4-subnormal not implies finite-conjugate-join-closed subnormal

This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., 4-subnormal subgroup) need not satisfy the second subgroup property (i.e., finite-conjugate-join-closed subnormal subgroup)
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Statement

It is possible to have a group $G$ and a finite collection of conjugate 4-subnormal subgroups of $G$ whose join is not subnormal. In fact, it is possible to have just two conjugate 4-subnormal subgroups whose join is not subnormal.

Facts used

1. 3-subnormal not implies finite-automorph-join-closed subnormal (in fact, it is possible to have just two automorphic 3-subnormal subgroups whose join is not subnormal.
2. Left residual of finite-conjugate-join-closed subnormal by normal is finite-automorph-join-closed subnormal

Proof

Hands-on proof

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Proof using facts

Suppose that every 4-subnormal subgroup of a group is finite-conjugate-join-closed subnormal. In particular, this means that if $H$ is a 3-subnormal subgroup of $L$, then whenever $L$ is a normal subgroup of a group $G$, $H$ is finite-conjugate-join-closed subnormal in $G$.

Fact (2) says that this implies that $H$ is finite-automorph-join-closed subnormal in $G$. Thus, we have shown that every 3-subnormal subgroup is finite-automorph-join-closed subnormal. This, however, contradicts fact (1).