Talk:Sylow theorems
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[edit] (old content uncategorized)
I removed this: From these observations, we can derive the following useful formula relating a group's order and its Sylow subgroups: which preceded the last statement. I don't think this last statement is a mere consequence of the first statements; it requires its own independent proof. AxelBoldt 18:04 Nov 6, 2002 (UTC)
- Yep. I was really thinking about highlighting the "meat" (i.e., the hard thing to remember exactly) about np. I separated these out and gave some examples. Chas zzz brown 19:36 Nov 6, 2002 (UTC)
Nitpick: The theorem is still true if p does not divide |G|; it's just that a Sylow p-subgroup then has order p0 = 1. Chas zzz brown 04:55 Nov 12, 2002 (UTC)
If we count the trivial group as a p group, and I don't know if that is common, or a good idea. The theorem that every p group has non-trivial center is then false. AxelBoldt 05:40 Nov 12, 2002 (UTC)
Good point :) Chas zzz brown 05:41 Nov 12, 2002 (UTC)
Added proofs of Sylow theorem based on W. R. Scott's Group Theory, Dover publications.
These proofs rotate more around the idea of conjugacy classes, normalizer, and centralizers; rather than the orbits and stabilizers from the concept of the group action of inner homomorphisms of G on G. I don't know if it's the current "standard" proof; Scott's book is from the 1960s. Hopefully the proof is not overly prolix. :) Chas zzz brown 10:15 Nov 12, 2002 (UTC)
About the first Sylow theorem, we can say a bit more. Namely:
Then for each 0 < i < n − 1, every subgroup of G of order pi is contained in a subgroup of order pi + 1, and is normal in it.
In particular: 1) every p-subgroup of G is contained in a Sylow p-subgroup of G; 2) for every i less or equal to n there exists a subgroup of order pi.
See for instance http://sps.nus.edu.sg/~limchuwe/mathweb/group13.html --Manta 20:09, 9 Mar 2005 (UTC)
- I think that these latter results should be separated out as corollaries at best. They are actually consequences of theorems on p-groups and for this reason I think they should be addressed separately. - Gauge 06:19, 22 May 2005 (UTC)
[edit] New proofs
I've written up some new proofs of the Sylow theorems on my talk page. The proofs are simpler than that appearing currently at this article, and they give more general results (these are probably "proofs from the Book"). The results are due to Wielandt. Please make suggestions and corrections as I plan to replace the proofs on this page with these new ones if there are no objections. - Gauge 20:52, 22 May 2005 (UTC)
- New proofs added with reference, page reorganized slightly for continuity. Note: the new proofs are also quite a bit more general and yield the Sylow theorems as special cases. - Gauge 07:09, 4 Jun 2005 (UTC)
[edit] Terminology
I don't like the terminology "p-Sylow subgroup" to describe these groups. Such a subgroup is in particular a p-group, so it makes sense to call them p-subgroups, and if they happen to be maximal, call them Sylow p-subgroups. It's not a huge deal to me, but since it was changed by an anonymous editor without stating rationale I figure I am entitled to change it back and see if anyone complains. - Gauge 23:56, 10 August 2006 (UTC)
- To weigh in a little late here, I agree with you about "Sylow p-subgroup" making more sense (and I think it's more prevalent), and I agree that we should use it in the article, but I don't see anything really wrong with the alternative phrasing. The object in question is a Sylow subgroup which in particular corresponds to the prime p. The advantage of your formulation is visual, since there's a space in "Sylow subgroup" that makes the p seem to refer only to "Sylow." The difference isn't huge, though. Personally, I have a slight bias toward "p-Sylow" since that's how the prof said it when I first encountered them... and if we're going to insist on math terms making sense, we'll have a lot of work to do before we hit group theory. ~ CZeke 20:18, 30 March 2007 (UTC)
[edit] Proof of Theorem 1
Note: In the proof of Theorem 1, |G| = pkm where p and m are NOT necessarily coprime. The theorem actually proves that a subgroup of order any prime power divisor of |G| exists, so we will not require p and m to be coprime. Part of why these proofs are so cool is that they prove much more than the usual Sylow theorems, and the proofs are shorter than the usual arguments. For instance, Theorem 1 already implies that any p-group has p-subgroups of every prime power order, which is usually a separate theorem. - Gauge 22:41, 11 August 2006 (UTC)
[edit] Sylow Theorems or Sylow Theorem?
I am tempted to move the article to the plural form; the whole article uses the plural, and it is more appropriate. Anyone object? - grubber 23:39, 26 December 2006 (UTC)
[edit] Third part of Theorem 3
I've always seen the third Sylow theorem presented with three parts, not two, so I've added the third. Any objections? ~ CZeke 20:00, 26 March 2007 (UTC)
