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04:00
ich bin niemand
du bist ziemlich weit dafür, dass du 18 bist
viel weiter als ich mit 18
okay, you can call me "Sylowmeister" if you want to
@MatheiBoulomenos :P
Anyway, back to $120$
@MatheiBoulomenos di bist nur 2 jahre elter wie mir
Woher weißt du, wie alt ich bin?
"als ich"
04:02
You love telling us which year you are
di hast gesugt dass di bist in der dritte jahre
@MatheiBoulomenos wie mir is what the yiddish translator tells me
I didn't know you were using yiddish
bad german, you mean
@MatheiBoulomenos lol
I'm 22 actually
04:04
in' in dem dritten jahr?
(in' = und)
dem
dritten
life is not always straightforward
@MatheiBoulomenos alright, Sylowmeister
Sr. Boulomenos
04:05
...
you should put this in your office
undergrads get offices?
I mean when he gets one
Well, I don't lol
you will
04:07
If you want to crack that $120$ nut, here are some hints
@MatheiBoulomenos no
but I think it's too difficult
sorry, I was not being a good Sylowmeister
try $180$ I think that's easier
Sylowmeister
Let $\varphi:G \to S_{15}$ be a monomorphism. Let $P_2$ be a 2-subgroup. Its normalizer has $120/15 = 8$ elements, so the normalizer is just itself (wait whaaaat)
that happens
04:17
ok gib mir einen Hinweis
Okay
you already constructed an embedding $G \to S_6$
what about the image of that embedding? Can it contain any odd permutations?
$180 = 2^2 \times 3^2 \times 5$. $n_2 \in \{9,15,45\}$, $n_3 \in \{10\}$, $n_5 \in \{6,36\}$ even more possibilities lol
@MatheiBoulomenos Let $\varphi: G \to S_6$ be a monomorphism given by the group action from $G$ by conjugation on the $6$ Sylow $5$-subgroups.
Why can't it contain any odd permutations?
If you have a subgroup $H \leq G$ and a normal subgroup $N \trianglelefteq G$, then what can you say about $H \cap N$?
it is also normal
$\varphi[G] \le G$ and $A_5 \trianglelefteq G$.
04:27
I wonder if we can generalize this result
So whenever you use a group action to define a monomorphism into $S_n$ for a simple group, you actually get a monomorphism to $A_n$
sometimes that's useful
Let $G$ be a simple group and $\varphi$ an embedding into $H$. Then, $\varphi[G]$ is contained in the smallest non-trivial normal subgroup of $H$.
correct
you might as well say contained in every non-trivial normal subgroup
04:30
Theorem: Simple groups are not simple at all.
it's not that simple (no pun intended)
$\varphi[G]$ might just intersect normal subgroups of $H$ trivially
hmm
I give up generalizing :P
the simple groups are classified
and then?
why are you people discussing a dead problem
04:31
it's fun
@MatheiBoulomenos what next?
I think you can figure out $180$
$A_n$ is simple for $n>4$ btw
I mean, for 120
Oh, for 120
you use one trivial fact
one trivial fact dass ich weiss nicht
04:34
if $A \subset B \subset C$ is a chain of subgroups, then $N_B(A) \subset N_C(A)$
that's what I did
are you even trying with the word order?
@0celo7 what?
In einem Nebensatz mit dass kommt das Verb ans Ende
oh and you need some combinatorics
@MatheiBoulomenos dass ich nicht weiss?
Combinatorics is trivial?
04:36
@MatheiBoulomenos oh Gottesgott
Well, that title is too much for sure :P
@MatheiBoulomenos no I was saying like omg
oh fur Gottesliebe
nobody says "oh für gottesliebe"
Just stay with "Oh mein Gott" to be safe
what do they say
@MatheiBoulomenos lol ok
Gregorian chants
04:38
