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I'm hoping that this is an easy question for someone.

How many elements can we expect to have multiplicative order at most $n^{1/c}$ in one of the finite fields $\mathbb{F}_p$ with $p$ prime with $n \le p \le n^2$, for $c=\frac{9\log{n}}{\log{\log{\log{n}}}}$?

Note that I'm trying to bound the expectation from below.

CLARIFICATION

We're taking the average order in one field, out of the set of finite fields with $n \le p \le n^2$.

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    $\begingroup$ I'm a little confused what set you're counting. Is it $\{(p, i) : \text{$p$ prime, $n \le p \le n^2$, $i \in \mathbb F_p$, and $i^m = 1$ for some $0 < m \le n^{1/c}$}\}$? $\endgroup$
    – LSpice
    Commented Aug 31, 2022 at 21:11
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    $\begingroup$ @LSpice: Oh, sorry... Yeah, you could count it that way. I was looking for the order for each finite field. Then take the average of this value over all of the finite fields $\mathbb{F}_p$, for $n \le p \le n^2$. $\endgroup$
    – Matt Groff
    Commented Aug 31, 2022 at 21:19
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    $\begingroup$ The multiplicative group of the field of $q$ elements is cyclic of order $q-1$, so the number of elements of order $d$ for $d$ dividing $q-1$ is $\phi(d)$ (where $\phi$ is the Euler phi-function). So this is a question about $\phi(d)$ for small values of $d$ dividing $q-1$. $\endgroup$
    – Gerry Myerson
    Commented Sep 1, 2022 at 0:47
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    $\begingroup$ To a first approximation, isn't it just $\lfloor n^{1/c} \rfloor$, pretty much independently of the range averaged over? Each prime equal to $1 \pmod d$ contributes $\varphi(d)$ elements of order $d$ as pointed out by Gerry Myerson; but we expect the primes to be equally distributed among the totients of $d$ so there's cancellation. $\endgroup$
    – Peter Taylor
    Commented Sep 1, 2022 at 8:48

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