6:41 AM

7:22 AM
Find the number of elements of orders $p$ and $q$ in a nonabelian group $G$ of order $pq$, with $p<q$ (both prime).

Approach: Let the number of elements of order $t$ with $n(t)$.

Let $n(p) = m$ and $n(q)=pq-1-m$. By Sylow's theorem, the number of subgroups of order $p$ is either $1$ or $p$. It cannot be $1$, so it must be $p$. ($q$ must be of the form $pk+1$ in order for $G$ to be nonabelian ).

Now, each of the sylow-p subgroups being of order $p$, it has $\phi(p)=p-1$ generators in it. There are $p$ of them. So, $n(p)=p(p-1)=m$, and $n(q)=pq-1-p(p-1)$.
@BalarkaSen Can you please verify this,...is this correct now?

7:58 AM
How come the room is so dead

8:31 AM
It's Sunday morning
3

Sooo
@AlessandroCodenotti, can you check that

no HKers here because they all go out to protest
right now

8:49 AM
Can any moderator give me the answer of this question

9:04 AM
Let's get the chat jumpin on a Sunday
anyone. anyone? Beuler?

9:47 AM
Wagwan

last sunday vs today

That's impressive

'At the end of the Umbrella Revolution we said that "We will be back". Today we are back.'

2 hours later…
11:39 AM

11:50 AM
Hi, what does $[x]$ represent? math.stackexchange.com/questions/3263724/…

Hey, is there a way to derive Cauchy's estimate for power series coefficients without complex analysis?
@Luyw The floor function, I'd assume

Shouldn't it have only the lower '_' in the brackets?

That's just a matter of convention, both versions are used in the literature.

Thank you

Actually, Kavi Rama Murthy specifies this definition in his answer too, now that I'm taking a closer look.

11:56 AM
Oh right, didn't notice.

12:49 PM

0

Let there be a set of N numbers. Now we are given two values K and X. Now we pick exactly K different numbers. And the each of that number with the XOR of X and that number. We can perform these operation as many times. Finally we have to find the maximum sum of the numbers (After performing th...

Someone help me in this

1 hour later…
1:52 PM
Hello. How would you turn the set of circles and lines in $\Bbb R^2$ into a smooth manifold?
If it's just the lines in $\Bbb R^2$ you can choose two charts, one for the non-vertical lines and one for the non-horizontal lines
You can send $y=mx+c$ to $(m,c)\in\Bbb R^2$ and $x=m'y+c'$ to $(m',c')\in\Bbb R^2$ in each case
With smooth chart transition on $U_0\cap U_1$ given by $(m,c)\mapsto (y=mx+c\iff x=ym^{-1}-cm^{-1})\mapsto (m^{-1},-cm^{-1})$.
It seems I might want to think of all these as being circles on the Riemann sphere or something
In which case it might just be the Riemann sphere that is the smooth manifold

Circles on S^2 are intersections with planes

On S^2?

lol yes
so you have an analogous construction to what you did for lines

Can anyone tell me how did the value of dy/dx come with 2h at the denominator in the second answer??

2:23 PM
aren't those the real lines in $\Bbb CP^1$

2:53 PM
@F.White Circles on S^2 are parametrized by a center and a radius (except (p, r) and (-p, pi-r) give the same circle)
So the manifold you want is S^2 x (0, $\pi$) modulo a diagonal $\Bbb Z/2$ action sending $(p, r)$ to $(-p, \pi - r)$ I suppose

Claim: If $N \triangleleft G$ and $G/N$ are $p$-groups, then $G$ is a $p$-group. Proof: Note that $G$ is a disjoint union of left cosets $N$. Since $N$ is finite, each left coset is finite; and since there are only finitely many distinct left cosets, it follows that $G$ is finite. Hence, $|G| = |H| \cdot |G:H| = p^k$ for some $k \in \Bbb{N}$...
Does this look right?

