Hey @Ted!

Hi Demonark

$ba=(b^3)^7a = ab^7$

@TedShifrin ugh

Well, for example, the dihedral group is given in that manner, Brody.

$a^n = b^2 = e$, $bab^{-1} = a^{-1}$.

$(a^m b^n) (a^p b^q) = a^{m+p} ~b^{7np+q}$

semi-direct product confirmed

activism.com/en_US/petition/…

Salut @GabrielRomon

Salut Gabriel

@TedShifrin can we conclude that its order is 100?

salut Ted et Leaky Nun

words are just finite products of those elements?

What is that about, GTR?

I would expect so, yes, Leaky. Semi-direct products are analogous to direct products in terms of the count.

@Gabriel: Now that I teach for AoPS, I'm curious. What is that petition?

@Brody yes

@TedShifrin interesting

@Brody: But you have enough information to make the words quite simple.

Leaky gave a formula up there ^^^ (which I've not checked).

hmm I skimmed over a chapter that mentioned defining equations, if that's related

@TedShifrin it's just the formula of a semidirect product

and it's for my example, not the dihedral group

@TedShifrin it's a joke essentially. But a few years ago he got on the nerves of a lot of users in the AOPS forums

He's gotten on my nerves in MSE.

I've never visited an AoPS forum.

then you should draft your own petition lol

how is it going with your teaching there ?

Hard to know so far. Only 4 students and 2 classes.

@TedShifrin how do you visualize semidirect products?

It's the first time in my life I've taught a class where I don't decide on homework, tests, or what's presented in class.

Experiment: $\langle a,b \mid a^{10}=b^{10}=1, a^2=b^2 \rangle$

ah that must be frustrating

@Leaky: I'm not sure I have a good answer. Intuitively, it's like a fiber bundle generalizing the product $X\times F$ ... the product structure twists. Here you're twisting by an automorphism of the normal subgroup.

@TedShifrin in English?

@TedShifrin Hello ! :D yes i did now :D

That was English. It's just geometry/topology, which you don't like :P But that's how I visualize things.

@KasmirKhaan you did what?

@TedShifrin thanks ! :) but did you understand what i said ><

oups texted worng guy

=p

what you said about what?

Oh ...

I meant

Can I read your book starting at any chapter?

like without losing continuity =p

Obviously math books aren't written to be read totally out of order, but you should be able to read somewhat out of order. You can read groups OK and part of chapter 7. But part of chapter 7 depends on rings, fields, etc.

I wrote it in the order I did on purpose. I explained earlier that it's easier to understand quotient constructions with rings than with groups.

Hmm I have the exam in a month and a litte more =p

so I think the best thing to do is to use your book after we introduced rings

that way It will be easy to follow :D

quotient groups is what we just started with

Not necessarily. I like the way I explained a lot of the group stuff, esp. group actions.

OMG

that is the topic I got problems with

let me see what page :D

That's the most important topic.

yes yes the way they explained it in dummit and foote is very criptic

Anyhow, I have a very early appointment in the morning, so good night, everyone.

I like drawing pictures of everything, and most algebra books draw none.

like read the whole text and count not get the notion of an action

See you Ted!

@TedShifrin Okay goodnight and thanks ! :D

Night.

G'n Ted

@TedShifrin in a finite group, how many elements of order 10 can there be?

In terms of n?

Find a set of irrational numbers (no singletons) that is closed under arithmetic means.

@Daminark no, in general

e.g. it can't be 3, but it can be 4

@Brody define arithmetic mean

or else I'm just going to go with $[\pi,2\pi] \cap \pi \Bbb Q$

@Daminark I'm wondering whether it can be 8

@LeakyNun in context of problem, $S\subset\Bbb{R}\setminus\Bbb{Q}$ and $AM:S^n\to S$ where $(x_1,\ldots,x_n)\mapsto\frac{x_1+\cdots+x_n}{n}$

@Brody ok, then I don't change my answer

@TedShifrin Right, but I didn't want to invoke Picard. Casorati-Weierstrauss is actually enough.

