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12:10 PM
@robjohn I think that inequality is not correct (or I miss something). How did you get that? Check for $k=1$.
 
Hi @DanielFischer!!! We have that $x+y=p^m(u_1+p^{n-m}u_2)$

Why when $n>m$ does it hold that $u_1+p^{n-m}u_2 \in \mathbb{Z}_p^{\star}$ ?
 
@Chris'ssis I was off by a factor of 2 since I was computing part of the integral for an interval of $\pi$ rather than $2\pi$
@Chris'ssis Now the formula is correct and asymptotically accurate
 
@evinda It holds if $u_1$ is a unit. Because $z \in \mathbb{Z}_p$ is a unit if and only if it is not a multiple of $p$. If you use the $\sum_{n=0}^\infty a_n p^n$ representation, that means $a_0 \neq 0$.
 
@DanielFischer I see... Thank you :)
@DanielFischer In order to prove that $\mathbb{Z}_p=\{ x \in \mathbb{Q}_p: w_p(x) \geq 0 \}$ can we just say that $x=p^m u \in \mathbb{Z}_p \Leftrightarrow m \geq 0$, i.e. $w_p(x) \geq 0$ ?
 
12:28 PM
@SayanChattopadhyay do set theory and calculus first.
 
Hi everyone
let's say I have $n$ points with successive integers as $x$-coordinates and arbitrary integers as $y$ coordinates. Is it possible in general to find a continuous periodic function (of period $n+1$) passing through those points?
 
@Chris'ssis $$ \begin{align} \left|\int_{2k\pi}^{2(k+1)\pi}\frac{\sin(x)}{x}\mathrm{d}x\right| &=\left|\int_{-\pi}^\pi\frac{-\sin(x)}{x+(2k+1)\pi}\mathrm{d}x\right|\\ &=\left|\int_{-\pi}^\pi\left(\frac{\sin(x)}{(2k+1)\pi} -\frac{\sin(x)}{x+(2k+1)\pi}\right)\mathrm{d}x\right|\\ &=\left|\int_{-\pi}^\pi\left(\frac{x\sin(x)}{(2k+1) \pi(x+(2k+1)\pi)}\right)\mathrm{d}x\right|\\ &\le\frac1{(2k+1)\pi(2k\pi)}\int_{-\pi}^\pi x\sin(x)\,\mathrm{d}x\\ &=\frac1{k(2k+1)\pi} \end{align} $$
 
@robjohn Nice. I think we might get something like that using the integration by parts too.
@robjohn Anyway, your way is very very nice, simple.
 
12:45 PM
@Chris'ssis As I have said; I am simple-minded :-)
 
@Chris'ssis did you think of a design for your book front cover?
@robjohn how do you do
 
@TheArtist Not really, not now. At the moment I only wanna write amazing problems with amazing proofs.
 
@Chris'ssis okay
 
I'll never be someone that published a book for the sake of publishing. If the book cannot reach a certain level, it will never be published (because I don't wanna publish any kind of book).
I work hard for this.
 
@TheArtist hi there
 
12:50 PM
@Chris'ssis cool :)
@robjohn hi
 
@evinda Not quite. There's always $0$ to be considered. Apart from that, if you know that every $x\in \mathbb{Q}_p\setminus \{0\}$ has a unique decomposition $x = p^mu$ with $u \in \mathbb{Z}_p^\times$ and $m\in\mathbb{Z}$, then that suffices for the $x\neq 0$ case.
 
@robjohn Simple integration by parts yields $$\left|\int_{2k\pi}^{2(k+1)\pi}\frac{\sin(x)}{x}\mathrm{d}x\right|\le\frac{1}{2‌​\pi k(k+1)}$$ Asymptotically, our results behave the same way.
 
@Chris'ssis I think that inequality may not hold... let me check again
 
@DanielFischer So, we say that every $x\in \mathbb{Q}_p\setminus \{0\}$ has a unique decomposition $x = p^mu$ with $u \in \mathbb{Z}_p^\times$ and $m\in\mathbb{Z}$. Then do we take cases for $m$ ?
 
