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?
Is there a non-cyclic group with every subgroup characteristic?
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0@Tobias: The automorphism group of $G$ maps to the automorphism group of $G/\Phi(G)$; $G/\Phi(G)$ is elementary abelian, and if it is not cyclic, you can lift a nontrivial automorphism to $G$. – 2012-06-01
2 Answers
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).
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0Why must such a group be abelian? – 2012-05-30
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".
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4Concerning the question posed by user1729: If $G$ is a groups whose subgroups are characteristic, then $G$ is abelian and the torsion subgroup of $G$ is locally cyclic. There are many torsion-free examples. See *Finiteness conditions on characteristic closures and cores of subgroups* by Giovanni Cutolo, Howard Smith and James Wiegold. – 2012-05-30