]

(0,2)

[somber bramble]

yep, and φ(3)?

[wind cypress]

(1,0)

[somber bramble]

aight but that won’t be a homomorphism now

[wind cypress]

Oh...

[somber bramble]

because you need to have φ(3) = φ(2+1) = φ(2) + φ(1) = (0,1) + (0,2) = (0,0)

agreed on every step?

does every = make sense?

[wind cypress]

Oh, damn

Yes

[somber bramble]

in particular what this means is that no matter what you choose φ(1) to be, that fully determines all values

because φ(n) = φ(1+1+…+1) = φ(1) + φ(1) + … + φ(1)

and further, you can’t have φ(n) = (0,0) for any n except 0

what does this tell you about the order of φ(1)?

[wind cypress]

hmm, you mean φ(0+1),

[somber bramble]

what?

you know what the order of an element is, right?

[wind cypress]

would need to be φ(0) + φ(1)

Not sure in this case

[somber bramble]

yea, but it is guaranteed because φ(0) is the identity of the target group

we ensured that bit

the order of an element is how many times you can add/multiply it to itself until you get the identity back

the order of 1 in ℤ/9ℤ is 9

because 1+1+…+1 (nine times) is 0, and that’s the first time it happens

what’s the order of (0,1) in the other group?

[wind cypress]

Uh, if i add (0,1) + (0,1) + (0,1) i get (0,0) so that's 3

[somber bramble]

aye

now, based on the discussion above, what should the order of φ(1) be?

[wind cypress]

Uh, three?

[somber bramble]

you sure?

[wind cypress]

I mean, in which group... if they both need to be 9

[somber bramble]

remember, we want φ(1+1…+1) to not be 0 until 1+1+…+1 itself is 0

[wind cypress]

But the first group it is 9, and the other, 3

[somber bramble]

well, some element in the other group has order 3

but we’ve already seen that that one wouldn’t work anyway

all that tells us is it was the wrong choice

but not whether a right choice exists

[wind cypress]

Right.

[somber bramble]

however, there are only so many elements in ℤ/3ℤ, so you can quickly list them all and find their orders

the general idea is this: an isomorphism must map elements of order k to elements of order k

because, in the more general case, if g is an element of order k, then you need to have this situation:
φ(g) = h
φ(g²) = h²
…
φ(g^{k-1}) = h^{k-1}
φ(g^k) = φ(id) = id

and none of the righthand sides must be the identity until the last

so g and h have the same order

[wind cypress]

Hmm

[somber bramble]

(I’ve written it as powers here, in the case above it was written with additions)

but does this statement make sense to you?

[wind cypress]

Well, some sense sure

[somber bramble]

but?

that doesn’t sound convinced

[wind cypress]

Heheheh

¨Hmm, if i mapped 1 to (1,1), and then 2 is (2,2), but that would lead to (0,0) too soon

Wait it's in the other group, where the order of the element is only three anyway

[somber bramble]

“the element number”?

[wind cypress]

order sorry 😃

[somber bramble]

yep, every element in ℤ/3ℤ × ℤ/3ℤ except for (0,0) has order 3 unless I’m mistaken

conclusion?

[wind cypress]

Then add (0,1) and (1,0) to (2,2)...

[somber bramble]

huh?

[wind cypress]

I mean, to reach all the points in that group. If it's allowed to chance what you are adding like that

[somber bramble]

well, φ(2) must be φ(1) + φ(1)

otherwise it’s not a homomorphism

[wind cypress]

That would work until we get to φ(3), which would give (0,0) again

[somber bramble]

what would work?

[wind cypress]

I mean, if i select (1,1) to be φ(1)

[somber bramble]

look, just look at the orders

what was the conclusion we made about the order of φ(1)?

what does it have to be if φ is to be an isomorphism?

[wind cypress]

Yes it has too little order compared to the other group

Homomorphism and bijection between those groups

[somber bramble]

that’s not an answer to my question

what does the order have to be

[wind cypress]

9?

[somber bramble]

yes

is there such an element?

[wind cypress]

Not in the other group

[somber bramble]

conclusion?

[wind cypress]

They are not isomorphic

[somber bramble]

correct

[wind cypress]

Huh... was simpler then what I thought...

[somber bramble]

looking at the orders of elements is a useful trick to decide two groups cannot be isomorphic

[wind cypress]

Was sidetracked since I can link the φ(n) to the other group, it has 9 points, but i guess I can't fill the homomorphism....

[somber bramble]

(however, as a warning, even if there are the same amount of elements for each order, they still might not be isomorphic)

(the orders can only tell you it’s not possible)

[wind cypress]

Well, sounds like this is going to be a long night, since that's kinda easy exercise, and I got way more until I got an exam in the morning.

[somber bramble]

why are you only studying now

[wind cypress]

I have been studying before, just was kinda hard course, and gotta do some last minute reading...

Are Z/6Z and S_3 isomorphic, S_3 mean group of permutations in the group {1,2,3}. This would be next. It's possible, sounds like it, both have 6 elements... but not sure yet.

