groups-rings-fields

[rapid cave]

I am starting to see whu do we care from things people here send

[tardy hedge]

because u like it

secretly ...

[quiet pelican]

My problem is more that it’s 90% meaningless character calculations by volume

[rapid cave]

Oh yeah there where a ton of characters

[knotty badger]

character theory was actually one of my favourite parts of rep theory

i just like doing calculations

and it was cool to figure out tricks for computing character tables

[rocky cloak]

Easy solution: attend a university that don't have any courses on representations of finite groups. Then your first rep course will be representations of something else

[delicate orchid]

no you just won't have a rep course at all

[rocky cloak]

My first rep course was on Artin algebras

[delicate orchid]

also who tf hates character theory? what's the problem topology boy, afraid you might have to count something?

[rocky cloak]

Characters are cool, but I'll let someone else compute them

[rapid cave]

My first repr theory course was on algebras, groups, lie algebras and quivers

All of the fun in one course

[delicate orchid]

I've just spent the past 4 years computing them I think it's your turn, jagr

[delicate orchid]

this goes hard

[rocky cloak]

Can't decide, let's just do all of them

[delicate orchid]

groups and quivers are basically two sides of the same coin tbf

[quiet pelican]

I’ve written a program which can compute the character table of any subgroup of S_n (given generators of it) but unfortunately I’ve still been forced to compute them by hand before

[delicate orchid]

if you squint really hard that is

[rocky cloak]

If you squint that hard everything is just algebras

[delicate orchid]

yunc... EVERY group is a subgroup of S_n

[quiet pelican]

The content of phrasing it like that is that the input is expected to be an explicit subgroup of S_n

[delicate orchid]

you gotta have a little fun finding the best embedding or it's boring

[quiet pelican]

And given a group in generator-relation form, the conversion is uncomputable in general
Probably similarly in most other forms that aren’t like set + multiplication table

[delicate orchid]

oh if you need the relators then yes you probably are fucked

or just do CharacterTable(G) in the online MAGMA calculator

[quiet pelican]

You don’t in my program
But <S | R> doesn’t convert into the desired form easily

[delicate orchid]

although the converstion is computable for finite groups

[delicate orchid]

yeah so find a nice group action

[quiet pelican]

But nice group actions are by isometries on geodesic spaces

[delicate orchid]

I think you've lost the plot

[kind temple]

what if you knew the order along with the presentation? is that enough to pin down a multiplication table?

[quiet pelican]

Am known to be insane

[delicate orchid]

yeah it would be

[quiet pelican]

Yes
Explicitly check every group of order n

[kind temple]

anything faster than that?

[delicate orchid]

Just transverse the cayley graph breadth first until you have enough distinct elements

[kind temple]

nice. problem solved for generators and relations

[delicate orchid]

then use the regular rep to write them as permutations

warning you may need more RAM than currently exists to do this

[elfin wraith]

Yeah my first rep theory course will be Lie algebras, I dont think the one problem sheet on modules of group rings in my noncom class counts nor does the lecture on characters for analytic number theory

[kind temple]

problem unsolved

[delicate orchid]

reminds me of how it is impossible to store the monster group in under like 5GB

[kind temple]

good lord

[delicate orchid]

no wait sorry

using the smallest faithful matrix representation it's 5GB

per element

196883 dimesional matrices go crazy

[elfin wraith]

Group theory scares me

[quiet pelican]

but groups

[delicate orchid]

it's ok I think you can drop the dimension down by 1 by taking matrices over F_2

that's a whole ~400KB saved!

[rapid cave]

Get more ram

Download some

[kind temple]

🐏

[elfin wraith]

Get this man into SWE right now!

[knotty badger]

3b1b makes a good point in that it's not just the size of the group that makes it difficult

there are far larger groups whose elements can be stored far more efficiently, like the symmetric groups

[delicate orchid]

yeah you just need a good action

unfortunately if the monster group had a good action we most likely wouldn't care about it

[knotty badger]

interesting, why is that

[delicate orchid]

heuristically the smaller the degree of the smallest faithful representation is the "less simple" the group in question is

it's not really a concrete thing just a vibe

[quiet pelican]

Nah SWEs learn to sacrifice consumer ram on the alter of the corporate gods

[elfin wraith]

Do we have any idea what the monster is the symmetries of or why it seems to be the largest one? Like do we have any handle on what the sporadic groups actually are or why they exist?

[noble nexus]

Something to do with modular forms

but I don't think anyone really understands why modular forms show up everywhere

also the leech lattice shows up in a lot of sporadic groups

the funniest "explanation" of why the leech lattice is special is because the sum of the first 24 squares is a square

I think most exceptional objects in math ultimately come down to coincidences in the natural numbers, but I don't know if that's entirely correct

[delicate orchid]

part of the study of finite simple groups was on their "involution" graphs, this post kind of goes into it https://mathoverflow.net/questions/355496/what-is-the-geometric-shape-of-the-monster-sporadic-group

[noble nexus]

and it's not entirely satisfying

[delicate orchid]

I'm a firm believer of this

[knotty badger]

that's interesting

[delicate orchid]

look me in the eye and tell me that PSL(2,4) \cong PSL(2,5) is due to some natural beauty and not luck

[noble nexus]

One of my favorite little coincidences in math is the surjective homomorphism from S4 to S3

and my favorite way of explaining it is that S3 is isomorphic to GL(2,2) and S4 is isomorphic to GA(2,2)

[delicate orchid]

what the CHUNGUS is a GA

[noble nexus]

Which is a coincidence because the dimension and field size are small enough to where all permutations are linear (the 3 nonzero elements of F2^2) or affine (the 4 elements of F2^2)

General affine group

[delicate orchid]

yeah that's true then

[noble nexus]

just general linear group + translations

[knotty badger]

what is it?

[delicate orchid]

canonical surjection S_4 -> S_4/V_4

[noble nexus]

Well there's a clear homomorphism from GA(2,2) to GL(2,2)

the kernel is the group of translations which is F_2^2 i.e. V_4

another way of explaining it is that any permutation of 4 points also permutes the 3 ways of partitioning 4 points into 2-element subsets

[elfin wraith]

The existance of the sporadic groups does just make me somewhat uncomfotable. Maybe it wouldnt so much if I knew anything about them, but knowing there is just exactly 26 groups which for some reason just happen to be finite simple makes me uncomfortable

Doesnt feel right at all

[knotty badger]

ah ok

[noble nexus]

yeah I agree, although if you think about them as small number coincidences that are washed away by general trends it doesn't seem as bad to me

[delicate orchid]

I feel much more uneasy about shit like the exceptional outer automorphism of S_6

that one is truly diabolical

[noble nexus]

Yeah that one is really weird and I haven't seen a satisfying explanation of it yet

[delicate orchid]

like this one is fine, S_1, ..., S_4 all have various normal subgroups and then S_n onwards is almost simple

but like, S_2 and S_6 are the weird ones for automorphism groups

wtf is going on in the middle

[noble nexus]

Oh if I remember it's because of a numerical coincidence around sylow subgroups

that gives an interesting embedding of S5

[delicate orchid]

sylow at which prime

[noble nexus]

At least that's the one I'm familiar with

[elfin wraith]

See these things I can usually just write off as a case of, thats just how the numbers happened to work out. And while im sure thats also the case for the sporadic groups, its just on a much larger scale, youre telling me of all possible ways to do this there just so happens to be exactly 26 exceptions? Crazy

[noble nexus]

S5 has 6 sylow 5 subgroups

and that gives an action of S5 on 6 points which gives a transitive embedding

a quick wikipedia browse shows that most constructions involve creating a "weird" subgroup of index 6

although I will say this doesn't really explain why that never happens again

[knotty badger]

one for each letter of the alphabet :P

[delicate orchid]

if it helps you sleep at night, 20 of them are subgroups of the monster

[noble nexus]

It does sound somewhat plausible that S(n) embedding transitively into S(n+1) should become vanishingly rare as n gets large, not sure how id quantify that

[elfin wraith]

This does help, I didnt know that. It still leaves another 5 weird ones

[rocky cloak]

The pariahs

[delicate orchid]

tag ur self I'm the O'Nan group

[vapid vale]

iirc the coincidence for S_6 is somewhat explained by (12) same size conjugacy class as (12)(34)(56)

which is true for no other order two conjugacy class of any other symmetric group

[delicate orchid]

I buy it. That's really odd

ah yes caue the outer automorphism is defined by permuting those two classes right

it's always the involutions 💔

[vapid vale]

is the verb for involutions to involve

oh involute

lol

[elfin wraith]

Truly insane we managed to categorise this shit

[delicate orchid]

it really is

group theory is solved everyone go home

[elfin wraith]

Infinite groups are a lie

Except maybe like Z and R

[vapid vale]

good pun

[delicate orchid]

more like a Lie

[delicate orchid]

those are rings, not groups!

a common blunder

[elfin wraith]

Exactly why theyre valid

[delicate orchid]

rings with non-commutative addition

[elfin wraith]

Throw back to the ANT homework where I got marked down for talking about C_n rather than Z/nZ, brother the latter is a ring

I was told that C_n isnt at all standard and I made that up

[noble nexus]

I write C_n a lot now

[elfin wraith]

Favourite intro ring theory exercise

[noble nexus]

Because I can't write Z_n in good conscience after learning p-adics

it just looks wrong to me

people write Z_2 and I have to convince my brain that's it's not the 2-adic numbers

[delicate orchid]

everyone fuckin uses C_n

[elfin wraith]

I still write Z_n sometimes because im lazy and will never touch the p-adics

[delicate orchid]

hell people just use n

[noble nexus]

also infinite groups are awesome

[elfin wraith]

I genuinely couldnt belive my eyes looking at the marking

I got fucked in that first hand in lol

[delicate orchid]

hope you contested that shit

[elfin wraith]

I did, got that specific mark back but I still got fucked

[noble nexus]

without infinite groups we wouldn't have the greatest group of all time (the lamplighter group)

[elfin wraith]

Did well in the rest of the course but it turned out that prof wanted every detail in every proof

[rocky cloak]

Where these graded by students one year above with too much power?

