Talk:Sylow theorems

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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 n_p. 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 $p^{i}$ 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)

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)[reply]

This is my first contribution. Forgive me if I violate any rules.

My comments regard the section "Example applications". The example of the order 15 group is fine.

However, the discussion of the simplicity of the groups of order 30 and 42 is a little bit weird.

First of all, there is a very elementary proof of the fact that any group G of order 2m with m odd, admits a normal subgroup of index 2: Let G act on itself by translation and show that the permutation induced by an element of order 2, is odd.

So, these are not very good examples of an application of Sylow theory. I also find the use of Burnside's difficult p^aq^b Theorem here rather out of proportion. Showing that all groups of order < 60 are not simple can easily be done without Burnside. It is also much easier than that.

It would perhaps make more sense to show that groups of order pq are not simple or that groups of order 24 or 36 are not simple. Stickelberger (talk) 14:26, 3 December 2009 (UTC)[reply]

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)[reply]

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)

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)[reply]

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)[reply]

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)[reply]

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)[reply]

Proof of Theorem 1

Note: In the proof of Theorem 1, |G| = p^km 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)[reply]

I would like to propose reorganizing the presentation of the proof to put the easy part first and add a few more hints to the part about counting Ω.

Since there was no objection, I have now moved the new wording to the main article as proposed two weeks ago.

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)[reply]

Page moved, per request at WP:RM. Cheers. -GTBacchus^(talk) 18:29, 3 January 2007 (UTC)[reply]

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)[reply]

I removed the following text from the introduction:

For any prime power pⁿ which divides the order of a given finite group G, there exists a subgroup of G of order pⁿ

At first I removed it thinking it was false, as n would have to divide the maximal power of p which divides the group. Then I realized that that is not actually the case (I think I was thinking about finite fields), but I still don't think the statement belongs in the introduction, since it's more a property of p-groups than a consequence of the Sylow theorems, and more importantly, could lead to confusion, thinking that any p-group (not necessarily maximal) which was a subgroup of a given group was a Sylow p-subgroup. 68.228.50.201 (talk) 01:22, 15 November 2010 (UTC)[reply]

Problem with the proof that groups of order 15 are cyclic?

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)[reply]

The product is a direct product because both subgroups have already been proved normal.—Emil J. 14:40, 13 February 2013 (UTC)[reply]

Ok I understand what went on in the proof now. I do think the wording could be better though, because the current wording does seem to imply that whenever you have two disjoint subgroups with sizes multiplying to the size of the group, then you have a direct product. This is not only wrong, but a common misconception, so might not be helpful in the first example of an introductory page. Shouldn't it be made clearer what is being used, to stop other beginners like me from getting confused? Also bearing in mind that it is the first example might it be simpler to use the less sophisticated argument that there must be an element of order 15 outside the two subgroups? Tobycrisford (talk) 23:11, 15 February 2013 (UTC)[reply]

Typography in proof of theorem 1

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)[reply]

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)[reply]

non-trivial statement without source or proof

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)[reply]

Article improvements

Here are some links whose content would improve this article

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)[reply]

Zorn's lemma

One defines a Sylow

p

-subgroup in an infinite group to be a p-subgroup (that is, every element in it has

p

-power order) that is maximal for inclusion among all

p

-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)[reply]

Wait — what is surprising?

The section Statement of the article contains this sentence:

"The surprising result here is that in the case of $\operatorname {Syl} _{p}(G)$ , all members are actually isomorphic to each other and have the largest possible order: if $|G|=p^{n}m$ with $n>0$ where $p$ does not divide $m$ , then every Sylow $p$ -subgroup $P$ has order $|P|=p^{n}$ ."