Ugh, I can't Sylow that

@Daminark that's the point :P

Why did you choose 10? You've taken away the only thing in group theory I know how to do

then let's deal with 9

@LeakyNun ok :D

there can be 6 elements with order 9

what else?

Oh I was kidding, you can't use Sylow anyway because that only tells you about the number of subgroups

But yeah which of these are you more interested in?

10 of course :P

but 9 if it's easy for you

I was thinking $x\Bbb{Q}_{>0}$ for any $x$ irrational, essentially the same @Leaky

@Brody right

Well in the 10 case I think that'd force it to have just 2 subgroups of order 10

@Daminark and I'm wondering where this is possible

Hey I'm back

And okay so the immediate thing to try would be $Z_{10} \oplus Z_{10}$

Which will probably end up being wrong

Oh yeah it is

the diagonal subgroup is cyclic of order 10

try Z/10*Z/10

What's star here? Semidirect product?

@Semi

free product. but i am not 100% if that works

it doesn't actually

there are millions of conjugate subgroups

Rip in subgroup

@LeakyNun You can have two subgroups of order $10$, but in the example I have in mind, only one of those will have elements of order $10$.

@TobiasKildetoft hmm, so how many elements of order 10 can we have?

@LeakyNun So an element of order $10$ is the same as a commuting pair of elements of orders $5$ and $2$

So the tricky part is the "commuting"

How do I put limits on the integral using math jax?

@TobiasKildetoft do you have the answer?

int_{lower}^{upper} ?

@Abcd \displaystyle \int_a^b

@LeakyNun I don't

$ \displaystyle \int_{to}^\pi$

@TobiasKildetoft oh ok

$\int\limits_{a}^{b}$

$\langle a,b \mid a^{10} = b^{10} = 1, ab=b^3a \rangle$

@TobiasKildetoft how many elements of order 10 are there in this group?

Also, does this group have order 100?

Isn't that group infinite?

@AlessandroCodenotti I don't think so...

@AlessandroCodenotti No, you can arrange the strings to always have first $a$'s then $b$'s, and the powers of those are limited

just put it in GAP and see :P

every element can be expressed in the form $a^p b^q$ with $0\le p,q < 10$

And then it is closed under product

GAP says 20 elements

@TobiasKildetoft ah right

@SteamyRoot :O

What's the order of $ab$? @Steamy

10, apparently

@SteamyRoot could you show us your code

Makes sense

Z/10*Z/10 modulo some relation of the form xyx^-1=y^N should produce a group with exactly two subgroups of order 10; someone needs to draw the cayley graph

@SteamyRoot can you list the elements of g in GAP?

Sorry for posting thrice.

the idea would be to identify the tons of conjugate subgroups in a way so that you don't produce a cycle of order 10

@SteamyRoot btw use Ctrl+K so that your code looks nice

press Ctrl+K after typing the code before pressing enter

Hey @Steamy, @Alessandro, and @Tobias!

Hi, Demonark

Morning

Yo @Leaky

Hi @Dami

gap> for x in g do
>   Print(x, " ");
> od;
<identity ...> a a^9 b a^2 a*b a^8 a^9*b a^3 a^2*b a^7 a^8*b a^4 a^3*b a^6 a^7\
*b a^5 a^4*b a^6*b a^5*b

Smells like a dihedral group kinda

Or not

gap> StructureDescription(g);
"C10 x C2"

lol

it cant be abelian...

@Daminark Hi

@SteamyRoot That can't be right.

The existence of this group follows from the axiom of choice. I choose to assert its existence.

Hmm

wait

What are C10 and C2?

Yeah, I may have forgotten an inverse on the relations somewhere, woops

Cyclic groups of those orders

@Daminark ... using choice again whenever you can @_@

@SteamyRoot they should be isomorphic

Hmmm, yeah, you're right.