[elfin wraith]

Just had to adjust, take a trip back to year 1

[delicate orchid]

funny way of spelling "dihedral of order 8" or "extraspecial of order p^3 for p odd"

[vapid vale]

basically every TDA textbook explains the relevant alg top. there is lots of interesting research done that requires more involved alg top/alg geo/rep theory (of quivers) but to understand and work with persistence modules you dont need very much – its taught here to undergraduates with no prerequisites beyond linear algebra

[elfin wraith]

How could you tell

[noble nexus]

also when you study infinite discrete groups you get to sound like a crazy person talking about groups that are "locally sometimes virtually maybe" abelian

[elfin wraith]

This is how all group theory sounds to me

[thorn jay]

yk what's funny? 2 and 6 are the only orders of which there exist no mutually orthogonal pair of Latin squares

[elfin wraith]

virtually polycyclic-byfinite ok bud

[vapid vale]

polycyclic-bifinite is there a flag for that

[delicate orchid]

unfortunately if you only consider outer automorphism groups then S_2 stops being weird

[karmic moat]

Group theorists spinning the prefix-root-suffix wheel to come up with some new type of group

[noble nexus]

The study of finite groups is virtually trivial

(literally)

[thorn jay]

well luckily that it doesnt exist for order 2 is a size issue. That jt doesnt exist for order 6 is a ??? issue

[delicate orchid]

skill issue

[noble nexus]

Yeah 2 is just weird

[vapid vale]

in my first college math class ever my (extremely brilliant) professor said that if we ever needed to guess an answer, to guess 0, 1, 2, or infinity. i could believe 0, 1, and infinity but 2 felt a little far fetched. now i somewhat understand

[next obsidian]

The question: what is 1+1

[delicate orchid]

0

[noble nexus]

2 = 1 + 1 is actually the most profound equation in math

[elfin wraith]

All fields are F_2

[noble nexus]

it's the cause of a lot of coincidences

[next obsidian]

If there’s a cause it isn’t a coincidence

[noble nexus]

true

[vapid vale]

disagree

[noble nexus]

order 2 symmetries also seem overwhelmingly cmmon in math

[vapid vale]

coincidence is the same as paradox

in that it can just mean counterintuitive/unapparent

[noble nexus]

I find it really crazy that finite (isomorphism classes of) groups have order a power of 2 with probability 1 in the limit

[vapid vale]

theyre 2 common

[delicate orchid]

it's ALWAYS the involutions

[noble nexus]

actually idk if that's proven but I think its conjectured

with very strong evidence

[delicate orchid]

I can explain why this should be true in a reasonable way if you want

[noble nexus]

sure

I do think that involutions seem a lot more common than any other number that's kinda how I've been thinking about it

[delicate orchid]

ok under the light assumption that the proportion of simple groups goes to 0 as the order goes to infinity

the idea is that if you have a group of order 2^n, then it is the smallest possible solvable group with n composition factors - in particular they all have to be C_2 (as subquotients of 2-groups are 2-groups and C_2 is the only simple 2-group). This gives us a lot of non-trivial choices to make as we reconstruct our group by recursively extending by a copy of C_2

I'm pretty sure we observe similar (much smaller) spikes in the number of groups at p^k for all k, it's just that 2 is the smallest prime so it overtakes everyone else

[noble nexus]

yeah that makes sense

but part of it is just that its so insane how much faster it overtakes

for ex in the first 50 billion groups apparently 99% of them have order 1024

[delicate orchid]

it's cause C_2 is SO small

like even if you extend by a single C_3 your group is suddenly 50% bigger

[noble nexus]

I guess 1024 is 2^10 but for even the next prime 3 you can only reach 3^6

for order less than 1000

[delicate orchid]

exactly

[noble nexus]

but for example you could have groups of order 2^8 x 3 and 2^7 x 5 etc and its kinda amazing that all of those contributions make up only 1%

[delicate orchid]

groups of order 1024 are the only ones with 10 composition factors, and the only ones with 9 are (I think...) order 512 and (potentially!) 1536

[noble nexus]

guess its just exponential growth though

[delicate orchid]

there may be something more to this story - the choices you make when extending aren't independent

[noble nexus]

I wonder if C2 also has more actions

yeah

like if each extension gives you more with C2 compared to a tower of C3

[delicate orchid]

yeah I was thinking about H^2(S ; F_2) for S a 2-group

[noble nexus]

I guess you could compare how many groups there are of say order 16 vs 81

let me look actually

this site is clutch https://people.maths.bris.ac.uk/~matyd/GroupNames/

[delicate orchid]

if you picked one prime power smaller I would've been able to tell you straight up 💔

although the ratio there is 1

there are actually MORE groups of order 3^k than 2^k

which I guess makes sense, C_3 has a non-trivial automorphism group for instance

[noble nexus]

2^3: 5 vs 3^3: 5
2^4: 14 vs 3^4: 15
2^5: 51 vs 3^5: 66

ya

oh I miscounted for 3^5

[delicate orchid]

not knowing the classification of groups of order p^5 💔

[noble nexus]

is anything known about the asymptotic amounts

or even conjectured

of how many groups of order p^k there are

[delicate orchid]

good question, this is known explicitly for k <= 8 (it may be known for k = 9 but I don't know that so it's not really known is it)

[noble nexus]

https://en.wikipedia.org/wiki/Higman–Sims_asymptotic_formula oh there we go

[thorn jay]

the only slight problem is that 3^k is wayy bigger than 2^k

[delicate orchid]

O(n^8/3) 🥀

[thorn jay]

lmao

[delicate orchid]

yeah exactly, which is why 2^k dominates

[thorn jay]

there exists one??

[delicate orchid]

there exists one for p^7

[thorn jay]

what

thats bullshit

as in, i do believe it

for what k is there a classification for p^k

[delicate orchid]

upon googling I can only find the p^6 one

[south patrol]

I can find them for p^67

[delicate orchid]

https://users.ox.ac.uk/~vlee/PORC/p8expp.pdf holy moly

[south patrol]

Ah retired

[thorn jay]

finite group theorists are insane

[noble nexus]

so if the number of groups of order p^n is approximately p^{2/27 n^3} then you should be able to visualize the approximate number of at least p-groups less than some order

[south patrol]

I've never met a group theorist who was infinite

[noble nexus]

2 growing far far faster than 3 or 5

which I guess is nice to see it that way

even for 100

its kinda striking how big of a difference it is

although for 2 this seems very wrong maybe thats a big O diff

or I entered something wrong

[delicate orchid]

is that meant to be 2^100 or

or just < 100

[noble nexus]

< 100 this is the function

probably big O

this is counting 2^18 groups of order 64 which is def wrong

[delicate orchid]

just worked out there's 340 2-groups of order < 100

[noble nexus]

should at least be asymptotically true though

for 10^7 lol

[vapid vale]

am i dumb, what is "order p^8 with exponent p"

[delicate orchid]

all elements are order p

[vapid vale]

oh

[delicate orchid]

exponent of the group not the order

[vapid vale]

got i

t

[delicate orchid]

CHALLENGE: classify the p-groups with order p^9 and exponent p when p = 2

[noble nexus]

no 👍

[delicate orchid]

exponent 2 groups are abelian

[vapid vale]

xd

[delicate orchid]

(xy)^2 = 1 => xy = (xy)^-1 => xy = y^-1x^-1 => xy = yx

that was the funny

[thorn jay]

okay that's cheating if youre classifying of exponent p

be a REAL man and claasify arbitrary exponents

[delicate orchid]

Best known is p^7 then

[thorn jay]

how do we even calculate that

[delicate orchid]

fuck knows

very carefully

https://arxiv.org/abs/1611.00461 this is cute

[tidal schooner]

I love this

[tidal schooner]

Geometric group theorists?

[elfin wraith]

https://tenor.com/view/perksofbeingawallflower-infinite-free-letgo-driving-gif-4158154

[astral ivy]

"Find all Abelian groups of order 2700 that don't contain elements of order 54 but do contain elements of order 50. Write their elementary and normal forms and find the number of elements of order 90 in each of them."

Okay so I have 2 issues with this task. First one is how do we find those that don't contain specific orders like 54 and 50 and my second issue is how do we find the number of elements of order 90.

1. As for all Abelian groups of order 2700:

$2700 = 3^3 \cdot 2^2 \cdot 5^2$
Further, for 3 we have: $Z_{27}$, $Z_3 \times Z_9$, $Z_3 \times Z_3 \times Z_3$.
For 2 we have: $Z_4$, $Z_2 \times Z_2$
For 5 we have: $Z_25$, $Z_5 \times Z_5$

In total, that's $3\cdot 2\cdot 2 = 12$. Now from these 12 groups I need to eliminate those that contain element of order 54 but I don't know how to do that. $54 = 2\cdot 27$ so it definitely has to do something with 27, I just don't know which one from $Z_{27}$, $Z_3 \times Z_9$, $Z_3 \times Z_3 \times Z_3$.

[cloud walrus]

danilojonic

[rocky cloak]

In a product of groups AxB, the order of an element (a, b) is the lcm of the order of a and b.

This means you can quickly identify what the order of elements are for a product of cyclic groups.

This is easiest when you write it as a product of groups of the form Z/p^k (as you have done)

[sly crescent]

In the specific case of p=2 we know up to p^10

(There are 49487367289)

[delicate orchid]

it's crazy that they've counted that many things correctly

[astral ivy]

So order of element 54 is 27x2 so I'm looking only at groups of that format right?

