Author 
Topic: Lagrange's Theorem (Read 851 times) 

Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Lagrange's Theorem
« on: Sep 20^{th}, 2007, 7:29pm » 
Quote Modify

Everybody knows that if a finite group G has a subgroup of order n, then n divides G. For which n is the converse true? That is, classify those integers n for which the following statement is true: If n divides G for some finite group G, then G has a subgroup of order n. (One direction is well known. The other took me a while!) An easier problem (though it doesn't seem to help in solving the first one) is to classify the n for which the following holds: if n divides the order of a finite group G, then G has a subgroup of index n.

« Last Edit: Sep 21^{st}, 2007, 4:59am by Eigenray » 
IP Logged 



JiNbOtAk
Uberpuzzler
Hana Hana No Mi
Gender:
Posts: 1187


Re: Lagrange's Theorem
« Reply #1 on: Sep 20^{th}, 2007, 9:35pm » 
Quote Modify

on Sep 20^{th}, 2007, 7:29pm, Eigenray wrote:Everybody knows that if a finite group G has a subgroup of order n, then n divides G. 
 Okay, I have to disagree. I for one, do not know that. ( I have trouble understanding it, never mind knowing it )


IP Logged 
Quis custodiet ipsos custodes?



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: Lagrange's Theorem
« Reply #2 on: Sep 20^{th}, 2007, 9:55pm » 
Quote Modify

on Sep 20^{th}, 2007, 9:35pm, JiNbOtAk wrote:Okay, I have to disagree. I for one, do not know that. ( I have trouble understanding it, never mind knowing it ) 
 This fact goes by the name Lagrange's Theorem, hence the title of the thread. It is one of the most basic theorems of group theory. Not to discourage you, but if you're not familiar with it, you probably won't make much progress on this problem without significant study. It took me three years


IP Logged 



Sameer
Uberpuzzler
Pie = pi * e
Gender:
Posts: 1261


Re: Lagrange's Theorem
« Reply #3 on: Sep 20^{th}, 2007, 10:09pm » 
Quote Modify

on Sep 20^{th}, 2007, 9:55pm, Eigenray wrote: This fact goes by the name Lagrange's Theorem, hence the title of the thread. It is one of the most basic theorems of group theory. Not to discourage you, but if you're not familiar with it, you probably won't make much progress on this problem without significant study. It took me three years 
 I was studying and was planning to do research in error coding... which also involves a lot of this subject and I still don't understand it properly... just to take a wild wild guess.. 2^n 1 has to be prime number?


IP Logged 
"Obvious" is the most dangerous word in mathematics. Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself. Sir Arthur Eddington, quoted in Bridges to Infinity



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: Lagrange's Theorem
« Reply #4 on: Sep 20^{th}, 2007, 10:22pm » 
Quote Modify

on Sep 20^{th}, 2007, 10:09pm, Sameer wrote:I was studying and was planning to do research in error coding... which also involves a lot of this subject and I still don't understand it properly... just to take a wild wild guess.. 2^n 1 has to be prime number? 
 That is sufficient. But it is not necessary. For example, it holds for n=4,8,9, and 11. Why?


IP Logged 



Sameer
Uberpuzzler
Pie = pi * e
Gender:
Posts: 1261


Re: Lagrange's Theorem
« Reply #5 on: Sep 20^{th}, 2007, 10:43pm » 
Quote Modify

I have a vague recollection of something involving a number with product of primes and the something about the power of those prime products.. gotta dig up the book and find out..


IP Logged 
"Obvious" is the most dangerous word in mathematics. Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself. Sir Arthur Eddington, quoted in Bridges to Infinity



JiNbOtAk
Uberpuzzler
Hana Hana No Mi
Gender:
Posts: 1187


Re: Lagrange's Theorem
« Reply #6 on: Sep 21^{st}, 2007, 12:20am » 
Quote Modify

I've heard of Lagrange, just not his theorem. Maybe I should have attended more of my Engineering Maths class back in uni. on Sep 20^{th}, 2007, 10:22pm, Eigenray wrote: That is sufficient. But it is not necessary. 
 Now what does this mean ? It's enough, but not needed ? <Note to self : It's never a good idea to break away from one's natural habitat.>


IP Logged 
Quis custodiet ipsos custodes?



