groups-rings-fields

[next obsidian]

obviously it has 0 because you add it in

if m,n in M_tor are annihalted by r,s non-zerodivisors then m + n is annihilated by rs, still not a zero divisor

let m in M_tor be annihilated by r, r not a zero divisor. Given any s in R, sm is either 0, or if it's non-zero it's still annihilated by r

since r(sm) = s(rm) = 0

[rotund aurora]

wait but, with this definition, Z/nZ over itself is always torsion-free??

because if r\in Z/nZ is such that there exists an s st sr=0, then s is a zero dvisior

[next obsidian]

everything ring will be torsion-free over itself. You make a tradeoff

you get actual submodules of torsion elements

but weird things like this happen

honestly, the real answer is that torsion doens't make that much sense if your ring isn't an integral domain haha

[rotund aurora]

hah xd

[next obsidian]

in that case the two definitions for torsion agree

to extend it for non-domains you either have to have weird stuff like p not being torsion in Z/pqZ, or you don't get submodules

[rotund aurora]

okay. Thanks

[burnt flower]

I basically understand none of my classes and go through them with the knowledge of high school math

[formal ermine]

for any homo. phi : G -> G', if H is normal in G, and H is a subset of the kernel of phi, then G' must be H?

[chilly radish]

I want you to think this through

first of all, is H a subgroup necessarily

second of all

what if we take H to be ker phi once, and antoher time we take it to be {e}

and the kernel is not trivial

if you stare at this critically you will see all of the holes with this question

this isn't even scratching the surface

for example, what if phi is the trivial homomorphism, G is not isomorphic to G', and we take H=ker phi = G

[formal ermine]

ok wait then I don't quite understand this
"We always have a homomorphism ..."
"Proposition. Let phi : G -> G' be a group homomorphism"
"(a) Is H normal in G with H being a subset of ker phi, then phi induces a unique homomorphism phi bar : G/H -> G', such that the diagram ... commutes"

are we also assuming that H is a subset of G'?

[coral spindle]

Why do you think this is true? Can you explain

[coral spindle]

No, what here would make you think we're assuming that?

[formal ermine]

it was my attempt at an understanding of the proposition above with the help of namington's messages in #discussion

[formal ermine]

because if phi(G) = H

but we have phi : G -> G'

[chilly radish]

it's not

[coral spindle]

phi(G) is not H

[chilly radish]

what makes you think that

[formal ermine]

the diagram

[chilly radish]

that is definitely a typo

but even if it wasn't, this doesn't mean that phi is surjective

[coral spindle]

Oh, well that makes a lot of sense

They meant G' there, not H

That is a simple typo

[formal ermine]

ohhhhh

[coral spindle]

And if it's not clear, this is just not true.

[formal ermine]

yeah, if H is a subset of G, G' can't be H unless G intersect G' is not empty

just realized

thanks

typos are horrifying

[chilly radish]

G intersect G' is meaningless unless G and G' are both subgroups of some bigger group anyways

or subsets of a bigger set

[formal ermine]

why is it meaningless?

[coral spindle]

It ain't gonna be a group

So group theoretically, it means nothing.

[formal ermine]

yeah I mean the elements in the group

[coral spindle]

There's no useful way of interpreting the intersection between arbitrary groups.

[formal ermine]

how do I say "G and G' must share some elements"

[coral spindle]

You would say their intersection is nonempty.

But we very rarely care about this

We would more often say that G and G' have a common subgroup H, for example.

[formal ermine]

but isn't that meaningless?

[formal ermine]

ah

[coral spindle]

You're missing the point of what we're saying here. It is meaningful to talk about the intersection of two arbitrary sets in a way that it isn't in group theory. Because of the typical set-theoretic construction of groups, yes indeed we would phrase this literally in terms of the sets involved. However if we are to be more precise (and in fact, more correct in some sense) we would really consider monomorphisms phi : H -> G and phi' : H -> G to describe this.

[formal ermine]

mh

how do you construct a group without a set?

[coral spindle]

You use some other foundations for mathematics.

[formal ermine]

eg?

[coral spindle]

Type theory. This is an abstract algebra channel; if you have questions about foundations, ask them in #foundations

[formal ermine]

oke

[cloud walrus]

potato

[south patrol]

i am confused about why this doesn't work; maybe I'm just misquoting the way direct limits interact with ext?

[coral spindle]

An injective resolution seems like a really hard way to do that haha

Surely you'd try a projective resolution first

[winter veldt]

I'm trying to figure out how to prove that 2^K - 1 is composite if K is composite. I'm not quite sure where to start.

[south patrol]

Okay fair enough that seems more sensible lol

I guess I just don't know any projective resolutions of Z[1/p] but I guess I can work one out

[wise mantle]

Does anyone here understand tensor products in terms of quotient spaces?

[south patrol]

in what way?

or do you mean the construction of the tensor products of vector spaces (which would sometimes use quotient spaces if in the general case)

[coral spindle]

This is a question more suited to #elementary-number-theory, but I'll give you a hint anyway. Consider k = mn and recall that x^a - 1 = (x-1)(x^{a-1} + x^{a-2} + ... + 1) for any x.

[wise mantle]

this way!

although

we have them constructed as the free abelian group on the cartesian product of the two sets

[south patrol]

Well the idea is essentially that you have "generators" given by F-linear combinations of pairs of elements and "relations" which make it behave as you'd like

[wise mantle]

so something like A tensor B is the quotient of the free abelian group generated by A X B on the elements

blah blah

[south patrol]

Oh okay tensor product of abelian groups

Sure

[wise mantle]

I don't think it helps that I don't fully understand quotient groups

But im struggling to picture what the space A tensor B looks like

[south patrol]

i really wouldn't worry about tensor products until you undrstand quotient groups

[wise mantle]

haha im working on questions on tensor products as part of a class im taking on a hilbert problem

so I am very much worrying 😦

I understand quotient groups simply as the set of cosets of one set in the other

[south patrol]

well take it rather as like a recommendation to make sure you work on quotient groups a bit more first

[wise mantle]

Yes ty I agree

But if i were to think of quotient groups as the set of cosets

how does that make sense with tensor products

[cloud walrus]

potato

[south patrol]

that seems much nicer! thanks

*btw, not about lol

[coral spindle]

Just taking a look now

Yeah looks about right to me

Obviously would need double checking but looks sane, you know

[south patrol]

well the idea just being we have an obvious surjection Z^2 -> Z[1/p] and we can identify the kernel with Z as in my map

[coral spindle]

Yeah I think so

[south patrol]

Noice

And this makes more sense actually

contravariance of Hom(-,Z) turning that limit for Z[1/p] into the limit giving you the padics; the result should be \hat(Z_p)/Z

[coral spindle]

I couldn't say; not terribly familiar with the p-adics myself

[south patrol]

Oh I mean dw

I was just saying like lol thanks for suggesting projective instead as more intuitive

[coral spindle]

No worries

But yeah the projective resolution tends to be easier to calculate for modules, definitely

[south patrol]

Oh yeah fair

Especially when stuff is fg and all

actually there is an error here lol

[hollow mica]

does anyone have an idea what Sym^k V means

[chilly radish]

k-th symmetric power of V

[tough raven]

I would guess the answer you're supposed to give is just that - the multiplicity of V in CG.

Unless you have seen some character theory by now

[lethal dune]

Can anyone gone an example of a monoid whose Grothendieck group is 0

Non zero monoid obviously

[tough raven]

(Wikipedia says, but I did check that) if the monoid has a zero element then its Grothendieck group is trivial

So non-negative integers under multiplication IG

[lethal dune]

it's Z for nnv integers

[gritty sparrow]

That is nnv integers under addition

For multiplication it is 0

[next obsidian]

:O

[gritty sparrow]

Omg, i’m your biggest fan! It’s an honor to meet you

[rustic crown]

[lethal dune]

Oh lol

[wooden ember]

Anyone know some good resources on Lie algebras actually explaining along the way why we care about what we’re doing? My course is building towards proving the semi simple classification but the work we’re doing along the way on nilpotent representations feels very ad hoc

It’s hard to see the point

[coral spindle]

I also felt that way about my first Lie algebras course. I'm not really sure about where a good place to get motivation is, but at the very least Erdmann–Wildon's book on Lie algebras might help you a bit

[rotund aurora]

oh wait I just realized

The reason why (1,0) is not linearly independent is because (0,1)(1,0)=(0,0), right?

