groups-rings-fields

[coral spindle]

Ofc this is technically false

Z/pZ for p prime has no nontrivial normal Abelian subgroup... for an appropriate definition of 'trivial' ;)

But you are not trying to say that it's a trivial subgroup but rather a trivial group

[cloud walrus]

longboard kayak

[ivory ore]

for 20 minutes im trying to figure out what is H'

oh well, for this we do not reach to 1 via derived series and just take the Gn=H right?

[ivory ore]

kinda lost here, brb i need food

[ivory ore]

why math has to abuse sign and language all the time, grrrr

[coral spindle]

G is a trivial subgroup of G

But it is typically not the trivial group

[coral spindle]

So G >= G^(1) >= G^(2) ...

And we're assuming that our group is not solvable

So...?

[ivory ore]

not it's gonna terminate before 1

or so to say it will terminate at a non-abelian simple group

so commutator of that terminal thing has to be itself, for commutator just helps you to get a abelian quotient, and here we can only have 1

[coral spindle]

This is false, but you have the right idea.

G^(1) = G does not mean the group is simple

We call such groups 'perfect'

[ivory ore]

are we doing topology now?

[coral spindle]

What?

[ivory ore]

poor joke, nvrmind

[coral spindle]

Clearly I'm missing something

[ivory ore]

"language abuse"

[ivory ore]

i can't see how is it false tho

would you walk me though that?

[ivory ore]

ohhhh, there has been a miscommunication by 1 i meant to indicate the trivial group = {e}

[coral spindle]

There exist groups G with [G, G] = G that are not simple

[ivory ore]

oh okay, actually i forgot for a moment that we are indeed working with the derived series(and was thinking of plain composition series)

yesh, we don't need to state whether it is simple or not

[worthy solar]

maybe i missed this but are there any conditions about when the scalars commute with the vectors

is that an extra condition or something that i can prove from this

because I dont necessarly see it

ie

$a\vb{v}=\vb{v}a, a\in \mathbb{F}, v\in V$

[cloud walrus]

Brandon7716

[worthy solar]

i guess i just would need to provide an example of such a vector spaces over a field where the inheirt left and right multiplation will give different outputs

[ashen heron]

va isnt exactly legal i dont think

[elfin wraith]

Vector spaces are modules over fields, which are commutative, so there’s no difference between left modules and right modules

[ashen heron]

the axioms are always stated in terms of F X V -> V

[elfin wraith]

It’s just customary to pick left action usually

[velvet hull]

Left vector space sounds cursed

[vocal pebble]

its a definition about notation(nothing much to worry about)

[vocal pebble]

they are defining both va and av to mean the image of (a,v)

this is completely fine as long as you give the notation the correct meaning

you could view it in the more general setting of the actions being the same but for an introductory definition that might just be unnecessary word salad, possibly

[rocky cloak]

The notation va is not define, but you could define it to equal av if you want.

[worthy solar]

that does make sense

[worthy solar]

yeah that is what i thought

[candid patrol]

Is there a non-computational way to show that H8 embeds into SL(2, F₃)?

[south patrol]

Lies

[coral spindle]

Oh really huh??? Is SL_2(5) a lie??!?!?!?

[thorn jay]

The trivial group

[south patrol]

I was thinking the trivial one too lol

[south patrol]

Well SL_2(R) is definitely a lie

I went to a talk on geometric group theory today

Fun times!

[thorn jay]

What's that?

[south patrol]

Well actually only half geometric group theory

[south patrol]

Sorry, forgot to reply. Tbh I know almost nothing about it but i believe a major point is studying groups (in particular "big" non-abelian, usually discrete but possibly w other topologies) by them acting on lots of stuff like trees etc, and by using some metric spaces theory

Like you can associate a graph (the Cayley graph) to a group and ask about its "geometry"

Like half of my office does this subject but admittedly I know almost nothing lol

[thorn jay]

Thanks, sounds interesting :0

[thorn jay]

Oh, mind explaining a little further?

The Cayley graph as directed graph? Or only the underlying no directed graph

[south patrol]

Uhh well viewing it as a space in the natural way

Like by gluing together copies of [0,1]

But then there is a metric you can put on it and stuffs

[thorn jay]

Alright, sure

And you let the group act on it too? By moving each vertex x to the one corresponding to xy?

Ah I should probably just look it up for myself instead of bothering you with these questions haha

[south patrol]

Yeahh

Which is nice

[south patrol]

Dw

Though tbh I don't know much anyway

[thorn jay]

I can imagine

It sounds a bit like how a quandle naturally acts on itself as topological space

(said topological space being either a partition topology or generated by some sequence of partition topologies)

[hot goblet]

having a group theory brain fart the statement "For all x, z in G, there exists a y in G s.t. x+y=z" is false right?

[wraith cargo]

Take y = -x+z

So it's true

[hot goblet]

oh fantastic you are right

that means the proof im working on about modules for like 3 hours is finally correct

thank you very much

[novel umbra]

Hi guys, could someone give me some advice?

I have a problem, I'm studying group theory. Well, I'm studying math, but the courses I'm studying, like probability and Lebegue's measurement, I understand the topics to a certain extent, but when it comes to the exercises, I get completely blocked, and my brain immediately searches the solutions and things like that. I really feel bad, and I think that's not what being a mathematician is all about. What do you think I should do?

It's like I'm too lazy to think

My thought has been like, "Bruh, are you really seeing the solution when you said you were going to do the exercise?" What I mean is, as a minimal mathematician, I should try it.

And if there are times when I try lately I have tried but as soon as I get stuck on something my brain gives up.

[chilly ocean]

must have a good mentor,

[novel umbra]

:c

[mossy slate]

Can there be multiple extension fields of the same cardinality, each using a different irreducible reduction polynomial?

In other words, is the irreducible reduction polynomial a parameter of an extension field along with the cardinality?

I am learning about some aspects of finite fields for the first time, so pardon me if my question is a dumb question 😅

[wraith cargo]

For finite fields, no
The extensions are unique

For infinite cardinalities the answer is yes

[velvet hull]

what I'm gathering is your question is essentially if there are more than one irreducible polynomials of the same degree, and the answer is yes

but any two finite field extensions of the same size are isomorphic, but not uniquely

[mossy slate]

For a more concrete example, let's say you have the Galois field GF(2^8). You can use x^8 + x^4 + x^3 + x + 1 as the reduction polynomial (as in the Advanced Encryption Standard). But are there other polynomials you could use, that would yield a different multiplication operation?

Thanks y'all for the help by the way 🫶

[frigid shard]

so your question amounts to whether two irreducible polynomials can have the same splitting field?

[velvet hull]

consider Q(sqrt(d)) for any d in Z.
this gives an infinite family of (mostly) non-isomorphic fields.
But, as the other person said, finite extensions of finite fields are isomorphic, though not necessarily equal

[ivory ore]

$H \text{ char } G$ is necessary and sufficient for the full automorphism group of $G$ to descend to automorphisms of $G/H$.

amirite?

[cloud walrus]

longboard kayak

[ivory ore]

better if a look at the hom and see what's going on here

[thorn jay]

Yeah it should be equivalent

[ivory ore]

yeah alr

also we need to take the fact into account that the subgroup is not unique upto order, or else it will be preserved without any option being left

idk what sort of groups to look at?

[velvet hull]

to empahsise the hint..... every subgroup of an abelian group is normal

[ivory ore]

oh yeah, i had that in mind, but i'm thinking if 2^n order would suffice or not

[velvet hull]

it's much more simple than that, just try constructing a group that has two copies of the same subgroup

[ivory ore]

V4

[velvet hull]

sure, that works, also ||Cn x Cn for any n||

[ivory ore]

got it, thanks

[thorn jay]

The humble switcharoo

[ivory ore]

i think taking multiple groups of same order and then taking semidirect product of it is the key to generally construct this?

[velvet hull]

no, you can probably get much more general than that

[ivory ore]

hmm one is acting on the nontrivial aut of the other twin, now i can "see"(in a handwavy way) why the whole aut is not gonna preserve them

[velvet hull]

the quaternion group is, I believe, the first nonsimple group that cannt be written as a semidirect product of groups

but every subgroup of Q8 is normal so it should work as an example

[ivory ore]

it has a centre C2

centre is char

also C2 is the only group of order 2 in Q8

OH, centre is not the only normal subgroup, 😝

[boreal inlet]

Is this false? One side is trivial, but the other side is definitely not obvious if it's true

I was thinking on these lines - there will be an s such that r_1 * s = r_2 and there will be another s' such that r_2 * s' = r_1

From here, we have r_1 * (ss' - 1) = 0

If R is an integral domain, we are done. So if it's false, R cannot be an integral domain

[rocky cloak]

Consider R = k[x, y]/(y(x^2 - 1))

The ideal generated by y and by xy are the same

[boreal inlet]

Oh my god

How can I forget about this example after a semester of AlgGeo

Skill issue fr

Thank you

[boreal inlet]

Wait, is x a unit in this ring?