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: Lagrange's Theorem
« Reply #7 on: Sep 21^{st}, 2007, 12:43am » 
Quote Modify

on Sep 21^{st}, 2007, 12:20am, JiNbOtAk wrote:Now what does this mean ? It's enough, but not needed ? 
 X sufficient for Y: X > Y X necessary for Y: not X > not Y


IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: Lagrange's Theorem
« Reply #8 on: Sep 21^{st}, 2007, 12:46am » 
Quote Modify

on Sep 20^{th}, 2007, 10:22pm, Eigenray wrote:That is sufficient. But it is not necessary. For example, it holds for n=4,8,9, and 11. Why? 
 I would guess n has to be a power of a single prime. But then, mathematics isn't a guessing game. I'm thinking about taking algebra next quarter; maybe I'll finally understand all this group theory a bit then.


IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



ThudnBlunder
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489


Re: Lagrange's Theorem
« Reply #9 on: Sep 21^{st}, 2007, 2:46am » 
Quote Modify

on Sep 21^{st}, 2007, 12:46am, towr wrote: I'm thinking about taking algebra next quarter; 
 Now that you have finished your thesis, aren't they kicking you out into the real world?


IP Logged 
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: Lagrange's Theorem
« Reply #10 on: Sep 21^{st}, 2007, 4:02am » 
Quote Modify

on Sep 21^{st}, 2007, 12:46am, towr wrote:I would guess n has to be a power of a single prime. 
 That is exactly correct. Now all you have to do is prove it!


IP Logged 



Barukh
Uberpuzzler
Gender:
Posts: 2276


Re: Lagrange's Theorem
« Reply #11 on: Sep 21^{st}, 2007, 4:28am » 
Quote Modify

I want to understand the question. Are we trying to find all n such that for every finite group G, if n  G, then G has a subgroup of order n?


IP Logged 



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: Lagrange's Theorem
« Reply #12 on: Sep 21^{st}, 2007, 4:34am » 
Quote Modify

on Sep 21^{st}, 2007, 2:46am, ThudanBlunder wrote:Now that you have finished your thesis, aren't they kicking you out into the real world? 
 No, not yet. I'm still working on a bachelor's degree in philosophy as well (which I hope to complete within 9 months or so).


IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: Lagrange's Theorem
« Reply #13 on: Sep 21^{st}, 2007, 4:57am » 
Quote Modify

on Sep 21^{st}, 2007, 4:28am, Barukh wrote:I want to understand the question. Are we trying to find all n such that for every finite group G, if n  G, then G has a subgroup of order n? 
 Yes that is it. And the same question with 'order n' replaced by 'index n' (although the proofs do not seem to have much in common).


IP Logged 



ThudnBlunder
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489


Re: Lagrange's Theorem
« Reply #14 on: Sep 21^{st}, 2007, 5:04am » 
Quote Modify

on Sep 21^{st}, 2007, 4:34am, towr wrote: No, not yet. I'm still working on a bachelor's degree in philosophy as well... 
 I know Abstract Algebra is useful for philosophy; but Group Theory? I suppose it could come in useful in the classification of your wacky, anythingispossible theories which may give your tutors a few grey hairs.

« Last Edit: Sep 21^{st}, 2007, 5:08am by ThudnBlunder » 
IP Logged 
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: Lagrange's Theorem
« Reply #15 on: Sep 21^{st}, 2007, 6:17am » 
Quote Modify

on Sep 21^{st}, 2007, 5:04am, ThudanBlunder wrote:I know Abstract Algebra is useful for philosophy; but Group Theory? 
 Well, I only need ~30 credits of philosophy, spread over a 60 credit year, so I have time left for some extracurricular courses.

« Last Edit: Sep 21^{st}, 2007, 6:18am by towr » 
IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



Michael Dagg
Senior Riddler
Gender:
Posts: 500


Re: Lagrange's Theorem
« Reply #16 on: Sep 21^{st}, 2007, 6:57am » 
Quote Modify

on Sep 20^{th}, 2007, 7:29pm, Eigenray wrote: (One direction is well known. The other took me a while!) 



IP Logged 
Regards, Michael Dagg



Barukh
Uberpuzzler
Gender:
Posts: 2276


Re: Lagrange's Theorem
« Reply #17 on: Sep 21^{st}, 2007, 7:09am » 
Quote Modify

After inspecting my home math library, seems like it's precisely the Sylow's theorem.


IP Logged 



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: Lagrange's Theorem
« Reply #18 on: Sep 21^{st}, 2007, 12:43pm » 
Quote Modify

on Sep 21^{st}, 2007, 7:09am, Barukh wrote:After inspecting my home math library, seems like it's precisely the Sylow's theorem. 
 That will only help in one direction. And even then, the statement of the theorem is not quite enough.