[wooden ember]

yep

[south patrol]

Is there a nice way to see that (assuming it's true!) complex (finite dimensional) representations of a compact group G with the same character are isomorphic? this is referred to in a paper i'm reading without proof

(We can also assume G is a Lie group for my purposes)

I know the proof for finite groups boils down to using orthogonality and that you can decompose into irreducible decompositions; does the same proof carry over somehow? probs uses Peter-Weyl

(I think I asked this before but still a bit puzzled why e.g. I can't find a reference to what looks like it should be a really important fact!)

[solar glacier]

I have a question in a proof that HK=KH iff HK is a subgroup of G given that H,K are subgroups. For showing HK =KH, it says the KH is clearly contained in HK by closure of subgroups, why is it KH \subset HK and not HK \subset KH for the trivial direction?

the proof is out od Dummit and Foote page 94 at the top

[wooden ember]

When HK is a subgroup, you can say that KH c HK because HK contains H and K so in particular contains every product of the form kh (since it is a subgroup)

you cant say that if HK is not a subgroup

[solar glacier]

then why cant you instnatly say HK c KH by the same reasoning

[wooden ember]

because KH isnt a priori a subgroup

[solar glacier]

HK c KH requires proof apparently

ahhhh

[wooden ember]

we only know that HK is a subgroup

[solar glacier]

gotcha gotcha

HK is the subgroup

not KH

thus KH by closure is contained in HK

[wooden ember]

(in the end so is KH but you need to show that)

[solar glacier]

yeah true, thanks! that clears it up!!

[south patrol]

i wonder if there some nicer perspective on these things with products of subgroups in the non-abelian case

i guess it is just a few checks though

[wooden ember]

wdym

[south patrol]

Eh dw

[rotund aurora]

How important is the distinction between an interal direct sum and an external direct sum?

[chilly ocean]

what do you mean? For me it's just that we want to know when a, say, module, is isomorphic to a direct sum of its submodules

[wet zodiac]

when do i learn this

early on or a bit into abstract alg

[gloomy night]

Hard to know for sure, but from the algebra courses I have seen, it will be early on.

[wet zodiac]

awesome

thank you

[coral spindle]

The first isomorphism theorem is what it's called.

[agile burrow]

Yeah, the first isomorphism theorem is a pretty fundamental concept and is typically taught early on

[south patrol]

This diagram also ignores like most important thing about it - the phi tilde is injective

[lapis trail]

I need to find the number of homomorphisms from GL_3(7) to A_3. I think I'm supposed to abelianize GL_3(7) by quotienting out SL_3(7), but I don't know what the elements of that quotient group would look like or if any will have order 3

[coral spindle]

Wow, that's a hard question

Fwiw I think you're right in your approach but I can't think of how to continue. Perhaps trying to characterise the quotient is helpful

[next obsidian]

Wtf lol

[lapis trail]

Wouldn't the quotient group need to be cyclic? Maybe the answer is 1

[coral spindle]

I'm leaning towards that, yeah

SL_3 is the derived subgroup after all

But my whole thought is that the kernel needs to contain it... is that true?

The kernel does have to be maximal

[lapis trail]

Oh

[coral spindle]

Or does it? A_3 is simple, right

[lapis trail]

Yes

It's Z3

[coral spindle]

right ofc, brain moment

[lapis trail]

It's gotta be 1

I gotta ask my prof tomorrow

[verbal harness]

I wonder if it is possible to illustrate why relations are associative in simple language? EG, without falling back on group theory given that I'm trying to explore some of the underpinnings of group theory to begin with and want to avoid circular reasoning. 🙂

[coral spindle]

Do you mean an operation rather than a relation? Relations are typically called transitive, not associative

As for the reason, it's just because composition of functions is associative; this is the real underpinning.

[verbal harness]

I'm ultimately reaching for "why is symmetry associative". And it's interchangeable to ask why Groups are associative (eg, why we require associativity to call something a group) which gets answered "group binary operation makes the most sense when it's a form of function composition, and that is always associative". So why are function applications associative? Apparently because functions are a subset of relations. Why are relations associative? Cue a bunch of latex that falls back on group theory. 😉

[coral spindle]

What? No it doesn't

You do not need any group theory to show that composition of relations is associative.

[verbal harness]

That's good to hear, and what I'm trying to learn up on. 🙂 Google searching and such has not yielded anything but latex that does rely on group theory though.

[coral spindle]

Neither do you need the fact that functions are a subset of relations to even argue why function composition is associative

[tribal moss]

On the contrary, you need to show it is associative before you can start using group theory to study it.

[verbal harness]

So how can the associativity of functions and/or transitivity of relations be described in simple language? Not something that has to be rigorous, just broadly illustrative. 🙂

[lapis trail]

That's how they figured out that associativity is important in the def of a group iirc, because function composition does it.

[coral spindle]

You're still using the wrong terminology. Functions are not associative; composition of functions is. Similarly composition of relations is.

Associativity communicates the following. If I want to put three pieces of pipe together, I can put the first two pieces together then put the last one on, or alternatively put the last two before putting on the first. Associativity states that these two methods will produce the same result.

[coral shale]

That's a great explanation, I like

[coral spindle]

I'm glad you like it

[verbal harness]

Yep, I understand what associativity is but I don't understand why that should always be a property of function or relation composition.
Commutivity is not guaranteed for example, even though that just "feels" simpler. And appears simpler to describe both in English and mathematical notation.
If I put on clothing, order of operations matters (not commutative) but still not order of composition (is associative).
Putting shoes on before socks does not result in the same outcome as putting on socks before shoes, but I can put socks into shoes and then apply that combination to my feet and get the same outcome as if I had put socks onto feet first.
It's like associativity says something about potential chronological perspectives, but it does so directly by subverting that perspective.

[lapis trail]

Ok maybe this is wrong because now I'm wondering if I can map everything but the center to zero, the center is Z6 since 7-1 is 6, that would give us 3 homomorphisms

[verbal harness]

You care about the order pipe pieces wind up in, because water is going to flow through it chronologically. But the statement about putting the pieces together specifically addresses that you can put the "last" bits together before you attach the "first" bits.
Commutivity also speaks to chronology, but somehow in a very different way.

[tribal moss]

Perhaps you're looking for something completely elementary:
(fog)oh = fo(goh) because they have the same value at all inputs: ((fog)oh)(x) = f(g(h(x))) = (fo(goh))(x)

[coral spindle]

I think this no longer has anything to do with group theory, to be frank with you

[coral spindle]

Oh drat yeah I think you're right there

Wait

No actually lol, I'm not sure how that map would actually work

[next obsidian]

Okay so

[lapis trail]

I'll find out

[next obsidian]

Any map from Gl_3(7) factors through the abelianization you’re right

And this is finite

[coral spindle]

Yeah I just confirmed this myself too

[next obsidian]

So you could theoretically write it down as a product of a bunch of Z/nZ

And then you’re reduced to just asking which of those admit maps to Z/3Z

And then you can take any product of those

So it really comes down to writing down the abelianizariin of your group in some normal form

[coral spindle]

Yeah as before, you should focus on that

[next obsidian]

Maybe everyone already said this

But I just thought of it and didn’t want to reread everything

[elder wave]

[verbal harness]

Associativity is an underpinning of group theory. Non-associative loops do not constitute symmetries. I could just use some help to wrap my head around what associativity actually brings to the table that makes it fundamental to symmetry.

[lapis trail]

I think it's because function composition does it, and symmetries are bijections

[verbal harness]

So Loops are not bijections?

[lapis trail]

Loops? Cycles?

[elder wave]

what is a loop for you here

[verbal harness]

https://upload.wikimedia.org/wikipedia/commons/3/3f/Magma_to_group4.svg

yeah.. but discord won't inline that, sorry. A Magma with divisibility and identity, but not necessarily associativity.

Groups obey four axioms. Loops are the superset of that which only obeys the first three.

[lapis trail]

Oh ok

Do you know much about what they do?

[chilly ocean]

@rustic crown

@coral spindle
MIT OCW has a course on Lie Algebras iirc.

[chilly ocean]

Multiplication by an element in a quasigroup is a bijection

The operation just doesn't correspond to function composition

[verbal harness]

Then it sounds like symmetries being bijections wouldn't play into it.
Group multiplication behaves like function composition in a way that other quasigroup multiplication is not guaranteed to, and somehow that is important to describe symmetries. But I don't know how.

