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Math Help - Normal subgroups

  1. #1
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    [SOLVED] Normal subgroups

    we know that if a group is abelian, all subgroups of it are normal. however - can you prove or disprove that if all subgroups of a group are normal, it is abelian?
    Last edited by shos; August 24th 2010 at 10:47 PM.
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  2. #2
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    We will disprove the claim.

    Consider the group Q_8 = \{\pm1,\pm i,\pm j,\pm k\} where i^2 = j^2 = k^2 = -1, (-1)^2 = 1. Also we need ijk = -1. Using the above relations you can show that Q_8 is a group. You can also show it is non-abelian by observing that ij = -ji.
    Note that Q_8 is generated by any two of i,j,k.

    Now show that if any subgroup H contains i then  \{\pm 1, \pm i\} \subseteq H. If  H = \{\pm 1, \pm i\}, then index of H in Q_8 is 2 and thus H is normal in Q_8. Suppose \{\pm 1, \pm i\} \subsetneq H, then one of j or k must belong to H. But then H = Q_8.

    We can proceed similarly for the case where a subgroup H contains j or k.
    The only remaining case is that H does not contain i, j or k. In this case H = \{1,-1\}, which is the center of Q_8 and thus H is normal in Q_8
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