@Chris'ssis Yes, they do, and I was just checking whether that is too small, but it appears not to be numerically
 
12:57 PM
@robjohn I used the integration by parts.
 
@Chris'ssis yes, but what term did you throw away? that cannot be exact
 
@evinda Has it been proved that every $x\neq 0$ has such a representation and that it is unique?
 
@robjohn Well, we have $1/(2k\pi)-1/(2(k+1) \pi)$- something. But it's absolute value of the whole thing.
 
@DanielFischer Yes, it has been proven.
 
@evinda Then you just need to know that $m \geqslant 0 \iff x\in\mathbb{Z}_p$.
 
1:02 PM
@Chris'ssis The part you are tossing is $$\int_{2\pi n}^{2\pi(n+1)}\frac{\cos(x)}{x^2}\mathrm{d}x$$ which turns out to be positive, I believe, and therefore your inequality holds.
 
@robjohn Yes. I'll try to explain this.
@robjohn that part with the absolute value is not OK (referring to the explanation).
 
@Chris'ssis It is positive because $x^2$ is convex
 
@robjohn Right.
 
@Chris'ssis That is an interesting exercise... show that $\int_0^{2\pi}f(x)\cos(x)\,\mathrm{d}x\ge0$ if $f$ is convex
 
@robjohn Yeah, true. I was thinking of a different way of tackling that part.
Let me think a bit.
 
1:12 PM
@DanielFischer So do use both of the following sentences or only the second one?


Each element $x \in \mathbb{Z}_p \setminus \{0\}$ has a unique decomposition $p^mu$ with $u \in \mathbb{Z}_p^{\ast}$ and $m \in \mathbb{N}$.

If $x \in \mathbb{Q}_p \setminus \{0\}$ then it has a unique decomposition $p^nu$ with $u \in \mathbb{Z}_p^{\ast}$ and $n \in \mathbb{Z}$
 
1:28 PM
@Chris'ssis are you saying there is something wrong with my explanation of the inequality, or something else?
 
@robjohn No, your way is OK. I was referring to my explanations, I did some explanations in a hurry and they were not proper for that situation. I retracted my saying at that point above.
 
@Chris'ssis ah, but the part about the positivity due to convexity makes it okay
 
@robjohn I met that before in some exercise. My different point was to use the integration by parts once or 2 times more.
Or, I just check now a new idea.
 
Hello!! Let $P(x, y) \in \mathbb{Q}[x, y]$. We consider the equation $P(x, y)=0 (1)$. If $a, b \in \mathbb{Q}$ such that $P(a, b)=0$ then $(a, b) \in \mathbb{Q}^2$, is called a rational solution. If $K$ a field, $\mathbb{Q} \leq K$, then if $(a, b) \in K \times K$ and $P(a, b)=0$ then $(a, b)$ is called a $K-$rational solution.

When $(a, b)$ is a $K-$rational solution does this mean that $(a, b) \in \mathbb{Q}$ ??

For example, $x^2-2$, is $\sqrt{2}$ a $\mathbb{R}-$solution ??
 
1:45 PM
@robjohn I perfectly understand what you mean. Thinking of explaining that mathematically (and I also think of other ways).
 
2:03 PM
@DanielFischer
I want to visualize CP^n as a quotient space of D^2n. Can you give me a hint?
The usual identify-upper-and-lower-hemisphere technique used for RP^n doesn't work, as there are too many elements with absolute value 1 in C.
 
2:23 PM
Hello @BalarkaSen!! If $(a, b) \in \mathbb{Q} \times \mathbb{Q}$ a rational solution of $P(x, y)=0 (1)$, where $P(x, y) \in \mathbb{Q}[x, y]$, then the equation $(1)$ has a $\mathbb{Q}_p-$rational solution $\forall p \leq \infty$.

The (rational) $x \in \mathbb{Q}$ is a perfect square of a rational $\Leftrightarrow$ when $x$ is a perfect square in $\mathbb{Q}_p, \forall p \leq \infty$.

