Derived subgroup not is divisibility-closed

Statement
It is possible to have a group $$G$$ and a prime number $$p$$ such that $$G$$ is a $$p$$-divisible group but the derived subgroup $$G' = [G,G]$$ is not a $$p$$-divisible group. In other words, for any $$g \in G$$, there exists $$x \in G$$ such that $$x^p = g$$, but there exists some $$g \in G'$$ for which there is no $$x \in G'$$ satisfying $$x^p = g$$.

In fact, for any prime number $$p$$, we can choose an example group $$G$$ specific to that prime. In fact, we can choose examples where $$G$$ is a divisible group for all primes, but the derived subgroup is not divisible by a specific prime of interest.

Facts used

 * 1) uses::General linear group over algebraically closed field is divisible
 * 2) uses::Derived subgroup of general linear group is special linear group
 * 3) uses::Special linear group over algebraically closed field is divisible precisely by those primes that do not divide its degree

For a single prime
Let $$K$$ be an algebraically closed field of characteristic zero. We could take $$K = \mathbb{C}$$ for concreteness. Let $$G = GL(p,K)$$.

By Fact (1), $$G$$ is divisible by all primes. In particular, it is divisible by $$p$$.

By Fact (2), the derived subgroup of $$G$$ is $$SL(p,K)$$.

By Fact (3), $$SL(p,K)$$ is not $$p$$-divisible.

For a collection of primes
Suppose $$\pi$$ is a finite collection of primes. We can construct an example of a group $$G$$ that is $$p$$-divisible for all $$p \in \pi$$, but such that the derived subgroup is not $$p$$-divisible for any $$p \in \pi$$. The idea is to take $$n$$ as the product of all the primes in $$\pi$$, and set $$G = GL(n,K)$$.

Fact (1) gives that $$G$$ is $$\pi$$-divisible.

Fact (2) gives that the derived subgroup of $$G$$ is $$SL(n,K)$$.

Fact (3) gives that $$SL(n,K)$$ is not $$p$$-divisible for any $$p \in \pi$$.