Characteristicity is not finite-relative-intersection-closed

Verbal statement
It is possible to have a group $$G$$ with subgroups $$H,K,L$$, such that $$H \le L, K \le L$$, $$H$$ characteristic in $$G$$ and $$K$$ characteristic in $$L$$, but $$H \cap K$$ not characteristic in $$G$$.

Other facts about finite-relative-intersection-closed

 * Subnormality is finite-relative-intersection-closed
 * Transitive and transfer condition implies finite-relative-intersection-closed

Related metaproperty satisfactions and dissatisfactions for characteristicity

 * Characteristicity is transitive
 * Characteristicity is strongly intersection-closed
 * Characteristicity does not satisfy intermediate subgroup condition
 * Characteristicity does not satisfy transfer condition

Example where $$L = HK$$
Suppose $$G$$ is the symmetric group on the set $$\{ 1,2,3,4 \}$$. Define:

$$H := \{, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}, K := \{ , (1,2,3,4), (1,3)(2,4), (1,4,3,2) \}$$

and define $$L = HK$$, explicitly, $$L$$ is a $$2$$-Sylow subgroup of $$G$$ isomorphic to the dihedral group:

$$L := \{, (1,2,3,4), (1,3)(2,4), (1,4,3,2), (1,2)(3,4), (1,4)(2,3), (1,3), (2,4) \}$$.

Then:


 * $$H$$ is characteristic in $$G$$: In fact, it is the only normal subgroup of order $$4$$ in $$G$$, and also equals the second commutator subgroup of $$G$$.
 * $$K$$ is characteristic in $$L$$: In fact, $$K$$ is an isomorph-free subgroup of $$L$$.
 * $$H \cap K$$ is not characteristic in $$G$$: The intersection $$H \cap K$$ is the two-element subgroup $$\{, (1,3)(2,4)\}$$, which is not characteristic -- in fact, it is not even normal in $$G$$, as it is not invariant under conjugation by $$(2,3)$$.

Example where $$K \le H$$
Suppose $$G$$ is the symmetric group on the set $$\{ 1,2,3,4 \}$$. Define:

$$H := \{, (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}, K := \{ , (1,3)(2,4) \}$$

and $$L$$ is a $$2$$-Sylow subgroup of $$G$$ isomorphic to the dihedral group, containing $$H$$:

$$L := \{, (1,2,3,4), (1,3)(2,4), (1,4,3,2), (1,2)(3,4), (1,4)(2,3), (1,3), (2,4) \}$$.

Then:


 * $$H$$ is characteristic in $$G$$: In fact, it is the only normal subgroup of order $$4$$ in $$G$$, and also equals the second commutator subgroup of $$G$$.
 * $$K$$ is characteristic in $$L$$: In fact, $$K$$ is the center of $$L$$.
 * $$H \cap K$$ is not characteristic in $$G$$: The intersection $$H \cap K$$ is the two-element subgroup $$K = \{, (1,3)(2,4)\}$$, which is not characteristic -- in fact, it is not even normal in $$G$$, as it is not invariant under conjugation by $$(2,3)$$.