Extension need not preserve powering

Statement
It is possible to have a group $$G$$, a normal subgroup $$H$$, and a prime number $$p$$ such that the following hold:


 * $$H$$ is powered over $$p$$.
 * The quotient group $$G/H$$ is also powered over $$p$$.
 * $$G$$ is not powered over $$p$$.

Note that it is not possible to construct finite examples, because in the finite case, being powered over a prime $$p$$ is equivalent to $$p$$ not dividing the order (see kth power map is bijective iff k is relatively prime to the order).

Proof
Below is an example where both $$H$$ and $$G/H$$ are rationally powered (i.e., powered over all primes), but $$G$$ is not powered over any prime. There may be simpler examples.

Let $$W$$ be the subgroup $$\exp(\mathbb{Q})$$ inside $$(\R^*)^+$$. Recall that the group particular group::GAPLus(1,R) is the group of affine maps from $$\R$$ to $$\R$$ where the multiplication is positive, i.e.:

$$GA^+(1,\R) = \R \rtimes (\R^*)^+$$

Let $$U$$ be the vector space over $$\mathbb{Q}$$ generated by $$W$$ as a subset of $$\R$$. Note that $$U$$ is a subring of $$\R$$, because its generating set is a multiplicative monoid:

$$G = U \rtimes W$$

Explicitly, it is the set of maps:

$$\{ x \mapsto ax + b \mid a \in W, b \in U \}$$

Let $$H$$ be the base of the semidirect product here, so it is isomorphic to $$U$$.


 * $$H$$ is powered over all primes: That's because it is isomorphic to $$U$$, a vector space over $$\mathbb{Q}$$.
 * $$G/H$$ is powered over all primes: That is because it is isomorphic to $$W = \exp(\mathbb{Q})$$, which is isomorphic to $$\mathbb{Q}$$.
 * $$G$$ is not powered over any prime. Consider the element of $$G$$ of the form $$x \mapsto ex + 1$$ (here, $$e = \exp(1)$$ is transcendental). For a prime $$p$$, the unique $$p^{th}$$ root of this in $$GA^+(1,\R)$$ is:

$$x \mapsto e^{1/p}x + \frac{1}{1 + e^{1/p} + e^{2/p} + \dots + e^{(p-1)/p}}$$

We would like to claim that the number $$\frac{1}{1 + e^{1/p} + e^{2/p} + \dots + e^{(p-1)/p}}$$ is not an element of $$U$$, so that this $$p^{th}$$ root is not in $$G$$. This can be deduced from the fact that $$e$$ is trancendental.