groups-rings-fields

[sacred wharf]

Bro just roasted all the lower lvl channels

[rocky cloak]

lower channels meaning #discussion ?

[knotty badger]

i have that channel unlisted

[vapid vale]

oh god

[sacred wharf]

Early university

Ode mvc etc

[rocky cloak]

Oh you don't get banned from those, makes sense I guess

[vapid vale]

i dont understand people who get anything out of cesspools like general discussion channels of thousands-of-member servers

[thorn jay]

probably the best course of action if you value your mental health

[knotty badger]

maybe you should ask the people who frequent discussy

personally i prefer distwossy, advancedussy, mathussy and studussy

[lean sail]

can anyone here explain the fourth isomorphism theorem in relatively simple terms?

[vapid vale]

i used to (well i still do) administrate a server called academic music and the general channels were often pretty bad (and we didn't even allow off topic discussion there)

[knotty badger]

-# i hated typing that

[karmic moat]

I have studying role its pretty awesome

[lean sail]

(lattice) isomorphism theorem

[thorn jay]

LATTICE
I've got a nice picture, hold on

[knotty badger]

oh is this the

subgroups of G/N corresponds to subgroups of G containing N

?

[vapid vale]

yez

[rocky cloak]

The subgroups of G/N are exactly of the form H/N for H a subgroup of G containing N

[thorn jay]

❤️❤️💕 ✨ correspondence theorem ✨ 💕 ❤️ ❤️

[vapid vale]

the most important one :)

[knotty badger]

so the way i view it is in terms of the quotient map $\pi : G \to G / N$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

you've met the result that the image of a subgroup is a subgroup

but did you know that the preimage of a subgroup is also a subgroup?

[sacred wharf]

Is that it,or are there like properties like inclusion preserving etc or ig its implied

[knotty badger]

you've actually already proven a special case of this when showing that kernels are subgroups

[vapid vale]

i think "exactly"in jagr's message carries that meaning

[knotty badger]

anyway, this is what lets you turn a subgroup of G/N into a subgroup of G containing N - apply pi^(-1)

[thorn jay]

this is whats happening, basically (Con A here is the lattice of what amounts to normal subgroups but more general)

[elfin wraith]

Too may cooks

[knotty badger]

the map $\pi$ is the other direction

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

so i would recommend you investigate $\pi \circ \pi^{-1}$ and $\pi^{-1} \circ \pi$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[vapid vale]

wait until my explanation, its the only good one

[twilit wraith]

Fourth isomorphism theorem my beloved

[astral ivy]

what is linkedin? circlejerk for the employed?

[rocky cloak]

If H < K then H/N < K/N yes.

Normaility is also persevered and H=K iff H/N = K/N

[karmic moat]

And those who are #OpenToWork

[thorn jay]

it says that, if you only consider everything ""above"" (containing) θ, then its exactly the same as what you get for A/θ, where the lattice isomorphisms are given by the preimage and quotient

[thorn jay]

well so far I've been the only one with a nice picture

checkmate

[vapid vale]

everyone in this channel likes to invoke my name

[rocky cloak]

What a beautiful name it is

[sacred wharf]

I mean ig to prove it you just look at homomorphical G to G/N and consider image of H

[lean sail]

okay, it says:
Let $N \trianglelefteq G$. Put $\overline{G} = G/N$. Then there is a bijection from the set of subgroups $A$ of $G$ which contain $N$ onto the set of subgroups
\begin{equation*}
\overline{A} = A/N
\end{equation*}
of the group $G/N$. In particular, every subgroup of $\overline{G}$ is of the form $A/N$ for some subgroup $A$ of $G$ with $N \subseteq A$ (namely, $A$ is the complete preimage of $\overline{A}$ in $G$).

[cloud walrus]

proofman

[knotty badger]

this is a nice visualisation of the statement of the theorem, but i don't see how to prove it from this

[lean sail]

is $A$ supposed to be a subgroup? or an entire set?

[cloud walrus]

proofman

[vapid vale]

just the set of subgroups

[sacred wharf]

If an element of H is in N then it just maps zero
Otherwise it maps to say h1 + N
If you look at all h + Ns they are a group since H is closed under addition, is that the idea?

[thorn jay]

I mean, from here it's the standard proof

[twilit wraith]

Its a set

[knotty badger]

what proof

[thorn jay]

this is just a visualisation to help you see what's going on

[knotty badger]

it's just a picture?

[vapid vale]

(enpeace i am assuming the A is the A in the question, not your A)

[rocky cloak]

Sort of unclear wording yeah.

A represents a subgroup, so it's the set of (subgroups A of G)

[thorn jay]

i know

[vapid vale]

oh oops

[thorn jay]

lol

[lean sail]

yeah, it's the profs writing, who i can't get a hold of

[vapid vale]

W reading

[lean sail]

to me, the A just seems like an individual subgroup, which contains N... \overline{A} can't really be a set of subgroups, because the elements of A/N are cosets, only one of which is a group...

[rocky cloak]

Which is the correct reading

They just don't give a name to the set of subgroups

[lean sail]

that's how i interpreted it, needed some reassurance

what are they doing with the \overline{A} and \overline{G}, like what's the point? just easier than A/N and G/N?

[thorn jay]

I do not like it

lol

[lean sail]

their writeup?

[rocky cloak]

No point as far as I can tell

[thorn jay]

yeah the overline is a little pointless

[lean sail]

confusing as hell

[thorn jay]

(39 buried, 0 found)

[vapid vale]

i really dislike the overline notation generally

[rocky cloak]

Like you wouldn't use it for anything, or in this context?

[thorn jay]

overline for a family of elements

[vapid vale]

i would try to avoid it unless it was clearly the right notation

[thorn jay]

🔛 🔝

[lean sail]

then, what do they mean when they say ``the set of subgroups \overline{A} = A/N of the group G/N''? as far as i can tell, there is only one element in A/N that's actually a subgroup, and that's N itself....

[rocky cloak]

I use $k = \overline{k}$ for algebraically closed

[vapid vale]

but like i hate \overline{k} for k in Z/nZ

[thorn jay]

A/N is a subgroup of G/N for every group A

[cloud walrus]

jagr2808

[thorn jay]

its just the image of A in the projection p : G -> G/N

[vapid vale]

oh yes i mean specifically for equivalence classes/images under a quotient

[thorn jay]

then I agree

[x] or x/~ are the only acceptable choices

[vapid vale]

obviously closures are fine

[rocky cloak]

Then I use [x] (if anything at all)

[thorn jay]

wdym "if anything at all"

[rocky cloak]

Consider 2 in Z/3

[lean sail]

ok, that makes sense. so we consider for each A, it's corresponding quotient group A/N, sort of?

[thorn jay]

that.. is acceptable

[knotty badger]

The image of a subgroup under a homomorphism is a subgroup, you see

[thorn jay]

yes

but really we only are interested in whenever A contains N

[knotty badger]

And the quotient map G -> G/N is a homomorphism

[thorn jay]

a.. canonical one, if you will

[knotty badger]

-# I won’t actually >:)

[balmy python]

first iso thm

[rocky cloak]

It's the set {A/N : A/N a subgroup of G/N}

[lean sail]

i think that comes up pretty early on right?

