groups-rings-fields

[oblique river]

it's more of a "bank" of examples I have rather than trying to come up with examples each time

[opal osprey]

I was trying to find online somewhere

Some kind of place that stores examples and counterexamples in algebra, or more specifically commutative algebra

There's books like

Counterexamples in topology

Which is great

But I couldn't find anything algebra related

[oblique river]

there might be things like that

but I'm also not convinced of the utility of such resources

[chilly ocean]

Show that a nonempty subset H of a group G is a subgroup of G if and only if ab^-1 belongs to H for all a,b in H. I understood the solution but they assume that b^-1 must be in H. How can they assume it?

This is the solution

What if H doesn’t have an element b^-1

[scarlet estuary]

subgroups are groups

so all elements must have inverses

b is in H, so b^-1 must be as well.

[chilly ocean]

Yes but we’re proving that the subset is a group we can’t use properties of it being a group

[scarlet estuary]

oh, i thought you were doing the other direction

my apologies

[chilly ocean]

np:)

[rustic crown]

since its non-empty, it contains an element... so if a in there then a*(a^-1) = e is there

now if b is any element, then e*(b^-1) is also there

[scarlet estuary]

det i dont follow

[chilly ocean]

What if the element is just the singleton a

[scarlet estuary]

oh i see what youre doing

yeah that works

we know ab^-1 in H for all a, b in H

[chilly ocean]

Well if b^-1 doesn’t belong to H then it automatically can’t be a group I presume? Can we justify it like this?

[rustic crown]

okie so we have a non-empty subset H of G. and the condition is if a and b are in H, then so is a*(b^-1)

[scarlet estuary]

set a = b

this gives that aa^-1 = e is in H

so eb^-1 in H

hence b^-1 in H

[chilly ocean]

Ohhhh

Thanks so much:)

[scarlet estuary]

this does require nonemptiness though, as det says

[chilly ocean]

Yes it does say to assume non emptiness

[scarlet estuary]

yeah.

[chilly ocean]

But is it not like, when you multiply e with b^-1 you automatically assume b^-1 is in H

[rustic crown]

so the operation is defined on G.

[chilly ocean]

Ahh got it thanks

[rustic crown]

you can multiply any 2 elements of G

[chilly ocean]

If b^-1 doesn’t belong in H then it’s not a group correct? So I can assume b^-1 does belong in H

[rustic crown]

(but we are trying to show H is a group)

.<

[chilly ocean]

I don’t understand why we can multiply e by b^-1 if it might not even be an element of H. If it’s not, then G is a group?

[rustic crown]

okie so lets start from the beginning

G is a group

[chilly ocean]

Yea

[rustic crown]

so you have a multiplication function G x G --> G

[chilly ocean]

Yea

[rustic crown]

this satisfies some nice properties

now the question is, if H is a subset of G. When can we guarantee that H is also a group with the operation induced from G.

[chilly ocean]

Yea I’m following so far

[rustic crown]

so if you pick 2 elements of H, call them a and b... then you want a*b to be an element of H. But apriori it is only an element of G.

[chilly ocean]

Yes

[rustic crown]

so one condition when H does become a group is when e is in H where e was the identity of G, for any 2 a, b in H we have a*b in H and lastly when a in H, then a^-1 in H.

[chilly ocean]

Yea exactly but they haven’t written it in the question so we had to assume it

Cos trivially, if it doesn’t then it’s not a group

So it must

[rustic crown]

the content of the proposition is that you can nicely package this into just one check. H should be non-empty and for any 2 elemetnts a and b in H, you have a*(b^-1) in H

[chilly ocean]

Ohhhh got it makes sense

Thank you

[rustic crown]

going from (1) => (2) is easy. Indeed, e in H, implies H is non-empty. And if a and b are in H, by the third condition b^-1 in H, and by the second condition a*b^-1 in H.

[shut halo]

By generators of a Lie Algebra, is it simply meant a basis for the Lie Algebra?

[sour plume]

Yeah, a vector space basis, usually also with specified relations for the Lie brackets between the basis elements

[shut halo]

usually also with specified relations for the Lie brackets between the basis elements

What do you mean by this? I'm getting confused on terminology between physics and maths

[sour plume]

If I have a vector space basis ${x_1,\dots,x_n}$ for my Lie algebra, people generally call them "generators" if they know how precisely they "generate" the complete structure of the Lie algebra; to do that, you need to know how all the Lie brackets look. Hence, you will generally only call such a set a set of "generators" if you have a precise expression of all Lie brackets $[x_i,x_j]$, expressed again in the basis ${x_1,\dots,x_n}$. That's basically it

[cloud walrus]

Lartomato

[sour plume]

So for example, people abstractly talk about the Lie algebra $\mathfrak{sl}_2$ as the 3-dimensional Lie algebra with generators ${e,h,f}$ with these Lie brackets between them

[cloud walrus]

Lartomato

[sour plume]

But that may be the "physics"-kind of definition, which isn't really a definition, lol. I'm now also seeing this definition on wikipedia

[shut halo]

Ok thanks a lot. I think I understand. It's just weird that my maths course never mention generators but my physics course always does.

[sour plume]

I think it's a matter of how many explicit calculations you want to do; in your math course, you probably don't often care that much about doing explicit calculations for any specific Lie algebra, but in your physics course, you very often care about specific Lie algebras liek su(2) and su(3). There, having "generators" of a Lie algebra -- so a vector space basis plus specified brackets -- is super important to make any precise statements

[latent anvil]

call a field extension $F/k$ separably generated if there's an intermediate field $F/E/k$ such that $E/k$ is purely transcendental and $F/E$ is separable and algebraic

[cloud walrus]

Sham "not a furry" rock

[latent anvil]

something I'm reading seems to claim that if $A$ is a finitely generated integral $k$-algebra then the field of fractions of $A$ is separably generated

[cloud walrus]

Sham "not a furry" rock

[latent anvil]

why is this the case?

i guess like, i can noether normalize to get $k[x_1,\ldots,x_d] \subseteq A$ integral

[cloud walrus]

Sham "not a furry" rock

[latent anvil]

and then we presumably take $E = k(x_1,\ldots,x_d)$

[cloud walrus]

Sham "not a furry" rock

[latent anvil]

where d is the krull dimension of A, i.e. the transcendence degree of F

but why is $F/E$ separable?

[cloud walrus]

Sham "not a furry" rock

[latent anvil]

wait wtf

this is basically saying that like, every variety has separable function field

which is not true(???)

im so confused lol

@next obsidian do you understand why the first sentence is true here

it seems like the idea of the proof is to do noether normalization to get a hypersurface in A^{d+1} which is birational to X, I think?

