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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0What is a characteristic subgroup? – 2012-05-29
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2@leo: We say that subgroup $H$ of $G$ is a characteristic subgroup if $\phi(H) = H$ for any automorphism $\phi$ of $G$ – 2012-05-29
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1Uh? A subgroup invariant under any automorphism of the group. This is a rather standard notion in group theory. – 2012-05-29
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0Yes, I believe that it is indeed the case that for a *finite* group, "every subgroup is characteristic" implies the group is cyclic. – 2012-05-29
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0@ArturoMagidin, ah, I didn't know that. See also http://groupprops.subwiki.org/wiki/Finite_group_implies_cyclic_iff_every_subgroup_is_characteristic – 2012-05-30
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0I once asked on MO about finite groups whose all _normal_ subgroups are characteristic: http://mathoverflow.net/questions/50864/groups-with-all-normal-subgroups-characteristic – 2012-05-30
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0If anyone knows a proof or at least a reference for the finite case (finite and every subgroup characteristic implies cyclic), you could post it as an answer. lhf's link doesn't have anything other than the statement in it really.. – 2012-05-30
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0@spin: I confess I'm having trouble showing it must be abelian in any case. For finite groups, though, you can reduce to the $p$-group case; then $G/\Phi(G)$ must be cyclic (otherwise not every cyclic subgroup is characteristic) and hence $G$ itself must be cyclic. – 2012-05-30
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0@ArturoMagidin: why must that quotient be cyclic for all cyclic subgroups to be characteristic? – 2012-06-01
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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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0+1 A great example! Having seen your example I am tempted to proffer $\mathbb{Q}/\mathbb{Z}$ as another. – 2012-05-29
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1@Jyrki: Those are almost the same example: $\mathbb{Z}_{p^\infty}$ is the $p$-Sylow of $\mathbb{Q}/\mathbb{Z}$. – 2012-05-29
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1Thanks. Nice example, $\mathbb{Z}_{p^{\infty}}$ seems to be an endless source of counterexamples in group theory.. – 2012-05-29
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0@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. – 2012-05-30
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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