Let $(\mathbb{Z}_+,+_p)$ be the integers modulo $p$ for a prime number $p$. Is every element a generator of the group for any prime $p$?

The question I am asking can be rephrased as follows: Let \begin{equation*} \mathcal{U}_n=\{i \mid hcf (i,n)=1,~1\leq i\leq n,~i\in\mathbb{N}\}. \end{equation*} Then $\forall u\in\mathcal{U}_n$, do we have $\langle u \rangle =\mathbb{Z}_p$ for any $p$?

This makes sense for some simple examples. Eg, for $(\mathbb{Z}_5,+_5)$, $3$ is generator since $3\to 1\to 4\to 2\to 0\to 3$, ie, $5$ times. Another example is $(\mathbb{Z}_7,+_7)$. The member $6$ is a generator since $6\to 5\to 4\to 3\to 2\to 1\to 0\to 6$. I am not sure if this holds for any member and any prime.

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    $\begingroup$ Yes, any non-zero element generates the additive group mod a prime. Your rephrasing does not seem to be a well-formed set. $\endgroup$ – Tobias Kildetoft Oct 4 '16 at 10:57
  • $\begingroup$ If $hcf = \gcd$ then $\mathcal{U}_n$ consists of all elements, which have a multiplicative inverse. So your group seems to be $(\mathbb{Z}_p,*_p)$. And of course are all $u \in \mathcal{U}_n$ generators of the additive group, because $u\ne 0.$ What is the relation between $n$ and $p$ or do you mean $ \mathcal{U}_p?$ $\endgroup$ – gammatester Oct 4 '16 at 11:03

The answer is yes. It follows from Lagrange's theorem that the order of any element (i.e. the size of the group generated by any element) of a group of size $p$ must be either $1$ or $p$.


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