[knotty badger]

Yep!

[knotty badger]

.

You also need this result

[lean sail]

aren't all of the A/N's subgroups of G/N?

[knotty badger]

In general, any function f : A -> B sets up a correspondence between subsets of A and subsets of B

It restricts to a bijection on certain nice subsets

[balmy python]

is there any tips for stuff that is very group specific

as in

maybe finding subgroups of a group

or finding which groups are normal

[vapid vale]

if you want to find subgroups you can just take an element and multiply it by itself a bunch

that'll give a few

[sacred wharf]

Sylow ofc

[balmy python]

what's sylow?

[balmy python]

that'll only give cyclic subgroups tbh

[sacred wharf]

O

[vapid vale]

yes

[thorn jay]

for normal subgroups, kernels of homomorphisms or the center are good ones

[sacred wharf]

I recommend reading up on this

[thorn jay]

try to construct a natural homomorphism and finding its kernel

[vapid vale]

you can construct homomorphisms and take their kernel

bra

[thorn jay]

I mean if its for an exam and it wasn't covered

[balmy python]

might do after my exam 😭 😭

[thorn jay]

I would recommend against that

so they don't accidentally use a result they aren't allowed to yet

[balmy python]

it's in a week and the guys making us do questions up to the lecture today

which is a decent amount of content

up to like orbit stabilisers and stuff

[sacred wharf]

Fun stuff

[balmy python]

yeah I just want the content to sink into my head more though

i feel like i haven't grasped group theory yet

compared to my other modules

[vapid vale]

modules

[balmy python]

subjects?

idk what to call them 😭

[balmy python]

anki!

[vapid vale]

no its just a math pun lol. there are objects called modules (which actually are groups with extra structure)

[thorn jay]

connected topological biquandle cohomology with coefficients in homogenous Beck-modules

[noble nexus]

if you've done group actions, a really natural way to find subgroups is to look for things your group acts on and then take the fixed point groups

(in fact every subgroup arises in this way ofc)

[knotty badger]

Yeah every subgroup arises as a stabiliser

[balmy python]

ohhh 😭

[sacred wharf]

Modules are sick until its over a pid

[noble nexus]

but often you can spot actions a lot easier than subgroups

[thorn jay]

wdym "until"

[vapid vale]

and wdym "sick"

[balmy python]

I will get back to you on this after i review todays lecture because that's what we did t

[thorn jay]

you're tryna tell me modules aren't awesome?

[sacred wharf]

Sick as in good

[sacred wharf]

Boooo pid

[thorn jay]

or are you an abstractionist elitist on their high horse merely wanting to work in the abstraction of abelian categories

[balmy python]

hmmm like this is one of the questions on my problem sheet but I don't even know how to tackle it. I get that the groups have the same size so if it weren't gonna be an isomorphism it'd probably be something to do with the order of the mapped elements not being the same

[rocky cloak]

PIDs aren't even artinian. Give me Nakayama algebras instead

[balmy python]

but how would i even approach this kinda question

[thorn jay]

sadge

[vapid vale]

nakayama means every module is uniserial?

[lean sail]

so when they say A is the complete preimage of \overline{A}, what exactly are they saying?

[knotty badger]

What do homomorphisms out of C_(nm) look like

[knotty badger]

I don’t actually know

[balmy python]

like a homomorphism from C_nm to another group?

[vapid vale]

i think you can just start with testing out some examples

[knotty badger]

Yes

[noble nexus]

they have the same order and one is cyclic, so if you want them to be isomorphic the other has to be cyclic

and in fact there's of course a unique cyclic group of each order, so that's enough

[balmy python]

cyclic?

[rocky cloak]

That A is everything that maps to A-over in G/N

[knotty badger]

Specifically I mean

Let G be another group

How can you produce a homomorphism C_(nm) -> G

[balmy python]

hmmm okay

[rocky cloak]

(as opposed to a preimage, which would just me something that maps onto A/N)

[vapid vale]

like if youre isomorphic to a cyclic group, you should just be able to take an element and repeatedly multiply it to get everything. try this with C_3 x C_5, C_2 x C_4, small examples. you'll at least get a feel for the problem statement before trying to apply group theory "methods"

[sacred wharf]

Diesnt this only gold if mn have gcd 1

[balmy python]

f(x)f(y) = f(xy) for elements x,y in C_nm and f(x),f(y), f(xy) in G?

or do you mean like

literally what is some homormorphisms

[knotty badger]

This is true, but there’s a simpler condition

[balmy python]

map the generator to a generator

[knotty badger]

It’s something you get familiar with when you work with group presentations a lot

[noble nexus]

yeah but we're not trying to give away the answer

[knotty badger]

So you’re half right, the homomorphism is determined by where you send the generator

[balmy python]

yeah

and then

[lean sail]

doesn't that follow directly from the canonical homomorphism?

[knotty badger]

So you have to map the generator “1” of C_(nm) to some g

Any other conditions?

[vapid vale]

(are multiple threads a possibility here lol)

[balmy python]

say we have h in C_(mn) which has generator g then
h = g^k for some k

f(h) = f(g^k) = f(g)^k

[knotty badger]

Yep

[rocky cloak]

Yes, most of what is being said is "obvious"

[knotty badger]

That shows the homomorphism is determined by f(g)

[balmy python]

so we map elements of the same power of the respective generators to each other

[knotty badger]

There’s one more condition you need

[balmy python]

bijection

[knotty badger]

We’re talking about just homomorphisms for now

The other condition you need is

[balmy python]

ah

[lean sail]

right, once you understand the write-up. so also what you said is basically also "the canonical homomorphism is a bijection between A and A-over?"

[balmy python]

hmmm

[knotty badger]

g^(nm) = e

[balmy python]

oh yeah ofc

😭

[knotty badger]

Thus, f(g)^(nm) = e

[balmy python]

yeah

[knotty badger]

In other words, to specify a homomorphism C_(nm) -> G

You have to pick an element of G, whose order divides nm

[sacred wharf]

Ye so maybe worth trying to prove this

[knotty badger]

And actually, doing this uniquely specifies a homomorphism

[balmy python]

we require the direct product to be cyclic with order that divides nm

[knotty badger]

You can read this off from the group presentation

C_(nm) = <x | x^(nm) = e>

This means that defining a homomorphism out of C_(nm) corresponds to picking some f(x) where f(x)^(nm) = e

[balmy python]

yeah

[knotty badger]

No more, no less

[elfin wraith]

I was about to say, are you taking intro algebra or something until I realised you probably are

[knotty badger]

So, given that, how can you define a homomorphism C_(nm) -> C_n x C_m?

[balmy python]

i guess a generator of C_n x C_m would be (g,h) where g,h are generators of each cyclic group

hmmm

[knotty badger]

That’s a good guess

[balmy python]

we want C_n x C_m = <x | x^(nm) = e> tbh

[knotty badger]

Though you’d have to check this actually is a generator

[balmy python]

yeah

[balmy python]

so then

(g,h)^nm = (g^nm, h^nm)

the question is if this is always (e,e)

[knotty badger]

What do you know about g and h

[balmy python]

well they're part of their respective cyclic groups

so g^m = e

h^n = e

but that feels off

[knotty badger]

mhm

it does?

