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12:00 AM
But you can get hourly parking that isn't too outrageous.
 
Lmao the timing.
 
@anakhro: Yeah, you can.
 
some books aren’t worth it to get for free even
 
I think most books generally have something worth it to read through.
Of course some are much better than others.
 
@ÉricoMeloSilva Kind of like Keisler's book...
 
12:01 AM
But there are gem comments and exercises through most.
 
@ÉricoMeloSilva You never know when you'll need to level a table
 
Yeah, I'm an old cynic, @anakhro. I disagree.
Or I have higher standards.
I know books with no gems.
 
Probably just higher standards.
I am easily entertained when it comes to mathematics.
 
My thought was that $\theta$ should be defined as $\theta (q')(n') = \phi_1(\phi_2(q') \phi_1 (n') \phi_2(q')^{-1})$, but I'm not sure that $\theta$ is a homomorphism...
 
@AlessandroCodenotti i already have way more books than tables in my apartment have legs
 
12:02 AM
The few things I read by Keisler where pretty good (and he has a lot of cool results in model theory)
 
LOL @Eric
 
@AlessandroCodenotti I'm referring to his Elementary Calculus...
 
@ÉricoMeloSilva You have more than 4 books? Impressive!
 
It's way too informal IMHO.
 
@user193319: It should just be a matter of using the isomorphisms to give a dictionary between one story and the other. I don't want to think about it :P
 
12:03 AM
i have more than one table so
 
Why do you need tables when you can get a Kindle?
 
Uhm are we talking about the same Keisler?
 
You could probably count other things with 4 legs, too ... not counting a dog.
 
Because you can't eat on your kindle?
 
12:04 AM
@AlessandroCodenotti maybe YOU can’t
 
@AlessandroCodenotti Well, you see, that's where you're wrong buddy.
 
losted
 
Yep, same Keisler
When my ebook reader broke I should have promoted it to cutting board instead of throwing it away I guess
 
That could be poisonous chopping.
 
gotta live life on the edge sometimes
 
12:07 AM
Anyway I'm leaving now. Bye everyone!
 
Night!
 
fare thee well
 
cya
 
luv u
 
0
Q: $(X,\mathscr T)$ is normal and each closed subsets of $X$ is a $G_{\delta}$ set.Then, $(X,\mathscr T)$ perfectly normal.

Math geekProve the following result without using Urysohn's lemma. $(X,\mathscr T)$ is normal and each closed subsets of $X$ is a $G_{\delta}$ set.Then, $(X,\mathscr T)$ perfectly normal. My effort: I have proved using Urysohn's lemma. How do I prove without the use of Urysohn's lemma? Let $A$ a...

 
12:08 AM
That escalated quickly
 
I can escalate it more if you want me to
 
You prefer elevators?
 
bullet trains
 
I dunno how escalators turn into bullet trains.
 
i'm literally listening to bullet train right now lmao.
but idk what bullet train has to do with elevators either
 
12:10 AM
please help me.
 
@Mathgeek I have called 911. They ask, "911, what is your emergency?"
 
1
Q: $(X,\mathscr T)$ is normal and each closed subsets of $X$ is a $G_{\delta}$ set.Then, $(X,\mathscr T)$ perfectly normal.

Math geekProve the following result without using Urysohn's lemma. $(X,\mathscr T)$ is normal and each closed subsets of $X$ is a $G_{\delta}$ set.Then, $(X,\mathscr T)$ perfectly normal. My effort: I have proved using Urysohn's lemma. How do I prove without the use of Urysohn's lemma? Let $A$ a...

 
Please don't keep spamming us with that.
 
@Mathgeek they don't seem to understand your question, so they sent all three, the fire and police department, and an ambulance.
 
@anakhro: I don't think humo(u)r is helping.
 
12:22 AM
i like how you put the parens on u...
 
@anakhro do you study math in police station or Fire force station? :(
 
i mean i know why you do, it just looks interesting...
 
@Dair why does he do it?
 
@anakhro To cover every base.
 
Every base would necessitate hundreds of languages.
 
