Mathematics - 2014-02-05 (page 1 of 3)

usukidoll

00:06

yo yo

anyone?? hears crickets

Alizter

chirp chirp

usukidoll

very funny. I'm trying to improve a short proof paper

it's from a question I asked earlier. Prove $(A \backslash B)' \backslash (B \backslash A)' = B \backslash A$

Definition 3.3.1 states that $A$ and $B$ are sets. The complement of $B$ relative to $A$, written $A \backslash B$ is the set
$ A \backslash B =[x:x \in A \land x \notin B]$

of course what I'm given are negation versions of this def.

$(A \backslash B)' =[x:x \notin A \lor x \in B]$

$ ( B \backslash A ) = [x: x \in B \land x \in A]$

$(B \backslash A)' =[x:x \notin B \lor x \in A]$

now apparently I have to apply def. 3.3.1 again to $(A \backslash B)' \backslash (B \backslash A)'$ to get $B \backslash A$, but I end up in a very nasty mess.

maybe I could use a set property which is $A \backslash B = A \cap B'$

-.-

so well let's work this one at a time. $(B \backslash A) = B \cap A'$

but you can clearly see that it's not the negation versions

robjohn

Darn. I got 5 upvotes just before the day change and they got eaten by the rep cap

usukidoll

@robjohn do you know how I can improve this proof that I posted ^

robjohn

@usukidoll $$ \begin{align} (A\setminus B)'\setminus(B\setminus A)' &=(A\cap B')'\cap(B\cap A')\\ &=(A'\cup B)\cap(B\cap A')\\ &=((A'\cup B)\cap B)\cap A'\\ &=B\cap A'\\ &=B\setminus A \end{align} $$

usukidoll

00:22

no umm ... I'm rewriting it

like a revision...

Pedro Tamaroff

@anon

anon

?

Pedro Tamaroff

Does SL/Z(SL) have a name?

Alizter

@PedroTamaroff Thanks for getting me excited.

Pedro Tamaroff

@Alizter Hehe, sorry.

I mean ${\rm SL}_n(\Bbb F)/Z({\rm SL}_n(\Bbb F))$

If the ascii wasn't clear enough.

usukidoll

00:30

@robjohn where did you get the intersection for $\backslash$

anon

Z(SL) is comprised of scalar multiples of the identity IIRC

robjohn

@usukidoll $A\setminus B=A\cap B'$

Pedro Tamaroff

@anon Then it is $\rm PSL$?

anon

yeah

usukidoll

not that one @robjohn the one for $(A \backslash B) \backslash$

Pedro Tamaroff

00:31

@anon OK.

usukidoll

that backslash after (A\B)'

robjohn

@usukidoll It is the same thing.

usukidoll

oh I see it much clearer this way.. thanks @robjohn

should've use set properties in the first place ugh

robjohn

01:06

@usukidoll I see that this is suggested in this hint

usukidoll

yeah

but I'm new at this so I didn't know that I could use set properties to make it easier to prove something...I've been doing definitions most of the time

like the prove that a is any set

that was just a definition.

and since it is a is a subset of ia

but the thing that irritates me sometimes is that I submit a proof assignment and it gets rejected like geez there are one liner proofs out there. What is there to write?

math.stackexchange.com/questions/658072/… this is a one or two liners MAX

a whole paragraph on something so small like this is insane

Pedro Tamaroff

01:21

@anon

@Mike YO.

Mike

wat

Pedro Tamaroff

I am proving erry finite abelian group has a subgroup of order $d$ for each divisor $d$ of $|G|$.

Now, I did a part, but I am missing another.

I prove it by induction on $|G|$. There is nothing to prove if $|G|=1$ or $=p$ a prime, so assume $|G|>1$ is composite.

Now choose a prime factor $p$, say $|G|=np$; so that $1<n<|G|$, and by Cauchy obtain an elt of order $p$.

Then $|G/\langle p\rangle|=n$. Moreover, every divisor of $|G|$ is of the form $dp^i,i=0,1$ for a divisor $d$ of $n$.

By induction, we have a subgroup $\bar H\leqslant G/\langle p\rangle$ of order $d$ for every $d\mid n$, and by the lattice iso, $\bar H=H/\langle p\rangle$ for $H$ a subgroup of $G$ containing $\langle g\rangle$.

Then $H$ has order $pd$.

