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If $A$ is an abelian group, there are a number of characteristic subgroups of it that immediately come to mind: Any of the groups $nA$, $A[n]$ (the $n$-torsion), the full torsion subgroup $T(A)$, the maximal divisible subgroup $D(A)$ etc. Any of these examples, however, can be obtained by (possibly transfinite) iteration of forming verbal $(nA)$ and marginal ($A[n]$) subgroups and taking sums and intersections of those. Do you know any (specific or general) examples of characteristic subgroups that are not of this form?

To be precise: Let $\mathcal{E}(A)$ denote the smallest (nontrivial) family of subgroups of $A$ that is closed under sums, intersections and formation of verbal and marginal subgroups $-$ these characteristic subgroups I call expected. What are some unexpected characteristic subgroups?

Here is an example of what I am talking about: Consider the group $A=\mathbb{Z}/2^n\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ for $n\geq 3$. It is straightforward to see that $A$ contains exactly three distinct subgroups of order $2^r$ for $1\leq r\leq n$. Now, if both of these inequalities are strict we have the expectedly characteristic $2^{n-r}A$ and $A[2^{r-1}]$ among them. Hence, the third group (which is cyclic generated by $(2^{n-r},1)$) must be characteristic as well and is easily seen to lie outside of $\mathcal{E}(A)$.

If you happen to know that some class of groups cannot have such subgroups, you are more than welcome to share these as well. (For instance, underlying additive groups of vector spaces form such a class.)

Lastly, even though I posed the question only for abelian groups, you may feel free to interpret it more liberally.

Shaun
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Tim Seifert
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    Given any (possibly non-abelian) group $G$ and any $x \in G$, one could consider the subgroup of $G$ generated by the elements of the form $f(x)$ where $f$ ranges over the automorphisms of $G$. It is clearly a characteristic subgroup (the characteristic closure of the cyclic subgroup of $G$ generated by $x$). But is that characteristic subgroup "expected" or "unexpected"? – Geoffrey Trang Dec 30 '23 at 21:50
  • Do you really want to have a fixed $A$ or rather want a natural construction for every $A$? By this I mean a functor from the category of abelian groups with isomorphisms. – Martin Brandenburg Dec 30 '23 at 21:53
  • @GeoffreyTrang Nice example! These subgroups are indeed (at least sometimes) of the unexpected form. Indeed, if we hold on to the condition that joins and meets of expected stuff should again be expected, we would find that every characteristic subgroup would then expected (since any such group is of course the join of groups of your suggested form). Do you happen to know some explicit instance where this construction is tractable while also giving an unexpected subgroup (in the sense of the question)? – Tim Seifert Dec 30 '23 at 22:36
  • @MartinBrandenburg Good point. I was asking for a specific example, but a general (functorial) construction (or indication of its impossibility) would of course be even more appreciated. I don't think abelian groups with isomorphisms would be the category of my choice though: This would allow, if I'm not mistaken, for "constructions" like forming the explicit examples in the question, while giving back the trivial group for everything else $-$ this does not seem to be a natural construction in a satisfactory way. Maybe a more interesting choice ... – Tim Seifert Dec 30 '23 at 22:48
  • ... would be to ask instead for a functorial construction of fully invariant subgroups that (at least for some groups) happen to give unexpected results (in the sense of the question). – Tim Seifert Dec 30 '23 at 22:51
  • If I'm not mistaken, a functorial construction would yield a fully invariant construction, and in free groups fully invariant implies verbal, so I suspect that a functorial construction would necessarily give verbal subgroups. – Arturo Magidin Dec 31 '23 at 08:51
  • @ArturoMagidin That is an interesting constraint, but there must be more to it, since (for instance) the perfect core or the torsion subgroup (for abelian groups) are functorial but not verbal. It seems to me that your objection only rules out functors that preserve epis (or maybe have some other form of exactness). – Tim Seifert Dec 31 '23 at 10:36
  • Yeah, I think I was glossing over the exactness issue. – Arturo Magidin Dec 31 '23 at 15:59
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    Characteristic subgroups of finite abelian groups are classified. This problem easily reduces to $p$-groups. For odd $p$, all the characteristic subgroups are expected (using your definition). For $p=2$ there are some unexpected ones, and they essentially all look like the example you gave. (There is a Klein section somewhere in the lattice and two of the intermediate groups are expected characteristic, so the third one is characteristic even though it may be unexpected.) – verret Dec 31 '23 at 19:38
  • @verret That is some very interesting information, thank you! Do you have a reference, where I can read more about these results? Do you happen to know if there can be (infinite) examples of a different flavour? – Tim Seifert Jan 04 '24 at 14:48
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    @TimSeifert You can look at Section 2 in https://arxiv.org/abs/0905.1885 (Characteristic Subgroups of Finite Abelian Groups by Brent Kerby and Emma Turner) for example. They credit some of these results to Miller-Baer and I think Baer dealt with more general cases than the finite one, so perhaps chasing references from that paper will lead to more interesting results. – verret Jan 04 '24 at 22:43

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