12:24 AM
to appease the English
 
As a Canadian, humour looks best.
 
Yeah, but the syntax of that sentence sucks. Humour is a Canadian?
 
@anakhro Did you ever play oboe?
 
I think policemen in Canada may be Mathematicians :P
 
@Dair no.
 
12:33 AM
Ok, nvm.
 
@TedShifrin Yes, humour is canadian.
 
The nerds have infiltrated!
 
Oh hell. Demonark is here.
 
Why can I never join this chat without Demonark crashing the party?
 
That's a long-standing problem.
 
12:36 AM
I think Daminark is my hero.
 
:D
 
I wish you still had that galaxy sloth profile pic.
I love that pic.
 
Hi Daminakhro.
 
@Ultradark did you work out your vector field?
 
Ask and you shall receive, lemme do it real quick
 
12:39 AM
You can do it with spherical or polar coordinates.
 
YAY!
 
@anakhro yeah $F=\{\sin(x),\sin(y)\}$
 
Guess I just gotta wait for a bit
Also I should update that to 4th year lmao
 
You're a turtle, eh?
 
What courses are you taking now?
 
12:42 AM
hmm that wasn't the one I remember but it's still awesome lol.
 
It's an older one, the first one I had on facebook in the "sloth photo" era
 
Demonark, you narrowing your decisions?
 
I had another one which just showed a sloth head at the center of a galaxy
 
hi chat
 
hi @Semiclassical
 
12:44 AM
Re courses: so I'm in final's week actually, this past quarter I had algorithms, commutative/AG, and differential topology. Though AG was a bit of a mess and tbh I need to just find a commutative book, work through it linearly, then an AG book, work through it linearly too. Next quarter, the plan is graph theory, ancient empires, algebraic number theory, and likely probability (heh) or maybe some compsci class
 
the line between "i have a result" and "i understand my result" is a frustrating one
 
Better not to draw it, @Semiclassic.
 
Re decisions: I'm gonna visit soon though at the moment I think there's basically a 75% chance I'm heading to Madison
 
take my favorite four-by-four matrix
 
12:45 AM
@Semiclassical I remember just brute algebraing far too many combinatorics problems and not actually understanding them...
 
Visiting is important. But great.
 
How was diff. topology, @Daminark? What did you cover?
 
I have a determinantal identity now which is neat
but i have no bloody idea why it comes out so nicely
 
@Semiclassical Life is so hard some times...
 
OK, have a good evening/night, all ...
 
12:47 AM
night
 
@TedShifrin cya
 
luv u ted
 
@anakhro i was waiting for this
 
I got Madison, Washington, and Notre Dame, I haven't relooked at Washington enough yet (which I should, I way underrated Madison when I applied and since realized it was really nice). Waiting on Michigan as a kind of high priority place, Rutgers is a waitlist because of the whole transcript business but I don't care about them much, Minnesota and Penn are also in the air but they're also very unlikely
 
@Dair waiting for what?
 
12:48 AM
that comment
 
here's hoping for Minnesota
 
@Daminark You got this my dude
 
Minnesota would make the decision perhaps a bit trickier. I still think I prefer Madison but if I remember right, Minnesota feels like it has a somewhat larger topology presence which is a bit of a plus, but the algebra/NT/AG at Madison is insane
 
@Daminark Sounds like regardless you've got at least one good option :D
 
Differential topology was basically, a couple weeks of technical definitions + Whitney embedding + tubular neighborhood theorem, then we did degree for a bit, then transversality and stuff along the lines of Poincare-Hopf. After that we did a bit of Morse theory, which I'll have to reread since I didn't absorb well, and then some de Rham cohomology
 
12:52 AM
Congrats! btw, I don't remember if I said it.
 
Also I did miss one of the De Rham cohomology classes so I'll be going through a bunch of Bott-Tu soon
@Dair thanks fam!
Not sure why it's not updating the picture, lemme close out of the browser and try again
 
@Daminark It's updated for me.
 
Yeah now it's goood
 
I was checking the profile pic out and was reminded that Area51 is a thing...
 
