groups-rings-fields

[south patrol]

Yup

But yeah uh this is not irreducible I think

Because you have an eigenvector

However, it is indecomposable

[chilly ocean]

oh i thought irreducible meant not a direct sum

[south patrol]

I think

[chilly ocean]

I confused irreducible with indecomposable

[south patrol]

Dw

[chilly ocean]

I'm guessing over char 2, C_2 has no irreducible non-trivial reps

[delicate orchid]

correct

[chilly ocean]

does it have other indecomposable ones other than this one and the trivial one?

[delicate orchid]

it has 2 indecomposables though

oh you've already got it

yes that's right

[south patrol]

gg

Oh yeah isn't this that like

[delicate orchid]

society if F_p contained pth roots of unity

[south patrol]

p-groups have only one (f.d.?) irrep over F_p

or is that false lol

[hidden wind]

mjau

[delicate orchid]

no that's true, a simple eigenvalue argument works

oh wait uh p-groups in general? not sure

[south patrol]

Maybe abelian p-groups

[hidden wind]

oop sorry wrong chat

[south patrol]

But that isn't much fo a generalisation then lol

[delicate orchid]

ok no it has to be true

[south patrol]

Is it true by like

[delicate orchid]

over a splitting field there's only one irreducible brauer character

[south patrol]

actions of p-groups having fixed points ish

Oh

Based

Sure

Oh yeah we can just work over an alg closed char p field and prove it there

[south patrol]

Idk what those are 😭

[delicate orchid]

the indecomposables are more interesting anyway

[south patrol]

When you help someone else and then get schooled...

always a bigger fish

[delicate orchid]

something something cde triangle...

something something p-modular system...

[south patrol]

Bobby Fusions

[delicate orchid]

no they're actually quite simple. You restrict to the p'-elements of your groups and pretend your roots of unity in ur alg closed char p field exist in C

then the brauer characters are the class functions from the p'-elements to these roots of unity such that the preimage in the char p field do geniunely corrispond to trace functions of irred. reps over ur field

[chilly ocean]

Are indecomposable decompositions analogous to abelian group extensions like irreducible decompositions are analogous to direct sums? So irreducibles are much simpler but you can't build as much stuff using just them

[south patrol]

Lol

[delicate orchid]

I'm not sure I follow this, indecomposable means you can't write it as a direct sum so the analogy would be split abelian group extenstions

and as you say irreducibles corrispond to simple modules and unless your group algebra is semisimple you can't build stuff out of them

[chilly ocean]

if the group algebra is semisimple then every rep is a direct sum of irreducibles?

what does it mean for it to be semisimple?

[delicate orchid]

exactly what you said

[chilly ocean]

lol. are there any nicer equivalent conditions or important special cases that imply it?

[delicate orchid]

for group algebras if the characteristic of the field doesn't divide the order of the group your algebra is semisimple

but you probably knew that one already

[chilly ocean]

can it divide it and still be semisimple?

[delicate orchid]

no

[delicate orchid]

it's equivalent to having zero jacobson and uhhhhhhhh being...

it's either noetherian or artinian

I'll say artinian because then it's definitely true

[chilly ocean]

i see

anyway, over any field, C_2 only has 2 indecomposable reps then

[delicate orchid]

yur

[chilly ocean]

this is too few reps to me, I want more! What if instead of over a field, we consider reps over a ring? Is there a well defined notion of representations on modules? Will there be rings where C_2 has more indecomposable reps?

[delicate orchid]

oh god you're actually going to make me think about Z/4Z[C_2]

[chilly ocean]

I was thinking about that

[delicate orchid]

it definitely has at least 2

there's 2 one dims

uhhh

is
(1 2)
(0,1)
still indecomposable?

[chilly ocean]

I hope so

[delicate orchid]

C_2 = <g>
v in Z/4Z^2 (ewwwwww)
g.v = g.(v_1, v_2) = (v_1+2v_2, v_2)

yeah I think it's indecomposable
<(0, 1)> isn't fixed, good enough for me

[chilly ocean]

i think it is decomposable

nvm

[delicate orchid]

lmk if you figure it out lol

[south patrol]

hm oka i will look this up further

[delicate orchid]

Navarro chapter 2 has everything you'll need

for now...

[south patrol]

Are you sying that to me or aaaaaaaaa

[delicate orchid]

u

[south patrol]

Okay thank

It's cool how like

modular rep theory turns up in topology / homotopy theory apparently

[chilly ocean]

Omg really?? how

oh homotopy

algebraic topology doesn't count as topology to me lol

[south patrol]

i'm honestly not completely sure, but representation theory comes up because of actions of symmetric groups being kinda inevitable

Oh lol bruh, do you have like pointset topology in mind

[chilly ocean]

yeah, I love pointset topology (don't judge!)

algebraic topology is cool too tho

[south patrol]

Yeah fair lol

[delicate orchid]

pointset topology is combinatorics not topology

[delicate orchid]

oh yeah it does but I don't remember why

do you still have the vector bundle - rep corrispondence over BG? questions questions

Benson 2 probably has a chapter on this

[austere prairie]

yo, anyone know what the K is?

[chilly ocean]

Wha- how? I feel like that about algebraic topology is the one who is like combinatorics since it uses simplexes and stuff like that

[delicate orchid]

a field

[austere prairie]

oh, wait is it just any arbitrary field

[delicate orchid]

ur literally looking at sets of sets how is that not combinatorial lol

[chilly ocean]

Depending on the context either an arbitrary field or either the real or complex numbers

[delicate orchid]

and the simplices are just in the background, you almost always blackbox them and just work with generalised homology theories or whatever

[chilly ocean]

Isn't that all of math lol

[delicate orchid]

but it wasn't a negative towards pointset, I love the combinatorial aspects

[delicate orchid]

yes but you're actually giving a fuck about them in pointset

[celest furnace]

Consider yourself judged.

[delicate orchid]

it's very "on the metal" to borrow a term from CompEng

[chilly ocean]

explodes

[celest furnace]

Lol

[south patrol]

isn't point set topology logic

[celest furnace]

I was thinking more like show that open connected subsets of R are path connected

Or show that Q is not a countable intersection of open intervals of R (?)

[tough raven]

So many interesting takes on point-set topology coming up here

[delicate orchid]

in an ideal world we wouldn't have to teach it ever but unfortunately we need to be """rigorous"""

[celest furnace]

This guy gets it

[acoustic turtle]

Can somebody check these two abstract algebra questions on ring modules?

[stark helm]

Given this condition and if I want to prove Sn is not solvable, is it true to argue that because An must a normal subgroup that contain all three cycles(even permutation), so there must exist H as normal subgroup that contain all three cycles, then all K as normal subgroup of H such that H/K abelian must contain all three cycles, which contradicts the sequence that e<G1<..<Sn

[crystal vale]

I take mapping from G-> G/M × G/N such that g->(gM, gN) and I think its kernel is M intersection N but I want to know where I am using G=MN condition?

[mighty kiln]

Surjectivity

[stark helm]

Is it correct to argue that since An is generated by three cycles, then if a group contains all three cycles, then the group is either An or Sn?

[tough raven]

For the first part:

I'm not sure what it means for L_α to be associative.
To be clear, since it's asked to show that L_α is a left R-module, you need to define addition and scalar multiplication R × L_α -> L_α operations on L_α and show that under addition, it is an abelian group and that the axioms of a left R-module are satisfied.

As I think you have realised, both of these maps are supposed to be the same as the ones on R, restricted to L_α, i.e., a + b in L_α is the same as a + b in R (you need to show that this lies in L_α), and for r in R, s in L_a, r . s is defined to be rs (you need to show that this lies in L_α and not just in R).
All that's left after that is to verify that L_α under addition is an abelian group and the scalar multiplication satisfies the axioms of a left R-module.

[tough raven]

For the second part, the statement is not true for any subgroup L of M. Try to figure out what condition L needs to satisfy for it to be true.

[crystal vale]

Thank you, if i am correct MN defined as finite sum of m_i•n_i, right?

[tough raven]

These computations seem to have ideas relevant to the solution but are conceptually off: it is not true that 1.a = 1 . aα, or r.(a+b) = r(aα+bα), or (rs) . a = rs(0) in the first part, or that (rsf, rsg) rsf - rsg, for example.

[tough raven]

I'm not sure if I've understood you correctly, but yes, you can use this to show that S_n is not solvable (for n >= 5).

Take H to be the subgroup of S_n generated by all 3-cycles (IDK if you know that this is A_n for n >= 5), which is non-trivial.

Then this statement shows that the commutator subgroup of H is H (by taking K = commutator subgroup), which shows that H is not solvable.
Alternatively, if you have a subnormal series for H with abelian quotients, you can take K to be the penultimate subgroup (just before H) to get that K = H, a contradiction.

