Let $p$ be a prime number, and GF$(p)$ its finite field. If $G=[x,x^2,x^3,...x^c]$ is a cyclic group with multiplicative order $c$, then $x$ is a generator of group $G$. That is, the multiplicative order of $x$ in GF$(p)$ is $c$. The elements $e$ in group $G$ are precisely those such that $e^c=1$. If $e$ is any element which cannot be written as $x^k$ for any positive integer $k$, then there exists an element $m$ such that $mx^k=e$. Let $G_2=[mx,mx^2,mx^3,...mx^c], G_3=[m_2x,m_2x^2,m_2x^3,...m_2x^c], G_4=[m_3x,m_3x^2,m_3x^3,...m_3x^c],...G_d=[m_{d-1}x,m_{d-1}x^2,m_{d-1}x^3,...m_{d-1}x^c]$
where $m, m_2,... m_d$ are distict elements. There are $(p-1)/c=d$ distinct groups such that each element $e$ is contained in exactly one group $G_n$. Let $D=[1,m,m_2,m_3,...m^{d-1}]$ be the group of distinct elements such that each of these elements will generate exactly one group (shown above).
It turns out that if $\gcd(c,d)=1$, then the elements contained in $D$ are those such that ${m_n}^d=1$.
What about when $\gcd(c,d)>1$ and:
- $c$ and $d$ do not have the same exact prime factors
- $c$ and $d$ have the same exact prime factors. That is, $c=a^b{a_2}^{b_2}{a_3}^{b_3}......{a_n}^{b_n}$ and $d$ = $c=a^c{a_2}^{c_2}{a_3}^{c_3}......{a_n}^{c_n}$.
Here is an example with the first case of $c$ and $d$ coprime:
$p=13, c=4, d=3$
Working in the field GF$(13)$ (the integers $\mod 13$) The first group $G$ is a cyclic group of order $4$: $[1, 5, 8, 12]$, and $1^4 = 5^4 = 8^4 = 12^4 = 1$. We have $d=3$, and $D=[1, 3, 9]$, it follows $1^3 = 3^3 = 9^3 = 1$. If every element in $D$ is multiplied with every element in $G$, we obtain the following groups:
$G: [1, 5, 8, 12]$, $G_2: [2, 3, 10, 12]$, and $G_3: [4, 6, 7, 10]$.
Every non-zero element in GF$(13)$ is included in exactly one of these groups.
If $c=2,d=6$, this approach doesn't work as $G$ is the cyclic group $[1, 3, 4, 9, 10, 12]$, $1^6 = 3^6 = 4^6 = 9^6 = 10^6 = 12^6 = 1$, and $D$ is the cyclic group $[1, 12]$, $1^2 = 12^2 = 1$. Multiplying every element in $D$ by every element in $G$ gives us two identical groups:
$G: [1, 3, 4, 9, 10, 12]$, and $G_2: [1, 3, 4, 9, 10, 12]$.
And not all non-zero elements in GF$(13)$ are included in one of these groups, unfortunately.
Does anybody have an approach around this?