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I have a group $G=(\mathbb{Z}_{251}^{*}, \cdot)$ with generator $g=71$ (so, having a generator, I'm given with the fact that is cyclic, right?)

further in the example of my study notes I read: "$n = |G| = 250$ ... $G$ is cyclic as it coincides with the multiplicative group of a finite field"

Could someone give me a justification for this?

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2 Answers 2

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$251$ is a prime number, thus $\mathbb{Z}_{251}$ is a field. I believe $\mathbb{Z}^{*}_{251}$ are the non zero elements of $\mathbb{Z}_{251}$.

Prove that the group, under multiplication, of all nonzero elements in a finite field must be a cyclic group.

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  • $\begingroup$ Yes, $\mathbb{Z_{251}^*}$ is defined as the set of the equivalence classes modulo $251$ smaller than $n$ and with no common factor with $n$ except for the neutral element $1$ $\endgroup$
    – ela
    Jul 15, 2016 at 12:46
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It is a basic theorem in ring theory that any finite subgroup of the group of units of an integral domain is cyclic. I remember the proof to be annoying, but it can be found in lots of places.

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