Since subgroups of solvable groups are solvable, if S_n were solvable, H would have to be.

(Incidentally, if S_n were solvable for arbitrarily large n, then every finite group would be solvable, being embeddable in all sufficiently large S_n.)

[stark helm]

Actually I don't know any stuffs about commutator subgroup or subnormal series, probably the stuff I can see is that we never get an identity

[crystal vale]

No I think MN is not defined as finite sum

[stark helm]

if the sequence require e<G1<...Sn, I never can get e here

[tough raven]

Then … what's your definition of solvable?

[crystal vale]

For any (sM, tN) what will be my g?

[tough raven]

That's what I'm calling a subnormal series (the sequence) with abelian quotients (because G_i/G_{i-1} is abelian).

[tough raven]

No, that's for subgroups of a ring usually.

MN := {mn | m in M, n in N}. This is a subgroup of G if M, N are subgroups, one of which is normal.

[stark helm]

Alternatively, if you have a subnormal series for H with abelian quotients, you can take K to be the penultimate subgroup (just before H) to get that K = H, a contradiction.
You mean take K<H and get K=H, where k has already contain all three cycles and H=An, then there should no be any group that contain all three cycles, so K does not exist?

[tough raven]

Well, you need gM = sM <=> g = sm for some m in M and gN = tN <=> g = tn. Thus sm = tn.
Conversely, if sm = tn for some m in M, n in N, then you can just take g = sm = tn.

So when can you find m in M, n in N such that sm = tn?

[tough raven]

I mean take K = G_{n-1}.
Then K/H is abelian, so K = H, which is a contradiction (first remove repetitions from your subnormal group if necessary so that G_0, …, G_n are all distinct).

[crystal vale]

Yes

[stark helm]

OK, so if K=G n-1, and H<K, (since H contain all three cycles which means An), then H=K

[crystal vale]

I get the answer online but I am not sure about when I will find m in M , n in N such that sm=tn? But here where we used G=MN so I think there must be a use of it

[tough raven]

Correct; there is.

[boreal inlet]

Uhm... let's say I have a field of characteristic zero, and L/K is a galois extension with Klien 4 as the galois group. How do I show L = K(√d1, √d2)?

[tough raven]

H < K because H is generated by the 3-cycles and K contains all the 3-cycles.

Do you see why K = H is a contradiction?

[stark helm]

because they should be An?

[tough raven]

No.

[tough raven]

In characteristic 0 (or not 2), any quadratic extension is obtained by adjoining a square root (the quadratic formula shows that a root of any irreducible quadratic polynomial can be expressed in terms of a square root).
Can you think of a way to use this?

[stark helm]

can you explain it?

[tough raven]

How did we choose K?

[stark helm]

by saying containing all three cycles?

[tough raven]

No.

We defined H to be the subgroup generated by the 3-cycles.

We showed that K contained all 3-cycles using the fact that you showed a screenshot of in your first post.

But how did we choose K?

[boreal inlet]

I do know the fact that for a field of char 0, K(a, b)/K has galois group of form Gal(K(a)/K) x Gal(K(b)/K).

So it boils down to find elements a and b which are not perfect squares because each of them will generate a extension of degree 2.

[crystal vale]

So how do we proceed?

[boreal inlet]

Problem is, I don't see how to

[stark helm]

in that case K<H, I only know K is normal subgroup of H, and H/k abelian implies all xyx^-1y^-1 is in intersection of subgroups of H?

[tough raven]

I'm not so sure about that.
I think the correct statement is that if L1, L2 are Galois extensions of K whose intersection is K, then the composite L1L2 is Galois over K and Gal(L1L2/K) is the product of Gal(L1/K) and Gal(L2/K).
So if K(a) and K(b) are both Galois over K and their intersection is K, then what you wrote is correct.

[tough raven]

OK, the second paragraph is the right approach though.

[tough raven]

Do you know about the Galois correspondence?

[tough raven]

I'm not sure what intersection you are talking about, but my question is much simpler. What was the definition of K?

[stark helm]

I actually have no idea

[tough raven]

You want to use that G = MN. So try to rewrite the desired condition sm = tn as something = product of something in m and something in n.

[crystal vale]

Okay thank you

[tough raven]

Stay tuned for next week's episode Read above and find out then.

[stark helm]

the intersection should be this

[cobalt heath]

Where did you bring K from

[boreal inlet]

Yes. What i see is, I can have an intermediate extension L/E such that the Galois group is the cyclic group

[stark helm]

I feel pretty confused if it means intersection of all the subgroups of G has every elements of the form xyx^-1y^-1. Or we just pick all subgroup that contains the form of xyx^-1y^-1?

[tough raven]

That doesn't matter for this.

[boreal inlet]

Then by tower formula E/K also has cyclic group of order 2 as the Galois group

[tough raven]

Seriously, you just need to recall how K was defined.

[stark helm]

K is a normal subgroup of H such that H/K is abelian, and it contain all three cycles?

[tough raven]

Yes, but how did I pick the K to apply this result to, to prove that S_n is not solvable?

[tough raven]

So you can find quadratic extensions of K if you can find subgroups of Gal(L/K) of index 2. And it's easy to find all of those.

[stark helm]

K=An or Sn?

[tough raven]

No.

[stark helm]

Will H=An?

[tough raven]

.

[tough raven]

Yes.

[stark helm]

I actually feel wondered that if H is generated by three cycles, isn't it represent that it contains all three cycles?

and also An contain all three cycles if H=An

[stark helm]

why K/H is abelian

[tough raven]

Because by assumption, the series has abelian quotients and H/K = G_n/G_{n-1}.

[crystal vale]

I want to prove that If G is an abelian simple group then G is isomorphic to Z_p for some p.

So it is given that G is an abelian simple group thus |G|>1 , hence there exists an element x, such that x ≠ 1 , and G is abelian , <x> is a normal subgroup but G is a simple group, hence <x>= G.

Now i will prove that G is finite, so if G is not finite that means x has infinite order , and <x>=<x^n> for all positive integers n , so x= x^(ns) then x^(ns-1)=1 it shows that x has finite order, so G is finite.
Let |x|=n
Now let G={1,x,.......,x^(n-1)} since |x^k|= n/gcd(n,k).

But for all <x^k>= G , when k≠1.
So for k≠1 , x^k has order n , it implies that gcd(n,k)=1 for all k<n, hence n is prime so it is isomorphic to Z_p.

Is it correct?

[boreal inlet]

Yess, I just did. I found an intermediate subfield which is literally just K(√d1). Then rest holds

Because L = E(√d2) and E = K(√d1), so, L = K(√d1, √d2)

[stark helm]

so we have H/K abelian and H=G_n, but how is it related that H/K abelian with K/H abelian?

[tough raven]

Where did K/H come from?

[stark helm]

come from three cycles?

[tough raven]

K/H doesn't matter here at all.

We have H/K abelian so your screenshot tells us that K contains all 3-cycles, so K = H.

[stark helm]

Then H=An and it means H is the smallest group containing all three cycles, is that true?

[tough raven]

H was defined as the smallest group containing all 3-cycles, so yes.

[stark helm]

Then one more question, since sequence of subnormal series never reaches e because all K=H, then it contradicts right?

[tough raven]

Yes.

[stark helm]

Got it, really appreciate your help

[crystal vale]

There is a question in which they want to prove that for a finite abelian group has a subgroup of order n for each positive divisor n of its order and use Cauchy theorem.

I thought if I show that there is always an element of order n then it will be proof, but I do not figure out how to show this.

[boreal inlet]

Isn't the first statement just Lagrange's Theorem

Or maybe I'm missing something

[crystal vale]

No

[dull marsh]

Its converse

[crystal vale]

Yes

Which book is good for the composition series concept?

[solemn garden]

This exercise requires Choice, right? Since you’d need to prove that every element of R - M is outside every proper ideal, and so you have to assume choice to say that any ideal is contained inside of a maximal ideal

[tough raven]

Not for the second part: any proper ideal must consist of nonunits and hence be contained within M.

[solemn garden]

I see

[tough raven]

For the first part, if r in R - M, then Rr + M => rs in 1 + M. If you can show that 1 + M consists of units then r is a unit.

Not sure if that can be done without AoC though.

[solemn garden]

Is it ok to just use choice if I see a method that uses choice first

[tough raven]

Yes.

(almost always?) you don't need to worry about whether Choice is necessary unless you want to.

[solemn garden]

Ok, thanks

[tough raven]

Let G be a group with finite exponent e (i.e., e is the smallest positive integer such that g^e = 1 for all g in G). Must G have an element of order e?

Note: this is easy to show for abelian groups, so one might as well focus on the non-abelian case.

[crystal vale]

In S(7) what will be the exponent?