fair enough
yes, combinatorics is trivial, but I didn't find it trivial to figure out that it is even relevant in this situation @0celo7
@MatheiBoulomenos I'm sure there's a... simpler explanation of how there are no simple groups with order $120$
no pun intended
I gave you a couple hints already
@MatheiBoulomenos combinatorics
and some normalizer
and some even permutations
nothing that I can touch
alright, I can touch combinatorics, but whatever
rigorous combinatorics is done using group actions :P
04:40
But you see why we have an embedding $G \to A_6$?
try to apply some trivial fact to that
sure
"trivial"
yeah, think about normalizers and combinatorics
lol
$A_6$ is a simple group
nope
you don't need the simplicity of $A_6$ for 120
You need to find a subgroup $H \leq G$ and then apply that $N_G(H) \leq N_{A_6}(H)$ thing
eh, the 5-subgroup?
so $N_G(H)$ has 20 elements
and God knows how many elements $N_{A_6}(\varphi[H])$ has
well, that's where the combinatorics come into play
...
"trivial"
by trivial I meant only the $N_B(A) \subset N_C(A)$ thing
what can you say about $\varphi[H]$ as a subgroup of $A_6$?
it has 20 elements
04:49
the image of a 5 element set has 20 elements?
oh you meant the normalizer
oh lol
it has 5 elements then
sure, so what kind of subgroup is it?
a 5-cycle
yes
and a $5$-Sylow subgroup
04:52
as $|A_6|=\frac{6!}{2}$
I'm sure you know some things about the normalizers of Sylowsubgroups
$|A_6| = 360 = 2^3 \times 3^2 \times 5$
no, I know nothing
the index of the normalizer of a Sylowsubgroup is equal to ...
$n_5 \in \{6,36\}$
@MatheiBoulomenos either $60$ or $10$
you don't even need Sylow theorems for that
but it isn't $10$ so it's $60$
04:55
why isn't it 10?
think about combinatorics
because of the chain thing
it's supposed to contain a group of 20 elements
so it can' be 10
every $5$-Sylow subgroup is made of $4$ $5$-cycles
mmhmm
so just count the number of $5$-cycles
oh, normalizer
eh
well
wait
what?
why? @MatheiBoulomenos
04:58
well, the index of the normalizer is the number of $5$-Sylow subgroups
@MatheiBoulomenos waaat
well $P^6_5/4 = 180$???
I mean, you used that before
to get the order of the normalizer, divide the group order by the number of Sylowsubgroups
we know the number of Sylowsubgroups if we know the number $5$-cycles
there are... $P^6_5/5 = 144$ 5-cycles... this is absurd enough
they form 36 sylow 5-subgroups
exactly as sylow predicted
so the normalizer has 10 elements, contradiction
that was hard enough
trust me, $180$ is easier
41 mins ago, by Leaky Nun
$180 = 2^2 \times 3^2 \times 5$. $n_2 \in \{9,15,45\}$, $n_3 \in \{10\}$, $n_5 \in \{6,36\}$ even more possibilities lol
you don't need to go through all combinations
I usually look at the largest prime first
Let $\varphi:G \to S_{36}$ be the group action induced by the $36$ Sylow $5$-subgroups. Let $P$ be a Sylow $5$-subgroup. Then, $N(P)$ contains $5$ elements.
@MatheiBoulomenos am I in the right direction?
05:07
Well yes, $N(P)$ is not crucial in the case that $n_5=36$
think about some things we talked about before
Well if $a \in P$ then $\varphi(a)$ permutes $35$ elements
and it is of order $5$
so it must be seven disjoint $5$-cycles
think about $\varphi(G)$
hmm
it is a subgroup of $A_{36}$?
I must be crazy if I'm dealing with $A_{36}$, lol
Oh no, I was in the wrong case :D
05:12
save that $\varphi(G)$ thing for later
sorry