[balmy python]

yeah because that would imply all positive integers are such that C_mn iso to C_m x C_n

[knotty badger]

So, be careful

[knotty badger]

If this is true, you get a homomorphism

From C_(nm) to C_n x C_m

[balmy python]

oh yeha

[knotty badger]

Sending k to (g^k, h^k)

[balmy python]

not necessarily an isomorphism

[knotty badger]

But you need more for it to be an isomorphism

Exactly

[balmy python]

hmmm okay

so let's check injectivity of the map

[knotty badger]

Though, this does show there are exactly n*m homomorphisms from C_(nm) to C_n x C_m

Which is kinda neat

[balmy python]

wait what?

oh homomorphisms

not isomorphisms

[knotty badger]

A homomorphism from C_(nm) -> G corresponds to picking a g in G with g^(nm) = e

[knotty badger]

Yeah

[balmy python]

don't all elements in g have that

[knotty badger]

Yes

[balmy python]

😭

oh wait

[knotty badger]

So every element of G defines a homomorphism

If G = C_n x C_m

[balmy python]

so we need to just pick a g in G with g^(nm) = e but it also generates every element as a g^k where 1 <= k < nm

[knotty badger]

Mhm

Another way to phrase this is

For a homomorphism, you need the order to divide nm

For an isomorphism, you need the order to equal nm

[balmy python]

for an iso u need it equal

yeah

hmmm

so we don't want g^k = e

[knotty badger]

We know this always holds

[balmy python]

can we have something like an original generator?

like say we already have a generator

[knotty badger]

of?

[balmy python]

where every element can be represented in terms of it

[balmy python]

C_m x C_n

[knotty badger]

Well if we had a generator of that, we’d know it’s cyclic

But that’s what we’re trying to determine

[balmy python]

ah shit

[knotty badger]

Here’s a suggestion

How about you try to figure out when they’re not isomorphic

[knotty badger]

I.e. when is this impossible?

[balmy python]

when each element doesn't have order equal to nm

[knotty badger]

This might be the time to try some small examples

[balmy python]

im so slow at understanding gt

[knotty badger]

Dw I am too

Still don’t really get it

[balmy python]

alright

probably something like C_4 or C_6

[white oxide]

Can somebody point me to why this is true? Suppose that $xy^{-1}, x^{-1}y \notin B$. Then $xy^{-1}$, $x^{-1}y \in \mathfrak{m} \subseteq B$, a contradiction. Is this the reasoning?

[cloud walrus]

okeyokay

[rapid cave]

Note (xy^-1) is the inverse of x^-1y

[knotty badger]

That would be y^-1 x right

Though ig if commutative you’re fine

[rapid cave]

But commutative

[knotty badger]

Okies

[rapid cave]

And the rest follows from the top sentence

This is the definition of a valuation ring they give

[balmy python]

did it after getting distracted a lot

uhh they're both not isomorphic

hmmm

we want the order of an element to be exactly g^nm

[sacred wharf]

ay there you go

[white oxide]

Yeah I implicitly used that in my proof ig

[rocky cloak]

By definitions either B contains xy^-1 or it contains (xy^-1)^-1 = yx^-1

[white oxide]

Oh yeah I guess that's simpler huh

[balmy python]

wait doesn't that just mean gcd(n,m) = 1 because say if we wanted a generator to only equal e when g^nm, then we want lcm(n,m) = nm which implies gcd(n,m) = 1?

[knotty badger]

You’ve got it

Indeed, for every g in C_n x C_m, g^(lcm(n, m)) = e

And lcm(n, m) <= nm

So our only chance is if they’re equal

[balmy python]

yep

ah and that's basically the whole question

😭

[knotty badger]

mhmmmm

[balmy python]

i'm trying to understand the orbit stabiliser theorem

so what it means is that if we have the stabiliser of an element and we have G/St(x), it is bijective to the orbit of x?

and not necessarily isomorphic because the orbit isn't a subgroup

so we have two elements in the same coset of G/St(x) lie within the same orbit?

[tribal moss]

An orbit does not consist of group elements at all.

[balmy python]

oh yeah forgot about that

it's a subset of the set the group is acting on

[tribal moss]

The basic idea is that if g and h are group elements in the same coset of St(x), then gx = hx.
And if g and h lie in different cosets of St(x), then gx != hx.

[balmy python]

ok yeah

and what really is good about actions on a group

like what i'm thinking in my head is that it's similar to matrices acting on a vector space or something

is it just associating permutations with group elements?

[tribal moss]

Yes. In many contexts the reason to care about groups at all is as a way to organize arguments about some permutations we're already interested in.

[balmy python]

ahhh

is there any nice examples other than matrix stuff?

I guess D8 acting on a set of possible square configurations?

[rapid cave]

burnside's theorem (lemma?)

[tribal moss]

Indeed.

[balmy python]

haven't come across that one yet

lemme have a look at it after

[tribal moss]

And in general a good alternative source of group actions is symmetry groups for geometric patterns and shapes.
Each element of the symmetry group acts on the set of all points in the plane.

[balmy python]

Each element of the symmetry group acts on the set of all points in the plane.

wdym?

[tribal moss]

Hmm, I think I worded that badly.

Each symmetry group has an action on the set of all points in the plane.

[rapid cave]

lets say you have a square, then C4 acts on it by rotating each point on the square 90 degrees

[balmy python]

yeah

[rapid cave]

but there can be other shapes with weirder symmetries

described by other groups

[balmy python]

why not just represent these rotations and flips as matrices acting on all points in a plane

[balmy python]

ahhh

[vapid vale]

well you can consider trying to find the order of the rotational symmetry group of, say, an icosahedron. so you let this group G act on the set of triangular faces, of which there are 20.

the order of this group G is the order of the stabilizer times the order of the orbit

i claim that its pretty easy to see what these are

[rapid cave]

this is representation theory!

[vapid vale]

what are the possible rotations that leave a certain triangle fixed

and how many of the 20 total triangles can you rotate that certain triangle to?

[balmy python]

like in D6?

[vapid vale]

in the icosahedron (picture attached)

[balmy python]

ohhh

hmmm

sending one corner to another corner determines a rotation?

[vapid vale]

wdym byb that

[balmy python]

so what you're saying is

how many times can i rotate the icosahedron?

[vapid vale]

im saying to fix a triangular face

[balmy python]

yeah

[vapid vale]

this is your element x \in X, where X is the set of faces

[balmy python]

ok yeah

[vapid vale]

and i'm asking how many ways can you rotate the icosahedron where x goes to x

[balmy python]

ahhh

3?

im defo wrong aren't i

😭 😭

[vapid vale]

no youre right

[balmy python]

yoooo no way

[vapid vale]

:D

ok, and how many faces can it be sent to?

(there are 20 total)

[balmy python]

the fixed triangle?

