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Classification of [itex]Z_7 \rtimes Z_6[/itex]

  1. Apr 12, 2013 #1
    1. The problem statement, all variables and given/known data

    I was trying to classify [itex] Z_7 \rtimes Z_6 \cong Z_2 \times Z_3[/itex], but I have questions about it...

    2. Relevant equations



    3. The attempt at a solution

    The possible homomorphisms are:

    1) [itex]\alpha[/itex] is trivial, so [itex]G \cong Z_7 \times Z_6[/itex]

    2) Kernel of [itex]\alpha[/itex] = [itex]Z_3[/itex]. So [itex]G \cong Z_7 \rtimes (Z_2 \times Z_3) \cong (Z_7 \rtimes Z_2) \times Z_3 = D_{14} \times Z_3/[/itex].

    3) Kernel of [itex]\alpha[/itex] = [itex]Z_2[/itex]. So [itex]G \cong Z_7 \rtimes (Z_2 \times Z_3) = (Z_7 \rtimes Z_3) \times Z_2 \cong Z_2 \times \text{(nonabelian group of order 21)}[/itex]. This nonabelian group of order 21 cannot be the dihedral or the quaternionic groups because the order is not even, right? It cannot be the symmetry or klien groups either. So do I just leave it this way, or do I need to continue?

    4) Kernel of [itex]\alpha[/itex] = [itex]\{e\}[/itex]. So [itex]G \cong Z_7 \rtimes Z_6[/itex]. This can't be isomorphic to the dihedral or quaternionic groups either, because it has a normal subgroup of order 7...and if it was dihedral or quaternionic, then the only normal subgroup would be of order 12, right? It can't be the symmetric or the klien groups either. So I'm also stuck with this. Is it ok to leave it this way?

    Thanks in advance
     
    Last edited: Apr 12, 2013
  2. jcsd
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