Mathematics - 2015-05-28 (page 1 of 3)

pjs36

12:11 AM

Ahoy, @AlexWertheim. I finally got some good news today, I may have a job teaching for the next year.

Mike Miller

12:21 AM

congrats!

pjs36

Thanks, Mike! I apparently scared Alex away, oops.

What are you up to, over the summer @MikeMiller ?

Mike Miller

12:43 AM

Sorry for the delay, I'm not paying much attention to chat right now, @pjs36. I'm doing a little travel to a summer school and a workshop and then when I get back from that I've got some teaching (and math, of course).

pjs36

1:01 AM

@MikeMiller Gotcha, no worries at all. It sounds like you'll be staying busy, at least!

Mike Miller

That's the goal!

pjs36

I can't say I've ever "workshopped" though, so that's definitely cool

Soham Chowdhury

@BalarkaSen But then his answer suffices?

Bye_World

2:07 AM

Is there such a thing as advanced trigonometry?

Maybe it's just Fourier analysis.

JMoravitz

2:28 AM

sigh. There's another round of grading out of the way.

Alex Wertheim

That's awesome, @pjs36! Glad to hear it :). Where at, if you don't mind my asking?

Remember me

3:04 AM

Hello!!

If there is something called advanced trigonometry I would love to study it

anon

you can try trigonometry on a sphere or in the hyperbolic plane instead of the Euclidean one

(it's a thing)

Remember me

Hmm...

Hello@anon

anon

hello

Remember me

So what have you been doing these days (in maths ) @anon

About solenoids?

anon

reskimming some hyperbolic stuff, beginning to look at "algebraic curves and riemann surfaces" by miranda but it's going slowly

Remember me

3:10 AM

Hmm....nice

anon

it only just occurred to me the other day that the cross-ratio [z,a;b,c] (or one definition of it at least - seems every source has its own definition) can be defined as the unique mobius transformation of z which sends (a,b,c) to (0,1,infinity).

Remember me

3:35 AM

Hey @Kaj

Kaj Hansen

Hey there

Remember me

Forgot my name again :p@Kaj

Kaj Hansen

haha, You're Sayan. Just kinda tired myself.

Soham Chowdhury

Heya @Kaj, what's cooking?

Kaj Hansen

Just checking in and seeing if there's anything interesting on Main

Soham Chowdhury

3:38 AM

Just got a little schoolwork over myself, now going to hit Aluffi.

Sayan?

Remember me

Nothing there ... The standard of questions are going very low

Yes..@Soham

Soham Chowdhury

Post some, then. :P

What are you doing?

Remember me

Doing physics homework having burgers then i will think of some topology

Kaj Hansen

Yeah, that's pretty standard. This one could use some more upvotes. It's so rare these days to see people having work in their OP: math.stackexchange.com/questions/1301995/…

Soham Chowdhury

I can't wait to have the field theory chapter done. :)

Remember me

3:40 AM

Well I just proved something in topology

You done open sets @Soham

Soham Chowdhury

A little, and not too well.

Kaj Hansen

Topology is open sets :P

2

Soham Chowdhury

Haha, I wouldn't know yet.

user147690

@Kaj Can you verify if what someone told me is true: Can the GCD of two polynomials be defined as the gcd of their degrees?

user147690

E.g. $p(x)=x^{15}+...$, $o(x)=x^{20},gcd(p,o)=5$

Kaj Hansen

3:53 AM

No no. Take the gcd of two polynomials to be the monic polynomial of largest possible degree dividing both of the given polynomials.

user147690

How would you find the degree of that monic polynomial in general?

