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Math Help - Properties of Z_{n}

  1. #1
    Junior Member platinumpimp68plus1's Avatar
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    Properties of Z_{n}

    I want to show that the elements of \mathbb{Z}_n under multiplication that have inverses are also the generators of \mathbb{Z}_n under addition.

    My answer so far: The only integers with multiplicative inverses are -1 and 1; and these are generators of ( \mathbb{Z}_n, +_{n}).

    I feel like this answer is too simple though... Can I show this more rigorously? Besides saying that (obviously) the gcd(1,n)=1, and \phi(x)=(1)x, \phi(x)=(-1)x, generates \mathbb{Z}_n? Or am I interpreting this question completely wrong?
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  2. #2
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    An element a \in \mathbb{Z}_n has multiplicative inverse iff (a,n)=1 which happens iff <a>= \mathbb{Z}_n, which means not only -1 and 1 have multiplicative inverses.
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  3. #3
    Junior Member platinumpimp68plus1's Avatar
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    Quote Originally Posted by Jose27 View Post
    An element a \in \mathbb{Z}_n has multiplicative inverse iff (a,n)=1 which happens iff <a>= \mathbb{Z}_n, which means not only -1 and 1 have multiplicative inverses.
    OOOOOOOOOH. Doh. Thank you. Bear with my stupidity, I'm in an abstract algebra course and did not take the recommended prereq so I'm a little lost at the moment. I was confusing the infinite \mathbb{Z} with \mathbb{Z}_n, I think.

    So does the fact that (a,n)=1 iff a generates \mathbb{Z}_n sufficiently answer this question? Or do you think there is more I need to show?
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  4. #4
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    Quote Originally Posted by platinumpimp68plus1 View Post

    So does the fact that (a,n)=1 iff a generates \mathbb{Z}_n sufficiently answer this question? Or do you think there is more I need to show?
    You would have to prove: a has inverse iff (a,n)=1 and then <a>= \mathbb{Z}_n iff (a,n)=1 (notice there are two iff's to be proven) but yes, this would finish it, since by transitivity you would be saying that a has an inverse iff a generates the whole additive group.
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  5. #5
    Junior Member platinumpimp68plus1's Avatar
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    I think I have the first statement proven, but I'm unsure about how to prove the second... any tips on where to start?
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  6. #6
    MHF Contributor Bruno J.'s Avatar
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    If (a,n)=1 then you can find integers x,y such that xa + yn = 1; clearly x is the multiplicative inverse of a (mod n).

    To show that a generates \mathbb{Z}_n as an additive group, consider the elements a,2a,3a,...,na. I claim that they are all distinct; because if ja\equiv ka \mod n, then a(j-k)\equiv 0 \mod n. Since a is invertible we can multiply by its inverse on both sides which yields j \equiv k \mod n. Hence \{a,2a,...,na\}=\mathbb{Z}_n.

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  7. #7
    Junior Member platinumpimp68plus1's Avatar
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    Thank you!!!
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