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Math Help - Groups (Z mod p)

  1. #1
    Member Jason Bourne's Avatar
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    Groups (Z mod p)

    (a) Show that if p is prime, all non-zero elements of \mathbb{Z}_{p} have a multiplicative inverse.

    (b) Deduce that  \mathbb{Z}^*_{p}, the set of non-zero elements of \mathbb{Z}_{p}, is a group under multiplication mod p.

    (c) Show that if n is not prime, then  \mathbb{Z}^*_{n} is not a multiplicative group.

    (d) Is  \mathbb{Z}^*_{17} cyclic? (prime order )

    Please help
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  2. #2
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    Quote Originally Posted by Jason Bourne View Post
    (a) Show that if p is prime, all non-zero elements of \mathbb{Z}_{p} have a multiplicative inverse.
    Let [x]_p \in \mathbb{Z}_p be a non-zero element (i.e. [x]_p \not = \{ 0,\pm p,\pm 2p,...\}). Now consider \{ [x]_p,[2x]_p,...,[(p-1)x]_p\}. None of these are equal to eachother, because otherwise x i \equiv x j (\bmod p) since \gcd (x,p)=1 it means i\equiv j (\bmod p) which is impossible since we are assuming that i\not = j. That means by pigeonhole principle that this set has to be a permutation of \{ [1]_p,...,[p-1]_p\} in some order. And so, there exists k such that kx]_p = [1]_p thus [k]_p[x]_p = [1]_p this means [k]_p is an inverse for [x]_p.

    (b) Deduce that  \mathbb{Z}^*_{p}, the set of non-zero elements of \mathbb{Z}_{p}, is a group under multiplication mod p.
    I did the hard part in (a). You should be able to do this now.

    (c) Show that if n is not prime, then  \mathbb{Z}^*_{n} is not a multiplicative group.
    Just show there exists an element that does not have an inverse. Hint: if n\geq 2 is not prime then there exists m\geq 2 such that \gcd (n,m)\not = 1, and argue that no such [x]_n can be found such that [mx]_n = [1]_n.

    (d) Is  \mathbb{Z}^*_{17} cyclic?
    Yes.
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