# Finite p-group

• June 10th 2011, 08:45 AM
zoek
Finite p-group
This is an exercise from a textbook with many mistakes –not only typos – so I am not sure if actually holds.

Let $G$ be a finite $p$-group ( $p$=prime). If $G/G’$ is cyclic then $G$ is cyclic too. ( $G’$ is the commutator subgroup or derived group of $G$).
• June 10th 2011, 03:00 PM
tonio
Quote:

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

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

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 $G$ is such a group, then $|Z(G)|>1$) ;

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

maximal sbgp. $M\leq G\, , \, G/M$ is abelian (even cyclic of order a prime) and thus $G'\leq M$);

3) As $G/G'=$ , we get that $G=G'\Longrightarrow G=$ 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)
• June 11th 2011, 02:56 AM
Swlabr
Quote:

Originally Posted by tonio
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 $G$ is such a group, then $|Z(G)|>1$) ;

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

maximal sbgp. $M\leq G\, , \, G/M$ is abelian (even cyclic of order a prime) and thus $G'\leq M$);

3) As $G/G'=$ , we get that $G=G'\Longrightarrow G=$ 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)

Is it true that $Z(G) \leq G^{\prime}$? 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).
• June 11th 2011, 04:15 AM
zoek
Thank you both!

I had not studied about Frattini subgroup, until today. I shall do it now!
• June 12th 2011, 10:18 PM
topspin1617
Quote:

Originally Posted by Swlabr
Is it true that $Z(G) \leq G^{\prime}$? Because that would make the question much easier!

No, that's not true in general.

For example, if $G$ is abelian, that certainly fails (though I suppose this would make the whole problem very easy... but still.).
• June 13th 2011, 12:40 AM
Swlabr
Quote:

Originally Posted by topspin1617
No, that's not true in general.

For example, if $G$ is abelian, that certainly fails (though I suppose this would make the whole problem very easy... but still.).

Sorry-that was meant to be $G^{\prime} \leq Z(G)$, as then $G/G^{\prime}\rightarrow G/Z(G)$ surjectively, and so $G/Z(G)$ is cyclic. It is true that if $G/Z(G)$ is cyclic then $G$ is abelian, which isn't too hard to prove.

However, this ( $G^{\prime} \leq Z(G)$) does not hold in general either. It holds if and only if $G$ has nilpotency class at most 2 (because if $G^{\prime}\leq Z(G)$ then $[[G, G], G]\leq [Z(G), G]=1$, and the converse is similar).

So, we want to prove that $G$ 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!
• June 13th 2011, 06:14 AM
Deveno
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).
• June 13th 2011, 06:22 AM
Swlabr
Quote:

Originally Posted by Deveno
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.

Yes, but the method is looking for a contradiction...so you can't give a real-life counter-example! (Also, $D_{16}$ has derived length 3, not 2...you'd need $D_8$ for that, I believe...)
• June 23rd 2011, 12:40 AM
zoek
Re: Finite p-group
Quote:

Originally Posted by zoek
Let $G$ be a finite $p$-group ( $p$=prime). If $G/G’$ is cyclic then $G$ is cyclic too. ( $G’$ is the commutator subgroup or derived group of $G$).

I think that finally I managed to solve this exercise:

1. $G$ finite $p$-group $\Rightarrow G$ nilpotent

2. $G$ nilpotent $\Rightarrow G’ \leq \Phi (G)$

3. $G/G’$cyclic $\Rightarrow G/\Phi (G)$ cyclic $\overset {G fin. p-grp }{\Longrightarrow }$ $G$ cyclic.

Quote:

Originally Posted by Swlabr
I mean, if the OP hadn't read the section on that yet, then there must be another proof!

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.