@MatheinBoulomenos Hello :D

oder guten morgen ._.

guten morgen

mathein ._.

on that question of characteristic

Strange fact but kinda cool

if M,N are char

I need to show that MN is also char

i need to show that phi( MN) is contained in MN

but can one consider this phi(M) phi(N)

these kind of polynomials, called eisenstein polynomials are important beyond just being irreducible

they appear in algebraic number theory

Yeah, this is from an ANT book

I see. It's also standard to do this in a Galois theory course, because it's a useful criterion to prove stuff is irreducible

I hope mathein can make some Group then we call it Matheinsgruppe

._.

or mahteinsalgebra

Yeah, this is one my algebra prof told me about some time back as one that sorta included some of the algebra it used

Stewart and Tall, Algebraic Number Theory and Fermat's Last Theorem

I see

Dami :D stop disturbing mathein and let him help me for once

-.-

"for once" :P

@MatheinBoulomenos did you see my Q?

haha

well my solution of that is to consider phi (M) phi (N)

that is contained in MN

It's true that for every group homomorphism phi(MN)=phi(M)phi(N)

also since phi is an automorphism we can write phi(MN)

but you have to prove that

aha

that is kinda why i got here ><

can we write stuff like that?

phi acting on a set

not just an element

ofc we can write it , but how do we do algebra on it

You basically have to plug in the definition of MN

that's all

hmm

@KasmirKhaan I'll gladly help you, but I'm not exclusively here to help you. I want to talk to other people, too

I know that was a joke with dami ><

okay ill keep working then and let you guys in Peace

sorry about this :)

no, you can ask me

What's the definition of MN?

well MN= {mn | m in M, n in N}

okay, so what happens if you apply phi to that set?

well I do get how it is done, but the thing is ,am not Confident that i do the right things

@MatheinBoulomenos have you come across the informal “proof” of the Borsuk-Ulam theorem on S^2, ie “there is a pair of antipodal points on earth with equal temperature and air pressure”? I have a serious doubt on the validity of that proof.

and I dont want to do something and Think it was done corect when it was not

my solution is let m in M, and n in N, phi(mn) = phi(m) phi(n)

@LeakyNun I don't think that statement is supposed to be an informal proof. It's a surprising consequence

but we know that phi(m) is an element in M , so is phi(n) in N

so their Product must be in MN

@MatheinBoulomenos that statement itself is not an informal proof. reparse my question

I think he's saying that he's found a proof that's really informal

And sketch

hence phi(MN) is contained in MN and thus MN is char

@Daminark unlike some other person, I do welcome your help

well, then I don't know the informal proof, no

have you by any chance read it? @Daminark

@KasmirKhaan yes, that's correct

@MatheinBoulomenos :D i need to write alot of proofs to get the experience

on class we did nothing ,so i dont really know if the proofs i do are well written or not

Nope @Leaky

is it just me on tired brain or is this guy not making sense?

challenge for leaky, find a normal subgroup that is not characteristic :D

@KasmirKhaan Z/2Z in V4

-__-

@Daminark @MatheinBoulomenos it goes like 1. pick a pair of antipodes A and B 2. pick a path of antipodes between A and B 3. there must be a zero 4. now do this infinitely often 5. they form a band of equitempertural antipodes which is continuous 6. apply the same argument; my doubt is on 6

nice it works

what if the band that is formed is actually S1 U S1?

I mean, S1 sqcup S1 can be closed under antipodes (it just takes points from one copy of S1 to another copy)

then you can’t actually travel from one point to the other

@KasmirKhaan have you proved that the inner automorphisms form a normal subgroup of the entirety of the automorphisms?

This is a tricky proof to think about

hmm

@KasmirKhaan it’s a good practice for abstract thinking

and symbol manipulations

that is what am very weak at

am trying now

ping me when you have a proof

okay :)

btw I don’t think you proved that the automorphisms themselves form a group

you may want to do that first

well i did

if sigma and tau are in Aut(G)

then their composition is there also

give a rigorous proof or none at all

hmm

don’t give me a proof sketch

but i had hard time starting the proof

so it may be a good practice to redo

like if i had a set of particualar elements i can show it is a Group or not

okay

@KasmirKhaan all the automorphisms of G

i know that the identity map is in Aut(G)

and also the conjugation map

but that is an inner aut(G)

@KasmirKhaan you don’t need it

seriously

the proof is straightforward from definition

Well i did do it starting that aut(G) being a subset of S_G

@KasmirKhaan don’t

but how does one assume elements in aut(G)

use only definition

nothing else

@KasmirKhaan why do you need to?

to prove associatitvy i need 3 elements

dont need exactly 3

phi sigma tau

yeah but ><

how do I know that such elements exist in first place

anyway let me keep thinking

:D

you don’t need to know

have you ever proved groups?

yes

closure is easy

because of composition of maps

if we have a 1-1 and onto map composed with also 1-1 and onto map then the Product is also a 1-1 and onto map

associativity says “for all phi, for all sigma, for all tau, (phi * sigma) * tau = phi * (sigma * tau)”. where do you need to prove that they exist??!!

