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If $p$ is a prime, $(\Bbb Z/p)^\times\simeq\Bbb Z/(p-1)$

Proof. Since $(\Bbb Z/p)^\times$ is abelian, $(\Bbb Z/p)^\times\simeq\prod_{i=1}^m\Bbb Z/n_i$ where $2\leq n_i\mid n_{i+1}$ for each $i$. Then for $a\in(\Bbb Z/p)^\times, a^{n_m}=1$ so the polynomial $x^{n_m}-1\in(\Bbb Z/p)[x]$ has at least $p-1 =|(\Bbb Z/p)^\times|$ many roots.

From this, I want to conclude $m=1$ so that $(\Bbb Z/p)^\times\simeq \Bbb Z/(p-1)$. Could you help?

Proof. If $(\Bbb Z/p)^\times$ is abelian, $(\Bbb Z/p)^\times\simeq \Bbb Z/p_1^{n_1}\times\cdots\times\Bbb Z/p_m^{n_m}$ and $p_i$ are distinct as if we let $k = \operatorname{lcm}(p_1,...,p_m)$, then $x^k =1$ for any $x\in (\Bbb Z/p)^\times$. If $p_i$ are not distinct, $|G|>k$ so the number of roots of $x^k-1$ has more element than $k$ elements.

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  • Does this answer your question? https://math.stackexchange.com/questions/989451/proving-that-the-multiplicative-group-mod-p-p-is-prime-is-cyclic – David Sheard Aug 29 '21 at 14:39
  • In fact, a more general statement holds: if $F$ is a field and $G\le F^\times$ is a finite subgroup of the multiplicative group over $F$, then $G$ is cyclic. – Hanul Jeon Aug 29 '21 at 14:58

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There is a proof linked in the comments, another quick proof is a follow : $\mathbb{F}_p^{\times}$ is cyclic because $\mathbb{F}_p$ is a commutative field, therefore it is isomorphic to $\mathbb{Z}/{\rm Card}(\mathbb{F}_p^{\times})\mathbb{Z}=\mathbb{Z}/(p-1)\mathbb{Z}$.

Tuvasbien
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