00:00 - 16:0016:00 - 00:00

12:08 AM
I don't know what to pick for $g(x)$

2 hours later…
2:25 AM
The plane spanned by r' and r'' is the osculating plane that a curve is "closest to lying in" at that point. This generalizes to higher dimensional hyperplanes for higher dimensional curves, right? Like, (by thinking about Taylor polynomials?) the 3d hyperplane spanned by the first three derivatives is the best fitting one.

2:38 AM
I got the answer for the same question. But I don't understand why (1) and (2) implies $f(e^{it})e^{-int}=1$

https://math.stackexchange.com/questions/696500/if-p-is-a-polynomial-then-either-pz-zn-or-max-z-1pz1

@N.Maneesh (1) says the integral of it is exactly $2\pi$, so the integral of the real part is $2\pi$. But (2) says the most the real part could be is 1, so it needs to be 1 almost everywhere on the interval. Since f is a polynomial, it must be identically 1.

@MarkS. Thank you very much :)

5 hours later…
7:30 AM
Morning all

heya

8:28 AM
rather quiet out here eh

not anymore
balarka in da house

hardly the party spotlight myself
which college are you in, @Subhasis, if i may ask

Durgapur Govt college. Lol

Morning everyone

8:36 AM
One from isi, one from Princeton and one from a noob college

Morning @Alessandro

Perfectly balanced
@BalarkaSen

my man isi is a shit college and u will hardly find anyone saying it out loud which makes it worse
im not too familiar with Durgapur Govt
Hey @Alessandro

i dont know what my level means; its just terrible on many different levels
subpar curriculum for a math undergrad, subpar students, ...

8:39 AM
explain

thats not true the faculty is fantastic

I don't know a thing that you utter while discussing topology
it was really an experience that day, you walking me step by step through the problem

i mean the point isnt that i have head start

then what is the point?

i just stated some of them; i dont think the course structure is good, thats one major problem
we had two analysis courses this year and i dont think it was actual analysis

8:43 AM
what makes you so good at mathematics?

i dont think anybody learnt any analysis in those courses

what is the course content?

analysis-2 eg was half a semester of metric spaces and half a semester of multivariable calculus done badly with riemann integration (which is one of the major things one should learn in a first analysis course imo) was sandwiched half-assedly in between
one learns more analysis by going through rudin than sitting in these courses

isn't multivariable at this level just calculation based?
@BalarkaSen rings some bell. I know very little. But I learnt it by digging it up step by step

i mean its not an engineering multivariable course

8:47 AM
well, things are worse here. We have basically an engineering multi
just computation

but i dont think any of the conceptual leaps from analysis in R to analysis in R^n was hit home in the course

no explanation why the heck the Jacobean appears in the change of variable
just mere computation

well it says something when one of the so-called best places for math in the country is, well, not that good

for the case of Jacobean: I believe it has something to do with the preservation of area, due to the change of scale factor

@SubhasisBiswas thats ok though, the proof of change of variables formula is involved. but one should have some intuitive grasp of why it appears
yes, thats why you scale by determinant of the Jacobian
its a measure of how much area is infinitisimally stretched

8:49 AM
yes.
a rectangle getting shaped into a parallelogram
or a bigger rectangular shape
I wish I could be more passionate towards this subject. it's very beautiful, but I just don't feel it

u dont have to take it as a job
just do what you want to do

i was trying to prepare myself for the isi mmath programme
entrance

oh shit dont come here lmao

midway, i gave up.

its worse than bmath

8:54 AM
well, I am not even worthy of that course, to be honest
mathematics course

@SubhasisBiswas i mean mmath is fundamentally a bad course and most of the students dont have proper mathematical foundation which is bad because ur a grad student who'll presumably persue a phd, so

@BalarkaSen well, still too much for a student who doesn't even know what $\mathbb{C}[X,Y]$ means

iisc faculty is great
i dont know about the student community
prolly worse

@BalarkaSen well, you have coached me for one day. You know enough to judge a student by his depth. What are your honest opinions about my abilities.
I don't mind if your opinion shatters my confidence
just be honest

@SubhasisBiswas its one thing to not know something and another thing to be paid to learn something but still not doing it, which is my criticism of the students in isi really
the student stipends are a total waste of money

9:00 AM
@BalarkaSen students are not devoted enough?