Kaj Hansen

There is an analog to the Euclidean algorithm in polynomial rings over a field @AlexClark

anon

gcd in the polynomial ring over which ring of scalars? Z? Q? R? C?

user147690

$\Bbb R[x]$

anon

in any case, given any three natural numbers $a,b,c$ with $a,b\ge c$ there exist polynomials $f(x)$ and $g(x)$ such that $\deg f=a$ and $\deg g=b$ and $\deg(\gcd(f,g))=c$. Thus, literally the only thing the degrees of $f$ and $g$ can tell us is an upper bound for the degree of their gcd.

user147690

3:58 AM

Fair enough

user147690

I suppose I better go find this E.A analog

user147690

Actually I would guess it really just is E.A with long division of polynomials?

anon

yes

Kaj Hansen

Indeed

anon

given any tuple $(f,g)$ one can replace either $f$ or $g$ by its remainder upon division by the other, then do it again and again until you get $(h,h)$. at least assuming your scalars are a field (this doesn't work in $\Bbb Z[x]$ for instance).

Remember me

4:21 AM

@Soham you can think of the powerset iterated n times of all real numbers

What do you basically mean by the question @Soham

Soham Chowdhury

5:29 AM

Anyone online right now?

When does a symmetric group have symmetric groups as subgroups?

For all $n > 3$?

Remember me

Yes I am But i cant help you with symmetric groups sorry @Soham

Mike Miller

if $n < m$ then there is a natural inclusion $S_n \subset S_m$

Soham Chowdhury

Wait, I think I didn't mean symmetric groups, exactly.

Why is there only one "multiplication table" for groups with $|G| \leq 3$?

Mike Miller

because you can write it down?

alternatively, because there is only one group of order $p$, $p$ prime, up to isomorphism

Soham Chowdhury

It seems like if the order is $4$, there is a sort of "free" subsquare in the table that can permute without destroying the "Sudoku property".

Oh, yes.

So: the trivial group is unique, and $2$ and $3$ are prime.

user147690

5:37 AM

@MikeM Irreducible polynomials in $\Bbb R[x]$ are only those with complex roots and degree 1 polynomials right?

Mike Miller

those quadratics with complex roots, yes

every polynomial in $\Bbb R[x]$ splits as a product of quadratic and linear factors

Soham Chowdhury

@Mike, so multiplication tables for groups of prime order have no subsquares that we can permute without losing "group-ness". Does that say anything about subgroups?

Mike Miller

google lagrange's theorem

Soham Chowdhury

Hmm.

Thanks.

user147690

@MikeMiller Oh I see now, thanks

Balarka Sen

6:06 AM

@SohamChowdhury almost. again, I am asking you to stop obsessing about getting to algebraic topology as fast as possible and to learn topology thoroughly.

@MikeMiller did you see what I pinged you on bordism?

Soham Chowdhury

@BalarkaSen what does that entail?

that's my question.

Balarka Sen

what does what entail?

Soham Chowdhury

"learning topology thoroughly"

Munkres part 1?

Balarka Sen

well, do a bit of Simmons and then study Munkres!

Soham Chowdhury

at what point am I ready, O master?

:P

Balarka Sen

6:08 AM

ready to do what?

Soham Chowdhury

to move on, as it were.

anyhow, I think I'll finish groups today.

which is cool.

Balarka Sen

Does Aluffi talk about classification of groups?

Soham Chowdhury

finitely-generated abelian groups?

that's in the second groups chapter, after rings and modules.

Balarka Sen

that and a bit about semidirect products, sylow's theorem, etc.

Soham Chowdhury

(I believe)

@BalarkaSen I haven't read that chapter yet.

All three Sylows are in there.

Balarka Sen

6:12 AM

eh, I prefer having all those in a single chapter.

user1667423

How can I regularize the sum from n = 1 to infinity of log(log(n))?

Soham Chowdhury

How far should I continue Simmons before starting with Munkres?

user1667423

I know how to do so for log(n) (take the derivative of the zeta function at s=0) but I'm stuck on this one.

Balarka Sen

make sure you do all those. those are the real group theory.

Soham Chowdhury

Hm, yes.

Balarka Sen

6:13 AM

@SohamChowdhury I have no idea. I've forgotten what Simmons does. I guess you will have the motivations after reading a bit about metric spaces.