So, something stands in $\mathbb{Q}$ iff it stands in $\mathbb{Q}_p, \forall p \in \mathbb{P}$ ??
 
@BalarkaSen $z \mapsto (z, \sqrt{1 - \lVert z\rVert^2})$ embeds $D^{2n}$ into $S^{2n+1} \subset \mathbb{C}^{n+1}$.
 
@Balarka!
 
@DanielFischer Eh, OK, but I can't visualize the embedding yet. But I guess I should figure that one out myself. Thanks.
Hi @Studentmath
 
@Balarka I have a nice question for you, after I solve it
 
@BalarkaSen There's something about the last coordinate that may help.
 
2:34 PM
@robjohn The general case is clear after integrating by parts and considering that a convex function has an increasing derivative.
@robjohn how did you prove that?
 
@Balarka okay I am done. If you want some fun group theory exercise let me know
 
@robjohn I mean it has an increasing first derivative. But still, this is only clear for some cases.
 
@Studentmath OK
 
Okay, you are given an abelian group of finite order $n$, $G$. We mark the element of largest order with $max_G$. We mark the number of elements with order $max_G$ in $G$ as $num_G$. Out of all the abelian groups of order $n$, find the one with the maximal $num_G$.
 
Huy
@DanielFischer: Does there exist a manifold with 12 orientations?
 
2:49 PM
I.e. express it as cyclic direct-product. Obviously prove it is the one with the maximal $num_G$
 
@Studentmath Noted. Will think about it later
@Huy How did the DiffGeo exam go?
 
It's nice.
 
Huy
@BalarkaSen: Does there exist a manifold with 12 orientations?
 
What is orientation of a manifold?
 
Huy
In mathematics, orientability is a property of surfaces in Euclidean space that measures whether it is possible to make a consistent choice of surface normal vector at every point. A choice of surface normal allows one to use the right-hand rule to define a "clockwise" direction of loops in the surface, as needed by Stokes' theorem for instance. More generally, orientability of an abstract surface, or manifold, measures whether one can consistently choose a "clockwise" orientation for all loops in the manifold. Equivalently, a surface is orientable if a two-dimensional figure such as in the space...
 
2:52 PM
@Huy No, every connected manifold (of dimension at least $1$) has either no orientation or two. You can combine orientations on different components as you like, so the number of orientations, if not $0$, is always a power of $2$ [or infinite, if you have infinitely many components].
 
Huy
I see.
 
@Huy Oh. No idea.
Way too much differential for me.
 
Huy
@BalarkaSen: It was better than I expected, but that was one question I didn't know the answer to.
@BalarkaSen: And now I can finally waste a day to play Sims.
 
Huy
@BalarkaSen: Do you like architecture, or building stuff, at all?
 
2:57 PM
Eh, why d'you ask?
 
Huy
Just making conversation.
If you're busy then nevermind.
 
No, I'm not busy. Taking a break.
 
Huy
(also because you always claim everything apart from maths [and sleep] is a waste of time)
 
Well, not much of architecture, but I like sculptures.
 
Huy
Cool.
 
2:59 PM
@robjohn Tnx, please look at my relative question
0
Q: What general function can I use to represent the next sequence: 2,2,2,2,2,3,3,2,2,2,2,2,3,3,2,2,2,2,2,3,3…?

Ilya_GazmanLook at this sequence: 2,2,2,2,2,3,3,2,2,2,2,2,3,3,2,2,2,2,2,3,3... It is defined as fallow: $$f(n)=\begin{cases} 3 &\text{if $n \bmod 7=6,0$}\\ 2 &\text{otherwise}\\ \end{cases}$$ David found a good representation for $f(n)$: $$g(n)=\frac{\sin\dfrac{2\pi(n+1)}{7}}{\sin\dfrac{2\pi}{7}} ...

 
Morning
 
Huy
Morning again, @MikeMiller.
 
@Huy Did you get an answer about your 12 orientations?
 