[rocky cloak]

It is not

I guess the easiest way to see it is to mod out (y), and use that x is not a unit in k[x]

[amber burrow]

is the following a valid proof that eveyr subgruop of a cylic group is cyclic:

consider $G = \langle g \rangle$

[cloud walrus]

donut123

[amber burrow]

and $H \subset G$

[cloud walrus]

donut123

[amber burrow]

choose minimal $a$ so that $g^a \in H$. We claim $H =\langle g^a\rangle$. Suppose not. Then there exists $b$ that is not a multiple of $a$, so $d=\gcd(a,b) \neq a \implies d < a$. Using bezout's lemma, there exists $m,n \in \mathbb{Z}$ such that $am + bn = d$. Then $(g^a)^m\cdot (g^b)^n = g^d \in H$. But this contradicts minimality of $a$. Done

[cloud walrus]

donut123

[amber burrow]

i know the usual method uses division algorithm

but this was on a final and i was being dumb (didnt have much time) and idk i thought of bezout's lemma first

[coral spindle]

Looks right to me. You do need to word things more carefully but I get what you mean.

It is indeed overly complicated

[amber burrow]

Let G be a finite group and let $\pi : G \to S_G$ be the left regular representation. Prove that
if $x$ is an element of $G$ of order $n$ and $|G| = mn$, then $\pi(x)$ is a product of $m$ $n$-cycles.
Deduce that $\pi(x)$ is an odd permutation if and only if $|x|$ is even and $\frac{|G|}{|x|}$ is odd.

(exercise from dummit and foote)

[cloud walrus]

donut123

[amber burrow]

was getting a bit stuck

so $\pi(x)$ partitions $G$ into disjoint orbits under left multiplication by $x$

[cloud walrus]

donut123

[delicate orchid]

G acts freely on itself, so nothing non-trivial in G can fix anything

[delicate orchid]

this is also a good approach

[amber burrow]

but im not really sure where to go from there

im guessing some kind of application of orbit stabilizer

[delicate orchid]

I presume you know what cosets are

[amber burrow]

yes

[delicate orchid]

so can you see the connection between those orbits and left cosets?

[amber burrow]

oh

so the orbits are different cosets of $\langle x \rangle$

[cloud walrus]

donut123

[delicate orchid]

exactly

[amber burrow]

ah

and the last part is kinda simple

alr that makes sense ty

[amber burrow]

Hint with this too pls: Let $G$ and $\pi$ be as in the preceding exercise. Prove that if $\pi(G)$ contains an odd permutation
then $G$ has a subgroup of index $2$.

[cloud walrus]

donut123

[amber burrow]

So there exists some $x \in G$ such that $\pi(x)$ is odd. By previous exercise, $|x|$ is even and $\frac{|G|}{|x|}$ is odd.

[cloud walrus]

donut123

[delicate orchid]

ok I've thought of a solution but it doesn't use the previous question at all lol

taking a permutation to it's sign (call this map s) is a group homomorphism ||with an index 2 kernel, A_n||, so ||the kernel of pi composed with s is G \cap A_n, because pi is injective||. Now ||use 2nd iso|| to conclude ||that G/G \cap A_5 is index 2||

I'm implicitly identifying G with pi(G) so I have less nonsense to write

[rocky cloak]

||G -> S_G -> C2 is surjective||

Would also be a short way of saying the same thing

[delicate orchid]

true and that avoids 2nd iso

I like using 2nd iso though it's fun

[rocky cloak]

I mean, 2nd iso is just composition of maps anyway

[amber burrow]

Yeah I thibj this is the intended way

The exercise referred to some previous exercise in a previous chapter

And it uses second iso

[delicate orchid]

in that case why wouldn't they accept jagr's solution lol

[chilly ocean]

Every Boolean ring is semi simple?

[amber burrow]

Didn’t completely pay attention, but someone mentioned smth about avoiding second iso

[delicate orchid]

it's pretty clear that the jacobson radical is 0, but I don't see why a boolean ring must be artinian

[chilly ocean]

We don't know the reverse Direction

[delicate orchid]

well if by "reverse direction" you mean that every semisimple ring is boolean that's very false, take C[G] for any finite group G

[chilly ocean]

Okay infinite product of Z_2 works i think

[molten bone]

Don't know how to construct an isomorphism here. Trying to find a isomorphism between (r^i s^j, k) ---> some element in D_6 where i = 1, ..., n; j = 1,2; and k = 0,1. I'd appreciate any advice.

[south patrol]

Probably the easiest imo is as follows: try to find two subgroups of D6 iso to D3 and Z/2Z

[potent condor]

If people keep writing D_{2n} instead of D_n I’m gonna start writing S_{n!} instead of S_n

[thorn jay]

no but you see

D_2n is good

S_n! not

[frigid shard]

You can also express $D_n$ in terms of semidirect products ($\mathbb{Z}_n \rtimes \mathbb{Z}_2$) and use that direct products are commutative

[cloud walrus]

n1lp0tence

[molten bone]

I tried listing out each element in the direct product and mapping it to different elements in D_6.

We haven't done a lot with semi-direct products to be honest.

[tough raven]

The ideals of the powerset R of a set X as a Boolean ring are ideals (closed under finite unions and subsets) of sets of X. If X is infinite, consider the ideal I of R consisting of all finite subsets of X. If J is a complement to I, then IJ ⊆ I ∩ J = 0 (this happens in every commutative semisimple ring). In particular applying this to {x} ∈ I and an arbitrary set A ∈ J, {x} ∩ A = ∅, so x ∉ A. But that just means that every set is empty, i.e., J = 0. But then R ≠ I + J. Thus, I has no complementary ideal.

[brave remnant]

I want to verify that F(a)/F an extension of F adjoining a with minimal polynomial f, we have F(a) either contains only a or it contains all the roots of f

can someone help please

[frigid shard]

This is certainly not true. Consider $\mathbb{Q}(\sqrt[6]{2})$.

[cloud walrus]

n1lp0tence

[brave remnant]

okay

but how do i know how many roots i get

there is a thm that says the number of roots i get by adjoining one root divides the degree of the extn

but i have no idea why?

Suppose [F(a):F] degree 6 extn then why can't i get 4 roots of m_a(x) in F(a) and just have them permuted sending one to the next until it cycles through all 4 under a field embedding F into F closure extending to F(a) into F closure

this is somehow a contraidction 😦

[velvet hull]

I can't think of an argument that doesn't use the Galois correspondence. but the very loose reason why this won't work is because if such an automorphism does exist, then it must also permute the remaining 2 roots in a way that you necessarily get more roots of the minimal polynomial

[brave remnant]

it is the case that i need to show this without using galois theory

i have an idea but im not sure if its correct

[frigid shard]

what extension are u referring to here, is it the one u get after the root is adjoined? do you mean this recursively, as in you adjoin one root at a time and count the number of new roots you get? If so there would be a silly counterexample: for any Galois extension adjoining one root would already give you all the roots, so adjoining another one just gives 0 which divides nothing.

[velvet hull]

presumably by adjoining to the base field only

[brave remnant]

There is a thm that E algebraic over F then take alpha in E-F then there exists a minimal polynomial of alpha over F called m_a then the number of roots of m_a in F(a) divides [F(a):F]

F(a)/F is the extn im referring to

it does just strictly divide no?

[frigid shard]

$Q(\sqrt[3]{2})$?

[cloud walrus]

n1lp0tence

[brave remnant]

like the example 6th root of 2 u get plus minus 6th root of 2 so only two roots in Q(6th root 2)

yes then u get one root only the real one since the other ones are complex

and it still divides

I am not sure why this is the case

ive been trying to figure this out with no avail for 5 hrs

[frigid shard]

I checked my notes from a while ago

I think this Corollary might help

the idea is probably to embed $\text{Gal}(k(\alpha))$ as a subgroup of something order $n$ and apply Lagrange

[cloud walrus]

n1lp0tence

[brave remnant]

not sure if I can use galois theory

although i can use field embeddings

[frigid shard]

well assuming Galois theory we can prove your theorem very easily from this: recall that given an extension $L/k$, $[L : L^G] = |G|$, and since $L^G$ is an intermediate field $[L : L^G]$ divides $[L : k]$, as desired.

[cloud walrus]

n1lp0tence

[frigid shard]

But $[L : L^G] = |G|$ is a highly non-trivial result tho, so idk how one would bypass this

[cloud walrus]

n1lp0tence

[frigid shard]

is this like on your homework?

