groups-rings-fields

[oblique river]

yes

and

because all the roots are contained in C

[chilly ocean]

oh yeah... I kept looking at it as x^7 - 11 = 0.

[oblique river]

that wouldnt' amtter

all the roots of that are still contained in C

[chilly ocean]

oh true

[oblique river]

and that polynomial has roots that aren't in R

namely all but 1

[chilly ocean]

but we couldn't adjoin the one real root of x^7 - 11 = 0, otherwise we'd just get back R wouldn't we?

[oblique river]

but that wouldn't be the splitting field

which is what you asked for

I mean

x^7 + 11 also has a real root

[chilly ocean]

right, just wanted to clarify...

[oblique river]

yeah

[chilly ocean]

oh yeah. but that one real root doesn't generate the other 6 imaginary roots

[oblique river]

x^7 + 11 isn't irreducible over R

correct

[raw moth]

Okay what I've gathered is I should sleep

[chilly ocean]

That's where I get confused actually, I know x^7 + 11 isn't irreducible over R since its degree higher than 2, but idk what conclusions to draw from that

[oblique river]

if a poly is irreducible then adjoining any root will giev isomorphic fields

like if f is irred. over a field K and if a and b are roots of f

the K(a) is iso. to K(b)

[chilly ocean]

Okay, so in this case, it isn't irreducible since if we adjoin the real root we just have R which isn't isomorphic to C

[oblique river]

yep

but adjoin any other root and you get C

[raw moth]

But this isn't iff right?

Eg x^4+1

[oblique river]

x^4 + 1 has no real roots, correct

[raw moth]

But is reducible in R

[chilly ocean]

Yeah I read about cyclotomic fields and didn't understand a thing

[oblique river]

correct

[raw moth]

Okay cool

[oblique river]

that's a good example of a non-irreducible poly that doesnt have any roots

[chilly ocean]

Okay that's exactly what I thought too, but there is this theorem in my book that seems to contradict it.. I will find it very quickly.

This: "Theorem. Any polynomial of positive degree with real coefficients can be
factored into a product of linear and quadratic terms with real coefficients."

[oblique river]

mhm

[chilly ocean]

How would we apply that to x^4 +1 or any cyclotomic type

[oblique river]

that poly can be factored as a product of two quadratics

[chilly ocean]

okay what about x^7 -1 = (x-1)(1 + x + x^2 + x^3 + x^4 +x^5 + x^6)

how can we factor the second term in that product?

[oblique river]

the statement of that theorem doesn't tell you how to do it

it just says that it's possible

but in this case it will factor into a product of 3 quadratics

if you're looking at things like x^n - 1

it'll be 2 linears + the rest quadratics if n is even

and 1 linear + the rest quadratics if n is odd

[chilly ocean]

Oh I see.. and those quadratics will be the terms "containing" all the nonreal roots?

[oblique river]

yep

over the real numbers, you just have R and C

[delicate bloom]

interesting, I think I accidentally proved that once while failing to try to prove the Eisenstein irreducibility criteria

[oblique river]

real roots will be linear

and nonreal roots will be quadratic

in pairs

[chilly ocean]

Right because complex also have their conjugate as a root

[oblique river]

yep

[chilly ocean]

Is there an efficient way to find this factorization given an arbitrary polynomial?

[oblique river]

like, to actually find the factors? or just to figure out if they're linear or quadratic

[chilly ocean]

To find the factors themselves... but I guess that would really just amount to solving the polynomial wouldnt it

[oblique river]

yeah

knowing the factors is equivalent to knowing all the roots

in this case

[chilly ocean]

Okay interesting. Thank you for clearing that up. That splitting field over R was getting to me

[oblique river]

in general you should expect questions "over R" to be easy

which is just to say that the only options are R or C

[gloomy garnet]

So I'm working with finding splitting fields over Q for given polynomials, and what I'm not sure how to do is find the degree of that field over Q, denoted $[E:\mathbb{Q}]$

[cloud walrus]

Sonicollin:

[gloomy garnet]

where E is the splitting field

for example if E = Q(zeta_6), what's the degree

[hot lake]

the degree [E:Q] is usually defined as the dimension of E as a Q-vector space

[gloomy garnet]

so I need a set of vectors that spans E where the coefficients are elements of Q?

[hot lake]

yes having a basis of E would be nice

[gloomy garnet]

ok that's where I get stuck with nth roots of unity (e.g. zeta_6)

I know that for Q(i), a basis over Q would be {1, i}

so for Q(zeta_6), would {1, zeta_6} be a basis over Q?

[hot lake]

well, would that be a field ?

what would be the coefficients of zeta6 * zeta6 ?

[gloomy garnet]

zeta6 squared is zeta3 right?

[hot lake]

hm yes

[gloomy garnet]

well zeta3 contains a sqrt(3)

so would that have to be thrown in as well?

[hot lake]

a sqrt(3) ??

[gloomy garnet]

right?