[somber bramble]

first try to do the same thing we just did

[wind cypress]

Order of elements is five in the first group

[somber bramble]

order ”of elements”?

also, no it’s not

there is no element of order 5 in either of those groups

[wind cypress]

Wait six of course 😃

[somber bramble]

there are six elements in either group

each has an order

[wind cypress]

Okay the 0 in first group has order of six

[somber bramble]

uh, no

0 always has order 1

cause 0 = 0

as in

the sum with one element

ya know

[wind cypress]

1+5 is congruent with 0, so 1 has order of 6

[somber bramble]

that’s not how it works

I mean, it does

but that’s still now how it works

it has order 6 because
1 ≠ 0
1+1 ≠ 0
…
1+1+1+1+1+1 = 0

[wind cypress]

Oh, okay

The second group has elements {1,2,3}, {1,3,2}, {2,1,3}, {2,3,1}, {3,2,1}, {3,1,2}

[somber bramble]

shitty notation but whatever

yes

[wind cypress]

Hmm, tedious to check by hand, which element could have order six

[somber bramble]

honestly if checking the orders on a six set group is too much effort then I cannot help you

[wind cypress]

Heheeh

Hmm, easier is to note that if group has 6 elements, it's isomorphic to Z/6Z

[somber bramble]

that you’ll have to prove

[wind cypress]

Time for a little break though, getting hungry.

[wind cypress]

Moving on for some other kinds of exercises, two of which are very similar:

1. Let p be a prime number, and let C be a cyclic group with an order of p^n with some positive integer n. Show that group C has an element with an order of p. (Order means how many elements are in the group. Element h has the same order as cyclic group <h>).

2. Let p be a prime number, and let G be a group with order of p^m with some positive integer m. Show that group G has an element with an order of p.

1. So, probably first thing to do, is type C = {e, a_1, a_2, ... a_ p, ... a_ (p^n-1) } which would be isomorphic to Z/p^nZ, if i understand correctly.

and a with index p would form a group of <e, a_1, a_2, ... a _p-1) which has p elements

But, now what...

[sharp sonnet]

I assume you did Lagrange's theorem?

[wind cypress]

I've heard of it

I really shouldn't have taken the abstract algebra yet, but next year, when I had more experience with math. But can't be helped now.

[sharp sonnet]

Well, it would be nice to know if you can use it to solve this question

[wind cypress]

Do you mean [G:H] = #G/#H ?

Where G is finite group and H < G?

[sharp sonnet]

yeah

well, for 1. it's even easier

what does it mean that C is cyclic?

[wind cypress]

Oh... man. Didn't think of that

Cyclic group can be generated by a single element in that group

[somber bramble]

since you didn't continue before let me just tell you: no, not all groups of order 6 are isomorphic

[sharp sonnet]

yeah, so just take the generator of that group

[wind cypress]

Oh, yeah, was away for a while, will get back to that though

[sharp sonnet]

and look at different powers of it

and what order they have

[stone fulcrum]

Oof, I've missed 2 hours of group theory convo

[sharp sonnet]

and well

[wind cypress]

Naah I kept a break

[sharp sonnet]

maybe once you've done that and think about lagranges theorem, you can do 2. yourself

[wind cypress]

But that just tells me p^n / p = p^(n-1)

[sharp sonnet]

Lagranges theorem tells you that the order of any subgroup divides the order of the original group

[wind cypress]

p^n = k*p ? with some k... ?

[sharp sonnet]

so if you take any non-identity element a of a group of order p^m, the subgroup <a> is of order p^n, where 0<n<=m

this reduces the problem in 2. to the problem in 1.

[wind cypress]

Oh you were already in the second exercise

[sharp sonnet]

well yeah, i thought i said that

in 1. you wont need lagrange

you know how your group looks like

if it's generated by a it's just {a^0, a^1, ..., a^(p^n-1)}

just check the order of every element

[wind cypress]

hmm with n=2 that would make a^p one of the elements in that group

Damn BSOD...

Anyway with m = n+1, that would make p^(m-n) p, and thus a^p in that group

[wind cypress]

Any calculator I can use to check something like this, permutations, like a = (26345) b = (1753), then (a^-1)*b?

there are numbers 1 to 7 available in the permutation group

About the earlier, S_3 permutation group, apparently it's generator is (1 2) (1 2 3), can't be done with one cycle.

[cerulean swan]

I need someone to help me with some algebra

can we get in a call

[sick acorn]

@wind cypress yeah, symmetric groups aren’t cyclic for n >= 3

[somber bramble]

there's a super easy algorithm for calculating the product of cycles

let's take your thing from above
(26345)^{-1}(1753)

first, inverting just means writing it backwards, so
(54362)(1753)

now, start at 1. go through all the cycles from right to left and just look what path 1 takes. So in this case, the second one takes it to 7, and the first leaves 7 alone. so 1 goes to 7. write
(17

now, look at what happens to 7. second one takes it to 5, then first one takes that to 4. so,
(174

continuing,
(1743
and then you see 3 goes to 1, so you close the bracket. Then, start a new one with the next number not yet listed, so 2, and continue until you've written all numbers. Then, in the end, delete all 1-cycles (things like (5)).
(1743)(256)

@wind cypress

[brisk granite]

Has anyone used dummit and foote for abstract algebra? Is it any good? Should I consider using more standard texts like Herstein?

[woven delta]

Lmao what

D&F is much more standard than Herstein

@brisk granite

[brisk granite]

ok

my bad

I think I was thinking of another book then

[onyx mirage]

Is there an O(1) algorithm for finding generators of cyclic groups with finite cardinalities?

@brisk granite i'm using D&F

[fringe nexus]

it should be the "smallest" element in the group

[onyx mirage]

I like it, but personally do think the amount of exercises are kind of overkill, and it doesn't help that some new theory like lagrange's theorem are only covered in exercises

@fringe nexus I'm talking about cyclic groups under the operation of taking the product

under addition it'd be the 2nd smallest element, 1, yea

[chilly ocean]

lagrange's theorem are only covered in exercises

[onyx mirage]

In chapter 1 I mean

[chilly ocean]

ah ok

flip to pp.89

[onyx mirage]

Yeah, I noticed sometimes they just like throwing in later theory into earlier exercises

[chilly ocean]

i like it

[onyx mirage]

The proof of lagrange's theorem is an exercise in chapter 1.3 if i'm not mistaken

[fringe nexus]

there is no algorithm for multiplicative

https://en.wikipedia.org/wiki/Primitive_root_modulo_n#Finding_primitive_roots

multiplicative group

besides just brute force

in general

[onyx mirage]

🤔 peculiar

[fringe nexus]

there are special cases i guess

where if you have more information

[onyx mirage]

No special cases where you could use things like fermat's little theorem, or anything?

