Subnormality of fixed depth satisfies intermediate subgroup condition

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., $$H \le K \le G$$), $$H$$ 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 $$k$$).

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

Hands-on proof
Given: A group $$G$$, a $$k$$-subnormal subgroup $$H$$, a subgroup $$K \le G$$ such that $$H \le K$$.

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

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

$$H = H_0 \le H_1 \le \dots \le H_k = G$$.

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

$$H = H_0 \le H_1 \cap K \le H_2 \cap K \le \dots \le H_k \cap K = K$$

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.