Talk:P-group

Dihedral notation

when you write $D_{4}$ do you mean the klein group or the eight isometries of a square?

Because the convention I know is that one denotes the group of all 2*n isometries on a regular n-gon by $D_{2n}$

Here it obviously has order 8. The conventions are at Wikipedia:WikiProject Mathematics/Conventions. So I'll make that Dih₄. Charles Matthews 19:49, 11 February 2006 (UTC)[reply]

Every profinite group contains a $p$ -group

The article says that every non-trivial group of finite order contains a $p$ -group. In fact, according to the definition here, the trivial group is itself a $p$ -group, so there's no need for the exception. Also, it's probably not inapposite to mention profinite groups here. JadeNB

I changed natural number to positive integer, so that the trivial group won't be a p-group.Rich 16:28, 19 September 2006 (UTC)[reply]

Is that wise? It would be better to correct the statement made to have content. Charles Matthews 18:22, 19 September 2006 (UTC)[reply]

I don't see what you mean but it's ok. You may revert or revise w/no hard feelings.I should mention I later added 'non-identity"Rich 21:06, 19 September 2006 (UTC)[reply]

The trivial group is NOT a p-group because 1 is not a prime number. We maybe should say that the definition of a finite p-group to be a group that has order pⁿ and let the other results follow. - grubber 20:54, 19 September 2006 (UTC)[reply]

That might be a more convenient definition but we're supposed to, as an encyclopedia, report on what the authorities say, and it's defined as every element has order a prime power by every text I can think of.Rich 21:11, 19 September 2006 (UTC)[reply]

That's right, and 1 is a prime power, namely p⁰. The trivial group is a p-group for all primes p. --Zundark 21:21, 19 September 2006 (UTC)[reply]

Ah, I was thinking 1ⁿ rather than p⁰. Looking at my Milne notes again, it doesn't admit or preclude n=0, and that's all I have nearby right now. Anyone have their Herstein book handy? I'm curious how he defines a p-group. - grubber 22:28, 19 September 2006 (UTC)[reply]

FYI, Herstein defines a p-Sylow subgroup, but I can't find any reference to a p-group at all. -grubber 14:57, 20 September 2006 (UTC)[reply]

In my haste to reply to your comment I mistakenly said that your definition was correct. Of course, it's not correct, as the symmetric group S₃ shows. A p-group is a group in which all elements are of p-power order. The rest of my reply stands: the trivial group is a p-group for every prime p (and, more generally, a π-group for every set π of primes). --Zundark 08:36, 20 September 2006 (UTC)[reply]

The def I provided is equivalent to that def (for finite G). What are you saying S₃ shows? S₃ isn't a p-group for any p. - grubber 14:48, 20 September 2006 (UTC)[reply]

Your definition is indeed a correct definition for finite p-groups. My comment wasn't addressed to you, it was addressed to Rich (as the indentation shows). As you say, S₃ isn't a p-group for any p. Yet it fits Rich's definition, since all its elements are of prime power order. --Zundark 15:34, 20 September 2006 (UTC)[reply]

Lol. Big sigh of relief! :) - grubber 18:19, 20 September 2006 (UTC)[reply]

The trivial group has to be a p-group, nilpotent group, solvable group, or else theorems are going to have pointless exceptions. Charles Matthews 15:29, 20 September 2006 (UTC)[reply]

I'll concede the point. Vacuously true statements are always a bit hairy, but I can understand the utility you say we gain by letting it be. - grubber 18:19, 20 September 2006 (UTC)[reply]

I went back to the old version except nonnegative integer instead of natural number because of ambiguity of whether 0 is a natural number. I hope this suits everyone better than my earlier edits. By the way, I believe Charles' information that the trivial group needs to be p-group, etc to make theorems easier to state, but isn't it ironic? We forbid 1 from being prime to avoid difficulty stating the the Fund Thm of Arith, but for similar reasons require {1} to be a p-group.Rich 18:46, 20 September 2006 (UTC)[reply]

Prime powers include 1, though. Not really paradoxical. Charles Matthews 20:42, 20 September 2006 (UTC)[reply]

-Yeah, good point.Rich 04:34, 22 September 2006 (UTC)[reply]

article is identical to...

