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Eigenray
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Lagrange's Theorem
« on: Sep 20th, 2007, 7:29pm » |
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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.
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| « Last Edit: Sep 21st, 2007, 4:59am by Eigenray » |
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JiNbOtAk
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Re: Lagrange's Theorem
« Reply #1 on: Sep 20th, 2007, 9:35pm » |
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on Sep 20th, 2007, 7:29pm, Eigenray wrote:| Everybody knows that if a finite group G has a subgroup of order n, then n divides |G|. |
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Okay, I have to disagree. I for one, do not know that. ( I have trouble understanding it, never mind knowing it )
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Eigenray
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Re: Lagrange's Theorem
« Reply #2 on: Sep 20th, 2007, 9:55pm » |
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on Sep 20th, 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 ) |
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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 
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Sameer
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Re: Lagrange's Theorem
« Reply #3 on: Sep 20th, 2007, 10:09pm » |
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on Sep 20th, 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 |
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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?
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"Obvious" is the most dangerous word in mathematics.
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Eigenray
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Re: Lagrange's Theorem
« Reply #4 on: Sep 20th, 2007, 10:22pm » |
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on Sep 20th, 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? |
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That is sufficient. But it is not necessary. For example, it holds for n=4,8,9, and 11. Why?
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Sameer
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Re: Lagrange's Theorem
« Reply #5 on: Sep 20th, 2007, 10:43pm » |
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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..
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Proof is an idol before which the mathematician tortures himself.
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JiNbOtAk
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Re: Lagrange's Theorem
« Reply #6 on: Sep 21st, 2007, 12:20am » |
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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 20th, 2007, 10:22pm, Eigenray wrote:
That is sufficient. But it is not necessary. |
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Now what does this mean ? It's enough, but not needed ? 
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towr
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Re: Lagrange's Theorem
« Reply #7 on: Sep 21st, 2007, 12:43am » |
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on Sep 21st, 2007, 12:20am, JiNbOtAk wrote:| Now what does this mean ? It's enough, but not needed ? |
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X sufficient for Y: X -> Y
X necessary for Y: not X -> not Y
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towr
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Re: Lagrange's Theorem
« Reply #8 on: Sep 21st, 2007, 12:46am » |
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on Sep 20th, 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? |
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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.
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ThudnBlunder
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Re: Lagrange's Theorem
« Reply #9 on: Sep 21st, 2007, 2:46am » |
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on Sep 21st, 2007, 12:46am, towr wrote:| I'm thinking about taking algebra next quarter; |
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Now that you have finished your thesis, aren't they kicking you out into the real world?
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Eigenray
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Re: Lagrange's Theorem
« Reply #10 on: Sep 21st, 2007, 4:02am » |
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on Sep 21st, 2007, 12:46am, towr wrote:| I would guess n has to be a power of a single prime. |
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That is exactly correct. Now all you have to do is prove it!
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Barukh
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Re: Lagrange's Theorem
« Reply #11 on: Sep 21st, 2007, 4:28am » |
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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?
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towr
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Re: Lagrange's Theorem
« Reply #12 on: Sep 21st, 2007, 4:34am » |
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on Sep 21st, 2007, 2:46am, ThudanBlunder wrote:| Now that you have finished your thesis, aren't they kicking you out into the real world? |
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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).
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Eigenray
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Re: Lagrange's Theorem
« Reply #13 on: Sep 21st, 2007, 4:57am » |
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on Sep 21st, 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? |
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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).
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ThudnBlunder
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Re: Lagrange's Theorem
« Reply #14 on: Sep 21st, 2007, 5:04am » |
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on Sep 21st, 2007, 4:34am, towr wrote:
No, not yet. I'm still working on a bachelor's degree in philosophy as well... |
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I know Abstract Algebra is useful for philosophy; but Group Theory?
I suppose it could come in useful in the classification of your wacky, anything-is-possible theories which may give your tutors a few grey hairs.