I feel like Euclid trying to work out a justification for an ugly fifth postulate. 🙂

[chilly ocean]

I don't know what to say

Symmetries work by having a state of an object, altering it, and obtaining another state of an object

[verbal harness]

In geometry, the fifth postulate relates to the Gaussian curvature of the space being considered, so altering or omitting it has a physically meaningful analog.

[chilly ocean]

So by the very nature they must behave like functions

Say, not symmetry but more like a change of a state

[verbal harness]

To me that sounds like a bijection though.

[rotund aurora]

So I suppose from external direct sums you get new modules from old (meaning you take the cartesian product, and define component-wise operations). Classic example: R^n. But I asked that because isomorphism relations is not all we care about, equality is also important. For example, Z^2=Z+Z (as an internal sum, and literal equality) is not exactly correct (Z is not even a submodule of Z^2). Sure, Z is isomorphic to (1,0)Z, but it is also isomorphic to (2,0)Z and (0,2)Z, and definitely, Z^2 is not equal to (2,0)Z+(0,2)Z. You may say this example is rather trivial, but sometimes, I think, this stuff may matter. That's why I asked.

[chilly ocean]

Yes? But quasi-groups aren't just bijections. This is a product Q x Q to Q which happens to be bijective coordinate-wise, but if you try taking x(yz) then it's not necessarily (xy)z so treating elements of Q as maps on Q with product inherited from Q doesn't give you that xy is composition of functions.

So while they are bijections, you don't compose them like bijections

So how an element of quasi-group acts on itself doesn't matter as much when composing them

Which is associativity said multiple times but in a different way

Redundancy coming from intuitive treatment of the subject, i. e. concept of symmetry

Symmetries are just bijections

But composing symmetries is to compose functions

Enough said

[delicate bloom]

is there any kind of easy criteria to check if a polynomial with rational coefficients F(x,y) can be decomposed as G(x,y)*(ax+b)+H(x)?

[coral shale]

calculus things, does x = b/a have to be some kind of minimum maybe? random thought

if u took the partial wrt y?

[delicate bloom]

dF/dy = dG/dy * (ax+b) so it should have the same linear term in it

maybe that's as good as it gets, which isn't terrible I suppose

[coral shale]

yh so partial y and find roots works?

where its always 0 for some x

[delicate bloom]

yeah, I could use rational root theorem I suppose too

[chilly ocean]

Am I missing something? Take G(x, y) = F(x, y), a = 0, b = 1, H(x) = 0

And here you go

[coral shale]

[delicate bloom]

well, a!=0

I suppose I should also clarify a,b are rational as well and the coeffs on G and H are rational too

[delicate bloom]

actually this is pretty decent for my case, I don't care which variable is which, so I have two shots at it. try to find a linear factor in y on F_x or a linear factor in x on F_y

[coral shale]

this is long division right

yh, think it always works. surprised ig

Does euclidean alg not work on the original thing. i assume u tried

[delicate bloom]

yeah, in the F_y case it'll be looking at coefficients which are polynomials in y, so it might not be so easy

[delicate bloom]

I considered it but didn't really get anywhere, maybe I didn't try hard enough lol

[next obsidian]

You can always divide with remainder if you’re dividing by a polynomial with a unital leading coefficient

So you can at least write F = G•(xa + b) + H(x,y)

I imagine that you can show this expression is unique in the same way division with reminader in a Euclidean domain is unique

And if that’s the case you just can do a divisor with remainder to get an H and then check if it’s a polynomial only in x

[delicate bloom]

well if I'm not misunderstanding you, I'm thinking once I get (ax+b) from F_y = G_y (ax+b) I already know H(x,y) doesn't depend on y because it got killed by the y derivative

[next obsidian]

I don’t understand. If you take the y-derivative of H you get F_y - G_y(ax + b)

I don’t see how you can conclude this is 0?

[delicate bloom]

I think I'm looking at something different than you

I don't know ax+b before hand

[next obsidian]

I took this expression

Wait what are a and b?

You aren’t fixing them?

[delicate bloom]

I'm starting from a polynomial F(x,y) with rational coefficients and wanting some criteria to show if I can factor it in this form G(x,y)*(ax+b)+H(x), I don't know what the G, H, or a,b will be

[next obsidian]

Oh

I assumed a and b were fixed already

Then you’re kinda pwned

[delicate bloom]

yeah it's sorta annoying yeah

also I should probably restate a != 0 so it's not trivial

also it doesn't matter if the variable is x or y, just that I can factor out a linear term in one variable out of it, F(x,y)=G(x,y)*(ay+b)+H(y) would be fine too

x and y are just arbitrary labels, and really for any change of variables I could probably be fine too, well, admittedly this question isn't that important for what I was doing so I hadn't thought about it too much lol

[rotund aurora]

With sets, a function f has a left inverse if and only if f is injective. This is not true if f has to preserve some structure on the sets (for example, in the case of modules). However, that still holds (unsurprisingly) in the case of vector spaces. I think this is category theory, but I write this in case someone has something interesting to say 😮

(I am asking about the question of necessary and sufficient conditions so that a morphism has a left-inverse)

[coral shale]

choice might lurk

[rotund aurora]

it doesn't lurk in modules

not even for right inverses, which is fun lol

I think the general answer is very similar for the case of modules, but I'm clueless as to how you would express it

[next obsidian]

Not if you restrict to finite dimensional vector spaces

The only thing you need is to make use of the fact t hat every vector space has a basis

And showing t hat every injection has a set theoretic retract doesn’t need choice, you need this to produce one-sided inverses to surjevtions because you have to pick one element in every fiber

[rotund aurora]

I suppose that question is equivalent to the following: Given a structure X and a substructure Y of X, when can we extend a morphism with domain Y and some fixed codomain to a morphism on the whole set X so that the restriction to Y is the same.

[ripe basalt]

https://math.stackexchange.com/a/3023502 can anyone tell me how this guy moved from the first summation to the second one (group theory conjugacy question)

[next obsidian]

It’s factoring

[coral spindle]

We're summing up |C(a)| for every element of the group, but we will collect elements together according to their having the same conjugacy class, in which case their centraliser has the same order (you should be able to prove this easily). So writing Cl(a) for the conjugacy class of a, then choosing representatives of each conjugacy class we can simply sum up |Cl(a)| * |C(a)| for each representative, instead of all the |C(a)| for each element of the group.

Hopefully that's the transformation you were talking about there.

[ripe basalt]

i am.. but

it feels like you wrote whats already there

[coral spindle]

Oh, right. They do explain that there.

Well if you can be more specific about what's confusing you, that would be helpful.

[ripe basalt]

C(a) is all elements g where ga = ag yes

i dont get how you're factoring it

[coral spindle]

There are |Cl(a)| elements which are conjugate to a

They all contribute |C(a)| to the sum

Hence in total they contribute |Cl(a)| * |C(a)|.

Is that clear?

[ripe basalt]

how does each thing in Cl(a) contribute |C(a)| to the sum?

[coral spindle]

Because if a and b are conjugate, then |C(a)| = |C(b)|.

[ripe basalt]

so if i pick something in Cl(a), this is just an element b where b = gag^-1 for some g

[coral spindle]

Yes, that is the definition.

[ripe basalt]

but this isn't an element of C(b)

[coral spindle]

No, who says so?

[ripe basalt]

C(b) is all g where gbg^-1 = b

[coral spindle]

Yes.

[ripe basalt]

we have gag^-1 = b

[coral spindle]

For some specific g in G, yes.

[ripe basalt]

so...

it's not fixing b under conjugation

[coral spindle]

And?

This is totally irrelevant

[ripe basalt]

that's C(b)

[coral spindle]

Yes what about it?

[ripe basalt]

is the elements which fix b under conjugation

[coral spindle]

Why is that relevant?

I think you're confusing yourself in some way that I cannot quite work out.

Let's review some definitions.

Cl(a) = {gag^-1 | g in G}

[ripe basalt]

yes

[coral spindle]

C(a) = {g in G | gag^-1 = a}.

[ripe basalt]

so all elements conjugate to a

[coral spindle]

Great.

[ripe basalt]

and all elements which fix a under conj, sure

[coral spindle]

Exercise: if b is in Cl(a), then |C(b)| = |C(a)|. Hint: conjugation.

Now let's look at the problem at hand

Hopefully you are aware that the conjugacy classes, Cl(a), partition G.

[ripe basalt]

yeah

[coral spindle]

Great.

[ripe basalt]

i understand the rest of everything

in the proof

just dont get the factoring piece

[coral spindle]

OK

Let's say each of a_1, a_2, ... a_n are conjugate, and b_1, b_2, ..., b_m are conjugate

So Cl(a_1) = {a_1, ..., a_n}

[ripe basalt]

wdym their 'conjugate'

oh

[ruby sundial]

If you have an inner product of vectorspace V does this induce a innerproduct on End(V)?

