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hi DogAteMy, orbit
12:21 AM
Hi. Any hints on how to find the asymptotic order of $P_K(1+K^{-\alpha})$ with $\alpha\in(0,2)$ if $P_n(x)$ denotes the Legendre polynomial? There are results when $n$ goes to infinity, but I couldn't find anything for my case, in which also $x>1$ rather than $x\in[-1,1]$.
@TedShifrin hello ted
@Daminark ive got a good group theory question
1:35 AM
Is there an easy way to show $|\text{Aut}(S_4 \oplus \mathbb Z_2)|=48$?
@Argon, what's the hard way?
@orbit-stabilizer The hard(est) way is to enumerate them all
@Argon ah yes
@Argon write down a 2x2 matrix with homomorphisms in each entry
@anon How do you mean? Any homomorphism?
1:45 AM
heya anon :)
not quite any
see the end of my answer here
hi ted
it's not terribly hard to prove the fact, and could be useful beyond just a single exercise
@anon Thanks, I'll take a look
the rest of my answer covers H^2 which is different from HxK mind you
Hi chat.
For anyone who's curious, the calculus group is officially over.
May the stooopid idea of classes on MSE rest in peace.
Applause for those who gave it a try.
1:49 AM
also kind of reminiscent of goursat's lemma, which Tobias explains nicely here
For anyone who's curious and insane, I wrote an explanation of SOAP, which makes really big numbers.
@Kasmir: Hello, but it's way past your bedtime.
Ah man.
I feel like I'm at the limit of the largest numbers I can write in a decently sized program. Spent so much time these past few weeks working on it.
At least I'm at the limit of what I'm comfortable with. That much is certain.
Why does a non-trivial homomorphism exist iff $q|p-1$? I know, given $H,K$, with orders $p,q$ respectively, and $q<p$, we have $Aut(H) \cong Z/(p-1)Z$. So for the homomorphism from $K $to$ Aut(H)$ to exist, q must divide $p-1$ since the image is a subgroup, but why are those sufficient conditions rather than just necessary conditions?
@SimplyBeautifulArt I think it would be useful to have a room going for people self-studying the same thing


Trying to understand extraordinarily large numbers.
2:04 AM
Hah, not my cup of tea
Q: How to disprove this fallacy that derivatives of $x^2$ and $x+x+x+\cdots\ (x\text{ times})$ are not same.

Bazinga Possible Duplicate: Where is the flaw in this argument of a proof that 1=2? (Derivative of repeated addition) $$x^2=\underbrace{{x+x+x+\cdots+x}}_{x \text{ times}}$$ $$\therefore \frac {dy}{ dx}\\ =\frac{d}{dx} \left( \underbrace{{x+x+x+\cdots+x}}_{x \text{ times}} \right) \\ = \u...

^ More ways to disprove your classmates.
Oy vey.
@orbit-stabilizer Also, check my profile.
Needs an update though
@SimplyBeautifulArt I did, what was I supposed to find?
@orbit-stabilizer I had thought your first comment to me was for a room for studying very large numbers.
Hence link to one chatroom as well as an off-stackexchange chat place for more.
2:06 AM
"I think it would be useful to have a room going for people self-studying the same thing"

