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Normal group contained in the center.

  1. Jan 21, 2012 #1
    I am doing exercises from Hungerford's text 'Algebra', and would appreciate if someone took the time to verify my write-up for me, and possibly provide me with tips how this could be done more efficiently (using less mathematical machinery)

    1. The problem statement, all variables and given/known data

    If a normal subgroup N of order p ( p prime) is contained in a group G of order [itex]p^n[/itex], then N is in the center of G.

    2. Relevant equations
    Orbit-Stabilizer Theorem: http://www.proofwiki.org/wiki/Orbit-Stabilizer_Theorem

    3. The attempt at a solution
    I construct the group action [itex]f: G \times H \rightarrow H[/itex] by this following rule: [itex]f((g,h)) = ghg^{-1}[/itex]. This is well-defined by normality of H of course.
    By Orbit-Stabilizer Theorem I know that the size of orbit of any [itex]h \in H[/itex] must have cardinality dividing [itex]|G| = p^n[/itex]. So cardinality of orbits is 1 or p (since anything bigger
    would imply more elements in an orbit than there are in H) and H is the disjoint union of orbits of it's elements, so all orbits cardinalities add up to p. But we see that the orbit of the identity in H must be of size 1 (itself), since for any [itex]g \in G[/itex] , [itex]geg^{-1} = e[/itex]. This means we have p-1 other elements in orbits, but orbit cardinalities have to divide p, so they are all of size 1.
    This means that for any [itex] h \in H[/itex] and all [itex] g \in G[/itex] we have [itex] ghg^{-1} = h[/itex], so all elements of H are in the center of G.
     
    Last edited: Jan 21, 2012
  2. jcsd
  3. Jan 21, 2012 #2

    Dick

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    Science Advisor
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    That's very nicely written. And the Orbit-Stabilizer Theorem is about the only piece of machinery you used. I can't think how you would do it more simply.
     
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