2 million!
@user193319 or just use Lagrange lol

3:10 PM
@LeakyNun So $|H|$ divides $|G|$. Why does that imply $|G|$ is a $p$-group?

|G| = |N| |G/N|

That's what I wrote.
$|G| = |N| \cdot |G : N|$ (except I actually had $H$ in place of $N$).

|N| is a power of p
|G/N| is a power of p
so naturally |G| is also a power of p

Yes.

You can even do this for infinite groups. $p$-group in the infinite context means every element has order some power of $p$. Then pick any $g \in G$; $gN$ has order $p^k$ as $G/N$ is a $p$-group, so $g^{p^k} \in N$ is an element of $N$. $N$ itself is a $p$-group, so $g^{p^k}$ itself must have order $p^\ell$.
Combine to get $g$ has order some power of $p$

3:15 PM
Oh, very nice.

3:33 PM
@BalarkaSen Is the orbit space of a smooth manifold with a $\Bbb Z/2\Bbb Z$-action still a smooth manifold or you have to go to stacks or something?

> Provided that the multiverse is negatively curved, one can even fit infinitely many infinitely large universes of the same dimensionality inside, each infinite universe is a horohyperball bounded by a horohypersphere. Using a Poincaré hyperball model to model the geometry of such a multiverse, every representation of an infinite universe would be a hyperball tangent to the outer boundary of the unit hyperball.

Also, something else I've been wondering for awhile: is it possible for each $g\in\Bbb N$ to find some $f:\Bbb R^3\to \Bbb R$ which has level-set as a genus g closed orientable surface
Isn't the notion of a multiverse just science-fiction? @Secret

There are finite bounded objects, finite unbounded objects, infinite unbounded objects. But what is rarely talked about, are infinite bounded objects. Horospheres are an example of these
Yes for now

I mean what does it even refer to? Spatially separated universes or something?

Check Mark Tegemark for an example of classification of multiverses

3:38 PM
I was hoping you could give me the quick overview since you're talking about it :P

ok briefly speaking:

@F.White No, nothing fancy like that. In this case the action on $S^2 \times (0, \pi)$ is free away from $S^2 \times \{\pi/2\}$, so the image of $S^2 \times (0, \pi/2) \sqcup (\pi/2, \pi)$ under the quotient map is $S^2 \times (0, 1)$ and $S^2 \times \{\pi/2\}$ quotients to $\Bbb{RP}^2 \times \{1\}$.

Level 1: Copies of our observable universes outside that repeats due to poncarie recurrence
Level 2: Universes with different laws of physics
Level 3: Quantum realities
Level 4: Different mathematical structures

So the quotient space is $(0, 1]$ with a $S^2$ stuck at every $0 < t < 1$ and a $\Bbb{RP}^2$ stuck at $t = 1$. A neighborhood of $\Bbb{RP}^2 \times \{1\}$ is the tautological line bundle on $\Bbb{RP}^2$
Aka, the configuration space of circles in $S^2$ is the unit disk bundle of the tautological line bundle on $\Bbb{RP}^2$.
This is the 2-dimensional analogue of a Mobius strip

What goes on in that mind of yours
What an impressively quick answer, thanks @BalarkaSen I'll have to work out exactly what this means

3:50 PM
@F.White Embed $\Sigma_g$ standardly in $\Bbb R^3$ and let $U$ be an $\epsilon$-neighborhood of it. By tubular neighborhood theorem $U \cong \Sigma_g \times \Bbb R$. Define $f : \Sigma_g \times \Bbb R \to \Bbb R$ by $f(x, t) = \rho(t)$ where $\rho: \Bbb R \to [0, 1]$ satisfies $\rho(0) = 0$, and $\rho \equiv 1$ outside $(-\epsilon, \epsilon)$. Then this defines a function $F : U \to \Bbb R$ such that $F^{-1}(0) = \Sigma_g$
Define $F \equiv 1$ outside of $U$. This gives a global function $F : \Bbb R^3 \to \Bbb R$ which is zero exactly on $\Sigma_g$ and $1$ outside of a small tubular neighborhood of it.

someone clarify my question plis
it's been bothering me

what is it

oh group theory
sorry I don't know what groups are

My professor gave this as an optional exercise like 6 weeks ago, although sometimes the questions are intentionally impossible, the exact phrasing was:

"Try to find an explicit smooth function $f : R^3 → R$ with a closed orientable surface of genus g as a level set."

