(i) Show that
$$ 7^1, 7^2, 7^3, \ldots , 7^{\phi(11)} \pmod {11} $$
produce a reduced residue system modulo 11 and complete the following table:
| Integer a | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| s where 7^s ≡ a (mod 11) |
(ii) By using your results of part (i), complete the following table:
| Integer a | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Order of a(mod 11) |
(i) The following calculations show that $7^n$ for $n =1, 2, \ldots, \phi(11)$ produces a reduced residue system. That is, the right-most column contains, just once, every factor of 11 that is co-prime to it.
| n | 7^n | 7^n mod 11 |
| 1 | 7 | 7 |
| 2 | 49 | 5 |
| 3 | 343 | 2 |
| 4 | 2401 | 3 |
| 5 | 16807 | 10 |
| 6 | 117649 | 4 |
| 7 | 823543 | 6 |
| 8 | 5764801 | 9 |
| 9 | 40353607 | 8 |
| 10 | 282475249 | 1 |
The following is the completed table.
| Integer a | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| s where 7^s ≡ a (mod 11) | 10 | 3 | 4 | 6 | 2 | 7 | 1 | 9 | 8 | 5 |
(ii) We will use the Order Formula (6.8) which states that if $a \pmod n$ has order $k$, then $a^s \pmod n$ has order $k/\gcd(s,k)$.
We can see from (i) above that the order of $7 \pmod {11}$ is 10, so $a=7, k=10$.
We consider $a^s \equiv 7^s \pmod n$. The table from (i) shows that $7^s \pmod {11}$ can be equivalent to every number from 1 to 10, for values of $s$ from 1 to 10. So the Order Formula as $10/\gcd(s,10)$ gives us the order for every number from 1 to 10 modulo 11.
The results of this calculation gives us the required table.
| Integer a | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Order of a(mod 11) | 1 | 10 | 5 | 5 | 5 | 10 | 10 | 10 | 5 | 2 |