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The Trouble with Normal Subgroups

  1. May 9, 2008 #1

    Tom Mattson

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    [SOLVED] The Trouble with Normal Subgroups

    1. The problem statement, all variables and given/known data
    Find an example of a group [itex]G[/itex] and subgroups [itex]H[/itex] and [itex]K[/itex] such that [itex]H[/itex] is normal in [itex]K[/itex], [itex]K[/itex] is normal in [itex]G[/itex], but [itex]H[/itex] is not normal in [itex]G[/itex].


    2. Relevant equations
    None.


    3. The attempt at a solution
    I have many attempts. :grumpy:

    In the beginning I was fixated on groups of orders 8 and 16, because I figured it would be easy to find nested subgroups with index 2 (that is, [itex][G:K] =2[/itex] and [itex][K:H]=2[/itex]). Finding such groups is fairly easy, but finding them with [itex]H[/itex] not normal in [itex]G[/itex] has proved elusive.

    Here are my false starts.

    Attempt 1:
    [itex]G=D_8[/itex], the dihedral group of order 8
    [itex]K=<b>[/itex], the subgroup of order 4 generated by [itex]b[/itex]
    [itex]H=<b^2>[/itex], the subgroup of order 2 generated by [itex]b^2[/itex]

    Result: No good, [itex]H[/itex] is normal in [itex]G[/itex]

    Attempt 2:
    [itex]G=Q_8[/itex], the quaternionic group of order 8
    [itex]K=\left<b\right>[/itex]
    [itex]H=\left<b^2\right>[/itex]

    I didn't even bother finishing this one. It became obvious that this attempt would fail for the same reason that the last one failed.

    Attempt 3:
    Here's where I tried to be more clever.

    [itex]G=GL_2\left(\mathbb{Z}_3\right)[/itex]
    [itex]K=SL_2\left(\mathbb{Z}_3\right)[/itex]
    [itex]H=[/itex] the subgroup generated by the elements of order 4 in [itex]K[/itex] (this is a subgroup of order 8).

    Once again, [itex]H[/itex] is normal in both [itex]K[/itex] and [itex]G[/itex].

    I also did some goofing around with [itex]S_n[/itex] and [itex]A_n[/itex], but to no avail.

    Little help?
     
  2. jcsd
  3. May 9, 2008 #2
    Try the dihedral group with order 8.
     
  4. May 13, 2008 #3

    Tom Mattson

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    That was my first attempt. It didn't work.
     
  5. May 13, 2008 #4

    Tom Mattson

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    Oh brother! I've got it. Like a dummy I was fixated on cyclic subgroups. I just considered a subgroup with two generators, and I got it.

    FYI, [itex]\left<ab^2\right>[/itex] is normal in [itex]D_8[/itex], [itex]\left<a,b^2\right>[/itex] is normal in [itex]\left<ab^2\right>[/itex], but [itex]\left<ab^2\right>[/itex] is not normal in [itex]D_8[/itex], as [itex]b^{-1}ab^2b=abb^3=ab^4=a[/itex], which is not in [itex]\left<ab^2\right>[/itex].

    So it was [itex]D_8[/itex] all along...
     
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