So I was expecting if it was possible that it'd be easy to write it out explicitly (my dream was really that it'd even be polynomial)

3:59 PM
You can write down a polynomial, but that's harder

Honestly, I was having trouble writing it down for the torus lol

explicit??

Sure, @Ryan, that's an exercise in Hirsch

@RyanUnger It wasn't intended as assessment, so sometimes the learning experience is meant to come from your failure :P

@BalarkaSen Hirsch has the hardest exercises in any book
some are unsolved

4:04 PM
@RyanUnger Wait, so this is open?
For some genera

Nah

This one is fine

@RyanUnger -_-
SOMEONE?

@SubhasisBiswas Ask more interesting problems :P
Hey @TedShifrin

4:06 PM
Hola all

Hi @ÍgjøgnumMeg
How does one say your name?

Kinda like "oy jugnom meh"

ig-yog-num?

I say I-go-no-meg

jug like the start of juggler?

4:07 PM
@BalarkaSen I played frisbee with that guy you mentioned

Yes :)

@RyanUnger Lol damn

some tiny chinese girl tackled me and my arm is still hurting

@ÍgjøgnumMeg Thanks :D

4:09 PM
@RyanUnger are you sure?

the pain is much better today

Frisbee is hipster sport

I spent 2 weeks playing ultimate frisbee at a summer school once
It was weird

4:23 PM
@F.White Here's an easier way to see your space is the tautological line bundle $\Bbb{RP}^2$. Every point in $\Bbb{RP}^2$ there's a line through origin in $\Bbb R^3$. For every such line there's a $(-1, 1)$'s worth of circles in $S^2$; slice the sphere by planes perpendicular to the line. So the configuration space is a $(-1, 1)$-bundle on $\Bbb{RP}^2$, and precisely the one which assigns to each point in $\Bbb{RP}^2$ the corresponding unit open interval in the line associated to that point.
The line bundle $E \to \Bbb{RP}^2$ where the fiber over $[\ell]$ is $\ell$ is the tautological line bundle. So that's just the open unit disk bundle of $E$.
In the earlier interpretation this is just the map $(S^2 \times (0, \pi))/\Bbb Z_2 \to S^2/\Bbb Z_2 = \Bbb{RP}^2$ which forgets the second factor.

Oops. Didn't realize I was logged in all night. Hi, @F.White.
Hi, a @Balarka

Hi @Ted!

As I always say, the exercises are what make (or not) a good book.

5:05 PM
hello
@ted
where can I find lists of groups of order up to 100

here's a good database by Tim Dokchitser at Bristol

hi @SubhasisBiswas
and hi @ÍgjøgnumMeg

Hey @Ted

I have a question. (I have been nagging about it the whole time)
@ÍgjøgnumMeg now that's awesome
Find the number of elements of orders $p$ and $q$ in a nonabelian group $G$ of order $pq$, with $p<q$ (both prime).

This one
i have written up a solution, don't know how good it is

Does it matter whether $q\equiv 1\pmod p$?

5:14 PM
I guess so

That's when there's going to be a nonabelian group of such an order

SSShhhhhh.

because a unique sylow subgroup (in this case) would imply the centre is G

trivial?

@SubhasisBiswas What you wrote didn't look good. There are $q - 1$ elements of order $q$, whereas you wrote some nonsense
@TedShifrin He's already written up a solution I didn't respond to

5:15 PM
@BalarkaSen i have corrected it

LOL, a @Balarka. I meant the necessary condition for non-abelian :)

Let the number of elements of order $t$ with $n(t)$.

Let $n(p) = m$ and $n(q)=pq-1-m$. By Sylow's theorem, the number of subgroups of order $p$ is either $1$ or $q$. It cannot be $1$, so it must be $q$. ($q$ must be of the form $pk+1$ in order for $G$ to be nonabelian ).

