Mathematics - 2016-11-25 (page 1 of 6)

DHMO

00:00

@meow-mix where?

SteamyRoot

It's not wrong. At all. $1/\Gamma(0) = 0$.

meow-mix

> 1/x is defined for all but x=0, but 1/1/x isnt undefined at 0

DHMO

@meow-mix who said it?

meow-mix

me

Null

eh, but 0!=1, so 1/0!=1, therefore the gammafunction doesn't help me at all, if it is like that

Semiclassical

00:01

$\Gamma(x)$ has poles, $1/\Gamma(x)$ has zeroes.

SteamyRoot

Actually, meow, that would be wrong.

Fargle

@SteamyRoot Only if you adopt the abuse of notation that $\frac{1}{\Gamma(x)}=\Lambda(x)$ (the reciprocal gamma function)

DHMO

@meow-mix and why should i believe you?

@Semiclassical abuse of notation

meow-mix

'well, i gave you a source

SteamyRoot

@Fargle It's not abuse of notation within the class of meromorphic functions

Semiclassical

00:02

Not really seeing why that'd be an abuse of notation.

DHMO

a source doesnt mean yourself @meow-mix

Akiva Weinberger

$\dfrac1{\Gamma(0)}$ is undefined but $\lim_{x\to0}\dfrac1{\Gamma(x)}=0$, no?

DHMO

@Semiclassical because Gamma(0) is undefined

the reciprocal of a pole is not zero

SteamyRoot

Except it is.

DHMO

@AkivaWeinberger yes, thank you

Fargle

00:02

$\frac{1}{\frac{1}{x}} = x$ only when $x \neq 0$. When $x = 0$ we adopt the convention that $\frac{1}{\frac{1}{x}} = 0$, but the function $\frac{1}{\frac{1}{x}}$ does in fact have a hole at $x = 0$.

Semiclassical

$\frac{1}{1/x}=x$ would disagree with you.

At most, it's a removable singularity.

meow-mix

yeah, i retract that comment

Semiclassical

And when it comes to complex analysis I don't much care about removable singularities.

meow-mix

well, it would be removable

DHMO

@Semiclassical thank you

people have interesting concepts about the meaning of 1/

SteamyRoot

00:04

Complex analysis has interesting concepts about that meaning :P

meow-mix

anyways, im not a complex analyst, im just reading from a source

Semiclassical

One does have to be a bit careful---the reciprocal of an essential singularity is not a zero.

DHMO

"interesting" was a polite way to say "wrong"

Fargle

Like, it's a perfectly acceptable thing to say, but it also begs a modicum of care.

Semiclassical

But the reciprocal of a pole is certainly a zero.

SteamyRoot

00:04

@DHMO It's not wrong at all.

DHMO

@Semiclassical nope

Semiclassical

Yes.

SteamyRoot

Seriously, read any book on complex analysis, chapter on "meromorphic functions"

Null

"You should absolutely believe everything on the internet" - Abraham Lincoln 1872

Akiva Weinberger

In a wheel, $//0=0$

SteamyRoot

00:05

The reciprocal of a pole is a zero. And in fact, the multiplicity of pole and multiplicity of zero match

meow-mix

here's a source

Wrench, J. W. Jr. "Concerning Two Series for the Gamma Function." Math. Comput. 22, 617-626, 1968.

DHMO

@SteamyRoot "1/Gamma(0)" does not mean "find the Gamma function, reciprocal it, fill the holes, evaluate at x=0"; it means "find the Gamma function, evaluate at x=0, reciprocal it"

meow-mix

in which the reciprocal gamma function is defined as entire

DHMO

at least it is an abuse of notation

SteamyRoot

Sigh...

Semiclassical

00:07

If it's an abuse of notation, it's one I could give a damn about.

DHMO

@meow-mix the "reiprocal gamma function" is not the "pointwise-reciprocal of the gamma function"

Semiclassical

For the purposes of actual computations, the reciprocal of a pole is a zero.

DHMO

@Semiclassical Gamma is C->C, not C->(C U {pole})

anyway, i have to go now

Null

@Semiclassical ah, now I atleast understand why one would do it like that.

SteamyRoot

@DHMO Please, just read a book on complex analysis.

meow-mix

00:08

and, @Null, incase you believe my source is fake, here is the full article, available at ams.org

ams.org/journals/mcom/1968-22-103/S0025-5718-1968-0237078-4/…

Null

@SteamyRoot any good recommendations?