Soham Chowdhury

I just wanted a clear set of prereqs for starting an algebraic topology book. I'm not obsessed with getting there as quick as possible, and I won't leave any ideas half-baked in my head.
$\hskip -0.5in \text{this} \rightarrow$"do a bit of Simmons and study Munkres" is enough on the point-set side of things then?

Remember me

@Soham If you are comfortable and know analysis then do till algebraic operators

Soham Chowdhury

Also, do you have any experience with Bredon, @BalarkaSen?

Balarka Sen

no

Soham Chowdhury

And my previous question?

Balarka Sen

6:16 AM

yes

Soham Chowdhury

good.

Remember me

@Balarka How much of algebra do i require for altop?

Anyways Hi@Balarka

Balarka Sen

meh, I am not going to suggest everyone how fast he can study everything and get to algebraic topology.

I have works to do

figure out yourself, I guess.

Remember me

Fine

Soham Chowdhury

yeah, I figured he would get pissed.

Algebra is beautiful.

Do as much as you can.

@Rem. ^

@BalarkaSen what have you done today? I did a bit of "olympiad"-ish math today morning and then started going through all the problems in this chapter.

Remember me

6:18 AM

Well homological algebra I guess Let me finish simmons first then I will go to DF and continue DF again

Balarka Sen

nothing much, yet. I have to think about a few problems I have been planning to think about for a few days.

Remember me

After that I suppose I will decide which to do Algebraic Topology or axiomatic set theory

Soham Chowdhury

I should go. Thanks for the help, @Balarka.

@Rememberme Learn set theory and then answer my question on Main :)

Remember me

I have learnt till that much

But i dont understand your question

@Soham

Balarka Sen

@Remember you don't need homological algebra to do algebraic topology. you won't get motivation for studying the former unless you do singular homology, on the contrary.

Remember me

6:20 AM

Ahh.. Thats at the far end of DF

@Soham You have left many statements undefined. When you say that In this Bijection $\Bbb{N}\rightarrow \Bbb{R}$ You say that it loses countability I dont understand what you mean by "loses countability"

And what loses countability, I tried thinking about that statement of yours and got a rough idea but i feel you cant work with that rough idea

Vrouvrou

@robjohn we have that $\Omega$ is bounded

robjohn

@Vrouvrou I know. So how does that help?

Remember me

Hello@robjohn

robjohn

@Rememberme hello

user147690

@Balarka do you think you could read over some proofs of mine later on? Just on finding all maximal ideals of $\Bbb R[x],\Bbb Z[x]$ and proving $\Bbb Z[i]$ is a PID

Vrouvrou

6:30 AM

@robjohn And $\mu$ is a finite radon measure

user147690

[the proofs aren't written yet, but I have my outlines mostly ready]

Vrouvrou

@robjohn so we have that $\mu(\Omega)<\infty $ no ?

robjohn

@Vrouvrou yes, I said that.

8 hours ago, by robjohn

I guess it would have to be...

Vrouvrou

@robjohn what is the center of mass of $\mu$

please

robjohn

Center of mass

In physics, the center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero. The distribution of mass is balanced around the center of mass and the average of the weighted position coordinates of the distributed mass defines its coordinates. Calculations in mechanics are often simplified when formulated with respect to the center of mass. In the case of a single rigid body, the center of mass is fixed in relation to the body, and if the body has uniform density, it will be located at the centroid. The center of...

Vrouvrou

6:36 AM

it is a constant

what is a constant?

center of mass ?

@AlexClark sure

@Vrouvrou $$\text{center of mass of }\Omega\text{ weighted by }\mu=\frac1{\mu(\Omega)}\int_\Omega x\,\mathrm{d}\mu$$

Balarka Sen

what were the maximal ideals of $\Bbb Z[x]$ you found?