@Mike I am missing the subgroups of order $d=p^0d$.

@anon @Mike has forsaken me.

Mike

01:40

@Pedro I do that all the time to you.

anon

just vary p

sorry, was shoveling

Mike

@Pedro Err... the subgroups of order $p^0d$? Those would be the trivial one. Let me make sure I'm not misreading you

Pedro Tamaroff

@anon Meaning I can now let $p$ vary throughout all prime factors of $|G|$?

Mike

Oh dih nvm

@Pedro Fix a d, pick a prime factor of d, do your argument

anon

actually pick a prime factor of n/d

Pedro Tamaroff

01:47

@anon I claimed that I can just repeat the process with all prime factors of $|G|$.

anon

sure

Pedro Tamaroff

And that gives me all divisors.

$$\|e^{it\Delta}u_0\|_{L^{q}_t L^{r}_x} = \sup_{G \in L^{q'}_tL^{r'}_x\backslash\{0\}} \left| \int_{\mathbb{R}^d} \int_\mathbb{R} G(x,t) e^{-it\Delta}u_0 \, dt \, dx \right| / \|G\|_{L^{q'}_tL^{r'}_x}.$$

DAAAAYUM.

This.

usukidoll

:O

how to be a successful proof writer

ughhhhhh I wish I knew da stepz

Pedro Tamaroff

@anon I think I understood a little about composition series, but... why are we interested in chains $1=G_0\lhd G_1\lhd\cdots\lhd G_s=G$ for which $G_{i+1}/G_i$ is abelian?

anon

it's the next best thing to being abelian

Pedro Tamaroff

02:01

I know that $G_{i+1}/G_i$ simple means $G_i$ is maximal normal in $G_{i+1}$.

$G_{i+1}/G_i$ abliean means $G_i$ contains $G_{i+1}'$.

Right?

anon

also, if a field extension has a galois group with abelian composition factors (i.e. solvable) then the extension is obtained through a tower of extensions by radical expressions

Pedro Tamaroff

@anon Heh, that comes latter. =P

I am proving subgroups and quotients of solvables are solvable.

Mike

be careful, you might summon the resident solvability by whatevers guy

Pedro Tamaroff

?

Jasper Loy

@pedro What is a phoenix complex?

Ah, someone unaccepted my answer.

I will now delete my answer, lol.

It's hard to earn a living on SE...

@usukidoll To become better at proofs, you read the proofs written by experts, so the more books you read, the better you are at writing.

usukidoll

02:26

yeah .. ....... this book I have is only the first -_-

too bad my prof goes all over the place

Jasper Loy

Actually, the course you are taking now should be done by first year math majors.

It should be the very first course they take, not the last.

But I know in many places, they are the last, lol.

usukidoll

I know :(

even my differential equations prof said that all the analysis courses are done in the first year and that was when he was a student in the Ukraine

Anthony

Helllllloooo

Can somebody explain the dedekind cuts section of this to me?

0.999...

In mathematics, the repeating decimal 0.999... (sometimes written with more or fewer 9s before the final ellipsis, or as 0.9, 0.(9), or {{nowrap|\scriptstyle\mathbf{0}.\mathbf{\dot{9}}}}) denotes a real number that can be shown to be the number one. In other words, the symbols "0.999..." and "1" represent the same number. Proofs of this equality have been formulated with varying degrees of mathematical rigor, taking into account preferred development of the real numbers, background assumptions, historical context, and target audience. Every nonzero, terminating decimal has an equal twi...

usukidoll

OH LAWLZ

Pedro Tamaroff

02:43

@Anthony WAT. We were talking about Dedekind cuts some days ago.

Anthony

It's all we're doing in my real analysis class right now.

usukidoll

barfs

Anthony

For the above link though, @PedroTamaroff, can you explain their logic from moving to a/b < 1-(1/10)^b?

Mike

dedekin cuts are for dweebs

Anthony

I really need help with this though :/

Mike

02:53

cauchy4lyfe

Pedro Tamaroff

@Mike Why not both? =D

Also, I have to prove subgroups and quotients of solvable groups are solvable.

@Mike Dat unnecessary work.

usukidoll

real analysis the worse undergrad math class ever! guest starring so many proofs and some strange notation

Pedro Tamaroff

@usukidoll One whinebulance coming up for you.

usukidoll

sends it back to @PedroTamaroff

I don't have to take that course ahahahahah but one of my classmates is and he said it sucks

lol

that's the class with 2 grad students

Pedro Tamaroff

What is the programme of the class?

