# Derived subgroup not is divisibility-closed

This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., derived subgroup) does not always satisfy a particular subgroup property (i.e., divisibility-closed subgroup)
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## Statement

It is possible to have a group $G$ and a prime number $p$ such that $G$ is a $p$-divisible group but the derived subgroup $G' = [G,G]$ is not a $p$-divisible group. In other words, for any $g \in G$, there exists $x \in G$ such that $x^p = g$, but there exists some $g \in G'$ for which there is no $x \in G'$ satisfying $x^p = g$.

In fact, for any prime number $p$, we can choose an example group $G$ specific to that prime. In fact, we can choose examples where $G$ is a divisible group for all primes, but the derived subgroup is not divisible by a specific prime of interest.

## Proof

### For a single prime

Let $K$ be an algebraically closed field of characteristic zero. We could take $K = \mathbb{C}$ for concreteness. Let $G = GL(p,K)$.

By Fact (1), $G$ is divisible by all primes. In particular, it is divisible by $p$.

By Fact (2), the derived subgroup of $G$ is $SL(p,K)$.

By Fact (3), $SL(p,K)$ is not $p$-divisible.

### For a collection of primes

As with the previous example, let $K$ be an algebraically closed field of characteristic zero, such as $\mathbb{C}$.

Suppose $\pi$ is a finite collection of primes. We can construct an example of a group $G$ that is $p$-divisible for all $p \in \pi$, but such that the derived subgroup is not $p$-divisible for any $p \in \pi$. The idea is to take $n$ as the product of all the primes in $\pi$, and set $G = GL(n,K)$.

Fact (1) gives that $G$ is $\pi$-divisible.

Fact (2) gives that the derived subgroup of $G$ is $SL(n,K)$.

Fact (3) gives that $SL(n,K)$ is not $p$-divisible for any $p \in \pi$.