[terse crystal]

Yes. {ord(x): x from G} is a finite set, whose elements are factors of e. Exponent is clearly the largest element of this set

[crystal vale]

In S(7) what will be the exponent? I think it is not true

[terse crystal]

You mean symmetry group S_7, permutations of 7 elements?

[crystal vale]

Yes

[terse crystal]

Oh my bad

[crystal vale]

e is LCM of elements of that set

[terse crystal]

Exponent is LCM of all orders

[crystal vale]

Yes

[terse crystal]

Yeah I am dumb

[crystal vale]

S(7) may be work

Because the exponent will be 84

[terse crystal]

S_3 too

[crystal vale]

Yeah

[terse crystal]

6

[rocky cloak]

In general it can be very hard to do algebra without choice, so I would recommend not ditching axioms.

[tough raven]

Oh, any non-cyclic group of squarefree order will work then 🤦 (or in general, I think a finite group has exponent equal to its order iff its Sylow subgroups are cyclic and such a group need not be cyclic/abelian/nilpotent).

[crystal vale]

I want to prove that if all the composition factors of G are of prime order then G is solvable, so all composition factors prime order means they are Abelian so it implies G is solvable, right?

[delicate orchid]

420, u forgot about the 5-cycles

[crystal vale]

But for Converse, if G is solvable then all composition factors of G are of prime order.
Let chain of subgroups
1=G_0 <.........<G_s=G so if I let this chain be the composition series the G(i+1)/G(i) is simple group and from G solvable it is abelian so we know that if K is abelian and simple group it is isomorphic to Z_p, so its composition factors will be cyclic, but this is not correct, please correct me

[crystal vale]

I thought I missed something, thank you

[delicate orchid]

The composition quotients don’t need to be abelian?

[crystal vale]

But I want to show that composition factors are of prime order so it is a simple group so if I have them are Abelian then they will isomorphic to Z_p

[delicate orchid]

Right so you’re doing a refinement thing

[crystal vale]

Refinement?

[delicate orchid]

So you start off with just a subnormal series and then keep inserting factors until you reach one of maximal length

[crystal vale]

I just picked a chain of solvable G then I assume that it is a composition series hence all factors are simple group

[tough raven]

I assume you meant to say simple?

[tough raven]

If I am understanding your definitions correctly then
solvable <=> there is a subnormal series with abelian quotients
and
all finite groups have composition series (subnormal series with simple quotients)
but you don't know that there is a single subnormal series with both properties.

You can show that there is one by doing as @ wewladstbh#0 suggesed and refining a subnormal series with abelian quotients by inserting factors until it has simple quotients.

[delicate orchid]

they definitely have to be simple, that's part of the definition right? We're showing they have to abelian so Notknow's proof seemed circular to me

in a general subnormal series they don't need to be simple though, that is correct

[tough raven]

Ah.

Well, in the end, I think we were saying the same thing.

[delicate orchid]

yur

[south patrol]

Nice

[crystal vale]

I don't know how to insert factor

[delicate orchid]

if your subnormal series isn't a composition series then somewhere there will be a H_i that's not a maximal normal subgroup of H_i+1, so you can put the maximal normal subgroup in between them and form a longer series

[delicate orchid]

I think I've recalled the standard proof of this though, you use induction on |G| and the fact that subgroups and quotients of solvable groups are solvable, assume your result holds for all H smaller than G and then do this

[crystal vale]

You use the fact that if H_i+1/H_i is not a simple group then H_i is not the maximal normal subgroup in H_i+1, right?

[delicate orchid]

yus

[crystal vale]

I want to prove that if T is a linear map from V to V then if T^m is injective then T is injective, where m is a positive integer so Iet T is not injective that means there exists u≠0 such that T(u)=0 so T^m(u)=0 , hence T^m is not injective, is it correct or I missing something?

[tawny pine]

its fine but we can do it without contradiction

[tough raven]

Eh, it's fine.

This is very benign as uses of contradiction go.

[boreal inlet]

Actually I'm curious, what are the other approaches?

[rocky cloak]

Definition of injective: T(x) = T(y) implies x=y.

Assume T(x) = T(y), then T^m(x) = T^m(y), which implies x=y

[tawny pine]

direct, the wording is just the same without the framing of contradiction

[delicate orchid]

contrapositive alsobeit

[crystal vale]

It's not contradiction

[rocky cloak]

Proving the contrapositive is implicitly a proof by contradiction I guess

[delicate orchid]

proof by truth table

[rocky cloak]

Like what you end up proving is that if
T^m is injective then T is not not injective. Then the final step that not not injective implies injective is a proof by contradiction.

[tawny pine]

doesnt matter if you call it contradiction or contrapositive. theyre essentially the same as jagr said

on top of that they even gave a proof that simply uses the def of injective

[rocky cloak]

Personally I'm fine with assuming LEM whenever, but I guess it's nice to avoid when not needed

[tawny pine]

based on this we see T need not be linear (nor V be a vector space). the claim holds for any function from a set to itself @crystal vale

[delicate orchid]

composition of monomorphisms is mono

[tawny pine]

in this case its more about fg mono implies g mono

[delicate orchid]

what's that silly theorem called

two-out-of-six, that's the one

[crystal vale]

Okay, thank you

[formal ermine]

why?

why am i being nerd emoji'd

[rocky cloak]

I'm not so concerned with the intricacies of what implies what, so I'm fine with just assuming whatever allows me to easily do cool math.

[south patrol]

To be fair there is a nice point I saw online about how proofs by contraposition are often nicer in terms of what you get out

Like, if you do a proof by contraposition then everything you prove in the course of your proof can be an interesting result in its own right

Whereas with a contradiction you can't take stuff out of it in the same way since you assumed nonsense

[delicate orchid]

yur contrapositive is nicer

[hidden haven]

Sometimes people assume ¬A but then prove A without using that assumption so might as well remove the assumption entirely

[fading field]

Honestly, if I have some statement that I think is true, but don't know how to prove it, I literally just go by contradiction and then if I don't use the hypothesis then I will just rewrite the proof

[hidden wind]

if i have some statement i think is true, i sometimes just don't bother proving it, MWAHAHA

[crystal turtle]

Also, if you have use contradiction then whatever you show in your proof is just nonsense and has to be thrown out. At least with contrapositive you can have meaningful steps along the way

[south patrol]

isn't that what I said

Jk i am not offended by agreement

This is also a stack exchange answer right aha

[stark helm]

I am wondering if there is a field A with four elements and a field B with eight elements, and i argue that because A/{0]=A*( multiplicative group) has three elements, B/{0}=B* multiplicative group has seven elements, then A* is not subgroup of B* by lagrange theorem, and therefore saying A is impossible to be the subfield of B, am I correct?

[south patrol]

Well

There is an assumption you have made there

That A is a subfield of B

[stark helm]

I am considering if A is possible to be isomorphic to the subfield of B

[delicate orchid]

It certainly seems like you’ve proven that it isn’t

There’s also the linear algebra argument

[stark helm]

I want to prove that a field with four elements can not be a subfield of a field with eight elements, my proof is valid right?

[coral spindle]

I don't think you are confused about the concept of proof by contradiction

Your proof is fine

[delicate orchid]

Actually this has got me thinking

[coral spindle]

If I were you (Wew) rn I would say something like "wholesome epic chungus tower law" idk

[delicate orchid]

Taking unit groups is a adjoint functor

So it should preserve inclusions I think

Unless I’ve got which adjoint it is backways

Anyway

[coral spindle]

It's uhhhhhhhhhhh

It's gonna be right adjoint to the group algebra functor

[delicate orchid]

Yur

[south patrol]

You need slightly more justification

[south patrol]

yes

Actually no I think I was being pedantic

Lol

[coral spindle]

Group algebras over F_(char B) too :3

How cute :3

[delicate orchid]

I just needed the word “right” tbh. Now is the tensor product right or left

[south patrol]

But I would say that if A is a subfield of B then A^x is a subgroup of B^x

[delicate orchid]

It’s right. Both “tensor” and “right” are red words so it has to be right

[south patrol]

That is obvious tbf

[coral spindle]

Tensor is left adjoint to hom. Think currying!