[brave remnant]

its like a part

of a larger

issue

i have an idea

[frigid shard]

yeah proving is same order of difficulty as proving the Galois correspondence

[frigid shard]

in that case you should probably not try to invoke this result

[brave remnant]

do you think i can use

my intuition is like this

F(a)/F gives less roots than the m_a / F

right

[frigid shard]

well less than is easy

[brave remnant]

but this is arbitrarily less than

it's some number

wait not sure thats going to work

i was going to say that (x-a)(x-b)(x-c)g(x)

then I just take the x-a x-b x-c and construct a new minimal polynomial

but that contradicts the minimality of the original

this sucks

[frigid shard]

lol

[frigid shard]

do you want to just say your larger issue

[brave remnant]

i think the larger issue gets resolved if I can find a way to show the F(a) problem

[frigid shard]

yeah but it could be that this is a stronger statement than your other issue, assuming that it doesn't require Galois theory

[brave remnant]

F(a)/F then there is a fixed field intermediate between F(a)/F

[frigid shard]

yes

[brave remnant]

I'm supposed to somehow use that?

[frigid shard]

that's essentially the argument i did above

[brave remnant]

i don't understand

[frigid shard]

the problem is if you don't know the order of the intermediate extension ur cooked

[frigid shard]

cf

[brave remnant]

but do we even care abt the order of the intermediate? by lagranges it must divide thats all we care about

[frigid shard]

you need to link it to order of Galois group

[brave remnant]

im very confused about the whole fixed field root permutation field embedding stuff

so my understanding is that if I have a phi:F to E where E is maybe a splittig field of F for example, then if I extend F to some larger field K where K algebraic over F, then there is a PHI such that PHI restricted to F is just phi. PHI is completely determined by where it sends the roots of an irredpolynomial over F

why can't i send those roots to any other root I want? Why am i constrained by the number of roots being permuted must divide the degree of the minimal polynomial

[frigid shard]

yeah this is the isomorphism extension theorem

[brave remnant]

an example is Q maps to C then f is x^4-2 then I extend phi that fixes Q to the splitting field of x^4-2 then I send fourth root of 2 to fourth root of 2 times i then to fourth root of 2 times -i and back to fourth root of 2

[frigid shard]

sure you can extend arbitrarily but you want the resultant diagram to commute, i.e. it be a morphism of extensions

[brave remnant]

in this case I send to 3 things but 3 does not divide 4 contradiction, but whyyyyyyyyy

[brave remnant]

where does the contradiction come from

[brave remnant]

is this true for fields of any characteristic?

[chilly ocean]

i

life is so good

[rocky cloak]

Any automorphism of F(a) is determined by where a is mapped and can map a to any root of the minimal polynomial.

Hence |Aut(F(a)/F)| equals the number of district roots in F(a).

Now calling this group G you have |G| = [F(a) : F(a)^G] which divides
[F(a) : F] = [F(a) : F(a)^G] * [F(a)^G : F]

[rocky cloak]

This is not a homomorphism, you must have f(-x) = -f(x) which you don't do here

[frigid shard]

This is same as the proof I gave, the problem is OP can’t use that |G| = [F(a):F(a)^G]

Do you know of a way to bypass this?

[rocky cloak]

Well, I don't know if this is any better, but if you let E be the splitting field of F(a) and G the automorphism group of E, then G acts transitively on the roots.

Take H to be the subgroup that fixes a and let H' be the subgroup that leaves F(a) invariant.

Then H' > H hence [H':H] divides [G:H]. The latter is the total number of roots while the furmer is the number of roots in F(a)

The only tricky think here is that G acts transitively on the roots.

You can prove by induction that any embedding F(a) -> E extends to an automorphism of E. This is a general fact for normal extensions

[frigid shard]

I feel like the root counting implicitly uses the Galois correspondence no?

[rocky cloak]

Not really no. It's just orbit stabilizer

[frigid shard]

Oh wait you’re so right

So this does solve their issue since showing G acts transitively is pretty straightforward

[hot goblet]

is there a concept analogous to group order for rings?

[delicate orchid]

the order of the ring?

[hot goblet]

shit i meant so say modules

is there a concept analogous to group order for modules

[delicate orchid]

ok they're still just sets so

just take the cardinality

in fact they're also groups so we don't even need to go that far down

[hot goblet]

mm cause Ive got a module generated by <m> and I've got another element with the same annihilator (hence same order) tryna show that this other element q also generates the module

[hot goblet]

true

i meant to say modules but typed rings sorry

[delicate orchid]

that would be the order of a group element not the group

[hot goblet]

uh huh yes

[delicate orchid]

but I think they meant element anyway so it's fine

[delicate orchid]

do we know that q is in <m>?

[hot goblet]

yes, <m>=M the whole module and so q in M means q in <m>

(by assumption q not m)

[delicate orchid]

oh I see yes you did say that sorry

[hot goblet]

all good

can I just say "oh if we think about these as groups then they have the same order and so <q> = M" that feels very handwavey

https://mathoverflow.net/questions/391369/elements-with-equal-annihilators

hmm

[delicate orchid]

then yeah I think you can conclude that <q> = <m>. My thoughts are: if we consider the multiplication map R -> End(M), the kernel of this map is exactly the annihilator of m (cause <m> = M), so if q is also annihilated by this kernel and only this kernel then we must have have <q> = M

if <q> < M, then there must be more elements of R which send q to 0, implying Ann(q) > Ann(m) which is a contradiction

[hot goblet]

when you say R -> End(M) your talking about sending scalars to endomorphism

hmm

so your suggesting a map p(a) that sends a in R to the map that sends n to an (n in M)

is that right?

fixed

[delicate orchid]

yeah

that's what modules are really, an abelian group with a map R -> End(M)

[hot goblet]

yeah nah yeah that makes sense thanks mate

[delicate orchid]

no worries

[hot goblet]

This stack exchange has a theorem
"If R is a quasi-Frobenius ring (which since R is commutative is the same a a Frobenius ring), then two elements have the same annihilator if and only if they are associates (differ by multiplication by a unit) if and only if they generate the same principal ideal."

well technically this is for rings but

[hot goblet]

this makes sense but I hadnt thought of it before, we only covered it in the context of a map RxM -> M being scalar multiplication

[delicate orchid]

yeah that also works - but it's just a nicely way to encode all of the scalar multiplication axioms into a short phrase

cause a ring homomorphism R -> End(M) automatically gives you a scalar multiplication, but not all maps RxM -> M do

[hot goblet]

ohh yeah thats really smart removes the need for some of the axioms

[hot goblet]

if you use this definition isnt the definition of a module just an abelian group with a map R->end(M) thats it?

[delicate orchid]

yup that's the full definition

[hot goblet]

huh that is a lot more elegant than the definition in my class

[amber burrow]

wait this doesn't make sense? how can you intersect G and A_G? G is just some group, not necessarily a permutation group

[rocky cloak]

Wew is identifying G with it's image in S_G

[amber burrow]

oh ok

yeah i just saw he mentioned that

i think you said exactly this, but i wanna make sureim doing the right thing

so you don't actually need second iso

since theres an odd permutation, pi compose f is surjective (cuz you can get -1)

so G/ker f is isomorphic to C_2

and then you're just done?

[rocky cloak]

Yup

[amber burrow]

alr - a little confusing since dummit and foote refers to a previous exercise about like second iso stuff

also technically ker pi compose s is not pi(G) intereset A_G, its just isomorphic

but actually you don't need any of that, you just need mpa surjective -> G/ker iso to C_2

[delicate orchid]

I'm gonna defend myself here, I'm 99% sure that it is straight up this intersection

[amber burrow]

but ker f is a subgruop of G

but pi(G) cap A_G are permutations

[delicate orchid]

as previously explained

we're identifying G with pi(G)

[amber burrow]

ok then i think ur right

[delicate orchid]

but yeah they're isomorphic anyway so who cares

[amber burrow]

yeah

Let $G$ be a finite group of composite order $n$ with the property that $G$ has a subgroup of
order $k$ for each positive integer $k$ dividing $n$. Prove that G is not simple

[cloud walrus]

donut123

[amber burrow]

Felt like this was too easy for being the last problem in the chapter

isn't it just

take a prime factor

by cauchy's theorem, you get a subgroup of order p

wait nvm lol

was gonna say cyclic and done lmao

[chilly ocean]

if p be the smallest prime dividing order of group then group is not simple (subgroup with index p is normal)

[amber burrow]

oh yeah lol forgot that

[chilly ocean]

but i didnt understand your solution

[amber burrow]

nah i was doing gibberish lol

There’s a subgroup of order n/p

So taking quotient gives index p

We know that one exists

And by what you said it’s normal

[round portal]

good morning guys it's ya boy texas

i need some help proving this assertion here

do y'all have any tips

if I could prove that every simple abelian group is cyclic it's home free right? but idk if that's necessarily true

[coral spindle]

Well every subgroup of an Abelian group is normal. Do you know of any nice examples of subgroups?