Now, each of the sylow-p subgroups being of order $p$, it has $\phi(p)=p-1$ generators in it. There are $q$ of them. So, $n(p)=q(p-1)=m$, and $n(q)=pq-1-q(p-1)=q-1$.

That's fine
Initially you wrote the number of Sylow $p$-subgroups is $p$ here which was nonsensical, which is why I didn't bother to respond.

@BalarkaSen yes. That is not what the sylow theorem says
feeling extremely foggy all day. Don't know what I am doing with my life
see you

@BalarkaSen That's awesome, thanks

5:20 PM
I mean you should learn to draw pictures of these things. A nonabelian group of order $pq$ looks like a flower with $q$ petals (Sylow $p$-subgroups) and center of the flower is the normal Sylow $q$-subgroup (assuming $q < p$ like in your setup)
Conjugating by a nontrivial element of the "center of the flower" has the action given by rotating the flower, i.e., the Sylow $p$-subgroups are conjugate

5:33 PM
@BalarkaSen lmao

6:29 PM
Yeah drawing pictures of these things really helps

Depends on what sort of thing you are looking to do
If you start getting used to only working with these things as pretty pictures, you will have no way to handle the more abstract things.

yeah
But a lot of geometric intuition guides me in algebraic geometry as well @TobiasKildetoft

sure, in some cases it is good to ground stuff in some geometric intuition
this is something I was always missing and felt made me have a much harder time with AG

I think a good entry point towards algebraic geometry is through complex analysis
then you will get good geometric grounding.
For instance I have a project I am working on I would like to publish it next year. But first I need to work on certain background mainly in complex analysis and Hartshorne and I will work on the problem next summer.
I think to enter AG and have good picture of what is going on it is through complex analysis and Riemann surfaces.

7:18 PM
@TobiasKildetoft Yeah that handicaps me in algebra sometimes
Most of the time good pictures come after you've done the hard algebraic work by getting your hands dirty
Math is strange
No particular formula works everytime, in fact any such formula fails most of the times
Hi @Karl

Hello @Balarka

How's it going?

Good, end of the quarter.

Hah that's always a relief for me (though it's a semester and not quarters on my end)

7:46 PM
Is there a site that lets people link to math formulas?
For example, a $25 hr job is$52K a year because ($25 an hour * 40 hrs a week * 52 weeks a year) but I would link a$25hr job is not enough for living in California at $60k a year cost and I would have a link on "$25hr" to a page that showed the math
the page would show "\$25 an hour * 40 hrs a week * 52 weeks a year = 52k a year"

1 hour later…
9:18 PM
Context: cryptography

-Choose two distinct primes p and q of approximately equal size so that their product n = pq is of the required length.
-Compute φ(n) = (p-1)(q-1).
-Choose a public exponent e, 1 < e < φ(n), which is coprime to φ(n), that is, gcd(e, φ(n))=1.
-Compute a private exponent d that satisfies the congruence ed ≡ 1 (mod φ(n)).
-Make the public key (n, e) available to others. Keep the private values d, p, q, and φ(n) secret.

**What does the trible equal sign indicate in this context?** [According to Wikipedia, it can mean different things depending on the context](https://en.

@SebastianNielsen it means equivalent modulo phi(n).
The idea is that it is "equality", but in a different sense than usual.

I am sorry, but how is that different from the "usual" equal sign (=)
?

You know like in elementary school how you did "congruent" shapes? Like triangles with the same angles, but not the same side lengths?

Uhm, I am not sure.

The notation "a ≡ b (mod p)" means that there exists an integer c such that "a = b + cp".
They could fully well use the notation "a = b (mod p)" instead, but it's clearer this way, since it's not your usual equality.

9:25 PM
Okay, thanks.

The rigorous reason is that they are defining a new notion of equality called an "equivalence relation".
Which respects all the properties of equality that you would normally expect, but often doesn't coincide with the "=" you are used to.