Akiva Weinberger

@DHMO You mean C\{nonpositive integers}->C

Semiclassical

Does it work as a pole for the purposes of the argument principle? If so, I don't really care about the abuse of notation.

Akiva Weinberger

Or $\bar{\Bbb C}\to\bar{\Bbb C}$

Null

@meow-mix hey, relax. i don't think it's fake. I'm just skeptical in general ;) Thanks nonetheless!

Akiva Weinberger

00:09

I think the word you're all missing is "Riemann sphere".

SteamyRoot

@Null Greene & Krantz - Function Theory of One Complex Variable for basic stuff (bachelor level), Miranda - Algebraic Curves and Riemann Surfaces for more advanced stuff (master level)

Semiclassical

There's probably an argument to be made for taking the reciprocal gamma function as the more essential object.

Since, properly defined, it is entire.

SteamyRoot

Well, you don't "need" the Riemann sphere to justify it, really. But certainly it works there; and there cannot be any discussion there either :P

Semiclassical

Though I think that the Gamma function itself is sorta unnecessary for this.

$f(x)=x$ is entire, and has a zero at zero.

$1/x$ has a pole at zero.

The only reason we're getting into knots is because we're treating the Gamma function as the more natural object, and it's not. Same as if we treated $(1/x)^{-1}$ as the definition of $x$.

Historically/pedagogically, it's probably more natural to start with the Gamma function. But at the level of complex analysis the opposite is true.

Fargle

@Semiclassical I actually rather like this perspective. Define $\Lambda(z) = \frac{i}{2\pi}\oint_{C} (-t)^z e^{-t}\, dt$ with $C$ as a curve enclosing the positive reals, starting and ending at positive infinity and going through $0$.

Then define $\Gamma(z) = \frac{1}{\Lambda(z)}$.

Semiclassical

00:16

Yeah, that Hankel integral (I saw it on Wikipedia)

Fargle

Same.

Semiclassical

Alternatively, start with the integral definition of $\Gamma(z)$ and take $\Lambda(z)=1/\Gamma(z)$ to be $1/\Gamma(z)$ with all singularities removed.

SteamyRoot

wait what

Going through $0$?

That can't be right - it should enclose $0$, but definitely not go through it

Semiclassical

That's not the wording Wikipedia has

"where C is a path encircling 0 in the positive direction, beginning at and returning to positive infinity with respect for the branch cut along the positive real axis. "

SteamyRoot

Okay, that makes more sense

meow-mix

00:19

im not sure why @DHMO got angry at me

Semiclassical

The general principle is that, in complex analysis, removable singularities are boring.

SteamyRoot

Actually, you could indeed take @Akiva's suggestion and use the Riemann Sphere. If you let $\Gamma(z)$ and its reciprocal be defined on $\mathbb{C}_\infty$, then there's no removable singularities to worry about at all

Semiclassical

Sounds legit.

SteamyRoot

And then just define $\Gamma$ and its reciprocal on $\mathbb{C}$ or $\mathbb{R}$ as the projection of the function from the Riemann sphere

Fargle

@meow-mix Nothing personal, I suspect. We here in MSE just care a lot about math. :3

Semiclassical

00:22

After all, $f(z)=1/z$ is a Mobius transform; it maps the Riemann sphere to its reflection across the real line (not a great way to say that, but w/e)

If one wanted to be truly precise, we work on $\mathbb{P}^1$

So you'd have homogeneous coordinates $p=[z:w]$ and $f(p)=[z^{-1}:w^{-1}]$

The point at infinity is then $p=[0:1]$, and near there we can represent $f(p)$ as $f(p)=[1:zw^{-1}]$ so that $f([0:1])=[1:0]$.

Near $[1:0]$ we can instead represent $f(p)$ as $f(p)=[wz^{-1}:1]$ and therefore $f([1:0])=[0:1]$

And done.

Null

@Semiclassical so $(3,+\infty)$ would make sense in that context as a point?

Semiclassical

you'd represent it as $[1:0]$.

But $[z:w]$ and $[\lambda z:\lambda w]$ are equivalent in P^1 if $\lambda\in\Bbb C$

heather

00:39

how important is it that I learn the theorems and how to prove things along with the actual math itself? will it significantly help my understanding?