But how do I distinguish further? I can't take out all Z_3 x Z_3 x Z_3, Z_3 x Z_9 and Z_27

I need something left to generate others right?

Like keep {Z_3}^3

[rocky cloak]

So the Z/27 is the problem, so you would take that one out

Next you need to figure out what's needed to get an element of order 50

[astral ivy]

AxB such that lcm(a, b) = 50?

That would be Z_2 x Z_25

[rocky cloak]

Yeah, so you need a Z/25 in there

So then what groups are you left with?

[astral ivy]

I see. Now I just rewrite them:
2: Z_2 x Z_2, Z_4
3: Z_3 x Z_3 x Z_3, Z_9 x Z_3
5: Z_5 x Z_5, Z_25.

Now connect only with Z_25 so:
Z_2 x Z_2 x Z_25, Z_3 x Z_3 x Z_3 x Z_25, Z_4 x Z_25, Z_9 x Z_3 x Z_25.

So there's 4 in total

Is this correct?

[wide spade]

idk if this is the right place to ask, but if i wanted to learn about infinite simple groups, is there a book or something that i can read ?

(ping me if u reply please)

[rocky cloak]

Maybe you wrote it a little weird, but you should have 2, 3 and 5 stuff in all of them

So Z/4 x Z/9 x Z/3 x Z/25 should be one of them

[tulip otter]

why is N either G or H?

[rocky cloak]

H has index p, which famously doesn't have many divisors

[astral ivy]

Isn't that just writing out the normal form though?

We take one of each (largest possible) and go down

until none are left

[rocky cloak]

You wrote
Z/4 x Z/25
which is not a group of order 2700

[tulip otter]

so since H has index p then H either has 1 or p orbits so that the index of the normalizer is one of these?

wait i think what i am saying doesnt make sense

[rocky cloak]

That's one way to think of it I guess.

More to the point Id say N(H) is a subgroup of G containing H.

So has index dividing the index of H

[astral ivy]

Okay yeah I see so 4 groups that don't contain elements of order 54 but contain elements of order 50 are:

1. (Z_2 x Z_2) x (Z_3 x Z_3 x Z_3) x (Z_25)
2. (Z_2 x Z_2) x ( Z_9 x Z_3) x (Z_25)
3. (Z_4) x ( Z_9 x Z_3) x (Z_25)
4. (Z_4) x ( Z_3 x Z_3 x Z_3) x (Z_25)

find the number of elements of order 90 in each of them

How about this?

[tulip otter]

ohhh yes, i was overthinking after all

tysm

[rocky cloak]

It might be easier to start by finding the number of elements with order dividing 90

Or just thinking about what numbers are needed to make an lcm of 90

[chilly ocean]

Idk much group theory, but I’ve been trying this for about thirty minutes and still can’t figure out how you get the last implication (xy=y^-1x^-1->xy=yx). How do you deduce that?

[knotty badger]

It’s from exponent 2

[rocky cloak]

x = x^-1

[astral ivy]

can you tell me what are the elementary forms? Are they literally these what I wrote:
(Z_2 x Z_2) x (Z_3 x Z_3 x Z_3) x (Z_25)
(Z_2 x Z_2) x ( Z_9 x Z_3) x (Z_25)
(Z_4) x ( Z_9 x Z_3) x (Z_25)
(Z_4) x ( Z_3 x Z_3 x Z_3) x (Z_25)

Or Z_9, Z_4 and Z_25 need to be simplified further (doesn't make sense to me if they should)

[chilly ocean]

Oh I’ve been trying to prove xyxy=1 implies xy=yx in general, is this only true when x=x^-1 for all elements of the group? It appears I have misread

[knotty badger]

Yeah it’s not true in general

[rocky cloak]

You can't simplify Z/9 any further no

[chilly ocean]

No wonder I couldn’t prove it haha

[astral ivy]

I do that by looking at their normal forms right?

a normal form will give me the largest group in each case

that's why I thought of that

[rocky cloak]

Whatever you find easier

[knotty badger]

For a good counterexample, consider x being a (nontrivial) rotation and y being a reflection in the dihedral group

xy is a reflection, so you definitely have xyxy = 1

However rotations and reflections don’t commute in general

[chilly ocean]

I see, thanks!

[knotty badger]

So you won’t usually have xy = yx

It’s nice to be useful in algebra sometimes…

[twilit wraith]

Semidirect products are giving me a rough time

On my homework I was asked to classify all groups of order 28 up to isomorphism which can indeed be done with semidirect products

[rocky cloak]

Only so many ways C4 can act on C7

[twilit wraith]

But part of my argument appealed to the presentation of a semidirect product knowing which generators get sent to which automorphisms

After talking with my professor though it seems like he doesn't want me to use presentations but im not sure how else I can do this

[twilit wraith]

Yes, given all possible maps from all groups of order 4 into C6, there can only be at most four isomorphism types

But im not sure how to find the nonabelian ones without using presentations

[rocky cloak]

So how are you using presentations exactly, or what do you mean by that?

Like you have your 4 groups, so you just need to argue they're not isomorphic right?

[twilit wraith]

I was assuming I find the isomorphism type as in "this semidirect product is isomorphic to this well known group"

[rocky cloak]

Is the semidirect product of C7 and C4 even a well known group?

I guess it's the binary dihedral group...

[vapid vale]

i recall those being called "dicyclic"

[rocky cloak]

Anyway, you can show that a group is a semidirect product by just noting how the 2-sylow group acts on the 7-sylow group.

[elfin wraith]

It’s certainly not one I know off the top of my head lol

[vapid vale]

binary dihedral is certainly a more useful group concept than dicyclic though

[twilit wraith]

Dicyclic group of order 28

[rocky cloak]

It's a finite subgroup of SL(2, C), so maybe you should

[vapid vale]

since its explicit from the SL2C SO3R double cover

[elfin wraith]

I’ve just never really looked at infinite groups tbf

[rocky cloak]

Yeah, I guess I assumed that wouldn't be a "well known" group in a class that asks you to classify semidirect products

[twilit wraith]

Fair enough

It could just be the case that im misunderstanding the goal

[rocky cloak]

But yeah if you know the dicyclic group you can just notice that it is a semidirect product of C7 and C4 and describe the action

Or I guess, since you've proven there's 4 groups, just list 4 groups of order 28

[twilit wraith]

I see

Theoretically I only have to check that the two semidirects given by a nontrivial homomorphism from C4/V4 into Aut(C7) are not isomorphic right?

This is assuming that trivial homomorphisms cause their related semidirect products to be abelian

At least i think

[knotty badger]

Omg same

[velvet hull]

oh my god

[kind temple]

i always saw you hiidostuff

[elfin wraith]

I always just kinda assumed it was some weeb shit

[merry harness]

I always thought it was hildostuff 😭

I am blind

[karmic moat]

i thought it was hildostuff too

[thorn jay]

I thought it was hiido stuff

[tardy hedge]

Same lol

[south patrol]

Oo lol

[tardy hedge]

mq blew everyones mind tonight

[elfin wraith]

It’s good that apparently none of us can read

[south patrol]

Or we just prefer separated words

Jk I cannot read

[thorn jay]

we would NOT survive in ancient Rome

💔

[twilit wraith]

Im popular now

But yes that is my name

I kinda have a new username now but I kept discord as this age old one

[sly rock]

The only thing I can fully keep up with in this chat
Just an aspiring math-knower after getting a 7 on AA HL

[thorn jay]

you're probably thinking of xyx^-1y^-1 = 1 => xy = yx

[chilly ocean]

ah yeah probably

Thanks!

[twilit wraith]

dude im just not seeing how to go about this semidirect product stuff

honestly the biggest struggle is with working with the presentations properly

im trying to show that $\langle x,a,b : x^7 = a^2 = b^2 = 1, axa^{-1} = x^{-1}, bxb^{-1} = x \rangle \cong D_{14}$

[cloud walrus]

hiidostuff

[twilit wraith]

my function between the two is $\psi(x) = r^2, \psi(a) = s, \psi(b) = r^7$

[cloud walrus]

hiidostuff

[twilit wraith]

but im struggling to show that $\psi$ is injective without going through a bunch of computation

[cloud walrus]

hiidostuff

[twilit wraith]

so i feel like im just doing it wrong

[vapid vale]

that doesn't look right

[thorn jay]

this presentation cannot be finite right?

it contains the free product of Z2 with itself as a subgroup

if you remove the generator b, then maybe it is easier to see how this gives a presentation of a semidirect product of C2 acting nontrivially on C7

@twilit wraith

[twilit wraith]

i mean idrk

i missed one of the lessons on semidirects and its showing

[thorn jay]

okay, do you know the intuition behind a semidirect product?

[twilit wraith]

not really

i mean i know its a way to extend the direct product to nonabelian groups in a way

[thorn jay]

Okay so, a group G is isomorphic to a nontrivial direct product if and only if G has normal subgroups N1 and N2 such that their intersection is trivial, and their join, N1N2, is G. Is this familiar?

going the other way, for any two groups G and H we just smush them together into a product G x H, very easily.

[thorn jay]

in some sense

[twilit wraith]

yes this makes sense

for semidirect product we only need one being normal

[thorn jay]

yes.

We relax the conditions a bit with G a group, N normal and H just a subgroup such that N and H have trivial intersection and NH = G. In the case where H is also normal, we can recover the group product in G very simply (as it must be isomorphic to the direct product). In general, however, the situation is more delicate. The normality of N yields a natural group action of H on N, by conjugation in G. Explicitly: r_h(n) = hnh^-1

[twilit wraith]

is this why the homomorphism from H to Aut(N) associated with a semidirect product determines what happens with conjugation?