[next obsidian]

I have no clue

[latent anvil]

l'oof

[next obsidian]

Idk why it’s separable, like it’s finite and algebraic

[latent anvil]

i mean yeah you can nn to get that

[next obsidian]

And I don’t think we’ve made characteristic assumptions

[latent anvil]

hnng

there are certainly non separable extensions of Fp(x)

[latent anvil]

does this proof sketch sound good tho

so like nn give us the intermediate field

and then field theory tells us there's sort of only extra relation element you need to get from k(T1,..,Td) to K(X)

so we can look at a hypersurface in A^{d+1} defined by the minimal polynomial of that thingy

or really in an open subset...

[next obsidian]

Right

I think I glanced at this and just said “sure” and didn’t think harder about it

Like algebraically this is saying that m_x always requires at least as many generators as the krull dimension and this is a fact I just know to be true

[latent anvil]

wait it is?

[next obsidian]

I think so, yeah

[latent anvil]

oh that gives the inequality dim m_x/m_x^2 >= dim X

[next obsidian]

Maybe you have to relate the krull dimension of the local ring to the dimension of X actually

[latent anvil]

since m_x/m_x^2 = m_x tensor A/m_x and then NAK

[next obsidian]

Right

[latent anvil]

I think nakayama does it here

[latent anvil]

so i agree this is equivalent to the inequality

but the inequality isn't the hard part imo

i guess it's not the interesting/surprising part to me

[next obsidian]

Sure

I’m just saying I glanced at this and went

“Okay m_x requires at least as many as the krull dim”

So whatever, the proof is whatever

[latent anvil]

sure

[next obsidian]

Haha

[latent anvil]

but like

that doesn't tell you the minimal number of generators = the krull dim for a dense open set of X, right?

[next obsidian]

Oh yeah

Not at all

[latent anvil]

that's what I mean by this triple of messages

[next obsidian]

Ah sure

[latent anvil]

because that's saying X is nonsingular except on some shitty closed subset

[next obsidian]

Right

Gross ewww

[latent anvil]

hahaha

[next obsidian]

You’re so small hahahahah

To the shitty singular part

[latent anvil]

lmfao

[next obsidian]

So like what you want is the separable degree to be 1 or whatever of that K(X)/L

So I think what it is is like, you consider the inclusion of L into an algebraic closure

[latent anvil]

I don;'t think I agree with this

[next obsidian]

And show there’s only a single extension of that to K(X)

[latent anvil]

so L is the field you get from nn right?

[next obsidian]

Right

you want K(X) separable over that

So it’s equivalent to showing the inclusion of L into an algebraic closure extends uniquely

[latent anvil]

Yes, I agree

wait do I

yes, an algebraic closure of L

(not of K)

because we know K(X)/L is finite, this is equivalent

[next obsidian]

Or err

I think this is measuring the wrong thing

[latent anvil]

oh dear

[next obsidian]

I think we want this equal to the degree of the extension

https://stacks.math.columbia.edu/tag/09HD

Read 14.7 and 14.8

The number of extensions into an algebraic closure is the degree of the separable closure over the base

[latent anvil]

ahhh yeah

[next obsidian]

So we want the separable closure to be everything

[latent anvil]

it should be equal to [K(X) : L]

not 1

[next obsidian]

Right

1 says purely inseparable

[latent anvil]

okay so

let's try and example

k[t] -> k[t] via frobenius

this is an integral extension

right?

$k[t^p] \subseteq k[t]$

[cloud walrus]

Sham "not a furry" rock

[latent anvil]

k is a field of char p, sorry

[next obsidian]

k being char p?

Okay

I agree

[latent anvil]

so we get $k(t^p) \subseteq k(t)$

[cloud walrus]

Sham "not a furry" rock

[latent anvil]

when we take fields of fractions

[next obsidian]

Yup

[latent anvil]

this k(t^P) is our intermediate extension

but the extension is not at all separable

so you can't just take any polynomial subring over which A is integral

right?

maybe we should ask baicheng

[next obsidian]

Yeah

Idk what is going on there lol

[latent anvil]

alright, sent

[ivory dust]

If ord(G) = 30, and H is normal subgroup of G isomorphic to Z6 how does one go about proving G is isomorphic to Z30

[scarlet estuary]

are you aware of the sylow theorems?

[ivory dust]

Noo

[scarlet estuary]

lagrange's theorem?

[ivory dust]

yess

i was given a hint to consider the homomorphism phi : G -> Aut(H) i.e. G->H->H phig(h) = ghg^-1, i proved its homomorphic

and i proved its kernel is order 30

[fair shard]

wait how would those help don't you need to look at automorphisms of Z6?

[ivory dust]

but not sure what that entails

[fair shard]

have you seen semidirect products

[ivory dust]

ive seen direct products, internal direct products

and external

idk what semi is

[fair shard]

maybe I'm not the best help actually I might be overcomplicating it

ok so H normal in g means left cosets are the same as right costs

cosets

[ivory dust]

since H normal, ghg^-1 is in H obviously thats what phi is mapping to tho, i think phig(h) = {e} order 1

[fair shard]

so here is where the homomorphism comes in

given g in G

gH=Hg meaning gh=h'g for some h'

and you can see that h' is ghg^-1

this is the homomorphism to aut(H)

each g is sent to the automorphism thar is conjugation by g

sorry I have to go now, try to classify the automorphisms and make a divisibility argument

[ivory dust]

Hm ok thank you

if order(phi(G)) = 1

well a thm we learned was that

phi(G) isomorphic to G/ker phi

so ord(G) = ord(ker phi) x ord(phi(G))

so 30 = 30 x 1

what would the

order of ker(phi) being 30 imply?

i think thats where im stuck

[fair shard]

where does this come from

[ivory dust]

phi(G) isomorphic to G/ker phi

and G finite order 30

it was a thm

for homorphic phi mapping G to a subgroup of G

[fair shard]

but it has || 1 only if |ker|=30

[ivory dust]

yea

i showed ord(ker) = 30

ghg^-1 = e for fixed g implies h = e

only one such h exists in H

and there are 30 cases for g

[terse crystal]

I just proved that G is abelian then by the structure of finite ( finitely generated) abelian groups we have that G is isomorphic to the direct products of Z / 2Z , Z / 3Z and Z / 5Z therefore isomorphic to Z / 30Z

First the order of G / H is 5 therefore a cyclic groups of order 5 generated by Hy, y ^5 belongs to H.choose a generator x of H we see that there is an isomorphism for H to itself defined by mapping x to yxy^-1

And this automorphism has order either 1 or 5 since y^5 belongs to H

And it can’t be 5 by direct check therefore G is an abelian group then the result follows

( yxy^-1=x^k then k^5 =1 mod 6)

[left sand]

there's a problem i thought of of the proof

$\sum_{k=1}^n \frac{f_k(t)}{p_k(t)^{r_k}} = 0$ but the problem is $p_i$ and $p_j$ could be $(t-1)^2$ and $(t-1)^6$.

and f_k is guaranteed to be a constant

[terse crystal]

P_i are all different irreducible polynomials

[cloud walrus]

Anticipation

[left sand]

hm, but how would you decompose $\frac{t^3+3}{(t-1)^3(t-3)^2}$

[cloud walrus]

Anticipation

[terse crystal]

In which field?

