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Core of a Group

  1. Feb 26, 2012 #1
    Let H be a subgroup of G and define the core of H as such
    core H={g[itex]\in[/itex]G| g[itex]\in[/itex]aHa^-1 for all a[itex]\in[/itex]G}= [itex]\bigcap[/itex]{aHa^-1|a[itex]\in[/itex]G}
    Prove that the core of H is normal in G and core H[itex]\subset[/itex]H.


    I am having a hard time proving this because isn't the definition of core H basically saying the the core is normal?
     
  2. jcsd
  3. Feb 26, 2012 #2

    Office_Shredder

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    It is pretty close to tautological, but as I like to say if it's so obvious it should be easy to prove. You need to do two things:

    1) Prove core(H) is actually a subgroup
    2) Prove that for a∈G, acore(H)a-1=core(H)
     
  4. Feb 26, 2012 #3
    Ok so would I say something like
    If e is in core H becaus aea^-1=e
    Let g and h^-1 be in the core then

    ag^a-1(aha^-1)=agh^-1a^-1 ..........is this right?
    so its a a sub group by the one step subgroup test.
     
  5. Feb 26, 2012 #4

    Office_Shredder

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    As long as you state that that's true for all a, that looks good to me
     
  6. Feb 26, 2012 #5
    Ok thanks, I thought there might have been something I was missing
     
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