Every abelian subgroup is contained in a maximal among abelian subgroups

Statement

Every abelian subgroup of a group is contained in an abelian subgroup that is Maximal among abelian subgroups (?): the bigger subgroup is not contained in any bigger abelian subgroup.

Facts used

Zorn's lemma
Abelianness is varietal

Proof

Given: A group $G$ , an Abelian subgroup $H$ .

To prove: There exists an Abelian subgroup $K\leq G$ containing $H$ that is not contained in any bigger Abelian subgroup.

Proof: Consider the collection of Abelian subgroups of $G$ containing $H$ , ordered by inclusion. We first show that in this partially ordered set, every chain has an upper bound.

Indeed, given an ascending chain of Abelian subgroups, consider their union. We want to show that this is again Abelian. Given an ascending chain of subgroups, any two elements in their union must belong to one of the subgroups. The condition that these two elements commute now follows from the fact that they commute in the smaller subgroup.

Thus, the partially ordered collection of Abelian subgroups containing $H$ satisfies the condition that every chain has an upper bound in the collection. Zorn's lemma thus yields that there exists a maximal element in the partially ordered collection. If $K$ is such a maximal element, then $K$ cannot be contained in a bigger Abelian subgroup, because that bigger Abelian subgroup would also be in the partially ordered set, contradicting maximality. Thus, $K$ is maximal among Abelian subgroups.

Note that the argument used to show that the union of an ascending chain of Abelian subgroups is Abelian, hinges on the fact that Abelian groups form a subvariety of the variety of groups. Indeed, for any varietal group property, the union of an ascending chain of subgroups each satisfying the property, also satisfies the property.