Akiva Weinberger

@heather I think it's very important

Null

i think only if you understand the "need" for the theorem

(just my humbly opinion)

@Brody hello, how are you? :)

Brody

Happy Thanksgiving fellow americaners

Akiva Weinberger

Happy Thanksgiving

Brody

Hi @Null. Doing well, thanks. You?

Null

00:44

@Brody good too!

Brody

Your question earlier spurred quite a discussion @Null

I didn't (and still don't) know a decent answer honestly

Null

@Brody im just drinking my beer solving excercises haha. I think the answer to my profs excercise lies in the prof...

(i didnt visit his seminar cuz of reasons)

sent him an email tho :D

Fargle

@heather Tremendously, at least as you ascend.

Null

now, just for the sake of it, assume that x! is defined always: Let $x\in\mathbb{R}$ be arbitary but fixed. Let $n=0$. $$\sum_{k=0}^{0}(-1)^k\binom{x}{k}=(-1)^0\binom{x}{0}=\frac{x!}{0!(x-0)!}=1=\fra‌c{(x-1)!}{0!(x-1-0)!}=(-1)^{0}\binom{x-1}{0}$$

mmh

lol screw that^^

Brody

01:11

@Null I see what you mean to write. What are you trying to show?

Null

just the basecase @brody of $$\sum_{k=0}^{n}(-1)^k\binom{x}{k}=(-1)^{n}\binom{x-1}{n}$$

Brody

Ohhh

Null

i think its easy to recognize what the red \frac should be

i mean, the excercise builds on the fact that its defined for all reals

so i guess my prof has a suiting definition of binomials

Brody

This is for all $n\in\{0,1,2,\ldots\}$?

Null

yep

Brody

01:15

I feel weird about allowing arbitrary $x\in\mathbb{R}$

Null

yeah

(-1)! is basicly the thing we talked for 2 hours^^

and I highly doubt that this is content for the first semester

Brody

What about integer $x\ge n+1$? (or no equality, if you prefer)

Null

well x is in that sense a constant

just that it is an arbitary constant

so it doesnt get replace by k in the sum

Brody

We can let $x$ vary; we simply restrict the domain

I see what you mean

Null

something like $x\in\mathbb{R}\setminus\mathbb{Z}$ imo

Brody

01:21

For whatever reason I doubt your prof. has you working with any extension of $x!$ beyond the non-negative integers

Null

But that's none of my buiseness^^. I just write below the proof "this only holds if x is not a negative integer under the usual definitions"

meow-mix

im back

Brody

Wb @meow-mix

meow-mix

when composition is a binary relation, does $(f,g)$ map to $g \circ f$ or $f \circ g$?

i would imagine $g \circ f$, im just making sure

Soscrates Fd

01:37

Hello all

meow-mix

hi, @soscrates

Soscrates Fd

What's the math topic being discussed if any

meow-mix

nothing really, im just wondering what the answer to my question is

Brody

@meow-mix I don't know what function this is

Soscrates Fd

Wish I could help, only have calculus2 understanding just a curious onlooker

meow-mix

01:39

composition of functions

in this case, the functions are elements of an automorphism set in a category

Soscrates Fd

Interesting

meow-mix

in fact

the category is $\mathsf{Grp}$

Brody

I don't know anything about categories

meow-mix

it doesnt even have to do with categories

was just mentioning the specifics

Null

@meow-mix does grp mean pseudogroup like rng for pseudoring?

meow-mix

01:41

basically, does the binary operation $\circ$ (composition) map $(f,g)$ to $f \circ g$ or $g \circ f$

@Null no, $\mathsf{Grp}$ is the category of groups whose morphisms are group homomorphisms

Akiva Weinberger

@Null Categories are usually TLAs (three-letter acronyms)

or, rather, three-letter abbreviations

meow-mix

@AkivaWeinberger counterexample to "Akiva's Conjecture": the category $\mathsf{Ring}$

Akiva Weinberger

"Usually"

meow-mix

i know, just thought id add in some humor :P

Akiva Weinberger

Also, someone once tried to use the category of Banach analytic manifolds in a paper, or $\sf BanAnaMan$.

7

meow-mix

01:43

anyways, akiva youre knowledgable. do you know?

