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Math Help - group of automorphisms

  1. #1
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    group of automorphisms

    Let p be odd prime number such that  2p+1 is prime as well. Let  n=4p+2. Identify Aut(\mathbb{Z}_n).

    Here's what I did.

    Aut(\mathbb{Z}_n) is isomorphic to the group U_{\phi(n)}=(numbers relatively prime to n, ._n), where \phi(n) denotes the number of these numbers relatively prime to n.
    But what is this group..?


    Also, \mathbb{Z}_n=\mathbb{Z}_{2(2p+1)} which is isomophic to \mathbb{Z}_2 \oplus \mathbb{Z}_{2p+1}, but that doesn't get me anywhere either.

    Please help, thank you.
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  2. #2
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    Quote Originally Posted by georgel View Post
    Let p be odd prime number such that  2p+1 is prime as well. Let  n=4p+2. Identify Aut(\mathbb{Z}_n).

    Here's what I did.

    Aut(\mathbb{Z}_n) is isomorphic to the group U_{\phi(n)}=(numbers relatively prime to n, ._n), where \phi(n) denotes the number of these numbers relatively prime to n.
    But what is this group..?


    Also, \mathbb{Z}_n=\mathbb{Z}_{2(2p+1)} which is isomophic to \mathbb{Z}_2 \oplus \mathbb{Z}_{2p+1}, but that doesn't get me anywhere either.

    Please help, thank you.
    Notice that \phi(n) = \phi(4p+2) = \phi(2)\phi(2p+1) = \phi(2p+1) = 2p.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Notice that \phi(n) = \phi(4p+2) = \phi(2)\phi(2p+1) = \phi(2p+1) = 2p.

    Thank you!

    So now the group is isomorphic to either  \mathbb{Z}_2\otimes\mathbb{Z}_p or \mathbb{Z}_{2p}. I check the order of the elements to determine which one.
    Is this okay?
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  4. #4
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    Quote Originally Posted by georgel View Post
    Thank you!

    So now the group is isomorphic to either  \mathbb{Z}_2\otimes\mathbb{Z}_p or \mathbb{Z}_{2p}. I check the order of the elements to determine which one.
    Is this okay?
    note that for any prime q and n \geq 1: \ Aut(\mathbb{Z}/q^n\mathbb{Z}) \cong \mathbb{Z}/(q^n-q^{n-1})\mathbb{Z}. in particular Aut(\mathbb{Z}/q\mathbb{Z}) \cong \mathbb{Z}/(q-1)\mathbb{Z}. also, if \gcd(a,b)=1, then: (\mathbb{Z}/ab\mathbb{Z})^{\times} \cong (\mathbb{Z}/a\mathbb{Z})^{\times} \times (\mathbb{Z}/b\mathbb{Z})^{\times}. thus:

    Aut(\mathbb{Z}/n\mathbb{Z}) \cong (\mathbb{Z}/n\mathbb{Z})^{\times} \cong (\mathbb{Z}/2\mathbb{Z})^{\times} \times (\mathbb{Z}/(2p+1)\mathbb{Z})^{\times} \cong \mathbb{Z}/2p \mathbb{Z}.
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