Vrouvrou

6:40 AM

@robjohn is this is in $\overline{\Omega}$ ?

robjohn

@Vrouvrou whatever domain you want

user147690

@BalarkaSen Hmmm haven't fully got this down yet, R[x] is ready to write up I imagine, and I have an idea how to write down the proof for $\Bbb Z[i]$ being a PID

Vrouvrou

@robjohn how ?

robjohn

@Vrouvrou what do you mean "how"?

Vrouvrou

i dont understand whatever domain you want

we work on omega

robjohn

6:56 AM

@Vrouvrou Fine... use $\Omega$ or $\overline{\Omega}$. Whatever domain you want to compute the center of mass of the measure.

Remember me

8:06 AM

I think this the best question with this tag:
http://math.stackexchange.com/questions/1302183/did-guinness-book-of-records-screw-this-up#1302183

Soham Chowdhury

8:53 AM

How can I prove that $A = \left( \begin{array}{ccc}
1 & 1 \\
0 & 1 \end{array} \right)$ has infinite order in $\text{GL}_2\mathbb{R}$?

nTuply

Hey everyone. Can someone please answer this question for me.

0

Q: Help understanding the range and kernel of a linear transformation

I'm having some trouble understanding the Range and Kernel of a linear transformation. The definition goes as follows: Let $T:V \longrightarrow W$ be a linear transformation. Define the sets $ker(T) = \{v\in V : T(v) =0\}$, $R(T) =\{w\in W : w=T(v)\; for\; some \; v\in V\}$. $ker(T)$ is the k...

Soham Chowdhury

I missed the brackets, sorry.

I tried using determinants, but that doesn't help in this case because $A$ has a det of 1.

Balarka Sen

@SohamChowdhury what does it mean to have infinite order?

Soham Chowdhury

No power is equal to the identity.

$I_2$, in this case.

Balarka Sen

right. so can you find an explicit formula for $A^n$?

using induction, maybe?

Soham Chowdhury

9:06 AM

Wait.

Balarka Sen

and then show that $A^n \neq I$

Soham Chowdhury

Is this the Fibonacci matrix?

Balarka Sen

who cares about big names

2

Soham Chowdhury

Shiiiit.

Balarka Sen

just find out an explicit formula for $A^n$. compute $A^2, A^3, A^4$ first.

and then induct something out.

Soham Chowdhury

9:09 AM

Oh. $\left(\begin{array}{ccc} 1 & n \\ 0 & 1\end{array}\right)$

Balarka Sen

mhm

Soham Chowdhury

So, not the Fibonacci matrix. (Which is cool btw, you can use it to prove Cassini's identity, I found.)

Thanks

For odd-order $g \in G$, does $g^2$ also have odd order?

I guess so.

Balarka Sen

prove it.

Soham Chowdhury

I'm doing that rn

Balarka Sen

@SohamChowdhury didn't even know that thing was called fibonacci matrix

Soham Chowdhury

9:15 AM

I'm not sure it is called that, but what else can you call it if not that?

Bonaccir chheler matrix?

Okay, I think I got a proof.

Balarka Sen

why would you need to call it anything in the first place? :P

Remember me

@BalarkaSen Right on the money :p

Soham Chowdhury

So let $0\leq r < |g|.$
$g^{2|g| - r} = g^{2|g|}\cdot g^{-r} = g^{-r} = (g^{-1})^r = e.$
This implies that $|g^{-1}| < |g|$.
This is a contradiction, except if $r = 0$. This implies $|g^2| = |g|$.

Something's wrong.

@AlexClark, online?

user147690

@SohamChowdhury Yes but depressed due to non-math things, so I am taking a break

user147690

But I'll have a coffee and get back to it in 10 I suppose

Soham Chowdhury

9:25 AM

Would it be too much to ask for you to look at the "proof"?

Okay, that's awesome.

That does not imply $|g^2| = |g|$.

Soham Chowdhury

9:47 AM

I seem to have lost my mind. I just made up a multiplication table for a group of order 3 with elements of order 2.