What does it cover?

usukidoll

03:00

what mine? just intro to advanced mathematics like truth tables set theory induction and all the other stuff in the mix

like trail mix

every major has to take it

but that real analysis stuff ........ depends on what math you're going for

Pedro Tamaroff

That's real analysis? Hm.

usukidoll

I don't have to take it! I heard it's hard and sucks anyway

I don't like signing up for courses that aren't required... waste of time and such...

so I rather focus my energy on Intro to Advanced Mathematics which is a proof writing class required for all math majors .. get it? ^^

Pedro Tamaroff

@El'endiaStarman Is there a problem, officer?

usukidoll

lol

El'endia Starman

@PedroTamaroff Not at all. Just taking a look to see if usukidoll was in here.

Pedro Tamaroff

03:04

@usukidoll The law's onto you.

usukidoll

dude

I met this person in some christian room

Pedro Tamaroff

So?

usukidoll

so?????

Pedro Tamaroff

@El'endiaStarman So are you christian or just want to know about christianity?

El'endia Starman

@PedroTamaroff ........I'm a moderator on the Christianity site. :P

Pedro Tamaroff

03:07

@El'endiaStarman Yes, that I can tell.

El'endia Starman

@PedroTamaroff What then was the point of your question?

usukidoll

: $\geq$

Pedro Tamaroff

@El'endiaStarman Well, my question was the point itself.

Are you a christian or do you study christianity? Or both?

El'endia Starman

@PedroTamaroff Both.

I also happen to be a math major, hence why I'm here.

Speaking of which, @usukidoll, how are you doing on those proofs? Still need help?

Ramana Venkata

@robjohn Can you look at this question

Victor

03:16

@PedroTamaroff : Do you understand math.stackexchange.com/questions/664107/… right now?

usukidoll

no I'm done for now .. I just gotta read more on induction @El'endiaStarman

prof thinks we know everything ha! WE'RE JUST NEWBIES!

it's going to be like last semester with my calc iv

Mike

@Pedro I'm trying to figure out a way to conjecture something for algebraic topology without computing the kernel of giant integral matrices.

usukidoll

super easy lecture hard a** homework everyone fails the exam everyone gets C's just to be happy

wth

Pedro Tamaroff

@Victor You're saying that any theorem in geometry is theoretically solvable by a supercomputer because geometry is axiomatized. But all mathematics is.

@Mike Details?

Victor

@PedroTamaroff - But is there infinite number of axiom or finite number of axiom

for integral equation?

Pedro Tamaroff

03:20

When you say integral equation you mean something like a differential equation but with integrals, yes?

Victor

yes

Mike

@Pedro I need to "compute the homology in general of the union of cones over X with the bases identified". I want to guess what this will be then prove it.

Victor

For diophaine equation there exist infinte number of algorithm in order to solve all the possible diophaine equation, but not sure if the integration is the same

Mike

But ever case I come up with is total bull.

Victor

@PedroTamaroff

Pedro Tamaroff

03:24

@Victor I have no idea. But there isn't an algorithm to solve the general Diophantine equation.

@Mike How can I help? =P

Anyone feeling cold? Here are some tropical maths.

Mike

@Pedro I'm just whining.

Pedro Tamaroff

@Mike I know what the cone over a space is, though.

Homology... not so much.

Victor

Oh sorry, My mistakes. @PedroTamaroff But will integral equation be axiomatized ?

And what is "Any theorem in geometry is theoretically solvable by a supercomputer because geometry is axiomatized. But all mathematics is" come from?

Pedro Tamaroff

I don't understand what you mean by "will integral equation be axiomatized?".

Where are you from?

Victor

Could there be one day the integration could be axiomized. I am Chinese

Pedro Tamaroff

03:31

I don't think I understand what you mean by "axiomatizing integration."

Victor

@PedroTamaroff

axiomatizing in oder to use computer to solve all the possible integration problems

Mike

@Victor The solution to Hilbert's 10th problem show that this isn't possible for Diophantine equations.

And those have finitely many axioms.

Ted Shifrin

Boo @Mike @Pedro

Pedro Tamaroff

@TedShifrin Heart trips?

How does one prove a graph is not planar?

