Permutable and 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., permutable subnormal subgroup) must also satisfy the second subgroup property (i.e., join-transitively subnormal subgroup)
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This article describes a computation relating the result of the Join operator (?) on two known subgroup properties (i.e., Permutable subnormal subgroup (?) and Subnormal subgroup (?)), to another known subgroup property (i.e., Subnormal subgroup (?))
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Statement with symbols

Suppose H, K \le G are subgroups such that H is a permutable subnormal subgroup -- it is both subnormal and permutable, and K is a subnormal subgroup. Then the join \langle H, K \rangle is also subnormal, and its subnormal depth is bounded by a function of the subnormal depths of H and K.

Facts used

  1. Subnormality is permuting join-closed