[left sand]

R

[terse crystal]

I have a general way solving the inverse of a matrix let me write it down

[left sand]

ok sorry i am busy for a bit so might ping you later again

[terse crystal]

Sure

[terse crystal]

You can always factorize polynomials in C first .when adding up to the final answer you always can get real rational functions again (since they are conjugated pairs)

[terse crystal]

Could you tell me which book it is from.I am looking for different materials to read

[latent anvil]

it's algebraic groups by humphreys

[terse crystal]

Thank you

[left sand]

got a few q..

i still dont know what b is

and what ur doing here

i think i got it nvm, but why is A invertible

also it seems you just decomposed into same irreducible polynomials again of higher powers

[left sand]

which is the problem of the proof i mentioned above,

so the proof that suppose a_k ≠ 0 then splitting and saying it doesnt divide left hand side wont work because it might divide

oh btw here is my solution so far, it may be im interpreting something wrong

[terse crystal]

A is invertible because those polynomials defined in the fifth row are linear independent. Proof is similar as what we discussed before

[terse crystal]

Those polynomials f_i show up If you add up those terms whose denominators are power of the same irreducible polynomial P_i

[left sand]

ah, ok, so would that be a basis? and the ones i wrote not necessarily a basis?

[terse crystal]

What you wrote forms a basis The only reason why all numerators in my pictures are 1 is because that in C irreducible polynomials have degree 1

[left sand]

ok, but how would i show it forms a basis, im stuck in the end at that part

cannot show LHS is not divisible by p_1(t)^r_1, after supposing a_1 ≠ 0 toward contradiction

i would like to show Q_1(t) is not divisible by p_1(t)^r_1 but can't because Q_1(t) could contain a higher power of p_1

[left sand]

ok how is the last implication true

im sorry my pea brain cant understand

[terse crystal]

f_i = Σ a_ijk x^j P_i(x)^(r_i-k)

Since j is smaller than the degree of P_i =d this polynomials can be divided

Into polynomials where the power of x is contained in [kd,(k+1)d-1] for all k

[left sand]

wait what do u mean contained

... every step get so hard for me this problem is insane

like what is kd and the stuff

[terse crystal]

Btw could you show me how to type (a_n)^r in latex, I want to write it clearer

[left sand]

$a_n^r$?

[cloud walrus]

Anticipation

[left sand]

or $(a_n)^r$

[cloud walrus]

Anticipation

[terse crystal]

$f_i=\sum a_ijk \cdot x^j \cdot (P_i)^(r_i -k)$

[cloud walrus]

Cogwheels of the mind

[left sand]

$f_i=\sum a_{ijk} \cdot x^j \cdot (P_i)^{(r_i -k)}$

[cloud walrus]

Anticipation

[terse crystal]

Thanks if the degree of P_i is d then you can see that this polynomials

[left sand]

where j < d

[terse crystal]

You can divide it into different sections where the power of x is contained in [kd,(k+1)d-1]

Because j is smaller than d

[left sand]

Wait so i will rewrite something

I believe I am trying to show $\sum_{k=0}^n \frac{a_k x^{s_k}}{p(x)^{r_k}} = 0$ implies $a_k = 0$

[cloud walrus]

Anticipation

[left sand]

where p is irreducible and $s_k < deg(p)$

[cloud walrus]

Anticipation

[terse crystal]

You are trying to show that

[left sand]

where does r_i - k come from, I don't understand how you rewrite to that form first of all

[terse crystal]

$\sum_{i,j,k} \frac{a_{ijk}x^j}{(p_i)^k}=0$ implies that $a_{ijk}=0$

[cloud walrus]

Cogwheels of the mind

[left sand]

wait cant we ignore the i,j's for now and solve this sub problem?

[terse crystal]

$1 \leq k \leq r_i$

[cloud walrus]

Cogwheels of the mind

[terse crystal]

It is insufficient. f_i doesn’t necessarily just contain one terms

[left sand]

i mean f_i is this

[terse crystal]

$\sum_{i,j,k} \frac{a_{ijk}x^j}{(p_i)^k}=\sum_i \frac{f_i (x)}{(p_i)^{r_i}$

[cloud walrus]

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

[left sand]

So let $f_i$ be $\sum_{k=0}^n \frac{a_k x^{s_k}}{p(x)^{r_k}} = 0$, how would the proof go?

[cloud walrus]

Anticipation

[left sand]

to show a_k = 0

where p is irreducible and s_k < deg p, r_k ≥ 1 distinct

[terse crystal]

Then it equals to $\frac{f}{p(x)^{r_n}$ where $f=\sum a_k x^{s_k} p(x)^{{r_n}-{r_k}}$

[cloud walrus]

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

[left sand]

Then it equals to $\frac{f}{p(x)^{r_n}}$ where $f=\sum a_k x^{s_k} p(x)^{{r_n}-{r_k}}$

[cloud walrus]

Anticipation

[terse crystal]

Yes but I can’t see why you have to discuss this particular form

[left sand]

i mean if i can show a_k = 0 then im done with the proof right

since thats the last implication you wrote

[left sand]

here

[terse crystal]

No like I said it’s not the normal case

Normally $f=\sum a_{jk} \cdot x^j \cdot (P)^{(r -k)}$

[cloud walrus]

Cogwheels of the mind

[left sand]

sorry what are the j,k,r i am completely lost

I agree up to this point, and yes f_k can be broken into monomial sum

but actually im lost already what is f_k

i think this problem is insane..

im just dumb

[terse crystal]

Smaller than degree p not smaller than or equal to sorry

[left sand]

ye

[terse crystal]

Oh so in other words

[left sand]

what did you say f_i was again?

