This is an exercise from a textbook with many mistakes –not only typos – so I am not sure if actually holds.

Let be a finite -group ( =prime). If is cyclic then is cyclic too. ( is the commutator subgroup or derived group of ).

Printable View

- Jun 10th 2011, 09:45 AMzoekFinite p-group
*This is an exercise from a textbook with many mistakes –not only typos – so I am not sure if actually holds.*

Let be a finite -group ( =prime). If is cyclic then is cyclic too. ( is the commutator subgroup or derived group of ). - Jun 10th 2011, 04:00 PMtonio

This is a very nice and non-trivial exercise, and we need to be sure we know/can prove the following:

1) A finite p-group is always nilpotent (hint: if is such a group, then ) ;

2) In a nilpotent group we always have that the Frattini subgroup of (hint: for any

maximal sbgp. is abelian (even cyclic of order a prime) and thus );

3) As , we get that and we're done (hint: the Frattini sbgp. of a

group is the set of all non-generators of the group. Apply now (2))

Tonio

3) - Jun 11th 2011, 03:56 AMSwlabr
Is it true that ? Because that would make the question much easier!

(Also, Tonio's approach is very similar to something called 'The Burnside Basis Theorem', which says that G/Frat(G) is abelian, and G is generated by n-elements, where G/Frat(G) is n-generated. See Robinson, A Course in the Theory of Groups). - Jun 11th 2011, 05:15 AMzoek
Thank you both!

I had not studied about Frattini subgroup, until today. I shall do it now! - Jun 12th 2011, 11:18 PMtopspin1617
- Jun 13th 2011, 01:40 AMSwlabr
Sorry-that was meant to be , as then surjectively, and so is cyclic. It is true that if is cyclic then is abelian, which isn't too hard to prove.

However, this ( ) does not hold in general either. It holds if and only if has nilpotency class at most 2 (because if then , and the converse is similar).

So, we want to prove that has nilpotency class at most two...however, this seems to be harder than I initially thought...but I am still uneasy about Tonio's use of the Frattini subgroup. I mean, if the OP hadn't read the section on that yet, then there must be another proof! - Jun 13th 2011, 07:14 AMDeveno
i don't think that approach will work. for example D16 has center {1,r^4} but [D16,D16] = <r^2>, which is cyclic of order 4.

(of course D16 has nilpotency class 3. so i suppose what we want to show is that G/[G,G] cyclic --> G is of nilpotency class 2.

this isn't obvious. perhaps one could get around using the Frattini subgroup by showing [G,G] is NOT maximal, and using induction). - Jun 13th 2011, 07:22 AMSwlabr
- Jun 23rd 2011, 01:40 AMzoekRe: Finite p-group
I think that finally I managed to solve this exercise:

1. finite -group nilpotent

2. nilpotent

3. cyclic cyclic cyclic.

It is only in my textbook that this exercise is located before Frattini subgroup and nilpotent groups (but the solution of this is totally incorrect). Everywhere else I found something about this (Rotman, Robinson and internet) it was after nilpotent groups and Frattini subgroup.