Characteristic not implies powering-invariant in solvable group

Statement
It is possible to construct a solvable group $$G$$ and a characteristic subgroup $$H$$ of $$G$$ such that $$H$$ is not a powering-invariant subgroup of $$G$$.

In fact, we can choose $$G$$ to be powered for any set of primes and $$H$$ to be powered over any choice of subset of the set of primes (and not over the others).

Related facts

 * Center is local powering-invariant, hence the center is powering-invariant
 * Fixed-point subgroup of a subgroup of the automorphism group implies local powering-invariant
 * Derived subgroup not is local powering-invariant

Similar facts for Lie rings

 * Characteristic not implies powering-invariant in solvable Lie ring

Example of a rationally powered group and a characteristic subgroup that is not powered for any prime
Suppose $$G$$ is the group $$GA^+(1,\R)$$, given explicitly as the group of linear maps with positive coefficient from $$\R$$ to itself under composition. Explicitly, the elements of $$G$$ are maps of the form:

$$x \mapsto ax + b, \qquad a,b \in \R, a > 0$$

We can think of $$G$$ as the semidirect product $$\R \rtimes (\R^\ast)^+$$, i.e., the semidirect product of the reals under addition by the action of the positive reals under multiplication with their natural action by multiplication.

Suppose $$H$$ is the subgroup of $$G$$ given by the semidirect product $$\R \rtimes (\mathbb{Q}^\ast)^+$$. Explicitly, $$H$$ is the subgroup comprising the maps of the form:

$$x \mapsto ax + b, a \in \mathbb{Q}, b \in \R, a > 0$$

Then, we note that:


 * $$G$$ is rationally powered: See GAPlus(1,R) is rationally powered
 * $$H$$ is not powered for any prime: For instance, the map $$x \mapsto 2x$$ is in $$H$$. For any prime $$p$$, its unique $$p^{th}$$ root in $$G$$ is $$x \mapsto 2^{1/p}x$$. But since $$2^{1/p} \notin \mathbb{Q}$$, this element is not in $$H$$.
 * $$H$$ is characteristic in $$G$$: First, note that the translation subgroup $$x \mapsto x + b$$ is precisely the derived subgroup $$[G,G]$$, hence it is characteristic. This subgroup is the base of the semidirect product and is isomorphic to the additive group of $$\R$$, hence is rationally powered. Now, note that $$H$$ is precisely the subgroup of $$G$$ of elements such that the automorphism of $$[G,G]$$ induced by its action by conjugation is a rational multiplication. Note that whether an automorphism is a rational multiplication is a purely group-theoretic condition, so the above specifies $$H$$ uniquely in an automorphism-invariant manner.

Example for suitably chosen sets of primes
Consider two prime sets $$\pi_1$$ and $$\pi_2$$ with $$\pi_2 \subset \pi_1$$ (as a proper subset). Define the following two subsets of $$\R$$:


 * $$T_1$$ is the subset of the set of positive reals comprising those numbers $$x$$ for which there is a natural number $$n$$ with all prime factors in $$\pi_1$$ for which $$x^n \in (\mathbb{Q}^\ast)^+$$.
 * $$T_2$$ is the subset of the set of positive reals comprising those numbers $$x$$ for which there is a natural number $$n$$ with all prime factors in $$\pi_2$$ for which $$x^n \in (\mathbb{Q}^\ast)^+$$.

Note that $$T_2 \subset T_1$$ because $$\pi_2 \subset \pi_1$$.

Now define $$G$$ and $$H$$ as the following groups of linear maps from $$\R$$ to $$\R$$:

$$G = \{ x \mapsto ax + b, a \in T_1, b \in \R \}$$

$$H = \{ x \mapsto ax + b, a \in T_2, b \in \R \}$$

Then, we can see that:


 * $$G$$ is powered precisely over the primes in $$\pi_1$$ and over no other primes
 * $$H$$ is powered precisely over the primes in $$\pi_2$$ and over no other primes
 * $$H$$ is a characteristic subgroup of $$G$$