[hot lake]

no

[gloomy garnet]

what about the zeta_n = cos(2pi/n) + i*sin(2pi/n) definition?

[hot lake]

would you happen to know the minimal polynomial of zeta6 by any chance ?

yes

[gloomy garnet]

then wouldn't zeta3 = (-1/2) + i*(sqrt(3)/2)?

[hot lake]

yes ?

[gloomy garnet]

so that's where I'm getting the sqrt(3)

[hot lake]

that doesn't mean sqrt(3) will be in Q(zeta3) or in Q(zeta6)

[gloomy garnet]

well that's confusing

[hot lake]

do you have computed the real parts and imaginary parts of zeta6 too ?

[gloomy garnet]

well let's see

isn't it (1/2) + i*(sqrt(3)/2)?

[hot lake]

yeah

so do you think it's possible to write zeta3 as a linear combination of 1 and zeta6 with rational coefficients ?

[gloomy garnet]

ahh I see

yes it is

wait

[hot lake]

if it is then it shows that the Q-span of 1 and zeta6 is a field, Q(zeta6)

and since zeta6 is not rational then you know that 1 and zeta6 are independant so it's a basis

[gloomy garnet]

ok nice

so why do I not have to check any other powers of zeta6?

[hot lake]

well you could

[gloomy garnet]

also I do happen to have the minimal polynomial of zeta6, and I know that it's degree is 2 which means [E:Q] is also 2

but I don't actually know the reasoning behind it very well

[hot lake]

well say x = zeta6

if x² wasn't in the Q-span

you would add it to your vector space and then try to write x³ as a linear combination of 1,x,x²

etc

[gloomy garnet]

ok

[hot lake]

and when you can write x^n as a combination of 1,x,x²,...,x^(n-1)

well it means you have found a polynomial f of degree n with rational coefficients such that f(x) = 0

[gloomy garnet]

hmm yeah you're right

[hot lake]

so back there I made you find the minimal polynomial of zeta6 again

[gloomy garnet]

linear algebra theory right?

[hot lake]

well yeah if you haven't done linear algebra, field theory is going to be nigh impossible

[gloomy garnet]

yeah I have haha thankfully

ok now it's coming together for me

[hot lake]

so once you have a vector space E = span(1,x,...,x^(n-1)) and you know that x^n is in E

then you know that multiplication by x is a map from E to E

because the image of each basis vector is in E

and multiplication by x is a linear map

so you don't need to check the higher powers any more

[gloomy garnet]

ok or in other words you can write x^n as a linear combination of the other vectors

[hot lake]

and in order to be a field you need not only stability with multiplication but also existence of inverses

and that's where the minimality of the polynomial comes in

[gloomy garnet]

hmm

[hot lake]

if y is some nonzero element of E

you can define a multiplication by y map

E -> E

and y has an inverse <=> 1 is in the image of it

<=> the image is E

<=> the map is injective (linear algebra !)

<=> the kernel is trivial

but now you ponder

what does it mean to have a nontrivial kernel

[gloomy garnet]

uh huh

[hot lake]

and it should imply somehow that your polynomial f(x) was reducible

I feel I'm missing a bit of the proof

say y = g(x) and you have found z = h(x) such that z*y = 0, you get g(x) * h(x) = 0 modulo f(x) (as polynomials in Q [x]) but that's not exactly what we want

[gloomy garnet]

ok

[hot lake]

anyway, g and h would have smaller degree, and one of g(x) and h(x) would need to evaluate to 0

which means you get a smaller polynomial than f that annihilates x

[gloomy garnet]

annihilates x?

[hot lake]

if x is in C, a polynomial f annihilates x if f(x)=0

[gloomy garnet]

ah

[hot lake]

and then you compute the gcd of f with (g or h) to get a factor of f which shows that f wasn't irreducible

i guess

so if you miss the fact that zeta3+1 = zeta6

[gloomy garnet]

I found that

[hot lake]

and go and find the polynomial f(x) = x^3+1 instead

well that polynomial isn't irreducible

so that's how you would know you have missed something

[gloomy garnet]

hmm right

you can only "mod out" by irreducible polynomials right?

[hot lake]

you can mod out by anything in Q[x]

except 0

[gloomy garnet]

oh in order to get a field I mean

[hot lake]

yeah

[junior quarry]

Hey @everyone, I'm trying to get into higher math and I like algebra because of its relationship to type theory and computer science. I have a couple of algebra books such as Arten (used in the Harvard online course), but I find it easier to learn things by making projects that utilize the concepts I want to learn (and I guess a lot of programmers are the same way). Are there any project-based abstract algebra courses online or elsewhere that any of you are aware of or have experience with?

[gloomy garnet]

oh btw thanks a ton @hot lake . You were a big help!

[outer monolith]

Hi anyone on here?