[vapid vale]

yeah

[balmy python]

19

[vapid vale]

well including itself

[balmy python]

20

[vapid vale]

right

[balmy python]

yep

[vapid vale]

so 3 was the stabilizer, 20 was the orbit, so we should believe this group has order 60

but we can double check

bringing the image back for reference

G also acts on the vertices

[balmy python]

wait quick check why is 20 the orbit

[vapid vale]

of which there are 12

[balmy python]

an orbit of x is {g(x) | g \in G}

ah got it

it's all the possible things it can be sent to

[vapid vale]

correct

[balmy python]

wait it's literally like the orbit around the sun

[vapid vale]

xd

[balmy python]

like a planet can only stick to its orbit

orbits are disjoint

[vapid vale]

so what is the order of the stabilizer of a vertex?

[balmy python]

3

[vapid vale]

well, look at a vertex, and look at a region surrounding the vertex

what does it look like?

[balmy python]

oh wait a vertex not a face 😭

[vapid vale]

xd

[balmy python]

ah so it's a pentagon so the stabilizer has order 5

[vapid vale]

right

[balmy python]

and then the orbit has order of the number of vertices

[vapid vale]

yes

so prety conclusively we believe this group has order 60 (if you are to believe there are 12 vertices)

so now, using orbit stabilizer and knowing the order of the icosahedral group, tell me how many edges there are

[balmy python]

wait hold on

[balmy python]

let me process this rq

[vapid vale]

sure

[balmy python]

oh yeah that's by os theorem right

|G| = |G(x)| |St(x)|

[vapid vale]

yep

[balmy python]

alright

and the group is just describing the rotations?

[vapid vale]

yes

[balmy python]

alright

[balmy python]

the stabilizer of the faces had order 3

wait no

we need to look at the stabiliser of edges

you can't really rotate it to map an edge to itself

unless it's the identity map

so i would say that |St(edges)| = 1

and then by os thm you get the |orbit(edges)| = 60

but it doesn't look like it has 60

[tribal moss]

You can rotate by 180° around the midpoint of the edge.

[balmy python]

but wouldn't that be different

[vapid vale]

if you look at it from the right angle you just have a vertical line segment: |

[balmy python]

360 deg around the midpoint of the line makes sense

wouldn't 180 mean that it has a line of symmetry down an edge?

[vapid vale]

you are not rotating around an edge in this image

if it helps try orienting it so the edge is facing vertically

[balmy python]

so if im gonna rotate around this right

180deg

[vapid vale]

yes

[balmy python]

wait 180 in which direction 😭

[vapid vale]

flipping it upside down

[balmy python]

this?

[vapid vale]

yes

[balmy python]

if i do that won't my edge be facing on the other side of the shape

[balmy python]

OH WAIT

you flip the edge itself

and not the shape with it 😭

i'm an idiot

[vapid vale]

well you are rotating the whole shape to be clear

you cant just rip off an edge and manipulate it - that wouldnt be a rotational symmetry

[balmy python]

ahhh i see

i shoulda just made origami

so that gives 30

fair enough

[vapid vale]

also random question, was this a joke

[balmy python]

no 😭

was that like the dumbest question ever

[vapid vale]

like esquie said, this is precisely what representation theory is

and you used the word represent

[balmy python]

ahhh

ive never knew what it's about

[vapid vale]

xd

[balmy python]

is it being able to represent many things in different ways?

[tribal moss]

Generally it's about representing arbitrary groups as matrices acting on e.g. R^n or C^n.

[balmy python]

ohhh

yeah like i learn group theory right

[knotty badger]

It’s often quite helpful to represent group elements as functions in some way

[balmy python]

but i don't really get what it's useful for 😭

[knotty badger]

Whether that’s set-theoretic functions (group actions)

Or linear maps (group representations)

[balmy python]

ahhh

[balmy python]

like with analysis i can tell

[knotty badger]

I think it can be helpful to view group theory as a separation of symmetry ideas

[balmy python]

it's got uses in probability a lot

and like odes

[knotty badger]

If you take the rigid symmetries of a cube

[balmy python]

rigid?

[knotty badger]

Distance-preserving

[knotty badger]

It permutes the edges, of which there are 12

It permutes the vertices, of which there are 8

It permutes the faces, of which there are 6

It permutes the diagonals, of which there are 4

And it can be thought of as linear maps on R^3, so as 3x3 matrices

All of these are rather different mathematical things

[balmy python]

yeah

[knotty badger]

You have permutations on sets of different sizes

And you have linear maps too

But they should have something in “common”

[balmy python]

so gt is more about the study of permutations?

[knotty badger]

Since they all came from the “symmetries of a cube”

[knotty badger]

Well, the way i like to think of it is as a separation of concerns

One half of the separation is the group of symmetries of the cube, abstractly

You focus just on how the group elements combine with each other

Rather than focusing on how they act on vertices or faces or 3D vectors

It’s a purely “syntactic” perspective, where you focus on symbols and how they combine

[knotty badger]

This helps you identify what’s in common to all the examples above

The other half is the “semantic” perspective

[balmy python]

hmmm

[knotty badger]

It’s all about interpreting or representing the group elements concretely

As permutations, or linear maps, or isometries

Having them actually act on something

It can really make life easier to cleanly separate these two parts

[balmy python]

what can be thought? the symmetries?

[knotty badger]

Yeah

[balmy python]

ahhh

so group theory mainly involves the abstration between similaritirs between different representations of symmetry?

[knotty badger]

Well group theory is a bit too large to have any one thing it mainly involves

But this is my perspective on it as a physicist

[balmy python]

😭

do you use it in qft?

[knotty badger]

You might start with an object that you want to investigate symmetries of

[knotty badger]

Extremely frequently

[balmy python]

damn

[vapid vale]

i should note that groups had been used purely as symmetries of things (roots of polynomials, euclidean space, general sets) 50-60 years before people decided to care about abstract groups in their modern definition thats taught in class, and group-like objects had been talked about much longer than that

[balmy python]

whats an example of things you used it for in qft

[knotty badger]

But it can often be useful to reinterpret these symmetries as acting on a different object

To do that, it’s very helpful to separate out the symmetries from the object

[knotty badger]

Writing down Lagrangians

[balmy python]

oh wait isnt noethers theorem proven by group theory

[knotty badger]

Mhm

[balmy python]

oooh

i did it last year for my project and forgor

[knotty badger]

It’s very natural to consider symmetries of objects

[balmy python]

was it like describinf forces with rotations and translations

[balmy python]

mhm

[knotty badger]

But one disadvantage of this approach is that you get very tied to the objects themselves

[balmy python]

even like rubiks cubes

[balmy python]

ahhh

[balmy python]

yeahhh

[knotty badger]

It can make it harder to see the things that two sets of symmetries “should” have in common

Like how the symmetries of a cube can be thought of as permutations of 4 diagonals, or of 6 faces

There should be something in common to these two perspectives

But it’s hard to see if all you know are the permutations

[balmy python]

can you represent this as a direfr producr or smth?

