Talk:Sylow theorems

message: 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)
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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)
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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)
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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 p^i is contained in a subgroup of order p^{i+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 p^i. 
See for instance http://sps.nus.edu.sg/~limchuwe/mathweb/group13.html --Manta 20:09, 9 Mar 2005 (UTC)
replies: text: 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)
thread_title:  (old content uncategorized) 

message: 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)
replies: text: 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)
replies: text: It should be noted on the page that the numbering for the proofs does not line up with that of the theorems.  These proofs are more general, and so it is not correct to label them the same way. 129.107.240.1 (talk) 03:06, 13 January 2009 (UTC)

text: Well, these proofs (based on Wielandts') are undoubtedly proofs from the "Book", thanks for sharing. I must admit, though, that I had quite a hard time understanding them, because they are written down in a very terse manner which makes them difficult to digest for people which do not have such deep insights into group theory and combinatorics. I'd say they sketch the idea, more or less, but that's probably ok within the scope of the article. Gzim75 (talk) 15:10, 27 November 2017 (UTC)
thread_title: New proofs

message: 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)
replies: text: 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)
thread_title: Terminology

message: 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)
replies: text: Page moved, per request at WP:RM.  Cheers. -GTBacchus(talk) 18:29, 3 January 2007 (UTC)
thread_title:  Sylow Theorem'''''s''''' or Sylow Theorem? 

message: 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)
thread_title:  Third part of Theorem 3 

message: I may be missing something here but isn't the proof that groups of order 15 are cyclic not quite right?  It's this bit that I'm not sure about:  "Since 3 and 5 are coprime, the intersection of these two subgroups is trivial, and so G must be the internal direct product of groups of order 3 and 5, that is the cyclic group of order 15. Thus, there is only one group of order 15 (up to isomorphism)."
Just because you have two subgroups of trivial intersection who's sizes multiply to give the size of the group, does NOT imply in general that the group is the direct product of these subgroups, this only holds if the subgroups commute.  I don't know if there is some other relevant theorem being implicitly assumed here which means you can make this step in this case, but if there is shouldn't it be stated?  I'd have thought the correct way to finish of the proof would be to consider an element not in either of these subgroups, which must then have order 15 since you've already found the elements of order 3 and 5, which therefore implies that the group is cyclic.  — Preceding unsigned comment added by Tobycrisford (talk • contribs) 14:23, 13 February 2013 (UTC)
replies: text: The product is a direct product because both subgroups have already been proved normal.—Emil J. 14:40, 13 February 2013 (UTC)
thread_title:  Problem with the proof that groups of order 15 are cyclic? 

message: In the proof of theorem 1, we fix an element \alpha \in \omega. We have only defined \omega as an element of the sets \Omega and G_\omega, not as a set in itself. I believe it should be changed to \Omega.65.112.10.195 (talk) 04:01, 24 October 2016 (UTC)
replies: text: You have misread this. \omega is a subset of a certain size, so this statement about an invariant point of it makes perfectly good sense. --Bill Cherowitzo  (talk) 04:35, 24 October 2016 (UTC)
thread_title:  Typography in proof of theorem 1 

message: The Consequences section currently says:
:Due to the maximality condition, if H is any p-subgroup of G, then H is a subgroup of a p-subgroup of order pn.
This isn't obvious to me. I think it should be explained, proved, sourced or removed.
--Joriki (talk) 04:51, 12 August 2018 (UTC)
thread_title:  non-trivial statement without source or proof 

message: Here are some links whose content would improve this article
* https://yutsumura.com/groups-of-order-100-200-is-it-simple/
* https://yutsumura.com/every-group-of-order-20449-is-an-abelian-group/
* https://web.archive.org/web/20210124212837/https://yutsumura.com/sylows-theorem-summary/
This would help readers get accustomed to using the theorems for proving results about groups using their cardinality. Also, there should be examples of groups in each of the special cases in the article, such as groups of order pq.
Wundzer (talk) 15:33, 12 February 2021 (UTC)
thread_title:  Article improvements 

message: : One defines a Sylow -subgroup in an infinite group to be a p-subgroup (that is, every element in it has -power order) that is maximal for inclusion among all -subgroups in the group. Such subgroups exist by Zorn's lemma.
I removed Zorn's lemma from this sentence. At the least it needs a cite and explanation. I don't see why all such chains have the upper bounds required by Zorn's lemma. let G be a group such that for all x in G, x != e, there's a y in G such that y ^ 3 = x, and exists an x in G s.t. x != e and x^3 = e. That should be a group that is infinite and has an infinite chain of 3-subgroups.--66.76.243.26 (talk) 18:15, 22 July 2022 (UTC)
thread_title:  Zorn's lemma