[thorn jay]

exactly!

because using only this group action we can completely recover the group structure on G

[twilit wraith]

wait, how come?

[cloud walrus]

.𝖊𝖓𝖕𝖊𝖆𝖈𝖊_𝖒𝖚𝖘𝖎𝖈

[thorn jay]

So there is a natural isomorphism from G to the set N x H with the group product (n1, h1) * (n2, h2) = (n1 r_h1(n2), h1h2)

[cloud walrus]

.𝖊𝖓𝖕𝖊𝖆𝖈𝖊_𝖒𝖚𝖘𝖎𝖈

[thorn jay]

all this to say that the map f defining the semidirect product can be thought of as some conjugation action of H on G, and in fact this is precisely what it is when you consider the semidirect product

[twilit wraith]

i guess i see why the direct product is the semidirect product but with conjugation

[thorn jay]

(without conjugation)

[twilit wraith]

without?

oh just that conjugation fixes N?

[thorn jay]

the direct product is the special case where f is the zero map

[twilit wraith]

mmm i see

[thorn jay]

this can be helpful, if you know that there are non nontrivial maps from H to Aut(G) then you immediately know that there cannot be any semidirect products besides the direct one

[twilit wraith]

i was understanding that nonabelian groups arise only from nontrivial maps from H to Aut(G)

i guess just whats really confusing me is that this nontrivial has to determine how conjugation works

[thorn jay]

if G and H are abelian, yes

[thorn jay]

in reply to this

[twilit wraith]

because from what i saw semidirect was simply just a weirder multiplication

[thorn jay]

wdym by this?

[twilit wraith]

i missed a word on accident

its still just not making sense to me why the homomorphism from H to Aut(G) has anything to do with the conjugation

[thorn jay]

well in a sense the homomorphism from H to Aut(G) is what defines the conjugation

[twilit wraith]

here it just seemed like conjugations always give you an isomorphism with a direct product

[thorn jay]

that is on the level of sets

[twilit wraith]

i might be misunderstanding the between-the-lines stuff with the equalities here

[twilit wraith]

do these last two equalities hold just by how semidirect products are constructed?

rather than the conjugation being exactly what r_h_1(n_2) is

[thorn jay]

we have not constructed any semidirect product here

[twilit wraith]

i mean by cancellation itll turn out to be that way

[thorn jay]

this is just showing that the group structures of N and H, and the particular map r : H -> Aut(N), which is determined by G, actually determine the group operation of G

[twilit wraith]

i guess im just not sure where r_h came from

[thorn jay]

here

[twilit wraith]

ah wait i might be getting it now

what matters is just where exactly an element of H sends the elements of N to under that conjugation

but we know the action of H on N by conjugation will always work by the fact that N is normal

[thorn jay]

yes, exactly

[twilit wraith]

to me though it just seems like conjugation is an easiest action rather than the only action we can possibly choose

[thorn jay]

conjugation is the most natural action to choose! We want to mimick the direct product, and the direct product works because you can commute the elements of your factor subgroups. In general, elements only commute "up to conjugation"

[twilit wraith]

i see

[thorn jay]

i.e. g h = (ghg^-1) g

[twilit wraith]

so its moreso that the semidirect product rises from the fact that we chose conjugation to be the action to work with

[thorn jay]

I wouldn't say it's a choice, rather that it's the thing that pops up when we look at wanting to move elements around in a word

anyways, that may be pedantics

[twilit wraith]

hmm

[twilit wraith]

so just to make sure im thinking about this right

we know just by multiplication that we can think of multiplying arbitrary elements of NH in the sense that gives a semidirect product later on

but in that multiplication a conjugation of n_2 by h_1 occurs

so then we can determine semidirect products by what that conjugation sends n_2 to

and thats what semidirect products are about

[twilit wraith]

and the homomorphism from H to Aut(N) comes because those homomorphisms determine all the possible cases of what n_2 can possibly be sent to

[thorn jay]

yes

that does sound about right

[twilit wraith]

yes

finally i think i get it

it makes a lot of sense in hindsight

[thorn jay]

always in hindsight

[twilit wraith]

truly lol

okay now im wondering what the general process of classifying groups using semidirect products should look like

because it feels like im all over the place with trying to classify groups of order 28

[thorn jay]

always start with sylow

[twilit wraith]

yes that part makes sense

the part where it stops making sense is where i need to know what to do with the homomorphisms i have

like i get that in the case of order 28, i get that no matter if my sylow 2-subgroup is Z4 or V4, i have only one nontrivial homomorphism from either group into Aut(C7) = C6

which then has to be inversion because thats a automorphism of order 2 that always exists

but then when im actually trying to determine whether or not a group i have is already isomorphic to another group i found in this process, i just dont know

i tried to use presentations as a start but it just got complicated real quick

[thorn jay]

(I'm just thinking outloud)
consider the sylow-7 subgroup. Then n_7 must divide 28/7 = 4, so either 1, 2, or 4. Furthermore it must be 1 mod 7, so n_7 = 1, i.e. G must have a normal subgroup N isomorphic to C7. Then obviously of course the sylow-4 subgroup H is disjoint from N and generates G along with N. Hence indeed G is isomorphic to C7 \rtimes C4 or C7 \rtimes V4.

As you remarked, both C4 and V4 have only one nontrivial homomorphism to C6 = Aut(C7), so for both there is one commutative and one noncommutative case. This leaves only to check the noncommutative G1 = C7 \rtimes C4 and G2 = C7 \rtimes V4

[twilit wraith]

wdym commutative and noncommutative

arent C4 and V4 both abelian

[thorn jay]

I meant the resulting semidirect product

commutative iff f is trivial

[twilit wraith]

oh i see

[thorn jay]

Now notice that by Sylow both G1 and G2 contain a unique copy of C7 which is normal. If they were isomorphic, this would give isomorphic factor groups, but one is C4 and the other is V4, so G1 and G2 cannot be isomorphic

[twilit wraith]

oh i see

so like G1/C7 = G2/C7

but its not

so no isomorphism exists

thats much simpler than what i was going with

[thorn jay]

I'm sure multiple approaches exist :3

[thorn jay]

yee

28 is nice and small, meaning that every factor in the composition series must be prime abelian by the classification of finite simple groups lol

[twilit wraith]

yeah

maybe from here i just need to recall some stuff i already know about groups

i was just overcomplicating it because of my lack of comfort with semidirect products

but now that i know i should be good

[thorn jay]

that's great!

hint: do not work with presentations. ever.

(/hj)

[twilit wraith]

yeah it just very unwieldy

apparently in rep theory its good but i imagine for much much different purposes

[thorn jay]

are you sure those aren't re\presentations?

[twilit wraith]

lol i suppose so

but i couldve sworn they were talking about group presentations

maybe i did genuinely just misinterpret though

[thorn jay]

maybe, ask wew or sm

[twilit wraith]

anywho i will rewrite my homework because i dont wanna make my prof read presentations

[thorn jay]

good luck o7

[twilit wraith]

thanks so much for the help

it feels very freeing to finally understand this

[thorn jay]

of course!

[twilit wraith]

last question, how does this proof look?

[thorn jay]

in the first paragraph: it must divide 4, as 4 is the index of a sylow-7 group in this case

[twilit wraith]

oh right oops

typo

[thorn jay]

for the rest looks good!

[twilit wraith]

awesome

[dense zinc]

9/10 minor mistake at the start but that was already pointed out

[thorn jay]

I wonder..

given an arbitrary congruence representable variety, is there a nice description of split short exact sequences?

for groups its semidirect products, for modules its direct products/sums

I think lie algebras have semidirect products too?

[twilit wraith]

ok groups of order 20 is a little harder

i end up getting that there are three homomorphisms from Z4 to Aut(Z5) obviously

[dense zinc]

Probably, I didn’t do a lot in group ring fields

[twilit wraith]

but idk how to distinguish these groups from each other

[thorn jay]

maybe it's nice to try and prove a sufficient condition for two semidirect products to be isomorphic

Here's a nice lemma: Let G be a p-group, and H1, H2 nonisomorphic groups of order not divisible by p. Then for any f1 : H1 -> Aut(G) and f2 : H2 -> Aut(G) we have that G \rtimes_f1 H1 and G \rtimes_f2 H2 are not isomorphic

see if you can prove this yourself

[twilit wraith]

wait doesnt this not relate to the problem im having though

[thorn jay]

it does, C5 is a p-group for p=5 and C4 and V4 are nonisomorphic groups with order not divisible by 5

[noble nexus]

No but in your case you should have a case where some of your semidirect products are isomorphic

for groups of order 20

[twilit wraith]

yes but the issue is that i need to show that some of the homomorphisms from Z4 to Aut(Z5) dont give me the same semidirect products

[thorn jay]

ah yeah the problem at hand

[noble nexus]

You want to think about symmetries that might relate your different actions

I believe that of the three, two should be isomorphic

[thorn jay]

that sounds reasonable to me

[twilit wraith]

im predicting that the homomorphisms which send 1 to the identity and inversion automorphism result in the same isomorphic type

[thorn jay]

I believe if one has two actions f, g : H -> Aut(G) such that they "differ" by an automorphism of H, then the resulting semidirect product are isomorphic

[thorn jay]

that would indeed imply this

[noble nexus]

yeah this is true

[dense zinc]

Me too

[noble nexus]

Although sending 1 to the identity is going to give you the trivial action no?

[twilit wraith]

no that would be 1 to 0

[noble nexus]

What do you mean the identity

[thorn jay]

they meant that the action is the identity

[twilit wraith]

oh oops my bad

[thorn jay]

i.e. we construct the holomorph of C5

[twilit wraith]

yeah nah ur right i read totally wrong there

[thorn jay]

but no one is based enough to know that term

[noble nexus]

Well the identity element in the automorphism group is the trivial transformation is what I mean

[noble nexus]

I know it!