Huy
@MikeMiller: Daniel gave me one. I mistakenly thought taking orientable, not-connected manifolds would add up in regards of orientation, and said that, and the prof said that I was on the right track but wasn't quite right yet. But I didn't realise my mistake.
 
Ok.
Seems to me Daniel's trying to take my job.
 
Huy
3:05 PM
Pretty sure my overall performance should have sufficed to pass though, felt a bit better than during my GR exam, but well, who knows.
 
bbl
 
Is anyone of you familiar with Hensel's Lemma??
 
@Huy I caught up above. I didn't know people still played the sims.
 
Huy
@MikeMiller: I do, before and after exams, traditionally.
 
Interesting.
 
3:15 PM
@MikeMiller Nobody can take my job of getting well, it's too tough.
 
@DanielFischer Could you take a look at my question? math.stackexchange.com/questions/1125049/…
 
Huy
@MikeMiller: One of my first houses. i.imgur.com/tb1Onk7.jpg I once did one I was really, really, really proud of but I can't find the screenshots anymore, just now.
 
@Huy How did you do it?
 
Huy
@evinda: I played Sims. :D
 
Aha! @Huy
 
Huy
3:22 PM
@evinda: I'm working on a new one just now.
 
Nice :D @Huy
 
Huy
I don't think I'll ever make another one remotely as gorgeous as the one I made 3 years ago, though.
@evinda: h2.abload.de/img/screenshotwldt.jpg and h2.abload.de/img/screenshot-3cxjc.jpg this is one of my favourite creations.
Even though I can tell I wasn't very experienced shaping houses at that time. But I put so much time and effort into the garden.
Or park.
 
Looks good!
 
I am looking at Mazur's theorem...

$$E(\mathbb{Q})_{\text{torsion}} \cong \mathbb{Z}/n\mathbb{Z}, \text{ for } n=1,2, \dots ,10,12$$
means that the torsion group $E(\mathbb{Q})_{\text{torsion}}$ contains the points $P$ which have the identity
$$nP=0, \text{ for } n=1,2, \dots ,10,12$$
and the coordinates are integers, right ??

What does
$$E(\mathbb{Q})_{\text{torsion}} \cong \mathbb{Z}/2m\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}, \text{ for } m=1,2,3,4$$
mean??
 
Huy
@MikeMiller: Can you name all the cars? h2.abload.de/img/screenshot-3cxjc.jpg
 
3:28 PM
If we have an equation $P(X,Y) \in \mathbb{Q}_p$ how do we deduce that it has to be $(x,y) \in \mathbb{Z}_p \times \mathbb{Z}_p$ , where $(x,y)$ is the solution of the equation?
 
I am reading the followng idea about Hensel's Lemma...
0
Q: Idea of Hensel's Lemma

Mary Star$$P(x, y)=0 \tag {1} \\ \text{ where }P(x, y) \in \mathbb{Q}[x, y]$$ How do we know that $(1)$ has a solution in a $\mathbb{Q}_p$ ? We will apply Hensel's Lemma. Idea: we begin from an element of $\mathbb{Q}_p$, that has some identities $(a_1)$. Then we create a sequence of values $a_1, ...

Could you explain to me which identities are meant??
 
@Huy No.
 
Huy
@MikeMiller: Not a car person. :-(
 
I've got a question
 
can someone refer me to a proof that for any ring $R$, $\sqrt{(0)}$ is a nil ideal?
 
3:42 PM
I am looking at $G$, $o(G)=888888$. It's not cyclic, and has an element of order $4$. So, I am wondering how many subgroups of order $24024$ it has, and if they are all isomorphic. Since I can say $G=C_2\times C_4 \times C_7 \times C_{11} \times C_{13} \times C_{37}$, I can say that there are 37 of such and are all isomorphic to $C_2\times C_4 \times C_7 \times C_{11} \times C_{13}$?
 
@ThomasAndrews please explain your comment
 
@Huy John, Edward, Richard, Amanda, Chloe, Ian, Karl, Lana and Julia.
 
Huy
@PedroTamaroff: Interesting.
 
