Join of subnormal subgroups is subnormal iff their commutator is subnormal

Statement
Suppose $$H,K$$ are fact about::subnormal subgroups of a group $$G$$. Then, the following are equivalent:


 * 1) The join of subgroups $$\langle H, K \rangle$$ is a subnormal subgroup of $$G$$.
 * 2) The subgroup $$H^K$$, i.e., the closure of $$H$$ under the action of $$K$$ by conjugation, is a subnormal subgroup of $$G$$.
 * 3) The commutator of two subgroups $$[H,K]$$ is a subnormal subgroup of $$G$$.

Moreover, if $$s_1, s_2, s_3$$ denote the subnormal depths of the subgroups in (1),(2),(3) respectively, and $$h,k$$ denote the subnormal depths of $$H,K$$, we have:

$$s_2 \le s_1 + 1, s_3 \le s_2 + 1, s_3 \le s_1 + 1, s_2 \le s_3h, s_1 \le s_2k$$.

Related facts

 * Normality is strongly join-closed
 * Normality is commutator-closed

Facts used

 * 1) uses::Commutator of two subgroups is normal in join: For $$H, K \le G$$, $$[H,K]$$ is normal in $$\langle H, K\rangle$$. Further, we have $$H[H,K] = H^K$$, $$K[H,K] = H^K$$, and $$KH^K = \langle H, K \rangle$$.
 * 2) uses::Subnormality is normalizing join-closed: For $$A, B \le G$$ subnormal of depths $$a,b$$, with $$B \le N_G(A)$$, we have that $$AB$$ is subnormal of depth at most $$ab$$.

Proof
Given: A group $$G$$ with subnormal subgroups $$H, K$$.

To prove: Consider the subgroups $$\langle H, K \rangle$$, $$H^K,$$, and $$[H,K]$$. If any one of them is subnormal, so are the others. Further, if $$s_1, s_2, s_3$$ are their subnormal depths, we have:

$$s_2 \le s_1 + 1, s_3 \le s_2 + 1, s_3 \le s_1 + 1, s_2 \le s_3h, s_1 \le s_2k$$.

Bounding the commutator in terms of the join
To prove: If $$\langle H, K \rangle$$ is subnormal of depth $$s_1$$, so is $$[H,K]$$, and if $$s_3$$ denotes its subnormal depth, we have $$s_3 \le s_1 + 1$$.

Proof: By fact (1), the commutator $$[H,K]$$ is a normal subgroup of $$\langle H, K \rangle$$, and thus, if $$\langle H, K \rangle$$ is subnormal of depth $$s_1$$, we have a subnormal series for $$[H,K]$$ of length $$s_1 + 1$$. This yields $$s_3 \le s_1 + 1$$.

Bounding the commutator in terms of the closure
To prove: If $$H^K$$ is subnormal of depth $$s_2$$, then $$[H,K]$$ is subnormal of depth $$s_3$$ with $$s_3 \le s_2 + 1$$.

Proof: By fact (1), $$[H,K]$$ is normal in $$\langle H, K \rangle$$, and in particular, is normal in the intermediate subgroup $$H^K$$. Thus, if $$H^K$$ has subnormal depth $$s_2$$, we obtain a subnormal series for $$[H,K]$$ of length $$s_2 + 1$$, yielding $$s_3 \le s_2 + 1$$.

Bounding the closure in terms of the join
To prove: If $$\langle H, K \rangle$$ is subnormal of depth $$s_1$$, then $$H^K$$ is subnormal of depth $$s_2$$ with $$s_2 \le s_1 + 1$$.

Proof: We have $$KH^K = \langle H, K \rangle$$, and by definition, $$K$$ normalizes $$H^K$$. Thus, $$H^K$$ is normal in $$\langle H, K \rangle$$. If $$\langle H, K \rangle$$ is subnormal of depth $$s_1$$, we get a subnormal series for $$H^K$$ of length $$s_1 + 1$$, yielding $$H^K$$ is subnormal with subnormal depth at most $$s_1 + 1$$.

Bounding the closure in terms of the commutator
To prove: If $$[H,K]$$ is subnormal of depth $$s_3$$ and $$H$$ is subnormal of depth $$h$$, then $$H^K$$ is subnormal of depth at most $$s_3h$$.

Proof: By fact (1), $$H$$ normalizes $$[H,K]$$ and $$H^K = H[H,K]$$. Thus, by fact (2), $$H^K$$ is subnormal of depth at most $$s_3h$$.

Bounding the join in terms of the closure
To prove: If $$H^K$$ is subnormal of depth $$s_2$$ and $$K$$ is subnormal of depth $$k$$, then $$\langle H, K \rangle$$ is subnormal of depth at most $$s_2k$$.

Proof: We have $$KH^K = \langle H, K \rangle$$, with $$K$$ normalizing $$H^K$$. Thus, by fact (2), $$\langle H, K \rangle$$ is subnormal of depth at most $$s_2k$$.

Textbook references

 * , Page 388, Theorem 13.1.6, Section 13.1 (Joins and intersections of subnormal subgroups)