Subnormality of fixed depth satisfies intermediate subgroup condition

This article gives the statement, and possibly proof, of a subgroup property (i.e., subnormal subgroup) satisfying a subgroup metaproperty (i.e., intermediate subgroup condition)
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Statement

Verbal statement

A subnormal subgroup of a group is also subnormal in every intermediate subgroup. In fact, its subnormal depth in any intermediate subgroup is bounded from above by the subnormal depth in the whole group.

Property-theoretic statement

The subgroup property of being a subnormal subgroup satisfies the subgroup metaproperty called the intermediate subgroup condition -- any subnormal subgroup of the whole group is also subnormal in every intermediate subgroup.

Statement with symbols

Suppose $H$ is a subnormal subgroup of a group $G$. Then, for any intermediate subgroup $K$ (i.e., $H\leq K\leq G$), $H$ is subnormal in $K$. Moreover, if $H$ is $k$-subnormal in $G$, $H$ is also $k$-subnormal in $K$. (Here, when we say $k$-subnormal, we mean the subnormal depth is at most $k$).

Related facts

Generalizations

• Subnormality satisfies inverse image condition
• Subnormality satisfies transfer condition

Related facts about normality and subnormality

• Normality is strongly UL-intersection-closed
• Normality satisfies transfer condition
• Normality satisfies inverse image condition
• Normality satisfies intermediate subgroup condition

Facts used

1. Normality satisfies transfer condition: If $H,K\leq G$ are subgroups such that $H$ is normal in $G$, then $H\cap K$ is normal in $K$.

Proof

Hands-on proof

Given: A group $G$, a $k$-subnormal subgroup $H$, a subgroup $K\leq G$ such that $H\leq K$.

To prove: $H$ is $k$-subnormal in $K$.

Proof: Consider a subnormal series for $H$ of length $k$:

$H=H_{0}\leq H_{1}\leq \dots \leq H_{k}=G$.

where $H_{i}$ is normal in $H_{i+1}$ for each $i$. We claim that the series:

$H=H_{0}\leq H_{1}\cap K\leq H_{2}\cap K\leq \dots \leq H_{k}\cap K=K$

is a subnormal series for $H$ in $K$. For this, observe that:

$H_{i}\cap K=H_{i}\cap (H_{i+1}\cap K)$.

We know that $H_{i}$ is normal in $H_{i+1}$, so by fact (1), $H_{i}\cap (H_{i+1}\cap K)$ is normal in $H_{i+1}\cap K$, yielding that $H_{i}\cap K$ is normal in $H_{i+1}\cap K$, as desired.