I am trying to define a congruence relation for the Hurwitz integers: https://en.wikipedia.org/wiki/Hurwitz_quaternion

Such that if you have Hurwitz integers like A, B and N,

A is congruent to B (mod N)

i'm not sure how that would look like, or how to define something like that and prove it

[oblique river]

In any reasonable definition of this, A being congruent to B mod N should be equivalent to (A-B) being divisible by N

that seems like a fine condition

[outer monolith]

thanks!

Sorry, my maths isn't that strong, and I have a difficult(for me) research task

so how would this work with other operations

like if i want to compute B = A^2 mod N?

[oblique river]

what do you mean? that's not a computation

that's a statement

which (according to our previous definition) means that B - A^2 is divisible by N

[outer monolith]

If B is unknown?

[oblique river]

I'm not sure what you mean then

[outer monolith]

I'm not sure either 😅

Let me think and try to form the right question

[smoky cypress]

So you're trying to calculate A^2 modulo N, when you are actually given A?

[oblique river]

oh, that's a different thing then

[outer monolith]

yes, and N is known

[oblique river]

why not just reduce all of the coefficients modulo N

[outer monolith]

err N is anytthing

[oblique river]

which is ok if N is odd at least

[outer monolith]

N is always odd in my case

[oblique river]

great

just reduce the coefficients mod N

[outer monolith]

coefficients of the quaternion correct?

[oblique river]

yeah

[outer monolith]

N is also a quaternion

[oblique river]

oh

what does it mean for N to be odd, then

[outer monolith]

the norm(N) is odd

[oblique river]

I see

if you know about rings and ideals

then you're just looking to describe the quotient ring H/(N)

(here I think we want (N) to be the two-sided ideal but I'm not totally sure since I don't deal with noncommutative rings all that much)

[outer monolith]

so primes form an ideal in Z correct?

[oblique river]

no

[outer monolith]

hmm

so

would the ring of odd norms form an ideal?

[oblique river]

no

[outer monolith]

I think yes, since H is an euclidean domain

[oblique river]

the sum of two odd norms need not be an odd norm

[outer monolith]

I think a similar argument can be made for H

https://en.wikipedia.org/wiki/Gaussian_integer#Principal_ideals

[oblique river]

the problem is that H isn't commutative which kinda screws things up

you're right that all left ideals are principal

as are all right ideals

but when you want to talk about "mod N" you need to determine if you mean left ideals or right ideals or two-sided ideals

[outer monolith]

my grasp on all of this is tenuous as it is 😦

[oblique river]

if you want to actually get a quotient ring though then you need a 2-sided ideal

yeah I don't know much noncommutative ring theory

sorry

[outer monolith]

haha its ok

thanks for your help

[oblique river]

np and gl

[chilly ocean]

so far i have done

How am I supposed to show part B?

[mild laurel]

Did you read the hint

[chilly ocean]

Yes

Am I reading this question correctly

From what I understand

The question is asking me for some power of (2+i) that equals an integer then it must be some multiple of 5

First I need to show that some power of (2+i) is 5

I have no clue how to do this

[oblique river]

no power of (2+i) is 5

[stone fulcrum]

<r> is all members of the form ar
So there's some number a ∈ Z[i] such that a(2 + i) = 5

[chilly ocean]

O

I thought that meant a generator

[stone fulcrum]

It does haha

But that's what it generates

Note that all powers of r is not usually an ideal

[solemn rain]

quick question

let G be a group , H be an abelian group , and f:G--->H is a group homomorphism. Show that [G,G] is contained in ker(f)

proof: let x be in [G,G]

x = [x_1,y_1][x_2,y_2] ..... for x_i,y_i in G for all i

f(x) = f([x_1,y_1][x_2,y_2]......) = f([x_1,y_1])f([x_2,y_2]).... = [f(x_1),f(y_1)][f(x_2),f(y_2)]......

= 1

hence x in ker(f)

[outer meadow]

you dont need to do this nonsense with so many x and y, just do it for one

[solemn rain]

yea it should be

x = [x'^a,y^b]

for some x and y generators

?

or is it

x = [x,y]^t

for some int t

i think its this 1 ^

[upper pivot]

just show it for arbritary x,y

[G,G] is generated by [x,y] for arbriatary x,y

so showing that is enough

[solemn rain]

cool

so am i right?

[upper pivot]

yeah

[solemn rain]

first GT problem ever woo

[chilly ocean]

i sort of did part b?

why do i need to show 5 is in <2+i>?

[mild laurel]

The picture's pretty hard to read

[chilly ocean]

uh click open original

should give a bigger clearer picture

[mild laurel]

Yeah, that's the right idea

you need to show that 5 is in <2 + i> since that's one inclusion

That shows that 5Z is a subset of <2 + i> intersect Z

The work afterwards shows the other direction

[chilly ocean]

?

so

wait

5 in <2+i> tells me

what again?

