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Math Help - prove not cyclic

  1. #1
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    prove not cyclic

    Hi All,

    How to I Claim: For any interger n> 1, the group u(n(n+2)) is not cyclic?

    Thanks.
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  2. #2
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    Quote Originally Posted by nguy0688 View Post
    Hi All,

    How to I Claim: For any interger n> 1, the group u(n(n+2)) is not cyclic?

    Thanks.
    If \gcd(n,m) = 1 then \text{U}(\mathbb{Z}/nm\mathbb{Z}) \simeq \text{U}(\mathbb{Z}/n\mathbb{Z}) \times \text{U}(\mathbb{Z}/m\mathbb{Z}) by the Chinese remainder theorem.

    Now if n is odd then \gcd(n,n+2) =1 and so \text{U}(\mathbb{Z}/n(n+2)\mathbb{Z}) \simeq \text{U}(\mathbb{Z}/n\mathbb{Z}) \times \text{U}(\mathbb{Z}/(n+2)\mathbb{Z}). Now this cannot be cyclic because the orders of those two groups are not relatively prime, since \phi(x) is for x>1.

    Similarly with n odd, just write, n(n+2) = 4(n/2)(n/2+1) and apply the argument.
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  3. #3
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    but. . .

    is there a way to prove this with out using the chinese remainder??? I am thinking about trying to prove it by showing that there is no element in u(n(n+2)) which generates the whole group. . . .but that's where i am stumped
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  4. #4
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    Quote Originally Posted by arsmath View Post
    is there a way to prove this with out using the chinese remainder??? I am thinking about trying to prove it by showing that there is no element in u(n(n+2)) which generates the whole group. . . .but that's where i am stumped
    Maybe, I am just not sure how easy that method is. What is so bad with what I posted?
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    Maybe, I am just not sure how easy that method is. What is so bad with what I posted?
    nothing wrong with your solution. . .just me trying to avoid the Chinese remainder thm.
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