1:13 AM
What's the norm of a point in a vector space?
is it 0
 
For the most part no, only 0 has norm 0
 
If so, then I think it's confusing to extend the notation of absolute value in $\mathbb{R}$ to norms in $\mathbb{R}^n$
 
So if you're referring to the standard Euclidean norm
Then the norm of a point $(x_1,\ldots,x_n)$ is $\sqrt{\sum_{i=1}^n x_i^2}$
 
@Daminark It's also the definition of the norm of a vector in that space too right
 
Point = vector
When you're in a vector space
 
1:16 AM
ooh didn't think of it like that
i just thought of vectors as things with direction & magnitude
and points don't really have direction or magnitude
 
So, if you wanna keep that interpretation, then a point can be identified with the arrow from the origin to that point
 
points have 0 measure...
 
What's a measure?
 
In mathematical analysis, a measure on a set is a systematic way to assign a number to each suitable subset of that set, intuitively interpreted as its size. In this sense, a measure is a generalization of the concepts of length, area, and volume. A particularly important example is the Lebesgue measure on a Euclidean space, which assigns the conventional length, area, and volume of Euclidean geometry to suitable subsets of the n-dimensional Euclidean space Rn. For instance, the Lebesgue measure of the interval [0, 1] in the real numbers is its length in the everyday sense of the word, specifically...
 
I just find it strange that the magnitude of the norm of a point can be larger than that of a vector with size
 
1:19 AM
well, at least for the standard way of measuring things...
 
Don't worry about measure for now, I'm not sure why that just game up. But yeah so, while there are some contexts which benefit from the "vectors have magnitude and direction" stuff but if you're gonna be doing math I recommend dropping that from your mind
 
@Daminark Points are points, vectors are vectors and there is a natural mapping between the two i guess is what i'm trying to get at...
 
@Daminark I feel like there should be a separate name for the norm of a point because it's the same word but one means the size of the vector the other means the distance from the origin
oh well I'll just try not to dwell on it
math was never really that precise in wording in my experience
 
I mean, the point is what I said earlier, for every point in $\mathbb{R}^n$ there's a unique arrow going from the origin to that point
And the norm of the point in the sense of the distance from the origin is exactly the length of that arrow
 
I'm just trying to see if obliv is confusing measure and norm...
 
1:22 AM
Okay so I should think of a point as a vector that starts from the origin?
 
So, I remember the business about "vectors as arrows" was a thing that we were taught in high school physics, is that what you're in right now?
 
@obliv I would say it depends on the context...
 
Yeah that's the only information I have of them @Daminark basically. But using that idea it makes way more sense. then $x \in \mathbb{R}^n$ that is $x = (x^1 - 0, x^2 - 0, ... , x^n - 0)$ is the length from the origin. lol very simple
 
Okay so, for now roll with this. Once you're out of that class find a potion at Walmart which permanently makes you forget about vectors as arrows/have length and direction
 
"potion at Walmart" wat.
 
1:33 AM
How do I construct a vector that isn't a point? Like, whats the formal notation
$\vec{xy} = $ I'm drawing a blank
 
So, here's the correct way to do vectors, for the sake of this conversation pretend you've never heard that word before
A (real) vector space is a set where you can add things together and you scale things by real numbers. For example, in $\mathbb{R}^n$, you can add by $(x_1,\ldots,x_n) + (y_1,\ldots,y_n) = (x_1 + y_1, \ldots, x_n + y_n)$, and scale by $r(x_1,\ldots,x_n) = (rx_1,\ldots, rx_n)$.
A vector is an element of a vector space. We also refer to vectors as points
 
Cool, so how would I write down a triangle for example
or even just a line in some space
 
You'd write it as a set of points. For example, if you think of the y-axis in the plane, you'd just take every point on the line. For example, $(0,1)$ is on the y-axis, $(1,1)$ is not, $(0,\frac{\pi}{47.99982})$ is, etc
 
So would you write it as like a sum or something? Is there some notation that lets me describe a horizontal line that spans from $(2,4) \to (3,4)$
 
So the y-axis is the set of points of the form $(0,x)$ where $x$ is any real number
 
1:41 AM
the typical construction is something like (2,4)*(1-t)+(3,4)*t
when t=0, that's (2,4). when t=1, it's (3,4)
and since it's a linear function of t, it'll interpolate between those two points as you go from t=0 to t=1
 
sorry what does interpolate mean
 
Hello, could someone help me with a example of a function that belongs to $H^2(\mathbb{T})$ where $\mathbb{T}$ is the one dimensional torus and $H^2$ is the sobolev space.
 