[terse crystal]

We replaced the final $\sum_{j,k}$ with $\sum_{k}$ $\sum_{j}$

[cloud walrus]

Cogwheels of the mind

[left sand]

yep my exact question is

why does f_i(x) = 0 as you wrote above mean that a_{ijk} = 0, i think im good once that is resolved

[terse crystal]

the power of x in each k doesn’t intersect at all

[left sand]

oh shoot i think i get it

wait

[rich ravine]

@shamrock#0598 I have not read the whole discussion, but there is a general theorem (Liu, Alg. geom and arithm. curves, prop 3.2.15) that integral varieties are geometrically reduced iff their function field is separable (in the sense of not necessarily algebraic extensions ; see Lang's Algebra, chap VIII.4 for general theory)

[terse crystal]

$(a+bx)+(cx^2+dx^3)+(ex^4+fx^5)=0$ then $(a+bx)=0$ $(cx^2+dx^3)=0$ $(ex^4+fx^5)=0$

[cloud walrus]

Cogwheels of the mind

[left sand]

oh do they really not intersect

wait ur right i see

right haha my brain

thanks

i cannot think

so much notation

[left sand]

just another thing to confirm

f_i is of degree less than r_i right

oh i mean

less than r_i*deg(p_i)

[terse crystal]

?I thought it was that “iff the intersection of the function field and the separate closure of k equals to k”

[left sand]

actually still dont get why it doesnt overlap

because im confused with indexing

[left sand]

i believe here f_i(x) is of degree < r_i*deg(p_i)

[terse crystal]

We proved that each f_i =0 like the degree of them don’t even matter

j being smaller than degree p is for proving each a is zero

[left sand]

here why can't you have

$x(x^2+x+1)^2 + (x^2+x+1)$ which is overlap

[cloud walrus]

Anticipation

[terse crystal]

Oh I was wrong I need to reconsider

[left sand]

So the problem is

First of all, if I got

[terse crystal]

Oh I figured it out

[left sand]

oh

im gonna summarize the problem again cuz maybe someone else can explain as well

so ignore what im saying for now

The question is to find a basis for R(t), and so far I claim that

[left sand]

$B = {t^n \vert n \geq 0} \cup {\frac{t^m}{p(t)^n}\vert n \in \mathbb{N} \text{P irreducible over} \mathbb{R} \text{and} m < deg(P)}}} is a basis

[mild laurel]

hm, one thing that might make things easier is to note that the irreducible polynomials over R are all degree 1 or 2

[left sand]

yea

i think cogwheel is proving in a general field F actually

[mild laurel]

why is the condition that m < deg(P) there? How do you write something like t^2/(t +1) in that basis?

[left sand]

that is equal to t+k+c/(t+1)

[mild laurel]

ah right

just divide

There's a proof on wikipedia, don't know if you've looked at that

[left sand]

nope

[left sand]

thx

[left sand]

why p|g_n?

[terse crystal]

the powers of p are greater than or equal to 1 in the previous terms

[left sand]

wait so then what?

[terse crystal]

? degree of g_n is smaller than P while it also can be divided by P so it can only be zero

[left sand]

nono i mean why can g_n be divided by p

?

[terse crystal]

This

[left sand]

yea but i dont get what you mean by this

or i mean how does that help

oh

is it because

[terse crystal]

$\sum_{k=1}^{n-1} g_k p^{n-k}=p (\sum_{k=1}^{n-1}g_k p^{n-k-1})$

[left sand]

wait

no

[cloud walrus]

Cogwheels of the mind

[terse crystal]

Which = -g_n

[left sand]

ohh

oh

thank you so much

i get the argument now

finally i can write the goddamn proof

[terse crystal]

Thank you actually latex is so interesting

I wonder why I kept taking pictures

[left sand]

its nice but usually i feel writing on ipad on notability or something is faster

but yea its professional

it can be decent fast too once u get used to it, which is not hard

[terse crystal]

Yeah hope that I can soon

[celest brook]

my work so far, following what professor did and it's all just a bit unclear to me. I always hear people say 'you can shorten' it, but that's usually if you assume 'blah blah's theorem or something else down the line I don't know about. I guess what i'm asking is if there is a faster way to convey it for a beginner rather than doing this

MT=multiplication table

cayley table

[mild laurel]

What's your definition of dihedral groups here?

[celest brook]

uhh whadya mean usual presentation

yesss

how?

yes

ohhh

that's where rs=sr^{-1} comes from

s^{2}=r^{n}

I guess it just feels weird writing out multplication tables

very nice when patterns appear though

:)

[mild laurel]

If your definition of the dihedral groups is "has this presentation", then you can just directly check that f(r) and f(J_3) give you generators for the presentation

[celest brook]

why do I only need to check the generators?

[mild laurel]

I'm saying that you check that those generators satisfy the relations for D_3

[celest brook]

well haven't I defined f(r) and f(J_3) to be the generators of D_3 anyway??

[mild laurel]

I mean sure

[celest brook]

how does that relate to an isomorphism

[mild laurel]

But this way you don't actually need to confirm this homomorphism

[celest brook]

why?

[mild laurel]

Like I said, if the way you've defined D_3 is by means of a presentation in terms of generators

Then you can just check that f(r) and f(J_3) satisfied this definition

[celest brook]

how come that will gurantee that there is a homomorphism? How does that deflect the need to check/

[mild laurel]

The group will just be D_3

[left sand]

lol theres another issue

you proven that a_jk = 0

but how do you know thats the only way to write r

as in r = $\sum_{k=1}^n \frac{f_k(t)}{p_k(t)^{r_k}}$

[cloud walrus]

Anticipation

[left sand]

which can be further decomposed into $\frac{f_k(t)}{p_k(t)^{r_k}} = \sum_{i < d, j \leq r_k} a_{ij} \frac{x^i}{p_k(x)^j}$ where $d = deg(p_k)$

[cloud walrus]

Anticipation

[left sand]

so here if r = 0, then a_ij = 0 can be shown but how do you prove no other lin combo so that b_ij ≠ 0

just say partial fraction decomposition is unique?

[terse crystal]

I don’t think there is any confusion here I directly proved that
$\frac{x^j}{p^k}$
is linear independent. All those f_i are just used in certain steps in the middle

[cloud walrus]

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

[left sand]

ok so first you say there is such a decomposition for every f/g right, which is the span part

[terse crystal]

You could check the last picture whether they are any specific mistake in it. I didn’t find any error...

[left sand]

no the picture has no error but im just still a bit conceptually confused

so for every f/g in R(t), you show it can be written uniquely as f/g = q+r

where r can be spanned by the t^n/p(x)

oh ok, i get the picture

the idea

[terse crystal]

From the beginning, also a good latex practice:

[cloud walrus]

the-last-knight

[median pawn]

How would you prove this?

The abelian part is easy, follows from fundamental theorem of finite abelian groups. The non abelian case seems tricky

[terse crystal]

The structure of groups having order p^k is clear when k is smaller than or equal to 7

[median pawn]

It's not clear to me 😦

[terse crystal]

Particularly I have an article here showing the case where k =3:

[median pawn]

Is it Keith Conrad's article?

[terse crystal]

Oh how to upload files here...

[median pawn]

That is pretty complicated, since it proves for p^3 in general

I'm looking for 8 and 27 in particular and hoping things will be easier

[terse crystal]

It’s not complicated when k=3 actually...