I just meant that in general. Like I would be interested in a room discussing Aluffi's Algebra
Oh haha
I wonder
What are the best quotes of the year from this chatroom?
Spend days sorting through the star board?
'my dad sees me reading a book on Galois Theory "Remember, it's just a theory." '
Lol, yeah I'm gonna do that I guess
2:09 AM
stop that, orbit
yes plz no gigantic chatjax text.
@orbit-stabilizer If you want to be showing some interesting starboard stuff, just do the links.
I dont know how :/
When you've got a chat message you wanna link, such as this one, click the gray arrow on the right
and copy the permalink.
2:13 AM
@anon I can understand until about the commutators. Why does that guarantee the automorphism is the matrix?
@Argon check that the matrix applied to an element equals the automorphism applied to an element
racks brain to remember answer written 2 years ago
Isn't it that matrix by definition?
assume $\phi$ starts out as an arbitrary automorphism, then associate a matrix to it, then check the matrix and $\phi$ evaluated at a tuple produce the same result
._. so much drama
Can I ask a question? Can I say a metric space (X,d) metrizable? Thank you!!
2:18 AM
One does not simply ask to ask a question and then ask away anyways before anyone has the chance to respond.
hmm .. "metrizable" = there is a metric. Hmm ...
@SimplyBeautifulArt One shouldn't even ask to ask.
Yeah, and I think you were the one that said that policy to me first :P
But why my professor said it is unclear. Cry.
Professors suck.
2:22 AM
Also, why are most of the highly starred chat messages about flags needing to stop? x.x
because there have been some infiltrators doing crazy flags in here
Oops didn't work
sup duds
2:28 AM
When you've got a chat message you wanna link, such as this one, click the gray arrow on the right
and copy the permalink.
@DHMO Man, I do miss you bud. I really do lol
I click the grey arrow. I click permalink. What happens next?
heya Eric
16 mins ago, by Simply Beautiful Art
When you've got a chat message you wanna link, such as this one, click the gray arrow on the right
and copy the permalink.
I did !
the permalink stuff here is highly confuzling.
2:30 AM
Now that's out of the way. How many elements of order 25 are there in $C_{5^{100}}$. How does one go about answering this?
@Ted I managed to acquire a copy of Bryant's EDS
On chrome:
will thumb through over break probably
Then you can copy/paste a link to a chat message.
Got it, thanks @Simply !
2:31 AM
Well, the real tôme is the Bryant/Chern/Griffiths/Goldschmidt EDS book, Eric.
ya that one
i didnt wanna write out all the names
ohhh ... great.
it takes more than thumbing
yeah i figured
it's one of the few books I didn't discard when I moved
but if you want to work through things, I'll be glad to participate
I'm off to cook dinner
have a nice meal
2:33 AM
I'll do that
Starred messages I personally noticed and found interesting:

["Math is hard, guys. Don't do it if you don't want to struggle to learn."](https://chat.stackexchange.com/transcript/36?m=39667233#39667233)

["A **tan** is good **cos** it's not a **sin**."](https://chat.stackexchange.com/transcript/36?m=39535988#39535988)

["We definitely need to encourage more women re math."](https://chat.stackexchange.com/transcript/36?m=38508407#38508407)