1:57 AM
@Daminark what did you cover of Morse theory?
 
2:08 AM
hey anyone here knows principle component analysis?
 
@obliv interpolate means to make a prediction within the range of data.
So if you hand a bunch of data points to someone so they can interpolate it, they will merely "connect the dots" in some fashion.
 
@anakhro have u done PCA?
 
Never heard of it. What is it?
Is it a drug?
 
I don't think that's the way Semiclassical was using it @anakhro but thank you
 
2:26 AM
In mathematics, linear interpolation is a method of curve fitting using linear polynomials to construct new data points within the range of a discrete set of known data points. == Linear interpolation between two known points == If the two known points are given by the coordinates ( x 0 , y 0 ) {\displaystyle (x_{0},y_{0})} and ( x 1 ...
 
$\substack{\text{for\\electron}}$
Heh
Doesn't work
 
So do linearly dependent vectors have to point in the same direction if that makes sense
@students yeah I don't know what half of those words mean but thanks I can search things on wiki as well.
 
Isa
Does someone know how to write a line through text ?
 
[s]a[/s]
nvm idk lol
 
2:44 AM
No idea.
 
Test
ah, there it is
@Isa put three dashes to the left and right of the text you want to strikethrough
 
Isa
@Semiclassical what are dashes ?
 
minus sign
-
 
Isa
nop, doesn't work
I meant in the ask window
 
3:02 AM
it's the same on the main site as here
--- text ---
except without the spaces
 
Isa
nop, I used \enclose but does not looks good
 
I see you have deleted your question How to write a line through text? before I was able to post something there.
@Isa In the posts on the main you can use <s>striked text</s>
In chat you can use ---striked text--- striked text.
There is an older post on meta about equations: Striking out equations
From the title "How to write a line through text?" I suppose that you're talking about text. (Not about equations.)
 
Isa
@MartinSleziak yes sorry I deleted bacause I found the solution so I didn't want to waste people's time. And yes I wanted to put a line through a word, not an equation.
 
3:18 AM
Well, possibly the same question (and answer to that question) could be useful to other users, too. It's certainly no big deal.
But since I have already started typing the answer, I wanted mention the information at least somewhere in chat - so that the answer gets to you at least in this way.
 
Isa
ah, I'm going to undeleted then
 
Although from your comment I see that you have already found that information elsewhere..
@Isa That's your call. It is certainly not a big deal.
 
Isa
yes, from the comments of my question on meta
 
I have posted the answer.
See you later!
 
 
1 hour later…
4:36 AM
How to see that $\Bbb Z_2[x] / \langle x^3 + x + 1 \rangle = \{ax^2 + bx + c + \langle x^3 + x + 1 \rangle: a, b, c \in \Bbb Z_2 \}$ has at least 8 elements? (I can see at least 8).
Do I have to calculate and check, or is their easier way?
 
4:47 AM
hi
is somebody here?
 
5:12 AM
@Silent Show that $ax^2+bx+c\ne0$ if $a$, $b$, and $c$ are not all zero
 
5:23 AM
some help
 
(is not identically zero, anyways)
 
how to prove that $f : [0,1] \rightarrow C $ is ameasurable function
C is the cantor set
and f(x) = \sum_{n=1}^{\inf}(2*b_{n}/3^{n})
 
6:09 AM
Random:
> The strengthened liar sentence is true if and only if false or undefined, so we have a new paradox, called the strengthened liar paradox. The problem with the strengthened liar paradox is known as a revenge problem: Given any solution to the liar, it seems we can come up with a new strengthened paradox, analogous to the liar, that remains unsolved.
Superresistance: Given any goal G and some formal system T, if G cannot be realised in T, then it cannot be realised in any extension of T
A concept X is said to be stubbornly nonexistent if X display superresistance in any consistent formal systems T
 
@AkivaWeinberger Here by $0$, do you mean $\langle x^3+x+1\rangle$, or zero polynomial in $Z_3[x]$? In later case that is obvious.
 