-.-

ill come up with good proof in a short

ill ping ya :D

would you stop the dramatic responses

I’m sure I’m not the first one to complain about that

i dont talk to many ppl

you Ted Mathein and Anon =p

sigh

so yeah just you who commented now but anyway, am thinking of it really

@LeakyNun okay

starting with closure

if sigma and tau are in Aut(G) then sigma composed with tau is also there

this comes from composition of functions

associativty also follows from composition of functions

stop

why

Aut(G) isn’t just the bijective maps

yes we know that if sigma nad tau are hom, then sigma tau is also a homorphism

t(ab) = t(a) t(b)

you need to state that

even if you don’t prove it

s(ab) =s(a) s(b)

so st(ab) = s (t(a) t(b) ) = st(a) st(b)

because both s,t are hom

the identity map is also in Aut(G)

go on

if tau is a hom so is tau inverse

why

because we showed Before that inverse of 1-1 and onto map is 1-1 and onto

also if tau is a hom, so is tau ^-1

@KasmirKhaan why

give a complete proof of that lemma

if t(ab) = t(a) t(b)

and I will be impressed

i take t (t'(a) t'(b)) = tt'(a) tt' (b)

=ab

not looking very nice

but i hope you see what i did with notation

now i take t' of all that

t'(ab) = t'(a) t'(b)

because t' (tt')

is that how you would write in an exam?

no but it is hard to type clear here

without latex

t (t'(a) t'(b)) = tt'(a) tt' (b) =ab

now

t^-1 ( t (t'(a) t'(b))) = t^-1 ( tt'(a) tt' (b)) = t^-1 (ab)

aha

@LeakyNun here leaky ? :D

what

so suppose sigma_a in inn(G)

and tau in Aut(G)

we look at tst'(g)

ts(t'(g)) = t (at'(g) a') = t(a) g t(a) '

because t is an auto morphism

´we can split that Product that way, and move the inverse sign up

tst'(g) is also an inner automorphsim

because we get an inner automorphism by conjugation by an element of G

did you come up with that proof yourself?

yes sir

why ?

good

:D

now complete the lemma earlier

what lemma ._.

inverse of automorphism is homomorphism

aha

t^-1 ( t (t'(a) t'(b))) = t^-1 ( tt'(a) tt' (b)) = t^-1 (ab)

i assume this one

so here since t is assumed to be an aut

LHS

( t (t'(a) t'(b))) = tt'(a) tt'(b)

i am lost in notation now ><

let me regroup

ah yes

(t^-1 t) ( t'(a) t'(b)) = ( t'(a) t'(b))

we do t' t first

to give us identity

so LHS ( t'(a) t'(b)) and RHS t' (ab)

and we are done :D

good

yeey :D

the key idea to recognize that inn(G)

its elements are of the form sigma_a

conjugation by sme element of G

yes

other than that, the rest is just manipulating composition of functions

@LeakyNun want screen shot of the problems im gonna do next ?

I allready having problem with the first ><

ok

well need to spend some time thinking of why the order was even mentioned there

in proiblem 30

also need to go eat something :D so ill ping ya when am back :)

thanks again leaky :D

@Daminark nice hat

Hint: how many normal subgroups of order $p$ does $G$ have?

Also merry Christmas to everyone!

@TheNotMe My understanding of partition numbers: oeis.org/A160096

the three isomorphism theorems in terms of category theory

@Daminark @MatheinBoulomenos

@MatsGranvik what are you trying to say exactly?

@LeakyNun hi

Wiwichu a merry crismas an a japi niu yir

$$\huge\overline{\underline {_{\color{red}\diamondsuit}^{ \color{green}\diamondsuit}\color{red}\diamondsuit \!\!\!\!\!\!\color{green}\diamondsuit^{ \color{red}\diamondsuit}_{\color{green}\diamondsuit }\quad\color{red}{\Bbb{Merry}}~\color{green}{\Bbb{Christmas!}}\quad^{ \color{red}\diamondsuit}_{\color{green}\diamondsuit} \color{green}\diamondsuit\!\!\!\!\!\! \color{red}\diamondsuit_{\color{red}\diamondsuit }^{\color{green}\diamondsuit}}}$$

@Liad hi

(redacted)

hi leaky

did you have a look at example 30?

@LeakyNun Wake up leaky

@LeakyNun Do you code in Java?

yes

@KasmirKhaan ?

well

Leaky

lets ignore some of the given facts of the question

So lets just say H is normal in G

and we look at phi(H) , for phi in Aut (G)

phi(H) = phi (gHg') = phi(g) phi(H) phi(g)' = x phi(H) x'

@LeakyNun leaky hope u still here ><

Ehm anyway, i want to know when does this fail to be contained in H

@Kasmir hint for question 30: how many normal subgroups of order $p$ does $G$ have?