Null

@AkivaWeinberger so if the number of letters is n. The number of categories is bounded by $n^3$? :-D

Akiva Weinberger

(Sadly, it was changed to $\sf BanMan$ before publication)

Null

hilarious^^

Akiva Weinberger

@meow-mix I think $f\circ g$, but I'm not quite sure why it matters

Brody

@meow-mix Not sure why you're talking about a mapping

Akiva Weinberger

01:45

unless you see $“\circ(f,g)”$ somewhere

It's like seeing $“+(3,5)”$ instead of $3+5$

meow-mix

because i have to prove that $g \mapsto \gamma_g$ is a homomorphism from $G$ to $\mathrm{Aut}(G)$

Fargle

@AkivaWeinberger Oh my God really?

Akiva Weinberger

I've related this story before

Fargle

I'm upset that didn't make it to print.

meow-mix

@AkivaWeinberger it matters because it has to do with homomorphisms

Akiva Weinberger

01:49

Is $\gamma_g$ conjugation? In any case, just choose whichever one makes the statement true :P

meow-mix

yes

Akiva Weinberger

Wait is it $x\mapsto gxg^{-1}$ or $x\mapsto g^{-1}xg$

Fargle

Usual composition of functions is taken $\circ(f,g)(x) = (f \circ g)(x) = f(g(x))$. That is, the composition is evaluated right through left--you map x first through $g$, then through $f$.

Akiva Weinberger

^, which is a confusing consequence of writing $“f(x)”$ - that is, the function first, and what the function acts on second

meow-mix

ok shit this is confusing

Akiva Weinberger

01:53

If I made the notation I probably would change it to $“(x)f”$

Fargle

@AkivaWeinberger I might do something like that, or $(x) \mapsto f$, read "x through f" rather than "f of x"

with a different arrow than $\mapsto$

@meow-mix What are you having trouble with in particular?

meow-mix

i have to prove that $g \mapsto \gamma_g$ is a homomorphism of groups

Brody

@AkivaWeinberger That might have some nice advantages, e.g. $(f(x))^2$ vs. $(x)f^2$

Fargle

@meow-mix Yes--what I mean is, where is your sticking point?

meow-mix

oh, im just "stuck" at proving that $(ab)x(ab)^{-1} = b(axa^{-1})b^{-1}$

heather

01:56

is there a book anyone knows about that just has things to try to prove? (explanations of the proofs would be a bonus)

meow-mix

try studying analysis

heather

analysis? okay...

meow-mix

hey @arctictern

arctic tern

hey

Semiclassical

@meow-mix you sure that shouldn't be $a(bxb^{-1})a^{-1}$?

Fargle

01:58

@meow-mix You're doing your composition wrong. $(\gamma_a \circ \gamma_b)(x) = \gamma_a(bxb^{-1}) = a(bxb^{-1})a^{-1}$.

Brody

Velleman's book How to Prove It comes to mind

meow-mix

i meant that, sorry

Brody

but this might not be what you want @heather

Semiclassical

That's pretty easy to prove, then. Just figure out what $(ab)^{-1}$ is.

Fargle

@meow-mix Well, just recall how to write $(ab)^{-1}$ in terms of $a^{-1}$ and $b^{-1}$.

@heather Are you looking for a sort of "intro to proofs"?

meow-mix

02:00

@Fargle how would i do that?

$a^{-1}b^{-1}$?

Semiclassical

Does $(ab)(a^{-1}b^{-1})=1$?

Fargle

@meow-mix Not quite: (ab)^{-1} should satisfy $(ab)(ab)^{-1} = 1$, but $aba^{-1}b^{-1}$ might not.

heather

@Fargle, yes, I suppose, but mainly I just want some practice proving things.

Semiclassical

If not, then that's not $(ab)^{-1}$.

meow-mix

let me ponder this

Fargle

02:01

@heather I can also recommend How to Prove It, as per what Brody said. I'll also see if I can dig up the text my prof used for that class--he wrote it himself and distributed it electronically.

Semiclassical

To put it in a more literal way: If I close a door and then lock it, would I undo that by opening the door and then unlocking it?

Fargle

Both have significant amounts of proofs to practice with.

heather

@Fargle, okay, thank you =)

that sounds great!

meow-mix

well no, because the group of "actions on the door" is not commutative

Semiclassical

Right. So what order would I do instead?

meow-mix

02:02

oh wait, $b^{-1}a^{-1}$?

Semiclassical

yup

(to complete the point: you'd undo them in the opposite order that you did them originally)

anyways. once you've got that, the result is immediate.

meow-mix

yeah

welp

time to prove the final thing in this exercise

that the homomorphism is trivial if and only if $G$ is abelian

Fargle

@heather I'll need your email for the latter.

meow-mix

(don't spoil it!)

