Divisibility is central extension-closed

Statement
Suppose $$G$$ is a group and $$H$$ is a central subgroup of $$G$$. Suppose $$p$$ is a prime number such that:


 * $$H$$ is $$p$$-divisible.
 * The quotient group $$G/H$$ is $$p$$-divisible.

Then, the whole group $$G$$ is $$p$$-divisible.

Dual fact
The dual fact is that dual::powering-injectivity is inherited by central extensions.

Other related facts

 * Powering is central extension-closed
 * Powering-injectivity is central extension-closed
 * Divisibility is inherited by extensions where the normal subgroup is contained in the hypercenter

Proof idea
The idea is that of successive approximation. We first obtain a $$p^{th}$$ root in the quotient group, then pick a representative. We then pick a representative, and measure the extent to which its $$p^{th}$$ power misses the mark. Then, we take a $$p^{th}$$ root of that, and use that as the "first-order correction" to our original choice of representative.

The subgroup being central is crucial in making sure that the product of the $$p^{th}$$ powers of the original representative and the correction is the $$p^{th}$$ power of the product.

Proof details
Given: A group $$G$$, a central subgroup $$H$$ of $$G$$, a prime $$p$$ such that both $$H$$ and $$G/H$$ are $$p$$-divisible. An element $$g \in G$$.

To prove: There exists $$x \in G$$ such that $$x^p = g$$.

Proof: Suppose $$\pi:G \to G/H$$ is the quotient map.