[fringe nexus]

i think you use the totient function actually

if you know it

and you can find orders of elements

but i am not 100% sure

[onyx mirage]

I think euler phi function doesn't rule out that your element could be a factor of your result I think

[fringe nexus]

this would make it more efficient

but still

[onyx mirage]

|(p+1)| = p^(n-1) in Z/p^n Z

[onyx mirage]

Had to prove that using binomium theorem, but don't know entirely how

[raw moth]

once you have one generator it's easy to find the rest

take g the generator, n the order of the group, then it's g^k where 1 <= k < n and gcd(k,n) = 1

[onyx mirage]

yeah, figured that

|g^a| = |g| / gcd(|g|,a) for cyclic groups

🤔 Is there a way to know HOW MANY generators a cyclic group should have?

I'm assuming all finite cyclic groups are isomorphic to Z/ p Z, so proving it for those would be sufficient

[somber bramble]

a cyclic group by definition has just one generator?

or do you have a different definition of cyclic

[onyx mirage]

They can have multiple potential singleton generators

[somber bramble]

oh, like that

[onyx mirage]

mhm

[somber bramble]

for finite groups it’s just the totient function

a k in ℤ/nℤ is a generator iff k and n are coprime in ℤ

[onyx mirage]

counterxample: let k=5 and n=7

Ah I'm talking about the generator under multiplication

[somber bramble]

5
5 + 5 = 10 ≡ 3
3 + 5 = 8 ≡ 1
1 + 5 = 6
6 + 5 = 11 ≡ 4
4 +5 = 9 ≡ 2
2 + 5 = 7 ≡ 0

and that’s all

[onyx mirage]

Yeah, under addition it's pretty trivial

[somber bramble]

wait so what is it you actually want?

[onyx mirage]

generators of (Z/p Z - {0}, *)

Z*

mhm

and perhaps an algorithm that can find a generator in constant time

[somber bramble]

well, at least for p prime, that is a cyclic group again, and so it would be isomorphic to ℤ/(p-1)ℤ with addition, and have φ(p-1) generators, right?

that doesn’t tell you the generators

but it should be that many

[onyx mirage]

That should help if it's possible

[mild laurel]

no known algorithm exists for finding generators

[somber bramble]

that’s false. you can brute force it :P

it’s not exactly a good algorithm

as it has complexity O(n²)

[onyx mirage]

It can run in O(n) if you use dynamic programming

[somber bramble]

I believe that

doesn’t even seem too difficult to write

[onyx mirage]

without it might even be worse than O(n^2)

[somber bramble]

why? you have n elements and have to do ≤n actions on each

(I’m assuming multiplication mod n to be O(1))

you can actually stop halfway through, though that doesn’t affect the complexity

[onyx mirage]

yeah it's just O(n^2) I was thinking of some other algo (interval matching) that kinda looked similar

but you don't have to consider other elements when calculating the order of 1

[mild laurel]

I don't see how dynamic programming helps here at all. If you do have a primitive root, you're never going to get back to the same position again

[raw moth]

if you have a cyclic group of order n, then there are phi(n) generators lol

[somber bramble]

I already said that read the convo afterwards

[onyx mirage]

It's still a matter of finding them

[raw moth]

@somber bramble ya I'm just confused why it was asked after what I said earlier :^)

anyway let me go and find my manners again because I'm tired as shit, sorry

[onyx mirage]

same

I somehow got tensed up muscle knots in my chin and it's killing me

[somber bramble]

I have today off and slept well

:)

[onyx mirage]

:))))))))))))

[somber bramble]

it’s 10AM, I’m sitting in my room, preparing for a day of playing Breath of the Wild

[onyx mirage]

ooh fun

I remember that

[somber bramble]

haven’t picked it up for a while

[onyx mirage]

Its content doesn't really scale with the difficulty

i.e. the game becomes too easy if you complete everything

[somber bramble]

I’m 3/5 of the way through, just need to beat the storm boss and then get to the castle i assume. the dungeons have been very underwhelming but I love the world

I hope the next installment focusses more on the dungeons again

well the next “proper” one, cadence of hyrule doesn’t count

super excited for that tho

[hollow peak]

I loved the game

but the dungeons were always the same

[onyx mirage]

that's why you go for calamity ganon as soon as the game starts :^)

[hollow peak]

speedruns are very fast in this game x)

[onyx mirage]

it's the first quest you get, so why wouldn't you

[somber bramble]

:P

[onyx mirage]

Maybe using shor could simplify the process of finding the generator 🤔

still no O(1), though

[rapid tusk]

@simple valley

[simple valley]

what

[rapid tusk]

How do

@simple valley

Got exam tomorrow and there are no solutions to anything

[mild laurel]

What is the definition of a partition?

[rapid tusk]

[mild laurel]

Do you understand this definition?

[rapid tusk]

not really :/

[simple valley]

what's the issue

[rapid tusk]

I don't get this Q at all and neither does anyone I know from my class

[simple valley]

also this is totally #proofs-and-logic territory

[rapid tusk]

[simple valley]

for the record

[rapid tusk]

Group theory tho no?