Fargle

SPOILER ALERT: it's true.

meow-mix

02:05

@Fargle WHY YOU GOTTA SPOIL IT

now im NEVER gonna return to math

im going back to chess crosses arms

Fargle

wow jeez I'm sorry

heather

@Fargle, I'm not sure I can just give out my email, sorry - not that I don't trust you or anything, just not sure my parents would be happy about it.

meow-mix

that's odd, my 3 messages lined up in a perfectly rectangular fashion

heather

@meow-mix, come on, math is much better than chess.

Fargle

Nah, I feel you, haha. In that case, How to Prove It should serve you quite well.

meow-mix

02:06

chess is fun :P

anyways

heather

@Fargle, okay, thanks for your help =)

meow-mix

lets prove this biconditional

Fargle

No problemo.

Brody

@meow-mix lol

Semiclassical

To show that it's trivial, you want $axa^{-1}=x$ regardless of $x$

meow-mix

02:07

hushhhh

thats the kind of spoilers i dont care for

im gonna pretend like that didnt happen

Semiclassical

pffft

Null

je n'ai pas le temps

remember that sentece if you ever face a duel at morning :P

heather

I do not have time?

meow-mix

@Null like Galois?

Null

Évariste Galois

Évariste Galois (* 25. Oktober 1811 in Bourg-la-Reine; † 31. Mai 1832 in Paris) war ein französischer Mathematiker. Er starb im Alter von nur 20 Jahren bei einem Duell, erlangte allerdings durch seine Arbeiten zur Lösung algebraischer Gleichungen, der so genannten Galoistheorie, postum Anerkennung. == Leben == Galois besuchte das College Louis-le-Grand in Paris, scheiterte zweimal an der Aufnahmeprüfung zur École polytechnique und begann ein Studium an der École normale supérieure. Mit 17 Jahren veröffentlichte er eine erste Arbeit über Kettenbrüche; wenig später reichte er bei der Académie des…

@meow-mix nice one^^

meow-mix

02:09

that feel when you link a german wikipedia article

Null

:D

heather

english translation:

Évariste Galois

Évariste Galois (French: [evaʁist ɡaˈlwa]; 25 October 1811 – 31 May 1832) was a French mathematician born in Bourg-la-Reine. While still in his teens, he was able to determine a necessary and sufficient condition for a polynomial to be solvable by radicals, thereby solving a problem standing for 350 years. His work laid the foundations for Galois theory and group theory, two major branches of abstract algebra, and the subfield of Galois connections. He died at age 20 from wounds suffered in a duel. == Life == === Early life === Galois was born on 25 October 1811 to Nicolas-Gabriel Galoi...

Null

he is legend :O

Fargle

I'm already older than he was when he died.

heather

he died from wounds suffered in a duel? geesh, didn't know that

Brody

02:10

I had no idea he died prematurely

meow-mix

dude, i hate when im just walking my dog, and my Aufnahmeprüfung gets tangled in a knot!

Semiclassical

Yeah, he died really young.

meow-mix

hopefully i wont die in a duel at 22

Brody

Shame

Null

@meow-mix it would give catch 22 a new meaning

Fargle

02:11

I'm a little late to have lasting contributions to algebra, I guess.

meow-mix

@Fargle maybe have lasting contributions to Inter-universal Teichmuller theory?

Fargle

@meow-mix If we're being real, probably not.

meow-mix

the identity of $\mathrm{Aut}(G)$ is $\gamma_1$ right?

Brody

@Fargle I'm sure there are plenty of famous names who did great things not far from their old deaths

Fargle

@meow-mix Test it to be sure.

@Brody Oh, absolutely. I'm merely being tongue-in-cheek.

meow-mix

02:13

well $\gamma_1$ maps $x \mapsto 1x1^{-1}$ which is just $x$

Fargle

My plan is to revolutionize our approach to math and then quit and live in isolation for close to 30 years.

meow-mix

so, yes?

Null

i want a heavy metal song about Galois.

