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Derived Subgroup (In particular Q8)

  1. Jul 16, 2011 #1
    So if G = Q8 = <a, b : a^4 = 1, b^2 = a^2, b^{-1}ab = a^{-1}>

    I'm fine with the notion of the derived subgroup G' = <[g,h] : g, h in G>

    (Where [g,h] = g^{-1}h^{-1}gh)

    But I can't see why G' = {1, a^2}, I can only seem to get everything to be 1!? i.e. g = a, h = a^3 ===> a^{-1}a^{-3}aa^3 = 1 etc.

    And given the conjugacy classes of Q8 are: {1}, {a^2}, {a, a^3}, {b, a^2b}, {ab, a^3b}

    Is it a case of just checking an element g and an element h from each of the 5 conjugacy classes, not all 8 elements? i.e. if we check g = b, we don't need to check g = a^2b as well.

    (Sorry if this should be in the homework area, but it's not a "homework" question, just a problem I've come across reading a textbook and is a more general derived subgroup problem too!)

    Thanks in advance!!! :-)
     
  2. jcsd
  3. Jul 16, 2011 #2

    micromass

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    Hi OMM! :smile:

    Just pick two elements that don't commute. For example, a and b.

    There is a very handy theorem for calculating derived subgroups:


    If G is a finite group, then the derived subgroup H of G is the smallest normal subgroup such that G/H is abelian.

    Q8 is not abelian, so {1} cannot be the derived subgroup. But Q8/{1,a2} is abelian (it contains 4 elements), so {1,a2} must be the derived subgroup. (it only needs to be checked that it's normal)
     
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