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Thread: modulo arithmetic proof

  1. #1
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    modulo arithmetic proof

    Afternoon guys, I'm stuck on this proof and I'd really appreciate it if someone could help solve it. ($\displaystyle Z $ = set of integers)

    Question/Theorem

    Let $\displaystyle Z^*_p$ be the set of nonzero integers modulo p. ($\displaystyle p \geq 2 $) Prove $\displaystyle Z^*_p $ is a group with multiplication mod $\displaystyle p$ if and only if $\displaystyle p$ is a prime number.

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    My (lame) effort so far:

    We need to show:
    (1) $\displaystyle Z^*_p $ is a group with multiplication mod $\displaystyle p$ $\displaystyle \Rightarrow $ $\displaystyle p $ is prime.

    (2) if $\displaystyle p$ is prime, $\displaystyle Z^*_p $ is a group with multiplication mod $\displaystyle p$
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    (1) $\displaystyle \forall x \in Z^*_p $, $\displaystyle \exists x^{-1} \in Z^*_p $ such that $\displaystyle xx^{-1} = 1 $

    so $\displaystyle xx^{-1} = 1 \mod p $
    $\displaystyle xx^{-1} -1 = 0 \mod p $
    $\displaystyle \Rightarrow $ p divides $\displaystyle xx^{-1} -1 $
    $\displaystyle \Rightarrow \exists a \in Z $ such that $\displaystyle xx^{-1} - pa =1 $

    I don't know where to go from here, or indeed if this is the right way of approaching this.
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    (2) if $\displaystyle p$ is prime, then any integer $\displaystyle q$ such that $\displaystyle 1 \leq q \leq p-1 $ is coprime to $\displaystyle p$ and so $\displaystyle \exists $ n,m $\displaystyle \in Z $ such that $\displaystyle np + mq =1 $

    Now using that, we need to show the properties of a group are satisfied.

    Showing associativity and identity (=1) are easy enough. But I don't know how to show an inverse exists. i.e. show that $\displaystyle \exists x^{-1} \in Z^*_p $ such that $\displaystyle xx^{-1} = 1 $ using the above property of a prime number.

    __________________________________________________ __________________

    Please help

    Thank you.
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  2. #2
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    Quote Originally Posted by WWTL@WHL View Post
    Afternoon guys, I'm stuck on this proof and I'd really appreciate it if someone could help solve it. ($\displaystyle Z $ = set of integers).
    Consider $\displaystyle \{ [0],[1],...,[p-1]\}$ we want to show if $\displaystyle [x]\not = [0]$ then there exists $\displaystyle [y]$ such that $\displaystyle [xy]=[1]$. This is equvalent to saying $\displaystyle xy\equiv 1(\bmod p)$. Since $\displaystyle p$ is prime it means $\displaystyle \gcd(x,p)=1$ since $\displaystyle [x]\not = [0]$. Thus, $\displaystyle ax+bp = 1$ for some integers $\displaystyle a,b$. This means that $\displaystyle ax\equiv 1(\bmod p)$ for some $\displaystyle a$. Thus, if we let $\displaystyle [y] = [a]$ then $\displaystyle [xy]=1$.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Consider $\displaystyle \{ [0],[1],...,[p-1]\}$ we want to show if $\displaystyle [x]\not = [0]$ then there exists $\displaystyle [y]$ such that $\displaystyle [xy]=[1]$. This is equvalent to saying $\displaystyle xy\equiv 1(\bmod p)$. Since $\displaystyle p$ is prime it means $\displaystyle \gcd(x,p)=1$ since $\displaystyle [x]\not = [0]$. Thus, $\displaystyle ax+bp = 1$ for some integers $\displaystyle a,b$. This means that $\displaystyle ax\equiv 1(\bmod p)$ for some $\displaystyle a$. Thus, if we let $\displaystyle [y] = [a]$ then $\displaystyle [xy]=1$.
    Thank you very much, perfecthacker! I totally forgot about the 'nonzero' part, so I guess I was barking up the wrong tree with that $\displaystyle xx^{-1} -1 = 0 (modp) $ stuff.

    Now I'm going to be very, very cheeky and ask for any points for the 2nd bit of the proof. i.e. we can't assume p is prime, and deduce it is from the fact that $\displaystyle Z^*_p $ satisying the properties of a group.

    I've outlined my thoughts in my original post, but admittedly, I don't have any concrete ideas. If you, or anyone else, can nudge me in the right direction, that'd be superb.
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    Suppose $\displaystyle n\geq 2$ (nobody ever cares for when $\displaystyle n=1$ because the group on only one element is really unintersing and never comes up anyway). We will show $\displaystyle \mathbb{Z}_n^{\text{x}}$ is not a field. If it is a field then $\displaystyle [a][b]\not = [0]$ if $\displaystyle [a],[b]\not = [0]$ (why?). But if $\displaystyle n$ is not a prime then we can write $\displaystyle n=ab$ where $\displaystyle 1<a,b<n$, thus $\displaystyle [a][b] = [n] = [0]$ while $\displaystyle [a],[b]\not = [0]$. This means it cannot be a field if $\displaystyle n$ has factors.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    If it is a field then $\displaystyle [a][b]\not = [0]$ if $\displaystyle [a],[b]\not = [0]$ (why?).
    Thanks for the reply, but I'm confused. Consider $\displaystyle Z^*_6 $.

    Let $\displaystyle [a] = [2] \not = [0] $ and let $\displaystyle [b] = [3] \not = [0] $

    Then surely $\displaystyle [ab] = [a][b] = (2mod6) . (3mod6) = 6mod6 = [0] $ ?
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  6. #6
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    Quote Originally Posted by WWTL@WHL View Post
    Thanks for the reply, but I'm confused. Consider $\displaystyle Z^*_6 $.

    Let $\displaystyle [a] = [2] \not = [0] $ and let $\displaystyle [b] = [3] \not = [0] $

    Then surely $\displaystyle [ab] = [a][b] = (2mod6) . (3mod6) = 6mod6 = [0] $ ?
    Yes [a]=[2] and [b]=[3] are non-zero and [2][3] = [6] = [0] so it means it cannot be a group under multiplication.
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  7. #7
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    Quote Originally Posted by ThePerfectHacker View Post
    Yes [a]=[2] and [b]=[3] are non-zero and [2][3] = [6] = [0] so it means it cannot be a group under multiplication.
    D'oh.

    Of course, yes.

    Thank you very much for all your help.
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