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Cyclic Group

  1. Dec 9, 2009 #1
    Hi everyone.

    How could I prove if something is a cyclic group? I was wondering because I can prove is something is a group, a subgroup, and a normal subgroup, but I have no Idea as to how to prove something is a cyclic group.

    Ex: Suppose K is a group with order 143. Prove K is cyclic.

    I read around and I kept seeing something about Sylow's theorems, but I never learned anything like that. Is there another approach?
  2. jcsd
  3. Dec 10, 2009 #2
    To prove that K is cyclic, it is sufficient to prove that it has normal subgroups of order 11 and 13. There's a theorem that states that, if we can find such normal subgroups and their intersection is trivial, then K is isomorphic to their direct product. The only groups of order 11 and 13 are Z_11 and Z_13. Since both are cyclic and 11 and 13 are coprime, their direct product is cyclic.

    By Sylow's theorems, there's exactly one of each in K.

    This works quite generally for prime pairs as long as p[itex]\not =[/itex] 1 mod q and q[itex]\not =[/itex]1 mod p. If p = 1 mod q (such would be the case, for example, if one of the primes is 2), the subgroup of order 2 is not unique and therefore not normal: therefore, for example, there's a non-cyclic group of order 6, [itex]S_3[/itex] (3 = 1 mod 2), and there's a nontrivial non-cyclic group of order 21 (7 = 1 mod 3).

    The same method goes if there are more than two primes in the decomposition.
    Last edited: Dec 10, 2009
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