@SteamyRoot Ahh, no, it is fine. The point is that the second generator becomes equal to its third power, so it really has order $2$

I meant, the inverse doesn't matter

I'm pretty sure GAP is right with its identification :P

gap> a*b=b*a;
true

this group is abelian

ab=ba=b^3a, so b^2=e

gap> Size(f/[f.1^10,f.2^10,f.1*f.2*f.1^-1*f.2^-1]);
100
gap> Size(f/[f.1^10,f.2^10,f.1*f.2*f.1^-1*f.2^-2]);
10
gap> Size(f/[f.1^10,f.2^10,f.1*f.2*f.1^-1*f.2^-3]);
20
gap> Size(f/[f.1^10,f.2^10,f.1*f.2*f.1^-1*f.2^-4]);
50
gap> Size(f/[f.1^10,f.2^10,f.1*f.2*f.1^-1*f.2^-5]);
20
gap> Size(f/[f.1^10,f.2^10,f.1*f.2*f.1^-1*f.2^-6]);
50
gap> Size(f/[f.1^10,f.2^10,f.1*f.2*f.1^-1*f.2^-7]);
20
gap> Size(f/[f.1^10,f.2^10,f.1*f.2*f.1^-1*f.2^-8]);
10
gap> Size(f/[f.1^10,f.2^10,f.1*f.2*f.1^-1*f.2^-9]);
100
gap> Size(f/[f.1^10,f.2^10,f.1*f.2*f.1^-1*f.2^-10]);

well, the symmetry isn't too surprising

gap> g := f/[f.1^10,f.2^10,f.1*f.2*f.1^-1*f.2^-4];
<fp group on the generators [ a, b ]>
gap> a := GeneratorsOfGroup(g)[1];
a
gap> b := GeneratorsOfGroup(g)[2];
b
gap> a*b=b*a;
false
gap> Size(g);
50

We should play with this group lol

gap> StructureDescription(g);
"C5 x D10"

@LeakyNun Well, there it is easy to count the number of elements of order $10$

gap> IsomorphismGroups(DirectProduct(CyclicGroup(5),DihedralGroup(10)),g);
[ f1, f2, f3 ] -> [ a^24, a^25, b^4 ]

what is a^24 doing lol

Heh :P

Things like IsomorphismGroups are great if you need any isomorphism, but it's rarely a natural or good-looking one :P

just checking whether they're isomorphic lol

gap> StructureDescription(f/[f.1^4,f.2^4,f.1*f.2*f.1^3*f.2]);
"C4 : C4"

what the hell is colon?

(quaternion group)

@LeakyNun the colon means semidirect product

hmm

and dot?

non-split extension

what does that mean?

$G.H$ then means that the group has a normal subgroup isomorphic to $H$ with quotient $G$ (or the other way around, I forgot)

and this quotient does not split

meaning that the group is not a semidirect product

@_@

is it well-defined?

no, but neither is the semidirect product

hmm

must Aut(G) be abelian if G is abelian?

Try $G = \mathbb{Z}^n$

nvm, Aut(Z2^3) is clearly not abelian

:c sniped

But if $\operatorname{Aut}(G)$ is nilpotent, then $G$ is nilpotent

And same goes for solvable.

must G be abelian if Aut(G) is abliean?

No, but $G/Z(G)$ must be

what's a counterexample?

Hmm, or maybe the answer is actually yes. I need to think a bit more

how come Aut(D8) = D8 ;_;

There are counterexamples, but they're not very easy afaik

Coincidence

@TobiasKildetoft heh?

gap> StructureDescription(AutomorphismGroup(DihedralGroup(8)));
"D8"
gap> StructureDescription(AutomorphismGroup(QuaternionGroup(8)));
"S4"

what is the rotation in Aut(D8)?

is there anyone here who thinks about a sheaf in terms of it's espace etale

i'm semi-confused about something

Hi

Any help is appreciated, plz keep it simple

@LeakyNun Aut(Z/2xZ/2) is not abelian, right?

