In NSA (ZFC+IST), what can we say about generators for $\mathbb{Z}/\nu\mathbb{Z}$ for unlimited $\nu$? Recently I've been going through a short text on Nonstandard Analysis that uses the axiomatic approach of Nelson (Internal Set Theory - IST). Its study has led me to be curious about the properties of the fields of integers modulo $p$ prime when $p$ is unlimited ($\nu$). This is interesting because, in a sense, it contains all the arithmetic of the standard $\mathbb{Z}$ in a finite set. Although I don't necessarily expect to find anything novel, I think that it is instructive with regard to NSA/IST.
As there are infinitely many primes, we know from Idealization there is an unlimited prime I will call $\nu$. We would like to look at $\mathbb{Z}/\nu\mathbb{Z}$, where $\nu$ is an unlimited prime number. This can take the usual definition (and properties) per the Transfer axiom.
$$\mathbb{Z}/\nu\mathbb{Z}=\{[0],[1],[2],\ldots ,[\nu -1]\}$$
Note that in IST this is finite (despite "containing" all standard $\mathbb{Z}$) since there's a bijection with $\{1,2,\ldots ,\nu\}\subseteq\mathbb{N}$. (In the IST axiomatic approach to NSA, $\mathbb{N}$ itself contains nonstandard elements, unlimited numbers, that are simply "revealed" by the axioms. This is in contrast to the other NSA approaches that actually extend $\mathbb{N}$.)
For $\mathbb{Z}/p\mathbb{Z}$ we would have the existence of a $k\in\mathbb{N}, k<p$ such that
$$\mathbb{Z}/p\mathbb{Z}=\{[0],[k^1],[k^2],\ldots ,[k^{p-1}]\}$$
Therefore, I think it is reasonable to ask if there exists a $\kappa\in\mathbb{N}, \kappa<\nu$ such that
$$\mathbb{Z}/\nu\mathbb{Z}=\{[0],[\kappa^1],[\kappa^2],\ldots ,[\kappa^{\nu-1}]\}$$
In particular, is there an unlimited $\kappa$? Is it necessarily unlimited? Necessarily limited?
If so, we would have an unlimited "generator" for all the standard elements of $\mathbb{Z}/\nu\mathbb{Z}$ (plus some nonstandard).
Although any solution is welcome, I'd particularly appreciate answers capable of elucidating the situation with the axioms of IST.
 A: You are asking if the multiplicative group of $\mathbb{Z}/\nu\mathbb{Z}$ is cyclic, and this is true: $\mathbb{Z}/\nu\mathbb{Z}$ is a finite field, and  multiplicative groups of finite fields are cyclic, nothing stops you carrying out the usual proof in IST.
Consider the set of primes $p$ for which 2 is a generator of the multiplicative group mod $p$.  This is conjectured to be infinite, in which case it contains a nonstandard element (that infinite sets contain nonstandard elements is proved in Robert's book, indeed it's Basic Principle number 1 on the inside cover of my copy).  Better, according to an  answer to this MO question the set of primes for which $2,3$ or $5$ is a primitive root is known to be infinite, so there are illimited primes whose multiplicative groups are generated by $2,3$ or $5$.
The number of generators of the multiplicative group mod $\nu$ is $\phi(\nu-1)$.  Since $\nu-1$ is illimited, so is $\phi(\nu-1)$ (the prime factorization of $\nu-1$ must have either an illimited prime or a limited prime to an illimited power or contain an illimited number of limited distinct primes, so use multiplicitivity of $\phi$).  
Lemma: a set of natural numbers whose size is illimited must contain an illimited element. Proof: the set is finite, so has a maximum. If the maximum was standard then the set would be contained in a standard interval, so have standard size.  
We can conclude there is always an illimited primitive root.
It is a theorem that there exists a (standard!) constant $C$ such that there are infinitely many primes $p$ for which the least positive primitive root has size $C \log p$. Thus there is an illimited prime $\nu$ satisfying this, so all the elements of $[0,\nu-1]$ which generate the multiplicative group are illimited. (This is exactly André Nicolas' comment above).
A: Assuming $\nu$ is 1 mod 4, if $\kappa$ is a generator then so is $\nu-\kappa$. One of the two is necessarily illimited. Hence there is always an illimited generator.
Otherwise, $\nu-\kappa$ is the square of a generator $\alpha$.  Here if $\alpha$ were finite, then $\nu-\kappa$ would be finite as well.
Thus in either case, there is necessarily an illimited generator.  If $\kappa$ itself is finite, then either $\nu-\kappa$ (in the 1 mod 4 case) or $\alpha$ (in the 3 mod 4 case) is an illimited generator.
A: If $\nu=2^k$ is a power of 2 then the multiplicative group modulo $\nu$ is generated by the coset of the number $5$ (for all $k$).  By transfer this remains true for infinite powers $k$.  Therefore in case case there is a finite generator for the multiplicative group.