Mike

@TedShifrin I want to show that... "If you take $n$ of the cones over the space $X$, and take the space obtained by the union at the bases, then this space is just the wedge of $n-1$ suspensions of $X$.

Any ideas?

Or is this even true?

Pedro Tamaroff

03:44

@Mike I also know what the suspension is! Yay.

CONFETTI.

Mike

OH! I don't mean the same.

I mean homotopy equivalent to.

Ramana Venkata

Can somebody look at this question

Ted Shifrin

What's $V-E+F$ @pedro?

Pedro Tamaroff

@TedShifrin Err... faces?

Ted Shifrin

Closed loops bound faces?

Mike

03:49

@Pedro $F$ is faces.

Pedro Tamaroff

@TedShifrin Right, that was my guess.

groupprops.subwiki.org/wiki/…

Ted Shifrin

He just has vertices and edges :)

Pedro Tamaroff

There you can see the lattice.

Mike

@TedShifrin I actually just want to compute the homology of that come thing. Am I doing the Wrong Thing here?

Ted Shifrin

Use Mayer-Vietoris?

Pedro Tamaroff

03:51

Suppose I have a chain $1=G_0\lhd G_1\lhd \cdots\lhd G_{s-1}\lhd G_s=G$ that proves $G$ is solvable. I want to obtain one for $H\leqslant G$.

What if I intersect the $G_i$s with $H$?

Ted Shifrin

I don't believe your conjecture, @Mike. Maybe.

@Pedro, my war with Mhenni continues.

Pedro Tamaroff

@TedShifrin I think you should stop wasting time with him.

Just downvote and/or delete his answers when needed.

Ted Shifrin

LOL, but he's wreaking havoc ...

ah, I need 20k to delete answers ...

Pedro Tamaroff

I think you have a bit of a fixation with him, though.

Cut him some slack.

Ted Shifrin

You started it ... Only when I stumble on him ... I don't go looking.

Pedro Tamaroff

03:55

I usually bother when he does really stupid stuff.

Ted Shifrin

so what if you do intersect?

Pedro Tamaroff

Well, I still have to prove $\dfrac{G_{i+1}\cap H}{G_i\cap H}$ is abelian.

Mike

@Ted I don't have that yet. Gimme a week or so.

Pedro Tamaroff

I guess I need to use the second isomorphism theorem?

Mike

Best I have is long exact sequences and excision.

Ted Shifrin

03:56

ok, @Mike, try excision and induction.

Pedro Tamaroff

Yay, this works.

Mike

oh, excise a cone at a time. Duh.

Thanks.

Pedro Tamaroff

You performing surgery Mike?

Fancy pansy.

Ted Shifrin

Surgery is more advanced.

Pedro Tamaroff

What is excision?

Sounds like surgery...

Ted Shifrin

03:59

Surgery involves handles :)

Mike

Excision is just removing points. Passing to sunspaces.

I may never know surgery...

anon

@PedroTamaroff consider $G_{i+1}\cap H\to G_{i+1}\to G_{i+1}/G_i$

Mike

@Ted Pretty sure my conjecture is true for path-connected spaces.

Pedro Tamaroff

@anon Right, that intersection quotient embeds in the G quotient.

Mike

Homotopy equivalent = same homotopy type. And I'm about to prove the latter.

No I'm not.

That was dumb.

Pedro Tamaroff

04:05

@anon For the quotient part, I was thinking about taking the quotient of the series by $N$... but first I have to enlarge each factor by $N$ to take the quotient.

Ted Shifrin

I'm confused, but let's try it with a circle and $n=3$.

Pedro Tamaroff

And I guess another iso theorem will hop in.

Ted Shifrin

It's not Easter yet, @Pedro, but almost my birthday :)

Pedro Tamaroff

@TedShifrin Pity, 61 is not prime.

anon

@PedroTamaroff even better: every subnormal series refines to a composition series (do you see how this is applicable)?

Pedro Tamaroff

04:07

@anon What does "subnormal" series mean?

anon

each term is normal in the next, but not necessarily normal beyond

Ted Shifrin

You sure?

Pedro Tamaroff

I take refines means we can find intermediate "subseries" to make the subnormal series normal.

