Derived subgroup is divisibility-closed in nilpotent group

Statement
Suppose $$G$$ is a nilpotent group and $$G'$$ is the derived subgroup of $$G$$. Then, $$G'$$ is a divisibility-closed subgroup of $$G$$, i.e., for any prime number $$p$$, if $$G$$ is $$p$$-divisible, so is $$G'$$.

Related facts

 * Lower central series members are divisibility-closed in nilpotent group: Essentially, the same proof works.
 * Derived series members are divisibility-closed in nilpotent group: Follows from this and divisibility-closedness is transitive.

Facts used

 * 1) [[uses::Equivalent of definitions of nilpotent group that is divisible for a set of primes]

Proof
The proof follows directly from Fact (1). Sspecifically, it is the (1) implies (4) implication of Fact (1) that we use. We make two cases:


 * $$G$$ has class one or less: In this case, the derived subgroup is trivial.
 * $$G$$ has class two or more: Using the (1) implies (4) implication within Fact (1), and setting $$i = 2, j = c + 1$$ (where $$c$$ is the nilpotency class of $$G$$) gives the result.