@AlessandroCodenotti thats the thing , we dont have sylow yet

You don't really need sylow here, it can be done without

hmm

ill save that hint for later

the other approach I had was since H of order p

H is cyclic

so it is generated by <a> say, g a^i g' = a^j

More generally suppose $|G|=ab$ with $\text{gcd}(a,b)=1$ and that $H$ is a normal subgroup of order $a$. Then how many normal subgroups of order $a$ are there in $G$?

hmm

we know that [G : H ] = b

but let me Think how many normal subgroups they can be

@AlessandroCodenotti aha ><

it can only be 1 such thing

yeah ?

Yes but why?

I cheated used results from next chapters, a p-sylow is normal iff n_p =1

cant find Another way to do it yet

Careful, in the $ab$ case you're not dealing with p-Sylow subgroups

hmm right ><

math.stackexchange.com/questions/2579674/…

Can you please help me to complete the proof.

Where will i get contradiction?

@AlessandroCodenotti ill keep working on the questions and come back later =P thanks for the hint :)

@ManeeshNarayanan I left short response in the set theory chatroom.

Realization: Complex Analytic Number Theory = CANT

I smell estimates in the air

epsilons and deltas and xis and etas and mus and gammas

@KasmirKhaan my philosophy is to avoid results that you can’t justify

there’s no problem with using sylow as long as you can prove it

but then you could just substitute the proof inside

I used the 3d Poincare conjecture once :3

I used ker CD = ker C oplus ker D once lol

I’m going to use Desargue a million times

I still have no idea how to prove it

whats Desargue

Are C and D matrices?

I think C and D are commuting matrices

@AkivaWeinberger commuting matrices

Descargues is that thingy

“O exists iff r exists”

I'm trying to remember how that proof goes. I vaguely remember it involving the third dimension

I don’t want any hint

Other picture

it’s a really neat theorem

recall that it says “p iff q”

it turns out that the two directions are duals of each other

so it only suffices to prove one direction

(the dual of a triangle is a triangle!)

Hm yeah makes sense

@BalarkaSen as you see i’m getting overexcited by duality again

Projective duality is neat tho

sup leaky

Well using that result

hi

I did this

Phi (H) has also same order as H

phi (H) is also normal in G

so it is equal to H and thus contained in H

because we have unique normal subgroup of same order as H

“it is equal to H and thus contained in H” you’re doing it backwards

we do have equality

that is stronger than contained in

just like normality

@KasmirKhaan they are actually equivalent

hmm so that proof is correct?

yes

i did not trust it alot tbh

AH

I think I got it

good thing that maths does not rely on your trust

Re Dezarg

Desargues

leaky stop being mean

@AkivaWeinberger good for you

MVT

so the image of a normal subgroup under automorphism is a normal subgroup

we used that fact here

@LeakyNun I once made a Galois theory for chains of characteristic subgroups

and the other fact was the uniqness of our normal subgroup of order H

@BalarkaSen what happened

so a characteristic subgroup is normal

but not the other way around

hmm

right

are they a generalized version of normal?

well

normal means fixed under inner automorphisms

they do seem to have Connection that i cant put my hand on

char means fixed under any automorphism

in that sense you can say one is generalizing other

neat

hmm leaky

by isomorphism

we dont mean tht the Groups as the same right

because if we dont have unique normal subgroup

sure

so we cant use that arguemnt

what do you mean?

phi(H) will be isomorphic to H

but phi(H) wont be contained in H

they are both groups of order p, of course they are isomorphic

well all we know that they are isomorphic

what i mean is

if we did not have the uniqness property

of the subgroup H

right

then we could not say that they are isomorphic hence phi(H) is contained in H

because they can be isomorphic with diffeent elements

@LeakyNun If $H$ is a subgroup of $G$ define $\text{Aut}_H(G)$ to be the subgroup of automorphisms of $G$ fixing $H$ pointwise. Say $N$ is a characteristic subgroup of $H$ is a characteristic subgroup of $G$. Then there is in fact a Galois left-exact sequence $1 \to \text{Aut}_H(G) \to \text{Aut}_N(G) \to \text{Aut}_N(H)$ similar to the Galois short exact sequence $1 \to \text{Gal}(K/L) \to \text{Gal}(K/F) \to \text{Gal}(L/F) \to 1$ for a chain of Galois extensions $K/L/F$.

nice

I didn't spend any more time thinking about this, but I suspect an interesting question to ask is if you can extend that left-exact sequence to a long exact sequence on the right

I suspect you'll end up with group cohomology of sorts

@Daminark triggered

leaky by fixed what do you mean

on the char

a subgroup H is normal if gHg' in contained H for all g in G, so for all inn (G)

but in case of a subgroup being char, means that phi(H) is contained in H for all phi in Aut(G)

by fixed here , what do u mean