[mild laurel]

that 5Z is a subset of <2 + i> intersect Z

[chilly ocean]

and then when i showed <2+i>intersectZ = 5Z

in the next part

just showed

[mild laurel]

No you didn't

You only showed that <2 + i> intersect Z is a subset of 5Z

[chilly ocean]

ah shit

so i still need to show

that 5Z is a subset of <2+i> intersect Z?

[mild laurel]

But that's what showing 5 is in <2 + i> shows

[chilly ocean]

was the 2ndd part pointless?

[mild laurel]

No, again, the second part shows that <2 + i> intersect Z is a subset of 5Z

This is how you show two sets A,B are equal. You show that A is a subset of B and B is a subset of A

[chilly ocean]

ok

so

what about part a and b shows c

Z[i]/<2+i>

the elements of it have the form a + bi + <2+i> correct?

[mild laurel]

yeah

so part (a) shows that every element of Z[i] can be represented by an integer

and part (b) shows which integers are enough

[chilly ocean]

Is this better?

[mild laurel]

yeah, that's the right idea

[tribal pasture]

Can anybody tell why g being a class function implies that it becomes 0 when the identity is satisfied

@oblique river

[oblique river]

are the chis the characters of irreps

[tribal pasture]

Yes

[oblique river]

that sum is the inner product of g and chi

[tribal pasture]

Yes

[oblique river]

or maybe |G| times that

[tribal pasture]

Yes yes |G| times

[oblique river]

the chis form an orthonormal basis with respect to that pairing

so if you pair to 0 with every basis element

you are 0

it's just linear algebra really

[tribal pasture]

Yes but we aren't summing over the chis

The chis form an orthonormal basis, yes, I agree but over here the sum is being evaluated over the group elements

[oblique river]

yes lol

that is the inner product

the inner product is not "sum over chi"

that wouldn't make sense

[tribal pasture]

But in this case then the basis should be X1(s_i) right for all i ?

[oblique river]

what?

chi(s_i) is a complex number

the set of class functions of G has the structure of a complex vector space

the characters of irreps form a basis

which is orthonormal with respect to that pairing

[tribal pasture]

YES Okay so the characters form a basis right of the class function space

[oblique river]

yes

[tribal pasture]

OHhh I think I got you. What you are saying is the fact that the sum is 0 implies that the inner product between g and X1 is 0. Correct?

[oblique river]

I mean like we agreed earlier, that sum is the inner product (up to a factor of |G|)

so if the sum is 0

[tribal pasture]

Yes yes

[oblique river]

then of course the inner product is 0

because they are the same

[tribal pasture]

And why does inner product being 0 implies g is 0 ?

[oblique river]

because the inner product is 0 for all chi

this is now just linear algebra

<g, chi> = 0 for all chi

and the chi form a basis

therefore g = 0

[tribal pasture]

I mean they only evaluated for chi1?

[oblique river]

it literally says "for all chi"

[tribal pasture]

Ahhhh Right right thanks! Sorry for the silly questions

[chilly ocean]

For part C

is this a suitable answer?

or am i missing something?

[vale coral]

Cam someone explain me the steps in the follow division? - concrete abstract algebra example 5.3.3? It is primarily the concept of dividying by both f1 and f2 at the same time that confuses me.

[woeful flint]

im trying to show that $1/p$ has decimal expansion with period that divides $p-1$. To do this I let $1/p = 0.\overline{a_1\cdots a_n}$ and so we must have that $p(\underline{a_1\cdots a_n}) = 10^n - 1$ i.e. $10^n = 1\pmod{p}$. Can we just use Lagrange's Theorem here or do we need to show something more?

[cloud walrus]

Kraft Macaroni:

[knotty mason]

From Fermat's little theorem you know that $10^p = 1 \pmod p$, and $p$ is the first time this occurs. You have $10^n = 1 \pmod p$ so you can use that to show that $n$ is a multiple of $p$. And you can also get some conditions about $p$ not being a multiple of $2$ or $5$.

[cloud walrus]

skymoo:

[compact plover]

How do you find the conjucacy classes of a sigma in S10?

[raw moth]

have you seen disjoint cycle decomposition?

[compact plover]

I figured it out! thank you though

[raw moth]

ah, no problem

if you haven't seen it before and have some spare time, you should think about what conjugacy classes look like in A_n

[patent girder]

Does there exists a finite set S and a partial order ≤ such that there is NO integer n such that S(n) has that type of partial order?

Can someone help me?

[woven delta]

what does S(n) mean here?

@patent girder

[patent girder]

its a function

[woven delta]

but S is a finite set

is S(n) the successor of n?

[patent girder]

does all partial orders of finite sets appear as the partial order in the divisor of some integer? is basically what its asking

[woven delta]

oh I see

well one thing is that there is exactly one maximal element

you can have lots of maximal elements in arbitrary partial orders on finite sets

assuming you are saying what I think you're saying

also I guess 1 is minimal?

