Derived subgroup not is divisibility-closed

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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.

Facts used

  1. General linear group over algebraically closed field is divisible
  2. Derived subgroup of general linear group is special linear group
  3. Special linear group over algebraically closed field is divisible precisely by those primes that do not divide its degree

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.