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Let $(\Bbb Z/p\Bbb Z)^\times$ be the multiplicative group of the integers modulo $p$, for $p$ a prime. It is well known that this group is cyclic (for a survey of the many proofs so far, see here). I'm trying with this other proof attempt, but (as long as everything is okay till there, which I'm unsure about) I'm stuck in the final step.

In the following, I'll assume as known the result $\operatorname{Aut}(\Bbb Z/p\Bbb Z)\cong(\Bbb Z/p\Bbb Z)^\times$.

All the nontrivial elements of $\Bbb Z/p\Bbb Z$ (the additive group of the integers modulo $p$) have order $p$. Automorphisms preserve the order of the elements, and hence every $\psi\in\operatorname{Aut}(\Bbb Z/p\Bbb Z)$ permutes the nontrivial elements of $\Bbb Z/p\Bbb Z$, namely: $$\psi\in\operatorname{Aut}(\Bbb Z/p\Bbb Z)\Longrightarrow \begin{cases} \psi(0) = 0 \\ \psi(i) = \sigma(i), \text{ for all }i=1,\dots,p-1 \\ \tag 1 \end{cases}$$ for some $\sigma\in S_{p-1}$. By the linearity of $\psi$, the permutation $\sigma$ in $(1)$ fulfils the condition: $$\sigma(k)\equiv\sigma(1)k\pmod p\tag 2$$ Every choice of $\sigma(1)$ gives rise to a distinct element of $\operatorname{Aut}(\Bbb Z/p\Bbb Z)$; in particular, the identity of $\operatorname{Aut}(\Bbb Z/p\Bbb Z)$ corresponds to $\sigma(1)=1$, and if $\sigma(1)\ne 1$, then $\sigma$ moves all the $p-1$ nontrivial elements of $\Bbb Z/p\Bbb Z$. Therefore, $\sigma\in S_{p-1}$ in $(1)$ is of the form (cycles in comma-notation): $$\small{\bigl(1,\sigma(1),\sigma(1)^2,\dots,\sigma(1)^{l-1}\bigr)\bigl(j_2,\sigma(1)j_2,\sigma(1)^2j_2,\dots,\sigma(1)^{l-1}j_2\bigr)\dots\bigl(j_r,\sigma(1)j_r,\sigma(1)^2j_r,\dots,\sigma(1)^{l-1}j_r\bigr)}\tag 3$$ for some $1\le r\le p-1$, where: $$1+\sigma(1)+\sigma(1)^2+\dots+\sigma(1)^{l-1}\equiv 0\pmod p\tag 4$$ and: $$lr=p-1\tag 5$$ The case $r=p-1$ corresponds to the trivial permutation/automorphism. For $r<p-1$, conditions $(4)$ and $(5)$ come from being $\sigma$ a fixed point-free permutation, and in turn they imply that there are at most $l_r-1=\frac{p-1}{r}-1$ automorphisms of order $l_r:=\frac{p-1}{r}$.

