Subnormality is permuting join-closed

Verbal statement
A join of two subnormal subgroups that permute is again a subnormal subgroup. Moreover, its subnormal depth is bounded by a function of the subnormal depths of the individual subnormal subgroups.

Statement with symbols
Suppose $$H$$ and $$K$$ are subnormal subgroups of a group $$G$$, and suppose further that $$HK = KH$$, i.e., the product of subgroups is again a subgroup. Then, $$HK$$ is a subnormal subgroup of $$G$$. Further, if the subnormal depth of $$H$$ is $$h$$ and the subnormal depth of $$K$$ is $$k$$, the subnormal depth of $$HK$$ is bounded by:

$$h[k]^h = hk(k+1)(k+2) \dots (k + h - 1)$$.

By symmetry, it is also bounded by the function $$k[h]^k$$. When $$h \le k$$, $$h[k]^h$$ is the smaller of the two numbers, hence the more relevant bound.

Related facts

 * Subnormality is normalizing join-closed
 * Normality is strongly join-closed
 * 2-subnormality is conjugate-join-closed
 * Join of normal and subnormal implies subnormal of same depth
 * Commutator subgroup satisfies ascending chain condition on subnormal subgroups implies subnormal join property
 * Nilpotent commutator subgroup implies subnormal join property

Applications

 * Permutable and subnormal implies join-transitively subnormal

Facts used

 * 1) uses::Modular property of groups: If $$A, B, C \le G$$ such that $$A \le C$$ we have $$A(B \cap C) = AB \cap C$$.
 * 2) uses::Subnormality is normalizing join-closed: If $$A, B \le G$$ are subnormal of depths $$a,b$$, and $$B \le N_G(A)$$, then $$AB$$ is a subnormal subgroup of depth at most $$ab$$.
 * 3) uses::Subnormality satisfies intermediate subgroup condition: If $$H \le L \le G$$ and $$H$$ is $$h$$-subnormal in $$G$$, $$H$$ is $$h$$-subnormal in $$L$$.
 * 4) uses::Subnormality is finite-relative-intersection-closed: Suppose $$A$$ is $$a$$-subnormal in $$C$$ and $$B \le C$$ such that $$B$$ is $$b$$-subnormal in some subgroup $$D$$ of $$C$$ containing both $$A$$ and $$B$$. Then, $$A \cap B$$ is $$(a + b)$$-subnormal in $$C$$.

Proof
Given: A group $$G$$, a subnormal subgroup $$H$$ of subnormal depth $$h$$ in $$G$$, a subnormal subgroup $$K$$ of subnormal depth $$k$$, such that $$HK = KH$$. Let $$J = HK$$.

To prove: $$J$$ is a subnormal subgroup of $$G$$ with subnormal depth at most $$hk(k+1) \dots (k + h - 1)$$.

Proof: We break the proof into many steps. Note that we use permutability where we use the fact that the product $$HK$$ is the subgroup generated by $$H$$ and $$K$$.

The key idea is to construct one subnormal series, first going down the series, and then going back up the series using some bounds. Our first few steps are in the ambient group $$J$$, and our later steps shift focus to the ambient group $$G$$.