Holomorphs are cool

[thorn jay]

thank you 🙏

[noble belfry]

Hey guys

[thorn jay]

they have an analogue for quandles which I find awesome

[noble belfry]

Can someone give a hint towards showing that $S_\infty$ has only 2 subgroups of finite index

[cloud walrus]

sudo

[noble nexus]

Does S infinity here mean all permutations of N or only finitely supported

there are differing conventions

[noble belfry]

Finitely supported

[thorn jay]

finitely supported meaning what here?

[noble nexus]

It only acts nonrivially on a finite zet

[noble belfry]

If p is an element then there is some M s.t p(n) = n for all n >= M

[noble nexus]

so basically just the union of Sn over all n

[thorn jay]

right okay

[twilit wraith]

is the set of all finitely supported permutations just countable?

[noble nexus]

Yea

[twilit wraith]

interesting

[noble nexus]

it's a countable union of finite sets (Sn)

[noble belfry]

Yeah

[twilit wraith]

oh right

[noble nexus]

one of the nicest groups to form a von neumann algebra out of in fact...

[noble belfry]

If you know to show this don’t tell me just maybe point me in the right direction 🙏

[noble nexus]

anyway for the proof I don't know the answer but I'd probably think about actions on finite sets I think

or at least that's one direction maybe

[thorn jay]

what is von Neumann doing here

[dense zinc]

I got a hint
Both subgroups are finite

[noble nexus]

any finite index subgroup certainly must contain permutations that are arbitrarily large

[noble nexus]

Well that hint is wrong

[noble belfry]

What

Yeah it’s not true i think the subgroups are $A_\infty$ and itself

[cloud walrus]

sudo

[noble nexus]

I also might think about the intersection of a finite subgroup with each Sn

Since it's just the union of Sn where Sn acts on the first n points

[twilit wraith]

oh wait for my earlier problem couldnt i argue non-isomorphism using kernels

[dense zinc]

Brainfart
I just lost all my knowledge

[dense zinc]

You could’ve

[noble nexus]

That sounds right to me

[noble belfry]

Oh

I know why

It’s because A_n is the only proper non trivial normal subgroup of S_n

For n > 4 maybe ?

Wait

[noble nexus]

You can form certain operator algebras using discrete groups, and the algebras you get are "irreducible" in a sense when the group is an infinite conjugacy class (ICC) group. S infinity is one of the simplest ICC groups. Also the fact that it is a union of increasing finite groups makes the structure of the von neumann group algebra particularly nice

[noble belfry]

That doesn’t help since finite index doesn’t imply normality 💀just that there’s a normal subgroup

[thorn jay]

does the union have to be a sequential direct limit?

[noble nexus]

Yeah more or less

[thorn jay]

awh

cuz else any locally finite group would be nice

[noble nexus]

it doesn't have to be in general but when it happens it's nice

[vapid vale]

my hint is that it is actually helpful

[noble belfry]

Wait this does actually help we’re guaranteed that the normal subgroup also has finite index

[vapid vale]

its not immediate from that

[noble belfry]

Yeah i think i see the line of reasoning

[vapid vale]

which is

[noble nexus]

I would think about how a finite index subgroup could lead you to a normal subgroup

(I don't actually know if that's the correct approach)

[vapid vale]

^ and prove that this normal subgroup also has finite index

[thorn jay]

oh take the core?

[vapid vale]

🙃

[noble belfry]

We have that if H has finite index, then there’s a normal subset N of H with finite index Then (S_n intersection N) is normal in S_n since for g intersection S_n,
gxg^-1 where x is in S_n is in S_n and also is in N by normality, which implies that S_n intersection N is a normal subgroup. So its either trivial or A_n or S_n for every n

[vapid vale]

if this is just a sketch, sure but you should prove N has finite index

also i personally would prefer to just work with A_infty rather than stating something for each S_n

[noble belfry]

That’s just because if we consider the homomorphism from G to Sym(G/H) induced by the action on the left cosets we get that |G/N| <= n!

[noble belfry]

That’s kind of an amazing consequence though

Group actions are really powerful and cool

[sonic coral]

b is not true as written, right? Don’t we need G and X to be the same size? Otherwise I think S_3 acting on {1,2,3} is a counter example

[velvet hull]

could you explain how it would be a counter example?

[sonic coral]

the stabilizer of each element in X is size 2.

so every x in X is fixed by some nonidentity element of G

like the claim is that there is an element in X with a trivial stabilizer, but i don’t think this is guaranteed under the conditions given

[rapid cave]

Looks like a counterexample yes

[cursive frost]

is there a footnote for (b)?

[rocky cloak]

Yeah, I'm not sure what this problem was trying to get at with b. It's just obviously false unless X is free.

[sonic coral]

the footnote says use a theorem from class. probably orbit stabilizer

[cursive frost]

ah weird

[sonic coral]

what do you mean by free?

[rocky cloak]

G under left multiplication essentially

[sonic coral]

oh i see

yeah we’d need a bijection from G to X and so they’d have to have the same size

[kind temple]

by orbit stabilizer yea

[vocal pebble]

Not to mention it contradicts part (a) for the example you gave

[swift tundra]

Does question 10 here have a typo? Either I think they meant a map f: M—>N/aN, or they forgot to add the condition that aM is contained in the kernel of f.

Actually perhaps is it well defined because projecting onto N/aN will definitely put aM in the kernel

[swift tundra]

I was wondering if I am headed in the right direction with my reasoning:

Since N is finitely generated and a is contained in the radical, then I can find a minimal generating set where none of the elements are in aN (similar to Nakayama). Therefore these n_i are not in the kernel of the induced map, and in particular we can find nonzero m_i such that f(m_i)=k, where k-n_i lies in aN. From here I want to argue why we can actually hit n_i, but I’m still working on that

[somber sleet]

I need some help with an exercise, I don't manage to finish it. Let G be a group acting faithfully on a set X with five elements. This action only has two orbits, one with 2 elements and the other with 3. How could G look like?

It's easy to achieve that G must be a subgroup of S2 x S3. Also if we consider the projections
p1: S2 x S3 \to S2 and p2: S2 x S3 \to S3, we get that p1(G) is a subgroup in S2 and p2(G) is a subgroup in S3. I do not come further, I am pretty confused

[errant quartz]

Let $G$ act on $X$, and let $x \in X$. Then $\psi: aStab(x) \rightarrow ax$ defines a bijection from $G/Stab(x)$ onto $O(x)$ (the orbit of $x$), which satisfies $\psi(g(aStab(x)))=g\psi(aStab(x))$ for all $g, a \in G.$

Question: Could this reasoning be considered similar to the reasoning that for a homomorphism $\phi$, $im(\phi) \cong G/ker(\phi)$? Specifically, something about the stabilizer of $x$ feels like the kernel of a homomorphism to me, and its orbit feels like the image of that homomorphism.

[cloud walrus]

muffin hamster

[rocky cloak]

Something that might be helpful is to think of the image of u as a submodule of N, and try to make a situation where you can apply Nakayama

[rocky cloak]

So you're pretty close. Next it would be smart to determine what p1(G) and p2(G) can be, and split into cases.

After that you just need to think about which elements of S2xS3 can be in your group.

[rocky cloak]

Yes, you have a homomorphism of G-sets
G -> X
g |-> gx
whose image is the orbit of x.

There isn't a superclear concept of kernel of G-sets, but the stabilizer fulfills that purpose here.

You can look at the universal algebra version of the isomorphism theorems
https://en.wikipedia.org/wiki/Isomorphism_theorems

[astral ivy]

To prove something is an isomorphism we first need to prove it's a homomorphism and then prove it's surjective and injective. I assume surjection and injection should go rather easy, all done following the definition but I'm not sure how proving a homomorphism would go and it's not because I couldn't apply the definition but rather because there's so many different examples that I get lost and can't connect how the definition impacts in that specific example

For example: Let G be a subgroup of Gl_2(R) generated by matrices (-1 0 0 1) and (0 - 1 1 0). Prove G is isomoprhic to D_4. First of all I don't even understand what "generated by" means, does it mean those 2 matrices are identities so we have 2 identity elements in a way?

What I can notice immediately is that from one generator matrix we can get the other one just with rotation which is an element of D_4, that's the only connection I can see

[elfin wraith]

Ok, so in general in algebra, we say that an element a generates an object if every element of that object can be written as a combination of the element. A little less abstractly, We see that D_4 is generated by 2 elements, a reflection and a rotation, thats because D_4 is the group of symmetries of a square, and we can get to all of these by taking some combination of rotations and reflections

[astral ivy]

Hmm okay but why is it generated with 2 matrices? (Which are identical if rotation operation from D4 is applied)

[elfin wraith]

Im not sure what you mean by why is it generated with 2 matrices. If you go about proving that its isomorphic to D_4 I guess you can argue that its because these matrices correspond to a rotation and a reflection throough some angle (or possibly just 2 reflections through different angles, the geometric interpretation isnt unique)

[astral ivy]

Okay, what should I do to prove G is isomoprhic to D_4? First show homomorphism right?

But here's where I'm lost because the examples are so varying and different.

By definition, a homomorphism is a function between groups that keeps the group operation. f(a * b) = f(a) • f(b)

This makes perfect sense when a and b are some algebraic expressions. But here they're not, they're matrices.

[elfin wraith]

Ok so lets say $f:G\to H$ is a function from a group $(G,\ast)$ to a group $(H,\times)$

[cloud walrus]

Nope

[elfin wraith]

Then $f$ is a homomorphism if for all $a,b\in G$, we have that $f(a\ast b) = f(a)\times f(b)$. So we need to ask ourselves, what is the operation on GL_2(R)?