You are on the right track

[amber burrow]

|testing|

||testing||

There

[amber burrow]

||center is either trivial or the whole group. It can’t be trivial, since every element commutes with powers of itself||

[fierce breach]

omg rachel goswell

[coral spindle]

I wasn’t talking about the centre

Indeed Z(G)=G for any Abelian group G

[chilly ocean]

every abelian shit is just a dirct product of littile shits, isomorphics to some prime clocks on wall.

[round portal]

alright bet

cool!

[round portal]

yup i think that seals it

[candid patrol]

It does but that’s a too strong theorem

[chilly ocean]

since group is abelian why dont you try some induction

without theorem

[candid patrol]

Suppose G abélien and |G| not prime, then take p prime which divides |G|, by Cauchy, there exists x with order p, and so, < x > is a normal proper subgroup of G

[round portal]

cool

[round portal]

I think there is an easier way

induction seems like overkill

[chilly ocean]

atleast it is basic

[amber burrow]

really general question, but when do you wanna think about group actions?

like on the surface, if you were the problem to show that if |G| = 2k where k is odd then G has a subgroup of index 2, i dont think i would think of group actions

[mighty kiln]

What are groups for, if not to act

[rocky cloak]

I think the answer here is basically anytime you want to think about a group.

Like, subgroups up to conjugation = transitive group actions. So if you're thinking about indexes of subgroups you probably want some kind of G/H group action.

Similarly the core of a subgroup H is the kernel of the group action G/H, and the normalizer is the automorphism group of G/H as a G-set. So this is often typically seen as group actions.

Then you have that homomorphisms to Sn and GLn are group, so understanding homomorphisms to them or related groups are best understood through that lense

And many of the nice theorems about finite groups are proven by finding some nice group action, like Cauchys theorem, the Sylow theorems, the center of p-group being nontrivial, burnsides complement theorem

[chilly ocean]

In my class group theory started from group actions , saying every groups comes from a action, and every group can be embedded in GL_n(F), AND YES cauchy and sylow theorem prooven using actions and embedding it was nice treat .

[rocky cloak]

Hmm, can every group be embedded in GLn(F) for some F and n?

[chilly ocean]

at least for finite group yes

[rocky cloak]

Well sure

[chilly ocean]

for infinite group not sure

[chilly ocean]

is it true for infinite?

[rocky cloak]

Idk, I would think not, but idk

[hazy reef]

why was the word ring chosen

[delicate orchid]

I dread to think about cardinality issues for trying to embed things bigger than uncountably infinite inside GL_n(F), even if F itself is larger than uncountably infinite (if such a thing is even possible for a field)

[delicate orchid]

I believe historically it's from the fact that Z/mZ has a "cycling" kind of feel to i t

https://mathoverflow.net/questions/117292/why-is-a-ring-called-a-ring

second answer seems mostly likely lol

[hazy reef]

but then only quotient rings should be called rings

[delicate orchid]

There are a myriad other explanations that are given in the thing I linked

[hazy reef]

by definition, isn't a ring any group where you can add and multiply elements together

[delicate orchid]

it could also just come from an archaic German interpretation of ring to be a collection of things

[delicate orchid]

Every ring is a quotient ring so

[chilly ocean]

why the word separable is used for the topological spaces having countable dense subset.

[elfin wraith]

Because in R you can separate any two reals with an element of Q

[chilly ocean]

why it is a general term then

[elfin wraith]

A lot of topology comes from generalisations of things you do on R. You’ve got to call it something, seems like a perfectly good things to call it.

[chilly ocean]

At begining it looks, but as I grew it was far from R, so I don't think this

[rocky cloak]

I mean, you growing doesn't really change that a lot of topology is motivated by R.

The word seperable comes from functional analysis anyway, where thing do look a lot like R

[delicate orchid]

95% of topologies I see are homotopy colimits of subsets of R so as far as I care all topology is quotients and subspaces of R^n

[stuck egret]

point set topology is just bigcap bigcup on steroids

[tardy hedge]

Poo and pee

Noo i dont want to lose my
Very active tag

I have in fact not been very active recently

[elfin wraith]

Ah, I didn’t know that, what’s the motivation in functional analysis?

I was always under the impression it was the thing I said but I’ve never actually looked it up

[rose prism]

wdym motivation

countable sets are much nicer to work with than uncountable ones

[elfin wraith]

Jagr mentioned separable comes from functional analysis, in what way does it show up there and how did that motivate the topological definition?

I don’t really know anything about FA

[rocky cloak]

Just like you said. R is a seperable Banach space

[south patrol]

Lol

[icy totem]

For coprime naturals m, n, consider a field extension L/K of
degree m. Then every element a ∈ K with an nth root in L has
an nth root in K.

I managed to solve it eventually, but I wanna see if someone can find a different proof from mine

Cause mine it's kinda random and I feel like it can be solved in other ways

[velvet hull]

Consider the tower of field extensions L - K[a^1/n] - K. Denote the intermediate field by F (and it is a subfield of L by assumption so such a tower exists).
Then [F:K] divides n (to see why notice that Aut(F/K) has an embedding into Z/nZ) such that m = [L:K] = [L:F] [F:K] by the tower formula.
But then as n,m are coprime by assumption [F:K] = 1, ie a^1/n is in K
There’s a problem with this proof, see below, I think I made some incorrect assumptions about the existence of roots of unity, I’ll come back to this later

[rocky cloak]

Consider K = Q, a=1 and n=3.

Then [Q(1^1/3):Q] has degree 2, so does not divide 3

Of course Q contains a different cube root of 1 (namely 1 itself)

[rocky cloak]

So the argument should work when K has an nth root of unity.

So maybe one can show separately that a root in K(1^1/n) implies a root in K.

[velvet hull]

1^1/n to denote an nth root of unity is cursed

[south patrol]

I think this is probably key

if you can reduce n to a prime, which seems reasonable

[sturdy spear]

\delta in an algebra book?

interesting

[chilly ocean]

kronecker deltas are used

[sturdy spear]

oh my bad true

brainfart sorry

[chilly ocean]

there is a triangular delta aslo

[lime warren]

how do yall vocalize these? for the first one obviously just wondering about the second statement

[eager willow]

Usually in context it's "Let f:A->B" to say "Let f be a function/morphism/map from A to B"

[eager willow]

Whereas the f labeling of the arrow, in context, often shows up in a diagram and is simply not vocalized

[chilly ocean]

can any body share a sevrer link for politics

[prisma bear]

For finitely generated Z modules A, B. If A, B does not have elements with order of some power of a fixed prime p, does it mean the statement hold for A tensor B?

This is from an exercise on Hatcher where he gave some complicated hints, but shouldn't this statement be obvious?

The same would hold true if I replace A with direct sum A_i, B with direct sum B_i right?

[chilly ocean]

please mention the page

chpater

[crystal vale]

Any hint? I know if p is prime ideal then R/p is domain but how can I verify it is integral domain or not.

I know I have to choose R and p such that R/p not finite

[chilly ocean]

see in C(R) and maximal ideal containing all cont function with compact suppport

[crystal vale]

But in C(R), there is an unity, right?

[chilly ocean]

try ring with continuosu func with compact support

[rocky cloak]

Take R = Z<x, y> to be the noncommutative polynomial ring and p=(x,y)

[thorn jay]

In commutator theory by Ralph McKenzie, deltas are used for congruences, as there they are denoted using Greek letters

[rocky cloak]

If I'm gonna label arrows I usually start from alpha and just go alphabetically

Reach delta pretty quick

[thorn jay]

Lots of arrows

[elfin wraith]

Wait until brother learns about differential algebra. Delta is common notation for derivations

[chilly ocean]

discriminant of polynomial

[crystal vale]

Is Z<x,y> exactly same as subring generated by x and y in Z[x,y] ?

Non-commutative ring, how?

[rocky cloak]

The non-commutative polynomial ring is not a subring of the commutative polynomial ring no

[tribal moss]

Do you require "integral domains" to have 1?

If not, then it appears that the definition of "prime ideal" would directly make R/p a domain.

On the other hand, if integral domains must have 1, then you could just take R to be the even integers, and p={0}...

[crystal vale]

yes

[crystal vale]

ahhhhh....

[crystal vale]

i didn't worked in non-commutative polynomial ring

[chilly ocean]

maybe he is thinking it as product ideal of Z and <x,y> inside Z[x,y]

[tribal moss]

I think Jagr simply misread the problem statement to have "noncommutative" instead of "commutative".

[chilly ocean]

ya maybe

but i also tried to work out with ring of continuous function with compact support, but not much aware about its prime ideal

[tribal moss]

Intuitively I think the only prime ideals there would be { f | f(a)=0 } for each a.

[chilly ocean]

or can we take rng of polynomial with 0 contant term

[chilly ocean]

0 function

[tribal moss]

{0} is not prime in the ring of continuous functions with compact support -- two bumps with disjoint supports have 0 as their product.

[chilly ocean]

ideal generated by bump functions?