A concept X is nullus sine fantasia if all weaker generalisations of it is stubbornly nonexistent
Objects that display superresistance include division by zero etc.
Objects that are nullus sine fantasia include impossibilities like squaring the circle, an infinite cardinal smaller than the natural numbers, and many others
Thus in my thought process, to prove that something does not exist requires proving that said thing is at least a nullus sine fantasia
Or in English:
 
6:25 AM
@AkivaWeinberger, is this reasoning correct: Since by division algorithm, we get $ax^2+bx+c=0(x^3+x+1)+ax^2+bx+c$, hence because division algorithm gives unique quotient and remainder, all $ax^2+bx+c+\langle x^3+x+1\rangle$ distinct?
Oh!!! So, I feel like concluding that polynomials with degree m mod (polynomial with degree m) are all distinct over field, where m<n. Am I right?
 
> I will not be convinced at a mathematical level that some X is impossible unless it will lead to a contradiction in all conceivable systems
2 years ago, I have demonstrated that division by zero is not a nullus sine fantasia, I will soon show more crazy concepts are also not
 
@MatheinBoulomenos, akiva is not here. Will you please check the validity of my above claim? Thanks
 
7:17 AM
I was thinking of, $a_1x^2+b_1x+c_1=a_2x^2+b_2x+c$ iff $(a_1-a_2)x^2+(b_1-b_2)x+(c_1-c_2)=0$
so it suffices to show that if $ax^2+bx+c=0$ then $a=b=c=0$
 
 
1 hour later…
8:20 AM
$ax = b(x')^ 2 + cxx''$
how do you solve such a de?
 
8:41 AM
@AkivaWeinberger Thank you very much.
Please verify this: polynomials with degree m mod (polynomial with degree n) are all distinct over field, where m<n. @AkivaWeinberger
That is two distinct polynomials of degree m remain distinct when we consider them mod (degree n polynomial) where m<n, right? (Here i am talking polynomials with field coefficients only)
 
9:39 AM
@Silent Yes
Notice however that you don't always get a field though
$\Bbb R[x]/\langle x^2\rangle$ for example is not a field because it has zero divisors
In that ring though the polynomials of the form $ax+b$ are all distinct, yeah
 
10:38 AM
@AkivaWeinberger he didn't say you get a field
 
11:34 AM
0
Q: Internal Semidirect with Factors Isomorphic to "Outside" Groups

user193319Here is a conjecture of mine: If $G$ is the internal semidirect product of $N \unlhd G$ and $Q \le G$, and $\phi_1 : N' \to N$ and $\phi_2 : Q' \to Q$ are isomorphisms, then there is some $\theta : Q' \to \text{Aut } N'$ such that $G \cong N' \rtimes_\theta Q'$ My thought was to take $\the...

 
Fair
So here's a question
Say we have $n+2$ vectors in $\Bbb R^n$
and say that, for every collection of $n+1$ vectors, there exists a vector whose dot product with everything in the collection is positive
Must there exist a vector whose dot product with all $n+2$ vectors is positive?
 
11:49 AM
I have this question, I can't figure out. If I have four coplanar vectors then can I prove that if they are linear dependent in a way x a + y b + z c + w d = 0 then x + y + z + w =0?
The converse is true.
I think no, because for the converse to be true pairs of two vectors have to be linearly dependent.
A b c d are position vectors.
 