http://experts.about.com/e/p/p/p-group.htm

word for word. though i can't say which is the original, or if its a copyright issue. kind of lame though. --Putgeminmouth

They copied Wikipedia. (It says so in the small print at the bottom of the page.) --Zundark 08:59, 10 November 2006 (UTC)[reply]

Hughes subgroup of a p-group

The Hughes subgroup of a finite p-group is the subgroup generated by all elements that are not of order p, according to I. D. MacDonald. That's all I know, but if someone knows more, by all means please put it in here.-Richard Peterson63.164.145.198 (talk) 07:24, 15 January 2008 (UTC)[reply]

There's a wiki page [1] about it. Charles Matthews (talk) 16:19, 15 January 2008 (UTC)[reply]

Great,thanksRich (talk) 21:53, 15 January 2008 (UTC)[reply]

Proof of the statement regarding normalizers

This is regarding a recent revert.

There are essentially four reasons why I tried to clean up the proof:

The statement "creating an infinite descent" is not really correct; at some point, the condition that Z(G) ≤ H is no longer true, and one cannot mod out Z.
When the "smaller example" was mentioned, this implies a relationship between G and G/Z when, in fact, they are completely different p-groups (as pointed out by User:Dto [Thank you Dto :) ]).
(Most importantly) Even if, at some point, one cannot mod out by the center, so what? There is no wording which leads the reader into believing that this really is a contradiction. To see this, there needs to be something else (for example, consideration of inverse images of the natural homomorphism from G to G/Z which implies something regarding H and N in G).
This is supposed to be an induction proof, but verification of the induction hypothesis is missing and it is hard to find the sequence of statements.

I am not necessarily advocating that the new wording be restored, because I agree that it is hard to follow, but I believe that it is correct and, as such, it would be a good start to better wording.

« D. Trebbien (talk) 19:22 2008 May 23 (UTC)

First let me make sure we are on the same page about this proof. The method of infinite descent is a type of proof by contradiction, so one should not be worried if there is an impossible conclusion (that is the point). If H really is a counterexample, then H contains the center Z of G, and H/Z=N/Z so H/Z really is another counterexample! Continuing, so H/Z contains the center Z2 of G/Z, so H/Z2 is another example, so H/Z2 contains the center Z3 of G/Z2, and H/Z3 is yet another counterexample, etc. Since the center is nontrivial, the order of H/Z is strictly smaller than the order of H, creating an infinite descending sequence of positive integers (their orders), a contradiction.

Now let me address the specific points:

The statement "creating an infinite descent" is correct. The normalizer of H/Z in G/Z is stated to be N/Z (perhaps adding a proof of this in a footnote like in your first version would be useful in a mathematical textbook, perhaps even in an encyclopedia article just giving a survey of the ideas), so the same argument shows that in all cases the center of G/Z is contained in H/Z. One can always mod out by Z, which is absurd given that Z is not the trivial subgroup, hence the contradiction.
The relationship is very commonly used and broadly understood. One set is said to be smaller than another set if its cardinality is smaller than the other set. In fact, when comparing groups, even better than comparing cardinality, is using a homomorphism to compare groups. In this case G and G/Z stand in an incredibly close relationship, not only because of the surjective homomorphism from G to G/Z, but because the kernel is precisely the center. Not only is G is a central extension, but it is a witness to the fact that G/Z is capable, and expresses G/Z as the group of inner automorphisms of G.
I think the contradiction is clear: given one example H there is a strictly smaller example H/Z, creating an infinite descent.
The beauty of infinite descent is it avoids irrelevant details of induction. As long as the set of counterexamples is well-ordered (downwardly), the contradiction is found. With typical induction arguments you are forced to consider a minimal counterexample, and develop its structure. This method is bolder and more courageous by developing the structure of all counterexamples.

If you still feel the argument is unclear, then consider adding a footnote explaining the N/Z=H/Z step which I omitted in order to have the article read more like an encyclopedia than a homework assignment. The rest of the proof is very standard and rather tidy. JackSchmidt (talk) 20:01, 23 May 2008 (UTC)[reply]

The problem was not the meaning of contradiction, induction, or cardinality, but that I was ignoring the important assumption that H was a counterexample (ie. equaled N). Oh well.

