Subnormal subgroup

Remarks

One writes $H \trianglelefteq\trianglelefteq G$. The least such $n$ is the defect of $H$ in $G$; the subgroups of defect at most $1$ are the normal subgroups. Example: in the alternating group $A_4$, let $V = \{e, (1\,2)(3\,4), (1\,3)(2\,4), (1\,4)(2\,3)\}$ and $H = \{e, (1\,2)(3\,4)\}$. Then $H \trianglelefteq V$ because $V$ is abelian, and $V \trianglelefteq A_4$, so $H$ is subnormal of defect $2$; it is not normal, since conjugation by $(1\,2\,3)$ sends $(1\,2)(3\,4)$ to $(2\,3)(1\,4)$. Non-example: in the symmetric group $S_3$, the subgroup $H = \{e, (1\,2)\}$ is not subnormal. The only subgroups of $S_3$ that contain $H$ are $H$ and $S_3$. So a chain from $H$ to $S_3$ must contain a step $H \trianglelefteq S_3$, but $H$ is not normal in $S_3$. In a nilpotent group every subgroup is subnormal.