# Characteristic subgroups and direct powers

Solved question: Suppose H is a characteristic subgroup of a group G. Is it then necessary that, for every natural number n, in the group $G^n$ (the external direct product of $G$ with itself $n$ times), the subgroup $H^n$ (embedded the obvious way) is characteristic?

Answer: No. A counterexample can be constructed where $G = \mathbb{Z}_8 \times \mathbb{Z}_2$ (here $\mathbb{Z}_n$ is the group of integers modulo $n$) with $H$ the subgroup

$$\{ (0,0), (2,1), (4,0), (6,1) \}$$

This subgroup sits in a weird diagonal sort of way and just happens to be characteristic (a quirk of the prime $2$ because there isn't enough space). We find that $H \times H$ is not characteristic in $G \times G$.

(ASIDE: The answer is yes, though, for many important characteristic subgroups, including fully invariant subgroups, members of the upper central series, and others that occur through typical definitions. Since for abelian groups of odd order, all characteristic subgroups are fully invariant, the answer is yes for abelian groups of odd order, so the example of order $2^4$ has no analogue in odd order abelian groups.)

My question is this:

1. Strongest: Is it true that if H is characteristic in G and $H \times H$ in $G \times G$, then each $H^n$ is characteristic in $G^n$ [NOTE: As Marty Isaacs points out in a comment to this question, $H \times H$ being characteristic in $G \times G$ implies H characteristic in G, so part of the condition is redundant -- as explained in (2)]?
2. Intermediate: Is there some finite $n_0$ such that it suffices to check $H^n$ characteristic in $G^n$ for $n = n_0$? Note that if $H^n$ is not characteristic in $G^n$ for any particular n, then characteristicity fails for all bigger $n$ as well. I'd allow $n_0$ to depend on the underlying prime of G if we are examining $p$-groups.
3. Weakest: Is there a test that would always terminate in finite time, that could tell, for a given H and G, whether $H^n$ is characteristic in $G^n$ for all n? The "try all n" terminates in finite time if the answer is no, but goes on forever if the answer is yes. In other words, is there a finite characterization of the property that each direct power of the subgroup is characteristic in the corresponding direct power of the group?

ADDED: My intuition, for what it's worth, is that those subgroups H of G that can be characterized through "positive" statements, i.e., those that do not make use of negations or $\ne$ symbols, would have the property that $H^n$ is characteristic in $G^n$. On the other hand, those whose characterization requires statements of exclusion (not a ...) would fail because negative statements are difficult to preserve on taking direct powers. But I don't know how to make this rigorous.

• That odd prime statement sounds handy, do you have a reference for "characteristic subgroups of abelian groups of odd order are fully invariant"? – Jack Schmidt Aug 16 '10 at 4:02
• No reference to a paper or book, but here's an online proof I wrote up: groupprops.subwiki.org/wiki/… It's rather long. What fails for the prime 2 is the first half of the proof: "it is the direct sum of its intersections with the direct summands" because in that part we use that the doubling map is an automorphism. The second half is bookkeeping that takes homomorphisms between cyclic subgroups and converts them to automorphisms by adding the identity map. – Vipul Naik Aug 16 '10 at 14:54
• There is a redundancy in Question 1 since if $H \times H$ is characteristic in $G \times G$, then $H$ is automatically characteristic in $G$. This is because every automorphism of $G$ extends to an automorphism of $G \times G$ stabilizing the first component. – Marty Isaacs Jan 28 '12 at 20:49
• @Marty Isaacs: I have added this note to Question 1. I had already incorporated a similar observation into Question 2, as you probably noticed. – Vipul Naik Mar 21 '12 at 15:26
• In the case of finite abelian groups, is it true that $\operatorname{Aut}(G \times G \times G)$ is generated by automorphisms stabilizing one copy of $G$ ? This holds at least when $G=(\mathbf{Z}/p^n\mathbf{Z})^k$. – François Brunault Sep 14 '12 at 7:16