Isomorph-freeness is not finite-upper join-closed

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

We can have a situation where a subgroup $H\leq G$ is isomorph-free in two intermediate subgroups $K$ and $L$, but is not isomorph-free in $\langle K,L\rangle$.

Proof

Example of a non-abelian prime-cubed order group

Further information: Semidirect product of cyclic group of prime-square order and cyclic group of prime order

Let $p$ be an odd prime, and let $G$ be a non-Abelian group of order $p^{3}$ obtained as the semidirect product of a cyclic group $A$ of order $p^{2}$ and a cyclic group $B$ of order $p$. Let $H$ be the center of $G$. Let $K$ and $L$ be two cyclic subgroups of order $p^{2}$. Then:

• $H$ is isomorph-free in $K$ as well as in $L$: it is the cyclic subgroup of order $p$ in the cyclic group of order $p$.
• $H$ is not isomorph-free in the join $\langle K,L\rangle =G$: The cyclic subgroup $B$ is isomorphic to $H$, for instance.