["mathematics is not about numbers, nor is painting about ink, nor is music about notes."](https://chat.stackexchange.com/transcript/36?m=35840237#35840
Oh, thanks for not working too.
Anyways, good night all
2:50 AM
3:15 AM
@anon What makes $\alpha(H)$ and $\beta(H)$ normal?
(Sorry for the trivial questions, this is new to me)
How can one show that the symmetries of the cube gives $S_4$ without doing the business about actions on opposite vertices?
Like I'm having an extremely bad time processing that picture, is there a sort of combinatorial way to execute this?
for your purposes you want to write $\alpha(H)$ and $\beta(K)$. because they are images of normal subgroups under automorphism / surjection
Oh of course thanks
@Daminark first figure out all the rotations of S_4 then see what the cycle types are of the corresponding actions on space diagonals
which doesn't avoid the diagonals like you want, but the diagonals are the best way to see it
like, the 90 degree face rotation has cycle type (abcd)
the 180 degree face rotations have cycle type (ab)(cd)
now do rotations around axes through vertices and edges
after explaining cycle types and drawing a nice picture I gave this exercise to a nursing student and she succeeded writing them all down
Wait I think we can embed the tetrahedron into the cube, and I've been convinced that rotations of that are A_4
3:21 AM
you're tackling the problem like an air bender, you need to be an earth bender and just do it :P
And I'm completely incompetent in visual stuff
So I'm trying to dodge that
invest in some piper cleaners
oh shit that sounds like a fun way to visualize stuff
I want a dry erase sphere
and a knobbly wobbly
that reminds me of a problem that came to mind thinking about S4 acting on the cube. Can a proper subgroup of a finite group contain a representative of every conjugacy class?
@anon that's a good question
3:27 AM
googled and MO says no apparently
Yes, makes sense
Yes, it does.
my guess is counting
Suppose there existed such a group, then let the elements of that group act on themselves by conjugation.
Mhmm.... no.
cause if you had such a subgroup $H \leq G$ youd have $\bigcup_{g \in G} gHg^{-1}$
3:29 AM
I'm getting confused around the section regarding "free products, semidirect products, and wreath products." It appears to me that in my case only diagonal matrices can be the automorphisms (?). Can I just then enumerate the number of automorphisms of $H$ and $K$ and find the number of matrices that can exist?
Oh yeah for a proper subgroup H that union can never be everything
@Argon I give the general form of matrices in ${\rm Aut}(H\times K)$, and whether or not they are all diagonal depends on if there are any group homomorphisms $H\to Z(K)$ or $K\to Z(H)$
if there are such homomorphisms, then the corresponding nondiagonal matrices still define automorphisms of $H\times K$
yeah counting gives you an inequality $|G| - 1 \leq (|H| - 1)[G:H]$
which implies $|G| = |H|$
if i ever taught people group theory id assign this as a prob
i rate it 7.5/10
It's p good, I know Frank assigned his class the stuff but Jack didn't
now how to generalize to the lie setting appropriately
3:33 AM
Urg this stuff makes no sense
^ I feel that
@Argon given $\phi$, define $\alpha,\beta,\gamma,\delta$ appropriately. prove $\alpha,\beta$ must be invertible, and $\beta,\gamma$ land in the centers of their codomains. prove $\phi$ evaluated at $(h,k)$ matches the result of evaluating the matrix at $(h,k)$. conversely, prove every such matrix defines an automorphism.
consider it an exercise to try on your own instead of someone else's writing to read.
(after defining what $\alpha,\beta,\gamma.\delta$ are anyway)
@anon what's a good place to read about rep theory of compact groups
I've got a final coming up and Farb never gave us references for the compact groups stuff
I wouldn't say I'm a good place to ask, as I just read pdfs and play fast and loose with convergence of integrals/sums
3:53 AM
Oh here is a fun problem
Prove the average degree of a graph is $< \sqrt{2m}$
what's m?
Number of edges
or is figuring out m part of the exercise
I feel like average degree should be 2m/n
where n is number of vertices
it is
4:07 AM
anon r u in hs too?
ha no
so I guess $2m/n<\sqrt{2m}\iff 2m<n^2$ means we look at the latter now
or $m<n^2/2$, but that follows from $m\le\binom{n}{2}$
so, done
im satisfied
im procrastinating work by doing less important work
what has become of my life
a rooted tree of procrastinations within procrastinations
@anon okay that's another way to do it
What I had in mind was using the adjacency matrix, which is why I liked it so much
easyboi counting mode
4:14 AM
The sum of the entries of the adjacency matrix is gonna be $2m$, but it has to be less than $n^2$
Because this is a 0-1 matrix, and the diagonal has to consist of zeroes
i give it a 6/10
no u
6/10 aint bad
Lol perhaps
Ds get degrees
4:17 AM
But yeah I dunno why but I was just cracking up so hard at this problem, like I put it in my "favorite problems ever" list
