Derived subgroup is divisibility-closed in nilpotent group

Statement

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

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. Equivalence 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.