Subnormality is not finite-upper join-closed

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

Suppose $G$ is a group, $H\leq G$ is a subgroup and $K,L\leq G$ 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

Proof

Example of the symmetric group

Further information: symmetric group:S5

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.