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Math Help - Cyclic group problem

  1. #1
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    Cyclic group problem

    Let G be a group of order 25. Prove that G is cyclic or g^5=e for all g in G.
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  2. #2
    Super Member Rebesques's Avatar
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    Well, it is either cyclic or not.

    Suppose it is not cyclic and choose an element g in G. The order of g, <g>, must divide the order of G, so <g> divides 25 or <g> is 1 or 5 or 25. In every case, the result holds true.
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  3. #3
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    Oh, man, this problem is easy. I really should have gotten it.

    Thanks.
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  4. #4
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    The group is not necessarity cyclic, how about \mathbb{Z}_5\times \mathbb{Z}_5.

    If G is cyclic then it is clearly true because G\simeq \mathbb{Z}_{25} and \left< 5 \right> = \{ 0,5,10,15,20\}.

    If G is not cyclic then choose x\in G so that x\not = e. Then \left< x\right> is a subgroup of G. It cannot (subgroup) have order 1 because it is not identity and it cannot be 25 (the full group) because we assumed it is not cyclic. Thus it must be 5 by Lagrange's theorem.
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  5. #5
    Super Member Rebesques's Avatar
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    I don't see where you disagree PH
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  6. #6
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    Quote Originally Posted by Rebesques View Post
    I don't see where you disagree PH
    I do not. I wanted to be more clear to tttcomrader.
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