Talk:P-group

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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 $D 2 n$

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)

[edit] 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)

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

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)

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)

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)

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)

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)

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)

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)

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)

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)

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

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)

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)

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)

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

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