[cloud walrus]

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

[elfin wraith]

The point is that the homomorphism takes you from the operation on G, to the operation on H, so this is how you answer your question about a and b being matrices

Are you happy with that? Do you know what the group operation on GL_2(R) is?

[astral ivy]

Yep

[astral ivy]

I don't. I think it's addition of matrices but I'm not sure (multiplication would be marked with an asterisk I think)

[knotty badger]

it's matrix multiplication, actually

[elfin wraith]

Ok, yeah this is where mathematicians get sloppy and why I kinda made the difference with the symbols for the operations. First point is unless its explicitly written with an addition sign, I woudlnt read into the actual symbol used. (and even then, all you can usually learn from additive notation is that the group is abelian, but this is more of a convention than an actual rule, in general you shouldnt read into the symbol)

Second point is that GL_n(R) is a group under multiplication, (see if you can spot which axiom fails for addition!)

But ok, if you take the operation to be matrix multiplication, are you now happy with the definiton of a homomorphism when theres matrices involved?

[glad osprey]

I dunno if you have learned about cyclic groups yet, but if it were generated by only 1 matrix, it would be cyclic. In which case it couldn't be isomorphic to D_4 which we know isn't cyclic (it should be intuitive that you cannot perform a rotation some number of times to get a reflection, and there is no reflection that gives you every possible rotation when repeated)

[astral ivy]

In this case it looks like the rotation is equal to reflection. Take the 2 generating matrices for example, rotate one once and we will get the 2nd one but also if we reflect the first one we will tet the 2nd one.

[elfin wraith]

So now, knowing that the operation is matrix multiplication, its easy, though somewhat tedious to actually just compute the isomorphism to $D_4$. $D_4$ is the group of symmetries of a square, so if $r$ is a rotation by $90^\circ$ and $s$ is a reflection along an axis, then we know that $D_4 = {e,r,r^2,r^3,s,sr,sr^2,sr^3}$

[cloud walrus]

Nope

[elfin wraith]

Now we want to show that we get exactly these same elements in our matrix group, because then the isomorphism just becomes a case of choosing what to call r and what to call s, do you think you could do that?

This is possibly easier if you know about group presentations and know that $D_4 \cong \langle a,b| a^4=b^2=(ab)^2 =e\rangle$

[cloud walrus]

Nope

[glad osprey]

Rotating a matrix is not a group operation in GL(R). You're mixing together things you can do outside a group with things you can do inside a group. For example, you can add two matrices, but addition is not an operation in GL(R), so you're not allowed to do that when studying GL(R) as a group

[astral ivy]

Well no I don't understand it. Let A and B be our generator 2x2 matrices. f(A*B) = f(A) • f(B) but what is • operation and how do I use it? Aren't operations in Dihedral group just rotation and reflexion which are both applied to only 1 element at a time so having them like this doesn't make sense to me. I'm having issues with understanding the notation.

f(A*B) would just be the product of those 2 matrices right?

[elfin wraith]

So A*B is the product of matrices yes, but f(A) and f(B) are not matrices, theyre rotations or reflections, and so composing them makes sense

[astral ivy]

Yep that's where my issue is, that's also why I can't properly understand homomorphism in this case

[elfin wraith]

Youre mapping out of matrices, so f(A) is not a matrix

[astral ivy]

So • is still multiplication but after the matrices get (individually)transformed in D_4

[elfin wraith]

Its the group opperation on D_4

[elfin wraith]

This is what I was saying here

[glad osprey]

IMO understanding D_4 abstractly is kinda tricky for a beginner, because on one level it's the symmetries of a square, but on another level it's just a set of symbols {e, r, s, ...} that combine in a certain way. The group presentation Nope mentioned above is probably the definition most mathematicians use in practice, but it's kinda abstract, and hard to connect with the symmetries of the square

[astral ivy]

Yes I know the identity property of D4 my issue is more about connecting it before and understanding how things work.

Operation on D_n is just composition of rotations and reflections?

[delicate orchid]

this presentation pmo. that will be all.

[wraith cargo]

Yes

[elfin wraith]

Yeah you can think of it this way, as sheddow mentions this can be kinda confusing. More accuratly, its what I wrote above, its the set of symbols $a,b$ subject to the relations that $a^4 = e, b^2 = e$ and $(ab)^2 = e$. And then multiplication here is just combining words and simplifying where you can apply these relations. But this is a bit more abstract

[cloud walrus]

Nope

[astral ivy]

Okay so let me understand this, f(A) • f(B) is roughly "Apply rotation/reflexion on matrix A, apply rotation/reflexion on matrix B, combine them". Except I don't really understand what their combination actually is.

[wraith cargo]

Arguably thinking about it geometrically is confusing because some of the compositions might not be very easy to visualize

[elfin wraith]

I would usually write <a,b| a^n,b^2> if that brings you any more joy

[delicate orchid]

I'm gonna slime u out like we playing superflat survival

that's just not even the right group that's an infinite group

[knotty badger]

yeah

[elfin wraith]

Youre getting the order wrong

[knotty badger]

you need (ab)^2 as well

[elfin wraith]

oops yeah lol

[astral ivy]

I see

[merry harness]

sorry if im interrupting smth, i kind of need a hint for ts...

[elfin wraith]

f(A) is not a matrix, it is some permutation of a square, which is sort of confusing i agree. But if you scroll up a little, I wrote out the elements of D_4 explicitly

[glad osprey]

No, this is the problem with thinking of D_4 as rotations/reflections. You're mapping matrix A and B into elements of D_4, so they're no longer matrices. You're never actually rotating matrices, you're mapping each matrix into a rotation/reflection, such that the multiplication of two matrices "correspond" to the composition of two rotations/reflections

[merry harness]

groups of homomorphism still kinda tricky for me ngl

[astral ivy]

Isn't it "apply rotation n-1 times, apply reflexion" (r^{n-1}s = e)

[elfin wraith]

Here, the group D_4 is explicitly the set mentioned, under the operation of just joining those words up

[wraith cargo]

They're cyclic groups so you just need to count where the generator of Z/mZ can be sent

[astral ivy]

I know what elements of D_4 are

[merry harness]

ok ok thanks

[elfin wraith]

So f(A) is one of those elements. What you need to do is choose one of them to send A to, this is how well build out isomorphism

[wraith cargo]

You have to show there only exist d homomorphisms + that it's a cyclic group

[knotty badger]

I don’t think this is true

[delicate orchid]

everyone knows u write semidirect product of <A|R> with <B|S> along phi as <A, B| R, S, {bab^{-1}phi(b)(a^{-1}) : a in A, b in B}> so D_n is <a,b | a^n, b^2, bab = a^-1>

[knotty badger]

I see

[elfin wraith]

I say take a step back for a miniute, lets just look at the subgroup of GL_2(R) that were considering

forget anything about D_4 for now

[astral ivy]

Okay

[elfin wraith]

I want you to just multiply those matrices together in a couple of different ways. Call one of them A, one of them B and compute AB, (AB)^2, B^2 etc

Just get a handle on how they can be combined

[astral ivy]

Alright done

[elfin wraith]

Ok, have you spotted anything? What can you tell me about (AB)^2?

[astral ivy]

It's A but with "inverted diagonal"

[wraith cargo]

That is not correct

[astral ivy]

We get the identity matrix I did it in head

Now I wrote it out

[elfin wraith]

Ok nice! What about A^2 and B^2?

[glad osprey]

I'm gonna take a page out of Pseudo's book and post a commutative diagram so the idea of a group homomorphism phi is that following either path in this diagram gives the same result: you can either apply phi to both elements then multiply, or you can multiply then apply phi. By multiplication here I mean the group operation, which is different for M and N (denoted by dot and circle respectively here)

[wraith cargo]

R u serious rn

[astral ivy]

A² is also identity matrix and B² is with negative diagonal

[glad osprey]

Dunno if it helps, maybe it's a bit easier seeing it visually than as phi(ab) = phi(a)phi(v)

[elfin wraith]

Ok nice, final thing, whats B^4?

Id like to hope you can make an educated guess about what it should be by now

[astral ivy]

Identity matrix

[wraith cargo]

https://tenor.com/view/hai-gif-10567033442484308038

Good job!

[elfin wraith]

Ok cool, so we now know that we have matrices A and B which satisfy A^4 = B^2 = (AB)^2 = I

[astral ivy]

Good job, I have some linear algebra knowledge for ts

[elfin wraith]

Does this remind you of anything?

[astral ivy]

Properties of D_4 identity

[wraith cargo]

Very nice

[elfin wraith]

Exactly!

[wraith cargo]

Here's a question for you tho
Do you understand why these matrices represent symmetries of a square?

[elfin wraith]

So now, going back to the isomorphism idea, could you make a guess about how you might want to relate these matrices with elements of D_4

[astral ivy]

You mean A corresponding to rotation and B corresponding to reflexion

[astral ivy]

Because they fit the properties for identity above?

[wraith cargo]

I mean but this is the wrong way around
D_4 as a group isn't random

God wait are we using the D_n notation here or D_2n

[elfin wraith]

D_n, were not heathens

[astral ivy]

Holy who uses D_2n

[wraith cargo]

A lot of people

[elfin wraith]

Dummit and Foote uses it I think hence a lot of people do too

[tulip otter]

at least everyone who uses D&F

[tulip otter]

oh sniped

[astral ivy]

I mean it doesn't really matter, it's more of would you like your order of the group cleaner

[glad osprey]

Just out of curiosity, does this make sense at all @astral ivy ? I want to know if this is helpful for a beginner or not

[astral ivy]

No I have no idea what you wrote out

[elfin wraith]

I would be amazed if it was, shit is barely helpful to me lmao

[wraith cargo]

I am validated in thinking that category theory is useless to beginners 🙏

[glad osprey]

It's just a visualization of the fact that multiplication followed by homomorphism is the same as homomorphism followed by multiplication

[elfin wraith]

Ok but in anycase, do you see what were getting at now? Do you belive that subgroup is isomorphic to D_4?

