Talk:P-group

message: 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)
thread_title:  Dihedral notation 

message: 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)
replies: text: 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)

text: 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)
replies: text:  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)
replies: text:  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)

text: 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)
replies: text:  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)
replies: text: 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)
replies: text:  Lol. Big sigh of relief! :) - grubber 18:19, 20 September 2006 (UTC)

text: 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)

text: Prime powers include 1, though. Not really paradoxical. Charles Matthews 20:42, 20 September 2006 (UTC)
thread_title:  Every profinite group contains a <math>p</math>-group 

message: 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)
thread_title:  article is identical to... 

message: 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)
replies: text: There's a wiki page  about it. Charles Matthews (talk) 16:19, 15 January 2008 (UTC)
thread_title: Hughes subgroup of a p-group

message: 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)
thread_title:  Properties is pretty slim 

message: 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)
replies: text: 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)
thread_title:  Definition 

message: . 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)
thread_title:  Is it reasonable to restrict this article to finite groups? 

message: 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)
replies: text: 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)
thread_title:  Number of groups of order 1024