Consider the integers of the form
$\quad pq + 1, \text{where 0 } \lt q \le p $
The corresponding set of residue classes $\{[pq + 1]\}$ form a cyclic group of order $p$ with generator $[p + 1]$.
Example: If $p = 11$ then $12$ generates a cyclic subgroup of order $11$ in $(\mathbb{Z}/{121}\mathbb{Z})^\times$:
$\; {[12]}^1 \equiv \;\;\, 12 \pmod {121}$
$\; {[12]}^2 \equiv \;\;\, 23 \pmod {121}$
$\; {[12]}^3 \equiv \;\;\, 34 \pmod {121}$
$\; {[12]}^4 \equiv \;\;\, 45 \pmod {121}$
$\; {[12]}^5 \equiv \;\;\, 56 \pmod {121}$
$\; {[12]}^6 \equiv \;\;\, 67 \pmod {121}$
$\; {[12]}^7 \equiv \;\;\, 78 \pmod {121}$
$\; {[12]}^8 \equiv \;\;\,89 \pmod {121}$
$\; {[12]}^9 \equiv\; 100 \pmod {121}$
$\; {[12]}^{10} \equiv 111 \pmod {121}$
$\; {[12]}^{11} \equiv\;\;\;\, 1 \pmod {121}$
I have a direct proof of the above using Euclidean division (representation) theory, but would be interested in seeing other proofs (or links/references). Also, the wikipedia link
$\quad$ Multiplicative group of integers modulo $n$
states
...though even for prime $n$ no general formula for finding generators is known.
So I am also interested in any partial progress made in this area, determining the order of elements in ${\displaystyle (\mathbb {Z} /n\mathbb {Z} )^{\times }}$.
Here we 'pattern construct' the larger cyclic group $K_{2p}$ generated by $[p-1]$ in $(\mathbb{Z}/{p^2}\mathbb{Z})^\times$ for $p \ge 5$.
The group $K_{2p}$ has $2p$ elements.
Set $k = p-1$, an even integer.
Define a list of numbers by starting at $p-1$ and incrementing by $2p$ while staying below $p^2 - 1$,
$\quad G_1: p-1, p-1+2p, p-1+4p, \dots, p-1+(k-1)p$
Now add $p$ to each number to create a second list,
$\quad G_2: 2p-1, 2p-1+2p, 2p-1+4p, \dots, 2p-1+kp$
The $\text{[.]}_{\, p^2}$ residues of the set of numbers in $G_1 \cup G_2$ are exactly the $k$ generators for $K_{2p}$ having order $2p$.
Continuing, we'll define another list of numbers by starting at $p+1$ and incrementing by $2p$
(equivalently, add $2$ to every number in $G_1 \cup G_2$),
$\quad H_1: p+1, p+1+2p, p+1+4p, \dots, p+1+(k-1)p$
Now add $p$ to each number to create a second list,
$\quad H_2: 2p+1, 2p+1+2p, 2p+1+4p, \dots, 2p+1+(k-1)p$
The $\text{[.]}_{\, p^2}$ residues of the set of numbers in $H_1 \cup H_2$ are exactly the $k$ elements in $K_{2p}$ having order $p$.
Since $2p - 2k = 2$ there are two elements that remain to be accounted for in $K_{2p}$. But those are the two elements $\{[1],[p^2-1]\}$ satisfying $x^2 = 1$.
Example: For $p = 11$ specify the proper subgroup $K_{22}$ of $(\mathbb{Z}/{121}\mathbb{Z})^\times$.
The elements of order $22$ consist of
$\quad [10], [32], [54], [76], [98],$
$\quad [21], [43], [65], [87], [109]$
The elements of order $11$ consist of
$\quad [12], [34], [56], [78], [100],$
$\quad [23], [45], [67], [89], [111]$
The elements of order $2$ consist of
$\quad [120]$
The elements of order $1$ consist of
$\quad [1]$