# 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 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.