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Math Help - a group that is a subgroup of Zn

  1. #1
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    a group that is a subgroup of Zn

    For each odd composite numbern=>3, any number satisfying a^(n-1)≡1(modn )
    is said to be a Fermat liar for n.
    F-Liar={ a/1<= a<= n and a^(n-1)≡1(modn }
    Show that F-Liar is a subgroup of Z^*n
    i tried to solve this by finding the size or order of F and it must divide the order of Zn to be F a subgroup of Zn.
    I think the size of Zn is n-1
    and the size of F is n
    but i could not get the whole prove
    the order of Zn is less than or equal n/2 and the order of F must be less than or equal the order of Zn but how can I show that.

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  2. #2
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    Quote Originally Posted by koko2009 View Post
    For each odd composite numbern=>3, any number satisfying a^(n-1)≡1(modn )
    is said to be a Fermat liar for n.
    F-Liar={ a/1<= a<= n and a^(n-1)≡1(modn }
    Show that F-Liar is a subgroup of Z^*n
    i tried to solve this by finding the size or order of F and it must divide the order of Zn to be F a subgroup of Zn.
    I think the size of Zn is n-1
    and the size of F is n
    but i could not get the whole prove
    the order of Zn is less than or equal n/2 and the order of F must be less than or equal the order of Zn but how can I show that.

    This is a mess: first, \mathbb{Z}_n^* is the multiplicative group of all units in \mathbb{Z}_n and its order is \phi(n) = Euler's Totient Function.
    Second, you need to show that the product of any two Fermat liars is again a F.L., which is pretty easy since everything's abelian here.

    Tonio
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