[simple valley]

at this stage there's no group theory involved

[mild laurel]

A partition is just a way of splitting up your set into separate parts

For example, one partition would be {{a}, {b,c}}

Since every element is in one of these partitions, and no element is in two

[simple valley]

every element is in at least one part: forall x in S, exists P such that x in P

this is the statement that S = union P

no element is in multiple parts: forall x in S, not exists P1 and P2 such that x in P1 and x in P2

the contrapositive of this statement is,

forall x in S, exists P1 and P2 such that x in P1 and x in P2 implies P1=P2

[rapid tusk]

[simple valley]

this can be restated as either, "different parts do not intersect"

or "parts either don't intersect or coincide"

equivalences of all of these formulations are basic set theory

[rapid tusk]

Why is this in my group theory notes then if this is just basic set theory

[mild laurel]

because you can't do group theory if you don't understand basic set theory

[rapid tusk]

.> yea true

cheers tho, we're just digesting what you all said lmao

[mild laurel]

Here's an analogy that might help or not

You're taking a pie (your set) and cutting it up into slices (your partition)

Every part of the pie is in some slice, and no part of the pie is in two different slices

This cutting up of the pie gives you a partition of the pie

[rapid tusk]

yea, that makes much more sense than their definition tbh

[simple valley]

to you perhaps

"their" definition is rigorous

meaning you can formulate precise objective statements bsed on it

with no leeway for misinterpretation

[rapid tusk]

yea true, i'm just dumb

[mild laurel]

I'm surprised that no one in your group theory class knew what a partition was and no one could figure it out through the definition

[rapid tusk]

Idk our class isn't the best at this module

[fringe nexus]

show that the operations on D_4 are permutations of the corners

[rapid tusk]

what are corners?

also D_4?

[fringe nexus]

Uh

have you not worked with the dihedral group

[rapid tusk]

no?

[fringe nexus]

is that not what S(square) means

ok nvm

[rapid tusk]

[fringe nexus]

i think it should be

because D_4 is isomorphic to a subgroup of s_4

[rapid tusk]

S(Square) is the symmetry group of the square

[fringe nexus]

yes

[rapid tusk]

reflections and rotations

[fringe nexus]

ok

thats the dihedral group

[rapid tusk]

oh? we never called it that

[fringe nexus]

ok so you have four points right

[rapid tusk]

yup

[fringe nexus]

say you label them 1, 2, 3, 4

now we know the set is generated by r,v

what happens to the points when you operate on it by r

1 - > 2 -> 3 - > 4 ->1 right

note that this is a permutation

in cycle notation its (1 2 3 4)

[rapid tusk]

(4,1,2,3)?

?

[fringe nexus]

uhh

cycle notation of (a b) means a->b->a

i guess 4123 works

but now you have an explicit way of constructing your isomorphism

since R, F generate your whole group

[rapid tusk]

hmm

.>

Group theory is hard smh

[fringe nexus]

[rapid tusk]

[fringe nexus]

[clear fiber]

What topics should I study in order to have the foundation to learn group theory?

I’ve taken a discrete mathematics course

And learned basic set theory

What else do I need?

[topaz solar]

I'd think you should be good with that, so long as you know how to do normie proofs

i always recommend linalg since linalg is great, but group theory doesn't take a whole lot of prereqs

[clear fiber]

Ye I can do normie proofs

And I’ve also taken linear algebra as well

[topaz solar]

Of course, consider your specific place's prerequisites and corequisites and such, but you'll be fine

So long as you can read a book and figure out what an equivalence relation is you're fine

[clear fiber]

Alright sweet

I think Imma dive headfirst into group theory this summer then

[void dragon]

What is normie proof?

[topaz solar]

A proof that isn't super difficult

Like proving the kernel of a homomorphism is a normal subgroup

[void dragon]

Oh I see, we can do that in one liner bar definition

[clear fiber]

Oh

I have no idea what that means lol

I was thinking more like, “prove the square root of 2 is irrational “

[void dragon]

I think it's about the same

[clear fiber]

Ok

Both linear algebra and group theory are sub fields of abstract algebra, correct?

[topaz solar]

yes

they are both algebra

And that's about the same level of proof, just with a bit more definitions behind it

[uncut girder]

Question: what are the prime ideals in Z[x]

[topaz solar]

@uncut girder if i remember correctly, it's irreducibles right?

(with some from from the prime ideals in Z too)

[uncut girder]

Yeah

[topaz solar]

can't forget everyone's favorite ideal (2)

[uncut girder]

What does this mean?

[topaz solar]

I have no idea on context, but looks like some weird diagram detailing some structural stuff about the ideals

The super dense scribbles seem to be more or less something like $\dots$

[cloud walrus]

Darkrifts:

[topaz solar]

But i don't know context so that's all I can offer @uncut girder

[uncut girder]

It's supposed to be a picture of Spec(Z[x])

[topaz solar]

Like each vertical line is a set of ideals which can be factored into (p) for the primes or some such

Ooh makes sense

[uncut girder]

It comes from Mumford's Red Book of Varieties and Schemes

[topaz solar]

Hm

I don't really know much on varieties and schemes so I can't offer much insight there, but it looks to me like it's showing relationships (possibly topological because I know Zariski is a thing though I know nothing about it) between the prime ideals?

[uncut girder]

Yes

V(E) is the set of prime ideals which contain E

[topaz solar]

Makes sense

[uncut girder]

Also

(x+3)(x+2)= x^2 ^5x +6 = x^2+1 (mod 5)

This would explain why the line for x^2+1 passes through (5, x+3) and (5, x+2) ??

[topaz solar]

It likely does

Since integers usually have some focus on mod, it wouldn't surprise me

The dotted circles are werid, but I guess that's to show it doesn't have the same kind of intersection?

