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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?

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What is a characteristic subgroup? – leo May 29 '12 at 20:08
@leo: We say that subgroup $H$ of $G$ is a characteristic subgroup if $\phi(H) = H$ for any automorphism $\phi$ of $G$ – spin May 29 '12 at 20:09
Uh? A subgroup invariant under any automorphism of the group. This is a rather standard notion in group theory. – DonAntonio May 29 '12 at 20:10
Yes, I believe that it is indeed the case that for a finite group, "every subgroup is characteristic" implies the group is cyclic. – Arturo Magidin May 29 '12 at 20:28
@ArturoMagidin, ah, I didn't know that. See also… – lhf May 30 '12 at 11:47
up vote 19 down vote accepted

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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+1 A great example! Having seen your example I am tempted to proffer $\mathbb{Q}/\mathbb{Z}$ as another. – Jyrki Lahtonen May 29 '12 at 20:24
@Jyrki: Those are almost the same example: $\mathbb{Z}_{p^\infty}$ is the $p$-Sylow of $\mathbb{Q}/\mathbb{Z}$. – Chris Eagle May 29 '12 at 20:35
Thanks. Nice example, $\mathbb{Z}_{p^{\infty}}$ seems to be an endless source of counterexamples in group theory.. – spin May 29 '12 at 20:40
@spin: A group in which every subgroup is characteristic must be abelian. For finite groups, you are correct that the group must be cyclic. For infinite torsion groups, their $p$-parts must be either finite cyclic or Prufer groups. But I believe that there are torsionfree groups in which every subgroup is characteristic and with reasonably complicated structure. – Arturo Magidin May 30 '12 at 3:27
Why must such a group be abelian? – Chris Eagle May 30 '12 at 9:34

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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Concerning 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. – Martin Brandenburg May 30 '12 at 11:44

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