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Math Help - Cyclic groups of order greater then 2 have more then one generator

  1. #1
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    Cyclic groups of order greater then 2 have more then one generator

    My problem is to prove that cyclic groups of order 3 or greater must have at least 2 generators.

    I think I found an answer but I'm worried that it's wrong.

    Here's my proof.

    In a group of order n generated by a there exists a "last element" a^{n-1}. This element can be shown to generate every inverse element and therefore every element in the group.

    For example a^{n-1}a^{n-1}=a^{2n}a^{-2}=ea^{-2}=a^{-2}
    a^{n-1}a^{n-1}a^{n-1}=a^{3n}a^{-3}=ea^{-3}=a^{-3}

    and so on until a^{n-1}a^{n-1}\cdots(n times)\cdots a^{n-1}=a^{n^2}a^{n}=e

    Thus it is shown that in every cyclic group with n>2 the element a^{n-1} generates the group as well as a.

    Is this valid?
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  2. #2
    Senior Member Tinyboss's Avatar
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    That's right--the inverse of a generator is a generator, and a generator is never its own inverse unless it's the only nonidentity element.
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  3. #3
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    Right, except I'm not sure if you're allowed to assume the cyclic group is finite. (I'm not sure if saying |G|>2 is supposed to mean that G is finite or not.) In this situation, though, just replace a^{n-1} with a^{-1} (which, in the case of a finite cyclic group, are the same thing).
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