[terse crystal]

Oh,exactly 😂

[terse crystal]

$\sum_{0 \leq t \leq T} c_t x^t + \sum_{1 \leq i \leq n, 0 \leq j < deg(p_i), 1 \leq k \leq r_i} \frac{a_{ijk} x^j}{{p_i}^k}=0$ then let $Q=\prod_{1 \leq i \leq n} (P_i)^{s_i}$, $Q_j = \prod_{1 \leq i \leq n, j \neq i } (P_i)^{s_i}$ $r=\sum_{0 \leq t \leq T} c_t x^t$, $\frac{f_i}{{p_i}^{s_i}}=\sum_{0 \leq j < deg(p_i), 1 \leq k \leq r_i} \frac{a_{ijk} x^j}{{p_i}^k}$ then we have that $\sum_{1 \leq i \leq n} f_i Q_i +rQ=0$ then if one of the $f_i \neq 0$ then $p_i | f_i$ which is a contradiction therefore all $f_i =0$ then $r=0$ so we can reduce it to the case where $\sum_{0 \leq j < deg(p_i), 1 \leq k \leq r_i} \frac{a_{ijk} x^j}{{p_i}^k}=0$ which we have proved in the last picture.

[cloud walrus]

Cogwheels of the mind

[left sand]

yea got it, thanks!

[terse crystal]

I edited the context but it didn’t change

$\sum_{0 \leq t \leq T} c_t x^t + \sum_{1 \leq i \leq n, 0 \leq j < deg(p_i), 1 \leq k \leq r_i} \frac{a_{ijk} x^j}{{p_i}^k}=0$ then let $Q=\prod_{1 \leq i \leq n} (P_i)^{s_i}$, $Q_j = \prod_{1 \leq i \leq n, j \neq i } (P_i)^{s_i}$ $r=\sum_{0 \leq t \leq T} c_t x^t$, $\frac{f_i}{{p_i}^{s_i}}=\sum_{0 \leq j < deg(p_i), 1 \leq k \leq r_i} \frac{a_{ijk} x^j}{{p_i}^k}$ then we have that $\sum_{1 \leq i \leq n} f_i Q_i +rQ=0$ then if one of the $f_i \neq 0$ then $p_i | f_i$ which is a contradiction therefore all $f_i =0$ then $r=0$ so we can reduce it to the case where $\sum_{0 \leq j < deg(p_i), 1 \leq k \leq r_i} \frac{a_{ijk} x^j}{{p_i}^k}=0$ which we have proved in the last picture.

[cloud walrus]

Cogwheels of the mind

[left sand]

lol

that sum looks monstrous

[terse crystal]

Where $p_i \not| f_i$.

[cloud walrus]

Cogwheels of the mind

[terse crystal]

Indeed😂

[chilly ocean]

,rotate

[chilly ocean]

I don’t quite get the sentence xN=eN iff x belongs to N.

In theorem 1.2

[viscid pewter]

what don't you get about it

[chilly ocean]

eN is the identity element of the quotient group G/N

Since you're trying to figure out the kernel

Yes but why should x belong to N

[viscid pewter]

so since N is a subgroup

if x is in N, then xN is just N again

or eN, alternatively

[chilly ocean]

Ohh yea it’s a subgroup

Overlooked that

Thanks:)

[viscid pewter]

i mean that's not quite rigorous yet but it's the intuition

[chilly ocean]

that just gives the only if part

[viscid pewter]

yeah, exactly

[chilly ocean]

but if xN = eN then x must be in N

so that's the other direction

What’s the reason for that?

Is it because NN=N?

[viscid pewter]

if xN = eN, then there's n in N such that xn = e, ie. x = n^-1, but N is closed under inverses so x is in N

that works i think

[chilly ocean]

if the left coset by N gives the whole of N then its just an element of N right? since N is a subgroup and is closed under the induced law of composition

[chilly ocean]

that works too!

I’m tripping rn ig if I come back to this later I’ll get it

Thanks:)

just look up the definition of what cosets mean ^^

[chilly ocean]

Ahh got it

since eN is just N and xN is just xn where n belongs to N, x must also belong to N because it’s a multiplicative subgroup

Correct?

[viscid pewter]

yeah

[chilly ocean]

Thanks again:)

[chilly ocean]

Give an example of a nonabelian group each of whose subgroups is normal.

[rustic crown]

Q8

[chilly ocean]

Yea that’s correct

How did u come up with the answer?

Did u guess and verify or was it a well defined method

Its a well known example ^^

the quaternions group

How are u supposed to just remember this i mean wow

[viscid pewter]

it's just like

what's the smallest perfect number? 6

it's just trivia

[chilly ocean]

So experience?

[viscid pewter]

ig?

it's like trivia really

[chilly ocean]

Hey is it ok if I leave out some things that I don’t understand in a math book during a first reading and come back to it later?

[viscid pewter]

yeah

[chilly ocean]

yay:)

[viscid pewter]

i mean it's really hard to understand everything the first time

[chilly ocean]

No cap bruh it’s so hard I can hardly do 20 pages in like 6 hours

don't expect to be able to master everything in a textbook

Yeah I’ve heard ppl say that before

its not a novel that you're reading ^^

[viscid pewter]

there was a thing i read where someone recommended like for one chapter, reading it three times: so first time you skim the thing just to get an idea of what's happening, then second time you go through it normally and try and understand the proofs a bit, and then third time you really get into it and nitpick all the proofs

and obviously it's not as cut-and-dried as that

different chapters, different readers, etc.

[chilly ocean]

Also, on top of that, group theory is that sort of thing one easily forgets in a couple of weeks

[viscid pewter]

but that's closer to it than just once and done i think

[chilly ocean]

I already forgot what I learned yesterday lmao

I took algebra this first semester and I barely remember anything

That’s rlly helpful thanks so much

[chilly ocean]

[rain crescent]

Is it possible to have 2 isomorphic groups in which one of them has a finite generator set and the other one doesnt?

[chilly ocean]

no

but both have finite generator set

[rain crescent]

I mean I think i wrote it wrong. I meant only one of them could be represented by a finite set

[chilly ocean]

Im interpreting his thing as in one group there doesnt exist a finite generating set

[rain crescent]

is that because the image of a generator by an isomorphism is a generator?

and is that a trivial fact or?

[viscid pewter]

it has to be like a surjective homomorphism i think

yeah

[chilly ocean]

hes saying you can weaken the assumption

but idk if its enough

[rain crescent]

ty

everyone

[chilly ocean]

mirza maybe repost your thing

uhh

sylov

yee

are my msgs sending?

[next obsidian]

P is a Sylow so it’s used up all the p

To make this more clear, if p divided the index of the normalized, then P doesn’t have maximal order

Like wrt p

This says that P has order p^k

But that p^k+1 divides |G|

Because

The index of the normalizer is the number of conjugates

By orbit stabilizer

Yeh

:(

Not gorilla

Monkey

[cursive temple]

Pedestal primate

[next obsidian]

I like that one

[chilly ocean]

Lets say π(x)=[x] (equivalence class of x), π: X -> X/~,
is it correct if we say for all a,b member of X, π(a,b)=[a][b]=π(b,a)=[b][a]=X to proof that is symmetric ?

because π only takes one argument

[carmine fossil]

Is there a binary operation defined on X/~ or smt?