[uncut girder]

(x+1)^2 =x^2 +2x + 1 = x^2+1 (mod 2)

[topaz solar]

yep seems right

[uncut girder]

Dotted circle means E is contained in that prime ideal

I think

[topaz solar]

They don't label it like they do the explicit intersections though, so idk

That part's weird to me, but I'm not the one with the book

[uncut girder]

I'm not reading that book either

[topaz solar]

oh lmao

[uncut girder]

It's an exercise in AM to draw Spec(Z[x])

[topaz solar]

am
oh boy

[uncut girder]

What

[topaz solar]

nothing

[uncut girder]

Well

x^2+1 is not contained in the ideal (5,x+3)

But x^2+1 is 0 in Z[x]/(5,x+3)

So I'm confused

[topaz solar]

So, x^2+1 isn't contained in 5, x+3, but it's a multiple of x+3 (mod 5) and you end up modding out the 5s, right?

[uncut girder]

Yeah

[topaz solar]

So you have (x+3)(x+2) = x^2+5x+6 = x^2 + 1 = 0 because you mod out (x+3), right?

[uncut girder]

Yeah

Wait a sec, x^2 +1 is actually in the ideal (5,x+3)
Since x^2 +1 = (x+2)(x+3) - (x+1)5

So those dark dots really do signify membership in V(E)

[topaz solar]

h m m, looks in there to me

[uncut girder]

V(E) are the closed sets in the Zariski topology

[topaz solar]

which means very little to me since i know basically nothing topology, other than Zariski and D(f) is a thing

[uncut girder]

Closed sets are complements of open sets

[topaz solar]

yes

i know that much

[uncut girder]

Open sets are notions of closeness

Do you know general topology?

[topaz solar]

The one thing I do know about zariski is the opens have the ideals that dont share a certain common element, which seems weird as far as closeness goes

And obviously no

I know super duper basics, but that's about it

[uncut girder]

Yeah

[topaz solar]

Whatever you were going to say, feel free to say it, it'll probably just go over my head

[uncut girder]

Zariski topology is very weak

In Spec(C[x]) an open set is the complement of a few points

the Zariski topology is coarser than the usual topology

[topaz solar]

i'm such an imbecile with topology that's almost meaningless to me

Makes sense

[uncut girder]

Why do you know CA when you dont know basic topology? 👀

[topaz solar]

I said almost

[uncut girder]

Commutative Algebra

[topaz solar]

oh
commutative algebra is much nicer than topo

And because I spent a year focusing on getting algebra

i've started this next cycle on analysis

which will give me some baby topo to use to learn topo better

[uncut girder]

Oh ok

[topaz solar]

My "learning cycles" are a bit inefficient i admit but I got most everything I know in math in a single year so can't be that bad

Anyhow, so those dots do mean membership, what I'm interested in is the particular meaning of the dotted circle (like around the V_3) and i feel i'm gonna be horribly disappointed because of some simple thing like "it's not contained in it"

[uncut girder]

Yeah so that's signifying the complement of V(x^2 +1) which is an open set

So its signifying closeness in the sense given by Zariski topology

[topaz solar]

ah, so literally is just "not contained in it" but with the additional zariski boie

Putting things that don't intersect over a given element together as being close seems a bit weird

though i guess having the things which do contain the same idea as closed sets does make a bit of sense because of limit points, which gives you the other ones by complementation

Again, sorry about not getting the topo stuff, dont bother explaining it since i'm about as mentally competent as a 3 year old when it comes to that stuff

[full blaze]

Abstract in 3 days

[uncut girder]

Question: in a general commutative ring, is 0 a prime ideal?

Or is the prime ideal containing 0 the nilradical?

Because if the ring has zero dividers, say fg=0, f, g non zero,

then fg being in the zero ideal does not imply f or g is 0.

[somber bramble]

an ideal is prime iff R/I is a domain ⇒ 0 is a prime ideal iff R/0 = R is a domain

[uncut girder]

Okay

[somber bramble]

so, as you said, depends on whether there are zero divisors

[uncut girder]

Is it true that the nilradical is prime

[somber bramble]

what is the nilradical?

[uncut girder]

The set of all nilpotent elements

x^n =0 for some n>0

[somber bramble]

is every zero divisor nilpotent?

[uncut girder]

Idk

[somber bramble]

in finite rings, every element is nilpotent, but R is not a prime ideal

(since R/R = 0, but in a domain, 0≠1)

[uncut girder]

How is every element nilpotent

[somber bramble]

oh wait, yea no

I was thinking in terms of groups

[uncut girder]

I know that the nilradical is the intersection of all prime ideals

[somber bramble]

https://math.stackexchange.com/questions/786393/when-is-nilradical-not-a-prime-ideal

[uncut girder]

And that R/N has no nilpotent elements, N being the nilradical

Nice

[uncut girder]

How to prove V(f) =X implies f is nilpotent
V(f) being the set of prime ideals containing f and X being the set of all prime ideals

We know that f is contained in every prime ideal

Oh shit so f is in the intersection of all prime ideals so f is in the nilradical, ie f is nilpotent. Ezpz

[uncut girder]

How to show X is quasi-compact?

[civic linden]

is an automorphism of a field F simply just creating a bijection of that field and preserving its structure?

[somber bramble]

yep

[coarse fern]

Is the symmetry group of an equilateral triangle isomorphic to the symmetry group of an isosceles triangle?

[somber bramble]

what do you think?

[coarse fern]

I think so

I think so

[somber bramble]

what symmetry of the isolesces corresponds to a rotation?

[coarse fern]

Is it not symmetric under rotations of n90 degrees?

[somber bramble]

n90?

what is n

[coarse fern]

Multiples of 90 until the identity permutation

Where j

Where n

[somber bramble]

a triangle, symmetric under 90 degree rotations?