[south patrol]

Tensor is a left adjoint

preserves surjections

[delicate orchid]

I hate this fucking bullshit

[delicate orchid]

YEAH SO ITS RIGHT EXACT

[south patrol]

yeah

[delicate orchid]

SO WHY IS IT ON THE LEFT

[coral spindle]

Hom(a x b, c) <-> Hom(a, Hom(b, c)) ergo tensor left, hom right epic chungus

[south patrol]

left adjoints are right exact

[coral spindle]

see bc it's on the left ya dingus

[south patrol]

maths is annoying

it's like how uh

[delicate orchid]

This is like the electron having negative charge

[south patrol]

left derived functors are right kan extensions

[delicate orchid]

Or the fuckin semidirect product symbol

[south patrol]

Lol

[delicate orchid]

Who knows which way round it goes just flip a coin

[south patrol]

With semidirect product i learnt a trick

it is like the normal subgroup symbol with extra decoration

[delicate orchid]

Yes I know the normal subgroup thing

[south patrol]

ye

sorry

[delicate orchid]

Stop MANSPLAINING TO ME

I might cry

Anyway

I prefer the linear algebra argument

[coral spindle]

You mean the tower law argument NERD

[south patrol]

sorry

Why would you need the tower law

[delicate orchid]

If A is a subfield of B then B is a - SHUT UP BOYTJIE - a A-vector space and thus has to have order equal to a power of 4

[south patrol]

yes exactly

[coral spindle]

Oh yh why did I call it the tower law

[south patrol]

I think it's like

[coral spindle]

Oh no wait I remember why

[south patrol]

if you're doing it relative to the prime field

[coral spindle]

Prime subfield

Yes

[south patrol]

ye

[coral spindle]

I am simply too good for you bitches

[south patrol]

This was fun because like

[delicate orchid]

[B : A] = [B:A][A:A] upon

[south patrol]

when i was taught about ifnite fields the prof didn't do this proof and it was way longer

he was proving like

F_(p^m) embeds in F_(p^n) iff m|n

with what you said being one direction lol

[hidden wind]

the name semidirect product gives me nightmares

[south patrol]

Actually what he said was the other argument here in retrospect lol

like p^m -1 | p^n -1

implies m|n

i think

[coral spindle]

They are all equivalent

All roads lead to the tower law over the prime subfield

[delicate orchid]

Semidirect product of fusion systems

[delicate orchid]

Tbf you have to show that if A is a subfield of B then B is always a vector space

That’s not strictly obvious at first glance

[south patrol]

True, but like

if you know any field theory you know this right lol

[delicate orchid]

It might even be an A-algebra

[south patrol]

The sus chungus...

[delicate orchid]

I wonder if anyone has tried to study the group of automorphisms of B as an A-algebra

[south patrol]

No.

[delicate orchid]

Probably not :letrollface:

[coral spindle]

Probably some really sensible people who would never fight a duel

If anyone at all

[delicate orchid]

Tbf whenever I say Galois group thays what I mean

I don’t care about the extension being galois

I can barely even remember what that means. Just give me the automorphisms you pendants

I must do… cohomology

[white oxide]

why are we allowed to let G be abelian

[delicate orchid]

Iirc the logic is you’ll eventually hit G/[G,G] anyway so just start there

Wait G is a general finite group?

Meh it probably still works by the same logic but this is weird

[white oxide]

[G, G] are the commutators of G right

clearly is crazy

[south patrol]

I think the point is just: say you have some inclusion H < G in your tower, where G/H is abelian. Then you can just find a cyclic tower of G/H and pull back

[delicate orchid]

Stop saying “pull back” to the undergrads

But yeah I think that’s more to the point

[crystal turtle]

pullbacks uwu

[white oxide]

at this point i forgot all group theory

[delicate orchid]

Remember when authors say “clearly” they don’t mean it’s clear

They mean they can’t be bothered to write out the details

[delicate bloom]

that's the charitable assumption, really they're gaslighting you 😛

[south patrol]

Where the author says "clearly" is where the reader spends 70% of their time

[celest furnace]

Disagree. Easily 90

Lol

[solemn kettle]

$K[V]$ is the coordinate ring of $V$ . Why can it be embedded in $L$?, I know that the polynomial ring in $n$ variables can be embedded in $L$ but I don't know how the quotient is embedded.

[cloud walrus]

eduardo291299

[tough raven]

You might get better answers for this in #algebraic-geometry.
(I'm guessing that it might also help to specify what V and L are.)

[solemn kettle]

$V$ is a K-variety and $L$ a field extension of $K$

[cloud walrus]

eduardo291299

[winter shore]

Hatcher says that the abelianization of this group is g copies of Z. How could I show this?

[a,b] is the commutator here

[delicate orchid]

That’s the abelianisation of the free group on g generators, which is the free abelian group on g generators which is Z^g

[coral spindle]

Not quite Wew, e.g. there is no relation [b_1, a_2] present

[delicate orchid]

Ah true

But when you abelainise you just add those in

It’s also on 2g variables

[coral spindle]

pi_1(M_g) looks like Z^2 * Z^2 ... * Z^2 (g times) where * denotes the free product. So I would actually expect that the Abelianisation looks like Z^2g?

[delicate orchid]

Yeah

Just comparing presentations they’re identical up to relabelling

[winter shore]

oops you're right

[coral spindle]

Anyway as wew alludes to, you can just use the correspondence theorem together with the knowledge that the Abelianisation of the free group of n variables is Z^n.

[winter shore]

my problem is that when abelianizing, there's some stuff that has already been made go to 0, e.g. [ai,bi]. It seems like then the abelianization of Pi_1(M_g) could be smaller than Z^2g

[delicate orchid]

Slam the universal property triangles together until it works

[delicate orchid]

No. If it’s already zero then sending it to zero doesn’t change anything

It’s the set of relators not the multiset!

[winter shore]

hmmmm I'll think a bit more about that, thank you

[dull ginkgo]

me when your diagram is a simplicial complex

[delicate orchid]

(You can also see this by using one of the elementary tietze transforms)

[south patrol]

I would just say that any map pi_1(M_g) into an abelian group factors uniquely through Z^2g

Like hopefully that is clear

[delicate orchid]

Beat you to it

You’re so category brained it’s unreal

But yes that would work

[winter shore]

I guess my problem is: Is the abelianization of G=< A | R> the same as K=<A | R \cup {[x,y] | x,y in A}>? These don't seem to be immeadiately the same

the one on the left can be constructed as < A | R> / normal subgroup generated by the [x,y]; but it's not clear to me that the thing on the right is a presentation for this

[delicate orchid]

R is a subset of your second thing. This is just set theory

[south patrol]

Lol

[winter shore]

I feel this is so simple and obvious yet I don't get it lol, I'll just return to it later

ty

[olive granite]

How did they derive the circled congruence?

I tried expanding the binomial but I don't see why p^α divides the omitted terms

[cloud walrus]

point9006

point9006

[olive granite]

That "p>2^2" is unclear to me anyway, as the purpose is proving the thing for p>2.
\alpha >= 3 instead is needed because this is an inductive step and the base case was \alpha = 2

Or just p^(\alpha-k) divides binom(p^(\alpha-2), k) is enough, but still I don't understand how to derive it eventually

[dire siren]

@olive granite
#elementary-number-theory is the right channel
you actually want binom(p^(a-2), k) to be divisible by p^(a-k); to prove this, use p-adic valuation and Legendre identity
so we want to prove that v_p of that binom is at leat a-k
now, v_p of that binom is v_p((p^(a-2))!)-v_p(k!)-v_p((p^(a-2))!)

and we have v_p(n!) = (n-s_p(n))/(p-1), where s_p(n) is the sum of the digits when n is written in base p

[olive granite]

Oh, thank you. If you want we can move to that channel.
Is it the legendre identity? Why did they omit these steps, anyway, they don't seem that trivial to me, or is it a well known fact?

[dire siren]

I don't know why they skipped the steps. By any change, are there any similar calculations done before?

also, there is a chance I might have made an error

[olive granite]

I checked but no, it doesn't seem. I'll try to see what I can do with your suggestion

[dire siren]

@olive granite maybe forget about that approach; I think I have something that works better

[olive granite]

Oh ok

[dire siren]

Write the binomial as $\frac{p^{\alpha-2}(p^{\alpha-2}-1)(p^{\alpha-2}-2)\ldots(p^{\alpha-2}-k+1)}{k!}$

[cloud walrus]

Filip

[dire siren]

so we want to prove that the exponent of p in this thing (notation v_p(...)) is at least a-k

v_p(nominator) is at least a-2

and v_p(k!), by Legendre, is (k-s_p(k-1))/(p-1) and this is at most (k-1)/(p-1)

now v_p(fraction)=v_p(nominator)-v_p(denominator) >= a-2 - (k-1)/(p-1)

and a-2 - (k-1)/(p-1) >= a-k, q.e.d.

[olive granite]

Awesome! I think I understand. So v_p(k!) is at most that quantity because s_p(k) is at least 1 right?

[dire siren]

yes

[olive granite]

Thank you very much, that's a very nice identity

[dire siren]

you're welcome
btw, as a remark, for k=p, the exponent of p in that binom should be a-3

[dire siren]

it's not relevant for the solution, but it answers this question

[olive granite]

Totally right

[covert fjord]

Hey guys are there collections of galois theory problems?