Brody

@Fargle Of course, just some encouragement nonetheless, lol

Null

that would be f***** brutal

heather

02:13

yeah, I'll try to avoid getting into a duel =P

meow-mix

i used to think Galois was pronounce "guh-loy" and i sounded like an idiot

Fargle

@meow-mix That doesn't mean it's the identity in Aut(G), though. You want to check that $\gamma_g \circ \gamma_1 = \gamma_1 \circ \gamma_g = \gamma_g$.

heather

i still am not completely sure how it's pronounced, to be honest

Fargle

@heather "gal-wah"

meow-mix

gal-wa, im pretty sure

Brody

02:13

OTOH, you could be laughed at by the community until old age, only to have your work finally accepted long after your death

meow-mix

that would be sad :(

Null

@meow-mix well, if you didn't learn french that's not unusual ;)

heather

@Fargle, hmm, yep, that's a whole lot different than how I thought it was pronounced =P

Fargle

@Brody Who is that supposed to be? Mine was Grothendieck >_>

Null

@Brody altho Nash got his work accepted before his death, i find his story very tragic

(insulin therapy ugh)

Brody

02:16

@Fargle No one in particular, but Cantor came to mind (not sure if his transfinite stuff made the mainstream or not prior to his last days)

Fargle

@Brody Oh, true.

meow-mix

@Fargle i dont understand, do you have a counter example? im trying to figure this out but i feel like $\gamma_1$ is the identity morphism

Semiclassical

It's not a question of whether it's right or not, it's whether you've shown it or not.

Fargle

@meow-mix It definitely is, but not because $1x1^{-1} = x$. For example, in the rationals, $2x2^{-1} = x$, but it is not the identity morphism.

meow-mix

how is $\gamma_2$ not the identity morphism?

Null

02:19

(ignore this) @Semiclassical unless we are talking about true but unprovable statements.

meow-mix

because it would map $x \mapsto 2x2^{-1} = x$ for all $x \in \mathbb{Q}$

Semiclassical

yes, well, this is group theory

Fargle

@meow-mix Because $\gamma_2 \circ \gamma_3 \neq \gamma_3$.

For example.

meow-mix

what? how?

Fargle

Or, well, this is a bad example. The conjugations of an abelian group are all trivial.

meow-mix

02:20

$\gamma_3(x) = 3x3^{-1} = x$

Semiclassical

This is the whole trivial thing. Need something which is more complicated.

Fargle

"identity morphism" $\neq$ "the identity map"

meow-mix

is there a counter example?

Semiclassical

Simplest nonabelian group is the Klein four-group.

Fargle

Take, for example, $\{\pm 1, \pm i, \pm j, \pm k\} \subset \Bbb H$.

@Semiclassical Definitely abelian.

Semiclassical

02:21

Probably could work out an example in there.

Am I being silly?

Fargle

This is a group, as you can check.

Also, $-1$ commutes with every element, so $\gamma_{-1}(x) = x$.

Akiva Weinberger

@Semiclassical Isn't that abelian

Semiclassical

Right, klein four-group is the first non-cyclic one.

Derp

Akiva Weinberger

I think you want S3

Fargle

But $\gamma_{-1} \circ \gamma_j = \gamma_{-j} \neq \gamma_j$.

meow-mix

02:23

so where is the hole in my reasoning?

Fargle

@meow-mix You're not proving that $\gamma_1$ is the identity on Aut(G).

You need to show that $\gamma_g \circ \gamma_1 = \gamma_g$.

Which you've already got a good first step for.

Mike Miller

what game are we playing

meow-mix

what first step would that be?

Semiclassical

Basically, it's just algebra that you need to verify

meow-mix

@MikeMiller the "meow sucks at math" game

Fargle

02:24

That $1x1^{-1} = x$.

Now take that composition, and show that it equals the RHS.

And do the same for $\gamma_1 \circ \gamma_g$.

Semiclassical

honestly, if you just write down that composition of conjugations, it's immediately obvious.

meow-mix

oh, so i show that $g(1x1^{-1})g^{-1} = gxg^{-1}$?

Fargle

Exactly.

Mary Star

We have that the dihedral group D_6 and the subgroup <σ>={σ, σ^2, σ^3, σ^4, σ^5, id}.
How can we find the elements of the quotient group D_6/<σ> ?

meow-mix

@MaryStar those are equivalence classes of $D_6$

which are related by

"symmetric"

wait

Mary Star

02:27

@meow-mix What do you mean?

meow-mix

what is $\sigma$?

or, if not using mathjax, σ

Mary Star

D_6=<σ,τ>

Semiclassical

Careful: There's two meanings of D_n

You presumably mean the one that contains 12 elements.

meow-mix

well yeah, they stated the subgroup consists of $6$ elements

Akiva Weinberger

Sigma = rotating the hexagon and tau = flipping it over?