[coarse fern]

Where n

Omg my phone

Omg my phone

Sorry, yes under integer multiples of 90

Oh wait

[somber bramble]

consider taking a triangle out of paper and rotating it 90 degrees

[coarse fern]

Okay

Wait would it be 180 degrees lol

[somber bramble]

like literally cut one out of paper and see if rotating works

if you can’t visualize it

[coarse fern]

I guess I’m lost on how to define symmetry because it does not look symmetric yet I can permit

I guess I’m lost on how to define symmetry because it does not look symmetric yet I can permit

Permute the set * uhhh

I thought if you rotate the triangle and if the distance between points is preserved then it checks out?

[somber bramble]

there’s very few symmetries in an isolesces triangle

by which I mean really just one

[coarse fern]

There’s a reflection through vertical and

[somber bramble]

yep

[coarse fern]

Wait

[somber bramble]

and the identity

[coarse fern]

Wat

I

[somber bramble]

which I forgot to count

[coarse fern]

Oh ok

Huh

Idk why but I can’t seem to wrap my head around verifying this isometric feature of symmetry

[somber bramble]

think about it. if you rotate the triangle, it’ll be in a very different shape

can’t get the corners to line up

isometries are rotations, translations and reflections, and their compositions

you don’t need to try anything but those things

[coarse fern]

Provided the distance is preserved right?

How do I verify that

[somber bramble]

they are preserved with those things

[coarse fern]

Well clearly most rotations do not preserve distance with this isosceles triangle so am still kind of iffy , I’ll do more examples and ask a more appropriate question

[somber bramble]

idk what oyu mean with “preserve distance with the triangle”

what you need is for the resulting shape to look identical to the original one

[coarse fern]

I get what you’re saying, which makes sense , I guess the formal definition is confusing me

I get what you’re saying, which makes sense , I guess the formal definition is confusing me

[somber bramble]

that’s the symmetries of a specific thing

namely the plane

oh, the thing below

[bleak abyss]

Just think of the shape as a subset of the plane, and consider isometries of the plane which map the shape to itself

[somber bramble]

as said, you just need isometries (rotations, reflections, translations and combinations thereof) which map each corner to another corner

well, or the same corner

ya know

you don’t have to think about it too abstractly

it’s a very tangible idea

[coarse fern]

Okay I see, yo

You guys out here making sense and my phone is tilting me

You guys out here making sense and my phone is tilting me

You guys out here making sense and my phone is tilting me

[bleak abyss]

For what it's worth, you can often dodge this. The main symmetry groups of shapes I've seen were:
Symmetries of cube = S_4
Symmetries of tetrahedron = A_4
Symmetries of n-gon = D_n

[somber bramble]

that the symmetries of the cube are S4 is, imo, nontrivial

[bleak abyss]

But actually in my mind I think of D_n as the automorphism group of the cycle graph

[somber bramble]

also that’s only rotation symmetries

imo “symmetries” is always very unclear

whether it also counts reflections or no

[coarse fern]

Does it not?

[bleak abyss]

Which is kinda the same thing but in my mind adjacency is way easier to process than "rigid motions"

[coarse fern]

For a regular pentagon you get symmetries at weird angles , how to calculate those angles ?

[coarse fern]

What a way to define equivalence relation 🌶

[solar vessel]

lmao

[chilly ocean]

nice groin area bro

[coarse fern]

Ty

[abstract sandal]

for an abelian group G, is the centre of G the same as G?

as all elements are in a class by themselves and hence the centre of G spans G?

[proven temple]

yes

[solar wyvern]

For K a field and G a finite abelian group, when is the group ring K(G) a field?

[oblique river]

only when G is trivial

@solar wyvern

[solar wyvern]

@oblique river This is true in general? i.e., a group ring can only be a field if G trivial?
e: and K a field ya

For distinct group elements a,b, will a+b ever be invertible?

[oblique river]

yes. if your ring R isn't a field then R[G] will never be

and as for an example, uhhhhhh

[raven tangle]

im so lost right now omds

[oblique river]

yeah it should be true

i mean like there should be an example where it's invertible

[raven tangle]

im 13 and confused

[oblique river]

even in something like C[Z/3Z]

well maybe not for cyclic groups...

rebecca if you're 13 then you probably shouldn't be asking for help in this channel.

[raven tangle]

yh

[oblique river]

@solar wyvern I can't think of an example or proof off the top of my head. Maybe section 3 of this could be useful: https://www.lakeheadu.ca/sites/default/files/uploads/77/docs/Cox Daniel.pdf

[solar wyvern]

hm, so you definitely have some less trivial units if K has r^2=0

and there's something called bicyclic units for R=ℤ

Theorem 3.22. Let G be an ordered group and R a ring without zero divisors. Then RG contains no nontrivial zero divisors and no nontrivial units

[woven delta]

What's the group ring R[Z/2Z]?

[oblique river]

When I said C earlier yes I meant the complex numbers

and what do you mean "what is it", liquid?

