Number of primative roots in finite fields of order p^n

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The number of primitive roots in finite fields of order p^n is indeed given by the Euler totient function φ(p^n - 1). In a finite field F = GF(p^n), the multiplicative group of units F* is cyclic with an order of p^n - 1. The number of generators of this cyclic group, which are the primitive elements, is φ(p^n - 1). Therefore, the statement about the number of primitive roots holds true for finite fields of order p^n. This confirms that the relationship between primitive roots and the Euler totient function applies consistently across these fields.
Bourbaki1123
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Is it true as it is for finite fields of order p^1, that the number of primitive roots of fields of order p^n is the euler totient of (P^n-1)? If not is there a different rule for the number?
 
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Well, say you have a finite field F = GF(pn). Then F*, the multiplicative group of units of F, is cyclic and has order pn - 1. But the number of generators of a cyclic group G is φ(|G|), so F* has φ(pn - 1) generators, i.e. F has φ(pn - 1) primitive elements.

Is that what you were asking?
 

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