[knotty badger]

There’s not an obvious mapping of diagonals to faces

[knotty badger]

Don’t think so

They’re not independent

[balmy python]

ahh

wait i just realised

i think a lot of what i study in linalg mirrors a lot of what i do in grouos

groups

like quotient vector spaces are obvious but

also like direct sums

if thats all about groups

what are rings even about 😭

like i know that it extends the axioms of a group to include multiplication

and theyre close to fields

[noble nexus]

rings are kinda about geometry

[knotty badger]

Rings are when you act on abelian groups

[noble nexus]

But explaining why takes a bit of effort

but yeah a more straight forward idea is that rings are symmetries of "linear objects"

[knotty badger]

Any abelian group has a ring of endomorphisms

[balmy python]

endomorphisms?

[knotty badger]

Whereas a general object just has a monoid of endomorphisms

[knotty badger]

Maps from the object to itself

[balmy python]

ahhh so not necessarily automorphisms

[knotty badger]

Mhm

Also, functions into a ring naturally form a ring

[balmy python]

so the preimage of a ring is a ring?

[knotty badger]

That’s not quite the statement

[balmy python]

oh 😭

what i also learnt in linalg was factorization in polynomials

and something about unique factorizarion domains

[knotty badger]

Polynomials are quite special for the following reason

[balmy python]

cant wait to learn about that

[knotty badger]

You can interpret the same polynomial in many different rings

[balmy python]

oooh

like isomorphisms?

[knotty badger]

All you need is a way to interpret the coefficients of your polynomial in your ring

For example

You can interpret the equation “x^2 + y^2 = 1” in any ring R

The set of points in R^2 that satisfy the equation could reasonably be called the “circle” in R

Moreover this is always a group

So every ring has its own version of the circle

[balmy python]

oooh

[balmy python]

can i have an example

or is it tedious

[knotty badger]

You can interpret this equation in the ring of rational numbers

This is essentially a ring of Pythagorean triples

In the reals you get the circle

[balmy python]

would still look like a circle if we graphed it though?

[knotty badger]

In the complex numbers i think you get C*?

[balmy python]

yeah

[knotty badger]

One with lots of gaps

[balmy python]

yep

[knotty badger]

You could even interpret it in the ring of polynomials

Or power series

[balmy python]

a polynomial in a polynomial

so a circle in polnomials would just be x,y as polynomials?

over some field

[knotty badger]

The point is, you can use info about solutions to a polynomial equation in one ring to help you understand stuff about polynomial equations in other rings

[balmy python]

ahhh

[knotty badger]

It would be two polynomials p and q such that p^2 + q^2 = 1

[balmy python]

yep

damn

[knotty badger]

For example, you can use the geometry of the circle in R to get info about the circle in Q

[balmy python]

i liked learning about polynomials in linalg a lot

especially irreducibles

[knotty badger]

Yeah, the ring of linear maps is another place you can interpret polynomials

So if V is a vector space

[balmy python]

isnt that more becayse Q is a subset of R?

[knotty badger]

You can consider pairs of linear maps L and K such that L^2 + K^2 = 1

[balmy python]

no way 😭

[knotty badger]

More than that, it’s a subring

[balmy python]

ahhh

[knotty badger]

This is a starting point for algebraic geometry

[balmy python]

think my uni offers it next year

[knotty badger]

Solutions of the same polynomial in different rings interact in interesting ways

[balmy python]

im liking analysis so far so i might wanna pick functional analysis

[knotty badger]

A big part of AG is figuring out exactly how

[balmy python]

might pick some probability and physics and if i have space maybe a bit of algebra

[balmy python]

ahhh

[knotty badger]

This is how AG can help solve things like fermat’s last theorem

You have the polynomial x^n + y^n = z^n

[balmy python]

because you can consider it in a different ring?

[knotty badger]

If you study this equation in tons of different rings

You can glean some info about how it behaves in integers

[balmy python]

ahhh

makes sense

maths lore is so cool

[knotty badger]

mhm

[balmy python]

do you use this stuff a lot in your phd?

[knotty badger]

Not much algebraic geometry really

But tons about rep theory

[south patrol]

O v nice

How does this interact w what u do

[balmy python]

oooh

[south patrol]

(I unfortunately do not rly know what you do aha)

But am interested

[knotty badger]

Important for understanding allowed terms in lagrangians

[balmy python]

i wanna do a phd but sometimes i feel like im too dumb for it 😭

[knotty badger]

There’s also fancier parts of rep theory for generalised symmetries

[knotty badger]

i can very much sympathise

[south patrol]

Ah ok

[balmy python]

like i really love math and would be quite sad if my job didnt involve lots of it in the future

[sacred wharf]

Phd seems daunting…

[south patrol]

As in to ensure they have the correct symmetries or smth else

[south patrol]

It is

[knotty badger]

That’s part of it

[balmy python]

at imperial i get impostor syndrome a lot 😭

[sacred wharf]

Not for u, u r prismatic potato

[knotty badger]

I’d be ok being a programmer

[karmic moat]

Someday I will understand wiles proof of flt

[south patrol]

Ah ur at imperial cool

[balmy python]

yeahhh

are u from the uk?

[south patrol]

Taylor btfo

[knotty badger]

Still don’t know if I really have what it takes to do a phd

[sacred wharf]

I mean i think thats in almost any math program

[balmy python]

woahhh

oxbridge potato?

[sacred wharf]

No they are in grocery store

Because they are a potato

[balmy python]

i sat in my friends lecture hall and felt like a genius

[knotty badger]

Oh like glados

[vapid vale]

idk if i can do a career in math

[south patrol]

I am a potato

[balmy python]

waitttt no way howd yk mister monster potato

[south patrol]

Discord is a small world

[balmy python]

😭

[sacred wharf]

When r u fighting chipper potato for #1 potato

[knotty badger]

idk either

[balmy python]

ive met him at oxford before

[karmic moat]

Preach one of my friends dmed me a few weeks ago and said “are u xyz” and i said yea

[south patrol]

Yeah

[south patrol]

Wait lol is mister monster at oxford

I thought he was elsewhere

[balmy python]

not anymore

hes graduated

wait what yr u in

[south patrol]

Wait maybe I know who he is and forgor

[south patrol]

Just started 2nd year phd, but I did undergrad there too

[balmy python]

ohhh you go oxford 😭

thats so cool

[sacred wharf]

Is getting phd at oxford harder than undergrad

[balmy python]

probably

[sacred wharf]

I still domt get how yall pay so much tuition fee

[south patrol]

Idk I mean phd is more chill in many ways than ug but then a phd is hard in its own ways (regardless of uni)

[balmy python]

i like oxford for the way they do tutorials

[south patrol]

Don't rub it in

[knotty badger]

PhD can be terrifyingly chill

[balmy python]

imperial doesnr have that so people tend to be behind on problem sheets

[south patrol]

I mean yeah if you are an international student then they torture you

[balmy python]

😭😭😭

home student 9k aint too bad though

[sacred wharf]

Fr 40k bru

[south patrol]

Yeah

[balmy python]

defo not worth going leeds 😂😂

[karmic moat]

Y’all dont get funding as phds?