[lapis latch]

I’m sure a textbook on Galois theory would lol

[delicate orchid]

pick an irreducible polynomial and work out the galois group, do this for 200 different polynomials

[rotund aurora]

Express cos(2pi/17) in radicals

[covert fjord]

Like problem book with different kinds of problems

[dull ginkgo]

Oh jesus fuck not this fucker again

[summer path]

inb4 there are 2 people asking how to compute galois groups for 3 years

[dull ginkgo]

This problem took me a long ass time to write out

[rotund aurora]

ok now do it for 65537

[dull ginkgo]

[rotund aurora]

reverse

[dull ginkgo]

Honestly the Gauss sum was the hardest part

[summer path]

why is there a role called "old abstract algebra"

as opposed to new abstract algebra?

[delicate orchid]

correct

the new roles are only avalible to grad+

[summer path]

where are the new roles

[delicate orchid]

https://discord.com/channels/268882317391429632/1204198027338711121

[rotund aurora]

First google search yields this https://link.springer.com/book/10.1007/978-3-319-72326-6

[covert fjord]

Thank

[rotund aurora]

but you can probably find most of the material in usual Galois theory books

a nice problem if you haven't seen it is to calculate the degree of Q(zeta_n)

[summer path]

waow big long list

[covert fjord]

I am learning it from notes of J.S Milne

[formal ermine]

He moved to L functions

[coral spindle]

Took the L

[formal ermine]

bro started computing galois groups before i even got into math

and only switched to computing L functions a couple of months ago

[clear fiber]

Hi, in this problem the book specifies that K is a subset of I. But it didnt seem like I needed that in the proof. Am I missing something?

[coral spindle]

And still has zero idea how to do it

[coral spindle]

Well the notation I/K is usually defined only to be when I is a superset of K. If you ignore that, you do actually get a similar result, but the ideal containing K you are actually looking at is I+K

[formal ermine]

i cant think of a non trivial way to define I/K if K is not a subset of I

[coral spindle]

Well it's immediately obvious.

It's just i + K for i in I

You get exactly the same quotient set in the end

[formal ermine]

nvm had a brainfart

yeah

[clear fiber]

@coral spindle Ok thank you that makes sense

[frank cosmos]

let K be of char=p>1. Why is x^p-t^p irred in K(t^p)[x]?

is it because the only root is (possible \pm) t which is not in K(t^p)?

im looking at a proof that K(t)/K(t^p) is a non-seperable extension

and the book says that x^p-t^p is irreducible over K(t^p)[x] but shares roots with its derivative in K(t) hence it is not seperable

but dont we have to show that the minimal polynomial is not seperable

and the minimal polynomial of t over K(t^p) is just x-t

which is coprime w/ 1

also, are the only field extensions of finite fields F_{p^m} \subset F_{p^n} where m<=n

[dire siren]

by Eisenstein, I think

[frank cosmos]

i see

[cobalt heath]

Ah so it is
x^p - s in K(s)[x]
Yea eisenstein (as always for irreducibility)

[frank cosmos]

also, why is K\subset L purely inseperable <-> every alpha \in L is a root of some polynomial x^(p^n)=a, n>0, p=char(K), a \in K

[cobalt heath]

Which direction are you having trouble with?

[wooden fulcrum]

can an action of a group ever be faithful on a set smaller than the group?

[coral spindle]

Yes

Think of nice groups which act on sets... you already know a very nice infinite family

[wooden fulcrum]

hmm I'm not sure, I don't know of many examples of groups acting on sets I don't know much of anything about actions yet

if I have an infinite group acting on a finite set then the function from the group to the endomorphisms can't be injective no?

[coral spindle]

You absolutely do know of an example of a group acting on sets. You know infinitely many examples.

Think for a bit of some of the first groups you learn about

[wooden fulcrum]

well I thought of adding or multiplication but there the target set can't be finite

[coral spindle]

What are some finite groups you know of.

[wooden fulcrum]

permutations, cyclic, matrices over cyclic with multiplication uhh

[coral spindle]

Tell me about permutations

What does the permutation group Sym(3) look like

And how does it relate to the set {1,2,3}

[wooden fulcrum]

is Sym S?

[coral spindle]

Yes

Symmetric group

[wooden fulcrum]

ah huh right you're right

xD

not that I was suspecting you weren't I mean

so the group can't be larger than the target set size factorial?

[coral spindle]

It cannot

This is because a group action of G on a set X is actually a group homomorphism G -> Sym(X)

[wooden fulcrum]

I initially got confused by this

[coral spindle]

Faithful actions are those for which the kernel of this map is 0

[wooden fulcrum]

just from the text given it does not seem to me that the second example is a counterexample of the first since both have cardinality 2^n I thought but idk I'm underslept

[wooden fulcrum]

thank you for the answer

[crude mason]

(a, b, c)(d, e, f) = (a + cd, b + e, cf).
does anyone know how to show this a binary operation in congruence classes of modulo 5 field

where c isnt 0

i dont even know how to show it commutes

[coral spindle]

Who says it commutes?

It's very unclear to me what your question is

[crude mason]

if its a binary operation it has to commute no?

[coral spindle]

No

[delicate orchid]

“A binary operation” is so vague it literally automatically satisfies it

[coral spindle]

A binary operation is just a function

So tadah, it's a binary operation

Elaborate on your question, there's context missing clearly.

[crude mason]

the question just says to show its a binary operation

where the group is (a,b,c) st a,b,c are in f5

[coral spindle]

Show the full question to us.

[crude mason]

nvm

i understand for some reason i was looking at defn of a ring

instead of binaery operation

this was the question if u care

[coral spindle]

c =/= 0
indeed it was nontrivial to show this precisely because you needed to specify the set being operated on more clearly.

A classic example of the xy problem

You should have shown the question from the start

[crude mason]

apologies

i did say c wasnt 0

[delicate orchid]

that makes it annoying though cause now I can't just immediately show everything by linear algebra

[crude mason]

how would you have shown it with lin alg if it was

bcus under the field it would be a vector space or something?

[coral spindle]

Wew is being facetious

[crude mason]

oh im not experienced enough for the memes ahahahaha

[delicate orchid]

Not really. You could write the right multiplication by a certain element as a matrix

[coral spindle]

Yes and this immediately shows everything

[cobalt heath]

Representation >.>

[coral spindle]

What representation

[cobalt heath]

Nevermind, likely I misread the joke

[crystal vale]

But if I and K are not related to each other and I want to find a set of cosets of I in K or K in I , then it makes sense or not?

[oblique shadow]

So do you prefer rings or groups first

[crystal vale]

Ring

[cobalt heath]

Morphisms

[celest furnace]

Groups

[still dew]

https://media.discordapp.net/attachments/867389710132707378/1208699411299958834/image.png?ex=65e43c0c&is=65d1c70c&hm=87bbeadf4f76296f270f195c5478f9ce91d563e5e9203ac49b526b161ebf81aa&

I am not able to find a counterexample of the second

For the first i think H intersection K \in {H,K} simply means one is a subset of the other so I was able to get through i guess

<S> here means subgroup generated by S

Ok nvm i just found an example using vector spaces over F_2 for the second

We consider ((0,0),(1,0)) ((0,0),(0,1)) ((0,0),(1,1)) right?

[quiet pelican]

Yeah, the second is just the non-cyclic group of order 4, and all its proper non-trivial subgroups (which is what you wrote)

[languid trellis]

The dihedral group D_n has order 2n because we can identify each rotation with a reflection. When we consider, say rotation by pi radians and reflection in y axis, these two function R^2 - > R^2 have exactly the same domain and image. As functions (subsets of R^2 \times R^2) they are equivalent. Why do we then say the dihedral group has order 2n and not order n?

[dull marsh]

Pretty sure rotation by pi degrees and reflection in y axis are not the same

(1, 1) goes to (1, -1) by the reflection and to (-1, -1) by the rotation

[languid trellis]

oh god i am braindead

my bad

[still dew]

Oh right!

[wooden fulcrum]

I think Sigma = {v, w} and G = Sigma^*, H = {v}^*, K = {w}^* and L = G would work also or anything where one of the subgroups contains the other 2 and the other 2 are disjoint

[wooden fulcrum]

oh this is a good way to put it nice

[astral fractal]

I'm a bit confused as to the geometric significance of conjugation is

Like if srs^-1 = r^-1 or something, what is this saying and does s being the "conjugator" mean anything

for dihedral groups

[delicate orchid]

a good example is the fact that to change basis in linear algebra you conjugate by a matrix

you can think of it as "going somewhere, doing something, then coming back" I guess

[delicate orchid]

and that we call conjugate matrices "similar"

[astral fractal]

yeah i thought it was similiar to the idea of similiar matrices

[delicate orchid]

continuing with this, diagonalisation writes one matrix as a conjugate of another, as does triangularisation

[astral fractal]

i guess geometrically i thought of it as you can restate a symmetry in terms of a composition

[delicate orchid]

but back to groups, the obvious example is that quotients are only well defined when you quotient by a normal subgroup, which is a union of conjugacy classes.