meow-mix

02:28

@AkivaWeinberger the opposite i believe

Fargle

@meow-mix $\sigma$ is one of the generators. The dihedral group with $2n$ elements is $D_n := \langle \sigma, \tau\;|\;\sigma^6 = \tau^2 = 1, \sigma\tau = \tau\sigma^{-1}\rangle$

@AkivaWeinberger Yes.

meow-mix

@MaryStar so, the quotient group consists of two sets

Akiva Weinberger

@meow-mix But sigma has order six

Fargle

@meow-mix No, because $\sigma$ has order $6$. A reflection has order $2$.

meow-mix

oh wait

hmm, thats weird

Semiclassical

02:29

One simple way to see what the answer has to be: D_6 contains 12 elements, and the subgroup contains 6 elements.

meow-mix

i would think sigma is symmetry

anyways

Fargle

@meow-mix Rotational symmetry is also a symmetry.

meow-mix

so it consists of two sets

Null

what you think about this? does it honor Galois sufficiently? img.memecdn.com/more-many-is-not-humanly-possible_o_7021901.jpg

Mary Star

@meow-mix How do we know that?

meow-mix

02:30

{σ, σ^2, ..., σ^6} and {tau σ, tau σ^2, ..., tau σ^6}

Semiclassical

D_6 contains 12 elements, and the subgroup contains 6 elements. So the quotient had better contain 12/6=2 elements.

Fargle

@MaryStar The elements of D6/ <sigma> are exactly the cosets of <sigma>. Since D6 has 12 elements, and <sigma> has 6, then there are two cosets of <sigma>: <sigma> itself, and g<sigma>, where g is some element outside <sigma>.

Semiclassical

@meow-mix Those two sets only contain 10 elements...

Mike Miller

tau is always an involution in my parlance

meow-mix

i meant to write 6

Semiclassical

02:31

Right. I'd have just included 1 and tau instead.

meow-mix

yeah, you could write {1<σ>, tau<σ>}

Semiclassical

Yeah. Those are the cosets.

meow-mix

anyways ill work on these exercises tomorrow

i want to play some chess :)

Semiclassical

Which if you think about this as the symmetries of the hexagon: You start with a hexagon with 6 labelled vertices. You can make this more symmetric by considering a hexagon colored red on one side and blue on the other. Then there are only two configurations: the hexagon faces red up or blue up.

meow-mix

how do you define divisibility in groups? is it that $g | k$ if $g^n = k$ for some $n$?

Semiclassical

02:36

Divisibility isn't really a concept in groups.

Mike Miller

You usually don't, but that would be acceptable.

One often says that an abelian group is "divisible" if every $g$ is of the form $nh$ for some $h$

though note we make no demands on uniqueness of h

Semiclassical

Wouldn't that just amount to being cyclic?

Oh, doesn't have to be the same h for each element.

Though. Couldn't you just say that $g=1g$ in that case?

I don't see how that can differ from being cyclic without it being trivial.

meow-mix

i guess that would apply more for rings? in which we define the quotient ring as a set of equivalence classes whose equivalence relation is $x \sim y$ if $x-y \in I$ (where $I$ is the ideal you're quotienting by)

Mary Star

Thank you for your help!! :)

Semiclassical

"An abelian group $(G,+)$ is divisible if, for every positive integer $n$ and every $g\in G$, there exists $y\in G$ such that $ny=g$." (wikipedia)

"for every positive integer n" is the key part

So 'divisible' by 1, by 2, by 3, etc.

Brody

02:51

@Semiclassical so $\Bbb Q$ is divisible but not $\Bbb Z$

Semiclassical

Right.

Brody

Is this like saying $G$ is closed under division by positive integers?

Well, this assumes $ny=g \Rightarrow y=\frac{g}{n}$ and "division" is defined

Mary Star

And when instead of $\langle\sigma\rangle$ we had $\langle\sigma^2\rangle$, we would have 4 cosets, right? Are they the following:
$\{\langle \sigma^2\rangle, \langle \sigma^2\rangle\sigma , \langle \sigma^2\rangle\tau , \langle \sigma^2\rangle\sigma \tau\}$
? @Semiclassical @meow-mix @Fargle

Semiclassical

$\langle \sigma^2\rangle$ would have 3 elements, so 12/3=4 cosets, yeah.

I think those are the right ones, as well.

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