[south patrol]

With loan ig ye

[sacred wharf]

https://tenor.com/view/i-mean-thats-bad-really-gif-20159382

[karmic moat]

Damn

[south patrol]

The above seems to all be about undergrad

[balmy python]

yeah we do but its very limited

[karmic moat]

Oh oops

[vapid vale]

doing a math phd is my "at least im making money while i stall" gambit

[south patrol]

At least here almost everyone has funding

[balmy python]

it used to be 3k a year tbh

back in my dads days

[south patrol]

Ig if I didn't have funding I would not do a phd lol

Well also wouldn't be able to afford it lol

[balmy python]

whatre your plans after phd

[sacred wharf]

I heard if no funding its basically free labour scam

[balmy python]

lecturing?

[sacred wharf]

They will win fields medal

[south patrol]

It already is in the sense that the pay is near minimum wage

[balmy python]

😭

[karmic moat]

No they call these “unpaid internships and you get paid in experience” god kids are so ungrateful these days

[balmy python]

potato medal

[south patrol]

The plan is to make it that far

[sacred wharf]

Number 1 vegetable medal fr

[balmy python]

whats your phd in 😭

[vapid vale]

do u like leeks

[south patrol]

Yeah hbu

[vapid vale]

yeah

[karmic moat]

Hell no potatoes are not the best vegetable

[balmy python]

whats your final form

[sacred wharf]

Imo not worth anywhere maybe part III with big maybe

[balmy python]

is it french fries

[south patrol]

Uh like derived algebraic geometry

[sacred wharf]

Why u hating on prismatic and chipper potato

[balmy python]

part iii sounds nice tbf

[karmic moat]

Never said potatoes are bad

[balmy python]

depending on my grade next year ill go for it

[karmic moat]

But potatoes def not a number 1 veggie

[balmy python]

its the most versatile veggie

[sacred wharf]

Aint it 40k even for yew kay

[karmic moat]

Hmm

[balmy python]

for the masters program?

nah

[balmy python]

think about it.

[sacred wharf]

9k?

[balmy python]

potatoes are everywhere around us

[karmic moat]

U ever had celtuce?

[balmy python]

yeah

[karmic moat]

That veggie is awesome

[sacred wharf]

O i thought that was undergrad only

[sacred wharf]

Idk have u seen a celtuce do phd in oxford

[balmy python]

but not as versatile

[karmic moat]

In chinese it’s called 莴笋

So god damn delicious

[karmic moat]

Better to be a master of one trade than to be a jack of all and master of none

[south patrol]

My brother convinced me a tiktok cat had died and now I am sad

[balmy python]

yeah be a master of potatoes

[south patrol]

But she is ok

[balmy python]

nooo

[karmic moat]

W

[south patrol]

https://tenor.com/view/metallica-master-of-puppets-metal-thrash-metal-james-hetfield-gif-5004481132394071709

[karmic moat]

Reinstalled tiktok today

[sacred wharf]

For now

[balmy python]

MPot 4 year course

[karmic moat]

WWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWWW

[balmy python]

£47k a year

[south patrol]

I am trying to get this up to speed on guitar

Pain.

[karmic moat]

Mastah mastah

[sacred wharf]

Top 3 deals in the UK

[south patrol]

Top 5 things to do in the uk

[balmy python]

the uk is just dayligjt robbery

[south patrol]

1. leave

[balmy python]

tube prices take up most my loan

and its not like it gets me to uni on time

[south patrol]

Remember when meal deals were £3

😭

[balmy python]

3.85 now.

EVEN MCDONALDS INFLATION

[sacred wharf]

Im applying for exchange in korea where they have 90 euro for 90 meals per month

[south patrol]

Bit expensive m8

[balmy python]

90 meals is crazy

u sure it aint north korea

[south patrol]

Go to Denmark where it is 90 euro for 1 meal

Yes I know they don't use euro before I get killed

[balmy python]

switzerland where it is 90 chf for a fork

[sacred wharf]

Mcdonalds is crazy here they have a deal 6 euro for double cheese burger and fries and nuggets and coke

[balmy python]

😭

[sacred wharf]

Idk i basically live in denmark and it aint that bad

[balmy python]

i have a friend that works at mcdonalds

so i get half price on all meals

[south patrol]

Idk but it funny to meme

[balmy python]

but back in the days a cheesburger was £1

[south patrol]

All meals or just meals at McDonalds

[balmy python]

mcdonalds 😭

[sacred wharf]

Bro is competing with manifold for most self reacts

[karmic moat]

Wendys got biggie bag meals

Goated tbh

[south patrol]

Wendys

[balmy python]

Oooh i have one near me

[karmic moat]

Oh grow up.

[balmy python]

but its mad expebsive

[balmy python]

😭😭

[karmic moat]

Like grow up by 6 or 7 years

[balmy python]

the best food in the uk is probably pepes

[karmic moat]

Like the frog?

[balmy python]

piri piri slaps

[karmic moat]

Hi walter

[balmy python]

have you tried it potato

[agile burrow]

Hi anamono

[karmic moat]

What’s up walter

[agile burrow]

I am also learning some algebraic groups

[sacred wharf]

Walter white to back to chemistry

[karmic moat]

What r u using to learn

[agile burrow]

Humphreys

[karmic moat]

W

[agile burrow]

I think like Borel and Springer and Humphreys all do the same things lol, I just picked one and stuck with it

[karmic moat]

Yea I use humphreys mostly for reference

Sometimes borel

[agile burrow]

And also I guess its nice that Humphreys restricts to an algebraically closed field for most of it and talks about rationality stuff at the end

[karmic moat]

Humphreys uses “d-group” tho where I havent really heard anywhere else

[agile burrow]

Yeah that's fair. But like it's just the diagonalizable stuff right?

[waxen bramble]

I need help, what does this mean?

[karmic moat]

Yea i think so

[velvet hull]

what part do you not understand?

[south patrol]

I haven't, what 's it

I think I have seen this but I am vegan so it does not seem ideal

[south patrol]

What is this about lol

[waxen bramble]

The number of automorphisms is the amount of different automorphisms of each group. I am doing the odd exercises, I dont understand why the answer is 2

[south patrol]

I was not correcting grammar but just repeating it nonsensically

[karmic moat]

I’m having fun rn bc I’m learning the same stuff but from the alg group perspective on one hand and the other from the lie group perspective so there’s small diffs

[sacred wharf]

It seems it is true that vegans only eat potatos

[agile burrow]

Lol yeah like characters of G_a

[karmic moat]

Yea I was kidding I thought u were making a deez nuts joke

[south patrol]

Oh lol

[karmic moat]

Then this

[velvet hull]

any automoprhism is determined by its action on a generating set, does that part make sense?

