# 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 $H \le L, K \le L$, $H$ characteristic in $G$ and $K$ characteristic in $L$, but $H \cap K$ not characteristic in $G$.

## 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 $K \le H$

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 $(2,3)$.