Divisibility is central extension-closed

This article gives the statement, and possibly proof, of a group property (i.e., divisible group for a set of primes) satisfying a group metaproperty (i.e., central extension-closed group property)
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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.

Related facts

Dual fact

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

Other related facts

Proof

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.

Step no.	Assertion/construction	Given data used	Previous steps used	Explanation	Commentary
1	Let $a=\pi (g)$ , so $a\in G/H$ .				moved to quotient group
2	There exists $b\in G/H$ such that $b^{p}=a$ .	$G/H$ is $p$ -divisible.	Step (1)	direct	took root in quotient group
3	Suppose $y\in \pi ^{-1}(b)$ , i.e., $y\in G$ is such that $\pi (y)=b$ . Then, $u=y^{-p}g$ is an element of $H$ .		Steps (1), (2)	We have that $\pi (y^{-p}g)=b^{-p}a$ which is the identity element of $G/H$ . Thus, $y^{-p}g\in H$ .	picked representative, this is our "first guess" $y$ , and measured how far it is, using $u$ .
4	There exists $v\in H$ such that $v^{p}=u$ .	$H$ is $p$ -divisible.	Step (3)	direct	picked $v$ as the "correction" for our first guess.
5	The element $x=yv$ works, i.e., $x^{p}=g$ .	$H$ is central.	Steps (3), (4)	We have $(yv)^{p}=y^{p}v^{p}$ because $v\in H$ and $H$ is central. Thus, $x^{p}=(yv)^{p}=y^{p}v^{p}=y^{p}u=y^{p}y^{-p}g=g$ , as desired.	multiplied the first guess and the correction to get a correct guess.