$\alpha \nmid n \Rightarrow \alpha$ is a generator of group $G$ Let $G=\left \langle x \right \rangle$ a cyclic group s.t. $ord(x)=n$. Let $a=x^\alpha$ s.t. $\alpha \nmid n$ and $0<\alpha<n$ . Prove that $x^\alpha$ is a generator of $G$.
My partial answer: Let $A:=\left \langle x^\alpha \right \rangle$. We'll assume in counterwise the oposite. Then, $|G|=n$ thus $|A|\leq n$. $|A|\ne n$ because elsewise we would get $\left \langle x^\alpha \right \rangle = G$. Thus, $|A|<n$. I would like for some help from here or to get another direction.

This is not true. For example, suppose $\text{ord}(x)=10$, and take $\alpha=6$. Then 
$$<x^\alpha>=\{e, x^6, x^{12}=x^2, x^8, x^{14}=x^4\}$$
which clearly is not the whole group. 
However, if $\gcd(\alpha,n)=1$ this is true, since then by the euclidean algorithm there exist $k,l \in \mathbb{Z}$ such that $k\alpha+ln=1$, so
$$x=x^{k\alpha+ln}=x^{k\alpha}x^{ln}=(x^{\alpha})^k (x^n)^l=(x^\alpha)^k$$
where the last equality follows from lagrange’s theorem (the order of $x$ divides the order of the group). The above equality now shows $x$ is in the subgroup generated by $x^{\alpha}$, so the subgroup contains all of $G$. 