[coral spindle]

OK

So

The sum we're looking at

is |C(a_1)| + ... + |C(a_n)| + |C(b_1)| + ... |C(b_m)|.

N.b. of course that n = |Cl(a_1)|, m = |Cl(b_1)|.

Now note

since |C(a_1)| = |C(a_i)| for any i

we have that this is equal to |C(a_1)| + ...(n times)... + |C(a_1)| + (the same for b_i)

Do you see that?

[ripe basalt]

sure

|C(a_1)| = |C(a_i)|
i guess i need to show this

it's the fact u gave before though

[coral spindle]

Then you're done. Because this is just |Cl(a_1)| * |C(a_1)| + |Cl(b_1)| * |C(b_1)|.

This should be clear now, yes?

We're summing over classes rather than elements of the group, now.

Is this clear?

[ripe basalt]

more clear, sure

[coral spindle]

Is it not totally clear?

OK well good luck.

[ripe basalt]

no it's not totally clear

[coral spindle]

OK well I don't know what else to say. I'm afraid I need to sleep now. Maybe someone else can help you with any specific questions you have.

[ripe basalt]

no one will i think

but thanks i guess

[ripe basalt]

i dont get why there's a_i and b_i now

i mean i just have 1 set of representatives, yes?

rather im choosing a set of reps

and you can't have Cl(a_1) = {a_1, a_2, ...., a_N}

because the representatives aren't supposed to sit in the same class..they define different classes

they are by definition the things not equivalent under the relation of conjugation

so no, this isn't clear at all

like if a_2 was conjugate to a_1 it wouldnt be a representative of a different conjugacy class, it'd just be something in Cl(a_1)

[next obsidian]

The point is

For all a_i in Cl(a_1)

[ripe basalt]

but i guess he had to leave immediately after the explanation

[next obsidian]

|C(a_i)| = |C(a_1)|

[ripe basalt]

there are no a_i in Cl(a_1)

are you mixing notation

a_i is a representative of a conjugacy class

[next obsidian]

Isn’t Cl(a_i) the conjugates of a_1?

Yeah, they’re elements my guy

[ripe basalt]

no they arent

[next obsidian]

Yes they are

Cl(a_1) is a set

[coral spindle]

I literally defined them as that

[ripe basalt]

if a_2 is in Cl(a_1) then a_2 is in the same conjugacy class as a_1 and so a_1 ~ a_2

[coral spindle]

The representatives are b_1 and a_1 in this miniature example

[ripe basalt]

so they cant be reps

[coral spindle]

You have just invented this idea that the a_i are representatives.

[ripe basalt]

no i didnt

it's in the proof, and you said it before

but ok if theyre not reps then fine

[next obsidian]

Lmfao

[coral spindle]

I'm not going to continue, I should be sleeping.

[next obsidian]

You can figure this out yourself

[ripe basalt]

why are you getting angry

i just said "if they're not representatives then this is fine"

anyways i think i get it now

[acoustic pine]

Let X ={0,...,n} for n a fixed number. Is there always a group G such that G acts freely and transitively on X?

[coral shale]

Cn or am i missing something

[acoustic pine]

is it really just Cn

am I just dumb

[coral shale]

i just googled free so idk

think so

[acoustic pine]

ok so I am just stupid

this is fine

[chilly radish]

Yes, cycles are fixed-point free

[acoustic pine]

I knew that much, I guess I just was wondering if there were more

Because for n=4, I could see the cases 3 cases of just "rotating" the set through, and swapping the first 2 and second two elements

so that is a group with 5 elements at least

[coral shale]

if n is composite u can chop it up into smaller cycles?

[chilly radish]

I doubt there is a full classification for general n, but there are surely more

[coral shale]

nvm not transitive

[acoustic pine]

ugh ok this is annoying

[chilly radish]

This is an interesting question tho

What's the context

[acoustic pine]

So about 5 days ago, at my colleges math club, my buddy posed the question: "What is the probablility that, if you drew n uno cards, that there is some order in which you can play those n cards, following normal uno rules"

We simplified the problem to the case of 4 colors, 10 cards. I represented this as a 4x10 graph

I started reading about Cayley graphs, and was curious of the graph of "playable cards from a point" is a cayley graph, and I believe it is, meaning that, for any L suits and W cards per suit, there is a group corresponding to the set of all legal moves from a given point

I wanted to do a combinatorial argument based off the cosets and compositions, because I found what I think is a strictly combinatorial solution

anyway, where this problem arises is looking at how to construct the graph itself. A cayley graph for GxH is the cartesian product graph of both. So, I wanted to find the group representing the free transitive permutations on L and W elements, and look at the product group

Very roundabout way of doing it, especially since the graph gets far more complex when you add wild cards and duplicates, but if the group structure idea holds, we're in business

[chilly radish]

That's pretty cool. Not sure if I have anything to add right now but those are all interesting problems

[acoustic pine]

"what happens when a statistician, an algebraist, and an actuary play uno" would be another appropriate setup

[chilly radish]

Lmao

In your graph, are the nodes numbers, suit of the next card coloured edges and "is a legal move" the rule that connects two vertices?

[acoustic pine]

so far we ignore coloring of nodes/edges, but ye

[chilly radish]

Well the edge colouring matters for cayley graphs

[acoustic pine]

Nodes are tuples of (C, N), for C a color and N a number

Oh it does?

I was focusing on using this theorem:

[chilly radish]

Cayley graphs as I know them are generally defined wrt a set of generators, edges represent multiplication by generators and each edge in coloured appropriately

Maybe you can disregard the edge colouring in some definition?

[acoustic pine]

According to this article, I can recover the generating set S, as well as the coloring from a graph, assuming it is Cayley

[chilly radish]

I see

Ok that's pretty cool I didn't know that

[acoustic pine]

Math wikipedia is full of surprises

[acoustic pine]

update: I'm dumb and that transposition makes the action not free

If you transpose then shift, it fixes some elements

[quiet pelican]

Do you have that every irreducible rep is a sub rep of the regular representation?

If so then you might be able to show that the regular rep is the sum of only the reps you’ve found

[quiet pelican]

and you can abuse that you know exactly what that sum is

[formal ermine]

what's a unit?

[lament dawn]

Of what

[formal ermine]

in Z/nZ

[lament dawn]

In a ring it generically refers to an invertible element

So here the same

[formal ermine]

ah ok thanks

[coral spindle]

I read this and the question of trying to characterise them is a very interesting one

Here's a small observation

If I'm not mistaken, the subgroup of S_4 generated by (1 2)(3 4) and (2 3)(4 1) acts freely and transitively on {1,2,3,4}.

I believe it is a subgroup of order 4 (again, I may be mistaken), and is isomorphic to V_4

This leads me to believe that characterising these groups will likely involve partitions of n, or perhaps more likely just divisors of n.

[coral shale]

Think I agree. If you have a product of unequally sized disjoint cycles, it cannot be free

[south patrol]

Oh cool lol this looks like one of those questions where it should be trivial or smth but it really isn't xd

[coral shale]

eg. ((12)(3456))^2 fixes 1, 2
More generally
((1...k)(k+1...n))^k fixes 1, ..., k

So any free permutation group on n elements needs to have elements that are products of n/k equally sized disjoint cycles for some k

[hollow mica]

why do people choose to keep everything on the same side

i.e.

why not write (v_1 + v_2) ⊗ w = v_1 ⊗ w + v_2 ⊗ w

[coral shale]

choice

but

= 0, ie. = e, the identity

[south patrol]

The point is that you mod out by those things

It is more natural to just give a list of relations - stuff we set to 0

[coral shale]

<a, b | aba^-1b^-1> or <a, b | ab=ba>, think most prefer the former

[hollow mica]

I see

[tribal moss]

If G acts freely and transitively on a nonempty set A, then choose and fix a0 in A. Then for each a in A there is exactly one g in G such that g(a0) = a, so we get a bijection between G and A. Through this bijection G acts on itself rather than on A, with an action that satisfies g(e) = g for all g. But then g(h) = g(h(e)) = gh(e) = gh, so the action is exactly the group's standard self-action by left multiplication.
So the free transitive group actions are, up to bijection, exactly the standard group self-actions.

And the free transitive actions on {1,...,n} are just the groups of order n.