[south patrol]

I liked Milne's book on alg groups provided you like

Translate everything from fake schemes into actual schemes

[sacred wharf]

yo hchan

[karmic moat]

Lol I tried reading his notes

[noble belfry]

In Z6 there are 2 elements with order 6 (1 and 5)

[karmic moat]

I have never seen someone treat something so concrete in such generality before

[south patrol]

This is like an actual book which I think is quite nice

Just yeah odd foundations

[karmic moat]

Granted they are really nice

[noble belfry]

If we construct an isomorphism from Z_6 to Z_6, since Z_6 is cyclic sending the generator 1 somewhere fixes everything else

[south patrol]

I just think it is nice that it covers non-algebraically-closed and positive characteristic fields well

[karmic moat]

Because I just replaced “let k be perfect” with “let k be C” and assumed everything to be semisimple and hell maybe simply connected

Oh

[noble belfry]

1 has to be sent to a generator so it can be sent to 1 or 5

[south patrol]

Which are what I mostly care about lol

[south patrol]

Lol

Mfw every char 0 field is perfect

[karmic moat]

W

Anyway yea rn i’m trying to get a grasp of it so im just doing simple simple type A stuff over C

[balmy python]

Hi, I know it's not my question but can you explain this compared to the definition of an isomorphism from a set to itself

[sacred wharf]

They nit the same ?

[tardy hedge]

what do you mean by this question?

bro is bored

[sacred wharf]

Bro its 2:38 am can you blame me

[tardy hedge]

why arent u sleepng

[sacred wharf]

Not particularly sleep tbh

I have to wake up at 11 tmrw so not too much in a rush

[velvet hull]

oh, the word isomorphism is dependent on what kind of objects you are working with

an isomorphism of sets usually means just a bijection

[vapid vale]

this isn't a definition per se, its a property of homomorphisms. homomorphism of a group requires f(ab) = f(a)f(b), so once you know what a homomorphism does to generators, you can multiply them and you know what it does to the whole group

[tough raven]

Prove that if G is a group whose composition series has length 2, then exactly one of the following is true: (i) G has a unique non-trivial proper normal subgroup (ii) G is a product of two simple groups.

[vapid vale]

we have 1 < N < G with N, G/N simple – if N is unique then youre done – otherwise, M \neq N is normal wth M \cap N = 1; by various isomorphism thms + simplicity, MN/N \cong M/(M \cap N) \cong G/N, so M is simple, they centralize each other, so G = M x N

sorry if you wanted hint not full answer, i liked that problem

[proud vapor]

For an essay I'm writing for my second year of uni I came across an open problem asking if the left Krull dimension always equals the right Krull dimension in a Neothrian ring dose anyone know if this is still an open problem.

[waxen bramble]

Ohh I see, what about Z?

[twilit wraith]

there arent any elements with order 6 in Z

[proud vapor]

I think if we map 1 to x in Z then the image is xZ since 1 generates Z so we need to send 1 to 1 or -1. So we have 2 automorphisms.

[knotty badger]

Conceptually, what makes modules behave differently to vector spaces

[rocky cloak]

I mean vector spaces are a type of module.

But the main thing that's special about vector spaces is that all vector spaces are free, and all vector spaces are semisimple.

[knotty badger]

Vector spaces being free requires choice in general

[rocky cloak]

That's true yes

[knotty badger]

What does it mean for a module to be semi simple…?

[knotty badger]

But it still seems that even infinite-dim vector spaces are better behaved than modules?

I could be wrong though

[rocky cloak]

A few equivalent definitions are
-Direct sum of simple modules
-Its the sum of its simple submodules
-Every submodule is a direct summand

[rocky cloak]

I mean without choice they get kinda crazy

[elfin wraith]

Well, they are modules. I guess it’s similar to how fields are just particularly nice rings, you assume a lot of structure so they’re somewhat forced to be “well behaved”

[rocky cloak]

But most people use choice

[rapid cave]

I choose you

[knotty badger]

Can you be more specific about how they’re more well-behaved than modules

[thorn jay]

sane people*

[knotty badger]

I guess I’m mostly thinking of functional analysis here

Where you use choice but not in the sense of using a hamel basis or anything

[elfin wraith]

Again, they are just modules. I think the main point is, as jagr said, that they’re free and semi simple which gives you a pretty good handle on them.

In general, any module over any ring could look like anything

[rocky cloak]

Functional analysis is all about topological vector spaces, which then has a lot more structure than just a vector space

[knotty badger]

But do people actually use that infinite dimensional vector spaces are free and semi simple

[elfin wraith]

I guess my best example about how they’re better behaved is just the fact that linear algebra is basically solved and taught to first years, and general module theory is

[knotty badger]

What I have in mind is like

A student who’s asking why modules are so much harder than vector spaces

I recognise that empirically they are viewed as harder

But this doesn’t tell me much at the conceptual level

[elfin wraith]

Well the fact that not all modules are free sets you up for them being harder

Having a basis is a massively helpful computational tool

[rocky cloak]

I mean in situations where you want to prove something is unique up to isomorphism you might try to reduce to vector spaces then use that they are uniquely determined by dimension.

For example in showing that a decomposition into modules with division rings as endomorphism rings is unique you would do this

[knotty badger]

Basis-free linalg still seems very well behaved though

More than you’d expect from just knowing that they’re modules

At least from what I’ve seen

[rocky cloak]

Well the semisimplicity would be a good example then.

Like you have a short exact sequence
Z/2 -> Z/4 -> Z/2

If these where vector spaces you could say Z/4 was the direct sum Z/2 (+) Z/2, but here it's more complicated

[delicate orchid]

I might say that anyway

[knotty badger]

Wew do you have opinions on this

[elfin wraith]

Fields are incredibly restrictive is certainly part of this though, like for a general ring you won’t have inverses nor commutativity

This leads to dimension being a bit of a bitch for arbitrary modules. A really useful thing for f.d. Vector spaces anyway is that they have a meaningful and easy notion of dimension. For modules there’s not generally an obvious way to define it, and in fact there’s like 4 or 5 different ways that you can off the top of my head (none of which give you as much) (edit- they certainly give you things, but relative to how far VS dimension can give you)

I think this is just that when you look at one specific object, there’s quite a lot you can say about it, but when you try to make a statement about all possible objects with only one loose structure in common, it’s hard to say a lot

[rocky cloak]

Another way to say it, that is maybe less helpful to this hypothetical student, is the the field of homological algebra exists.

[knotty badger]

Maybe I’m just not getting this

[rocky cloak]

Did you have any specific results in mind?

[elfin wraith]

Would you agree that linear algebra is nice?

It’s perhaps not “easy” but it’s certainly well understood

[knotty badger]

I’m not sure because this feels like a loaded Q

[elfin wraith]

Ok

[sacred wharf]

At least its nice to learn

[rocky cloak]

Like when I think about linear algebra being nice I think about things like
If T:V -> W is a linear map
V is the direct sum of kerT and ImT.

If T:V -> V, then T has an eigenvalue, and V breaks into a direct sum of (generalized) eigenspaces. (V finite dim)

For finite dimensional spaces you have the dual, and V = V**.

[knotty badger]

Like my notion of “nice” doesn’t seem to match up with your notion of “nice”

[rocky cloak]

All of these are things that are not true for modules in general

Even things like the subspace of a finite dimensional vector space is finite dimensional doesn't necessarily translate

[south patrol]

Well having an eigenvalue isn't true without more assumptions ig but ye

[rocky cloak]

Well, you did call them "well behaved", so maybe you can expand on what you mean by that

That maps can be represented by matrices and that you have these nice relations between invertibly, determinant, adjugate matrix is something that is true for free modules (over commutative rings), but not for modules in general.

