I am working on a cryptography example problem. The problem is the following:
For the group $G = \langle \mathbb{Z}_{26}^*, x\rangle$
a) find the order of the group
b) find the order of each element in the group
c) Is the group a cyclic group? Prove the answer and find the generator(s).
So far, I've started on a and b. This is what I have so far:
a) Since $\mathbb{Z}_{26}^*={1,3,5,7,9,11,15,17,19,21,23,25}$; the order of the group is $12$.
b)
$|1| = 1$,
$|3| = 3$,
$|5| = 4$,
$|7| = ?$,
$|9| = 3$,
$|11| = ?$,
$|15| = ?$,
$|17| = 6$,
$|19| = ?$,
$|21| = 4$,
$|23| = 6$,
$|25| = 2$.
I am not sure about a or b. Are $7,11,15,19$ part of the group?
Am I on the right track? Any help/advice would be greatly appreciated. Thanks!
(a) is exactly correct; you want exactly those numbers which do not have $2$ or $13$ as factors.
(b) is also correct where you have answered it. Now note that $$7^2 = 23, 11^2 = 17, 15^2 = 17, 19^2 = 23,$$ so that all of these elements $(7,11,15,19)$ have order $12$ (as all the right-hand sides have order $6$).
Edit: (c) The existence of an element of order $12$ automatically implies that the group is cyclic; for example, the powers of $7$ exhaust the group: $$1,7,23,5,9,11,25,19,3,21,17,15.$$