Subnormality is not finite-upper join-closed

Statement
Suppose $$G$$ is a group, $$H \le G$$ is a subgroup and $$K,L \le G$$ are subgroups containing $$H$$. Then, it can happen that $$H$$ is a subnormal subgroup of $$K$$ and of $$L$$, but $$H$$ is not a subnormal subgroup of the join of subgroups $$\langle K, L \rangle$$.

Related facts

 * 2-subnormality is not upper join-closed: Note that the same example works.
 * Normality is upper join-closed
 * Subnormality is permuting upper join-closed in finite: This result is also known as the Maier-Wielandt theorem.
 * Subnormality is not permuting upper join-closed

Example of the symmetric group
Let $$G$$ be the symmetric group on the set $$\{ 1,2,3,4,5 \}$$. Let $$K$$ and $$L$$ be the dihedral groups given as follows:

$$K = \langle (1,3,2,4), (1,2) \rangle; \qquad L = \langle (1,3,2,5), (1,2) \rangle$$

Define $$H = K \cap L$$. Then, $$H$$ is a two-element subgroup comprising $$(1,2)$$ and the identity permutation.

Observe that:


 * $$H$$ is a 2-subnormal subgroup in both $$K$$ and $$L$$. In particular, $$H$$ is subnormal in both $$K$$ and $$L$$.
 * The join of $$K$$ and $$L$$ is $$G$$. This follows from some straightforward computation.
 * $$H$$ is not a subnormal subgroup of $$G$$. In fact, $$H$$ is a contranormal subgroup of $$G$$: the normal closure of $$H$$ in $$G$$ is the whole of $$G$$. This follows from the fact that transpositions generate the finitary symmetric group.