Subnormality is normalizing join-closed

Statement
Suppose $$H,K \le G$$ are fact about::subnormal subgroups, with the property that $$K \le N_G(H)$$: in other words, $$K$$ normalizes $$H$$. Then the join of subgroups $$\langle H, K \rangle$$ is also subnormal. Moreover, the fact about::subnormal depth of $$\langle H, K \rangle$$ is bounded from above by the products of subnormal depths of $$H$$ and $$K$$.

Related facts

 * Join of normal and subnormal implies subnormal of same depth
 * 2-subnormality is conjugate-join-closed
 * Subnormality is permuting join-closed

Facts used

 * 1) uses::Join of normal and subnormal implies subnormal of same depth: If $$L$$ is normal in $$G$$ and $$K$$ is $$k$$-subnormal in $$G$$, then $$KL$$ is subnormal in $$G$$ with subnormal depth at most $$k$$.
 * 2) uses::Normality is upper join-closed: If a subgroup is normal in two intermediate subgroups, it is normal in their join.

Proof
Given: A group $$G$$, subnormal subgroups $$H, K \le G$$ such that $$K \le N_G(H)$$, i.e., $$K$$ normalizes $$H$$. $$H$$ has subnormal depth $$h$$ and $$K$$ has subnormal depth $$k$$.

To prove: $$HK = \langle H, K \rangle$$ is a subnormal subgroup, with subnormal depth at most $$hk$$.

Proof: Consider the descending chain $$G_i$$ defined by $$G_0 = G$$, and $$G_{i+1}$$ is the normal closure of $$H$$ in $$G_i$$. This is the fastest descending subnormal series for $$H$$, and thus, $$G_h = H$$.

First, observe that since conjugation by any element of $$K$$ preserves $$H$$, it also preserves all the subgroups $$G_i$$, which are defined in terms of $$H$$. Thus, $$K$$ normalizes $$G_i$$ for each $$i$$.

We claim that $$G_{i+1}K$$ is subnormal of depth at most $$k$$ in $$G_iK$$.

Let's see why. First, note that $$G_{i+1}$$ is normal in $$G_i$$, and $$K$$ normalizes $$G_{i+1}$$, so $$G_{i+1}$$ is normal in $$G_iK$$ (fact (2)). Also, since $$K$$ is $$k$$-subnormal in $$G$$, $$K$$ is $$k$$-subnormal in $$G_{i+1}K$$. Applying fact (1) with $$G_iK$$ as the ambient group, $$G_{i+1}$$ as the normal subgroup, and $$K$$ as the subnormal subgroup, yields that $$G_{i+1}K$$ is $$k$$-subnormal in $$G_iK$$.

Thus, we have a chain:

$$HK = G_hK \le G_{h-1}K \le \dots \le G_1K \le G_0K = G$$

where each member is $$k$$-subnormal in its successor. This tells us that $$HK$$ is $$hk$$-subnormal in $$G$$.

Textbook references

 * , Page 387, Section 13.1 (Joins and intersections of subnormal subgroups)