[tribal moss]

What do you mean? Principal ideals? Those are generally not prime, I think.

[chilly ocean]

please give some nice prime ideal

[tribal moss]

{ f | f(42) = 0 }.

I think all of them have that form.

[chilly ocean]

okay

this isnt working though

[tribal moss]

"Working" in which sense? Notknow's problem?

[chilly ocean]

yes

[tribal moss]

Right, because then the quotient ring is just the reals.

[chilly ocean]

maybe result holds

[undone ledge]

i think there are more

[tribal moss]

Hmm?

[chilly ocean]

this ring without unity is an ideal of some ring, can we use that

[tribal moss]

You did see my suggestion of 2Z/{0}, right?

[undone ledge]

take an infinite compact set S and consider the multiplicative submonoid of functions that vanish at finitely many points in S. then a prime ideal that avoids this multiplicative submonoid cannot be of the form {f | f(a) = 0}

[chilly ocean]

No

[tribal moss]

I think that's a good counterexample to Notknow's problem.

(Or even any proper nontrivial ideal in a domain, quotiented by {0}).

[crystal vale]

yes

[chilly ocean]

at the end it was most simple

[tough raven]

IIRC on a compact Hausdorff space X, those are the maximal ideals; the closed (wrt uniform convergence) ideals are {f : f(Z) = 0} for each closed subset Z, and this ideal should be prime iff X\Z is connected(?).

[tough raven]

Yes, I think it is obvious using the classification of fg ℤ-modules.

[amber burrow]

Ok I don’t know this stuff, but I was just recognizing some stuff: isn’t Hausdorff a topology or thing, but maximal ideal is a ring theory thing?

[tough raven]

Yes. (Compact and continuous are also topology things.)

[stuck egret]

Hello

[stuck egret]

If they were both related wouldn't they be an example of a prescheme or small scheme?

I'm talking about Hausdorff and ideal thing donut123 said

[tribal moss]

I don't think { f | f(Z) = 0 } is prime if Z is an extended closed subset.
Choose f with support outside Z, and g and h with disjoint supports both inside Z. Then (f+g)(f+h) = f² is in the ideal, but neither f+g nor f+h is.

[chilly ocean]

maybe his host space is compact housdorff not R

[tribal moss]

What I describe could equally well take place on, say, the unit circle, which is compact Hausdorff: If Z has nonempty interior there will still be space for disjoint bumps g and h on it.

[undone ledge]

@tribal moss

[chilly ocean]

life is sweet chin, jst bang it

[tribal moss]

I agree there's such a multiplicative submonoid. But it's not obvious to me that there's a prime ideal that avoids it?
(It doesn't seem to be a general ring fact at play, at least: in Z, the nonzero squares form a multiplicative submonoid, but there's no nonzero prime ideal that avoids them).

[undone ledge]

{0} is prime there though

[tribal moss]

Yeah, but it's still not clear to me which general argument you're appealing to here.

[undone ledge]

i would assume that (a maximal ideal that avoids the submonoid => that ideal is prime) doesn't rely on the ring having a unit

0 is not in the submonoid

[tribal moss]

I may be forgetting something basic, but how do you know there's a maximal ideal that avoids it? And is it [maximal ideal] that avoids it or maximal [ideal that avoids it]? I can get the latter by Zorn, but it's not clear that it would necessarily be prime.

[undone ledge]

maximal [ideal that avoids it]

[tribal moss]

Okay, I agree such an ideal must exist. Why must it be prime, though?

[undone ledge]

if you have two elements outside that multiply together to go inside then one of them must not be in the submonoid, but then the ideal generated by including that element is also disjoint from the submonoid

[tribal moss]

if you have two elements outside that multiply together to go inside then one of them must not be in the submonoid
Okay so far.
but then the ideal generated by including that element is also disjoint from the submonoid
Why? Perhaps there's a way to get into the submonoid by adding our new element to something already in the ideal.

[undone ledge]

oops i made a mistake

consider the elements where you can multiply by something in the submonoid to go inside

this is an ideal and contains everything inside

and also avoids the submonoid

[tribal moss]

Wait, why?

How can "contains everything inside" and "avoids the submonoid" even be true about the same set, ideal or not?

[undone ledge]

sorry let me actually use letters

M multiplicative submonoid, P maximal wrt the ideals that avoid M

[tribal moss]

And, just to be sure, "avoid M" means has empty intersection with M, right?

[undone ledge]

consider {x in our ring | exists m in M such that mx is in P}

yeah

[amber burrow]

I get that - I’ve taken some topology - what I’m confused about is that you are using a ring theory thing to describe a topology thing

Aren’t they pretty separate?

[undone ledge]

this is an ideal and avoids M

and contains P

assume 0 is not in M and M is nonempty

[tribal moss]

Hmm, I'll need to think about why that is an ideal first.

[undone ledge]

when you sum two you can multiply by the two ms that exist

[tribal moss]

Okay if we have a x and y in your claimed ideal (which has no name yet!) with mx in P, yn in P, then (x+y)mn is in P too.
So I accept it's an ideal.

And it cannot contain anything in M.

So it must actually be P itself, or it wouldn't be maximal.

[undone ledge]

exactly

(that's why i didn't give it a name)

[tribal moss]

Actually, could it be smaller than P?

[undone ledge]

it contains P

[tribal moss]

Ah, because your entire space is compact so 1 is in M.

[undone ledge]

i meant abstractly, it contains P

[tribal moss]

Now I'm lost again.

[undone ledge]

take any p from P. then any m in M works, since mp is in P

[tribal moss]

Ah, right, P is an ideal. Sorry.

[undone ledge]

i'm definitely overcomplicating this so you shouldn't apologize

for any element x in our ring consider {y in our ring | xy is in P}

we want to show that this set is P itself

that would show that P is prime

[tribal moss]

Yes it would.

[undone ledge]

the set is an ideal

it also contains P

[tribal moss]

Right.

[undone ledge]

well assume x is outside P

[tribal moss]

Outside what?

[undone ledge]

otherwise the set is just the ring itself

[tribal moss]

Huh?

[undone ledge]

P is an ideal, so if x was in P then any element multiplied with x goes inside P

[tribal moss]

Ah, I see, "the set" was {y | xy in P }.

[undone ledge]

this ({y | xy in P}) is also eventually P but call it P_x

sorry

P_x also avoids M, because x was outside P to start with

like if there was an element m in M such that mx is in P then x is in P

by what we said earlier

[tribal moss]

Okay, so putting it all together I think we have shown that if 0 not in M, then there's a prime ideal P disjoint from M

In a general ring.

So now back to function rings, this shows there's a prime ideal whose every member is a function with infinitely many zeroes.

[undone ledge]

this is the case for {f | f(a) = 0} as well

because our functions have compact support and we're working in C(ℝ, ℝ)

[tribal moss]

Oh, I thought you had switched to using a compact domain, like S^1.

[undone ledge]

well there we would have a unit

idk i was responding to your initial comment

[undone ledge]

this one

[tribal moss]

Okay.

So we have a prime ideal in which every member has infintely many zeroes in, say [0,1].

[undone ledge]

yeah

[tribal moss]

And that won't be { f | f(a)=0 } no matter whether a is in [0,1] or not.

Okay, I'm convinced.

Thanks for the explanation.

[undone ledge]

it's very nonconstructive but i actually don't think there's a constructive way to show there are other prime ideals

[tribal moss]

Sounds like fair game. ;-)

[tough raven]

It's true if the saturation (closed under divisors) of the submonoid is proper (does not contain 0).

[tough raven]

Why not?

[tough raven]

They're different, but they can be connected. The ring under consideration depends on the original topological space, so it's reasonable for its algebraic properties to be related to the space's topological ones.

[tardy hedge]

the set of associated primes of an ideal I of R is the same as Ass_R(R/I)?

an element of Ass_R(R/I) is an ideal Ann(m) for m in R/I?

in atiyah macdonald it says the asociated primes of I are the prime ideals that occur in the set r(I:x) for x in R. So from that it looks like radicals of annihilators of elements in R/I

[candid patrol]

Is is true that for a surjective $f: G \rightarrow H$, we have:
$D(G) = f^{-1}(D(H)) \cdot \operatorname{Ker}(f)$?

[cloud walrus]

UGOBEL

[tardy hedge]

What is D(G)

[delicate orchid]

Deez gnuts

[tardy hedge]

ong what is D(G)

[trim wind]

probably the subgroup generatd by all divisible subgroups

[chilly ocean]

Hi everyone I have a question related to the Banach-Tarski Paradox. Is G-equidecomposability a bijection?

[candid patrol]

D(G) = { [a,b] = abab^{-1}b^{-1} | a,b € G }

Also know as [G,G] (commutator subgroup)

[coral spindle]

D for Derived subgroup

[candid patrol]

yes

[storm kiln]

No right? If H is abelian, then D(H) = {1} and f^-1(D(H)) = ker f as well

[balmy wraith]

Is there is a definition/term for the set of all pairs elements (x, y) in a ring (R, *, +) such that x*y = x+y?