[patent girder]

i didn't really understand the question which is what i was asking about

[woven delta]

so given a number n, you can make a factor tree of n

this tree is a partial order if you say j < k if j|k

factor tree here means all the elements m with m|n

[patent girder]

ok

i get that

[woven delta]

so 1 is minimal and n is maximal

so there is exactly one minimal and exactly one maximal element

[patent girder]

and the remaining elements are the factors of n, right?

[woven delta]

all the elements are the factors of n

its pretty easy to construct other invariants of the factor tree

for example the number of elements a, b, c where a<c, b<c, but a and b are not comparable

if you look at prime powers it is easy to get a lower bound

[patent girder]

ok thankyou

[woven delta]

but yeah the maximal and minimal elements are the easiest invariants

[solemn rain]

how can i prove

that a group mod maximal subgroup is simple

[woven delta]

lol

4th isomorphism theorem

[willow forum]

cyclic subgroups can be infinite right?

[woven delta]

only if they're Z

[solemn rain]

yes ntky

[willow forum]

dont mean the intrustion, just looking for a definition clarification

yepyep that's what I meant

some <a> in Z

thanks !

[solemn rain]

oh okay so

the group mod maximal is not just simple

it has no subgroups

at all

?

right?

[woven delta]

yes

[solemn rain]

cool

ty

[woven delta]

this is immediate

[solemn rain]

yea yea fuck this shit i am just going to write like umm

fourth isomoprhism theorem

lol

qed

t

y

[woven delta]

assuming the maximal subgroup is normal obviously

[solemn rain]

yes

didnt mention that mb

Let G be a finite group, let H be a normal subgroup ofG and let P be a Sylow p-subgroup of H. Then G = HN_G(P)

do i prove this using G ~ HxK

if H and K are normal and intersect at 1?

[woven delta]

it's wrong

[solemn rain]

what

whats wrong

Let G be a finite group, let H be a normal subgroup ofG and let P be a Sylow p-subgroup of H. Then G = HN_G(P)
?>

[woven delta]

you need more conditions

[solemn rain]

1 sec

this is frattini argument imma just google

[woven delta]

oh ok

[solemn rain]

fuck i wnated to prove this on my own

now i cant lmao

[woven delta]

isn't the trivial subgroup normal though?

[solemn rain]

it is but it doesnt have a sylow p subgroup

right?

[woven delta]

oh I read that as a sylow p subgroup of G

fuck

[solemn rain]

okay cool

now cani prove this using what i said

wtf

who deleted

wtf

[woven delta]

I deleted

[solemn rain]

why

[woven delta]

cause I didn't want to see it

[solemn rain]

what

this is my msgn ot yours

but okay

can i prove this

using G ~ HxK

?

where H and K are normal

and intersect at 1

[woven delta]

I don't see why that's relevant

[solemn rain]

G ~ HK* sorry

[woven delta]

N_G(P) is not normal

[solemn rain]

yea

so no

okay

wait why

[woven delta]

also they intersect nontrivially

why would it be normal

[solemn rain]

yea yea

sorry

[woven delta]

normalizer is the largest subgroup of G st P is normal in it

[solemn rain]

ye ayea

i got the proof

thank you

[woven delta]

lol what

[solemn rain]

i gotg the proof

[woven delta]

did you read the proof online?

[solemn rain]

yes

[woven delta]

smh

[solemn rain]

no

it was you

i had good intentions

bro

[woven delta]

you didn't have to read the proof

just look at the statement

smh

[solemn rain]

oh wow

i just lok at the statement

and im done

?

damn ur good

[woven delta]

I mean to confirm that the problem was correct

[solemn rain]

okay now

1 more problem im sorry

this is the big one

i am shit with nilpotent groups

i cant even remember their definition lmaoa

now how can i prove

that A finite group is nilpotent if and only if every maximal subgroup is normal

so ig a nilpotent gruop G is a any group such that

Z_c(G) = G for some c

now i really dk what Z_i(G) mean

i think Z_i(G) = Z_i-1(G/Z_i) something like thjat

[woven delta]

Z(G/G_i)

[solemn rain]

Z_i(G) = Z(G/G_i)

and G_i is the group of index i in the

umm

composition

series right?

[woven delta]

G_0 < G_1 <...<G_n=G

where < is normal subgroup

[solemn rain]

yea thats the composiition series

oh

[woven delta]

and G_{i+1}/G_i subgroup of Z(G/G_i)

[solemn rain]

okay

so this is called the lower central series

G_0 < G_1 <...<G_n=G
^

right?

[woven delta]

I am reading this off wiki lol

[solemn rain]

yea nilpotent groups suck right?

[woven delta]

this is just central series

idk if they suck

[solemn rain]

okay so u dont know about them

?

[woven delta]

since idk why they are useful

I can still do this kind of problem

[solemn rain]

yea okay any hints?

[woven delta]

we'll see once I do it

[solemn rain]

okay

[woven delta]

probably maximal subgroup is normal implies nilpotent is trivial

We know G_n/G_{n-1} is contained inside Z(G_n/G_{n-1}). Obviously this tells us G_n/G_{n-1} is abelian, so every subgroup is normal.

this gives us the nilpotent implies maximal subgroup is normal direction

[solemn rain]

okay

[woven delta]

do you see why?

