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Groups and Subgroups!

  1. Jul 10, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove that if a group G has no subgroup other than G and {e}, then G is cyclic....

    2. Relevant equations



    3. The attempt at a solution

    we could say that, let a E G - {e} then we construct <a>...
     
  2. jcsd
  3. Jul 10, 2008 #2

    quasar987

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    Sounds good. Why are you having doubts?
     
  4. Jul 10, 2008 #3
    how do I construct <a> ??
     
  5. Jul 10, 2008 #4

    HallsofIvy

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    Well, you were the one who used that notation- what is the definition of "<a>"?
     
  6. Jul 10, 2008 #5

    quasar987

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    Huh? Well by definition, <a>={..., a^-1,e,a,a²,a³,...}. Since <a> is a subgroup by construction and since it is not equal to {e}, it is equal to G.
     
  7. Jul 10, 2008 #6
    oh yeah sure! I meant how would say, after constructing it! I forgot it is equal to {e} it is equal to G...this is what I get!
    We could say that, let a E G - {e} then we construct <a> =
    a, a^2, a^3... a^n=1 <- cyclic subgroup generated by element a.
    Since <a> is subgroup, it must coincide with G (because G is the only subgroup), but then a generates G: <a> = G, which is definition of G being cyclic....
    are we done after saying this statement or should we have to mention something else?
    thanks for offering help!
     
  8. Jul 10, 2008 #7

    quasar987

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    Well what you wrote for the definition of <a> is incorrect, and don't forget to say "since <a> is not {e}, it is G".
     
  9. Jul 10, 2008 #8
    ohh yeah I am sorry about that...I was in hurry or something to write it down! but yeah I believe everything else is fine and I would not forget to mention that <a> is not equal to {e}..it is G!
    Thanks!
     
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