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Normal Subgroup

  1. Apr 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Let Dn = {1,a,..an-1, b, ba,...ban-1} with |a|=n, |b|=2,
    and aba = b.
    show that every subgroup K of <a> is normal in Dn.


    3. The attempt at a solution
    First, we show <a> is normal in Dn.
    <a> = {1,a,...an-1} has index 2 in Dn and so is normal
    by Thm (If H is a subgroup of index 2 in G, then H is normal in G.)
    Next, Since <a> is cyclic, K is also cyclic and abelian.
    Let k in K, x in G and 1 in G. ( G = Dn)
    k = 1k = (xx-1)k = x(kx-1) because K is abelian.
    k in K => xkx-1 in K for all x in G
    => K is a normal in G = Dn.
     
  2. jcsd
  3. Apr 10, 2009 #2
    are you sure the relation isn't meant to be [itex]aba^{-1}=b[/itex]?

    also i think it suffices to show that you have an index of 2 OR that for [itex]n \in N, \forall g \in G, gng^{-1} \in N[/itex] in order to imply [itex]N \triangleleft G[/itex].
     
  4. Apr 10, 2009 #3

    matt grime

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    This doesn't follow. K is abelian, so elements of K commute with themselves; they do not commute with arbitrary elements of G. But then they do not need to - you're trying to show that K is normal, not central.
     
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