Is a group cyclic with its generators A group (S, $\odot$) is called cyclic if there exists g $\in$ S such that for every a $\in$ S there exists an integer n such that a = g $\odot$ g  ...  $\odot$ g (n times). If such a g $\in$ S exists, it is called a generator.
Is the group $\mathbb{Z}^{*}_{13}$= {1, 2 ... 11, 12} together with multiplication modulo 13 then cyclic? I can't seem to find the generators
 A: Without some background in number theory, all you can do is try various numbers. Obviously $1$ doesn't work. So try $2$. We get lucky, $2$ works. For let us find the various positive powers of $2$, reduced modulo $13$. We get, in order,
$$2,4,8,3,6,12,11,9,5,10,7,1.$$
Remark: There are other generators, a total of $4$ of them. It turns out that they are $2^1$, $2^5$, $2^7$, and $2^{11}$ (modulo $13$). 
At this early stage, if you want all the generators, it is best to compute. Let us test $3$. The various powers, modulo $13$, are $3$, $9$, $1$, and now things start all over again, so we certainly won't get everything. 
Next let's try $4$. The powers of $4$ will be the even powers of $2$, so we can look back on the work we did with $2$ and see that we will only get $4$, $3$, $12$, $9$, $10$, and $1$. 
Quite a few left. Let's try $5$. It turns out that $5$ is no good, because $5^4$ gives $1$. Continue.
A: Try to determine the order of each element. Iff there is no element of order 12, the group isn't cyclic.
A: It is cyclic. You can take for example $2$ as generator:
$$1=2^{11},2=2^1,3=2^4,4=2^2,5=2^8,6=2^5$$
$$7=2^{10},8=2^3,9=2^7,10=2^9,11,12=2^6$$
In fact $\mathbb{Z}_p^*$ is prime for every prime $p$.