[woven delta]

Idk, I guess I want it to be C

[oblique river]

oh, it's certainly not

haha sorry

[woven delta]

That's good

[oblique river]

it's isomorphic to R[x]/(x^2 - 1)

whereas C is isomorphic to R[x]/(x^2 + 1)

[woven delta]

Oh ok

[oblique river]

as a ring, it's isomorphic to the direct sum R + R

which isn't an integral domain

[woven delta]

I'm on the train so I wasn't able to work it out by hand lol

[oblique river]

as I told flim earlier, group rings are never fields unless the group is trivial and the ring was already a field

[solar wyvern]

We now arrive at a conjecture which is very well known among those in the field:
was this intentional

[oblique river]

what is that from?

also i dont know what you mean by intentional

is there some pun i'm missing or something

[solar wyvern]

yeah just bad pun

[oblique river]

oh i dont get it sry :(

[bleak abyss]

in the "field"

[oblique river]

oh lol

eh

[solar wyvern]

@oblique river a little confused about solvable & (solvable by) radical extensions--lang defines these in the finite (and separable) case:
solvable extension E/k as one whose minimal Galois extension K/k has solvable galois group whereas a
(solvable by) radical extension E/k is defined wrt towers of fields, i.e., for k=E_0⊂...⊂E_i⊂E_{i+1}⊂...⊂E_m=E
E_{i+1}/E_i is a splitting field for one of X^n-1, X^p-X-a, or X^n-a

these aren't necessarily normal extensions are they? rather, for the tower stuff, do you always consider the maximal tower or do you need to consider every one--example for nested fields k<F<E<K<Ω does K/F need to be a splitting field of that form, or does it suffice that Ω/K,K/E, E/F etc all are?

[oblique river]

yeah they're not necessarily normal

for example Q(cbrt(2)) / Q is solvable

and solvable by radicals

well maybe not with lang's convention because it's not a splitting field

but Q(cbrt(2), zeta_3) / Q is solvable by radicals

[solar wyvern]

@oblique river maybe I phrased it awkwardly but Q(cbrt2)/Q would be a solvable extension because Gal(Q(cbrt2,zeta_3)/Q) = S_3 is solvable and has Q(cbrt2)/Q as a subextension?

[oblique river]

yeah it is solvable

but your definition of "solvable by radicals" seems to not fit it because it requires E_{i+1}/E_i to be a splitting field

[solar wyvern]

https://puu.sh/DAVVe/3ef014f6cf.png

yeah I think i fucked up @oblique river

[oblique river]

can you give some context

like what should i give feedback on

[solar wyvern]

@oblique river where'd I go wrong (with my definitions)

also this might be a bit vague but: unlike galois/normal extensions solvable--or equivalently, radical--extensions are a "distinguished class": that is closed under pushouts (F/k, F'/k solvable implies FF'/k solvable) and subextensions of solvable extensions are solvable.

my question is whether this the maximal subcategory of galois extensions so that you have such a "distinguished class"

[oblique river]

i think the answer is probably "no" to that question

like just take "the smallest subcategory generated by the solvable extensions and the splitting field of x^5 - x - 1 over Q"

which almost certainly shouldn't give you all extensions

and the definitions seem fine?

[somber bramble]

Algebra Prof in last class: “Should we put rep theory on the exam? I don’t know… I’ll have to think about it”
Also Algebra Prof:

[void dragon]

[quiet cave]

Unilateral decision making

[somber bramble]

For this question, is the following a valid solution?

No. Assume $f$ was a generator. Then $\deg(f) = 5$, as no nonzero polynomial of smaller degree is in the ideal. Further, $f \mid x^5 + 1$, so we see that $f = x^5 + 1$. Polynomial division then shows that $x^7 + 1 = (x^5 + 1)(x^2) + x^2 + 1$, so $x^7 + 1 \notin \langle x^5 + 1 \rangle$, contradicting the assumption.

[cloud walrus]

Sascha Baer:

[fickle brook]

uhhhhhhhh

aren't all ideals in K[x] principal for K a field tho

besides, just because f generates <x^5 + 1, x^7 + 1> doesn't mean deg(f) = 5

@somber bramble

[somber bramble]

oh fuck right I’m dumb there are elements of smaller degree in there

because $(x^2)(x^5 + 1) + (x^7 + 1) = x^2 + 1$

[cloud walrus]

Sascha Baer:

[somber bramble]

and yea you’re right

fields are PID, and if R is a PID then R[x] is one too

so I have to find a generator

[fickle brook]

second statement isn't true

polynomial ring over a PID need not itself be a PID

[somber bramble]

oh right that was for UFD

dammit it’s so hard to keep all these terms straight

[fickle brook]

K[x] is a euclidean domain for K a field

[somber bramble]

meaning I can find the generator as the gcd

so I have to find the gcd of those two elements

…let’s see if I can do the euclidean algorithm from memory cause I’ve not done it in a while

I always mess up which two things I have to pick to continue

if I write
a = bq + r
is gcd(a,b) = gcd(q,r) or gcd(b,r)

and is there a sensible way to remember which one

[fickle brook]

b and r

ok so how i remember it

is this

suppose we want to find gcd(a,b). let a_0 = a, a_1 = b

then perform a series of euclidean divisions as follows

$a_0 = a_1 q_1 + a_2 \ a_1 = a_2 q_2 + a_3 \ a_2 = a_3 q_3 + a_4 \ \vdots$

until the sequence $(a_i)$ hits zero

[cloud walrus]

Ann:

Ann:

[somber bramble]

and then the result was the last nonzero a_i right

cause gcd(x,0) = x

so I get X+1 as the gcd and that’s the generator

[fickle brook]

ye

[somber bramble]

and I’m already stuck on the next question yay

so d = 2, because A8 has index 2. but now I need to somehow look at S8 and see if it has any other subgroups with half the elements and I don’t really know any tools that could be helpful. any hints (e.g. relevant theorems)

hm, let’s say there exists another subgroup H of index 2 of S8 and it is not A8. then there is an element with odd sign in H.
further, all elements of S8 without A8 but plus the identity can’t be a subgroup (it has one element too much), so there is also an element of even sign in H. call it h
subgroups of index 2 are always normal I think.
Thus H is closed under conjugation, but the orbit of h under conjugation is A8. So A8 is a proper subgroup of H, contradiciton

does this line of reasoning work out?

okay yea subgroups with index 2 are normal: ${gH | g \in G} = {H, G \setminus H} = {Hg | g \in G}$ and $gH = G \setminus H \iff g \notin H$ and same from the other side

[cloud walrus]

Sascha Baer:

[uncut girder]

Given s.e.s 0->A->B->C->0 of modules
Show that it is not ness true that
0->Hom(X,A)->Hom(X,B)->Hom(X,C)->0
Is exact. (Specifically, exactness will fail at Hom(X,C) )

The maps in the second sequence are just post composition of the corresponding map in the first sequence

[uncut girder]

<@&286206848099549185>

[marble wagon]

take modules which aren't free or projective

[covert vector]

@uncut girder did you get it yet?