More for being cute than being hard, but still
i dont get excited by counting graph things really
Lol yeah, I've gotten the vibe that we have... pretty different taste
i just kinda get bored doing it
i get excited if it shows up in connection with something else
like turning something into combinatorics gets me intrigued
I've got my group theory exam tmr morning...
4:32 AM
@orbit good luck
@Eric yeah I haven't thought about connections with other things so much
I just sorta found that the stuff in algebra that I've enjoyed most was the Sylow counting, and mostly because I just liked the exercises themselves
@Daminark Why is this true?
Why must a group of order 182 have a subgroup of order 91?
Okay so if you have a group of order 182
You know by Sylow that you must have a normal subgroup of order 7, right?
i think about most math in terms of connections with other things
Call that $K$. Now you also know by Sylow that you have /some/ group of order 13, which may not be normal, yeah?
4:35 AM
Yeah, that's true.
Either 1 of them, or 14
Call that $H$. Then because $K$ is normal, you have $HK \le G$
Okay. I don't see why it has to be a subgroup, lemme prove it
gimmie a sec
Can you prove that if $H$ and $K$ are subgroups of $G$, that $HK \le G \iff HK = KH$?
Oh boy this does not bode well.
Gimmie a few more seconds
One direction is easy. Suppose $HK = KH$. Then $HK \leq G$ because $e \in HK$ obvs. And since $hk = k'h'$, we have that $HK$ is closed under multiplication. And ofc inverses exist because $(hk)^-1 = k^{-1}h^{-1}$.
That direction suffices for this problem
Because $K$ is normal so ofc $HK = KH$
4:50 AM
Okay, how do you do the other direction?
Well, assume $HK \le G$
Then for $hk \in HK$, you know $k^{-1}h^{-1} \in HK$
So $k^{-1}h^{-1} = h'k'$ for some $h'\in H$ and $k'\in K$
But then $hk = k'^{-1}h'^{-1} \in KH$
So that does it. So since in our above question, $K$ was normal by virtue of $n_7 = 1$, we know that $HK \le G$
Okay, so $HK \leq G$. Let's continue.
Well, we know that the size of $HK$ has to be $7\times 13 = 91$
(This is because $H\cap K = \{e\}$)
4:53 AM
So that's why $G$ has a subgroup of order 91, meaning index 2
So that's normal
If you have a group of order 91, just abstractly, let's look at it
Ah, and we don't have to consider any semidirect product to find what group of order 91 there is because it's cyclic.
We know $n_7 = 1$ and $n_{13} = 1$ easily
So then we have a normal, cyclic subgroup of order 91
Right. Got it.
Here is where you have to use the semidirect product
Let's call $HK = N$
4:55 AM
You mean G?
$HK = G$
No... $G$ is the group of order 182, $HK$ is the subgroup of order 91
Oh I see, I thought we were relabelling K to be the group of order 91. Sorry
Oh no it's the same as before
Okay, let's call $HK = N$.
So yeah we now know that there is some subgroup $T$ of order 2
4:57 AM
By sylow right
We know that $TN = G$
And $N \cap T = \{e\}$
So this is the internal semidirect product of $T$ and $N$
And $|NT| = |N||T| =|G|$ which means $NT = G$.
Huh, that NT looks off
This means that $G$ is isomorphic to some external semidirect product of $N = C_2$ and $T = C_{91}$
4:58 AM
So now we need to classify the automorphisms?
Yeah, first we find all the automorphisms, and then we see which ones give isomorphic groups
We want to find all homomorphisms from $C_2$ to $\text{Aut}(C_{91}) \cong C_6\times C_{12}$
Why can't we look at automorphisms from C_91 to C_2?
$C_{91}$ is the normal subgroup
Right. my bad.
So you have to take a map from $C_2$ to the automorphism group of $C_{91}$
5:01 AM
Now, why is $Aut(C_{91}) \cong C_6 \times C_{12}$.
But yeah so you can either take the trivial automorphism, in which case the group is $C_{182}$, or you can map to a subgroup of order 2 of $C_6\times C_{12}$
Well, you know $C_{91}\cong C_7\times C_{13}$
And we know there's no homomorphism from $C_7$ to $C_{13}$
You have 6 degrees of freedom in the first one and 12 in the second and so it works?
So $\text{Aut}(C_7\times C_{13}) \cong \text{Aut}(C_7) \times \text{Aut}(C_{13})$
5:03 AM
Oh I see. We disregard the trivial homomorphism?
I mean the other way to do it is to just find $C_{91}^{\times}$
Right, which is just $\phi(91)$.
Well, the order is $\phi(91)$
Ah yes.
@Daminark Okay so we want to a homomorphism from C_2 to C_6 x C_12
But it's a general fact that if $G$ and $H$ have coprime orders, then $\text{Aut}(G\times H) \cong \text{Aut}(G) \times \text{Aut}(H)$, which is easy to do when you note you don't have homomorphisms between them, so I'll move on
5:07 AM
That's good to know
So either the homomorphism is trivial or it sticks you in a subgroup of order 2, and I think there are just 3 of those in $C_6\times C_{12}$
$(3,0)$, $(0,6)$ and $(3,6)$ should be it
So now we have at most 4 groups of order 182, and we can find which are isomorphic
How did you come up with the subgroups of order 2 in C_6 x C_12?
A subgroup of order 2 is an element of order 2
But you have to be order dividing 2 in each coordinate
