Abelian normal is not join-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., Abelian normal subgroup) not satisfying a subgroup metaproperty (i.e., join-closed subgroup property).
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Statement

It is possible to have a group $G$ with Abelian normal subgroups $H,K$ such that the join $⟨H,K⟩$ is not an Abelian normal subgroup.

Related facts

• Cyclic normal is not join-closed
• Abelian characteristic is not join-closed

Proof

Example of the dihedral group

{{further|[[Particular example::dihedral group:D8]}}

Let $G$ be the dihedral group of order eight:

$G=⟨a,x\mid a^4=x^2=e,ax{a}^{-1}=x^{-1}⟩$.

Let $H,K$ be subgroups of $G$ given as follows:

$H=⟨a⟩,,K=⟨a^2,x⟩$.

Both $H$ and $K$ are Abelian normal subgroups, but the join of $H$ and $K$, which is the whole group $G$, is not an Abelian normal subgroup.

Example of the quaternion group

Further information: quaternion group

In the quaternion group, the cyclic subgroups generated by $i$ and $j$ are both Abelian normal of order four, but their join, which is the whole group, is not Abelian.

Any non-Abelian group of prime-cubed order

Further information: prime-cube order group:p2byp, prime-cube order group:U3p

If $p$ is an odd prime, the two non-Abelian $p$-groups of order $p^3$ again offer examples of groups with Abelian normal subgroups whose join is not normal. In both cases, there are multiple subgroups of order $p^2$, that are Abelian and normal, and whose join is the whole group, which is not normal.