# 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 $H \le K \le G$ 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.

## Proof

### Hands-on proof

Given: $H \le K \le G$, 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).