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Let $p\geq 2$ be a prime and let g be an element of order $p-1$ in $\Bbb Z_p$. Prove that every non-zero element of $\Bbb Z_p$ can be written as a power of $g$.

So i wanted to start this proof by proving the the elements $[g],[g]^2,[g]^3,...,[g]^{p-1}$ are all distinct. But im a bit uncertain on how do this . I thought it could be something with inverses since we are in $\Bbb Z_p$ but that didn't really workout.

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    $\begingroup$ If those elements are not distinct then two of them are the same. Deduce that the order of $g$ must then be lower than $p-1$. $\endgroup$
    – lulu
    May 21 '19 at 18:20
  • $\begingroup$ @lulu well the order is the smallest number m such that $g^m \equiv\ 1\ mod\ p$. So if two elements are the same then there exist a k and y such that $[g]^y = [g]^k$ but these y and k don't say anything about the order, correct? they just say that two powers give the same result with the base g. $\endgroup$
    – johnnyB
    May 21 '19 at 18:32
  • $\begingroup$ If $g^y=g^k$ with $y>k$ then $g^{y-k}=1$. $\endgroup$
    – lulu
    May 21 '19 at 18:36
  • $\begingroup$ @lulu Ok so i think i'm with both of you only one thing remains. There was a post here that got removed it seems that asked why the inverse of $[g]^k$ is $[g]^{-k}$ in modular arithmetic the inverse is usually noted as $[g]^{-1}$ the thing that got me thinking is that $[g]^{-k} will be a fraction and not an integer. So is this really correct? $\endgroup$
    – johnnyB
    May 23 '19 at 14:03
  • $\begingroup$ Fraction? There are no fractions in modular arithmetic, not in the usual sense. If $g$ is invertible, then we can speak of $g^{-1}$ and we can raise that inverse to the $k^{th}$ power. That's what we mean by $g^{-k}$. $\endgroup$
    – lulu
    May 23 '19 at 14:14
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Hint Suppose the powers $g, g^2, \ldots, g^{p - 1}$ are not distinct, so that $g^k = g^l$ for some $1 \leq k < l \leq p - 1$. Multiply both sides by $g^{-k}$.

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  • $\begingroup$ Oh $g^{-k}$ is the inverse?If we multipliy we get that $g^k = g^l$ multiplying by $g^{-k}$ we get $g^k * g^{-k} = g^l * g^{-k}$ which is equal to $1 = g^{l-k}$ So since k < l the power will be positive. But how do we know that $l-k \not =0?$ if that is the case then this is not a problem. Since $g^0 = 1$ $\endgroup$
    – johnnyB
    May 21 '19 at 18:49
  • $\begingroup$ Since $1 \leq k < l \leq p - 1$ we have $1 \leq l - k < p - 1$. $\endgroup$ May 21 '19 at 21:23
  • $\begingroup$ Is the fact that [1] is not in $Z_p$ the contradiction then? That is we can write non-primes as a power of g. I'm sorry but i am having some difficulty seeing the contradiction. $\endgroup$
    – johnnyB
    May 23 '19 at 5:43
  • $\begingroup$ Since $g^{l-k} = 1$, the order of $g$ divides $l-k$ and so cannot be $p-1$. $\endgroup$ May 23 '19 at 6:40

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