Cyclic normal is not join-closed

This article gives the statement, and possibly proof, of a subgroup property (i.e., cyclic normal subgroup) not satisfying a subgroup metaproperty (i.e., join-closed subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about cyclic normal subgroup|Get more facts about join-closed subgroup property|

Statement

It is possible to have a group $G$ with two cyclic normal subgroups (i.e., subgroups that are both cyclic and normal) whose join is not a cyclic normal subgroup.

Since the join is always a normal subgroup, the particular thing that fails is that the join is cyclic.

Related facts

• Abelian normal is not join-closed
• Abelian characteristic is not join-closed
• Normality is strongly join-closed

Proof

Example of the quaternion group

Further information: quaternion group

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

Example of a prime-cube order group for odd prime

Further information: prime-cube order group:p2byp

In the non-Abelian group of order $p^{3}$ obtained as a semidirect product of a cyclic group of order $p^{2}$ and a cyclic group of order $p$, there are many different cyclic normal subgroups of order $p^{2}$ (in fact, there are $p$ of them). The join of any two of these is the whole group, which is not cyclic.