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Group is a union of proper subgroups iff. it is non-cyclic

  1. Dec 8, 2016 #1
    1. The problem statement, all variables and given/known data

    Prove that a finite group is the union of proper subgroups if and only if the group is not cyclic.

    2. Relevant equations

    None

    3. The attempt at a solution

    " => "
    If the group, call it G, is a union of proper subgroups, then, for every subgroup, there is at least one element of G that is not in that particular subgroup. But then we know that none of the subgroups can represent all the elements of G. Therefore, G is not cyclic.

    " <= "
    If the group is not cyclic, then no element a in G generates G. That means that G is the union of all the <a> subgroups for all a in G.

    Is this correct?
     
  2. jcsd
  3. Dec 8, 2016 #2

    fresh_42

    Staff: Mentor

    Looks fine to me.
     
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