[frigid shard]

well representations over rings are just modules, which is just linear algebra xD

[undone ledge]

these will exactly be 1 + a unit

[frigid shard]

seriously tho any graduate algebra book should have a section on module theory

check dummit and foote or aluffi

well this is still group representation

GL_2(R) -> Aut(k)

so the keyword is group representation theory

classical representation theory usually still studies functors into k-Vect, only more modern literature would contain study of representations over modules

but free modules are really not that different from vector spaces

what is ur current source

ic

i dont think any definition would change work over R-mod so if ur comfortable with classical representation theory just translate over the same definitions

lol

well i suppose the point of the paper (partially) is to work with more general representations

[arctic trail]

linear algebra is either over division rings or commutative rings (usually over something that's both)

if you venture outside that you're dead

[frigid shard]

well i meant that as a joke but at least R-mod is still an abelian category even when R is bad

[lyric turtle]

hey i got the answer for the point group as D1 but im a bit confused how i could go about finding the index for this. Anyone have any suggestions?

[minor fulcrum]

do you know what the group G is, because its subgroup of translations is of course just Z

[tough raven]

I think that's exaggerating. Things like Smith normal form are over PIDs and lemmas like the centre or classification of ideals of matrix rings are both valid over general rings and useful in module theory over those rings.

[crystal vale]

Can I say like that, since if H is a subgroup of index n in G then there is a unique group homomorphism from G to S_n?

I know there is a group homomorphism G to S_n and kernel \subset H, but I am not sure about uniqueness.

And if we can say there is bijection between subgroup of index n and homomorphisms then we can say G has finite subgroup of index n, because G is finitely generated so there are finite group homomorphism G to S_n, right?

[rocky cloak]

I mean it's not unique, but you don't need uniqueness for the exercise.

If there are more maps G -> Sn than index n subgroups that's still a finite amount.

[crystal vale]

Yes G to S_n there are finitely group homomorphism because G is finitely generated

[hard hearth]

this doesnt seem right to me... the fact that the r(x) that the division algo in Z[x] gives us might be nonzero is precisely what we are trying to eliminate right? i rather have the following proof:
the division algorithm in Q[x] gives us that g = fq' --(eq. 1) . now if we apply the division algo in Z[x], we get that g = fq + r => g-fq = r but we know f divides the lhs (in Z[x]) by equation (1), thus f divides r (in Z[x]) thus f divides fq + r = g (in Z[x]) as desired

1. does my concern make sense
   2)if 1), is my "alternate" proof fine?

[trim wind]

The proof in the book is fine. The point is that deg r < deg f, so it has to be the same in Q[x], since it's unique with w.r.t. this property

[rocky cloak]

The proof in the picture is saying that you do division with remainder works the same over Q and Z, so when you performed the computation you will get r=0.

I don't quite follow your proof. You say that f divides g-fq by equation 1. But wouldn't that mean q' is in Z[x], in which case you would already be done.

[hard hearth]

🤦‍♂️

ignore whatever i said

[round portal]

guys how tf do I show that A_n is simple for all n >= 5

fr

i'm kinda going insane over simple groups n shit

helps if I use that

$A_{n} = <\begin{pmatrix} 1 & 2 & i \end{pmatrix} | i \in {1, 2, ..., n}>$?

[cloud walrus]

scarpa

[chilly ocean]

as fas as i know about a proof using induction, and using that any non trivial conjugacy class of A_n has at least n element

[round portal]

this is true

my prof proved in class

i'm just asking if i can use it to prove that A_n is simple for all n >= 5

[chilly ocean]

yes

on stackexchange i saw a proof

[chilly ocean]

which uses this fact

[round portal]

cool

chatgpt told me that i could show that every normal subgroup of An contains a 3-cycle

does that work?

[tough raven]

As @ jagr2808 mentioned, it's enough that {subgroups} → {homororphisms} is injective, but that isn't entirely trivial either. ||Make sure to represent the cosets by 1, ..., n so that H is represented by 1; then H = Stab(1) recovers the subgroup from the homomorphism.||

[south patrol]

Btw you can use \langle, \rangle for the angle brackets

[round portal]

thanks buddy

im still struggling

with this problem

😦

[south patrol]

;;

Ngl I have forgotten how to do this lol

[round portal]

like

I know the strategy

pick a normal subgroup of A_n, ok

show that it contains a 3-cycle

since N is normal, it contains all 3-cycles

I'm struggling to show that N contains a 3-cycle

[south patrol]

If you've dealt with n= 5 then for n > 5 you can try intersecting with A_(n-1) and stuffs

[round portal]

that sounds like a whole lotta work

[undone ledge]

N nontrivial

intuitively you have enough room to move around things

[round portal]

surely

but is there a way to explicitly show

[undone ledge]

you have all the conjugations to play with

[round portal]

ooh

that's true eh

[undone ledge]

like give yourself a random bunch of even permutations and see if you can produce a 3-cycle

(using those permutations and their conjugates)

[round portal]

alright!

you helped a lot actually

thanks bud

[undone ledge]

actually what i said isn't quite correct

you can only conjugate by even permutations

e.g. a 3-cycle

[mossy slate]

Is it really true that all finite fields of the same order are isomorphic? That would be wild...

It seems like, given that the operations can be anything as long as they satisfy the necessary conditions of commutativity, associativity, etc... you could make two finite fields of the same order that are not isomorphic

[somber badge]

what are some methods to count |Hom(G,S_n)|? can I consider permutation representations and use stuff like orbit-stabilizer and burnside's lemma? i'm not sure how to go about these two though, or if there are any other methods

[south patrol]

Yea

The point is that if K is a finite field then it has p^n elements for a prime p and integer n >= 1. So every element of K is a root of the polynomial x^(p^n) - x by Lagrange's theorem. This means K is a splitting field of this polynomial over Fp, and splitting fields are unique up to isomorphism

Note that these are not unique up to a canonical/unique isomorphism though

[mossy slate]

Hopefully one day soon I will learn the math required to understand all that (less than 3 weeks till my first college semester starts!)

[south patrol]

Ah nice, sorry I wasn't sure of your background - I suppose the stuff here may technically be a little bit away but probably not too far if you read ahead (and you seem to have read ahead to be asking this aha)

[elfin wraith]

(ive finished my degree and dont know what a splitting field is)

(which is my fault for not taking galois theory)

[south patrol]

Valid

What do you do nope actually lol idk if I have talked to u much

heh

(Hello)

[elfin wraith]

Went to one lecture, the man was insanely dry, took categories for quantum theory instead

[south patrol]

Chad

[elfin wraith]

Uhhhhh algebra vaguely, I like rings, generally happy if theyre around. I fuck with topology too, my UG thesis was about Betti numbers in algebra and topology, I liked the homological algebra I learned there

[south patrol]

Ok I forgot we have talked lol

Now I remember it all

[south patrol]

V cool ye

[elfin wraith]

Im hoping I actually get into a masters somewhere and the idea becomes more clear

I think K theory seems cool maybe some noncom geometry, maybe some rep theory

[south patrol]

V nice

[south patrol]

I would expect this to be very hard in general

[mossy slate]

Haha yeah, I know some stuff about really "advanced" math, but there are a lot of gaps in my knowledge. I've been studying cryptography, and right now I'm learning about AES, which uses Galois field arithmetic. That's what got me onto the subject lately.

[south patrol]

Or do you mean for specified examples

[south patrol]

Ah yeah sure that is a natural way into finite fields

[velvet hull]

here's another interpretation: a finite field of size n will contain exactly the n unique solutions to the polynomial x^n-x / x^(n-1)-1 (the proof of this is nontrivial).
but then any 2 solution sets to a polynomial are necessarily "the same", so two fields are isomorphic

[south patrol]

Isn't that what I said aha

But should be x^(n-1) - 1

[velvet hull]

oh you did lol

[south patrol]

Nws tho lol like just funny

[velvet hull]

happens a lot here

i know it annoys me when someone reposts my answer lol

[rocky cloak]

It can be useful to connect it with G-sets of size n I guess.

There's a formula connecting it to index n subgroups here
https://qchu.wordpress.com/2015/11/15/finite-index-subgroups-of-the-modular-group/

[thorn jay]

rings are great. I hate them, and especially commutative rings make me angry at how many miracles they perform

[somber badge]

I was thinking of Hom(Q_8, S_4) in particular, since someone said there's a really short solution with orbit-stabilizer a while ago

[somber badge]

is a G-set the same thing as a permutation rep?

[south patrol]

A G-set is a set equipped with an action of the group G on it

[rocky cloak]

A map Q8 -> S4 is the same as an action of Q8 on {1,2,3,4}.