So that only gives 3 options
Oh, that's right
So, 4 homomorphisms. 1 being trivial.
So now I recommend just bashing them out to see which are isomorphic. You'll get a $D_{182}$ in there, wouldn't surprise me if there weren't any others tbh
Anyway, I'm dead tired and so imma slep
5:11 AM
I see, and by bashing them out, you mean computing them by applying the autmorphism?
Alright, thank you! Goodnight!
This was extremely helpful!
Yeah, like see how the automorphism acts on $C_{91}$ and find if any of them do the same thing. And yeah no problem, good night!
lol, welcome back
Lol I came to log out from my phone :P
See you!
3 hours later…
8:42 AM
$\Bbb Q(x,y) \cong \Bbb Q(x+y,x-y)$?
@LeakyNun Yes
as fields?
long time no see @TobiasKildetoft
@LeakyNun Yes, as fields. In fact, one can argue that they are equal
indeed, one can
$\overline{\Bbb C(x)} \cong \Bbb C$?
9:03 AM
So to continue where I left off yesterday...
First to recap the definition of a measure:
Let $\mu : X \to \Bbb{R} \cup \{\pm \infty\}$ be a measure. Then
1. $\mu (\varnothing) = 0$
2. $\mu (X) \geq 0$
3. $\mu \left( \bigcup_{i \in I} \mu (X_i)\right) \leq \sum_{i \in I} \mu (X_i)$
You need $I$ to be countable and the $X_i$ taken from some fixed $\sigma$-algebra of subsets of $X$ (which is why I find it more intuitive to begin with an outer measure and construct a measure from it)
9:18 AM
is the trouble of dealing with an uncountable sum the only reason why we cannot define a measure with uncountable subadditivity, or is there more fundamental reasons to that?
If you want points to have measure 0 and some set to have positive measure you can't have uncountable additivity (anyway an uncountable sum can be finite iff all but countably many factors are 0)
@AlessandroCodenotti is it false when $I$ is uncountable?
wait, yes it is
I see, since all elements in a set are points and points have zero measure thus allowing uncountable additivity will result in basically a sum of zeros which is always zero in the end, thus making all sets have measure zero
ok, now to revise on outer measure...
(also typo in the very first line, I should be saying the positive extended real ray, otherwise I will have negative values in the measure)
Points don't have to have measure zero, but it's nice when they do
@Secret for example here's an uncountably subadditive measure: enumerate the rationals $f:\Bbb N \to \Bbb Q$. Define $\mu(A) = \displaystyle \sum_{n=0}^\infty \begin{cases} 2^{-n} & f(n) \in A \\ 0 & f(n) \notin A \end{cases}$
it's even defined everywhere
it fails translation-invariance
9:31 AM
@LeakyNun hmm..., since $f$ is an enumeration, does it mean it is well ordering the rationals and thus under this measure, $\mu (A) = \frac{1}{2^n}$ if only the nth rational is in $A$?
yes, and it does not follow
I mean, if only the nth rational (and no other rationals) are in A, then by that formula, $\mu (A)$ should only have $\frac{1}{2^n}$ contributed to it
Outer measure: Let $m : 2^X \to \Bbb{R}_{\geq 0} \cup \{\infty\}$ be an outer measure. Then:
1. $m (\varnothing) = 0$
2. $A \subseteq B \implies m(A) \leq m(B)$
3. $m\left( \bigcup_{i=1}^{\aleph_0}A_i \leq \sum_{i=1}^{\aleph_0}m(A_i)\right)$
Given some $E \subseteq X$. $E$ is measurable under $m$ if $\forall A \in 2^X$:
$m(A) = m(E \cap A) + m(E^c \cap A)$
Therefore the Lebesgue outer measure is the infimum of the covers of $E$ formed by closed intervals
Let $\lambda^*$ be the Lebesgue outer measure.
Now to figure out how to prove $\lambda^*[a,b] = b-a$
10:32 AM
A patient is sitting in the waiting room of a doctor's office. We assume that its waiting time in minutes is exponentially distributed with parameter $\lambda = 0.2$. Within what time will the patient be treated with probability $0.9$?
The patient waited $5$ minutes without being called. How long does he have to wait with probability $0.9$?

I have done the following:
$$P(X\leq x)=1-e^{-\lambda x}\Rightarrow 0.9=1-e^{-0.2x} \Rightarrow e^{-0.2x} =0.1 \Rightarrow x \approx 11.513$$
Within the first $11.5$ minutes the patient will be treated with probability $0.9$.
2 hours later…
12:39 PM
Q: Abelian monoids and integer polynomial composition?

mickConsider distinct integer polynomials of distinct degree. Let $*$ denote composition. Consider when $a,b,c,d$ forms an abelian monoid. Many questions arise naturally. For instance 1) can we classify them ? 2) Is every finitely generated abelian monoid a submonoid of a nonfinitely generated a...

Any ideas ?
@mick Most of that seems to not make any sense to me
Why not ??
what do you mean by "Consider when a,b,c,d forms an abelian monoid"?
abc's are polynomial , They form abelian monoid by composition
but no set of four integer polynomials will form a monoid under composition.
at least not when they have distinct degrees
I think there is only one way to have four polynomials form a monoid under composition, and that is by choosing $x$, $-x$, $-x+1$ and $-x -1$

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