[astral fractal]

our class hasnt gotten that far yet

we only introduced normal subgroups so far

[delicate orchid]

yeah so once you learn group actions and formalise this notion there's loads of nice things to do with conjugation. For example, if two group elements are conjugate then they fix the same number of points

D_n isn't the best example of this because things either fix everything, 2 points, or nothing lol

[astral fractal]

i see

[icy totem]

Sorry for the stupid question, but are group actions and group operations synonyms?

[delicate orchid]

no

[icy totem]

What is the difference?

[delicate orchid]

well for one group operators are G x G -> G, group actions are G x X -> X for any set X

for example let G = S_n and X be a set of size n

then there's a canonical group action S_n x X -> X given by permuting the elements of X

[icy totem]

I see thanks

[thin stream]

hello :)

im brand new to group theory (literally started my course last week) and came across this q

parts i, ii and iii done with no problem and i showed isomorphism between S1 and SO(2)

(if my course's notation is different from yours, S1 is the complex numbers w/ modulus 1, SO(2)=O(2)=2x2 orthogonal matrices)

and my idea for showing the second isomorphism isn't true was to compose isomorphisms so that by contradiction it would mean $S^1\cong S^1\times {\pm 1}$

[cloud walrus]

george clooney real account

[thin stream]

now this should be the easy part: prove that claim is obviously false

but i'm not sure how to, in general, prove two things are not isomorphic, all i did was make the "not isomorphic" problem easier lol

[celest furnace]

You need to look for a property preserved under isomorphism that one group has and the other doesn’t

[thin stream]

i can't use cardinality because theyre both uncountable ;-;

[celest furnace]

For finite groups you usually compare elements of a certain order

[thin stream]

i havent learnt what those properties are

this is the very first isomorphism question i have ever been exposed to

[celest furnace]

Maybe look at the number of elements of order 5 (solutions to z^5=1) for example

[thin stream]

so S^1 has e^2api/5 with a =0,1,2,3,4

and S^1 i guess has (e^2api/5, 1) and (e^2api/5, -1) for a=0,1,2,3,4

so LHS has 5 of them, RHS has 10 of them, so not isomorphic?

[celest furnace]

Yeah this technique works but this example doesn’t

Since -1^5 = -1 not 1

[thin stream]

oh lol oops

[celest furnace]

If you use an even number instead of 5 this works though!

[thin stream]

no ur right

okay cool lol i'm just doing baby math mistakes like -1^5=1 lmaoo

so just choose elements of order 2 basically

[celest furnace]

Prove that if x has order n and phi is an isomorphism then phi(x) also has order n most importantly

[thin stream]

S^1 has 1,-1 and S^1 x +-1 has (1,1),(1,-1),(-1,1),(-1,-1)

as elements of order 2

and then unequal amount so not isomorphic?

coolcool

[celest furnace]

Yeah that works

Nice job

[thin stream]

okay cool that dioesn't look too bad

tysm !!

[celest furnace]

👍

[thin stream]

loving group theory so far, my second fav course after set theory hehe

[celest furnace]

It gets better! It’s awesome

[celest furnace]

Also this is crazy lol

[thin stream]

oxford university🤷‍♂️

they hate their students

[celest furnace]

Wow Oxford! Nice

[thin stream]

my professor is rly good at explaining it tho, we kinda whizzed past all of the preliminary definitions in like 15min without making anyone feel too out-of-breath

so ik that when we get to higher group theory stuff it'll be pretty palatable

[thin stream]

would u say this is a reasonable proof?

forgive the weird sideways inequality signs i tried to make a primitive kind of equality diagram to make the connection simpler

[celest furnace]

Not quite

You’ve shown that |phi(g)| <= n (technically |phi(g)| | n) but you haven’t ruled out that it could be less

[thin stream]

ahh true

okay cool lemme try the second bit

wait doesn't it just follow from inverse mapping

[celest furnace]

Yep it would follow by running this in reverse

Nice job

[thin stream]

okay cool

problem sheet complete now i can sleep

🫡🫡

[celest furnace]

💤💤💤

[frank cosmos]

Does the same proof show that if K=Q(zeta), zeta primitive n-th root of unity, then if L=K(Sqrtn) for some a \in K, Gal(L,K) is cyclic?

because Gal is a subgroup of Z/nZ, hence cyclic itself

assume my claim/proof is right. then is the converse true? a cyclic extension is of this form?

[south patrol]

Do you mean replacing p with general n?

[celest furnace]

I think adjoining the 8th root of unity leads to a non cyclic Galois group(?) since (Z/8)^x is not cyclic

[south patrol]

that isn't what they're asking though

[frank cosmos]

yes, sorry

[celest furnace]

Pointing out that Gal is a subgroup of (Z/n)^x, not Z/n

[south patrol]

No they are correct

[south patrol]

You are doing Q(zeta)/Q right

[celest furnace]

Yeah

[south patrol]

Hello1 is doing Q(zeta, nth root)/Q(zeta)

[frank cosmos]

then its multiplicative yeah

[celest furnace]

Oooh

[frank cosmos]

proving abel

[south patrol]

Anyway, no the same proof does not work, since they critically used the fact p was prime

[frank cosmos]

i dont need to be Z/nZ, just cyclic

[south patrol]

Ah then agreed yes

[frank cosmos]

this wasnt expliclty mentioned so wasnt sure

[south patrol]

It's cyclic and you can just identify what it is using the splitting field

[frank cosmos]

how

oh, so like the subgroup {0,1,2} is sqrt(a)*zeta^0, sqrt(a)*zeta^1, etc?

by the fund theorem

[south patrol]

The proposition is not quite correct technically as written since you should force x^p - a to have no roots but yes

[south patrol]

Idk what you mean

[south patrol]

Just calculate the order of the splitting field of x^n - a over Q(zeta)

I need to remember how you do this here though

[frank cosmos]

is this because then the extension isnt normal?

[south patrol]

No I mean lol if we take a = 1 say

then clearly K = L

[frank cosmos]

and then Gal=0

[south patrol]

Sure

Though yeah I mean uh

[frank cosmos]

where is this hypothesis used though in the proof

[south patrol]

Well they say the Galois group is the trivial group or Z/pZ itself

And the Galois group has the order of the extension

[frank cosmos]

i see

[south patrol]

So they are tacitly assuming the extension is not of degree 1 lol

But yeah slightly pedantic point but kinda important when we wanna generalise to other n

Anyway like you can check that x^n - a is irreducible if a is not an nth root

By Eisenstein basically

[frank cosmos]

ye

[south patrol]

i fink

[frank cosmos]

how do they get that the quotients are cyclic

this confuses me

[south patrol]

Isn't that what you have just computed

[frank cosmos]

im also confused on the notation, n, a_0 etc

is n the deg of the poly?

are the a_i roots?

[south patrol]

Okay yeah I agree that is kinda badly written

[frank cosmos]

do u know whats going on here

[south patrol]

Well it's solvable by radicals

[frank cosmos]

also i think this only shows (<-)

[south patrol]

so you can just write it where each field is obtained from the last by adjoining nth roots ig

[south patrol]

oh yeah lol

[frank cosmos]

what are the n and ai?

[south patrol]

n is just random and a_0 an element of K_0

[frank cosmos]

are the ai and n just arbitary elements that we adjoin the n-th root of until we get L

[south patrol]

you are just adjoining an nth root of an element to form K_1

yes

[frank cosmos]

do you mean K

[south patrol]

It's badly written though

Sorry yes

[frank cosmos]

ok so start with K, add some (n) roots of unity, add sqrtn, repeat process until L. guaranteed to eventually terminate since roots given by radicals. Then the corresponding subgroups are all cyclic, the quotient is then cyclic (idk why prime order), and terminates at 1 thus solvable

i dont get why K(sqrtn) \subset L

or even why adjoining roots of unity to K is a subset of L

ah, i think this may be because the extension is normal? x^n-1 has a root in L thus all roots are in L

[frank cosmos]

need to show that x^n-a_i=0 has some root in L

or maybe just put L in a radical extension of K, and then take the a_i guaranteed by defn

[pastel condor]

In the solution (second image), can someone explain the step $\phi(M) + I\phi(M) + I^2N = \phi(M) + I^2N$?

[cloud walrus]

Henry

[delicate orchid]

phi(M) is some sub module of N, so it’s closed under multiplication from R, in particular Iphi(M) \subseteq phi(M)

[frank cosmos]

this is unaswered btw

[pastel condor]

ohhhhh i see

I was trying see how $I\phi(M)$ is a subset of $I^2N$, but I came to the conclusion that $I^2$ is a subset of $I$, not the other way around

[cloud walrus]

Henry

[boreal inlet]

Let f be a degree n monic irreducible seperable polynomial in K[x] where K is a field. Let L be the splitting field of f over K. I know that L/K is Galois. But, would I be right to say that n = |Gal(L/K)|?