Thinking about this as a G-set it decomposes into orbits of the form Q8/H for subgroups H.

These subgroups must have size at least 2, since otherwise Q8/H would be bigger than 4, so they are nontrivial.

But any nontrivial subgroup contains {±1}.

So any such map factors
Q8 -> Q8/{±1} -> S4.

Counting maps C2xC2 -> S4 is much easier.

[south patrol]

Nice

[somber badge]

sorry i'm kind of new to group actions and stuff, but what does this mean? "Thinking about this as a G-set it decomposes into orbits of the form Q8/H for subgroups H". why do the orbits of the action on {1,2,3,4} have to be of the form Q8/H

[south patrol]

This is essentially orbit stabiliser

Any G-set decomposes as the disjoint union of the orbits

[somber badge]

yea

[south patrol]

So you can reduce to the case where G acts transitively on the set

But then in that case it's of the form Orb(x) = G/Stab(x) for any x

[somber badge]

oh ic ic

thank you!

[somber badge]

what does it mean for the map to factor?

[rocky cloak]

It means that it's the composition of two maps

[somber badge]

oh i see thank you!

[chilly ocean]

that works for A5

[crystal vale]

i have to think about the example of countable group which has uncountable subgroup, any hint?

[chilly ocean]

yes, power set is group under symmetric operation

[crystal vale]

i see

[chilly ocean]

you can make one now

[crystal vale]

but how power set is countable?

infinite countable

[chilly ocean]

you can take all the subests of natural with finite cardinality

[crystal vale]

i see

[chilly ocean]

this set is countable

now it has uncountable many subgroups

[crystal vale]

yes

[chilly ocean]

now you can make subgroups containing singletons of the form {n}

[crystal vale]

yes i can because every non-identity element has order 2

[chilly ocean]

this shows there are atleast aleph 0

Subgroups

[crystal vale]

yes

[chilly ocean]

what about subgroups containg {x,y}

[crystal vale]

yes because {x,y} also has order 2

so total number of subgroups containing {x,y} at least | N \times N |, right?

[chilly ocean]

we can repeat this for arbitrary for large cardinality

[crystal vale]

yes

i don't know much about infinite cardinalities, so how they will implies there are uncountable subgroup?

[chilly ocean]

do you know basic operation on aleph 0

aleph 0*ALEPH 0=aleph 0

[crystal vale]

no

[crystal vale]

is it because N \times N countable?

[chilly ocean]

you can say but it is countable union of countable set is countable

[crystal vale]

okay

[chilly ocean]

you can easyliy now see that there is a bijextion from set of subgroups that we are identifying of the form {x,y,.....} to $\mathbb{N}\times\mathbb{N}\times......}$

[crystal vale]

yes

[cloud walrus]

Akhi Mishra(Riemman Slayer)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[chilly ocean]

the last product is infinite

for {x} you have aleph 0 subgroups

for {x,y} you have again aleph 0*aleph 0=aleph 0

so there is bijection to $\mathbb{N}\times\mathbb{N}\times......}$

[cloud walrus]

Akhi Mishra(Riemman Slayer)
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[chilly ocean]

so you have 2^(aleph_0) subgroups atleast

hence uncountable

[crystal vale]

i don't get this part

[chilly ocean]

thats the claim think about it

[crystal vale]

okay

so are you map subgroups {x} -> (x,0,0,...), {x,y} -> (x,y, 0,0,0,..), {x1,..,xN} -> ( x1,.., xN, 0, 0,..)?

yes this is not well - defined

[chilly ocean]

You can be more precise

[crystal vale]

ordered (x,y) maps to (x,y, 0,...)

[chilly ocean]

Yes

[crystal vale]

so we get injection

oh there are more subgroups

[chilly ocean]

You have to be more precise you map creats some problems

[crystal vale]

{ e, {x}, {x,y}, {y} }

[chilly ocean]

Yes but our 2 order subgroups are sufficient

[crystal vale]

but i don't see how it makes bijection with N\times N\times....

thank you, i will look at it

[velvet hull]

quick sanity check, is the arbitrary union of Galois extensions Galois? (assuming that the extensions are compatible of course)

[twilit creek]

Hey can someone explain what are the éléments of a gallois field F_q and F_q^m ?

[dim widget]

You mean the compositum?

The Union of fields inside a given algebraic closure is unlikely to be a field even

[velvet hull]

assuming it's a field, I guess is the implicit assumption

but for instance if you fix a prime p, the union over all finite fields of characteristic p (ignoring the compatibility issues) is isomorphic to the algebraic closure of F_p

[dim widget]

So then are you assuming the union is over subfields which are contained within each other?

[velvet hull]

not necessarily all contained within each other, but it induces a poset with a maximal element, yes

[dim widget]

So all fields are contained within some other field? Or you just mean that the union is contained within a field?

Anyway you have to ask a precise question to get a real answer

[velvet hull]

okay, let me rephrase.

[chilly ocean]

cyclic group

[velvet hull]

we have an arbitrary family of Galois extensions, all of which are embedded in a field - take it to be the closure or whatever.
is the union of the family of fields, assuming it is a well-defined subfield, Galois?

do tell if it's not precise enough

[dim widget]

Then the union is equal to the compositum, and indeed one can show that the compositum of galois extensions is galois with galois group a subgroup of the product of the various galois groups in the union.

But it’s good to ask questions without implicit assumptions.

[dim widget]

This follows because one can check by hand that it’s normal and separable because any element lies inside of one of the extensions which are known to be galois

[velvet hull]

what's a sketch of the proof?

is it just via the embedding into the direct product of the Galois groups

[dim widget]

Yes if a galois element is trivial on every extension in the union it is per definition trivial

And since all of the extensions are normal the galois group of the union preserves each of them

So the map g \to \prod_i g|_{F_i} is injective homomorphism of groups

[velvet hull]

yes, that tracks

[rocky cloak]

Consider any element, it is contained in one of the extensions in the union. Hence its minimal polynomial splits and is seperable

This is for a literal union though

[velvet hull]

i just needed the verification lol I am literally writing down what yall describing right now

[chilly ocean]

i applied for postgrad tag and getting undergrad wow, how it is given

[frigid shard]

bro is unworthy

[chilly ocean]

i have a degree but

[frigid shard]

im joking lol

[chilly ocean]

no i just want to know what is the crietria for it

[frigid shard]

perhaps u can try modmail

or just ask in #advanced-lounge many mods are active there

[chilly ocean]

i think still it is fine, will it change anything

[frigid shard]

idk iirc the form just asks for like textbooks/papers uve read recently

[frigid shard]

no

[chilly ocean]

okay

[frigid shard]

like the role doesn't really matter

[rocky cloak]

It grants you acces to the graduate lounge, not that its necessarily a very intersting place to be

[chilly ocean]

as i have completed algebra coursework, i will shift to other channels soon(alg top, adv analysis ) , will i miss anything else related if i dont have postgraduate roll

[rocky cloak]

no I think thats it

[thorn jay]

You get cool roles to tell the world what kinda math you're interested in

[chilly ocean]

how to get it

[thorn jay]

.. you apply for postgrad

[chilly ocean]

i did that but got something else

[thorn jay]

Then, idk what to tell you man

You can always try again ig

[chilly ocean]

i still have amazing group theory problems, for now i will be fine

i and @chilly ocean have some sets of nice problems which are still unsolved so we will post it here. for now undergrad is fine, as i will stay here only

[thorn jay]

You, won't get problem sets if you have the postgrad role

[chilly ocean]

i have own problems left to work out

you wait i will post after 24 hour

[thorn jay]

I just didn't see how it relates at all to the previous conversation

So I got kinda confused

[chilly ocean]

i mean i dont need go anywhere else for now , for problems

[thorn jay]

Yes, how does that relate to getting the postgrad role

[chilly ocean]

i mean even if i am not getting it i am surviving here

[amber burrow]

ok this is an assertion used in a proof of a bigger problem

in an abelian group G order pq, where p and q are primes, if every nonidentity element is of order p, the index of centralizer for every nonidentity element is q

(they contradict this later)

but why si that true

oh wait

for some element x of order p

Z(<x>) is a subgroup of G

p <= Z(<x>), since x commutes with powers of itself

if equals index is q

it cant be of order q, since <x> is a subgroup of Z(<x>)

and p cant divide q (both primes)

ok i got it

[chilly ocean]

abelian group of order pq is Z_pq

[amber burrow]

yeah the problem was to show that if |G| = pq distinct primes, p <=q, and p does not divide q-1, then G is abelian

[chilly ocean]

without sylow?\

[amber burrow]

yup

its like Lagrange + Class equation

if Z(G) = p,q, or pq, G/Z(G) is cyclic so abelian

so suppose Z(G) = 1

use what i just mentioned to get that theres an element of order q

so index p

let H = <x>

[rocky cloak]