I don't think ability really matters at the end of the day, what matters is sticking to it, whatever you are doing. You spent a whole day and solved a problem that you wanted to solve; that's how one does math
I'm quite impressed by the perseverance

but I lack the intellect.
that's a key factor

meh i dont care about cleverness

can you get me started on some other topic today, like group theory?
starting from a basic level to higher up

just pick up some book, like Artin, and read through the chapter on group theory
thats the best approach

9:03 AM
well, I started studying in the morning. Then started scrolling though the election results. Haven't touched that book since.

lol
set a goal like "finish this specific chapter by today"
that usually helps
how much group theory do you know

@BalarkaSen very basic actually. I don't even have much grasp on $D_8$.
I know small things about cyclic groups, even lesser amount of knowledge on permutation groups, then comes normal and quotient groups.
very little actually. And yes, no idea about sylow theorem and conjugacy classes. I won't say that's because of lack of cleverness. That's because of lack of hard work and discipline

do you know what an isomorphism is

yes
sort of "property preserving"
if even a single isomorphism exists b/w two groups, then they are just identical, except for the symbols involved

Prove that if $G$ is a group of order a prime $p$ then $G$ is isomorphic to the cyclic group $\Bbb Z_p$

9:10 AM
well, if a group has a prime order, then it must be cyclic. all cyclic groups of order $n$ are isomorphic to $Z_n$

why is it cyclic

group must be cyclic due to lagrange's theorem

yup
Ok, prove that if $G$ is a group of order $4$ then $G$ is isomorphic to $\Bbb Z_2 \times \Bbb Z_2$ or $\Bbb Z_4$

"either" ?

Prove that as well :)
i.e., $\Bbb Z_2 \times \Bbb Z_2$ and $\Bbb Z_4$ are not isomorphic

9:12 AM
okay, if a group is of order $4$, then it is either a cyclic group, or isomorphic to Klein's 4 group
only two groups upto order $4$

I mean that's basically the question. Why is that so?

well, the cyclic thing can be excluded. we have to prove that only one non cyclic group exists of order $4$.

mhm

order of an element cannot be $3$
neither can be $1$ ( absurd case, since every element becomes an identity)
so, only option is where the elements are of order $1$ and $2$
now, a group with an even order must have an odd number of elements of order $2$

by "neither can be 1" you mean "there is an element of order not 1"

9:16 AM
it must be the case here, that the group has three elements of order $2$
@BalarkaSen sorry, ignore that sentence, it's just a gibberish (regarding 1).
Now, three elements of order two.
is the only possibility

thats correct

now let the elements be $a, b, c, e$. Being of order two, it must be $ab=c$ ( and so on)
so, we get Klein's four group

OK. Can you generalize your result to groups of order $p^2$?

well, that's a problem from conjugacy classes i guess
wait up
gotta eat

Bon apetit

9:23 AM
@BalarkaSen you will see that most people in most math colleges in India are "sub-par". Everyone just wants to solve Olympiad problems. It's frustrating after I point.

9:34 AM
i mean people don't care about learning
its a serious problem because on one hand you have thousands of students cramming for engineering colleges and the scarce few who do get in science just want to enjoy college life or whatever
we have broken up our education cycle beyond repair

@BalarkaSen tru dat.
@BalarkaSen thanks :D

i have been reading this book called "Academically Adrift" by Arum & Roksa which explores similar situations in colleges in US throughout decades

@BalarkaSen well, what do I prove? that it is isomorphic to $Z_p \times Z_p$?

@SubhasisBiswas Not all groups of order $p^2$ is isomorphic to $\Bbb Z_p \times \Bbb Z_p$

for example?