What does [a][b] even mean?

[chilly ocean]

it means two equivalence classes from a and b. since both of them are member of X, therefore both are part of X/~.

[chilly ocean]

no

this is the actual question

[carmine fossil]

By ~_π, they mean the equivalence a (~_π) b iff π(a)=π(b)

[chilly ocean]

ah got it

then it will become π(a)=π(b) then π(b)=π(a) and [a]=[b] is equal to [b]=[a]

same thing goes for transitivity

thanks @carmine fossil

one more question, if I want to proof that is a surjective map, I have to describe since all members of X/~ are part of X, then there always exist an equivalence class such that π(x)=[x]. should I proof it differently ? because it's too obvious.

[carmine fossil]

That's it

Any representative of the equivalence class gets mapped to the equivalence class

[chilly ocean]

we can also use the definition of fibers, and say fibers are the same as equivalence classes, therefore, there always exist y s.t. f(x)=y

and y = [x]

[molten silo]

wOULD THIS WORK

[next obsidian]

How is this algebra

But also that makes sense

[molten silo]

shit wrong chat

[next obsidian]

Never thought about the indicator function being square integrable

[oblique river]

I'm pretty sure that here, f^2(x) is referring to [f(x)]^2

and not the compositional square

[molten silo]

i see

thanks

[oblique river]

you can still do something with the indicator function though

but you'll have to modify it in some way

[scarlet estuary]

yeah you can fix that

[oblique river]

i'd say it's bad form to give an answer to a question someone else is asking for help on

unless they're specifically asking for that

but yes, you are right

[cursive temple]

Pretty cool problem though

[next obsidian]

OH LMFAO

IM A DUMBASS

I just went “yup that makes sense to me”

myself

[molten silo]

me to

[celest brook]

If know H=x^{-1}Hx can I automatically left and right actions on both sides to get that x^{-1}Hx=x^{-1}Hx?

[scarlet estuary]

erm

x^{-1}Hx=x^{-1}Hx follows from the definition of equality

[celest brook]

no i asusme

huh

[scarlet estuary]

a = a for all a

[celest brook]

???

[viscid pewter]

they're the same thing

[scarlet estuary]

if you have two things

[viscid pewter]

you have the same thing on the right and left

[scarlet estuary]

and theyre the same

theyre equal

[viscid pewter]

like, literally the same thing spelt the same

[celest brook]

oh wait

dgoidjkog

I meant

if

H=xHx^{-1} can i use operations to imply that H=x^{-1}Hx as well

I guess I am more asking how would I show they are the same

[mild laurel]

yes, take your equation and multiply both sides on the right by x

and then multiply both sides by x^{-1} on the left

you get what you want

[celest brook]

so I am allowed to do that with this?

[mild laurel]

I mean, you should probably say why its true, since this is equality of sets, and not equality of elements

[celest brook]

i'm trying to use subset arguements and it's very annoying

prof is just like 'write it in reverse order' for other case

eiojgk,s

[mild laurel]

well, I guess it depends what you know

[mild laurel]

do you know that H = xHx^{-1} for a single x in G, or do you know that this holds for all x in G?

[celest brook]

I could see why (xy)H(xy)^{-1} was in H clearly, but you could just simplify that down into one case

I'm a bit confused why I can't do the same with inverses

i'm sorry I'm so stupid

[mild laurel]

No you're fine

so how did how do the (xy)H(xy)^{-1} case?

[celest brook]

you know that xHx^{-1} in H, and yHy^{-1} in H

you can also rewrite (xy)^{-1} as y^{-1}x^{-1}

so you get

x(yHy^{-1})x^{-1}

you know yHy^{-1} will be in H

so you can rewrite above as x^{-1}Hx

which you know is also in H

so therefore (xy) in H

[mild laurel]

I think some of those should be lower case h's but yes

[celest brook]

confusing

[chilly ocean]

Alright class today we will be learning lower case vs upper case letters

[next obsidian]

I would just never include a lowercase h

You can just do it on the level of the entire H

[celest brook]

oops

yeah i didnt mean to do that on top

[mild laurel]

I think on the level of entire H, it's unclear that you can move from (xy)H(y^{-1}x^{-1}) to x(yHy^{-1})x^{-1}

[next obsidian]

Wat

[mild laurel]

like you need to appeal to elements and their associativity here

[next obsidian]

I mean you definitely can’t turn the x^-1 on the right to an x

[celest brook]

group

[next obsidian]

[viscid pewter]

that's not so bad surely

[mild laurel]

sure, but I'm just saying doing it on the level of H isn't enough

[next obsidian]

He’s saying that you’ll need to appeal to individual elements

[viscid pewter]

can take that as given

no?

[mild laurel]

ehhhhhhhhhh

[viscid pewter]

aw come on

ok fine

[next obsidian]

To prove that you can use associativity here

Which I ageee

But also

I don’t care

[viscid pewter]

well to prove yes but it just feels obvious

like i can run through the 3 lines in my head

[mild laurel]

this is an intro algebra class

anyways sorry

[viscid pewter]

oh, ok then

[celest brook]

i am confused

[viscid pewter]

ignore me

[celest brook]

oh wait, I forgot to say that x,y in K

but I think that's implied

[mild laurel]

you can do a similar thing for the other case. If you know that H = xHx^{-1}, then you can show H = x^{-1}Hx by doing the subset arguments

[celest brook]

yes

[mild laurel]

then what are you having trouble with?

[mild laurel]

seems fine

[celest brook]

i'm confused in what to write b=

would it just be b=kh or smth like that?

[mild laurel]

okay I'm not sure why you switched to y

[celest brook]

o

[mild laurel]

so you want to say that given some element of the form xhx^{-1} in xHx^{-1}, that this is equal to h' for some h' in H

and then rearrange just like you did the other case

[celest brook]

okk

hmm

wait I'm confused

ohhh

ugh my brain

b=k*xHx^{-1}?

why can't I get this???

[rigid cave]

I've just learned about finite fields (Galois fields) from a PDF I've found, but how do you construct them? For example, what are the elements in F_(2^2)? Are there any good PDF's that explain this?

[mild laurel]

Do you know what polynomial rings and quotient rings are?

[rigid cave]

yes

[mild laurel]

You can think of F_(2^2) as being F_2[x]/(x^2 + x + 1)

Since the latter is a field and has order 4

It's a field because I specifically chose x^2 + x + 1 to be irreducible

[rigid cave]

Wait, but how does the latter have order 4?

[mild laurel]

Because given any polynomial f in F_2[x], you can reduce it down to be linear or constant in F_2[x]/(x^2 + x + 1)

by basically dividing by x^2 + x + 1

[rigid cave]

Oh, right!

[rigid cave]

And this is true because two finite fields of the same order are isomorphic, right?

