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  1. #1
    ynj
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    isomorphism

    prove:U(p)is isomorphic to Z(p-1),where p is prime.Z(p-1)is the additive group of modulo p-1.U(p) means all the positive intergers that is relatively prime to p.
    Thanks a lot!!
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  2. #2
    Super Member Gamma's Avatar
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    More General result

    In fact this is a consequence of a more general theorem.

    A finite subgroup of the multiplicative group of a field is cyclic. In particular, if F is a finite field, then the multiplicative group F^{\times} of nonzero elements of F is a cyclic group.

    By The Fundamental Theorem, the finite subgroup can be written as a direct product of cyclic groups

    \mathbb{Z}/\mathbb{Z}_{n_1}\times \mathbb{Z}/\mathbb{Z}_{n_2} \times ... \times \mathbb{Z}/\mathbb{Z}_{n_k}

    Where n_k|...|n_2|n_1.

    In general, if G is a cyclic group and d||G|, then G contains precisely d elements of order dividing d. Since n_k divides the order of each of the cyclic groups in the direct product, it follows that each direct factor contains n_k elements of order dividing n_k. If k were greater than 1, there would therefore be a total of more than n_k such elements. But then there would be more than n_k roots of the polynomial p(x)=x^{n_k}-1 in the field F, which contradicts the fact that a polynomial of degree n can have at most n roots. Thus k=1 and the group is cyclic.

    Notice \mathbb{Z}_p is a finite field, so your question is a trivial corollary of the above theorem.
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  3. #3
    ynj
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    Thank you for your solution! In your solution, there are some theorems that I still don't know. I am a newcomer of abstract algebra. So many things to be studied.....
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