Now, $\operatorname{Aut}(\Bbb Z/p\Bbb Z)$($\cong(\Bbb Z/p\Bbb Z)^\times$) is a finite Abelian group; so, if $n$ is the maximal order among the elements in the group, then the order of every element divides $n$. Let $l_r$ be the maximal order, for some $r\ge1$, and $\{1,d_2,\dots,d_{D_r}\}$ the set of the proper divisors of $l_r$. Then, from $(4)$, an upper bound to the number of automorphisms is given by: \begin{alignat}{1} m_r &= \underbrace{l_r-1}_{\text{max num. aut. order }l_r}+\underbrace{\frac{l_r}{d_2}-1}_{\text{max num. aut. order }l_r/d_2}+\space\dots\space+\underbrace{\frac{l_r}{d_{D_r}}-1}_{\text{max num. aut. order }l_r/d_{D_r}}+1 \\ \tag 6 \end{alignat} where the final "$+1$" accounts for the trivial automorphism. From $(6)$: \begin{alignat}{1} m_r &= \frac{l_r}{1}+\frac{l_r}{d_2}+\dots+\frac{l_r}{d_{D_r}}+\frac{l_r}{l_r}-D_r \\ &= \frac{l_r}{1}+\frac{l_r}{d_2}+\dots+\frac{l_r}{d_{D_r}}+\frac{l_r}{l_r}-(D_r+1)+1 \\ &= \text{sum of the divisors of }l_r-\text{number of the divisors of }l_r\space+\space 1 \\ \tag 7 \end{alignat} For $p$ such that $q:=\frac{p-1}{2}$ is prime, the maximal order $l_r$ is either $l_1=p-1$, or $l_2=q$, or $l_q=2$. In the latter two cases, $(7)$ yields, respectively: \begin{alignat}{2} m_2 &= q+1-2+1=q &&<p-1 \\ m_q &= 3-2+1=2 &&<p-1, \text{ as soon as }p>3 \\ \tag 8 \end{alignat} which both contradict the fact that there are $p-1$ automorphisms. Therefore, there must be some automorphism of (maximal) order $p-1$, and hence $(\Bbb Z/(2q+1)\Bbb Z)^\times$ is cyclic for every prime $q$ such that $2q+1$ is prime either, such as: $(\Bbb Z/5\Bbb Z)^\times$, $(\Bbb Z/7\Bbb Z)^\times$, $(\Bbb Z/11\Bbb Z)^\times$, $(\Bbb Z/23\Bbb Z)^\times$, $(\Bbb Z/47\Bbb Z)^\times$, $(\Bbb Z/59\Bbb Z)^\times$, $(\Bbb Z/83\Bbb Z)^\times$, etc.

Likewise, if $\frac{p-1}{2}$ is composite, I think I should use $(7)$ to firstly prove that $r>2\Rightarrow m_2>m_r$, and then that $m_2<p-1$. This is where I'm stuck. (But it's far more probable that the upper bound $(7)$ is simply too weak to prove the cyclicity of $(\Bbb Z/p\Bbb Z)^\times$ for $p$'s different from those already accounted for here above.)

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    Another (shorter) proof is, that the multiplicative group of a finite field is cyclic, see here. So for $\Bbb F_p$ this gives that $(\Bbb Z/p)^{\times}$ is cyclic. This is on page $5$ of Conrad's note you have linked above. It is this MO-question. – Dietrich Burde Dec 07 '21 at 12:41
  • Well your aproach is very simmilar to the standard aproach for proving cyclicity ,but your estimates for the number of automorphism of order $l$ is far from sharp, and so it does not have any chance of sucsess. you obtain this estimate from $4$, but if $l$ is composite, then $x^l+x^{l-1}+...+1$ is not ireducible, in other word of $\gamma$ has order $2k$, then $\gamma^k$ has order $2$ not $2k$. – ali Dec 14 '21 at 13:05
  • but if you think more you can give the exact number of automorphism of order $l$ (spoiler alert:the answer is $\phi(l)$) and then you get something simmilar to 7, which need a little work to prove the theorem. – ali Dec 14 '21 at 13:07
  • @ali, do you mean that $\sum_{d\mid l_{max}}\phi(l_{max}/d)<p-1$ for $l_{max}<p-1$? Why? –  Dec 15 '21 at 06:22
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    yes in fact we have $\sum_{d\mid l_{max}}\phi(l_{max}/d)=l_{max}$! you can prove this by induction on the number of primes dividing $l_{max}$, you can also prove this by noting that every $n\le l_{max}$ can written in a unique way as $n=dm$ where $d|l$ and $(m,l/d)=1$ – ali Dec 15 '21 at 09:08
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    in fact this also shows that you don't have room for eror in your estimates – ali Dec 15 '21 at 09:10
  • @ali. Thanks for your time. Now I see that this is actually the proof #1 of the resource linked in my post. –  Dec 15 '21 at 10:42

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