Actually, there's a theorem you can apply here
if $n_5=6$, then $\varphi:G \to A_6$ monomorphic, then $|N_G(H)| = 180/6 = 30$, then there is an element of order $3$, then there is an element of order $15$, and have fun embedding it into $A_6$
yes that works
you can also say that $\varphi(G)$ is a subgroup of $A_6$ of index $2$
but $A_6$ is simple
@MatheiBoulomenos :D
waaaat
you win
05:14
well, for the $n_5=36$ case
this follows from Burnside's transfer theorem
but that's a bit advanced
please quote the theorem
I think it's also possible without it
@MatheiBoulomenos then how?
by counting arguments
again
05:17
Morning everyone!
Morning @Perturbative
@MatheiBoulomenos oh well, $\varphi[N(H)]$ only has 5 elements, contradiction
wait, it isn't
Morning @MatheiBoulomenos
@MatheiBoulomenos wie
Does there exist a vector space $V$ isomorphic to $\mathbb{R}^m$, where $V$ isn't an $m$-dimensional manifold?
05:21
@MatheiBoulomenos I'm going to sleep after this
I'm thinking lol
I know I solved this once
@Perturbative dimension is preserved under isomorphism, no?
@LeakyNun Yep
then how can it be different?
I think I got it
05:23
@MatheiBoulomenos wie
look at the $3$-Sylow groups
@LeakyNun There's a difference between isomorphism, homeomorphism and diffeomorphism
@MatheiBoulomenos ok wait
Let $\varphi: G \to A_{10}$ monomorphic with $|P|=9$. Then, $|N(P)| = 180/10 = 18$.
you have to do some case work
well either $P=C_3^2$ or $C_9$
wait
$P$ is abelian isn't it
so $N(P)$ has to be all of $G$?
05:24
it is
no
not every abelian subgroup is normal
oh, abelian with itself only
consider the intersection of two 3-sylow subgroups
if this is always trivial
count elements in $G$
if not, consider the normalizer of the intersection
there you need to apply that $P$ is abelian
$80+144 > 180$ lol
let $A = H \cap K$
and how?
Well, any subgroup of an abelian group is normal
thus $H, K \leq N_G(A)$
why?
oh wait, yes
and then
?
05:30
what could the order of $N_G(A)$ be?
9 or 18
well, we know it's not 9
why not?
oh, ok
but 18 is not possible as well
right
gute nacht
05:31
why?
@MatheiBoulomenos because otherwise $N_G(A)$ would be $N_G(H)$
but it contains $K$
you don't know that $N_G(A) \subset N_G(H)$
But we know that $H,K \subset N_G(A)$
this means that $N_G(A)$ contains at least $2$ $3$-Sylowsubgroups
But this is impossible
by Sylow
$18=2\times3^2$
there can't be $2$ Sylow subgroups
So there are still the possibilities that $N_G(A)$ has order $36$
or $180$
if it's $180$, $A$ is normal and we win
not 180, or else A would be normal
why not 45?
05:35
But we know that there are $144$ elements of order $5$
...
lol
so there are only $180-144=36$ left
so $A$ can't have any non-trivial conjugates
ok, next?
as it's the whole rest
we're done
we won against $180$
how?
why not 36?
05:36
well, there are only $36$ elements not of order $5$
and we have subgroup of order $36$
then $N_G(A)$ itself must be normal
as it just consists of every element which is not of order $5$
conjugation preserves the order
oh ok
how did you come up with these arguments all on your own
Well, I'm the Sylowmeister
gute Nacht
05:39
gute Nacht
05:57
@LeakyNun 1st order Peano axiom remove the induction axioms
Finally got the review for the paper we sent to Transactions of the AMS more than a year ago. Very positive comments in general.
How does $\frac{f(d_2+d_3)-f(d_2)}{(d_2+d_3)-d_2}\leq \frac{f(d_3)-f(0)}{d_3-0}$ hold here
 
