Find all generators of $(\mathbb Z \setminus 11 \mathbb Z)^\times$ I need to find all generators of this group:  $$(\mathbb Z \setminus 11 \mathbb Z)^\times$$
I know that this group is cyclic, because $11$ is a prime number. It also has exactly $10$ elements and so each generator has to have the order of $10$. 
One approach (a very manual one, unfortunately) would be to try all possible elements and check which of them have the order of $10$, but I do not think that this is a wise choice.  
I also tried to use the fact that, for finite groups,
$$\text{A is a subgroup of B } \Rightarrow \text{card(A)} \mid \text{  card(B})$$
And since the cardinality of our group is $10$, it is enough to check these numbers: $1, 2, 5, 10$ and, obviously, $1$ does not do the job. 
And so we need to check these: $2, 5, 10$.
Is it a correct way to solve this? Is there an easier approach?
 A: I don't really follow your argument. You're not looking for subgroups, you're looking for the whole group.
All cyclic groups of a fixed size are isomorphic to all other cyclic groups of the same size. So once you find one generator, you can find all other generators easily, because given any generator $g$, the morhpism $g^i \rightarrow i$ is an isomorphism to the additive group Z$_{10}$. Finding elements of order 10 in Z$_{10}$ consists of merely finding elements that are coprime with 10: 1, 3, 7, and 9. And once you have elements of order 10 in Z$_{10}$, you can then pull them back to your original group.
Taking the powers of 2, we get:
1, 2, 4, 8, 5, 10, 9 ,7, 3, 6.
Since those are 10 distinct values, 2 is a generator. This list not only confirms that 2 is a generator, but gives the other generators: we can take the 1st power of 2 (which is just 2), the 3rd power (which is 8), the 7th (which is 7, and the 9th (which is 6). So the generators are 2, 8, 7, 6.
A: I don't think this will improve the search alot, but maybe a little.


*

*Find all $a$ such that $a^2\equiv_{11}1$ and eliminate them from your search. Notice that once $a^2\equiv_{11}1$ also $(-a)^2\equiv_{11}1$, so you can eliminate $-a$ aswell.

*If $a^{5}\equiv_{11}-1$, and $a\not\equiv_{11}-1$ then $a$ is a generator. This is because the order of $a$ divides $10$, and the only possible order for $a$ in this case is $10$.
Also, suppose you manage to find a generator $a$, then $a^3,a^7,a^9$ will be generators aswell, since $\{1,3,7,9\}$ are integers less than, and co-prime, to  $10$.
A: Each element except identity is a generator since group order is prime. 
Edit: I did not notice the the star at the top of group
