Solvable group with abelianization that is divisible by a prime need not be divisible by that prime

Statement
It is possible to have a solvable group $$G$$ and a prime number $$p$$ such that the abelianization of $$G$$ is $$p$$-divisible, but $$G$$ itself is not $$p$$-divisible.

Related facts

 * Residually nilpotent group with abelianization that is divisible by a prime need not be divisible by that prime
 * Equivalence of definitions of nilpotent group that is divisible for a set of primes

Proof
Let $$G$$ be a infinite dihedral group:

$$\langle a,x \mid xax^{-1} = a^{-1}, x = x^{-1} \rangle$$

and let $$p$$ be any prime number other than 2. Then:


 * $$G$$ is solvable: In fact, $$G$$ is metacyclic. It has a cyclic normal subgroup $$\langle a \rangle$$ with a cyclic quotient group isomorphic to cyclic group:Z2 (generated by the image of $$x$$).
 * The abelianization of $$G$$ is $$p$$-divisible: The abelianization of $$G$$ is a Klein four-group, which is divisible for all odd primes.
 * $$G$$ is not $$p$$-divisible: The element $$a \in G$$ has no $$p^{th}$$ roots.