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Math Help - Isomorphic

  1. #1
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    Isomorphic

    Let G be any group with no proper , nontrivial subgroups and assume abs value(G)>1. Prove that G must be isomorphic to Z_p for some prime p.



    I know we have an isomorphism if a group is 1-1, onto, and the homomorphism property holds.

    G is cyclic means G=<a>, where a is a generator.

    I know a cyclic group of order n has exactly one subgroup of order m for each positive divisor m of n.
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  2. #2
    Senior Member roninpro's Avatar
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    Write |G|=n>1. Let a\ne e. What can be said about the order of \langle a \rangle?
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  3. #3
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    has exactly one subgroup of order m for each positive divisor m of n
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  4. #4
    Senior Member roninpro's Avatar
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    First, you don't know that G is cyclic. Second, you didn't answer the question. What is the order/size of \langle a\rangle?
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  5. #5
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    is it infinite order since there is no n such that a^n=e
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  6. #6
    Senior Member roninpro's Avatar
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    I took that your group G was finite. If this is so, then \langle a\rangle cannot be infinite!
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  7. #7
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    then it has finite order
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  8. #8
    Senior Member roninpro's Avatar
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    You should be able to pin it down, though. In particular, can 1<\langle a\rangle<|G|?
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  9. #9
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    I guess i"m confused on how to pin it down.
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  10. #10
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    order is n
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  11. #11
    Senior Member roninpro's Avatar
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    Precisely. If the order of \langle a\rangle were in between 1 and n, it would be a proper subgroup, contrary to your assumption. This means that \langle a\rangle=G; in other words, G is cyclic. Now you just need to show that |G| is prime.
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