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

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

Proof
Let $$G$$ be the 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 residually nilpotent: The $$c^{th}$$ member of lower central series of $$G$$ for $$c \ge 1$$, is $$\langle a^{2^c} \rangle$$. The intersection of these is trivial.
 * 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.