9:40 AM
Think

@BalarkaSen well, same applies to me too, kind of. I wasted so many countless hours gaming and on social media (and chasing girls), that nothing can fix it
it's not for everyone.
@BalarkaSen will, but what do I do with this problem? (generalisation)?

its not an individual problem but of course you can project the problem within you etc

@BalarkaSen which result , I mean

@BalarkaSen True. I mean I don't really see a problem with enjoying college life. You need to know your priorities if you truly want to learn something.

that can wait; first give me an example of a group of order $p^2$ which is not isomorphic to $\Bbb Z_p \times \Bbb Z_p$

9:47 AM
I believe that if you truly want to develop academically you should develop as a person as well and that implies developing yourself socially and academically. That will happen I guess if you interact outside our comfort zones such as academics etc.

@Albas the notion of enjoying college life is synonymous to lowkey adolescent hedonism to be honest so i dont think those two are compatible. i mean the process of learning doesn't pay off in the short term, it's painful and unpleasant because you realize how pathetic you are and there's so little you actually understand and even though you understand it how slow you are to internalize it
2
its sort of depressing if you linger on it really
but ofc you can learn and socialize at the same time its just that the postmodern culture equates socialization with partying

okay, I am taking a wild guess
I haven't calculated yet
$U_{32}$ and $\mathbb{Z}_4 \times \mathbb{Z}_4$ are not isomorphic (!?)

What's $U_{32}$
$(\Bbb Z_{32})^\times$?

Yup that's the issue isn't it, conforming people to certain norms which are mostly hollow. But again I don't have much of a problem with partying even though I don't really prefer it. If people are partying, drinking etc it's their thing. If they know what their priorities are then they know how to get themselves back. I know many quite distinguished people in science (mathematics specifically) who have done "partying" in their undergraduate/graduate studies but are no less in learning

$U_{n}= \{\overline{k} \in \mathbb{Z}_n: gcd(n,k)=1\}$

9:58 AM
Yea certain books like Gallian use that notation for Z_n*

@Albas ultimately, we need to be happy. We'll all be dead within a few decades

@SubhasisBiswas That's true, $U_{32}$ is $\Bbb Z_2 \times \Bbb Z_8$. But this is not an example to what I was asking
$p$ was prime
also, you're thinking WAY too hard

dude, give me some times
time*
don't think that I am same as you.

I think that it's more of a problem of how mathematics is put forward as a subject. It's said to be a subject of prodigies and geniuses. A subject if you don't have anything unique in yourself you cannot succeed. This kind of an outlook forces people to build deep insecurities in themselves about their own knowledge and they start shying away from learning anything new, experimenting anything new.
Whereas it's actually a lot of hard work, a lot of failures. There aren't any moments. There are culminations.
Rant over lol

sure, but i dont think its a result of the social view of math. anyone with minimum working knowledge in math can experience this learning mechanism, where you see some technique, try to apply it to solve problems, fail innumerably many times, and then culminate in to a proper understanding of the technique
the idea is that's how learning always ever works, this should not be a point of demotivation

10:07 AM
$A_3 \times A_3$?

@SubhasisBiswas $A_3$ is isomorphic to $\Bbb Z_3$ (Exercise)

yeah
cyclic
group of order $3$

Try $p = 2$. What's a group of order $4$ not isomorphic to $\Bbb Z_2 \times \Bbb Z_2$? :P

$A_3 \simeq Z_3$
@BalarkaSen dude, this is cheating
I was trying to find a group of order $p^2$, that is not cyclic and not isomorphic to $Z_p \times Z_p$
it is an obvious fact.

@SubhasisBiswas Well, I was objecting to this, in which you didn't mention the cyclic case at all
I warned you you were thinking too hard :P

10:11 AM
fak u

So what is the statement of the problem that you want to prove for groups of order $p^2$?

to find a group of order $p^2$, that is not cyclic and not isomorphic to $Z_p \times Z_p$
I guess it is not possible to find a counterex

Do you think there is such a group?
@SubhasisBiswas OK. Prove it.

