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I have a question about properties of the multiplicative groups.

Let we have finite field of prime order $2^k$ -1. $k$ is co-prime with $$\frac{2^k-2}{k}$$.

It is clear that multiplicative group of such field has subgroup of order.

$$\frac{2^k-2}{k}$$.

How it is possible to find formula for generator $g$ of this subgroup (this subgroup is always cyclic) ?

For example for $k=5$ $g=6$ For example for $k=7$ $g=24$

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    $\begingroup$ The equality $k = \frac{2^k - 2}k$ is obviously false. (For example, as you say, for $k = 5$ it does not work.) What do you mean? $\endgroup$
    – LSpice
    Commented Jun 11, 2020 at 13:58
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    $\begingroup$ How can the question be the same? You've made a conjecture, and shown that it is false. $\endgroup$
    – LSpice
    Commented Jun 11, 2020 at 14:37
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    $\begingroup$ OK, I see that you have changed the question to remove the incorrect conjecture; now it makes sense. The subgroup is cyclic—all finite subgroups of multiplicative groups of fields are—so there is no need to gather evidence for the existence of $g$. My suspicion is that this is probably not much easier than finding a primitive generator for a random $\mathbb Z/p\mathbb Z$, which is hard, but I have no evidence for that. It may be interesting even to see if you can come up with an $\mathrm o(2^k/k)$ algorithm for testing membership in this subgroup (but I'm no expert; it may be known). $\endgroup$
    – LSpice
    Commented Jun 11, 2020 at 14:45
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    $\begingroup$ You might as well ask for a generator $x$ for the whole multiplicative group of the field, and then take $x^{k}$. $\endgroup$
    – Geoff Robinson
    Commented Jun 11, 2020 at 15:31
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    $\begingroup$ I can’t give you an example, as only two Wieferich primes are known. But anyway, the multiplicative group is $C_{2^k-2}$, which is isomorphic to $C_k\times C_{(2^k-2)/k}$ only if $k$ is coprime to $(2^k-2)/k$, and there is no a priori reason this should be the case. $\endgroup$
    – Emil Jeřábek
    Commented Jun 11, 2020 at 16:26

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