@TedShifrin Heh, I was thinking 51.

anon

@TedShifrin depends which ring :)

Ted Shifrin

Thanks @anon :D

anon

04:09

@PedroTamaroff to make the subnormal series a composition series (which is basically just an unrefinable / maximally refined subnormal series)

Pedro Tamaroff

@anon Yes, composition. Hehehe.

Ted Shifrin

@Mike: So what is the space for $X=S^1$, $n=3$?

Pedro Tamaroff

@anon I'm thinking how that is applicable.

I am thinking $N$ has to appear somewhere in the series, or inbetween.

anon

1<N<G is a subnormal series. just refine it to a composition series, delete the terms appearing strictly before N, and then quotient everything by N: you will obtain a composition series for the quotient G/N

Pedro Tamaroff

And I can quotient above $N$.

@anon Right, that proves every normal subgroup of $G$ is part of a composition series.

Dis for you, @Mike

Ted Shifrin

04:12

hmm, @Mike, up to homotopy type mayhaps you win ....

Mike

@Ted No, I win for spheres. I know it's true for spheres.

($S^1$ is where I formulated the conjecture.)

anon

hmm, is that from contagion pedro?

Mike

It's probably false for pathological spaces.

Ted Shifrin

hawaiian earrings, maybe not.

Pedro Tamaroff

@anon Maybe, I know it from MUSE.

Let me check.

Mike

04:14

Another source might be someplace homology disagrees with homotopy.

Klein bottle, say.

Pedro Tamaroff

@anon Seems not.

anon

ah, wwz, that's where

Pedro Tamaroff

Oh! WWZ used MUSE, now I recall.

I watched it... the whole movie was Brad Pitt saying "I am fucking awesome though I am fifty."

I LOL'd pretty hard when the Harvard doctor died.

No offense towards Harvard intended.

anon

haha. it's like they were setting him up to be some young savior and then he gets offed straight after introduction

Pedro Tamaroff

Yes, that was absolutely hilarious, and slightly realistic.

I also watched Tom Cruise's "Oblivion", and liked the story. But again the movie was too Hollywoodish. If they'd dwelled more into the story, things would have been more sci-fi like and waaay more interesting.

@anon After I refine that, I just quotient above $N$ by $N$ yes?

Mike

04:22

@TedShifrin Excision won't work in the obvious way. At least, I can't pick my excised sets to be one of the cones and its inferior.

Ted Shifrin

No, you need the upper part of the cone only?

Pedro Tamaroff

@anon Hmm, but how do I show $G/N$ is solvable though? I have a composition series, alright.

Mike

ah, so I want Z = the top point, A = whole cone.

in hatchets notations. I can explain if you want.

kiasy

Hey, any idea how to use Master Theorem on this: T(n-1)+2, I'm so confused ?? Please help.

Ted Shifrin

Nah ... I can fake it. And it's almost bedtime.

Mike

04:25

@kiasy You're gonna need to tell us more than that. I don't know what T is or the master theorem. In addition, I'd like to hear what you tried first.

@Ted Alright. Anyway thy excision should work.

kiasy

@Mike I've tried doing a=1 b=-1 d=2 but log-1 or 1 doesn't make sense...

log-1 of 1*

Mike

Ok, that's still completely useless to me, but meh.

Pedro Tamaroff

@kiasy Ask on main.

Mike

@Ted Actually I don't think this approach will work at all, I can't recover the homology of the full space from this.

anon

@PedroTamaroff the composition factors of G/N are quotients of composition factors of G

Pedro Tamaroff

04:29

@anon Ah.

Mike

I'm dumb. I don't need excision.

This is a simple application of the long exact sequence.

anon

@PedroTamaroff :)

Pedro Tamaroff

You mean to use $(G/N)/(H/N)\simeq G/H$, yes?

@anon But, wait.

anon

err, no, not quotients, what am I saying. I just showed the composition factors of G/N are a submultiset of the factors of G

Ted Shifrin

For a pair, you mean, @Mike?

anon

04:30

@PedroTamaroff There's more?