[mild laurel]

yeah

[rigid cave]

Okay, thank you so much! One more thing: I know that $F(\alpha) \cong F[X]/(f(X))$ where $f(X)$ is the minimal polynomial to $\alpha$ because I've seen the proof. However, is there any intuition behind this?

[cloud walrus]

older sister

[rigid cave]

For example, I can understand why Q(i) is isomophic to Q[X]/(X^2 +1) just because they behave the same, but I can't see why this is true intuitively for any other value

[mild laurel]

how do you think about F(\alpha)? Or how have you defined F(\alpha)?

[rigid cave]

I think of it as the smallest field containing alpha that also contains F

[mild laurel]

okay sure

and you've seen the idea that if the degree of f is d, then 1, \alpha, \alpha^2, ..., \alpha^{d-1} gives you a basis for F(\alpha) over F?

[rigid cave]

Yes

[mild laurel]

Then that's kind of how I think of it

Any element of F(\alpha) can be written as a_0 + a_1 \alpha + a_2 \alpha^2 + ... + a_{d-1} \alpha^{d-1} since thats a basis

and you can also think of this as a polynomial a_0 + a_1x + a_2x^2 + ... + a_{d-1}x^{d-1}

and thats basically how the isomorphism between those two fields works

[rigid cave]

Ohhhh

Well that was quite nice! Thank you so much!

[mild laurel]

np

[viscid pewter]

ok this is going to be so trivial but how can i prove that, with K normal in H and G and H a subgroup of G and H/K iso to Zp, H is normal in G

i've been stuck on this far longer than i care to admit

[chilly ocean]

I think this is more of less "just do it" assuming the proof I have works

You know, like start with ghg^-1 and show that this is in H

[viscid pewter]

hmm

i was trying homomorphism stuffs

[chilly ocean]

Actually wait a second

[viscid pewter]

probs worth trying symbols...

[chilly ocean]

NVM, maybe this doesn't work

[viscid pewter]

i really hope it's a true statement

i'm like 95% sure it is but

i mean i think the heart of my problem is maybe i don't understand what H/K being iso to Zp really says about H?

i think if H/K is iso to Zp then there exists j in H such that j^p is in K, j isn't in K and H = jK

but i don't know what i can do with that

[chilly ocean]

(Anyway yeah sorry my scratch work was too scratch that it didn't make sense, but anyway) my thoughts so far are, so I think it is sufficient to show that gxg^-1 is in H where x is such that xK generates H/K. Notably (gxg^-1)^p is in K (and thus in H). Then maybe this problem is about taking pth roots eg if y^p is in K then y is in H. As a backwards direction, at least it is the case that the map H to H that maps y to y^p actually maps into K, does this imply gxg^-1 is in K somehow?

[next obsidian]

What’s the problem?

[chilly ocean]

He posted it above

[next obsidian]

Not sure whose problem it is

[viscid pewter]

mine

[next obsidian]

Ohhh

[viscid pewter]

with K normal in H and G and H a subgroup of G and H/K iso to Zp:
P: H is normal in G

[next obsidian]

Wait

Uh

p is prime?

[viscid pewter]

yes

[next obsidian]

Finite groups?

[viscid pewter]

yes

I think it is sufficient to show that gxg^-1 is in H where x is such that xK generates H/K
i got this far, i think it's right

[chilly ocean]

So I guess I have: y in H implies y^p in K. And I want to turn this around to y^p in K implies y in H

[viscid pewter]

that doesn't feel true tho

[next obsidian]

Well that’s simply every single x which isn’t in K

But is in H

So I don’t think it says all that much about x tbh

[viscid pewter]

like with G = Z2 x Z2 x Z2 = {(a, b, c)} where a, b, c are all 0 or 1, {(0, b, c} is a H, {0, 0, c)} is a K
then (1, 0, 0)^2 is in K but (1, 0, 0) is not in H

so it's just false?

[chilly ocean]

You're right

Hm

[viscid pewter]

can we do anything with the fact that (G/K)/(H/K) is iso to G/H??

i wish that meant G/H was automatically well defined

... just to check, it doesn't, right?

[next obsidian]

Maybe I’m dumb as hell but

I think I have a proof even when H/K isn’t Z_p lol

So N_G(K) = G right? And N_H(K) = H

[viscid pewter]

yyyes

[next obsidian]

So it suffices to show that N_G(K) is contained in N_G(N_H(K))

Because then you have G = N_G(K) < N_G(N_H(K)) = N_G(H)

So the normalizer of H is G

[viscid pewter]

yes

[next obsidian]

So take a g in N_G(K) so that gKg^-1 = K

And then take an element in N_H(K) so an h such that hKh^-1 = K

Then look at ghg^-1

We want this to also be in N_H(K)

Ahhhh i see where I went wrong

So ghg^-1 does fix K

Under conjugation

You get ghg^-1Kgh^-1g^-1

And then you just like collapse this in 3 steps

What you need to show is that ghg^-1 is IN H

Which I’m gonna try to use the fact that H/K is Z_p now to do that

[viscid pewter]

ok

i kinda get that ig lol

[next obsidian]

Oh but

I think this isn’t very useful because

I doubt this is true in general

And if I try to use special information this is literally just saying that

ghg^-1 is in H when h is in H

But idk, maybe something brilliant happens

[chilly ocean]

Wait what the fuck?

This is false right? K=1, H = Zp, G = something that contains Zp

[next obsidian]

Yeah

[viscid pewter]

hrnnn

is 1 not normal in everything

[next obsidian]

It is normal in everything

But H wouldn’t be normal in G or at least you can make a G where it isn’t

[viscid pewter]

oh right

fuck

no, wait

uhhhh

fuck

[next obsidian]

Did you just hope this was true?

Or is this a problem from a book or something

[viscid pewter]

i hoped it was true in order to solve a problem from a book

[next obsidian]

[viscid pewter]

look

it seemed reasonable at the time

gimme a sec

[next obsidian]

At least say you think it’s true at the start sad face

[viscid pewter]

i literally did tho

and also throughout

[next obsidian]

Oof

What’s the different problem tho

That u wanted this to be true for

[viscid pewter]

oh wait i should have been thinking about all this other structure

so it's like, G is solvable

prove all the subgroups in the chain are normal in G

and there was some stuff about how Nk+1/Nk is prime cyclic

but i think it would have to be the fact that above H but below G, there's more normal subgroups?

the chain goes up all the way?

[next obsidian]

I asked a friend

[viscid pewter]

is this too mangled for you to understand

[next obsidian]

And they said this isn’t even true either

[viscid pewter]

f me

[next obsidian]

https://en.m.wikipedia.org/wiki/Supersolvable_group

[viscid pewter]

ok so i'll write out all the details of the problem that i have access to

[next obsidian]

Such a group is super solvable I guess

And A_4 is solvable but not super solvable

[viscid pewter]

show these are all equivalent:
G is solvable
there's a chain 1 norm in H1 norm in ... norm in G such that the composition factors are cyclic
the composition factors are prime
there's a chain 1 norm in N1 norm in ... norm in G such that all the Ni are normal in G and Ni+1/Ni is abelian

[next obsidian]

Eh?