[south patrol]

Oh fair enough

[tribal moss]

Also @coral spindle

[quiet pelican]

DMd for semi dox reasons

[coral spindle]

Yeah I realised this, that’s a very cool observation. I guess this means that this is as hard as classifying all groups haha

[chilly ocean]

Can you explain this..

[coral shale]

<a, b> must have order a multiple of lcm(ord(a), ord(b)) = 2, but also divides 3! by lagrange. It's not 2 as it has to contain HK, so it's 6. idk if that's the argument

what does not seem straightforward?

[chilly ocean]

Might be the wrong channel: the question is from representation theory. Say I have the group $G=\left\langle\begin{pmatrix} a & b \ 0 & d \end{pmatrix}\right\rangle$ and the represntation $\rho$ in $\mathbb{C}^2$ is just left-multiplication. If I am not mistaken, this representation falls apart as two subrepresentations with $\text{diag}(a,1)$ and $\begin{pmatrix} 1 & b/a \ 0 & c \end{pmatrix}$ whenever $a,b,c\neq 0$. I ought to compute $\text{Hom}G(\rho,\rho)$ (the intertwiners) or at least its dimension, which I have a hunch is of dimension 2 (I am not sure how to calculate this). This should be a counterexample to Schur's lemma, in the sense that it shows that the factors of the decomposition from before not being semisimple means that $\dim\mathbb{C} \text{Hom}_G(\rho,\rho)\neq 1$. Can anyone help me out with this? The calculation of the intertwiners. Thanks in advance.

[cloud walrus]

miqq

[chilly ocean]

The problem in question, book is An introduction to representation theory by Kowalski.

[midnight summit]

Hello!

can anyone recommend a good book for lie algebra?

one that is not written scientifically but mostly for intuitive understanding

[acoustic pine]

I knew the problem was more interesting than Z/nZ!

[coral spindle]

Well you'll like Tropo's observation here @acoustic pine

[acoustic pine]

Very well written too

[coral spindle]

So as I mentioned, this problem is as hard as finding all groups of order n. Which is to say... very hard indeed lol

But I guess it's easy when n is prime!!! 🤪

[lethal dune]

Or 1

[chilly ocean]

I think I figured it out if anyone is interested. Let $\langle e_1, e_2\rangle$ be a basis for $\mathbb{C}^2$. Then the subspace $\langle e_1 \rangle$ is invariant under the action of $\rho$ and is a 1-dimensional subrepresentation $\tilde{\rho}$ say. Being 1-dimensional means it's irreducible. If $\rho$ were semisimple, then there should be one other subrepresentation $\pi$ which should also be on some invariant subspace and that $\tilde{\rho}\oplus\pi\cong\rho$. However, any other possible subspace (so the two generated by $e_2$ or $e_1+e_2$ are not invariant, hence $\rho$ can't be semisimple. What $\text{hom}G(\rho,\rho)$ ought to be still eludes me, but at least I've established $\rho$ not to be semisimple. I do think that $\rho$ decomposes somehow like $\rho \cong \tilde{\rho}\bigoplus{i=1}^\infty \rho$ hence the intertwiner set is actually just injections/projections. I could be wrong though.

[cloud walrus]

miqq

[main needle]

The smallest $n$ such that $Z_2 \times Z_4$ is a subgroup of $S_n$ is n = 6 ?

[cloud walrus]

ru0xffian

[coral spindle]

Hm interesting question

I am inclined to think yes but it's not easy to prove

I think you'll have to argue via generators

[chilly ocean]

We can see that n is at most 6

[next obsidian]

Can you somehow reduce to A_n

Idk, I think I would try to check for all lower n

And verify it’s not possible

[main needle]

I just tried to work with group presentation it's generated by an element of order 4 and element of order 2 such that they commute

[quiet pelican]

Can’t be <4 by number of elements, can’t be 4 as a 4 cycle can’t commute with a product of 1 or 2 disjoint transpositions

And I think a similar argument rules out 5

[coral spindle]

Very nice

Yeah I think that does it

[chilly ocean]

(1, 1) needs to be a permuation consisting of disjoint 2 and 4-cycles

[coral spindle]

So it does have to be 6

[main needle]

Yeah I argued similarly

[coral spindle]

I have worked through some small examples of this thing before

Like, finding minimal faithful actions

Finding the number for the cyclic groups is quite interesting.

I won't spoil in case you're interested in working it out in general

[chilly ocean]

yeah, so (1, 1) needs to be multiple of at least one 4-cycle and at least one 2-cycle

disjoint*

[quiet pelican]

Is it ||sum of prime factors, counting multiplicities (eg. Z/4Z would be 2 + 2 = 4)||?

[coral spindle]

Ah not quite

I mean for example what would Z/8Z be? It's not too hard to see this must be embedded in Sym(8), and nothing smaller

An instructive example is working out Z/6Z

[quiet pelican]

🤦 If I’d actually tried to work out my example through the proof I had, I would have seen it didn’t work ||but I believe that the same proof shows that it’s the sum of the highest powers of each prime dividing it, and that this follows quite easily from the fact that Z/p^k Z minimally embeds in S_{p^k}, and the Chinese remainder theorem (and the fact that if two cycles of different (coprime?) length are either disjoint, or don’t commute)||

[coral spindle]

I agree with you completely

That was also my reasoning

neat, isn't it?

[quiet pelican]

Yeah, that was nice

[coral spindle]

I figured out you can also embed Dih(2n) in the same way. Like the minimal faithful action of Dih(2n) and Z/nZ has the same size

Which I thought was really cool

The only exception is n=2.

[chilly ocean]

what's more immediately useful to study: multi-linear algebra or universal algebra?

I would say it depends on what you'd like to do after.

For me, multilinear algebra would be more useful, because I like geometry.

It's hard to rate one subject over another as more useful in general.

[coral spindle]

Neither of those are particularly widely applicable, to be quite frank with you

There are better places to look. If you're personally interested though, this is your choice

[chilly ocean]

< (12),(23) >
Is the set of all finite products of (12), (23), and their inverses.

So that gets you:
(12),
(12)(12) = e
(23)
(12)(23) = ( 123 )

Anything else?

[pastel cliff]

in general angle brackets mean the group generated by x, or in this case (1 2) and (2 3)

[next obsidian]

Yeah multi linear algebra is lowkey just subsumed into linear algebra via tensor product and related things

I think the theory of bilinear forms and stuff is a notable exception where there’s useful and interesting stuff to say which doesn’t readily fall into that category

But besides that, not really

[chilly ocean]

idk wym that they're not "widely applicable".

Mmm... good to know on the multi-linear stuff.
Thanks.

[pastel cliff]

[coral spindle]

Multilinear algebra is just a small part of linear algebra, and universal algebra doesn't really have any results that can be used elsewhere.

[chilly ocean]

Universal algebra btfo.

[coral spindle]

It's a nice little bit of perspective, and it's fun

[next obsidian]

Universal algebra is something you learn because you like it IMO

[coral spindle]

Yeah as I said, if you're personally interested then that's your choice

[next obsidian]

It just isn’t a very good payout if you’re looking at like usefulness to time

[chilly ocean]

So... you all have studied universal algebra then?

[coral spindle]

I've done a little.

If you're wanting a top-down view of mathematical structures, model theory would be more worthwhile.

[chilly ocean]

Multilinear algebra is, in my experience, the kind of thing you learn because you're using it for something else. You don't really do it for its own sake.

[coral spindle]

It is easy to think that because it says "multi" on front, it's an upgrade!

[leaden rock]

I'd argue the top-down view of mathematical structures is category theory

[chilly ocean]

(23)(12)=(132)

[delicate bloom]

do differential geometry

[next obsidian]

No, but everyone that ever studies universal algebra that I’ve talked to never has anything to really show for it. All I’ve seen is that they say words like quasi ring, semi groups, semi ring, etc and then they can’t say anything useful you can do with them

[leaden rock]

They do have applications in symplectic geometry

[next obsidian]

Like if you want to describe algebraic structures like this, I think operads also do that

And operads are very important in higher categorical stuff

[chilly ocean]

I'm interested. Go on.

[next obsidian]

Yeah I mean, that’s the first I’ve heard of it, but I don’t think it makes sense to study these weird esoteric structures just because you might eventually use it when you become a symplectic geometer. It seems like something you should just pickup when you find yourself needing it, which is also the case with multi linear algebra

This is all again just assuming you don’t want to study these things for their own sake

[leaden rock]

I see what you mean. I am inclined to agree with you. It's a nice tool to have in the toolbox, but certainly not something to delve too deep in

[chilly ocean]

Where does universal algebra see application in symplectic geometry?

