Abelian normal is not join-closed

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

Related facts

 * Cyclic normal is not join-closed
 * Abelian characteristic is not join-closed
 * Nilpotent normal is finite-join-closed

A group obtained as a join of Abelian normal subgroups is termed a group generated by Abelian normal subgroups. Such groups have a number of nice properties.

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

$$G = \langle a, x \mid a^4 = x^2 = e, axa^{-1} = x^{-1}\rangle$$.

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

$$H = \langle a \rangle, \qquad, K = \langle a^2, x \rangle$$.

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
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
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.