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

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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.

Related facts

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 \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).