Grigori Perelman ftw

[chilly ocean]

Is there anything else?
The claim is that you can get all of S_3.

Are you able to make a 2-way containment argument?

[leaden rock]

Showing isomorfisms between symplectic manifolds, and quoting (I think? translating from Swedish) certain tangent spaces. I know it has certain applications in the knot theory part of symplectic geometry

[leaden rock]

[chilly ocean]

Any references?

[leaden rock]

Are you looking for books or pdf:s on the topic?

[chilly ocean]

On the applications of universal algebra you're mentioning.

I've studied symplectic geometry, but have never seen universal algebra in it.

[leaden rock]

Maybe I have the wrong idea of what universal algebra is

I interpreted it as abstract algebra

Yes, you're right. I was thinking of abstract algebra for some reason. My bad, ignore my gibberish

[chilly ocean]

All good, it happens.

[chilly ocean]

universal algebra can give you an eagle-eye view on some things

that said, multi-linear algebra 100%

[chilly ocean]

I did study universal algebra before

[chilly ocean]

Hmm.... what's more useful...
Graph Theory or Multi-linear algebra?

depends on where tbh

in applications graphs are really common

algorithms etc

but like if you study physics for example then probably multilinear algebra

but you could dive into manifolds at that point

well, not that multilinear algebra is only used there, it has applications in geometric measure theory too for example

in algebra in general

tensor products of modules, so I guess, it kind of incorporates some fundamentals for you to do homological algebra

[chilly ocean]

I agree, at least with the first part

[chilly ocean]

yep. There is some things that you learn while doing universal algebra but that's a time that could be probably better spent doing something like further studying ring theory or number theory, if you're into that kind of stuff

Btw, a modern book about universal algebra is Bergman's Universal algebra

I picked up Grazter's from the library.
🤷

it does mention Hilbert nullstelensatz few times so I guess it's better to dive into it with some algebra knowledge already

personally I read Sankappanavar

[chilly ocean]

https://www.amazon.com/Course-Universal-Algebra-Graduate-Mathematics/dp/0387905782

?

[main needle]

Is the the Dihedral group of order 24 subgroup of $S_n$ for $n < 12 $ ?

[cloud walrus]

ru0xffian

[quiet pelican]

Yes. See a lot of the discussion above, or consider the rotation subgroup’s embedding

[chilly ocean]

Are you asking if its a subgroup for all such symmetric groups? Well, no, because it's order exceeds those of S_2, S_3, etc.

But, iirc, there's a theorem that all groups are isomorphic to some subgroup of a corresponding symmetric group.

So probably since S_4 has 24 elements, good chance it's a subgroup of that or at least S_n>5.

[quiet pelican]

Starting here #groups-rings-fields message

[chilly ocean]

Yes

[main needle]

Got it thanks

[quiet pelican]

It embeds into ||S7|| by something like ||r -> (123)(4567) and s -> (23)(57)||

[solar glacier]

Is this the correct set of iso classes of group order 72 using versions I and II of fundamental theorem

Using that z_mn = z_m x z_n iff GCD of m n is 1

[void cosmos]

yes

you are correct

if by the arrows you mean isomoprhic

[solar glacier]

yes by arrows i mean iso @void cosmos

[void cosmos]

then yea

[solar glacier]

sweet thanks! preparing for an algebra exam lol

[void cosmos]

yea good luck

any questions just ask here

[solar glacier]

just trying groups of diff orders and their decomposition

thanks!!

[void cosmos]

btw

this is not the classification of groups of order 72

its the classification of abelian groups of order 72

up to isomorphism

[solar glacier]

no just abelian

yes

up to iso

im dealing w abeilan groups using the fundmentla theorem for abelian groups

[void cosmos]

yea

[solar glacier]

version 1 requires the group be finitely generated

version 2 just requires ableian of order n>1

[void cosmos]

yea

[solar glacier]

cool thanks!!!!

[trail stump]

am I understanding this correctly. let's say it's finite, and a group is a simple group, but it doesn't guarantee any subgroup of it simple?

[next obsidian]

Yeah

Like, if your group has an element of order nm, where these aren’t 1

Then the subgroup generated by it has a normal subgroup

Of order say, n

So any element of composite order not equal to |G| will break this

[trail stump]

gotya thank you

[tender bough]

screenshot is from my lecture notes of Lie Algebra class. Why is the eigenvalue non-zero?

[wind steeple]

ig adₓ is [x,.]

[tender bough]

yep

[wind steeple]

bc it is non abelian there is a x st [x,.] is non zero

bc it is ad-diagonalizable, every ad_x is diagonalizable (by def ig, I don't know the def of ad-diagonalizable lol)

so [x,.] has an eigenvector

with non zero eigenvalue

bcp if every eigenvalue of [x,.] were 0, it would be 0 since it is diagonalizable

[tender bough]

Thanks! That makes sense!

[tender bough]

Help, I can't see this. I can see that [x, [y, -]] = [y, [x, -]] since x and y commute and the adjoint representation is a homomorphism from g to gl(g). But why do [x, -] and [y, -] have common eigenvectors?

[chilly ocean]

"Simultaneously diagonalizable" means that there is a basis which diagonalizes both of them, i.e., which consists of eigenvectors of each. It's a general linear algebra fact that commuting diagonalizable operators are simultaneously diagonalizable.

I'm not completely familiar with the Lie algebra stuff being thrown around here, but I think that'll do it for you.

[tender bough]

ahh thanks!

yeah it doesn't seem to have anything to do with Lie algebra

[hollow mica]

this is the definition of the tensor product

what exactly is addition in the vector space L

I am confused because it says V x W is a basis

[next obsidian]

It’s formal sums of simple tensors v (x) w

Except you can like factor according to the rules described there

[hollow mica]

wait idk what tensors are yet

this is the definition

[next obsidian]

So like v (x) w + v’ (x) w = (v + v’) (x) w

These are symbols

In that they call these (v,w)

Anyway, just look this up in a book

[hollow mica]

this is (v + v', w) ?

[next obsidian]

Don’t learn it from Wikipedia

[next obsidian]

Sure

[hollow mica]

any recs

[next obsidian]

Like any book on abstract algebra lol

[hollow mica]

ok

I know what a quotient space is

but "modding over these relations" is confusing to me

[next obsidian]

Well then

I don’t think you should be learning what a tensor product is

Unless you’re only doing it for vector spaces

Or you need it for a specific purpose

[hollow mica]

only for vector spaces

[next obsidian]

These are a mildly advanced construction

Oh well then like

Idk look it up in Intro to Smooth manifolds or something

They describe it for vector spaces

It’s way easier there

[hollow mica]

okay sure

this is so much better lol

[ripe basalt]

Let H be cyclic and N some arbitrary subgroup. If \phi \psi are injective homomorphisms from H --> Aut(N) such that phi(H) = psi(H) show that N ⋉_phi H ~= N ⋉_psi H

would the easiest thing to do here be construct an isomorphism?

oh the ascii is backwards

sec

Let $H$ be cyclic and $N$ some arbitrary somegroup. If $\phi, \psi$ are injective homomorphisms from $H$ to $Aut(N)$ with $\phi(H) = \psi(H)$ so that $N \rtimes_{\phi} H \cong N \rtimes_{\psi} H$

[cloud walrus]

*-algebra

[ripe basalt]

heh..help

should the identity map between these two define an isomorphism

oh no it's something a little more careful like $(n,h) \mapsto (n,h^a)$

[cloud walrus]

*-algebra

[ripe basalt]

ok i think this does it actually

ok that was easier than i thought

[tiny jolt]

I need some help on this question

I believe that the composition series of $G$ is $F_{20} \supset D_5\supset \ZZ_5\supset{1}$, but then I found that the commutator subgroup of $G$ has order $5$

[cloud walrus]

pramana

[ruby sundial]

Okay

I need to show every submodule of M/N can be written as L+N/N for some submodule L of M

My proof

There is a correspondence between submodules of M containing N and submodules of M/N

Let S be a submodule of M/N

then there is corresponding submodule L in M that contains N

L contains N implies N=L cap N

Using second isomorphism theorem we have that L/L cap N isomorphic to L+N/N

S is isomorphic to L/L cap N?

[next obsidian]

Do you know the proof that like

Every ideal of R/I is of the form J/I where J contains I?