[wraith cargo]

I think it's important to realize that if someone has trouble with abstraction, even more abstraction isn't really helpful

[twilit wraith]

sorry to interrupt, but i have to prove that maximal subgroups of finite nilpotent groups have prime index

[knotty badger]

Uh no

[astral ivy]

I can see it but how do I "show" it?

[glad osprey]

This equivariance of the order of operations is maybe not clear to a beginner when you write f(ab) = f(a)f(b)

[knotty badger]

Yeah exactly

[glad osprey]

It's not more abstraction, it's just "following this arrow or this arrow" is the same. You don't need to know a shred of cat theory, and I'm not trying to teach cat theory

[wraith cargo]

I genuinely can't tell if you're ragebaiting

[knotty badger]

Flowcharts are not exactly a super abstract concept

[glad osprey]

Yes, exactly, it's just a flow chart

[twilit wraith]

i was thinking of letting G be a direct product of its sylow p-subgroups, letting M be some arbitrary direct product of subgroups of those sylow p-subgroups, and then showing that we can find a subgroup that M is contained in if we go deeper than "removing one section of some p-subgroup"

[elfin wraith]

You simply need to choose somewhere to send A and somwhere to send B which are going to preserve the identity

[wraith cargo]

Yes they are because you add extra steps to a simple enough concept
Why do you need to draw a diagram to represent f(ab)=f(a)f(b)
That already is equivariance it's literally the definition
You're just drawing out the same equation in a diagram that's harder to read

[astral ivy]

I genuinely admire the Dummit and Foote literature, our 2 semester abstract algebra course recommends only about 80 pages of literature for problem solving, it's just not enough to learn on your own.

[glad osprey]

Btw, wtf is with people shouting about rage baiting immediately when they're arguing with someone they don't agree with?

[twilit wraith]

i.e. if M is a direct product where two of the factors are proper, then its a subgroup of smth where one of the factors are proper, or if its a direct product where one of its factors has more than prime index in its respective sylow p-subgroup, then that factor is contained in a p-group of higher order

[elfin wraith]

Notice that $f(a) = f(e\ast a) = f(e)\times f(a)$ and $f(a) = f(a\ast e) = f(a)\times f(e)$ if $f$ is a homomorphism, so you know it needs to send the identity to the identity

[cloud walrus]

Nope

[elfin wraith]

This combined with the calculations you just did you should you where things need to be sent

[wraith cargo]

I shouldn't have said that in hindsight but I come from an advisor that has for years taught me that using category theory too early leads to students who read too much of the nlab and just regurgitate the words they read over there without understanding what they're reading and in the long term end up hurting themselves and others they introduce to this style of thinking

[elfin wraith]

Theres some people I want to tag but I wont...

I dont know that category theory "too early" is nessicarily harmful, but I also dont think its helpful

[noble nexus]

Category theory makes simple things complicated so that in advanced contexts it can make complicated things simple

but if you aren't in those contexts it just makes simple things complicated for no reason

[glad osprey]

Yep, I agree, too much cat theory early on isn't the best idea 👍 and I found out what I wanted - the commutative diagram didn't make any sense to danilo, so I probably won't use that explanation again

[wraith cargo]

I probably should have said something akin to "students who lack a dose of mathematical maturity fall into this problem"

[merry harness]

I look forward to learning category theory

[wraith cargo]

Yeah sorry that I'm rabid about this but I've seen many people who fell into this hole and I enter like a fight or flight response lmao
Sorry once again

[noble nexus]

Ive been guilty of over categorifying in explanations sometimes

I have sinned...

It's tempting

[glad osprey]

Lol, it's okay, no worries

[elfin wraith]

I will say, just as a generic point, if someone is struggling with the concept of the image of a function (which was fundamentallty the issue there) adding any more complexity to the situation is likely not the call in any capacity

[wraith cargo]

This isn't a priori bad but you have to know who your audience is and if they're willing and mature enough to learn this style of thinking
When I was learning category theory I was in an algebraic topology class and I was constantly asking if X concept was actually Y concept from category theory

[rapid cave]

I once tried to introduce cat theory into a seminar on GT

a professor not even from that course stopped me

once I asked him about something related to that

[elfin wraith]

I think this is the point where it makes sense to introduce categories

[rapid cave]

and AG right?

[elfin wraith]

and you can get to the opposite siutation like in my AG course where activly avoiding category theory just got annoying

[glad osprey]

Yeah, in this case it wasn't the right explanation. But the hard thing is trying to figure out what people struggle with as beginners: sometimes it's very concrete, like what an operation is, or what the elements of a set is, and sometimes it's more abstract, like "why are we doing this", how do these things fit together etc

[elfin wraith]

Yeah

[tardy hedge]

I liked this btw lol

[wraith cargo]

Like in my opinion if you wanna learn category theory, first you learn the examples classically to have some preset intuition and then you're ready to understand why we abstract them in the ways we do using categories
Like you can do it the other way around where you treat category theory as a very fundamental explanation, but for many students I think the definitions of category theory can feel very unmotivated

[glad osprey]

I think when teaching math, you often just have to guess, and throw a lot of explanation at the student in the hope that something sticks

[elfin wraith]

My issue is also that if you learn category theory much before the natural paces you learn it anyway, you just dont have any examples to actually work with

[noble nexus]

Yeah examples become really contrived

[elfin wraith]

You can maybe drip feed some parts through intro algebra or whatever, but like, if you dont have Ring and Group and Top and pi_1 what are you even talking about?

You can do a decent ammount with Set sure, but its not the most interesting course

[noble nexus]

it's why I've never really been a fan of resources that try and make category theory super simple or do category theory for people without a lot of the other math because like why

[wraith cargo]

I think it's good that you tried at least to have a very exotic explanation that might give someone unexpected intuition

[noble nexus]

Not to say it's bad for those to exist but

[wraith cargo]

Like it's not bad to experiment pedagogically

[knotty badger]

Wait why

[wraith cargo]

But if you're really going off the beaten path things get very subtle

And you have to be very careful and very carefully listen to the feedback you get from students

[noble nexus]

Just because without a lot of the interesting math examples I think it's hard to really give people a picture of why it matters

[wraith cargo]

Because how do you motivate category theory definitions without existing examples other than "this is cool :3 UwU"

[noble nexus]

and it encourages undergrads to (no offense) waste their time on it before having the background. Not to say it isn't interesting, and if people want to learn it then by all means it's great

[elfin wraith]

Even as someone who very much loves algebra just because its fun, when I tried to learn category theory before any algtop I was just bored to tears, its a bunch of stuff you can do, and I kept thinking sure ok, but I never really cared

[noble nexus]

(I was that undergrad to some extent, and category theory made so much more sense to me once I had examples of it everywhere)

[knotty badger]

Hm this hasn’t really been my experience

[elfin wraith]

and even if I had cared I dont really gain anything from it, because im not doing anything where its conceviably helpful

[noble nexus]

someone with background will see the definition of a functor and immediately be able to come up with 10 examples

[elfin wraith]

H_1,H_2,H_3,.....

[languid trellis]

Spec!!!

i think

[glad osprey]

Yeah, a few years ago I tried to learn from Category Theory for Programmers by Bartosz Milewsky, and while I kinda got some intuition for what a functor was, most of it was way too much handwaving to actually understand anything. I realized that I needed to go through the "normal" math curriculum before this could make any sense to me

[elfin wraith]

Yeah, its a functor from CRing to Top

[knotty badger]

There are lots of very concrete examples you can use for these

[knotty badger]

Hm bartosz’s stuff was handwavy?

In what way

[noble nexus]

Category theory for FPs I think is ok because again they have some relevant background that lets them care and come up with examples

Granted I don't really know anything about FP so maybe there are other issues there

[glad osprey]

From what I can remember, it was mostly just a bunch of pigs and arrows. A monad is just a pig in the category of endopigs (he might not have said this, it's been a long time since I read it)

[knotty badger]

I mean I’ve watched the lectures associated with the book and they did not seem handwavy

As if having diagrams is a bad thing

[glad osprey]

Maybe it's just that Haskell represents such a narrow part of cat theory that I felt there wasn't enough examples to justify the abstraction

[knotty badger]

Haskell has lots of useful examples of things from cat theory though

[glad osprey]

I just remember that when I got to natural transformations, I fell off pretty hard

[wraith cargo]

[knotty badger]

This is a common experience

Part of why I dedicated so much time to figuring out how to explain naturality

[knotty badger]

Do you understand them now

[quiet pelican]

Someone with the background will probably have heard the word functor be name dropped by books/lectures without defn before /hj

[marble hinge]

I am one of those guys who became interested in category theory at some point just because I was a Haskell dev 🙂

[glad osprey]

Hmm, kind of... There's a whole bunch of natural transformations in diff geo, so I've been exposed to it a lot, but I don't have a good sense of why we care, or how to use the fact that something is a natural transformation

[marble hinge]

You see and use all those words: Functor, Monoid, Monad, etc right in your code, so at some point you want to dig deeper 🙂

[quiet pelican]

You need to either say contravariant, or drop an op in there 🙃

[glad osprey]

Same, it's actually part of the reason I started studying math

[marble hinge]

But normally people coming from that way don’t have enough background in abstract algebra

My colleague had a PhD in category theory though, so not everyone is as clueless as me 😄

Isomorphism is cool! 😎 I like it when I can just relabel things and get to something that I understand already:)

