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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 generator $g$ of this subgroup (this subgroup is always cyclic) ?

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

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 $k=7$ $g=24$

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 generator $g$ of this subgroup (this subgroup is always cyclic).

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

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 generator $g$ of this subgroup (this subgroup is always cyclic) ?

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

I have a question about properties of the multiplicative groups.

Let we have finite field of prime order $2^k$ -1.

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

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

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

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

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 generator $g$ of this subgroup (this subgroup is always cyclic).

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