@BalarkaSen ok. Watch this
If not cyclic, then the elements must have order either $1$ or $p$.
there can be only one element of order $1$, so there must be $p^2-1$ number of elements of order $p$
amirit so far?

That's correct

10:16 AM
ok
now, we choose individual elements of $G$ and map one by one to an element of $Z_p \times Z_p$
sorry
proving $ker \phi = \{e\}$ does it

Can someone give me the definition of a real subspace of a complex vector space? Is it the one which is closed under the scalar multiplication by real numbers?

You chose some arbitrary bijection $\phi : G \to \Bbb Z_p \times \Bbb Z_p$. There is no reason this is a homomorphism.

yes yes.
wait
the group must be commutative
being of order $p^2$

@ThomasShelby That sounds correct to my ears. Any $\Bbb C$-vector space $V$ can be thought, by restriction of scalars, as a $\Bbb R$-vector space. A subspace of that real vector space is what you want
@SubhasisBiswas Why?

groups of order $p^r$ has a non trivial centre (had to look it up, I forgot). So, in case of $p^2$, there are only a few possibilities. Centre is of order $1$ (impossibility); of order $p$ and of order $p^2$.
suppose, $o(Z) \neq p^2$
then $Z$ is cyclic. But, $G/Z$ cannot be non trivial cyclic.

10:27 AM
That is false.
Right. Why?

@BalarkaSen Okay, thank you! Do you happen to know any theorem which states that a real subspace of $M_n(\Bbb C)$ is (topologically) closed? $M_n(\Bbb C)$ is the space of all n by n matrices.

@Thomas There is nothing specific about $M_n(\Bbb C)$ here... any subspace of a finite-dimensional real vector space is closed.

@BalarkaSen because, for a non commutative group, $G/Z$ is non cyclic.
But if commutative, then $G =Z$.

Yeah but like, why?
You just restated the question

10:33 AM
OK :)

Btw @BalarkaSen, where do you plan to do your summer project ? You were saying you might be doing it on the h principle.

I'm visiting TIFR in July
It's not a project; just going to learn something

Ah I see. It's a bad habit of calling it a project. Everybody calls it that here.

Project usually means you do something out of which some kind of publication comes out
Nothing of that sort

@BalarkaSen Thank you!!

10:43 AM
I know. The problem is that here I get a scholarship which kind of forces me to do a "summer project" each summer where I have to write up a report and give a seminar in the end. It's mostly learning stuff from a book but that's what they call it.
The questions though that you gave @SubhasisBiswas is a very nice exercise. I remember it troubling me a little in my group theory course.

Oh that's nice. I wish we had that as well.

Yea it's something that atleast makes people learn something new.
Though we have to kind of make sure that these projects are somewhat interdisciplinary. The institute kind of pushes the entire notion that interdisciplinary work is very important.

That's not false

Yea that is not. I am completely for it. But it just becomes a lot of work.
Like I have to juggle QFT and geometry at the moment which is a lot of work. It's my fault I took QFT but I find it extremely fascinating

10:51 AM
I am just studying the two at the moment and let's see where that takes me. There was this person here who did his project on the topic of the atiyah singer index theorem and it's relations to the heat equations. That was kinda cool.

Index theorems are really cool but I don't know the first thing

Interdisciplinary works are quite common here. Physicists taking mathematics classes and mathematics students taking physics classes.

I only learnt the Toeplitz index theorem and some basic things about Fredholm operators because I had to write a report on the works of an Indian mathematician for a non-credit course on intro to mathematics in my first sem, and I chose Patodi
I didn't actually write anything about the super general Atiyah-Patod-Singer index theorem, but just mentioned Toeplitz and Gauss-Bonnet as examples of index theorems
'cuz those are the only two things I understand

I just understand Gauss Bonnet that too only for surfaces. I just have heard this but isn't atiyah singer index some generalisation of the Riemann Roch theorem. The guy who did his project was talking about it in his presentation.