Then also Cayley-Hamilton is nice, but this also has an analog for modules (over commutative rings) by reducing to the free case

[south patrol]

To me I feel I get the most mileage out of the fact that everything is free. There is not much interesting "internal structure" (although understanding maps in a systematic way can still be complicated of course (though they are still simpler than they would be for general modules)) and being able to work with dimensions in fd case is extremely powerful

[vapid vale]

modules are like families of vector spaces

[rocky cloak]

Quiver-gang unite!

[vapid vale]

:D

[tribal moss]

Huh, how does that work?

[vapid vale]

M (x) K(A/p)

[tribal moss]

Can you elaborate on what you mean by that?

[rocky cloak]

How much can you actually recover about a module from it's residues?

[vapid vale]

for a module M over a commutative ring A, you have the residue field at each prime ideal. the fiber of M at p is tensoring with the residue field. perhaps its some intuition that a module carries the information of many vector spaces, so should intuitively be more "complicated"

[vapid vale]

certainly not the module itself; i was just responding to the conceptual question (although this is not a good answer to the conceptual student asking the question)

[vocal pebble]

i always thought modules ought to be more complicated because they can have free resolutions that are more than just the trivial one unlike vector spaces which means there can be relations among generators, relations among relations of generators and so on

[rocky cloak]

Indeed, not all modules are free

[vocal pebble]

well i guess the answer to the original question is just that

[vocal pebble]

but i guess this is saying more than just that not all modules are free. they can be more and more complex in some sense?

[vapid vale]

but also this yes. the general point i was making is that you at the very least need many many vector spaces to be as complicated as a module, and even still you need lots of compatibility/further information/limitations to dream of recovering the module

which to me feels like a direct comparison between the two

[knotty badger]

ok i have a much more elementary question

let $G$ be a subgroup and $N$ be a normal subgroup

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

suppose you define a multiplication on $G /N$ by $(gN) * (g' N) := gg' N$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

what would be your preferred strategy to show this is "well-defined"?

from what i can tell, there are two strategies you can take

1. take different coset representatives $g_1$ and $g_1'$, and show $g g' N = g_1 g_1' N$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[knotty badger]

2. show that this agrees with the setwise product of the subsets, defined by AB = {ab | a in A, b in B}, and inherit the well-definedness from the setwise product

perhaps there are others

[vocal pebble]

define normal subgroups to be the ones which are kernels of homomorphisms, so the group structure on the target gives you the wel definedness for G/N

[knotty badger]

wouldn't you still need to argue that the group structure on the target agrees with this coset multiplication rule?

also that's not the def of normal subgroup i'm taking

[vocal pebble]

arguing that it agrees is not hard because you have a homomorphism no?

[rocky cloak]

I like strategy 2, but I feel students are sometimes confused about setwise products, so then it can be best to just deal directly with elements.

[knotty badger]

i'm not convinced

[vocal pebble]

what

[knotty badger]

yeah for this, what do students get confused by wrt setwise products

[vocal pebble]

you have phi(g1)phi(g2) = phi(g1g2), so you have g1 N * g2N = g1g2N. kind of similar to the setwise product argument

[rocky cloak]

Well, I'm not sure exactly what trips them up tbh. Just the idea of manipulating a whole set at the time I guess

[rocky cloak]

You can't really show that every normal subgroup is a kernel from this though

[knotty badger]

i see i see

[vocal pebble]

thats why at the very beginning i said "define normal subgroups to be the ones which are kernels of homomorphisms"

with the widely accepted definitions you will probably be using one of the two methods mentioned yeah

[somber sleet]

Do you know where I can find a proof that calculates the unit group of Z/p^kZ

[rocky cloak]

For a third method, that's probably not very helpful:

You could look at the image of G in Sym(G/N), then show that the kernel is N iff N is normal.

[supple ice]

try to show that a is a unit in Z/p^kZ iff gcd(a,p^k)=1

[vapid vale]

if you were teaching it i’d walk through 2 and use 1 as an exercise

[knotty badger]

oh, why is that?

[elfin wraith]

I think my first instinct would be to do 1), but I don’t have any deep reason to do so, it’s just my first thought

I guess one of the nice things about that method is you can easily then work out a very explicit example of how it fails for non normal subgroups but I think both methods are equally good

[knotty badger]

Yeah for me I don’t see a clear winner between the two

Nor do I think there necessarily should be one

I’ll keep this in mind

One interesting thing I've found is that students mix up setwise commutativity with elementwise commutativity

i.e. they think $gn = ng$ rather than $gN = Ng$

[cloud walrus]

Pseudo (Cat theory #1 Fan)

[vapid vale]

well you’re writing things as gN which implies you’re talking about cosets i.e. honest sets. so in that language it’s good to actually explain via set wise products bc they are unintuitive otherwise. but if they conclude themselves that representative multiplication works, they will take away the multiplicative structure on G/N as honest elements

that’s a good notion to feel like you have intellectual ownership of, hence exercise

[somber sleet]

IS there some way to compute Z[i]? Like how do I know which elements this contains?

[quiet pelican]

In what sense?
Like, how is the definition, “a + bi, where a, b are integers”, not explicit enough?

(Well it’s not quite the definition, but you can eliminate higher powers of i with i^2 = -1)

[somber sleet]

do you see the "a+bi for a,b \in Z"? Yes, how do you explicitely get that

[quiet pelican]

If you have an element ai^n \in Z[i], then this is equal to (-1)^{floor(n/2)} a i^{n mod 2}

And then do this to every term for general linear combinations

[knotty badger]

do you think it could be at all helpful to bring up monoids to my students?

when i was learning group theory, i found it weird that:

• for group homomorphisms, you only require f(g h) = f(g) f(h)
• for group actions and representations, you additionally require that f(e) is the identity map

once i learned a bit more about monoids this made sense

[delicate orchid]

gNg'N = gg'NN = gg'N BOOM!!! how you like THEM apples

[knotty badger]

yes that's approach 2

[vocal pebble]

i was always convinced with "you want homomorphisms to preserve some structure, so make f(hg) = f(g)f(h)", and then defined actions as homomorphisms, so that f(e) = id was forced

[knotty badger]

right but the definition of a group homomorphism doesn't require identities to be sent to identities

it's something you can deduce

however for group actions and representations you have to explicitly check identity preservation

[delicate orchid]

could also go the congurences route if they know about quotient sets

[knotty badger]

mhm, one way of thinking about it is you can either go the "partition" route or the "equivalence relation" route

[delicate orchid]

yeah. Honestly I think I like the "equivalence relation" route more

cause like, every subgroup partitions the group into cosets that's not special

[knotty badger]

this is true

[elfin wraith]

Tell me how i knew who the reactions were from without checking

[thorn jay]

it was probably the amount of reactions

[knotty badger]

heyyy enpeace

[thorn jay]

haiii

[knotty badger]

i am currently marking the first problem sheet for my groups and reps students

hence why all these elementary Qs

[elfin wraith]

I think the word congruence just summons you

[elfin wraith]

Oh interesting, is this both an intro to group theory and to representations?

[thorn jay]

most of the time in UA that's not category theoretical you spend thinking about congruences im ngl, so its like a sleeper agent