[solemn rain]

yea cuz G is nilpotent

[woven delta]

it's cause 4th isomorphism theorem

[solemn rain]

yes

yea yea

i thought We know Gn/G{n-1} is contained inside Z(Gn/G{n-1}).

was cuz of nilpotent

lmao

[woven delta]

yeah

I was asking about why this implies a maximal subgroup is normal

[solemn rain]

if G_n/G_n-1 is abelian

does this mean G is abelian

[woven delta]

no

[solemn rain]

is this statemtn for all n

[woven delta]

no

G_n = G

[solemn rain]

oh

n is the class irght

nilpotency class?

or no no wait wtf

ugh sorry i dont get it now

i am really not that good with the definitions

u assumed G is nilpotent

so there exists c such that Z_c(G) = G

Z(G/G_c) = G

[woven delta]

I'm using the wikipedia definitions lol

[solemn rain]

bro im not that good to just

go from def to results

so please slow on me

okay so Z(G/G_c) = G

[woven delta]

okay so we are looking at G_n/G_{n-1}, where G_n is G

[solemn rain]

for some c

okay now lets considier G_n/G_(n-1)

now whats n

okay well we have this

[woven delta]

it's the n st G_n=G

[solemn rain]

composition series

G_0 < G_1 < .... G_n-1<G_n = G

now

[woven delta]

this isn't a composition series

[solemn rain]

norm al

composition series

< means normal yea

okay so

[woven delta]

yeah

[solemn rain]

this is nilpotent

so G_n/G_n-1 is abelian?

isnt this for solvable?

lmao

sorry

wait

G_n/G_n-1

is in Z(G/G_c)

[woven delta]

okay so we have $G_n/G_{n-1}$ is contained in $Z(G_n/G_{n-1})$

[cloud walrus]

Liquid:

[woven delta]

so it is abelian

[solemn rain]

okay

[woven delta]

which gives us a correspondence

between normal subgroups in $G_n$ that contain $G_{n-1}$

[solemn rain]

yes

yea

[cloud walrus]

Liquid:

[solemn rain]

thatshio fourthi so

[woven delta]

and subgroups in $G_n$ that contain $G_{n-1}$

[solemn rain]

okay now i get why G_n/G_n-1 is abelian

[cloud walrus]

Liquid:

[solemn rain]

how do u say all subgroups are normal then

[woven delta]

not sure, only seeing at least one is normal so far.

maybe you can do some third isomorphism theorem stuff

hmm

[solemn rain]

tbh idk at all really

i give up

[woven delta]

okay let's try the other direction instead. if every maximal subgroup is normal then the quotient by a maximal subgroup is in fact a group of order p

[solemn rain]

yea

tahts by some theorem actually

lmao how did u know this xd

some bigass theorem

[woven delta]

sylow

[solemn rain]

no not sylow

[woven delta]

it's immediate from sylow

[solemn rain]

oh yeah nvm

mi mxied with

with the theorme on p groups

that all maximla subgroups are of index p

[woven delta]

so by choosing a maximal subgroup N consider 0 < N < G

[solemn rain]

and normal

[woven delta]

I think that works

[solemn rain]

thats for p groups tho

[woven delta]

so maximal implies only subgroups of quotient is 0 and G/N

apply sylow to get existence of a subgroup of order p for any prime factor p

and you're done

[solemn rain]

okay i get the first line

[woven delta]

wait I think I'm wrong

which is alright

[solemn rain]

i dont get how im done

[woven delta]

okay so the first terms in the sequence should be N < G

[solemn rain]

yea

and N is normal

[woven delta]

I think you have to use all the maximal subgroups

[solemn rain]

i dont see how u can get from this

to proviing there is c such that Z_c(G) is G

[woven delta]

okay so I am building the sequence backwards from how you are

[solemn rain]

wait

[woven delta]

can you post your definition?

[solemn rain]

let P be asylow p subgroup of G

if P is normal then im done

cuz

[woven delta]

lol

[solemn rain]

if every sylow

is normal then G is nilpotent

[woven delta]

this seems unrelated

[solemn rain]

no bro

if i let P be asylow p subgroup

and show its normal im done

[woven delta]

why would that be true?

[solemn rain]

okaay so

suppose P_i is normal in G for all i

like suppose all sylow p subgroups are normal

[woven delta]

okay but why would all sylow p subgroups be normal

[solemn rain]

cuz its nilpotent

[woven delta]

that seems much stronger than nilpotent

that implies you are a product of your sylow p subgroups

[solemn rain]

yea ur right

now that implies its nilpotent

[woven delta]

okay so we start with N < G

now take a different normal subgroup N' which intersects N nontrivially.

