Subnormality satisfies transfer condition

Verbal statement
The intersection of a subnormal subgroup with any subgroup is subnormal in that subgroup. Moreover, the subnormal depth of the intersection is bounded from above by the subnormal depth of the original subgroup.

Statement with symbols
Suppose $$H$$ is a subnormal subgroup of a group $$G$$, and $$K \le G$$ is a subgroup. Then, $$H \cap K$$ is a subnormal subgroup of $$K$$. Moreover, if $$H$$ is $$k$$-subnormal in $$G$$, $$H \cap K$$ is $$k$$-subnormal in $$K$$ (i.e., its subnormal depth is at most $$k$$.

Related facts about subnormality

 * Subnormality satisfies intermediate subgroup condition
 * Subnormality satisfies inverse image condition
 * Subnormality satisfies image condition

Related facts about normality

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

Facts used

 * 1) uses::Normality satisfies transfer condition: If $$H$$ is a normal subgroup of a group $$G$$, and $$K \le G$$, then $$H \cap K$$ is a normal subgroup of $$K$$.
 * 2) uses::Transfer condition is subordination-closed: If $$p$$ is a subgroup property satisfying the transfer condition, the subordination of $$p$$.

Hands-on proof
Given: A group $$G$$, a $$k$$-subnormal subgroup $$H$$ of $$G$$, and a subgroup $$K$$ of $$G$$.

To prove: $$H \cap K$$ is a $$k$$-subnormal subgroup of $$K$$.

Proof: Since $$H$$ is $$k$$-subnormal subgroup of $$G$$, we have a subnormal series:

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

We claim that the following is a subnormal series for $$H \cap K$$ in $$K$$:

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

For this, we need to show that each $$H_i \cap K$$ is normal in $$H_{i+1} \cap K$$. For this, note that:

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

Since $$H_i$$ is normal in $$H_{i+1}$$, fact (1) tells us that $$H_i \cap (H_{i+1} \cap K) = H_i \cap K$$ is normal in $$H_{i+1} \cap K$$, completing the proof.

Property-theoretic proof
This follows directly from facts (1) and (2).