The proof will be exactly the same, you basically just need to note that if J > I, then J + I = J

Except for modules it’s like, if L > N, then L + N = L

[next obsidian]

Also same result for groups!

[ruby sundial]

oh okay

[next obsidian]

I don’t think you need to use a second iso here

Rather, second iso sort of is predicated on this result in some sense

[ruby sundial]

i guess I dont

[next obsidian]

As is 3rd iso

[ruby sundial]

So S in M/N is in form L/N for some L in M containing N and L=L+N

[next obsidian]

Right, but that last condition

Is equivalent to saying L > N

[ruby sundial]

N is subgroup of L?

[next obsidian]

Sub module

[ruby sundial]

Is that what geq means here

ok

[next obsidian]

But that really is just a set theoretic inclusion

So like in general, if you have L < M, and look at L’s imagr inside of M/N

This will be equal to the set (L + N)/N

But what you really want is a bijevtion between submodules of M/N

[ruby sundial]

yes

[next obsidian]

And some set of submodules of M

[ruby sundial]

So start with a submodule of M

[next obsidian]

This is why you stipulate that L > N

Because both L and L+N correspond to (L+N)/N

But among all the submodules of M which correspond to (L+N)/N

Aka those whose image is that set

Only one of them contains N

Namely that’s like, L + N

So if we look at the map
{L a sub module of M, L > N} -> {submodules of M/N}

Where the map sends L to L/N

This is a bijection

The inverse takes a sub module of M/N, and sends it to the inverse image

Under the map M -> M/N

[ruby sundial]

yea

okay

I was thinking of just using that fact first

[next obsidian]

Wait you knew that fact already?

[ruby sundial]

the correspondence of submodules from the quotient map

yeah

[next obsidian]

[ruby sundial]

I only did for rings though

[next obsidian]

That automatically implies your result

Ah okay

[ruby sundial]

lol woo

[next obsidian]

Yeah the proof works exactly the same

[cloud walrus]

walter

[agile burrow]

This can be ignored 🙂 I forgot how the diagonal action works and made a goofy mistake

[analog pumice]

Can someone check a proof: If $R$ is a commutative ring and $I,J\leq R$ are ideals such that $I\subseteq\sqrt J$, then $\sqrt I\subseteq\sqrt J$.

Proof: $I\subseteq\sqrt J\implies\sqrt I\subseteq\sqrt{\sqrt J}=\sqrt J$.

[cloud walrus]

cucaracha

[agile burrow]

Seems fine to me if you've shown that radicals preserve inclusion and that the radical of the radical of J is just the radical of J

[south patrol]

Ye

[pine patio]

if K and F are both finite fields and G = Gal(K/F) acts on K then does there always exist an element z of K with Stab_G(z) = {1}?

[wind steeple]

why ? If you take a subextension K'/F, then Gal(K/K') is a subgroup of Gal(K/F) and it's containing the stabiliser of points of K'

and it is probably non trivial

[chilly radish]

Anyone know a tool for generating Cayley Graphs for known groups?

[south patrol]

Does this question make sense like

Stabilisers would normally be defined for elements of K

[wintry pasture]

Given that a finite Group G acts transitively and faithfully on a finite set $\Omega$, with $|\Omega|>1$
Show that the fixed point free permutations generate a transitive normal subgroup of G.

I am sitting on this for about 2 hours now and just managed to show that there exists at least one fix point free element, would anyone mind giving me a bit of help with the approach?

[cloud walrus]

Philip

[rotund aurora]

How many units can a ring have?

[wintry pasture]

|R|-1, where R is your ring

[rotund aurora]

this doesn't answer the question

i think not at least

so if F is a fintie field, then there are F-1 units, but this number is of the form p^k-1

and Z[i] has four units

oh well I guess it's in that form. Are those all the possibilities?

also, I forgot to mention that I'm more interested in the case that our ring has characteristic zero

[pine patio]

i mean if K and F are both finite fields and G = Gal(K/F) acts on K then does there always exist an element z of K with Stab_G(z) = {1}?

[wintry pasture]

Ah, you mean if the number of units a ring can have underlies some restrictions other than maximum and minimum?

[rotund aurora]

I'm not following you ?

If you adjoin an inverse of 2 in Z, then you will also get an inverse of 4, 8, 16 and so on, so you will get infintiely many units

[wintry pasture]

Im not quite sure what exactly you are asking for

[rotund aurora]

Given a ring R of characteristic 0, how many units can R contain?

[south patrol]

Oh okay so the homework problem lol

Well hint: what do you know about the group of units of a finite field?

[wintry pasture]

I think you generally can have 2k units for any k\in \N

[pine patio]

it’s generated by 1?

[gritty sparrow]

This is true for any (finite) galois extension, interestingly enough

[pine patio]

i know why it’s true when the fields are infinite but not when they’re finite

[gritty sparrow]

there is a proof that works in the infinite case but not the finite one? very interesting

[pine patio]

yeah it uses fixed fields of elements of G

[gritty sparrow]

also the group of units is obviously not generated by 1, that would just be the trivial group

[pine patio]

i mean cyclic

[gritty sparrow]

yeah, that's right

btw what is the proof for the infinite field case?

[pine patio]

it assumes that if K/F is a finite extension then G is finite

[south patrol]

Yes it is cyclic

Once you know it is cyclic this is a one liner

So think about it a bit more

You don't need to use the fact G is finite

[pine patio]

so G acts on K\ {0} which is cyclic

so if a is the generator of this cyclic group then any automorphism of K that fixes a would also fix all K, so no non trivial automorphism fixes a right? so stab_G(a) = {id}

[rotund aurora]

But this would imply that the 2p-th cyclotomic ring has only 2p units, I don't think that is the case?

I don't know about cyclotomic rings, so idk

[solar glacier]

Question about proof verification

For showing a group of order 112 is not simple can o suppose that it is simple then n2, n7 can’t be 1

112 = 7•2^4 where

n7 = 1 or 8

And n2 = 1 or 7

Assuming simplicity we get

n7=8 and n2=7

Thus there’s 6 times 8 elements of order 7

And 7 times 15 elements of order 2 contradicting the group having order 112

Does this work ?

As 105 +48 > 112

@ me with either a verify or critique

Anyone??

[coral spindle]

Here is some info on this question: https://math.stackexchange.com/questions/3367423/is-every-group-the-unit-group-of-some-ring
Turns out it's fairly hard to answer this. For instance, there is no ring with precisely 5 units.

The first answer is the one I refer to for this fact.

[coral spindle]

This is false.

It is true when R is a finite field, certainly, but there are plenty of finite rings which are not fields.

[rotund aurora]

What is an example of a minimal spanning set that contains no torsion elements?

and that it is not linearly independent *

[coral spindle]

Sure

Consider the Z-module Q

Well actually now that I say that, that has no minimal spanning set

[rotund aurora]

I just read that modules need not have minimal spanning sets 😮 crazyyy algebra is blowing my mind today

[coral spindle]

Certainly one of minimal cardinality but not minimal by inclusion

[rotund aurora]

that was probably expected, but didn't think about it

I mean minimal with set inclusion, not cardinality

[coral spindle]

I know.

I'm thinking.

OK so consider the ring $\mathbb Z[X]$, and consider the ideal $I=(2, X)$. Clearly as a $\mathbb Z[X]$-module, this is spanned by these non-torsion elements $2$ and $X$, but it's not too hard to see that $I$ is nonprincipal, hence this is a minimal spanning set. However these are not $\mathbb Z[X]$-linearly independent, since $X\cdot2 + (-2)\cdot X = 0$.

Come on TeX...

[cloud walrus]

Boytjie

[rotund aurora]

I think $\mathbb Z[\sqrt 2, 1/2 \sqrt 2]$ over $\mathbb Z$ is an example. The set being ${1,\sqrt 2,\frac 12\sqrt 2}$

[cloud walrus]

Croqueta

[coral spindle]

Uh that's not right, you can just omit sqrt(2) and you get a smaller spanning set.

[rotund aurora]

oh right

[coral spindle]

I think my example works just fine :)

[rotund aurora]

thanks for the example 👍

I think you can also take the span of (6,15) over Z, then (6,15) is a minimal spanning set, but not linearly independent

[coral spindle]

That doesn't work either. In fact any finitely generated torsion-free Z-module will not work.

I urge you to find a smaller spanning set there.

[rotund aurora]

that's 3Z no?

[coral spindle]

It is indeed. So it is generated by 3.

[rotund aurora]

yes, but I meant the set {6,15}, which doesn't contain 3

[coral spindle]

Ah right, I see.