[solemn rain]

contradiction

suppose P is not normal in G

N_G(P) is in M

cuz P is a subgroup of N_G(P)

and M is maximal

M is normal by assumption

boom

farttini go brr

now what

XD

G= MN_G(P)

now what

[woven delta]

sorry what?

[solemn rain]

frattinis argument

the thing i wanted to prove

earlier

[woven delta]

nope

you seem to be claiming that N_G(P) is inside M

which would give us a contradiction

but it's not necessarily

[solemn rain]

it should bro

cuz M is maximal

[woven delta]

nope

[solemn rain]

and P is a subgroup of N_G(P)

^

by induction

[woven delta]

there are other maximal subgroups though

yeah sorry your argument is bad

[solemn rain]

😢

[woven delta]

lol

[solemn rain]

bro listen

[woven delta]

tbh idk if the statement you made is true

[solemn rain]

bro

bro

if H < G

then H < N_G(H)

H is asubgroup of G

[woven delta]

lol

[solemn rain]

that is

if G is nilpotent ofc

[woven delta]

i'm not sure what you're saying

[solemn rain]

let G be anilpotent group

and H be a subgroup of G

then H is a subgroup of N_G(H)

good?

[woven delta]

this is always true

not sure what your point is

[solemn rain]

okay

so now if M is maximal

then M contains N_G(H) by default

as H is a subgroup of M

isnt that why u said my argument is wrong

[woven delta]

apparently the statement you made is true btw (the nilpotent implies sylow p subgroups are normal)

[solemn rain]

as M should not always ocntian N_G(H)

yea cool

[woven delta]

why does M contain N_G(H)

[solemn rain]

cuz H is in M

[woven delta]

so?

[solemn rain]

N_G(H) is a subgroup

M is maximal

[woven delta]

lol

there are many maximal subgroups

why can't some other one contain N_G(H)

[solemn rain]

dont they all do

any subgroup is contained in a maximal subgroup no?

how wouldnt a maximal subgroup not contian

another smaller subgroup

[woven delta]

what

[solemn rain]

not contain*

ur implying that there can exist am aximal

subgroup that cant contain N_G(H) right?

how tho i dont understand sorry

im slow

[woven delta]

are you saying that every maximal subgroup contains any group?

[solemn rain]

any subgroup

is contained in a maximal subgroup

thats what im saying

[woven delta]

okay, your proof is actually almost good

[solemn rain]

oh really

why almost

[woven delta]

yeah

[solemn rain]

whats wrong

[woven delta]

cause you may need to switch up the maximal subgroup you're looking at

you have to say "let M be a maximal subgroup that contains P and N_G(P)"

[solemn rain]

it ocntains P by definition

i dont know why it doesnt contain

[woven delta]

not just "let M be a maximal subgroup that contains P"

[solemn rain]

N_G(P) by definitio

[woven delta]

not every maximal subgroup contains N_G(P)

that would be dumb

[solemn rain]

xd

ty

okay

so what happens

if M doesnt contain N_G(P)

[woven delta]

then you just find some M' that does

lol

[solemn rain]

bro

are u sure bro?

[woven delta]

this is why you need all maximal subgroups to be normal

[solemn rain]

u remember G = MN_G(P)

[woven delta]

not just one

[solemn rain]

right?

by frattini

[woven delta]

lol

you can apply frattini to M' and P

cause P is inside M'

[solemn rain]

can i say this

let M be a maximal subgroup that contains N_G(P)

G = MN_G(P)

[woven delta]

yeah

[solemn rain]

N_G(P) is in M tho

so MN_G(P) is M

now what XDDD

[woven delta]

so that gives us a contradiction

[solemn rain]

what contradiction lmfao

[woven delta]

it contradicts frattini

[solemn rain]

mmm yea

okay so im done herre?

[woven delta]

or you could say "suppose P not normal"

[solemn rain]

yea i did that

ye ayea i rememebr now

okay did we do the other part?

the other direction?

[woven delta]

well you're not done

[solemn rain]

i didnt do the other part

did i

[woven delta]

you have to use the sylow p subgroups to construct the series

[solemn rain]

yea

i have this proved already from b4

if G is a direct produtct

[woven delta]

okay cool

[solemn rain]

of syulows

[woven delta]

then you're done

[solemn rain]

then its ni;lpotent

okay

did we do

the other direction

and ifits nilpotent its also a direct product of ..

aswell

[woven delta]

no

[solemn rain]

man this group shit is getting confusing as fuck

too many things

lmao

its like cuphead

[woven delta]

eh it's not too bad

[solemn rain]

its bdsm tho

okay anwyays

did we do

the other direction

[woven delta]

we showed there's at least one normal maximal guy

but you have to show every maximal subgroup is normal

or every sylow p subgroup is normal

[solemn rain]

wait

i said

did is ay that

P is as ubgroup of N_G(P)

cant we do that on M

and then do a contradiction

[woven delta]

how would that work?

