Is there a non-cyclic group with every subgroup characteristic? Suppose $G$ is a group such that every subgroup of $G$ is a characteristic subgroup. Does this mean that $G$ is cyclic? I remember reading that this is true in the finite case, is that right? What about the infinite case?
 A: Arturo has proffered the Prüfer $p$-group. However, this is infinitely generated, which leaves us with the following question:

Does there exist a finitely generated example?

Answer: No, there does not.
Proof: As has already been pointed out, our group is necessarily abelian, and it is well-known that every finitely generated abelian group is of the form $\mathbb{Z}^n\times C_{m_1} \times\ldots\times C_{m_i}$ for some finite list of natural numbers $(n, m_1, \ldots, m_i)$ with $n$ possible zero. We can assume $n>0$ here, as we are looking for an infinite example.
Take the given generators for $G$ in terms of its direct-product decomposition, $G=\langle a_1, \ldots, a_n, b_{m_1}, \ldots, b_{m_i}\rangle$. Clearly the map which sends $a_1$ to $a_1c$, for $c$ some generator other than $a_1$, and keeps every other generator fixed is an automorphism of $G$, but $\langle a_1\rangle\neq\langle a_1c\rangle$, so not every subgroup of $G$ is characteristic.
Thus, there can be no generator other than $a_1$ and so $G=\mathbb{Z}$ is cyclic.
We thus have another question:

If every subgroup of $G$ is characteristic, is $G$ locally cyclic?

I am not sure of the answer to this, but I suspect it is "yes".
A: Consider the Prüfer $p$-group, $\mathbb{Z}_{p^{\infty}}$. Since every proper subgroup of $G$ is finite, and there is one and only one subgroup of each finite order $p^k$, every subgroup of $G$ is characteristic. But $G$ is not cyclic. (It is, of course, quasicyclic: every finitely generated subgroup is cyclic).