12:32 PM
Does someone know of a function that belong to $L^2(\mathbb{T})$, where $\mathbb{T}$ is the one dimension torus
 
1:10 PM
@PolarBear Hint: $a-d$, $b-d$, and $c-d$ are linearly dependent
 
 
1 hour later…
2:22 PM
I am working on classifying all noncommutative groups of order $p^3$. So I know that $G$ must contain a normal subgroup $N$ of order $p^2$; since it is of order $p^2$, it must be abelian. The hint I was given is to find an element $c$ of order $p$ that is not in $N$. I know that $|Z(G)| = p$, so $Z(G) = \langle z \rangle$ with $|z| = p$. However, the problem is that $Z(G) \le N$...
How do I show that $G$ contains another element of order $p$ that isn't in $N$?
 
2:39 PM
@user193319 that's not true at least for $p=2$ (not sure about odd $p$)
 
Oh, I forgot to add that $p$ is odd.
And then from there I am suppose to deduce that $G = N \rtimes \langle c \rangle$.
 
2:57 PM
@AkivaWeinberger They aren't. How are they?
 
@PolarBear They're three coplanar vectors
 
Also, I would like to add, the points represented by the position vectors a b c d are coplanar
I wasn't clear then.
 
Yes I know
That's why $a-d$, $b-d$, and $c-d$ are linearly dependent
You have three vectors in a two-dimensional space
 
@AkivaWeinberger I meant they are all linearly dependent. Three. But not two right?
Oh, I see. Sorry misunderstood.
I can't think further. What to do next?
 
What does linearly dependent mean? You get an equation.
 
3:01 PM
Yes, we do.
 
What equation do you get from the fact that they're linearly dependent?
 
So, I can get equation finally like xa+yb+cz=(x+y+z)(d)
 
In other words, $xa+yb+cz+(-x-y-z)d=0$
(You meant to write (x+y+z)d, not x+y+z(d) )
 
Yeah, I am sorry about that. So, w = -x-y-z
Oh yeah, got it.
 
@user193319 I think it is also false for odd $p$
 
3:04 PM
Oh, really? Do you have a counterexample?
 
Now $x+y+z+(-x-y-z)=0$, so we've found our coefficients that add to zero.
 
@user193319 in a group of order $p^3$, the elements of order $p$ form a subgroup and that subgroup can have order $p^2$
 
@AkivaWeinberger Exactly! Thanks. I have one more doubt. For example a and d are collinear with the origin and b and c the same in opp way and |a|=|c|=|d|=1 and |b|=2 then -a+d+2c-b = 0 but -1+1+2-1≠0
What's wrong with this example
 
1
Q: Finite chain and finite antichain implies that the poset is finite

user 170039The problem with which I am struggling is the following, Let $(P,\le)$ be a poset. If every chain and antichain of $P$ are of finite then $P$ is also finite. My Attempt Notice that by Hausdorff Maximality Principle every chain is contained in a maximal chain. By hypothesis the number of e...

 
@MatheinBoulomenos Why is the set of all elements killed by $p$ closed under products? Also, I don't see why the existence of a such a subgroup is a problem.
 
3:13 PM
well, if the set of the elements of order p form a subgroup of order p^2, then there can't be any element of order p not in that subgroup
 
@PolarBear Like this?
 
it's closed under products due to the formula $x^py^p=[x,y]^{p(p-1)/2}(xy)^p$
and because $[G,G]$ has order $p$
so $[x,y]^{p(p-1)/2}=1$
 
Sure $a+d=0$ and $b+2c=0$, but we also have $3a-2b-4c+3d=0$
 
@AkivaWeinberger Yes, like that.
@AkivaWeinberger Oh, I see. So there exists one such quadruple
@AkivaWeinberger Thanks a lot! :)
 
3:19 PM
$x^x$
$x^x$
 
@AkivaWeinberger, when dealing with $Z[\sqrt d]$, I encounter, e.g., $Z[\sqrt{-3}]$. What does $\sqrt{-3}$ mean here? Because, $x^2+3=0$ has two roots right? Which one is $\sqrt{-3}$ here?
 
By $Z$ you mean the integers $\Bbb Z$ or some other ring?
 
You can't add one to $\Bbb Z$ without adding the other
 
Yeah, you get the same ring either way
Besides, there's even an isomorphism that maps one square root to the other
(complex conjugation)
 
@MatheinBoulomenos So you are saying that this set lives inside of $N$? I tried proving this but I don't see it.
 