Oh I don't know Chern-Gauss-Bonnet. When I say GB I mean GB for surfaces.
I don't know anything about AS. Could be so

11:05 AM
I am taking this course on algebraic curves next semester(will Riemann Roch pop up there?)I know like algebra till rings and modules. Is that going to be enough? Do I need to do some more stuff? I think they will follow Fulton or Shafarevich

I don't know too much algebraic geometry. I learnt whatever I did from Shafarevich and I had some background in commutative algebra before that

Okhay

1 hour later…
user131753
12:37 PM
@BalarkaSen: Did you receive my message above where I mentioned my email? I just checked my email and saw that there has been no email from you. In any case, if you send me those images via email, please give me a ping.

@user170039 Ah I forgot, let me do that

user131753
23 hours ago, by user 170039
@BalarkaSen I will be glad to receive them. My email id is user170039@gmail.com.

Oh you have mentioned in there in the message.
@user170039 I shared a dropbox folder with you

user131753
@BalarkaSen Yeah I got it. Thanks.

No problem. Sorry if it's not super readable at times

user131753
12:44 PM
@BalarkaSen $\ddot\smile$

1:25 PM
@Balar

@BalarkaSen well said

@BalarkaSen, if two quotient groups $G/Z$ and $J/Z$ are isomorphic, are $G$ and $J$ isomorphic?
or a bit generalised version

What does it mean to quotient by the same subgroup when the two groups are different? Do you mean if $G/H \cong G'/H'$, $H \cong H'$, then $G \cong G'$?

Let $G/P \simeq J/Q$ and $P \simeq Q$

That's what I wrote.

1:28 PM
hmmm...
@BalarkaSen how do you come up with an answer so fast
@BalarkaSen I was typing. Didn't want to delete the sentence lol
@BalarkaSen is there any restrictive case, where this holds. Like, $o(G)=o(J)$.
?
or, when the group is commutative and the quotient groups are cyclic.

@SubhasisBiswas Well, at this level of generality, there's no hope of having this. Take $G = G' = S_n \times A_n$, $H = A_n \times \{e\}$ and $H' = \{e\} \times A_n$.

and I have never given it a thought until this very day.

Oh, wait, you asked for something else.
I am giving an example of $G \cong G'$, $H \cong H'$ but $G/H \not\cong G'/H'$

yes, the other way.

Ok, a counterexample for you is easier. Take $G = G' = \Bbb Z_4 \times \Bbb Z_2$. Quotient by $H = \Bbb Z_2 \times \Bbb Z_2$ and $H' = \Bbb Z_4 \times \{0\}$.
@SubhasisBiswas Everything's abelian here, and the quotient $G/H \cong G'/H' \cong \Bbb Z_2$, so... :)

1:39 PM
hmmmm

Here's something which is true. If there is a homomorphism $\varphi : G \to G'$ such that $\varphi(H) \subset H'$ so that $\varphi|_H : H \to H'$ is an isomorphism and the induced homomorphism $G/H \to G'/H'$ from $\varphi$ is an isomorphism, then $\varphi$ is an isomorphism between $G$ and $G'$.
This is the "corrected" statement for the statement I was giving a counterexample of at first
For yours, the correct statement is simply that if there is an isomorphism $\varphi : G \to G'$ such that $\varphi(H) \subset H'$ such that $\varphi|_H : H \to H'$ is also an isomorphism, then the induced homomorphism $G/H \to G'/H'$ is an isomorphism

@BalarkaSen wtf they had you doing the index theorem as a freshman?
american universities are the worst

The second corrected statement follows from the first isomorphism theorem. The first corrected statement follows from the short five lemma