[wraith cargo]

The whole point of category theory is to formalize the notion of what it means for something to be compatible in some general sense
The reason we care, vaguely, is think about what happens in homology
You get maps from H_n(X) to H_{n+1}(X)
But the n-th homology and the n+1-st homology are different functions, so what's going on here
Well this is where natural transformations come into play
The start of the definition makes enough sense, it's any arbitrary morphism between functors with the same domain and codomain so we abstract the idea of these maps between homology groups for example
But the rest of the definition is a bit more mysterious
You want some diagram to commute, but why?
Well this is a bit of a more philosophical question
The additional data of a commutative diagram is a compatibility requirement that allows natural transformations to respect the data that functors carry in a more powerful form
You abstractify the notion of a map preserving the structure of your objects

God I am like

Fucked in the head

I said "this is the point" like 5 times

Now the way you use this is to say

If something is a natural transformation

Automatically you get that it respects whatever algebraic operation you have on your object

If u have something like this for example

The fact that it's a natural transformation literally automatically allows you to know that the R-module operations are compatible on both functors

And this is what you'd ideally want from a philosophical point of view right

If operations wouldn't be compatible under a transformation like this something is morally wrong

[delicate orchid]

The whole point of category theory is to glue triangles together

[wraith cargo]

And maybe this even speaks to a wider philosophy of what we want algebra to look like in the sense that in an ideal world we build all the operations we put on objects interact in expected and we'll behaved ways

[glad osprey]

That makes a lot of sense, thanks

My copy of Category Theory in Context came in the mail today, so I'll start learning more about it soon

[thorn jay]

I've got that book on the shelf next to me

[elfin wraith]

My category theory course next sem ends with Kan complexes, remind me to come back and laugh at this

[thorn jay]

Kan is a name I'm seemingly unable to escape

[elfin wraith]

Just kant do it

[rapid cave]

lets say I have a month of free time, what should I do in that month (in terms of math)

[thorn jay]

learn universal algebra

[quiet pelican]

Physics 🙃 /j
what maths are you generally interested in?

[rapid cave]

NT, and things around it

[elfin wraith]

Commutative algebra seems wise

[rapid cave]

I have small list, AG, CFT

modular forms

I probably don't need more things

[elfin wraith]

universal algebras number one salesman

Also very much a thing, persoanlly my prefernce

[thorn jay]

I am truly Spamton G. Spamton

[elfin wraith]

https://assets.cambridge.org/97805218/36876/sample/9780521836876ws.pdf

[thorn jay]

how much do you know about sheaves?

[rapid cave]

little

[thorn jay]

because maybe it is of interest to look at a book just covering sheaves

[wraith cargo]

Lol it's kind of a cop out answer but IMHO it's good to understand that this is a trend in basically all math
A lot of definitions come from a kind of vibe based approach where you try to frame things in the most idealized fashion possible
Especially in research this is very important to understand and it's also why definitions get updated a lot and evolve over time
You'll rarely land on the correct idea the first time because you might not have accounted for all the compatibilities and things you'd want to have in future work

[thorn jay]

you'll need it

[rapid cave]

hopefully will learn a bit in the next month

[elfin wraith]

I need to learn about sheaves

[rapid cave]

its a name?

[thorn jay]

Bosch has a book on it, it's probably pretty good

I've got another book of his and at least his exposition is good

[rapid cave]

your explanation was really clear

and interesting

thank you

[thorn jay]

wait no, Bredon

Bredon has a book on Sheaves

I looked through it a bit and it looks pretty good

[rapid cave]

prerequisites?

[proud vigil]

would a question about monoids belong here

[thorn jay]

yes

[thorn jay]

let me look

[thorn jay]

looks like its a book mainly about sheaf cohomology, which is good for NT and AG I believe

many things are sheaf cohomology

[rapid cave]

oh great

I am going towards sheaf cohomology

[thorn jay]

I'm pretty sure the main prereqs are the same as for AG

[rapid cave]

I told you I have a course on it. and I can go to lectures

[thorn jay]

then learning about sheaves and getting comfortable with them is great I think

[rapid cave]

whats the difference between that and hartshorne ch2,3

[thorn jay]

hartshorne is about schemes, not sheaves

I believe that working with sheaves in greater generality is good

[rapid cave]

so its a good idea to learn both

[thorn jay]

yes, but schemes come after sheaves

[rapid cave]

https://link.springer.com/book/10.1007/978-1-4612-0647-7 this?

[thorn jay]

yus

[vapid vale]

why

[proud vigil]

consider S^S, where S is a set with n elements.

1. how many subsets of S^S are monoids of size n!, with function composition as the operation?

2. how many subsets of S^S are monoids, with the same operation?

for 1), i can think of n! + 1 monoids:

Sn is one

the others come from defining an order on the n elements (with n! total possible orderings, so n! different monoids)) and letting each function map each element to a >= element (with n! total possibilities for assignments, meaning n! elements in a monoid)

but im not sure what the cleanest way to prove that these are the only ones would be, or if there are actually others. the fact that there's already 2 distinct types of monoids here is throwing me off

and i have ideas for 2, but they're sort of contingent on how 1) plays out

[rapid cave]

Oh yeah someone recommended me a book on Ergodic Theory for NT

forgot

I probably have enough things to do

[thorn jay]

idk I feel like the setting of schemes sometimes obscures the more general phenomena going on with sheaves

meh, not out of experience just in my opinion its better to learn about things in general

having to learn what a sheaf is and immediately after having to come to terms with the definition of an affine scheme was not fun so maybe that's why I think this way

[vapid vale]

this has a very strong assumption that the intention is to just work with the general phenomena going on with sheaves

[thorn jay]

I mean, if there's a course about sheaf cohomology that they're attending

[proud vigil]

i guess 2 is maybe a lot uglier actually, it doesnt seem that consistent how many elements you need to satisfy closure

[rapid cave]

its more on the AG side

[thorn jay]

mm, I see

[rapid cave]

so yeah maybe having a more complete picture is nice

but man this will require a ton of effort

[dim wind]

why must the number of element of order d in a finite group be a multiple of phi(d)?

[wraith cargo]

For any element of order d, it generates a cyclic subgroup of order d, and for every such element you can find phi(d) such other subgroups by exponentiating the element of integers relatively prime to d

To write this out if an element g has order d, then g^n has order d iff GCD(n,d)=1

[dim wind]

that I'm ok with

[proud vigil]

okay i kind of brute forced it and i think i found a counterexample for n=3 :( so theres a lot more monoids than i thought

[wraith cargo]

Okay let me rephrase what I said

[thorn jay]

submonoids are in general very not well-behaved

[wraith cargo]

Every cyclic subgroup of order d has phi(d) elements of order d in it
And every element of order d generates a cyclic subgroup of order d (these need not be distinct)
Ergo the number of elements of order d is (number of cyclic subgroups of order d)*phi(d)

So phi(d) | number of elements of order d

[dim wind]

I think I see it

[wraith cargo]

UwU

[proud vigil]

do you have to check that there's no repeated elements between cyclic subgroups?

[wraith cargo]

No because if there are then they're the same subgroup since they're all generators of order d

[proud vigil]

oh true

[dim wind]

thx

[delicate orchid]

Aye aye captain

[sonic coral]

wait is this not true? I pointed out the issue and what condition should be added, but he said that is too strong and we just need |X| > 1…

[rocky cloak]

What you said initially is correct.

Though if I'm allowed to fanfiction what the problem should say:

"For |X|>1, show that there exists a g in G such that gz =/= z for all z in X"

This statement is kinda similar, and is at least true

[chilly ocean]

This is incredibly obvious yet feels... enlightening. Thank you.

[thorn jay]

lwk this is how categorize algebraic structures and homomorphisms

[glad osprey]

Nice, love that it helped someone at least

[mint seal]

the essence of category theory

[glad osprey]

Just found this in my mailbox btw

[mint seal]

sweeet

I'm sure you'll have a Riehl good time with it

[proud vigil]

omg wait riehl is a professor here

thats so cool

[mint seal]

lucky!

[tidal schooner]

Great book, enjoy

[twilit wraith]

I have to prove that G/Z(G) is nilpotent implies G is nilpotent

Honestly tho idrk where to start

I thought that their central series would intersect at some point but doesn't seem like it

[chilly ocean]

just wanted to add now that after thinking some more about this I’m really starting to grasp the idea of a group object and things of that nature way better than I did before!

Thank you again

[thorn jay]

do you know about upper central series?

if you know that then the result is immediate basically

[twilit wraith]

Yes but it didnt seem like Z(G/Z(G)) intersects the upper central series of G at any point

Nilpotent groups

Intro group theory

my class is just going through a lot of group stuff ig

but this is the intro alg course for the grad program at my uni

also this is the last topic for group theory iirc

at least in dummit and foote the next chapter is on rings

[velvet hull]

Intro group theory (graduate)

[twilit wraith]

its been fun but stressful

im not fully sure ill get As in all my grad courses this semester which is worrying

idk how much that matters for grad school but i imagine a good amount

wdym

yes

i know in europe you have to do a masters before a phd

the first two years of a phd in the US is essentially doing masters in europe

yeah

i may go to europe for a phd but idrk

id like to stay in the US because everyone i love is here but you know

its kinda getting rough

not sure

[mint seal]

quals are exams that grad students take after 1 to 2 years of study. The most common ones are in algebra, real analysis, complex analysis, and topology, but some schools offer quals in other topics (probability, PDEs, etc.) They usually follow the content of the corresponding grad courses pretty closely

[twilit wraith]

theres a diffgeo qual at my school which is pretty cool

well the topology and diffgeo qual are the same but still

[mint seal]

nice

oh, interesting

that's kind of surprising

[twilit wraith]

yeah