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Take the integers coprime to $p$ (all but multiples of $p$). Does there always exist an integer (generator) $a$ coprime to $p$ that generates the entire group of coprime integers under powers of $a$?

For example $3^k\equiv 3, 2, 6, 4, 5, 1 \mod 7$ therefore $3$ is the generator of the group because it generates all coprime integers.

I'm not to familar with group theory, so I don't know where to start, any hints? Or maybe its more advanced than I think.

Dane Bouchie
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  • Yes, such an element is called primitive. I am not aware of a completely elementary proof that does not use at least some amount of group theory. – Tobias Kildetoft Feb 18 '15 at 09:22
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    If you mean you're investigating the group mod $p$, the answer is yes, they're called primitive roots. In particular, you have $\varphi(p-1)$ primitive roots mod $p$. – PITTALUGA Feb 18 '15 at 09:23
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    The reason this works is that the multiplicative group $K^$ of any finite field $K$ is cyclic. For if not, then there would be an $n< |K|-1$ with $g^n=1$ for all $g \in K^$, but then the equation $x^n-1$ would have more than $n$ solutions in $K$, which is impossible. – Derek Holt Feb 18 '15 at 09:26
  • You can see my solution here: https://math.stackexchange.com/a/4590349/575114 – Ganesh Chowdhary Sadanala Dec 03 '22 at 03:02

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