I'm trying to complete a standard proof concerning "ruler and compass constructibility" and it boils down to this result which I've seen stated in some books but haven't seen or been able to come up with a proof:
"If is a finite group of order then there exist subgroups such that ."
Can anyone help with this please?
I see, thanks. I've had a look at the Sylow theorems and they will solve my problem. From what I gather, my original question seems to be a particular case of a standard result. Just to check though, is there a quick and simple proof to show that when is a group of order , then there exist subgroups
such that ? Or really is the use of the Sylow theorems about the simplest argument we can give? Just because if there is indeed something simple that I could use instead, I'd much rather do it that way. I should have said: we can also use the fact that the centre of any group of order (p prime) is not trivial, and we can use Cauchy's theorem (if this helps, but I get the feeling they only help to prove the Sylow theorems anyway...)
we'll prove this by induction over it's clear for now suppose the claim is true for all groups of order with and let be a group of order consider two cases:
case 1. is abelian: by Cauchy, has an element of order 2. let and then and so, by the induction hypothesis, has a subgroup of order
i.e. there exists a subgroup of such that and thus
case 2. is non-abelian: so for some integer let then thus, by the induction hypothesis, has a subgroup of
order i.e. there exists a subgroup of such that and hence
Also, the result is proven in the literature, for example on page 139 of Robinson, A Course in the Theory of Groups.
well, Sylow theorems have been stated in different ways. i don't have access to any group theory textbook right now and so i googled the Sylow theorems and the first noteable thing was a
lecture note of some professor in the university of Glasgow and this guy agrees with me. see theorem 4.10. anyway, the proof is the same as the proof i gave for p = 2.
I suppose it is just my own opinion though.
It may well be seen that way; nevertheless, most text books I know don't include it as part of Sylow theorems but as something that can more or less easily be proven without any reference to Sylow theorems, using only say the class equation and some induction.
Even in the case of the link you produced that part of what that author calls Sylow theorem is not proved within that theorem but apart, as lemma 4.11, without any reference to the actual Sylow theorem but only using precisely the fact that the center of finite p-groups is non-trivial! Thus, I'd say that including this fact within the Sylow theorem there is a pretty artificial thing.
It'd be interesting to know, me thinks, if Sylow included this result as part of his theorems or if it was known already.
Thanks very much for this. Without worrying too much about exactly what the Sylow theorems say, I was looking at your more elementary proof and it's precisely the kind of argument I was after. One thing confuses me though: Cauchy's theorem doesn't require a group to be abelian, right? So why doesn't the proof you give for "case 1: G abelian" hold even when G isn't abelian?
On a related note, whilst I was searching for an answer to this problem I came across the comment, "The centre of any group of order (p prime) is not trivial. Then by Cauchy's theorem, we can pick an element in the centre of the group of order 2." So clearly there must be something to do with "commutativity" going on here, but I don't really understand what it is! Thanks.
On the contrary: thanx to the fact that the center of any finite p-group isn't trivial one doesn't need to worry about commutativity. For the induction step one only takes a non-trivial element of the center of the group and then looks at the subgroup generated by that element, which is normal for being central, and then passes to the factor group and etc.