Prove that a group $G$ is not cyclic if and only if $G$ is a union of proper subgroups.

Question

Can anyone please help me prove this question?

Question: So first we can assume that $G$ is not cyclic. Then we need to show that $G$ is a union of proper subgroups. How can I do this?

Answer 1

HINT: If $G$ is not cyclic, $\langle g\rangle\ne G$ for any $g\in G$. If $G$ is cyclic, and $g$ is a generator of $G$, then $G$ is the only subgroup of $G$ containing $g$.

Answer 2

Take any $\,g_1\in G\,$ . Since $\,G\neq\langle g_1\rangle\,$ ,we have that there exists $\,g_2\in G-\langle g\rangle\,$ . So now look at $\,\langle g_1\rangle\cup\langle g_2\rangle\,$ , If this is $\,G\,$ we're done, otherwise there exists $\,g_3\in G-\left(\langle g_1\rangle\cup\langle g_2\rangle\right)\,$ , so look at

$$\bigcup_{i=1}^3\langle g_i\rangle$$

and etc.

Added on request: Suppose

$$G=\bigcup_{H\lneq G} H$$

If $\,G\,$ is cyclic then $\,G=\langle x\rangle\,$ , for some $\,x\in G\,$ , but then

$$\color{red}{\forall\;H\lneq G\;,\;\;x\notin H}\;\Longrightarrow \bigcup_{H\lneq G}H\neq G\,\,\,\,\text{...contradiction!}$$

The gist of the above slick proof is the red part: can you see why it is true?