No of distinct roots of f in an algebraic closure of K should be the no of K-algebra embeddings of K(alpha) into the closure, and the Cardinality of the automorphism group in general is less than this as it's mostly a subgroup. But for Galois group, do we achieve equality?

Oh right

For every K-algebra embedding of L to a fixed algebraic closure of K, as L/K is normal (because splitting field), it induces an automorphism on L.

So it should be equality in this case

My bad, should have figured this out a long time ago

[cobalt heath]

Uhm

How does that work for e.g. f(x) = x^3 - 2 over Q?

[south patrol]

[L:K] = |Gal(L/K)|, but that needn't be the degree of f

The example Absta gives is an example

[south patrol]

I should point out that this is only true for Galois extensions, but if you take the splitting field of a separable poly then that gives you a Galois extension

[cobalt heath]

Yeah, permutation of roots can be wonky

You can switch them around without anyone noticing, in more ways than cyclics

[boreal inlet]

Isn't that Galois?

[cobalt heath]

Yes, the extension is Galois

But what is the extension field?

[boreal inlet]

Oh shit.

Degree 6

Wait then where did I do wrong

Every K-alg homomorphism from L to K closure should be an automorphism

L is normal

That means |Hom(L, K bar)| <= |Gal(L/K)|, and also in general, |Gal(L/K)| <= |Hom(L, K bar)| as every K-automorphism of L is also an embedding of L into K bar

This is infact true for all finite normal extensions

Cardinality of Hom(L, K bar) is simply the seperable degree, and is equal in this particular case as L/K is also seperable.

But there's the irreducible condition right? In this case, Gal(L/K) acts transitively over R_L(f), that is the set of roots of f in L.

f can have at most n roots, so it acts as some transitive subgroup of S_n

...oh right. This doesn't say it needs to be equal.

But how do we say n divides this?

If there's an order n element we are done

[cobalt heath]

Do we say this

[glossy crag]

Is the rational canonical form good for anything besides proving Cayley-Hamilton and that similarity of matrices does not depend on the base field? I feel like I've never seen it used for anything besides this and it's nowhere near as powerful as e.g. Jordan form.

[boreal inlet]

An exercise literally asks me to prove this when f is irreducible.

[cobalt heath]

Hmm

[boreal inlet]

Even in your example, 3 divides 6

[cobalt heath]

Maybe if alpha is a root of the function

K(alpha) : K has degree n or something?

[boreal inlet]

If it has the same exact degree as the degree of the polynomial, then yeah we are done

But your counterexample said that is not always the case

[coral spindle]

You are aware that the size of the Galois group is the degree of the splitting field, yes? And you know the tower law?

Hopefully you also know the degree of the extension achieved by adjoining the root of an irreducible polynomial

[delicate orchid]

pizza tower law

[cloud solar]

Let G be a group with 2023 elements. How can I find the number of elements x with the property that x^7=e. How can I find the number of elements x with x^(17^2)=e

[rotund aurora]

Isnt 2023 prime

,w 2023 factor

[cloud walrus]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by croqueta3385

[rotund aurora]

Lmao

[cloud solar]

It has one 7-sylow subgroup

And one 17^2-sylow subgroup

[rotund aurora]

Nice

[cloud solar]

I know the number of elements with x^7=e is 1(mod7)

But I dont think this helps

And the group may be non abelian

[boreal inlet]

Yes.

So, you want to mean that

[L:K] = [L:K(alpha)] [K(alpha):K] = n * [L : K(alpha)].

If L was somehow a simple extension and alpha was that particular adjoined element

Then only [L:K] = n

[delicate orchid]

The order of a subgroup has to divide the order of the group, so your 7-sylow subgroups are cyclic of order 7

likewise your Sylow 17-subgroups have to be order 17^2, which is a square of a prime, and so are also abelian

[cloud solar]

Ok right

But for example

All the elements of order 7

Are in that 7-Sylow subgroup?

Yes I am idiot this is true

[delicate orchid]

I don't buy that there's only one of them

maybe

[cloud solar]

n7 divides 17^2

[delicate orchid]

one sec

[cloud solar]

And n7=1 mod7

[delicate orchid]

yes I don't need the sylow theorems to be mansplained to me thanks

just hard to think about 289 and multiples of 7 in my head

yeah ok I can see it now. Strange

oh yeah duh, your entire group is abelian hahaha

[cloud solar]

How?

[delicate orchid]

all of your sylow subgroups are normal and abelian

[cloud solar]

And that implies the entire group is abelian?

[delicate orchid]

all your sylow subgroups are normal iff your group is nilpotent, and thus is the product of it's sylow subgroups

there's also some theorem where |G| = p^2q with q not 1 mod p^2 and p not 1 mod q then you conclude this but it's literally the same thing

[coral spindle]

But you’re trying to prove that n divides the order of the galois group yes?? Think about how to use this fact.

You’re basically there, I have nothing else to add really

[cloud solar]

I know this result

But I think it was p^2q with q not 1 mod p

And q>p

For abelian

And pq with p<q and q not 1 mod p

For cyclic

[delicate orchid]

I've just googled it and it agrees with me, but yours still works with p = 17 and q = 7 so w/e

[cloud solar]

Oh

[south arrow]

How to find out if Z2 x Z5 x Z8 and Z4 x Z20 are isomorphic or not?

[boreal inlet]

@coral spindle

[boreal inlet]

I think I was there

[rocky cloak]

One thing you can do is try to figure out the maximal order of an element in each. If it's different they can't be isomorphic.

[pure arrow]

Hi, I was reading a bit of rings,
I came across the Ring of real quaternions, In the book(Topics in Algebra, Herstein) it is stated that
ijk = -1
However, when I did vector calculus, I have used ijk = 1, are these two different?

Please clarify!

[coral spindle]

It is standard to have ijk=-1 but you could also choose a different basis for the quaternions — substituting -i for i, for example — and you would get the same formulae but with ijk = 1. I have never seen anyone actually do this though.

[pure arrow]

What is basis for quaternions?

[coral spindle]

1, i, j, k.

If you don't know what a basis is in terms of linear algebra I can't really explain more than that.

[pure arrow]

I know basis in linear algebra, just wasn't sure if it was the same here

[coral spindle]

It's the same here.

[pure arrow]

Are quaternions and the direction unit vectors the same thing?

[coral spindle]

What do you mean by this

What are the direction unit vectors

Is 1 a direction unit vector?

[pure arrow]

I don't know how to rigorously define these things

The unit vectors pointing in the x, y and z directions are defined as i,j,k respectively.

[coral spindle]

Is 1 a direction unit vector then?

Because then no, clearly not

Because 1 is in the quaternions

I want to convey to you a very important idea that will help you understand algebra.

It does not matter what things in some algebraic structure are, it matters how they relate to each other.

The quaternions would still behave the same if they were bits of shoestring.

As long as you had elements you called 1, i, j, and k, and you had all the equations hold, they would be the quaternions.

[delicate orchid]

that being said, if you take the standard inner product on R^4 and just slap it on the quaternions, then i j k all have norm 1 and are orthogonal

this is of course implicitly choosing your basis to include i, j, k but as boytjie said earlier who chooses a different basis lol

[rocky cloak]

Historically the names i, j, k for the unit vectors derives from the quaternions. The quaternions predate linear algebra, and inner product and cross product where formulated in terms of quaternion multiplication.

[coral spindle]

dang I didn't know that

I did imagine that right-handedness of the coordinate system aligned with ijk=-1 but this explains it

neato

Funny considering you can get the cross product fairly directly from determinants

[winter shore]

Is this a common construction on groups? Does it have a name?

[timber pollen]

Hi, can anyone help me find a ring with an element that is irreducible but not prime?

[delicate orchid]

This looks like a standard biset construction but more generally it’s just a pullback

I don’t know how this relates to G-principle bundle nonsense though sorry

[coral spindle]

It's a good idea to look at extensions of Z by some square root, so try looking at rings like Z[sqrt(n)] for some integer n, which may be negative! You will be able to find irreducible elements, and prove they are irreducible, by looking at their norms. Give it a shot!

[timber pollen]

I'm unfamiliar with the notation, what do you mean by Z[sqrt(n)]?

I only know Z[X] as ring of polynoms with coefficients from Z

[coral spindle]

I mean the subring {x + y sqrt(n) | x, y in Z} of C.

If you've never seen this, prove it's a ring now. It's nontrivial

[timber pollen]

Ah, we used Z(sqrt(n)) for that 👍

[coral spindle]

This is iffy notation.

Z[sqrt(n)] is much more standard.