But the last one isn’t equivalent

[viscid pewter]

apparently not

[next obsidian]

Where did u get this problem

[viscid pewter]

d&f

[next obsidian]

Tf?

What number?

[viscid pewter]

i can't get a picture directly rn

uhhh look it's the chapter that first covers composition series

4. something?

[next obsidian]

@old lava yo

Can u confirm that this was an exercise in D&F?

[viscid pewter]

i bet this is gonna be me misinterpreting/mistranscribing somehow

[chilly ocean]

I have done quite a few exercises in D and F, this does vaguely sound familiar

But eg abelian factor -> cyclic factor sounds possible, by refining the chain (abelian groups are very nice)

And eg prime/cyclic -> abelian is trivial

[next obsidian]

Yes

[chilly ocean]

I guess it is just the last one

[next obsidian]

If you remove normal in G

All are equivalent

[viscid pewter]

aaa

i am like 80% sure it said Ni was normal in G

what if

d&f uses 'solvable' to mean supersolvable

or

no, i'm so confused, i don't see where i've gone wrong

[celest brook]

what does k and g being in the same conjugacy class mean? I understand Cl(g), but not exactly what Cl(k) means? Or is that k different than the k the definition?

ohh

wai

[viscid pewter]

if k is in Cl(g), g is in Cl(k)

[celest brook]

because k=hgh^{-1} -> h^{-1}kh=g ?

[viscid pewter]

uhhh basically

yeah

+- a typo

[celest brook]

o

that better?

[viscid pewter]

yes

[celest brook]

for reflexivity, would it simply be to show that g in Cl(g)?

[viscid pewter]

well yes, that's the condition

[celest brook]

and that's obvious from that fact of letting h=e

e identity

[viscid pewter]

yes

[old lava]

yo, I'm here

what do I need to confirm

[next obsidian]

This

The last bit isn’t equivalent

You can’t always get Ni normal in G

Do you recall an exercise like this in D&F?

[old lava]

this one

[viscid pewter]

yes

[next obsidian]

What in the fuck

[old lava]

I did do it

it was one of the hardest ones

[viscid pewter]

?!

[old lava]

(iii) -> (iv) took forever

[next obsidian]

you can’t ensure that Ni is normal!?

[old lava]

I might've made a mistake then

lmfao

[next obsidian]

Take A4, it has a unique normal subgroup, the Klein 4

Wtf?!

[viscid pewter]

ok so like

[next obsidian]

What exercise number is that

[old lava]

ya, but A_4 / klein_4 isn't abelian

each N_i must be normal and each N_i / N_{i+1} is abelian

[next obsidian]

Yes, but the thing is A4 is solvable

[old lava]

oh wait monkaS

[next obsidian]

But you can’t get a composition series with each thing in the chain normal

[old lava]

I was thinking A_5

[next obsidian]

Which exercise is this?

[old lava]

exercise 8, section 3.4

[viscid pewter]

wait

ok i knew there was a 4 in there somewhere lol

just now remembering that 4.x is the group actions, not this

[next obsidian]

Okay wait...

[viscid pewter]

i think the trick is right at the start

[next obsidian]

Maybe you can get an abelian series with each Ni normal

[viscid pewter]

i think they might use solvable where everyone else uses supersolvable

[next obsidian]

But it isn’t a composition series

Like you’ll have non simple quotients

In which case I suppose it could be true?

Maybe...

[old lava]

wait, it does work for A_4, 1 \triangle V_4 \triangle A_4 does satisfy it

A_4 / V_4 is cyclic group of order 3

[viscid pewter]

hnnngh

[old lava]

And V_4 / 1 is obviously abelian

[next obsidian]

But you can’t get a composition series

With each factor normal

So it has to be an abelian series only

Wtf

[viscid pewter]

you'd have to go Z2 in V4

and Z2 isn't normal in A4

[next obsidian]

Right, but Z2 isn’t normal in A4

[viscid pewter]

yeah

[next obsidian]

But if you only want an abelian series

Then you’re fine with the filtration Poros gave

Tfw

[old lava]

we don't need to do Z_2 in A_4

that's not necessary by the condition they gave

1 \triangle V_4 \triangle A_4 exists is all the condition says

[viscid pewter]

yes but like

[old lava]

they don't say it's a composition series

[viscid pewter]

well the series would have to be 1 in Z2 in V4 in A4

wait

[old lava]

no?

[next obsidian]

No

[viscid pewter]

wait

[next obsidian]

It’s an abelian series

Abelian quotients

Not simple quotients

[viscid pewter]

yeah, ii and iii say the factors are cyclic prime abelian

[old lava]

you have to show that the minimal nontrivial normal subgroup is necessarily abelian, and then use induction

to prove 3 -> 4

and I'm pretty sure it works

[viscid pewter]

wait, fuck

different chains!

oh my fucking god

[old lava]

different chains indeed

[viscid pewter]

fuuuck

the destroyer of hope

i just assumed all those other things were leading up to showing those chains were actually equal...

ok well thanks all for not holding a grudge over me wasting your time lmao

i am shooketh

i'm just gonna do something else now, i can't even

[old lava]

lol, why would I hold a grudge

[celest brook]

we don't want G to be abelian because the inverse won't be in the conjugacy class?

[viscid pewter]

uhhhhhhh

[celest brook]

oh wait

ok, that's not even helpful for what the question is asking

hmm

I've already proven that Cl(e) makes the trivial group

[chilly ocean]

Hi, I'm working on part (iii). I'm having trouble figuring this out.

What have you done so far?

a is a complex root of f(x), so I know that $a^3+2a+1=0$. This was used in part (ii).

[cloud walrus]

Mega Euler

[chilly ocean]

Indeed, $\frac{u}{v} = \frac{a^2+1}{a^3+2a+1}$

[cloud walrus]

Mega Euler

[chilly ocean]

Not sure how to proceed from here

I've gone in circles with the denominator

You messed up the formula, but doesn't really matter

I would try to write 1/v as a polynomial in a

Oh hmmm

You know, like write v times q(a) = 1 and solve coefficients of q

Or maybe directly doing it with u/v is better, not sure

q(a) being a multiplicative identity of v?

Yeah, ie 1/v

1/v*

Er, multiplicative inverse , I think you mean

inverse*

Yes

What is the point in finding coefficients in q? I thought that Q[x] is the polynomial ring with coefficients coming from the rationals.

Well, finding coefficients of q will tell you what 1/v is (in Q(a)=Q[x]/(f(x))) in the basis 1,a, a^2.
I'm not sure what you mean about Q[x]

Q[x] is the set of all polynomials with rational coefficients