Prove or disprove that if every proper subgroup $G$ is cyclic, then $G$ is cyclic.
If I used group $U(8)$ as an example to disprove the statement, does it work? I just say that every subgroup of $U(8)$ is of order $2$ and that they are cyclic: $\langle 3\rangle = \{1,3\}, \langle 5\rangle = \{1,5\}$ and $\langle 7\rangle= \{1,7\}.$
I saw an other post saying that $C_2 \times C_2$ disproved the statement, but I want to check if the way I did it works.
In your explanation, we can also say that if a cyclic group of order $n$ must have exactly $\phi(n)$ of order $n$, So in your case, if group $G$ is cyclic of order $4$ then it must have $\phi(4)=2$ elements of order $4$ which are not there, So your example will also work, This explanation by the definition of a cyclic group. For more reading, you can see in 'cyclic group' in "Contemporary Abstract Algebra".
Note: Even if every subgroup is cyclic then the group may not be abelian so it is not cyclic, $Q_8$ and $S_3$ are the examples.
Here is a very quick (dis)proof: There exist non-cyclic finite groups.
(If the statement held then every finite group would be cyclic, because if $G$ is a finite group of order $n$ then all proper subgroups are of smaller order, hence cyclic by induction. If the statement held then $G$ would be cyclic, and hence all finite groups would be cyclic.)