[delicate orchid]

Typically (…) is used for field extensions

[…] for ring extensions

[timber pollen]

You're right, I got confused 🙂

[winter shore]

The notation made me think it was a pullback but what is the ambient category? E is a right G-set while F is a left one

[delicate orchid]

good question, apart from the obvious boring answer that we can just consider a left/right G-set as a right/left one by acting on the other side by g^-1 I think it's a pull back in the category of (G,G)-bisets

where we consider a left/right G-set as a (G,G)-biset with a trivial right/left action

[winter shore]

Thanks, I'll think about that

Could you say where do bisets arise? First time I meet this concept, I found https://link.springer.com/book/10.1007/978-3-642-11297-3 but they don't talk about its motivation

[delicate orchid]

well the simpliest case I know of is double cosets K\G/H

which come up in like, Mackey's theorem

I only know about them because my phd research uses them so they're not a very common construction it seems

[winter shore]

I see, thanks

[dull ginkgo]

Let F be a field, G a finite automorphism subgroup of Aut(F), and F^G the fixed-point subfield. Is there an automorphism of F^G[X_1...X_N] that extends each of G's elements that just "permutes" the indeterminates?

[reef smelt]

do fields have the Schröder–Bernstein property ?

[dull ginkgo]

I keep trying to think about it, I know G embeds into S_|G|, and S_|G| is an automorphism subgroup of F^G[X_1...X_N] that fixes F^G and permutes the indeterminates

but I'm unsure if that guarantees that for a valuation map and an automorphism s (and it's isomorphism-image) s* that v(s*(p)) = s(v(p))

I am a moron nevermind

[dim widget]

no

algebraically closed fields probably do

but for instance C(continuum many variables) embeds into C

or just C(t)

[coral spindle]

The theory of acfs (in a particular characteristic) is categorical in all cardinals, right?

So indeed yes, since they would have equal cardinalities

[dim widget]

i do know some of these words

but yes algebraically closed fields are classified by prime field and transcendence basis size

so they are basically fancy sets, and so schroder bernstein holds

[coral spindle]

Yes so indeed they're categorical in all cardinals for that reason lol

Categorical here just means that if a model has a particular cardinality, then it is unique with that cardinality up to isomorphism

[dull ginkgo]

Hey so, going off this again

Let F be a field, G a finite automorphism subgroup of Aut(F), and F^G the fixed-point subfield. If we have an element of F, x with a full orbit under G, then is F^G(s_1(x)…s_n(x)) always iso to F, where s_n is an enumeration of G

[rocky cloak]

If "full orbit" means the orbit has size |G|, then yes.

[F:F^G] = |G| and G permutes the roots of the minimal polynomial of x, so is the minimal polynomial of x has |G| roots, then F^G(x) = F.

[dull ginkgo]

I was going to try to prove [F:D^G] = |G| through that & integrality/algebraicity

[rocky cloak]

One thing you can do is prove that F/F^G is normal and seperable, hence a Galois extension.

[dull ginkgo]

I was trying to find a weird back route to Artin’s lemma by proving F is a finitely generated F^G algebra then proving it’s a finitely generated module by virtue of Noether Normalization

Instead of the usual Artin’s lemma proof which is vector space magic

[dull ginkgo]

I am trying to come up with a contradiction by showing there can’t be an intermediate field that it gets mapped to

[minor wraith]

Does "symmetry" in group theory refer to the actual bijections themselves?

else, what is a "symmetry" referring to precisely?

i.e., mirroring a pentagon vertically preserves the figure in a plane

so is this referred to as a "symmetry"?

[tender wharf]

maybe D_n

i guess

that makes sense

[fading field]

the symmetry is an isometry from R^2 -> R^2 that fixes the image (eg a pentagon will "stay the same" after applying the map)

[tender wharf]

but you're probably looking for the idea of "permutation"

[fading field]

these will form a group under composition

[tender wharf]

did you watch a 3blue1brown video perchance lol

[dull marsh]

Most of the times you can think of symmetries as bijections that preserve some property/structure of an object (or nothing at all, that is the case in Sn)

[fading field]

With this, you can get all the "possible symmetries" of a shape in the plane (note that this is a VERY SPECIFIC VERSION of the notion of symmetry, but it will work for now), which are translations, reflections, and rotations. Note that for any particular "thing" in the plane you don't get translations in your symmetry group unless your thing is unbounded, so that's why D_n for example is only rotations and reflections, since polygons are bounded (ex. suppose you have some set X a subset of R^2 such that there it's bounded on the right and there is a single point that has the biggest x-coordinate: where is the "rightmost" point going to go if you translate to the right? What will map to the "rightmost point" if you translate to the left?) (an example of where translations would be in your group is if you have a "tiling pattern", like the set {(x,y) | x or y is an integer}, then we can say that translations "up" or "down" or "to the right" or "to the left" or some Z-linear combination of these is in our symmetry group)

[fading field]

you can expand this notion of symmetry to mean "a group acts on a set which has structure", which is sort of the most "general" notion of symmetry. When X is a finite set, you get the symmetric group S_n as the group of symmetries of X. When X is a "riemannian manifold with the additional data of a set inclusion" you get the usual notion of "symmetries" that you're familiar with. Symmetry is literally everywhere in group theory. You can rephrase "normal subgroups" and "cosets" in the language of symmetry if you wanted to. You can have groups act on manifolds. You can have groups act on groups. You can even have groups act on themselves. In fact, when you let a group act on itself and explore the consequences, you get a nice theorem that says "every finite group is a subgroup of S_n"

[tender wharf]

(cayley)

[fading field]

@minor wraith I'll probably leave it there because I have to sleep, but you should take away the lesson that "symmetry is literally intrinsic to what it means to be a group" and that you could literally define a group as "a thing that acts on a space" and people would more or less be happy with that definition, and you can expand the notion of "symmetries of a thing" to mean "a group acting on that thing in a nice way" and people would be happy with that as well.

if you have any questions about it I'll probably respond just ping me or something

[crystal turtle]

I never really cared much for the "groups are symmetries" philosophy

[rose prism]

you mean the “geometry and representation theory” philosophy?

[fading field]

I really do think that the perspective is helpful pedagogically, and it has the ability to stay rigorous, as long as one is willing to expand on what they consider to be "symmetry"

[rose prism]

thats an understatement

it is like

[fading field]

yeah

i agree

[rose prism]

the entire point lol

[fading field]

I am trying to convince someone who is unconvinced

[rose prism]

“i dont agree with [standard thing]” followed by zero elaboration doesnt warrant a thoughtful response imo

[fading field]

maybe not

[rose prism]

but that is up to you ofc

pedagogically i agree as well

but it can be tricky

[crystal turtle]

maybe it doesn't I was just thinking out loud I guess

[rose prism]

i think saying this as a slogan and not explaining doesnt help that much

like intro group theory courses often dont do group actions which is bad i think

[vagrant zinc]

Guys, could someone give me a hand finding the neutral?

[rose prism]

in part bc omitting that language frustrates the instructor’s attempt to explain the geometry

[vagrant zinc]

[rose prism]

diagonal matrix with 1s on diagonal

i can imagine a group theory course built around some central example would be good

but maybe too slow

[fading field]

I had the opportunity to substitute for an professor who was teaching algebra

I'm upset that I didn't do it

would have been a really good experience

[rose prism]

its also hard to find good examples that are suitable for the level of a student who needs to be learning group theory

yeah ive never taught it either

seems interesting

[fading field]

I think studying A_n does a disservice to introductory AA students

[rose prism]

yes

[ivory trail]

the undergrad course used the wallpaper groups as examples

[rose prism]

for sure

actually maybe a course could do linear algebra with group theory flavor and then the rep theory of S_n

maybe too hard

[fading field]

I think there are like three or four groups that ought to be studied in-depth: S_n, D_n, the basics about GL(n,R), SL(n,R), PSL(n,R) i think are probably good ones

uh there are probably more

[rose prism]

well cyclic groups lol

[fading field]

oh yeah

after you do all the introductory stuff

[rose prism]

maybe also GL(n, Fq)

[fading field]

like you can talk about cyclic groups and basics for like 10 weeks and then just do examples for the last 5

or however long it would take to do the very basics

[rose prism]

i think its a function of what basics are important

[fading field]

idk I'm not a pedagogy-head

[rose prism]

like

i will say that ring theory and module theory seem harder to motivate

[ivory trail]

it's nice to use the sylow theorems to classify small groups or whatever

[rose prism]

but they need to be done

[fading field]

I think this should be left for the qual course and if people are interested in it they can do it

i hate the theory of finite groups

just like

[rose prism]

maybe some basic covering space theory is doable

[fading field]

to study finite groups becuase they're finite is a waste of time

to me

[crystal turtle]

You know what in retrospect maybe I just didn't care much for group theory

[fading field]

lol

[rose prism]

or like playing with teichmuller space

[rose prism]

that is much more reasonable lol

i wonder how often people teach group theory following the historical order

and do galois theory first or whatever

[fading field]

lmao