Doing some practice for sets and groups, when this question has me absolutely stumped! The set S consists of the eight elements 9^1, 9^2, 9^3, ..., 9^8 where the operation is modulo 64. Determine each of the elements of S as an integer between 0 and 63. Under multiplication modulo 64, the set S forms a group G with identity 1.
- Write down the inverses of each of the remaining elements of G.
Using the same information as above, know that a group in which each element x can be written as a power of a particular element g of the group is said to be a cyclic group; and such an element g is called a generator of the group.
- Write down all the possible generators of G
Apply the binomial theorem: $$ 9^n=(1+8)^n=1+8n,\qquad n=1,\cdots,8. $$ Note that we can ignore the higher order terms since they are divisible by $64$. All the terms are in $[0,63]$, except for the $n=8$ term which reduces to $1$.
Now observe you have a cyclic group of order $8$. Hence it is isomorphic to the integers mod 8, which has the 4 generators $1,3,5,7$. These correspond to $$\{9^1,9^3,9^5,9^7\}=\{9,25,41,57\}.$$