2 hours later…
07:45
Hmm, date on the review is written in French, and the reviewer points out a sole reference that we should add, being a paper in French from 2008 (which is really late to still be publishing in French). My reviewer-recognition sense is tingling.
08:03
I am reading a proof and I have a question.

We have that $y-x>0$. By the Archimidean property we have that $n>\frac{1}{y-x}$. There also exists $a,b\in \mathbb{Z}$ such that $a<xn$ and $b>yn$. Why does the last sentence hold? From which property does this follow?
@MaryStar That is just the same property used twice
So, the Archimidean property? So, do we have that for $n\in \mathbb{N}$ and $\frac{a}{x}\in\mathbb{R}$ we get that $\frac{a}{x}<n\Rightarrow a<xn$ ?
But how do we get then $b>yn$ ?
@MaryStar The archimedean property is just that for any real there are both larger and smaller integers
and since $xn$ and $yn$ are reals that is all there is to it
08:44
From the Archimedean Axiom we have that $\forall x\in \mathbb{R} \exists n\in \mathbb{N}: n>x$. The negation is $\exists x\in \mathbb{R}\forall n\in\mathbb{N}:n\leq x$, right?

From the Archimedean Axiom we have that $\forall X\in \mathbb{R} \exists n\in \mathbb{N}: n>X$. At the negation we have that for this $n$ there exists a $Y\in\mathbb{R}$ such that $n\leq Y$.

So, we have that $X<n\leq Y$, right?

At the proof we have: There also exists $a,b\in \mathbb{Z}$ such that $a<xn$ and $b>yn$, where $n\in \mathbb{N}$ and $x,y\in \mathbb{R}.
From $X<n\leq Y$ we also get $-Y<-n\leq -X$, therefore an integer is between two reals, right?

So, since $a,b\in\mathbb{Z}$, we have that $a<xn$ and $b>yn$, where $xn, yn\in\mathbb{R}$.

Is this correct? @TobiasKildetoft
@MaryStar Most of it is unreadable
And no, there is no way a statement starting with "since $a,b\in\mathbb{Z}$" can be the correct reason why those numbers exist
09:13
[Some summary of the recent finding in Mathworks: Topic 1]
Predicative mathematics have injections and surjections linked to different things: injection measures size while surjection measures complexity. In ZF type set theories however, both injection and surjection are tied to size, which becomes cardinality for well ordered sets
Also it is consistent with predicative mathematics that the universal object is countable
what's predicative mathematics ?
Uh it basically ensures no objects being defined or constructed is circular. Let me see if I can find a better explanation:
For example, the Hartogs number proof of $\omega_1$ relies on that the collection of countable ordinals is a set, but that requires the powerset axiom, which assumes @user21820 (can you help me to fill in the details?)
@Secret: Lol, instead of posting in the main room, you could just ask those interested to join the relevant chat-rooms.

 Mathworks (Not the main chat!)

Maths department of SecretLabs SE Branch (chat.stackexchange.c...
From the Archimedean Axiom we have that $\forall x\in \mathbb{R} \exists n\in \mathbb{N}: n>x$.

Can can we show that for any real there are both larger and smaller integers?