Okay, this is what I have so far

@SubhasisBiswas I just realized I again gave a counterexample to something different: $G \cong G'$, $G/H \cong G'/H'$ but $H \not \cong H'$. Sigh. You are asking $G/H \cong G'/H'$, $H \cong H'$ but $G \not \cong G'$. Take $G = \Bbb Z_6$, $G' = S_3$ and $H = H' = \Bbb Z_3$. Both quotients $G/H$ and $G'/H'$ are $\Bbb Z_2$.
Tons of example, eg with $\Bbb Z_2 \times \Bbb Z_2$ and $\Bbb Z_4$, with the diagonally embedded $\Bbb Z_2$ in the first and the cyclic subgroup $\Bbb Z_2$ in the latter (this is even abelian)
This is known as the "extension problem": Given $N$ a normal subgroup of $G$, and the quotient $G/N$, if you know the isomorphism type of $N$ and $G/N$, how well do you understand the isomorphism type of $G$?
Sorry for the confusion. Basically there are tons of counterexamples to all possible variations to the problem, is the takeaway

1:59 PM
What we have so far:

$o(G)=p^2$. Let $a \in G$. $G/<a> \cong \mathbb{Z}_p \times \mathbb{Z}_p/<(\overline{1},\overline{1}) \cong \mathbb{Z}_p$

Again, $<(\overline{1},\overline{1}) \cong <a> \cong \mathbb{Z}_p$
can we conclude anything from this?

Right, that's not enough to conclude $G \cong \Bbb Z_p \times \Bbb Z_p$

We need at least one homomorphism
b/w $G$ and ...

I mean, $\Bbb Z_4/\Bbb Z_2 \cong (\Bbb Z_2 \times \Bbb Z_2)/\langle (1, 1) \rangle \cong \Bbb Z_2$, but $\Bbb Z_4 \not\cong \Bbb Z_2 \times \Bbb Z_2$.
You're not really using the fact that $G$ is not cyclic anywhere in your approach

Yes, I need to use it.

But this is a good observation.

2:02 PM
not enough.
@BalarkaSen tbh, i kinda used it to define the quotient group. But only a mild application

I don't understand; how? All you needed was an element $a$ of order $p$ in $G$.
That will exist even if $G$ is cyclic.

yes. being cyclic also ensures commutativity. That's why I said, this is not a proper application of the property. Way too loose

I liked your original approach with the center
You should try to work on that

@BalarkaSen which one.

2:19 PM
@SubhasisBiswas You wanted to show if $G$ is a group of order $p^2$, then $G = Z(G)$.

it is a book proof
a standard theorem

Oh, you have seen this before?

@BalarkaSen yes.

Guys, can you answer my question? Median is 50th percentile. But there are two kinds of definition of percentiles, exclusive and inclusive. Which definition of a percentile do we use when calculating median? Or can be just choose arbitrary definition and use it to calculate the median?

@SubhasisBiswas OK. What about groups of order $pq$, $p \neq q$? :P
(How many groups of order $6$ do you know? Can you prove they exhaust all isomorphism classes?)

2:28 PM
$\mathbb{Z}_6$, $S_3$

Prove that every group of order $6$ is isomorphic to one of the two you wrote down.

@BalarkaSen is that all? or does there exist some more?

None.

Umm..only two groups upto isomorphism of order $6$?

That's right

2:31 PM
okay. Learnt something new today
I am trying to prove it. watch it step by step
excluding the cyclic case right away

Don't prove it on this chatbox, get a pen and paper and work it out fully before posting it.

we tryna prove that the only non-cyclic group of order $6$ is $S_3$?

That's a restatement, yes

@BalarkaSen I cannot stress enough how important this is. Still I ignore it oftentimes
*almost all the time

It's very easy to get confused while writing things out here, happens to me all the time

2:40 PM
I am getting $\frac{f((0,0)+t\frac{(1,10^-6)}{(\sqrt{1+10^{-12}}}-f(0,0)}{t}$
Which I got on simplification $\frac{-1}{t}$ as $t \to 0$, I will get limit goes to infinity. But answer given as $-1$
$\frac{f((0,0)+t\frac{(1,10^-6)}{(\sqrt{1+10^{-12}}}})$ $=f(t(\frac{10^{12}}{\sqrt{1+10^{12}}},\frac{10^{6}}{\sqrt{1+10^{12}}})$
$f((0,0)+t\frac{(1,10^-6)}{\sqrt{1+10^{-12}}})$ $=f(t(\frac{10^{12}}{\sqrt{1+10^{12}}},\frac{10^{6}}{\sqrt{1+10^{12}}})$
Here $|y|\leq x^2$

3:10 PM
@BalarkaSen, I have a question

@N.Maneesh I believe the formula is $f(0+th, 0+t k)-f(0,0)/t$
@BalarkaSen do the number of elements of a given order say $m$ always divide the order of a group?