@huy I am wondering what you look like now.
 
Huy
@ABeautifulMind: Unshaved.
 
3:49 PM
@Huy I see. Which part? LOL.
 
Huy
@ABeautifulMind: That escalated quickly.
 
@Chris'ssis If $f$ is convex, then $$ \begin{align} f(\pi-x)&\le\frac\pi{2\pi-2x}f(x)+\frac{\pi-2x}{2\pi-2x}f(2\pi-x)\\ f(\pi+x)&\le\frac{\pi-2x}{2\pi-2x}f(x)+\frac\pi{2\pi-2x}f(2\pi-x)\\ f(\pi-x)+f(\pi+x)&\le f(x)+f(2\pi-x) \end{align} $$ Therefore, $$ \begin{align} \int_0^{2\pi}f(x)\cos(x)\,\mathrm{d}x &=\int_0^{\pi/2}[f(x)-f(\pi-x)-f(\pi+x)+f(2\pi-x)]\cos(x)\,\mathrm{d}x\\ &\ge0 \end{align} $$
 
@Huy Now all the mods will come in after getting flags.
 
Hey, @robjohn can you help me with my last asked question please?
 
@Samuel What's the definition of a nil ideal?
 
3:53 PM
@Huy From left to right: Mercedes-Benz SLS AMG, Bentley Continental, Pre-2008 Lamborghini Gallardo, a Jaguar (resolution makes it impossible to tell imo), Porsche Panamera, BMW M6 (unsure, resolution is poor), and a Rolls Royce Phantom (I believe) on the far right
 
Huy
@ABeautifulMind: I look a tad older than on other pictures that you can find of me.
 
@Ilya_Gazman Didn't I give two answers in chat already?
 
Oops, didn't see the other two outside the garage. Those two, from left to right, are the Ferrari 458 Italia and an Aston Martin One-77
 
Huy
@teadawg1337: It's just a 325i coupé, IIRC, the one my father currently owns.
 
@teadawg1337 Good try, but they're all named Clarence.
 
3:55 PM
@MikeMiller: good morning
 
Morning
 
Huy
@teadawg1337: I'm not sure actually about the Jag. Are you sure? I don't recall ever having a Jag in my garage. I can't make it out just now either, but I try to remember (and to recognise).
 
Hrmpf, or is there only one subgroup of that order? I am confused.
 
@Huy Hmm, it might be a Maserati... I'm not entirely sure tbh
 
Huy
@teadawg1337: I should make a Sims house with 78 One-77s.
@teadawg1337: No, 99% not a Maserati either. I don't like Maseratis.
 
3:58 PM
@robjohn Thanks for your previous help! Now the question is different. I don't want to spam it, but can you please look at my last posted link for the question. It is based on your answer
 
@Huy It really looks quite similar to a Jaguar XK to me
 
Would I be right to say that if $G$ is Abelian, so that $G=C_2\times C_4\times C_7\times C_{11}$ it would have one and only one subgroup of order $56$, as that subgroup would have one sylow-p subgroup of order $2^3$, and one of order $7$ (as it is abelian too), and they are precisely the same ones as in $G$. So there is only one way to form it? Or am I messing things up?
 
Wait... No, the back isn't rounded like a Jaguar.... I have no idea
.... Definitely a Jaguar XK
 
Huy
@teadawg1337: I think it's actually the Maserati after all. Maybe I was drunk when I bought the cars.
@teadawg1337: Which cars do you think I should buy for my new house? I'll only want to have 3-4, because I have become so humble.
 
@Huy Derp, didn't even look at the exhaust........... Yeah, it's definitely a Maserati
@Huy I haven't the slightest idea
 
Huy
4:08 PM
@teadawg1337: What is your favourite car?
 
@robjohn Back. Nice.
 
@Huy Far too many to choose from.... Probably the Koenigsegg Agera
Maybe the Enzo Ferrari or McLaren 12C
 
Huy
@teadawg1337: Do you have a car?
 
@Huy I have a Toyota Camry with my name on it (figuratively), but I don't have my driver's license yet
 
Huy
@teadawg1337: That's rather odd.
 