[solemn rain]

idk D

let G be a nilpotent group

now by all these theorems i am reading about

if H is a subgroup then H is a proper subgroup of normaliser of H

let M be am ximal that is not normal

[woven delta]

obviously

[solemn rain]

boom M is not maximal no more

[woven delta]

why do you keep stating that like its a big fact

[solemn rain]

cuz N_G is not M

[woven delta]

what are you proving

[solemn rain]

that if G is nilpotent then

all maximals

are nmormals

[woven delta]

how did you use nilpotent

[solemn rain]

bro

i swear bro

are u sure

its notab ig fact?

that H is a subgroup]

of N_G(H)
that

that uses nilpotent

[woven delta]

no

[solemn rain]

okay

u know what

prove it for me please

[woven delta]

obviously H normalizes itself

suppose g \in H, h \in H

then hgh^{-1} \in H

[solemn rain]

N_G(H) = {h | ghg^-1 is in G}

right?

the set of elements of H that satisify normalityh

in G

[woven delta]

no

N_G(H) = {g | ghg^-1 is in H}

for all h in H

lol

[solemn rain]

uh oh

yea

ur right

sry xd

[woven delta]

you kept invoking that as a big fact

[solemn rain]

this shit is confusing as fuck

[woven delta]

I was like wtf

[solemn rain]

yea mb

but ey

hey

i am really confsued with Normalizer right now but ik this

if H is normal in G

then N_G(H) = H

right?

now

[woven delta]

no

[solemn rain]

wtf

no

G

[woven delta]

N_G(H)=G

[solemn rain]

N_G(H) =G

yea

okay so is N_G(H) = { g | gH=Hg} ?

[woven delta]

yeah

[solemn rain]

now how is H a subgroup of that

[woven delta]

it's the largest subgroup in which H is normal

what

[solemn rain]

XD

[woven delta]

wtf are you saying

[solemn rain]

how is H < N_G(H)

[woven delta]

because every element of N_G(H) normalizes H

by definition

and every element of H normalizes H

[solemn rain]

okay

my brain got fried

okaay

i goti t

now

back to this

suppose G is nilpotent

we want to show M a maximal subgroup is normal

arbitrary

so let M be a maximal subgroup that is not normal

okay?"

u with me

?

[woven delta]

yeah

lol

[solemn rain]

okay so

by the easy fact that im too bad

to know

M is a subgroup of N_G(M)

now

N_G(M) is not G bro

cuz M is not normal

contradiction

fuck nilpotency idfc

whats wrong with this

[woven delta]

N_G(M)=M

M is normal in itself

[solemn rain]

okay

now

[woven delta]

lol if what you said was true

[solemn rain]

oh fuck

ur going to make me ruin maths

by my hand now

arent u

[woven delta]

then every sylow p subgroup is normal

[solemn rain]

yes

thats true for nilpotent

[woven delta]

looks like you fucked group theory

[solemn rain]

no no listen bor

thats true for nilpotent

i assumed G is nilpotent

[woven delta]

you didn't use nilpotent though

at all

[solemn rain]

thats what im saying bro

that H is as ubgropu of normaliseri n G

cuz of nilpotency

somehow u proved me wrong

and im super sad

[woven delta]

lol

I'm sad too

[solemn rain]

wait

why are us ad now

[woven delta]

cause you ruined group theory

it's over

[solemn rain]

its not like its ur wife or whatever

fuck this shit

its uselss anyways

[woven delta]

lol

[solemn rain]

listen bro

listen

im sure

i have htis math sense

that nobody else has

wait

if G is nilpotent

this implies ( if H < G then H < N_G(H) )

this implies P_i <| G for all i

( P_i is sylow )

[woven delta]

that's always true though

[solemn rain]

this implies G is a direct prodcut of all sylows

[woven delta]

the first implication

[solemn rain]

yea

i meant

the implication

that this implies all sylows subgroups are normal

is true

[woven delta]

wtf

[solemn rain]

all sylwos
Xd

dk

[woven delta]

wtf are you saying anymore?

[solemn rain]

u know these umm

statements in textbooks

that say

all these conditions are equivalent

tahts what ims aying

[woven delta]

so you're saying that all groups are nilpotent

good to know

[solemn rain]

okay bro give me a moment please

im going to fix all of thsi shit

[woven delta]

lol

is this the definition you are using?

[solemn rain]

I FIXEDD IT BRO

listen

listen

if H < G then H < N_G(H)

this is true

for all groups

[woven delta]

yes

[solemn rain]

but not proper tho

[woven delta]

no

[solemn rain]

yes

[woven delta]

oh

[solemn rain]

nilpotency

saysa

proper

so htis

[woven delta]

oh ok

[solemn rain]

makes

[woven delta]

yeah

[solemn rain]

N_G(M) not M

must be G

no

[woven delta]

that does it

[solemn rain]

yea

so im done right

[woven delta]

alright great

yeah

[solemn rain]

yea

tysm for taking all this time

again