$f(0+th, 0+t k)-f(0,0)/t=f((0,0)+t(h,k))-f(0,0)/t$

@SubhasisBiswas Look at $\Bbb Z/3\Bbb Z$

@SubhasisBiswas right?

3:17 PM
@N.Maneesh ah yes

I have normalize the unit vector $(1,10^-6)$

@Astyx finite groups

$\Bbb Z_3$ is a finite group

Oof, I am stupid
let's go the problem manish gave us
I am trying
now, $|10^{-6}t| < |t^2|$ only upto a certain $t$ , right?
as $t \to 0$, the inequality inverts after $|t| \leq 10^{-6}$
am I right so far?

$10^{-12}\leq 1$. isn't it $|y|\ge x^2$?
here it is not the case. so, $f(h,k)=0$. am I correct?

3:22 PM
@BalarkaSen, do something

I'm not here to solve random problems, I have work to do

ok ok.
I gotta appreciate the time you are putting to improve me :)

It is $|10^{-6}t|\ge t^2$ and that's surely true for sufficiently small $t$, no?

@Thorgott exactly my point
after a certain $t$, inequality inverts
hence, the numerator becomes $0$

But the inequality is satisfied, so the numerator should be $1$
Oh wait, $f(0,0)=1$, so it cancels out, nvm

3:41 PM
@SubhasisBiswas I have a follow up to your question. How many elements of order $d$ are there in $\Bbb Z_n$?

@BalarkaSen $\phi(d)$ ?

When $d$ divides $n$, yes.

If it is, then my question was beyond stupid. It would have implied $\phi (d) \mid n$, whenever $d|n$.

Suppose $G$ is a group of order $n$ such that for every divisor $d$ of $n$, $G$ contains a unique subgroup of order $d$. Can you prove $G$ is cyclic?

@BalarkaSen wait up
This is what I have done so far (the prev question).

@BalarkaSen, although I have missed a short part of the proof, here is what I have done far. I am trying to fully complete it (the missed part).

Let $a \in G$, with $o(G)=6$. Then either $o(a)=3$, or $o(a)=2$. The number of elements of order $2$ in $G$ must be odd. Therefore, the number of elements of order $2$ is either $3$ or $1$.

It cannot be the case where $|\{a \in G: o(a)=2\}|=1$ (I missed the proof here, will come back to it). So, let the elements be $p, q, r , x, y, e$, with $o(p)=o(q)=o(r)=2$ and $o(x)=o(y)=3$. Now, we ass

3:48 PM
Good.

Suppose that $V,W$ are normed linear spaces such that $\mathcal{L}(V,W)$, the space of bounded linear operators $V\rightarrow W$ with the operator norm, is complete. Is $W$ necessarily complete? The converse of this is well-known to be true.

@BalarkaSen means?
@Thorgott well @BalarkaSen, here's your kind of problem.

Your line of thought is correct I mean

is this some sort of Banach space?
I don't know.

A complete normed linear space is also called a Banach space, yes

3:51 PM
@BalarkaSen :D

@Thorgott How do you conclude? From the definition of limit. Right? By taking $\epsilon=10^{-6}$

@N.Maneesh yee haw
apply some intuition, @N.Maneesh. $O(x^2)$ has a much faster converging rate.

Yes, that would be the formal way to do it.

user131753
@SubhasisBiswas Disregard intuition. Be explicit $\ddot\smile$.

user131753
(That's the advise I am usually given.)

3:57 PM
Thank you @SubhasisBiswas @Thorgott

@user170039 yes, intuition for normies like me is often misleading.

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