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Group center properties

  1. Dec 2, 2011 #1
    I am aware of the theorem |G/Z(G)|=p with p prime implies G/Z(G) is cyclic and thus G is abelian, but I do not understand why. Is there not a theorem that says G abelian [itex]\Leftrightarrow[/itex] Z(G)=G? So what if |G|=p[itex]^{3}[/itex] and |Z(G)|=p[itex]^{2}[/itex]? This implies |G/Z(G)|=p implying G is abelian however G[itex]\neq[/itex]Z(G). What is the ambiguity here?
     
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  3. Dec 2, 2011 #2

    micromass

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    There is no ambiguity. The situation you mention can just never occur. You have just proven that it can never oocur that |G|=p3 and |Z(G)|=p2.

    You can use this, for example to prove that all groups of order p2 are abelian. Indeed, Z(G) is a subgroup of G, so there are three possibilities:
    • |Z(G)|=1 It can be proven that this can never happen (a group with prime order must always have nontrivial center)
    • |Z(G)|=p Then your theorem says the group is abelian, thus |Z(G)|=p2, which is a contradiction.
    • |Z(G)|=p2 is the only remaining possibility.
     
  4. Dec 2, 2011 #3
    Ah I see! Thanks for clearing that up.
     
  5. Dec 3, 2011 #4

    lavinia

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    Try looking at the exact sequence of groups,

    0 -> Z(G) -> G -> G/Z(G) -> 1

    - Any group of prime order is cyclic so G/Z(G) is cyclic.
    - If the sequence is split then the group is isomorphic to the direct product Z(G) x G/Z(G)
    so G is abelian and Z(G) = G
    - If the sequence is not split what happens?
     
  6. Dec 3, 2011 #5
    I do not follow, I don't believe I've ever looked at this sequence or talked about what a split sequence is. Micromass cleared up the problem I was having though.
     
  7. Dec 3, 2011 #6

    Deveno

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    lavina, do you mean not left-split (because right-splits need not be direct products)?
     
  8. Dec 5, 2011 #7

    lavinia

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    To me split means that there is an inverse of the projection onto G/Z(G)

    Since Z(G) is in the center of the group split means direct product - I think.

    What you get is that there is a subgroup isomorphic to G/Z(G) that commutes with every other element of the group. this should be a direct product.

    If the sequence is not split then a lift of a generator of the quotient has some power that lies in Z(G) but again it commutes with everything so G is abelian.
     
    Last edited: Dec 5, 2011
  9. Dec 5, 2011 #8

    lavinia

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    I am sorry. I thought he only cleared up a special case. Anyway. This is a different way of looking at the problem that I thought you might find interesting.
     
  10. Dec 6, 2011 #9

    Deveno

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    ah, so you DID mean right-split. let me assure you that right-split ≠ direct product.

    consider the following short exact sequence:

    0→Cn→Dn→Dn/Cn→0.

    (where we map a generator of Cn, to a generating rotation of Dn).

    define:

    f:Dn/Cn→Dn by:

    f(Cn) = 1
    f(Cns) = s

    then if p is the canonical projection: p(f(Cnx)) = Cnx.

    but Dn is certainly not a direct product of abelian groups.
     
  11. Dec 6, 2011 #10

    lavinia

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    If The kernel is in the center of the group then split means direct product. Otherwise not necessarily.I think the proof is straight forward.
     
    Last edited: Dec 6, 2011
  12. Dec 7, 2011 #11

    Deveno

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    i see how this is true, because if H acts on N by conjugation, and N is in Z(G), the action has to be trivial.
     
  13. Dec 8, 2011 #12

    lavinia

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    yes. that's it.
     
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