Isn't this impossible?

Remember me

Well I am happy today!!

I dont know how

Soham Chowdhury

Good.

Oh, never mind.

It was wrong.

user147690

@Rememberme Why are you happy today?

Remember me

I have got a 10 in my school results@AlexC

Soham Chowdhury

@Rememberme Awesome.

@AlexClark Help me, I seem to be going mad.

user147690

9:53 AM

@Rememberme 10 out of what?

user147690

@SohamChowdhury I can try, but maybe madness is contagious!

Soham Chowdhury

Shit.

user147690

Sure you don't mean \hrule

user147690

Nvm they are both things

Soham Chowdhury

So, essentially this is a group multiplication table.

$\begin{array} - & e & a & b \\ e & e & a & b \\ a & b & e & a \\ b & a & b & e\end{array}$

All the elements seem to have order 2.

What's wrong?

user147690

9:56 AM

So it's non commutative

Soham Chowdhury

@AlexC?

But a group of order 3 can't have elements of order 2.

user147690

What is your identity?

Soham Chowdhury

Lagrange's theorem.

e.

user147690

Looks like you don't have one

user147690

Check your identity again

Soham Chowdhury

9:57 AM

Oho.

Right.

Told you I was going mad.

I need a coffee too, I guess.

user147690

Haha that's fine

Soham Chowdhury

"If $|g|$ is odd, what can you say about $|g^2|$?"

I'll work on this now.

user147690

What is $g$?

Soham Chowdhury

An element.

user147690

Enjoying group theory? I think it is one of the better topics. Are you working from Aluffi for this?

Soham Chowdhury

10:01 AM

Yes, I am. Yes, I am.

But brief periods of incompetence are painful and disturbing. You'd know, from the other day.

Oh, it's an easy proof.
$$|g^2| = \frac{|g|}{\gcd(2, |g|)} \overset{|g| \;\text{is odd}}{=} \frac{|g|}{1} = |g| $$.

@BalarkaSen

user147690

10:23 AM

Wow so empty here, 6 members

Balarka Sen

Now it's 7.

Soham Chowdhury

10:36 AM

Is that correct?

Paging @BalarkaSen

Balarka Sen

yep

Soham Chowdhury

Mmm.

"If g and h commute and the gcd of their orders is 1, the order of their product is the product of their orders."

I got till $|gh| = k|g||h|$.

user147690

So the only maximal ideals of $\Bbb R[x]$ are of the form $(x-a)$ and $(x^2-2ax+(a^2+b^2))$

Balarka Sen

Fact : If $g, h$ commutes then $(gh)^n = h^ng^n$

Prove this.

@AlexClark Yes.

Soham Chowdhury

I can't do $(gh)^n = g^nh^n$, obviously.

user147690

10:39 AM

@BalarkaSen Feels kinda anticlimatic

Balarka Sen

Wait, no, what about $x^4 + 1$?

Irreducible polynomials over $\Bbb R$ are surely not just quadratics and linear things.

Soham Chowdhury

Can I use that, @Balarka?

user147690

@BalarkaSen Over $\Bbb R[x]$ the maximals seem to be

Balarka Sen

@SohamChowdhury Well, why not?

Soham Chowdhury

Then $(gh)^n = (hg)^n = h^ng^n$.

Balarka Sen

10:41 AM

$(x^4 + 1)$ is maximal, but yet not what you say it is.

Yes, @Soham. What's the contradiction?

user147690

@BalarkaSen $x^4+1=-(x^2+\sqrt{2}x-1)(x^2+\sqrt{2}x+1)$ so it is generated by $(x^2+\sqrt{2}x+1)$

Soham Chowdhury

No, I thought I would have to prove that separately.

Balarka Sen

Yikes, yes, need coffee.

@SohamChowdhury You're not making sense. I said that $g$ and $h$ commutes. So $(gh)^n = g^nh^n = h^ng^n$

Soham Chowdhury

And that's easy too. $(gh)^n = ghghgh\cdots = gggg\cdots hhhh\cdots = g^nh^n$ if gh commute.

