Does there exist a characteristically simple group, which is not a direct product of simple groups?

Question

Does there exist a characteristically simple group, which is not a direct product of simple groups? A characteristically simple group is a group without non-trivial proper characteristic subgroups.

The only thing I know, that if such group $G$ exists, it should not be Artinian:

Suppose it is. If it has no non-trivial proper normal subgroups, then it is simple. If we have one, we can choose the minimal one (let’s denote it as $N$), according to the Zorn lemma (which can be applied, as our group is Artinian). It is not hard to see that $N$ will be simple. Then, $\langle \Pi_{\phi \in Aut(G)} \phi(N) \rangle = G$ as it is characteristic. And as all $\phi(N)$ are normal in it and either are the same subgroup or intersect trivially (because of their minimality), that results in $G$ being the direct product of $[Aut(G):Stab_{Aut(G)}(N)]$ isomorphic copies of $N$.

However, not all groups are Artinian. And I do not know how to deal with non-Artinian case here.

Answer 1

$\blacktriangleright$ Abelian characteristically simple groups: these are precisely

• abelian groups $G$ of prime exponent (i.e., nonzero vector spaces over $\mathbf{Z}/p\mathbf{Z}$ for some prime $p$)
• torsion-free divisible nontrivial abelian groups (i.e., nonzero vector spaces over $\mathbf{Q}$). This includes the case of $\mathbf{Q}$ already mentioned in a comment.
• "the" trivial group.

$\blacktriangleright$ A characteristically simple group with a minimal nontrivial normal subgroup (e.g., artinian) is necessarily a restricted direct power $S^{(I)}$ of a simple group $S$ (and $I$ being any nonempty index set). In the abelian case, this corresponds to the case with prime exponent. In the nonabelian case, the decomposition is unique, in the sense that the group is precisely restricted direct product of its minimal nontrivial normal subgroups. (Note however that such powers fail to be artinian, and even fail to satisfy min-n, as soon as the index set $I$ is infinite.)

$\blacktriangleright$ there also exist non-abelian characteristically simple groups with no minimal nontrivial normal subgroup. A simple example is obtained by fixing a nonabelian finite simple group $S$, a Cantor space $K$ and considering the subgroup of $S^K$ consisting of continuous (= locally constant) maps $K\to S$: this is a countable, nonabelian characteristically simple group, with no minimal nontrivial normal subgroup. (Edit: Here, the Cantor space can be replaced by any infinite compact, totally disconnected Hausdorff space on which the self-homeomorphism group acts minimally, which provides uncountable examples.)

$\blacktriangleright$ there also exist infinite finitely generated characteristically simple groups with no minimal nontrivial normal subgroup. These are necessarily nonabelian, so fall in the previous category, but are harder to construct: J. Wilson produced in 1976 such examples that are in addition finitely generated.

Answer 2

Actually, as it was pointed out in the comments $(\mathbb{Q}, +)$ is an example of such group.

It is non-simple $(\mathbb{Q}, +)$ as $(\mathbb{Z}, +) \triangleleft (\mathbb{Q}, +)$.

It is characteristically simple, as for any field $\mathbb{F}$, $Aut(\mathbb{F}, +)$ acts transitively on $\mathbb{F}\setminus\{0\}$.

It is not a direct product of any two non-trivial groups, as was proved in four different ways here: Is $(\mathbb{Q},+)$ the direct product of two non-trivial subgroups?