Stem extension preserves divisibility for nilpotent groups

Statement
Suppose $$G$$ is a nilpotent group and $$\pi$$ is a set of prime numbers such that $$G$$ is $$\pi$$-divisible (i.e., $$G$$ is a $$\pi$$-divisible nilpotent group). Suppose $$E$$ is a stem extension of $$G$$ with quotient map $$\varphi:E \to G$$. Then, $$E$$ is also a $$\pi$$-divisible group.

Related facts

 * Schur multiplier of divisible nilpotent group need not be divisible by any prime

Facts used

 * 1) uses::Divisibility is quotient-closed
 * 2) uses::Equivalence of definitions of nilpotent group that is divisible for a set of primes

Proof
Let $$A$$ be the kernel of $$\varphi$$. Since the extension is a stem extension, we have $$A \le E' \cap Z(E)$$. In particular, $$A \le E'$$. Thus, $$E/E'$$ is a quotient of the group $$E/A \cong G$$. By Fact (1), $$E/E'$$ is $$\pi$$-divisible. Since $$E$$ is nilpotent, Fact (2) yields that $$E$$ is $$\pi$$-divisible.