[rotund aurora]

yeah

[coral spindle]

In that case you can find arbitrarily large examples of that within Z itself.

[rotund aurora]

this works as before because there are we can solve 6x=15y but x and y will be non-units

and their gcd is 3, which is strictly smaller than any of them

So in general, you just want a set of vectors such that there exists a non-trivial linear combination equal to zero, but such that all the coefficients are all non-units

Then, this set will be a minimal spanning set of the set they span but it will also be linearly dependent

I should move on, but whatever. Let $R$ be a ring, and put $X=\bigcup_{i\in I}{x_i}$ and consider the ring of polynomials $R[X]$ as a module over itself. Let $Y$ be a proper subset of $X$. Then $Y$ is linearly dependent, but it is a minimal spanning set of $\text{Span}(Y)$.

[cloud walrus]

Croqueta

[rotund aurora]

Is this true? I think I came up with a proof for the case when R is an integral domain, but I don't know if I'm being completelly delusional

[coral spindle]

The second part certainly isn't true, and in fact you pointed out a counterexample earlier

[rotund aurora]

no

oh yes

[coral spindle]

:)

[rotund aurora]

oh yeah the second part is stupid

[coral spindle]

As for the first part, I'm thinking

[rotund aurora]

the second part doesn't follow from my "proof"

This should be the correct statement I think

[coral spindle]

The first part is also not true (I think I have an example) but I think if you require that L be spanning as well as L.I. then it is true – haven't checked.

[coral spindle]

This is still not right.

I can make minimal spanning sets of Z that are arbitrarily large

[rotund aurora]

yeah lol

[coral spindle]

Let me demonstrate

Let p_1, ..., p_n be distinct primes

Then set x_i = the product of all the p_j, except when i = j – we omit that

[rotund aurora]

you take a bunch of integers, that are all coprime, but every subset is not coprime

[coral spindle]

Yes that is right

This set x_1, ..., x_n spans z and is minimal

So anyway

For a counterexample for the first one, consider the ring R = k<X,Y> of non-commuting polynomials over some field k. Doesn't matter which field.

Then the set {X, Y} is l.i. but is of larger cardinality than the spanning set {1}.

(this is within the module R over itself)

(left module, of course)

[rotund aurora]

My idea was to "close" the R-module M into a vector space V over the field of fractions F of R in a natural manner. So that a subset of M was independent if and only if it was independent in the vector space closure over F

[coral spindle]

I think that does work for domains R, yes. You could make this construction explicit with localisation or the tensor product.

[rotund aurora]

This is sketchy, but that was my idea. I didn't bother to check the details

[coral spindle]

This is the localisation, yes.

[rotund aurora]

Then I wrote this

pi(S) should be pi(L) lol

[coral spindle]

It would be best if you specified that the set L is going to be R-lin ind in M, but Frac(R)-lin ind in V

[rotund aurora]

pi is the map that sends v to (v,1), etc

[coral spindle]

This is clear, yes

[rotund aurora]

I thought about that

but shouldn't that be true?

[coral spindle]

It is, yes

You only really need to prove the 2nd part. The 1st part follows immediately

[rotund aurora]

yes

But then, if $S$ spans $M$ and $L$ is linearly independent (over $R$), then $\pi(S)$ spans $V$ and $\pi(L)$ is linearly independent (over Frac$(R)$), therefore $|\pi(L)|\leq |\pi(S)|$

[cloud walrus]

Croqueta

[rotund aurora]

and pi is injective

[coral spindle]

Yes

[rotund aurora]

so then |L|\leq |S| ?

[coral spindle]

For domains R, yes

Although actually

It is not true in general that pi is injective

But your conclusion that |L| \leq |S| is correct, just for slightly more subtle reasons.

[rotund aurora]

If S is minimal spanning set in $M$ then $|S|-1=\dim(V)$ and so for any other spanning set $S'$, $|S|-1\leq |S'|$.

[cloud walrus]

Croqueta

[rotund aurora]

The problem is that pi is not injective?

[coral spindle]

But this is wrong, we've seen this before

Again, minimal spanning sets can be arbitrarily large

[solar glacier]

Can someone verify if I did this correct

[rotund aurora]

yes, that's why I ask if the problem is my assumption that pi was injective

[coral spindle]

Well indeed it is not true in general that pi is injective, so that is indeed wrong

I don't know how you've concluded that |S|-1 = dim(V) even in that case

But dim(V) means there is a l.i. spanning set of size dim(V), so that's just wrong

You can choose a basis like that

[rotund aurora]

My idea was that if $S$ is a minimal spanning set in $M$, then $\pi(S)$ is a minimal spanning set in $V$, the vector space closure of $M$

[cloud walrus]

Croqueta

[coral spindle]

Well again that's not true, and we've seen this

{2, 3} is a minimal spanning set for Z.

[rotund aurora]

you are right

[coral spindle]

Yup

You can notice something about how pi behaves on S and L to show that |L| <= |S|.

[rotund aurora]

I think in that case I mixed up the notions of linear independence and span

[coral spindle]

It is certainly true that |pi(L)| <= |pi(S)|. You should recall this.

[rotund aurora]

Because it is not true that if pi(S) spans V then S spans M

so S can not span M while pi(S) spans V

[coral spindle]

That's true as well

[rotund aurora]

Example: {2} over Z does not span Z, but {2} over Q spans Q

[coral spindle]

I urge you to think about this as you're very close

[rotund aurora]

I'm lost at how pi is not injective. Recall that (v,a)=(w,b) if and only if bv=aw, therefore, pi(v)=pi(w) iff (v,1)=(w,1) iff v=w

[coral spindle]

Set M = Z/2Z.

Calculate V

Admittedly, I think this is due to an oversight with the way you constructed V. I need to think about what's missing.

The point is that since every element of M is torsion, everything gets sent to 0 under this construction

Yup indeed, (1 + 2Z, 1) ~ (0 + 2Z, 2) here

Could the issue be that this relation ~ isn't an equivalence? Let me think quickly

Yes indeed this is the issue.

You need to consider the transitive closure of ~.

[rotund aurora]

mmh

actually

[coral spindle]

Put simply, it is not the case that bv = aw iff (v,a)=(w,b) if you complete ~ to be an equivalence relation.

Whoops typo

[rotund aurora]

yes I see

[coral spindle]

Would you like some hints for showing |L| <= |S|?

[rotund aurora]

If the module is torsion free, then that relation is transitive tho

[coral spindle]

Indeed well done, you've noticed pi is injective on non-torsion elements

[rotund aurora]

[coral spindle]

Of course it is not sufficient to look only at torsion-free modules, but you can use this fact

You should be very close now...

[rotund aurora]

So every module is a direct sum of a torsion-free module and a torsion module

[coral spindle]

Sure, a bit unexpected but I'm interested

OK well

Hint: ||Elements of a basis are not torsion||

[rotund aurora]

Suppose $M=M_\text{Tors}\oplus M/M_{Tors}$, then $V=\pi(M_{Tors})\oplus \pi(M/M_{tors})$

[cloud walrus]

Croqueta

[rotund aurora]

that's was what I was thinking

and in a vector space yyou cannot have torsion

[coral spindle]

Yup

[rotund aurora]

so torsion gets killed, is that what you were saying? :p

[coral spindle]

No that wasn't my hint

If you want to use this, you'll need of course to argue why this construction respects direct sums

Hopefully it's obvious why torsion modules go to 0

[rotund aurora]

yes

If v is not torsion, then pi(v) cannot be torsion, because if a/b pi(v)=0 then a pi(v)=0 and so pi(av)=0, but pi is injective in the torsion free part

[coral spindle]

Nice

[rotund aurora]

aren't we done? xd

like, a linearly independent set contains no torsion elements

[coral spindle]

Yup

And of course, |pi(S)| <= |S|.

So that does it

[rotund aurora]

thanks for the patience haha

[coral spindle]

No worries

It's a very neat problem

[rotund aurora]

its crazy that you cannot really say anything in general about minimal spanning sets tho

[coral spindle]

Well they are kinda weird objects

I think we can be fooled by vector spaces into thinking that minimal spanning sets are nice

But no, it's just vector spaces that are nice haha

[rotund aurora]

yeah, this is very tricky

[hollow parrot]

If G is a group and H is a subgroup and for all h in H, there exists h' in H such that gh =g'h' for some g, g' in G, is this sufficient information to conclude gH = g'H, ie, that g, g' are representatives for the same coset?

[coral spindle]

Yes, that is enough.