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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 ${\displaystyle (\mathbb {Z} /n\mathbb {Z} )^{\times }}$.

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  • Did you mean $0\color{red}\lt q\lt p$? – J. W. Tanner Aug 16 '20 at 17:09
  • Using the binomial theorem, $(pq+1)^p\equiv1\pmod {p^2}$. Is that an answer to your question? – J. W. Tanner Aug 16 '20 at 17:11
  • @J.W.Tanner We've identified a cyclic subgroup along with a generator. – CopyPasteIt Aug 16 '20 at 17:16
  • You have identified a generator of a cyclic subgroup, but not a generator of the entire group – J. W. Tanner Aug 16 '20 at 17:18
  • @J.W.Tanner Yes, but I figure other work has been done for subgroups/generators of ${\displaystyle (\mathbb {Z} /n\mathbb {Z} )^{\times }}$ (more general). – CopyPasteIt Aug 16 '20 at 17:20
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    $(1+p Z)/(1+p^2 Z)$ is a group with $p$ elements which is prime so it is automatically cyclic, any element other than $1$ is a generator. For $p$ odd with $g$ a generator of $Z/pZ^\times$ then $p-1$ divides the order of $g\bmod p^2$ which must be either $p-1$ or $(p-1)p$, in the former case $g (1+p)$ is a generator of $Z/p^2 Z^\times$. Something similar holds with $Z/p^k Z^\times$ and $Z/nZ^\times\cong \prod_{p^k | n} Z/p^k Z^\times$ is understood from there (for $n$ odd as the case $p$ even is a bit different) – reuns Aug 16 '20 at 17:37

1 Answers1

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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

$\quad [1]$

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