Is there a group theoretic proof that $(\mathbf Z/(p))^\times$ is cyclic? Theorem: The group $(\mathbf Z/(p))^\times$ is cyclic for any prime $p$.

Most proofs make use of the fact that for $r\geq 1$, there are at most $r$ solutions to the equation $x^r=1$ in $\mathbf Z/(p)$, a result which doesn't seem — understandably — to have any group theoretic proofs.
K. Conrad gives seven different proofs — and hints at some others — in his paper here. The first six make use of the previously mentioned fact, while the seventh proof makes extensive use of cyclotomic polynomials and is thus still not group-theoretic.
I was also able to find a linear algebra based proof in the second chapter of Teoría Elemental de Grupos by Emilio Bujalance García, but still, no group theoretic proof to be found.
 A: There's no such a proof, at least to my knowledge, and the following (very partial) argument is no exception (the off-group theory step will be remarked). Rather, it is just an attempt to try giving some "group-theoretic flavour" to the matter, by using (as suggested in the comments) the isomorphism $\operatorname{Aut}({\bf Z}/(p))\cong ({\bf Z}/(p))^\times$, moreover in the very special case $p=2q+1$, where $q$ is another prime.
All the nontrivial elements of ${\bf Z}/(p)$ (the additive group of the integers modulo $p$) have order $p$. Now, automorphisms preserve the order of the elements, and hence every $\psi\in\operatorname{Aut}({\bf Z}/(p))$ permutes the nontrivial elements of  ${\bf Z}/(p)$, namely:
$$\psi\in\operatorname{Aut}({\bf Z}/(p))\Longrightarrow \begin{cases}
\psi(0) = 0 \\
\psi(i) = \sigma(i), \text{ for all }i=1,\dots,p-1 \\
\tag 1
\end{cases}$$
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 $\operatorname{Aut}({\bf Z}/(p))$; in particular, the identity of $\operatorname{Aut}({\bf Z}/(p))$ corresponds to $\sigma(1)=1$, and if $\sigma(1)\ne 1$, then $\sigma$ moves all the $p-1$ nontrivial elements of ${\bf Z}/(p)$. 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 identity. 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 $l_r:=\frac{p-1}{r}$ (this is the off-group theory step).
Now, $\operatorname{Aut}({\bf Z}/(p))$($\cong({\bf Z}/(p))^\times$) 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 identity. 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 for $p>3$, respectively:
\begin{alignat}{2}
m_2 &= q+1-2+1=q &&<p-1 \\
m_q &= 2+1-2+1=2 &&<p-1 \\
\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 $({\bf Z}/(2q+1))^\times$ is cyclic for every prime $q$ such that $2q+1$ is prime either, such as: $({\bf Z}/(5))^\times$, $({\bf Z}/(7))^\times$, $({\bf Z}/(11))^\times$, $({\bf Z}/(23))^\times$, $({\bf Z}/(47))^\times$, $({\bf Z}/(59))^\times$, $({\bf Z}/(83))^\times$, etc.
