10 page paper is perfect length :)

i published 1 paper and I've got 2 more in waiting

cool

yes but the other 2 are not as powerful so thats why I'm waiting

If I write two papers one will basically be just a combinatorial argument and an algebraic topology argument. I think I'd have to wait to post the first one until I did the second because the arguments would be too easy to scoop.

I need to prove more

@PVAL-inactive well in that case you should put it all together

or just put both on Arxiv at the same time

so that nobody can scoop it

The second one is what will probably happen.

do you have any other applications of the combinatorial stuff?

Though I'd like to write the first paper, then post it so its out there. Then spend time on the combinatorial stuff.

@ForeverMozart I don't know that means.

The combinatorial stuff is likely very easy for a lot of people to do.

hey @BalarkaSen I would like to discuss something

in group theory

I will only help if I can help.

Assume K is finite cyclic group, H is an arbitrarily subgroup. Suppose $\phi_1$ and $\phi_2$ are homomorphism from $K \rightarrow Aut(H)$ such that $\phi_1(K)$ and $\phi_2(K)$ are conjugate subgroups in Aut(H). I want to prove that $H \rtimes_{\phi_1} K$ and $H \rtimes_{\phi_2} K$ are isomorphic subgroups.

The way I did this is as follows

So we know K is cyclic right so it is generated by some element k i.e K = <k>. By Hypothesis $\sigma \phi_1(k) \sigma^{-1} = \phi_2(k)^a$ for some $a \in \mathbb{Z}$

so we can define the following homomorphism

$(h,k) \mapsto (\sigma(h),k^a)$

I proved that this is a group homomorphism between $H \rtimes_{\phi_1} K$ to the other one.

Since K is finite so it has some order n. Since $\phi_1(K)$ and $\phi_2(K)$ are conjugate so they have same order m, they are also cyclic.

@PVAL I think a lot of people post two papers at once. I think that's what I'm going to try to do.

@MikeMiller It's becoming rather bittersweet to continue to be proving things while trying to write things up.

@Adeek Sounds good to me so far.

I think me now and myself a year ago would get into a fight over that statement.

I am stuck here though

So all you have to do is to write down an inverse for your homomorphism?

I was thinking to prove that a has relative order to n.

@PVAL I hate writing.

if a is relatively prime to n then I am just taking a generator to a generator

then I will be done

right @BalarkaSen ?

You want to show $\sigma$ sends a generator to a generator?

yeah

@MikeMiller Are you convinced you have results now.

?

i'm a compulsive editor

@Adeek There's no reason this is true though.

Think in a seminar today someone announced a result and someone else crushed their proof.

Yikes, @PVAL. That sucks.

lol

hm

@PVAL-inactive lol crushed

just a sec @BalarkaSen

@Adeek What you probably want is to say there is a representative in [a modulo m] which is prime to n. That would suffice, nope?

@PVAL Yes. Paper 2 I'm not sure about. But i think it will be a weekend to check if I have results and months to write them up.

@BalarkaSen why would it suffice ?

$\phi_2(k)^a$ only depends on the mod m class of $a$, because $\phi_2(k)^m = 0$.

If someone crushed my proof I think I would quit math

By assumption $m$ is the order of $\phi_2(K)$.

@PVAL Can you email me details?

oh ok yeah so what we would like to show is that we have that gcd(|a|,|K| = n) = 1 right ?

yeah

|a| is bad notation, but yeah. You want to show (b, n) = 1 where b = a modulo m.

@MikeMiller I'd rather not. It was an internal seminar (internal to the grad students here).

since k is generator of k so $\phi_1(k)$ and $\phi_2(k)$ are both generators of $\phi_1(K)$ and $\phi_2(K)$

That sounds like a straightforward number theory exercise now.

so $\sigma \phi_1(k) \sigma^{-1} = \phi_2(k)^a$ tells us that gcd(m,a) = 1

eyyo

does m | n ?

@PVAL misunderstood. agree it would be best not to share.

@Adeek Yes, because it's a subgroup.

oh oke

yes then I am done

but wait now why is my orginal maps between the semi-direct an isomorphism ?

You now know that $\sigma$ is an isomorphism. Write down an inverse for your homomorphism.

alright

@MikeMiller Okay so if $b/4-1/2 \geq 5$ and $p$ is prime. There is a unique filling of $L(p,q)$ up to homeomorphism.

just a sec

oh ok I got it now @BalarkaSen since we know that this particular a is relatively prime mod n so we know it is invertible i.e we must have b such that ab = 1.

Right.

so we can define the following map $(h,k) \mapsto (\sigma^{-1}(h),k^b)$

this will give us the required inverse map

@PVAL What is b? Some number parameterizing the contact structure?

@MikeMiller b is the lower bound on the terms of the continued fraction decomposition, as in my first comment.

ok.

This is true for any contact structure on those manifolds

that was a nice problem thank you @BalarkaSen

sure

just a sec @BalarkaSen we have $\phi_i : K \rightarrow Aut(H)$ right so how do we know that we have |K| is divisible by the $|\phi_i|$?

is it just first isomorphism theorem ?

@Adeek Yes.

yeah nvm it is trivial it is lagrange theorem and first isomorphism theorem I guess

Just the 1st isom theorem suffices. No Lagrange needed.

Hi all

yeah