Do you mean in a non-abelian group? Otherwise the centralizer is always just everything

[amber burrow]

oops i just shouldn't have specificed anything

the problem is to prove G is abelian

[elfin wraith]

One of those new fangled non-commutative Abelian groups

[rocky cloak]

Not every group of order pq is abelian though, but I guess you just have an extra condition

[chilly ocean]

non abelian shit, hanging out with two prime ladies

[amber burrow]

Yeah if p does not divide q-1

[amber burrow]

is this a joke? i thought commutative and abelian meant the same thing 😭

[elfin wraith]

It was a joke, they are the same

As jagr mentioned the centraliser isnt very interesting for abelian groups because everything commutes

[amber burrow]

yeah

i was framing the question badly

[chilly ocean]

thats the indian system of teaching grp theory

[rocky cloak]

I feel like this proof is trying very hard not to prove Cauchy's theorem

[amber burrow]

yeah in the exercises you use cauchy's theorem to show it is cyclic

i think they were kinda trying to show off techniques from the chapter

mainly that N_G(H)/C_G(H) is a subgroup of Aut(H)

[rocky cloak]

I guess that's pedagogical

[amber burrow]

had to google what that word meant 🥲

but yeah i guess so

[rocky cloak]

Welcome to English 802

[elfin wraith]

I just flicked through my group theory homeworks from last semester because I thought I vaguely remembered doing this problem and im realsing I have 0 recollection of doing any of those homeworks

Im sure that bodes well for the exam next week

[rocky cloak]

I really think it's bad to have too good a memory if you want to be a mathematician. You need a slightly bad memory because you need to forget the way you approached [a problem] the previous time because it's a bit like evolution, DNA.

-Andrew Wiles

[elfin wraith]

This was the only thing I remembered, because I was very salty about this being the only mark I lost on the entire homework when it literally was by construction

[chilly ocean]

jagr2808-

[rocky cloak]

Something poetic in me remembering this quote isn't there

[elfin wraith]

Its a nice quote though, what use are all these new tools and ideas we learn if we can't then return and apply them to motivating problems. Also just nice to see a problem which you struggled with before comes naturally now, even with the same tools

[chilly ocean]

same goes for problems of life

[glad osprey]

What was the construction?

[elfin wraith]

I mean I haven't looked at this in a while so potentially I am mistaken and there was something else to say, but thats the problem

I was mildly irritated at the time but I got 19/20 i wasn't losing sleep over it lol

[glad osprey]

Seems like you wrote pretty explicitly that N_n < N_{n+1} follows because of the correspondence theorem

[harsh citrus]

Was reading Milne's Galois theory notes. Have a doubt as to why $$\frac{Gal(E_2 /F)}{Gal (E_2 / E_1 \cap E_2)} \cong Gal(E_1 \cap E_2 /F)$$
holds

[cloud walrus]

lajok000_92503

[chilly ocean]

as it says fundamental theorem

[harsh citrus]

Yeah fundamental theorem of Galois theory but I don't see t

It

@chilly ocean I don't see it

[chilly ocean]

are you sure its saying fundamental theorem of galois theory

for me it looks like first isomorphism theorem of groups

[harsh citrus]

Yes

[chilly ocean]

the by (d) check it

[harsh citrus]

Got it
Cam you explain the next line of counting as well @chilly ocean

[harsh citrus]

Each sigma_1 in Gal(E1/F) restricted to E1 cap E2...

[elfin wraith]

That was my feeling lol

[chilly ocean]

Each sigma_1 in Gal(E1/F) restricted to E1 cap E2... then you want sigma 2 such that sigma_1 = sigma 2 on intersection of E1 and E2 that is sigma 1^{-1} sigma _2 is identity on E1 cap E2

[harsh citrus]

Yes
But I don't see why why said that the extensions of Sigma 1 restricted to E1 intersection E2 gas exactly [E2: E1 intersection E2] Extensions to an element of Gal(E2/F)

[chilly ocean]

but we know there are [E_2: E_1 cap E_2] of them which are trivial on E_1 cap E_2 and element of Gal(E2/F) so sigma 1^{-1} sigma _2 must be choosen in [E_2: E_1 cap E_2] ways

[rocky cloak]

In general [E:K] = |Gal(E/K)|

[harsh citrus]

Ok
I got it
Thanks guys
I hope I do good tomorrow

[chilly ocean]

thats what the line is saying just before the computations of degree extensions

[harsh citrus]

Thanks bro

Man why do I take up algebra when I suck at it.

[chilly ocean]

all the best bro, please share something intresting , from ur ppr

[harsh citrus]

I will
My Prof always puts interesting problems which is why I am stressed
In 15 hours from now I will post

Gn. Will sleep 4 hours then grind again

I am genuinely feeling tensed. I had studied before but forgot everything due to lack of revising

Gn

[hard hearth]

is it because they dont share the same identity?

[rocky cloak]

Yes

[hard hearth]

if i define subrings differently so that Rx{0} becomes a subring always, what problems arise ?

say i remove the condition that subring needs to have same identity

it still feels reasonable to me

subrings share identity also feels reasonable.. hmm not sure

[elfin wraith]

There are differing conventions afaik

[thorn jay]

what you then do is define a "rng" which does not require a multiplicative identity at all

[hard hearth]

a funny example, if we take nxn matrices, then we can say that the subset of matrices with identical entries everywhere is a subring if we use this alternate definition (here multiplicative identity will be the matrix with 1/n entry everywhere)

[thorn jay]

So R x {0} is a subrng of R x R, but not a subring

[hard hearth]

but that is not what i am saying

i am saying they can have different identities

[thorn jay]

they do

what you're defining is a subrng which is a ring

[hard hearth]

i am not exactly defining a subrng, since i require identity, just not the same identity. but you said subrng does not ask for identity at all. so its between subrng and subring?

[thorn jay]

"subrng which is a ring"

[hard hearth]

yes

[rocky cloak]

Then the intersection of subrings need not be a subring for example

[elfin wraith]

You can take like {0,3} inside Z/6Z

[hard hearth]

ohh i see we want this propert

makes sense

sure thank

[elfin wraith]

I could also be mistaken but I don’t really think subrings are that interesting, at least not in the same way we care about subgroups in group theory

Modules are the more interesting objects when you look at rings

[stiff gull]

looking for a tutor to help me with my final exam review

[hard hearth]

so in this example, under my definition, the intesection of this subring and the subring of multiples of identity will be empty, which is not a ring

oh no wait it will be trivial ring

[rocky cloak]

Consider for example ZxZ with the "subrings"
R1 = Zx0
R2 = {(a, b) : a = b mod 2}

The intersection is 2Zx0

[hard hearth]

R2 is 2Zx2Z U (2Z+1)x(2Z+1), yeah? their intersection is as you describe, but 2Zx0 is not a "subring" because no multiplicative identity

cool

[thorn jay]

groups are werid

[hard hearth]

makes sense why subrngs dont ask for 1 (no 1 here)

[thorn jay]

weird*

the fact you can embed the lattice of normal subgroups into that of regular subgroups leads to all sorts of strange constructions like subnormal series which you simply dont see anywhere else in algebra (modules dont count, theyre groups)

[hard hearth]

whats a regular subgroup?

[thorn jay]

just a subgroup

[hard hearth]

oh xd

[white oxide]

Is $\sqrt{(x, y)} \subset k[x, y]$ prime for the following reason? If $fg \in \sqrt{(x, y)}$, then $(fg)^r \in (x, y)$ for some $r \in \mathbb{Z}^+$. Suppose that neither $f$ nor $g$ is in $\sqrt{(x, y)}$. Then $f$ and $g$ must be constant, which contradicts the fact that $(fg)^r \in (x, y)$.

[cloud walrus]

okeyokay

[thorn jay]

(x, y) is itself prime, so the radical of (x, y) is simply (x, y), so prime :)

[white oxide]

Ah yeah makes sense

Thanks

[thorn jay]

( (x, y) is prime as k[x, y]/(x, y) is isomorphic to k)

[white oxide]

Oh ya

[amber burrow]

this is probably the wrong channel (because i don't know what this really is), but where/when does one typically learn "Sylvester's Criterion"

smth about positive definiteness

[elfin wraith]

I’m not aware of it but positive definiteness would lead me to believe linear algebra

Knowing the content of the statement should probably give you a hint

[amber burrow]

yeah i dont know why i didnt look it up- pretty obviously linear algebra since it uses matrices

but do you know when one learns what positive definiteness is?

i saw it used as the solution to some problem, but ive taken a linear algebra class and never heard of it

[elfin wraith]

I learned about it in my intro linear algebra class, yours possibly just didn’t bring it up

It’s just kinda a general idea that appears a lot though

[amber burrow]

really, an intro class?

i mean my class didn't even define self-adjoint (which i think is the same as hermetian?) until the last month or so of class, and that was basically only to prove the spectral theorem