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| « Last Edit: Sep 21st, 2007, 5:08am by ThudnBlunder » |
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towr
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Re: Lagrange's Theorem
« Reply #15 on: Sep 21st, 2007, 6:17am » |
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on Sep 21st, 2007, 5:04am, ThudanBlunder wrote:| I know Abstract Algebra is useful for philosophy; but Group Theory? |
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Well, I only need ~30 credits of philosophy, spread over a 60 credit year, so I have time left for some extracurricular courses.
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| « Last Edit: Sep 21st, 2007, 6:18am by towr » |
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Michael Dagg
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Re: Lagrange's Theorem
« Reply #16 on: Sep 21st, 2007, 6:57am » |
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on Sep 20th, 2007, 7:29pm, Eigenray wrote:
(One direction is well known. The other took me a while!)
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Barukh
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Re: Lagrange's Theorem
« Reply #17 on: Sep 21st, 2007, 7:09am » |
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After inspecting my home math library, seems like it's precisely the Sylow's theorem.
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Eigenray
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Re: Lagrange's Theorem
« Reply #18 on: Sep 21st, 2007, 12:43pm » |
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on Sep 21st, 2007, 7:09am, Barukh wrote:| After inspecting my home math library, seems like it's precisely the Sylow's theorem. |
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That will only help in one direction. And even then, the statement of the theorem is not quite enough.
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Barukh
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Re: Lagrange's Theorem
« Reply #19 on: Sep 22nd, 2007, 3:15am » |
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on Sep 21st, 2007, 12:43pm, Eigenray wrote:
That will only help in one direction. |
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Correct.
Quote:| And even then, the statement of the theorem is not quite enough. |
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How come? Sylow theorem states: "If n is a power of a prime, and n | |G|, then G has a subgroup of order n."
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| « Last Edit: Sep 22nd, 2007, 3:15am by Barukh » |
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Eigenray
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Re: Lagrange's Theorem
« Reply #20 on: Sep 22nd, 2007, 6:21am » |
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It guarantees a subgroup of order pk, where pk is the largest power of p dividing |G|. But you also need subgroups of order pi for all i < k. (Of course, this may follow from the proof of the theorem.)
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Obob
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Re: Lagrange's Theorem
« Reply #21 on: Sep 22nd, 2007, 7:00am » |
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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 pk, it is enough to show that any p-group 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.
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Barukh
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Re: Lagrange's Theorem
« Reply #22 on: Sep 22nd, 2007, 9:41am » |
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on Sep 22nd, 2007, 6:21am, Eigenray wrote:| It guarantees a subgroup of order pk, where pk is the largest power of p dividing |G|. But you also need subgroups of order pi for all i < k. (Of course, this may follow from the proof of the theorem.) |
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This is a statement of the existence of Sylow p-subgroup.
All three books on Abstract Algebra (Johnson, Herstein, Fraleigh) state and prove Sylow's first theorem as I described it before.
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| « Last Edit: Sep 22nd, 2007, 9:42am by Barukh » |
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ecoist
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Re: Lagrange's Theorem
« Reply #23 on: Sep 22nd, 2007, 3:40pm » |
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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 Zp extended by a primitive m-th root of unity, say x. Then m divides |F|-1; 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 semi-direct 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.
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| « Last Edit: Sep 22nd, 2007, 3:43pm by ecoist » |
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Eigenray
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Re: Lagrange's Theorem
« Reply #24 on: Sep 22nd, 2007, 6:35pm » |
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on Sep 22nd, 2007, 7:00am, Obob wrote:
But this is not a group.
on Sep 22nd, 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. |
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The usual statement of Sylow's theorem is about maximal p-subgroups.
on Sep 22nd, 2007, 3:40pm, ecoist wrote:| I hope this is a partial proof and not swiss cheese. |
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That's exactly what I had! Note that this is a special case.
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