Characteristicity is not finite-relative-intersection-closed

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

Verbal statement

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

Related facts

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

Example where $L=HK$

Further information: symmetric group:S4, dihedral group:D8, subgroup structure of symmetric group:S4

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 $K\leq H$

Further information: symmetric group:S4, dihedral group:D8, subgroup structure of symmetric group:S4

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 $(2,3)$.