Conjugate-join-closed subnormal implies join-transitively subnormal

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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.

Related facts

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.