No, I thought I couldn't assume $(gh)^n = g^nh^n$. That's why I said that.

Balarka Sen

Right. So if $n$ is the order of $gh$, then $n$ is the order of $g$ and $h$ too.

Done.

@SohamChowdhury What's to assume? It's a provable fact.

Soham Chowdhury

10:44 AM

How is that, now?

@BalarkaSen Yes, I thought I had to supply the proof.

Balarka Sen

I don't get whatever you're trying to say.

Soham Chowdhury

No, it's like what happened yesterday with norms.

I was sort of assuming the answer and saying something like "we know that x is true, so x."

Anyway, miscommunication can't be helped sometimes.

@BalarkaSen "How is that?" was in reference to the fact that I didn't understand this message. (linked)

How does that follow from the previous fact?

Oh, I see.

Balarka Sen

no, that was a typo.

OK, so what was the thing we wanted to prove again?

user147690

Usually when my conversations get so muddled I start again lol

Soham Chowdhury

@BalarkaSen $(gh)^n = g^nh^n$ for commuting $g,h$.

Balarka Sen

10:48 AM

No, no that was the hint I have you.

Soham Chowdhury

$|gh| = |g||h|$.

Balarka Sen

What was your problem?

Right.

So assume $|g| = n$, $|h| = m$

We know that $(gh)^k = g^kh^k$.

Soham Chowdhury

Should I continue?

Balarka Sen

Sure.

Soham Chowdhury

I'm not really sure what to do. I can show that $(gh)^{mn} = e$: $(gh)^n = g^nh^n = h^n \implies (gh)^{mn} = (h^n)^m = e$.

But how is it minimal?

Balarka Sen

10:52 AM

Start by assuming it's not.

Soham Chowdhury

The division algorithm trick?

Let $0\leq k<mn$.

Balarka Sen

We already know that $(gh)^{mn} = 1$. If $(gh)^k$ for some $k$ smaller than $mn$, then $k | mn$ must hold.

Recall that you have some restriction on $m, n$

Use that.

Soham Chowdhury

Um, then, wlog, $k | m$? Because $\gcd(m,n) = 1$?

I have no clue what to do from there.

Balarka Sen

Think, think.

Soham Chowdhury

Am I moving in the right direction?

Balarka Sen

11:00 AM

Yes, you are.

Soham Chowdhury

I had copied this text from a while back. Don't laugh.
$(gh)^{mn-k} = e \implies gh^{mn}\cdot((gh^{-1})^k) = e \implies |gh| = k$. I can replace ${mn}$ by $k$ and find $k_1$, and so on. Infinite descent?

I'm thinking.

btw, we haven't even proved Lagrange's theorem yet in the text.

Balarka Sen

You don't need Lagrange to do this.

Soham Chowdhury

Not Lagrange, but isn't $|g|$ divides $|G|$ a corollary or something?

We haven't proved it yet, anyhow.

Balarka Sen

Where have I used that?

Soham Chowdhury

Oh, never mind.

$k|m$. What next?

Balarka Sen

11:09 AM

Anyway, you're thinking too hard. $k | m$, so $e = (gh)^k = g^k h^k$. This means $gh = 1$. However, that contradicts $(|g|, |h|) = 1$

Remember me

@AlexClark 10 ... This is something we call cgpa cumalative grade point nonsense I don't know the last word so that is out of 10 which we get when we finish a whole year which is out of 10 in that I got a 10

user147690

@Rememberme average lol(I mean it is cumulative grade point average). Good work!

Remember me

Yes some shit

Soham Chowdhury

@AlexClark Whoa, you risked being misunderstood there lol

user147690

@SohamChowdhury Yes I realised when I finished reading whoops!

