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Normal subgroup of prime order in the center

  1. Oct 25, 2007 #1
    1. The problem statement, all variables and given/known data

    Let H be a normal subgroup of prime order p in a finite group G. Suppose that p is the smallest prime dividing |G|. Prove that H is in the center Z(G).

    2. Relevant equations

    the Class Equation?
    Sylow theorems are in the next section, so presumably this is to be done without them.

    3. The attempt at a solution

    Not completely sure of a solution, but here's (at least some of) what we know:

    1. Since H is normal, [tex]ghg^{-1} \in H[/tex].
    2. Since [tex]|H|[/tex] is prime, [tex]H[/tex] is cyclic and abelian.
    3. [tex]G[/tex] is finite, with order [tex]|G| = p^nq[/tex].
    4. The normalizer [tex]N(H)[/tex] (stabilizer under conjugation) is all of [tex]G[/tex]...
    5. ...so [tex]|G| = |N(H)|[/tex] ??
    6. Probably some more relevant properties.

    And we want to show that [tex]H \subseteq Z(G)[/tex], i.e. [tex]H \subseteq \{g \in G | gx = xg \forall x \in G\}[/tex]
     
    Last edited: Oct 25, 2007
  2. jcsd
  3. Oct 25, 2007 #2
    So would these things imply that H = G is cyclic, thus abelian and is the center?
     
  4. Oct 25, 2007 #3
    Well, from the facts you've given, since H has prime order, it's cyclic, and every cyclic group is abelian. Now since H is normal, like you've also shown we have

    [tex]ghg^{-1} \in H[/tex]

    It can quite easily be shown that this is true if

    [tex] \forall g \in G[/tex] [tex] \forall h \in H, gh \in Hg[/tex]

    that is, [tex] \exists h' \in H[/tex] such that

    [tex] gh = h'g[/tex]

    but H is abelian so...make some conclusion.

    Since this holds for every member of H when applied to every member of G, the result follows
     
  5. Oct 25, 2007 #4
    I misinterpreted the problem. i was thinking of G acting on H, as a subgroup, and not of G acting on H element-wise, by just permuting the elements around in H.

    this means if H was not in Z(G), then its orbit would have order 2,...,p-1, none of which divide |H| and |G| since p is prime and it's the least prime that divides |G|.

    i believe this does it.
     
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