Ok, this may look familiar. For example, here you find a very close question, but not using the normal part of it. This is a exercise from a Brazillian book, Paulo A. Martin's "Grupos, Corpos e Teoria de Galois" (the translation is quite easy). It says:
If every proper subgroup of $G$ is cyclic and normal then is $G$ finite cyclic?
The answer is, probably, no. The book puts a sugestion, wich is to set $G$ as $GL_{2}(\mathbb{C})$, with generators $$ a = \left( \begin{matrix} 0 & i \\ i & 0 \end{matrix} \right), b= \left( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right) $$ and study its subgroups. Well, this seems quite hard. I've found this paper online, but it doesn't help, since this is supposed to be a easy/medium level question. There is another example of group that we can use as an counter example? Any help would be appreciated! Thanks in advance!
There is a very elementary counterexample, namely the Kleinian $4$-group $$ G=C_2\times C_2. $$ Every proper subgroup has order $1$ or $2$, and hence is cyclic. Since $G$ is abelian, every subgroup is normal. Of course $G$ is not cyclic.
This example already has been given at this duplicate, where normality holds, since $C_p\times C_p$ is abelian.