$G$ is cyclic $iff$ there is an element of order $|G|$. I'm trying to prove that if there is an element of order $|G|=n$ there the group is cyclic. But I have some problem. Let $a\in G\backslash \{1\}$. And consider $\left<a\right>$ which is cyclic. Let $b\in G$. Then, $b^n=1=a^n\in \left<a\right>$. Does the fact that $b^n\in \left<a\right>$ implies that $b\in G$, and if yes, why ?

By definition $G$ is cyclic iff $G=\langle a\rangle$ and so order of $a$ should be $|G|$. And if the order of $a$ is $|G|=n$ then $a$ generate $n$ distinct elements, and so $a$ generate whole of $G$.