4:19 PM
I'm turning 19 in a few months, plus I'm terrified of being severely injured in a crash
 
hey yall
I like math
 
Huy
@teadawg1337: Is there any reason for that? Also, is it usual on your country to only being able to obtain a driver's license at age 19?
 
I live in America, most people have a driver's license at 16. I'm just profoundly lazy
 
Huy
I see.
And why are you terrified of that unlikely situation?
 
I don't have a driver's license either, @teadawg1337
 
4:22 PM
Hi guys
 
@MikeMiller: one where every element is nilpotent
 
@SamuelYusim But that's true of that ideal by its very definition.
 
@robjohn It a result known in classical analysis, right?
 
Do you have good public transportation, that you can handle without a car?
 
$\sqrt{(0)}$ is the intersection of all prime ideals in the ring
 
4:24 PM
That's silly. That should be a theorem, not a definition.
 
Hi @MikeMiller
 
$\sqrt{I}$, to me, means $\{ f \in A : f^k \in I \text{ for some } k\}$ :P
(At least, that should be a theorem if that's the notation you're using.)
 
oh, well obviously it works that way then
 
hence the confusion
anyway, you've got one inclusion, yes? every nilpotent element is in that intersection.
 
Is anyone familiar with the Foundation series by Isaac Asimov? DON'T SPOIL ANYTHING, I've only started reading it yesterday
 
4:26 PM
@teadawg1337 Everyone dies
 
i have heard issac assimovs name@teadawg1337
I robot@teadawg1337
 
@robjohn I read some books like Problems in mathematical analysis and curiously I didn't meet it there. It's too nice.
 
gtg
 
@Chris'ssis the convex functions and cosine?
 
hi again
 
4:28 PM
@Chris'ssis I've never seen it before, but I doubt it's new
 
@robjohn I mean that generalization.
 
@teadawg supposed to be good
 
@robjohn That way of yours there is amazing.
 
@Chris'ssis have you stayed on the chat since 4am ?
 
@MikeMiller sure, since if $n$ is nilpotent, $0 = n^k$ for some $k$ and so either $n$ or $n^{k-1}$ is in the prime ideal. If it's $n^{k-1}$ we can use induction to peel off powers of $n$ until we do have $n$ is in the ideal
 
4:30 PM
Oh, it's fantastic. I love his idea of psychohistory
The idea of using mathematics to predict the future with pinpoint accuracy is fascinating beyond description
Anyways, I've an appointment to attend to
 
@robjohn some problems are simply like hell (when needed to go into details). I mean that limit is just a tiny bit of a problem I've beed working on.
 
@MikeM I am confused by a certain question, mind giving a quick look?
 
@Ramanewbie Are you spying on me? :-)
 
@Studentmath Sure, just a second
 
@MikeM cheers, take your time :)
 
4:33 PM
Great, I have just found out how CP^n/CP^{n-1} looks like.
 
@Chris'ssis -_- lol no, I just realize you're here, as 5 hours ago !
 
Now there should be no problem in computing homology.
 
@Ramanewbie Not really. I was away for 45 minutes or so. I just returned, but I'm not on chat, I'm working on my book.
 
@Chris'ssis ok... hope it's ready soon !
 
(although I use the chat, yes)
 
4:36 PM
@Samuel Now if $s$ is not nilpotent you need to show there's some prime ideal that doesn't contain it. The method I would use is a Zorn's lemma argument.
@Studentmath What's up?
 
@Huy Could you maybe take a look at my question? math.stackexchange.com/questions/1125049/…
 
Huy
@evinda: Sorry, I'm done with work until Tuesday.
 