I assume by the "N/Z=H/Z step" you mean showing that "H/Z = N_G/Z(H/Z)". ~~This probably will be helpful to state somewhere...~~

« D. Trebbien (talk) 21:16 2008 May 23 (UTC)

Already stated. Oh bother. « D. Trebbien (talk) 21:20 2008 May 23 (UTC)

I should probably mention that this proof is just supposed to be an example of central induction. The nilpotent group article is probably the best place for a very clear exposition of the so called "normalizer condition" (every proper subgroup is a proper subgroup of its normalizer). It is a very nice result that groups whose upper central series reach the whole group have the normalizer condition, and at least for some restricted class of groups, these are equivalent to the group being nilpotent. For p-groups we just want to say "having a nontrivial center is useful, because arguments like this one work out". Since none of this is really proved at nilpotent group#Properties, it might be wise to put *another* correct proof there. If you want to really go to town on that article, the textbooks by Kurosh are the basic source for that section, but I think Robinson's textbook is also just fine (as in, tells you more than you wanted to know about how nilpotent groups fit into the grand scheme of things).

Would it help to have other examples of central induction in this article? It might make it clearer that the normalizer condition isn't the focus of that section. Often in computational group theory the problem for G is hard, but easier for G/N, and in favorable cases the answer for G/N can be lifted to G (this is called the homomorphism principle). When G is a p-group, one usually takes N to be contained in the center. JackSchmidt (talk) 20:30, 23 May 2008 (UTC)[reply]

Properties is pretty slim

Actually looking at the properties section, it doesn't really seem to support the "much is known about p-groups", as there is only a single subsection. This should probably be fixed. There are lots of results (even on arxiv) on their conjugacy classes (upper and lower bounds on how many and how large each is), their representations (in char p and otherwise), counting their subgroups, normal subgroups, etc.

Probably this article should link to most of the articles on special sorts of p-groups as well. I think many articles are not linked. JackSchmidt (talk) 20:36, 23 May 2008 (UTC)[reply]

Should you prove any properties of p-groups listed? I could probably write up a few good properties, but whether or not I could prove them in a succinct way is a different story Triangl (talk) 22:16, 1 December 2008 (UTC)[reply]

Dispute: Prevalence of p-groups

The comment in the "Prevalence" section that almost all groups are 2-groups or even p-groups requires a proof or references. The reference given there doesn't have a proof of such fact or further reference for this fact. I think it is still a conjecture.

Definition

The definition is given as

"The elements of a p-group all have order a power of p." (i've paraphrased it)

and then goes on to say a Group is a p-group iff it has order a power of p (ie the number of elements in the group is a power of p)

It is probably more intuitive to have the second bit as the definition and the first as a corollary. It seems to be more immediately obvious that the latter implies the former. I won't change it unless someone else agrees though. Triangl (talk) 22:19, 1 December 2008 (UTC)[reply]

The point is that the "order is a power of p" definition only works for finite groups. The definition given in the article is needed to cover the case of infinite p-groups. --Zundark (talk) 22:40, 1 December 2008 (UTC)[reply]

Is it reasonable to restrict this article to finite groups?

Find sources: Google (books · news · scholar · free images · WP refs) · FENS · JSTOR · TWL. That finds plenty of textbooks that at least mention the infinite ones. I'm not sure how much is known in general about the infinite ones though. 86.127.138.67 (talk) 21:44, 10 April 2015 (UTC)[reply]

Number of groups of order 1024

This page (2021) states that there are 49487367289 groups of order 1024. Should the number 49487365422 in the section "Prevalence" be corrected? See also this Math Stack Exchange question.

Also, This Math Stack Exchange answer cited a better estimation ( $O(n^{5/2})$ in the exponent instead of $O(n^{8/3})$ ). Should this be taken into account for an improvement?

I'd like to point out that, it should be noted the constant served as a bound for $O(n^{8/3})$ does not depend on $p$ . 129.104.240.174 (talk) 00:26, 28 August 2023 (UTC)[reply]

I have changed the incorrect number per my comment at Talk:1024_(number)#Number_of_groups_of_order_1024, but I have not touched the formula. We cannot cite Stackexchange as it is a self-published source, and I don't have access to the cited sources to verify whether they give the revised estimate directly, or whether this is innovation from the Stackexchange answerer. Barnards.tar.gz (talk) 15:30, 28 August 2023 (UTC)[reply]