It's S3

@BalarkaSen Or a better description: $2\times 2$ matrices.

yep, Aut(Z/nxZ/n) is GL_2(Z/n)

nice

Hi guys, can anyone here help me to solve this puzzle?

no

smells like hexadecimal

apart from that, no time

#2busywatchingmemes

Ah, I finally got it

guys I need a help of a moderator

can any one guide me how to reach one

@Mour_Ka There is an email I think. If it is related to a specific post, you can raise a flag on that post

well I did raised a flag actually 4 days ago.

What sort of flag?

Flag for Abusive comment. I have a person who used two accounts to reply to me doing two negative votes and was agrressive in his response on meta. so I want to check if I violated the rules to receive all of that or he is just doing mistakes.

he deleted some comments to look like he is different person but I copied all so lets say he is caught. I am not sure also if deleted comments can be seen by moderators.

@Kari

@Dodsy Usually a chat ban lasts for a while only, so don't worry about it. I have been banned many times too. =D

Hello. May I recieve help with this linear algebra question?

@DennisKim just ask; don't ask to ask

I had difficulties with generalizing the inverse matrix in this case

@Semi semisemisemisemisemisemisemisemisemisemisemisemisemisemisemi. Finally finally finally, after so much agony last year, I understand (the basics of) waves and oscillations and Fourier TRANSFORMS and and and.... just so much! I think a million quarters fell yesterday.

And I revised the quantum chapters from last year, and I just understand it soooooo much better, like when they mention group velocity or what not (I finally understand why $v_{\operatorname{group}}=\dfrac{d\omega}{dk}$) Sorry I had to share this!:P I really increased my physical understanding in under 24 hours. Can you image how hopeless I was when I didn't understand Fourier transforms last year, while we were doing optics and quantum mechanics.

Or when they would mention time upon time the "dispersion relation", and I just wouldn't have any idea what they were really talking about. It is so bittersweet:P

Btw!!! I've chosen to do condensed matter this year, but it seems to be a bit of a "hand wavy" subject. Like, we do get "rigourous things", but they told us that not everything will be clear, because we’re not advanced enough, lolz.

Anyhow, I think in theory it would be relatively doable if we had more time, because I've noticed they sometimes skip a chapter or paragraphs (by Hook&Hall, if you are familiar with it, donno if it’s really standard literature), that I think actually provide some fundamental notions of the formulas used. So I don't know. We have basically gotten a 'physical-like' treatment of lattices (like what they are and how to describe them), and now we're considering the free electron model.

I think that's alright - but we'll also be doing some classical mechanics formalisms these weeks!!! WHICH IS SO COOL!!! (I’ve been waiting for this FOREVER!!:P) And we'll also be doing some more proper quantum mechanical FORMALISMSSSSS, WHUUUUUUU!!!! I'm so excited for this year:P Haha, anyhow, I hope you don't mind I share this:P But I had to update you, even though you know the undergraduate programme already:P But still, EPICNESS is around the corner. (actually it's already here) Bya!

(sorry for the spam, peeps:P)

but I just feel very excited and had to share this

but other than that I have no idea lol

come later when Ted is here :P

@DennisKim

@DennisKim Hello Dennis, you misspelled 'receive'.

Thank you for now :) sorry for the typo

@Daminark You'll love this. A few sections forward in Ch 1 Forster explains how to do Galois theory with Riemann surfaces

That's the gateway to Grothendieck's algebro-geometric point of view of Galois theory

@Mour_Ka Deleted comments can be seen by the moderators of the relevant site.

@BalarkaSen You reading Forster? ^^

Forster seems to be the most popular book on Riemann surfaces.

It's pretty good, I'd say

Not so geometric though

There's also Springer, Farkas/Kra, Donaldson, Ahlfors/Sario, Napier/Ramachandran, Varolin, Miranda, lol.