Remember me

11:10 AM

Right

user147690

Sorry if I offended you for a second there Sayan haha

Balarka Sen

Equivalently, note that $g^k \neq 1$ as $k < n$, and $h^k \neq 1$ as $(|h|, |g|) = 1$. So $g^kh^k = 1$ is absurd.

Remember me

This might seem idiotic but o will call two open spheres the same of the radius of both of them are same

Right

Soham Chowdhury

@BalarkaSen Because $gh = 1$ implies both have equal orders?

Remember me

Or nonsense

Balarka Sen

11:12 AM

@SohamChowdhury You really should think about a problem (for hours, yes) before asking questions in here.

yes, @Soham

user147690

I have no idea what you just said @Rem

Soham Chowdhury

@BalarkaSen I know, I know :(

user147690

"This might seem idiotic but o will call two open spheres the same of the radius of both of them are same"

Soham Chowdhury

"o" $\mapsto$ "we"

Remember me

I am on my phone so sorry for bad grammar

Soham Chowdhury

11:13 AM

Hammer?

user147690

" Of" $\mapsto$ "if"?

Balarka Sen

yes, your hrammer has always been mucked up

user147690

What do you mean by open spheres? Open balls?

user147690

Are they located at the exact same point?

Remember me

Nope

user147690

11:14 AM

They are at different points, then by what means do you mean call them the same?

Remember me

Well I thought this because we call two circles equal if they are of the same radius

Soham Chowdhury

@BalarkaSen What if $g$ and $h$ are inverses of each other?

Remember me

So I thought that might imply to open spheres

user147690

@Rememberme I don't even know what it means to name them haha

Balarka Sen

@SohamChowdhury Can't be : we assumed $(|g|, |h|) = 1$. That's the whole point.

user147690

11:16 AM

In my mind they should only be 'equal' if they are the same, e.g. same centre and same radius

Remember me

Well I will ask you again when I am on my pc

Soham Chowdhury

@BalarkaSen What if they're both = 1?

We assume they're different, then?

user147690

@Rememberme Ok

Remember me

Typing on the phone is literally difficult

user147690

@Rememberme I know haha, always frustrates me

Balarka Sen

11:17 AM

Everything gets muddled up when you assume they are identity :P

Remember me

Again wrong grammar

Balarka Sen

Treat that case separately.

Soham Chowdhury

@BalarkaSen Ah, now I feel like I understand. Thanks a ton, man.

Remember me

Okay well my battery is going to go dead talk later

Balarka Sen

Just a suggestion : it'd be better if you do a bit of elementary number theory before doing algebra deeply.

Not a lot : pick up Niven-Zuckerman-Montgomery and do exercises from the first few chapters.

Remember me

11:19 AM

Yes @SohamChowdhury balarka is right over there

Soham Chowdhury

@BalarkaSen I have Burton.

Will that do?

Balarka Sen

doesn't have good exercises, I think

Remember me

Well exercises are very easy in it

Soham Chowdhury

Okay.

Remember me

Also when you go ahead in it it feels like burton rushes

Soham Chowdhury

11:25 AM

I really need to brush up on elementary NT anyhow, so that's actually a very good thing you reminded me of.
I have a lot of trouble following that kind of proof, even though I'm familiar with the material to some extent.

Soham Chowdhury

11:50 AM

Last exercise!
Let $G$ be an abelian group, and let $g$ be an element of max finite order. Show that for any other $h$ with finite order, $|h|$ divides $g$.
@BalarkaSen, here's my proof.

The hint is this: "calculate $|g^{p^m}h^s|$".

$$|g^{p^m}h^s| = |g^{p^m}||h^s| \\
= \frac{|g|}{\gcd(|g|,p^m)} \times \frac{|h|}{\gcd(|h|,s)} \\
= \frac{|g||h|}{p^ms} = p^nr$$.
$r > s \implies p^nr > p^mr$, so this element has order greater than $g$, contradiction.

We are considering $g=p^mr, h=p^ms$ for prime p and relatively prime $r,s$ and $m<n$.