3:26 PM
@user193319 I'm saying that the set of elements killed by $p$ can be a valid choice for $N$
and if that happens, no element oustide of $N$ will have order $p$
 
So, $N$ actually equals the set of all elements killed by $p$? What do you mean by "valid choice for $N$"? I can't control what $N$ is; it's just some group guaranteed to exist by a theorem.
 
yes, you can't control what it is and so it might be the set of all elements killed by $p$
 
@AkivaWeinberger, @AlessandroCodenotti, Thsnk you!
 
But what if it isn't possible? Don't I need an example of such a group to know that it is possible? Otherwise we are saying that JS Milne made a huge error in his book.
 
@user193319 consider $G= \left\{ \begin{pmatrix}1+pa & b \\ 0 & 1 \end{pmatrix} \mid a,b \in \Bbb Z/p^2 \Bbb Z \right\}$
this actually has order $p^3$ because it only depends on $a \pmod{p}$
 
3:36 PM
How then do I show there are only two noncommutative groups of order $p^3$? I was just going off the claim/hint made my Milne, which you say is false. I know what the two groups look like; I just need to show they are the only two.
 
You have to first realize that such a group must be a central extension of $\Bbb Z/p \times \Bbb Z/p$ by $\Bbb Z/p$. There are many ways to proceed from there.
 
@MatheinBoulomenos See page 48 jmilne.org/math/CourseNotes/GT.pdf
 
yeah, I'd proceed with group cohomology
 
The easiest, though not the most elementary, way is to compute $H^2(\Bbb Z/p \times \Bbb Z/p; \Bbb Z/p)$
 
How does this not contradict what you are saying?
 
3:40 PM
@MatheinBoulomenos Sure, same.
 
In both examples (3.14 and 3.15) there is an element of order $p$ not in some subgroup of order $p^2$.
I don't know anything about cohomology.
 
@user193319 but that isn't what you said, you said that for any subgroup of order $p^2$; there is an element of order $p$ in that subgroup
 
No, I didn't say any subgroup of order $p^2$. I just want to show that there is an element of order $p$ outside of $N$.
 
you just said there exists a subgroup of order $p^2$
 
Yeah, a normal subgroup of order $p^2$ that must also be commutative; and I called it $N$. Then JS Milne's hint is to find an element $c$ of order $p$ not in $N$ and argue that $G = N \rtimes \langle c \rangle$.
 
3:44 PM
Can someone please tell whether "m-l divides k" will be "m-l divides n" in the 3rd line of the proof of Proposition 3.
 
@user193319 I can't find that claim
 
@MatheinBoulomenos See page 125.
 
@user193319 yeah that's wrong
 
How else can I proceed, if not with cohomology?
 
Pullback the generators of $G/Z(G) \cong (\Bbb Z/p)^2$ to $G$, and work out the relations.
 
3:52 PM
Does $G$ contain a subgroup $H$ such that $H \cong G/Z(G)$ via the projection map?
 
not necessarily
 
That'd force the extension to split, i.e., be a semidirect product. You don't know this apriori.
 
Yeah, I know. That's what I am trying to prove, and that would be one way of proving it.
 
if $G$ has an element of order $p^2$, then this won't be true
 
It's not true if $p = 2$ (think $Q_8$).
But the odd prime vs p = 2 cases are kind of different
 
3:55 PM
@BalarkaSen Why is $G/Z(G) \cong (\Bbb{Z}/p)^2$? Isn't it possibly isomorphic to $\Bbb{Z}/p^2$? What do you mean by pullback generators?
 
