union of proper subgroup is proper? 
Question:
Prove that a finite group is the union of proper subgroups IFF the group is not cyclic.

Let G be a finite group.
Suppose G is the union of proper subgroups $b_{i}$.
This means that there is an element in G that is not in $b_{1}$.
Iterating this reasoning, we see that there is at least one element in G that is not in the union of the proper subgroups $b_{i}$.
Then what?
This feels like those days where no questions can be solved.
Any hint is appreciated. Thanks in advance.
Edit: If G were cyclic, then, there exists an element, say $a$, that generates G. But $G=\cup _{i=1}^{n}b_{i}$ implies that $a$ generates $\cup _{i=1}^{n}b_{i}$ too. Hence, $\cup _{i=1}^{n}b_{i}$ is not a proper union of subgroup since every element in G is in $\cup _{i=1}^{n}b_{i}$.

Let $G$ be a group, finite or infinite. Observe that the following statements are equivalent.


*

*$G$ is the union of some proper subgroups.

*$G$ is the union of all of its proper subgroups.

*Each element of $G$ belongs to some proper subgroup of $G.$

*For each element $g\in G,$ $\langle g\rangle$ is a proper subgroup of $G.$

*There is no element $g\in G$ such that $\langle g\rangle=G.$

*$G$ is not cyclic.

