# Conjugate-join-closed subnormal implies join-transitively subnormal

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., conjugate-join-closed subnormal subgroup) must also satisfy the second subgroup property (i.e., join-transitively subnormal subgroup)
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## Statement

Suppose $H$ is a conjugate-join-closed subnormal subgroup of a group $G$: for any subset $S$ of $G$, the subgroup $H^S = \langle H^s \mid s \in S \rangle$ is a subnormal subgroup of $G$.

Then, $H$ is a join-transitively subnormal subgroup of $G$: for any subnormal subgroup $K$ of $G$, the join of subgroups $\langle H, K \rangle$ is also a subnormal subgroup of $G$.

## Facts used

1. Join of subnormal subgroups is subnormal iff their commutator is subnormal: Suppose $H, K \le G$ are subnormal subgroups. Then $[H,K]$ is subnormal if and only if $H^K$ is subnormal, if and only if $\langle H, K \rangle$ is subnormal.

## Proof

Given: A conjugate-join-closed subnormal subgroup $H$ of a group $G$, a subnormal subgroup $K$ of $G$.

To prove: $\langle H, K \rangle$ is subnormal.

Proof:

1. (Given data used: $H$ is conjugate-join-closed subnormal in $G$): Since $H$ is conjugate-join-closed subnormal in $G$, $H^K$ is subnormal in $G$.
2. (Given data used: $H, K$ are subnormal in $G$; Fact used: fact (1)): By fact (1), we get that $\langle H, K \rangle$ is subnormal in $G$.