Assume $a\ne 0$ and $b \ne 0$ and there is no zero divisor. Can $6a^2b^2=0$?

I know

@ShaVuklia haha, nice

I heard of mathworld.wolfram.com/AbelsImpossibilityTheorem.html a long time ago . But i didn't think a lot about it, simply said myself 'I understand it.'. Recently It became complicated to me. If we can't find a formula for the roots, is that mean it is impossible to find the roots exactly?

Depends what you mean by exactlyy.

If you've got a specific polynomial with known coefficients, then the roots can be found numerically to arbitrary accuracy. so in that sense there's no impediment.

on the other hand, if you have symbolic parameters then what you'd need is a formula that works for all such choices. and that's exactly what you won't have in general.

i though it was like the inexpressibility of pi in terms of $1,2,\dots\, \sqrt{2} $ etc

eh, that's a bit higher level.

suppose you had the degree five polynomial $x^5-x+1$

i guess i should ask you 'Can we have polynomial that have a root neither $1$ nor $\sqrt{2}$ but like pi?'

this has exactly one real root. however, this real root cannot be expressed in closed-form in terms of radicals

oh, that's the one

however, this real root is still the root of a polynomial with rational coefficients, and therefore is said to be algebraic

so, my definition for algebraic is pathetic

by contrast, there's no polynomial with rational coefficients which has $\pi$ as a root

So $\pi$ is not algebraic and is instead said to be transcendental

The terminology is, to my mind, unfortunately a bit limited here

There are irrational numbers which can be expressed via radicals; there are irrational numbers which can't be expressed via radicals but are nevertheless algebraic; and there are irrational numbers which are transcendental.

I'm not sure there's a simple name for the second category beyond "algebraic numbers which can't be written using radicals"

understood, well, can we express the root of $x^5-x+1$ via other things?

depends what you mean by 'other things.'

@AbdullahUYU there's a "bring radical" function (denoted BR) that can solve quintic equations

ehhh

bring radical isn't so much telling you "how to solve the quintic"

kk

for example, a series that involves radicals maybe @Semiclassical

as much as saying "how can I reduce solving the quintic to a standard case"

(that's a bit of a simplification as well)

one route would be: "Can I obtain this root to arbitrary accuracy by numerical algorithms?"

and to that the answer would definitely be yes

You can express a root as an infinite series via the Lagrange inversion formula

yeah, that works

though you probably have to be careful to avoid convergence headaches.

mathoverflow.net/questions/32099/… that's the relevant MO question

hmm, additionally i guess we can do that via newton method.

@AbdullahUYU yeah, that's in the 'solve via numerical algorithms' category

yeah i mean't it

kk

i.e. people think I do hard maths while I actually do intermediate maths

(because I never get the hard maths right)

so now i interested in @AlessandroCodenotti 's response. thank you all btw

apparently you can also write the root in terms of the hypergeometric 4F3 function, but uh

that's not really a pleasant endeavor

(and it's specific to the quintic case I was mentioning.)

there's also a connection to Jacobi theta functions and elliptic function theory: en.wikipedia.org/wiki/…

but, again, this is hardly a pleasant task

there's some other methods in that vein in the sections after that link

to clarify, do you mean expressing the root in terms of something isn't worth generally, or specifically in terms of hypergeometric 4F3 function?

in general, it's not worth expressing the roots of a nonsolvable quintic in terms of special functions.

much better to use some simple numerical algorithm like Newton's method

(especially given how fast Newton's method converges. # of correct digits doubling after each step? hard to beat)

Hi. I got a stupid doubt.. suppose I have an average of expectations $1/N^2 \sum_{i,j=1}^N E[A_{i,j}]$. I'd like instead to write just $E[A_{i,j}]$ as if indices $i$ and $j$ also are random/arbitrary and are chosen uniformly, as to simplify notation. Is that ok or is it too nonstandard?