Mike

@Ted Yeah. X will be our full coned thing, its sunspace will be... uh...

whatever, I can figure it out

Pedro Tamaroff

Suppose I refine $1\lhd N\lhd G$ to $1=G_0\lhd G_1\lhd \cdots \lhd G_j= N\lhd \cdots \lhd G_s=G$, @anon

anon

mmhmm

kiasy

@PedroTamaroff

Mike

The subspace should be open so that it's a "good pair" in general.

kiasy

04:32

0

Q: Master theorem questn

I need to solve the following: T(n)=T(n-1)+8 I've tried doing a=1 b=-1 and d=8 but log(base -1) of 1 doesn't make sense. Any suggestions?

ramanujans-master-theorem

Pedro Tamaroff

I have no guarantee $G_{i+1}/G_i$ will be abelian.

anon

G was assumed solvable no? I thought you were trying to show G solv => G/N solv

Ted Shifrin

you can usually thicken closed things just fine, @Mike.

Pedro Tamaroff

@kiasy T(n)=T(n-1)+8 gives T(n)-T(n-1)=8. This means by telescopy that T(N)-T(0)=8N, dear @kiasy

@anon Yes.

Mike

@Ted The theorem doesn't say X a CW complex or anything pleasant. So I'd better not make assumptions.

Pedro Tamaroff

04:34

But being solvable means every composition series has abelian factors? @anon As I have it, solvable means there is a subnormal series with abelian factors.

Ted Shifrin

What theorem?

Mike

Brain typo. The problem.

Ted Shifrin

Oh ...

Mike

It's also not assumed our spaces are hausdorff or anything. Hatcher clarifies.

So we make no assumptions about this hellish space.

Ted Shifrin

nah, it's a neighborhood in the cone that deformation retracts fine, methinks.

Pedro Tamaroff

04:36

@Mike Damn. That's fucked up.

anon

@PedroTamaroff if there is a subnormal series with abelian factors then any composition series it refines to will have cyclic factors. since every composition series has the same multiset of composition factors, we're done.

that invokes jordan-holder of course

Mike

@Ted I can probably just pick my pair to be X, a cone minus its base.

That's open, the only issue is that X/that seems like it would be ba.

Ted Shifrin

hatcher drives me nuts with his covering map definitions. No one can figure out why he's obtuse.

Pedro Tamaroff

@anon Hmm... maybe I'll stick to multiplying by $N$ and taking the quotient.

anon

ugh

Mike

04:37

oh, I'm dumb.

kiasy

thanks @ped

Pedro Tamaroff

@anon Don't hate me. =)

kiasy

@PedroTamaroff ill look at that

Mike

I can pick a closed subset of the clone. There's an obvious deformation retract.

Hmm... no. That gives me Nothing. Whatever. I can figure it out.

Ted Shifrin

Night @Mike, @Pedro, @anon

Pedro Tamaroff

04:42

Bubyes.

Mike

How the hell can you people tell what T is in that question?

kiasy

we r cool

anorton

@Mike recurrence solving comes with practice. :-) Also: For people who want to slowly work on retagging a few more questions: ramanujans-master-theorem is a very misused tag.

(The Master Theorem is used for asymptotic behavior of recurrence relations. But, the Ramanujan Master Theorem is something entirely different.)

Jasper Loy

@pedro Now I know what a phoenix is, lol.

Pedro Tamaroff

@anon I honestly don't follow this.

=/

anon

04:55

which part?

Pedro Tamaroff

Well, we're assuming we have a subnormal series 1 = G_0 < G_1 < ... < G_s = G with abelian factors, yes? And then we take a normal subgroup N of G; and we want to show G/N admits such a subnormal series.

Mike

@anorton Master theorem is a stupid name for a theorem.

Pedro Tamaroff

You told me first to take the subnormal seriess 1 < N < G and refine it to a composition series.

Jasper Loy

So I gave a proof in the answer, and then OP asks for a proof not knowing that my proof is a proof, sigh........

Pedro Tamaroff

Say I refine it to 1 < N_0 < N_1 < ... < N_s=N< H_0 < H_1< ... < H_s=G

I am trying to see how "if there is a subnormal series with abelian factors then any composition series it refines to will have cyclic factors. since every composition series has the same multiset of composition factors, we're done." takes the wheel after I've done this.

That is, I don't know how to apply your comment to my problem.

anon

04:59

(1) If 1<...<G is subnormal with abelian factors it refines to a comp series (call it A) with cyclic factors.
(2) 1<N<G refines to a composition series (call it B)
(3) the factors in A and B are the same (Jordan-Holder)
(4) the factors in N<...<G are the same as in N/N<...<G/N
(5) conclusion: the series N/N<...<G/N have cyclic factors

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$Mathematics$ Mathematics