# 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.

## 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 Facts used 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 $(yv)^p = y^pv^p$ because $v \in H$ and $H$ is central. Thus, $x^p = (yv)^p = y^pv^p = y^pu = y^py^{-p}g = g$, as desired. multiplied the first guess and the correction to get a correct guess.