We have that $n>x$. Do we maybe multipliply that inquality by -1? Then we get $-n<-x$. Does this help us? @TobiasKildetoft
@MaryStar yes, that helps, because if $x$ is negative, you just replace it by $-x$ and do it again.
09:33
@TobiasKildetoft, Can you please provide an example for some metric space $X$, a point $p\in X$ and a positive real number $r$ such that all three of the following different: a) open ball centered at $p$ with radius $r$, call it $N_r(p)$, (b) closure of $N_r(p)$ and (c) closed ball centered at $p$ with radius $r$
@Silent Why are you asking me that?
@TobiasKildetoft oh. you answered mary star some Archimedean property question, so I assumed that you know real analysis.
Sorry.
{-1} u [0;1], p=0, r=1
@mercio,wow!! such a nice example! thank you so much.
Do you mean the following?
We have that $\forall x\in \mathbb{R} \exists n\in\mathbb{N}: n>x$. For $x=-y>0$, $\exists n\in \mathbb{N}: n>-y \Rightarrow y>-n$.
09:41
ich habe ungeschlafen
@LeakyNun ausgeschlafen?
unslept
Ah du hast nicht geschlafen... @LeakyNun Warum nicht?
@LeakyNun I had a look at your solutions. Looks good in general. A few places where you seem to go into unnecessary detail (such as apparently proving the isomorphism theorem rather than just using it), and a few where the level of detail seemed slightly too low, but not by much
@Evinda lol I just meant I woke up
@TobiasKildetoft would you give me feedback?
09:43
Aha :D @LeakyNun
like, "where?"
@LeakyNun I am going to lunch now. I will give you some more details later
@TobiasKildetoft ok thanks
are you going to eat roedgroed med floed?
10:03
Good morning! Which math problems are you wrestling today, channel?
I am trying to morph functions to other functions smoothly. But I have something of a bug which makes my solution hoppy. Maybe I just forgot some cost term in my optimization, I am not sure.
so we are called "channel" now?
Sorry, I haven't been on these chats here too much so I did not know what to call it.
lolll I'm just kidding
[Random]
Mathematics is the study of formal objects and rules, whose properties and truth value can be unambiguously determined. Using this we can define the following:
10:10
hmm.. that statistics is not real math?
not pure math, or at least less than 90 percent pure?
sorry for the bad joke, had a "breaking bad" marathon the other week.
well there exists modal logic and fuzzy logic, thus truth values can be probabilistic
(and we have beyesian and frequentist interpretations)
Also experimental mathematics and applied mathematics count because we do have a formal system plus rules that suits the applied or algorithmic situations
yep
bayesians and frequentists, the arch enemies, no? lol.
Not to me, I like the updating property of beyesianists and also the (PDF fixed but unknown) property of frequentists
beyesian is useful if we knew very little or is impkractical to sample the full parameter space that controls the probabilities
frequentist is useful if the pdf is sufficiently known
(Also I hate autocorrect)
by the way, why do people say Montreohhl, should it not be pronounced "moon ree al". dammit there is no good way to write this with english sounds.
That word is not English I think...
10:18
i would suspect it to be french, but i am not so good at french to say for sure how to pronounce it.
I agree with your take on bayesian and frequentist.
They are useful in different situations.
Nowadays I am quite useful-focusing, for being on a math forum.
So what have you done today Secret? Are you doing statistics?
I am doing some philosophy of mathematics stuff
ah, that is interesting. i liked reading popular scientific things about that when I was younger. but it was of course very basic.
like constructivism vs formalism vs intuitionism
that kind of philosophy?
but so far it's haphazard as I know little about the topic.
i and some users are discussing foundations of non constructive, constructive and predicative mathematics
though for me, I am more strongly interested on all possible weird properties of infinity in order to better understand it
i digged this "Godel Escher Bach" book when I was younger also although i never studied much music theory so I probably did not understand all the bachy things, although i love listening to music.
Ah, I am really bad at infinities. Do you mean like cardinalities in set theory?
10:26
Not just that, but things like infinite dedekind finite sets, amorphous sets, induction, ordinals etc.
Ah okay, I dont know much about it, but I like fractals and similar things.
However, it seems in predicative mathematics, we don't even have uncountability, though complexity and cardinality corresponds to surjectivity and injectivity
Many fractals have some infinite recursion or iteration, but I don't know if that is the kind of infinity you are interested in.
Fractals and recursive maps are also interesting. Though they might come later in the foundational perspective
for recursive maps, we can talk about fixed points and limit ordinals
I probably lack too much of the foundations of set theory to be speculating too much about it. I like programming pretty images. Maybe I know like the most basic algebra to investigate some things about it. But not higher level stuff.
10:30
Let $(x_n)$ be a sequence in some topological space $X$. Initially, I suspected that if $A := \{x_n ~|~ n \in \Bbb{N} \}$ is not closed, then $(x_n)$ must be a convergent sequence. But I think this is wrong. Let $(x_n)$ be some enumeration of the rational numbers; surely such a sequence does not converge. But $A = \Bbb{Q}$ is not closed in $\Bbb{R}$. Does this sound right?
Is there a button to turn on MathJax in chat?
Sorry this: tinyurl.com/cfqcvpc
@user193319 thanks!
@mathreadler You're welcome. I just recently discovered it myself--and boy was I pleased!
(May not be correct stuff) Given a formal object A, a subobject B is defined to be one which inherits the properties of A. If B derives A, then B is stronger than A.
10:34
of course a countable sunset makes sense. you can count on it! :D
to not be clouded if you are waiting for it
@user193319 That's correct
@AlessandroCodenotti Thanks!
still havent managed to get my chatjax working, and the \Bbb{Q} command got me hungry.
A formal object and rules (usually a formal system) is said to be a foundation of mathematics if it derives mathematics itself. Therefore a submathematics is a subobject of mathematics. For example, constructive mathematics is a submathematics of mathrmatics
(Note that the term "submathematics" does not really exists in the literature, thus all of this is rambles)
I'm gonna go grab some pizza slices and see if I can find this studenty cafe-bar again.

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