[uncut girder]

Yeah

[covert vector]

ok coo

something finite like Z/nZ should do the trick, cuz Z has no nonzero finite order elements

[uncut girder]

Yepp

[full blaze]

just checkin lets say you have H and K which are subgroups of a group G, The intersection between the two subgroups H intersection K is another subgroup of G, not a group of G

[fickle brook]

what would "a group of G" even mean

[solar wyvern]

group of groups

[fickle brook]

[wary sphinx]

Can groups be group elements?

[somber bramble]

you’d have to find some interesting way of doing that
obviously you could just take any group and label its elements with names of other groups

but there’s no reason to do that

there is some ways you can combine groups with one another tho

the free product of groups comes to mind

or the direct sum

[wary sphinx]

invertibility seems like the tricky thing

[marble wagon]

there's no reason to do that

?????

what are classifying spaces

[fickle brook]

yeah what are classifying spaces

[somber bramble]

arbitrarily naming group elements as other groups does seem pointless. if there’s some reason to label group elements as groups systematically, then it won’t be arbitrary anymore

[marble wagon]

yeah and red things are red

classifying spaces are when you have some kind of geometric object, parametrized by another geometric object

to be more concrete, for example, a kind of smooth manifold parametrized by another smooth manifold

so for example we can parametrize circles by their radius

giving the space of circles a topology where circles with similar radius are closer

etc

[marble wagon]

if you look it up you'll find the classifying space BG instead which is something else

so maybe a good first place to read is https://en.wikipedia.org/wiki/Moduli_space

although it's rather technical

[bleak abyss]

Oh you weren't talking about BG

I was so confused

[marble wagon]

ye

I meant classifying spaces of geometric objects

[bleak abyss]

Yeah I've always heard them referred to as either parameter spaces or moduli spaces

Well one professor of mine once called something a moduli space and was like meh I should call it a parameter space

And that was the only time I heard of parameter space

So idk if that's morally/technically different or whatever he was going for

[marble wagon]

I think classifying spaces in the sense of BG are named as such because they classify bundles

[bleak abyss]

Yeah I think principal G-bundles on X are given by [X,BG]

[marble wagon]

but what do I know

I need to learn topological k theory eventually

[bleak abyss]

Yeah same I basically just know some words Peter told me 😛

[marble wagon]

nice

[bleak abyss]

He was more gunning for cohomology with this

Since BG is a K(G,1) if G is discrete

[marble wagon]

group cohomology sucks

I don't get it

[bleak abyss]

Yeah it seems tough to deal with, bits and pieces have come up since H^2 somehow deals with group extensions and one time I had an AT problem which required computing cohomology of a space and nobody (not even the TA) could figure out how to do it without appealing to group cohomology

[marble wagon]

do you mean the other way around?

like

the one time I had to do group cohomology I was like

nah dude fuck groups, I'll just calculate the usual cohomology of BG

or something

admittedly the presentation of the material I read was awful

I should revisit it eventually

written by someone who knows what they're doing and gets to the point

instead of wanking over endless combinatorial descriptions that don't mean anything

[bleak abyss]

Nah we had to compute H_3(K(Z/p,1))

And none of us were sure how to do it

Our TA eventually gave in and Googled and was like welp I guess time for group homology

And we just cited the fact that H_3(Z/p) = Z/p

The context was this

So let's say you have a compact 3-manifold with finite fundamental group

The universal cover is a homotopy 3-sphere

That's pretty easy

[marble wagon]

okay that makes sense, for cyclic groups you can get very explicit stuff

are homotopy 3 spheres just 3 spheres?

[bleak abyss]

Yeah but that's Poincare conjecture I think

[marble wagon]

ye I thought so

okay keep going

[bleak abyss]

But yeah this isn't bad to see, if you have a compact space with finite pi_1 then its universal cover is also compact

So the universal cover of this guy has pi_1 = 0, thus H_1 = 0, by UCT also H^1 = 0. By Poincare duality H_2 = 0. It's oriented since it covers the oriented double cover of M so H_3 = Z

[marble wagon]

yeah that's fine, but why did you want to compute H_3(K(Z/p,1))?

[bleak abyss]

So next problem was to show that if pi_1 is abelian, then it's cyclic

[marble wagon]

I remember this problem lol

it has come up like 4 times

[bleak abyss]

You can pass to a cover and it basically just boils down to showing that (Z/p)^2 is impossible

[marble wagon]

ye

[bleak abyss]

And yeah you basically have to say that you can stack on a bunch of cells and make a K((Z/p)^2,1) out of it which must have H_4 of dimension 0 or 1

And use Kunneth

But yeah my solution was handwavy as fuck

[marble wagon]

okay I see

[bleak abyss]

Eh I mean maybe not that handwavy actually, the construction of the K((Z/p)^2,1) makes sense. pi_3(M) = Z since it's covered by S^3 so it is true you just need a single 4-cell

Kill the generator and you're done

And then focus only on high-dim cells

[marble wagon]

yea it looks fine just concise