Join-transitively subnormal of normal implies finite-conjugate-join-closed subnormal

This article describes a computation relating the result of the Composition operator (?) on two known subgroup properties (i.e., Join-transitively subnormal subgroup (?) and Normal subgroup (?)), to another known subgroup property (i.e., Finite-conjugate-join-closed subnormal subgroup (?))
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Statement

Statement with symbols

Suppose $H\leq K\leq G$ are subgroups such that $K$ is a normal subgroup of $G$ and $H$ is a join-transitively subnormal subgroup of $K$ (in other words, the join of $H$ with any subnormal subgroup of $G$ is subnormal). Then, $H$ is a finite-conjugate-join-closed subnormal subgroup of $G$: a join of finitely many conjugate subgroups of $H$ in $G$ is again subnormal.

Related facts

• 2-subnormal implies join-transitively subnormal
• 3-subnormal implies finite-conjugate-join-closed subnormal
• 2-subnormality is conjugate-join-closed
• Join-transitively subnormal of normal satisfies finite-conjugate-join-closed

Facts used

1. Join-transitively subnormal implies finite-automorph-join-closed subnormal
2. Finite-automorph-join-closed subnormal of normal implies finite-conjugate-join-closed subnormal

Proof

Hands-on proof

Given: $H\leq K\leq G$, such that $K$ is normal in $G$ and $H$ is join-transitively subnormal in $K$.

To prove: For any finite set $S\subseteq G$, the subgroup $H^{S}$, defined as the join of subgroups $H^{g},g\in S$, is subnormal in $G$.

Proof:

1. Each $H^{g}$ is join-transitively subnormal in $K$: Conjugation by $g$ defines an automorphism of $K$, since $K$ is normal in $G$. Thus, each $H^{g}$ is the image of $H$ under an automorphism of $K$. Since automorphism preserve subgroup properties, each $H^{g}$ is join-transitively subnormal in $K$.
2. $H^{S}$ is subnormal in $K$: Since $S$ is a finite set, $H^{S}$ is the join of finitely many subgroups $H^{g}$, each of which is join-transitively subnormal in $K$. By induction, we see that $H^{S}$ is also join-transitively subnormal in $K$; in particular, it is subnormal in $K$.
3. $H^{S}$ is subnormal in $G$: Since $H^{S}$ is subnormal in $K$ and $K$ is normal in $G$, we obtain that $H^{S}$ is subnormal in $G$.

Proof using given facts

The proof follows directly by piecing together facts (1) and (2).