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On the group of the invertible elements in $\mathbb{Z}_p$, the question asks to show that the group is cyclic.

This must have something to do with the representation of $G$ as a product of groups with prime power order, and I think that I should be able to find an element that generates the group, but so far no idea on how to do that.

By the way, $G$ is abelian.

Shaun
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José
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    There are lots of proofs of this. Many depend on Euler's $\varphi$ function that counts the number of integers less than $n$ and relatively prime to $n$. You can use it to count the number of generators of the cyclic subgroup generated by an element of $\mathbb{Z}_p^*$. That's the start of one proof. – Ethan Bolker Feb 21 '19 at 00:05
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    See https://mathoverflow.net/questions/54735/collecting-proofs-that-finite-multiplicative-subgroups-of-fields-are-cyclic and https://math.stackexchange.com/questions/59903/finite-subgroups-of-the-multiplicative-group-of-a-field-are-cyclic – lhf Feb 21 '19 at 00:08
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    See also https://en.wikipedia.org/wiki/Primitive_root_modulo_n#Finding_primitive_roots – lhf Feb 21 '19 at 00:08
  • It is cyclic because $p$ prime implies $\Bbb{Z}_p$ is a field so that $x^d-1$ has at most $d$ roots in it, since $\Bbb{Z}_p^*$ has $p-1$ elements it means they are exactly the roots of $x^{p-1}-1$ and one of them is a root of $x^{p-1}-1$ but not of $x^d-1$ for any $d < p-1$ thus it is a generator – reuns Sep 04 '19 at 22:12

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