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Math Help - Finite p-group

  1. #1
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    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).
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  2. #2
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    Quote Originally Posted by zoek View Post
    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'<Frat(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'=<xG'> , we get that G=<x>G'\Longrightarrow G=<x> 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)
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by tonio View Post
    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'<Frat(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'=<xG'> , we get that G=<x>G'\Longrightarrow G=<x> 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).
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    Thank you both!

    I had not studied about Frattini subgroup, until today. I shall do it now!
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    Quote Originally Posted by Swlabr View Post
    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.).
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by topspin1617 View Post
    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!
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  7. #7
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    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).
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Deveno View Post
    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...)
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  9. #9
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    Re: Finite p-group

    Quote Originally Posted by zoek View Post
    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 View Post
    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.
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