Every non-cyclic group $G$ has at least $2$ non-trivial cyclic subgroups I tried to prove it but got stuck. Here's my attempt:
$G$ isn't cyclic, then it's order musn't be prime, so $|G|\geq 4$
Now, pick two elements: $a, b\in G$ so that $e\ne a \ne b$. I can do this becuase there at least 3 elements other than $e$.
Now I try to prove that $<a> \ne <b>$. How can I proceed from here?

Choose $a\in G$ first. Then, consider $<a>$. Since $G$ is non-cyclic, $G\not=<a>$. Take $b\in G\setminus<a>$.
