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Normal subgroups and Abelian property

  1. Sep 16, 2008 #1
    If all subgroups of a group are normal, is the group abelian?
    I know that the answer is NO...

    Can you give a counter-example..???

    Better still can you logically deduce some property the counter-example must have, which will ease our way to finding it...
  2. jcsd
  3. Sep 16, 2008 #2


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    Whenever you want to look for a nonabelian counterexample, it's usually good to try the quaternion group Q first (and then the symmetric and dihedral groups!). In the case at hand this strategy is particularly fruitful, because
    (1) Q does have this property (all its subgroups are normal), and
    (2) every group which has this property contains a copy of Q as a subgroup.

    The idea is that if the group G which we're after is nonabelian, then it contains two noncommuting elements x and y in G. If we let Q0 be the subgroup of G generated by x and y, then it can be shown that Q0 is isomorphic to the quaternion group. For a proof, see Rotman's Theory of Groups.
  4. Sep 17, 2008 #3
    Thanks for the example and the suggestion...

    Firs thing I tried with was dihedral groups and was unsuccessful, quaternion groups never crossed my mind...
    I tried with the Klein four subgroup (dihedral group with [tex]x^2=e,y^2=e[/tex]) which satisfies the condition, but is Abelian and thus obviously ahould satisfy the conditions ,of no use...

    A little more patience may have helped because the quaternion group is similar to Klein four group...

    For example,
    In Klein four subgroup, [tex]x^2=y^2=z^2=e[/tex] and [tex]xy=z,yz=x,xz=y[/tex]
    And in quaternion,[tex]x^2=y^2=z^2=-e[/tex] and [tex]xy=z,yz=x,zx=y[/tex]
    Is there some general quaternion group??
    I only know of the quaternion group of order 8...

    In that case, how can we have bijection between the subgroup generated by [tex]x,y[/tex] and quaternion group if their cardinality is not the same and thus, isomorphism??...

    EDIT: In Wikipedia, I see something called the generalised quaternion group, did you mean that??
    Last edited: Sep 17, 2008
  5. Sep 17, 2008 #4


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    Maybe my outline of the proof was overly simplistic. I don't remember much of it, but what I do remember is that you can always choose two appropriate noncommuting elements x and y such that the group generated by {x,y} is isomorphic to the quaternion group of order 8. The proof isn't exactly easy though.
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