IP Logged 



Barukh
Uberpuzzler
Gender:
Posts: 2276


Re: Lagrange's Theorem
« Reply #19 on: Sep 22^{nd}, 2007, 3:15am » 
Quote Modify

on Sep 21^{st}, 2007, 12:43pm, Eigenray wrote: That will only help in one direction. 
 Correct. Quote:And even then, the statement of the theorem is not quite enough. 
 How come? Sylow theorem states: "If n is a power of a prime, and n  G, then G has a subgroup of order n."

« Last Edit: Sep 22^{nd}, 2007, 3:15am by Barukh » 
IP Logged 



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: Lagrange's Theorem
« Reply #20 on: Sep 22^{nd}, 2007, 6:21am » 
Quote Modify

It guarantees a subgroup of order p^{k}, where p^{k} is the largest power of p dividing G. But you also need subgroups of order p^{i} for all i < k. (Of course, this may follow from the proof of the theorem.)


IP Logged 



Obob
Senior Riddler
Gender:
Posts: 489


Re: Lagrange's Theorem
« Reply #21 on: Sep 22^{nd}, 2007, 7:00am » 
Quote Modify

Let us prove that if a prime power divides the order of a group G, then G has a subgroup of that order. Since the Sylow theorems guarantee a subgroup of order p^{k}, it is enough to show that any pgroup G has subgroups of each prime power order. Let g be any nonidentity element whatsoever. By Lagrange, the order of g is a prime power. Then some suitable power of g has order p. Now consider G/<g>. By induction on the order of G, G/<g> has subgroups of every order dividing G/<g>. But subgroups of G/<g> are in bijective correspondence with subgroups of G containing <g>. So G has subgroups of each prime power order.


IP Logged 



Barukh
Uberpuzzler
Gender:
Posts: 2276


Re: Lagrange's Theorem
« Reply #22 on: Sep 22^{nd}, 2007, 9:41am » 
Quote Modify

on Sep 22^{nd}, 2007, 6:21am, Eigenray wrote:It guarantees a subgroup of order p^{k}, where p^{k} is the largest power of p dividing G. But you also need subgroups of order p^{i} for all i < k. (Of course, this may follow from the proof of the theorem.) 
 This is a statement of the existence of Sylow psubgroup. All three books on Abstract Algebra (Johnson, Herstein, Fraleigh) state and prove Sylow's first theorem as I described it before.

« Last Edit: Sep 22^{nd}, 2007, 9:42am by Barukh » 
IP Logged 



ecoist
Senior Riddler
Gender:
Posts: 405


Re: Lagrange's Theorem
« Reply #23 on: Sep 22^{nd}, 2007, 3:40pm » 
Quote Modify

I hope this is a partial proof and not swiss cheese. Let n be a positive integer not a power of a prime. We construct a group G of order a multiple of n which contains no subgroup of order n. There exists a prime p dividing n such that n=am, where a is a power of p, m is prime to p, and a < m. Let F be the field Z_{p} extended by a primitive mth root of unity, say x. Then m divides F1; whence F>a. Let V be the additive structure of F and let M=<x>. Then M acts faithfully and irreducibly on V. Let G be the corresponding semidirect product of M and V. Suppose that H is a subgroup of G of order n. Then, W = (V intersect H), is a proper subgroup of V and a nontrivial normal subgroup of H. Thus the normalizer of W in G contains a Sylow subgroup of G, for each prime dividing the order of G. Hence W is normal in G. Therefore, M fixes W, a nontrivial proper subgroup of V, contradicting that M acts irreducibly on V. Hence G has no subgroup of order n.

« Last Edit: Sep 22^{nd}, 2007, 3:43pm by ecoist » 
IP Logged 



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: Lagrange's Theorem
« Reply #24 on: Sep 22^{nd}, 2007, 6:35pm » 
Quote Modify

on Sep 22^{nd}, 2007, 7:00am, Obob wrote: But this is not a group. on Sep 22^{nd}, 2007, 9:41am, Barukh wrote:All three books on Abstract Algebra (Johnson, Herstein, Fraleigh) state and prove Sylow's first theorem as I described it before. 
 The usual statement of Sylow's theorem is about maximal psubgroups. on Sep 22^{nd}, 2007, 3:40pm, ecoist wrote:I hope this is a partial proof and not swiss cheese. 
 That's exactly what I had! Note that this is a special case.


IP Logged 