A ok... No problem... @Huy
 
(To be more clear: use Zorn's to pick an ideal maximal w.r.t. not containing $s$. Show this ideal is prime.)
 
yeah we did it that way in class
 
4:41 PM
@MikeM so I have $G$ of order $888888$, I know it is not cyclic and has an element of order 4. So $G=C_2\times C_4\times C_7\times C_{11} \times C_{13} \times C_{37}$
 
but I think the last part of the proof we did is wrong
 
Hi @BalarkaSen
 
Now, I wonder how many subgroups of order $24024$ it has. I think only one,
Since, let such subgroup be $H$, then it is the direct-product of the sylow-subgroups of $G$, besides $C_{37}$. As $G$ is Abelian, there is only one sylow-subgroup of each kind...
But something along that argument sounds wrong, or weak. I just don't know.
 
Hi @Studentmath
 
oh wait I think I figured out the problem
 
4:43 PM
Hey @Sayan
 
my professor's exposition doesn't seem to carry over to paper very well, and his typed notes are twice as bad as what I copy down in class
 
@Studentmath "So $G=...". I don't buy that.
 
Are u guys talking about group
Theory
 
Oh! Yes I do, if you assume $G$ Abelian.
 
HARRO.
 
4:44 PM
Yes, I do. Also, I opened it up wrong.
But that's besides the point
I can take it to be $999999$ and $C_3\times C_9 \times...$ instead. And then ask the question about how many subgroups of order 999999/37 it has
It would be only one, right?
 
@PedroTamaroff Why.....why....hello
 
@KevinDriscoll Hey. What have you been doing?
 
@Studentmath Quite right. The projection onto the $C_{37}$ factor is trivial, so must just be the produxt of the first few factors.
 
Okay, but would it be, if, say, I asked about subgroup of order $999999/(37\cdot 3)=9009$. So it only holds the $C_9$ factor of the sylow-3 subgroup of $G$. How can I go along investigating how many subgroups of such may be?
Obviously no more than 3, and at least one.
 
Hi everyone! Could it be that the space of end points of a free group on 2 or more generatiors is homeomorphic to the circle?
 
4:53 PM
@Studentmath Are you asking about a general group, or just the Abelian one you already know about, with a factor of $C_3 \times C_9$?
 
@PedroTamaroff Quantum field theory! Been going pretty hardcore in learning mode. Lots of long and tedious computations. Have you had your exams yet?
 
@KevinDriscoll I have to sit for two finals in about a month.
 
If the latter: there are two, depending on whether the subgroup of that factor is $C_9$ or $C_3 \times C_3$.
 
Going to study tomorrow a while.
 
@MikeM the Abelian group I already know about.
 
4:54 PM
@Pedro Oh not for a month!? I thought it was sooner than that. Are you otherwise on a break or whaaaat?
 
that's what I thought, but how could it be $C_3\times C_3$?
 
@KevinDriscoll Yes, I'm on summer vacations. Finished Dec 23rd and will go back on March 16th.
 
@PedroTamaroff So you're sitting for these exams to be exempted from taking those classes? Or did you already take the classes?
 
@KevinDriscoll Oh, no. I never took an exam to be exempted of anything. =P
Just sitting for the final exams of courses I've taken.
 
@Pedro That sounds miserable then. They do that here in some secondary schools, sit for exams during/after the break following that class
I've never understood this approach
 
4:57 PM
Any $9009$ group factors to either of these two, but considering it is a subgroup of a group with sylow-3 subgroup being $C_3 \times C_9$
 
@KevinDriscoll Well, you can sit for them right after you finish. One can choose the date.
 
I want to take the exam and be DONE with it. Having that hanging over your break seems miserable.
 
I chose not to sit for them in december.
 
@Ilya_Gazman: are you looking for more than $$b+(a-b)\left(\left\lfloor\frac{n+d}c\right\rfloor -\left\lfloor\frac{n-1}c\right\rfloor\right)$$
 
@Pedro Oh!? Are the exams different to prevent students from cheating through collaboration?
 
4:58 PM
Is anyone here familiar with ends of topological spaces?
 
@KevinDriscoll Sure, there is not a fixed exam. Every new date there's a different exam.
Sometimes the exam is oral, so the professor decides what to ask more freely.
 
@Pedro Wow, how do the get away with that? Most professors I know hate making 1 exam, much less several.
 

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