If it's isomorphic to $\Bbb Z/p^2$ then $G$ would be abelian.
I mean, look at preimages of the generators of $(\Bbb Z/p)^2$ by $G \to G/\Bbb Z$. Call them $x$ and $y$, say. Then $G$ is generates by $x, y$ and a generator of $Z \cong \Bbb Z/p$. But now notice that as $xZ$ and $yZ$ commutes, $[x, y] \in Z$. Assuming $G$ is nonabelian, $[x, y]$ is non-identity, hence generates $Z$.
So $G$ is in fact generated by $x$ and $y$ and $Z$ is generated by the commutator $[x, y]$.
You need some extra trick to work out the relators at this point. I think the correct identity is $x^ny^m = y^m x^n [x, y]^{nm}$.
Every element of $G$ can be written as $x^a y^b [x, y]^c$, and using that identity you can check that multiplication on the indices work like $(a_1, b_1, c_1) "+" (a_2, b_2, c_2) = (a_1+a_2, b_1+b_2, c_1 + c_2 + a_1c_2)$.
This gives a homomorphism from the Heisenberg group of order $p^3$ to $G$. If $x$ and $y$ have order $p$ you're done; this is an isomorphism.
Well, I mean, if $x$ and $y$ don't have order $p$ then this is not a map.
I don't recall how to deal with that issue but in that case the group is what @MatheinBoulomenos wrote down
*should be $c_1 + c_2 + a_1b_2$.
 
@BalarkaSen Using your method, is it possible to show that the group is isomorphic to the group in example 3.14 or 3.15 in jmilne.org/math/CourseNotes/GT.pdf (see page 48)?
 
4:19 PM
Wait...what does $G \to G/\Bbb{Z}$ mean?
Why are you modding out by the integers?
 
Hi, a @Balarka
 
A function $f(x)$ of a variable $x$ has a discontinuity point at $x = 2$. Then which of the following can be said?

(a) $f(x) + 2$ also has the same discontinuity point (b) $f(x)$ is continuous for all $x<2$ and for all $x >2$
(c) $[f(x)]^2$ is continuous for all $x$ (d) $f(x)$ cannot have any other discontinuity point
 
$F_m|F_n$ iff $m|n$, and that's so nice
(for $m\ne2$)
 
So what do you think, @MrAP?
hi, DogAteMy.
 
Hi
Herbert Federer (July 23, 1920 – April 21, 2010) was an American mathematician. He is one of the creators of geometric measure theory, at the meeting point of differential geometry and mathematical analysis. == Career == Federer was born July 23, 1920, in Vienna, Austria. After emigrating to the US in 1938, he studied mathematics and physics at the University of California, Berkeley, earning the Ph.D. as a student of Anthony Morse in 1944. He then spent virtually his entire career as a member of the Brown University Mathematics Department, where he eventually retired with the title of Professor...
 
4:33 PM
I think that the answer is option (a)
 
> Nevertheless, the book's unique style exhibits a rare and artistic economy that still inspires admiration, respect—and exasperation.
 
@TedShifrin
 
Correct, @MrAP.
 
You don't usually see Wikipedia talking like that
 
But I do not completely understand why. I got it by eliminating the other options. Can you explain make me understand?
 
4:34 PM
Well, it's accurate. I sat through a series of lectures he gave at an AMS meeting years ago, commenting repeatedly to a professor next to me that I wished someone else (particular) were giving the lectures.
Well, @MrAP, all the others are wrong. Do you see that?
 
Yes. It cannot be option (b) and (d). The ones left are (a) and (c).
(a) seemed more meaningful.
 
(c) will work only for very special functions.
Can you give an example where (c) is correct and one where it is not?
 
$f(x)=\sqrt(-x)$ I guess is where (c) is correct.
 
No, that's not going to work.
Well, I guess it depends on the definition of discontinuity in your book/course.
 
and $f(x)=\frac{1}{x-2}$ is where it is not correct.
Why is that not going to work?
 
4:40 PM
For me, a point must be in the domain of the function to talk about continuity and discontinuity.
But some books are different.
If your textbook's definition is that any point not in the domain is a point of discontinuity, then you're OK on both.
At any rate, if $\lim_{x\to 2} f(x)\ne f(2)$, then it follows that $\lim_{x\to 2} (f(x)+2) \ne f(2)+2$, so (a) is correct.
 
Oh. So what do you think is an example of a function where (c) is correct?
 
What about $f(x)=\begin{cases} 1, & x\ge 2 \\ -1, & x<2 \end{cases}$?
 
Oh Right!
 
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