# Join of subnormal subgroups is subnormal iff their commutator is subnormal

## Statement

Suppose $H,K$ are Subnormal subgroup (?)s 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: $s_2 \le s_1 + 1, s_3 \le s_2 + 1, s_3 \le s_1 + 1, s_2 \le s_3h, s_1 \le s_2k$.

## Facts used

1. 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. 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 $s_2k$.